Precalculus: Concepts Through Functions, a Unit Circle Approach to Trigonometry 9780321931047, 1292058749, 9781292058740, 5535575615, 0321931041

Table of contents :
Cover......Page 1
Title......Page 2
Copyright......Page 3
Contents......Page 6
To the Student......Page 16
Preface to the Instructor......Page 17
Prepare for Class ‘‘Read the Book’’......Page 22
Practice ‘‘Work the Problems’’......Page 23
Review ‘‘Study for Quizzes and Tests’’......Page 24
Resources for Success......Page 25
Applications Index......Page 26
F Foundations: A Prelude to Functions......Page 34
F.1 The Distance and Midpoint Formulas......Page 35
Use the Distance Formula......Page 36
Use the Midpoint Formula......Page 38
Graph Equations by Plotting Points......Page 42
Find Intercepts from a Graph......Page 44
Test an Equation for Symmetry......Page 45
Know How to Graph Key Equations......Page 47
Calculate and Interpret the Slope of a Line......Page 52
Find the Equation of a Vertical Line......Page 55
Use the Point-Slope Form of a Line; Identify Horizontal Lines......Page 56
Write the Equation of a Line in Slope-Intercept Form......Page 57
Identify the Slope and y-Intercept of a Line from Its Equation......Page 58
Graph Lines Written in General Form Using Intercepts......Page 59
Find Equations of Parallel Lines......Page 60
Find Equations of Perpendicular Lines......Page 61
Write the Standard Form of the Equation of a Circle......Page 67
Graph a Circle......Page 68
Work with the General Form of the Equation of a Circle......Page 69
Chapter Project......Page 74
1 Functions and Their Graphs......Page 75
Determine Whether a Relation Represents a Function......Page 76
Find the Value of a Function......Page 79
Find the Domain of a Function Defined by an Equation......Page 82
Formthe Sum, Difference, Product, and Quotient of Two Functions......Page 84
1.2 The Graph of a Function......Page 89
Identify the Graph of a Function......Page 90
Obtain Information from or about the Graph of a Function......Page 91
Determine Even and Odd Functions from a Graph......Page 99
Identify Even and Odd Functions from the Equation......Page 100
Use a Graph to Determine Where a Function is Increasing, Decreasing, or Constant......Page 101
Use a Graph to Locate Local Maxima and Local Minima......Page 102
Use a Graph to Locate the Absolute Maximum and the Absolute Minimum......Page 103
Use a Graphing Utility to Approximate Local Maxima and Local Minima and to Determine Where a Function is Increasing or Decreasing......Page 104
Find the Average Rate of Change of a Function......Page 105
Graph the Functions Listed in the Library of Functions......Page 111
Graph Piecewise-defined Functions......Page 116
Graph Functions Using Vertical and Horizontal Shifts......Page 122
Graph Functions Using Compressions and Stretches......Page 125
Graph Functions Using Reflections about the x-Axis and the y-Axis......Page 127
Build and Analyze Functions......Page 134
Construct a Model Using Direct Variation......Page 139
Construct a Model Using Joint or Combined Variation......Page 140
Chapter Review......Page 144
Chapter Test......Page 148
Chapter Projects......Page 149
2 Linear and Quadratic Functions......Page 151
Use Average Rate of Change to Identify Linear Functions......Page 152
Determine Whether a Linear Function Is Increasing, Decreasing or Constant......Page 155
Find the Zero of a Linear Function......Page 156
Build Linear Models from Verbal Descriptions......Page 157
Draw and Interpret Scatter Diagrams......Page 163
Distinguish between Linear and Nonlinear Relations......Page 164
Use a Graphing Utility to Find the Line of Best Fit......Page 165
2.3 Quadratic Functions and Their Zeros......Page 170
Find the Zeros of a Quadratic Function Using the Square Root Method......Page 171
Find the Zeros of a Quadratic Function by Completing the Square......Page 173
Find the Zeros of a Quadratic Function Using the Quadratic Formula......Page 174
Find the Point of Intersection of Two Functions......Page 176
Solve Equations That Are Quadratic in Form......Page 177
2.4 Properties of Quadratic Functions......Page 181
Graph a Quadratic Function Using Transformations......Page 182
Identify the Vertexand Axis of Symmetry of a Quadratic Function......Page 184
Graph a Quadratic Function Using Its Vertex, Axis, and Intercepts......Page 185
Find a Quadratic Function Given Its Vertex and One Other Point......Page 188
Find the Maximum or Minimum Value of a Quadratic Function......Page 189
Solve Inequalities Involving a Quadratic Function......Page 193
2.6 Building Quadratic Models from Verbal Descriptions and from Data......Page 197
Build Quadratic Models from Verbal Descriptions......Page 198
Build Quadratic Models from Data......Page 202
Find the Complex Zeros of a Quadratic Function......Page 208
2.8 Equations and Inequalities Involving the Absolute Value Function......Page 211
Solve Absolute Value Inequalities......Page 212
Chapter Review......Page 217
Chapter Test......Page 220
Cumulative Review......Page 221
Chapter Projects......Page 222
3 Polynomial and Rational Functions......Page 224
Identify Polynomial Functions and Their Degree......Page 225
Graph Polynomial Functions Using Transformations......Page 229
Identify the Real Zeros of a Polynomial Function and Their Multiplicity......Page 230
Analyze the Graph of a Polynomial Function......Page 237
Build Cubic Models from Data......Page 239
3.2 The Real Zeros of a Polynomial Function......Page 245
Use the Remainder and Factor Theorems......Page 246
Use Descartes’ Rule of Signs to Determine the Number of Positive and the Number of Negative Real Zeros of a Polynomial Function......Page 248
Use the Rational Zeros Theorem to List the Potential Rational Zeros of a Polynomial Function......Page 249
Find the Real Zeros ofa Polynomial Function......Page 250
Solve Polynomial Equations......Page 252
Use the Theorem for Bounds on Zeros......Page 253
Use the Intermediate Value Theorem......Page 254
3.3 Complex Zeros; Fundamental Theorem of Algebra......Page 259
Use the Conjugate Pairs Theorem......Page 260
Find a Polynomial Function with Specified Zeros......Page 261
Find the Complex Zeros of a Polynomial Function......Page 262
3.4 Properties of Rational Functions......Page 265
Find the Domain of a Rational Function......Page 266
Find the Vertical Asymptotes of a Rational Function......Page 269
Find the Horizontal or Oblique Asymptote of a Rational Function......Page 270
Analyze the Graph of a Rational Function......Page 276
Solve Applied Problems Involving Rational Functions......Page 287
3.6 Polynomial and Rational Inequalities......Page 291
Solve Polynomial Inequalities......Page 292
Solve Rational Inequalities......Page 293
Chapter Review......Page 299
Cumulative Review......Page 303
Chapter Projects......Page 304
4 Exponential and Logarithmic Functions......Page 306
Form a Composite Function......Page 307
Find the Domain of a Composite Function......Page 308
Determine Whether a Function Is One-to-One......Page 315
Determine the Inverse of a Function Defined by a Map or a Set of Ordered Pairs......Page 317
Obtain the Graph ofthe Inverse Function from the Graph of the Function......Page 319
Find the Inverse of a Function Defined by an Equation......Page 320
Evaluate Exponential Functions......Page 327
Graph Exponential Functions......Page 331
Define the Number e......Page 334
Solve Exponential Equations......Page 336
4.4 Logarithmic Functions......Page 344
Evaluate Logarithmic Expressions......Page 345
Graph Logarithmic Functions......Page 346
Solve Logarithmic Equations......Page 351
Work with Properties of Logarithms......Page 357
Write a Logarithmic Expression as a Single Logarithm......Page 360
Evaluate a Logarithm Whose Base Is Neither 10 Nor e......Page 361
Graph a Logarithmic Function Whose Base Is Neither 10 Nor e......Page 363
Solve Logarithmic Equations......Page 366
Solve Exponential Equations......Page 368
Solve Logarithmic and Exponential Equations Using a Graphing Utility......Page 369
4.7 Financial Models......Page 372
Determine the Future Value of a Lump Sum of Money......Page 373
Calculate Effective Rates of Return......Page 376
Determine the Present Value of a Lump Sum of Money......Page 377
Determine the Rate of Interest or the Time Required to Double a Lump Sum of Money......Page 378
Find Equations of Populations That Obey the Law of Uninhibited Growth......Page 382
Find Equations of Populations That Obey the Law of Decay......Page 384
Use Newton’s Law of Cooling......Page 385
Use Logistic Models......Page 387
4.9 Building Exponential, Logarithmic, and Logistic Models from Data......Page 392
Build an Exponential Model from Data......Page 393
Build a Logarithmic Model from Data......Page 394
Build a Logistic Model from Data......Page 395
Chapter Review......Page 400
Chapter Test......Page 405
Cumulative Review......Page 406
Chapter Projects......Page 407
5 Trigonometric Functions......Page 408
5.1 Angles and Their Measures......Page 409
Convert between Decimals and Degrees, Minutes, Seconds Measures for Angles......Page 411
Find the Length if an Arc of a Circle......Page 412
Convert from Degreesto Radians and from Radians to Degrees......Page 413
Find the Area of a Sector of a Circle......Page 416
Find the Linear Speed of an Object Traveling in Circular Motion......Page 417
5.2 Trigonometric Functions: Unit Circle Approach......Page 423
Find the Exact Values of the Trigonometric Functions Using a Point on the Unit Circle......Page 425
Find the Exact Values of the Trigonometric Functions of Quadrantal Angles......Page 426
Find the Exact Values of the Trigonometric Functions......Page 428
Find the Exact Values of the Trigonometric Functions......Page 429
Find the Exact Values of the Trigonometric Functions for Integer Multiples......Page 431
Use a Circle of Radius r to Evaluate the Trigonometric Functions......Page 433
Determine the Domain and the Range of the Trigonometric Functions......Page 440
Determine the Period of the Trigonometric Functions......Page 442
Find the Values of the Trigonometric Functions Using Fundamental Identities......Page 444
Find the ExactValues of the Trigonometric Functions of an Angle......Page 446
Use Even-Odd Properties to Find the Exact Values of the Trigonometric Functions......Page 449
5.4 Graphs of the Sine and Cosine Functions......Page 453
Graph Functions of the Form y = A sin......Page 454
Graph Functions of the Form y = A cos......Page 456
Determine the Amplitude and Period of Sinusoidal Functions......Page 457
Graph Sinusoidal Functions Using Key Points......Page 459
Find an Equation for a Sinusoidal Graph......Page 463
5.5 Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions......Page 468
Graph Functions of the Form......Page 470
Graph Functions of the Form......Page 473
Graph Sinusoidal Functions of the Form......Page 476
Build Sinusoidal Models from Data......Page 480
Chapter Review......Page 487
Cumulative Review......Page 493
Chapter Projects......Page 494
6 Analytic Trigonometry......Page 496
6.1 The Inverse Sine, Cosine, and Tangent Functions......Page 497
Find the Exact Value of an Inverse Sine Function......Page 498
Find an Approximate Value of an Inverse Sine Function......Page 499
Use Properties of Inverse Functions to Find Exact Values of Certain Composite Functions......Page 500
Find the Inverse Function of a Trigonometric Function......Page 505
Solve Equations Involving Inverse Trigonometric Functions......Page 506
6.2 The Inverse Trigonometric Functions (Continued)......Page 509
Find the Exact Value of Expressions Involving the Inverse Sine, Cosine, and Tangent Functions......Page 510
Define the Inverse Secant, Cosecant, and Cotangent Functions......Page 511
Write a Trigonometric Expression as an Algebraic Expression......Page 512
Solve Equations Involving a Single Trigonometric Function......Page 515
Solve Trigonometric Equations Quadratic in Form......Page 518
Solve Trigonometric Equations Using Fundamental Identities......Page 519
Solve Trigonometric Equations Using a Graphing Utility......Page 520
6.4 Trigonometric Identities......Page 524
Establish Identities......Page 526
6.5 Sum and Difference Formulas......Page 532
Use Sum and Difference Formulas to Find Exact Values......Page 533
Use Sum and Difference Formulas to Establish Identities......Page 534
Use Sum and Difference Formulas Involving Inverse Trigonometric Functions......Page 538
Solve Trigonometric Equations Linear in Sine and Cosine......Page 539
6.6 Double-angle and Half-angle Formulas......Page 544
Use Double-angle Formulas to Establish Identities......Page 545
Use Half-angle Formulas to Find Exact Values......Page 548
Express Products as Sums......Page 554
Express Sums as Products......Page 556
Chapter Review......Page 558
Cumulative Review......Page 562
Chapter Projects......Page 563
7 Applications of Trigonometric Functions......Page 564
Find the Value of Trigonometric Functions of Acute Angles Using Right Triangles......Page 565
Solve Right Triangles......Page 567
Solve Applied Problems......Page 568
7.2 The Law of Sines......Page 577
Solve SSA Triangles......Page 579
Solve Applied Problems......Page 582
7.3 The Law of Cosines......Page 588
Solve SAS Triangles......Page 589
Solve Applied Problems......Page 590
7.4 Area of a Triangle......Page 594
Find the Area of SAS Triangles......Page 595
Find the Area of SSS Triangles......Page 596
Build a Model for an Object in Simple Harmonic Motion......Page 601
Analyze Simple Harmonic Motion......Page 603
Analyze an Object in Damped Motion......Page 604
Graph the Sum of Two Functions......Page 605
Chapter Review......Page 610
Chapter Test......Page 613
Chapter Projects......Page 614
8 Polar Coordinates; Vectors......Page 616
Plot Points Using Polar Coordinates......Page 617
Convert from Polar Coordinates to Rectangular Coordinates......Page 619
Convert from Rectangular Coordinates to Polar Coordinates......Page 621
Transform Equations between Polar and Rectangular Forms......Page 623
8.2 Polar Equations and Graphs......Page 626
Identify and Graph Polar Equations by Converting to Rectangular Equations......Page 627
Test Polar Equations for Symmetry......Page 630
Graph Polar Equations by Plotting Points......Page 631
Plot Points in the Complex Plane......Page 641
Convert a Complex Number between Rectangular Form and Polar Form......Page 642
Find Products and Quotients of Complex Numbers in Polar Form......Page 643
Use De Moivre’s Theorem......Page 644
Find Complex Roots......Page 645
8.4 Vectors......Page 649
Graph Vectors......Page 651
Find a Position Vector......Page 652
Add and Subtract Vectors Algebraically......Page 653
Find a Scalar Multiple and the Magnitude of a Vector......Page 654
Find a Vector from Its Direction and Magnitude......Page 655
Model with Vectors......Page 656
Find the Dot Product of Two Vectors......Page 663
Find the Angle between Two Vectors......Page 664
Determine Whether Two Vectors Are Orthogonal......Page 665
Decompose a Vector into Two Orthogonal Vectors......Page 666
Compute Work......Page 668
8.6 Vectors in Space......Page 671
Find Position Vectors in Space......Page 672
Perform Operations on Vectors......Page 673
Find the Dot Product......Page 674
Find the Angle between Two Vectors......Page 675
Find the Direction Angles of a Vector......Page 676
Find the Cross Product of Two Vectors......Page 680
Know Algebraic Properties of the Cross Product......Page 682
Find aVector Orthogonal to Two Given Vectors......Page 683
Find the Area of a Parallelogram......Page 684
Chapter Review......Page 686
Chapter Test......Page 689
Chapter Projects......Page 690
9 Analytic Geometry......Page 692
Know the Names of the Conics......Page 693
Analyze Parabolas with Vertex at the Origin......Page 694
Analyze Parabolas with Vertex at (h, k)......Page 697
Solve Applied Problems Involving Parabolas......Page 698
9.3 The Ellipse......Page 702
Analyze Ellipses with Center at the Origin......Page 703
Analyze Ellipses with Center at (h, k)......Page 707
Solve Applied Problems Involving Ellipses......Page 708
9.4 The Hyperbola......Page 712
Analyze Hyperbolas with Center at the Origin......Page 713
Find Asymptotes of Hyperbola......Page 717
Analyze Hyperbolas with Center at (h, k)......Page 719
Solve Applied Problems Involving Hyperbolas......Page 720
Identify a Conic......Page 725
Use a Rotation of Axes to Transform Equations......Page 726
Analyze an Equation Using Rotation of Axes......Page 728
Identify Conics without a Rotation of Axes......Page 730
Analyze and Graph Polar Equations of Conics......Page 732
Convert the Polar Equation of a Conic to a Rectangular Equation......Page 736
9.7 Plane Curves and Parametric Equations......Page 738
Find a Rectangular Equation for a Curve Defined Parametrically......Page 739
Use Time as a Parameter in Parametric Equations......Page 741
Find Parametric Equations for Curves Defined by Rectangular Equations......Page 744
Chapter Review......Page 750
Cumulative Review......Page 753
Chapter Projects......Page 754
10 Systems of Equations and Inequalities......Page 755
10.1 Systems of Linear Equations: Substitution and Elimination......Page 756
Solve Systems of Equations by Substitution......Page 758
Solve Systems of Equations by Elimination......Page 759
Identify Inconsistent Systems of Equations Containing Two Variables......Page 761
Solve Systems of Three Equations Containing Three Variables......Page 762
Identify Inconsistent Systems of Equations Containing Three Variables......Page 764
Express the Solution of a System of Dependent Equations Containing Three Variables......Page 765
Write the Augmented Matrix of a System of Linear Equations......Page 771
Write the System of Equations from the Augmented Matrix......Page 772
Perform Row Operationson a Matrix......Page 773
Solve a System of Linear Equations Using Matrices......Page 774
10.3 Systems of Linear Equations: Determinants......Page 785
Use Cramer’s Rule to Solve a System of Two Equations Containing Two Variables......Page 786
Evaluate 3 by 3 Determinants......Page 789
Use Cramer’s Rule to Solve a System of Three Equations Containing Three Variables......Page 790
Know Properties of Determinants......Page 792
10.4 Matrix Algebra......Page 795
Find the Sum and Difference of Two Matrices......Page 797
Find Scalar Multiples of a Matrix......Page 799
Find the Product of Two Matrices......Page 800
Find the Inverse of a Matrix......Page 804
Solve a System of Linear Equations Using an Inverse Matrix......Page 808
10.5 Partial Fraction Decomposition......Page 814
Decompose P/Q, Where Q Has Only Nonrepeated Linear Factors......Page 815
Decompose P/Q, Where Q Has Repeated Linear Factors......Page 817
Decompose P/Q, Where Q Has a Nonrepeated Irreducible Quadratic Factor......Page 819
Decompose P/Q, Where Q Has a Repeated Irreducible Quadratic Factor......Page 820
Solve a System of Nonlinear Equations Using Substitution......Page 822
Solve a System of Nonlinear Equations Using Elimination......Page 823
Graph an Inequality......Page 831
Graph a System of Inequalities......Page 833
Set up a Linear Programming Problem......Page 839
Solve a Linear Programming Problem......Page 840
Chapter Review......Page 846
Chapter Test......Page 849
Cumulative Review......Page 850
Chapter Projects......Page 851
11 Sequences; Induction; the Binomial Theorem......Page 852
Write the First Several Terms of a Sequence......Page 853
Write the Terms of a Sequence Defined by a Recursive Formula......Page 856
Use Summation Notation......Page 857
Find the Sum of a Sequence......Page 858
Determine Whether a Sequence Is Arithmetic......Page 863
Find a Formula for an Arithmetic Sequence......Page 864
Find the Sum of an Arithmetic Sequence......Page 865
Determine Whether a Sequence Is Geometric......Page 869
Find a Formula for a Geometric Sequence......Page 870
Find the Sum of a Geometric Sequence......Page 871
Determine Whether a Geometric Series Converges or Diverges......Page 872
Solve Annuity Problems......Page 875
Prove Statements Using Mathematical Induction......Page 880
11.5 The Binomial Theorem......Page 884
Evaluate n/j......Page 885
Use the Binomial Theorem......Page 887
Chapter Review......Page 891
Cumulative Review......Page 893
Chapter Projects......Page 894
12 Counting and Probability......Page 895
Count the Number of Elements in a Set......Page 896
Solve Counting Problems Using the Multiplication Principle......Page 898
12.2 Permutations and Combinations......Page 901
Solve Counting Problems Using Permutations Involving n Distinct Objects......Page 902
Solve Counting Problems Using Combinations......Page 904
Solve Counting Problems Using Permutations Involving n Nondistinct Objects......Page 906
Construct Probability Models......Page 910
Compute Probabilities of Equally Likely Outcomes......Page 912
Find Probabilities of the Union of Two Events......Page 914
Use the Complement Rule to Find Probabilities......Page 915
Chapter Review......Page 920
Chapter Test......Page 922
Chapter Projects......Page 923
13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function......Page 924
Find a Limit Using a Table......Page 925
Find a Limit Using a Graph......Page 927
13.2 Algebra Techniques for Finding Limits......Page 930
Find the Limit of a Sum, a Difference, and a Product......Page 931
Find the Limit ofa Polynomial......Page 932
Find the Limit of a Power or a Root......Page 933
Find the Limit of a Quotient......Page 934
Find the Limit of an Average Rate of Change......Page 935
Find the One-sided Limits of a Function......Page 937
Determine Whether a FunctionIs Continuous......Page 939
Find an Equation of the Tangent Line to Graph a Function......Page 944
Find the Derivative of a Function......Page 946
Find the Instantaneous Speed of a Particle......Page 947
13.5 The Area Problem; The Integral......Page 951
Approximate the Area Under the Graph of a Function......Page 952
Approximate Integrals Using a Graphing Utility......Page 956
Chapter Review......Page 957
Chapter Test......Page 960
Chapter Projects......Page 961
Work with Sets......Page 962
Graph Inequalities......Page 965
Find Distance on the Real Number Line......Page 966
Evaluate Algebraic Expressions......Page 967
Use the Laws of Exponents......Page 968
Evaluate Square Roots......Page 970
Use a Calculator to Evaluate Exponents......Page 971
Use the Pythagorean Theorem and Its Converse......Page 974
Know Geometry Formulas......Page 976
Understand Congruent Triangles and Similar Triangles......Page 977
Recognize Monomials......Page 983
Recognize Polynomials......Page 984
Add and Subtract Polynomials......Page 985
Multiply Polynomials......Page 986
Know Formulas for Special Products......Page 987
Divide Polynomials Using Long Division......Page 988
Work with Polynomials in Two Variables......Page 991
A.4 Factoring Polynomials......Page 993
Factoring the Difference of Two Squares and the Sum and Difference of Two Cubes......Page 994
Factor a Second-Degree Polynomial......Page 995
Factor by Grouping......Page 997
Factor a Second-Degree Polynomial......Page 998
Complete the Square......Page 999
Divide Polynomials Using Synthetic Division......Page 1002
Reduce a Rational Expression to Lowest Terms......Page 1006
Multiply and Divide Rational Expressions......Page 1007
Add and Subtract Rational Expressions......Page 1008
Use the Least Common Multiple Method......Page 1009
Simplify Complex Rational Expressions......Page 1011
Work with nth Roots......Page 1016
Simplify Radicals......Page 1017
Rationalize Denominators......Page 1018
Simplify Expressions with Rational Exponents......Page 1019
A.8 Solving Equations......Page 1024
Solve Linear Equations......Page 1025
Solve Rational Equations......Page 1027
Solve Equations by Factoring......Page 1028
Solve Radical Equations......Page 1029
Translate Verbal Descriptions into Mathematical Expressions......Page 1033
Solve Interest Problems......Page 1034
Solve Mixture Problems......Page 1035
Solve Uniform Motion Problems......Page 1036
Solve Constant Rate Job Problems......Page 1038
Use Interval Notation......Page 1042
Use Properties of Inequalities......Page 1043
Solve Inequalities......Page 1045
Solve Combined Inequalities......Page 1046
A.11 Complex Numbers......Page 1050
Add, Subtract, Multiply, and Divide Complex Numbers......Page 1051
B.1 The Viewing Rectangle......Page 1058
B.2 Using a Graphing Utility to Graph Equations......Page 1060
B.3 Using a Graphing Utility to Locate Intercepts and Check for Symmetry......Page 1062
B.4 Using a Graphing Utility to Solve Equations......Page 1063
B.5 Square Screens......Page 1065
B.7 Using a Graphing Utility to Solve Systems of Linear Equations......Page 1066
B.9 Using a Graphing Utility to Graph Parametric Equations......Page 1068
Answers......Page 1070
Photo Credits......Page 1176
Index......Page 1178

Citation preview

GLOBAL EDITION

Precalculus

Concepts Through Functions A Unit Circle Approach to Trigonometry THIRD EDITION

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Precalculus Concepts Through Functions

A Unit Circle Approach To Trigonometry

Third Edition Global Edition Michael Sullivan Chicago State University

Michael Sullivan, lll Joliet Junior College

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Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsonglobaleditions.com © Pearson Education Limited 2015 The rights of Michael Sullivan and Michael Sullivan, III to be identified as the authors of this work has been asserted by them in accordance with the Copyright, Designs and Patents Act 1988. Authorized adaptation from the United States edition, entitled Precalculus: Concepts through Functions, A Unit Circle Approach to Trigonometry, 3rd edition, ISBN 978-0-321-93104-7, by Michael Sullivan and Michael Sullivan, III, published by Pearson Education © 2015. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a license permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6–10 Kirby Street, London EC1N 8TS. All trademarks used herein are the property of their respective owners. The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks, nor does the use of such trademarks imply any affiliation with or endorsement of this book by such owners. The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. ISBN 10: 1-292-05874-9 ISBN 13: 978-1-292-05874-0 British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library 10 9 8 7 6 5 4 3 2 1

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Typeset in Times Ten by Cenveo® Publisher Services Printed and bound by Courier Kendallville in The United States of America

For Michael S., Kevin, and Marissa (Sullivan) Shannon, Patrick, and Ryan (Murphy) Maeve, Sean, and Nolan (Sullivan) Kaleigh, Billy, and Timmy (O’Hara) The Next Generation 

Contents

F

To the Student

15

Preface to the Instructor

17

Prepare for Class ‘‘Read the Book’’

21

Practice ‘‘Work the Problems’’

22

Review ‘‘Study for Quizzes and Tests’’

23

Resources for Success

24

Applications Index

25

Foundations: A Prelude to Functions F.1 The Distance and Midpoint Formulas

33 34

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F. 4 Circles

66

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Chapter Project

1

Functions and Their Graphs 1.1 Functions

73

74 75

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1.2 The Graph of a Function

88

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1.3 Properties of Functions

98

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5

6

CONTENTS

1.4 Library of Functions; Piecewise-defined Functions

110

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1.5 Graphing Techniques: Transformations

121

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1.6 Mathematical Models: Building Functions

133

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1.7 Building Mathematical Models Using Variation

138

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2

Chapter Review

143

Chapter Test

147

Chapter Projects

148

Linear and Quadratic Functions 2.1 Properties of Linear Functions and Linear Models

150 151

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2.2 Building Linear Models from Data

162

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2.3 Quadratic Functions and Their Zeros

169

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2.4 Properties of Quadratic Functions

180

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2.5 Inequalities Involving Quadratic Functions

192

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2.6 Building Quadratic Models from Verbal Descriptions and from Data

196

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2.7 Complex Zeros of a Quadratic Function

207

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210

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Chapter Review

216

Chapter Test

219

CONTENTS

3

Cumulative Review

220

Chapter Projects

221

Polynomial and Rational Functions

223

3.1 Polynomial Functions and Models

224

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3.2 The Real Zeros of a Polynomial Function

244

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3.3 Complex Zeros; Fundamental Theorem of Algebra

258

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3.4 Properties of Rational Functions

264

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3.5 The Graph of a Rational Function

275

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3.6 Polynomial and Rational Inequalities

290

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4

Chapter Review

298

Chapter Test

302

Cumulative Review

302

Chapter Projects

303

Exponential and Logarithmic Functions 4.1 Composite Functions

305 306

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4.2 One-to-One Functions; Inverse Functions

314

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4.3 Exponential Functions

326

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343

7

8

CONTENTS

4.5 Properties of Logarithms

356

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4.6 Logarithmic and Exponential Equations

365

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Logarithmic and Exponential Equations Using a Graphing Utility

4.7 Financial Models

371

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Sum of Money

4.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models

381

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4.9 Building Exponential, Logarithmic, and Logistic Models from Data

391

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5

Chapter Review

399

Chapter Test

404

Cumulative Review

405

Chapter Projects

406

Trigonometric Functions

407

5.1 Angles and Their Measures

408

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5.2 Trigonometric Functions: Unit Circle Approach

422

Find the Exact Values of the Trigonometric Functions Using a Point on the 6OJU
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Functions of p/4 = 45°r
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Functions of p/6 = 30° and p/3 = 60°r
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5.3 Properties of the Trigonometric Functions

439

Determine the Domain and the Range of the Trigonometric Functions r
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y = A cos (vx) Using Transformations

452

CONTENTS

r
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5.5 Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions

467

Graph Functions of the Form y = A tan(vx) + B and y = A cot (vx) + B r
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5.6 Phase Shift; Sinusoidal Curve Fitting

475

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6

Chapter Review

486

Chapter Test

492

Cumulative Review

492

Chapter Projects

493

Analytic Trigonometry 6.1 The Inverse Sine, Cosine, and Tangent Functions

495 496

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6.2 The Inverse Trigonometric Functions (Continued)

508

Find the Exact Value of Expressions Involving the Inverse Sine, Cosine, BOE
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6.3 Trigonometric Equations

514

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6.4 Trigonometric Identities

523

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6.5 Sum and Difference Formulas

531

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Trigonometric Equations Linear in Sine and Cosine

6.6 Double-angle and Half-angle Formulas

543

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6.7 Product-to-Sum and Sum-to-Product Formulas

553

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Chapter Review

557

Chapter Test

561

Cumulative Review

561

Chapter Projects

562

9

10

CONTENTS

7

Applications of Trigonometric Functions

563

7.1 Right Triangle Trigonometry; Applications

564

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7.2 The Law of Sines

576

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7.3 The Law of Cosines

587

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7.4 Area of a Triangle

593

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7.5 Simple Harmonic Motion; Damped Motion; Combining Waves

600

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of Two Functions

8

Chapter Review

609

Chapter Test

612

Cumulative Review

613

Chapter Projects

613

Polar Coordinates; Vectors 8.1 Polar Coordinates

615 616

1MPU
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1PMBS
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'PSNT

8.2 Polar Equations and Graphs

625

Identify and Graph Polar Equations by Converting to Rectangular &RVBUJPOT
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by Plotting Points

8.3 The Complex Plane; De Moivre’s Theorem

640

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Complex Roots

8.4 Vectors

648

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with Vectors

8.5 The Dot Product

662

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Angles of a Vector

670

CONTENTS

8.7 The Cross Product

11

679

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Parallelogram

9

Chapter Review

685

Chapter Test

688

Cumulative Review

689

Chapter Projects

689

Analytic Geometry

691

9.1 Conics

692

Know the Names of the Conics

9.2 The Parabola

693

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9.3 The Ellipse

701

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9.4 The Hyperbola

711

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9.5 Rotation of Axes; General Form of a Conic

724

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9.6 Polar Equations of Conics

731

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9.7 Plane Curves and Parametric Equations

737

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Equations

10

Chapter Review

749

Chapter Test

752

Cumulative Review

752

Chapter Projects

753

Systems of Equations and Inequalities

754

10.1 Systems of Linear Equations: Substitution and Elimination 4PMWF
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Containing Three Variables

755

12

CONTENTS

10.2 Systems of Linear Equations: Matrices

770

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10.3 Systems of Linear Equations: Determinants

784

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10.4 Matrix Algebra

794

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10.5 Partial Fraction Decomposition

813

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10.6 Systems of Nonlinear Equations

821

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of Nonlinear Equations Using Elimination

10.7 Systems of Inequalities

830

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10.8 Linear Programming

838

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11

Chapter Review

845

Chapter Test

848

Cumulative Review

849

Chapter Projects

850

Sequences; Induction; the Binomial Theorem 11.1 Sequences

851 852

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11.2 Arithmetic Sequences

862

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11.3 Geometric Sequences; Geometric Series

868

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11.4 Mathematical Induction

879

Prove Statements Using Mathematical Induction

11.5 The Binomial Theorem

883

n Evaluate a b
r
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Chapter Review

890

Chapter Test

892

Cumulative Review

892

Chapter Projects

893

CONTENTS

12

Counting and Probability 12.1 Counting

13

894 895

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12.2 Permutations and Combinations

900

Solve Counting Problems Using Permutations Involving n
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12.3 Probability

909

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Complement Rule to Find Probabilities

13

Chapter Review

919

Chapter Test

921

Cumulative Review

922

Chapter Projects

922

A Preview of Calculus: The Limit, Derivative, and Integral of a Function 13.1 Finding Limits Using Tables and Graphs

923 924

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13.2 Algebra Techniques for Finding Limits

929

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13.3 One-sided Limits; Continuous Functions

936

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13.4 The Tangent Problem; The Derivative

943

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Instantaneous Speed of a Particle

13.5 The Area Problem; The Integral

950

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Integrals Using a Graphing Utility

A

Chapter Review

956

Chapter Test

959

Chapter Projects

960

Review A.1 Algebra Essentials

A1 A1

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A13

14

CONTENTS

A.3 Polynomials

A22

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Variables

A.4 Factoring Polynomials

A32

Factoring the Difference of Two Squares and the Sum and Difference of 5XP
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x2 + Bx + C
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A.5 Synthetic Division

A41

Divide Polynomials Using Synthetic Division

A.6 Rational Expressions

A45

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A.7 nth Roots; Rational Exponents

A55

Work with nUI
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A.8 Solving Equations

A63

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A.9 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications

A72

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A.10 Interval Notation; Solving Inequalities

A81

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A.11 Complex Numbers

A89

Add, Subtract, Multiply, and Divide Complex Numbers

B

Graphing Utilities

B1

B.1 The Viewing Rectangle

B1

B.2 Using a Graphing Utility to Graph Equations

B3

B.3 Using a Graphing Utility to Locate Intercepts and Check for Symmetry

B5

B.4 Using a Graphing Utility to Solve Equations

B6

B.5 Square Screens

B8

B.6 Using a Graphing Utility to Graph Inequalities

B9

B.7 Using a Graphing Utility to Solve Systems of Linear Equations

B9

B.8 Using a Graphing Utility to Graph a Polar Equation

B11

B.9 Using a Graphing Utility to Graph Parametric Equations

B11

Answers

AN1

Photo Credits Index

C1 I1

To the Student As you begin, you may feel anxious about the number of theorems, definitions, QSPDFEVSFT
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your concerns are normal. This textbook was written with you in mind. If you attend class, work hard, and read and study this book, you will build the knowledge and TLJMMT
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the book has a large number of examples and clear explanations to help you break down the mathematics into easy-to-understand steps. Reading will provide you with a clearer understanding, beyond simple memorization. Read before class (not after) TP
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Use the Features We use many different methods in the classroom to communicate. Those methods, when incorporated into the book, are called “features.” The features serve many purposes, from providing timely review of material you learned before (just when you need it), to providing organized review sessions to help you prepare for quizzes and tests. Take advantage of the features and you will master the material. 5P
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book. Refer to the “Prepare for Class,” “Practice,” and “Review” on pages 21–23. Spend fifteen minutes reviewing the guide and familiarizing yourself with the features by flipping to the page numbers provided. Then, as you read, use them. This is the best way to make the most of your textbook. Please do not hesitate to contact us, through Pearson Education, with any questions, suggestions, or comments that would improve this text. We look forward to hearing from you, and good luck with all of your studies.

Best Wishes! Michael Sullivan Michael Sullivan, III

15

Preface to the Instructor

A

s professors at both an urban university and a community college, Michael Sullivan and Michael Sullivan, III, are aware of the varied needs of Precalculus students, ranging from those who have little mathematical background and a fear of mathematics courses, to those having a strong mathematical education and a high level of motivation. For some of your students, this will be their last course in mathematics, whereas others will further their mathematical education. This text is written for both groups. As a teacher, and as an author of precalculus, engineering calculus, finite mathematics, and business calculus texts, Michael Sullivan understands what students must know if they are to be focused and successful in upperlevel math courses. However, as a father of four, he also understands the realities of college life. As an author of B
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and son, Michael Sullivan, III, understands the trepidations and skills students bring to the Precalculus course. Michael, III also believes in the value of technology as a tool for learning that enhances understanding without sacrificing math skills. Together, both authors have taken great pains to ensure that the text contains solid, studentfriendly examples and problems, as well as a clear and seamless writing style. A tremendous benefit of authoring a successful series is the broad-based feedback we receive from teachers and students. We are sincerely grateful for their support. Virtually every change in this edition is the result of their thoughtful comments and suggestions. We are sincerely grateful for this support and hope that we have been able to take these ideas and, building upon a successful first edition, make this series an even better tool for learning and teaching. We continue to encourage you to share with us your experiences teaching from this text.

this course. By introducing functions early in the course, students are less likely to develop bad habits. Another advantage of the early introduction of functions is that the discussion of equations and inequalities can focus around the concept of a function. For example, rather than asking students to solve an equation such as 2x2 + 5x + 2 = 0, we ask students to find the zeros of f1x2 = 2x2 + 5x + 2 or solve f1x2 = 0 when f1x2 = 2x2 + 5x + 2. While the technique used to solve this type of problem is the same, the fact that the problem looks different to the student means the student is less apt to say, i0I
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how to solve it.” In addition, in Calculus students are going to be asked to solve equations such as f ′1x2 = 0, so solving f1x2 = 0 is a logical prerequisite skill to practice in Precalculus. Another advantage to solving equations through the eyes of a function is that the properties of functions can be included in the solution. For example, the linear function f 1x2 = 2x - 3 has one real zero because the function f is increasing on its domain.

About This Book

r Retain Your Knowledge This new category of problems in the exercise set are based on the article “To Retain New Learning, Do the Math” published in the Edurati Review in which author Kevin Washburn suggests that “the more students are required to recall new content or skills, the better their memory will be.” It is frustrating when students cannot recall skills learned earlier in the course. To alleviate this recall problem, we have created “Retain Your Knowledge” problems. These are problems considered to be “final exam material” that students must complete to maintain their skills. All the answers to these problems appear in the back of the book. r Guided Lecture Notes Ideal for online, emporium/redesign courses, inverted classrooms or traditional lecture classrooms. These lecture notes assist students in taking thorough, organized, and understandable notes as they watch the Author in Action videos by asking students to complete definitions, procedures, and examples based

This book utilizes a functions approach to Precalculus. Functions are introduced early (Chapter 1) in various formats: maps, tables, sets of ordered pairs, equations, and HSBQIT
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numeric, graphic, and verbal representations of functions. This allows students to make connections between the visual representation of a function and its algebraic representation. It is our belief that students need to “hit the ground running” so that they do not become complacent in their studies. After all, it is highly likely that students have been exposed to solving equations and inequalities prior to entering this class. By spending precious time reviewing these concepts, students are likely to think of the course as a rehash of material learned in other courses and say to themTFMWFT
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16

Features in the Third Edition Rather than provide a list of new features here, that information can be found on pages 21–23. This places the new features in their proper context, as building blocks of an overall learning system that has been carefully crafted over the years to help students get the most out of the time they put into studying. Please take the time to review the features listed on pages 21–23 and to discuss them with your students at the beginning of your DPVSTF
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these features, they are more successful in the course.

New to the Third Edition

PREFACE

on the content of the videos and book. In addition, experience suggests that students learn by doing and understanding the why/how of the concept or property. Therefore, many sections will have an exploration activity to motivate student learning. These explorations will introduce the topic and/or connect it somehow to either a real world application or previous section. For example, when teaching about the vertical line test in Section 1.2, after the theorem statement, the notes ask the students to explain why the vertical line test works by using the definition of a function. This helps students process the information at a higher level of understanding. r Chapter Projects, which apply the concepts of each chapter to a real-world situation, have been enhanced to give students an up-to-the-minute experience. Many projects are new and Internet-based, requiring the student to research information online in order to solve problems. r Exercise Sets at the end of each section remain classified according to purpose. The “Are You Prepared?” exercises have been expanded to better serve the student who needs a just-in-time review of concepts utilized in the section. The Concepts and Vocabulary exercises have been updated. These fill-in-the-blank and True/False problems have been written to serve as reading quizzes. Skill Building
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DPNQVUBtional skills and are often grouped by objective. Mixed Practice exercises have been added where appropriate. These problems offer a comprehensive assessment of the skills learned in the section by asking problems that relate to more than one objective. Sometimes these require information from previous sections so students must utilize skills learned throughout the course. Applications and Extension problems have been updated and many new problems involving sourced information and data have been added to bring relevance and timeliness to the exercises. The Explaining Concepts: Discussion and Writing exercises have been updated and reworded to stimulate discussion of concepts in online discussion forums. These can also be used to spark classroom discussion. r The Chapter Review now includes answers to all the problems. We have created a separate review worksheet for each chapter to help students review and practice key skills to prepare for exams. The worksheets can CF
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17

Changes in the Third Edition r CONTENT r Chapter 2, Section 4 A new objective “Find a quadratic function given its vertex and one point” has been added. r Chapter 2, Section 5 A new example was added to illustrate that quadratic inequalities may have the empty set or all real numbers as a solution. r Chapter 3, Sections 1 and 4 The content related to describing the behavior of the graph of a polynomial or rational function near a zero has been removed. r Chapter 3, Section 4 Content has been added that discusses the role of multiplicity and behavior of the graph of rational function as the graph approaches a vertical asymptote. r ORGANIZATION r Chapter 3, Sections 5 and 6 Section 5, The Real Zeros of a Polynomial Function and Section 6, Complex Zeros, Fundamental Theorem of Algebra have been moved to Sections 2 and 3, respectively. This was done in response to reviewer requests that “everything involving polynomials” be located sequentially. Skipping the new Sections 2 and 3 and proceeding to Section 4 Properties of Rational Functions can be done without loss of continuity.

Using this Book Effectively and Efficiently with Your Syllabus To meet the varied needs of diverse syllabi, this book contains more content than is likely to be covered in a typical Precalculus course. As the chart illustrates, this book has been organized with flexibility of use in mind. Even within a given chapter, certain sections are optional and can be omitted without loss of continuity. See the detail following the flow chart. F

APPENDIX A

APPENDIX B

1

2

3

4

5

9.1-9.4

6

10

11

12

9.5-9.7

13 7

8.1-8.3

8.4-8.7

Foundations A Prelude to Functions Quick coverage of this chapter, which is mainly review material, will enable you to get to Chapter 1, Functions and Their Graphs, earlier.

18

PREFACE

Chapter 1 Functions and Their Graphs Perhaps the most important chapter. Sections 1.6 and 1.7 are optional.

Chapter 10 Systems of Equations and Inequalities Sections 10.2–10.7 may be covered in any order. Section 10.8 requires Section 10.7.

Chapter 2 Linear and Quadratic Functions Topic selection depends on your syllabus. Sections 2.2, 2.6, and 2.7 may be omitted without a loss of continuity.

Chapter 11 Sequences; Induction; the Binomial Theorem There are three independent parts: Sections 11.1–11.3, Section 11.4, and Section 11.5.

Chapter 3 Polynomial and Rational Functions Topic selection depends on your syllabus. Section 3.6 is optional. Chapter 4 Exponential and Logarithmic Functions Sections 4.1–4.6 follow in sequence. Sections 4.7–4.9 are optional. Chapter 5 Trigonometric Functions The sections follow in sequence. Section 5.6 is optional.

Chapter 12 Counting and Probability The sections follow in sequence. Chapter 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function If time permits, coverage of this chapter will provide your students with a beneficial head-start in calculus. The sections follow in sequence.

Chapter 7 Applications of Trigonometric Functions Sections 7.4 and 7.5 may be omitted in a brief course.

Appendix A Review This review material may be covered at the start of a course or used as a just-in-time review. Specific references to this material occur throughout the text to assist in the review process.

Chapter 8 Polar Coordinates; Vectors Sections 8.1–8.3 and Sections 8.4–8.7 are independent and may be covered separately.

Appendix B Graphing Utilities Reference is made to these sections at the appropriate place in the text.

Chapter 6 Analytic Trigonometry Sections 6.2 and 6.7 may be omitted in a brief course.

Chapter 9 Analytic Geometry Sections 9.1–9.4 follow in sequence. Sections 9.5, 9.6, and 9.7, are independent of each other, but each requires Sections 9.1–9.4.

Acknowledgments Third Edition Textbooks are written by authors, but evolve from an idea to final form through the efforts of many people. It was Don Dellen who first suggested this book and series. Don is remembered for his extensive contributions to publishing and mathematics. Thanks are due to the following people for their assistance and encouragement to the preparation of this edition: r From Pearson Education: Anne Kelly for her substantial contributions, ideas, and enthusiasm; Peggy Lucas, who is a huge fan and works tirelessly to get the word out; Dawn Murrin, for her unmatched talent at getting the details right; Peggy McMahon for her organizational skills and leadership in overseeing production; Chris Hoag for her continued support and genuine interest; Greg Tobin for his leadership and commitment to excellence; and the Pearson Math and Science Sales team, for their continued confidence and personal support of our books. r Bob Walters, Production Manager, who passed away after a long and valiant battle fighting lung disease. He was an old and dear friend—a true professional in every sense of the word. r Accuracy checkers: C. Brad Davis, who read the entire manuscript and accuracy checked answers. His attention to detail is amazing; Timothy Britt, for creating the Solutions Manuals and accuracy checking answers. r Michael Sullivan, III would like to thank his colleagues at Joliet Junior College for their support and feedback. Finally, we offer our grateful thanks to the dedicated users and reviewers of our books, whose collective insights form the backbone of each textbook revision. 0VS
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please accept our apology. Thank you all. Gary Amara—South Maine Community College Richard Andrews—Florida A&M University Jay Araas—Sheridan College Jessica Bernards—Portland Community college Rebecca Berthiaume—Edison State College Susan Bradley—Angelina College Michael Brook—University of Delaware Tim Chappell—Penn Valley Community College Christine Cole—Moorpark College Alicia Collins—Mesa Community College Rebecca Cosner—Spokane Community College Jerry DeGroot—Purdue North Central Joanna DelMonaco—Middlesex Community College Stephanie Deacon—Liberty University Jerrett Dumouchel—Florida Community College at Jacksonville Vaden Fitton—North Virginia Community College Carrie Rose Gibson—North Idaho College Nina Girard—University of Pittsburgh at Johnstown Mary Beth Grayson—Liberty University Scott Greenleaf—South Maine Community College Donna Harbin—University of Hawaii-Maui Celeste Hernandez—Richland College Gloria P. Hernandez—Louisiana State University at Eunice .BSJU[B
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19

20

ACKNOWLEDGMENTS

Susitha Karunaratne—Purdue University North Central Debra Kopcso—Louisiana State University Yelena Kravchuk—University of Alabama at Birmingham Mary Krohn—Butler University Lynn Marecek—Santa Ana College James McLaughlin—West Chester University ,BUIMFFO
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.JBNJ Laura Pyzdrowski—West Virginia University Mike Rosenthal—Florida International University Phoebe Rouse—Louisiana State University Brenda Santistevan—Salt Lake Community College Catherine Sausville—George Mason University Ingrid Scott—Montgomery College Charlotte Smedberg—University of Tampa Leslie Soltis—Mercyhurst College Katrina Staley—North Carolina Agricultural and Technical State University Sonya Stephens—Florida A&M University John Sumner—University of Tampa Steve Szabo—Eastern Kentucky University Marilyn Toscano—University of Wisconsin, Superior Timothy L. Warkentin—Cloud County Community College Hayat Weiss—Middlesex Community College Larissa Williamson—University of Florida 4IBSZO
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Michael Sullivan, III Joliet Junior College

Global Edition Pearson would like to thank and acknowledge the following people for their work on the Global Edition: r Contributor:
r Sunila Sharma, Delhi University r Reviewers: r Yosum Kurtulmaz, Bilkent University
r Mohd. Hasan Shahid, Jamia Millia Islamia

Feature

Description

Benefit

Page

Every Chapter Opener begins with...

Chapter Opening Article & Project

Each chapter begins with a current article and ends with a related project. The article describes a real situation.

The Article describes a real situation. The 305, 406 Project lets you apply what you learned to solve a related problem.

NEW! Internet Based Projects

The projects allow for the integration of spreadsheet technology that students will need to be a productive member of the workforce.

The projects allow the opportunity for students to collaborate and use mathematics to deal with issues that come up in their lives.

305, 406

Every Section begins with...

Each section begins with a list of objectives. Objectives also appear in the text where the objective is covered.

These focus your studying by emphasizing what’s most important and where to find it.

326

Preparing for this Section

Most sections begin with a list of key concepts to review with page numbers.

Ever forget what you’ve learned? This feature highlights previously learned material to be used in this section. Review it, and you’ll always be prepared to move forward.

326

Now Work the ‘Are You Prepared?’  Problems

Problems that assess whether you have the Not sure you need the Preparing for This 326, 337 prerequisite knowledge for the upcoming Section review? Work the ‘Are You section. Prepared?’ problems. If you get one wrong, you’ll know exactly what you need to review and where to review it!

Now Work

These follow most examples and direct you to a related exercise.

Learning Objectives 2 Sections contain...

PROBLEMS

WARNING

Warnings are provided in the text.

We learn best by doing. You’ll solidify 333, 338 your understanding of examples if you try a similar problem right away, to be sure you understand what you’ve just read. These point out common mistakes and help you to avoid them.

360 232, 347

These represent graphing utility activities to foreshadow a concept or solidify a concept just presented.

You will obtain a deeper and more intuitive understanding of theorems and definition.

These provide alternative descriptions of select definitions and theorems.

Does math ever look foreign to you? This feature translates math into plain English.

CALCULUS

These appear next to information essential for the study of calculus.

Pay attention–if you spend extra time now, 102, 334 you’ll do better later!

SHOWCASE EXAMPLES

These examples provide “how-to” instruction by offering a guided, step-by-step approach to solving a problem.

With each step presented on the left and the mathematics displayed on the right, students can immediately see how each step is employed.

These are examples and problems that require you to build a mathematical model from either a verbal description or data. The homework Model It! problems are marked by purple headings.

351, 379 It is rare for a problem to come in the form, “Solve the following equation”. Rather, the equation must be developed based on an explanation of the problem. These problems require you to develop models that will allow you to describe the problem mathematically and suggest a solution to the problem.

Exploration and Seeing the Concept

In Words

Model It! Examples and Problems

343

236

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Feature

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Benefit

Page

‘Are You Prepared?’ Problems

These assess your retention of the prerequisite material you’ll need. Answers are given at the end of the section exercises. This feature is related to the Preparing for This Section feature.

326, 337 Do you always remember what you’ve learned? Working these problems is the best way to find out. If you get one wrong, you’ll know exactly what you need to review and where to review it!

Concepts and Vocabulary

These short-answer questions, mainly Fillin-the-Blank and True/False items, assess your understanding of key definitions and concepts in the current section.

It is difficult to learn math without knowing the language of mathematics. These problems test your understanding of the formulas and vocabulary.

Skill Building

Correlated to section examples, these problems provide straightforward practice.

It’s important to dig in and develop your skills. These problems provide you with ample practice to do so.

337–339

Mixed Practice

These problems offer comprehensive assessment of the skills learned in the section by asking problems that relate to more than one concept or objective. These problems may also require you to utilize skills learned in previous sections.

Learning mathematics is a building process. Many concepts are interrelated. These problems help you see how mathematics builds on itself and also see how the concepts tie together.

339–340

Applications and Extensions

These problems allow you to apply your skills to real-world problems. They also allow you to extend concepts learned in the section.

You will see that the material learned within the section has many uses in everyday life.

340–342

Discussion and Writing

“Discussion and Writing” problems are colored red. These support class discussion, verbalization of mathematical ideas, and writing and research projects.

To verbalize an idea, or to describe it clearly in writing, shows real understanding. These problems nurture that understanding. Many are challenging but you’ll get out what you put in.

342

NEW! Retain Your Knowledge

These problems allow you to practice content learned earlier in the course.

The ability to remember how to solve all the different problems learned throughout the course is difficult. These help you remember.

342

Now Work

Many examples refer you to a related homework problem. These related problems are marked by a pencil and orange numbers.

If you get stuck while working problems, look for the closest Now Work problem and refer back to the related example to see if it helps.

336, 339

Chapter Review Problems

Every chapter concludes with a comprehensive list of exercises to practice. Use the list of objectives to determine the objective and examples that correspond to the problems.

401–404 Work these problems to verify you understand all the skills and concepts of the chapter. Think of it as a comprehensive review of the chapter.

PROBLEMS

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337

Feature

Description

Benefit

Page

Chapter Review at the end of each chapter contains...

Things to Know

A detailed list of important theorems, formulas, and definitions from the chapter.

Review these and you’ll know the most 399–400 important material in the chapter!

You Should Be able to...

Contains a complete list of objectives by section, examples that illustrate the objective, and practice exercises that test your understanding of the objective.

Do the recommended exercises and you’ll have mastery over the key material. If you get something wrong, review the suggested page numbers and try again.

Review Exercises

These provide comprehensive review Practice makes perfect. These problems 401–404 and practice of key skills, matched to the combine exercises from all sections, giving you a comprehensive review in Learning Objectives for each section. one place.

Chapter Test

About 15-20 problems that can be taken as Be prepared. Take the sample practice 404–405 a Chapter Test. Be sure to take the Chapter test under test conditions. This will get you ready for your instructor’s test. If you Test under test conditions—no notes! get a problem wrong, you can watch the Chapter Test Prep Video.

Cumulative Review

These problem sets appear at the end of each chapter, beginning with Chapter 2. They combine problems from previous chapters, providing an ongoing cumulative review.

These are really important. They will ensure that you are not forgetting anything as you go. These will go a long way toward keeping you primed for the final exam.

405

Chapter Project

The Chapter Project applies to what you’ve learned in the chapter. Additional projects are available on the Instructor’s Resource Center (IRC).

The Project gives you an opportunity to apply what you’ve learned in the chapter to the opening article.If your instructor allows, these make excellent opportunities to work in a group, which is often the best way of learning math.

406

In selected chapters, a web-based project The projects allow the opportunity for students to collaborate and use mathematics is given. to deal with issues that come up in their lives.

406

NEW! Internet Based Projects

401

23

Resources for Success Instructor Resources

Student Resources

Additional resources can be downloaded from www.pearsonglobaleditions.com/sullivan.

Additional resources to help student success:

TestGen®

Chapter Test Prep Videos

®

TestGen (www.pearsonglobaleditions.com/sullivan)

enables instructors to build, edit, print, and administer tests using a computerized bank of questions developed to cover all the objectives of the text.

Students can watch instructors work through step-by-step solutions to all chapter test exercises from the textbook. These are available on YouTube.

PowerPoint® Lecture Slides Fully editable slides that correlate to the textbook.

Instructor Solutions Manual Includes fully worked solutions to all textbook exercises.

Mini Lecture Notes Includes additional examples and helpful teaching tips, by section.

Online Chapter Projects Additional projects that let students apply what was learned in the chapter.

24

Algebra Review Four Chapters of Intermediate Algebra review. Perfect for a slower-paced course or for individual review.

Applications Index Acoustics

Astronomy

amplifying sound, 403 loudness of sound, 355, 405 loudspeaker, 607 tuning fork, 607 whispering galleries, 708

angle of elevation of Sun, 573 distances of planets from Sun, 861 planetary orbits, 707–08, 711 Earth, 710 Jupiter, 710 Mars, 710 Mercury, 737 Pluto, 711 radius of the Moon, 437

Aerodynamics modeling aircraft motion, 689–90

Aeronautics Challenger disaster, 390

Aviation

Agriculture

modeling aircraft motion, 689–90 orbital launches, 767

farm management, 843 farm workers in U.S., 390–91 field enclosure, 828 grazing area for cow, 598 minimizing cost, 843 removing stump, 661

Biology

bearing of aircraft, 571, 574 flight time and ticket price, 167 frequent flyer miles, 584–85 holding pattern, 521 intersection point of two planes, 134–35 parking at O’Hare International Airport, 118 revising a flight plan, 592 speed and direction of aircraft, 656, 660

age versus total cholesterol, 398 alcohol and driving, 351, 356 bacterial growth, 382–83, 395 E-coli, 109, 152–53 blood pressure, 521 blood types, 899 bone length, 219 cricket chirping, 206 gestation period, 215 healing of wounds, 340, 355 maternal age versus Down syndrome, 168 muscle force, 661 yeast biomass as function of time, 394–95

Archaeology

Business

age of ancient tools, 384 age of fossil, 388, 389 age of tree, 389 date of prehistoric man’s death, 403

advertising, 167, 219 automobile production, 312, 783 blending coffee, A78 car rentals, 159 checkout lines, 918 clothing store, 920 cookie orders, 848 cost of can, 286–87, 289 of commodity, 313 of manufacturing, 243, 297, 837, A13, A78 marginal, 191, 218 minimizing, 218, 843, 848 of production, 108, 313, 810–11, 848 of theater ticket per student, 297 of transporting goods, 119 weekly, 239 cost equation, 63, 142 cost function, 160 average, 92 demand for candy, 142 for jeans, 167 for PCs, 396 demand equation, 218, 303 discounts, 313

Air travel

Architecture brick staircase, 867, 891 Burj Khalifa building, A14–A15 floor design, 865–66, 891 football stadium seating, 867 mosaic design, 867, 891 Norman window, 204, A20 parabolic arch, 204 racetrack design, 710 special window, 204 stadium construction, 867 window design, 204

Area. See also Geometry of Bermuda Triangle, 598 under a curve, 507 of isosceles triangle, 552 of sector of circle, 415–16, 419 of segment of circle, 610

Art fine decorative pieces, 437

drive-thru rate at Burger King, 336 at Citibank, 340, 355 earnings of young adults, 754 equipment depreciation, 877 ethanol production, 395 expense computation, A79 Jiffy Lube’s car arrival rate, 340–41, 355 managing a meat market, 844 milk production, 396 mixing candy, A78 mixing nuts, A78 new-car markup, A88 orange juice production, 783 personal computer price and demand, 396 precision ball bearings, A13 presale order, 767 product design, 844 production scheduling, 843 product promotion, 64 profit, 811 cigar company, 132 maximizing, 841–42, 843–44 profit function, 88, 191–92 rate of return on, 379 restaurant management, 767 revenue, 191, 195-64, A78 advertising and, 167 airline, 844 from calculator sales, 181 of clothing store, 799–800 daily, 192 from digital music, 132 instantaneous rate of change of, 950, 958 maximizing, 191–92, 203 monthly, 191–92 from seating, 878 theater, 768 revenue equation, 142 salary, 867 gross, 87 increases in, 877, 891 sales commission on, A88 of movie theater ticket, 755, 759–60, 767 net, 40 salvage value, 403 straight-line depreciation, 156–57, 160 supply and demand, 157–58, 160 tax, 297 toy truck manufacturing, 837 transporting goods, 837 truck rentals, 63, 161 unemployment, 921 wages of car salesperson, 63

25

26

Applications Index

Calculus

Computers and computing

area under a curve, 507 carrying a ladder around a corner, 522 maximizing rain gutter construction, 552 projectile motion, 522 Simpson’s rule, 204

Dell PCs, 396 graphics, 659, 812–13 households without personal computers, 390 JPEG image compression, 812 laser printers, A79 LCD monitors, 218 social media, 812 three-click rule for web design, 812 website map, 811 Word users, 390

Carpentry. See also Construction pitch, 65

Chemistry alpha particles, 723 decomposition reactions, 389 drug concentration, 288 gas laws, 142 pH, 354 purity of gold, A79 radioactive decay, 388, 389, 396, 404 radioactivity from Chernobyl, 389 reactions, 204 salt solutions, A79 sugar molecules, A80 volume of gas, A88

Combinatorics airport codes, 901 binary codes, 920 birthday permutations, 903, 907, 914–15, 918, 920 blouses and skirts combinations, 899 book arrangements, 907 box stacking, 907 code formation, 907 combination locks, 908 committee formation, 905, 907, 908, 920 Senate committees, 908 flag arrangement, 906, 920 letter codes, 901, 921 letter combinations, 920 license plate possibilities, 907, 920, 921 lining people up, 902, 907 number formation, 899, 907, 908, 921 objects selection, 908 seating arrangements, 920 shirts and ties combinations, 899 telephone numbers, 920 two-symbol codewords, 898 word formation, 905–06, 907, 921

Communications cell phone plan, 74 cell phone service, 118, 148, 161 cell phone towers, 397 installing cable TV, 137 long distance comparing phone companies, 218 international call plan, 161 phone charges, 160 satellite dish, 698, 700 spreading of rumors, 340, 355 Touch-Tone phones, 556–57, 608

Construction of box, 828 closed, 147 open, 137 of brick staircase, 891 of can, 301 of coffee can, A80 of cylindrical tube, 828 of enclosures around garden, A79 around pond, A79 maximizing area of, 198–99, 203 of fencing, 198–99, 203, 828 minimum cost for, 289 of flashlight, 700 of headlight, 700 of highway, 574, 585, 611 installing cable TV, 137 of open box, 179 pitch of roof, 575 of rain gutter, 204, 430, 552, 565–66 of ramp, 584 access ramp, 64 of rectangular field enclosure, 203 sidewalk area, 438 of stadium, 204, 867 of steel drum, 289 of swimming pool, A21 of swing set, 593 of tent, 597 TV dish, 700 vent pipe installation, 710

Crime income and, 398

Cryptography matrices in, 811

Decorating Christmas tree, A15–A16

Demographics birth rate mother’s age and, 206 of unmarried women, 191 diversity index, 354–55 life expectancy, A88 marital status, 900 mosquito colony growth, 388–89

population. See Population rabbit colony growth, 860

Design of awning, 585–86 of box with minimum surface area, 289 of fine decorative pieces, 437 of Little League Field, 421 of water sprinkler, 419

Direction of aircraft, 660 compass heading, 660 for crossing a river, 660, 661 of fireworks display, 722 of lightning strikes, 722 of motorboat, 660 of swimmer, 688

Distance Bermuda Triangle, A21 bicycle riding, 97 from Chicago to Honolulu, 508 circumference of Earth, 421 between cities, 414–15, 419 between Earth and Mercury, 586 between Earth and Venus, 586 from Earth to a star, 573–74 of explosion, 723 height of aircraft, 584, 586 of bouncing ball, 877, 891 of bridge, 584 of building, 574, 575, 576 of cloud, 569 of Eiffel Tower, 573 of embankment, 574 of Ferris Wheel rider, 521 of Great Pyramid of Cheops, 586, A21 of helicopter, 582, 611 of hot-air balloon, 574 of Lincoln’s caricature on Mt. Rushmore, 574 of monument, 574 of mountain, 581, 584 of statue on building, 569–70 of tree, 437 of Willis Tower, 574 from home, 97 from Honolulu to Melbourne, Australia, 508 of hot-air balloon to airport, 612 from intersection, 40 from intersection, 40, 136 length of guy wire, 592 of mountain trail, 574 of ski lift, 584 limiting magnitude of telescope, 403 to the Moon, 584 nautical miles, 420 pendulum swings, 873, 877 to plateau, 573

Applications Index

across pond, 573 range of airplane, A80 reach of ladder, 573 of rotating beacon, 474–75 at sea, 581–82, 585 of ship from shore, 573 to shore, 585, 610 between skyscrapers, 575 sound to measure, A71 to tower, 586 traveled by wheel, A20 between two moving vehicles, 40 toward intersection, 136 between two objects, 574 viewing, 437 visibility of Gibb’s Hill Lighthouse beam, 570–71, 576, A22 visual, A21 walking, 97 of water tower to building, 900 width of gorge, 572 of Mississippi River, 575 of river, 568, 610

Economics Consumer Price Index (CPI), 380 demand equations, 303 earnings of young adults, 754 federal deficit, 379, 403 income versus crime rate, 398 inflation, 379–80 IS-LM model in, 768 marginal propensity to consume, 878 multiplier, 878 participation rate, 88 personal computer price and demand, 396 poverty rates, 301 poverty threshold, 40 relative income of child, 811 unemployment, 921

Education admission probabilities, 921 age distribution of community college, 921 college costs, 379, 877 college tuition and fees, 810 degrees awarded, 897 doctoral degrees awarded, 918 faculty composition, 918 funding a college education, 403 grade computation, A88 grade-point average and video games, 167 IQ tests, A88 learning curve, 341, 355 maximum level achieved, 850 multiple-choice test, 907 Spring break, 843 student loan, 148 interest on, 810 true/false test, 907

Electricity alternating current (ac), 491, 542 alternating current (ac) circuits, 466, 484 alternating current (ac) generators, 466 charging a capacitor, 608 cost of, 116–17 current in RC circuit, 341 current in RL circuit, 341, 355 impedance, A95 Kirchhoff’s Rules, 768, 783 parallel circuits, A95 resistance in, 274 rates for, 63–64, A88 resistance, 142, 143, 274, A52, A54 due to a conductor, 148 voltage foreign, A13 U.S., A13

Electronics loudspeakers, 607 microphones, 50 sawtooth curve, 552, 608

Energy ethanol production, 395 heat loss through wall, 140 through window, 147 nuclear power plant, 722–23 solar, 50, 668, 700 thermostat control, 131–32

Engineering bridges clearance, 466 Golden Gate, 200–01 parabolic arch, 218, 700–01 semielliptical arch, 710, 751 suspension, 204, 700 crushing load, A71 drive wheel, 611 Gateway Arch (St. Louis), 701 grade of road, 65 horsepower, 142 lean of Leaning Tower of Pisa, 585 maximum weight supportable by pine, 139 moment of inertia, 557 piston engines, 436–37 product of inertia, 552 road system, 624 robotic arm, 678 rods and pistons, 593 rod tolerance, 215 safe load for a beam, 143 searchlight, 530, 700, 751 whispering galleries, 710

Entertainment banquet hall rental, 843 cable subscribers, 398 Demon Roller Coaster customer rate, 341

27

movie theater, 507 theater revenues, 768

Environment endangered species population, 340 lake pollution control laws, 860 oil leakage, 312

Exercise and fitness. See also Sports heartbeats during exercise, 153–54 for weight loss, A88

Finance. See also Investment(s) balancing a checkbook, A13 bills in wallet, 921 calculator sales revenue, 181 clothes shopping, 849 college costs, 379, 877 computer system purchase, 379 cost of car rental, 119 of driving a car, 63 of electricity, 116–17 of fast food, 767 minimizing, 218, 289 of natural gas, 118–19 of RV rental, 220 of tattoo, 685 of trans-Atlantic travel, 87, 95–96 of triangular lot, 597 cost equation, 142 cost function, 160 cost minimization, 191 credit cards balance on, 820 debt, 860 interest on, 379 minimum payments for, 119–20 payment, 860 demand equation, 203, 220 depreciation, 340, 399 of car, 371, 406 division of money, A73–A74, A78 electricity rates, 63–64 federal income tax, A88 financial planning, 767, 780, 783–84, 834–35, 836, 838, 844, A73–A74, A78 foreign exchange, 313 future value of money, 243 gross salary, 87 international call plan, 161 life cycle hypothesis, 205 loans, A78 car, 860 interest on, 148, 810, A73 repayment of, 379 student, 810 mortgages fees, 119 interest rates on, 379 payments, 138, 141, 147 second, 379 national debt, 108–09 price appreciation of homes, 379

28

Applications Index

prices demand vs., 218 of fast food, 769 for soda and hot dog combinations, 161 refunds, 767 rents and square footage, 205 revenue equation, 142 revenue maximization, 191, 197–98 rich man’s promise, 878 salary calculation, 313 salary options, 878–79 saving for a car, 379 for a home, 877 savings accounts interest, 379 sinking fund, 877 taxes, 160 e-filing returns, 119 federal income, 119, 325 luxury, 160 used-car purchase, 379 water bills, A88

Food and nutrition animal, 844 calories in fast foods, 76–77 candy, 166 colored candies, 909–10, 921 cooler contents, 921 cooling time of pizza, 389 fast food, 767, 769 Girl Scout cookies, 918 hospital diet, 768, 783 “light” foods, A88 milk production, 396 number of possible meals, 897–98 pig roasts, 390 raisins, 166 warming time of Beer stein, 389

Forestry wood product classification, 388

Games die rolling, 910–11, 912, 921 grains of wheat on a chess board, 878 Powerball, 921

Gardens and gardening. See also Landscaping enclosure for, A79

Geography area of Bermuda Triangle, 598 area of lake, 597, 611 grade of a mountain trail, 829 inclination of hill, 669 inclination of mountain trail, 568–69, 610 width of a river, 568

Geology earthquakes, 355–56

Geometry

Health. See also Medicine

angle between two lines, 542 balloon volume, 312 circle area of, 597, A78 area of sector of, 415–16, 419 circumference of, A7, A12, A78 equation of, 794 inscribed, 135–36, 599 length of chord of, 593 radius of, 827 collinear points, 793 cone volume, 142, 313 cube length of edge of, 257 surface area of, A13 volume of, A13 cylinder inscribing in cone, 137 inscribing in sphere, 136 volume of, 142, 313 Descartes’s method of equal roots, 828 equation of line, 793 ladder angle, 612 polygon area of, 794 number of sides of, 179 quadrilateral area, 612 rectangle area of, 87, 134, 218, 420, 711, A12 dimensions of, 218, 827 inscribed in semicircle, 136, 553 perimeter of, A12 semicircle inscribed in, 136 semicircle area, 597, 612 sphere surface area of, A12 volume of, A12 square area of, A20, A78 perimeter of, A78 surface area of balloon, 312 of cube, A13 of sphere, A12 triangle area of, 597, 612, 794, A12 circumscribing, 587 equilateral, A12 inscribed in circle, 136 isosceles, 87, 827, 828 Pascal’s, 860 perimeter of, A12 right, 572 sides of, 613 volume of paralleliped, 684

breast cancer survival rate, 396 cigarette use among teens, 64 expenditures on, 88 ideal body weight, 325 life cycle hypothesis, 205 pancreatic cancer survival rate, 340

Government federal deficit, 403 federal income tax, 88, 119, 325, A88 first-class mail charge, 120 national debt, 108–09 stimulus package (2009), 379

Home improvement. See also Construction painting a house, 769 painting a room, 475

Investment(s) annuity, 874–75, 877 in bonds, 844 EE Series, 379 Treasuries, 783, 784, 834–35, 836, 838 Treasury notes vs. Treasury bonds, 780 zero-coupon, 376, 380 in CDs, 375, 844 compound interest on, 372–75, 379, 467, 929 diversified, 768–69 division among instruments, A78 doubling of, 377, 380 in fixed-income securities, 844 433(K), 877, 891 growth rate for, 379 IRA, 379, 874–75, 877 in mutual fund, 392–93 return on, 379, 843, 844 in stock appreciation, 379 beta, 150, 221–22 NASDAQ stocks, 907 NYSE stocks, 907 portfolios of, 900 price of, 878 time to reach goal, 379, 380 tripling of, 377, 380

Landscaping. See also Gardens and gardening pond enclosure, 218 removing stump, 661 tree cutting, 584, 783 watering lawn, 419

Law and law enforcement motor vehicle thefts, 918 violent crimes, 88

Leisure and recreation cable TV, 137 centrifugal force ride, 419 community skating rink, 148 Ferris wheel, 71, 420, 521, 586, 607 gondola, 419 swing displacement, 613 video games and grade-point average, 167

Applications Index

Marketing. See Business Measurement optical methods of, 530 of rainfall, 668

Mechanics. See Physics Medicine. See also Health blood pressure, 521 breast cancer survival rate, 396 drug concentration, 108, 288 drug medication, 340, 355 healing of wounds, 340, 355 pancreatic cancer, 340 spreading of disease, 404

Meteorology weather balloon height and atmospheric pressure, 393–94

Miscellaneous bending wire, 828 biorhythms, 466 carrying a ladder around a corner, 474, 522 citrus ladders, 867 cross-sectional area of beam, 87–88, 95 curve fitting, 768, 782, 847 drafting error, 40 pet ownership, 918 rescue at sea, 581–82 rooms in housing units, 87 surface area of balloon, 312 surveillance satellites, 575–76 volume of balloon, 312 window dimensions, 179 wire enclosure area, 136

Mixtures. See also Chemistry blending coffees, 837, 848, A74–A75, A78 blending teas, A78 cement, A80 mixed nuts, 767, 837, 848, A78 mixing candy, A78 solution, 767 water and antifreeze, A79

Motion. See also Physics catching a train, 752 on a circle, 419 of Ferris Wheel rider, 521 of golf ball, 95, 522 minute hand of clock, 418 objects approaching intersection, 748 of pendulum, 608 revolutions of circular disk, A20 simulating, 742–43 tortoise and the hare race, 827 uniform, 136, 748, A75–A77, A78–A79

Motor vehicles

Pets

alcohol and driving, 351, 356 approaching intersection, 748 automobile production, 312, 783 automobile theft, 918 average car speed, A80 brake repair with tune-up, 921 braking load, 669, 688 crankshafts, 585 depreciation of, 305, 371, 399, 406 distance between, 437 with Global Positioning System (GPS), 403 loans for, 860 miles per gallon, 205–06 new-car markup, A88 RV rental cost, 220 spin balancing tires, 420 stopping distance, 88, 191, 325 used-car purchase, 379 windshield wiper, 419

dog roaming area, 420

Music iPod storage capacity for, 161 revenues from, 132

Navigation avoiding a tropical storm, 592 bearing, 571, 591 of aircraft, 571, 574 of ship, 574, 611 charting a course, 661 commercial, 584–85 compass heading, 660 correct direction for crossing river, 660 error in correcting, 589–90, 611 time lost due to, 585 rescue at sea, 581–82, 584 revising a flight plan, 592

Oceanography tides, 485

Optics angle of incidence, 522–23 angle of refraction, 522–23 bending light, 523 index of refraction, 522–23 intensity of light, 142 laser beam, 573 laser projection, 552 lensmaker’s equation, A54 light obliterated through glass, 340 magnitude of telescope, 403 measurements using, 530 mirrors, 723 reflecting telescope, 700

Pediatrics height vs. head circumference, 196, 325

29

Pharmacy vitamin intake, 768, 784

Photography camera distance for full-body shot, 574

Physics angle of elevation of Sun, 573 bouncing balls, 891 braking load, 669, 688 damped motion, 603–04 direction of aircraft, 660 Doppler effect, 289 falling objects, 141 force, 659, A78 muscle, 661 resultant, 659 of wind on a window, 140, 142 gravity, 274, 297 on Earth, 87, 325 on Jupiter, 87 harmonic motion, 602, 607, 611 heat loss through a wall, 140 heat transfer, 522 horsepower, 142 inclination of mountain trail, 568–69 inclination of ramp, 661 intensity of light, 142 kinetic energy, 143, A78 maximum weight supportable by pine, 139 moment of inertia, 557 motion of object, 602, 744 Newton’s law, 141 pendulum motion, 419, 607, 608, 873, A62, A71 period, 132, 326 simple pendulum, 141 pressure, 142, A78 product of inertia, 552 projectile motion, 181, 199–200, 203–04, 436, 437–38, 522, 547, 552, 557, 741–42, 747, 748, 749, 752 artillery, 513 hit object, 748 thrown object, 747 rate of change average, 960 instantaneous, 946, 949 safe load for a beam, 143 simulating motion, 742–43 sound to measure distance, A71 static equilibrium, 657, 660, 661, 688, 689 static friction, 661 stopping distance, 191 stress of materials, 143 stretching a spring, 142 tension, 657, 660, 688, 689, 883 thrown object, 195, 205, 655, 947–48, 949 truck pulls, 660

30

Applications Index

uniform motion, 136, 748, 752, A75–A77, A78–A79 velocity down inclined planes, A62 vertically propelled object, 179, 195 vibrating string, 142 weight, 142, 147, 656, 660 effect of elevation on, 96 work, 678, A78

Play wagon pulling, 659, 666, 667

Population. See also Demographics bacterial, 388, 389, 390, 395 decline in, 389 of divorced people, 201–02 E-coli growth, 109, 152–53 of endangered species, 340, 390 of fruit fly, 387 as function of age, 87 growth in, 388, 389 insect, 274, 388 of rabbit colony, 860 of trout, 860 of United States, 370, 396–97, 893 of world, 371, 397, 403, 851 future of, 960

Probability checkout lines, 918 classroom composition, 918 “Deal or No Deal” TV show, 894 exponential, 336, 340, 355 gender composition of 35-child family, 912 household annual income, 918 Monty Hall Game, 922 Poisson, 341 “Price is Right” games, 918 of shared birthdays in room of n people, 391 tossing a fair coin, 909, 911 of winning a lottery, 919

Psychometrics IQ tests, 215

Pyrotechnics fireworks display, 722

Rate. See also Speed of car, 419 catching a bus, 747–48 catching a train, 747 current of stream, 768 of emptying oil tankers, A80 a pool, A80 a tub, A80 to keep up with the Sun, 420 miles per gallon, 205–06 revolutions per minute of bicycle wheels, 419 of pulleys, 421

speed average, A80 of current, A78–A79 of motorboat, A78–A79 of moving walkways, A79 of plane, A80

Real estate commission schedule, A88 cost of triangular lot, 597 ground area covered by building, 597–98 price appreciation of homes, 379 rents and square footage, 205 valuing a home, 33, 73

Recreation bungee jumping, 297 Demon Roller Coaster customer rate, 341 online gambling, 918

Security

basketball, 908 free throws, 95, 575 granny shots, 95 biathlon, A80 bungee jumping, 297 exacta betting, 921 football, 710, 752, 908, A79 golf, 918 distance to the green, 591 putts, 398–99 sand bunkers, 513 hammer throw, 492 Olympic heroes, A80 pool shots, 576 races, 825, 827–28, A80 relay runners, 920 swimming, 613, 688 tennis, A79

Statistics. See Probability

security cameras, 573

Surveys

Seismology

of appliance purchases, 899 data analysis, 896, 899 stock portfolios, 899 of summer session attendance, 899 of TV sets in a house, 918

calibrating instruments, 751

Sequences. See also Combinatorics ceramic tile floor design, 865–66 Drury Lane Theater, 867 Fibonacci, 860 football stadium seating, 867 seats in amphitheater, 867

Speed of aircraft, 660 angular, 419, 491 of current, 420, 848 of cyclists moving in opposite directions, A80 as function of time, 97, 136 of glider, 610 ground, 660 instantaneous of ball, 947–48, 949, 958 on the Moon, 949–50 linear, 416–17 on Earth, 419, 420 of Moon, 420 of motorboat, 660, A76–A77 revolutions per minute of pulley, 420 of rotation of lighthouse beacons, 491 of swimmer, 688 of truck, 573 of wheel pulling cable cars, 420 wind, 767

Sports baseball, 748–49, 908, 920 diamond, 39 dimensions of home plate, 597 field, 592 Little League, 40, 421 on-base percentage, 162–63 stadium, 592 World Series, 908

Technology. See also Computers and computing Blu-ray drive, 419 DVD drive, 419 iPod storage capacity for music, 161

Temperature of air parcel, 867 body, A13 conversion of, 313, 325 cooling time of pizza, 389 cricket chirping and, 206 measuring, 64, 132 after midnight, 243 monthly, 484–85, 491 of portable heater, 403 relationship between scales, 132 sinusoidal function from, 480–81 of skillet, 403 warming time of Beer stein, 389 wind chill factor, 404

Tests and testing IQ, A88

Time for Beer stein to warm, 389 for block to slide down inclined plane, 436 Ferris Wheel rider height as function of, 521 to go from an island to a town, 137 hours of daylight, 482–83, 485–86, 506–07 for pizza to cool, 389 of sunrise, 420, 507 of trip, 437, 451

Applications Index

Transportation. See also Air travel; Motor vehicles de-icing salt, 513 Falls Incline Railway, 574

Travel. See also Air travel; Navigation bearing, 611 drivers stopped by the police, 405 driving to school, 142 parking at O’Hare International Airport, 118

Volume of box, 669 of gasoline in tank, A62 of water in cone, 137

artillery, 513

weather satellites, 71 wind chill, 120, 404

Weather

Work, 667, 678

atmospheric pressure, 340, 355 avoiding a tropical storm, 592 cooling air, 867 hurricanes, 242, 484 lightning strikes, 719–20, 722 rainfall measurement, 668 relative humidity, 341

computing, 667, 668, 688 constant rate jobs, 848 pulling a wagon, 666, 667 ramp angle, 669 wheel barrow push, 659 working together to do a job, A77, A79

Weapons

31

Foundations: A Prelude to Functions How to Value a House

F

Two things to consider in valuing a home are, first, how does it compare to similar homes that have sold recently? Is the asking price fair? And second, what value do you place on the advertised features and amenities? Yes, other people might value them highly, but do you? Zestimate home valuation, RealEstateABC.com, and Reply.com are among the many algorithmic (generated by a computer model) starting points in figuring out the value of a home. They show you how the home is priced relative to other homes in the area, but you need to add in all the things that only someone who has seen the house knows. You can do that using My Estimator, and then you create your own estimate and see how it stacks up against the asking price.

Looking at “Comps” Knowing whether an asking price is fair will be important when you’re ready to make an offer on a house. It will be even more important when your mortgage lender hires an appraiser to determine whether the house is worth the loan you’re after. Check with your agent, Zillow.com, propertyshark.com, or other websites to see recent sales of homes in the area that are similar, or comparable, to what you’re looking for. Print them out and keep these “comps” in a three-ring binder; you’ll be referring to them quite a bit. Note that “recent sales” usually means within the last six months. A sales price from a year ago may bear little or no relation to what is going on in your area right now. In fact, some lenders will not accept comps older than three months. Market activity also determines how easy or difficult it is to find accurate comps. In a “hot” or busy market, with sales happening all the time, you’re likely to have lots of comps to choose from. In a less active market, finding reasonable comps becomes harder. And if the home you’re looking at has special design features, finding a comparable property is harder still. It’s also necessary to know what’s going on in a given sub-segment. Maybe large, high-end homes are selling like hotcakes, but owners of smaller houses are staying put, or vice versa. Source: http://realestate.yahoo.com/Homevalues/How_to_Value_a_House.html

—See the Internet-based Chapter Project—

Here we connect algebra and geometry using the rectangular coordinate system. In the 1600s, algebra had developed to the point that René Descartes (1596–1650) and Pierre de Fermat (1601–1665) were able to use rectangular coordinates to translate geometry problems into algebra problems, and vice versa. This allowed both geometers and algebraists to gain new insights into their subjects, which had been thought to be separate but now were seen as connected.

Intercepts; Symmetry F.3 Lines F.4 Circles

Chapter Project

33

34

CHAPTER F  Foundations: A Prelude to Functions

F.1 The Distance and Midpoint Formulas PREPARING FOR THIS SECTION  Before getting started, review the following: r
"MHFCSB
&TTFOUJBMT
"QQFOEJY
"
4FDUJPO
"
pp. A1–A10)

r
(FPNFUSZ
&TTFOUJBMT
"QQFOEJY
"
4FDUJPO
"
pp. A13–A19)

Now Work the ‘Are You Prepared?’ problems on page 38.

OBJECTIVES 1 Use the Distance Formula (p. 35) 2 Use the Midpoint Formula (p. 37)

Rectangular Coordinates

Figure 1 y 4 2 –4

–2

O

2

4

x

–2 –4

Figure 2 y 4 3 (–3, 1) 1 –4 3 (–2, –3)

3

(3, 2) 2

O 3

x 4 2 (3, –2)

2

Figure 3 y Quadrant II x < 0, y > 0

Quadrant I x > 0, y > 0

Quadrant III x < 0, y < 0

Quadrant IV x > 0, y < 0

x

A point on the real number line is located by a single real number called the coordinate of the point. For work in a two-dimensional plane, points are located by using two numbers. Begin with two real number lines located in the same plane: one horizontal and the other vertical. The horizontal line is called the x-axis, the vertical line the y-axis, and the point of intersection the origin O. See Figure 1. Assign coordinates to every point on these number lines using a convenient scale. Recall that the scale of a number line is the distance between 0 and 1. In mathematics, we usually use the same scale on each axis, but in applications, a different scale is often used. The origin O has a value of 0 on both the x-axis and the y-axis. Points on the x-axis to the right of O are associated with positive real numbers, and those to the left of O are associated with negative real numbers. Points on the y-axis above O are associated with positive real numbers, and those below O are associated with negative real numbers. In Figure 1, the x-axis and y-axis are labeled as x and y, respectively, and an arrow at the end of each axis is used to denote the positive direction. The coordinate system described here is called a rectangular or Cartesian* coordinate system. The plane formed by the x-axis and y-axis is sometimes called the xy-plane, and the x-axis and y-axis are referred to as the coordinate axes. Any point P in the xy-plane can be located by using an ordered pair 1x, y2 of real numbers. Let x denote the signed distance of P from the y-axis (signed means that, if P is to the right of the y-axis, then x 7 0, and if P is to the left of the y-axis, then x 6 0); and let y denote the signed distance of P from the x-axis. The ordered pair 1x, y2 , also called the coordinates of P, then gives us enough information to locate the point P in the plane. For example, to locate the point whose coordinates are 1 - 3, 12 , go 3 units along the x-axis to the left of O and then go straight up 1 unit. We plot this point by placing a dot at this location. See Figure 2, in which the points with coordinates 1 - 3, 12 , 1 - 2, - 32 , 13, - 22 , and 13, 22 are plotted. The origin has coordinates 10, 02 . Any point on the x-axis has coordinates of the form 1x, 02 , and any point on the y-axis has coordinates of the form 10, y2 . If 1x, y2 are the coordinates of a point P, then x is called the x-coordinate, or abscissa, of P and y is the y-coordinate, or ordinate, of P. We identify the point P by its coordinates 1x, y2 by writing P = 1x, y2 . Usually, we will simply say, “the point 1x, y2 ” rather than “the point whose coordinates are 1x, y2 .” The coordinate axes divide the xy-plane into four sections called quadrants, as shown in Figure 3. In quadrant I, both the x-coordinate and the y-coordinate of all points are positive; in quadrant II, x is negative and y is positive; in quadrant III, both x and y are negative; and in quadrant IV, x is positive and y is negative. Points on the coordinate axes belong to no quadrant.

Now Work

PROBLEM

11

*Named after René Descartes (1596–1650), a French mathematician, philosopher, and theologian.

SECTION F.1  The Distance and Midpoint Formulas

Figure 4

35

COMMENT On a graphing calculator, you can set the scale on each axis. Once this has been done, you obtain the viewing rectangle. See Figure 4 for a typical viewing rectangle. You should now read Section B.1, The Viewing Rectangle, in Appendix B. ■

1 Use the Distance Formula If the same units of measurement (such as inches, centimeters, and so on) are used for both the x-axis and y-axis, then all distances in the xy-plane can be measured using this unit of measurement.

EXAMPLE 1 Solution

Finding the Distance between Two Points

Find the distance d between the points 11, 32 and 15, 62 .

First plot the points 11, 32 and 15, 62 and connect them with a straight line. See Figure 5(a). To find the length d, begin by drawing a horizontal line from 11, 32 to 15, 32 and a vertical line from 15, 32 to 15, 62 , forming a right triangle, as shown in Figure 5(b). One leg of the triangle is of length 4 (since 0 5 - 1 0 = 4), and the other is of length 3 (since 0 6 - 3 0 = 3). By the Pythagorean Theorem, the square of the distance d that we seek is d 2 = 42 + 32 = 16 + 9 = 25 d = 225 = 5

Figure 5

y 6

y 6

(5, 6)

(5, 6) d

d 3

3 (1, 3)

0

(1, 3) 4 (5, 3)

0

6 x

3

3

3

6 x

(b)

(a)

r

The distance formula provides a straightforward method for computing the distance between two points.

THEOREM

Distance Formula The distance between two points P1 = 1x1 , y1 2 and P2 = 1x2 , y2 2 , denoted by d 1P1 , P2 2 , is

In Words To compute the distance between two points, find the difference of the x-coordinates, square it, and add this to the square of the difference of the y-coordinates. The square root of this sum is the distance.

Figure 6

d1P1 , P2 2 = 2 1x2 - x1 2 2 + 1y2 - y1 2 2

(1)

Proof of the Distance Formula Let 1x1 , y1 2 denote the coordinates of point P1 and let 1x2 , y2 2 denote the coordinates of point P2 . Assume that the line joining P1 and P2 is neither horizontal nor vertical. Refer to Figure 6(a). The coordinates of P3 are 1x2 , y1 2 . The horizontal distance from P1 to P3 is the absolute value of the difference of the x-coordinates, 0 x2 - x1 0 . The vertical distance from P3 to P2 is the

y

y P2  (x2, y2)

y2 y1

P1  (x1, y1)

P3  (x2, y1) x2

x1 (a)

x

P2  (x2, y2)

y2

d(P1, P2)

y1 P1  (x1, y1)

⏐y2  y1⏐

⏐x2  x1⏐ x2

x1 (b)

P3  (x2, y1) x

36

CHAPTER F  Foundations: A Prelude to Functions

absolute value of the difference of the y-coordinates, 0 y2 - y1 0 . See Figure 6(b). The distance d 1P1 , P2 2 that we seek is the length of the hypotenuse of the right triangle, so, by the Pythagorean Theorem, it follows that 3 d 1P1 , P2 2 4 2 = 0 x2 - x1 0 2 + 0 y2 - y1 0 2 = 1x2 - x1 2 2 + 1y2 - y1 2 2 d 1P1 , P2 2 = 2 1x2 - x1 2 2 + 1y2 - y1 2 2

Now, if the line joining P1 and P2 is horizontal, then the y-coordinate of P1 equals the y-coordinate of P2; that is, y1 = y2 . Refer to Figure 7(a). In this case, the distance formula (1) still works, because, for y1 = y2 , it reduces to d 1P1 , P2 2 = 2 1x2 - x1 2 2 + 02 = 2 1x2 - x1 2 2 = 0 x2 - x1 0

Figure 7

y

y y2 P1  (x1, y1)

y1

d (P1, P2)

P2  (x2, y1)

⏐y2  y1⏐ d (P1, P2) P1  (x1, y1)

y1

⏐x2  x1⏐ x2

x1

P2  (x1, y2)

x1

x

(a)

x

(b)

A similar argument holds if the line joining P1 and P2 is vertical. See Figure 7(b).

EX A MPL E 2

Solution

■

Using the Distance Formula

Find the distance d between the points 1 - 3, 52 and (3, 2). Use the distance formula, equation (1), with P1 = (x1, y1) = ( - 3, 5) and P2 = (x2, y2) = (3, 2). Then d = 2[3 - 1 - 32]2 + 12 - 52 2 = = = =

Now Work

PROBLEMS

15

AND

262 + 1 - 32 2 236 + 9 245 325 ≈ 6.71

r

19

The distance between two points P1 = 1x1, y1 2 and P2 = 1x2, y2 2 is never a negative number. Furthermore, the distance between two points is 0 only when the points are identical—that is, when x1 = x2 and y1 = y2. Also, because 1x2 - x1 2 2 = 1x1 - x2 2 2 and 1y2 - y1 2 2 = 1y1 - y2 2 2, it makes no difference whether the distance is computed from P1 to P2 or from P2 to P1; that is, d 1P1 , P2 2 = d 1P2 , P1 2 . The introduction to this chapter mentioned that rectangular coordinates enable us to translate geometry problems into algebra problems, and vice versa. The next example shows how algebra (the distance formula) can be used to solve geometry problems.

EX A MPL E 3

Using Algebra to Solve Geometry Problems

Consider the three points A = 1 - 2, 12 , B = 12, 32 , and C = 13, 12 . (a) (b) (c) (d)

Plot each point and form the triangle ABC. Find the length of each side of the triangle. Verify that the triangle is a right triangle. Find the area of the triangle.

SECTION F.1  The Distance and Midpoint Formulas

Solution

37

(a) Figure 8 shows the points A, B, C and the triangle ABC. (b) To find the length of each side of the triangle, use the distance formula, equation (1).

d 1A, B2 = 2 3 2 - 1 - 22 4 2 + 13 - 12 2 = 216 + 4 = 220 = 225 d 1B, C2 = 2 13 - 22 2 + 11 - 32 2 = 21 + 4 = 25 d 1A, C2 = 2 3 3 - 1 - 22 4 2 + 11 - 12 2 = 225 + 0 = 5

(c) If the triangle is a right triangle, then the sum of the squares of the lengths of two of the sides will equal the square of the length of the third side. (Why is this sufficient?) Looking at Figure 8, it seems reasonable to conjecture that the right angle is at vertex B. We shall check to see whether

Figure 8 y B = (2, 3)

3 A = (–2, 1)

3 d 1A, B2 4 2 + 3 d 1B, C2 4 2 = 3 d 1A, C2 4 2

C = (3, 1)

–3

Using the results from part (b) yields

x

3

3 d 1A, B2 4 2 + 3 d 1B, C2 4 2 = 1 225 2 + 1 25 2 2 = 20 + 5 = 25 = 3 d 1A, C2 4 2 2

It follows from the converse of the Pythagorean Theorem that triangle ABC is a right triangle. (d) Because the right angle is at vertex B, the sides AB and BC form the base and height of the triangle. Its area is Area =

1 1 1Base2 1Height2 = 1 225 2 1 25 2 = 5 square units 2 2

Now Work

29

2 Use the Midpoint Formula

Figure 9 y P2 = (x 2, y2) y2 M = (x, y) y

y1

PROBLEM

r

y – y1

x – x1 P1 = (x1, y1) x1

y2 – y x2 – x B = (x 2, y)

A = (x, y1) x

x2

x

THEOREM

In Words To find the midpoint of a line segment, average the x-coordinates of the endpoints, and average the y-coordinates of the endpoints.

We now derive a formula for the coordinates of the midpoint of a line segment. Let P1 = 1x1 , y1 2 and P2 = 1x2 , y2 2 be the endpoints of a line segment, and let M = 1x, y2 be the point on the line segment that is the same distance from P1 as it is from P2 . See Figure 9. The triangles P1 AM and MBP2 are congruent. [Do you see why? Angle AP1 M = angle BMP2 ,* angle P1 MA = angle MP2 B, and d 1P1 , M2 = d 1M, P2 2 is given. Thus we have angle–side–angle.] Hence, corresponding sides are equal in length. That is, x - x1 = x2 - x 2x = x1 + x2 x1 + x2 x = 2

and

y - y1 = y2 - y 2y = y1 + y2 y1 + y2 y = 2

Midpoint Formula The midpoint M = 1x, y2 of the line segment from P1 = 1x1 , y1 2 to P2 = 1x2 , y2 2 is M = 1x, y2 = ¢

x1 + x2 y1 + y2 , ≤ 2 2

(2)

*A postulate from geometry states that the transversal P1 P2 forms congruent corresponding angles with the parallel line segments P1 A and MB.

38

CHAPTER F  Foundations: A Prelude to Functions

Finding the Midpoint of a Line Segment

EX A MPL E 4

Find the midpoint of the line segment from P1 = 1 - 5, 52 to P2 = 13, 12 . Plot the points P1 and P2 and the midpoint. Apply the midpoint formula (2) using x1 = - 5, y1 = 5, x2 = 3, and y2 = 1. Then the coordinates 1x, y2 of the midpoint M are

Solution

Figure 10 y P1  (–5, 5)

x =

5

M  (–1, 3)

That is, M = 1 - 1, 32 . See Figure 10.

P2  (3, 1)

–5

5

y1 + y2 x1 + x2 -5 + 3 5 + 1 = = - 1 and y = = = 3 2 2 2 2

Now Work

x

PROBLEM

r

35

F.1 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. On the real number line the origin is assigned the number 0 . (p. A4) 2. If - 3 and 5 are the coordinates of two points on the real number line, the distance between these points is 8 . (p. A6) 3. If 3 and 4 are the legs of a right triangle, the hypotenuse is 5 . (pp. A13–A14)

4. Use the converse of the Pythagorean Theorem to show that a triangle whose sides are of lengths 11, 60, and 61 is a right triangle. (p. A14) 112 + 602 = 612 5. State the formula for the area A of a triangle whose base is b and whose altitude is h. (p. A15) A = 1 bh 2 6. State the three cases for which two triangles are congruent. (p. A16) ASA, SSS, SAS

Concepts and Vocabulary 7. If 1x, y2 are the coordinates of a point P in the xy-plane, then x is called the x-coordinate or abscissa of P, and y y-coordinate or ordinate is the of P. 8. The coordinate axes divide the xy-plane into four sections quadrants called .

9. The distance d between two points P1 = (x1, y1) and P2 = (x2, y2) is d = 2(x2 - x1)2 + (y2 - y1)2. 10. If three distinct points P, Q, and R all lie on a line, and if midpoint d1P, Q2 = d1Q, R2 , then Q is called the of the line segment from P to R.

Skill Building In Problems 11 and 12, plot each point in the xy-plane. Tell in which quadrant or on what coordinate axis each point lies.

*11. (a) A = 1 - 3, 22

*12. (a) A = 11, 42

(d) D = 16, 52

(b) B = 16, 02

(b) B = 1 - 3, - 42

(e) E = 10, - 32

(c) C = 1 - 2, - 22

(d) D = 14, 12 (e) E = 10, 12

(c) C = 1 - 3, 42

(f) F = 16, - 32

(f) F = 1 - 3, 02

*13. Plot the points 12, 02 , 12, - 32 , 12, 42 , 12, 12 , and 12, - 12 . Describe the set of all points of the form 12, y2 , where y is a real number.

*14. Plot the points 10, 32 , 11, 32 , 1 - 2, 32 , 15, 32 , and 1 - 4, 32 . Describe the set of all points of the form 1x, 32 , where x is a real number.

In Problems 15–28, find the distance d1P1 , P2 2 between the points P1 and P2 . 15. 25

16. 25

y 2 P = (2, 1) 2 P1 = (0, 0) –2

–1

2

x

17. 210

y P2 = (–2, 1) 2 P = (0, 0) 1 –2

–1

2

x

P2  (–2, 2)

–2

18. 210

y 2

P1  (1, 1) 2 x

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

y P1 = (–1, 1) 2 –2

–1

P2 = (2, 2)

2

x

SECTION F.1  The Distance and Midpoint Formulas

19. P1 = 13, - 42; P2 = 15, 42 21. P1 = 1 - 3, 22; P2 = 16, 02

2217 285

23. P1 = 14, - 22; P2 = 1 - 2, - 52

325

25. P1 = 1 - 0.2, 0.32; P2 = 12.3, 1.12 27. P1 = 1a, b2; P2 = 10, 02

26.89 ≈ 2.62

2a2 + b2

39

20. P1 = 1 - 1, 02; P2 = 12, 42 5 22. P1 = 12, - 32; P2 = 14, 22 229 24. P1 = 1 - 4, - 32; P2 = 16, 22 525 26. P1 = 11.2, 2.32; P2 = 1 - 0.3, 1.12 28. P1 = 1a, a2; P2 = 10, 02

23.69 ≈ 1.92

 a 22

In Problems 29–34, plot each point and form the triangle ABC. Verify that the triangle is a right triangle. Find its area. *29. A = 1 - 2, 52; B = 11, 32; C = 1 - 1, 02

*30. A = 1 - 2, 52; B = 112, 32; C = 110, - 112

*33. A = 14, - 32; B = 10, - 32; C = 14, 22

*34. A = 14, - 32; B = 14, 12; C = 12, 12

*31. A = 1 - 5, 32; B = 16, 02; C = 15, 52

*32. A = 1 - 6, 32; B = 13, - 52; C = 1 - 1, 52

In Problems 35–44, find the midpoint of the line segment joining the points P1 and P2 . 35. P1 = 13, - 42; P2 = 15, 42 37. P1 = 1 - 3, 22; P2 = 16, 02

(4, 0)

39. P1 = 14, - 22; P2 = 1 - 2, - 52

3 a , 1b 2

41. P1 = 1 - 0.2, 0.32; P2 = 12.3, 1.12 43. P1 = 1a, b2; P2 = 10, 02 a a, b b 2 2

7 a1, - b 2 (1.05, 0.7)

36. P1 = 1 - 2, 02; P2 = 12, 42 38. P1 = 12, - 32; P2 = 14, 22

40. P1 = 1 - 4, - 32; P2 = 12, 22

1 a3, - b 2

42. P1 = 11.2, 2.32; P2 = 1 - 0.3, 1.12 44. P1 = 1a, a2; P2 = 10, 02 a a, a b 2 2

(0, 2)

1 a - 1, - b 2 (0.45, 1.7)

Applications and Extensions 45. Find all points having an x-coordinate of 2 whose distance from the point 1 - 2, - 12 is 5. (2, 2); (2, - 4) 46. Find all points having a y-coordinate of - 3 whose distance from the point 11, 22 is 13. (13, - 3); ( - 11, - 3) 47. Find all points on the x-axis that are 5 units from the point 14, - 32 . (0, 0); (8, 0) 48. Find all points on the y-axis that are 5 units from the point 14, 42 . (0, 1); (0, 7) 49. Geometry The medians of a triangle are the line segments from each vertex to the midpoint of the opposite side (see the figure). Find the lengths of the medians of the triangle with vertices at A = 10, 02, B = 16, 02 , and C = 14, 42 . 217; 225; 229

C

Median Midpoint

A

B

50. Geometry An equilateral triangle is one in which all three sides are of equal length. If two vertices of an equilateral triangle are 10, 42 and 10, 02, find the third vertex. How many of these triangles are possible? Two triangles are possible; vertex: 1 223, 2 2 or 1 - 223, 2 2 s

a 23 a *52. Geometry Verify that the points (0, 0), (a, 0), and a , b 2 2 are the vertices of an equilateral triangle. Then show that the midpoints of the three sides are the vertices of a second equilateral triangle (refer to Problem 50). In Problems 53–56, find the length of each side of the triangle determined by the three points P1 , P2 , and P3 . State whether the triangle is an isosceles triangle, a right triangle, neither of these, or both. (An isosceles triangle is one in which at least two of the sides are of equal length.) *53. P1 = 12, 12; P2 = 1 - 4, 12; P3 = 1 - 4, - 32

*54. P1 = 1 - 1, 42; P2 = 16, 22; P3 = 14, - 52 *55. P1 = 1 - 2,- 12; P2 = 10, 72; P3 = 13, 22 *56. P1 = 17, 22; P2 = 1 - 4, 02; P3 = 14, 62

57. Baseball A major league baseball “diamond” is actually a square 90 feet on a side (see the figure). What is the distance directly from home plate to second base (the diagonal of the square)? 9022 ft ≈ 127.28 ft 2nd base

s

90 ft

s

51. Geometry Find the midpoint of each diagonal of a square with side of length s. Draw the conclusion that the diagonals of a square intersect at their midpoints. [Hint: Use (0, 0), (0, s), (s, 0), and (s, s) as the vertices of the square.] s s a , b 2 2

3rd base

Pitching rubber 1st base

90 ft

Home plate

40

CHAPTER F  Foundations: A Prelude to Functions

58. Little League Baseball The layout of a Little League playing field is a square 60 feet on a side. How far is it directly from home plate to second base (the diagonal of the square)? Source: Little League Baseball, Official Regulations and Playing Rules, 2012. 6022 ft ≈ 84.85 ft

(a) Find an estimate for the desired intersection point. (b) Find the length of the median for the midpoint found in part (a). See Problem 49. ≈1.285 units Source: www.uwgb.edu/DutchS/structge/s100.htm 63. (a) (2.65, 1.6) y

59. Baseball Refer to Problem 57. Overlay a rectangular coordinate system on a major league baseball diamond so that the origin is at home plate, the positive x-axis lies in the direction from home plate to first base, and the positive y-axis lies in the direction from home plate to third base.

1.5 1.3

*(a) What are the coordinates of first base, second base, and third base? Use feet as the unit of measurement.

(c) If the center fielder is located at 1300, 3002 , how far is it from there to third base? 302149 ft ≈ 366.20 ft

60. Little League Baseball Refer to Problem 58. Overlay a rectangular coordinate system on a Little League baseball diamond so that the origin is at home plate, the positive x-axis lies in the direction from home plate to first base, and the positive y-axis lies in the direction from home plate to third base. *(a) What are the coordinates of first base, second base, and third base? Use feet as the unit of measurement. (b) If the right fielder is located at 1180, 202 , how far is it from there to second base? 40210 ft ≈ 126.49 ft

(c) If the center fielder is located at 1220, 2202 , how far is it from there to third base? 202185 ft ≈ 272.03 ft 61. Distance between Moving Objects A Ford Focus and a Mack truck leave an intersection at the same time. The Focus heads east at an average speed of 30 miles per hour, while the truck heads south at an average speed of 40 miles per hour. Find an expression for their distance apart d (in miles) at the end of t hours. d = 50t mi 62. Distance of a Moving Object from a Fixed Point A hot-air balloon, headed due east at an average speed of 15 miles per hour and at a constant altitude of 100 feet, passes over an intersection (see the figure). Find an expression for the distance d (measured in feet) from the balloon to the intersection t seconds later. d = 210,000 + 484t 2 ft

15 mph 100 ft

(2.6, 1.5) (1.4, 1.3) 1.4

2.6 2.7

x

64. Net Sales The figure illustrates how net sales of Wal-Mart Stores, Inc., grew from 2008 through 2012. Use the midpoint formula to estimate the net sales of Wal-Mart Stores, Inc., in 2010. How does your result compare to the reported value of $405 billion? $409 billion Source: Wal-Mart Stores, Inc., 2012 Annual Report Wal-Mart Stores, Inc. Net sales ($ billions)

Net sales ($ billions)

(b) If the right fielder is located at 1310, 152 , how far is it from there to second base? 522161 ft ≈ 232.43 ft

(2.7, 1.7)

1.7

500 450 400 374 350 300 250 200 150 100 50 0 2008

444

2009

2010 Year

2011

2012

65. Poverty Threshold Poverty thresholds are determined by the U.S. Census Bureau. A poverty threshold represents the minimum annual household income for a family not to be considered poor. In 2004, the poverty threshold for a family of four with two children under the age of 18 years was $19,157. In 2012, the poverty threshold for a family of four with two children under the age of 18 years was $23,283. Assuming poverty thresholds increase in a straight-line fashion, use the midpoint formula to estimate the poverty threshold of a family of four with two children under the age of 18 in 2008. How does your result compare to the actual poverty threshold in 2008 of $21,834? $21,220 Source: U.S. Census Bureau

66. Horizontal and Vertical Shifts Suppose that A = 12, 52 are the coordinates of a point in the xy-plane. (a) Find the coordinates of the point if A is shifted 3 units to the right and 2 units down. (5, 3) (b) Find the coordinates of the point if A is shifted 2 units to the left and 8 units up. (0, 13)

63. Drafting Error When a draftsman draws three lines that are to intersect at one point, the lines may not intersect as intended and subsequently will form an error triangle. If this error triangle is long and thin, one estimate for the location of the desired point is the midpoint of the shortest side. The figure shows one such error triangle.

67. Completing a Line Segment Plot the points A = 1 - 1, 82 and M = 12, 32 in the xy-plane. If M is the midpoint of a line segment AB, find the coordinates of B. (5, - 2)

SECTION F.2  Graphs of Equations in Two Variables; Intercepts; Symmetry

41

Discussion and Writing the terms coordinate axes, ordered pair, coordinates, plot, x-coordinate, and y-coordinate.

68. Write a paragraph that describes a Cartesian plane. Then write a second paragraph that describes how to plot points in the Cartesian plane. Your paragraphs should include

‘Are You Prepared?’ Answers 1. 0

2. 8

3. 5

4. 112 + 602 = 612

5. A =

1 bh 2

6. Angle–side–angle; side–side–side; side–angle–side

F.2 Graphs of Equations in Two Variables; Intercepts; Symmetry PREPARING FOR THIS SECTION Before getting started, review the following: r
4PMWJOH
&RVBUJPOT
"QQFOEJY
"
4FDUJPO
"
QQ
"m"

Now Work the ‘Are You Prepared?’ problems on page 48.

OBJECTIVES 1 2 3 4 5

Graph Equations by Plotting Points (p. 41) Find Intercepts from a Graph (p. 43) Find Intercepts from an Equation (p. 44) Test an Equation for Symmetry (p. 44) Know How to Graph Key Equations (p. 46)

1 Graph Equations by Plotting Points An equation in two variables, say x and y, is a statement in which two expressions involving x and y are equal. The expressions are called the sides of the equation. Since an equation is a statement, it may be true or false, depending on the value of the variables. Any values of x and y that result in a true statement are said to satisfy the equation. For example, the following are all equations in two variables x and y: x2 + y 2 = 5

2x - y = 6

y = 2x + 5

x2 = y

The first of these, x2 + y2 = 5, is satisfied for x = 1, y = 2, since 12 + 22 = 1 + 4 = 5. Other choices of x and y, such as x = - 1, y = - 2, also satisfy this equation. It is not satisfied for x = 2 and y = 3, since 22 + 32 = 4 + 9 = 13 ≠ 5. The graph of an equation in two variables x and y consists of the set of points in the xy-plane whose coordinates 1x, y2 satisfy the equation.

EXAM PL E 1

Determining Whether a Point Is on the Graph of an Equation Determine if the following points are on the graph of the equation 2x - y = 6. (a) 12, 32

Solution

(b) 12, - 22

(a) For the point 12, 32, check to see whether x = 2, y = 3 satisfies the equation 2x - y = 6. 2x - y = 2122 - 3 = 4 - 3 = 1 ≠ 6

The equation is not satisfied, so the point 12, 32 is not on the graph.

42

CHAPTER F  Foundations: A Prelude to Functions

(b) For the point 12, - 22, 2x - y = 2122 - 1 - 22 = 4 + 2 = 6 The equation is satisfied, so the point 12, - 22 is on the graph.

Now Work

EX AM PL E 2

r

9

PROBLEM

How to Graph an Equation by Plotting Points (SBQI
UIF
FRVBUJPO
y = - 2x + 3

Step-by-Step Solution Step 1: Find points (x, y) that satisfy the equation. To determine these points, choose values of x and use the equation to find the corresponding values for y. See Table 1.

Step 2: Plot the points found in the table as shown in Figure 11(a). Now connect the points to obtain the graph of the equation (a line), as shown in Figure 11(b).

Table 1

y = −2x + 3

(x, y)

-2

- 2( - 2) + 3 = 7

( - 2, 7)

-1

- 2( - 1) + 3 = 5

( - 1, 5)

0

- 2(0) + 3 = 3

(0, 3)

1

- 2(1) + 3 = 1

(1, 1)

2

- 2(2) + 3 = - 1

(2, - 1)

Figure 11 y 8

y 8 (–2, 7) (–1, 5)

(–2, 7)

6 4

–4

6

(–1, 5) (0, 3)

2

(0, 3) 2

(1, 1) 2

–2

4 x (2, –1)

–4

(1, 1) 2

–2

4 x (2, –1)

–2

–2 (a)

EX A MPL E 3

x

r

(b)

Graphing an Equation by Plotting Points (SBQI
UIF
FRVBUJPO
y = x2

Solution

Table 2

Table 2 provides several points on the graph. Plotting these points and connecting them with a smooth curve gives the graph (a parabola) shown in Figure 12.

x

y = x2

(x, y)

-4

16

( - 4, 16)

-3

9

( - 3, 9)

-2

4

( - 2, 4)

-1

1

( - 1, 1)

0

0

(0, 0)

1

1

(1, 1)

2

4

(2, 4)

3

9

(3, 9)

4

16

(4, 16)

Figure 12 y 20 (– 4, 16) (–3, 9)

(4, 16)

15 10

(3, 9)

5 (–2, 4) (2, 4) (1, 1) (–1, 1) (0, 0) –4 4

x

r

SECTION F.2  Graphs of Equations in Two Variables; Intercepts; Symmetry

43

The graphs of the equations shown in Figures 11 and 12 do not show all the points that are on the graph. For example, in Figure 11 the point 120, - 372 is a part of the graph of y = - 2x + 3, but it is not shown. Since the graph of y = - 2x + 3 could be extended out as far as we please, we use arrows to indicate that the pattern shown continues. It is important when illustrating a graph to present enough of the graph so that any viewer of the illustration will “see” the rest of it as an obvious continuation of what is actually there. This is referred to as a complete graph. One way to obtain a complete graph of an equation is to plot a sufficient number of points on the graph for a pattern to become evident. Then these points are connected with a smooth curve following the suggested pattern. But how many points are sufficient? Sometimes knowledge about the equation tells us. For example, we will learn in the next section that if an equation is of the form y = mx + b, then its graph is a line. In this case, only two points are needed to obtain the graph. One purpose of this book is to investigate the properties of equations in order to decide whether a graph is complete. Sometimes we shall graph equations by plotting points. Shortly, we shall investigate various techniques that will enable us to graph an equation without plotting so many points. COMMENT Another way to obtain the graph of an equation is to use a graphing utility. Read Section B.2, Using a Graphing Utility to Graph Equations, in Appendix B. ■ Figure 13 Graph crosses y-axis

Two techniques that sometimes reduce the number of points required to graph an equation involve finding intercepts and checking for symmetry.

y Graph crosses x-axis

2 Find Intercepts from a Graph x

Graph touches x-axis

Intercepts

EXAM PL E 4

The points, if any, at which a graph crosses or touches the coordinate axes are called the intercepts. See Figure 13. The x-coordinate of a point at which the graph crosses or touches the x-axis is an x-intercept, and the y-coordinate of a point at which the graph crosses or touches the y-axis is a y-intercept. For a graph to be complete, all its intercepts must be displayed.

Finding Intercepts from a Graph Find the intercepts of the graph in Figure 14. What are its x-intercepts? What are its y-intercepts?

Solution Figure 14

1 - 3, 02,

y 4

3 a , 0b , 2

4 a0, - b , 3

10, - 3.52,

14.5, 02

3 4 The x-intercepts are - 3, , and 4.5; the y-intercepts are - 3.5, - , and 3. 2 3

(4.5, 0) 5 x 4

(0, –3 ) (0, 3.5)

10, 32,

(0, 3)

( 3–2 , 0) 4 (3, 0)

The intercepts of the graph are the points

r

In Example 4, note the following usage: If the type of intercept (x- versus y-) is not specified, then report the intercept as an ordered pair. However, if the type of intercept is specified, then report only the coordinate of the specified intercept. For x-intercepts, report the x-coordinate of the intercept; for y-intercepts, report the y-coordinate of the intercept.

Now Work

PROBLEM

37(A)

44

CHAPTER F  Foundations: A Prelude to Functions

3 Find Intercepts from an Equation The intercepts of a graph can be found from its equation by using the fact that points on the x-axis have y-coordinates equal to 0 and points on the y-axis have x-coordinates equal to 0.

Procedure for Finding Intercepts 1. To find the x-intercept(s), if any, of the graph of an equation, let y = 0 in the equation and solve for x. 2. To find the y-intercept(s), if any, of the graph of an equation, let x = 0 in the equation and solve for y.

Because the x-intercepts of the graph of an equation are those x-values for which y = 0, they are also called the zeros (or roots) of the equation.

EX A MPL E 5

Finding Intercepts from an Equation Find the x-intercept(s) and the y-intercept(s) of the graph of y = x2 - 4.

Solution

To find the x-intercept(s), let y = 0 and obtain the equation

x + 2 = 0 x = -2

x2 - 4 1x + 22 1x - 22 or x - 2 or x

= = = =

0 0 0 2

Factor. Zero-Product Property Solve.

The equation has two solutions, - 2 and 2. The x-intercepts (the zeros) are - 2 and 2. To find the y-intercept(s), let x = 0 in the equation. y = x2 - 4 = 02 - 4 = -4 The y-intercept is - 4.

Now Work

PROBLEM

19

r

COMMENT For many equations, finding intercepts may not be so easy. In such cases, a graphing utility can be used. Read the first part of Section B.3, Using a Graphing Utility to Locate Intercepts and Check for Symmetry, in Appendix B to find out how a graphing utility locates intercepts. ■

4 Test an Equation for Symmetry Another helpful tool for graphing equations by hand involves symmetry, particularly symmetry with respect to the x-axis, the y-axis, and the origin. Symmetry often occurs in nature. Consider the picture of the butterfly. Do you see the symmetry?

DEFINITION

A graph is said to be symmetric with respect to the x-axis if, for every point 1x, y2 on the graph, the point 1x, - y2 is also on the graph. A graph is said to be symmetric with respect to the y-axis if, for every point 1x, y2 on the graph, the point 1 - x, y2 is also on the graph. A graph is said to be symmetric with respect to the origin if, for every point 1x, y2 on the graph, the point 1 - x, - y2 is also on the graph.

SECTION F.2  Graphs of Equations in Two Variables; Intercepts; Symmetry

45

Figure 15 illustrates the definition. Note that when a graph is symmetric with respect to the x-axis, the part of the graph above the x-axis is a reflection or mirror image of the part below it, and vice versa. When a graph is symmetric with respect to the y-axis, the part of the graph to the right of the y-axis is a reflection of the part to the left of it, and vice versa. Symmetry with respect to the origin may be viewed in two ways: 1. As a reflection about the y-axis, followed by a reflection about the x-axis 2. As a projection along a line through the origin so that the distances from the origin are equal

Figure 15

y

y

y (–x1, y1)

(x2, y2) (x1, y1)

(x3, y3)

(x1, –y1)

x (x3, –y3)

(–x2, y2)

(x2, y2)

(x2, –y2)

x

(x2, y2)

x

(–x2, –y2) (–x1, –y1)

Symmetry with respect to the x-axis

EXAM PL E 6

(x1, y1)

(x1, y1)

Symmetry with respect to the y-axis

Symmetry with respect to the origin

Symmetric Points

(a) If a graph is symmetric with respect to the x-axis, and the point 14, 22 is on the graph, then the point 14, - 22 is also on the graph. (b) If a graph is symmetric with respect to the y-axis, and the point 14, 22 is on the graph, then the point 1 - 4, 22 is also on the graph. (c) If a graph is symmetric with respect to the origin, and the point 14, 22 is on the graph, then the point 1 - 4, - 22 is also on the graph.

Now Work

r

PROBLEM

27

When the graph of an equation is symmetric with respect to the x-axis, the y-axis, or the origin, the number of required points to plot in order to see the pattern is reduced. For example, if the graph of an equation is symmetric with respect to the y-axis, then once points to the right of the y-axis are plotted, an equal number of points on the graph can be obtained by reflecting them about the y-axis. Because of this, before graphing an equation, it is wise to determine whether any symmetry exists. The following tests are used for this purpose.

Tests for Symmetry To test the graph of an equation for symmetry with respect to the X-AXIS

Y-AXIS

ORIGIN

Replace y by - y in the equation. If an equivalent equation results, the graph of the equation is symmetric with respect to the x-axis. Replace x by - x in the equation. If an equivalent equation results, the graph of the equation is symmetric with respect to the y-axis. Replace x by - x and y by - y in the equation. If an equivalent equation results, the graph of the equation is symmetric with respect to the origin.

46

CHAPTER F  Foundations: A Prelude to Functions

Testing an Equation for Symmetry

EX A MPL E 7

Test y =

Solution

x-Axis:

y-Axis:

Origin:

4x2 for symmetry. x + 1 2

To test for symmetry with respect to the x-axis, replace y by - y. Since 4x2 4x2 -y = 2 is not equivalent to y = 2 , the graph of the equation x + 1 x + 1 is not symmetric with respect to the x-axis. To test for symmetry with respect to the y-axis, replace x by - x. Since 41 - x2 2 4x2 4x2 y = = is equivalent to y = , the graph of the 1 - x2 2 + 1 x2 + 1 x2 + 1 equation is symmetric with respect to the y-axis. To test for symmetry with respect to the origin, replace x by - x and y by - y. 41 - x2 2

-y =

1 - x2 2 + 1

Replace x by - x and y by - y.

-y =

4x2 x2 + 1

Simplify.

y = -

4x2 x2 + 1

Multiply both sides by - 1.

Since the result is not equivalent to the original equation, the graph of the 4x2 equation y = 2 is not symmetric with respect to the origin. x + 1

Figure 16

r

5

−5

5

Seeing the Concept Figure 16 shows the graph of y = with respect to the y-axis?

4x2 x + 1 2

using a graphing utility. Do you see the symmetry

−5

Now Work

PROBLEM

57

5 Know How to Graph Key Equations The next three examples use intercepts, symmetry, and point plotting to obtain the graphs of key equations. It is important to know the graphs of these key equations because they will be used later. The first of these is y = x3.

EX A MPL E 8

Graphing the Equation y = x 3 by Finding Intercepts and Checking for Symmetry (SBQI
UIF
FRVBUJPO
y = x3 by plotting points. Find any intercepts and check for symmetry first.

Solution

First, find the intercepts. When x = 0, then y = 0; and when y = 0, then x = 0. The origin 10, 02 is the only intercept. Now test for symmetry. x-Axis: y-Axis: Origin:

Replace y by - y. Since - y = x3 is not equivalent to y = x3, the graph is not symmetric with respect to the x-axis. Replace x by - x. Since y = 1 - x2 3 = - x3 is not equivalent to y = x3, the graph is not symmetric with respect to the y-axis. Replace x by - x and y by - y. Since - y = 1 - x2 3 = - x3 is equivalent to y = x3 (multiply both sides by - 1), the graph is symmetric with respect to the origin.

To graph y = x3, use the equation to obtain several points on the graph. Because of the symmetry, we only need to locate points on the graph for which x Ú 0.

SECTION F.2  Graphs of Equations in Two Variables; Intercepts; Symmetry

47

See Table 3. Since (1, 1) is on the graph, and the graph is symmetric with respect to the origin, the point ( - 1, - 1) is also on the graph. Figure 17 shows the graph.

Table 3

x

y = x3

(x, y)

0

0

(0, 0)

1

1

(1, 1)

2

8

(2, 8)

3

27

(3, 27)

Figure 17 y

(0, 0) –6

(1, 1) 6

(– 1, – 1)

(– 2, – 8)

EXAM PL E 9

(2, 8)

8

–8

x

r

Graphing the Equation x = y 2 B
(SBQI
UIF
FRVBUJPO
x = y2. Find any intercepts and check for symmetry first. C
(SBQI
x = y2, y Ú 0.

Solution

Table 4 y

x = y2

(x, y)

0

0

(0, 0)

1

1

(1, 1)

2

4

(4, 2)

3

9

(9, 3)

Figure 20 6

(a) The lone intercept is 10, 02 . The graph is symmetric with respect to the x-axis since x = 1 - y2 2 is equivalent to x = y2. The graph is not symmetric with respect to the y-axis or the origin. To graph x = y2, use the equation to obtain several points on the graph. Because the equation is solved for x, it is easier to assign values to y and use the equation to determine the corresponding values of x. Because of the symmetry, start by finding points whose y-coordinates are non-negative. Then use the symmetry to find additional points on the graph. See Table 4. For example, since 11, 12 is on the graph, so is 11, - 12. Since 14, 22 is on the graph, so is 14, - 22, and so on. Plot these points and connect them with a smooth curve to obtain Figure 18. (b) If we restrict y so that y Ú 0, the equation x = y2, y Ú 0, may be written equivalently as y = 1x. The portion of the graph of x = y2 in quadrant I is therefore the graph of y = 1x. See Figure 19. Figure 18

Y1  x

Figure 19

y 6 (9, 3) (1, 1)

(4, 2)

(1, 1)

(0, 0)

2

10

6

y 6

Y2   x

EX AM PL E 10

2 (1, 1)

(9, 3)

(0, 0) 10 x

5 (4, 2)

(9, 3)

–2

5

10 x

r

COMMENT To see the graph of the equation x = y2 on a graphing calculator, graph two equations: Y1 = 1x and Y2 = - 1x. See Figure 20. We discuss why in Chapter 1. ■

Graphing the Equation y =

1 x

1 . Find any intercepts and check for symmetry first. x Check for intercepts first. If we let x = 0, we obtain 0 in the denominator, which makes y undefined. We conclude that there is no y-intercept. If we let y = 0, 1 we get the equation = 0, which has no solution. We conclude that there is no x 1 x-intercept. The graph of y = does not cross or touch the coordinate axes. x (SBQI
UIF
FRVBUJPO
y =

Solution

(4, 2)

48

CHAPTER F  Foundations: A Prelude to Functions

Next check for symmetry:

Table 5 1 y = x

x 1 10 1 3 1 2 1

(x, y) a

1

1 , 10b 10 1 a , 3b 3 1 a , 2b 2 11, 12

1 2 1 3 1 10

1 a2, b 2 1 a3, b 3 1 a10, b 10

10 3 2

2 3 10

y-Axis: Origin:

Use the equation to form Table 5, and obtain some points on the graph. Because of the symmetry, we only find points (x, y) for which x is positive. From 1 Table 5 we infer that if x is a large and positive number, then y = is a positive x number close to 0. We also infer that if x is a positive number close to 0, then 1 y = is a large and positive number. Armed with this information, we can graph x the equation. 1 Figure 21 illustrates some of these points and the graph of y = . Observe how x the absence of intercepts and the existence of symmetry with respect to the origin are utilized.

y

Figure 21

3

(––12 , 2) (1, 1) 3

r

(2, ––12 ) x

3

(2,  ––12 )

(1, 1)

( ––12 , 2)

1 1 , which is not equivalent to y = . x x 1 1 1 Replacing x by - x yields y = = - , which is not equivalent to y = . -x x x 1 Replacing x by - x and y by - y yields - y = - , which is equivalent to x 1 y = . The graph is symmetric with respect to the origin. x Replacing y by - y yields - y =

x-Axis:

1 COMMENT Refer to Example 2 in Appendix B, Section B.3, for the graph of y = found x using a graphing utility. ■

3

F.2 Assess Your Understanding ‘Are You Prepared?’ 

Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.

1. Solve the equation 2(x + 3) - 1 = - 7. (pp. A64–A66)

5 - 66 2. Solve the equation x 2 - 4x - 12 = 0. (pp. A67–A68)

5 - 2, 66

Concepts and Vocabulary  3. The points, if any, at which a graph crosses or touches the coordinate axes are called intercepts .

6. True or False To find the y-intercept(s) of the graph of an equation, let x = 0 and solve for y. True

4. Because the x-intercepts of the graph of an equation are those x-values for which y = 0, they are also called zeros roots or .

7. True or False The graph of an equation in two variables x and y consists of the set of points in the xy-plane whose coordinates (x, y) satisfy the equation. True

5. If, for every point 1x, y2 on the graph of an equation, the point 1 - x, y2 is also on the graph, then the graph is y-axis symmetric with respect to the .

8. True or False If a graph is symmetric with respect to the x-axis, then it cannot be symmetric with respect to the y-axis. False

Skill Building In Problems 9–14, tell whether the given points are on the graph of the equation. 9. Equation: y = x4 - 1x Points: 10, 02; 11, 12; 1 - 1, 02 12. Equation: y = x + 1 3

Points: 11, 22; 10, 12; 1 - 1, 02 (0, 1); 1 - 1, 02

10. Equation: y = x3 - 21x

11. Equation: y2 = x2 + 9

13. Equation: x2 + y2 = 4

14. Equation: x2 + 4y2 = 4

Points: 10, 02; 11, 12; 11, - 12 10, 02; 11, - 12 Points: 10, 32; 13, 02; 1 - 3, 02 (0, 3)

(0, 0)

Points: 10, 22; 1 - 2, 22; 1 22, 22 2 10, 22; 1 22,22 2

1 Points: 10, 12; 12, 02; a2, b 2 (0, 1); (2, 0) In Problems 15–26, find the intercepts and graph each equation by plotting points. Be sure to label the intercepts. *15. y = x + 2

*16. y = x - 6

*17. y = 2x + 8

*18. y = 3x - 9

*19. y = x - 1

*20. y = x - 9

*21. y = - x + 4

*22. y = - x2 + 1

*23. 2x + 3y = 6

*24. 5x + 2y = 10

*25. 9x2 + 4y = 36

*26. 4x2 + y = 4

2

2

2

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

49

SECTION F.2  Graphs of Equations in Two Variables; Intercepts; Symmetry

In Problems 27–36, plot each point. Then plot the point that is symmetric to it with respect to (a) the x-axis; (b) the y-axis; (c) the origin. *27. 13, 42

*28. 15, 32

*29. 1 - 2, 12

*30. 14, - 22

*31. 15, - 22

*32. 1 - 1, - 12

*33. 1 - 3, - 42

*34. 14, 02

*35. 10, - 32

*36. 1 - 3, 02

In Problems 37–48, the graph of an equation is given. (a) Find the intercepts. (b) Indicate whether the graph is symmetric with respect to the x-axis, the y-axis, or the origin. 37. 38. 39. 40. y y y y 3

3

–3

–3

(a) 1 - 1, 02; 11, 02 (b) x-axis, y-axis, origin 41.

3x

–3

y 3

45.

y 3

4

(a) 1 - 2, 02, 10, - 32, 12, 02 (b) y-axis 44.

y

x

3x

6

6

6 40

(a) 1 - 4, 02, 10, 02, 14, 02

(b) x-axis, y-axis, origin

(b) origin

(b) origin 47.

y 3

4

48.

8

2 3x

3 x

(a) 1 - 2, 02, 10, 22, 12, 02, 10, - 22 (a) 1 - 2, 02, 10, 02, 12, 02 46.

3

3

40

3

3x –3

(a) (0, 0) (b) x-axis

x

6

–3

–3

 –– 2

–1

p p (a) a - , 0b, 10, 12, a , 0b 2 2 (b) y-axis 43. y

(a) (0, 1) (b) no symmetry 42.

y 3

 –  – –– 2

3x

–3

3x

–3

4

1

4

2

4

3x

3

4

8 3

3

(a) x-intercepts: 3 - 2, 14 y-intercept: 0 (b) no symmetry

(a) x-intercepts: 3 - 1, 24 y-intercept: 0 (b) no symmetry

(a) no intercepts (b) origin

(a) no intercepts (b) x-axis

In Problems 49–52, draw a complete graph so that it has the type of symmetry indicated. *49. y-axis

*50. x-axis y

y

5

9

(0, 2)

*51. Origin

(0, 0)

–9

 , 2) (–– 2

5x

–5 9x (2, –5)

–9

4

(5, 3)

(–4, 0)

*52. y-axis

y

(2, 2)

(, 0) x

(0, 0)

–2

–4

–5

(0, –9)

*53. y2 = x + 4

*54. y2 = x + 9

3 *55. y = 2 x

*56.

*57. y = x4 - 8x2 - 9

*58. y = x4 - 2x2 - 8

*59. 9x2 + 4y2 = 36

*60. 4x2 + y2 = 4

*61. y = x3 + x2 - 9x - 9

*62. y = x3 + 2x2 - 4x - 8

*63. y = |x| - 4

*64.

y = |x 0 - 2

3x *65. y = 2 x + 9

x2 - 4 *66. y = 2x

- x3 *67. y = 2 x - 9

*68.

y =

5 y = 2 x

*71. y = 1x

*72.

y =

x4 + 1 2x5

In Problems 69–72, draw a quick sketch of each equation. *70. x = y2

3x

–3

In Problems 53–68, list the intercepts and test for symmetry.

*69. y = x3

y (0, 4) 4

1 x

50

CHAPTER F  Foundations: A Prelude to Functions

73. If 13, b2 is a point on the graph of y = 4x + 1, what is b?

13

75. If 1a, 42 is a point on the graph of y = x2 + 3x, what is a? - 4 or 1

74. If 1 - 2, b2 is a point on the graph of 2x + 3y = 2, what is b? 2

76. If 1a, - 52 is a point on the graph of y = x2 + 6x, what is a? - 5 or - 1

Mixed Practice In Problems 77–84, (a) find the intercepts of the graph of each equation, (b) test each equation for symmetry with respect to the x-axis, the y-axis, and the origin, and (c) graph each equation by plotting points. Be sure to label the intercepts on the graph and use any symmetry to assist in drawing the graph. *77. y = x2 - 5

*78. y = x2 - 8

*79. x - y2 = - 9

*80. x + y2 = 4

*81. x + y = 9

*82. x + y = 16

*83. y = x - 4x

*84. y = x3 - x

2

2

2

2

3

Applications and Extensions 85. (JWFO
UIBU
UIF
QPJOU


JT
PO
UIF
HSBQI
PG
BO
FRVBUJPO
that is symmetric with respect to the origin, what other point is on the graph? 1 - 1, - 22 86. If the graph of an equation is symmetric with respect to the y-axis, and 6 is an x-intercept of this graph, name another x-intercept. - 6

90. Solar Energy The solar electric generating systems at Kramer Junction, California, use parabolic troughs to heat a heat-transfer fluid to a high temperature. This fluid is used to generate steam that drives a power conversion system to produce electricity. For troughs 7.5 feet wide, an equation for the cross section is 16y2 = 120x - 225.

87. If the graph of an equation is symmetric with respect to the origin, and - 4 is an x-intercept of this graph, name another x-intercept. 4 88. If the graph of an equation is symmetric with respect to the x-axis, and 2 is a y-intercept, name another y-intercept. - 2 89. Microphones In studios and on stages, cardioid microphones are often preferred for the richness they add to voices and for their ability to reduce the level of sound from the sides and rear of the microphone. Suppose one such cardioid pattern is given by the equation 1x2 + y2 - x2 2 = x2 + y2. (a) Find the intercepts of the graph of the equation. (b) Test for symmetry with respect to the x-axis, the y-axis, and the origin. x-axis symmetry 89. (a) (0, 0), (2, 0), (0, 1), 10, - 12

(a) Find the intercepts of the graph of the equation. a

15 , 0b 8 (b) Test for symmetry with respect to the x-axis, the y-axis, and the origin. x-axis symmetry Source: U.S. Department of Energy

Discussion and Writing In Problem 91, use a graphing utility.

91. B

(SBQI
y = 2x2 , y = x, y = 0 x 0 , and y = 1 1x2 2, noting which graphs are the same. same: y = 2x2 and y = |x| (b) Explain why the graphs of y = 2x2 and y = 0 x 0 are the same. 2x2 = 0 x 0 (c) Explain why the graphs of y = x and y = 1 1x2 are not the same. 2

(d) Explain why the graphs of y = 2x2 and y = x are not the same. y Ú 0 for y = 2x2 92. Explain what is meant by a complete graph. 93. Draw a graph of an equation that contains two x-intercepts; at one the graph crosses the x-axis, and at the other the graph touches the x-axis. 91. (c) y = 1 2x2 2 has domain [0, q 2 ; y = x has domain 1 - q, q 2

‘Are You Prepared?’ Answers 1. 5 - 66

2. 5 - 2, 66

94. Make up an equation with the intercepts 12, 02, 14, 02, and 10, 12. Compare your equation with a friend’s equation. Comment on any similarities. 95. Draw a graph that contains the points 1 - 2, - 12, 10, 12, 11, 32, and 13, 52. Compare your graph with those of other students. Are most of the graphs almost straight lines? How many are “curved”? Discuss the various ways that these points might be connected.

96. An equation is being tested for symmetry with respect to the x-axis, the y-axis, and the origin. Explain why, if two of these symmetries are present, the remaining one must also be present. 97. Draw a graph that contains the points ( - 2, 5), ( - 1, 3), and 10, 22 that is symmetric with respect to the y-axis. Compare your graph with those of other students; comment on any similarities. Can a graph contain these points and be symmetric with respect to the x-axis? the origin? Why or why not?

SECTION F.3  Lines

51

F.3 Lines OBJECTIVES 1 2 3 4 5 6 7 8 9 10

Instructor Note For many students, this material is review. Consider your audience when setting the pace.

Calculate and Interpret the Slope of a Line (p. 51) Graph Lines Given a Point and the Slope (p. 54) Find the Equation of a Vertical Line (p. 54) Use the Point–Slope Form of a Line; Identify Horizontal Lines (p. 55) Find the Equation of a Line Given Two Points (p. 56) Write the Equation of a Line in Slope–Intercept Form (p. 56) Identify the Slope and y-Intercept of a Line from Its Equation (p. 57) Graph Lines Written in General Form Using Intercepts (p. 58) Find Equations of Parallel Lines (p. 59) Find Equations of Perpendicular Lines (p. 60)

In this section we study a certain type of equation that contains two variables, called a linear equation, and its graph, a line.

1 Calculate and Interpret the Slope of a Line Figure 22 Line

Rise

Consider the staircase illustrated in Figure 22. Each step contains exactly the same horizontal run and the same vertical rise. The ratio of the rise to the run, called the slope, is a numerical measure of the steepness of the staircase. For example, if the run is increased and the rise remains the same, the staircase becomes less steep. If the run is kept the same but the rise is increased, the staircase becomes more steep. This important characteristic of a line is best defined using rectangular coordinates.

Run

DEFINITION

Let P = (x1 , y1 2 and Q = (x2 , y2 2 be two distinct points. If x1 ≠ x2 , the slope m of the nonvertical line L containing P and Q is defined by the formula m =

y2 - y1 x2 - x1

x1 ≠ x2

(1)

If x1 = x2 , L is a vertical line and the slope m of L is undefined (since this results in division by 0).

Figure 23(a) provides an illustration of the slope of a nonvertical line; Figure 23(b) illustrates a vertical line. Figure 23 L Q = (x 2, y2)

L

y2

Q = (x 1, y2)

y1

P = (x 1, y1)

Rise = y2 – y1

P = (x 1, y1) y1

y2

Run = x2 – x1 x1

(a) Slope of L is m =

x2 y2 – y1 _______ x2 – x1

x1

(b) Slope is undefined; L is vertical

52

CHAPTER F  Foundations: A Prelude to Functions

As Figure 23(a) illustrates, the slope m of a nonvertical line may be viewed as

In Words

The symbol ∆ is the Greek letter delta. In mathematics, ∆ is read ∆y “change in,” so is read “change ∆x in y divided by change in x.”

m =

y2 - y1 Rise = x2 - x1 Run

or as m =

Change in y y2 - y1 ∆y = = x2 - x1 Change in x ∆x

That is, the slope m of a nonvertical line measures the amount y changes when x ∆y changes from x1 to x2. The expression is called the average rate of change of y ∆x with respect to x. Two comments about computing the slope of a nonvertical line may prove helpful: 1. Any two distinct points on the line can be used to compute the slope of the line. (See Figure 24 for justification.) Figure 24 Triangles ABC and PQR are similar (equal angles), so ratios of corresponding sides are equal. Then y2 - y1 Slope using P and Q = = x2 - x1 d(B, C) = Slope using A and B d(A, C)

y

Q = (x 2, y2) y2 – y1

P = (x 1, y1) B

x2 – x1

R x

A

C

Since any two distinct points can be used to compute the slope of a line, the average rate of change of a line is always the same number. 2. The slope of a line may be computed from P = (x1 , y1 2 to Q = (x2 , y2 2 or from Q to P because y2 - y1 y1 - y2 = x2 - x1 x1 - x2

EX A MPL E 1

Finding and Interpreting the Slope of a Line Given Two Points

The slope m of the line containing the points 11, 22 and 15, - 32 may be computed as m =

-3 - 2 -5 5 = = 5 - 1 4 4

or as m =

2 - 1 - 32 5 5 = = 1 - 5 -4 4

For every 4-unit change in x, y will change by - 5 units. That is, if x increases by 4 units, then y will decrease by 5 units. The average rate of change of y with respect 5 to x is - . 4

Now Work

PROBLEMS

11

AND

r

17

To get a better idea of the meaning of the slope m of a line L, consider the following example.

EX A MPL E 2

Finding the Slopes of Various Lines Containing the Same Point (2, 3) Compute the slopes of the lines L1 , L2 , L3 , and L4 containing the following pairs of QPJOUT
(SBQI
BMM
GPVS
MJOFT
PO
UIF
TBNF
TFU
PG
DPPSEJOBUF
BYFT L1 : L2 : L3 : L4 :

P P P P

= = = =

12, 32 12, 32 12, 32 12, 32

Q1 Q2 Q3 Q4

= = = =

1 - 1, - 22 13, - 12 15, 32 12, 52

SECTION F.3  Lines

Solution

Let m1 , m2 , m3 , and m4 denote the slopes of the lines L1 , L2 , L3 , and L4 , respectively. Then -2 - 3 -5 5 A rise of 5 divided by a run of 3 = = -1 - 2 -3 3 -1 - 3 -4 m2 = = = -4 3 - 2 1 3 - 3 0 m3 = = = 0 5 - 2 3 m4 is undefined because x1 = x2 = 2

Figure 25 L2

m1 =

L4

L1

y 5

Q4  (2, 5) P  (2, 3)

m3  0

Q3  (5, 3)

5

m1 

5– 3

L3

5 x Q2  (3, 1)

Q1  (1, 2)

53

3 m4 undefined

The graphs of these lines are given in Figure 25. Figure 25 illustrates the following facts:

m2  4

r

1. When the slope of a line is positive, the line slants upward from left to right 1L1 2. 2. When the slope of a line is negative, the line slants downward from left to right 1L2 2. 3. When the slope is 0, the line is horizontal 1L3 2. 4. When the slope is undefined, the line is vertical 1L4 2.

Seeing the Concept On the same screen, graph the following equations: Y1 = 0

Slope of line is 0.

Y2 =

1 x 4 1 Y3 = x 2

1 Slope of line is . 4 1 Slope of line is . 2

Y2  14 x

Y4 = x

Slope of line is 1.

3

Y5 = 2x

Slope of line is 2.

Y6 = 6x

Slope of line is 6.

Figure 26 Y6  6x Y5  2x Y4  x 2 Y3  12 x 3

Y1  0

See Figure 26.

2

Seeing the Concept On the same screen, graph the following equations:

Figure 27 Y6  6x Y5  2x Y4  x

2

Y3   12 x Y2   14 x 3

3

Y1  0 2

Y1 = 0

Slope of line is 0.

1 Y2 = - x 4 1 Y3 = - x 2 Y4 = - x

1 Slope of line is - . 4 1 Slope of line is - . 2 Slope of line is - 1.

Y5 = - 2x

Slope of line is - 2.

Y6 = - 6x

Slope of line is - 6.

See Figure 27.

Figures 26 and 27 illustrate that the closer the line is to the vertical position, the greater the absolute value of the slope.

54

CHAPTER F  Foundations: A Prelude to Functions

2 Graph Lines Given a Point and the Slope EX A MPL E 3

Graphing a Line Given a Point and the Slope

Draw a graph of the line that contains the point 13, 22 and has a slope of: (a)

Solution Figure 28 y 6 (7, 5) Rise = 3

3 4

(b) -

Rise 3 . The slope means that for every horizontal movement (run) of Run 4 4 units to the right, there will be a vertical movement (rise) of 3 units. Start at the given point, 13, 22 and move 4 units to the right and 3 units up arriving at the point 17, 52. Drawing the line through this point and the point 13, 22 gives the graph. See Figure 28. (b) The fact that the slope is (a) Slope =

(3, 2)

-

Run = 4 –2

10 x

5

4 -4 Rise = = 5 5 Run

means that for every horizontal movement of 5 units to the right, there will be a corresponding vertical movement of - 4 units (a downward movement). Start at the given point 13, 22 and move 5 units to the right and then 4 units down, arriving at the point 18, - 22. Drawing the line through these points gives the graph. See Figure 29.

Figure 29 y

Alternatively, consider

(–2, 6) 6

-

Rise = 4 Rise = – 4 10 x

–2

4 4 Rise = = 5 -5 Run

so that for every horizontal movement of - 5 units (a movement to the left), there will be a corresponding vertical movement of 4 units (upward). This approach leads to the point 1 - 2, 62, which is also on the graph of the line in Figure 29.

(3, 2) Run = 5 Run = –5 –2

4 5

r

(8, –2)

Now Work

PROBLEM

23

3 Find the Equation of a Vertical Line EX A MPL E 4

Graphing a Line (SBQI
UIF
FRVBUJPO
x = 3

Solution

To graph x = 3, find all points 1x, y2 in the plane for which x = 3. No matter what y-coordinate is used, the corresponding x-coordinate always equals 3. Consequently, the graph of the equation x = 3 is a vertical line with x-intercept 3 and undefined slope. See Figure 30.

Figure 30

y 4 (3, 3) (3, 2) (3, 1) 1 1

(3, 0) (3, 1)

5 x

r

SECTION F.3  Lines

55

Example 4 suggests the following result:

THEOREM

Equation of a Vertical Line A vertical line is given by an equation of the form x = a where a is the x-intercept.

COMMENT In order for an equation to be graphed using a graphing utility, the equation must be expressed in the form y = 5expression in x6 . But x = 3 cannot be put in this form. To overcome this, most graphing utilities have special commands for drawing vertical lines. LINE, PLOT, and VERT are among the more common ones. Consult your manual to determine the correct methodology for your graphing utility. ■ Figure 31

4 Use the Point–Slope Form of a Line; Identify Horizontal Lines

y

L (x, y )

Now let L be a nonvertical line with slope m that contains the point 1x1 , y1 2. See Figure 31. Any other point 1x, y2 on L gives

y – y1

(x 1, y1)

m =

x – x1

y - y1 x - x1

or y - y1 = m1x - x1 2

x

THEOREM

Point–Slope Form of an Equation of a Line An equation of a nonvertical line with slope m that contains the point 1x1 , y1 2 is y - y1 = m1x - x1 2

EXAM PL E 5 Figure 32 y 6

(2, 6) Rise  4

(1, 2)

Using the Point–Slope Form of a Line

An equation of the line with slope 4 that contains the point 11, 22 can be found by using the point–slope form with m = 4, x1 = 1, and y1 = 2. y - y1 = m1x - x1 2 y - 2 = 41x - 12

Run  1 2

y = 4x - 2 5 x

2

EXAM PL E 6

Solution

m = 4, x1 = 1, y1 = 2 Solve for y.

r

See Figure 32 for the graph.

Finding the Equation of a Horizontal Line

Find an equation of the horizontal line containing the point 13, 22. Because all the y-values are equal on a horizontal line, the slope of a horizontal line is 0. To get an equation, use the point–slope form with m = 0, x1 = 3, and y1 = 2. y - y1 = m1x - x1 2

Figure 33 y

y - 2 = 0 # 1x - 32

4 (3, 2)

–1

(2)

1

3

m = 0, x1 = 3, and y1 = 2

y - 2 = 0 y = 2

5 x

See Figure 33 for the graph.

r

56

CHAPTER F  Foundations: A Prelude to Functions

Example 6 suggests the following result:

THEOREM

Equation of a Horizontal Line A horizontal line is given by an equation of the form y = b where b is the y-intercept.

5 Find the Equation of a Line Given Two Points (JWFO
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find the equation.

Finding an Equation of a Line Given Two Points

EX A MPL E 7

Find an equation of the line containing the points 12, 32 and 1 - 4, 52.
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line.

Solution

First compute the slope of the line with 1x1, y1 2 = (2, 3) and 1x2, y2 2 = ( - 4, 5). m =

5 - 3 2 1 = = -4 - 2 -6 3

m =

y2 - y1 x2 - x1

1 Use the point 1x1, y1 2 = 12, 32 and the slope m = - to get the point–slope form 3 of the equation of the line.

Figure 34 y (–4, 5)

y - 3 = -

(2, 3)

1 1x - 22 3

y - y1 = m(x - x1)

2 –4

–2

10

x

r

See Figure 34 for the graph.

In the solution to Example 7, we could have used the other point, 1 - 4, 52 , instead of the point 12, 32. The equation that results, although it looks different, is equivalent to the equation that we obtained in the example. (Try it for yourself.)

Now Work

PROBLEM

37

6 Write the Equation of a Line in Slope–Intercept Form Another useful equation of a line is obtained when the slope m and y-intercept b are known. In this event, both the slope m of the line and a point 10, b2 on the line are known; then use the point–slope form, equation (2), to obtain the following equation: y - b = m1x - 02

THEOREM

or y = mx + b

Slope–Intercept Form of an Equation of a Line An equation of a line with slope m and y-intercept b is y = mx + b

Now Work

PROBLEMS

45

AND

51 (EXPRESS

SLOPE–INTERCEPT FORM)

(3)

ANSWERS IN

SECTION F.3  Lines

Figure 35 y = mx + 2

Seeing the Concept

Y4  3x  2 Y2  x  2

Y5  3x  2 Y3  x  2 4

57

To see the role that the slope m plays, graph the following lines on the same square screen.

Y1  2 6

6

Y1 = 2

m = 0

Y2 = x + 2

m = 1

Y3 = - x + 2

m = -1

Y4 = 3x + 2

m = 3

Y5 = - 3x + 2

m = -3

See Figure 35. What do you conclude about the lines y = mx + 2?

4

Seeing the Concept Figure 36 y = 2x + b Y4  2x  4 4

To see the role of the y-intercept b, graph the following lines on the same square screen.

Y2  2x  1 Y1  2x Y3  2x  1 Y5  2x  4

6

Y1 = 2x

b = 0

Y2 = 2x + 1

b = 1

Y3 = 2x - 1

b = -1

Y4 = 2x + 4

b = 4

Y5 = 2x - 4

b = -4

6

See Figure 36. What do you conclude about the lines y = 2x + b?

4

7 Identify the Slope and y-Intercept of a Line from Its Equation When the equation of a line is written in slope–intercept form, it is easy to find the slope m and y-intercept b of the line. For example, suppose that the equation of a line is y = - 2x + 3 Compare this equation to y = mx + b. y = - 2x + 3 y =

c

c

mx + b

The slope of this line is - 2 and its y-intercept is 3.

Now Work

EXAM PL E 8

PROBLEM

71

Finding the Slope and y-Intercept Find the slope m and y-intercept b of the equation 2x + 4y = 8.
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equation.

Solution

To obtain the slope and y-intercept, write the equation in slope–intercept form by solving for y. 2x + 4y = 8 4y = - 2x + 8 1 y = - x + 2 2

Figure 37 2x + 4y = 8 y 4 (0, 2) –3

2 1 (2, 1) 3

x

y = mx + b

1 The coefficient of x, - , is the slope, and the constant, 2, is the yJOUFSDFQU
(SBQI 2 1 the line with the y-intercept 2 and the slope - . Starting at the point 10, 22, go to 2 the right 2 units and then down 1 unit to the point 12, 12. Draw the line through these points. See Figure 37.

r

Now Work

PROBLEM

77

58

CHAPTER F  Foundations: A Prelude to Functions

8 Graph Lines Written in General Form Using Intercepts Refer to Example 8. The form of the equation of the line 2x + 4y = 8 is called the general form.

DEFINITION

The equation of a line is in general form* when it is written as Ax + By = C

(4)

where A, B, and C are real numbers and A and B are not both 0. If B = 0 in equation (4), then A ≠ 0 and the graph of the equation is a vertical C line: x = . If B ≠ 0 in equation (4), then we can solve the equation for y and write A the equation in slope–intercept form as we did in Example 8. Another approach to graphing equation (4) would be to find its intercepts. Remember, the intercepts of the graph of an equation are the points where the graph crosses or touches a coordinate axis.

EX A MPL E 9

Graphing an Equation in General Form Using Its Intercepts (SBQI
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2x + 4y = 8 by finding its intercepts.

Solution

To obtain the x-intercept, let y = 0 in the equation and solve for x. 2x + 4y = 8 2x + 4(0) = 8

Let y = 0.

2x = 8 x = 4

Divide both sides by 2.

The x-intercept is 4 and the point 14, 02 is on the graph of the equation. To obtain the y-intercept, let x = 0 in the equation and solve for y. 2x + 4y = 8 2(0) + 4y = 8 4y = 8

Figure 38

y = 2

y 4

(4, 0) 3

Divide both sides by 4.

The y-intercept is 2 and the point 10, 22 is on the graph of the equation. Plot the points 14, 02 and 10, 22 and draw the line through the points. See Figure 38.

(0, 2)

–3

Let x = 0.

x

r

Now Work P R O B L E M 9 1 Every line has an equation that is equivalent to an equation written in general form. For example, a vertical line whose equation is x = a can be written in the general form

1#x + 0#y = a

A = 1, B = 0, C = a

A horizontal line whose equation is y = b can be written in the general form

*Some books use the term standard form.

0#x + 1#y = b

A = 0, B = 1, C = b

SECTION F.3  Lines

59

Lines that are neither vertical nor horizontal have general equations of the form Ax + By = C

A ≠ 0 and B ≠ 0

Because the equation of every line can be written in general form, any equation equivalent to equation (4) is called a linear equation. Figure 39

9 Find Equations of Parallel Lines

y

Rise Run

Rise Run x

THEOREM

When two lines (in the same plane) do not intersect (that is, they have no points in common), they are parallel. Look at Figure 39. There we have drawn two lines and have constructed two right triangles by drawing sides parallel to the coordinate axes. These lines are parallel if and only if the right triangles are similar. (Do you see why? Two angles are equal.) And the triangles are similar if and only if the ratios of corresponding sides are equal.

Criteria for Parallel Lines Two nonvertical lines are parallel if and only if their slopes are equal and they have different y-intercepts. The use of the words “if and only if” in the preceding theorem means that actually two statements are being made, one the converse of the other. If two nonvertical lines are parallel, then their slopes are equal and they have different y-intercepts. If two nonvertical lines have equal slopes and they have different y-intercepts, then they are parallel.

EX AM PL E 10

Showing That Two Lines Are Parallel Show that the lines given by the following equations are parallel. L1 : 2x + 3y = 6,

Solution Figure 40

To determine whether these lines have equal slopes and different y-intercepts, write each equation in slope–intercept form. L1 : 2x + 3y = 6

y 5

5 x L1 3

L2 : 4x + 6y = 0

3y = - 2x + 6 y = -

5

L2 : 4x + 6y = 0

L2

EX AM PL E 11

Solution

6y = - 4x

2 x + 2 3

2 Slope = - ; y@intercept = 2 3

y = -

2 x 3

2 Slope = - ; y@intercept = 0 3

2 Because these lines have the same slope, - , but different y-intercepts, the lines are 3 parallel. See Figure 40.

r

Finding a Line That Is Parallel to a Given Line

Find an equation for the line that contains the point 12, - 32 and is parallel to the line 2x + y = 6. Since the two lines are to be parallel, the slope of the line being sought equals the slope of the line 2x + y = 6. Begin by writing the equation of the line 2x + y = 6 in slope–intercept form. 2x + y = 6 y = - 2x + 6

60

CHAPTER F  Foundations: A Prelude to Functions

The slope is - 2. Since the line being sought contains the point 12, - 32, use the point–slope form to obtain

Figure 41 y

y - y1 y - 1 - 32 y + 3 y 2x + y

6

6

6 x 2x  y  6 (2, 3) 5

m1x - x1 2 - 21x - 22 - 2x + 4 - 2x + 1 1

Point–slope form m = - 2, x1 = 2, y1 = - 3 Simplify. Slope–intercept form General form

This line is parallel to the line 2x + y = 6 and contains the point 12, - 32 . See Figure 41.

r

Now Work

2x  y  1

= = = = =

PROBLEM

59

10 Find Equations of Perpendicular Lines When two lines intersect at a right angle (90°), they are perpendicular. See Figure 42.

THEOREM

Criterion for Perpendicular Lines Two nonvertical lines are perpendicular if and only if the product of their slopes is - 1.

Figure 42 y

Here we shall prove the “only if” part of the statement: If two nonvertical lines are perpendicular, then the product of their slopes is - 1.

90° x

In Problem 128, you are asked to prove the “if” part of the theorem: If two nonvertical lines have slopes whose product is - 1, then the lines are perpendicular.

Figure 43 y Slope m2 A = (1, m2) Slope m1

Rise = m 2 x

Run = 1 O

1

Rise = m1

B = (1, m1)

Proof Let m1 and m2 denote the slopes of the two lines. There is no loss in generality (that is, neither the angle nor the slopes are affected) if we situate the lines so that they meet at the origin. See Figure 43. The point A = 11, m2 2 is on the line having slope m2 , and the point B = 11, m1 2 is on the line having slope m1 . (Do you see why this must be true?) Suppose that the lines are perpendicular. Then triangle OAB is a right triangle. As a result of the Pythagorean Theorem, it follows that 3 d 1O, A2 4 2 + 3 d 1O, B2 4 2 = 3 d 1A, B2 4 2

(5)

By the distance formula, we can write the squares of these distances as

3 d 1O, A2 4 2 = 11 - 02 2 + 1m2 - 02 2 = 1 + m22 3 d 1O, B2 4 2 = 11 - 02 2 + 1m1 - 02 2 = 1 + m21 3 d 1A, B2 4 2 = 11 - 12 2 + 1m2 - m1 2 2 = m22 - 2m1 m2 + m21

Using these facts in equation (5), we get

11 + m22 2 + 11 + m21 2 = m22 - 2m1 m2 + m21

which, upon simplification, can be written as m1 m2 = - 1 If the lines are perpendicular, the product of their slopes is - 1.

■

You may find it easier to remember the condition for two nonvertical lines to be perpendicular by observing that the equality m1 m2 = - 1 means that m1 and m2 are 1 1 negative reciprocals of each other; that is, either m1 = or m2 = . m2 m1

SECTION F.3  Lines

61

Finding the Slope of a Line Perpendicular to Another Line

EX AM PL E 12

3 2 If a line has slope , any line having slope - is perpendicular to it. 2 3

Finding the Equation of a Line Perpendicular to a Given Line

EX AM PL E 13

Find an equation of the line that contains the point 11, - 22 and is perpendicular to the line x + 3y = 6.
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Solution

First write the equation of the given line in slope–intercept form to find its slope. x + 3y = 6 3y = - x + 6 1 y = - x + 2 3

y x  3y  6

Proceed to solve for y. Place in the form y = mx + b.

1 The given line has slope - . Any line perpendicular to this line will have slope 3. 3 Because the point 11, - 22 is on this line with slope 3, use the point–slope form of the equation of a line.

Figure 44 y  3x  5

6

y - y1 = m1x - x1 2 y - 1 - 22 = 31x - 12

4 2 x

2

2

4

Point–slope form m = 3, x1 = 1, y1 = - 2

To obtain other forms of the equation, proceed as follows:

6

y + 2 y + 2 y 3x - y

(1, 2)

2

r

4

= = = =

31x - 12 3x - 3 3x - 5 5

Simplify. Slope–intercept form General form

r

Figure 44 shows the graphs.

Now Work

PROBLEM

65

F.3 Assess Your Understanding Concepts and Vocabulary 1. The slope of a vertical line is horizontal line is 0 .

7. Two nonvertical lines have slopes m1 and m2, respectively. The lines are parallel if m1 = m2 and the y-intercepts are unequal; the lines are perpendicular if m1m2 = - 1 .

undefined ; the slope of a

2. For the line 2x + 3y = 6, the x-intercept is y-intercept is 2 .

3

and the

8. The lines y = 2x + 3 and y = ax + 5 are parallel if 2 . a =

3. A horizontal line is given by an equation of the form y = b , where b is the y-intercept .

9. The lines y = 2x - 1 and y = ax + 2 are perpendicular if a = −½ .

4. True or False Vertical lines have an undefined slope. True

10. True or False Perpendicular lines have slopes that are reciprocals of one another. False

5. True or False The slope of the line 2y = 3x + 5 is 3. False 6. True or False The point 11, 22 is on the line 2x + y = 4. True

Skill Building In Problems 11–14, (a) find the slope of the line and (b) interpret the slope. *11.

*12.

y 2

(–2, 1) 2

(2, 1)

(0, 0) –2

–1

*13.

y

*14.

y (–2, 2)

2

(1, 1)

y (–1, 1)

2

(2, 2)

(0, 0) 2

x

–2

–1

2

x

–2

–1

2

x

–2

–1

2

In Problems 15–22, plot each pair of points and determine the slope of the line containing them. Graph the line. *15. 12, 32; 14, 02

*19. 1 - 3, - 12; 12, - 12

*16. 14, 22; 13, 42

*20. 14, 22; 1 - 5, 22

*17. 1 - 2, 32; 12, 12

*21. 1 - 1, 22; 1 - 1, - 22

*18. 1 - 1, 12; 12, 32

*22. 12, 02; 12, 22

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

x

62

CHAPTER F  Foundations: A Prelude to Functions

In Problems 23–30, graph the line containing the point P and having slope m. *23. P = 11, 22; m = 3

*27. P = 1 - 1, 32; m = 0

3 2 *26. P = 11, 32; m = 4 5 *29. P = 10, 32 ; slope undefined *30. P = 1 - 2, 02; slope undefined

*24. P = 12, 12; m = 4

*25. P = 12, 42; m = -

*28. P = 12, - 42; m = 0

In Problems 31–36, the slope and a point on a line are given. Use this information to locate three additional points on the line. Answers may vary. [Hint: It is not necessary to find the equation of the line. See Example 2.] 3 *31. Slope 4; point 11, 22 *33. Slope - ; point 12, - 42 *32. Slope 2; point 1 - 2, 32 2 4 *36. Slope - 1; point 14, 12 *35. Slope - 2; point 1 - 2, - 32 *34. Slope ; point 1 - 3, 22 3 In Problems 37–44, find an equation of the line L. *37.

*38.

y 2

(2, 1)

*39. L

y

L

(–2, 1) 2

(–1, 3)

(0, 0) –2

–1

x

y

*41.

3

–2

*42.

–1

3

(3, 3)

–2

*43.

y

3

–2

L

(1, 2)

–1

x

2

–1 L

3 x

L is parallel to y = 2x

–1

L

(2, 2)

2

–1

x

y 3

*44.

y 3 (1, 2) L

L y = 2x

2

(–1, 1)

x

2

y

(1, 1)

(0, 0) 2

*40.

y

3 x y = –x

L is parallel to y = –x

–1 y = 2x

3 x

L is perpendicular to y = 2x

(–1, 1) –3 L

1 x y = –x

L is perpendicular to y = –x

In Problems 45–70, find an equation for the line with the given properties. Express your answer using either the general form or the slope–intercept form of the equation of a line, whichever you prefer. *45. Slope = 3; containing the point 1 - 2, 32 2 *47. Slope = - ; containing the point 11, - 12 3 *49. Containing the points 11, 32 and 1 - 1, 22

*46. Slope = 2; containing the point 14, - 32 1 *48. Slope = ; containing the point 13, 12 2 *50. Containing the points 1 - 3, 42 and 12, 52

*51. Slope = - 3; y@intercept = 3

*52. Slope = - 2; y@intercept = - 2

*53. x@intercept = 2; y@intercept = - 1

*54. x@intercept = - 4; y@intercept = 4

*55. Slope undefined; containing the point 12, 42 57. Horizontal; containing the point 1 - 3, 22

y = 2

*59. Parallel to the line y = 2x; containing the point 1 - 1, 22

*61. Parallel to the line 2x - y = - 2; containing the point 10, 02

*63. Parallel to the line x = 5; containing the point 14, 22

1 65. Perpendicular to the line y = x + 4; containing the point 2 11, - 22 2x + y = 0 or y = - 2x 67. Perpendicular to the line 2x + y = 2; containing the point 1 - 3, 02 x - 2y = - 3 or y = 1x + 3 2 2 *69. Perpendicular to the line x = 8; containing the point 13, 42

*56. Slope undefined; containing the point 13, 82

*58. Vertical; containing the point 14, - 52

*60. Parallel to the line y = - 3x; containing the point 1 - 1, 22

*62. Parallel to the line x - 2y = - 5; containing the point 10, 02 64. Parallel to the line y = 5; containing the point 14, 22

y = 2

66. Perpendicular to the line y = 2x - 3; containing the point 11, - 22 x + 2y = - 3 or y = - 1x - 3 2 2 68. Perpendicular to the line x - 2y = - 5; containing the point 10, 42 2x + y = 4 or y = - 2x + 4

*70. Perpendicular to the line y = 8; containing the point 13, 42

SECTION F.3  Lines

63

In Problems 71–90, find the slope and y-intercept of each line. Graph the line. *71. y = 2x + 3 *76. y = 2x + *81. x + y = 1

1 2

*86. x = 2

*72. y = - 3x + 4

*73.

1 y = x - 1 2

*74.

1 x + y = 2 3

*75. y =

1 x + 2 2

*77. x + 2y = 4

*78. - x + 3y = 6

*79. 2x - 3y = 6

*80. 3x + 2y = 6

*82. x - y = 2

*83. x = - 4

*84. y = - 1

*85. y = 5

*87. y - x = 0

*88. x + y = 0

*89. 2y - 3x = 0

*90. 3x + 2y = 0

In Problems 91–100, (a) find the intercepts of the graph of each equation and (b) graph the equation. *91. 2x + 3y = 6

*92. 3x - 2y = 6

1 1 *97. x + y = 1 2 3 *100. - 0.3x + 0.4y = 1.2

*96. 5x + 3y = 18

*95. 7x + 2y = 21

*94. 6x - 4y = 24

*93. - 4x + 5y = 40

*99. 0.2x - 0.5y = 1

*98. x -

2 y = 4 3

102. Find an equation of the y-axis. x = 0

101. Find an equation of the x-axis. y = 0

In Problems 103–106, the equations of two lines are given. Determine whether the lines are parallel, perpendicular, or neither. 1 103. y = 2x - 3 105. y = 4x + 5 Neither 104. y = x - 3 106. y = - 2x + 3 Neither 2 y = 2x + 4 Parallel y = - 4x + 2 1 y = - 2x + 4 y = - x + 2 2 Perpendicular In Problems 107–110, match each graph with the correct equation: (a) y = x 107. (b)

(b) y = 2x 108. (c)

8

6

6

(c) y = 4

12

8

109. (d)

12

4

x 2

(d) y = 4x 110. (a)

8

3

3

8

2

3

3

2

Applications and Extensions *111. Geometry Use slopes to show that the triangle whose vertices are 1 - 2, 52 , 11, 32 , and 1 - 1, 02 is a right triangle. *112. Geometry Use slopes to show that the quadrilateral whose vertices are 11, - 12 , 14, 12 , 12, 22 , and 15, 42 is a parallelogram. *113. Geometry Use slopes to show that the quadrilateral whose vertices are 1 - 1, 02 , 12, 32 , 11, - 22 , and 14, 12 is a rectangle. *114. Geometry Use slopes and the distance formula to show that the quadrilateral whose vertices are 10, 02 , 11, 32 , 14, 22 , and 13, - 12 is a square. 115. Truck Rentals A truck rental company rents a moving truck for one day by charging $39 plus $0.60 per mile. Write a linear equation that relates the cost C, in dollars, of renting the truck to the number x of miles driven. What is the cost of renting the truck if the truck is driven 110 miles? 230 miles? C = 0.60x + 39; $105.00; $177.00 116. Cost Equation The fixed costs of operating a business are the costs incurred regardless of the level of production. Fixed costs include rent, fixed salaries, and costs of leasing machinery. The variable costs of operating a business are the costs that change with the level of output. Variable costs include raw materials, hourly wages, and

electricity. Suppose that a manufacturer of jeans has fixed daily costs of $500 and variable costs of $8 for each pair of jeans manufactured. Write a linear equation that relates the daily cost C, in dollars, of manufacturing the jeans to the number x of jeans manufactured. What is the cost of manufacturing 400 pairs of jeans? 740 pairs? C = 8x + 500; $3700; $6420 117. Cost of Driving a Car The annual fixed costs for owning a small sedan are $6735, assuming the car is completely paid for. The cost to drive the car is approximately $0.45 per mile. Write a linear equation that relates the cost C and the number x of miles driven annually. C = 0.45x + 6735 Source: AAA, April 2012 118. Wages of a Car Salesperson Dan receives $375 per week for selling new and used cars at a car dealership in Oak Lawn, Illinois. In addition, he receives 5% of the profit on any sales that he generates. Write a linear equation that represents Dan’s weekly salary S when he has sales that generate a profit of x dollars. S = 0.05x + 375 119. Electricity Rates in Illinois Commonwealth Edison Company supplies electricity to residential customers for a monthly customer charge of $13.04 plus 10.62 cents per kilowatt-hour for up to 800 kilowatt-hours (kw-hr).

64

CHAPTER F  Foundations: A Prelude to Functions

*123. Access Ramp A wooden access ramp is being built to reach a platform that sits 30 inches above the floor. The ramp drops 2 inches for every 25-inch run. y Platform 30"

Ramp x

(a) Write a linear equation that relates the monthly charge C, in dollars, to the number x of kilowatt-hours used in a month, 0 … x … 800. C = 0.1062x + 13.04 * C
(SBQI
UIJT
FRVBUJPO (c) What is the monthly charge for using 200 kilowatthours? $34.28 (d) What is the monthly charge for using 500 kilowatthours? $66.14 *(e) Interpret the slope of the line. Source: Commonwealth Edison Company, January 2013. 120. Electricity Rates in Florida Florida Power & Light Company supplies electricity to residential customers for a monthly customer charge of $7.00 plus 8.55 cents per kilowatt-hour for up to 1000 kilowatt-hours (kw-hr). (a) Write a linear equation that relates the monthly charge C, in dollars, to the number x of kilowatt-hours used in a month, 0 … x … 1000. C = 0.0855x + 7.00 * C
(SBQI
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FRVBUJPO (c) What is the monthly charge for using 200 kilowatthours? $24.10 (d) What is the monthly charge for using 500 kilowatthours? $49.75 *(e) Interpret the slope of the line. Source: Florida Power & Light Company, January 2013. *121. Measuring Temperature The relationship between Celsius (°C) and Fahrenheit (°F) degrees of measuring temperature is linear. Find a linear equation relating °C and °F if 0°C corresponds to 32°F and 100°C corresponds to 212°F. Use the equation to find the Celsius measure of 70°F. *122. Measuring Temperature The Kelvin (K) scale for measuring temperature is obtained by adding 273 to the Celsius temperature. (a) Write a linear equation relating K and °C. (b) Write a linear equation relating K and °F (see Problem 121).

(a) Write a linear equation that relates the height y of the ramp above the floor to the horizontal distance x from the platform. (b) Find and interpret the x-intercept of the graph of your equation. (c) Design requirements stipulate that the maximum run be 30 feet and that the maximum slope be a drop of 1 inch for every 12 inches of run. Will this ramp meet the requirements? Explain. (d) What slopes could be used to obtain the 30-inch rise and still meet design requirements? Source: www.adaptiveaccess.com/wood_ramps.php 124. Cigarette Use A report in the Child Trends DataBase indicated that in 2000, 20.6% of twelfth grade students reported daily use of cigarettes. In 2011, 10.3% of twelfth grade students reported daily use of cigarettes. *(a) Write a linear equation that relates the percent y of twelfth grade students who smoke cigarettes daily to the number x of years after 2000. *(b) Find the intercepts of the graph of your equation. *(c) Do these intercepts have any meaningful interpretation? (d) Use your equation to predict the percent for the year 2025. Is this result reasonable? - 2.8%; No Source: www.childtrends.org/databank 125. Product Promotion A cereal company finds that the number of people who will buy one of its products in the first month that the product is introduced is linearly related to the amount of money it spends on advertising. If it spends $40,000 on advertising, then 100,000 boxes of cereal will be sold, and if it spends $60,000, then 200,000 boxes will be sold. *(a) Write a linear equation that relates the amount A spent on advertising to the number x of boxes the company aims to sell. (b) How much expenditure on advertising is needed to sell 300,000 boxes of cereal? $80,000 *(c) Interpret the slope.

*126. Show that the line containing the points 1a, b2 and 1b, a2 , a ≠ b, is perpendicular to the line y = x. Also show that the midpoint of (a, b) and 1b, a2 lies on the line y = x. *127. The equation 2x - y = C defines a family of lines, one line for each value of C. On one set of coordinate axes, graph the members of the family when C = - 4, C = 0, and C = 2. Can you draw a conclusion from the graph about each member of the family? *128. Prove that if two nonvertical lines have slopes whose product is - 1, then the lines are perpendicular. [Hint: Refer to Figure 43 and use the converse of the Pythagorean Theorem.]

SECTION F.3  Lines

65

Discussion and Writing 129. Which of the following equations might have the graph shown? (More than one answer is possible.) (a) 2x + 3y = 6 (b) - 2x + 3y = 6

y

(c) 3x - 4y = - 12 (d) x - y = 1 (e) x - y = - 1

x

(f) y = 3x - 5 (g) y = 2x + 3 (h) y = - 3x + 3

133. m Is for Slope The accepted symbol used to denote the slope of a line is the letter m. Investigate the origin of this symbolism. Begin by consulting a French dictionary and looking up the French word monter. Write a brief essay on your findings. 134. Grade of a Road The term grade is used to describe the inclination of a road. How is this term related to the notion of slope of a line? Is a 4% grade very steep? Investigate the grades of some mountainous roads and determine their slopes. Write a brief essay on your findings.

(b), (c), (e), (g)

130. Which of the following equations might have the graph shown? (More than one answer is possible.) (a) 2x + 3y = 6 (b) 2x - 3y = 6

Steep 7% Grade

(c) 3x + 4y = 12

y

(d) x - y = 1 (e) x - y = - 1

x

(f) y = - 2x - 1 (g) y = -

1 x + 10 2

(h) y = x + 4

(a), (c), (g)

131. The figure shows the graph of two parallel lines. Which of the following pairs of equations might have such a graph? (c) (a) x - 2y = 3 x + 2y = 7 (b) x + y = 2 x + y = -1

135. Carpentry Carpenters use the term pitch to describe the steepness of staircases and roofs. How is pitch related to slope? Investigate typical pitches used for stairs and for roofs. Write a brief essay on your findings. 136. Can the equation of every line be written in slope–intercept form? Why? No, consider vertical lines. *137. Does every line have exactly one x-intercept and one y-intercept? Are there any lines that have no intercepts?

y

138. What can you say about two lines that have equal slopes and equal y-intercepts? They are the same.

(c) x - y = - 2 x - y = 1

x

(d) x - y = - 2 2x - 2y = - 4

139. What can you say about two lines with the same x-intercept and the same y-intercept? Assume that the x-intercept is not 0. They are the same line.

(e) x + 2y = 2 x + 2y = - 1

140. If two distinct lines have the same slope but different x-intercepts, can they have the same y-intercept? No

132. The figure shows the graph of two perpendicular lines. Which of the following pairs of equations might have such a graph? (d) (a) y - 2x = 2 y + 2x = - 1 (b) y - 2x = 0 2y + x = 0 (c) 2y - x = 2 2y + x = - 2 (d) y - 2x = 2 x + 2y = - 1 (e) 2x + y = - 2 2y + x = - 2

y

x

*141. If two distinct lines have the same y-intercept but different slopes, can they have the same x-intercept? 142. Which form of the equation of a line do you prefer to use? Justify your position with an example that shows that your choice is better than another. Have reasons. *143. What Went Wrong? A student is asked to find the slope of the line joining 1 - 3, 22 and 11, - 42 . He states that the 3 slope is . Is he correct? If not, what went wrong? 2

66

CHAPTER F  Foundations: A Prelude to Functions

F.4 Circles PREPARING FOR THIS SECTION  Before getting started, review the following: r
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pp. A38–A39)

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pp. A63–A69)

Now Work the ‘Are You Prepared?’ problems on page 70.

OBJECTIVES 1 Write the Standard Form of the Equation of a Circle (p. 66) 2 Graph a Circle (p. 67) 3 Work with the General Form of the Equation of a Circle (p. 68)

1 Write the Standard Form of the Equation of a Circle One advantage of a coordinate system is that it enables us to translate a geometric statement into an algebraic statement, and vice versa. Consider, for example, the following geometric statement that defines a circle.

DEFINITION

Figure 45 shows the graph of a circle. To find the equation, let 1x, y2 represent the coordinates of any point on a circle with radius r and center 1h, k). Then the distance between the points 1x, y2 and 1h, k2 must always equal r. That is, by the distance formula,

Figure 45 y

A circle is a set of points in the xy-plane that are a fixed distance r from a fixed point 1h, k2. The fixed distance r is called the radius, and the fixed point 1h, k2 is called the center of the circle.

(x, y) r

2 1x - h2 2 + 1y - k2 2 = r

(h, k ) x

or, equivalently, 1x - h2 2 + 1y - k2 2 = r 2

DEFINITION

The standard form of an equation of a circle with radius r and center 1h, k2 is 1x - h2 2 + 1y - k2 2 = r 2

THEOREM

(1)

The standard form of an equation of a circle of radius r with center at the origin 10, 02 is x2 + y 2 = r 2

DEFINITION

If the radius r = 1, the circle whose center is at the origin is called the unit circle and has the equation x2 + y 2 = 1

See Figure 46. Note that the graph of the unit circle is symmetric with respect to the x-axis, the y-axis, and the origin.

SECTION F.4   Circles

Figure 46 Unit circle x2 + y2 = 1

67

y 1

1

1x

(0,0) 1

EXAM PL E 1

Solution

Writing the Standard Form of the Equation of a Circle

Write the standard form of the equation of the circle with radius 5 and center 1 - 3, 62.

Substitute the values r = 5, h = - 3, and k = 6, into equation (1). 1x - h2 2 + 1y - k2 2 = r 2

(x - ( - 3))2 + (y - 6)2 = 52

1x + 32 2 + 1y - 62 2 = 25

Now Work

PROBLEM

7

r

2 Graph a Circle

The graph of any equation of the form 1x - h)2 + (y - k)2 = r 2 is that of a circle with radius r and center 1h, k2.

EXAM PL E 2

Solution

Graphing a Circle

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1x + 32 2 + 1y - 22 2 = 16 Since the equation is in the form of equation (1), its graph is a circle. To graph the equation, compare the given equation to the standard form of the equation of a circle. The comparison yields information about the circle. 1x + 32 2 + 1y - 22 2 = 16 1x - 1 - 32 2 2 + 1y - 22 2 = 42

Figure 47 (–3, 6)

y 6

c

–10

(–3, 2)

c

1x - h2 + 1y - k2 = r 2

4 (–7, 2)

c

2

(1, 2) 2 x

–5 (–3, –2)

We see that h = - 3, k = 2, and r = 4. The circle has center 1 - 3, 22 and a radius of 4 units. To graph this circle, first plot the center 1 - 3, 22. Since the radius is 4, locate four points on the circle by plotting points 4 units to the left, to the right, and up and down from the center. Then use these four points as guides to obtain the graph. See Figure 47.

Now Work

EXAM PL E 3

Solution

2

PROBLEMS

23(A)

r

AND

(B)

Finding the Intercepts of a Circle

For the circle 1x + 32 2 + 1y - 22 2 = 16, find the intercepts, if any, of its graph. This is the equation discussed and graphed in Example 2. To find the x-intercepts, if any, let y = 0 and solve for x. Then 1x + 32 2 + 1y - 22 2 = 16

In Words

The symbol { is read “plus or minus.” It means to add and subtract the quantity following the { symbol. For example, 5 { 2 means “5 - 2 = 3 or 5 + 2 = 7.”

1x + 32 2 + 10 - 22 2 = 16 1x + 32 + 4 = 16 2

1x + 32 2 = 12 x + 3 = { 212 x = - 3 { 223

y = 0 Simplify. Subtract 4 from both sides. Solve for x + 3. Solve for x

The x-intercepts are - 3 - 223 ≈ - 6.46 and - 3 + 223 ≈ 0.46.

68

CHAPTER F  Foundations: A Prelude to Functions

To find the y-intercepts, if any, let x = 0 and solve for y. Then 1x + 32 2 + 1y 10 + 32 2 + 1y 9 + 1y 1y y

22 2 22 2 22 2 22 2 - 2 y

= = = = = =

16 16 16 7 { 27 2 { 27

x = 0

Solve for y - 2. Solve for y.

The y-intercepts are 2 - 27 ≈ - 0.65 and 2 + 27 ≈ 4.65. Look back at Figure 47 to verify the approximate locations of the intercepts.

Now Work

PROBLEM

23(C)

r

3 Work with the General Form of the Equation of a Circle Eliminate the parentheses from the standard form of the equation of the circle given in Example 3 to obtain 1x + 32 2 + 1y - 22 2 = 16 x + 6x + 9 + y2 - 4y + 4 = 16 2

which simplifies to x2 + y2 + 6x - 4y - 3 = 0 It can be shown that any equation of the form x2 + y2 + ax + by + c = 0 has a graph that is a circle, or has a graph that is a point, or has no graph at all. For example, the graph of the equation x2 + y2 = 0 is the single point 10, 02 . The equation x2 + y2 + 5 = 0, or x2 + y2 = - 5, has no graph, because sums of squares of real numbers are never negative.

DEFINITION

When its graph is a circle, the equation x2 + y2 + ax + by + c = 0 is the general form of the equation of a circle.

If an equation of a circle is in the general form, the method of completing the square can be used to put the equation in standard form so that its center and radius can be identified.

EX A MPL E 4

Graphing a Circle Whose Equation Is in General Form (SBQI
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x2 + y2 + 4x - 6y + 12 = 0

Solution

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x, group the terms involving y, and put the constant on the right side of the equation. The result is 1x2 + 4x2 + 1y2 - 6y2 = - 12 Next, complete the square of each expression in parentheses. Remember that any number added on the left side of the equation must be added on the right.

SECTION F.4   Circles

1x2 + 4x + 42 + 1y2 - 6y + 92 = - 12 + 4 + 9

Figure 48

c

y

(2, 4)

c

6

(1, 3)

4

1 (3, 3)

69

6

4 a b = 4 2 2

a

-6 2 b = 9 2

1x + 22 2 + 1y - 32 2 = 1

(2, 3) (2, 2)

3

1 x

This equation is the standard form of the equation of a circle with radius 1 and center 1 - 2, 32. To graph the equation, use the center 1 - 2, 32 and the radius 1. See Figure 48.

Now Work

EXAM PL E 5

Solution Figure 49

Factor

PROBLEM

r

27

Finding the General Equation of a Circle

Find the general equation of the circle whose center is 11, - 22 and whose graph contains the point 14, - 22. To find the equation of a circle, we need to know its center and its radius. Here, the center is 11, - 22. Since the point 14, - 22 is on the graph, the radius r will equal the distance from 14, - 22 to the center 11, - 22. See Figure 49. Thus,

y 3

r = 2 14 - 12 2 + 3 - 2 - 1 - 22 4 2 = 29 = 3 r (1, 2)

x

5 (4, 2)

5

The standard form of the equation of the circle is

1x - 12 2 + 1y + 22 2 = 9

Eliminate the parentheses and rearrange the terms to get the general equation x2 + y2 - 2x + 4y - 4 = 0

Now Work

PROBLEM

13

COMMENT Example 6 requires information from Section B.5, Square Screens, in Appendix B.

EXAM PL E 6

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Using a Graphing Utility to Graph a Circle (SBQI
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x2 + y2 = 4

Solution Figure 50 2

This is the equation of a circle with center at the origin and radius 2. To graph this equation, first solve for y. x2 + y 2 = 4

Y1 = 4 − x 2

y 2 = 4 - x2 y = { 24 - x2

−3

3

−2

Y2 = − 4 − x 2

Subtract x 2 from each side. Solve for y.

There are two equations to graph on the same square screen: Y1 = 24 - x2 and Y2 = - 24 - x2. (Your circle will appear oval if you do not use a square screen.) See Figure 50. The graph is “disconnected” because of the resolution of the calculator.

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Overview The discussion in Sections F.3 and F.4 about lines and circles dealt with two main types of problems that can be generalized as follows: 1. (JWFO
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70

CHAPTER F  Foundations: A Prelude to Functions

F.4 Assess Your Understanding ‘Are You Prepared?’  Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 2. Solve the equation 1x - 2)2 = 9. (pp. A63–A69)

1. To complete the square of x2 + 10x, you would (add/ subtract) the number 25 . (pp. A38–A39) add

{ - 1, 5}

Concepts and Vocabulary 3. True or False Every equation of the form

5. The standard form of an equation of a circle with radius r and center (h, k) is 1x - h2 2 + 1y - k2 2 = r 2.

x + y + ax + by + c = 0 2

2

6. The circle whose equation is given by x2 + y2 = 1 is called unit the circle.

has a circle as its graph. False radius is the distance from the center 4. For a circle, the to any point on the circle.

Skill Building In Problems 7–10, find the center and radius of each circle. Write the standard form of the equation. 8.

7. y

9. y

y

(4, 2)

(2, 3) (h, k)

(h, k)

(1, 2) (0, 1)

10. y

(2, 1) (0, 1) (1, 2)

x (1, 0)

center: (2, 1); radius: 2 1x - 22 2 + 1y - 12 2 = 4

x

center: (1, 2); radius: 2

1x - 12 2 + 1y - 22 2 = 4

x

x

3 5 center: a , 2b ; radius: 2 2 9 5 2 ax - b + 1y - 22 2 = 2 4

center: (1, 2); radius: 22 1x - 12 2 + 1y - 22 2 = 2

In Problems 11–20, write the standard form of the equation and the general form of the equation of each circle of radius r and center 1h, k2. Graph each circle. *11. r = 2; *15. r = 5; *19. r =

1 ; 2

1h, k2 = 10, 02

*12. r = 3;

1h, k2 = 14, - 32 *16. r = 4;

1h, k2 = 10, 02

*13. r = 2;

1h, k2 = 12, - 32 *17. r = 4;

1 1h, k2 = a , 0b 2

*20. r =

1 ; 2

1h, k2 = 10, 22 *14. r = 3;

1h, k2 = 1 - 2, 12 *18. r = 7;

1h, k2 = 11, 02

1h, k2 = 1 - 5, - 22

1 1h, k2 = a0, - b 2

In Problems 21–34, (a) find the center 1h, k2 and radius r of each circle; (b) graph each circle; (c) find the intercepts, if any. *21. x2 + y2 = 4

*22. x2 + 1y - 12 2 = 1

*23. 21x - 32 2 + 2y2 = 8

*24. 31x + 12 2 + 31y - 12 2 = 6

*25. x2 + y2 - 2x - 4y - 4 = 0

*26. x2 + y2 + 4x + 2y - 20 = 0

*27. x2 + y2 + 4x - 4y - 1 = 0

*28. x2 + y2 - 6x + 2y + 9 = 0

*29. x2 + y2 - x + 2y + 1 = 0

1 = 0 2 2 2 *33. 2x + 8x + 2y = 0

*31. 2x2 + 2y2 - 12x + 8y - 24 = 0

*32. 2x2 + 2y2 + 8x + 7 = 0

*30. x2 + y2 + x + y -

*34. 3x2 + 3y2 - 12y = 0

In Problems 35– 42, find the standard form of the equation of each circle. 36. Center 11, 02 and containing the point 1 - 3, 22 35. Center at the origin and containing the point 1 - 2, 32 x2 + y2 = 13 1x - 12 2 + y2 = 20 38. Center 1 - 3, 12 and tangent to the y-axis 37. Center 12, 32 and tangent to the x-axis 1x - 22 2 + 1y - 32 2 = 9 1x + 32 2 + 1y - 12 2 = 9 40. With endpoints of a diameter at (4, 3) and (0, 1) 39. With endpoints of a diameter at (1, 4) and ( - 3, 2) 1x - 22 2 + 1y - 22 2 = 5 1x + 12 2 + 1y - 32 2 = 5 41. Center ( - 1, 3) and tangent to the line y = 2 1x + 12 2 + 1y - 32 2 = 1

42. Center (4, - 2) and tangent to the line x = 1 1x - 42 2 + 1y + 22 2 = 9

* Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

SECTION F.4   Circles

In Problems 43–46, match each graph with the correct equation. (a) 1x - 32 2 + 1y + 32 2 = 9 43. (c)

44. (d)

4

6

(b) 1x + 12 2 + 1y - 22 2 = 4

45. (b)

6

9

6

9

6

4

(c) 1x - 12 2 + 1y + 22 2 = 4

(d) 1x + 32 2 + 1y - 32 2 = 9 46. (a)

4

6

71

6

9

6

9

4

6

Applications and Extensions 47. Find the area of the square in the figure. 18 units2 y

x2  y2  9

x

and a diameter of 520 feet, with one full rotation taking approximately 30 minutes. Find an equation for the wheel if the center of the wheel is on the y-axis. Source: LasVegas Review Journal x2 + 1y - 2902 2 = 67,600 51. Weather Satellites Earth is represented on a map of a portion of the solar system so that its surface is the circle with equation x2 + y2 + 2x + 4y - 4091 = 0. A weather satellite circles 0.6 unit above Earth with the center of its circular orbit at the center of Earth. Find the equation for the orbit of the satellite on this map.

48. Find the area of the blue shaded region in the figure, assuming the quadrilateral inside the circle is a square. 36p - 72 units2 y

r

x 2  y 2  36

x

x2 + y2 + 2x + 4y - 4168.16 = 0 *52. The tangent line to a circle may be defined as the line that intersects the circle in a single point called the point of tangency. See the figure. y

49. Ferris Wheel The original Ferris wheel was built in 1893 by 1JUUTCVSHI
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r x

If the equation of the circle is x2 + y2 = r 2 and the equation of the tangent line is y = mx + b, show that:

(a) r 2 11 + m2 2 = b2 [Hint: The quadratic equation x2 + 1mx + b2 2 = r 2 has exactly one solution.] - r 2m r 2 (b) The point of tangency is ¢ , ≤. b b (c) The tangent line is perpendicular to the line containing the center of the circle and the point of tangency.

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72

CHAPTER F  Foundations: A Prelude to Functions

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to find an equation of the tangent line to the circle x2 + y2 - 4x + 6y + 4 = 0 at the point 13, 212 - 32. 22 x + 4y = 1122 - 12 55. Refer to Problem 52. The line x - 2y + 4 = 0 is tangent to a circle at 10, 22. The line y = 2x - 7 is tangent to the same circle at 13, - 12. Find the center of the circle. (1, 0)

56. Find an equation of the line containing the centers of the two circles x2 + y2 - 4x + 6y + 4 = 0 x + 5y = - 13

and x2 + y2 + 6x + 4y + 9 = 0

57. If a circle of radius 2 is made to roll along the x-axis, what is an equation for the path of the center of the circle? y = 2 58. If the circumference of a circle is 6p, what is its radius? 3

Discussion and Writing 59. Which of the following equations might have the graph shown? (More than one answer is possible.) (b), (c), (e), (g) (a) 1x - 22 2 + 1y + 32 2 = 13 (b) 1x - 22 + 1y - 22 = 8 2

2

(a) 1x - 22 2 + y2 = 3

y

(b) 1x + 22 2 + y2 = 3

(c) 1x - 22 2 + 1y - 32 2 = 13 x

(f) x + y + 4x - 2y = 0 2

2

(e) x2 + y2 + 10x + 16 = 0

x

(f) x2 + y2 + 10x - 2y = 1

(g) x2 + y2 - 9x - 4y = 0

(g) x2 + y2 + 9x + 10 = 0

(h) x + y - 4x - 4y = 4 2

(d) 1x + 22 2 + y2 = 4

2

(e) x2 + y2 - 4x - 9y = 0

y

(c) x2 + 1y - 22 2 = 3

(d) 1x + 22 + 1y - 22 = 8 2

60. Which of the following equations might have the graph shown? (More than one answer is possible.) (b), (e), (g)

2

(h) x2 + y2 - 9x - 10 = 0

61. Explain how the center and radius of a circle can be used to graph a circle.

62. What Went Wrong? A student stated that the center and radius of the graph whose equation is 1x + 32 2 + 1y - 22 2 = 16 are (3, - 2) and 4, respectively. Why is this incorrect? The center is 1 - 3, 22 .

‘Are You Prepared?’ Answers 1. add; 25

2. 5 - 1, 56

Chapter Project

73

Chapter Project The graph below, called a scatter diagram, shows the points (291.5, 268), (320, 305), . . . , (368, 385) in a Cartesian plane. From the graph, it appears that the data follow a linear relation.

Sale Price ($ thousands)

Zestimate vs. Sale Price in Oak Park, IL

Internet-based Project Determining the Selling Price of a Home Determining how much to pay for a home is one of the more difficult decisions that must be made when purchasing a home. There are many factors that play a role in a home’s value. Location, size, number of bedrooms, number of bathrooms, lot size, and building materials are just a few. Fortunately, the website Zillow.com has developed its own formula for predicting the selling price of a home. This information is a great tool for predicting the actual sale price. For example, the data below show the “zestimate”—the selling price of a home as predicted by the folks at Zillow and the actual selling price of the home, for homes, in Oak Park, Illinois.

380 360 340 320 300 280 300 320 340 360 Zestimate ($ thousands)

1. Imagine drawing a line through the data that appears to fit the data well. Do you believe the slope of the line would be positive, negative, or close to zero? Why? 2. Pick two points from the scatter diagram. Treat the zestimate as the value of x and treat the sale price as the corresponding value of y. Find the equation of the line through the two points you selected. 3. Interpret the slope of the line. 4. Use your equation to predict the selling price of a home whose zestimate is $335,000.

Zestimate ($ thousands)

Sale Price ($ thousands)

291.5

268

320

305

371.5

375

303.5

283

351.5

350

314

275

(b) Select two points from the scatter diagram and find the equation of the line through the points.

332.5

356

(c) Interpret the slope.

295

300

313

285

368

385

(d) Find a home from the Zillow website that interests you under the “Make Me Move” option for which a zestimate is available. Use your equation to predict the sale price based on the zestimate.

5. Do you believe it would be a good idea to use the equation you found in part 2 if the zestimate were $950,000? Why or why not? 6. $IPPTF
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www.zillow.com and randomly select at least ten homes that have recently sold. (a) Draw a scatter diagram of your data.

1

Functions and Their Graphs Choosing a Cellular Telephone Plan Most consumers choose a cellular telephone provider first and then select lect an appropriate plan from that provider. What type of plan to select depends ds on your use of the phone. For example, is text messaging important? How many minutes do you plan to use the phone? Do you desire a data plan to browse owse the Web? The mathematics learned in this chapter can help you decide cide which plan is best suited for your particular needs.

—See the Internet-based Chapter Project—

In this chapter, we look at a special type of equation involving two variables called a function. This chapter deals with what a function is, how to graph functions, properties of functions, and how functions are used in applications. The word function apparently was introduced by René Descartes in 1637. For him, a function was simply any positive integral power of a variable x. Gottfried Wilhelm Leibniz (1646–1716), who always emphasized the geometric side of mathematics, used the word function to denote any quantity associated with a curve, such as the coordinates of a point on the curve. Leonhard Euler (1707–1783) employed the word to mean any equation or formula involving variables and constants. His idea of a function is similar to the one most often seen in courses that precede calculus. Later, the use of functions in investigating heat flow equations led to a very broad definition that originated with Lejeune Dirichlet (1805–1859), which describes a function as a correspondence between two sets. That is the definition used in this text.

74

Functions The Graph of a Function Properties of Functions Library of Functions; Piecewise-defined Functions 1.5 Graphing Techniques: Transformations 1.6 Mathematical Models: Building Functions 1.7 Building Mathematical Models Using Variation Chapter Review Chapter Test Chapter Projects

SECTION 1.1   Functions

75

1.1 Functions PREPARING FOR THIS SECTION Before getting started, review the following: r
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OBJECTIVES 1 2 3 4

Determine Whether a Relation Represents a Function (p. 75) Find the Value of a Function (p. 78) Find the Domain of a Function Defined by an Equation (p. 81) Form the Sum, Difference, Product, and Quotient of Two Functions (p. 83)

1 Determine Whether a Relation Represents a Function

Figure 1 y 5 y  3x 1 (1, 2) 4

2 (0, 1)

2

4

x

5

EXAM PL E 1

Often there are situations where the value of one variable is somehow linked to the value of another variable. For example, an individual’s level of education is linked to annual income. Engine size is linked to gas mileage. When the value of one variable is related to the value of a second variable, we have a relation. A relation is a correspondence between two sets. If x and y are two elements in these sets, and if a relation exists between x and y, then we say that x corresponds to y or that y depends on x, and we write x S y. There are a number of ways to express relations between two sets. For example, the equation y = 3x - 1 shows a relation between x and y. It says that if we take some number x, multiply it by 3, and then subtract 1, we obtain the corresponding value of y. In this sense, x serves as the input to the relation and y is the output of the relation. This relation, expressed as a graph is shown in Figure 1. In addition to being expressed in equations and graphs, relations can be expressed through a technique called mapping. A map illustrates a relation as a set of inputs with an arrow drawn from each element in the set of inputs to the corresponding element in the set of outputs. Ordered pairs can be used to represent x S y as 1x, y2.

Maps and Ordered Pairs as Relations Figure 2 shows a relation between states and the number of representatives each state has in the House of Representatives (Source: www.house.gov). The relation might be named “number of representatives.”

Figure 2 State Alaska Arizona California Colorado Florida North Dakota

Number of Representatives 1 7 9 27 53

In this relation, Alaska corresponds to 1, Arizona corresponds to 9, and so on. Using ordered pairs, this relation would be expressed as 5 (Alaska, 1), (Arizona, 9), (California, 53), (Colorado, ), (Florida, 2), (North Dakota, 1) 6

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One of the most important concepts in algebra is the function. A function is a special type of relation. To understand the idea behind a function, let’s revisit the relation presented in Example 1. If we were to ask, “How many representatives does

76

CHAPTER 1  Functions and Their Graphs

Figure 3 Person

Phone Number

Dan

555 – 2345

Gizmo

549 – 9402 930 – 3956

Colleen

555 – 8294

Phoebe

839 – 9013

Alaska have?,” you would respond “1.” In fact, each input state corresponds to a single output number of representatives. Let’s consider a second relation, one that involves a correspondence between four people and their phone numbers. See Figure 3. Notice that Colleen has two telephone numbers. There is no single answer to the question “What is Colleen’s phone number?” -FUT
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DEFINITION

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EX A MPL E 2

Let X and Y be two nonempty sets.* A function from X into Y is a relation that associates with each element of X exactly one element of Y. The set X is called the domain of the function. For each element x in X, the corresponding element y in Y is called the value of the function at x, or the image of x. The set of all images of the elements in the domain is called the range of the function. See Figure 5. Since there may be some elements in Y that are not the image of some x in X, it follows that the range of a function may be a subset of Y, as shown in Figure 5. For example, consider the function y = x2. Since x2 Ú  for all real numbers x, the range of y = x2 is 5 y y Ú 6 , which is a subset of the set of all real numbers, Y. Not all relations between two sets are functions. The next example shows how to determine whether a relation is a function.

Determining Whether a Relation Represents a Function Determine which of the following relations represent a function. If the relation is a function, then state its domain and range. B
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Figure 6 Source: Each company’s Web site.

Calories

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*The sets X and Y will usually be sets of real numbers, in which case a (real) function results. The two sets can also be sets of complex numbers, and then we have defined a complex function. In the broad definition (proposed by Lejeune Dirichlet), X and Y can be any two sets.

SECTION 1.1   Functions

77

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Price

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In Words For a function, no input has more than one output. The domain of a function is the set of all inputs; the range is the set of all outputs.

EXAM PL E 3

PROBLEM

15

The idea behind a function is its predictability. If the input is known, we can use the function to determine the output. With “nonfunctions,” we don’t have this QSFEJDUBCJMJUZ
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Determining Whether a Relation Represents a Function Determine whether each relation represents a function. If it is a function, state the domain and range.

(a) 5 (1, ), (2, 5), (3, ), (, ) 6 (b) 5 (1, ), (2, ), (3, 5), (, 1) 6 (c) 5 ( - 3, 9), ( - 2, ), (, ), (1, 1), ( - 3, ) 6

Solution

(a) This relation is a function because there are no ordered pairs with the same first element and different second elements. The domain of this function is 5 1, 2, 3, 6 , and its range is 5 , 5, , 6 .

78

CHAPTER 1  Functions and Their Graphs

(b) This relation is a function because there are no ordered pairs with the same first element and different second elements. The domain of this function is 5 1, 2, 3, 6 , and its range is 5 , 5, 16 . (c) This relation is not a function because there are two ordered pairs, 1 - 3, 92 and 1 - 3, ), that have the same first element and different second elements.

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In Example 3(b), notice that 1 and 2 in the domain both have the same image in the range. This does not violate the definition of a function; two different first elements can have the same second element. A violation of the definition occurs when two ordered pairs have the same first element and different second elements, as in Example 3(c).

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PROBLEM

19

Up to now we have shown how to identify when a relation is a function for relations defined by mappings (Example 2) and ordered pairs (Example 3). But relations can also be expressed as equations. The circumstances under which equations are functions are discussed next. To determine whether an equation, where y depends on x, is a function, it is often easiest to solve the equation for y. If any value of x in the domain corresponds to more than one y, the equation does not define a function; otherwise, it does define a function.

EX A MPL E 4

Determining Whether an Equation Is a Function Determine whether the equation y = 2x - 5 defines y as a function of x.

Solution

The equation tells us to take an input x, multiply it by 2, and then subtract 5. For any input x, these operations yield only one output y. For example, if x = 1, then y = 2(1) - 5 = - 3. If x = 3, then y = 2(3) - 5 = 1. For this reason, the equation is a function.

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EX A MPL E 5

Determining Whether an Equation Is a Function Determine whether the equation x2 + y2 = 1 defines y as a function of x.

Solution

To determine whether the equation x2 + y2 = 1, which defines the unit circle, is a function, solve the equation for y. x2 + y 2 = 1 y 2 = 1 - x2 y = { 21 - x2

Instructor Note: We postpone the vertical-line test until the next section when we introduce the graph of a function.

For values of x between - 1 and 1, two values of y result. For example, if x = , then y = {1, so two different outputs result from the same input. This means that the equation x2 + y2 = 1 does not define a function.

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PROBLEM

33

2 Find the Value of a Function Functions are often denoted by letters such, as f, F, g, G, and others. If f is a function, then for each number x in its domain, the corresponding image in the range is designated by the symbol f 1x2, read as “f of x” or as “f at x.” We refer to f1x2 as the value of f at the number x; f1x2 is the number that results when x is given and the function f is applied; f 1x2 is the output corresponding to x or the image of x; f1x2 does not mean “f times x.” For example, the function HJWFO
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SECTION 1.1   Functions

79

Figure 9 illustrates some other functions. Notice that, in every function, for each x in the domain there is one value in the range. Figure 9 1 5 f (1) 5 f (21)

1 21 0

0 5 f (0)

2

2 5 f ( 2)

22

2 1–2 5 F (22)

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21 5 F (21)

f(x) 5 x 2

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Domain

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Range (a) f (x) 5 x

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4

1. It accepts only numbers from the domain of the function. 2. For each input, there is exactly one output (which may be repeated for different inputs).

f

Output y  f(x)

EXAM PL E 6

Solution

For a function y = f1x2, the variable x is called the independent variable, because it can be assigned any of the permissible numbers from the domain. The variable y is called the dependent variable, because its value depends on x. Any symbols can be used to represent the independent and dependent variables. For example, if f is the cube function, then f can be given by f1x2 = x3 or f1t2 = t 3 or f1z2 = z3. All three functions are the same. Each says to cube the independent variable to get the output. In practice, the symbols used for the independent and dependent variables are based on common usage, such as using C for cost in business. The independent variable is also called the argument of the function. Thinking of the independent variable as an argument can sometimes make it easier to find the value of a function. For example, if f is the function defined by f1x2 = x3, then f tells us to cube the argument. Thus, f122 means to cube 2, f1a2 means to cube the number a, and f1x + h2 means to cube the quantity x + h.

Finding Values of a Function

For the function f defined by f 1x2 = 2x2 - 3x, evaluate (a) f132

(b) f1x2 + f132

(c) 3f1x2

(e) - f1x2

(f) f13x2

(g) f1x + 32

(d) f1 - x2 f1x + h2 - f1x2 (h) h

(a) Substitute 3 for x in the equation for f , f1x2 = 2x2 - 3x, to get f 132 = 2132 2 - 3132 = 1 - 9 = 9

The image of 3 is 9.

h ≠ 

80

CHAPTER 1  Functions and Their Graphs

(b) f1x2 + f132 = 12x2 - 3x2 + 192 = 2x2 - 3x + 9 (c) Multiply the equation for f by 3. 3f1x2 = 312x2 - 3x2 = x2 - 9x (d) Substitute - x for x in the equation for f and simplify. f1 - x2 = 21 - x2 2 - 31 - x2 = 2x2 + 3x

Notice the use of parentheses here.

(e) - f1x2 = - 12x2 - 3x2 = - 2x2 + 3x (f) Substitute 3x for x in the equation for f and simplify. f 13x2 = 213x2 2 - 313x2 = 219x2 2 - 9x = 1x2 - 9x (g) Substitute x + 3 for x in the equation for f and simplify. f1x + 32 = 21x + 32 2 - 31x + 32 = 21x2 + x + 92 - 3x - 9 = 2x2 + 12x + 1 - 3x - 9 = 2x2 + 9x + 9 (h)

f1x + h2 - f1x2 3 21x + h2 2 - 31x + h2 4 - 3 2x2 - 3x4 = h h c f(x + h) = 2(x + h)2 - 3(x + h)

21x2 + 2xh + h2 2 - 3x - 3h - 2x2 + 3x Simplify. h 2x2 + xh + 2h2 - 3h - 2x2 Distribute and combine = like terms. h =

xh + 2h2 - 3h h h(x + 2h - 3) = h =

= x + 2h - 3

Combine like terms. Factor out h.

Divide out the h’s.

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Notice in this example that f 1x + 32 ≠ f 1x2 + f132 , f 1 - x2 ≠ - f1x2 , and 3f1x2 ≠ f13x2. The expression in part (h) is called the difference quotient of f, an important expression in calculus.

Now Work

PROBLEMS

39

AND

75

Most calculators have special keys that allow you to find the value of certain commonly used functions. For example, you should be able to find the square function f1x2 = x2, the square root function f 1x2 = 1x, the reciprocal function 1 f1x2 = = x -1, and many others that will be discussed later in this text (such as x ln x and log x 
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Finding Values of a Function on a Calculator (a) f1x2 = x2 1 (b) F1x2 = x (c) g1x2 = 1x

f11.232 = 1.232 = 1.5225 1 F11.232 = ≈ .13215 1.23 g11.232 = 11.23 ≈ 1.115552

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SECTION 1.1   Functions

81

COMMENT Graphing calculators can be used to evaluate any function. Figure 11 shows the SFTVMU
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COMMENT The explicit form of a function is the form required by a graphing calculator. ■

Implicit Form of a Function In general, when a function f is defined by an equation in x and y, we say that the function f is given implicitly. If it is possible to solve the equation for y in terms of x, then we write y = f1x2 and say that the function is given explicitly. For example, Implicit Form 3x + y = 5 x2 - y =  xy = 

Explicit Form y = f 1x2 = - 3x + 5 y = f 1x2 = x2 -   y = f 1x2 = x

SUMMARY Important Facts about Functions (a) For each x in the domain of a function f, there is exactly one image f1x2 in the range; however, an element in the range can result from more than one x in the domain. (b) f is the symbol that we use to denote the function. It is symbolic of the equation (rule) that we use to get from an x in the domain to f1x2 in the range. (c) If y = f1x2, then x is called the independent variable or argument of f, and y is called the dependent variable or the value of f at x.

3 Find the Domain of a Function Defined by an Equation Often the domain of a function f is not specified; instead, only the equation defining the function is given. In such cases, we agree that the domain of f is the largest set of real numbers for which the value f 1x2 is a real number. The domain of a function f is the same as the domain of the variable x in the expression f1x2.

EXAM PL E 8

Finding the Domain of a Function Find the domain of each of the following functions.

Solution

(a) f1x2 = x2 + 5x

(b) g1x2 =

3x x - 

(c) h 1t2 = 2 - 3t

(d) F1x2 =

23x + 12 x - 5

2

(a) The function says to square a number and then add five times the number. Since these operations can be performed on any real number, the domain of f is the set of all real numbers. (b) The function g says to divide 3x by x2 - .
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82

CHAPTER 1  Functions and Their Graphs

In Words The domain of g found in Example 8(b) is {x 0 x ≠ - 2, x ≠ 2}. This notation is read, “The domain of the function g is the set of all real numbers x such that x does not equal - 2 and x does not equal 2.”

(c) The function h says to take the square root of  - 3t. But only nonnegative numbers have real square roots, so the expression under the square root (the radicand) must be nonnegative (greater than or equal to zero). This requires that  - 3t Ú  - 3t Ú -   t … 3 The domain of h is e t ` t …

  f or the interval a - q , d . 3 3

(d) The function F says to take the square root of 3x + 12 and divide this result by x - 5. This requires that 3x + 12 Ú , so x Ú - , and that x - 5 ≠ , so x ≠ 5. Combining these two restrictions, the domain of F is {x 0 x Ú - , x ≠ 5}.

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The following steps may prove helpful for finding the domain of a function that is defined by an equation and whose domain is a subset of the real numbers.

Finding the Domain of a Function Defined by an Equation 1. Start with the domain as the set of real numbers. 2. If the equation has a denominator, exclude any numbers that give a zero denominator. 3. If the equation has a radical of even index, exclude any numbers that cause the expression inside the radical (the radicand) to be negative. That is, solve radicand Ú .

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PROBLEM

51

If x is in the domain of a function f, we shall say that f is defined at x, or f 1 x2 exists. If x is not in the domain of f, we say that f is not defined at x, or f 1 x2 does x not exist. For example, if f1x2 = 2 , then f12 exists, but f112 and f1 - 12 do x - 1 not exist. (Do you see why?) We will say more about finding the range when we look at the graph of a function in the next section. When a function is defined by an equation, it can be difficult to find the range. Therefore, we shall usually be content to find just the domain of a function when the function is defined by an equation. We shall express the domain of a function using inequalities, interval notation, set notation, or words, whichever is most convenient. When functions are used in applications, the domain may be restricted by physical or geometric considerations. For example, the domain of the function f defined by f1x2 = x2 is the set of all real numbers. However, if f is used to obtain the area of a square when the length x of a side is known, then we must restrict the domain of f
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EX A MPL E 9

Finding the Domain in an Application Express the area of a circle as a function of its radius. Find the domain.

Solution Figure 12 A

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See Figure 12. The formula for the area A of a circle of radius r is A = pr 2. Using r to represent the independent variable and A to represent the dependent variable, the function expressing this relationship is A(r) = pr 2

In this setting, the domain is 5 r 0 r 7 6 . (Do you see why?)

r

SECTION 1.1   Functions

83

Observe, in the solution to Example 9, that the symbol A is used in two ways: It is used to name the function, and it is used to symbolize the dependent variable. This double use is common in applications and should not cause any difficulty.

Now Work

PROBLEM

89

4 Form the Sum, Difference, Product, and Quotient of Two Functions Next we introduce some operations on functions. Functions, like numbers, can be added, subtracted, multiplied, and divided. For example, if f1x2 = x2 + 9 and g1x2 = 3x + 5, then f 1x2 + g1x2 = 1x2 + 92 + 13x + 52 = x2 + 3x + 1

The new function y = x2 + 3x + 1 is called the sum function f + g. Similarly, f 1x2 # g1x2 = 1x2 + 92 13x + 52 = 3x3 + 5x2 + 2x + 5

The new function y = 3x3 + 5x2 + 2x + 5 is called the product function f # g. The general definitions are given next.

DEFINITION

If f and g are functions: The sum f + g is the function defined by 1 f + g2 1x2 = f1x2 + g1x2

In Words

Remember, the symbol x stands for intersection. It means you should find the elements that are common to two sets.

DEFINITION

The domain of f + g consists of the numbers x that are in the domains of both f and g. That is, domain of f + g = domain of f ∩ domain of g.

The difference f − g is the function defined by 1f - g2 1x2 = f1x2 - g1x2

The domain of f - g consists of the numbers x that are in the domains of both f and g. That is, domain of f - g = domain of f ∩ domain of g.

DEFINITION

The product f ~ g is the function defined by 1f # g2 1x2 = f 1x2 # g1x2

The domain of f # g consists of the numbers x that are in the domains of both f and g. That is, domain of f # g = domain of f ∩ domain of g.

DEFINITION

The quotient

f is the function defined by g f 1x2 f a b 1x2 = g g1x2

g1x2 ≠ 

84

CHAPTER 1  Functions and Their Graphs

f consists of the numbers x for which g1x2 ≠  and that are in the g domains of both f and g. That is, f domain of = 5 x 0 g(x) ≠ 6 ∩ domain of f ∩ domain of g g The domain of

EX AM PL E 10

Operations on Functions Let f and g be two functions defined as f 1x2 =

1 x + 2

and g1x2 =

x x - 1

Find the following, and determine the domain in each case. (a) 1f + g2 1x2

Solution

(b) 1f - g2 1x2

(c) 1f # g2 1x2

f (d) a b (x2 g

The domain of f is 5 x 0 x ≠ - 26 and the domain of g is 5 x 0 x ≠ 16 . (a) 1f + g2 1x2 = f 1x2 + g1x2 = =

1 x + x + 2 x - 1

x1x + 22 x - 1 x2 + 3x - 1 + = 1x + 22 1x - 12 1x + 22 1x - 12 1x + 22 1x - 12

The domain of f + g consists of those numbers x that are in the domains of both f and g. Therefore, the domain of f + g is 5 x 0 x ≠ - 2, x ≠ 16 . (b) 1f - g2 1x2 = f1x2 - g1x2 = =

1 x x + 2 x - 1

x1x + 22 - 1x2 + x + 12 x - 1 = 1x + 22 1x - 12 1x + 22 1x - 12 1x + 22 1x - 12

The domain of f - g consists of those numbers x that are in the domains of both f and g. Therefore, the domain of f - g is 5 x 0 x ≠ - 2, x ≠ 16 . (c) 1f # g2 1x2 = f1x2 # g1x2 =

x 1 # x = x + 2 x - 1 1x + 22 1x - 12

The domain of f # g consists of those numbers x that are in the domains of both f and g. Therefore, the domain of f # g is 5 x 0 x ≠ - 2, x ≠ 16 . 1 f 1x2 f x + 2 1 #x - 1 x - 1 (d) a b 1x2 = = = = g x x g1x2 x + 2 x1x + 22 x - 1 f consists of the numbers x for which g(x) ≠  and that are in g the domains of both f and g. Since g(x) =  when x = ,
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f as - 2 and 1 from the domain. The domain of is 5 x 0 x ≠ - 2, x ≠ , x ≠ 16 . g

The domain of

r

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PROBLEM

63

In calculus, it is sometimes helpful to view a complicated function as the sum, difference, product, or quotient of simpler functions. For example, F1x2 = x2 + 1x is the sum of f1x2 = x2 and g(x) = 1x. x2 - 1 H1x2 = 2 is the quotient of f1x2 = x2 - 1 and g1x2 = x2 + 1. x + 1

SECTION 1.1   Functions

85

SUMMARY A relation between two sets of real numbers so that each number x in the first set, the domain, has corresponding to it exactly one number y in the second set. A set of ordered pairs 1x, y2 or (x, f (x)) in which no first element is paired with two different second elements. The range is the set of y-values of the function that are the images of the x-values in the domain. A function f may be defined implicitly by an equation involving x and y or explicitly by writing y = f1x2. If a function f is defined by an equation and no domain is specified, then the domain will be taken to be the largest set of real numbers for which the equation defines a real number. y = f1x2 f is a symbol for the function. x is the independent variable or argument. y is the dependent variable. f1x2 is the value of the function at x, or the image of x.

Function

Unspecified domain

Function notation

1.1 Assess Your Understanding ‘Are You Prepared?’

Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.

1. The inequality - 1 6 x 6 3 can be written in interval notation as ( - 1, 32 . QQ
"m"

1 is 2. If x = - 2, the value of the expression 3x2 - 5x + x 21.5 . QQ
"m"

x - 3 3. The domain of the variable in the expression x +  is {x 0 x ≠ - } . Q
"

4. Solve the inequality 3 - 2x 7 5. Graph the solution set. QQ
"m" {x 0 x 6 - 1}

Concepts and Vocabulary 5. If f is a function defined by the equation y = f 1x2, then x is called the independent variable and y is the dependent variable. 6. The set of all images of the elements in the domain of a function is called the range .

7. If the domain of f is all real numbers in the interval 3, 4 and the domain of g is all real numbers in the interval 3 - 2, 54, then the domain of f + g is all real numbers in the interval [, 5] . f consists of numbers x for which 8. The domain of g g(x2 ≠

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f and g .

9. If f 1x2 = x + 1 = x3 - 1x + 12.

and

g(x) = x3,

then

(g - f )(x)

10. True or False Every relation is a function. False

11. True or False The domain of 1f # g2 1x2 consists of the numbers x that are in the domains of both f and g. True 12. True or False The independent variable is sometimes referred to as the argument of the function. True 13. True or False If no domain is specified for a function f, then the domain of f is taken to be the set of real numbers. False x2 -  14. True or False The domain of the function f 1x2 = is x 5 x 0 x ≠ {26. False

Skill Building In Problems 15–26, determine whether each relation represents a function. For each function, state the domain and range. 15.

Person Elvis

Birthday Jan. 8

Function; Domain: {Elvis, Colleen, Kaleigh, Marissa}; 3BOHF
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^

16.

Father Bob

Colleen

Daughter Beth Diane

Kaleigh

Mar. 15

John

Linda

Marissa

Sept. 17

Chuck

Marcia

Not a function

86

CHAPTER 1  Functions and Their Graphs

17. Hours Worked

Not a function

Salary

*18.

$200

20 Hours

$300 30 Hours

$350

40 Hours

$425

*19. 5 (2, ), ( - 3, ), (, 9), (2, 1) 6

Level of Education

Average Income

Less than 9th grade 9th–12th grade High School Graduate Some College College Graduate

$18,120 $23,251 $36,055

*20. 5( - 2, 5), ( - 1, 3), (3, ), (, 12) 6

*22. 5 (, - 2), (1, 3), (2, 3), (3, ) 6

*25. 5( - 2, ), ( - 1, 1), (, ), (1, 1) 6

*23. 5 ( - 2, ), ( - 2, ), (, 3), (3, ) 6

*26. 5 ( - 2, 1), ( - 1, ), (, 3), (1, ) 6

$45,810 $67,165

*21. 5 (1, 3), (2, 3), (3, 3), (, 3) 6

*24. 5( - , ), ( - 3, 3), ( - 2, 2), ( - 1, 1), ( - , ) 6

In Problems 27–38, determine whether the equation defines y as a function of x.

31. y2 =  - x2

29. y =

32. y = { 21 - 2x

No

3x - 1 36. y = x + 2

35. y = 2x2 - 3x +  Yes

30. y = 0 x 0

1 Yes x 33. x = y2 No

28. y = x3 Yes

27. y = x2 Yes

No

37. 2x2 + 3y2 = 1

Yes

Yes

34. x + y2 = 1

No

38. x2 - y2 = 1

No

No

In Problems 39– 46, find the following for each function: (a) f 12

(b) f 112

(c) f 1 - 12

(d) f 1 - x2

(e) - f 1x2

*39. f 1x2 = 3x2 + 2x - 

*40. f 1x2 = - 2x2 + x - 1

*43. f 1x2 = 0 x 0 + 

*44. f 1x2 = 2x2 + x

(f) f 1x + 12

*41. f 1x2 =

(g) f 12x2

(h) f 1x + h2 *42. f 1x2 =

x x + 1 2x + 1 *45. f 1x2 = 3x - 5 2

x2 - 1 x + 

*46. f 1x2 = 1-

1 1x + 22 2

In Problems 47–62, find the domain of each function. *47. f 1x2 = - 5x +  *51. g1x2 =

*48. f 1x2 = x2 + 2

x x - 1

*52. h1x2 =

2

55. h1x2 = 23x - 12 58. f 1x2 = 61. P 1t2 =

x 2x -  2t -  3t - 21

2x x -  2

{x 0 x Ú } 56. G1x2 = 21 - x

{x 0 x 7 }

59. p1x2 =

2 Ax - 1

x2 x *50. f 1x2 = x2 + 1 x2 + 1 x - 2 x +  53. F1x2 = 3 {x 0 x ≠ } *54. G1x2 = 3 x + x x - x  57. f 1x2 = {x 0 x 7 9} 2x - 9

*49. f 1x2 =

{x 0 x … 1} {x 0 x 7 1}

{t 0 t Ú , t ≠ }

60. q1x2 = 2- x - 2 62. h1z2 =

2z + 3 z - 2

{x 0 x … - 2}

{z 0 z Ú - 3, z ≠ 2}

In Problems 63–72, for the given functions f and g, find the following. For parts (a)–(d), also find the domain. (a) 1f + g2 1x2

(b) 1f - g2 1x2

(c) 1f # g2 1x2

(e) 1f + g2 132

(f) 1f - g2 12

(g) 1 f # g)(22

f (d) a b 1x2 g f (h) a b 112 g

*63. f 1x2 = 3x + ; g1x2 = 2x - 3

*64. f 1x2 = 2x + 1; g1x2 = 3x - 2

*67. f1x2 = 1x; g1x2 = 3x - 5

*68. f 1x2 = 0 x 0 ; g1x2 = x

*65. f 1x2 = x - 1; g1x2 = 2x2

1 1 *69. f 1x2 = 1 + ; g1x2 = x x 2x + 3 x *71. f 1x2 = ; g1x2 = 3x - 2 3x - 2 1 73. Given f 1x2 = 3x + 1 and 1f + g2 1x2 =  - x, find the 2  function g. g1x2 = 5 - x 2

*66. f 1x2 = 2x2 + 3; g(x) = x3 + 1 *70. f 1x2 = 2x - 1; g1x2 = 2 - x *72. f 1x2 = 2x + 1; g1x2 =

2 x

f 1 x + 1 , find the and a b 1x2 = 2 x g x - x x - 1 function g. g(x) = x + 1

74. Given f 1x2 =

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

SECTION 1.1   Functions

In Problems 75– 82, find the difference quotient of f ; that is, find 75. f 1x2 = x + 3 *79. f 1x2 =

1 x2



76. f 1x2 = - 3x + 1 *80. f 1x2 =

f 1x + h2 - f 1x2

-3

1 x + 3

h

87

, h ≠ , for each function. Be sure to simplify.

*77. f 1x2 = x2 - x + 

*78. f 1x2 = 3x2 - 2x + 

*81. f 1x2 = 2x

*82. f 1x2 = 2x + 1

[Hint: Rationalize the numerator.]

Applications and Extensions 83. If f 1x2 = 2x3 + Ax2 + x - 5 and f 122 = 5, what is the value of A? -  2 84. If f 1x2 = 3x2 - Bx +  and f 1 - 12 = 12, what is the value of B? 5 3x +  and f 12 = 2, what is the value of A? -  85. If f 1x2 = 2x - A 2x - B 1 86. If f 1x2 = and f 122 = , what is the value of B? - 1 3x +  2 2x - A and f 12 = , what is the value of A? 87. If f 1x2 = x - 3 Where is f not defined? A = ; undefined at x = 3 x - B 88. If f 1x2 = , f 122 = , and f 112 is undefined, what x - A are the values of A and B? A = 1, B = 2

*(b) When is the height of the rock 15 meters? When is it 
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*89. Geometry Express the area A of a rectangle as a function of the length x if the length of the rectangle is twice its width. *90. Geometry Express the area A of an isosceles right triangle as a function of the length x of one of the two equal sides. 91. Constructing Functions Express the gross salary G of a QFSTPO
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her gross salary G as a function of the number x of items sold. 93. Population as a Function of Age The function P(a) = .a2 - 3.92a + 31.9 represents the population P (in millions) of Americans that were a
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97. Cost of Trans-Atlantic Travel
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3, x + 1 x

where x is the ground speed (airspeed { wind). (a) What is the cost per passenger for quiescent (no wind) conditions? $222 C
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miles per hour?  98. Cross-sectional Area The cross-sectional area of a beam cut from a log with radius 1 foot is given by the function A1x2 = x21 - x2 , where x represents the length, in feet, of half the base of the beam. See the figure on the next page. Determine the cross-sectional area of the beam if the length of half the base of the beam is as follows: (a) One-third of a foot 
GU2 (b) One-half of a foot 
GU2 (c) Two-thirds of a foot 1.99 ft2

88

CHAPTER 1  Functions and Their Graphs

The cost C, in dollars, of selling x cell phones, in hundreds, is C 1x2 = .5x3 - 2x2 + 5x + 5. *(a) Find the profit function, P 1x2 = R 1x2 - C 1x2. *(b) Find the profit if x = 15 hundred cell phones are sold. *(c) Interpret P(15).

A(x ) ⫽ 4x 公1 ⫺ x 2 1

x

*99. Economics The participation rate is the number of people in the labor force divided by the civilian population (excludes military). Let L 1x2 represent the size of the labor force in year x and P 1x2 represent the civilian population in year x. Determine a function that represents the participation rate R as a function of x. 100. Crimes Suppose that V 1x2 represents the number of violent crimes committed in year x and P 1x2 represents the number of property crimes committed in year x. Determine a function T that represents the combined total of violent crimes and property crimes in year x. T(x) = V 1x2 + P 1x2

101. Health Care Suppose that P 1x2 represents the percentage of income spent on health care in year x and I 1x2 represents income in year x. Determine a function H that represents total health care expenditures in year x. H(x) = P(x) # I(x)

102. Income Tax Suppose that I 1x2 represents the income of an individual in year x before taxes and T 1x2 represents the individual’s tax bill in year x. Determine a function N that represents the individual’s net income (income after taxes) in year x. N(x) = I(x) - T(x) 103. Profit Function Suppose that the revenue R, in dollars, from selling x cell phones, in hundreds, is R 1x2 = - 1.2x2 + 22x.

Discussion and Writing x2 - 1 *107. Are the functions f 1x2 = x - 1 and g1x2 = the x + 1 same? Explain. 108. Investigate when, historically, the use of the function notation y = f 1x2 first appeared.

‘Are You Prepared?’ Answers 1. 1 - 1, 32

2. 21.5

104. Profit Function Suppose that the revenue R, in dollars, from selling x clocks is R 1x2 = 3x. The cost C, in dollars, of selling x clocks is C 1x2 = .1x2 + x + . *(a) Find the profit function, P 1x2 = R 1x2 - C 1x2. (b) Find the profit if x = 3 clocks are sold.  *(c) Interpret P   105. Stopping Distance When the driver of a vehicle observes an impediment, the total stopping distance involves both the reaction distance (the distance the vehicle travels while the driver moves his or her foot to the brake pedal) and the braking distance (the distance the vehicle travels once the brakes are applied). For a car traveling at a speed of v miles per hour, the reaction distance R, in feet, can be estimated by R 1v2 = 2.2v. Suppose that the braking distance B, in feet, for a car is given by B 1v2 = .5v2 + .v - 15. *(a) Find the stopping distance function D(v) = R(v) + B(v). (b) Find the stopping distance if the car is traveling at a TQFFE
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106. Some functions f have the property that f 1a + b2 = f 1a2 + f 1b2 for all real numbers a and b. Which of the following functions have this property? Only h(x) = 2x (a) h1x2 = 2x (b) g1x2 = x2 1 (c) F 1x2 = 5x - 2 (d) G1x2 = x 109. Find a function H that multiplies a number x by 3 and then subtracts the cube of x and divides the result by your age. 3x - x3 H(x) = age

3. 5 x 0 x ≠ - 6

4. 5x 0 x 6 - 16

⫺1

0

1

1.2 The Graph of a Function PREPARING FOR THIS SECTION Before getting started, review the following: r
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OBJECTIVES 1 Identify the Graph of a Function (p. 89) 2 Obtain Information from or about the Graph of a Function (p. 90)

In applications, a graph often demonstrates more clearly the relationship between two variables than, say, an equation or table would. For example, Table 1 on the next page shows the average price of gasoline at a particular gas station in Texas (for the ZFBST
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SECTION 1.2  The Graph of a Function

Table 1

Year

Price

Year

Price

Year

Price

1983

3.10

1993

1.93

2003

2.13

1984

2.79

1994

1.94

2004

2.36

1985

2.64

1995

1.98

2005

2.92

1986

2.00

1996

2.02

2006

3.13

1987

2.12

1997

1.94

2007

3.31

1988

2.00

1998

1.70

2008

3.61

1989

2.01

1999

1.83

2009

2.70

1990

2.25

2000

2.19

2010

3.08

1991

2.07

2001

2.02

2011

3.78

1992

2.02

2002

1.99

2012

3.84

89

Source: http://www.randomuseless.info/gasprice/gasprice.html

Figure 13 Average retail price of gasoline (2012 dollars) 4.50 Price (dollars per gallon)

4.00 3.50 3.00 2.50 2.00 1.50 1.00

2012

2010

2008

2006

2004

2002

2000

1998

1996

1994

1992

1990

1988

1986

0.00

1984

0.50

Source: http://www.randomuseless.info/gasprice/gasprice.html

We can see from the graph that the price of gasoline (adjusted for inflation) fell GSPN

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information such as this from an equation requires that some calculations be made. Look again at Figure 13. The graph shows that for each date on the horizontal axis there is only one price on the vertical axis. The graph represents a function, although the exact rule for getting from date to price is not given. When a function is defined by an equation in x and y, the graph of the function is the graph of the equation; that is, it is the set of points 1x, y2 in the xy-plane that satisfy the equation.

1 Identify the Graph of a Function In Words If any vertical line intersects a graph at more than one point, the graph is not the graph of a function.

THEOREM

Not every collection of points in the xy-plane represents the graph of a function. Remember, for a function, each number x in the domain has exactly one image y in the range. This means that the graph of a function cannot contain two points with the same x-coordinate and different y-coordinates. Therefore, the graph of a function must satisfy the following vertical-line test.

Vertical-Line Test A set of points in the xy-plane is the graph of a function if and only if every vertical line intersects the graph in at most one point.

EXAM PL E 1

Identifying the Graph of a Function 8IJDI
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90

CHAPTER 1  Functions and Their Graphs y 6

Figure 14

y 4

y

y 3

1 (1, 1)

4 3

3x (a) y  x 2

Solution

4x 4

(1, 1)

6 x

1 1

3

(b) y  x 3

1 x

(c) x  y 2

(d) x 2  y 2  1

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r

PROBLEM

15

2 Obtain Information from or about the Graph of a Function

If 1x, y2 is a point on the graph of a function f, then y is the value of f at x; that is, y = f1x2 . Also if y = f 1x2, then 1x, y2 is a point on the graph of f. For example, if 1 - 2, 2 is on the graph of f, then f1 - 22 = , and if f152 = , then the point 15, 2 is on the graph of y = f1x2 . The next example illustrates how to obtain information about a function if its graph is given.

Obtaining Information from the Graph of a Function

EX A MPL E 2 Figure 15 y 4 2

⫺2 ⫺4

(4␲, 4)

(2␲, 4)

(0, 4)

(5––2␲, 0)

(––␲2 , 0)

(7––2␲, 0)

(3––2␲, 0) (␲, ⫺4)

x

(3␲, ⫺4)

Solution

Let f be the function whose graph is given in Figure 15. (The graph of f might represent the distance y that the bob of a pendulum is from its at-rest position at time x. Negative values of y mean that the pendulum is to the left of the at-rest position, and positive values of y mean that the pendulum is to the right of the at-rest position.) 3p (a) What are f 12, f a b , and f13p2? 2 (b) What is the domain of f ? (c) What is the range of f ? (d) List the intercepts. (Recall that these are the points, if any, where the graph crosses or touches the coordinate axes.) (e) How many times does the line y = 2 intersect the graph? (f) For what values of x does f1x2 = - ? (g) For what values of x is f1x2 7 ? (a) Since 1, 2 is on the graph of f, the yDPPSEJOBUF

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and for each such number y, there is at least one corresponding number x in the domain. The range of f is 5 y 0 -  … y … 6 or the interval 3 - , 4 . (d) The intercepts are the points p 3p 5p (, ), a , b , a , b , a , b , and 2 2 2

a

p , b 2

SECTION 1.2  The Graph of a Function

91

(e) Draw the horizontal line y = 2 on the graph in Figure 15. Notice that the line intersects the graph four times. (f) Since 1p, - 2 and 13p, - 2 are the only points on the graph for which y = f1x2 = - , we have f 1x2 = -  when x = p and x = 3p. (g) To determine where f1x2 7 , look at Figure 15 and determine the xWBMVFT
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p for which the y-coordinate is positive. This occurs on 3p 5p p p b ∪ a , b ∪ a , p d . Using inequality notation, f1x2 7  for 2 2 2 2 p 3p 5p p  … x 6 or 6 x 6 or 6 x … p. 2 2 2 2 When the graph of a function is given, its domain may be viewed as the shadow created by the graph on the x-axis by vertical beams of light. Its range can be viewed as the shadow created by the graph on the y-axis by horizontal beams of light. Try this technique with the graph given in Figure 15. c ,

r

The Zeros of a Function If f1r2 =  for a real number r, then r is called a real zero of f . For example, the zeros of the function f1x2 = x2 -  are - 2 and 2 since f1 - 22 =  and f122 = . We can identify the zeros of a function from its graph. To see how, we must remember that y = f1x2 means that the point 1x, y2 is on the graph of f . So, if r is a zero of f , then f1r2 = , which means the point 1r, 2 is on the graph of f . That is, r is an x@intercept. We conclude that the x@intercepts of the graph of a function are also the zeros of the function.

EXAM PL E 3

Finding the Zeros of a Function from Its Graph Find the zeros of the function f whose graph is shown in Figure 15.

Solution

The zeros of a function are the x@intercepts of the graph of the function. Because the p 3p 5p p x@intercepts of the function shown in Figure 15 are , , , and , the zeros of 2 2 2 2 p 3p 5p p f are , , , and . 2 2 2 2

Now Work

EXAM PL E 4

PROBLEMS

9

AND

13

r

Obtaining Information about the Graph of a Function Consider the function: f1x2 =

x + 1 x + 2

(a) Find the domain of f. 1 (b) Is the point a1, b on the graph of f ? 2 (c) If x = 2, what is f1x2 ? What point is on the graph of f ? (d) If f1x2 = 2, what is x? What point is on the graph of f ? (e) What are the x-intercepts of the graph of f (if any)? What point(s) are on the graph of f ?

Solution

(a) The domain of f is {x 0 x ≠ - 2}, since x = - 2
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 (b) When x = 1, f112 =

1 + 1 2 = 1 + 2 3

2 1 The point a1, b is on the graph of f; the point a1, b is not. 3 2 (c) If x = 2, then 2 + 1 3 f122 = = 2 + 2  3 The point ¢ 2, ≤ is on the graph of f. 

92

CHAPTER 1  Functions and Their Graphs

(d) If f1x2 = 2, then x x x x

+ + + +

1 2 1 1 x

= 2 = 2(x + 22 = 2x +  = -3

Multiply both sides by x + 2. Distribute. Solve for x.

If f1x2 = 2, then x = - 3. The point 1 - 3, 22 is on the graph of f. (e) The x-intercepts of the graph of f are the real solutions of the equation f1x2 =  that are in the domain of f. x + 1 =  x + 2 x + 1 =  x = -1

Multiply both sides by x + 2. Solve for x.

x + 1 =  is x = - 1, so - 1 is the x + 2 only x-intercept. Since f 1 - 12 = , the point ( - 1

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f.

The only real solution of the equation f1x2 =

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EX A MPL E 5

r

PROBLEM

25

Average Cost Function The average cost C (per computer) of manufacturing x computers per day is given by the function C(x) = .5x2 - 3.39x + 1212.5 +

2, x

Determine the average cost of manufacturing: B

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C

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D

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Solution

(a) The average cost of manufacturing x = 3 computers is 2, = $1351.5 3 (b) The average cost of manufacturing x =  computers is C(3) = .5(3)2 - 3.39(3) + 1212.5 +

2, = $1232.9  (c) The average cost of manufacturing x = 5 computers is C() = .5()2 - 3.39() + 1212.5 +

C(5) = .5(5)2 - 3.39(5) + 1212.5 +

2, = $1293. 5

(d)
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PROBLEM

31

Table 3

r

SECTION 1.2  The Graph of a Function

93

SUMMARY Graph of a Function Vertical-Line Test

The collection of points 1x, y2 that satisfies the equation y = f1x2. A collection of points is the graph of a function provided that every vertical line intersects the graph in at most one point.

1.2 Assess Your Understanding ‘Are You Prepared?’

Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.

1. The intercepts of the equation x2 + y2 = 1 ( - , ), (, ), (, - 2), (, 22 . Q


2. True or False The point 1 - 2, - 2 is on the graph of the equation x = 2y - 2. QQ
m False

are

Concepts and Vocabulary 3. A set of points in the xy-plane is the graph of a function if and only if every vertical line intersects the graph in at most one point.

6. True or False A function can have more than one yintercept. False

5. Find a such that the point 1 - 1, 22 is on the graph of f 1x2 = ax2 + . a = - 2

8. True or False The y-intercept of the graph of the function y = f 1x2, whose domain is all real numbers, is f 12. True

7. True or False The graph of a function y = f 1x2 always crosses the y-axis. False

4. If the point 15, - 32 is a point on the graph of f, then f( 5 ) = -3 .

Skill Building 9. Use the given graph of the function f to answer parts B
m
P  y (0, 3) 4

10. Use the given graph of the function f to answer parts B
m
P  y

(2, 4) 4

(4, 3)

(5, 3)

(–4, 2)

(10, 0) (11, 1)

(–2, 1) 2

(–3, 0)

(4, 0) 5

–5

(6, 0)

(–5, –2) (–6, –3)

–3

11 x

–2 (0, 0) –2

(8, – 2)

Find f 12 and f 1 - 2. f 12 = 3; f 1 - 2 = - 3 Find f 12 and f 1112. f 12 = ; f 1112 = 1 Is f 132 positive or negative? 1PTJUJWF Is f 1 - 2 positive or negative? Negative For what values of x is f 1x2 = ? - 3, , and 1 For what values of x is f 1x2 7 ? - 3 6 x 6 ; 1 6 x … 11 What is the domain of f ? 5x 0 -  … x … 116 What is the range of f ? 5y 0 - 3 … y … 6 What are the x-intercepts? - 3, , and 1 What is the y-intercept? 3 1 How many times does the line y = intersect the 2 graph? Three times (l) How many times does the line x = 5 intersect the graph? Once (m) For what values of x does f 1x2 = 3? 
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 (n) For what values of x does f 1x2 = - 2? - 5 and  (o) What are the zeros of f? ¢



(a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k)

–4

2

4

(6, 0) 6

x

(2, –2)

Find f 12 and f 12. f 12 = ; f 12 =  Find f 122 and f 1 - 22. f 122 = - 2; f 1 - 22 = 1 Is f 132 positive or negative? Negative Is f 1 - 12 positive or negative? 1PTJUJWF For what values of x is f 1x2 = ? 

BOE
 For what values of x is f 1x2 6 ?  6 x 6  What is the domain of f ? 5 x 0 -  … x … 6 What is the range of ƒ? 5y 0 - 2 … y … 36 What are the x-intercepts? 

BOE
 What is the y-intercept?  How many times does the line y = - 1 intersect the graph? Twice (l) How many times does the line x = 1 intersect the graph? Once (m) For what value of x does f 1x2 = 3? 5 (n) For what value of x does f 1x2 = - 2? 2 (o) What are the zeros of f? 



(a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k)

94

CHAPTER 1  Functions and Their Graphs

In Problems 11–22, determine whether the graph is that of a function by using the vertical-line test. If it is, use the graph to find: (a) The domain and range *11.

(b) The intercepts, if any *12.

y 3

⫺3

*16.

y 3

⫺3

*19.

y (⫺1, 2) 3

*20.

3 ⫺3

y 3

1 ␲ –– 2

⫺␲

␲x

*21.

y 3

*18.

–– 2

y 4

(3, 2)

3x

⫺3

x

4x 4

*22.

y

y

9

3x

␲

(4, 3)

4

(1–2 , 5)

4

6 ⫺3

␲

␲ ⫺ –– 2 ⫺1

3

⫺3

(1, 2)

␲ ⫺–– 2 ⫺1

3

3x

x ⫺3

⫺␲

*17.

y 3

⫺3

y

1

3x

⫺3

3x

*14.

y

⫺3

⫺3

*15.

*13.

y 3

⫺3

3x

(c) Any symmetry with respect to the x-axis, the y-axis, or the origin

3 ⫺1 ⫺3

3 x (2, ⫺3)

⫺3

3 x

In Problems 23–28, answer the questions about the given function. 23. f 1x2 = 2x2 - x - 1 (a) Is the point 1 - 1, 22 on the graph of ƒ? Yes (b) If x = - 2, what is f 1x2? What point is on the graph of ƒ? f ( - 2) = 9; ( - 2, 9) *(c) If f 1x2 = - 1, what is x? What point(s) is (are) on the graph of ƒ? 1 (d) What is the domain of ƒ? All real numbers - ,1 (e) List the x-intercepts, if any, of the graph of f. 2 (f) List the y-intercept, if there is one, of the graph of f. - 1 1 (g) What are the zeros of f ? - , 1 2 24. f 1x2 = - 3x2 + 5x (a) Is the point 1 - 1, 22 on the graph of ƒ? No (b) If x = - 2, what is f 1x2? What point is on the graph of ƒ? f ( - 2) = - 22; ( - 2, - 22) *(c) If f 1x2 = - 2, what is x? What point(s) is (are) on the graph of ƒ? 5 (d) What is the domain of ƒ? All real numbers , (e) List the x-intercepts, if any, of the graph of f. 3 (f) List the y-intercept, if there is one, of the graph of f.  5 (g) What are the zeros of f ? , 3 x + 2 25. f 1x2 = x -  (a) Is the point 13, 12 on the graph of ƒ? No *(b) If x = , what is f 1x2? What point is on the graph of ƒ? (c) If f 1x2 = 2, what is x? What point(s) is (are) on the graph of ƒ? 1; (1, 22 (d) What is the domain of ƒ? {x x ≠ } (e) List the x-intercepts, if any, of the graph of f. - 2 1 (f) List the y-intercept, if there is one, of the graph of f. 3 (g) What are the zeros of f ? - 2

x2 + 2 x +  3 (a) Is the point a1, b on the graph of ƒ? Yes 5 *(b) If x = , what is f 1x2? What point is on the graph of ƒ? 1 *(c) If f 1x2 = , what is x? What point(s) is (are) on the graph of ƒ? 2

26. f 1x2 =

What is the domain of ƒ? {x 0 x ≠ - } List the x-intercepts, if any, of the graph of f. None 1 List the y-intercept, if there is one, of the graph of f. 2 What are the zeros of f ? None 2x2 27. f 1x2 =  x + 1 (a) Is the point 1 - 1, 12 on the graph of ƒ? Yes *(b) If x = 2, what is f 1x2? What point is on the graph of ƒ? (c) If f 1x2 = 1, what is x? What point(s) is (are) on the graph of ƒ? - 1, 1; ( - 1, 1), (1, 12 (d) What is the domain of ƒ? All real numbers (e) List the x-intercepts, if any, of the graph of f.  (f) List the y-intercept, if there is one, of the graph of f.  (g) What are the zeros of f ?  2x 28. f 1x2 = x - 2 1 2 (a) Is the point a , - b on the graph of ƒ? Yes 2 3 *(b) If x = , what is f 1x2? What point is on the graph of ƒ? (c) If f 1x2 = 1, what is x? What point(s) is (are) on the graph of ƒ? - 2; ( - 2, 12 (d) What is the domain of ƒ? {x 0 x ≠ 2} (e) List the x-intercepts, if any, of the graph of f.  (f) List the y-intercept, if there is one, of the graph of f.  (g) What are the zeros of f ?  (d) (e) (f) (g)

* Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

SECTION 1.2  The Graph of a Function

95

Applications and Extensions 29. Free-throw Shots
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the key to a successful foul shot in basketball lies in the arc of the shot. Brancazio determined the optimal angle of UIF
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the ball can be modeled by the function h(x) = -

where x is the horizontal distance that the golf ball has traveled.

x2 + x +  v2

where h is the height of the ball above the floor, x is the forward distance of the ball in front of the foul line, and v is the initial velocity with which the ball is shot in feet per second. Suppose a player shoots a ball with an initial velocity PG

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x

(a) Find the domain of A. 5x 0  6 x 6 16 *(b) Use a graphing utility to graph the function A = A(x2. (c) Create a TABLE with TblStart =  and ∆Tbl = .1. Which value of x in the domain found in part (a) maximizes the cross-sectional area? What should be the length of the base of the beam to maximize the crosssectional area? 
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GU 33. Cost of Trans-Atlantic Travel
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per hour. The cost C (in dollars) per passenger is given by C 1x2 = 1 +

3, x + 1 x

CHAPTER 1  Functions and Their Graphs

where x is the ground speed 1airspeed { wind). B
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 (b) Find the domain of C. {x 0 x 7 } *(c) Use a graphing utility to graph the function C = C(x). *(d) Create a TABLE with TblStart =  and ∆Tbl = 5. F
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minimizes the cost per passenger? 
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C (100, 280000) 250,000

Cost (dollars per day)

96

150,000

100,000

50,000

(50, 51000) (30, 32000)

(0, 5000)

(10, 19000) 10

20

30

40 50 60 70 80 Number of computers

90 100 q

37. Reading and Interpreting Graphs Let C be the function whose graph is given below. This graph represents the cost C of using m anytime cell phone minutes in a month for a five-person family plan.

yg(x) C 2000

(2, 2) (4, 1) 2 2

(6, 1) (6, 0) x yf(x)

4

(3, 2)

(5, 2)

(4, 3)

4

(a) 1 f + g)(22 (c) 1 f - g)(2

3 -1 2

(14400, 1822)

(b) 1 f + g)(2 - 2 (d) 1g - ƒ2 12 1 f 1 (f) a b(2 g 3

36. Reading and Interpreting Graphs Let C be the function whose graph is given in the next column. This graph represents the cost C of manufacturing q computers in a day. *(a) Determine C  
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inflection point. Describe the behavior of the graph around the inflection point.

1500 Cost (dollars)

2 (2, 1)

(e) 1 f # g)(22

200,000

1000

500 (2000, 210) (0, 80)

(1000, 80) 5000

10000 Number of minutes

15000 m

Determine C  
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*(a) *(b) *(c) *(d)

Discussion and Writing 38. Describe how you would proceed to find the domain and range of a function if you were given its graph. How would your strategy change if you were given the equation defining the function instead of its graph? *39. How many x-intercepts can the graph of a function have? How many y-intercepts can the graph of a function have? *40. Is a graph that consists of a single point the graph of a function? Can you write the equation of such a function? 41. Match each of the following functions with the graph on the next page that best describes the situation. (a) The cost of building a house as a function of its square footage III C
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IV (c) The height of a human as a function of time I (d) The demand for Big Macs as a function of price V (e) The height of a child on a swing as a function of time? II

SECTION 1.2  The Graph of a Function y

y

y

y

x

x

y

x

x

x

(III)

(II)

(I)

97

(V)

(IV)

42. Match each of the following functions with the graph that best describes the situation. (a) The temperature of a bowl of soup as a function of time II (b) The number of hours of daylight per day over a 2-year period V (c) The population of Texas as a function of time IV (d) The distance traveled by a car going at a constant velocity as a function of time III F
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*44. Consider the following scenario: Jayne enjoys riding her bicycle through the woods. At the forest preserve, she gets PO
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She then travels down the incline in 3 minutes. The next 
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(in feet) as a function of time. 45. The following sketch represents the distance d (in miles) that Kevin was from home as a function of time t (in hours). Answer the questions by referring to the graph. In parts B
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(4.2, 2.8)

(5.3, 0)

t

x (V)

*(d) From t = 2. to t = 3 *(e) From t = 3 to t = 3.9 *(f) From t = 3.9 to t = .2 *(g) From t = .2 to t = 5.3 *(h) What is the farthest distance that Kevin was from home? (i) How many times did Kevin return home? Twice 46. The following sketch represents the speed v (in miles per hour) of Michael’s car as a function of time t (in minutes). v (t ) (7, 50)

(2, 30)

(8, 38)

(4, 30)

(4.2, 0)

(7.4, 50)

(7.6, 38)

(6, 0)

(9.1, 0)

t

*(a) Over what interval of time was Michael traveling fastest? *(b) Over what interval(s) of time was Michael’s speed zero? * D
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*(a) From t =  to t = 2 *(b) From t = 2 to t = 2.5 *(c) From t = 2.5 to t = 2.

49. Explain why the vertical-line test works.

‘Are You Prepared?’ Answers 1. 1 - , ), 1, ), 1, - 2), 1, 22

x (IV)

(III)

*43. Consider the following scenario: Barbara decides to take a walk. She leaves home, walks 2 blocks in 5 minutes at a constant speed, and realizes that she forgot to lock the door. So Barbara runs home in 1 minute. While at her doorstep, it takes her 1 minute to find her keys and lock the door. Barbara walks 5 blocks in 15 minutes and then decides to jog home. It UBLFT
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2. False

98

CHAPTER 1  Functions and Their Graphs

1.3 Properties of Functions PREPARING FOR THIS SECTION Before getting started, review the following: r
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OBJECTIVES 1 Determine Even and Odd Functions from a Graph (p. 98) 2 Identify Even and Odd Functions from the Equation (p. 99) 3 Use a Graph to Determine Where a Function is Increasing, Decreasing, or Constant (p. 100) 4 Use a Graph to Locate Local Maxima and Local Minima (p. 101) 5 Use a Graph to Locate the Absolute Maximum and the Absolute Minimum (p. 102) 6 Use a Graphing Utility to Approximate Local Maxima and Local Minima and to Determine Where a Function is Increasing or Decreasing (p. 103) 7 Find the Average Rate of Change of a Function (p. 104)

To obtain the graph of a function y = f1x2 , it is often helpful to know certain properties that the function has and the impact of these properties on the way that the graph will look.

1 Determine Even and Odd Functions from a Graph The words even and odd, when applied to a function f, describe the symmetry that exists for the graph of the function. A function f is even if and only if, whenever the point 1x, y2 is on the graph of f , the point 1 - x, y2 is also on the graph. Using function notation, we define an even function as follows:

DEFINITION

A function f is even if, for every number x in its domain, the number - x is also in the domain and f 1 - x2 = f 1x2 A function f is odd if and only if, whenever the point 1x, y2 is on the graph of f , the point 1 - x, - y2 is also on the graph. Using function notation, we define an odd function as follows:

DEFINITION

A function f is odd if, for every number x in its domain, the number - x is also in the domain and f1 - x2 = - f 1x2 3FGFS
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THEOREM

A function is even if and only if its graph is symmetric with respect to the y-axis. A function is odd if and only if its graph is symmetric with respect to the origin.

SECTION 1.3   Properties of Functions

EXAM PL E 1

99

Determining Even and Odd Functions from the Graph %FUFSNJOF
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an odd function, or a function that is neither even nor odd.

Figure 17

y

y

x

x

x

(b)

(a)

Solution

y

(c)

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because the graph is neither symmetric with respect to the y-axis nor symmetric with respect to the origin. D
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symmetric with respect to the origin.

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Now Work

PROBLEMS

21(a), (b),

AND

(d)

2 Identify Even and Odd Functions from the Equation In the next example, we use algebraic techniques to verify whether a given function is even, odd, or neither.

EXAM PL E 2

Identifying Even and Odd Functions Algebraically Determine whether each of the following functions is even, odd, or neither. Then determine whether the graph is symmetric with respect to the y-axis, with respect to the origin, or neither. (a) f1x2 = x2 - 5 (c) h 1x2 = 5x3 - x

Solution

(b) g1x2 = x3 - 1 (d) F1x2 = 0 x 0

(a) To determine whether f is even, odd, or neither, replace x by - x in f1x2 = x2 - 5. Then f1 - x2 = 1 - x2 2 - 5 = x2 - 5 = f 1x2 Since f1 - x2 = f1x2, the function is even, and the graph of f is symmetric with respect to the y-axis. (b) Replace x by - x in g1x2 = x3 - 1. g1 - x2 = 1 - x2 3 - 1 = - x3 - 1 Since g1 - x2 ≠ g1x2 and g1 - x2 ≠ - g1x2 = - 1x3 - 12 = - x3 + 1, the function is neither even nor odd. The graph of g is not symmetric with respect to the y-axis, nor is it symmetric with respect to the origin. (c) Replace x by - x in h 1x2 = 5x3 - x. h 1 - x2 = 51 - x2 3 - 1 - x2 = - 5x3 + x = - 15x3 - x2 = - h 1x2 Since h 1 - x2 = - h 1x2 , h is an odd function, and the graph of h is symmetric with respect to the origin.

100

CHAPTER 1  Functions and Their Graphs

(d) Replace x by - x in F1x2 = 0 x 0 . Then

Figure 18 y

F1 - x2 = 0 - x 0 = 0 - 1 0 # 0 x 0 = 0 x 0 = F1x2

5 (0, 4)

(⫺6, 0)

(3, 4) y = f (x ) (6, 1)

(⫺2, 0) ⫺4

Now Work

6 x

0

(⫺4, ⫺2)

Since F1 - x2 = F1x2, F is an even function, and the graph of F is symmetric with respect to the y-axis.

⫺2

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PROBLEM

33

3 Use a Graph to Determine Where a Function Is Increasing, Decreasing, or Constant $POTJEFS
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function, and note that parts of the graph are going up, parts are going down, and parts are horizontal. In such cases, the function is described as increasing, decreasing, or constant, respectively.

Determining Where a Function Is Increasing, Decreasing, or Constant from Its Graph

EX A MPL E 3

Determine the values of x
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it decreasing? Where is it constant?

Solution

To determine where a function is increasing, where it is decreasing, and where it is constant, we use strict inequalities involving the independent variable x, or we use open intervals* of xDPPSEJOBUFT
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WARNING Describe the behavior of a graph in terms of its x-values. Do OPU
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increasing from the point 1 - , - 22 to the point 1, 2. Rather, say it is increasing on the interval ( - , ). j

DEFINITIONS

A function f is increasing on an open interval I if, for any choice of x1 and x2 in I, with x1 6 x2 , we have f1x1 2 6 f 1x2 2. A function f is decreasing on an open interval I if, for any choice of x1 and x2 in I, with x1 6 x2 , we have f1x1 2 7 f 1x2 2. A function f is constant on an open interval I if, for all choices of x in I, the values of f1x2 are equal.

In Words If a function is decreasing, then, as the values of x get bigger, the values of the function get smaller. If a function is increasing, then, as the values of x get bigger, the values of the function also get bigger. If a function is constant, then, as the values of x get bigger, the values of the function remain unchanged. Figure 19

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More precise definitions follow:

Figure 19 illustrates the definitions. Graphs are read like a book—from left to right. Thus, the graph of an increasing function goes up from left to right, the graph of a decreasing function goes down from left to right, and the graph of a constant function remains at a fixed height. y

y

y

f (x 1)

f (x 2) f (x 1)

x1

x2

x

f (x 1)

f (x 2)

x1

x2

x

x1

(b) For x 1 < x 2 in l, f(x 1) > f (x 2); f is decreasing on I.

(a) For x 1 < x 2 in l, f (x 1) < f (x 2); f is increasing on I.

Now Work

PROBLEMS

x2

l

l

l

f(x 2)

11, 13, 15,

(c) For all x in I, the values of f are equal; f is constant on I. AND

21(C)

*The open interval 1a, b2 consists of all real numbers x for which a 6 x 6 b.

x

SECTION 1.3   Properties of Functions

101

4 Use a Graph to Locate Local Maxima and Local Minima Suppose f is a function defined on an open interval I containing c. If the value of f at c is greater than or equal to the values of f on I, then f has a local maximum at c
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y

f(c)

y

f has a local maximum at c. (c, f(c))

c l

f(c)

x

(a)

DEFINITIONS

(c, f(c))

c l

f has a local minimum at c.

x

(b)

A function f has a local maximum at c if there is an open interval I containing c so that, for all x in I, f1x2 … f1c2. We call f1c2 a local maximum value of f. A function f has a local minimum at c if there is an open interval I containing c so that, for all x in I, f 1x2 Ú f1c2. We call f1c2 a local minimum value of f. If f has a local maximum at c, then the value of f at c is greater than or equal to the values of f near c. If f has a local minimum at c, then the value of f at c is less than or equal to the values of f near c. The word local is used to suggest that it is only near c, not necessarily over the entire domain, that the value f1c2 has these properties.

EXAM PL E 4 Figure 21 y

2

y ⫽ f(x)

Figure 21 shows the graph of a function f.

(1, 2)

(–1, 1) –2

3

Finding Local Maxima and Local Minima from the Graph of a Function and Determining Where the Function Is Increasing, Decreasing, or Constant

x

Solution

WARNING The y-value is the local maximum value or local minimum value, and it occurs at some x-value. For example, in Figure 21, we say f has a local maximum at 1 and the local maximum value is 2. j

(a) At what value(s) of x, if any, does f have a local maximum? List the local maximum values. (b) At what value(s) of x, if any, does f have a local minimum? List the local minimum values. (c) Find the intervals on which f is increasing. Find the intervals on which f is decreasing. The domain of f is the set of real numbers. (a) f has a local maximum at 1, since for all x close to 1, we have f1x2 … f112. The local maximum value is f112 = 2. (b) f has local minima at - 1 and at 3. The local minimum values are f1 - 12 = 1 and f132 = . (c) The function whose graph is given in Figure 21 is increasing for all values of x between - 1 and 1 and for all values of x greater than 3. That is, the function is increasing on the intervals 1 - 1, 12 and 13, q 2 , or for - 1 6 x 6 1 and x 7 3. The function is decreasing for all values of x less than - 1 and for all values of x *Some texts use the term relative instead of local.

102

CHAPTER 1  Functions and Their Graphs

between 1 and 3. That is, the function is decreasing on the intervals 1 - q , - 12 and 11, 32 , or for x 6 - 1 and 1 6 x 6 3.

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Now Work

PROBLEMS

17

AND

19

5 Use a Graph to Locate the Absolute Maximum and the Absolute Minimum Look at the graph of the function f given in Figure 22. The domain of f is the closed interval 3 a, b4 . Also, the largest value of f is f1u2 and the smallest value of f is f1v2 . These are called, respectively, the absolute maximum and the absolute minimum of f on 3 a, b4 .

Figure 22 y (u, f(u)) y = f (x)

DEFINITION

Let f denote a function defined on some interval I. If there is a number u in I for which f 1x2 … f1u2 for all x in I, then f1u2 is the absolute maximum of f on I, and we say the absolute maximum of f occurs at u.

(b, f (b))

(a, f(a))

(y, f (y)) a

u

x

y b

If there is a number v in I for which f1x2 Ú f1v2 for all x in I, then f1v2 is the absolute minimum of f on I, and we say the absolute minimum of f occurs at v.

domain: [a, b] for all x in [a, b], f(x) ≤ f (u) for all x in [a, b], f(x) ≥ f (y) absolute maximum: f(u) absolute minimum: f(y)

The absolute maximum and absolute minimum of a function f are sometimes called the extreme values of f on I. The absolute maximum or absolute minimum of a function f may not exist. Let’s look at some examples.

Finding the Absolute Maximum and the Absolute Minimum from the Graph of a Function

EX A MPL E 5

For each graph of a function y = f1x2 in Figure 23, find the absolute maximum and the absolute minimum, if they exist. Also, find any local maxima or local minima. Figure 23 y

y

y

y

y (3, 6) 6

6

6

(5, 5)

6

6

4

4

(5, 4) 4

4

4

(4, 4)

(5, 3)

(4, 3)

(0, 3)

(1, 2) (0, 1) 1

(3, 1) 3

5

x

1

3

(b)

(a)

Solution

WARNING A function may have an absolute maximum or an absolute minimum at an endpoint, but not a local maximum or a local minimum. Why? Local maxima and local minima are found over some open interval I, and this interval cannot be created around an endpoint. ■

(1, 1) (2, 1) 5

x

2

2

2

2

2

(1, 4)

1

3

(c)

(2, 2)

(0, 0) 5

x

1

3

(d)

5

x

1

3

5

(e)

(a) The function f whose graph is given in Figure 23(a) has the closed interval
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x = . (b) The function f whose graph is given in Figure 23(b) has the domain 5 x|1 … x … 5, x ≠ 36 . Note that we exclude 3 from the domain because of the “hole” at (3, 1). The largest value of f on its domain is f152 = 3, the absolute maximum. There is no absolute minimum. Do you see why? As you trace the graph, getting closer to the point (3, 1), there is no single smallest value. [As soon as you claim a smallest value, we can trace closer to (3, 1) and get a smaller value!] The function has no local maxima or minima.

x

SECTION 1.3   Properties of Functions

103

(c) The function f whose graph is given in Figure 23(c) has the interval 3 , 54 as its domain. The absolute maximum of f is f152 = . The absolute minimum is 1. Notice that the absolute minimum 1 occurs at any number in the interval 3 1, 24 . The function has a local minimum of 1 at every x in the interval 3 1, 24 , but it has no local maxima. (d) The graph of the function f given in Figure 23(d) has the interval 3 , q 2 as its domain. The function has no absolute maximum; the absolute minimum is f12 = . The function has no local maxima or local minima. (e) The graph of the function f in Figure 23(e) has the domain 5 x 0 1 6 x 6 5, x ≠ 26 . The function f has no absolute maximum and no absolute minimum. Do you see why? The function has a local maximum of 3 at x = , but no local minima.

r

In calculus, there is a theorem with conditions that guarantee a function will have an absolute maximum and an absolute minimum.

THEOREM

Extreme Value Theorem If f is a continuous function* whose domain is a closed interval 3 a, b 4 , then f has an absolute maximum and an absolute minimum on 3 a, b4 . The absolute maximum (minimum) can be found by selecting the largest (smallest) value of f from the following list: 1. The value of f at any local maxima and local minima of f in 1a, b2 . 2. The value of f at each endpoint of 3 a, b4 —that is, f (a) and f (b). 3. The value of f on any interval in 3 a, b4 on which f is constant.

For example, the graph of the function f given in Figure 23(a) is continuous on the closed interval 3 , 54 . The Extreme Value Theorem guarantees that f has extreme values on 3 , 54 . To find them, we list 1. The value of f at the local extrema: f (3) = , f () =  2. The value of f at the endpoints: f () = 1, f (5) = 5 5IF
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PROBLEM

45

6 Use a Graphing Utility to Approximate Local Maxima and Local Minima and to Determine Where a Function Is Increasing or Decreasing To locate the exact value at which a function f has a local maximum or a local minimum usually requires calculus. However, a graphing utility may be used to approximate these values by using the MAXIMUM and MINIMUM features.

EXAM PL E 6

Using a Graphing Utility to Approximate Local Maxima and Minima and to Determine Where a Function Is Increasing or Decreasing (a) Use a graphing utility to graph f1x2 = x3 - 12x + 5 for - 2 6 x 6 2. Approximate where f has a local maximum and where f has a local minimum. (b) Determine where f is increasing and where it is decreasing.

*Although a precise definition requires calculus, we’ll agree for now that a continuous function is one whose graph has no gaps or holes and can be traced without lifting the pencil from the paper.

104

CHAPTER 1  Functions and Their Graphs

Solution

(a) Graphing utilities have a feature that finds the maximum or minimum point of a graph within a given interval. Graph the function f for - 2 6 x 6 2. The MAXIMUM and MINIMUM commands require us to first determine the open interval I. The graphing utility will then approximate the maximum or minimum value in the interval. Using MAXIMUM, we find that the local maximum is 11.53 and that it occurs at x = - .2, rounded to two decimal places. See Figure  B 
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30

30

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f is increasing from x = - 2 to x = - .2 and from x = .2 to x = 2, so f is increasing on the intervals 1 - 2, - .22 and 1.2, 22 , or for - 2 6 x 6 - .2 and .2 6 x 6 2. The graph is decreasing from x = - .2 to x = .2, so f is decreasing on the interval 1 - .2, .22 , or for - .2 6 x 6 .2.

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Now Work

PROBLEM

53

7 Find the Average Rate of Change of a Function In Foundations, Section 3, we said that the slope of a line could be interpreted as the average rate of change. To find the average rate of change of a function between any two points on its graph, calculate the slope of the line containing the two points.

DEFINITION In Words

The symbol ∆ is the Greek capital letter delta and is read “change in.”

If a and b, a ≠ b, are in the domain of a function y = f1x2, the average rate of change of f from a to b is defined as Average rate of change =

f1b2 - f1a2 ∆y = ∆x b - a

a ≠ b

(1)

The symbol ∆y in equation (1) is the “change in y,” and ∆x is the “change in x.” The average rate of change of f is the change in y divided by the change in x.

EX A MPL E 7

Finding the Average Rate of Change Find the average rate of change of f1x2 = 3x2: B
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Solution

(a) The average rate of change of f1x2 = 3x2 from 1 to 3 is f132 - f112 ∆y 2 - 3 2 = = = = 12 ∆x 3 - 1 3 - 1 2 (b) The average rate of change of f1x2 = 3x2 from 1 to 5 is f152 - f112 ∆y 5 - 3 2 = = = = 1 ∆x 5 - 1 5 - 1  (c) The average rate of change of f1x2 = 3x2
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r

SECTION 1.3   Properties of Functions

See Figure 25 for a graph of f1x2 = 3x2. The function f is increasing for x 7 . The fact that the average rate of change is positive for any x1, x2, x1 ≠ x2, in the interval 11, 2 indicates that the graph is increasing on 1 6 x 6 . Further, the average rate of change is consistently getting larger for 1 6 x 6 , which indicates that the graph is increasing at an increasing rate.

Figure 25 y 160

Now Work

(7, 147) 120

Average rate of change 5 24

80

Average rate of change 5 18 (3, 27)

(1, 3) 2

4

PROBLEM

61

The Secant Line (5, 75)

40

(0, 0)

105

The average rate of change of a function has an important geometric interpretation. Look at the graph of y = f1x2
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(a, f(a)) and (b, f(b)). The line containing these two points is called the secant line; its slope is msec =

Average rate of change 5 12 6

f 1b2 - f1a2 f1a + h2 - f 1a2 = b - a h

x

Figure 26

y

y  f (x ) Secant line (b, f (b ))  (a  h, f (a  h)) f (b )  f (a )  f (a  h)  f (a )

(a, f (a ))

Instructor Note: You may want to draw the connection between msec f(x + h) - f(x) and . h

THEOREM

bah

a

b ah

x

Slope of the Secant Line The average rate of change of a function f from a to b equals the slope of the secant line containing the two points (a, f (a)) and (b, f (b)) on its graph.

EXAM PL E 8

Finding the Equation of a Secant Line Suppose that g(x) = 3x2 - 2x + 3. (a) Find the average rate of change of g from - 2 to 1. (b) Find an equation of the secant line containing 1 - 2, g( - 2)2 and 11, g(1)2 . (c) Using a graphing utility, draw the graph of g and the secant line obtained in part (b) on the same screen.

Solution

(a) The average rate of change of g1x2 = 3x2 - 2x + 3 from - 2 to 1 is g112 - g1 - 22 1 - 1 - 22  - 19 = 3 15 = = -5 3

Average rate of change =

g(1) = 3(1)2 - 2(1) + 3 = 4 g(- 2) = 3(- 2)2 - 2(- 2) + 3 = 19

(b) The slope of the secant line containing 1 - 2, g( - 2)) = ( - 2, 19) and (1, g(1)) = 11, 2 is msec = - 5
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the secant line.

106

CHAPTER 1  Functions and Their Graphs

y - y1 y - 19 y - 19 y

Figure 27 24

= = = =

msec 1x - x1 2 - 5(x - ( - 2)) - 5x - 1 - 5x + 9

Point–slope form of the secant line x1 = - 2, y1 = g(- 2) = 19, m sec = - 5 Distribute. Slope–intercept form of the secant line

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67

PROBLEM

1.3 Assess Your Understanding ‘Are You Prepared?’

Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.

1. The interval 12, 52 can be written as the inequality 2 6 x 6 5 . QQ
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the point 13, - 22. (p. 55) y + 2 = 5(x - 3)

2. The slope of the line containing the points 1 - 2, 32 and 13, 2 is 1 . QQ
m

5. The intercepts of the equation ( - 3, ), (3, ), (, - 9) . QQ
m

y = x2 - 9

are

3. Test the equation y = 5x2 - 1 for symmetry with respect to the x-axis, the y-axis, and the origin. QQ
m y-axis

Concepts and Vocabulary 9. True or False A function f has a local maximum at c if there is an open interval I containing c so that for all x in I, f 1x2 … f 1c2. True

6. A function f is increasing on an open interval I if, for any choice of x1 and x2 in I, with x1 6 x2 , we have f 1x1 2 6 f 1x2 2. 7. A(n) even function f is one for which f 1 - x2 = f 1x2 for every x in the domain of ƒ; a(n) odd function f is one for which f 1 - x2 = - f 1x2 for every x in the domain of f.

10. True or False Even functions have graphs that are symmetric with respect to the origin. False

8. True or False A function f is decreasing on an open interval I if, for any choice of x1 and x2 in I, with x1 6 x2 , we have f 1x1 2 7 f 1x2 2. True

Skill Building In Problems 11–20, use the graph of the function f given.

y

11. Is f increasing on the interval 1 - , - 2)? Yes 12. Is f decreasing on the interval 1 - , - )?

13. Is f increasing on the interval 12, 1)?

10

No

(2, 10)

(⫺2, 6)

No

14. Is f decreasing on the interval 12, 5)? Yes

(5, 0)

(⫺5, 0)

15. List the interval(s) on which f is increasing. ( - , - 2), (, 2), (5, q ) ⫺10 16. List the interval(s) on which f is decreasing. ( - q , - ), ( - 2, ), (2, 5) 17. Is there a local maximum value at 2? If yes, what is it? :FT


⫺5

(0, 0)

(⫺8, ⫺4)

10 x

5

⫺6

18. Is there a local maximum value at 5? If yes, what is it? No 19. List the number(s) at which f has a local maximum. What are the local maximum values? 20. List the number(s) at which f has a local minimum. What are the local minimum values?

- 2, 2; , 1 - , , 5; - , , 

In Problems 21–28, the graph of a function is given. Use the graph to find: (a) The intercepts, if any (b) The domain and range (c) The intervals on which the function is increasing, decreasing, or constant (d) Whether the function is even, odd, or neither *21.

*22.

y 4 (⫺4, 2)

(⫺3, 3)

y

(3, 3)

*23.

3 (0, 3)

*24.

y 3

y 3

(0, 2)

(4, 2)

(0, 1) ⫺4 (⫺2, 0)

(2, 0)

4x

⫺3

(⫺1, 0)

(1, 0)

3 x

⫺3

3 x

⫺3

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

(1, 0)

3x

107

SECTION 1.3   Properties of Functions

*25.

*26.

y 2

*27.

y 2

(––␲2 , 1)

( – 3, 2)

(0, 1)

*28.

y 3

1 0, – 2

( ) (3, 1)

( –1, 2) ⫺␲

␲ ⫺–– 2

␲

␲ –– 2

⫺␲

x

⫺ ␲2

(⫺␲, ⫺1)

(⫺ ––␲2 , ⫺1)

␲

␲ 2

x

1 –, 0 3

( )

(␲, ⫺1)

(2, 2)

(⫺2, 1)

(⫺2.3, 0)

(0, 1)

⫺3

3 x (1, –1) (2, –1)

–3

⫺2

⫺2

y 3

3 x

(⫺3, ⫺2)

–3

(3, 0)

⫺2

In Problems 29–32, the graph of a function f is given. Use the graph to find: (a) The numbers, if any, at which f has a local maximum value. What are the local maximum values? (b) The numbers, if any, at which f has a local minimum value. What are the local minimum values? *29.

*30.

y 4

*31.

y

*32.

y

(

3 (0, 3)

⫺4 (⫺2, 0)

1

(0, 2) ⫺␲

4x

(2, 0)

⫺3 (⫺1, 0)

(1, 0)

3 x

␲ –– , 2

␲ ⫺–– 2

1)

␲

␲ –– 2

(⫺ ––␲2 , ⫺1)

y 2

⫺ π2

⫺π

x

π 2

π x

(⫺π, ⫺1)

⫺1

(π, ⫺1) ⫺2

In Problems 33–44, determine algebraically whether each function is even, odd, or neither. 33. f 1x2 = x3

37. F 1x2 = 2x

34. f 1x2 = 2x - x2

Odd

3

38. G1x2 = 1x

Odd

x2 + 3 41. g1x2 = 2 x - 1

Even

Neither

x 42. h1x2 = 2 x - 1

Even

35. g1x2 = - 3x2 - 5

39. f 1x2 = x + 0 x 0

- x3 43. h1x2 = 3x2 - 9

Odd

36. h1x2 = 3x3 + 5

Even

40. f 1x2 = 22x + 1 3

Neither

2x 44. F 1x2 = 0x0

Odd

Neither

2

Even

Odd

In Problems 45–52, for each graph of a function y = f 1x2, find the absolute maximum and the absolute minimum, if they exist. Identify any local maxima or local minima. *45.

*46.

y

*47.

y

(1, 4)

4 2

2

(2, 2) (5, 1) 1

*49.

3

1

*50.

2 (0, 1)

(1, 1) 1

x

5

*51.

3

(3, 2) (0, 0)

1

1

1

x

3

1

3

x

x

(1, 3) 2

(1, 1)

(4, 1)

1

3

y

(3, 2)

2

1

1

*52.

y

(2, 4)

2 (0, 2)

x

5

(2, 0) 2 (1, 1)

(1, 3)

(4, 3)

(5, 0) 3

(2, 4)

4

2

y 4 (1, 3)

(2, 3)

(3, 4)

4

(0, 2)

x

5

y

(0, 3)

(1, 1)

y 4

(4, 4)

4

(3, 3)

*48.

y

(0, 2)

(3, 1)

x

3

1

(2, 2) 2

1 (2, 0) 3

x

2

(1, 3)

In Problems 53–60, use a graphing utility to graph each function over the indicated interval and approximate any local maximum values and local minimum values. Determine where the function is increasing and where it is decreasing. Round answers to two decimal places. 1 - 2, 22

*53. f 1x2 = x3 - 3x + 2

*55. f 1x2 = x - x 5

3

1 - 2, 22

*59. f 1x2 = .25x + .3x - .9x + 3 3

*56. f1x2 = x - x 

*57. f 1x2 = - .2x3 - .x2 + x -  

*54. f 1x2 = x3 - 3x2 + 5

2

1 - , 2

1 - 3, 22

61. Find the average rate of change of f 1x2 = - 2x2 + . B
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-  (b) From 1 to 3 -  D
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- 1

2

1 - 1, 32

1 - 2, 22

*58. f1x2 = - .x3 + .x2 + 3x - 2

1 - , 52

*60. f 1x2 = - .x - .5x + .x - 2 

3

2

1 - 3, 22

62. Find the average rate of change of f 1x2 = - x3 + 1.
B
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-  (b) From 1 to 3 - 13 (c) From - 1 to 1 - 1

108

CHAPTER 1  Functions and Their Graphs

63. Find the average rate of change of g1x2 = x3 - 2x + 1. (a) From - 3 to - 2 1 (b) From - 1 to 1 - 1 (c) From 1 to 3 11

67. g(x) = x2 - 2 (a) Find the average rate of change from - 2 to 1. - 1 (b) Find an equation of the secant line containing 1 - 2, g( - 2)2 and 11, g(1)2. y = - x

64. Find the average rate of change of h1x2 = x2 - 2x + 3. (a) From - 1 to 1 - 2 C
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 (c) From 2 to 5 5

68. g(x) = x2 + 1 (a) Find the average rate of change from - 1 to 2. 1 (b) Find an equation of the secant line containing 1 - 1, g( - 12 and 12, g(2)2. y = x + 3

65. f 1x2 = 5x - 2 (a) Find the average rate of change from 1 to 3. 5 (b) Find an equation of the secant line containing 11, f 112 2 and 13, f 132 2. y = 5x - 2

66. f 1x2 = - x + 1 (a) Find the average rate of change from 2 to 5. -  (b) Find an equation of the secant line containing 12, f 122 2 and 15, f 152 2. y = - x + 1

69. h(x) = x2 - 2x B
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 (b) Find an equation of the secant line containing 12, h(2)2 and 1, h()2. y = x -  70. h(x) = - 2x2 + x B
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 - 5 (b) Find an equation of the secant line containing 1, h()2 and 13, h(3)2. y = - 5x

Mixed Practice 71. g(x) = x3 - 2x (a) Determine whether g is even, odd, or neither. Odd (b) There is a local minimum value of - 5 at 3. Determine a local maximum value. 
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x = - 3

72. f 1x2 = - x3 + 12x (a) Determine whether f is even, odd, or neither. Odd C
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73. F 1x2 = - x + x2 +  (a) Determine whether F is even, odd, or neither. Even C
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(c) Suppose the area under the graph of F between x =  and x = 3 that is bounded below by the xBYJT
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square units. Using the result from part (a), determine the area under the graph of F between x = - 3 and x =  bounded below by the x-axis. 
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Applications and Extensions 75. Minimum Average Cost The average cost per hour in dollars, C, of producing x riding lawn mowers can be modeled by the function C(x) = .3x2 + 21x - 251 +

25 x

*(a) Use a graphing utility to graph C = C(x2 . (b) Determine the number of riding lawn mowers to produce in order to minimize average cost. 
NPXFSTIS (c) What is the minimum average cost? $239/mower 76. Medicine Concentration The concentration C of a medication in the bloodstream t hours after being administered is modeled by the function C(t) = - .2t  + .39t 3 - .25t 2 + .t + .5 (a) After how many hours will the concentration be highest? 
IS (b) A woman nursing a child must wait until the DPODFOUSBUJPO
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him. After taking the medication, how long must she wait before feeding her child? 
IS 77. National Debt The size of the total debt owed by the United States federal government continues to grow. In fact, according to the Department of the Treasury, the debt per QFSTPO
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debt D depends on the year y, and each input corresponds to exactly one output, the debt is a function of the year. Thus D(y), represents the debt for each year y. Year

Debt (billions of dollars)

Year

Debt (billions of dollars)

2001

5807

2007

9008

2002

6228

2008

10,025

2003

6783

2009

11,910

2004

7379

2010

13,562

2005

7933

2011

14,790

2006

8507

2012

16,066

Source: www.treasurydirect.gov

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SECTION 1.3   Properties of Functions

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$1252 billion per year *(f) What is happening to the average rate of change as time passes?

*(e) What is happening to the average rate of change as time passes? 79. For the function f1x2 = x2, compute each average rate of change: B
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1 C
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 D
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 E
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 *(f) Use a graphing utility to graph each of the secant lines along with f . *(g) What do you think is happening to the secant lines? (h) What is happening to the slopes of the secant lines? Is there some number that they are getting closer to? What is that number? 5IFZ
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.

78. E.coli Growth A strain of E.coli
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The data shown in the table are collected. The population is measured in grams and the time in hours. Since population P depends on time t, and each input corresponds to exactly one output, we can say that population is a function of time. Thus P(t) represents the population at time t. Time (hours), t

Population (grams), P

0

0.09

2.5

0.18

3.5

0.26

4.5

0.35

6

0.50

109

80. For the function f 1x2 = x2, compute each average rate of change: (a) From 1 to 2 3 (b) From 1 to 1.5 2.5 (c) From 1 to 1.1 2.1 E
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 F
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 *(f) Use a graphing utility to graph each of the secant lines along with f . *(g) What do you think is happening to the secant lines? (h) What is happening to the slopes of the secant lines? Is there some number that they are getting closer to? What is that number? They are getting closer to 2.

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What does the slope of this line segment represent? (c) Find the average rate of change of the population from 
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Problems 81–88 require the following discussion of a secant line. The slope of the secant line containing the two points 1x, f (x)2 and (x + h, f (x + h)) on the graph of a function y = f 1x2 may be given as msec =

f 1x + h2 - f 1x2 1x + h2 - x

=

f 1x + h2 - f 1x2 h

h ≠ 

In calculus, this expression is called the difference quotient of f. (a) Express the slope of the secant line of each function in terms of x and h. Be sure to simplify your answer. (b) Find msec for h = .5,  and  at x = 1. What value does msec approach as h approaches ? (c) Find the equation for the secant line at x = 1 with h = .1. (d) Use a graphing utility to graph f and the secant line found in part (c) on the same viewing window.

*81. f 1x2 = 2x + 5

*82. f 1x2 = - 3x + 2

*83. f 1x2 = x2 + 2x

*84. f 1x2 = 2x2 + x

*85. f 1x2 = 2x2 - 3x + 1

*86. f 1x2 = - x2 + 3x - 2

*87. f 1x2 =

*88. f 1x2 =

1 x

1 x2

Discussion and Writing 89. Draw the graph of a function that has the following properties: domain: all real numbers; range: all real numbers; intercepts: 1, - 32 and 13, ); a local maximum value of - 2 is at - 1; a local minimum value of -  is at 2. Compare your graph with those of others. Comment on any differences. 90. 3FEP
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increasing on 1 - q , - 1), (2, q ); decreasing on 1 - 1, 22. Again compare your graph with others and comment on any differences. 91. How many x-intercepts can a function defined on an interval have if it is increasing on that interval? Explain. At most one

92. Suppose that a friend of yours does not understand the JEFB
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explanation, complete with graphs, that clarifies the idea. *93. Can a function be both even and odd? Explain.

94. Using a graphing utility, graph y = 5 on the interval 1 - 3, 32. Use MAXIMUM to find the local maximum values on 1 - 3, 32. Comment on the result provided by the calculator.

*95. A function f has a positive average rate of change on the interval 3 2, 5 4 . Is f increasing on 3 2, 5 4 ? Explain. *96. Show that a constant function f(x) = b has an average SBUF
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PG y = 2 - x2 on the interval 3 - 2, 24. Explain how this can happen.

110

CHAPTER 1  Functions and Their Graphs

‘Are You Prepared?’ Answers 1. 2 6 x 6 5

2. 1

4. y + 2 = 5(x - 32

3. symmetric with respect to the y-axis

5. 1 - 3, ), (3, ), (, - 92

1.4 Library of Functions; Piecewise-defined Functions PREPARING FOR THIS SECTION Before getting started, review the following: r
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Now Work the ‘Are You Prepared?’ problems on page 117.

OBJECTIVES 1 Graph the Functions Listed in the Library of Functions (p. 110) 2 Graph Piecewise-defined Functions (p. 115)

1 Graph the Functions Listed in the Library of Functions Figure 28

First we introduce a few more functions, beginning with the square root function. 0O
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y = 1x
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function f1x2 = 1x. Based on the graph, we have the following properties:

y 6 (1, 1)

(4, 2)

(9, 3)

Properties of f1 x 2 = 1x

(0, 0) –2

5

10 x

EX A MPL E 1

Solution

1. The domain and the range are the set of nonnegative real numbers. 2. The x-intercept of the graph of f 1x2 = 1x
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y-intercept of the graph of f1x2 = 1x
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x = .

Graphing the Cube Root Function

3 (a) Determine whether f 1x2 = 2 x is even, odd, or neither. State whether the graph of f is symmetric with respect to the y-axis or symmetric with respect to the origin. 3 (b) Determine the intercepts, if any, of the graph of f1x2 = 2 x. 3 (c) Graph f1x2 = 2x.

(a) Because 3 3 -x = - 2 x = - f1x2 f1 - x2 = 2

the function is odd. The graph of f is symmetric with respect to the origin. 3 (b) The y-intercept is f12 = 2  = . The x-intercept is found by solving the equation f1x2 = . f1x2 =  3 2 x =  x = 

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f (x) = 2x Cube both sides of the equation.

111

SECTION 1.4  Library of Functions; Piecewise-defined Functions

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#FDBVTF of the symmetry with respect to the origin, we find only points 1x, y2 for which 3 x Ú . Figure 29 shows the graph of f1x2 = 2 x.

Table 4

3

Figure 29

x

y =f1x2 = 2x

(x, y)

0

0

10, 02

1 8

1 2

1 1 a , b 8 2

1

1

11, 12

2

22 ≈ 1.26

3 12, 222

8

2

3

y 3 3

(1, 1)

( 1–8, 1–2)

(2, 2 )

( 1–8 , 1–2)

3

3

x

(0, 0)

18, 22

3

(1, 1)

(2, 2 )

r

3

From the results of Example 1 and Figure 29, we have the following properties of the cube root function.

3 Properties of f1 x 2 = 2x

1. The domain and the range are the set of all real numbers. 3 2. The x-intercept of the graph of f 1x2 = 2 x
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 3. The graph is symmetric with respect to the origin. The function is odd. 4. The function is increasing on the interval 1 - q , q 2. 5. The function does not have any local minima or any local maxima.

EXAM PL E 2

Solution

Graphing the Absolute Value Function

(a) Determine whether f1x2 = 0 x 0 is even, odd, or neither. State whether the graph of f is symmetric with respect to the y-axis or symmetric with respect to the origin. (b) Determine the intercepts, if any, of the graph of f1x2 = 0 x 0 . (c) Graph f1x2 = 0 x 0 . (a) Because

f1 - x2 = 0 - x 0 = 0 x 0 = f 1x2

the function is even. The graph of f is symmetric with respect to the y-axis. (b) The y-intercept is f12 = 0  0 = . The x-intercept is found by solving the equation f1x2 =  or 0 x 0 = . So the xJOUFSDFQU
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 (c) Use the function to form Table 5 and obtain some points on the graph. Because of the symmetry with respect to the y-axis, we need to find only points 1x, y2 for which x Ú .
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f1x2 = 0 x 0 . Figure 30

Table 5

x

y = f(x) = ∣ x ∣

1x, y 2

0

0

(0, 0)

1

1

(1, 1)

2

2

(2, 2)

3

3

(3, 3)

y 3

(3, 3) (2, 2) (1, 1)

3 2 1

(3, 3) (2, 2)

2 1

1

(1, 1) 1 2 (0, 0)

3

x

r

112

CHAPTER 1  Functions and Their Graphs

Instructor Note: Emphasize the importance of knowing the properties of the key functions presented here. Being able to visualize basic functions will be helpful later with topics such as nonlinear systems and graphing using transformations.

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of the absolute value function.

Properties of f(x)= ∣ x ∣

1. The domain is the set of all real numbers. The range of f is 5 y 0 y Ú 6 . 2. The x-intercept of the graph of f 1x2 = 0 x 0
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y-intercept of the graph of f1x2 = 0 x 0
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x = .

Seeing the Concept

Graph y = 0 x 0 on a square screen and compare what you see with Figure 30. Note that some graphing calculators use abs 1x2 for absolute value.

Below is a list of the key functions that we have discussed. In going through this list, pay special attention to the properties of each function, particularly to the shape of each graph. Knowing these graphs, along with key points on each graph, will lay the foundation for further graphing techniques.

Figure 31 Constant Function

Constant Function

y

f1x2 = b

f(x) = b

b is a real number

(0,b) x

Figure 32 Identity Function

See Figure 31. The domain of a constant function is the set of all real numbers; its range is the set consisting of a single number b. Its graph is a horizontal line whose y-intercept is b. The constant function is an even function.

Identity Function

f(x) = x

y 3

f1x2 = x (1, 1)

–3 (–1, –1)

(0, 0)

3 x

See Figure 32. The domain and the range of the identity function are the set of all real numbers. Its graph is a line whose slope is 1 and whose yJOUFSDFQU
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Square Function f1x2 = x2

SECTION 1.4  Library of Functions; Piecewise-defined Functions

See Figure 33. The domain of the square function is the set of all real numbers; its range is the set of nonnegative real numbers. The graph of this function is a parabola whose vertex is at 1, 2, which is also the only intercept. The square function is an even function that is decreasing on the interval 1 - q , 2 and increasing on the interval 1, q 2 .

Figure 33 Square Function f (x ) =

y (–2, 4)

x2

(2, 4)

4

(–1, 1)

(1, 1) 4 x

(0, 0)

–4

113

Cube Function f1x2 = x3

Figure 34 Cube Function y 4

f (x ) = x 3

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(1, 1) ⫺4 (⫺1, ⫺1)

(0, 0)

x

4

⫺4

Figure 35 Square Root Function y

f(x) =

2

⫺1

Square Root Function

x

f1x2 = 1x

(1, 1)

(4, 2) 5 x

(0, 0)

Figure 36 Cube Root Function y 3

f (x ) =

3

See Figure 35. The domain and the range of the square root function are the set of nonnegative real numbers. The intercept of the graph is at 1, 2. The square root function is neither even nor odd and is increasing on the interval 1, q 2.

Cube Root Function

x

3 f1x2 = 2 x

3

(1, 1)

(⫺ 1–8,⫺ 1–2)

(2, 2 )

( 1–8 , 1–2)

⫺3

3 x (0, 0) 3

(⫺1, ⫺1)

(⫺2,⫺ 2 )

⫺3

Figure 37 Reciprocal Function y 2

Reciprocal Function

1– , 2

( 2) f (x ) =

(⫺2, ⫺ 1–2 )

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1 –– x

f1x2 =

1 x

(1, 1)

⫺2

2 x

(⫺1, ⫺1) ⫺2

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114

CHAPTER 1  Functions and Their Graphs

Figure 38 Absolute Value Function y

f(x) = ⏐x ⏐

Absolute Value Function f1x2 = 0 x 0

3 (2, 2)

(⫺2, 2)

(1, 1)

(⫺1, 1) ⫺3

(0, 0)

3 x

Instructor Note: Students often apply the greatest integer function by just dropping the decimal portion of a number. Be sure to show them why this won’t work for negative arguments.

DEFINITION

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 The domain of the absolute value function is the set of all real numbers; its range is the set of nonnegative real numbers. The intercept of the graph is at 1, 2 . If x Ú , then f1x2 = x, and this part of the graph of f is the line y = x; if x 6 , then f1x2 = - x, and this part of the graph of f is the line y = - x. The absolute value function is an even function; it is decreasing on the interval 1 - q , 2 and increasing on the interval 1, q 2. The notation int1x2 stands for the largest integer less than or equal to x. For example, 1 3 int112 = 1, int12.52 = 2, inta b = , inta - b = - 1, int1p2 = 3 2  This type of correspondence occurs frequently enough in mathematics that we give it a name.

Greatest Integer Function f1x2 = int1x2* = greatest integer less than or equal to x

Table 6 x

y =f(x) =int(x)

-1

-1

( - 1, - 1)

1 2

-1

1 a - , - 1b 2

1 4

-1

1 a - , - 1b 4

0

0

(0, 0)

0

1 a , b 4

0

1 a , b 2

0

3 a , b 4

-

1 4 1 2 3 4

(x, y)

We obtain the graph of f 1x2 = int1x2
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at x = 1. Although a precise definition requires the idea of a limit, discussed in calculus, in a rough sense, a function is said to be continuous if its graph has no gaps or holes and can be drawn without lifting a pencil from the paper on which the graph is drawn. We contrast this with a discontinuous function. A function is discontinuous if its graph has gaps or holes so that its graph cannot be drawn without lifting a pencil from the paper. *Some books use the notation f 1x2 = 3 x 4 instead of int 1x2.

SECTION 1.4  Library of Functions; Piecewise-defined Functions

Figure 40 f 1x2 = int 1x2 6

⫺2

6

From the graph of the greatest integer function, we can see why it is also called a step function. At x = , x = {1, x = { 2, and so on, this function is discontinuous because, at integer values, the graph suddenly “steps” from one value to another without taking on any of the intermediate values. For example, to the immediate left of x = 3, the y-coordinates of the points on the graph are 2, and at x = 3 and to the immediate right of x = 3, the y-coordinates of the points on the graph are 3. Consequently, the graph has gaps in it. COMMENT When graphing a function using a graphing utility, you can choose either the connected mode, in which points plotted on the screen are connected, making the graph appear without any breaks, or the dot mode, in which only the points plotted appear. When graphing the greatest integer function with a graphing utility, it may be necessary to be in the dot mode. This is to prevent the utility from “connecting the dots” when f 1x2 changes from POF
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gaps even when in “connected” mode. ■

⫺2 (a) Connected mode 6

⫺2

115

6 ⫺2 (b) Dot mode

The functions discussed so far are basic. Whenever you encounter one of them, you should see a mental picture of its graph. For example, if you encounter the function f1x2 = x2, you should see in your mind’s eye a picture like Figure 33.

Now Work

PROBLEMS

9

THROUGH

16

2 Graph Piecewise-defined Functions Sometimes a function is defined using different equations on different parts of its domain. For example, the absolute value function f1x2 = 0 x 0 is defined by two equations: f 1x2 = x if x Ú  and f1x2 = - x if x 6 . For convenience, these equations are generally combined into one expression as f1x2 = 0 x 0 = e

x if x Ú  - x if x 6 

When a function is defined by different equations on different parts of its domain, it is called a piecewise-defined function.

EXAM PL E 3

Analyzing a Piecewise-defined Function The function f is defined as - 2x + 1 if - 3 … x 6 1 f1x2 = c 2 if x = 1 x2 if x 7 1 (a) Find f1 - 22, f112, and f122. (c) Locate any intercepts. (e) Use the graph to find the range of f.

Solution

(b) Determine the domain of f. (d) Graph f . (f) Is f continuous on its domain?

(a) To find f1 - 22, observe that when x = - 2, the equation for f is given by f1x2 = - 2x + 1, so f 1 - 22 = - 2( - 2) + 1 = 5 When x = 1, the equation for f is f1x2 = 2. That is, f112 = 2 When x = 2, the equation for f is f1x2 = x2, so f122 = 22 =  (b) To find the domain of f, look at its definition. Since f is defined for all x greater than or equal to - 3, the domain of f is {x 0 x Ú - 3}, or the interval 3 - 3, q 2 . (c) The y-intercept of the graph of the function is f 12. Because the equation for f when x =  is f1x2 = - 2x + 1, the y-intercept is f12 = - 212 + 1 = 1.

116

CHAPTER 1  Functions and Their Graphs

The x-intercepts of the graph of a function f are the real solutions to the equation f1x2 = . To find the x-intercepts of f, solve f1x2 =  for each “piece” of the function, and then determine whether the values of x, if any, satisfy the condition that defines the piece. f1x2 =  - 2x + 1 =  - 3 … x 6 1 - 2x = - 1 1 x = 2

Figure 41 y 8

4

(1,2)

(2,4)

(0,1)

( 1–2 , 0)

(1, ⫺1)

4

x

f1x2 =  x2 =  x 7 1 x = 

1 The first potential x-intercept, x = , satisfies the condition - 3 … x 6 1, so 2 1 x = is an x-intercept. The second potential x-intercept, x = , does not satisfy 2 1 the condition x 7 1, so x =  is not an x-intercept. The only x-intercept is . The 2 1 JOUFSDFQUT
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a , b . 2 (d) To graph f , graph each “piece.” First graph the line y = - 2x + 1 and keep only the part for which - 3 … x 6 1. Then plot the point 11, 22 because, when x = 1, f1x2 = 2. Finally, graph the parabola y = x2 and keep only the part for which x 7 1.
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 (e) From the graph, we conclude that the range of f is 5 y y 7 - 16 , or the interval 1 - 1, q 2. (f) The function f is not continuous because there is a “jump” in the graph at x = 1.

Now Work

EX A MPL E 4

f 1x2 =  2 =  x = 1 No solution

PROBLEM

29

r

Cost of Electricity *O
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NPOUI (c) If C is the monthly charge for x kWhr, develop a model relating the monthly charge and kilowatt-hours used. That is, express C as a function of x. Source: Duke Energy Progress, 2013

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per kWhr. That is, if x 7 , then C 1x2 = .9512 + .5 + .51x - 2 = . + .5 + .5x - . = .5x + 1.5

117

SECTION 1.4  Library of Functions; Piecewise-defined Functions

The rule for computing C follows two equations:

Figure 42 C

C 1x2 = e

.95x + .5 if  … x …  .5x + 1.5 if x 7 

The Model

Charge (dollars)

180 (1500, 142.90)

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120 60

(800, 82.98)

6.50 (300, 35.18)

0

400

x

800 1200 Usage (kWhr)

1.4 Assess Your Understanding ‘Are You Prepared?’

Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.

1. Sketch the graph of y = 1x. Q


2. Sketch the graph of y =

3. List the intercepts of the equation y = x3 - . Q


1, - 2, 12, 2

1 .
QQ
m

x

Concepts and Vocabulary 4. The function f 1x2 = x2 is decreasing on the interval 1 - q , 2 .

7. True or False The cube root function is odd and is decreasing on the interval 1 - q , q 2. False

5. When functions are defined by more than one equation, they are called piecewise-defined functions.

8. True or False The domain and the range of the reciprocal function are the set of all real numbers. False

6. True or False The cube function is odd and is increasing on the interval 1 - q , q 2. True

Skill Building In Problems 9 –16, match each graph to its function. A. Constant function B. Identity function E. Square root function F. Reciprocal function

C. Square function G. Absolute value function

D. Cube function H. Cube root function

9.

C

10.

A

11.

E

12.

G

13.

B

14.

D

15.

F

16.

H

In Problems 17–24, sketch the graph of each function. Be sure to label three points on the graph. *17. f 1x2 = x

*18. f 1x2 = x2

*19. f 1x2 = x3

*20. f 1x2 = 1x

*21. f 1x2 =

*22. f 1x2 = 0 x 0

3 *23. f 1x2 = 2 x

*24. f 1x2 = 3

1 x

x2 25. If f 1x2 = c 2 2x + 1 find: (a) f 1 - 22

27. If f 1x2 = e

if x 6  if x =  if x 7 





(b) f 12

2x -  x3 - 2

2

if - 1 … x … 2 if 2 6 x … 3

find: (a) f 12 -  (b) f 112 - 2

(c) f 122

- 3x 26. If f 1x2 = c  2x2 + 1 5

(c) f 122 
(d) f 132 25

find: (a) f 1 - 22

28. If f 1x2 = e

if x 6 - 1 if x = - 1 if x 7 - 1


(b) f 1 - 12

x3 3x + 2

find: (a) f 1 - 12 - 1




if - 2 … x 6 1 if 1 … x … 

(c) f 12

1

(b) f 12 
(c) f 112 5 (d) f 132 11

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

118

CHAPTER 1  Functions and Their Graphs

In Problems 29–40: (a) Find the domain of each function.

(b) Locate any intercepts.

(d) Based on the graph, find the range.

(e) Is f continuous on its domain?

if x ≠  if x = 

*29. f 1x2 = b

2x 1

*32. f1x2 = b

x + 3 - 2x - 3

1 + x *35. f1x2 = b 2 x *38. f 1x2 = b

2 - x 2x

*30. f 1x2 = b

if x ≠  if x = 

3x 

x + 3 *33. f 1x2 = c 5 -x + 2

if x 6 - 2 if x Ú - 2

1 *36. f 1x2 = c x 3 2 x

if x 6  if x Ú 

(c) Graph each function. *31. f 1x2 = b

- 2x + 3 3x - 2

if x 6 1 if x Ú 1

if - 2 … x 6 1 2x + 5 if x = 1 *34. f 1x2 = c - 3 if x 7 1 - 5x if x 6 

*37. f 1x2 = b

if x Ú 

if - 3 … x 6 1 *39. f 1x2 = 2 int 1x2 if x Ú 1

0x0 x3

if - 3 … x 6  if x =  if x 7 

if - 2 … x 6  if x Ú 

*40. f 1x2 = int 12x2

In Problems 41–44, the graph of a piecewise-defined function is given. Write a definition for each function. *41.

*42.

y

*43.

y 2

2

(2, 2) (1, 1)

(⫺1, 1) ⫺2

⫺2

2 x

(0, 0)

y (0, 2)

2 (2, 1)

(2, 1)

(⫺1, 1)

*44.

y

2 x

(0, 0)

⫺2

(⫺1, 0) (0, 0)

(2, 0) x

(1, 1)

⫺2

2 x

(⫺1, ⫺1)

45. If f 1x2 = int 12x2, find (a) f 11.22

2

(b) f 11.2

3

(c) f 1 - 1.2

-

x 46. If f 1x2 = int a b, find 2 (a) f 11.22  (b) f 11.2



(c) f 1 - 1.2

-1

Applications and Extensions 47. Cell Phone Service
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following function is used to compute the monthly cost for a subscriber: C 1x2 = b

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if  … x … 5 if x 7 5

where x is the number of anytime minutes used. Compute the monthly cost of the cellular phone for use of the following numbers of anytime minutes: B

$39.99
C


D

 Source: Sprint PCS 48. Parking at O’Hare International Airport The short-term OP
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if  if 1 if 3 if  if 9

6 6 6 6 …

x x x x x

… … … 6 …

1 3  9 2

Determine the fee for parking in the short-term parking garage for (a) 2 hours 
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 Source: O’Hare International Airport 49. Cost of Natural Gas
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following rate schedule for natural gas usage in single-family residences: Monthly service charge $22.25 1FS
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NPOUI *(c) Develop a function that models the monthly charge C for x therms of gas. *(d) Graph the function found in part (c). Source: Peoples Energy, Chicago, Illinois, 2013

SECTION 1.4  Library of Functions; Piecewise-defined Functions

50. Cost of Natural Gas
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*52. Federal Income Tax
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x equals taxable income and y equals the tax due, construct a function y = f 1x2 for Schedule Y-1. 2013 Tax Rate Schedules Schedule X-Single If Taxable Income Is Over

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The Tax Is This Amount

Schedule Y-l - Married Filing Jointly or Qualified Widow(er) Plus This %

Of the Excess Over

If Taxable Income Is Over

But Not Over

The Tax Is This Amount

Plus This %

Of the Excess Over

$0

$8,925

$0

+

10%

$0

$0

$17,850

$0

+

10%

$0

8,925

36,250

892.50

+

15%

8,925

17,850

72,500

1,785

+

15%

17,850

36,250

87,850

4,991.25

+

25%

36,250

72,500

146,400

9,982.50

+

25%

72,500

87,850

183,250

17,891.25

+

28%

87,850

146,400

223,050

28,457.50

+

28%

146,400

183,250

398,350

44,603.25

+

33%

183,250

223,050

398,350

49,919.50

+

33%

223,050

398,350

400,000

115,586.25

+

35%

398,350

398,350

450,000

107,768.50

+

35%

398,350

400,000

–

116,163.75

+

39.6%

400,000

450,000

–

125,846

+

39.6%

450,000

Source: Internal Revenue Service

53. Cost of Transporting Goods A trucking company transports HPPET
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Credit Score

Loan-Level Price Adjustment Rate

… 659

3.00%

660–679

2.50%

680–699

1.75%

700–719

1%

720–739

0.5%

Ú 740

0.25%

Source: Fannie Mae.

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 *56. Minimum Payments for Credit Cards Holders of credit cards issued by banks, department stores, oil companies, and so on receive bills each month that state minimum amounts that must be paid by a certain due date. The minimum due depends on the total amount owed. One such credit card

120

CHAPTER 1  Functions and Their Graphs

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57. Wind Chill The wind chill factor represents the equivalent air temperature at a standard wind speed that would produce the same heat loss as the given temperature and wind speed. One formula for computing the equivalent temperature is t W = d 33 -

11.5 + 11v - v2 133 - t2

22. 33 - 1.595133 - t2

 … v 6 1.9 1.9 … v … 2 v 7 2

where v represents the wind speed (in meters per second) and t
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*58. Wind Chill
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each additional ounce up to 13 ounces. First-class rates do not apply to flats weighing more than 13 ounces. Develop a model that relates C, the first-class postage charged, for a flat weighing x ounces. Graph the function. Source: United States Postal Service

Discussion and Writing *65. Exploration Graph y = x3. Then on the same screen graph y = 1x - 12 3 + 2. Could you have predicted the result?

In Problems 60–67, use a graphing utility. *60. Exploration Graph y = x . Then on the same screen graph y = x2 + 2, followed by y = x2 + , followed by y = x2 - 2. What pattern do you observe? Can you predict the graph of y = x2 - ? Of y = x2 + 5? 2

*66. Exploration Graph y = x2, y = x, and y = x on the same screen. What do you notice is the same about each graph? What do you notice is different?

*61. Exploration Graph y = x2. Then on the same screen graph y = 1x - 22 2, followed by y = 1x - 2 2, followed by y = 1x + 22 2. What pattern do you observe? Can you predict the graph of y = 1x + 2 2? Of y = 1x - 52 2?

*67. Exploration Graph y = x3, y = x5, and y = x on the same screen. What do you notice is the same about each graph? What do you notice is different? *68. Consider the equation

*62. Exploration Graph y = 0 x 0 . Then on the same screen graph 1 y = 2 0 x 0 , followed by y =  0 x 0 , followed by y = 0 x 0 . 2 What pattern do you observe? Can you predict the graph of 1 y = 0 x 0 ? Of y = 5 0 x 0 ?  *63. Exploration Graph y = x2. Then on the same screen graph y = - x2. What pattern do you observe? Now try y = 0 x 0 and y = - 0 x 0 . What do you conclude?

y = b

1 

if x is rational if x is irrational

Is this a function? What is its domain? What is its range? What is its y-intercept, if any? What are its x-intercepts, if any? Is it even, odd, or neither? How would you describe its graph?

69. Define some functions that pass through 1, 2 and 11, 12 and are increasing for x Ú . Begin your list with y = 1x, y = x, and y = x2. Can you propose a general result about such functions?

*64. Exploration Graph y = 1x. Then on the same screen graph y = 1- x. What pattern do you observe? Now try y = 2x + 1 and y = 21 - x2 + 1. What do you conclude?

‘Are You Prepared?’ Answers 1. y

2.

2 (1, 1)

y 2

(4, 2)

3. 1, - 2, 12, 2 (1, 1) 2 x

(0, 0)

4

x

(⫺1, ⫺1)

SECTION 1.5  Graphing Techniques: Transformations

121

1.5 Graphing Techniques: Transformations OBJECTIVES 1 Graph Functions Using Vertical and Horizontal Shifts (p. 121) 2 Graph Functions Using Compressions and Stretches (p. 124) 3 Graph Functions Using Reflections about the x-Axis and the y-Axis (p. 126)

At this stage, if you were asked to graph any of the functions defined by 1 3 y = x, y = x2, y = x3, y = 1x, y = 2 x, y = , or y = 0 x 0 , your response should x be “Yes, I recognize these functions and know the general shapes of their graphs.” *G
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Sometimes we are asked to graph a function that is “almost” like one that we already know how to graph. In this section, we develop techniques for graphing such functions. Collectively, these techniques are referred to as transformations.

1 Graph Functions Using Vertical and Horizontal Shifts EXAM PL E 1

Vertical Shift Up Use the graph of f1x2 = x2 to obtain the graph of g1x2 = x2 + 3. Find the domain and range of g.

Solution

Table 7

Begin by obtaining some points on the graphs of f and g. For example, when x = , then y = f12 =  and y = g12 = 3. When x = 1, then y = f112 = 1 and y = g112 = .
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x

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y = g1x2 = x2 + 3

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4

7

(⫺2, 7)

-1

1

4

(⫺1, 4)

0

0

3

1

1

4

2

4

7

y = x2 + 3 y (2, 7) (1, 4) 5 (⫺2, 4)

Up 3 units

(2, 4) (0, 3) y = x 2

(⫺1, 1) ⫺3

(1, 1) (0, 0)

3

x

The domain of g is all real numbers, or ( - q , q ). The range of g is [3, q ).

EXAM PL E 2

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Vertical Shift Down Use the graph of f1x2 = x2 to obtain the graph of g1x2 = x2 - . Find the domain and range of g.

Solution

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122

CHAPTER 1  Functions and Their Graphs

Table 8

y = f1 x2 = x2

x

Figure 44

y = g1x2 = x2 − 4

-2

4

0

-1

1

-3

0

0

-4

1

1

-3

2

4

0

y (–2, 4)

y = x2

4

(2, 4)

Down 4 units

Down 4 units (2, 0) 4 x

(⫺2, 0) (0, 0)

y = x2 ⫺ 4 ⫺5

(0, ⫺4)

The domain of g is all real numbers, or 1 - q , q 2 . The range of g is [ - , q ).

r

Note that a vertical shift affects only the range of a function, not the domain. For example, the range of f1x2 = x2 is 3 , q 2 . In Example 1 the range of g is 3 3, q 2 , whereas in Example 2 the range of g is 3 - , q 2 . The domain for all three functions is all real numbers. We are led to the following conclusions: If a positive real number k is added to the output of a function y = f1x2, the graph of the new function y = f1x2 + k is the graph of f shifted vertically up k units. If a positive real number k is subtracted from the output of a function y = f1x2, the graph of the new function y = f1x2 - k is the graph of f shifted vertically down k units.

Now Work

PROBLEM

39

Horizontal Shift to the Right

EX A MPL E 3

Use the graph of f1x2 = 2x to obtain the graph of g1x2 = 1x - 2. Find the domain and range of g.

Solution

Table 9

The function g1x2 = 1x - 2 is basically a square root function. Table 9 lists some points on the graphs of f and g. Note that when f1x2 =  then x = , and when g1x2 = , then x = 2. Also, when f1x2 = 2, then x = , and when g1x2 = 2, then x = . Notice that the x-coordinates on the graph of g are 2 units larger than the corresponding x-coordinates on the graph of f for any given y-coordinate. We conclude that the graph of g is identical to that of f, except that it is shifted horizontally 2 units to the right. See 'JHVSF
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y = g1x2

x

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x

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0

0

2

0

1

1

3

1

4

2

6

2

9

3

11

3

Figure 45

y 5

y = "x

Right 2 units (4, 2)

y = "x − 2

(6, 2) (0, 0)

(2, 0)

9

x

Right 2 units

The domain of g is 3 2, q 2 and the range is 3 , q 2 .

EX A MPL E 4

r

Horizontal Shift to the Left Use the graph of f1x2 = 1x to obtain the graph of g1x2 = 1x + . Find the domain and range of g.

Solution

Again, the function g1x2 = 1x +  is basically a square root function. Its graph is the same as that of f,
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123

SECTION 1.5  Graphing Techniques: Transformations

Figure 46

y 5

y = "x + 4

Left 4 units

(0, 2) (4, 2)

(−4, 0) −5

(0, 0)

y = "x x

5

Left 4 units

The domain of g is 3 - , q 2 and the range is 3 , q 2 .

Now Work

PROBLEM

r

43

Note that a horizontal shift affects only the domain of a function, not the range. For example, the domain of f1x2 = 1x is 3 , q 2 . In Example 3 the domain of g is 3 2, q 2
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g is 3 - , q 2 . The range for all three functions is 3 , q 2 . We are led to the following conclusion.

In Words

For y = f1x - h2, add h to each x-coordinate on the graph of y = f1x2. For y = f1x + h2, subtract h from each x-coordinate on the graph of y = f1x2.

EXAM PL E 5

Solution

Figure 47

If the argument x of a function f is replaced by x - h, h 7 , the graph of the new function y = f1x - h2 is the graph of f shifted horizontally right h units. If the argument x of a function f is replaced by x + h, h 7 , the graph of the new function y = f1x + h2 is the graph of f shifted horizontally left h units. Observe the distinction between vertical and horizontal shifts. The graph of f1x2 = x3 + 2 is obtained by shifting the graph of y = x3 up 2 units, because we evaluate the cube function first and then add 2. The graph of g1x2 = 1x + 22 3 is obtained by shifting the graph of y = x3 left 2 units, because we add 2 to x before we evaluate the cube function. Vertical and horizontal shifts are sometimes combined.

Combining Vertical and Horizontal Shifts

Graph the function f1x2 = 0 x + 3 0 - 5. Find the domain and range of f. We graph f in steps. First, note that the rule for f is basically an absolute value function, so begin with the graph of y = 0 x 0
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y

y

y

5

5

5 (−1, 2)

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(2, 2) (0, 0)

(−5, 2) 5

x

(−3, 0)

2

x

(−1, −3)

(−5, −3)

y = 0x 0 (a)

Replace x by x + 3; Horizontal shift left 3 units

y = 0 x + 30 (b)

Subtract 5: Vertical shift down 5 units

x

2

(−3, −5) y = 0 x + 30 − 5 (c)

The domain of f is all real numbers, or 1 - q , q 2 . The range of f is 3 - 5, q 2 .

Check: Graph Y1 = f 1x2 = 0 x + 3 0 - 5
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124

CHAPTER 1  Functions and Their Graphs

In Example 5, if the vertical shift had been done first, followed by the horizontal shift, the final graph would have been the same. Try it for yourself.

Now Work

PROBLEMS

45

AND

75

2 Graph Functions Using Compressions and Stretches EX A MPL E 6

Vertical Stretch Use the graph of f1x2 = 2x to obtain the graph of g1x2 = 22x.

Solution

Table 10

EX A MPL E 7

To see the relationship between the graphs of f and g
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y = g1x2

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= 22x

0

0

0

1

1

2

4

2

4

9

3

6

y = 2"x

Figure 48 y 5 (1, 2)

y = "x

(4, 4) (4, 2)

(0, 0)

10 x

5

(1, 1)

r

Vertical Compression 1 Use the graph of f1x2 = 0 x 0 to obtain the graph of g1x2 = 0 x 0 . 2

Solution

Table 11

1 as large as the 2 corresponding y-coordinate on the graph of f. The graph of f1x2 = 0 x 0 is vertically 1 1 compressed by a factor of to obtain the graph of g1x2 = 0 x 0 . For example, 12, 22 2 2 is on the graph of f, but 12, 12 is on the graph of g
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For each x, the y-coordinate of a point on the graph of g is

y = f1 x2 x -2

In Words

For y = af1x2 , the factor a is “outside” the function, so it affects the y-coordinates. Multiply each y-coordinate on the graph of y = f1x2 by a.

= 0x0 2

-1

1

0

0

1

1

2

2

Figure 49

y = g1x2 1 = 0x0 2

y

y =⏐x⏐

4

1 1 2 0 1 2 1

y= (2, 2)

(2, 2)

(– 2, 1) 4

1– x ⏐ ⏐ 2

(2, 1) (0, 0)

4 x

r

When the right side of a function y = f1x2 is multiplied by a positive number a, the graph of the new function y = af1x2 is obtained by multiplying each y-coordinate on the graph of y = f1x2 by a. The new graph is a vertically compressed (if  6 a 6 1) or a vertically stretched (if a 7 1) version of the graph of y = f1x2.

Now Work

PROBLEM

47

What happens if the argument x of a function y = f1x2 is multiplied by a positive number a, creating a new function y = f1ax2? To find the answer, look at the following Exploration.

SECTION 1.5  Graphing Techniques: Transformations

125

Exploration On the same screen, graph each of the following functions:

Figure 50 3 Y2 =

2x

Y1 =

x

Y3 =

–x 2

4

0 0

1 1 x Y3 = f a x b = x = 2 A2 A2 Create a table of values to explore the relation between the x- and y-coordinates of each function. Y1 = f(x) = 1x

Y2 = f(2x) = 12x

Result You should have obtained the graphs in Figure 50. Look at Table 12(a). Notice that (1, 1), (4, 2), and (9, 3) are points on the graph of Y1 = 1x. Also, (0.5, 1), (2, 2), and (4.5, 3) are points on the graph of 1 Y2 = 22x. For a given y-coordinate, the x-coordinate on the graph of Y2 is of the x-coordinate on Y1. 2

Table 12

We conclude that the graph of Y2 = 22x is obtained by multiplying the x-coordinate of each point on 1 the graph of Y1 = 1x by . The graph of Y2 = 22x is the graph of Y1 = 1x compressed horizontally. 2 Look at Table 12(b). Notice that (1, 1), (4, 2), and (9, 3) are points on the graph of Y1 = 1x. Also x . For a given y-coordinate, notice that (2, 1), (8, 2), and (18, 3) are points on the graph of Y3 = A2 the x-coordinate on the graph of Y3 is 2 times the x-coordinate on Y1. We conclude that the graph of x is obtained by multiplying the x-coordinate of each point on the graph of Y1 = 1x by 2. The Y3 = A2 x is the graph of Y1 = 1x stretched horizontally. graph of Y3 = A2

Based on the Exploration, we have the following result:

In Words

For y = f (ax), the factor a is “inside” the function, so it affects the x-coordinates. Multiply each x-coordinate on the graph of 1 y = f (x) by . a

If the argument x of a function y = f1x2 is multiplied by a positive number a, then the graph of the new function y = f 1ax2 is obtained by multiplying each 1 x-coordinate of y = f 1x2 by . A horizontal compression results if a 7 1, and a a horizontal stretch results if  6 a 6 1. Let’s look at an example.

EXAM PL E 8

Graphing Using Stretches and Compressions

The graph of y = f 1x2 is given in Figure 51. Use this graph to find the graphs of (a) y = 2f1x2

Solution

(a) The graph of y = 2f1x2 is obtained by multiplying each y-coordinate of y = f1x2 by 2. See Figure 52. (b) The graph of y = f 13x2 is obtained from the graph of y = f1x2 by multiplying 1 each x-coordinate of y = f1x2 by . See Figure 53. 3 Figure 52

Figure 51

1

1

( 2 , 1(

 3 2 5 3 2 2 3 , 1 2

(

Figure 53

( 2 , 2(

(

y

( 52 , 2(

2

( 52 , 1(

y  f(x)

 2

y 3

y

(b) y = f13x2

2 1

1 

x

( 6 , 1( ( 56 , 1(

2



3 2

2 52 3

1 

(

3

(

3 , 2 2

2

PROBLEMS

y  f(3x)

69(e)

AND

(g)

3

x

2 3

( 2 , 1 (

y  2f(x)

Now Work

2



x

1

r

126

CHAPTER 1  Functions and Their Graphs

3 Graph Functions Using Reflections about the x-Axis and the y-Axis Reflection about the x-Axis

EX A MPL E 9

Graph the function f1x2 = - x2. Find the domain and range of f.

Solution

Begin with the graph of y = x2,
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y = x2

y = −x2

-2

4

-4

-1

1

-1

0

0

0

1

1

-1

2

4

-4

Table 13 Figure 54 y 4

(⫺2, 4)

(⫺1, 1) ⫺4

y = x2 (2, 4)

The domain of f is all real numbers, or 1 - q , q 2 . The range of f is 1 - q , 4 .

(1, 1)

(–1, –1)

(1, – 1)

(–2, –4)

–4

4

r

x

When the right side of the function y = f 1x2 is multiplied by - 1, the graph of the new function y = - f1x2 is the reflection about the x-axis of the graph of the function y = f1x2.

(2, – 4) y = –x 2

Now Work

EX AM PL E 10

PROBLEM

51

Reflection about the y-Axis Graph the function f1x2 = 1- x. Find the domain and range of f.

Solution

To get the graph of f1x2 = 1- x, begin with the graph of y = 1x, as shown in Figure 55. For each point 1x, y2 on the graph of y = 1x, the point 1 - x, y2 is on the graph of y = 1- x. Obtain the graph of y = 1- x by reflecting the graph of y = 1x about the y-axis. See Figure 55.

Figure 55

y 4 y=

y=

–x ( – 1, 1) –5

x

(4, 2)

( – 4, 2) (0, 0)

(1, 1) 5

x

The domain of f is 1 - q , 4 . The range of f is 3 , q 2.

In Words

For y = - f (x), multiply each y-coordinate on the graph of y = f1x2 by - 1. For y = f1 - x2 , multiply each x-coordinate by - 1.

When the graph of the function y = f 1x2 is known, the graph of the new function y = f1 - x2 is the reflection about the y-axis of the graph of the function y = f1x2.

r

SECTION 1.5  Graphing Techniques: Transformations

127

SUMMARY OF GRAPHING TECHNIQUES To Graph:

Draw the Graph of f and:

Functional Change to f(x)

Vertical shifts

 

y = f1x2 + k, k 7  y = f1x2 - k, k 7 

Raise the graph of f by k units. Lower the graph of f by k units.

Horizontal shifts

 

y = f1x + h2 , h 7  y = f1x - h2, h 7 

Shift the graph of f to the left h units. Shift the graph of f to the right h units.

Compressing or stretching

 

 

Add k to f 1x2. Subtract k from f1x2 .  

 

   

Multiply each y-coordinate of y = f 1x2 by a. Stretch the graph of f vertically if a 7 1. Compress the graph of f vertically if  6 a 6 1. 1 Multiply each x-coordinate of y = f 1x2 by . a Stretch the graph of f horizontally if  6 a 6 1. Compress the graph of f horizontally if a 7 1.

Reflection about the x-axis

 

y = af 1x2, a 7      y = f1ax2 , a 7 

y = - f 1x2

 

y = f1 - x2

Reflect the graph of f about the y-axis.

Solution

Replace x by ax.     Multiply f1x2 by - 1.  

Reflection about the y-axis

Instructor Note: Encourage students to write the current function after each transformation, and remind them that transformations are cumulative.

Multiply f1x2 by a.    

 

Reflect the graph of f about the x-axis.

EX AM PL E 11

Replace x by x + h. Replace x by x - h.

Replace x by - x.

Determining the Function Obtained from a Series of Transformations Find the function that is finally graphed after the following three transformations are applied to the graph of y = 0 x 0 . 1. Shift left 2 units 2. Shift up 3 units 3. Reflect about the y-axis 1. Shift left 2 units: Replace x by x + 2. 2. Shift up 3 units: Add 3. 3. Reflect about the y-axis: Replace x by - x.

Now Work

PROBLEM

y = 0x + 20 y = 0x + 20 + 3 y = 0 -x + 20 + 3

27

r

The examples that follow combine some of the procedures outlined in this section to get the required graph.

EX AM PL E 12

Combining Graphing Procedures Graph the function f1x2 =

3 + 1. Find the domain and range of f . x - 2

128

CHAPTER 1  Functions and Their Graphs

Solution

1 It is helpful to write f as f(x) = 3 a b + 1. Now use the following steps to x - 2 obtain the graph of f : STEP 1: y =

1 x

Reciprocal function

1 3 STEP 2: y = 3 # a b = x x

1 Multiply by 3; vertical stretch of the graph of y = by x a factor of 3.

3 x - 2 3 STEP 4: y = + 1 x - 2

Replace x by x - 2; horizontal shift to the right 2 units.

STEP 3: y =

Add 1; vertical shift up 1 unit.

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y 4 (1, 1)

24

(1, 3)

(3, 3)

3 2, – 2

( )

(2, 12–) 4 x

y 4

y 4

24

(3, 4)

(4, 3–2 )

4 x

x

4

(21, 21)

(4, 5–2 ) 4 x

24 (1, 22)

(21, 23)

24 1 x

D  y 5 ––

Multiply by 3; Vertical stretch

(1, 23) 24

24

24 3 x

E  y 5 ––

Replace x by x 2 2; Horizontal shift right 2 units

3 x –2

F  y 5 –––

Add 1; Vertical shift up 1 unit

3 x –2

G  y 5 ––– 1 1

1 is {x 0 x ≠ } and its range is {y 0 y ≠ }. Because we x shifted right 2 units and up 1 unit to obtain f , the domain of f is {x 0 x ≠ 2} and its The domain of y =

range is {y 0 y ≠ 1}.

Hint: Although the order in which transformations are performed can be altered, you may consider using the following order for consistency: 1. Reflections 2. Compressions and stretches 3. Shifts ■

EX AM PL E 13

r

Other orderings of the steps shown in Example 12 would also result in the graph of f. For example, try this one: STEP 1: y =

1 x

1 x - 2 3 STEP 3: y = x - 2 3 STEP 4: y = + 1 x - 2 STEP 2: y =

Reciprocal function Replace x by x - 2; horizontal shift to the right 2 units. 1 Multiply by 3; vertical stretch of the graph of y = x - 2 by a factor of 3. Add 1; vertical shift up 1 unit.

Combining Graphing Procedures Graph the function f1x2 = 21 - x + 2. Find the domain and range of f .

Solution

Because horizontal shifts require the form x - h, begin by rewriting f1x2 as f1x2 = 21 - x + 2 = 2 - 1x - 12 + 2. Now use the following steps:

129

SECTION 1.5  Graphing Techniques: Transformations

STEP 1: y = 1x

Square root function

STEP 2: y = 2 - x

Replace x by - x; reflect about the y-axis.

STEP 3: y = 2 - (x - 1) = 21 - x

Replace x by x - 1; horizontal shift to the right 1 unit.

STEP 4: y = 21 - x + 2

Add 2; vertical shift up 2 units.

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(1, 1)

(4, 2)

(a) y 

(4, 2)

y (3, 4) 5 (0, 3)

(3, 2)

(1, 2)

(0, 1) 5 x 5

(0, 0)

5

y 5

y 5

y 5

(1, 1) (0, 0)

5 x 5

(b) y  x

x Replace x by x; Reflect about y-axis

5 x 5

(1, 0)

Replace x by x  1; (c) y  Horizontal shift  right 1 unit 

5 x

(x  1) Add 2; (d) y   x  1 Vertical shift up 2 units 1x

1x2

The domain of f is ( - q , 1] and the range is [2, q ).

Now Work

PROBLEM

r

61

1.5 Assess Your Understanding Concepts and Vocabulary 4. True or False The graph of y = - f 1x2 is the reflection about the x-axis of the graph of y = f 1x2. True

1. Suppose that the graph of a function f is known. Then the graph of y = f 1x - 22 may be obtained by a(n) horizontal shift of the graph of f to the right

5. True or False To obtain the graph of f(x) = 2x + 2, shift the graph of y = 2x horizontally to the right 2 units. False

a distance of 2 units. 2. Suppose that the graph of a function f is known. Then the graph of y = f 1 - x2 may be obtained by a reflection y about the -axis of the graph of the function y = f 1x2.

6. True or False To obtain the graph of f(x) = x3 + 5, shift the graph of y = x3 vertically up 5 units. True

3. Suppose that the graph of a function g is known. The graph vertical of y = g(x) + 2 may be obtained by a up shift of the graph of g a distance of 2 units.

Skill Building In Problems 7–18, match each graph to one of the following functions: A. y = x2 + 2

E. y = 1x - 22

F. y = - 1x + 22

2

I. y = 2x2 7.

C. y = 0 x 0 + 2

B. y = - x2 + 2

B

8.

E

y 3

9.

3 x

3

3 x

L. y = - 2 0 x 0 H

y 1 3

3

H. y = - 0 x + 2 0

K. y = 2 0 x 0

J. y = - 2x2 y 3

D. y = - 0 x 0 + 2

G. y = 0 x - 2 0

2

10.

D

y 3

3x 3

3 x

130

CHAPTER 1  Functions and Their Graphs

11.

I

y 3

12.

A

y 5

13.

3 x

3

L

y 3

3

14.

3 x 6

3

3

15.

F

y 4

16.

3 x

3

4

17.

y 4

G

4 x

4

6 x 4

3

J

y 3

3

4 x

4

1

3 x

C

y 8

18.

K

y 3

3 x

3

4

3

In Problems 19–26, write the function whose graph is the graph of y = x3, but is: 19. 4IJGUFE
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23. Reflected about the y-axis

y = -x

3

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y = x3

24. Reflected about the x-axis

y = - x3

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y = a xb = x  

In Problems 27–30, find the function that is finally graphed after each of the following transformations is applied to the graph of y = 1x in the order stated. 27. (1) Shift up 2 units (2) Reflect about the x-axis (3) Reflect about the y-axis

y = - ( 2- x + 2)

29. (1) Reflect about the x-axis (2) Shift up 2 units (3) Shift left 3 units y = - 2x + 3 + 2

28. (1) Reflect about the x-axis (2) Shift right 3 units (3) Shift down 2 units y = - 2x - 3 - 2 30. (1) Shift up 2 units (2) Reflect about the y-axis (3) Shift left 3 units y = 2- (x + 3) + 2 = 2- x - 3 + 2

31. If 13, 2 is a point on the graph of y = f 1x2, which of the following points must be on the graph of y = - f 1x2? (c) (a) 1, 32 (b) 1, - 32 (c) 13, - 2 (d) 1 - 3, 2

32. If 13, 2 is a point on the graph of y = f 1x2, which of the following points must be on the graph of y = f 1 - x2? (d) (a) 1, 32 (b) 1, - 32 (c) 13, - 2 (d) 1 - 3, 2

33. If 11, 32 is a point on the graph of y = f 1x2, which of the following points must be on the graph of y = 2f 1x2? (c) (a) 11, 32 (b) 12, 32

34. If 1, 22 is a point on the graph of y = f 1x2, which of the following points must be on the graph of y = f(2x)? (c) (a) 1, 12 (b) 1, 22 (c) 12, - 22 (d) 1, 2

(c) 11, 2

1 (d) a , 3b 2 35. Suppose that the x-intercepts of the graph of y = f 1x2 are - 5 and 3. *(a) What are the x-intercepts of the graph of y = f 1x + 22? *(b) What are the x-intercepts of the graph of y = f 1x - 22? *(c) What are the x-intercepts of the graph of y = f 1x2? *(d) What are the x-intercepts of the graph of y = f 1 - x2? 37. Suppose that the function y = f 1x2 is increasing on the interval 1 - 1, 52. (a) Over what interval is the graph of y = f 1x + 22 increasing? ( - 3, 32 (b) Over what interval is the graph of y = f 1x - 52 increasing? (, 12 *(c) What can be said about the graph of y = - f 1x2? *(d) What can be said about the graph of y = f 1 - x2?

36. Suppose that the x-intercepts of the graph of y = f 1x2 are -  and 1. *(a) What are the x-intercepts of the graph of y = f 1x + 2? *(b) What are the x-intercepts of the graph of y = f 1x - 32? *(c) What are the x-intercepts of the graph of y = 2f 1x2? *(d) What are the x-intercepts of the graph of y = f 1 - x2? 38. Suppose that the function y = f 1x2 is decreasing on the interval 1 - 2, 2. (a) Over what interval is the graph of y = f 1x + 22 decreasing? ( - , 52 (b) Over what interval is the graph of y = f 1x - 52 decreasing? (3, 122 *(c) What can be said about the graph of y = - f 1x2? *(d) What can be said about the graph of y = f 1 - x2?

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

SECTION 1.5  Graphing Techniques: Transformations

131

In Problems 39–68, graph each function using the techniques of shifting, compressing, stretching, and/or reflecting. Start with the graph of the basic function (for example, y = x2) and show all stages. Be sure to show at least three key points. Find the domain and the range of each function. *40. f 1x2 = x2 + 

*39. f 1x2 = x2 - 1 *42. g1x2 = x - 1

*45. f 1x2 = 1x - 12 3 + 2 *48. g1x2 =

*41. g1x2 = x3 + 1

*43. h1x2 = 2x + 2

3

*44. h1x2 = 2x + 1

*46. f 1x2 = 1x + 22 3 - 3

1 1x 2

*49. h1x2 =

3 *51. f 1x2 = - 2 x

*47. g1x2 = 1x

1 2x

3 *50. h1x2 = 22x

*52. f 1x2 = - 1x

3 *53. g1x2 = 2 -x

*55. h1x2 = - x3 + 2

*56. h1x2 =

*57. f 1x2 = 21x + 12 2 - 3

*58. f 1x2 = 31x - 22 2 + 1

*59. g1x2 = 22x - 2 + 1

*60. g1x2 = 3 0 x + 1 0 - 3

*61. h1x2 = 1- x - 2

*62. h1x2 =

*63. f 1x2 = - 1x + 12 3 - 1

*64. f 1x2 = - 2x - 1

*65. g1x2 = 2 0 1 - x 0

*54. g1x2 =

1 -x

*67. h1x2 = 2 int 1x - 12

*66. g1x2 = 22 - x

1 + 2 -x  + 2 x

*68. h1x2 = int 1 - x2

In Problems 69–72, the graph of a function f is illustrated. Use the graph of f as the first step toward graphing each of the following functions: (a) F 1x2 = f 1x2 + 3 (e) Q1x2 = 69.

1 f 1x2 2

y 4 (0, 2)

(b) G1x2 = f 1x + 22

(c) P 1x2 = - f 1x2

(f) g1x2 = f 1 - x2

(g) h1x2 = f 12x2 *71.

y 4

*70. (2, 2)

1

(2, 2)

2 (4, 0)

4 (4, 2)

2

x

4 2 (4, 2)

2

y

4 x

2

2 (2, 2)

(4, 2)

π

(d) H1x2 = f 1x + 12 - 2

π –2

π– 2

1 π ( –2 , 1)

y

*72.

(π–2 , 1)

1

π

x

π

π –2

(π, 1)

π– 1

2

π

x

(π, 1)

Mixed Practice In Problems 73–82, complete the square of each quadratic expression. Then graph each function using the technique of shifting. (If necessary, refer to Appendix A, Section A.4 to review completing the square.)

*73. f 1x2 = x2 + 2x

*77. f 1x2 = x + x + 1 2

*81. f 1x2 = - 3x2 - 12x - 1

*74. f 1x2 = x2 - x

*78. f 1x2 = x - x + 1 2

*82. f 1x2 = - 2x2 - 12x - 13

83. (a) Graph f(x) =  x - 3 - 3 using transformations. (b) Find the area of the region bounded by f and the x-axis that lies below the x-axis.

*75. f 1x2 = x2 - x + 1

*79. f 1x2 = 2x - 12x + 19 2

*76. f 1x2 = x2 + x + 2

*80. f 1x2 = 3x2 + x + 1

84. (a) Graph f(x) = - 2 x -  +  using transformations. (b) Find the area of the region bounded by f and the x-axis that lies above the x-axis.

Applications and Extensions

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*85. The equation y = 1x - c2 2 defines a family of parabolas, one parabola for each value of c. On one set of coordinate axes, graph the members of the family for c = , c = 3, and c = - 2.

76 72 68 64 60 56 0

t 4 8 12 16 20 24 Time (hours after midnight)

132

CHAPTER 1  Functions and Their Graphs

(a) At what temperature is the thermostat set during daytime hours? At what temperature is the thermostat set overnight? ž'
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y 2 (1, 1) 3 (2, 1)

(2, 0) 3 x (1, 1) 2

93. The graph of a function f is illustrated in the figure. *(a) Draw the graph of y = 0 f 1x2 0 . *(b) Draw the graph of y = f 1 0 x 0 2.

l Ag

y 2

where g ≈ 32.2 feet per second per second is the acceleration of gravity.

(1, 1)

(22, 0) 23

(2, 0) 3 x

(21, 21) (0, 21) 22

94. Suppose (1, 3) is a point on the graph of y = f 1x2 . (a) What point is on the graph of y = f 1x + 32 - 5? 1 - 2, - 22 (b) What point is on the graph of y = - 2f 1x - 22 + 1? (3, - 5) (c) What point is on the graph of y = f 12x + 32? ( - 1, 3)

95. Suppose 1 - 3, 52 is a point on the graph of y = g1x2 . (a) What point is on the graph of y = g1x + 12 - 3? ( - , 2) (b) What point is on the graph of y = - 3g1x - 2 + 3? (1, - 12) (c) What point is on the graph of y = g13x + 92? ( - , 5)

SECTION 1.6  Mathematical Models: Building Functions

133

Discussion and Writing *96. Suppose that the graph of a function f is known. Explain how the graph of y = f 1x2 differs from the graph of y = f 1x2. *97. Suppose that the graph of a function f is known. Explain how the graph of y = f 1x2 - 2 differs from the graph of y = f 1x - 22. 98. The area under the curve y = 1x bounded below by the 1 square units. Using the x-axis and on the right by x =  is 3

ideas presented in this section, what do you think is the area under the curve of y = 1- x bounded below by the x-axis and on the left by x = - ? Justify your answer. *99. Explain how the range of the function f1x2 = x2 compares to the range of g1x2 = f 1x2 + k. *100. Explain how the domain of g1x2 = 2x compares to the domain of g1x - k2, where k Ú . 1 
sq. units 3

1.6 Mathematical Models: Building Functions OBJECTIVE 1 Build and Analyze Functions (p. 133)

1 Build and Analyze Functions Real-world problems often result in mathematical models that involve functions. These functions need to be constructed or built based on the information given. In building functions, we must be able to translate the verbal description into the language of mathematics. This is done by assigning symbols to represent the independent and dependent variables and then finding the function or rule that relates these variables.

EXAM PL E 1

Finding the Distance from the Origin to a Point on a Graph Let P = 1x, y2 be a point on the graph of y = x2 - 1.

(a) Express the distance d from P to the origin O as a function of x. (b) What is d if x = ? (c) What is d if x = 1? 22 (d) What is d if x = ? 2 (e) Use a graphing utility to graph the function d = d 1x2, x Ú . Rounded to two decimal places, find the value(s) of x at which d has a local minimum. [This gives the point(s) on the graph of y = x2 - 1 closest to the origin.]

Solution

Figure 58

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d = 2 1x - 2 2 + 1y - 2 2 = 2x2 + y2

2 y ⫽ x2 ⫺ 1 1 P⫽ (x, y) (0, 0) d 1 ⫺1 2 x

Since P is a point on the graph of y = x2 - 1, substitute x2 - 1 for y. Then d 1x2 = 2x2 + 1x2 - 12 2 = 2x - x2 + 1 The distance d is expressed as a function of x. (b) If x = , the distance d is

⫺1

d 12 = 2 - 2 + 1 = 21 = 1 (c) If x = 1, the distance d is

Figure 59

d 112 = 21 - 12 + 1 = 1

2

22 , the distance d is 2 22 22  22 2 1 1 23 da b = a b - a b + 1 = - + 1 = 2 2 2 B 2 B 2

(d) If x =

2

0 0

(e) Figure 59 shows the graph of Y1 = 2x - x2 + 1. Using the MINIMUM feature on a graphing utility, we find that when x ≈ .1, the value of d is

134

CHAPTER 1  Functions and Their Graphs

smallest. The local minimum is d ≈ . rounded to two decimal places. Since d 1x2 is even, it follows by symmetry that when x ≈ - .1, the value of d is also a local minimum. Since 1 {.12 2 - 1 ≈ - .5, the points 1 - .1, - .52 and 1.1, - .52 on the graph of y = x2 - 1 are closest to the origin.

r

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EX A MPL E 2

PROBLEM

1

Area of a Rectangle A rectangle has one corner in quadrant I on the graph of y = 25 - x2, another at the origin, a third on the positive y-axis, and the fourth on the positive x-axis. See 'JHVSF
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Solution

Figure 60 y 30 20

(x, y) y ⫽ 25 ⫺ x 2

10 ⫺1

1 2 3 4 5

x

Express the area A of the rectangle as a function of x. What is the domain of A? Graph A = A 1x2. For what value of x is the area largest?

(a) The area A of the rectangle is A = xy, where y = 25 - x2. Substituting this expression for y, we obtain A 1x2 = x125 - x2 2 = 25x - x3. (b) Since 1x, y2 is in quadrant I, we have x 7 . Also, y = 25 - x2 7 , which implies that x2 6 25, so - 5 6 x 6 5. Combining these restrictions, we have the domain of A as 5 x  6 x 6 56 , or 1, 52 using interval notation. (c)
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5 0

PROBLEM

7

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Solution

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SECTION 1.6  Mathematical Models: Building Functions

(b)
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Figure 63 N Plane

135

400 mph

Figure 64 500 600 miles

d Plane 250 mph E 400 miles

0

2

⫺50

Now Work

r

(b)

(a) PROBLEM

19

1.6 Assess Your Understanding Applications and Extensions 1. Let P = 1x, y2 be a point on the graph of y = x2 - . (a) Express the distance d from P to the origin as a function of x. d(x) = 2x - 15x2 +  (b) What is d if x = ?  (c) What is d if x = 1? 25 ≈ . *(d) Use a graphing utility to graph d = d1x2. (e) For what values of x is d smallest? ≈ - 2. or ≈2. 2. Let P = 1x, y2 be a point on the graph of y = x2 - . (a) Express the distance d from P to the point 1, - 12 as a function of x. d1x2 = 2x - 13x2 + 9 (b) What is d if x = ?  (c) What is d if x = - 1? 23 ≈ . *(d) Use a graphing utility to graph d = d1x2. (e) For what values of x is d smallest? ≈ - 2.55 or ≈2.55

3. Let P = 1x, y2 be a point on the graph of y = 1x. (a) Express the distance d from P to the point 11, 2 as a function of x. d1x2 = 2x2 - x + 1 *(b) Use a graphing utility to graph d = d1x2. (c) For what values of x is d smallest? .5 1 4. Let P = 1x, y2 be a point on the graph of y = . x *(a) Express the distance d from P to the origin as a function of x. *(b) Use a graphing utility to graph d = d1x2. (c) For what values of x is d smallest? 1 or - 1 5. A right triangle has one vertex on the graph of y = x , x 7 , at 1x, y2, another at the origin, and the third on the positive y-axis at 1, y2, as shown in the figure. Express the area A of 1 the triangle as a function of x. A1x2 = x 2 3 3

y

(0, y)

y⫽x

(x, y)

*6. A right triangle has one vertex on the graph of y = 9 - x2, x 7 , at 1x, y2, another at the origin, and the third on the positive x-axis at 1x, 2. Express the area A of the triangle as a function of x. 7. A rectangle has one corner in quadrant I on the graph of y = 1 - x2, another at the origin, a third on the positive y-axis, and the fourth on the positive x-axis. See the figure. (a) Express the area A of the rectangle as a function of x. A(x) = x(1 - x2) (b) What is the domain of A? {x  6 x 6 } *(c) Graph A = A1x2. For what value of x is A largest? y 16

y ⫽ 16 ⫺ x 2 (x, y)

8

4

(0,0)

x

8. A rectangle is inscribed in a semicircle of radius 2. See the figure. Let P = 1x, y2 be the point in quadrant I that is a vertex of the rectangle and is on the circle. y y ⫽ 4 ⫺ x2

⫺2

P ⫽ (x, y )

2

x

(a) Express the area A of the rectangle as a function of x. A(x) = 2x2 - x2 (b) Express the perimeter p of the rectangle as a function of x. p(x) = x + 22 - x2 *(c) Graph A = A1x2. For what value of x is A largest? *(d) Graph p = p1x2. For what value of x is p largest?

9. A rectangle is inscribed in a circle of radius 2. See the figure on the next page. Let P = 1x, y2 be the point in quadrant I x (0, 0) that is a vertex of the rectangle and is on the circle. (a) Express the area A of the rectangle as a function of x. A(x) = x2 - x2 *Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

136

CHAPTER 1  Functions and Their Graphs y 2

P ⫽ (x, y)

⫺2

x

2 ⫺2

16. Geometry An equilateral triangle is inscribed in a circle of radius r. See the figure. Express the circumference C of the circle as a function of the length x of a side of the triangle. x2 [Hint: First show that r 2 = .] C 1x2 = 223 px 3 3 x

x r

2 2 x ⫹y ⫽4

(b) Express the perimeter p of the rectangle as a function of x. p(x) = x + 2 - x2 *(c) Graph A = A1x2. For what value of x is A largest? *(d) Graph p = p1x2. For what value of x is p largest? 10. A circle of radius r is inscribed in a square. See the figure. r

(a) Express the area A of the square as a function of the radius r of the circle. A(r) = r 2 (b) Express the perimeter p of the square as a function of r. p(r) = r 11. Geometry
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18. Uniform Motion Two cars leave an intersection at the TBNF
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d

19. Uniform Motion Two cars are approaching an intersection. One is 2 miles south of the intersection and is moving at a DPOTUBOU
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15. Geometry A semicircle of radius r is inscribed in a rectangle so that the diameter of the semicircle is the length of the rectangle. See the figure. R r

(a) Express the area A of the rectangle as a function of the radius r of the semicircle. A(r) = 2r 2 (b) Express the perimeter p of the rectangle as a function of r. 15. (b) p1r2 = r

Sphere

h

SECTION 1.6  Mathematical Models: Building Functions

21. Inscribing a Cylinder in a Cone Inscribe a right circular cylinder of height h and radius r in a cone of fixed radius R and fixed height H. See the illustration. Express the volume 2 pH1R - r2r V of the cylinder as a function of r. V 1r2 = 2 R [Hint: V = pr h. Note also the similar triangles.] r

H

137

*(a) If a person can row a boat at an average speed of 3 miles per hour and the same person can walk 5 miles per hour, build a model that expresses the time T that it takes to go from the island to town as a function of the distance x from P to where the person lands the boat. (b) What is the domain of T ? 5 x  … x … 126 (c) How long will it take to travel from the island to town if UIF
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Cone

22. Installing Cable TV MetroMedia Cable is asked to provide service to a customer whose house is located 2 miles from the road along which the cable is buried. The nearest connection box for the cable is located 5 miles down the road. See the figure.

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House

h

Stream 2 mi

Box 5 mi

x

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P

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Island

x

x

x

24 in.

x

x x

x 24 in.

(a) Express the volume V of the box as a function of the length x of the side of the square cut from each corner. V(x) = x(2 - 2x)2 (b) What is the volume if a 3-inch square is cut out? 92 in.3 D
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138

CHAPTER 1  Functions and Their Graphs

1.7 Building Mathematical Models Using Variation OBJECTIVES 1 Construct a Model Using Direct Variation (p. 138) 2 Construct a Model Using Inverse Variation (p. 139) 3 Construct a Model Using Joint or Combined Variation (p. 139)

When a mathematical model is developed for a real-world problem, it often involves relationships between quantities that are expressed in terms of proportionality: Force is proportional to acceleration. When an ideal gas is held at a constant temperature, pressure and volume are inversely proportional. The force of attraction between two objects is inversely proportional to the square of the distance between them. Revenue is directly proportional to sales.

Instructor Note: The material in this section is optional and can be skipped without loss of continuity.

Each of these statements illustrates the idea of variation, or how one quantity varies in relation to another quantity. Quantities may vary directly, inversely, or jointly.

1 Construct a Model Using Direct Variation DEFINITION

Let x and y denote two quantities. Then y varies directly with x, or y is directly proportional to x, if there is a nonzero number k such that y = kx

Figure 65 y

The number k is called the constant of proportionality. 5IF
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y and x if y varies directly with x and k 7 , x Ú . Note that the constant of proportionality is, in fact, the slope of the line. If two quantities vary directly, then knowing the value of each quantity in one instance enables us to write a model that is true in all cases.

x

Mortgage Payments

EX A MPL E 1

The monthly payment p on a mortgage varies directly with the amount B borrowed. If UIF
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model that relates the monthly payment p to the amount B borrowed for a mortgage with these terms. Then find the monthly payment p when the amount B
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Solution

Because p varies directly with B, we know that p = kB for some constant k. Because p = .5 when B = 1, it follows that

Figure 66

.5 = k 112 k = .5

Monthly payment

p 800

(120, 798)

600

Since p = kB, we have p = .5B

400

The model

In particular, when B = $12,, we find that

200 0

Solve for k.

p = .51$12,2 = $9 40 80 120 160 Amount borrowed (000's)

B

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p and the amount B borrowed.

Now Work

PROBLEMS

3

AND

21

r

SECTION 1.7  Building Mathematical Models Using Variation

139

2 Construct a Model Using Inverse Variation DEFINITION Figure 67

Let x and y denote two quantities. Then y varies inversely with x, or y is inversely proportional to x, if there is a nonzero constant k such that

y

y =

k x

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EXAM PL E 2

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pounds. Write a model relating the maximum weight W (in pounds) to length l (in feet). Find the maximum weight W that can be safely supported by a length of 25 feet.

Solution

Because W varies inversely with l, we know that

Figure 68

W =

k l

for some constant k. Because W = 5 when l = 1, we have k 1 k = 5

5 =

Since W =

k , we have l W =

5 l

The Model

In particular, the maximum weight W that can be safely supported by a piece of pine 25 feet in length is W = Figure 69

5 = 2 pounds 25

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W and the length l.

W 600

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(10, 500)

500 400

31

3 Construct a Model Using Joint or Combined Variation

300 200

(25, 200)

100 0

PROBLEM

r

5

10

15

20

25

l

When a variable quantity Q is proportional to the product of two or more other variables, we say that Q varies jointly with these quantities. Combinations of direct and/or inverse variation may also occur. This is usually referred to as combined variation. Let’s look at an example.

140

CHAPTER 1  Functions and Their Graphs

EX A MPL E 3

Loss of Heat through a Wall The loss of heat through a wall varies jointly with the area of the wall and the difference between the inside and outside temperatures, and inversely with the thickness of the wall. Write an equation that relates these quantities.

Solution

Begin by assigning symbols to represent the quantities: L = Heat loss

T = Temperature difference d = Thickness of wall

A = Area of wall Then

L = k

AT d

where k is the constant of proportionality.

r

In direct or inverse variation, the quantities that vary may be raised to powers. 'PS
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EX A MPL E 4

Force of the Wind on a Window The force F of the wind on a flat surface positioned at a right angle to the direction of the wind varies jointly with the area A of the surface and the square of the speed v
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Since F varies jointly with A and v2, we have

Figure 70

F = kAv2 Wind

where k is the constant of proportionality. We are told that F = 15 pounds when A =  # 5 = 2 feet 2 and v = 3 miles per hour. This gives 15 = k 122 192 1 k = 12

F = kAv 2, F = 150, A = 20, v = 30

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1 1122 1252 = 25 pounds 12

PROBLEM

39

r

141

SECTION 1.7  Building Mathematical Models Using Variation

1.7 Assess Your Understanding Concepts and Vocabulary k 2. True or False If y varies directly with x, then y = , where k x is a constant. False

1. If x and y are two quantities, then y is directly proportional to x if there is a nonzero number k such that y = kx.

Skill Building In Problems 3–14, write a general formula to describe each variation. 1 3. y varies directly with x; y = 2 when x = 1 y = x 5 5. A varies directly with x2; A = p when x = 2

A = px2

7. F varies inversely with d 2; F = 1 when d = 5

F =

25 d2

4. v varies directly with t; v = 1 when t = 2

v = t

6. V varies directly with x3; V = 3p when x = 3

V =

y varies inversely with 1x; y =  when x = 9

y =

8.

9. z varies directly with the sum of the squares of x and y; z = 5 when x = 3 and y = 

1 z = (x2 + y2 2 5

10. T varies jointly with the cube root of x and the square of d; T = 1 when x =  and d = 3

12. z varies directly with the sum of the cube of x and the square of y; z = 1 when x = 2 and y = 3

M =

1 3 z = (x + y2 2 1

13. The square of T varies directly with the cube of a and inversely with the square of d; T = 2 when a = 2 and d =  14. The cube of z varies directly with the sum of the squares of x and y; z = 2 when x = 9 and y = 

Applications and Extensions

p 3 r 3

19. F = . * 1 - 11 a

2x

3 T = d2 2 x

11. M varies directly with the square of d and inversely with the square root of x; M = 2 when x = 9 and d = 

15. V =

p 3 x 3 12

z3 =

9d 2 22x

T2 =

 2 (x + y2 2 9

a3 d2

mM b d2

In Problems 15–20, write an equation that relates the quantities. 15. Geometry The volume V of a sphere varies directly with the p cube of its radius r. The constant of proportionality is . 3 16. Geometry The square of the length of the hypotenuse c of a right triangle varies jointly with the sum of the squares of the lengths of its legs a and b. The constant of proportionality is 1. c 2 = a2 + b2

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p = .9B; $91.5

17. Geometry The area A of a triangle varies jointly with the lengths of the base b and the height h. The constant of 1 1 proportionality is . A = bh 2 2 18. Geometry The perimeter p of a rectangle varies jointly with the sum of the lengths of its sides l and w. The constant of proportionality is 2. p = 21l + w2

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p = .99B; $153.25

19. Physics: Newton’s Law The force F (in newtons) of attraction between two objects varies jointly with their masses m and M (in kilograms) and inversely with the square of the distance d (in meters) between them. The constant of proportionality is G = . * 1 - 11.

23. Physics: Falling Objects The distance s that an object falls is directly proportional to the square of the time t of the fall. If BO
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24. Physics: Falling Objects The velocity v of a falling object is directly proportional to the time t of the fall. If, after 
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142

CHAPTER 1  Functions and Their Graphs

25. Physics: Stretching a Spring The elongation E of a spring balance varies directly with the applied weight W (see the figure). If E = 3 when W = 2, find E when W = 15. 2.25

29. (a) D =

29    p

E

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 p 2 3 equation for V. V = r h 28. Cost Equation The cost C of chocolate-covered almonds 3 varies directly with the number A of pounds of almonds QVSDIBTFE
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30. (a) t =

29. Demand Suppose that the demand D for candy at the movie theater is inversely related to the price p. B
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Chapter Review

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Discussion and Writing 47. Using a situation that has not been discussed in the text, write a real-world problem that you think involves two variables that vary inversely. Exchange your problem with another student to solve and critique.

45. In the early seventeenth century, Johannes Kepler discovered that the square of the period T for a planet to orbit the Sun varies directly with the cube of its mean distance a from the Sun. Go to the library and research this law and Kepler’s other two laws. Write a brief paper about these laws and Kepler’s place in history. 46. Using a situation that has not been discussed in the text, write a real-world problem that you think involves two variables that vary directly. Exchange your problem with another student to solve and critique.

48. Using a situation that has not been discussed in the text, write a real-world problem that you think involves three variables that vary jointly. Exchange your problem with another student to solve and critique.

Chapter Review Library of Functions Constant function (p. 112) f 1x2 = b The graph is a horizontal line with y-intercept b.

Identity function (p. 112) f 1x2 = x The graph is a line with slope 1 and yJOUFSDFQU


y

y 3

y f (x ) = b

Square function (pp. 112–113) f 1x2 = x2 The graph is a parabola with intercept at 1, 2. ( – 2, 4)

(2, 4)

4

(0,b) (1, 1) x

–3 ( – 1, – 1)

(0, 0)

(– 1, 1)

3 x

Square root function (pp. 110 and 113) f 1x2 = 1x

y

y

4

2 (1, 1)

⫺4 (⫺1, ⫺1)

(0, 0)

4

x

⫺1

4 x

(0, 0)

–4

Cube function (p. 113) f 1x2 = x3

(1, 1)

Cube root function (pp. 111 and 113) 3 f 1x2 = 2 x y 3

(1, 1)

(0, 0)

(4, 2) 5 x

3

(1, 1)

(⫺ 1–8,⫺ 1–2)

(2, 2 )

( 1–8 , 1–2)

⫺3

3 x (0, 0)

⫺4

3

(⫺1, ⫺1)

(⫺2,⫺ 2 )

⫺3

144

CHAPTER 1  Functions and Their Graphs

Reciprocal function (p. 113) 1 f1x2 = x

Absolute value function (pp. 112 and 114) f 1x2 = 0 x 0

y

y

2

3

⫺2

(⫺1, 1) 2 x

⫺3

y 4 (2, 2)

(⫺2, 2) (1, 1)

Greatest integer function (p. 114) f 1x2 = int 1x2

(0, 0)

2

(1, 1) 3 x

⫺2

(⫺1, ⫺1)

2

4

x

⫺3 ⫺2

Things to Know Function (pp. 75–78)

  Function notation (pp. 78–81)

A relation between two sets such that each element x in the first set, the domain, has corresponding to it exactly one element y in the second set. The range is the set of y-values of the function for the x-values in the domain. A function can also be characterized as a set of ordered pairs 1x, y2 in which no first element is paired with two different second elements. y = f 1x2

f is a symbol for the function.  

x is the argument, or independent variable.

 

y is the dependent variable.

   

Difference quotient of f (pp. 80 and 109) Domain (pp. 81–83) Vertical-line test (p. 89) Even function f (p. 98) Odd function f (p. 98)

f 1x2 is the value of the function at x, or the image of x.

A function f may be defined implicitly by an equation involving x and y or explicitly by writing y = f 1x2. f 1x + h2 - f 1x2

h ≠  h If unspecified, the domain of a function f defined by an equation is the largest set of real numbers for which f 1x2 is a real number. A set of points in the plane is the graph of a function if and only if every vertical line intersects the graph in at most one point. f 1 - x2 = f 1x2 for every x in the domain ( - x must also be in the domain).

f 1 - x2 = - f 1x2 for every x in the domain ( - x must also be in the domain).

Increasing function (p. 100)

A function f is increasing on an open interval I if, for any choice of x1 and x2 in I, with x1 6 x2 , we have f 1x1 2 6 f 1x2 2.

Decreasing function (p. 100)

A function f is decreasing on an open interval I if, for any choice of x1 and x2 in I, with x1 6 x2 , we have f 1x1 2 7 f 1x2 2.

Constant function (p. 100)

A function f is constant on an open interval I if, for all choices of x in I, the values of f 1x2 are equal.

Local maximum (p. 101)

A function f has a local maximum at c if there is an open interval I containing c so that, for all x in I, f 1x2 … f 1c2.

Local minimum (p. 101)

A function f has a local minimum at c if there is an open interval I containing c so that, for all x in I, f 1x2 Ú f 1c2.

Absolute maximum and Absolute minimum (p. 102)

Let f denote a function defined on some interval I.

If there is a number u in I for which f 1x2 … f 1u2 for all x in I, then f 1u2 is the absolute maximum of f on I and we say the absolute maximum of f occurs at u. If there is a number v in I for which f 1x2 Ú f 1v2 , for all x in I, then f 1v2 is the absolute minimum of f on I and we say the absolute minimum of f occurs at v.

Average rate of change of a function (p. 104)

Direct variation (p. 138) Inverse variation (p. 139)

The average rate of change of f from a to b is f 1b2 - f 1a2 ∆y = ∆x b - a y = kx, k ≠  k y = ,k ≠  x

a ≠ b

Chapter Review

145

Objectives Section

You should be able to...

Examples

Review Exercises

Determine whether a relation represents a function (p. 75) Find the value of a function (p. 78) Find the domain of a function defined by an equation (p. 81) Form the sum, difference, product, and quotient of two functions (p. 83)

1–5 6, 7 8, 9 10

1, 2 3–5, 15, 39 6–11 12–14

Identify the graph of a function (p. 89) Obtain information from or about the graph of a function (p. 90)

1 2–5

27, 28 16(a)–(e), 17(a), 17(e), 17(g)

1 2

17(f) 18–21

3 4

17(b) 17(c)

5

17(d)

2

Determine even and odd functions from a graph (p. 98) Identify even and odd functions from the equation (p. 99) Use a graph to determine where a function is increasing, decreasing, or constant (p. 100) Use a graph to locate local maxima and local minima (p. 101) Use a graph to locate the absolute maximum and the absolute minimum (p. 102) Use a graphing utility to approximate local maxima and local minima and to determine where a function is increasing or decreasing (p. 103) Find the average rate of change of a function (p. 104) Graph the functions listed in the library of functions (p. 110) Graph piecewise-defined functions (p. 115)

6 7, 8 1, 2 3, 4

22, 23, 40(d), 41(b) 24–26 29, 30 37, 38

1.5

1

Graph functions using vertical and horizontal shifts (p. 121)

1–5, 11, 12, 13

   

2 3

Graph functions using compressions and stretches (p. 124) Graph functions using reflections about the x-axis or y-axis (p. 126)

6–8, 12 9–11, 13

16(f), 31, 33, 34, 35, 36 16(g), 32, 36 16(h), 32, 34, 36

1.6

1

Build and analyze functions (p. 133)

1–3

40, 41

1.7    

1

Construct a model using direct variation (p. 138) Construct a model using inverse variation (p. 139) Construct a model using joint or combined variation (p. 139)

1 2 3,4

42 43 44

1.1      

1

1.2  

1

1.3    

1

   

4

 

6

  1.4  

7

2 3 4

2

2 3

5

1

2 3

Review Exercises In Problems 1 and 2, determine whether each relation represents a function. For each function, state the domain and range. 2. 5 15, 32, 12, 52, 15, - 32 6 Not a function

1. 5 13, 22, 1 - 2, 52, 11, 22 6 Function; domain { - 2, 1, 3}; range {2, 5}

In Problems 3–5, find the following for each function: (a) f 132 (b) f 1 - 32 (c) f 1 - x2 *3. f 1x2 =

3x x2 - 1

(d) - f 1x2

*4. f 1x2 = 2x2 - 3x

(e) f 1x - 12 *5. f 1x2 =

In Problems 6–11, find the domain of each function. x - 1 {x 0 x ≠ 2, x ≠ - 2} x2 - 4 x *9. f 1x2 = 2 x + 2x - 3 6. f 1x2 =

7. f 1x2 = 22 - x 10. f 1x2 =

1x x2 - 1

{x x … 2} {x 0 x Ú 0, x ≠ 1}

8. g(x) = 11. g(x) =

7 - x x2

0x0 x

{x x ≠ 0}

x 1x + 8

In Problems 12–14, find f + g, f - g, f # g, and

5x x 7 - 86

f for each pair of functions. State the domain of each of these functions. g x + 1 1 *13. f 1x2 = 3x2 + x + 1; g(x) = 3x *14. f 1x2 = ; g(x) = *12. f 1x2 = 2 - x; g(x) = 3x + 1 x - 1 x f 1x + h2 f 1x2 15. Find the difference quotient of f 1x2 = - 2x2 + x + 1; that is, find , h ≠ 0. - 4x + 1 - 2h h

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

(f) f 13x2

2

146

CHAPTER 1  Functions and Their Graphs

16. Use the graph of the function f shown to: *(a) Find the domain and the range of f. (b) List the intercepts. (0, 0) (c) Find f 1 - 22. - 1 (d) Find the value(s) of x for which f 1x2 = - 3. (e) Solve f 1x2 7 0. {x 0 6 x … 3} *(f) Graph y = f 1x - 32. 1 *(g) Graph y = f a xb. 2 *(h) Graph y = - f 1x2.

17. Use the graph of the function f shown to find: *(a) The domain and the range of f . *(b) The intervals on which f is increasing, decreasing, or constant. *(c) The local minimum values and local maximum values. *(d) The absolute maximum and absolute minimum. (e) Whether the graph is symmetric with respect to the x-axis, the y-axis, or the origin. No symmetry (f) Whether the function is even, odd, or neither. Neither *(g) The intercepts, if any.

-4

y 4

y 4

(3, 3)

(3, 0)

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(0, 0)

⫺6 (⫺4,⫺3) (⫺3, 0)

x

5

(4, 3)

6 x (2, ⫺1)

⫺4

(⫺4, ⫺3)

⫺4

In Problems 18–21, determine (algebraically) whether the given function is even, odd, or neither. 18. f 1x2 = 2x5 - x

19. g(x) =

Odd

4 + x2 1 + x4

21. f 1x2 =

*20. G(x) = x2 - 3x + 4

Even

x 1 + x2

Odd

In Problems 22 and 23, use a graphing utility to graph each function over the indicated interval. Approximate any local maximum values and local minimum values. Determine where the function is increasing and where it is decreasing. *22. f 1x2 = 2x3 - 5x + 1

1 - 3, 32

*23. f 1x2 = 2x4 - 5x3 + 2x + 1

24. Find the average rate of change of f 1x2 = 8x - x. (a) From 1 to 2 23 (b) From 0 to 1 7

1 - 2, 32

2

(c) From 2 to 4

47

In Problems 25 and 26, find the average rate of change from 3 to 5 for each function f. Be sure to simplify. 25. f 1x2 = 2x - 3

26. f 1x2 = x3 - 5x

2

44

In Problems 27 and 28, is the graph shown the graph of a function? 27. No

28. Yes

y

y

x

x

In Problems 29 and 30, sketch the graph of each function. Be sure to label at least three points. *29. f 1x2 = 0 x 0

*30. f 1x2 = 1x

In Problems 31–36, graph each function using the techniques of shifting, compressing or stretching, and reflections. Identify any intercepts on the graph. State the domain and, based on the graph, find the range. *31. F 1x2 = 0 x 0 - 4

*32. g1x2 = - 2 0 x 0

*34. f 1x2 = 21 - x

*35. h(x) = (x - 1) + 2

In Problems 37 and 38, (a) Find the domain of each function. (d) Based on the graph, find the range. *37. f1x2 = b

3x x + 1

x *38. f 1x2 = c 1 3x

*33. h(x) = 2x - 1 *36. g(x) = - 2(x + 2)3 - 8

2

if - 2 6 x … 1 if x 7 1 if - 4 … x 6 0 if x = 0 if x 7 0

(b) Locate any intercepts. (e) Is f continuous on its domain?

(c) Graph each function.

39. A function f is defined by f 1x2 =

Ax + 5 6x - 2

If f 112 = 4, find A. A = 11

42. p =

427 B; $1083.92 65,000

Chapter Test

40. Constructing a Closed Box A closed box with a square base is required to have a volume of 10 cubic feet. *(a) Build a model that expresses the amount A of material used to make such a box as a function of the length x of a side of the square base. (b) How much material is required for a base 1 foot by 1 foot? 42 ft 2 (c) How much material is required for a base 2 feet by 2 feet? 28 ft 2 *(d) Graph A = A1x2. For what value of x is A smallest? 41. Area of a Rectangle A rectangle has one vertex in quadrant I on the graph of y = 10 - x2, another at the origin, one on the positive x-axis, and one on the positive y-axis. *(a) Express the area A of the rectangle as a function of x. (b) Find the largest area A that can be enclosed by the rectangle. ≈12.17 sq. units

147

terms. Then find the monthly payment p when the amount borrowed is $165,000 43. Weight of a Body The weight of a body varies inversely with the square of its distance from the center of Earth. Assuming that the radius of Earth is 3960 miles, how much would a man weigh at an altitude of 1 mile above Earth’s surface if he weighs 200 pounds on Earth’s surface? 199.9 lb 44. Heat Loss The amount of heat transferred per hour through a glass window varies jointly with the surface area of the window and the difference in temperature between the areas separated by the glass. A window with a surface area of 7.5 square feet loses 135 BTU per hour when the temperature difference is 40°F. How much heat is lost per hour for a similar window with a surface area of 12 square feet when the temperature difference is 35°F? 189 BTU

42. Mortgage Payments The monthly payment p on a mortgage varies directly with the amount B borrowed. If the monthly payment on a 30-year mortgage is $854.00 when $130,000 is borrowed, find a model that relates the monthly payment p to the amount B borrowed for a mortgage with the same

Chapter Test Prep Videos include step-by-step solutions to all chapter test exercises and can be found on this text’s Channel. (search “SullivanPrecalcUC3e”)

Chapter Test

1. Determine whether each relation represents a function. For each function, state the domain and the range. *(a) 5 12, 52, 14, 62, 16, 72, 18, 82 6 Function (b) 5 11, 32, 14, - 22, 1 - 3, 52, 11, 72 6 Not a function (c) Not a function y 6

In Problems 2– 4, find the domain of each function and evaluate each function at x = - 1. x + 2 *3. g1x2 = *2. f 1x2 = 24 - 5x 0x + 20 x - 4 *4. h1x2 = 2 x + 5x - 36 5. Using the graph of the function f : y

4

4 (1, 3)

2

(0, 2)

x ⫺4

⫺2

2

4

⫺2

(5, 3)

x 4

2

⫺4

(d)

(2, 0)

(2, 0)

4

4

(5, 2) (3, 3)

y

*(a) Find the domain and the range of f. (b) List the intercepts. (0, 2), ( - 2, 0), (2, 0) (c) Find f 112. f 112 = 3 (d) For what value(s) of x does f 1x2 = - 3? - 5 and 3 *(e) Solve f 1x2 6 0.

6 4 2 x 4

2

2

4

2

Function; domain: all real numbers; range: {y 0 y Ú 2}

*6. Use a graphing utility to graph the function f 1x2 = - x4 + 2x3 + 4x2 - 2 on the interval 1 - 5, 52. Approximate any local maximum values and local minimum values rounded to two decimal places. Determine where the function is increasing and where it is decreasing. 2x + 1 7. Consider the function g(x) = b x - 4 *(a) Graph the function. (b) List the intercepts. (0, - 4), (4, 0) (c) Find g1 - 52. g( - 5) = - 9 (d) Find g122. g(2) = - 2

if x 6 - 1 if x Ú - 1

148

CHAPTER 1  Functions and Their Graphs

8. For the function f 1x2 = 3x2 - 2x + , find the average rate of change of f
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Chapter Projects

A 1 2 Monthly fee 3 Allotted number of anytime minutes 4 5 6 7 8 9 10 11 12 13 14 15 16

Number of anytime minutes used Cost per additional minute Monthly cost of text messaging Monthly cost of data plan Allotted data per month (MB) Data used Cost per additional MB of data

B $

C

149

D

39.99 700 700

$ $ $

0.40 20.00 9.99 25 30

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Cost of phone minutes Cost of data

=IF(B42; g1x2 =

3 x + 2 2

2 x + 1

In Problems 21–30, find the domain of the composite function f ∘ g. 21. f 1x2 =

2 3 ; g1x2 = x - 1 x

5 x x ≠ 0, x ≠ 26

22. f 1x2 =

1 2 ; g1x2 = x + 3 x

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312

CHAPTER 4  Exponential and Logarithmic Functions

23. f 1x2 =

x 4 ; g1x2 = x - 1 x

5 x x≠ - 4, x ≠ 06

25. f 1x2 = 1x ; g1x2 = 2x + 3

3 e x x Ú - f 2 27. f 1x2 = x2 + 1; g1x2 = 2x - 1 5 x x Ú 16

29. f 1x2 =

x - 6 ; g1x2 = 2x x - 2

5 x 0 x Ú 0, x ≠ 46

24. f 1x2 =

x 2 ; g1x2 = x + 3 x

2 e x x≠ - , x≠0 f 3

26. f 1x2 = x - 2; g1x2 = 21 - x

28. f 1x2 = x2 + 4; g1x2 = 2x - 2 30. f 1x2 = 2x; g1x2 =

x x - 3

5x x … 16 5x x Ú 26

5x 0 x … 0 or x 7 36

In Problems 31–46, for the given functions f and g, find: B
f ∘ g

C
g ∘ f

D
f ∘ f

E
g ∘ g State the domain of each composite function. *31. f 1x2 = 2x + 3; g1x2 = 3x *33. f 1x2 = 3x + 1; g1x2 = x *35. f 1x2 = x2; 3 *37. f 1x2 = x x *39. f 1x2 = x *41. f 1x2 = 1x;

*32. f 1x2 = - x; g1x2 = 2x - 4

*34. f 1x2 = x + 1; g1x2 = x2 + 4

2

*36. f 1x2 = x2 + 1; g1x2 = 2x2 + 3 1 2 *38. f 1x2 = ; g1x2 = x + 3 x x 2 ; g1x2 = *40. f 1x2 = x + 3 x *42. f 1x2 = 2x - 2 ; g1x2 = 1 - 2x

g1x2 = x2 + 4 2 ; g1x2 = 1 x 1

4 x g1x2 = 2x + 3

; g1x2 = -

*44. f 1x2 = x2 + 4 ; g1x2 = 2x - 2

*43. f1x2 = x2 + 1; g1x2 = 2x - 1 *45. f 1x2 =

x - 5 x + 2 ; g1x2 = x + 1 x - 3

*46. f 1x2 =

In Problems 47–54, show that 1f ∘ g2 1x2 = 1g ∘ f2 1x2 = x.

2x - 1 x + 4 ; g1x2 = x - 2 2x - 5

*47. f 1x2 = 2x; g1x2 =

1 x 2 3 *49. f 1x2 = x3; g1x2 = 1x

*48. f 1x2 = 4x; g1x2 =

1 x 4 *50. f 1x2 = x + 5; g1x2 = x - 5

1 1x + 62 2 1 *53. f 1x2 = ax + b; g1x2 = 1x - b2 a

*52. f 1x2 = 4 - 3x; g1x2 =

*51. f 1x2 = 2x - 6; g1x2 =

a ≠ 0

*54. f 1x2 =

1 4 - x

3

1 1 ; g1x2 = x x

In Problems 55–60, find functions f and g so that f ∘ g = H. For 55–60, other answers are possible. 55. H x = 12x + 32 4 57. H x = 2x2 + 1 59. H x = 0 2x + 1 0

f 1x2 = x4; g1x2 = 2x + 3

f 1x2 = 2x ; g1x2 = x2 + 1

f 1x2 = |x|; g1x2 = 2x + 1

56. H1x2 = 11 + x2 2

58. H x = 21 - x2

60. H x = 0 2x + 3 0 2

3

f 1x2 = x3; g1x2 = 1 + x2

f 1x2 = 2x ; g1x2 = 1 - x2 f 1x2 = |x|; g1x2 = 2x2 + 3

Applications and Extensions 61. If f 1x2 = 2x3 - 3x2 + 4x - 1 and g1x2 = 2, find 1f ∘ g2 1x2 and 1g ∘ f2 1x2. 1f ∘ g2 1x2 = 11; 1g ∘ f2 1x2 = 2 x + 1 62. If f 1x2 = , find 1f ∘ f2 1x2. 1f ∘ f2 1x2 = x, x ≠ 1 x - 1 2 63. If f 1x2 = 2x + 5 and g1x2 = 3x + a, find a so that the graph of f ∘ g crosses the y-axis at 23. - 3, 3

64. If f 1x2 = 3x2 -  and g1x2 = 2x + a, find a so that the graph of f ∘ g crosses the yBYJT
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- 5, 5 In Problems 65 and 66, use the functions f and g to find: (a) f ∘ g (b) g ∘ f (c) the domain of f ∘ g and of g ∘ f (d) the conditions for which f ∘ g = g ∘ f

*65. f 1x2 = ax + b; g1x2 = cx + d ax + b ; g1x2 = mx *66. f 1x2 = cx + d 67. Surface Area of a Balloon The surface area S
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SECTION 4.1   Composite Functions

*71. Production Cost The price p, in dollars, of a certain product and the quantity x sold obey the demand equation p = -

1 x + 100 0 … x … 400 4

Suppose that the cost C, in dollars, of producing x units is C =

1x + 600 25

Assuming that all items produced are sold, find the cost C as a function of the price p. [Hint: Solve for x in the demand equation and then form the composite.] *72. Cost of a Commodity The price p, in dollars, of a certain commodity and the quantity x sold obey the demand equation p = -

1 x + 200 0 … x … 1000 5

Suppose that the cost C, in dollars, of producing x units is C =

1x + 400 10

Assuming that all items produced are sold, find the cost C as a function of the price p. 73. Volume of a Cylinder The volume V of a right circular cylinder of height h and radius r is V = pr 2h. If the height is twice the radius, express the volume V as a function of r. V r = 2pr 3

313

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74. Volume of a Cone The volume V of a right circular cone 1 is V = pr 2h. If the height is twice the radius, express the 79. 3 2 volume V as a function of r. V r = pr 3 3 75. Foreign Exchange Traders often buy foreign currency 80. in the hope of making money when the currency’s value DIBOHFT
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Suppose that f x = x3 + x2 - 16x - 16 and g x = x2 - 4. Find the zeros of f ∘ g x . - 222, - 23, 0, 222, 23 Suppose that f x = 2x3 - 3x2 - x + 12 and g x = x + 5. Find the zeros of f ∘ g x . - , - , - 3 2 If f and g are odd functions, show that the composite function f ∘ g is also odd. If f is an odd function and g is an even function, show that the composite functions f ∘ g and g ∘ f are both even. Let f x = ax + b and g x = bx + a, where a and b are integers. If f 1 =  and f g 20

- g f 20

= - 14, find the product of a and b.* 15

Retain Your Knowledge Problems 84–87 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. *84. Given f x = 3x +  and g x = x - 5, find 1f + g2 x , f 1f - g2 x , 1f # g2 x , and a b x . State the domain of g each. 85. Find the real zeros of f 1x2 = 2x - 52x + 2.

1 ,4 4

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‘Are You Prepared?’ Answers 1. - 21

2. 4 - 1x2

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3. 5x 0 x ≠ - 5, x ≠ 56

314

CHAPTER 4  Exponential and Logarithmic Functions

4.2 One-to-One Functions; Inverse Functions PREPARING FOR THIS SECTION Before getting started, review the following: r
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OBJECTIVES 1 Determine Whether a Function Is One-to-One (p. 314) 2 Determine the Inverse of a Function Defined by a Map or a Set of Ordered Pairs (p. 316) 3 Obtain the Graph of the Inverse Function from the Graph of the Function (p. 318) 4 Find the Inverse of a Function Defined by an Equation (p. 319)

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Figure 7 Population (in millions)

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6.5

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Washington

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South Dakota

0.8

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North Carolina

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Oklahoma

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Life Expectancy (in years) 11

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in the range. Functions such as the one in Figure 6 are given a special name.

DEFINITION

In Words A function is not one-to-one if two different inputs correspond to the same output.

A function is one-to-one if any two different inputs in the domain correspond to two different outputs in the range. That is, if x1 and x2 are any two different inputs of a function f, then f is one-to-one if f1x1 2 ≠ f1x2 2 . 1VU BOPUIFS
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functions, functions that are not one-to-one, and relations that are not functions.

SECTION 4.2   One-to-One Functions; Inverse Functions

315

Figure 8 y1

x1

y2

x2

Domain

EXAM PL E 1

y3

x3

Range

Domain

(a) One-to-one function: Each x in the domain has one and only one image in the range.

x1

y1 y2

x2

y3

x3

y1

x1

x3

Range

Domain

(b) Not a one-to-one function: y1 is the image of both x 1 and x 2.

y3

Range

(c) Not a function: x 1 has two images, y1 and y2.

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C
{ - 2, 6 , - 1, 3 , 0, 2 , 1, 5 , 2,  }

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PROBLEMS

11

AND

15

r

For functions defined by an equation y = f 1x2 and for which the graph of f is known, there is a simple test, called the horizontal-line test, to determine whether f is one-to-one.

THEOREM Figure 9 f 1x1 2 = f 1x2 2 = h and x1 ≠ x2; f is not a one-to-one function.

Horizontal-line Test If every horizontal line intersects the graph of a function f in at most one point, then f is one-to-one.

y y ⫽ f (x) (x 1, h)

(x 2, h) h

x1

x2

y⫽h x

EXAM PL E 2

The reason why this test works can be seen in Figure 9, where the horizontal line y = h intersects the graph at two distinct points, 1x1 , h2 and 1x2 , h2 . Since h is the image of both x1 , and x2 , and x1 ≠ x2 , f is not one-to-one. Based on Figure 9, the horizontal-line test can be stated in another way: If the graph of any horizontal line intersects the graph of a function f at more than one point, then f is not one-to-one.

Using the Horizontal-line Test For each function, use its graph to determine whether the function is one-to-one. B
f1x2 = x2


Solution

C
g1x2 = x3

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316

CHAPTER 4  Exponential and Logarithmic Functions

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y

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y

(1, 1)

23

y51 23

3 x

23

3 x

23

(a) A horizontal line intersects the graph twice; f is not one-to-one.

Now Work

3

3

3 (21, 1)

y5x

PROBLEM

(b) Every horizontal line intersects the graph exactly once; g is one-to-one.

r

19

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THEOREM

A function that is increasing on an interval I is a one-to-one function on I. A function that is decreasing on an interval I is a one-to-one function on I.

2 Determine the Inverse of a Function Defined by a Map or a Set of Ordered Pairs DEFINITION

In Words Suppose that f is a one-to-one function such that the input 5 corresponds to the output 10. In the inverse function f -1, the input 10 will correspond to the output 5.

EX AMPL E 3

Suppose that f is a one-to-one function. Then, corresponding to each x in the domain of f, there is exactly one y
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the domain and the range of the inverse function. State

Population (in millions)

Oklahoma

3.8

Indiana

6.5

Washington

6.9

South Dakota

0.8

North Carolina

9.8

SECTION 4.2   One-to-One Functions; Inverse Functions

Solution

317

The function is one-to-one. To find the inverse function, interchange the elements in the domain with the elements in the range. For example, the function receives as input Indiana and outputs 6.5 million. So the inverse receives as input 6.5 million and outputs Indiana. The inverse function is shown next. Population (in millions)

State

3.8

Oklahoma

6.5

Indiana

6.9

Washington

0.8

South Dakota

9.8

North Carolina

The domain of the inverse function is {3., 6.5, 6.9, 0., 9.}. The range of the inverse function is {Oklahoma, Indiana,
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If the function f is a set of ordered pairs 1x, y2 , then the inverse of f , denoted f -1, is the set of ordered pairs 1y, x2 .

Finding the Inverse of a Function Defined by a Set of Ordered Pairs

EXAM PL E 4

Find the inverse of the following one-to-one function: {1 - 3, - 22, 1 - 2, - 2, 1 - 1, - 12, 10, 02, 11, 12, 12, 2, 13, 22} State the domain and the range of the function and its inverse.

Solution

The inverse of the given function is found by interchanging the entries in each ordered pair and so is given by {1 - 2, - 32, 1 - , - 22, 1 - 1, - 12, 10, 02, 11, 12, 1, 22, 12, 32 } The domain of the function is { - 3, - 2, - 1, 0, 1, 2, 3}. The range of the function is {- 2, - , - 1, 0, 1, , 2}. The domain of the inverse function is {- 2, - , - 1, 0, 1, , 2}. The range of the inverse function is {- 3, - 2, - 1, 0, 1, 2, 3}.

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Now Work

Figure 11 Domain of f

Range of f f

f Range of f ⫺1

PROBLEMS

25

AND

29

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f is a one-to-one function, it has an inverse function, f -1. See Figure 11. The results of Example 4 and Figure 11 suggest two facts about a one-to-one function f and its inverse f -1.

⫺1

Domain of f

⫺1

WARNING Be careful! f -1 is a symbol for the inverse function of f. The - 1 used in f -1 is not an exponent. That is, f -1 does not mean the reciprocal 1 . j of f; f -1 x is not equal to f x

%omain of f = 3ange of f -1 3ange of f = %omain of f -1

Look again at Figure 11 to visualize the relationship. Starting with x, applying f, and then applying f -1 gets x back again. Starting with x, applying f -1, and then applying f, gets x back again. To put it simply, what f does, f -1 undoes, and vice versa. See the illustration that follows. Input x from domain of f Input x from domain of f - 1

Apply f Apply f -1

f1x2 f - 1 1x2

Apply f -1

f - 1 1f 1x2 2 = x

Apply f

f 1 f - 1 1x2 2 = x

318

CHAPTER 4  Exponential and Logarithmic Functions

In other words, f -1 1f1x2 2 = x, where x is in the domain of f

f1f -1 1x2 2 = x, where x is in the domain of f -1

Figure 12 f f −1

x

f(x ) = 2x

f −1(2x) = 1–2 (2x) = x

Consider the function f1x2 = 2x, which multiplies the argument x by 2. The inverse function f -1 undoes whatever f does. So the inverse function of f is 1 f -1 1x2 = x, which divides the argument by 2. For example, f132 = 2132 = 6 2 1 and f - 1 162 = 162 = 3, so f - 1 undoes what f did. This is verified by showing that 2 1 1 1 f -1 1 f1x2 2 = f -1 12x2 = 12x2 = x and f1 f -1 1x2 2 = f a xb = 2a xb = x 2 2 2 See Figure 12.

EX A MPL E 5

Verifying Inverse Functions

3 B
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g1x2 = x3 is g -1 1x2 = 1 x.

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f1x2 = 2x + 3 is f -1 1x2 =

Solution

3 3 B
g -1 1g1x2 2 = g -1 1x3 2 = 2 x = x

1 1x - 32 . 2

for all x in the domain of g

g1g 1x2 2 = g1 2x2 = 1 2x2 = x for all x in the domain of g -1 for all x in the 1 1 C
f -1 1 f1x2 2 = f -1 12x + 32 = 3 12x + 32 - 34 = 12x2 = x domain of f 2 2 1 1 for all x in the f1 f -1 1x2 2 = f a 1x - 32 b = 2c 1x - 32 d + 3 = 1x - 32 + 3 = x domain of f -1 2 2 -1

3

3

3

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EX A MPL E 6

Verifying Inverse Functions

1 1 is f -1 1x2 = + 1. For what values of x is x x - 1 f -1 1 f1x2 2 = x For what values of x is f1 f -1 1x2 2 = x

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f1x2 =

Solution

The domain of f is {x x ≠ 1} and the domain of f -1 is {x x ≠ 0}. Now f -1 1 f1x2 2 = f -1 a

1 b = x - 1

1 + 1 = 1x - 12 + 1 = x provided x ≠ 1 1 x - 1 1 1 1 f1 f -1 1x2 2 = f a + 1b = = = x provided x ≠ 0 x 1 1 a + 1b - 1 x x

Now Work

Figure 13 y

y⫽x

(a, b)

b a

(b, a) a

b

x

PROBLEMS

33

AND

r

37

3 Obtain the Graph of the Inverse Function from the Graph of the Function

Suppose that 1a, b2 is a point on the graph of a one-to-one function f defined by y = f1x2 . Then b = f1a2. This means that a = f -1 1b2, so 1b, a2 is a point on the graph of the inverse function f -1. The relationship between the point 1a, b2 on f and the point 1b, a2 on f -1 is shown in Figure 13. The line segment with endpoints 1a, b2 and 1b, a2 is perpendicular to the line y = x and is bisected by the line y = x.
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SECTION 4.2   One-to-One Functions; Inverse Functions

THEOREM

319

The graph of a one-to-one function f and the graph of its inverse f -1 are symmetric with respect to the line y = x.

Figure 14 illustrates this result. Once the graph of f is known, the graph of f -1 may be obtained by reflecting the graph of f about the line y = x. Figure 14

y

y ⫽ f (x)

y⫽x

(a 3, b 3) y ⫽ f ⫺1(x)

(a 2, b 2)

(b 3, a 3) x

(a 1, b 1) (b 2, a 2) (b 1, a 1)

EXAMPL E 7

Graphing the Inverse Function 5IF
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of its inverse.

Solution

Begin by adding the graph of y = x
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1 - 2, - 12, 1 - 1, 02, and 12, 12 are on the graph of f, the points 1 - 1, - 22, 10, - 12, and 11, 22 must be on the graph of f -1.
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f -1 is the reflection about the line y = x of the graph of f, draw f -1.
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Figure 15

y 3

y 3

y⫽x (1, 2)

Exploration Simultaneously graph Y1 = x, Y2 = x3, 3 and Y3 = 2x on a square screen with - 3 … x … 3. What do you observe about the graphs of Y2 = x3, its inverse 3 Y3 = 2x, and the line Y1 = x? Repeat this experiment by simultaneously graphing Y1 = x, 1 Y2 = 2x + 3, and Y3 = (x - 3) on a 2 square screen with - 6 … x … 3. Do you see the symmetry of the graph of Y2 and its inverse Y3 with respect to the line Y1 = x?

y ⫽ f (x) (⫺1, 0) ⫺3

y ⫽ f(x) (2, 1) 3 x

(⫺2, ⫺1)

⫺3 (⫺2, ⫺1)

3 x (0, ⫺1) (⫺1, ⫺2) ⫺3

⫺3

y ⫽ f ⫺1(x)

(b)

(a)

Now Work

(2, 1)

(⫺1, 0)

PROBLEM

r

43

4 Find the Inverse of a Function Defined by an Equation The fact that the graphs of a one-to-one function f and its inverse function f -1 are symmetric with respect to the line y = x tells us more. It says that we can obtain f -1 by interchanging the roles of x and y in f. Look again at Figure 14. If f is defined by the equation y = f1x2 then f -1 is defined by the equation x = f1y2

320

CHAPTER 4  Exponential and Logarithmic Functions

The equation x = f 1y2 defines f -1 implicitly. Solving this equation for y yields the explicit form of f -1, that is, y = f -1 x

Let’s use this procedure to find the inverse of f1x2 = 2x + 3.
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How to Find the Inverse Function

EX A MPL E 8

Find the inverse of f1x2 = 2x + 3. Graph f and f -1 on the same coordinate axes.

Step-by-Step Solution

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f1x2 with y in f 1x2 = 2x + 3 and obtain y = 2x + 3. Now interchange the variables x and y to obtain

Step 1 Replace f(x) with y. In y = f(x), interchange the variables x and y to obtain x = f(y). This equation defines the inverse function f -1 implicitly.

x = 2y + 3 This equation defines the inverse f -1 implicitly. To find the explicit form of the inverse, solve x = 2y + 3 for y.

Step 2 If possible, solve the implicit equation for y in terms of x to obtain the explicit form of f -1, y = f -1(x).

x = 2y + 3 2y + 3 = x 2y = x - 3 1 y = 1x - 32 2

Reflexive Property; If a = b, then b = a. Subtract 3 from both sides. 1 Multiply both sides by . 2

The explicit form of the inverse f - 1 is f -1 1x2 =

1 1x - 32 2

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f and f - 1 are inverses. 1 The graphs of f1x2 = 2x + 3 and its inverse f -1 1x2 = 1x - 32 are shown in 2 Figure 16. Note the symmetry of the graphs with respect to the line y = x.

Step 3 Check the result by showing that f -1(f(x)) = x and f(f -1(x)) = x.

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Figure 16 y

Procedure for Finding the Inverse of a One-to-One Function STEP 1: In y = f 1x2, interchange the variables x and y to obtain

f(x)  2x  3 5

yx f

5

x = f1y2

1

(x)  1–2(x  3) 5

x

This equation defines the inverse function f -1 implicitly. STEP 2: If possible, solve the implicit equation for y in terms of x to obtain the explicit form of f -1: y = f -1 1x2

5

EX A MPL E 9

STEP 3: Check the result by showing that

f -1 1 f x2 2 = x and f1 f -1 1x2 2 = x

Finding the Inverse Function The function f 1x2 =

2x + 1 x - 1

x ≠ 1

is one-to-one. Find its inverse and check the result.

SECTION 4.2   One-to-One Functions; Inverse Functions

Solution

321

STEP 1:
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f1x2 with y and interchange the variables x and y in y =

2x + 1 x - 1

x =

2y + 1 y - 1

to obtain

STEP 2: Solve for y. x = x1y - 12 xy - x xy - 2y 1x - 22y

= = = =

y =

2y + 1 y - 1 2y + 1 2y + 1 x + 1 x + 1

Multiply both sides by y - 1. Apply the Distributive Property. Subtract 2y from both sides; add x to both sides. Factor.

x + 1 x - 2

Divide by x - 2.

The inverse is f -1 1x2 = STEP 3:

x + 1 , x ≠ 2 x - 2

Replace y by f -1(x).

Check:

2x + 1 + 1 2x + 1 x - 1 2x + 1 + x - 1 3x f -1 1 f1x2 2 = f -1 a b = = = = x, x ≠ 1 x - 1 2x + 1 2x + 1 - 21x - 12 3 - 2 x - 1 f1 f -1 1x2 2 = f a

x + 1 b = x - 2

x + 1 b + 1 21x + 12 + x - 2 x - 2 3x = = = x, x ≠ 2 x + 1 x + 1 - 1x - 22 3 - 1 x - 2

2a

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Exploration

2x + 1 x + 1 In Example 9, it was found that, if f(x) = , then f-1(x) = . Compare the vertical and x - 1 x - 2 -1 horizontal asymptotes of f and f . Result The vertical asymptote of f is x = 1, and the horizontal asymptote is y = 2. The vertical asymptote of f-1 is x = 2, and the horizontal asymptote is y = 1.

Now Work

PROBLEMS

51

AND

65

If a function is not one-to-one, it has no inverse function. Sometimes, though, an appropriate restriction on the domain of such a function will yield a new function that is one-to-one. Then the function defined on the restricted domain has an inverse function. Let’s look at an example of this common practice.

EX AM PL E 10

Solution

Finding the Inverse of a Domain-restricted Function Find the inverse of y = f 1x2 = x2 if x Ú 0. Graph f and f - 1.

The function y = x2
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restricting the domain of this function to x Ú 0, as indicated, yields a new function that is increasing and, therefore, is one-to-one. As a result, the function defined by y = f1x2 = x2, x Ú 0, has an inverse function, f -1. Follow the steps given previously to find f -1.

322

CHAPTER 4  Exponential and Logarithmic Functions

STEP 1: In the equation y = x2, x Ú 0, interchange the variables x and y. The result is

Figure 17 y 2

f (x )  x 2, x  0

x = y2

yx f 1(x )  x

2

x




y Ú 0

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STEP 2: Solve for y to get the explicit form of the inverse. Since y Ú 0, only one solution for y is obtained: y = 1x. So f -1 1x2 = 1x. STEP 3:

Check:

f -1 1 f1x2 2 = f -1 1x2 2 = 2x2 = 0 x 0 = x since x Ú 0

f 1 f -1 1x2 2 = f 1 1x2 = 1 1x2 2 = x

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f1x2 = x2, x Ú 0, and f -1 1x2 = 1x. Note that the domain of f = range of f - 1 = [0, q and the domain of f - 1 = range of f = [0, q .

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SUMMARY 1. If a function f is one-to-one, then it has an inverse function f -1. 2. %PNBJO
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f = 3ange of f -1;
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f = %omain of f -1. 3. To verify that f -1 is the inverse of f, show that f -1 1 f1x2 2 = x for every x in the domain of f and that f1 f -1 1x2 2 = x for every x in the domain of f -1. 4. The graphs of f and f -1 are symmetric with respect to the line y = x.

4.2 Assess Your Understanding ‘Are You Prepared?’

Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.

1. Is the set of ordered pairs { 11, 32, 12, 32, 1 - 1, 22} a GVODUJPO
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f 1x2 = 2 x + 3x - 1 {x 0 x ≠ - 6, x ≠ 3}

1 + 1 x 4. Simplify: QQ
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1 1 x2 x , x ≠ 0, x ≠ - 1, x ≠ 1 1-x

Concepts and Vocabulary 5. If x1 and x2 are any two different inputs of a function f, then f is one-to-one if f 1x1 2 ≠ f 1x2 2 . 6. If every horizontal line intersects the graph of a function f at one@to@one function. no more than one point, f
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7. If f is a one-to-one function and f 132 = , then f -1 12 = 3 .

8. If f -1 denotes the inverse of a function f, then the graphs of f and f -1 are symmetric with respect to the line y = x . 9. If the domain of a one-to-one function f is 34, q 2, then the 34, q 2 . range of its inverse, f -1, is 10. True or False If f and g are inverse functions, the domain of f is the same as the range of g. True

Skill Building In Problems 11–18, determine whether the function is one-to-one. 11.

Domain

Range

20 Hours 25 Hours

one-to-one

12.

Domain

Range

$200

Bob

Karla

$300

Dave

Debra

30 Hours

$350

John

Dawn

40 Hours

$425

Chuck

Phoebe

one-to-one

SECTION 4.2   One-to-One Functions; Inverse Functions

13.

Domain

Range

20 Hours

$200

not one-to-one

14.

Domain

25 Hours

Karla

Dave

Debra Phoebe

$350

John

40 Hours

$425

Chuck

15. { 12, 62, 1 - 3, 62, 14, 92, 11, 102} 17. { 10, 02, 11, 12, 12, 162, 13, 12}

16. { 1 - 2, 52, 1 - 1, 32, 13, 2, 14, 122}

not one-to-one

18. { 11, 22, 12, 2, 13, 12, 14, 322}

one-to-one

not one-to-one

Range

Bob

30 Hours

323

one-to-one

one-to-one

In Problems 19–24, the graph of a function f is given. Use the horizontal-line test to determine whether f is one-to-one. 19.

one-to-one

y 3

⫺3

20.

⫺3

3 x

⫺3

22.

one-to-one

y 3

21.

y 3

⫺3

3 x

3 x

⫺3

⫺3

not one-to-one 23.

y

not one-to-one

one-to-one

y 3

24.

y 3

not one-to-one

2 ⫺3

3 ⫺3

x

⫺3

3 x

3 x

In Problems 25–32, find the inverse of each one-to-one function. State the domain and the range of each inverse function. *25. Location

Annual Precipitation (inches)

*26.

Domestic Gross (in millions)

Title

Atlanta, GA

49.7

Avatar

$761

Boston, MA

43.8

Titanic

$659

Las Vegas, NV

4.2

Marvel’s The Avengers

$623

Miami, FL

61.9

The Dark Knight

$535

Los Angeles, CA

12.8

Star Wars: Episode One – The Phantom Menace

$475

Source: currentresults.com

*27. Age 30 40

Source: boxofficemojo.com Monthly Cost of Life Insurance $10.55 $12.89 $19.29

45 Source: acequotes.com

*29. { 1 - 3, 52, 1 - 2, 92, 1 - 1, 22, 10, 112, 11, - 52}

*31. { 1 - 2, 12, 1 - 3, 22, 1 - 10, 02, 11, 92, 12, 42}

*28.

State

Unemployment Rate

Virginia Nevada Tennessee Texas

5.6% 9.7% 7.7% 6.3%

Source: United States Bureau of Labor Statistics, Jan. 2013

*30. { 1 - 2, 22, 1 - 1, 62, 10, 2, 11, - 32, 12, 92}

*32. { 1 - 2, - 2, 1 - 1, - 12, 10, 02, 11, 12, 12, 2}

In Problems 33–42, verify that the functions f and g are inverses of each other by showing that f 1 g1x2 2 = x and g 1 f 1x2 2 = x. Give any values of x that need to be excluded from the domain of f and the domain of g. 1 1x - 42 3 x *35. f 1x2 = 4x - ; g1x2 = + 2 4

*33. f 1x2 = 3x + 4; g1x2 =

*34. f 1x2 = 3 - 2x; g1x2 = *36. f 1x2 = 2x + 6; g1x2 =

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1 1x - 32 2

1 x - 3 2

324

CHAPTER 4  Exponential and Logarithmic Functions

3 *37. f 1x2 = x3 - ; g1x2 = 2 x + 

*38. f 1x2 = 1x - 22 2, x Ú 2; g1x2 = 1x + 2

*39. f 1x2 =

*40. f 1x2 = x; g1x2 = x

1 1 ; g1x2 = x x 4x - 3 2x + 3 ; g1x2 = *41. f 1x2 = x + 4 2 - x

*42. f 1x2 =

x - 5 3x + 5 ; g1x2 = 2x + 3 1 - 2x

In Problems 43–48, the graph of a one-to-one function f is given. Draw the graph of the inverse function f -1. For convenience (and as a hint), the graph of y = x is also given. *43.

y 3

*44.

y⫽x

y⫽x

y 3

y 3

y⫽x

(1, 2)

(0, 1)

(2, 1–2)

(⫺1, 0) ⫺3

⫺3

3 x

(1, 0)

(2, 1) ⫺3 (⫺1, ⫺1)

3 x

(0, ⫺1)

(⫺2, ⫺2)

(⫺2, ⫺2)

⫺3

⫺3

*46.

*45.

y⫽x

y 3

*47.

y 3

3 x

⫺3 y⫽x

*48.

y 2

y⫽x

(⫺2, 1) ⫺3

⫺3

3 x (1, ⫺1) ⫺3

⫺2

3 x

2 x

⫺2

⫺3

In Problems 49–60, the function f is one-to-one. Find its inverse and check your answer. Graph f, f -1, and y = x on the same coordinate axes. *49. f 1x2 = 3x

f -1 1x2 =

*51. f 1x2 = 4x + 2 *53. f 1x2 = x3 - 1

*50. f 1x2 = - 4x

1 x 3

f -1 1x2 =

x 1 4 2 3 -1 f 1x2 = 1x + 1

*55. f 1x2 = x2 + 4, x Ú 0 f -1 1x2 = 2x - 4, x Ú 4 4 4 *57. f 1x2 = f -1 1x2 = x x 1 2x + 1 f -1 1x2 = *59. f 1x2 = x - 2 x

1 f -1 1x2 = - x 4 1 - x -1 *52. f 1x2 = 1 - 3x f 1x2 = 3 3 *54. f 1x2 = x3 + 1 f -1 1x2 = 1 x - 1

*56. f 1x2 = x2 + 9, x Ú 0 f -1 1x2 = 2x - 9, x Ú 9 3 3 *58. f 1x2 = f -1 1x2 = x x 4 4 *60. f 1x2 = f -1 1x2 = - 2 x + 2 x

In Problems 61–72, the function f is one-to-one. Find its inverse and check your answer. *61. f 1x2 = *63. *65. *67. *69. *71.

2 3 + x 3x f 1x2 = x + 2 2x f 1x2 = 3x - 1 3x + 4 f 1x2 = 2x - 3 2x + 3 f 1x2 = x + 2 x2 - 4 f 1x2 = , 2x2

2 - 3x x 2x f -1 1x2 = x - 3 x f -1 1x2 = 3x - 2 3x + 4 f -1 1x2 = 2x - 3 - 2x + 3 f -1 1x2 = x - 2 f -1 1x2 =

x 7 0

f -1 1x2 =

2 11 - 2x

*62. f 1x2 = *64. f 1x2 = *66. f 1x2 = *68. f 1x2 = *70. f 1x2 = *72. f 1x2 =

4 4 f -1 1x2 = 2 2 - x x 2x x f -1 1x2 = x - 1 x + 2 3x + 1 -1 f -1 1x2 = x x + 3 2x - 3 3 + 4x f -1 1x2 = x + 4 2 - x - 3x - 4 2x - 4 f -1 1x2 = x - 2 x + 3 x2 + 3 3 , x 7 0 f -1 1x2 = A 3x - 1 3x2

Applications and Extensions 73. 6TF
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y = f 1x2
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the following: B
f 1 - 12 0 C
f 112 2 D
f -1 112 0 E
f -1 122 1

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f 122 C
f 112 0 D
f -1 102 1 E
f -1 1 - 12 0 2

SECTION 4.2   One-to-One Functions; Inverse Functions

75. If f 12 = 13 and f is one-to-one, what is f -1 1132



-5 76. If g1 - 52 = 3 and g is one-to-one, what is g -1 132 *77. The domain of a one-to-one function f is [5, q 2, and its range is [ - 2, q 2. State the domain and the range of f -1. *78. The domain of a one-to-one function f is [0, q 2, and its range is [5, q 2. State the domain and the range of f -1. *79. The domain of a one-to-one function g is 1 - q , 04 , and its range is [0, q 2. State the domain and the range of g -1. *80. The domain of a one-to-one function g is [0, 15], and its SBOHF
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*83. Find the inverse of the linear function f 1x2 = mx + b, m ≠ 0 *84. Find the inverse of the function f 1x2 = 2r 2 - x2, 0 … x … r 85. A function f has an inverse function. If the graph of f lies in quadrant I, in which quadrant does the graph of f -1
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*87. The function f 1x2 = 0 x 0 is not one-to-one. Find a suitable restriction on the domain of f so that the new function that results is one-to-one. Then find the inverse of f.

*88. The function f 1x2 = x4 is not one-to-one. Find a suitable restriction on the domain of f so that the new function that results is one-to-one. Then find the inverse of f.

In applications, the symbols used for the independent and dependent variables are often based on common usage. So, rather than using y = f 1x2 to represent a function, an applied problem might use C = C 1q2 to represent the cost C of manufacturing q units of a good since, in economics, q is used for output. Because of this, the inverse notation f -1 used in a pure mathematics problem is not used when finding inverses of applied problems. Rather, the inverse of a function such as C = C 1q2 will be q = q1C2. So C = C 1q2 is a function that represents the cost C as a function of the output q, and q = q1C2 is a function that represents the output q as a function of the cost C. Problems 89–92 illustrate this idea. * B
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W h = 45.5 + 2.3 h - 60 .] 9 92. Temperature Conversion The function F 1C2 = C + 32 5 converts a temperature from C degrees Celsius to F degrees Fahrenheit.

*93. Income Taxes The function

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326

CHAPTER 4  Exponential and Logarithmic Functions

96. Period of a Pendulum The period T (in seconds) of a simple pendulum as a function of its length l (in feet) is given by

97. Given f 1x2 =

l T 1l2 = 2p A 32.2

ax + b cx + d

find f -1 1x2. If c ≠ 0, under what conditions on a, b, c, and d - dx + b ; f = f -1if a = - d is f = f -1? f - 1 1x2 = cx - a

*(a) Express the length l as a function of the period T. (b) How long is a pendulum whose period is 3 seconds? 7.34 ft

Discussion and Writing *98. Can a one-to-one function and its inverse be equal? What must be true about the graph of f for this to happen? Give some examples to support your conclusion. 99. Draw the graph of a one-to-one function that contains the points ( - 2, - 32, 10, 02 , and 11, 52 . Now draw the graph of its inverse. Compare your graph to those of other students. Discuss any similarities. What differences do you see? *100. Give an example of a function whose domain is the set of real numbers and that is neither increasing nor decreasing on its domain, but is one-to-one.

101. Is every odd function one-to-one? Explain. No

*102. Suppose that C 1g2 represents the cost C, in dollars, of manufacturing g cars. Explain what C -1 1800,0002 represents. 103. Explain why the horizontal-line test can be used to identify one-to-one functions from a graph. 104. Explain why a function must be one-to-one in order to have an inverse that is a function. Use the function y = x2 to support your explanation.

[Hint: Use a piecewise-defined function.]

Retain Your Knowledge Problems 105–108 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 6x2 - 11x - 2 105. If f 1x2 = 3x2 - 7x, find f 1x + h2 - f 1x2 6xh + 3h2 - 7h *108. Find the domain of R(x) = . Find any 2x2 - x - 6 *106. Use the techniques of shifting, compressing or stretching, horizontal, vertical, or oblique asymptotes. and reflections to graph f(x) = -  x + 2 0 + 3. 3 Domain: {x 0 x ≠ - , x ≠ 2}; *107. Find the zeros of the quadratic function f(x) = 3x2 + 5x + 1. 2 3 What are the x-intercepts, if any, of the graph of the function? Vertical asymptote: x = - ; Horizontal asymptote: y = 3 2

‘Are You Prepared?’ Answers 1. Yes; for each input x there is one output y. 2. Increasing on 10, q 2; decreasing on 1 - q , 02

3. {x x ≠ - 6, x ≠ 3} x , x ≠ 0, x ≠ - 1, x ≠ 1 4. 1 - x

4.3 Exponential Functions PREPARING FOR THIS SECTION Before getting started, review the following: r
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OBJECTIVES 1 2 3 4

Evaluate Exponential Functions (p. 326) Graph Exponential Functions (p. 330) Define the Number e (p. 333) Solve Exponential Equations (p. 335)

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SECTION 4.3   Exponential Functions

327

ar where the base a is a positive real number and the exponent r is a rational number. But what is the meaning of ax, where the base a is a positive real number and the exponent x
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C
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15

It can be shown that the familiar laws for rational exponents hold for real exponents.

THEOREM

Laws of Exponents If s, t, a, and b are real numbers with a 7 0 and b 7 0, then as # at = as + t 1s = 1

1as 2 = ast 1ab2 s = as # bs s 1 1 a-s = s = a b a0 = 1 a a t

(1)

Introduction to Exponential Growth Suppose a function f has the following two properties:

Table 1 x

f(x)

0

5

1

10

2

20

3

40

4

80

1. The value of f doubles with every 1-unit increase in the independent variable x. 2. The value of f at x = 0 is 5, so f 102 = 5. Table 1 shows values of the function f for x = 0, 1, 2, 3, and 4. Let’s find an equation y = f1x2 that describes this function f. The key fact is that the value of f doubles for every 1-unit increase in x. f102 = 5 f112 = 2f 102 = 2 # 5 = 5 # 21 f122 = 2f 112 = 215 # 22 = 5 # 22

Double the value of f at 0 to get the value at 1. Double the value of f at 1 to get the value at 2.

328

CHAPTER 4  Exponential and Logarithmic Functions

f 132 = 2f 122 = 215 # 22 2 = 5 # 23 f142 = 2f 132 = 215 # 23 2 = 5 # 24 The pattern leads to f1x2 = 2f1x - 12 = 215 # 2x-1 2 = 5 # 2x

DEFINITION

An exponential function is a function of the form f1x2 = Cax

WARNING It is important to distinguish a power function, g x = axn, n Ú 2, an integer, from an exponential function, f x = C # ax, a ≠ 1, a 7 0. In a power function, the base is a variable and the exponent is a constant. In an exponential function, the base is a constant and the exponent is a variable. j

where a is a positive real number a 7 0 , a ≠ 1, and C ≠ 0 is a real number. The domain of f is the set of all real numbers. The base a is the growth factor, and because f 0 = Ca0 = C, C is called the initial value. In the definition of an exponential function, the base a = 1 is excluded because this function is simply the constant function f1x2 = C # 1x = C. Bases that are negative also are excluded; otherwise, many values of x would have to be excluded 3 1 and x = .
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1 x F1x2 = a b + 5 3

G1x2 = 2 # 3x - 3

For each function, note that the base of the exponential expression is a constant and the exponent contains a variable. In the function, f1x2 = 5 # 2x, notice that the ratio of consecutive outputs is constant for 1-unit increases in the input. This ratio equals the constant 2, the base of the exponential function. In other words, f112 5 # 21 = = 2 f102 5

f122 5 # 22 = = 2 f112 5 # 21

f132 5 # 23 = = 2 and so on f122 5 # 22

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THEOREM

For an exponential function f1x2 = Cax, where a 7 0 and a ≠ 1, if x is any real number, then f1x + 12 = a f1x2

In Words For 1-unit changes in the input x of an exponential function f(x) = C # a x , the ratio of consecutive outputs is the constant a.

or f1x + 12 = af1x2

Proof f1x + 12 Cax + 1 = = ax + 1 - x = a1 = a f1x2 Cax

EX AMPL E 2

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SECTION 4.3   Exponential Functions

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y

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0

2

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329

For each function, compute the average rate of change of y with respect to x and the ratio of consecutive outputs. If the average rate of change is constant, then the function is linear. If the ratio of consecutive outputs is constant, then the function is exponential.

Table 2 (a) x

y

-1

5

Average Rate of Change

Ratio of Consecutive Outputs

∆y

2 5

∆x

(b)

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1 -1 = 2 2

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- 4 - ( - 1) = -3 2 - 1

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2

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3

-7

Average Rate of Change

Ratio of Consecutive Outputs

0

2

1

x

y

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32

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16

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2

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8 1 = 16 2

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2

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4

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7

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Ratio of Consecutive Outputs

330

CHAPTER 4  Exponential and Logarithmic Functions

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2 Graph Exponential Functions If we know how to graph an exponential function of the form f x = ax, then we DBO
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EX A MPL E 3

Solution Table 3 x

f(x) = 2x

−10

2-10 ≈ 0.00098 1 2-3 = 8 1 2-2 = 4 1 2-1 = 2

−3 −2 −1 0

20 = 1

1

21 = 2

2

22 = 4

3

23 = 8

10

Graphing an Exponential Function

Graph the exponential function: f 1x2 = 2x The domain of f1x2 = 2x is the set of all real numbers. Begin by locating some points on the graph of f1x2 = 2x, as listed in Table 3. Because 2x 7 0 for all x, the range of f is 0, q . Therefore, the graph has no x-intercepts, and in fact, the graph will lie above the x-axis for all x. As Table 3 indicates, the y-intercept is 1. Table 3 also indicates that as x S - q , the value of f1x2 = 2x gets closer and closer to 0. Thus, the x-axis 1y = 02 is a horizontal asymptote to the graph as x S - q . This provides the end behavior for x large and negative. To determine the end behavior for x large and positive, look again at Table 3. As x S q , f1x2 = 2x grows very quickly, causing the graph of f 1x2 = 2x to rise very rapidly. It is apparent that f is an increasing function and, hence, is one-to-one. 6TJOH
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(1, 2) (0, 1) 3

x

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SECTION 4.3   Exponential Functions

331

Graphs that look like the one in Figure 19 occur very frequently in a variety of situations. For example, the graph in Figure 20 illustrates the number of Facebook subscribers by year from 2004 to 2012. One might conclude from this graph that the number of Facebook subscribers is growing exponentially.

Subscribers (in millions)

Figure 20

1100 1000 900 800 700 600 500 400 300 200 100 0 2004 2005 2006 2007 2008 2009 2010 2011 2012 Year

Later in this chapter, more will be said about situations that lead to exponential growth. For now, let’s continue to explore properties of exponential functions. The graph of f1x2 = 2x in Figure 19 is typical of all exponential functions of the form f1x2 = ax with a 7 1. Such functions are increasing functions and hence are one-to-one. Their graphs lie above the x-axis, pass through the point 10, 12 , and thereafter rise rapidly as x S q . As x S - q , the x-axis 1y = 02 is a horizontal asymptote. There are no vertical asymptotes. Finally, the graphs are smooth and continuous with no corners or gaps. Figure 21 illustrates the graphs of two more exponential functions whose bases are larger than 1. Notice that the larger the base, the steeper the graph is when x 7 0, and when x 6 0, the larger the base, the closer the graph of the equation is to the x-axis.

Figure 21 y

y=6

x

(1, 6)

6

y = 3x 3 1– 3

(–1, ) y⫽0

–3

(1, 3)

Seeing the Concept Graph Y1 = 2x and compare what you see to Figure 19. Clear the screen, graph Y1 = 3x and Y2 = 6x , and compare what you see to Figure 21. Clear the screen and graph Y1 = 10x and Y2 = 100x.

(0, 1)

(–1, 1–6 )

3

x

The following display summarizes information about f1x2 = ax, a 7 1.

Properties of the Exponential Function f(x) = ax, a + 1

Figure 22 y

(1, a)

(−1, a1 ) y⫽0

(0, 1) x

1. The domain is the set of all real numbers, or - q , q using interval notation; the range is the set of positive real numbers, or 0, q using interval notation. 2. There are no x-intercepts; the y-intercept is 1. 3. The x-axis 1y = 02 is a horizontal asymptote as x S - q 3 lim ax = 0 4 . xS - q 4. f1x2 = ax, a 7 1, is an increasing function and is one-to-one. 1 5. The graph of f contains the points 10, 12, 11, a2, and a - 1, b . a 6. The graph of f is smooth and continuous, with no corners or gaps. See Figure 22.

Now consider f1x2 = ax when 0 6 a 6 1.

332

CHAPTER 4  Exponential and Logarithmic Functions

EX A MPL E 4

Graphing an Exponential Function 1 x Graph the exponential function: f 1x2 = a b 2

Solution Table 4 x

1 x f(x) = a b 2

−10

1 - 10 = 1024 a b 2

−3

1 -3 = 8 a b 2

−2

1 -2 = 4 a b 2

−1

1 -1 a b = 2 2

0

1 0 a b = 1 2

1 x The domain of f1x2 = a b consists of all real numbers. As before, locate some 2 1 x points on the graph as shown in Table 4. Because, a b 7 0 for all x, the range of f is 2 the interval 10, q 2. The graph lies above the x-axis and has no x-intercepts. The 1 x y-intercept is 1. As x S - q , f1x2 = a b grows very quickly. As x S q , the 2 values of f1x2 approach 0. The x-axis 1y = 02 is a horizontal asymptote as x S q . It is apparent that f is a decreasing function and hence is one-to-one. Figure 23 illustrates the graph.

Figure 23

y 6

(–2, 4)

1

1

1 1 a b = 2 2

2

1 1 2 a b = 2 4

3

1 1 3 a b = 2 8

10

1 10 a b ≈ 0.00098 2

3 (–1, 2) (0, 1)

(1, 1–2)

(2, 1–4 ) (3, 1– ) 8

–3

3

x y=0

r

1 x The graph of y = a b also can be obtained from the graph of y = 2x 2 1 x using transformations. The graph of y = a b = 2-x is a reflection about the y-axis 2 of the graph of y = 2x
SFQMBDF
x by - x 
4FF
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 B
BOE
C 

y

Figure 24

y y = 2x

6

6 1 x

y = ( –2) (2, 4)

(–2, 4)

3

(–2, 1–4 ) (–1, 1–2 ) (–3, 1–8 )

3 (1, 2)

Using a graphing utility, simultaneously graph: 1 x (a) Y1 = 3x, Y2 = a b 3 1 x (b) Y1 = 6x, Y2 = a b 6

1 x Conclude that the graph of Y2 = a b , a for a 7 0, is the reflection about the y-axis of the graph of Y1 = ax.

(0, 1)

(0, 1)

y=0

Seeing the Concept

(–1, 2)

3 (a) y = 2x

x

(1, 1–2)

(2, 1–4 ) (3, 1– ) 8

–3

Replace x by −x ; Reflect about the y-axis.

3

x y=0

1 x

(b) y = 2−x = ( –2)

1 x The graph of f 1x2 = a b in Figure 23 is typical of all exponential functions of 2 the form f x = ax with 0 6 a 6 1. Such functions are decreasing and one-to-one. Their graphs lie above the x-axis and pass through the point 10, 12. The graphs rise rapidly as x S - q . As x S q , the x-axis 1y = 02 is a horizontal asymptote. There are no vertical asymptotes. Finally, the graphs are smooth and continuous, with no corners or gaps.

SECTION 4.3   Exponential Functions

Figure 25 illustrates the graphs of two more exponential functions whose bases are between 0 and 1. Notice that the smaller base results in a graph that is steeper when x 6 0.
8IFO
x 7 0, the graph of the equation with the smaller base is closer to the x-axis. The following display summarizes information about the function f1x2 = ax, 0 6 a 6 1.

Figure 25 y=(

1– x 6 y

)

6

(–1, 6)

333

y = ( 1– ) x 3

3

(–1, 3)

(0, 1)

(1, 1–6 )

–3

Properties of the Exponential Function f(x) = ax, 0 * a * 1

(1, 1–3) 3

x y⫽0

Figure 26 y

(–1, 1–a) (0, 1)

(1, a) x y⫽0

EXAM PL E 5

1. The domain is the set of all real numbers, or 1 - q , q 2 using interval notation; the range is the set of positive real numbers, or 10, q 2 using interval notation. 2. There are no x-intercepts; the y-intercept is 1. 3. The x-axis 1y = 02 is a horizontal asymptote as x S q 3 lim ax = 0 4 . x Sq 4. f1x2 = ax, 0 6 a 6 1, is a decreasing function and is one-to-one. 1 5. The graph of f contains the points a - 1, b , 10, 12, and 11, a2 . a 6. The graph of f is smooth and continuous, with no corners or gaps. See Figure 26.

Graphing Exponential Functions Using Transformations Graph f1x2 = 2-x - 3 and determine the domain, range, and horizontal asymptote of f.

Solution

Begin with the graph of y = 2x.
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Figure 27

y

y

y

10

10

10

(3, 8)

(23, 8)

(23, 5) (2, 4)

(21, 1–2)

(22, 4) (21, 2) (0, 1)

(1, 2) (0, 1)

y50

3

x

23

(1, 1–2 ) 1

(22, 1) x y50

(21, 21) (0, 22) 24

(a) y 5 2 x

Replace x by 2 x ; reflect about y-axis.

(b) y 5 22x

Subtract 3; shift down 3 units.

(

x 2 5– 1, 2 ) 2 y 5 23

(c) y 5 22x 2 3

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f1x2 = 2-x - 3 is the interval 1 - q , q 2 and the range is the interval 1 - 3, q 2. The horizontal asymptote of f is the line y = - 3.

r

Now Work

PROBLEM

41

3 Define the Number e .BOZ QSPCMFNT
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334

CHAPTER 4  Exponential and Logarithmic Functions

One way of arriving at this important number e is given next.

DEFINITION

The number e is defined as the number that the expression a1 +

1 n b n

(2)

approaches as n S q . In calculus, this is expressed, using limit notation, as e = lim a1 + n Sq

1 n b n

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BT
n takes on increasingly large values. The last number in the right column in the table is correct to nine decimal places. Therefore, e = 2.1212c

3FNFNCFS
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indicate that the decimal places continue. Because these decimal places continue but do not repeat, e is an irrational number. The number e is often expressed as a decimal rounded to a specific number of places. For example, e ≈ 2.12 is rounded to five decimal places. The exponential function f1x2 = e x, whose base is the number e, occurs with such frequency in applications that it is usually referred to as the exponential function. Indeed, most calculators have the key e x or exp 1x2 , which may be used to evaluate the exponential function for a given value of x.*

Table 6

Table 5 1 n

n

y

1 n

a1 +

1 b n

x

ex

-2

e - 2 ≈ 0.14

-1

e - 1 ≈ 0.37

1

1

2

2

2

0.5

1.5

2.25

0

e0 ≈ 1

5

0.2

1.2

2.48832

1

e1 ≈ 2.72

10

0.1

1.1

2.59374246

2

100

0.01

1.01

2.704813829

e2 ≈ 7.39

1,000

0.001

1.001

2.716923932

10,000

0.0001

1.0001

2.718145927

100,000

0.00001

1.00001

2.718268237

0.000001

1.000001

2.718280469

-9

2.718281827

1,000,000

Figure 28

1 +

n

1,000,000,000

10

-9

1 + 10

(2, e 2)

Now use your calculator to approximate e x for x = - 2, x = - 1, x = 0, x = 1, and x = 2. See Table 6. The graph of the exponential function f1x2 = e x is given in 'JHVSF

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2 6 e 6 3, the graph of y = e x lies between the graphs of y = 2x and y = 3x.
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6

3

1– e

(–1, ) y⫽0

(–2, e1–2)

Seeing the Concept

(1, e)

Graph Y1 = ex and compare what you see to Figure 28. Use eVALUEate or TABLE to verify the points on the graph shown in Figure 28. Now graph Y2 = 2x and Y3 = 3x on the same screen as Y1 = ex. Notice that the graph of Y1 = ex lies between these two graphs.

(0, 1) 0

3

x

EX A MPL E 6

Graphing Exponential Functions Using Transformations Graph f1x2 = - e x - 3 and determine the domain, range, and horizontal asymptote of f.

Solution

Begin with the graph of y = e x. Figure 29 shows the stages. *If your calculator does not have one of these keys, refer to your owner’s manual.

SECTION 4.3   Exponential Functions

Figure 29

y

(0, 21)

(–1, – 1–e )

6

y50

(1, 2e)

23

3

x

3

(–2, – e1–2)

(2, e 2 )

y

y

y50

y50

( 1, – e1–2) 23

(2, – 1–e )

x

(3, 21) (4, 2e)

(1, e) 26

26

(–1, 1–e )

335

(2, 2e 2 )

(0, 1) 3

(–2, e1–2) (a) y 5 e x

(5, 2e 2 )

x (b) y 5 2e x

Multiply by −1; Reflect about the x-axis.

Replace x by x − 3; Shift right 3 units.

(c) y 5 2e x23

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f 1x2 = - e x - 3 is the interval 1 - q , q 2, and the range is the interval 1 - q , 02. The horizontal asymptote is the line y = 0.

Now Work

PROBLEM

r

53

4 Solve Exponential Equations Equations that involve terms of the form ax, a 7 0, a ≠ 1, are referred to as exponential equations. Such equations can sometimes be solved by appropriately BQQMZJOH
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  If au = av, then u = v.

In Words When two exponential expressions with the same base are equal, then their exponents are equal.

EXAM PL E 7

(3)

1SPQFSUZ

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B
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QSPQFSUZ



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CF
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UIF
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CBTF

Solving Exponential Equations Solve each exponential equation. B
3x + 1 = 1



C
42x - 1 = x + 3

Solution

B
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1 = 34, write the equation as 3x + 1 = 1 = 34 Now the expressions on each side of the equation have the same base, 3. Set the exponents equal to each other to obtain x + 1 = 4 x = 3 The solution set is {3}. C

42x - 1 = x + 3 2 3 122 2 2x - 1 = 123 2 x + 3 4 = 2 ; 8 = 2 r s rs 2 2x - 1

3 x + 3

1a 2 = a 2 = 2 2 2x - 1 = 3 x + 3 If a u = a v, then u = v. 4x - 2 = 3x + 9 x = 11 The solution set is {11}.

Now Work

PROBLEMS

63

AND

73

r

336

CHAPTER 4  Exponential and Logarithmic Functions

EX A MPL E 8

Solving an Exponential Equation Solve: e -x = 1e x 2 2

Solution

2

#

1 e3

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e on the right side. 1e x 2

2

#

1 = e 2x # e -3 = e 2x - 3 e3

As a result, 2

e -x - x2 2 x + 2x - 3 1x + 32 1x - 12 x = - 3 or x

= = = = =

e 2x - 3 2x - 3 Apply property (3). Place the quadratic equation in standard form. 0 Factor. 0 Use the Zero-Product Property. 1

The solution set is 5 - 3, 16 .

Now Work

EX A MPL E 9

PROBLEM

r 79

Exponential Probability Between 9:00 pm and 10:00 pm
DBST
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used to determine the probability that a car will arrive within t minutes of 9:00 pm. F1t2 = 1 - e -0.2t B
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B
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pm
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before 9:05 pm  C
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B
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PG

pm CFGPSF

pm  D Graph F using your graphing utility. E
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F approach as t increases without bound in the positive EJSFDUJPO

Solution

B
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B
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NJOVUFT
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F t

at t = 5. F 5 = 1 - e -0.2 5 ≈ 0.63212 c

Use a calculator. Figure 30

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5*
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63% probability that a car will arrive within 5 minutes. C
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F t at t = 30. F 30 = 1 - e -0.2 30 ≈ 0.995

c

Use a calculator. Figure 31

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F 1 approaches can be found by letting t S q . Since e -0.2t = 0.2t , it follows that e e -0.2t S 0 as t S q . Therefore, F approaches 1 as t gets large. The algebraic analysis is confirmed by Figure 31.

1

0

30 0

Now Work

r

PROBLEM

109

SECTION 4.3   Exponential Functions

337

SUMMARY Properties of the Exponential Function
%PNBJO
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- q , q ; range: the interval 0, q

x-intercepts: none; y-intercept: 1 Horizontal asymptote: x-axis y = 0 as x S - q Increasing; one-to-one; smooth; continuous See Figure 22 for a typical graph. %PNBJO
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- q , q ; range: the interval 0, q

x-intercepts: none; y-intercept: 1 Horizontal asymptote: x-axis y = 0 as x S q %FDSFBTJOH
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DPOUJOVPVT See Figure 26 for a typical graph.  

f1x2 = ax, a 7 1


f1x2 = ax, 0 6 a 6 1



If au = av, then u = v.

4.3 Assess Your Understanding ‘Are You Prepared?’

Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1 1. 43 = 64 ; 2>3 = 4 ; 3-2 = . QQ
"m"
BOE
4. Find the average rate of change of f 1x2 = 3x - 5 from 9 QQ
"m"

x = 0 to x = 4. QQ
m
m
3 2x 2 2. Solve: x + 3x = 4 QQ
"m"
{ - 4, 1} has y = 2 as a 5. True or False The function f 1x2 = x - 3 3. True or False To graph y = 1x - 22 3, shift the graph of horizontal asymptote. QQ
m
True y = x3 to the left 2 units. QQ
m
False

Concepts and Vocabulary 6. " O
exponential function is a function of the form f 1x2 = Cax, where a 7 0, a ≠ 1, and C ≠ 0 are real numbers. The base a is the growth factor and C is the initial value. f 1x + 12 = a . 7. For an exponential function f 1x2 = Cax, f 1x2 8. True or False The domain of the exponential function f 1x2 = ax, where a 7 0 and a ≠ 1, is the set of all real numbers. True 9. True or False The range of the exponential function f x = ax, where a 7 0 and a ≠ 1, is the set of all real numbers. False

10. True or False The graph of the exponential function f 1x2 = ax, where a 7 0 and a ≠ 1,has no x-intercept. True

*11. The graph of every exponential function f 1x2 = ax, where a 7 0 and a ≠ 1, passes through three points: , . , and 12. If the graph of the exponential function f 1x2 = ax, where a 7 0 and a ≠ 1, is decreasing, then a must be less than 1 . 13. If 3x = 34, then x =

4 . 1 x 14. True or False The graphs of y = 3x and y = a b are 3 identical. False

Skill Building In Problems 15–24, approximate each number using a calculator. Express your answer rounded to three decimal places. *15. B
32.2
*17. B
23.14


C
32.23
C
23.141


D
32.236
D
23.1415


E
315 E
2p

*16. B
51.
*18. B
22.


C
51.3
C
22.1


D
51.32
D
22.1


E
513 E
2e

*19. B
3.12.


C
3.142.1


D
3.1412.1


E
pe

*20. B
2.3.1


C
2.13.14


D
2.13.141


E
e p

21. e

1.2

3.320

22. e

-1.3



23. e

-0.5



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24. e

2.1



338

CHAPTER 4  Exponential and Logarithmic Functions

In Problems 25–32, determine whether the given function is linear, exponential, or neither. For those that are linear functions, find a linear function that models the data; for those that are exponential, find an exponential function that models the data. 25.

26.

x

f1 x 2

-1

3

-1

2

0

6

0

5

1

12

1

8

2

18

2

11

3

30

27.

g1x2

3

x

H 1x 2

-1

30.

x

g1x2

2

-1

6

0

4

0

1

1

6

1

0

2

8

2

3

3

10

3

10

Linear; H1x2 = 2x + 4

x

f1x 2

1 4 1

-1

3 2 3

1

4

1

6

2

16

2

12

3

64

3

24

-1 0

14

28.

H 1 x2

x

Linear; g1x2 = 3x + 5

Neither 29.

x

0

Exponential; f 1x2 = 3 # 2x

Exponential; H1x2 = 4x 31.

x

F 1x 2

-1

2 3 1 3 2 9 4 27 8

0 1 2

Neither 3

32.

x

F 1x 2

-1

1 2 1 4 1 8 1 16 1 32

0 1 2 3

3 x Exponential; F x = a b 2

Exponential; F x =

1# 1 x a b 4 2

In Problems 33–40, the graph of an exponential function is given. Match each graph to one of the following functions. B
y = 3x
C
y = 3-x
D
y = - 3x
E
y = - 3-x G
y = 3x - 1


F
y = 3x - 1
33.

B

y 3

H
y = 31 - x


34.

F

y 3

35.

I
y = 1 - 3x %

y 1

y⫽0 2x

⫺2 y⫽0 ⫺2

y⫽0 ⫺2

2x ⫺1

37.

A

y 3

H

y 1

⫺2

2x

38.

⫺3

C

y

⫺3

39.

E 40.

y 3

1

y

G

3

2x

2x ⫺1

y⫽1

2x ⫺1

y⫽0 ⫺2 y⫽0 ⫺2

36.

⫺2 y ⫽ ⫺1

⫺3

2x ⫺1

⫺2

2x

y⫽0

⫺1

In Problems 41–52, use transformations to graph each function. Determine the domain, range, and horizontal asymptote of each function. *42. f 1x2 = 3x - 2

*41. f 1x2 = 2x + 1 *45. f 1x2 = 3

#

1 a b 2

x

*49. f 1x2 = 2 + 4x-1

*43. f 1x2 = 3x - 1

*44. f 1x2 = 2x + 2

x

*47. f 1x2 = 3-x - 2

*48. f 1x2 = - 3x + 1

*50. f 1x2 = 1 - 2x + 3

*51. f 1x2 = 2 + 3x>2

*52. f 1x2 = 1 - 2-x>3

*46. f 1x2 = 4

#

1 a b 3

In Problems 53–60, begin with the graph of y = e x [Figure 28] and use transformations to graph each function. Determine the domain, range, and horizontal asymptote of each function. *53. f 1x2 = e -x

*57. f 1x2 = 5 - e -x

*54. f 1x2 = - e x

*58. f 1x2 = 9 - 3e -x

*55. f 1x2 = e x + 2

*59. f 1x2 = 2 - e -x>2

*56. f 1x2 = e x - 1

*60. f 1x2 =  - 3e 2x

SECTION 4.3   Exponential Functions

339

In Problems 61–80, solve each equation. 61. x = 3

5 36

62. 5x = 5-6

1 x 1 5 26 65. a b = 5 25 3 69. 3x = 9x 5 - 22, 0, 226 2

73. 3x

-

= 22x

1 1 x 66. a b = 536 4 64 2 1 70. 4x = 2x e 0, f 2 2 74. 5x +  = 1252x 52, 46

5 - 1, 6 5 - 46

77. e x = e 3x + 

5 - 66

78. e 3x = e 2 - x 1 49 1 4

81. If 4x = , what does 4 - 2x equal 83. If 3 - x = 2, what does 32x equal

1 e f 2

5 - 46

63. 2-x = 16

64. 3-x = 1

1 68. 5x + 3 = 5 - 46 5 72. 9-x + 15 = 2x 566

75. 4x # 2x = 162

76. 92x # 2x = 3-1

2

5 - 4, 26

2

1 2 51, 26 80. e 4 x # e x = e 12 e2 1 82. If 2x = 3, what does 4 - x equal 9 1 -x 3x 84. If 5 = 3, what does 5 equal 2 2

79. e x = e 3x

#

In Problems 85–88, determine the exponential function whose graph is given. 85.

f 1x2 = 3x

y

86.

20

16

16

12

y⫽0

87.

(–1, 1–3 ) –3

y (–1, – 1–6 ) –1 –10

–2

5 - 6, 26

12 (2, 9)

8

(0, 1)

4

–1 –2

(–1, 1–5 )

(1, 3) 1

2

3 x

–3

f 1x2 = - 6x

(0, –1) 1

1 e - 1, - f 3

f 1x2 = 5x

y

20

8

5 - 46

3 67. 22x - 1 = 4 e f 2 71. -x + 14 = 16x 566

2 3 x (1, –6)

y⫽0

88.

–2

4

–1 –2

y 2

(1, 5) (0, 1) 1

2

3 x

y⫽0

f 1x2 = - e x (0, –1) 3 x

y⫽0

(1, –e) (–1, – 1–e ) –4

–20 –8

–30 (2, –36) –40

*89. Find an exponential function with horizontal asymptote y = 2 whose graph contains the points 10, 32 and 11, 52 .

(2, –e 2)

–12

*90. Find an exponential function with horizontal asymptote y = - 3 whose graph contains the points 10, - 22 and 1 - 2, 12 .

Mixed Practice *91. Suppose that f 1x2 = 2x. B
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f 142
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f
1 C
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f 1x2 = , what is x
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*92. Suppose that f 1x2 = 3x. B
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f 142
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f 1 C
*G
f 1x2 = , what is x
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9 of f

93. Suppose that g1x2 = 4x + 2. * B
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g1 - 12
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g C
*G
g1x2 = 66, what is x
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of g
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94. Suppose that g1x2 = 5x - 3. * B
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of g
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1 x 95. Suppose that H x = a b - 4. 2 * B
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1 x 96. Suppose that F 1x2 = a b - 3. 3 * B
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F. - 1

340

CHAPTER 4  Exponential and Logarithmic Functions

In Problems 97–100, graph each function. Based on the graph, state the domain and the range, and find any intercepts. *97. f 1x2 = e

e -x ex

*99. f 1x2 = e

- ex - e -x

if x 6 0 if x Ú 0 if x 6 0 if x Ú 0

*98. f 1x2 = e *100. f 1x2 = e

ex e -x - e -x - ex

if x 6 0 if x Ú 0 if x 6 0 if x Ú 0

Applications and Extensions 101. Optics If a single pane of glass obliterates 3% of the light passing through it, the percent p of light that passes through n successive panes is given approximately by the function p1n2 = 10010.92 n B
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B
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where P is the total population of the community and d is the number of days that have elapsed since the rumor began. In a community of 1000 students, how many students will IBWF
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probability can be used to determine the probability that a car will arrive within t minutes of 12:00 pm:

A1n2 = A0 e -0.35n

F1t2 = 1 - e -0.1t

describes the area of a wound after n days following an injury when no infection is present to retard the healing. Suppose that a wound initially had an area of 100 square millimeters. B
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105. Advanced-Stage Pancreatic Cancer The percentage of patients P who have survived t years after initial diagnosis of advanced-stage pancreatic cancer is modeled by the function P t = 100 0.3 t Source: Cancer Treatment Centers of America B
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110. Exponential Probability Between 5:00 pm and 6:00 pm, cars BSSJWF
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B
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1

SECTION 4.3   Exponential Functions

* E Graph F using a graphing utility. F
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hour. The following formula from probability can be used to determine the probability that x cars will arrive between 5:00 pm and 6:00 pm.

where x! = x

#

1x - 12

#

#g#3#2#1

B
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x = 15 cars will arrive between 5:00 pm and 6:00 pm. 0.0516 C
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x = 20 cars will arrive between 5:00 pm and 6:00 pm.  112. Poisson Probability
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Demon Roller Coaster at the rate of 4 per minute. The following formula from probability can be used to determine the probability that x people will arrive within the next minute.

where x! = x

#

1x - 12

4xe -4 x!

# x - 2 # g # 3 # 2 # 1

B
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the air to the maximum amount that it can hold at a specific temperature. The relative humidity, R, is found using the following formula: 4221

I =

E 31 - e - R>L t 4 R I

20xe -20 x!

1x - 22

P 1x2 =

115. Current in a RL Circuit The equation governing the amount of current I
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approximates the number of words L that the student will have learned after t minutes. B
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114. Learning Curve Suppose that a student has 500 vocabulary words to learn. If the student learns 15 words after 5 minutes, the function

L

⫺

R = 10 T + 459.4 - D + 459.4 + 2

R ⫹

where T
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341

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117. If f is an exponential function of the form f 1x2 = C # ax with growth factor 3 and f 162 = 12, what is f 12 36

342

CHAPTER 4  Exponential and Logarithmic Functions

*118. Another Formula for e
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values of 2 +

Problems 124 and 125 provide definitions for two other transcendental functions. *124. The hyperbolic sine function, designated by sinh x, is defined as

1 1 1 + + g + 2! 3! n!

for n = 4, 6, , and 10. Compare each result with e. [Hint: 1! = 1, 2! = 2 # 1, 3! = 3 # 2 # 1, n! = n1n - 12

#g#

sinh x =

B
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f 1x2 = sinh x is an odd function. C Graph f 1x2 = sinh x using a graphing utility.

132 122 112.]

*119. Another Formula for e
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various values of the expression. Compare the values to e.

*125. The hyperbolic cosine function, designated by cosh x, is defined as

2 + 1 1 + 1 2 + 2 3 + 3 4 + 4 etc.

cosh x =

f 1x + h2 - f 1x2 h

= ax

#a

- 1 h

1cosh x2 2 - 1sinh x2 2 = 1 *126. Historical Problem
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conjectured that the function

h ≠ 0

*121. If f 1x2 = ax, show that f 1A + B2 = f 1A2 *122. If f 1x2 = ax, show that f 1 - x2 =

1 x 1e + e -x 2 2

B
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1 x 1e - e -x 2 2

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f1x2 = 212 2 + 1 x

# f 1B2.

for x = 1, 2, 3, c , would always have a value equal to a prime number. But Leonhard Euler 1109132 showed that this formula fails for x = 5.
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the prime numbers produced by f for x = 1, 2, 3, 4. Then show that f 152 = 641 * 6,00,41, which is not prime.

1 . f 1x2

*123. If f 1x2 = ax, show that f 1ax2 = 3f 1x2 4 a.

Discussion and Writing 127. The bacteria in a 4-liter container double every minute. After 60 minutes the container is full. How long did it take UP
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130. As the base a of an exponential function f 1x2 = ax, where a 7 1, increases, what happens to the behavior of its graph for x 7 0
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Retain Your Knowledge Problems 132–135 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 132. Solve the inequality: x3 + 5x2 … 4x + 20 *135. - q , - 5] h [ - 2, 2] x + 1 133. Solve the inequality: Ú 1 2, q

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-3

-3

-4

0

-5

3

-6

6

-7

Consider the quadratic function f x = x2 + 2x - 3. B
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‘Are You Prepared?’ Answers 1. 64; 4;

1 9

2. { - 4, 1}

3. False

4. 3

5. True

SECTION 4.4   Logarithmic Functions

343

4.4 Logarithmic Functions PREPARING FOR THIS SECTION Before getting started, review the following: r
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Now Work the ‘Are You Prepared?’ problems on page 352.

OBJECTIVES 1 Change Exponential Statements to Logarithmic Statements and Logarithmic Statements to Exponential Statements (p. 344) 2 Evaluate Logarithmic Expressions (p. 344) 3 Determine the Domain of a Logarithmic Function (p. 345) 4 Graph Logarithmic Functions (p. 345) 5 Solve Logarithmic Equations (p. 350)

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y = f1x2 has an inverse function that is defined JNQMJDJUMZ
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x = f 1y2. In particular, the exponential function y = f1x2 = a x, where a 7 0 and a ≠ 1, is one-to-one and, hence, has an inverse function that is defined implicitly by the equation x = ay, a 7 0, a ≠ 1 This inverse function is so important that it is given a name, the logarithmic function.

DEFINITION

The logarithmic function to the base a, where a 7 0 and a ≠ 1, is denoted by y = log a x
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iy is the logarithm to the base a of xu
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CZ y = log a x if and only if x = ay

In Words When you read logax, think to yourself “a raised to what power gives me x? ”

The domain of the logarithmic function y = log a x is x 7 0.

As this definition illustrates, a logarithm is a name for a certain exponent. So log ax represents the exponent to which a must be raised to obtain x.

EXAM PL E 1

Relating Logarithms to Exponents B
*G
y = log 3 x, then x = 3y. For example, the logarithmic statement 4 = log 3 1 is equivalent to the exponential statement 1 = 34. 1 C
*G
y = log 5 x, then x = 5y. For example, - 1 = log 5 a b is equivalent to 5 1 = 5-1. 5

r

344

CHAPTER 4  Exponential and Logarithmic Functions

1 Change Exponential Statements to Logarithmic Statements and Logarithmic Statements to Exponential Statements The definition of a logarithm can be used to convert from exponential form to logarithmic form, and vice versa, as the following two examples illustrate.

EX A MPL E 2

Changing Exponential Statements to Logarithmic Statements Change each exponential statement to an equivalent statement involving a logarithm. B
1.23 = m



C
e b = 9



D
a4 = 24

Solution

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y = log a x and x = ay, where a 7 0 and a ≠ 1, are equivalent. B
*G
1.23 = m, then 3 = log 1.2 m. C
*G
e b = 9, then b = log e 9. D
*G
a4 = 24, then 4 = log a 24.

Now Work

EX A MPL E 3

PROBLEM

r

9

Changing Logarithmic Statements to Exponential Statements Change each logarithmic statement to an equivalent statement involving an exponent. B
log a 4 = 5



C
log e b = - 3



D
log 3 5 = c

Solution

B
*G
log a 4 = 5, then a5 = 4. C
*G
log e b = - 3, then e -3 = b. D
*G
log 3 5 = c, then 3c = 5.

Now Work

PROBLEM

r

17

2 Evaluate Logarithmic Expressions To find the exact value of a logarithm, write the logarithm in exponential notation using the fact that y = log a x is equivalent to ay = x, and use the fact that if au = ay, then u = y.

EX A MPL E 4

Finding the Exact Value of a Logarithmic Expression Find the exact value of: B
log 2 16

Solution

C
log 3

1 2

B
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log 216, think “2 SBJTFE
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 u
y = log 2 16 2y = 16 2y = 24 y = 4

Change to exponential form. 16 = 24 Equate exponents.

1 , think “3 2 1 raised to what power yields u
2 1 y = log 3 2

C
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log 3

3y =

1 2

3y = 3-3

Therefore, log 2 16 = 4.

y = -3 Therefore, log 3

Now Work

PROBLEM

25

Change to exponential form. 1 1 = 3 = 3-3 27 3 Equate exponents.

1 = - 3. 2

r

SECTION 4.4   Logarithmic Functions

345

3 Determine the Domain of a Logarithmic Function The logarithmic function y = log a x has been defined as the inverse of the exponential function y = a x. That is, if f1x2 = a x, then f -1 1x2 = log a x. Based on the discussion given in Section 4.2 on inverse functions, for a function f and its inverse f -1, %omain of f -1 = 3ange of f and 3ange of f -1 = %omain of f Consequently, it follows that %omain of the logarithmic function = 3ange of the exponential function = 10, q 2 3ange of the logarithmic function = %omain of the exponential function = 1 - q , q 2

The next box summarizes some properties of the logarithmic function:

y = log a x %omain:

1defining equation: x = ay 2 10, q 2 3ange: 1 - q , q 2

The domain of a logarithmic function consists of the positive real numbers, so the argument of a logarithmic function must be greater than zero.

EXAMPL E 5

Finding the Domain of a Logarithmic Function Find the domain of each logarithmic function.

B
F1x2 = log 2 1x + 32 1 + x C
g1x2 = log 5 a b 1 - x D
h 1x2 = log 1>2 0 x 0

Solution

B
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F consists of all x for which x + 3 7 0, that is, x 7 - 3.
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interval notation, the domain of F is 1 - 3, q 2 . C
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g is restricted to 1 + x 7 0 1 - x Solve this inequality to find that the domain of g consists of all x between - 1 and 1, that is, - 1 6 x 6 1 or, using interval notation, 1 - 1, 12. D
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0 x 0 7 0, provided that x ≠ 0, the domain of h consists of all real numbers except zero or, using interval notation, 1 - q , 02 ∪ 10, q 2.

r

Now Work

PROBLEMS

39

AND

45

4 Graph Logarithmic Functions Because exponential functions and logarithmic functions are inverses of each other, the graph of the logarithmic function y = log a x is the reflection about the line y = x of the graph of the exponential function y = a x, as shown in Figure 32 on the following page.

346

CHAPTER 4  Exponential and Logarithmic Functions

Figure 32

y

y ⫽ ax ( ⫺1,

1– a

y

y⫽x

3

y ⫽ ax y ⫽ x

3 (1, a)

)

(a, 1) (⫺1, 1–a

(1, a)

(0, 1) ⫺3

(1, 0)

3 x

1– a,

⫺1) y ⫽ loga x

(

(0, 1) )

⫺3

y ⫽ loga x (a, 1)

(1, 0)

3 x

( 1–a ,⫺1)

⫺3

⫺3

(a) 0 ⬍ a ⬍ 1

(b) a ⬎ 1

For example, to graph y = log 2 x, graph y = 2x and reflect it about the line 1 x y = x. See Figure 33. To graph y = log 1>3 x, graph y = a b and reflect it about 3 the line y = x. See Figure 34.

Figure 33

Figure 34 y 2

(⫺1, 12 )

y⫽ (1, 2)

2x

y⫽ 1

y⫽x

y

(⫺1, 3)

3

y ⫽ log2x

(0, 1)

(2, 1)

⫺2

x

(3)

(0, 1) ⫺3

(1, 0) 2 x

( 13 , 1) (1, 13 ) 3 x (1, 0) (3, ⫺1) y ⫽ log1/3x

( 12 , ⫺1) ⫺3

⫺2

Now Work

y⫽x

PROBLEM

59

The graphs of y = log a x
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Properties of the Logarithmic Function f1 x 2 = loga x

1. The domain is the set of positive real numbers, or 10, q 2 using interval notation; the range is the set of all real numbers, or 1 - q , q 2 using interval notation. 2. The x-intercept of the graph is 1. There is no y-intercept. 3. The y-axis x = 0 is a vertical asymptote of the graph. 4. A logarithmic function is decreasing if 0 6 a 6 1 and is increasing if a 7 1. 1 5. The graph of f contains the points 11, 02, 1a, 12, and a , - 1b . a 6. The graph is smooth and continuous, with no corners or gaps.

In Words

y = logex is written y = ln x.

If the base of a logarithmic function is the number e, the result is the natural logarithm function. This function occurs so frequently in applications that it is given a special symbol, ln
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SECTION 4.4   Logarithmic Functions

y = ln x if and only if x = e y

347

(1)

Because y = ln x and the exponential function y = e x are inverse functions, the graph of y = ln x can be obtained by reflecting the graph of y = e x about the line y = x. See Figure 35. Other points on the graph of f 1x2 = ln x can be found by using a calculator with an  ln 
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 Figure 35

y

Table 7

5

x

ln x

1 2

- 0.69

2

0.69

3

1.10

ye x

yx

(1, e) ( 0, 1) (1, 1–e )

Seeing the Concept Graph Y1 = ex and Y2 = ln x on the same square screen. Use eVALUEate to verify the points on the graph given in Figure 35. Do you see the symmetry of the two graphs with respect to the line y = x?

y  0 3

( 1, 0) 1

( 1–e ,1)

Graphing a Logarithmic Function and Its Inverse

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f, find the range of f - 1. G
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y = - ln 1x - 22, begin with the graph of y = ln x and use transformations. See Figure 36.

Solution

y

3

3 x0

x0

1

1

(e, 1) (1, 0)

1

(

1– , e

D y  In x

x2

y

y

3

1

3 x

x0

E XAM PL E 6

Figure 36

yIn x

(e, 1)

3 x

1

1)

( 1–e , 1)

1

Multiply by  1; reflect about the x-axis.

E y  In x

(3, 0) 3 x

(1, 0)

( 1–e  2,1)

1 1 1

(e, 1) Replace x by x  2; shift right 2 units.

3

5

x

(e2, 1)

F y  In (x 2)

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f1x2 = - ln 1x - 22 is the set of all real numbers. The vertical asymptote is x = 2.
L t 4 R

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Loudness of Sound Problems 127–130 use the following discussion: The loudness L x , measured in decibels (dB), of a sound of x intensity x, measured in watts per square meter, is defined as L x = 10 log , where I0 = 10-12 watt per square meter is the least intense I0 sound that a human ear can detect. Determine the loudness, in decibels, of each of the following sounds. 130. %JFTFM
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E# -1 128. Amplified rock music: intensity of 10 watt per square meter. 110 dB 129. Heavy city traffic: intensity of x = 10-3 watt per square meter. 90 dB The Richter Scale Problems 131 and 132 use the following discussion: The Richter scale is one way of converting seismographic readings into numbers that provide an easy reference for measuring the magnitude M of an earthquake. All earthquakes are compared to a zero-level earthquake whose seismographic reading measures 0.001 millimeter at a distance of 100 kilometers from the epicenter. An earthquake whose seismographic reading measures x millimeters has magnitude M x , given by M x = log¢

x ≤ x0

356

CHAPTER 4  Exponential and Logarithmic Functions

where x0 = 10-3 is the reading of a zero-level earthquake the same distance from its epicenter. In Problems 131 and 132, determine the magnitude of each earthquake. 131. Magnitude of an Earthquake
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having an accident of 5 or more should not have driving privileges, at what concentration of alcohol in the bloodstream should a driver be arrested and charged XJUI
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Discussion and Writing 134. Is there any function of the form y = xa, 0 6 a 6 1, that increases more slowly than a logarithmic function whose CBTF
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8IZ 136. Critical Thinking In buying a new car, one consideration might be how well the price of the car holds up over time. %JGGFSFOU
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Age in Years New

1

2

3

4

5

$38,000

$36,600

$32,400

$28,750

$25,400

$21,200

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Retain Your Knowledge Problems 137–140 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. *139. 6TF
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*140. A complex polynomial function f of degree 4 has the zeros 1 to 1. 12 - 1, 2, and 3 - i.
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f. Then find a 2 polynomial function with real coefficients that has the zeros.

‘Are You Prepared?’ Answers 1. B
{x 0 x … 3}

C
{x 0 x 6 - 2 or x 7 3}

2. {x 0 x 6 - 4 or x 7 1}

3. {3}

4.5 Properties of Logarithms OBJECTIVES 1 Work with the Properties of Logarithms (p. 356) 2 Write a Logarithmic Expression as a Sum or Difference of Logarithms (p. 359) 3 Write a Logarithmic Expression as a Single Logarithm (p. 359) 4 Evaluate a Logarithm Whose Base Is Neither 10 Nor e (p. 360) 5 Graph a Logarithmic Function Whose Base Is Neither 10 Nor e (p. 362)

1 Work with the Properties of Logarithms Logarithms have some very useful properties that can be derived directly from the definition and the laws of exponents.

SECTION 4.5   Properties of Logarithms

EXAMPL E 1

Establishing Properties of Logarithms B
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log a 1 = 0.


Solution

357

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log a a = 1.

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y = log a x
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y = log a 1. Then y ay ay y log a 1

= = = = =

log a 1 1 a0 0 0

= = = = =

log a a a a1 1 1

Change to an exponential statement. a 0 = 1 since a 7 0, a ≠ 1 Solve for y. y = loga 1

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y = log a a. Then y ay ay y log a a

Change to an exponential statement. a = a1 Solve for y. y = loga a

r

To summarize: log a 1 = 0

THEOREM

log a a = 1

Properties of Logarithms In the properties given next, M and a are positive real numbers, a ≠ 1, and r is any real number. The number log a M is the exponent to which a must be raised to obtain M. That is, aloga M = M

(1)

The logarithm to the base a of a raised to a power equals that power. That is, log a ar = r

(2)

The proofs use the fact that y = ax and y = log a x are inverses.

Proof of Property (1) For inverse functions, f 1f -1 1x2 2 = x for all x in the domain of f - 1 6TF
f1x2 = ax and f -1 1x2 = log a x to find f1f -1 1x2 2 = aloga x = x for x 7 0 Now let x = M to obtain aloga M = M, where M 7 0.

■

Proof of Property (2) For inverse functions, f -1 1f1x2 2 = x for all x in the domain of f 6TF
f1x2 = ax and f -1 1x2 = log a x to find f -1 1f1x2 2 = log a ax = x for all real numbers x Now let x = r to obtain log a ar = r, where r is any real number.

■

358

CHAPTER 4  Exponential and Logarithmic Functions

EX A MPL E 2

Using Properties (1) and (2) B
2log2 p = p



C
log 0.2 0.2-22 = - 22



D
ln e kt = kt

Now Work

PROBLEM

r

15

Other useful properties of logarithms are given next.

THEOREM

Properties of Logarithms In the following properties, M, N, and a are positive real numbers, a ≠ 1, and r is any real number.

The Log of a Product Equals the Sum of the Logs log a 1MN2 = log a M + log a N

(3)

The Log of a Quotient Equals the Difference of the Logs log a a

M b = log a M - log a N N

(4)

The Log of a Power Equals the Product of the Power and the Log log a Mr = r log a M

(5)

ax = e x ln a

(6)

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Proof of Property (3) Let A = log a M and let B = log a N. These expressions are equivalent to the exponential expressions aA = M and aB = N Now

log a 1MN = log a 1aA aB 2 = log a aA + B = A + B = log a M + log a N

Law of Exponents Property (2) of logarithms ■

Proof of Property (5) Let A = log a M. This expression is equivalent to aA = M Now

log a Mr = log a 1aA 2 = log a arA = rA = r log a M r

Law of Exponents Property (2) of logarithms ■

Proof of Property (6) 1SPQFSUZ



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a = e, yields e ln M = M Now let M = ax
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e ln a = e x ln a = 1e ln a 2 x = ax x

Now Work

PROBLEM

19

■

SECTION 4.5   Properties of Logarithms

359

2 Write a Logarithmic Expression as a Sum or Difference of Logarithms Logarithms can be used to transform products into sums, quotients into differences, and powers into factors. Such transformations prove useful in certain types of calculus problems.

Writing a Logarithmic Expression as a Sum of Logarithms

EXAM PL E 3

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log a 1x2x2 + 12, x 7 0, as a sum of logarithms. Express all powers as factors. log a 1x2x2 + 12 = log a x + log a 2x2 + 1 1>2 = log a x + log a 1x2 + 12 1 = log a x + log a 1x2 + 12 2

Solution

loga 1M # N2 = loga M + loga N

loga M r = r loga M

r

Writing a Logarithmic Expression as a Difference of Logarithms

EXAM PL E 4

8SJUF ln

x2 , 1x - 12 3

x 7 1

as a difference of logarithms. Express all powers as factors.

Solution

ln

x2 = ln x2 - ln 1x - 12 3 = 2 ln x - 3 ln1x - 12 1x - 12 3 c c M loga a b = loga M - loga N N

r

logaM r = r logaM

Writing a Logarithmic Expression as a Sum and Difference of Logarithms

EXAM PL E 5

8SJUF log a

2x2 + 1 , x 7 0 x3 1x + 12 4

as a sum and difference of logarithms. Express all powers as factors.

Solution

log a

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that the variable may assume. For example, the domain of the variable for log a x is x 7 0 and for log a x - 1 is x 7 1. If these functions are added, the domain is x 7 1. That is, the equality log a x + log a x - 1 = log a[x x - 1 ] is true only for x 7 1. j

2x2 + 1 = log a 2x2 + 1 - log a 3 x3 1x + 12 4 4 x3 1x + 12 4 = log a 2x2 + 1 - 3 log a x3 + log a 1x + 12 4 4

= log a 1x + 12 - log a x - log a 1x + 12 1 = log a 1x2 + 12 - 3 log a x - 4 log a 1x + 12 2 2

Now Work

PROBLEM

1>2

3

Property (4) Property (3)

4

Property (5)

r

51

3 Write a Logarithmic Expression as a Single Logarithm "OPUIFS
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logarithms with the same base as a single logarithm. This skill will be needed to solve certain logarithmic equations discussed in the next section.

EXAM PL E 6

Writing Expressions as a Single Logarithm 8SJUF
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3 D
log a x + log a 9 + log a 1x2 + 12 - log a 5 B
log a  + 4 log a 3

C


360

CHAPTER 4  Exponential and Logarithmic Functions

Solution

B
log a  + 4 log a 3 = log a  + log a 34 r logaM = logaM r = log a  + log a 1 logaM + logaN = loga 1M # N2 = log a 1 # 12 = log a 56 2 C
ln  - ln 152 - 12 = ln 2>3 - ln 125 - 12 r logaM = logaM r 3 3 82/3 = 1 1 82 2 = 22 = 4 = ln 4 - ln 24 4 M = ln a b logaM - logaN = loga a b N 24 1 = ln a b 6 = ln 1 - ln 6 = - ln 6 ln1 = 0

D
log a x + log a 9 + log a 1x2 + 12 - log a 5 = log a 19x2 + log a 1x2 + 12 - log a 5 = log a 3 9x1x2 + 12 4 - log a 5 = log a J

9x1x2 + 12 R 5

r

WARNING A common error made by some students is to express the logarithm of a sum as the sum of logarithms. log a 1M + N2

Correct statement

is not equal to

log a 1MN2 = log a M + log a N

log a M + log a N Property (3)

Another common error is to express the difference of logarithms as the quotient of logarithms. log a M - log a N is not equal to

log a M log a N

M Correct statement log a M - log a N = log a a b Property (4) N A third common error is to express a logarithm raised to a power as the product of the power times the logarithm. 1log a M2 r is not equal to r log a M

Correct statement

Now Work

log a Mr = r log a M PROBLEMS

57

AND

j

Property (5)

63

Two other important properties of logarithms are consequences of the fact that the logarithmic function y = log a x is a one-to-one function.

THEOREM

Properties of Logarithms In the following properties, M, N, and a are positive real numbers, a ≠ 1. If M = N, then log a M = log a N. If log a M = log a N, then M = N.

(7) (8)

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exponential and logarithmic equations, a topic discussed in the next section.

4 Evaluate a Logarithm Whose Base Is Neither 10 Nor e Logarithms to the base 10, common logarithms, were used to facilitate arithmetic DPNQVUBUJPOT
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SECTION 4.5   Properties of Logarithms

361

the number e remain very important because they arise frequently in the study of natural phenomena. Common logarithms are usually abbreviated by writing log, with the base understood to be 10, just as natural logarithms are abbreviated by ln, with the base understood to be e. .PTU
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log and ln keys to calculate the common logarithm and the natural logarithm of a number, respectively. Let’s look at an example to see how to approximate logarithms having a base other than 10 or e.

EXAMPL E 7

Approximating a Logarithm Whose Base Is Neither 10 Nor e Approximate log 2 .
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Solution

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log 2 
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y = log 2 . Then 2y = . Because 22 = 4 and 23 = , the value of log 2  is between 2 and 3. 2y =  ln 2y = ln  y ln 2 = ln  ln  y = ln 2 y ≈ 2.04

Property (7) Property (5) Exact value

r

Approximate value rounded to four decimal places

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to logarithms involving the base e. In general, the Change-of-Base Formula is used.

THEOREM

Change-of-Base Formula If a ≠ 1, b ≠ 1, and M are positive real numbers, then log a M =

log b M log b a

(9)

Proof Let y = log a M. Then ay = M log b ay = log b M

Property (7)

y log b a = log b M log b M y = log b a log b M log a M = log b a

Property (5) Solve for y.

r

y = loga M

Because calculators have keys only for log and ln , in practice, the Change-of-Base Formula uses either b = 10 or b = e. That is, log a M =

EXAMPL E 8

log M log a

and log a M =

ln M ln a

Using the Change-of-Base Formula Approximate: B
log 5 9
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C
log 22 25

(10)

362

CHAPTER 4  Exponential and Logarithmic Functions

Solution

1 log 5 log 25 log 9 2 1.94939000 = C
log 22 25 = B
log 5 9 = ≈ 1 log 5 0.69900043 log 22 log 2 2 ≈ 2.9 log 5 = ≈ 2.3219 log 2 or or ln 9 4.46363 ≈ log 5 9 = ln 5 1.60943912 1 ln 5 ln 25 2 ≈ 2.9 log 22 25 = = 1 ln 22 ln 2 2 ln 5 = ≈ 2.3219 ln 2

r

Now Work

PROBLEMS

23

AND

71

5 Graph a Logarithmic Function Whose Base Is Neither 10 Nor e The Change-of-Base Formula also can be used to graph logarithmic functions whose base is neither 10 nor e.

EX AMPL E 9

Graphing a Logarithmic Function Whose Base Is Neither 10 Nor e 6TF
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y = log 2 x.

Figure 42

Solution

2

0

2

4

Because graphing utilities have only logarithms with the base 10 or the base e, use the Change-of-Base Formula to express y = log 2 x in terms of logarithms with base log x ln x 10 or base e. Graph either y = or y = to obtain the graph of y = log 2 x. ln 2 log 2 See Figure 42. log x ln x Check:

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Now Work

PROBLEM

79

SUMMARY Properties of Logarithms In the list that follows, a, b, M, N, and r are real numbers. Also, a 7 0, a ≠ 1, b 7 0, b ≠ 1, M 7 0, and N 7 0. Definition

y = log a x means x = ay

 

Properties of logarithms

log a 1 = 0; log a a = 1 aloga M = M; log a ar = r log a 1MN2 = log a M + log a N M log a a b = log a M - log a N N log b M log a M = log b a

log a Mr = r log a M ax = e x ln a If M = N, then log a M = log a N.

Change-of-Base Formula

If log a M = log a N, then M = N.

 

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363

SECTION 4.5   Properties of Logarithms

Historical Feature

L

tables, published in 1614, listed what would now be called natural logarithms of sines and were rather difficult to use. A London professor, Henry Briggs, became interested in the tables and visited Napier. In their conversations, they developed the idea of common logarithms, which were published in 1617. Their importance for calculation was immediately recognized, and by 1650 they were being printed as far away as China. They remained an important calculation tool until the advent of the inexpensive handheld calculator about 1972, which has decreased their calculational, but not their theoretical, importance. A side effect of the invention of logarithms was the popularization of the decimal system of notation for real numbers.

ogarithms were invented about 1590 by John Napier (1550–1617) and Joost Bürgi (1552–1632), working independently. Napier, whose work had the greater influence, was a Scottish lord, a secretive man whose neighbors were inclined to believe him to be in league with the devil. His approach to logarithms was very John Napier different from ours; it was based on the relationship (1550–1617) between arithmetic and geometric sequences, discussed in a later chapter, and not on the inverse function relationship of logarithms to exponential functions (described in Section 4.4). Napier’s

4.5 Assess Your Understanding Concepts and Vocabulary 1. log a1 =

2. log aa =

0

5. log a 1MN2 =

log aM

+

3. alogaM =

1

M 6. log a a b = log aM N 8. If log a x = log a6, then x =

log aN

7. log aM = r

r log a M log 5  , then M = 9. If log  M = log 5 

4. log aar =

M

log aN 6 .

10. True or False ln1x + 32 - ln12x2 =

 .

11. True or False log 2 13x4 2 = 4 log 2 13x2

12. True or False

False

ln  = 2 ln 4

r

ln1x + 32

False

ln12x2

False

Skill Building In Problems 13–28, use properties of logarithms to find the exact value of each expression. Do not use a calculator. 13. log 3 31 log2 

17. 2

14. log 2 2-13





18. e

ln 

15. ln e -4

- 13

19. log  2 + log  4



16. ln e 22

-4

22

20. log 6 9 + log 6 4

1

2

22. log  16 - log  2 1 23. log 2 6 # log 6  3 24. log 3  # log  9 2 21. log 6 1 - log 6 3 1 log3 5 - log3 4 5 log5 6 + log5  loge2 16 42 27. e 4 28. e loge2 9 3 26. 5 25. 3 4 In Problems 29–36, suppose that ln 2 = a and ln 3 = b. Use properties of logarithms to write each logarithm in terms of a and b. 2 29. ln 6 a + b 30. ln a - b 31. ln 1.5 b - a 32. ln 0.5 - a 3 1 2 1 5 35. ln 2 6 1a + b2 36. ln 4 1a - b2 33. MO

3a 34. MO

3b 5 A3 4 In Problems 37–56, write each expression as a sum and/or difference of logarithms. Express powers as factors. x 38. log 3 log 3 x - 2 39. log 2 z3 3 log 2z 40. log x5 5 log  x 37. log 5 125x2 2 + log 5 x 9 e x 42. ln 1 - ln x 43. ln x ln x - x 44. ln1xe x 2 ln x + x 41. ln1ex2 1 + ln x x e a *45. log a 1u2 v3 2 u 7 0, v 7 0 *46. log 2 a 2 b a 7 0, b 7 0 *47. ln1x2 21 - x2 0 6 x 6 1 *48. ln1x21 + x2 2 x 7 0 b 3 2 x1x + 22 x3 2 x + 1 x3 2x + 1 *49. log 2 ¢ b x 7 1 *51. logJ ≤ x 7 3 *50. log 5 a 2 R x 7 0 *52. logJ R x 7 2 2 x - 3 x - 1 1x + 32 1x - 22 2 *53. lnJ

x2 - x - 2 R 1x + 42 2

1>3

x 7 2*54. lnJ

1x - 42 2 x - 1 2

2>3

R

x 7 4 *55. ln

5x21 + 3x 1x - 42 3

x 7 4

*56. lnc

3 5x2 2 1 - x d 41x + 12 2

0 6 x 6 1

In Problems 57–70, write each expression as a single logarithm. 57. 3 log 5 u + 4 log 5 v

log 5 1u3v4 2

1 1 60. log 2 a b + log 2 ¢ 2 ≤ x x *63. lna

log 2 a

1 b x3

58. 2 log 3 u - log 3 v

log 3 a

u2 b v

*61. log 4 1x2 - 12 - 5 log 4 1x + 12

59. log 3 2x - log 3 x3

log 3 a

1 x5>2

b

*62. log1x2 + 3x + 22 - 2 log1x + 12

x x2 + x + 6 x + 1 x2 + 2x - 3 4 ≤ - log¢ b + lna b - ln1x2 - 12*64. log¢ ≤ *65.  log 2 23x - 2 - log 2 a b + log 2 4 2 x - 1 x x + 2 x x - 4

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364

CHAPTER 4  Exponential and Logarithmic Functions

3 *66. 21 log 3 1 x + log 3 19x2 2 - log 3 9

*67. 2 log a 15x3 2 -

*69. 2 log 2 1x + 12 - log 2 1x + 32 - log 2 1x - 12

1 1 1 *68. log1x3 + 12 + log1x2 + 12 log a 12x + 32 2 3 2 *70. 3 log 5 13x + 12 - 2 log 5 12x - 12 - log 5 x

In Problems 71–78, use the Change-of-Base Formula and a calculator to evaluate each logarithm. Round your answer to three decimal places. 71. log 3 21



72. log 5 1



73. log 1>3 1

75. log 12 

5.615

76. log 15 



77. log p e

- 3.0



74. log 1>2 15

- 3.90

78. log p 22

0.303

In Problems 79–84, graph each function using a graphing utility and the Change-of-Base Formula.

*79. y = log 4 x * 80. y = log 5 x *81. y = log 2 1x + 22 * 82. y = log 4 1x - 32 * 83. y = log x - 1 1x + 12 *84. y = log x + 2 1x - 22

Mixed Practice 86. If f 1x2 = log 2 x, g1x2 = 2x, and h1x2 = 4x, find: * B
1f ∘ g2 1x2.
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1g ∘ f2 1x2.
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85. If f 1x2 = ln x, g1x2 = e x, and h1x2 = x2, find: * B
1f ∘ g2 1x2.
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Applications and Extensions In Problems 87–96, express y as a function of x. The constant C is a positive number. 87. ln y = ln x + ln C

89. ln y = ln x + ln1x + 12 + ln C 91. ln y = 3x + ln C

88. ln y = ln1x + C2

y = Cx y = Cx1x + 12

y = x + C

90. ln y = 2 ln x - ln1x + 12 + ln C

y = Ce 3x

92. ln y = - 2x + ln C

93. ln1y - 32 = - 4x + ln C y = Ce - 4x + 3 1 1 *95. 3 ln y = ln12x + 12 - ln1x + 42 + ln C 2 3 97. Find the value of log 2 3 # log 3 4 # log 4 5 # log 5 6 # log 6  # log  . 3

99. Find the value of log 2 3 # log 3 4 # g # log n 1n + 12 # log n + 1 2. 1 *101. Show that log a 1x + 2x2 - 12 + log a 1x - 2x2 - 12 = 0.

y =

y = Ce - 2x

Cx2 x + 1

94. ln1y + 42 = 5x + ln C y = Ce 5x - 4 1 1 *96. 2 ln y = - ln x + ln1x2 + 12 + ln C 2 3 98. Find the value of log 2 4 # log 4 6 # log 6 . 3

100. Find the value of log 2 2 # log 2 4 # g # log 2 2n.

n!

*102. Show that log a 1 2x + 2x - 12 + log a 1 2x - 2x - 12 = 0. *103. Show that ln11 + e 2x 2 = 2x + ln11 + e -2x 2. f 1x + h2 - f 1x2 h 1>h = log a a1 + b , h ≠ 0. *104. Difference Quotient If f 1x2 = log a x, show that h x *106. If f 1x2 = log a x, show that f 1AB2 = f 1A2 + f 1B2. *105. If f 1x2 = log a x, show that - f 1x2 = log 1>a x. 1 * 107. If f 1x2 = log a x, show that f a b = - f 1x2. x M *109. Show that log a a b = log a M - log a N, where a, M, and N N are positive real numbers and a ≠ 1.

*108. If f 1x2 = log a x, show that f 1xa 2 = af 1x2. *110. Show that log a a

1 b = - log a N, where a and N are N positive real numbers and a ≠ 1.

Discussion and Writing 111. Graph Y1 = log1x2 2 and Y2 = 2 log1x2 using a graphing VUJMJUZ
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Retain Your Knowledge Problems 115–118 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. *115. 6TF
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1 - 5 { 221 - 2, , 5 2

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365

SECTION 4.6   Logarithmic and Exponential Equations

4.6 Logarithmic and Exponential Equations PREPARING FOR THIS SECTION Before getting started, review the following: r
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OBJECTIVES 1 Solve Logarithmic Equations (p. 365) 2 Solve Exponential Equations (p. 367) 3 Solve Logarithmic and Exponential Equations Using a Graphing Utility (p. 368)

1 Solve Logarithmic Equations In Section 4.4 logarithmic equations were solved by changing a logarithmic expression to an exponential expression. That is, they were solved by using the definition of a logarithm: y = log a x is equivalent to x = ay

a 7 0, a ≠ 1

For example, to solve the equation log 2 11 - 2x2 = 3, write the logarithmic equation as an equivalent exponential equation 1 - 2x = 23 and solve for x. log 2 11 - 2x2 = 3

1 - 2x = 23 - 2x =  x = -

Change to an exponential equation. Simplify.

 2

Solve.

 Check: log 2 11 - 2x2 = log 2 a1 - 2a - b b = log 2 11 + 2 = log 2  = 3 2

23 = 

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EXAM PL E 1

M, N, and a are positive and a ≠ 1.

Solving a Logarithmic Equation Solve: 2 log 5 x = log 5 9

Solution

The domain of the variable in this equation is x 7 0. Note that each logarithm is to the same base, 5. Then find the exact solution as follows: 2 log 5 x = log 5 9 loga M r = r loga M log 5 x2 = log 5 9 If loga M = loga N, then M = N. x2 = 9 x = 3 or x = - 3 3FDBMM
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366

CHAPTER 4  Exponential and Logarithmic Functions

Check: 2 log 5 3 ≟ log 5 9 log 5 32 ≟ log 5 9 log 5 9 = log 5 9

r loga M = loga M r

The solution set is 5 36 .

Now Work

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PROBLEM

13

Often one or more properties of logarithms are needed to rewrite the equation as a single logarithm. In the next example, the log of a product property is used.

EX A MPL E 2

Solution

WARNING A negative solution is not automatically extraneous. Check whether the potential solution causes the argument of any logarithmic expression in the equation to be negative. j

Solving a Logarithmic Equation

Solve: log 5 1x + 62 + log 5 1x + 22 = 1 The domain of the variable requires that x + 6 7 0 and x + 2 7 0, so x 7 - 6 and x 7 - 2. This means any solution must satisfy x 7 - 2. To obtain an exact solution, express the left side as a single logarithm. Then change the equation to an equivalent exponential equation. log 5 1x + 62 + log 5 1x + 22 log 5 3 1x + 62 1x + 22 4 1x + 62 1x + 22 x2 + x + 12 x2 + x +  1x + 2 1x + 12 x = -  or x

1 1 51 5 0 0 -1

loga M + loga N = loga(MN) Change to an exponential equation. Simplify. Place the quadratic equation in standard form. Factor. Zero-Product Property

Only x = - 1 satisfies the restriction that x 7 - 2, so x = -  is extraneous. The solution set is 5 - 16 , which you should check.

Now Work

EX A MPL E 3

= = = = = = =

PROBLEM

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21

Solving a Logarithmic Equation Solve: ln x = ln x + 6 - ln x - 4

Solution

The domain of the variable requires that x 7 0, x + 6 7 0, and x - 4 7 0. As a result, the domain of the variable here is x 7 4. Begin the solution using the log of a difference property. ln x = ln x + 6 - ln x - 4

x + 6 ln x = ln a b x - 4 x + 6 x = x - 4 x x - 4 = x + 6 x2 - 4x = x + 6 x2 - 5x - 6 = 0 x - 6 x + 1 = 0 x = 6 or x = - 1

M In M - ln N = ln a b N If ln M = ln N, then M = N. Multiply both sides by x - 4. Simplify. Place the quadratic equation in standard form. Factor. Zero-Product Property

Because the domain of the variable is x 7 4, discard - 1 as extraneous. The solution set is {6}, which you should check.

Now Work

PROBLEM

31

r

SECTION 4.6   Logarithmic and Exponential Equations

367

2 Solve Exponential Equations In Sections 4.3 and 4.4, exponential equations were solved algebraically by expressing each side of the equation using the same base. That is, they were solved by using the one-to-one property of the exponential function: If au = av, then u = v

a 7 0, a ≠ 1

For example, to solve the exponential equation 42x + 1 = 16, notice that 16 = 42 and apply the property above to obtain the equation 2x + 1 = 2, and the solution is 1 x = . 2 Not all exponential equations can be readily expressed so that each side of the equation has the same base. For such equations, algebraic techniques often can be used to obtain exact solutions. In the next example two exponential equations are solved by changing the exponential expression to a logarithmic expression.

EXAMPL E 4

Solving Exponential Equations 4PMWF
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Solution

C
 # 3x = 5

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22 = 4 and 23 =  , write the exponential equation as the equivalent logarithmic equation. 2x = 5 x = log 2 5 =

c

ln 5 ln 2

Change@of@Base Formula (9), Section 4 .5

Alternatively, the equation 2x = 5 can be solved by taking the natural logarithm PS
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 # 3x = 5 5 3x = 

Exact solution Approximate solution

ln 5 f. ln 2

5 ln a b  ln 3

Exact solution

≈ - 0.42

Now Work

In M r = r ln M

Solve for 3x.

5 x = log 3 a b = 

The solution set is

If M = N, then ln M = ln N.

W

Approximate solution

5 ln a b  ¶ . ln 3

PROBLEM

43

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368

CHAPTER 4  Exponential and Logarithmic Functions

EX A MPL E 5

Solving an Exponential Equation Solve: 5x - 2 = 33x + 2

Solution

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in x that can be solved. 5x - 2 = 33x + 2 ln 5x - 2 = ln 33x + 2

If M = N, ln M = ln N.

1x - 22 ln 5 = 13x + 22 ln 3

1ln 52 x - 2 ln 5 = 13 ln 32x + 2 ln 3

NOTE: Because of the properties of logarithms, exact solutions involving logarithms often can be expressed in multiple ways. For example, the solution to 5x - 2 = 33x + 2 from Example 5 can be expressed equivalently 2 ln 15 ln 225 as or among others. ln 5 - ln 27 ln( 5/27) j Do you see why?

1ln 52x - 13 ln 32x = 2 ln 3 + 2 ln 5

1ln 5 - 3 ln 32x = 21ln 3 + ln 52 21ln 3 + ln 52 x = ln 5 - 3 ln 3 ≈ - 3.212

The solution set is e

Now Work

ln M r = r ln M Distribute. Place terms involving x on the left. Factor. Exact solution Approximate solution

21ln 3 + ln 52 f. ln 5 - 3 ln 3

PROBLEM

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53

The next example deals with an exponential equation that is quadratic in form.

EX A MPL E 6

Solving an Exponential Equation That Is Quadratic in Form Solve: 4x - 2x - 12 = 0

Solution

Note that 4x = 22 x = 2 2x = 2x 2, so the equation is quadratic in form and can be written as 2x 2 - 2x - 12 = 0

Let u = 2x; then u 2 - u - 12 = 0.

Now factor as usual. 12x - 42 12x + 32 = 0 2 - 4 = 0 or 2x + 3 = 0 2x = 4 2x = - 3 x

(u - 4)(u + 3) = 0 u - 4 = 0 or u + 3 = 0 u = 2x = 4

u = 2x = - 3

The equation on the left has the solution x = 2, since 2x = 4 = 22; the equation on the right has no solution, since 2x 7 0 for all x. The only solution is 2. The solution set is 5 26 .

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PROBLEM

61

3 Solve Logarithmic and Exponential Equations Using a Graphing Utility The algebraic techniques introduced in this section to obtain exact solutions apply only to certain types of logarithmic and exponential equations. Solutions for other types are usually studied in calculus, using numerical methods. For such types, a graphing utility can be used to approximate the solution.

369

SECTION 4.6   Logarithmic and Exponential Equations

Solving Equations Using a Graphing Utility

EXAM PL E 7

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4

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1

PROBLEM

71

0

4.6 Assess Your Understanding ‘Are You Prepared?’

Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 5 - 3, 106

1. Solve x2 - x - 30 = 0. QQ
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Skill Building In Problems 5–40, solve each logarithmic equation. Express any irrational solution in exact form and as a decimal rounded to three decimal places. 16 5. log 4 x = 2 {16} 6. log 1x + 62 = 1 {4} 7. log 2 15x2 = 4 e f 5 10 8. log 3 13x - 12 = 2 e f 9. log 4 1x + 22 = log 4  {6} 10. log 5 12x + 32 = log 5 3 {0} 3 1 1 1 11. log 3 x = 2 log 3 2 {16} 12. - 2 log 4 x = log 4 9 e f 13. 3 log 2 x = - log 2 2 e f 2 3 3 14. 2 log 5 x = 3 log 5 4 {} 15. 3 log 2 1x - 12 + log 2 4 = 5 {3} 16. 2 log 3 1x + 42 - log 3 9 = 2 {5} 21 17. log x + log1x + 152 = 2 {5} 18. log x + log 1x - 212 = 2 {25} 19. log12x + 12 = 1 + log1x - 22 e f  15 21. log 2 1x + 2 + log 2 1x + 2 = 1 { - 6} 22. log 6 1x + 42 + log 6 1x + 32 = 1 { - 1} 20. log12x2 - log1x - 32 = 1 e f 4 23. log  1x + 62 = 1 - log  1x + 42 { - 2} 24. log 5 1x + 32 = 1 - log 5 1x - 12 5 26 *25. ln x + ln1x + 22 = 4 *27. log 3 x + 1 + log 3 1x + 42 = 2

*26. ln1x + 12 - ln x = 2

29. log 1>3 1x2 + x2 - log 1>3 1x2 - x2 = - 1 5 26

*28. log 2 x + 1 + log 2 1x + 2 = 3

30. log 4 1x2 - 92 - log 4 1x + 32 = 3

{6} 9 31. log a 1x - 12 - log a 1x + 62 = log a 1x - 22 - log a 1x + 32 e f 32. log a x + log a 1x - 22 = log a 1x + 42 {4} 2 33. 2 log 5 x - 3 - log 5  = log 5 2 \^ 34. log 3 x - 2 log 3 5 = log 3 x + 1 - 2 log 3 10 35. 2 log 6 x + 2 = 3 log 6 2 + log 6 4

5 - 2 + 4226

3 36. 3 log  x - log  2 = 2 log  4 5 42 26 1 3 38. log x - 1 = log 2 5 1 + 226 3 40. ln x - 32 ln x + 2 = 0 5e, e 4 6

37. 2 log 13 x + 2 = log 13 4x +  5 - 23, 236 1 39. log 3 x 2 - 5 log 3 x = 6 e , 29 f 3

1 e f 3

In Problems 41–68, solve each exponential equation. Express any irrational solution in exact form and as a decimal rounded to three decimal places. 41. 2x - 5 = 

42. 5-x = 25

{}

*45. -x = 1.2 53. 1.2x = 10.52 -x

{0}

*61. 16 + 4

- 3 = 0

*65. 3 # 4x + 4 # 2x +  = 0

58. 32x + 3x - 2 = 0 x+1

*62. 9 - 3

*48. 0.3140.2x 2 = 0.2 4 1-x *52. a b = 5x 3 *56. e x + 3 = px

x

3 *51. a b = 1 - x 5 *55. p1 - x = e x

*54. 0.31 + x = 1.2x - 1 x

*44. 3x = 14

*47. 5123x 2 = 

*50. 2x + 1 = 51 - 2x

*57. 22x + 2x - 12 = 0 x+1

*43. 2x = 10

*46. 2-x = 1.5

*49. 31 - 2x = 4x

x

{ - 2}

+ 1 = 0

{0}

*66. 2 # 49x + 11 # x + 5 = 0

59. 32x + 3x + 1 - 4 = 0

*63. 25 -  # 5 = - 16 x

x

*67. 4x - 10 # 4-x = 3

{0}

60. 22x + 2x + 2 - 12 = 0

*64. 36x - 6 # 6x = - 9

*68. 3x - 14 # 3-x = 5

In Problems 69–82, use a graphing utility to solve each equation. Express your answer rounded to two decimal places. 69. log 5 1x + 12 - log 4 1x - 22 = 1

71. e x = - x

{ - 0.5}

{2.9}

72. e 2x = x + 2

70. log 2 1x - 12 - log 6 1x + 22 = 2

{ - 1.9, 0.45} 73. e x = x2

{ - 0.0}

{12.15}

74. e x = x3

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{1.6, 4.54}

{1}

370

CHAPTER 4  Exponential and Logarithmic Functions

75. ln x = - x

{0.5}

79. e x + ln x = 4

{1.32}

76. ln12x2 = - x + 2 80. e x - ln x = 4

{1.16}

77. ln x = x3 - 1

{0.05, 1.4} 81. e -x = ln x

{0.39, 1}

{1.31}

78. ln x = - x2 82. e -x = - ln x

{0.65} {0.5}

Mixed Practice In Problems 83–96, solve each equation. Express irrational solutions in exact form and as a decimal rounded to three decimal places. 83. log 9 1x - 52 = log 3 1x + 12 {2, 3} [Hint: Change log 9 1x - 52 to base 3.]

84. log 2 1x + 12 - log 4 x = 1

86. log 16 x + log 4 x + log 2 x = 

87. log 9 x + 3 log 3 x = 14 {1} 1 90. log 2 xlog2 x = 4 e 4, f 4

{1} * 85. log 2 13x + 22 - log 4 x = 3

1 e , 4f 88. 21log 4x2 2 + 3log x = log 2 16 16 2 e x + e -x 2 3 e - 1, f 91. = 1 {0} 89. 1 222 2 - x = 2x 3 2 [Hint:
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e x.] {ln13 + 2222, ln 3 - 222 } e x - e -x e x + e -x e x - e -x 92. = 3 93. = 2 94. = -2 ≈{1.63, - 1.63} 2 2 2 {ln1 - 2 + 252} ≈ { - 1.444} {ln12 + 252} ≈ {1.444} 96. log 2 x + log 6 x = 3 *95. log 5 x + log 3 x = 1 # 3 ln 2 ln 6 [Hint:
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f 1x2 = 3x and g x = 10 on the same Cartesian 97. f 1x2 = log 2 1x + 32 and g1x2 = log 2 13x + 12. B
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Applications and Extensions 107. A Population Model
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continues, the model P 1t = 31611.00 t - 2013 represents the population P
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 Source: U.S. Census Bureau

SECTION 4.7   Financial Models

371

108. A Population Model The population of the world in 2013 XBT

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2062 Source: U.S. Census Bureau 109. Depreciation The value V of a Chevy Cruze LT that is t years old can be modeled by V 1t2 = 1,0010.42 t. B
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110. Depreciation The value V of a Honda Civic LX that is t years old can be modeled by V 1t2 = 1,95510.9052 t. B
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After 9.3 years Source: Kelley Blue Book

Discussion and Writing 111. Fill in a reason for each step in the following two solutions. Solve: log 3 1x - 12 2 = 2

Solution A

Solution B

log 3 1x - 1 2 = 2

1x - 12 2 = 32 = 9 ______ 1x - 12 = { 3 ______ x - 1 = - 3 or x - 1 = 3 ______

log 3 1x - 12 2 = 2 2 log 3 1x - 12 = 2 ______ log 3 1x - 12 = 1 ______

x - 1 = 31 = 3 ______ x = - 2 or x = 4 ______ x = 4 ______ Both solutions given in Solution A check. Explain what caused the solution x = - 2 to be lost in Solution B.

Retain Your Knowledge Problems 112–115 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. x + 5 x and g x = , find f ∘ g. Then find *112. Solve: 4x3 + 3x2 - 25x + 6 = 0. *114. For f x = x - 2 x - 3 113. Find the domain of the domain of f ∘ g. f 1x2 = 2x + 3 + 2x - 1. {x x Ú 1} 115. %FUFSNJOF
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{ 10, - 42, 12, - 22, 14, 02, 16, 22} is one-to-one. one-to-one

‘Are You Prepared?’ Answers 1. { - 3, 10}

2. { - 2, 0}

3. { - 1.43}

4. { - 1.}

4.7 Financial Models PREPARING FOR THIS SECTION Before getting started, review the following: r
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Now Work the ‘Are You Prepared?’ problems on page 381.

OBJECTIVES 1 2 3 4

Determine the Future Value of a Lump Sum of Money (p. 372) Calculate Effective Rates of Return (p. 375) Determine the Present Value of a Lump Sum of Money (p. 376) Determine the Rate of Interest or the Time Required to Double a Lump Sum of Money (p. 377)

372

CHAPTER 4  Exponential and Logarithmic Functions

1 Determine the Future Value of a Lump Sum of Money * *OUFSFTU
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THEOREM

Simple Interest Formula If a principal of P dollars is borrowed for a period of t years at a per annum interest rate r, expressed as a decimal, the interest I charged is I = Prt

(1)

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simple interest. In problems involving interest, the term payment period is defined as follows: Annually: Semiannually: Quarterly:

Once per year Twice per year Four times per year

Monthly: 12 times per year Daily: 365 times per year*

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EX A MPL E 1

Computing Compound Interest A credit union pays interest of 2% per annum compounded quarterly on a certain savings plan. If $1000 is deposited in such a plan and the interest is left to accumulate, IPX
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I = Prt. The principal P is $1000 and the rate of 1 interest is 2, = 0.02. After the first quarter of a year, the time t is year, so the 4 interest earned is 1 I = Prt = 1$10002 10.022 a b = $5 4 The new principal is P + I = $1000 + $5 = $1005. At the end of the second quarter, the interest on this principal is 1 I = 1$10052 10.022 a b = $5.03 4 At the end of the third quarter, the interest on the new principal of $1005 + $5.03 = $1010.03 is 1 I = 1$1010.032 10.022 a b = $5.05 4 Finally, after the fourth quarter, the interest is 1 I = 1$1015.02 10.022 a b = $5.0 4 After 1 year the account contains $1015.0 + $5.0 = $1020.16. The pattern of the calculations performed in Example 1 leads to a general formula for compound interest. For this purpose, let P represent the principal to be invested at a per annum interest rate r that is compounded n times per year, so the time of each 1 compounding period is years.
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SECTION 4.7   Financial Models

Interest = principal * rate * time = P # r

#

1 = P n

#

373

r a b n

The amount A after one compounding period is A = P + P

#

r a b = P n

#

a1 +

r b n

After two compounding periods, the amount A, based on the new principal r P # a1 + b , is n A = P # a1 +

r r r r 2 r r b + P # a1 + b a b = P # a1 + b a1 + b = P # a1 + b n n n n n n

5

5 New principal

c

Factor out P # 11 + nr 2

Interest on new principal

After three compounding periods, the amount A is A = P

#

a1 +

r 2 b + P n

#

r 2 r b a b = P n n

a1 +

#

a1 +

r 2 b n

#

a1 +

r b = P n

#

a1 +

r 3 b n

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Because t years will contain n # t compounding periods, the amount after t years is A = P

THEOREM

Exploration To see the effects of compounding interest monthly on an initial deposit r 12x b with of $1, graph Y1 = a1 + 12 r = 0.06 and r = 0.12 for 0 … x … 30. What is the future value of $1 in 30 years when the interest rate per annum is r = 0.06 (6%)? What is the future value of $1 in 30 years when the interest rate per annum is r = 0.12 (12%)? Does doubling the interest rate double the future value?

NOTE The results shown here differ from those in Example 1 due to rounding. j

EXAM PL E 2

#

a1 +

r nt b n

Compound Interest Formula The amount A after t years due to a principal P invested at an annual interest rate r compounded n times per year is A = P

#

a1 +

r nt b n

(2)

For example, to rework Example 1, use P = $1000, r = 0.02, n = 4
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t = 1 year to obtain

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r nt 0.02 4 1 b A = $1020.15 b = 1000a1 + n 4 *O
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A is typically referred to as the future value of the account, and P is called the present value.

Now Work

PROBLEM

7

Comparing Investments Using Different Compounding Periods Investing $1000 at an annual rate of 10% compounded annually, semiannually, quarterly, monthly, and daily will yield the following amounts after 1 year: Annual compounding 1n = 12:

Semiannual compounding 1n = 22: 2VBSUFSMZ
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1n = 42:

A = P # 11 + r2 = 1$10002 11 + 0.102 = $1100.00 r 2 A = P # a1 + b 2 = 1$10002 11 + 0.052 2 = $1102.50 r 4 A = P # a1 + b 4 = 1$10002 11 + 0.0252 4 = $1103.1

374

CHAPTER 4  Exponential and Logarithmic Functions

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A = P

#

a1 +

r 12 b 12

= 1$10002 a1 + %BJMZ
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A = P

#

a1 +

0.10 12 b = $1104.1 12

r 365 b 365

= 1$10002 a1 +

0.10 365 b = $1105.16 365

r From Example 2, note that the effect of compounding more frequently is that the amount after 1 year is higher: $1000 compounded 4 times a year at 10% results JO


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CPVOE Let’s find the answer. Suppose that P is the principal, r is the per annum interest rate, and n is the number of times that the interest is compounded each year. The amount after 1 year is A = P

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Table 8

(1  nr )n n = 100

n = 1000

n = 10,000

er

r = 0.05

1.0512580

1.0512698

1.0512710

1.0512711

r = 0.10

1.1051157

1.1051654

1.1051704

1.1051709

r = 0.15

1.1617037

1.1618212

1.1618329

1.1618342

r = 1

2.7048138

2.7169239

2.7181459

2.7182818

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Pe r, the interest is said to be compounded continuously.

THEOREM

Continuous Compounding The amount A after t years due to a principal P invested at an annual interest rate r compounded continuously is A = Pe rt

(4)

SECTION 4.7   Financial Models

EXAMPL E 3

375

Using Continuous Compounding The amount A that results from investing a principal P of $1000 at an annual rate r of 10% compounded continuously for a time t of 1 year is A = $1000e 0.10 = 1$10002 11.10512 = $1105.1

Now Work

PROBLEM

r

13

2 Calculate Effective Rates of Return Suppose that you have $1000 and a bank offers to pay 3% annual interest on a TBWJOHT
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be earned for you to have the same amount at the end of the year as if the interest IBE
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determine the value of the $1000 in the account that earns 3% compounded monthly. A = +1000a1 +

0.03 12 b 12

r n Use A = P a1 + b with P = $1000, r = 0.03, n = 12. n

= $1030.42 4P
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I = Prt with t = 1, I = +30.42, and P = $1000, the annual simple interest rate is 0.03042 = 3.042,. This interest rate is known as the effective rate of interest. The effective rate of interest is the equivalent annual simple interest rate that would yield the same amount as compounding n times per year, or continuously, after 1 year.

THEOREM

Effective Rate of Interest The effective rate of interest re of an investment earning an annual interest rate r is given by r n Compounding n times per year: re = a1 + b - 1 n r Continuous compounding: re = e - 1

EXAMPL E 4

Computing the Effective Rate of Interest—Which Is the Best Deal? 4VQQPTF
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offering the best deal.

Solution

The bank that offers the best deal is the one with the highest effective interest rate. American Express

First Internet Bank

0.022 4 0.0215 12 b - 1 re = a1 + b - 1 re = a1 + 4 12 ≈ 1.0211 - 1 ≈ 1.0221 - 1

Discover

0.0212 365 b - 1 365 ≈ 1.02143 - 1

re = a1 +

= 0.0211

= 0.0221

= 0.02143

= 2.11,

= 2.21,

= 2.143,

The effective rate of interest is highest for First Internet Bank, so First Internet Bank is offering the best deal.

r

Now Work

PROBLEM

23

376

CHAPTER 4  Exponential and Logarithmic Functions

3 Determine the Present Value of a Lump Sum of Money 8IFO
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referring to the present value of money. The present value of A dollars to be received at a future date is the principal that you would need to invest now so that it will grow to A dollars in the specified time period. The present value of money to be received at a future date is always less than the amount to be received, since the amount to CF
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A = P

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To solve for P, divide both sides by a1 + A a1 +

THEOREM

r nt b n

r nt b n

r nt b . The result is n

= P or P = A # a1 +

r -nt b n

Present Value Formulas The present value P of A dollars to be received after t years, assuming a per annum interest rate r compounded n times per year, is P = A

#

a1 +

r -nt b n

(5)

If the interest is compounded continuously, then P = Ae -rt

(6)

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EX AMPL E 5

Computing the Value of a Zero-Coupon Bond "
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How much should you be willing to pay for it now if you want a return of B

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r -nt 0.0 -121102 = $1000a1 + b = $450.52 b n 12

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PROBLEM

15

r

SECTION 4.7   Financial Models

377

4 Determine the Rate of Interest or the Time Required to Double a Lump Sum of Money EXAM PL E 6

Rate of Interest Required to Double an Investment 8IBU
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interest formula with n = 1 and t = 5 to find r.

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r nt b n 11 + r2 5 a1 +

2 = 11 + r2

5

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5

1 + r = 22

Take the fifth root of each side.

5

r = 22 - 1 ≈ 1.1469 - 1 = 0.1469 5IF
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EXAM PL E 7

PROBLEM

r

31

Time Required to Double or Triple an Investment B
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P is the initial investment and P is to double, the amount A will be 2P.
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It will take about 22 years to triple the investment.

Now Work

PROBLEM

35

r

378

CHAPTER 4  Exponential and Logarithmic Functions

4.7 Assess Your Understanding ‘Are You Prepared?’

Answers are given at the end of these exercises. If you get a wrong answer, read the page listed in red.

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"m" $15

2. If you borrow $5000 and, after 9 months, pay off the loan in the amount of $5500, what per annum rate of interest was DIBSHFE
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Concepts and Vocabulary 3. 5IF
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a bank in the form of a loan or by a bank from an individual JO
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principal.

5. In working problems involving interest, if the payment period of the interest is quarterly, then interest is paid 4 times per year.

4. If a principal of P dollars is borrowed for a period of t years at a per annum interest rate r, expressed as a decimal, the interest I charged is I = Prt . Interest charged according to this formula is called simple interest.

6. The effective rate of interest is the equivalent annual simple interest rate that would yield the same amount as compounding n times per year, or continuously, after 1 year.

Skill Building In Problems 7–14, find the amount that results from each investment. 7. $100 invested at 4% compounded quarterly after a period of 2 years 

11. $600 invested at 5% compounded daily after a period of 3 years 

8. $50 invested at 6% compounded monthly after a period of 3 years 

12. 
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2 years 

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13. $1000 invested at 11% compounded continuously after a period of 2 years  14. 
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period of 3 years 

In Problems 15–22, find the principal needed now to get each amount; that is, find the present value. 15. To get $100 after 2 years at 6% compounded monthly 

19. To get $600 after quarterly $554.09

2

years

at

4%

compounded

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25. For 5% compounded continuously  26. For 6% compounded continuously 

In Problems 27–30, determine the rate that represents the better deal. 1 *27. 6% compounded quarterly or 6 , compounded annually 4 1 *28. 9% compounded quarterly or 9 , compounded annually 4 *29. 
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35.* B

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SECTION 4.7   Financial Models

379

Applications and Extensions *51. Comparing IRA Investments
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39. Time Required to Reach a Goal If Tanisha has $100 to JOWFTU
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she can take advantage of this investment opportunity or to CF
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Assume a rate of interest of 6% compounded continuously. *42. Time Required to Reach a Goal How many years will it UBLF
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year. This was a 4.2% increase from the previous year. Source: The College Board B
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individuals to put money aside now to help pay for college later. If one such plan offers a rate of 4% compounded continuously, how much should be put in a college savings plan in 2015 to pay for 1 year of the cost of college at a 4-year private college for an incoming GSFTINBO
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$33,441

43. Price Appreciation of Homes
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*54. Analyzing Interest Rates on a Mortgage Colleen and Bill have just purchased a house for $650,000, with the seller holding a second mortgage of $100,000. They promise to pay the seller $100,000 plus all accrued interest 5 years from now. The seller offers them three interest options on the second mortgage: B
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46. Paying off a Loan John requires $3000 in 6 months to pay off a loan that has no prepayment privileges. If he has the $3000 now, how much of it should he save in an account paying 3% compounded monthly so that in 6 months he will IBWF
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$2955.39 47. Return on a Stock George contemplates the purchase of 100 shares of a stock selling for $15 per share. The stock pays no dividends. The history of the stock indicates that it should grow at an annual rate of 15% per year. How much should UIF

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55. 2009 Federal Stimulus Package
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≈$1020 billion; $233 billion Source: U.S. Treasury Department

*49. Comparing Savings Plans Jim places $1000 in a bank account that pays 5.6% compounded continuously. After 1 year, will he have enough money to buy a computer TZTUFN
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56. Per Capita Federal Debt In 2013, the federal debt was about 
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Inflation Problems 57–62 require the following discussion. Inflation is a term used to describe the erosion of the purchasing power of money. For example, if the annual inflation rate is 3%, then $1000 worth of purchasing power now will have only $970 worth of purchasing power in 1 year because 3% of the original $1000 (0.03 * 1000 = 30) has been eroded due to inflation. In general, if the rate of inflation averages r per annum over n years, the amount A that $P will purchase after n years is A = P where r is expressed as a decimal.

#

11 - r2 n

380

CHAPTER 4  Exponential and Logarithmic Functions

57. Inflation If the inflation rate averages 3%, how much will 
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60. Inflation If the amount that $1000 will purchase is only $930 BGUFS

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58. Inflation If the inflation rate averages 2%, how much will 
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Problems 63–66 involve zero-coupon bonds. A zero-coupon bond is a bond that is sold now at a discount and will pay its face value at the time when it matures; no interest payments are made. 63. Zero-Coupon Bonds A zero-coupon bond can be redeemed in 20 years for $10,000. How much should you be willing to pay for it now if you want a return of: B

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 67. Time to Double or Triple an Investment The formula ln m

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Problems 69–72 require the following discussion. The Consumer Price Index (CPI) indicates the relative change in price over time for a fixed basket of goods and services. It is a cost-of-living index that helps measure the effect of inflation on the cost of goods and services. The CPI uses the base period 1982–1984 for comparison (the CPI for this period is 100). The CPI for January 2013 was 230.28. This means that $100 in the period 1982–1984 had the same purchasing power as $230.28 in January 2013. In general, if the rate of inflation averages r percent per annum over n years, then the CPI index after n years is C1I = C1I 0 a1 +

r n b 100

where CPI0 is the CPI index at the beginning of the n-year period. Source: U.S. Bureau of Labor Statistics 69. Consumer Price Index B
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Discussion and Writing 73. Explain in your own words what the term compound interest NFBOT
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continuous compounding
NFBO 74. Explain in your own words the meaning of present value. 75. Critical Thinking You have just contracted to buy a house and will seek financing in the amount of $100,000. You go to several banks. Bank 1 will lend you $100,000

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origination fee of 1.5%. Bank 3 will lend you $100,000 at the rate of 9.125% amortized over 30 years with no loan origination fee. Bank 4 will lend you $100,000 at the rate of 
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SECTION 4.8   Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models

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table to assist you. If the amount of the monthly payment EJE
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have sound reasons for your choice. Compare your final EFDJTJPO
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381

Monthly Payment

Loan Origination Fee

Bank 1

$786.70

$1,750.00

Bank 2

$977.42

$1,500.00

Bank 3

$813.63

$0.00

Bank 4

$992.08

$0.00

Retain Your Knowledge Problems 76–79 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. x 76. Find the remainder R when f x = 6x3 + 3x2 + 2x - 11 is *78. The function f x = is one-to-one. Find f -1. x 2 divided by g x = x - 1. Is g a factor of f

R = 0; yes 79. Solve: log 2 x + 3 = 2 log 2 x - 3 {6} *77. Find the real zeros of f x = x5 - x4 - 15x3 - 21x2 - 16x - 20. Then write f in factored form.

‘Are You Prepared?’ Answers 1 2. 13 % 3

1. $15

4.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models OBJECTIVES 1 Find Equations of Populations That Obey the Law of Uninhibited Growth (p. 381) 2 Find Equations of Populations That Obey the Law of Decay (p. 383) 3 Use Newton’s Law of Cooling (p. 384) 4 Use Logistic Models (p. 386)

1 Find Equations of Populations That Obey the Law of Uninhibited Growth Figure 44

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A varies with time t according to the function

A

A 1t2 = A0e kt A0 t (a) A(t ) ⫽ A0 e k t , k ⬎ 0 A A0

t (b) A(t ) ⫽ A0 e kt , k ⬍ 0

(1)

Here A0 is the original amount 1t = 02 and k ≠ 0 is a constant. If k 7 0,
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A is increasing over time; if k 6 0, the amount A is decreasing over time. In either case, when an amount A WBSJFT
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exponential law, or the law of uninhibited growth 1k 7 02 or decay 1k 6 02. See Figure 44. 'PS
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was shown to follow the law of uninhibited growth. In this section, additional phenomena that follow the exponential law will be studied. Cell division is the growth process of many living organisms, such as amoebas, plants, and human skin cells. Based on an ideal situation in which no cells die and no by-products are produced, the number of cells present at a given time follows the law of uninhibited growth. Actually, however, after enough time has passed, growth at an exponential rate will cease due to the influence of factors such as lack of living space and dwindling food supply. The law of uninhibited growth accurately models only the early stages of the cell division process. The cell division process begins with a culture containing N0 cells. Each cell in the culture grows for a certain period of time and then divides into two identical

382

CHAPTER 4  Exponential and Logarithmic Functions

cells. Assume that the time needed for each cell to divide in two is constant and does not change as the number of cells increases. These new cells then grow, and eventually each divides in two, and so on.

Uninhibited Growth of Cells A model that gives the number N of cells in a culture after a time t has passed JO
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k 7 0

(2)

where N0 is the initial number of cells and k is a positive constant that represents the growth rate of the cells. 6TJOH
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positive real numbers, even though the number of cells being counted must be an integer. This is a common practice in many applications.

EX A MPL E 1

Bacterial Growth A colony of bacteria grows according to the law of uninhibited growth according to the function N 1t2 = 100e 0.045t, where N is measured in grams and t is measured in days. B
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Solution

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N0 , is obtained when t = 0, so N0 = N 102 = 100e 0.045102 = 100 grams

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N 1t2 = 100e 0.045t to N 1t2 = N0e kt. The value of k, 0.045, indicates a growth rate of 4.5%. D
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N 152 = 100e 0.045152 ≈ 125.2 grams. E
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equation N 1t2 = 140. 100e 0.045t = 140 e 0.045t = 1.4 0.045t = ln 1.4 ln 1.4 t = 0.045 ≈ .5 days

Divide both sides of the equation by 100. Rewrite as a logarithm. Divide both sides of the equation by 0.045.

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N 1t2 = 200 grams, so the doubling time can be found by solving the equation 200 = 100e 0.045t for t. 200 = 100e 0.045t 2 = e 0.045t ln 2 = 0.045t ln 2 t = 0.045 ≈ 15.4 days

Divide both sides of the equation by 100. Rewrite as a logarithm. Divide both sides of the equation by 0.045.

The population doubles approximately every 15.4 days.

Now Work

PROBLEM

1

r

SECTION 4.8   Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models

EXAMPL E 2

383

Bacterial Growth A colony of bacteria increases according to the law of uninhibited growth. B
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N is the number of cells and t is the time in hours, express N as a function of t. C
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number of cells in the culture. D
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Solution

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N of cells at time t is N 1t2 = N0e kt

where N0 is the initial number of bacteria present and k is a positive number. C
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k, note that the number of cells doubles in 3 hours. Hence But N 132 = N0e

N 132 = 2N0

k132

, so

N0e k132 = 2N0 e 3k = 2 Divide both sides by N0 3k = ln 2 Write the exponential equation as a logarithm. 1 k = ln 2 ≈ 0.23105 3 The function that models this growth process is therefore N 1t2 = N0e 0.23105t D
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t needed for the size of the colony to triple requires that N = 3N0 . Substitute 3N0 for N to get 3N0 = N0e 0.23105t Divide both sides by N0 3 = e 0.23105t Write the exponential equation as a logarithm. 0.23105t = ln 3 ln 3 t = ≈ 4.55 hours 0.23105 *U
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for a total time of 6 hours.

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2 Find Equations of Populations That Obey the Law of Decay 3BEJPBDUJWF
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Uninhibited Radioactive Decay The amount A of a radioactive material present at time t is given by A 1t2 = A0e kt

k 6 0

(3)

where A0 is the original amount of radioactive material and k is a negative number that represents the rate of decay. All radioactive substances have a specific half-life, which is the time required for half of the radioactive substance to decay. Carbon dating uses the fact that all living PSHBOJTNT
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384

CHAPTER 4  Exponential and Logarithmic Functions

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ratio of carbon 12 to carbon 14 is constant. But when an organism dies, the original amount of carbon 12 present remains unchanged, whereas the amount of carbon 14 begins to decrease. This change in the amount of carbon 14 present relative to the amount of carbon 12 present makes it possible to calculate when the organism died.

EX AMPL E 3

Estimating the Age of Ancient Tools Traces of burned wood along with ancient stone tools in an archeological dig in Chile XFSF
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Solution

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A of carbon 14 present at time t is A 1t2 = A0e kt

where A0 is the original amount of carbon 14 present and k
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first seek the number k
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UIF 1 original amount of carbon 14 remains, so A 15302 = A0 . Then 2 1 A = A0e k15302 2 0 1 Divide both sides of the equation by A 0. = e 530k 2 1 Rewrite as a logarithm. 530k = ln 2 1 1 k = ln ≈ - 0.00012096 530 2 'PSNVMB



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If the amount A
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follows that 0.016A0 = A0e - 0.00012096t Divide both sides of the equation by A 0. 0.016 = e - 0.00012096t Rewrite as a logarithm. - 0.00012096t = ln 0.016 ln 0.016 t = ≈ 33,30 years - 0.00012096 5IF
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conclusion to argue that humans lived in the Americas nearly 34,000 years ago, much earlier than is generally accepted.

Now Work

r

PROBLEM

3

3 Use Newton’s Law of Cooling Newton’s Law of Cooling* states that the temperature of a heated object decreases exponentially over time toward the temperature of the surrounding medium.

Newton’s Law of Cooling The temperature u of a heated object at a given time t can be modeled by the following function: u 1t2 = T + 1u 0 - T2e kt

k 6 0

(4)

where T is the constant temperature of the surrounding medium, u 0 is the initial temperature of the heated object, and k is a negative constant. /BNFE
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SECTION 4.8   Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models

EXAM PL E 4

385

Using Newton’s Law of Cooling "O
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whose air temperature is 30°C. B
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T = 30 and u 0 = 100, the temperature u t
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t
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where k is a negative constant. To find k, use the fact that u = 0 when t = 5. Then u 1t2 = 30 + 0e kt 0 = 30 + 0e k152 50 = 0e 5k 50 e 5k = 0 5 5k = ln  1 5 k = ln ≈ - 0.063 5 

u(5) = 80 Simplify. Solve for e 5k. Take ln of both sides. Solve for k.

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(5)

Find t when u = 50°C. 50 = 30 + 0e - 0.063t 20 = 0e - 0.063t 20 e - 0.063t = 0 2 - 0.063t = ln  2 ln  t = ≈ 1.6 minutes - 0.063

Simplify.

Take ln of both sides.

Solve for t.

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36 seconds. C
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t when u = 35°C. 35 = 30 + 0e - 0.063t 5 = 0e - 0.063t 5 e - 0.063t = 0 5 - 0.063t = ln 0 5 ln 0 t = ≈ 39.2 minutes - 0.063

Simplify.

Take ln of both sides.

Solve for t.

The object will reach a temperature of 35°C after about 39.2 minutes. D
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t increases, the exponent - 0.063t becomes unbounded in the negative direction. As a result, the value of e - 0.063t approaches zero, so the

386

CHAPTER 4  Exponential and Logarithmic Functions

value of u, the temperature of the object, approaches 30°C, the air temperature of the room.

r

Now Work

PROBLEM

13

4 Use Logistic Models

The exponential growth model A 1t2 = A0e kt, k 7 0, assumes uninhibited growth, NFBOJOH
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could be modeled using this function, assuming that no cells die and no by-products are produced. However, cell division eventually is limited by factors such as living space and food supply. The logistic model, given next, can describe situations where the growth or decay of the dependent variable is limited.

Logistic Model In a logistic model, the population P after time t is given by the function P1t2 =

c 1 + ae -bt

(6)

where a, b, and c are constants with a 7 0 and c 7 0. The model is a growth model if b 7 0; the model is a decay model if b 6 0. The number c is called the carrying capacity
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value P1t2 approaches c as t approaches infinity; that is, lim P1t2 = c. The number t Sq 0 b 0 is the growth rate for b 7 0 and the decay rate for b 6 0.
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a typical logistic decay function. Figure 45

P(t )

yc

P(t )

yc

(0, P(0))

1– c 2

1– c 2

Inflection point

Inflection point

(0, P(0)) t

t (a)

(b)

From the figures, the following properties of logistic growth functions emerge. Properties of the Logistic Model, Equation (6) 1. The domain is the set of all real numbers. The range is the interval 10, c2, where c is the carrying capacity. 2. There are no x-intercepts; the y-intercept is P102. 3. There are two horizontal asymptotes: y = 0 and y = c. 4. P1t2 is an increasing function if b 7 0 and a decreasing function if b 6 0. 1 5. There is an inflection point where P1t2 equals of the carrying capacity. 2 The inflection point is the point on the graph where the graph changes from being curved upward to being curved downward for growth functions, and the point where the graph changes from being curved downward to being curved upward for decay functions. 6. The graph is smooth and continuous, with no corners or gaps.

SECTION 4.8   Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models

EXAM PL E 5

387

Fruit Fly Population 'SVJU
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population after t days is given by 230 P 1t2 = 1 + 56.5e -0.3t B
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one-half of the carrying capacity.

Solution

230 . The carrying capacity of the half-pint 1 bottle is 230 fruit flies. The growth rate is 0 b 0 = 0 0.3 0 = 3, per day. C
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P102.

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t S q , e -0.3t S 0 and P1t2 S

P102 =

230

1 + 56.5e -0.3102 230 = 1 + 56.5 = 4

Thus, initially, there were 4 fruit flies in the half-pint bottle. D
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P152. P152 =

230

≈ 23 fruit flies 1 + 56.5e -0.3152 After 5 days, there are approximately 23 fruit flies in the bottle. E
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P1t2 = 10. 230 1 + 56.5e -0.3t 230 1.2 0.2 0.0049 ln 10.00492 t

Figure 46

= 10 = = = = = ≈

1011 + 56.5e -0.3t 2 1 + 56.5e -0.3t 56.5e -0.3t e -0.3t - 0.3t 14.4 days

Divide both sides by 180. Subtract 1 from both sides. Divide both sides by 56.5. Rewrite as a logarithmic expression. Divide both sides by - 0.37.

230

250

Y1 5 1 1 56.5e20.37x

Y2 5 115 0

25

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5IF 1 + 56.5e -0.3x population will reach one-half of the carrying capacity in about 10.9 days 
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Look back at Figure 46. Notice the point where the graph reaches 115 fruit flies POFIBMG
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increasing at an increasing rate to increasing at a decreasing rate. For any logistic growth function, when the population reaches one-half the carrying capacity, the population growth starts to slow down.

Now Work

PROBLEM

23

388

CHAPTER 4  Exponential and Logarithmic Functions

EX A MPL E 6

Exploration On the same viewing rectangle, graph Y1 =

500 1 + 24e-0.03x

and Y2 =

500

Wood Products The EFISCEN wood product model classifies wood products according to their life-span. 5IFSF
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Institute, the percentage of remaining wood products after t years for wood products XJUI
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1 + 24e-0.08x

What effect does the growth rate 0 b 0 have on the logistic growth function?

P1t2 =

100.3952 1 + 0.0316e 0.051t

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 E Explain why the numerator given in the model is reasonable.

Solution

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0 b 0 = 0 - 0.051 0 = 5.1,. C
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P1102. 100.3952 ≈ 95.0 P1102 = 1 + 0.0316e 0.0511102 So 95% of long-life-span wood products remain after 10 years. D
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P1t2 = 50. 100.3952 1 + 0.0316e 0.051t 100.3952 2.009 1.009 31.956 ln 131.9562 t

= 50 = = = = = ≈

5011 + 0.0316e 0.051t 2 1 + 0.0316e 0.051t 0.0316e 0.051t e 0.051t 0.051t 59.6 years

Divide both sides by 50. Subtract 1 from both sides. Divide both sides by 0.0316. Rewrite as a logarithmic expression. Divide both sides by 0.0581.

It will take approximately 59.6 years for the percentage of long-life-span wood products remaining to reach 50%. E The numerator of 100.3952 is reasonable because the maximum percentage of wood products remaining that is possible is 100%.

Now Work

r

PROBLEM

27

4.8 Assess Your Understanding Applications and Extensions 1. Growth of an Insect Population The size P of a certain insect population at time t
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26.6 days *Author’s Note: Surprisingly, the chemical formulas for glucose and fructose are the same: This is not a typo.

390

CHAPTER 4  Exponential and Logarithmic Functions

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 Source: Linda Tappin, “Analyzing Data Relating to the Challenger Disaster,” .BUIFNBUJDT
5FBDIFS Vol. 87, No. 6, September 1994, pp. 423–426.

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6

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represents the percentage of households that do not own a personal computer t
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2011 Source: U.S. Department of Commerce 28. Farmers The logistic model W 1t2 =

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SECTION 4.9   Building Exponential, Logarithmic, and Logistic Models from Data

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8IZ Source: U.S. Department of Agriculture 29. Birthdays The logistic model P 1n2 =

113.319

391

models the probability that in a room of n people, no two people share the same birthday. * B
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Retain Your Knowledge Problems 30–33 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. x2 2y 30. Find the equation of the linear function f that passes *32. 8SJUF
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4.9 Building Exponential, Logarithmic, and Logistic Models from Data PREPARING FOR THIS SECTION Before getting started, review the following: r
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1 Build an Exponential Model from Data (p. 392) 2 Build a Logarithmic Model from Data (p. 393) 3 Build a Logistic Model from Data (p. 394)

Finding the linear function of best fit 1y = ax + b2 for a set of data was discussed in Section 2.2. Likewise, finding the quadratic function of best fit 1y = ax2 + bx + c2 and finding the cubic function of best fit 1y = ax3 + bx2 + cx + d2 were discussed in Sections 2.6 and 3.1, respectively. In this section we discuss how to use a graphing utility to find equations of best fit that describe the relation between two variables when the relation is thought to be exponential 1y = abx 2, logarithmic 1y = a + b ln x2 , or logistic c ¢y = ≤. As before, a scatter diagram of the data is drawn to help 1 + ae -bx determine the appropriate model to use. 'JHVSF

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Figure 47 y

y

x

y

y

x

y

x

x

y  ab x, a  0, b  1

y  ab x, 0  b  1, a  0

y  a b In x, a  0, b  0

Exponential

Exponential

Logarithmic

y  a b In x, a  0, b  0 y  Logarithmic

x c

bx , a  0, b  0, c  0

1 ae

Logistic

392

CHAPTER 4  Exponential and Logarithmic Functions

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OFYU
example shows how data can lead to an exponential model.

Table 9

EX A MPL E 1

Year, x

Account Value, y

0

20,000

1

21,516

2

23,355

3

24,885

4

27,484

5

30,053

6

32,622

Solution

Fitting an Exponential Function to Data .BSJBI
EFQPTJUFE
 
JOUP
B
XFMMEJWFSTJGJFE
NVUVBM
GVOE

ZFBST
BHP
5IF
EBUB
JO
5BCMF

SFQSFTFOU
UIF
WBMVF
PG
UIF
BDDPVOU
FBDI
ZFBS
GPS
UIF
MBTU

ZFBST B
6TJOH
B
HSBQIJOH
VUJMJUZ
ESBX
B
TDBUUFS
EJBHSBN
XJUI
ZFBS
BT
UIF
JOEFQFOEFOU
variable. C
6TJOH
B
HSBQIJOH
VUJMJUZ
CVJME
BO
FYQPOFOUJBM
NPEFM
GSPN
UIF
EBUB D
&YQSFTT
UIF
GVODUJPO
GPVOE
JO
QBSU
C
JO
UIF
GPSN
A = A0 e kt. E
(SBQI
UIF
FYQPOFOUJBM
GVODUJPO
GPVOE
JO
QBSU
C
PS
D
PO
UIF
TDBUUFS
EJBHSBN F
6TJOH
UIF
TPMVUJPO
UP
QBSU
C
PS
D


QSFEJDU
UIF
WBMVF
PG
UIF
BDDPVOU
BGUFS

ZFBST G Interpret the value of k
GPVOE
JO
QBSU
D  B
&OUFS
UIF
EBUB
JOUP
UIF
HSBQIJOH
VUJMJUZ
BOE
ESBX
UIF
TDBUUFS
EJBHSBN
BT
TIPXO
JO
'JHVSF
 C
"
HSBQIJOH
VUJMJUZ
àUT
UIF
EBUB
JO
5BCMF

UP
BO
FYQPOFOUJBM
NPEFM
PG
UIF
GPSN
y = abx
VTJOH
UIF
&91POFOUJBM
3&(SFTTJPO
PQUJPO
'JHVSF

TIPXT
UIBU
y = abx = 19,20.4311.05562 x. Notice that 0 r 0 = 0.999, which is close to 1, indicating a good fit. Figure 48

Figure 49

40,000

⫺1

7 0

D
5P
FYQSFTT
y = abx in the form A = A0 e kt, where x = t and y = A, proceed as follows: abx = A0 e kt, x = t If x = t = 0, then a = A0 . This leads to a = A0, bx = e kt t bx = 1e k 2 b = ek x = t x x Because y = ab = 19,20.4311.05562 , this means that a = 19,20.43 and b = 1.0556. a = A0 = 19,20.43 and b = e k = 1.0556 To find k, rewrite e k = 1.0556 as a logarithm to obtain k = ln 11.05562 ≈ 0.0210 As a result, A = A0 e kt = 19,20.43e 0.0210t.

SECTION 4.9   Building Exponential, Logarithmic, and Logistic Models from Data

393

E
4FF
'JHVSF

GPS
UIF
HSBQI
PG
UIF
FYQPOFOUJBM
GVODUJPO
PG
CFTU
àU Figure 50 40,000

⫺1

7 0

F
-FU
t = 10
JO
UIF
GVODUJPO
GPVOE
JO
QBSU
D 
5IF
QSFEJDUFE
WBMVF
PG
UIF
BDDPVOU
after 10 years is A = A0e kt = 19,20.43e 0.02101102 ≈ +45,04 G The value of k = 0.0210 = .210% represents the growth rate of the account. It represents the rate of interest earned, assuming the account is growing continuously.

r

Now Work

PROBLEM

1

2 Build a Logarithmic Model from Data .BOZ
SFMBUJPOT
CFUXFFO
WBSJBCMFT
EP
OPU
GPMMPX
BO
FYQPOFOUJBM
NPEFM
JOTUFBE
UIF
independent variable is related to the dependent variable using a logarithmic model.

Table 10

EXAM PL E 2

Atmospheric Pressure, p

Height, h

760

0

740

0.184

725

0.328

700

0.565

650

1.079

630

1.291

600

1.634

580

1.862

550

2.235

Fitting a Logarithmic Function to Data Jodi, a meteorologist, is interested in finding a function that explains the relation CFUXFFO
UIF
IFJHIU
PG
B
XFBUIFS
CBMMPPO
JO
LJMPNFUFST
BOE
UIF
BUNPTQIFSJD
QSFTTVSF
NFBTVSFE
JO
NJMMJNFUFST
PG
NFSDVSZ
PO
UIF
CBMMPPO
4IF
DPMMFDUT
UIF
EBUB
shown in Table 10. B
6TJOH
B
HSBQIJOH
VUJMJUZ
ESBX
B
TDBUUFS
EJBHSBN
PG
UIF
EBUB
XJUI
BUNPTQIFSJD
pressure as the independent variable. C
*U
JT
LOPXO
UIBU
UIF
SFMBUJPO
CFUXFFO
BUNPTQIFSJD
QSFTTVSF
BOE
IFJHIU
GPMMPXT
B
MPHBSJUINJD
NPEFM
6TJOH
B
HSBQIJOH
VUJMJUZ
CVJME
B
MPHBSJUINJD
NPEFM
GSPN
UIF
data. D
%SBX
UIF
MPHBSJUINJD
GVODUJPO
GPVOE
JO
QBSU
C
PO
UIF
TDBUUFS
EJBHSBN E
6TF
UIF
GVODUJPO
GPVOE
JO
QBSU
C
UP
QSFEJDU
UIF
IFJHIU
PG
UIF
XFBUIFS
CBMMPPO
if the atmospheric pressure is 560 millimeters of mercury.

Solution

Figure 51 2.4

B
&OUFS
UIF
EBUB
JOUP
UIF
HSBQIJOH
VUJMJUZ
BOE
ESBX
UIF
TDBUUFS
EJBHSBN
4FF
Figure 51. C
"
HSBQIJOH
VUJMJUZ
àUT
UIF
EBUB
JO
5BCMF

UP
B
MPHBSJUINJD
GVODUJPO
PG
UIF
GPSN
y = a + b ln x
CZ
VTJOH
UIF
-0(BSJUIN
3&(SFTTJPO
PQUJPO
4FF
'JHVSF

PO
the next page. The logarithmic model from the data is h 1p2 = 45.63 - 6.9025 ln p

525 ⫺0.2

775

where h is the height of the weather balloon and p is the atmospheric pressure. Notice that 0 r 0 is close to 1, indicating a good fit. D
'JHVSF

TIPXT
UIF
HSBQI
PG
h 1p2 = 45.63 - 6.9025 ln p on the scatter diagram.

394

CHAPTER 4  Exponential and Logarithmic Functions

Figure 53

Figure 52

2.4

775

525 ⫺0.2

E
6TJOH
UIF
GVODUJPO
GPVOE
JO
QBSU
C


+PEJ
QSFEJDUT
UIF
IFJHIU
PG
UIF
XFBUIFS
balloon when the atmospheric pressure is 560 to be h 15602 = 45.63 - 6.9025 ln 560 ≈ 2.10 kilometers

Now Work

PROBLEM

r

5

3 Build a Logistic Model from Data Logistic growth models can be used to model situations for which the value of the EFQFOEFOU
WBSJBCMF
JT
MJNJUFE
.BOZ
SFBMXPSME
TJUVBUJPOT
DPOGPSN
UP
UIJT
TDFOBSJP
For example, the population of the human race is limited by the availability of OBUVSBM
SFTPVSDFT
TVDI
BT
GPPE
BOE
TIFMUFS
8IFO
UIF
WBMVF
PG
UIF
EFQFOEFOU
variable is limited, a logistic growth model is often appropriate.

EX A MPL E 3

Fitting a Logistic Function to Data The data in Table 11 represent the amount of yeast biomass in a culture after t hours.

Table 11

700

20 0

Figure 55

Yeast Biomass

Time (hours)

Yeast Biomass

0

9.6

10

513.3

1

18.3

11

559.7

2

29.0

12

594.8

3

47.2

13

629.4

4

71.1

14

640.8

5

119.1

15

651.1

6

174.6

16

655.9

7

257.3

17

659.6

8

350.7

18

661.8

9

441.0

Source:
5PS
$BSMTPO
¾CFS
(FTDIXJOEJHLFJU
VOE
(SÕTTF
EFS
)FGFWFSNFISVOH
JO
8ÛS[F
#JPDIFNJTDIF
;FJUTDISJGU
#E

QQ
m


Figure 54

⫺2

Time (hours)

Solution

B
6TJOH
B
HSBQIJOH
VUJMJUZ
ESBX
B
TDBUUFS
EJBHSBN
PG
UIF
EBUB
XJUI
UJNF
BT
UIF
independent variable. C
6TJOH
B
HSBQIJOH
VUJMJUZ
CVJME
B
MPHJTUJD
NPEFM
GSPN
UIF
EBUB D
6TJOH
B
HSBQIJOH
VUJMJUZ
HSBQI
UIF
GVODUJPO
GPVOE
JO
QBSU
C
PO
UIF
TDBUUFS
diagram. E
8IBU
JT
UIF
QSFEJDUFE
DBSSZJOH
DBQBDJUZ
PG
UIF
DVMUVSF F
6TF
UIF
GVODUJPO
GPVOE
JO
QBSU
C
UP
QSFEJDU
UIF
QPQVMBUJPO
PG
UIF
DVMUVSF
BU
t = 19 hours. B
4FF
'JHVSF

GPS
B
TDBUUFS
EJBHSBN
PG
UIF
EBUB C
"
HSBQIJOH
VUJMJUZ
àUT
B
MPHJTUJD
HSPXUI
NPEFM
PG
UIF
GPSN
y =

c by 1 + ae -bx using the LOGISTIC regression option. See Figure 55. The logistic model from the data is 663.0 y = 1 + 1.6e -0.540x where y is the amount of yeast biomass in the culture and x is the time.

SECTION 4.9   Building Exponential, Logarithmic, and Logistic Models from Data

395

D
4FF
'JHVSF

GPS
UIF
HSBQI
PG
UIF
MPHJTUJD
NPEFM Figure 56 700

⫺2

20 0

E
#BTFE
PO
UIF
MPHJTUJD
HSPXUI
NPEFM
GPVOE
JO
QBSU
C


UIF
DBSSZJOH
DBQBDJUZ
PG
the culture is 663. F
6TJOH
UIF
MPHJTUJD
HSPXUI
NPEFM
GPVOE
JO
QBSU
C


UIF
QSFEJDUFE
BNPVOU
PG
ZFBTU
biomass at t = 19 hours is y =

Now Work

663.0 1 + 1.6e -0.540 19

PROBLEM

≈ 661.5

r

7

4.9 Assess Your Understanding Applications and Extensions 1. Biology A strain of E. coli
#FV
SFD"
JT
QMBDFE
JOUP
a nutrient broth at 30° Celsius and allowed to grow. The following data are collected. Theory states that the number of bacteria in the petri dish will initially grow according to the law of uninhibited growth. The population is measured using an optical device in which the amount of light that passes through the petri dish is measured.

2. Ethanol Production The data in the table below represent FUIBOPM
QSPEVDUJPO
JO
CJMMJPOT
PG
HBMMPOT
JO
UIF
6OJUFE
States from 2000 to 2012.

Year

Ethanol Produced (billion gallons)

2000 (x = 0)

1.6

2007 (x = 7)

Population, y

2001 (x = 1)

1.8

2008 (x = 8)

9.0

0

0.09

2002 (x = 2)

2.1

2009 (x = 9)

10.6

2.5

0.18

2003 (x = 3)

2.8

2010 (x = 10)

13.2

3.5

0.26

2004 (x = 4)

3.4

2011 (x = 11)

13.9

4.5

0.35

2005 (x = 5)

3.9

2012 (x = 12)

13.3

6

0.50

2006 (x = 6)

4.9

Time (hours), x

Source:
%S
1PMMZ
-BWFSZ
+PMJFU
Junior College

* B
%SBX
B
TDBUUFS
EJBHSBN
USFBUJOH
UJNF
BT
UIF
JOEFQFOEFOU
variable. C
6TJOH
B
HSBQIJOH
VUJMJUZ
CVJME
BO
FYQPOFOUJBM
NPEFM
from the data. y = 0.090311.3342 x D
&YQSFTT
UIF
GVODUJPO
GPVOE
JO
QBSU
C
JO
UIF
GPSN
N1t2 = N0 e kt. N t = 0.0903e 0.2915t * E
(SBQI
UIF
FYQPOFOUJBM
GVODUJPO
GPVOE
JO
QBSU
C
PS
D
on the scatter diagram. F
6TF
UIF
FYQPOFOUJBM
GVODUJPO
GSPN
QBSU
C
PS
D
UP
predict the population at x =  hours. 0.69 G
6TF
UIF
FYQPOFOUJBM
GVODUJPO
GSPN
QBSU
C
PS
D
UP
QSFEJDU
XIFO
UIF
QPQVMBUJPO
XJMM
SFBDI


I

Year

Ethanol Produced (billion gallons) 6.5

Source:
3FOFXBCMF
'VFMT
"TTPDJBUJPO


* B
6TJOH
B
HSBQIJOH
VUJMJUZ
ESBX
B
TDBUUFS
EJBHSBN
PG
the data using 0 for 2000, 1 for 2001, and so on as the independent variable. C
6TJOH
B
HSBQIJOH
VUJMJUZ
CVJME
BO
FYQPOFOUJBM
NPEFM
from the data. y = 1.5256 1.2236 x D
&YQSFTT
UIF
GVODUJPO
GPVOE
JO
QBSU
C
JO
UIF
GPSN
A t = A0e kt. A t = 1.5256e 0.201t * E
(SBQI
UIF
FYQPOFOUJBM
GVODUJPO
GPVOE
JO
QBSU
C
PS
D
on the scatter diagram F
6TF
UIF
NPEFM
UP
QSFEJDU
UIF
BNPVOU
PG
FUIBOPM
UIBU
XJMM
be produced in 2015. 31.5 billion gallons * G Interpret the meaning of k in the function found in QBSU
D 

%VF
UP
TQBDF
SFTUSJDUJPOT
BOTXFST
UP
UIFTF
FYFSDJTFT
NBZ
CF
GPVOE
JO
UIF
"OTXFST
JO
UIF
CBDL
PG
UIF
CPPL

396

CHAPTER 4  Exponential and Logarithmic Functions

3. Advanced-Stage Breast Cancer The data in the table below represent the percentage of patients who have survived after diagnosis of advanced-stage breast cancer at 6-month intervals of time. Time after Diagnosis (years)

5. Milk Production The data in the table below represent the OVNCFS
PG
EBJSZ
GBSNT
JO
UIPVTBOET
BOE
UIF
BNPVOU
PG
NJML
QSPEVDFE
JO
CJMMJPOT
PG
QPVOET
JO
UIF
6OJUFE
4UBUFT
for various years.

Percentage Surviving

Dairy Farms (thousands)

Milk Produced (billion pounds)

0.5

95.7

Year

1

83.6

1980

334

128

1.5

74.0

1985

269

143

2

58.6

1990

193

148

2.5

47.4

1995

140

155

3

41.9

2000

105

167

33.6

2005

78

177

2010

63

193

3.5 Source: Cancer Treatment Centers of America

Source: Statistical Abstract of the United States, 2012 * B
6TJOH
B
HSBQIJOH
VUJMJUZ
ESBX
B
TDBUUFS
EJBHSBN
PG
UIF
data, with time after diagnosis as the independent variable. C
6TJOH
B
HSBQIJOH
VUJMJUZ
CVJME
BO
FYQPOFOUJBM
NPEFM
from the data. y = 11.226 0.013 x D
&YQSFTT
UIF
GVODUJPO
GPVOE
JO
QBSU
C
JO
UIF
GPSN
A t = A0e kt. A t = 11.226e -0354t * E
(SBQI
UIF
FYQPOFOUJBM
GVODUJPO
GPVOE
JO
QBSU
C
PS
D
on the scatter diagram. F
8IBU
QFSDFOUBHF
PG
QBUJFOUT
EJBHOPTFE
XJUI
BEWBODFE stage cancer are expected to survive for 4 years after JOJUJBM
EJBHOPTJT
2.% * G Interpret the meaning of k in the function found in part D 

* B
6TJOH
B
HSBQIJOH
VUJMJUZ
ESBX
B
TDBUUFS
EJBHSBN
PG
UIF
data with the number of dairy farms as the independent variable. C 
6TJOH
B
HSBQIJOH
VUJMJUZ
CVJME
B
MPHBSJUINJD
NPEFM
GSPN
the data. y = 330.0549 - 34.500 ln x * D
(SBQI
UIF
MPHBSJUINJD
GVODUJPO
GPVOE
JO
QBSU
C
PO
UIF
scatter diagram. E
*O

UIFSF
XFSF

UIPVTBOE
EBJSZ
GBSNT
JO
UIF
6OJUFE
4UBUFT
6TF
UIF
GVODUJPO
JO
QBSU
C
UP
QSFEJDU
UIF
BNPVOU
PG
NJML
QSPEVDFE
JO


CJMMJPO
QPVOET F
5IF
BDUVBM
BNPVOU
PG
NJML
QSPEVDFE
JO

XBT

CJMMJPO
QPVOET
)PX
EPFT
ZPVS
QSFEJDUJPO
JO
QBSU
E
DPNQBSF
UP
UIJT
6OEFS
CZ

CJMMJPO
QPVOET

4. Chemistry A chemist has a 100-gram sample of a radioactive material. He records the amount of radioactive material FWFSZ
XFFL
GPS

XFFLT
BOE
PCUBJOT
UIF
GPMMPXJOH
EBUB

6. Economics and Marketing The following data represent the QSJDF
BOE
RVBOUJUZ
EFNBOEFE
GPS
%FMM
QFSTPOBM
DPNQVUFST
in a recent year.

Week

Weight (in grams)

0

100.0

Price ($/Computer)

Quantity Demanded

2300

152 159

1

88.3

2000

2

75.9

1700

164

3

69.4

1500

171

4

59.1

1300

176

5

51.8

1200

180

6

45.5

1000

189

* B
6TJOH
B
HSBQIJOH
VUJMJUZ
ESBX
B
TDBUUFS
EJBHSBN
XJUI
week as the independent variable. C
6TJOH
B
HSBQIJOH
VUJMJUZ
CVJME
BO
FYQPOFOUJBM
NPEFM
from the data. y = 100.32610.692 x D
&YQSFTT
UIF
GVODUJPO
GPVOE
JO
QBSU
C
JO
UIF
GPSN
A1t2 = A0 e kt. A1t2 = 100.326e - 0.1314t * E
(SBQI
UIF
FYQPOFOUJBM
GVODUJPO
GPVOE
JO
QBSU
C
PS
D
on the scatter diagram. F
'SPN
UIF
SFTVMU
GPVOE
JO
QBSU
C


EFUFSNJOF
UIF
IBMGMJGF
of the radioactive material. 5.3 weeks G
)PX
NVDI
SBEJPBDUJWF
NBUFSJBM
XJMM
CF
MFGU
BGUFS

XFFLT
0.14 g * H
8IFO
XJMM
UIFSF
CF

HSBNT
PG
SBEJPBDUJWF
NBUFSJBM

* B
6TJOH
B
HSBQIJOH
VUJMJUZ
ESBX
B
TDBUUFS
EJBHSBN
PG
UIF
data with price as the dependent variable. C
6TJOH
B
HSBQIJOH
VUJMJUZ
CVJME
B
MPHBSJUINJD
NPEFM
GSPN
the data. y = 32,41.02 - 600.96 ln x * D
6TJOH
B
HSBQIJOH
VUJMJUZ
ESBX
UIF
MPHBSJUINJD
GVODUJPO
GPVOE
JO
QBSU
C
PO
UIF
TDBUUFS
EJBHSBN E
6TF
UIF
GVODUJPO
GPVOE
JO
QBSU
C
UP
QSFEJDU
UIF
OVNCFS
PG
%FMM
QFSTPOBM
DPNQVUFST
UIBU
XJMM
CF
EFNBOEFE
JG
UIF
price is $1650. ≈16 computers 7. Population Model The following data represent the QPQVMBUJPO
PG
UIF
6OJUFE
4UBUFT
"O
FDPMPHJTU
JT
JOUFSFTUFE
JO
CVJMEJOH
B
NPEFM
UIBU
EFTDSJCFT
UIF
QPQVMBUJPO
PG
UIF
6OJUFE
States.

SECTION 4.9   Building Exponential, Logarithmic, and Logistic Models from Data

397

8. Population Model The following data represent the world population. An ecologist is interested in building a model that describes the world population.

Year

Population

1900

76,212,168

1910

92,228,496

1920

106,021,537

Year

Population (billions)

1930

123,202,624

2001

6.17

1940

132,164,569

2002

6.24

1950

151,325,798

2003

6.32

1960

179,323,175

2004

6.40

1970

203,302,031

2005

6.47

1980

226,542,203

2006

6.55

1990

248,709,873

2007

6.63

2000

281,421,906

2008

6.71

2010

308,745,538

2009

6.79

2010

6.86

2011

6.94

2012

7.02

Source: U.S. Census Bureau

* B
6TJOH
B
HSBQIJOH
VUJMJUZ
ESBX
B
TDBUUFS
EJBHSBN
PG
UIF
data using years since 1900 as the independent variable and population as the dependent variable. * C
6TJOH
B
HSBQIJOH
VUJMJUZ
CVJME
B
MPHJTUJD
NPEFM
GSPN
UIF
data. * D
6TJOH
B
HSBQIJOH
VUJMJUZ
ESBX
UIF
GVODUJPO
GPVOE
JO
QBSU
C
PO
UIF
TDBUUFS
EJBHSBN E
#BTFE
PO
UIF
GVODUJPO
GPVOE
JO
QBSU
C


XIBU
JT
UIF
DBSSZJOH
DBQBDJUZ
PG
UIF
6OJUFE
4UBUFT
   F
6TF
UIF
GVODUJPO
GPVOE
JO
QBSU
C
UP
QSFEJDU
UIF
QPQVMBUJPO
PG
UIF
6OJUFE
4UBUFT
JO

≈315,203,2 * G
8IFO
XJMM
UIF
6OJUFE
4UBUFT
QPQVMBUJPO
CF
   H
$PNQBSF
BDUVBM
64
$FOTVT
GJHVSFT
UP
UIF
QSFEJDUJPOT
GPVOE
JO
QBSUT
F
BOE
G 
%JTDVTT
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Source: U.S. Census Bureau

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population of the world in 2020. ≈.65 billion G
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end of June each year. Year

Cell Sites (thousands)

Year

Cell Sites (thousands)

1985 (x = 1)

0.6

1999 (x = 15)

74.2

1986 (x = 2)

1.2

2000 (x = 16)

95.7

1987 (x = 3)

1.7

2001 (x = 17)

114.1

1988 (x = 4)

2.8

2002 (x = 18)

131.4

1989 (x = 5)

3.6

2003 (x = 19)

147.7

1990 (x = 6)

4.8

2004 (x = 20)

174.4

1991 (x = 7)

6.7

2005 (x = 21)

178.0

1992 (x = 8)

8.9

2006 (x = 22)

197.6

1993 (x = 9)

11.6

2007 (x = 23)

210.4

1994 (x = 10)

14.7

2008 (x = 24)

220.5

1995 (x = 11)

19.8

2009 (x = 25)

245.9

1996 (x = 12)

24.8

2010 (x = 26)

251.6

1997 (x = 13)

38.7

2011 (x = 27)

256.9

1998 (x = 14)

57.7

2012 (x = 28)

285.6

Source:
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398

CHAPTER 4  Exponential and Logarithmic Functions

10. Cable Rates The following date represents the average monthly rate charged for a basic cable television subscription JO
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maximum possible average monthly rate for basic DBCMF
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Average Monthly Rate (dollars)

Year 1980 (x = 0)

7.69

1985 (x = 5)

9.73

1990 (x = 10)

16.78

1995 (x = 15)

23.07

1997 (x = 17)

26.48

1998 (x = 18)

27.81

1999 (x = 19)

28.92

2000 (x = 20)

30.37

2001 (x = 21)

32.87

2002 (x = 22)

34.71

2003 (x = 23)

36.59

2004 (x = 24)

38.14

2005 (x = 25)

39.63

2006 (x = 26)

41.17

2007 (x = 27)

42.72

2008 (x = 28)

44.28

2009 (x = 29)

46.13

2010 (x = 30)

47.89

Source: Statistical Abstract of the United States, 2012

Mixed Practice 11. Age versus Total Cholesterol The following data represent the age and average total cholesterol for adult males at various ages.

Age

Total Cholesterol

27

189

40

205

50

215

60

210

70

210

80

194

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whose income is $55,000. 121.4 crimes per 1000 households Income Level

Property Crime Rate

5000

201.1

11,250

157.0

20,000

141.6

30,000

134.1

42,500

139.7

62,500

120.0

Source: Statistical Abstract of the United States, 2012

13. Golfing The data on the following page represent the expected percentage of putts that will be made by QSPGFTTJPOBM
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Chapter Review

114.. Depreciation of a Chevrolet Impala The following data 14 represent the asking price and age of a Chevrolet Impala SS.

Distance (feet)

Expected Percentage

Distance (feet)

Expected Percentage

2

99.3

14

25.0

Age

Asking Price

3

94.8

15

22.0

1

$27,417

4

85.8

16

20.0

1

$26,750

5

74.7

17

19.0

2

$22,995

6

64.7

18

17.0

2

$23,195

7

55.6

19

16.0

3

$17,999

8

48.5

20

14.0

4

$16,995

9

43.4

21

13.0

4

$16,490

10

38.3

22

12.0

11

34.2

23

11.0

12

30.1

24

11.0

13

27.0

25

10.0

Source: TheSandTrap.com

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percentage of 30-foot putts will be made. 5.1%

399

Source: cars.com

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Impala SS that is 5 years old.  

Retain Your Knowledge Problems 15–18 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 3 322 15. Construct a polynomial function that might have the graph 16. 3BUJPOBMJ[F
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17. 1 y f x = x + 3 x + 1 2 x - 2

the unlabeled side in the given right triangle. 23 3 4 2 2 –4

–2

2

4 x

–2 1 –4

*18. Graph the equation x - 3 2 + y2 = 25.

Chapter Review Things to Know Composite function (p. 306)

f ∘ g x = f g x

; The domain of f ∘ g is the set of all numbers x in the domain of g for which g1x2 is in the domain of f.

One-to-one function f (p. 314)

A function for which any two different inputs in the domain correspond to two different outputs in the range For any choice of elements x1 , x2 in the domain of f, if x1 ≠ x2 , then f 1x1 2 ≠ f 1x2 2.

Horizontal-line test (p. 315)

If every horizontal line intersects the graph of a function f in at most one point, f is one-to-one.

400

CHAPTER 4  Exponential and Logarithmic Functions

Inverse function f −1 of f (pp. 316–319)

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f 1f -1 1x2 2 = x for all x in the domain of f -1

The graphs of f and f -1 are symmetric with respect to the line y = x. Properties of the exponential function (pp. 328, 331, 333)

f 1x2 = Cax, a 7 1, C 7 0





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f 1x2 = Cax, 0 6 a 6 1, C 7 0
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Number e (p. 334)

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If au = av, then u = v.

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f 1x2 = log a x, a 7 1


1y = log a x means x = a 2
y

1 n 1 n b as n S q ; that is, lim a1 + b = e. n Sq n n

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Natural logarithm (p. 347)

y = ln x means x = e .

Properties of logarithms (pp. 357, 358, 360)

log a1 = 0

y

log a a = 1 aloga M = M

log a ar = r

M log a 1MN2 = log a M + log a N log a a b = log a M - log a N N r log a M = r log a M x If M = N, then log a M = log a N a = e x ln a

Formulas

If log a M = log a N, then M = N  

Change-of-Base Formula (p. 361)

log a M =

log b M

Continuous compounding (p. 374)

log b a r nt A = P # a1 + b n A = Pe rt

Effective rate of interest (p. 375)

Compounding n times per year: re = a1 +

Present Value Formulas (p. 376)

Continuous compounding: re = e r - 1 r -nt P = A # a1 + b or P = Ae -rt n A1t2 = A0 e kt

Compound Interest Formula (p. 373)

Growth and decay (p. 382, 383) Newton’s Law of Cooling (p. 384) Logistic model (p. 386)

u1t2 = T + u 0 - T e kt k 6 0 c P 1t2 = 1 + ae -bt

r n b - 1 n

Chapter Review

401

Objectives Section

You should be able to . . .

Example(s)

Review Exercises

1, 2, 4, 5 2–4

1–7 5–7

1, 2 3, 4

8(a), 9 8(b)

3 Obtain the graph of the inverse function from the graph

7

9

of the function (p. 318) 4 Find the inverse of a function defined by an equation (p. 319)

8, 9, 10

10–13 14(a), (c), 47(a) 31–33   35, 36, 39, 41

4.1  

1 Form a composite function (p. 306)

4.2  

1 Determine whether a function is one-to-one (p. 314)

   

2 Find the domain of a composite function (p. 307) 2 Determine the inverse of a function defined by a map or

a set of ordered pairs (p. 316)

4.3      

1 Evaluate exponential functions (p. 326)

4 Solve exponential equations (p. 335)

1 3–6 p. 334 7, 8

4.4

1 Change exponential statements to logarithmic statements

2, 3

15, 16

 

and logarithmic statements to exponential statements (p. 344) 2 Evaluate logarithmic expressions (p. 344)

4

14(b), (d), 19, 46(b), 48(a), 49 17, 18, 34(a) 34(b), 46(a) 37, 40, 46(c), 48(b) 20, 21

2 Graph exponential functions (p. 330) 3 Define the number e (p. 333)

     

3 Determine the domain of a logarithmic function (p. 345)

5 6, 7 8, 9

4.5        

1 Work with the properties of logarithms (p. 356)

4.6    

1 Solve logarithmic equations (p. 365)

4.7      

1 Determine the future value of a lump sum of money (p. 372)

4 Graph logarithmic functions (p. 345) 5 Solve logarithmic equations (p. 350) 2 Write a logarithmic expression as a sum or difference of logarithms (p. 359) 3 Write a logarithmic expression as a single logarithm (p. 359) 4 Evaluate a logarithm whose base is neither 10 nor e (p. 360) 5 Graph a logarithmic function whose base is neither 10 nor e (p. 362) 2 Solve exponential equations (p. 367) 3 Solve logarithmic and exponential equations using a

1, 2 3–5 6 7, 8 9

22–25 26–28 29 30

1–3 4–6 7

37, 43 38, 42, 44, 45 35–45

1–3 4 5 6, 7

50, 55 50 51 50

graphing utility (p. 368) 2 Calculate effective rates of return (p. 375) 3 Determine the present value of a lump sum of money (p. 376) 4 Determine the rate of interest or the time required to double

a lump sum of money (p. 377) 4.8

1 Find equations of populations that obey the law

1, 2

54

     

of uninhibited growth (p. 381) 2 Find equations of populations that obey the law of decay (p. 383) 3 Use Newton’s Law of Cooling (p. 384) 4 Use logistic models (p. 386)

3 4 5, 6

52 53 56

1 2 3

57 58 59

1 Build an exponential model from data (p. 392)

4.9    

2 Build a logarithmic model from data (p. 393) 3 Build a logistic model from data (p. 394)

Review Exercises 1. Evaluate each expression using the graphs of y = f 1x2 and y = g1x2 shown in the figure. y

(–5, 4) (–8, 2) (–2, 1)

(7, 6)

6 4 2

–10 –8 –6 –4 –2 –2 (0, –3) (–8, –2) –4 (–5, –4) –6

y = f (x) (4, 3) (10, 3) (2, 0) (7, 0) 2

4

(2, –4)

6

8 10 x (10, –2) y = g(x)

(0, –6) (4, –6)

(a) (b) (c) (d)

1g ∘ f2 1 - 82 - 4 1f ∘ g2 1 - 82 1 1g ∘ g2 172 - 6 1g ∘ f2 1 - 52 - 6

In Problems 2–4, for the given functions f and g find: (a) (b) (c) (d)

1f ∘ g2 132 - 65 1g ∘ f2 1 - 32 665 1f ∘ f2 122 23 1g ∘ g2 1 - 22 2

*2. f 1x2 = 3 - 4x; g1x2 = 3x2 - 10

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

402

CHAPTER 4  Exponential and Logarithmic Functions

*3. f 1x2 = 2x + 2: g1x2 = 2x2 + 1

*9. State why the graph of the function is one-to-one. Then draw the graph of the inverse function f -1. For convenience (and as a hint), the graph of y = x is also given.

*4. f 1x2 = 2x2 - 1; g1x2 = 3x - 5 In Problems 5–7, find f ∘ g, g ∘ f, f ∘ f, and g ∘ g for each pair of functions. State the domain of each composite function.

y 4

*5. f 1x2 = 2 - x; g1x2 = 3x + 1

y=x (3, 3)

*6. f 1x2 = 2x - 1; g1x2 = x2 + 3x + 1 *7. f 1x2 =

1 x + 1 ; g1x2 = x - 1 x

(2, 0) 4 x

–4

8. For the function { 13, 22, 15, 102, 12, 52, 17, 32}

(0, –2) (–1, –3) –4

(a) Verify that the function is one-to-one. (b) Find the inverse of the given function. (a) one-to-one (b) {(2, 3), (10, 5), (5, 2), (3, 7)}

In Problems 10–13, the function f is one-to-one. Find the inverse of each function and check your answer. State the domain and the range of f and f - 1 . 2x + 3 2x + 3 f - 1 1x2 = 5x - 2 5x - 2 *12. f 1x2 = 2x - 2 f - 1 1x2 = x2 + 2, x Ú 2

1 x + 1 f - 1 1x2 = x - 1 x *13. f 1x2 = x1>3 + 1 f - 1 1x2 = 1x - 12 3 *11. f 1x2 =

*10. f 1x2 =

14. If f 1x2 = 4x and g1x2 = log 2 x, evaluate each of the following. (a) f 122

1 1 (d) ga b - 3 64 8 15. Convert 35 = z to an equivalent statement involving a logarithm. log 3 z = 5 16

(b) g1162

4

(c) ƒ( - 32

16. Convert log3 z = 7 to an equivalent statement involving an exponent. 37 = z In Problems 17 and 18, find the domain of each logarithmic function. 2 18. H1x2 = log 2 1x2 - 3x + 22 a , qb 3 In Problems 19–21, evaluate each expression. Do not use a calculator. 17. f 1x2 = log13x - 22

19. log 3 27

20. ln e 22

3

22

( - q , 1) ∪ (2, q )

21. 5log5 0.3

0.3

In Problems 22–25, write each expression as the sum and/or difference of logarithms. Express powers as factors. xy 23. log 2 1a2 2b2 4, a 7 0, b 7 0 8 log 2 a + 2 log 2 6 *22. log 5 ¢ 2 ≤, x 7 0, y 7 0, z 7 0 z *24. lna

2x2 + 1 y 1x

2

b, x 7 0, y 7 0

25. ln¢

2x + 3 ≤ , x 7 2 2 ln(2x + 3) - 2 ln(x - 1) - 2 ln(x - 2) x2 - 3x + 2

In Problems 26–28, write each expression as a single logarithm. 3

26.

1 log 2 x3 - 3 log 2 1x2 + 12, x 7 0 2

28.

1 1 1 ln1x2 + 12 - 4 ln - 3ln1x - 42 + ln x4 2 2 2

x2 1 log 2 2 27. 3 ln x2 - ln 3 - 43ln1x2 + 32 - ln1x4 3 2 (x + 1) lnc

162x2 + 1 2x1x - 42

ln

x8 13(x2 + 3)4

d

29. Use the Change-of-Base Formula and a calculator to evaluate log 4 19 . Round your answer to three decimal places. 2.124 *30. Graph y = log 3 x using a graphing utility and the Change-of-Base Formula. In Problems 31–34, use the given function f to: (a) Find the domain of f. (b) Graph f. (c) From the graph, determine the range and any asymptotes of f. (d) Find f -1. (e) Find the domain and the range of f -1 . (f) Graph f -1 . *31. f 1x2 = 2x - 3

*32. f 1x2 = 1 + 3 - x

*33. f 1x2 = 3e x - 2

*34. f 1x2 =

1 ln1x + 32 2

In Problems 35–45, solve each equation. Express any irrational solution in exact form and as a decimal rounded to 3 decimal places. 5 1 2 35. 53x + 7 = 25 e - f *36. 3x + x = 23 37. log x 64 = - 3 e f 3 4 2 39. 252x = 5x - 12 5 - 2, 66 40. log 3 2x - 2 = 2 {83} *38. 5x = 3x + 2 1 x2 # 5x x# 41. 8 = 4 2 e , -3f 43. log 7 1x + 22 + log 7 1x - 42 = 1 {5} 42. 2 5 = 10x 516 2 *45. 9x + 4 # 3x - 3 = 0 44. e 1 - x = 5 {1 - ln 5} ≈ { - 0.609}

Chapter Review

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Temperature (°F)

0

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1

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2

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3

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4

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5

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6

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7

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8

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9

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10

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11

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12

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13

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14

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15

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404

CHAPTER 4  Exponential and Logarithmic Functions

*(c) Graph the exponential function found in part (b) on the scatter diagram. (d) Predict how long it will take for the probe to reach a temperature of 110°F. ≈83 s

cold. Those who come in contact with someone who has this cold will themselves catch the cold. The following data represent the number of people in the small town who have caught the cold after t days.

58. Wind Chill Factor The data below represent the wind speed (mph) and wind chill factor at an air temperature of 15°F. Days, t Wind Speed (mph)

Wind Chill Factor (°F)

5

7

10

3

15

0

20

⫺2

25

⫺4

30

⫺5

35 ⫺7 Source: U.S. National Weather Service

*(a) Using a graphing utility, draw a scatter diagram with wind speed as the independent variable. (b) Using a graphing utility, build a logarithmic model from the data. y = 18.921 - 7.096 ln x *(c) Using a graphing utility, draw the logarithmic function found in part (b) on the scatter diagram. (d) Use the function found in part (b) to predict the wind chill factor if the air temperature is 15°F and the wind speed is 23 mph. ≈ - 3°F 59. Spreading of a Disease Jack and Diane live in a small town of 50 people. Unfortunately, both Jack and Diane have a

1. Given f 1x2 =

5

2. Determine whether the function is one-to-one. (a) y = 4x2 + 3 not one-to-one (b) y = 2x + 3 - 5 one-to-one 2 and check your answer. *3. Find the inverse of f 1x2 = 3x - 5 State the domain and the range of f and f -1. 4. If the point 13, - 52 is on the graph of a one-to-one function f, what point must be on the graph of f -1? 1 - 5, 32

In Problems 5–7, solve each equation. 5. 3x = 243

{5}

6. log b16 = 2

{4}

7. log 5 x = 4

{625}

In Problems 8–11, use a calculator to evaluate each expression. Round your answer to three decimal places. 8. e + 2

22.086

3

9. log 20

1.301

10. log 321

2.771

11. ln 133

4.890

2 4 8 14 22 30 37 42 44

*(a) Using a graphing utility, draw a scatter diagram of the data. Comment on the type of relation that appears to exist between the day and the number of people with a cold. *(b) Using a graphing utility, build a logistic model from the data. *(c) Graph the function found in part (b) on the scatter diagram. (d) According to the function found in part (b), what is the maximum number of people who will catch the cold? In reality, what is the maximum number of people who could catch the cold? ≈47; 50 *(e) Sometime between the second and third day, 10 people in the town had a cold. According to the model found in part (b), when did 10 people have a cold? *(f) How long will it take for 46 people to catch the cold?

Chapter Test Prep Videos include step-by-step solutions to all chapter test exercises and can be found on this text’s Channel. (search “SullivanPrecalcUC3e”)

Chapter Test x + 2 and g1x2 = 2x + 5, find: x - 2 (b) 1g ∘ f2 1 - 22 *(a) f ∘ g and state its domain (c) 1f ∘ g2 1 - 22 - 3

Number of People with Cold, C

0 1 2 3 4 5 6 7 8

In Problems 12 and 13, use the given function f to: (a) Find the domain of f. (b) Graph f. (c) From the graph, determine the range and any asymptotes of f. (d) Find f - 1 , the inverse of f. (e) Find the domain and the range of f - 1 . (f) Graph f - 1 .

* 12. f 1x2 = 4x + 1 - 2 *13. f 1x2 = 1 - log 5 1x - 22 In Problems 14–19, solve each equation. 14. 5x + 2 = 125 * 16. 8 - 2e

-x

{1}

= 4

* 18. 7x + 3 = e x

15. log1x + 92 = 2

{91}

* 17. log1x + 32 = log1x + 62 2

*19. log 2 1x - 42 + log 2 1x + 42 = 3

4x3 ≤ as the sum and/or difference of x - 3x - 18 logarithms. Express powers as factors. 21. A 50-mg sample of a radioactive substance decays to 34 mg after 30 days. How long will it take for there to be 2 mg remaining? ≈250.39 days

* 20. Write log 2 ¢

2

22. (a) If $1000 is invested at 5% compounded monthly, how much is there after 8 months? $1033.82 (b) If you want to have $1000 in 9 months, how much do you need to place in a savings account now that pays 5% compounded quarterly? $963.42 (c) How long does it take to double your money if you can invest it at 6% compounded annually? 11.9 yr

Cumulative Review

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16. Data Analysis The following data represent the percent of all drivers by age who have been stopped by the police for any reason within the past year. The median age represents the midpoint of the upper and lower limit for the age range.

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406

CHAPTER 4  Exponential and Logarithmic Functions

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Internet-based Project I.

Depreciation of Cars
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The following projects are available on the Instructor’s Resource Center (IRC): II. Hot Coffee A fast-food restaurant wants a special container to hold coffee. The restaurant wishes the container to quickly cool the coffee from 200° to 130°F and keep the liquid between 110° and 130°F as long as possible. The restaurant has three containers to TFMFDU
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5

Trigonometric Functions Length of Day Revisited The length of a day depends on the day of the year as well as on the latitude of the location. Latitude gives the location of a point on Earth north or south of the equator. In Chapter 3 we found a model that describes the relation between the length of day and latitude for a specific day of the year. In the Internet Project at the end of this chapter, we will find a model that describes the relation between the length of day and day of the year for a specific latitude.

—See the Internet-based Chapter Project I—

In this chapter we define the trigonometric functions, six functions that have wide application. We shall talk about their domain and range, see how to find values for them, graph them, and develop a list of their properties. There are two widely accepted approaches to the development of the trigonometric functions: one uses right triangles; the other uses circles, especially the unit circle. In this book, we develop the trigonometric functions using the unit circle. In Section 7.1, we present right triangle trigonometry.

5.3 5.4 5.5 5.6

Angles and Their Measure Trigonometric Functions: Unit Circle Approach Properties of the Trigonometric Functions Graphs of the Sine and Cosine Functions Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions Phase Shift; Sinusoidal Curve Fitting Chapter Review Chapter Test Cumulative Review Chapter Projects

407

408

CHAPTER 5  Trigonometric Functions

5.1 Angles and Their Measure PREPARING FOR THIS SECTION Before getting started, review the following: r
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Now Work the ‘Are You Prepared?’ problems on page 417.

OBJECTIVES 1 Convert between Decimals and Degrees, Minutes, Seconds Measures for Angles (p. 410) 2 Find the Length of an Arc of a Circle (p. 411) 3 Convert from Degrees to Radians and from Radians to Degrees (p. 412) 4 Find the Area of a Sector of a Circle (p. 415) 5 Find the Linear Speed of an Object Traveling in Circular Motion (p. 416)

Figure 1 V

Ray

Line

"
ray, or half-line, is that portion of a line that starts at a point V on the line and FYUFOET
JOEFGJOJUFMZ
JO
POF
EJSFDUJPO
5IF
TUBSUJOH
QPJOU
V of a ray is called its vertex. 4FF
'JHVSF
 *G
UXP
SBZT
BSF
ESBXO
XJUI
B
DPNNPO
WFSUFY
UIFZ
GPSN
BO
angle. We call one ray of an angle the initial side and the other the terminal side. The angle formed is identified by showing the direction and amount of rotation from the initial side to the terminal side. If the rotation is in the counterclockwise direction, the angle is positive; if the rotation is clockwise, the angle is negative.
4FF
'JHVSF


Figure 2 ide al s

al min

Vertex Initial side

e

al min

Vertex

e sid

Ter

Ter

␤

n mi Ter ␣

sid

Initial side

␥

Vertex

Initial side

Counterclockwise rotation Positive angle

Clockwise rotation Negative angle

Counterclockwise rotation Positive angle

(a)

(b)

(c)

Lowercase Greek letters, such as a
BMQIB


b
CFUB


g
HBNNB


BOE
u
UIFUB


BSF
PGUFO
VTFE
UP
EFOPUF
BOHMFT
/PUJDF
JO
'JHVSF
 B
UIBU
UIF
BOHMF
a is positive because the direction of the rotation from the initial side to the terminal side is counterclockwise. The angle b
JO
'JHVSF
 C
JT
OFHBUJWF
CFDBVTF
UIF
SPUBUJPO
JT
DMPDLXJTF
5IF
BOHMF
g in 'JHVSF
 D
JT
QPTJUJWF
/PUJDF
UIBU
UIF
BOHMF
a
JO
'JHVSF
 B
BOE
UIF
BOHMF
g in 'JHVSF
 D
IBWF
UIF
TBNF
JOJUJBM
TJEF
BOE
UIF
TBNF
UFSNJOBM
TJEF
)PXFWFS
UIF
measures of a and g are unequal, because the amount of rotation required to go from the initial side to the terminal side is greater for angle g than for angle a. "O
BOHMF
u is said to be in standard position
JG
JUT
WFSUFY
JT
BU
UIF
PSJHJO
PG
B
rectangular coordinate system and its initial side coincides with the positive xBYJT
See Figure 3. y

Figure 3

Terminal side Vertex

y

␪ Initial side

x Terminal side

(a) ␪ is in standard position; ␪ is positive

Vertex Initial side x ␪

(b) ␪ is in standard position; ␪ is negative

SECTION 5.1   Angles and Their Measure

409

When an angle u is in standard position, the terminal side will lie either in a quadrant, in which case we say that u lies in that quadrant, or the terminal side will lie on the xBYJT
PS
UIF
yBYJT
JO
XIJDI
DBTF
XF
TBZ
UIBU
u is a quadrantal angle. For FYBNQMF
UIF
BOHMF
u
JO
'JHVSF
 B
MJFT
JO
RVBESBOU
**
UIF
BOHMF
u
JO
'JHVSF
 C
MJFT
in quadrant IV, and the angle u
JO
'JHVSF
 D
JT
B
RVBESBOUBM
BOHMF y

Figure 4

y

y



 x



(a)  lies in quadrant II

x

x

(b)  lies in quadrant IV

(c)  is a quadrantal angle

"OHMFT
BSF
NFBTVSFE
CZ
EFUFSNJOJOH
UIF
BNPVOU
PG
SPUBUJPO
OFFEFE
GPS
UIF
initial side to become coincident with the terminal side. The two commonly used measures for angles are degrees and radians.

Degrees HISTORICAL NOTE One counterclockwise rotation is 360° due to the Babylonian year, which had 360 days. j

Figure 5

5IF
BOHMF
GPSNFE
CZ
SPUBUJOH
UIF
JOJUJBM
TJEF
FYBDUMZ
PODF
JO
UIF
DPVOUFSDMPDLXJTF
EJSFDUJPO
VOUJM
JU
DPJODJEFT
XJUI
JUTFMG

SFWPMVUJPO
JT
TBJE
UP
NFBTVSF

EFHSFFT
 BCCSFWJBUFE
ž
One degree, 1°, is
SFWPMVUJPO
"
right angle is an angle that 3  NFBTVSFT
ž
PS
revolution; a straight angle
JT
BO
BOHMF
UIBU
NFBTVSFT
ž
PS
4 
SFWPMVUJPO
4FF
'JHVSF

"T
'JHVSF
 C
TIPXT
JU
JT
DVTUPNBSZ
UP
JOEJDBUF  a right angle by using the symbol . y

y

y

Terminal side

Terminal side x

Initial side Vertex

x

Vertex Initial side (b) right angle, 1–4 revolution counterclockwise, 90°

(a) 1 revolution counterclockwise, 360°

Terminal side Vertex

x

Initial side 1

(c) straight angle, –2 revolution counterclockwise, 180°

It is also customary to refer to an angle that measures u degrees as an angle of u degrees.

EXAM PL E 1

Drawing an Angle Draw each angle in standard position. B
ž




C
- °




D
ž




E
ž  B
"O
BOHMF
PG
ž
JT
of a right angle.  4FF
'JHVSF
 Figure 6

 revolution in 4 UIF
DMPDLXJTF
EJSFDUJPO
4FF
'JHVSF


C
"O
BOHMF
PG
- ° is

Figure 7

y Te rm in al sid e

Solution

Vertex

Initial side x

Terminal side

45°

Vertex Initial side

y

x

90°

CHAPTER 5  Trigonometric Functions

D
"O
BOHMF
PG
ž
DPOTJTUT
PG
B
SPUBUJPO
UISPVHI
ž
GPMMPXFE
CZ
B
SPUBUJPO
UISPVHI
ž
4FF
'JHVSF


E
"O
BOHMF
PG
ž
DPOTJTUT
PG

SFWPMVUJPO
ž
GPMMPXFE
CZ
B
SPUBUJPO
UISPVHI
ž
4FF
'JHVSF


Figure 8

Figure 9

y

y

225° Initial side x

405°

sid e

Vertex

Te rm in al sid e

410

al rm

in

x

Initial side

Vertex

Te

r

Now Work

PROBLEM

11

1 Convert between Decimals and Degrees, Minutes, Seconds Measures for Angles "MUIPVHI
TVCEJWJTJPOT
PG
B
EFHSFF
NBZ
CF
FYQSFTTFE
VTJOH
EFDJNBMT
UIF
OPUJPO
PG
minutes and seconds may also be used. One minute, denoted by 1′, is defined as   degree. One second, denoted by 1″, is defined as minute, or, equivalently,   
EFHSFF
"O
BOHMF
PG
TBZ

EFHSFFT

NJOVUFT

TFDPOET
JT
XSJUUFO
DPNQBDUMZ 3 as 3°4′″. To summarize:  counterclockwise revolution = 3° ° = ′ ′ = ″

(1)

It is sometimes necessary to convert from the degree, minute, second notation 1D°.′S″2 to a decimal form, and vice versa.

EX A MPL E 2

Converting between Degrees, Minutes, Seconds, and Decimal Forms B
$POWFSU
°′″ to a decimal in degrees. Round the answer to four decimal places. C
$POWFSU
ž
UP
UIF
EFHSFF
NJOVUF
TFDPOE
D°.′S″
OPUBUJPO
3PVOE
UIF
answer to the nearest second.

Solution

B
#FDBVTF
′ = a

 5  =  b and ″ = a b = a   

#

 5 b , convert as follows: 

a

 

°′″ = ° + ′ + ″

= ° +  # ′ +  # ″

COMMENT Graphing calculators BOE
TPNF
TDJFOUJGJD
DBMDVMBUPST
IBWF
the ability to convert from degrees, minutes, seconds to decimal form, and vice versa. Consult your owner’s manual. ■

= ° + 

#

a

 5 b +  

≈ ° + .° + .° = .°

#

#

 5 b 

Convert minutes and seconds to degrees.

SECTION 5.1   Angles and Their Measure

411

C
#FDBVTF
° = ′ and ′ = ″, proceed as follows: .° = = = = = = = = ≈

Now Work

° + .° ° + . # ° ° + 1.2 1′2 ° + .3′ ° + ′ + .3′ ° + ′ + .3 # ′ ° + ′ + 1.32 1″2 ° + ′ + .″ °′″

PROBLEMS

23

AND

Convert fraction of degree to minutes;1° = 60′.

Convert fraction of minute to seconds; 1′ = 60″. Round to the nearest second.

r

29

In
NBOZ
BQQMJDBUJPOT
TVDI
BT
EFTDSJCJOH
UIF
FYBDU
MPDBUJPO
PG
B
TUBS
PS
UIF
precise position of a ship at sea, angles measured in degrees, minutes, and even seconds are used. For calculation purposes, these are transformed to decimal form. In other applications, especially those in calculus, angles are measured using radians.

Radians "
central angle
JT
B
QPTJUJWF
BOHMF
XIPTF
WFSUFY
JT
BU
UIF
DFOUFS
PG
B
DJSDMF
5IF
SBZT
PG
B
DFOUSBM
BOHMF
TVCUFOE
JOUFSTFDU
BO
BSD
PO
UIF
DJSDMF
*G
UIF
SBEJVT
PG
UIF
DJSDMF
JT
r and the length of the arc subtended by the central angle is also r, then the measure of the angle is 1 radian.
4FF
'JHVSF
 B 

Ter mi

Ter

nal si

de

mi nal sid e

Figure 10

r

1 radian

3

r r

1 Initial side

Initial side

1 1 radian

(a)

3

(b)

'PS
B
DJSDMF
PG
SBEJVT

UIF
SBZT
PG
B
DFOUSBM
BOHMF
XJUI
NFBTVSF

SBEJBO
TVCUFOE
BO
BSD
PG
MFOHUI

'PS
B
DJSDMF
PG
SBEJVT

UIF
SBZT
PG
B
DFOUSBM
BOHMF
XJUI
NFBTVSF

SBEJBO
TVCUFOE
BO
BSD
PG
MFOHUI

4FF
'JHVSF
 C  Figure 11 u s = u1 s1

2 Find the Length of an Arc of a Circle s 

1 s1

r

Now consider a circle of radius r and two central angles, u and u, measured in radians. Suppose that these central angles subtend arcs of lengths s and s, SFTQFDUJWFMZ
BT
TIPXO
JO
'JHVSF

'SPN
HFPNFUSZ
UIF
SBUJP
PG
UIF
NFBTVSFT
PG
the angles equals the ratio of the corresponding lengths of the arcs subtended by these angles; that is, u s = s u 

(2)

412

CHAPTER 5  Trigonometric Functions

Suppose that u = 
SBEJBO
3FGFS
BHBJO
UP
'JHVSF
 B 
5IF
MFOHUI
s of the arc subtended by the central angle u =  radian equals the radius r of the circle. Then s = r
TP
FRVBUJPO

SFEVDFT
UP u s = r 

THEOREM

or s = ru

(3)

Arc Length For a circle of radius r, a central angle of u radians subtends an arc whose length s is (4)

NOTE Formulas must be consistent with regard to the units used. In equation (4), we write s = ru To see the units, however, go back to equation (3) and write s length units u radians = 1 radian r length units s length units = r length units

u radians 1 radian

The radians divide out, leaving s length units = 1r length units2u

s = ru

where u appears to be “dimensionless” but, in fact, is measured in radians. So, in using the formula s = r u, the dimension for u is radians, and any convenient unit of length (such as inches or meters) may be used for s and r. j

EX A MPL E 3

Finding the Length of an Arc of a Circle 'JOE
UIF
MFOHUI
PG
UIF
BSD
PG
B
DJSDMF
PG
SBEJVT

NFUFST
TVCUFOEFE
CZ
B
DFOUSBM
BOHMF
PG

SBEJBO

Solution

6TF
FRVBUJPO

XJUI
r =  meters and u = .. The length s of the arc is

r

s = ru = 1.2 = . meter

Now Work

PROBLEM

71

3 Convert from Degrees to Radians and from Radians to Degrees Figure 12 1 revolution = 2p radians s  2r 1 revolution

With two ways to measure angles, it is important to be able to convert from one to the other. Consider a circle of radius r
"
DFOUSBM
BOHMF
PG
 SFWPMVUJPO
XJMM
TVCUFOE
BO
BSD
FRVBM
UP
UIF
DJSDVNGFSFODF
PG
UIF
DJSDMF
'JHVSF
 
#FDBVTF
UIF
DJSDVNGFSFODF
PG
a circle of radius r equals pr, we substitute pr for s
JO
FRVBUJPO

UP
GJOE
UIBU
GPS
an angle u
PG

SFWPMVUJPO s = ru

r

pr = ru u = p radians

u = 1 revolution; s = 2pr Solve for u.

From this, we have



SFWPMVUJPO = p radians

(5)

SECTION 5.1   Angles and Their Measure

413

4JODF

SFWPMVUJPO = 3°, we have 3° = p radians %JWJEJOH
CPUI
TJEFT
CZ

ZJFMET ° = p radians

(6)

%JWJEF
CPUI
TJEFT
PG
FRVBUJPO

CZ

5IFO  degree =

p radian 

%JWJEF
CPUI
TJEFT
PG

CZ
p. Then  degrees =  radian p We have the following two conversion formulas:*  degree =

EXAM PL E 4

p radian 

 radian =

 degrees p

(7)

Converting from Degrees to Radians Convert each angle in degrees to radians. B
ž


C
ž


D
- 4°


E
ž


F
ž

Solution

B
° =  #  degree = 

#

p p radian = radians  3 p p C
° =  # ° =  # radian = radians   p p D
- 4° = - 4 # radian = - radian  4 p p E
° =  # radian = radians   p F
° =  # radian ≈ . radians 

r

&YBNQMF

QBSUT
B m E


JMMVTUSBUFT
UIBU
BOHMFT
UIBU
BSF
iOJDFu
GSBDUJPOT
PG
B
SFWPMVUJPO
BSF
FYQSFTTFE
JO
SBEJBO
NFBTVSF
BT
GSBDUJPOBM
NVMUJQMFT
PG
p, rather than as p EFDJNBMT
'PS
FYBNQMF
B
SJHIU
BOHMF
BT
JO
&YBNQMF
 E


JT
MFGU
JO
UIF
GPSN
radians,  p 3.4 XIJDI
JT
FYBDU
SBUIFS
UIBO
VTJOH
UIF
BQQSPYJNBUJPO
≈ = . radians.   8IFO
UIF
GSBDUJPOT
BSF
OPU
iOJDF u
VTF
UIF
EFDJNBM
BQQSPYJNBUJPO
PG
UIF
BOHMF
BT
JO
&YBNQMF
 F 

Now Work

EXAM PL E 5

PROBLEMS

35

AND

61

Converting Radians to Degrees Convert each angle in radians to degrees. 3p 3p p radian



C
radians


D
radians   4 p E
radians F

SBEJBOT 3 B


*

Some students prefer instead to use the proportion

given and solve for the measurement sought.

Degree °

=

Radian . Then substitute for what is p

414

CHAPTER 5  Trigonometric Functions

Solution

p p # p #  radian =  radian = degrees = 3° p    3p #  3p radians = C
degrees = ° p   3p #  3p radians = D
degrees = - 3° p 4 4 p #  p radians = E
degrees = 4° p 3 3  F
4 radians = 4 # degrees ≈ .° p B


Now Work

PROBLEM

r

47

5BCMF

MJTUT
UIF
EFHSFF
BOE
SBEJBO
NFBTVSFT
PG
TPNF
DPNNPOMZ
FODPVOUFSFE
angles. You should learn to feel equally comfortable using degree or radian measure for these angles.

Table 1

EX A MPL E 6

Degrees

0°

30° p 6

45° p 4

60° p 3

90° p 2

120° 2p 3

135° 3p 4

150° 5p 6

180°

Radians

0

Degrees

210°

225°

240°

270°

300°

315°

330°

360°

Radians

7p 6

5p 4

4p 3

3p 2

5p 3

7p 4

11p 6

2p

p

Finding the Distance between Two Cities The latitude of a location L is the measure of the angle formed by a ray drawn from the center of Earth to the equator and a ray drawn from the center of Earth to L
4FF
'JHVSF
 B 
4JPVY
'BMMT
4PVUI
%BLPUB
JT
EVF
OPSUI
PG
%BMMBT
5FYBT
'JOE
UIF
EJTUBODF
CFUXFFO
4JPVY
'BMMT
ž
¿
OPSUI
MBUJUVEF
BOE
%BMMBT
ž
¿ north MBUJUVEF 
4FF
'JHVSF
 C 
"TTVNF
UIBU
UIF
SBEJVT
PG
&BSUI
JT

NJMFT

Figure 13

North Pole

North Pole

L

Sioux Falls 43°33' Dallas

θ° Equator

South Pole

South Pole (a)

Solution

Equator

32°46'

(b)

5IF
NFBTVSF
PG
UIF
DFOUSBM
BOHMF
CFUXFFO
UIF
UXP
DJUJFT
JT
ž
¿ - ž
¿ =
ž¿. 6TF
FRVBUJPO



s = ru
#VU
SFNFNCFS
UP
àSTU
DPOWFSU
UIF
BOHMF
PG
ž¿ to radians. u = °4′ ≈ .33° = .33 c 47′ = 47a

1 b° 60

#

p radian ≈ . radian 

SECTION 5.1   Angles and Their Measure

415

6TF
u = . radian and r = 3 miles
JO
FRVBUJPO
 
5IF
EJTUBODF
CFUXFFO
the two cities is s = ru = 3 # . ≈ 44 miles

NOTE If the measure of an angle is given as 5, it is understood to mean 5 radians; if the measure of an angle is given as 5°, it means 5 degrees.

j

r

When an angle is measured in degrees, the degree symbol will always be shown. )PXFWFS
XIFO
BO
BOHMF
JT
NFBTVSFE
JO
SBEJBOT
UIF
VTVBM
QSBDUJDF
JT
UP
PNJU
UIF
p word radians. Thus, if the measure of an angle is given as , it is understood to mean  p radian. 

Now Work

PROBLEM

107

4 Find the Area of a Sector of a Circle Consider a circle of radius r. Suppose that u, measured in radians, is a central angle PG
UIJT
DJSDMF
4FF
'JHVSF

8F
TFFL
B
GPSNVMB
GPS
UIF
BSFB
A
PG
UIF
TFDUPS
TIPXO
JO
CMVF
GPSNFE
CZ
UIF
BOHMF
u. Now consider a circle of radius r and two central angles u and u, both measured JO
SBEJBOT
4FF
'JHVSF

'SPN
HFPNFUSZ
UIF
SBUJP
PG
UIF
NFBTVSFT
PG
UIF
BOHMFT
equals the ratio of the corresponding areas of the sectors formed by these angles. That is,

Figure 14 A θ r

u A = u A

Figure 15 u A = u1 A1

Suppose that u = p radians. Then A = area of the circle = pr . Solving for A, we find A

A = A

θ

u u  = pr  = r u u p  c

r

A1

A1 = pr2

θ1

u1 = 2p

THEOREM

Area of a Sector The area A of the sector of a circle of radius r formed by a central angle of u radians is A =

EXAM PL E 7

  r u 

(8)

Finding the Area of a Sector of a Circle 'JOE
UIF
BSFB
PG
UIF
TFDUPS
PG
B
DJSDMF
PG
SBEJVT

GFFU
GPSNFE
CZ
BO
BOHMF
PG
ž
Round the answer to two decimal places.

Solution

6TF
FRVBUJPO

XJUI
r =  feet and u = 3° = u must be in radians.] A =

p
SBEJBO
  - 2 3 - 2 = =  C  

r

*U
JT
JOUFSFTUJOH
UP
DPNQBSF
UIF
BOTXFS
GPVOE
JO
&YBNQMF
 B
XJUI
UIF
BOTXFS
UP
&YBNQMF

PG
4FDUJPO

5IFSF
JU
XBT
GPVOE
UIBU cos

p  = cos ° = 1 2 + 2 2  

#BTFE
PO
UIJT
BOE
UIF
SFTVMU
PG
&YBNQMF
 B


UIJT
NFBOT
UIBU  1 2 + 2 2 

and

3 + 2 

BSF
FRVBM
4JODF
FBDI
FYQSFTTJPO
JT
QPTJUJWF
ZPV
DBO
WFSJGZ
UIJT
FRVBMJUZ
CZ
TRVBSJOH
FBDI
FYQSFTTJPO
5XP
WFSZ
EJGGFSFOUMPPLJOH
ZFU
DPSSFDU
BOTXFST
DBO
CF
PCUBJOFE
depending on the approach taken to solve a problem.

Now Work

EX A MPL E 7

PROBLEM

19

Finding Exact Values Using Half-angle Formulas  p If cos a = - , p 6 a 6 ,
GJOE
UIF
FYBDU
WBMVF
PG   a a a B
sin



C
cos



D
tan   

Solution

'JSTU
PCTFSWF
UIBU
JG
p 6 a 6

p p a p a , then 6 6 . As a result, lies in quadrant II.     

a a B
#FDBVTF
lies in quadrant II, sin 7 , so use the +
TJHO
JO
GPSNVMB
B
UP
HFU     - a- b  a  - cos a sin = =  B  R      2 = = = = H A  2

SECTION 6.6  Double-angle and Half-angle Formulas

549

a a C
#FDBVTF
lies in quadrant II, cos 6 , so use the -
TJHO
JO
GPSNVMB
C
UP
HFU     + a- b  a  + cos a cos = =  B  R     2 = = = R  2 a a D
#FDBVTF
lies in quadrant II, tan 6 , so use the -
TJHO
JO
GPSNVMB
D
UP
HFU  

a  - cos a tan = =  B  + cos a




   - a- b   = = -    + a- b b b 

r

"OPUIFS
XBZ
UP
TPMWF
&YBNQMF
 D
JT
UP
VTF
UIF
SFTVMUT
PG
QBSUT
B
BOE
C 

a tan = 

Now Work

PROBLEM

a 2   = - = a 2 cos   sin

7(C), (D),

AND

(F)

a that does not contain + and - signs, making it  NPSF
VTFGVM
UIBO
GPSNVMB
 D 
5P
EFSJWF
JU
VTF
UIF
GPSNVMBT There is a formula for tan

 - cos a =  sin

a 

Formula (9)

and a a a sin a = sin c a b d =  sin cos   

Double-angle Formula

Then  - cos a = sin a

a 

a  a = = tan a a a   sin cos cos     sin

sin

Because it also can be shown that  - cos a sin a = sin a  + cos a UIJT
SFTVMUT
JO
UIF
GPMMPXJOH
UXP
)BMGBOHMF
'PSNVMBT

Half-angle Formulas for tan

tan

A 2

a  - cos a sin a = =  sin a  + cos a

(11)

550

CHAPTER 6  Analytic Trigonometry

8JUI
UIJT
GPSNVMB
UIF
TPMVUJPO
UP
&YBNQMF
 D
DBO
CF
PCUBJOFE
BT
GPMMPXT cos a = -

 

p 

p 6 a 6

sin a = - 2 - cos a = -

B

 -

   = =  B  

5IFO
CZ
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   - a- b  a  - cos a  tan = = = = -  sin a    

6.6 Assess Your Understanding Concepts and Vocabulary 1. cos 1u2 = cos u - sin u =  cos u -  =  -  sin u.

4. True or False tan1u2 =

 - cos u u 2. sin =   3. tan

 tan u  - tan u

True

5. True or False sin1u2
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6. True or False tan1u2 + tan1u2 = tan1u2

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Skill Building In Problems 7–18, use the information given about the angle u,  … u 6 p, to find the exact value of: u u u (a) sin1u2 (b) cos 1u2 (c) sin (d) cos (e) tan1u2 (f) tan     p  p  p *7. sin u = ,  6 u 6 *8. cos u = ,  6 u 6 *9. tan u = , p 6 u 6       *10. tan u =

p  , p 6 u 6  

*11. cos u = -

2 , 

p 6 u 6 p 

*12. sin u = -

2 , 

p 6 u 6 p 

*13. sec u = , sin u 7 

*14. csc u = - 2, cos u 6 

*15. cot u = - , sec u 6 

*16. sec u = , csc u 6 

*17. tan u = - , sin u 6 

*18. cot u = , cos u 6 

In Problems 19–28, use the Half-angle Formulas to find the exact value of each expression. 19. sin .°

3 - 2 

23. DPT
ž
27. sina -

p b 

3 + 2  -

20. DPT
ž


3 + 2 

24. TJO
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3 - 2 

21. tan

p 

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p 

28. cos a -

 - 2

p b 

22. tan

p 

*26. csc

p 

3 - 2 

In Problems 29–40, use the figures to evaluate each function given that f1x2 = sin x, g1x2 = cos x, and h1x2 = tan x. y

y x2 ⫹ y2 ⫽ 1

x2 ⫹ y2 ⫽ 5

(x, 2)

␣

␪ x

x (⫺1–4 , y)

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

-  + 2

551

SECTION 6.6  Double-angle and Half-angle Formulas

29. f 1u2

-

 

u 31. ga b 

 

-

33. h1u2

 

u 34. ha b 

 + 2 

35. g1a2

 

a 37. ƒa b 

2 

a 38. ga b 

-

2 

a 39. ha b 

-

   - cos 1u2 + cos 1u2.   

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31 - 22

30. g1u2



2 

36. f 1a2

2 

40. h1a2

-

2 

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*43. Develop a formula for cos 1u2 as a third-degree polynomial in the variable cos u.

*44. Develop a formula for cos 1u2 as a fourth-degree polynomial in the variable cos u.

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sin1u2 as a fifth-degree polynomial *46. 'JOE
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cos 1u2 as a fifth-degree polynomial in the variable sin u. in the variable cos u. In Problems 47–68, establish each identity. cot  u -  cot u - tan u *47. cos u - sin u = cos 1u2 *48. = cos 1u2 *49. cot 1u2 = cot u + tan u  cot u *50. cot 1u2 =

 1cot u - tan u2 

*53. cos 1u2 - sin 1u2 = cos 1u2 *56. sin u cos u =

 3 - cos 1u2 4 

y sec y +  *59. cot  =  sec y -  *62.  *64.

*51. sec1u2 =

sec u  - sec u

*54. 1 sin u cos u2 1 -  sin u2 = sin1u2 *57. sec

*55.

u  =   + cos u

*63.

cos u + sin u cos u - sin u =  tan1u2 cos u - sin u cos u + sin u

cos 1u2

 sec u csc u 

 + sin1u2

*58. csc

=

cot u -  cot u + 

u  =   - cos u

u  *61. cos u = u  + tan   - tan

y *60. tan = csc y - cot y 

 sin u + cos u sin1u2 =  sin u + cos u

sin1u2 sin u

-

cos 1u2 cos u

= 

 tan u - tan u  -  tan u  *67. ln 0 sin u 0 = 1ln 0  - cos 1u2 0 - ln 2 

*65. tan1u2 =

*66. tan u + tan1u + °2 + tan1u + °2 =  tan1u2 *68. ln 0 cos u 0 =

*52. csc1u2 =

 1ln 0  + cos 1u2 0 - ln 2 

In Problems 69–78, solve each equation on the interval  … u 6 p. *69. cos 1u2 +  sin u = 

*70. cos 1u2 =  -  sin u

p p p p e , , , f    

72. sin1u2 = cos u

75.  - sin u = cos 1u2

76. cos 1u2 +  cos u +  = 

p p e , , p, f  

e ,

p p , f  

*74. cos 1u2 + cos 1u2 = 

*73. sin1u2 + sin1u2 = 

No real solution

77. tan1u2 +  sin u = 

71. cos 1u2 = cos u

78. tan1u2 +  cos u = 

e

p p , f  

p p p p e , , , f    

Mixed Practice In Problems 79–90, find the exact value of each expression.  79. sina sin- b 

2 

 83. tanc  cos- a - b d    87. sin a cos- b  

 

80. sinc  sin-  

2 d 

2 

 81. cos a sin- b 

 

 82. cos a cos- b 

 

 84. tana tan- b 

 

 85. sina cos- b 

 

 86. cos c  tan- a - b d 

-

 

  88. cos a sin- b  

 

 89. seca tan- b 

 

 90. cscc  sin- a - b d 

-

 

In Problems 91–93, find the real zeros of each trigonometric function on the interval  … u 6 p. p p p p 91. f 1x2 = sin1x2 - sin x , , p, 92. f 1x2 = cos 1x2 + cos x , p, 93. f 1x2 = cos 1x2 + sin x    

p p ,  

552

CHAPTER 6  Analytic Trigonometry

Applications and Extensions 94. Constructing a Rain Gutter A rain gutter is to be constructed PG
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95. Laser Projection In a laser projection system, the optical angle or scanning angle u is related to the throw distance D from the scanner to the screen and the projected image width W by the equation  W  D = csc u - cot u

v 2 3sin1u2 - cos 1u2 - 4 

*98. Sawtooth Curve An oscilloscope often displays a sawtooth curve. This curve can be approximated by sinusoidal curves of varying periods and amplitudes. A first approximation to the sawtooth curve is given by y =


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(b) At what times t does the graph of V touch the graph of y = e -t>3? When does the graph of V touch the graph of y = - e -t>3? (c) When is the voltage V between - 0.4 and 0.4 volt? *56. The Sawtooth Curve An oscilloscope often displays a sawtooth curve. This curve can be approximated by sinusoidal curves of varying periods and amplitudes. (a) Use a graphing utility to graph the following function, which can be used to approximate the sawtooth curve. 1 1 f 1x2 = sin12px2 + sin 14px2, 2 4

and y = sin12pht2

0 … x … 4

y = sin32p18522t4 + sin32p112092t4 Use a graphing utility to graph the sound emitted by touching 7. Touch-Tone phone

1

2

3

697 cycles/sec

4

5

6

770 cycles/sec

7

8

9

852 cycles/sec

*

0

#

941 cycles/sec

(b) A better approximation to the sawtooth curve is given by f 1x2 =

1 1 1 sin12px2 + sin 14px2 + sin 18px2 2 4 8

Use a graphing utility to graph this function for 0 … x … 4 and compare the result to the graph obtained in part (a). (c) A third and even better approximation to the sawtooth curve is given by 1 1 1 1 f 1x2 = sin12px2 + sin 14px2 + sin 18px2 + sin11px2 2 4 8 1

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1209 cycles/sec

1477 cycles/sec

1336 cycles/sec

*58. Use a graphing utility to graph the sound emitted by the * key on a Touch-Tone phone. See Problem 57. 59. CBL Experiment Pendulum motion is analyzed to estimate simple harmonic motion. A plot is generated with the position of the pendulum over time. The graph is used to find a sinusoidal curve of the form y = A cos 3B 1x - C2 4 + D. %FUFSNJOF
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Discussion and Writing 61. Use

a

graphing utility to graph the function sin x f 1x2 = , x 7 0. Based on the graph, what do you x sin x conjecture about the value of for x close to 0? x 62. Use a graphing utility to graph y = x sin x, y = x2 sin x, and y = x3 sin x for x 7 0. What patterns do you observe?

1 1 63. Use a graphing utility to graph y = sin x, y = 2 sin x, x x 1 and y = 3 sin x for x 7 0. What patterns do you observe? x 64. How would you explain to a friend what simple harmonic motion is? How would you explain damped motion?

Chapter Review

609

Retain Your Knowledge Problems 65–68 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your xy3 4x - 3 mind so that you are better prepared for the final exam. 65. f - 1(x) = 66. log 7 a b x - 1 x + y x - 3 4 p 65. The function f(x) = , x ≠ 4, is one-to-one. Find its 68. Given cos a = , 0 6 a 6 , find the exact value of: x - 4 5 2 inverse function. a 3210 a 210 a 1 66. Write as a single logarithm: log 7 x + 3 log 7 y - log 7 (x + y) (a) cos (b) sin (c) tan 2 10 2 10 2 3 67. Solve: log (x + 1) + log (x - 2) = 1 {4}

‘Are You Prepared?’ Answer 1. A = 5 T =

p 2

Chapter Review Things to Know Formulas Law of Sines (p. 577)

sin B sin C sin A = = a b c

Law of Cosines (p. 587–588)

c 2 = a2 + b2 - 2ab cos C b2 = a2 + c 2 - 2ac cos B a2 = b2 + c 2 - 2bc cos A

Area of a triangle (pp. 594–595)

K =

1 1 1 1 bh K = ab sin C K = bc sin A K = ac sin B 2 2 2 2

K = 2s 1s - a2 1s - b2 1s - c2

where s =

1 1a + b + c2 2

Objectives You should be able to N

Example(s)

Review Exercises

7.1

1

Find the value of trigonometric functions of acute BOHMFT
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Solve SAA or ASA triangles (p. 578)

1, 2

8, 19

 

2

Solve SSA triangles (p. 578)

3–5







 

3

Solve applied problems (p. 581)




32, 33

7.3

1

Solve SAS triangles (p. 588)

1

11, 15, 20

 

2

Solve SSS triangles (p. 589)

2

13, 14, 17

 

3

Solve applied problems (p. 589)

3

34

7.4

1

Find the area of SAS triangles (p. 594)

1






 

2

Find the area of SSS triangles (p. 595)

2

23, 24

7.5

1

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Section

610

CHAPTER 7  Applications of Trigonometric Functions

Review Exercises In Problems 1 and 2, find the exact value of the six trigonometric functions of the angle u in each figure. *2.

*1.

4

4





2

3

In Problems 3–5, find the exact value of each expression. Do not use a calculator. 3. cos 62° - sin 28°

4.

0

tan 70° cot 20°

5. cos2 40° + cos2 50°

1

1

In Problems 6 and 7, solve each triangle. *6.

10

A

20° a

*7.

5

b

A 2

B a

In Problems 8–20, find the remaining angle(s) and side(s) of each triangle, if it (they) exists. If no triangle exists, say “No triangle.” *8. A = 30°, B = 75°, b = 5

*9. A = 100°, a = 5, c = 2

*10. a = 3, c = 1, C = 110°

*11. a = 3, c = 1, B = 100°

*12. a = 4, b = 6, A = 50°

*13. a = 2, b = 3, c = 1

*14. a = 6, b = 8, c = 4

*15. a = 1, b = 3, C = 40°

*16. a = 3, b = 5, C = 40°

1 4 *17. a = 1, b = , c = 2 3 *20. c = 5, b = 4, A = 70°

*18. a = 3, A = 10°, b = 4

*19. a = 4, A = 20°, B = 100°

In Problems 21–25, find the area of each triangle. 21. a = 2, b = 3, C = 40° 24. a = 4, b = 2, c = 5

1.93

3.80

22. a = 7, c = 10, B = 60°

30.31 23. a = 4, b = 3, c = 5

25. B = 40°, C = 60°, b = 4

6

10.57

26. Area of a Segment Find the area of the segment of a circle whose radius is 8 inches formed by a central angle of 30°. 0.76 in2 27. Geometry The hypotenuse of a right triangle is 7 feet. If one leg is 4 feet, find the degree measure of each angle. 34.85° and 55.15° 28. Finding the Width of a River Find the distance from A to C across the river illustrated in the figure. 23.32 ft

1454 ft 5¯ Lake Michigan

1 mi

30. Finding the Speed of a Glider From a glider 200 feet above the ground, two sightings of a stationary object directly in front are taken 1 minute apart (see the figure). What is the speed of the glider? 132.55 ft/min

A

25¯ C

50 ft

40¯

B 10¯

29. Finding the Distance to Shore The Willis Tower in Chicago is 1454 feet tall and is situated about 1 mile inland from the shore of Lake Michigan, as indicated in the figure on the top right. An observer in a pleasure boat on the lake directly in front of the Willis Tower looks at the top of the tower and measures the angle of elevation as 5°. How far offshore is the boat? 2.15 mi

200 ft

31. Finding the Grade of a Mountain Trail A straight trail with a uniform inclination leads from a hotel, elevation 5000 feet, to a lake in a valley, elevation 4100 feet. The length of the trail is 4100 feet. What is the inclination (grade) of the trail? 12.7°

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

Chapter Review

32. Finding the Height of a Helicopter Two observers simultaneously measure the angle of elevation of a IFMJDPQUFS
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611

35. Approximating the Area of a Lake To approximate the area of a lake, Cindy walks around the perimeter of the lake, taking the measurements shown in the illustration. Using UIJT
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100'

70' 100°

50°

50' 125' 50'

33. Constructing a Highway A highway whose primary directions are north–south is being constructed along the west coast of Florida. Near Naples, a bay obstructs the straight path of the road. Since the cost of a bridge is prohibitive, engineers decide to go around the bay. The illustration shows the path that they decide on and the measurements taken. What is the length of highway needed to go around the bay? 
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36. Finding the Bearing of a Ship The Majesty leaves the Port at #PTUPO
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34. Correcting a Navigational Error A sailboat leaves St. Thomas bound for an island in the British West Indies, 200 miles away. Maintaining a constant speed of 18 miles per hour, but encountering heavy crosswinds and strong currents, the crew finds, after 4 hours, that the sailboat is off DPVSTF
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St. Thomas

200 mi

15°

38. Rework Problem 37 if the belt is crossed, as shown in the figure. 
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39. An object attached to a coiled spring is pulled down a distance a = 3 units from its rest position and then released. Assuming that the motion is simple harmonic with period T = 4 seconds, develop a model that relates the displacement d of the object from its rest position after t seconds. Also assume that the positive direction of the motion is up. p d = - 3 cos a tb 2

612

CHAPTER 7  Applications of Trigonometric Functions

In Problems 40 and 41, the displacement d (in feet) that an object travels in time t (in seconds) is given. (a) Describe the motion of the object. (b) What is the maximum displacement from its rest position? (c) What is the time required for one oscillation? (d) What is the frequency? *40. d = 6 sin 12t2

*41. d = - 2 cos 1pt2

*42. An object of mass m = 40 grams attached to a coiled spring with damping factor b = 0.75 gram/second is pulled down a distance a = 15 centimeters from its rest position and then released. Assume that the positive direction of the motion is up and the period is T = 5 seconds under simple harmonic motion. (a) Develop a model that relates the displacement d of the object from its rest position after t seconds. (b) Graph the equation found in part (a) for 5 oscillations. *43. The displacement d (in meters) of the bob of a pendulum of mass m (in kilograms) from its rest position at time t (in seconds) is given as d = - 15e -0.6t>40 cos ¢ (a) (b) (c) (d) (e)

2p 2 0.36 b t≤ C 5 1600 a

Describe the motion of the object. What is the initial displacement of the bob? That is, what is the displacement at t = 0? Graph the motion using a graphing utility. What is the displacement of the bob at the start of the second oscillation? What happens to the displacement of the bob as time increases without bound?

*44. Use the method of adding y-coordinates to graph y = 2 sin x + cos 12x2 .

Chapter Test Prep Videos include step-by-step solutions to all chapter test exercises and can be found on this text’s Channel. (search “SullivanPrecalcUC3e”)

Chapter Test

2. Find the exact value of sin 40° - cos 50°. 0

*1. Find the exact value of the six trigonometric functions of the angle u in the figure. 3  6

In Problems 3–5, use the given information to determine the three remaining parts of each triangle. *3.

*4.

C 17

b a

*5.

12

C

5

41⬚

22⬚

8

A

c

B 10

B

52⬚

C

19

In Problems 6–8, solve each triangle. *6. A = 55°, C = 20°, a = 4

*7. a = 3, b = 7, A = 40°

*9. Find the area of the triangle described in Problem 8.

*8. a = 8, b = 4, C = 70°

[Hint: Triangle ABC is a right triangle.]

*10. Find the area of the triangle described in Problem 5.

B

11. A 12-foot ladder leans against a building. The top of the ladder leans against the wall 10.5 feet from the ground. What is the angle formed by the ground and the ladder? 61.0° 12. A hot-air balloon is flying at a height of 600 feet and is directly above the Marshall Space Flight Center in Huntsville, Alabama. The pilot of the balloon looks down at the airport that is known to be 5 miles from the Marshall Space Flight Center. What is the angle of depression from the balloon to the airport? 1.3° 13. Find the area of the shaded region enclosed in a semicircle of diameter 8 centimeters. The length of the chord AB is 6 centimeters. 9.26 cm2

6 A

C

8

14. Find the area of the quadrilateral shown. 54.15 square units 5 72

11

7

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

8

Chapter Projects

15. Madison wants to swim across Lake William from the fishing lodge (A) to the boat ramp (B), but she wants to know the distance first. Highway 20 goes right past the boat ramp and County Road 3 goes to the lodge. The two roads intersect at point (C), 4.2 miles from the ramp and 3.5 miles from the lodge. Madison uses a transit to measure the angle of intersection of the two roads to be 32°. How far will she need to swim? ≈2.23 mi

613

B

A

5 40⬚

Lake William

B

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3.5

6x

3

5x 32⬚

CR

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20 Hwy

A

17. The area of the triangle shown below is 542
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18. Logan is playing on her swing. One full swing (front to back UP
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at an angle of 42° with the vertical. If her swing is 5 feet long, and we ignore all resistive forces, build a model that relates her horizontal displacement (from the rest position) after time t. pt d = 5(sin 42°) sina b 3

Cumulative Review *1. 'JOE
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1 - 5, 12 and radius 3. Graph this circle. *3. Determine the domain of the function f 1x2 = 2x2 - 3x - 4 *4. Graph the function y = 3 sin 1px2.

*5. Graph the function y = - 2 cos 12x - p2. 3p 6 u 6 2p, find the exact value of: 2 *(b) cos u *(c) sin12u2 3 1 1 *(e) sina u b *(f) cos a u b 5 2 2

6. If tan u = - 2 and *(a) sin u (d) cos 12u2

*7. Graph each of the following functions on the interval 30, 44: (a) y = e x (b) y = sin x (c) y = e x sin x (d) y = 2x + sin x *8. Sketch the graph of each of the following functions: (a) y = x (b) y = x2 (c) y = 1x (d) y = x3

(e) y = e x (g) y = sin x (i) y = tan x

(f) y = ln x (h) y = cos x

*9. Solve the triangle in which side a is 20, side c is 15, and angle C
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2x2 - 7x - 4 x2 + 2x - 15

*12. Solve 3x = 12. Round your answer to two decimal places. 13. Solve log 3 1x + 82 + log 3 x = 2.

{1}

14. Suppose that f 1x2 = 4x + 5 and g1x2 = x2 + 5x - 24. (b) Solve f 1x2 = 13. {2} *(a) Solve f 1x2 = 0. *(c) Solve f 1x2 = g1x2. *(d) Solve f 1x2 7 0. *(e) Solve g1x2 … 0. *(f) Graph y = f 1x2. *(g) Graph y = g1x2.

Chapter Projects I.

Spherical Trigonometry When the distance between two MPDBUJPOT
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OA and

614

CHAPTER 7  Applications of Trigonometric Functions

OB to intersect the tangent lines at P and Q, respectively. See the figure. List the plane right triangles. Determine the measures of the central angles. C a

b O

Q

B

A

c

2. Apply the Law of Cosines to triangles OPQ and CPQ to find two expressions for the length of PQ. 3. Subtract the expressions in part (2) from each other. Solve for the term containing cos c. 4. Use the Pythagorean Theorem to find another value for OQ2 - CQ2 and OP 2 - CP 2. Now solve for cos c. 5. Replacing the ratios in part (4) by the cosines of the sides of the spherical triangle, you should now have the Law of Cosines for spherical triangles: cos c = cos a cos b + sin a sin b cos C Source: 'PS
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II. The Lewis and Clark Expedition Lewis and Clark followed several rivers in their trek from what is now Great Falls, Montana, to the Pacific coast. First, they went down the Missouri and Jefferson rivers from Great Falls to Lemhi, Idaho. Because the two cities are at different longitudes and EJGGFSFOU
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angles are found by using the differences in the latitudes and longitudes of the towns. See the diagram.) Then find the length of the arc joining the two towns. (Recall s = ru.) 2. From Lemhi, they went up the Bitteroot River and the Snake River to what is now Lewiston and Clarkston on the border of Idaho and Washington. Although this is not really a side to a triangle, make a side that goes from Lemhi to Lewiston and Clarkston. If Lewiston and Clarkston are at about 4.5°N 117.0°W, find the distance from Lemhi using the Law of Cosines for a spherical triangle and the arc length. 3. How far did the explorers travel just to get that far?

1. Great Falls is at approximately 47.5°N and 111.3°W. Lemhi is at approximately 45.5°N and 113.5°W. (Assume that the rivers flow straight from Great Falls to Lemhi on UIF
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4. Draw a plane triangle connecting the three towns. If the distance from Lewiston to Great Falls is 282 miles, and if the angle at Great Falls is 42° and the angle at Lewiston is 48.5°, find the distance from Great Falls to Lemhi and from Lemhi to Lewiston. How do these distances compare with the ones computed in parts (1) and (2)? Source: 'PS
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&YQFEJUJPO American Journey: The Quest for Liberty to 1877, Texas Edition. Prentice Hall, 1992, p. 345. Source: For map coordinates: National Geographic Atlas of the World, published by National Geographic Society, 1981, pp. 74–75.The following projects are available at the Instructor’s Resource Center (IRC): III. Project at Motorola: How Can You Build or Analyze a Vibration Profile? Fourier functions not only are important to analyze vibrations, but they are also what a mathematician would call interesting. Complete the project to see why. IV. Leaning Tower of Pisa Trigonometry is used to analyze the apparent height and tilt of the Leaning Tower of Pisa. V. Locating Lost Treasure Clever treasure seekers who know the Law of Sines are able to find a buried treasure efficiently.

South

VI. Jacob’s Field Angles of elevation and the Law of Sines are used to determine the height of the stadium wall and the distance from home plate to the top of the wall.

Polar Coordinates; Vectors How Airplanes Fly Essentially there are 4 aerodynamic forces that act on an airplane in flight; these are lift, drag, thrust and weight (i.e. gravity). In simple terms, drag is the resistance of air molecules hitting the airplane (the backward force), thrust is the power of the airplane’s engine (the forward force), lift is the upward force and weight is the downward force. So for airplanes to fly and stay airborne, the thrust must be greater than the drag and the lift must be greater than the weight (so as you can see, drag opposes thrust and lift opposes weight). This is certainly the case when an airplane takes off or climbs. However, when it is in straight and level flight the opposing forces of lift and weight are balanced. During a descent, weight exceeds lift and to slow an airplane drag has to overcome thrust. Thrust is generated by the airplane’s engine (propeller or jet), weight is created by the natural force of gravity acting upon the airplane and drag comes from friction as the plane moves through air molecules. Drag is also a reaction to lift, and this lift must be generated by the airplane flight. This is done by the wings of the airplane. . . The generation of lift is a widely discussed and sometimes disputed theory, but there are some key factors that nobody argues. A cross section of a typical airplane wing will show the top surface to be more curved than the bottom surface. This shaped profile is called an ‘airfoil’ (or ‘aerofoil’) and the shape exists because it’s long been proven (since the dawn of flight) that an airfoil generates significantly more lift than opposing drag i.e. it’s very efficient at generating lift. During flight air naturally flows over and beneath the wing and is deflected upwards over the top surface and downwards beneath the lower surface. Any difference in deflection causes a difference in air pressure (‘pressure gradient’) and because of the airfoil shape the pressure of the deflected air is lower above the airfoil than below it. As a result the wing is ‘pushed’ upwards by the higher pressure beneath or, you can argue, it is ‘sucked’ upwards by the lower pressure above.

8

Source: Pete Carpenter. How Airplanes Fly—the basic principles of flight.

http://www.rc-airplane-world.com/how-airplanes-fly.html, accessed July 2013. © rc-airplane-world.com

—See Chapter Project I—

Outline

This chapter is in two parts: Polar Coordinates, Sections 8.1–8.3, and Vectors, Sections 8.4–8.7. They are independent of each other and may be covered in either order. Sections 8.1–8.3: In Foundations we introduced rectangular coordinates x and y and discussed the graph of an equation in two variables involving x and y. In Sections 8.1 and 8.2, we introduce polar coordinates, an alternative to rectangular coordinates, and discuss graphing equations that involve polar coordinates. In Section 4.3, we discussed raising a real number to a real power. In Section 8.3, we extend this idea by raising a complex number to a real power. As it turns out, polar coordinates are useful for the discussion. Sections 8.4–8.7: We have seen in many chapters that we are often required to solve an equation to obtain a solution to applied problems. In the last four sections of this chapter, we develop the notion of a vector and show how it can be used to model applied problems in physics and engineering.

8.1 8.2 8.3 8.4 8.5 8.6 8.7

Polar Coordinates Polar Equations and Graphs The Complex Plane; De Moivre’s Theorem Vectors The Dot Product Vectors in Space The Cross Product Chapter Review Chapter Test Cumulative Review Chapter Projects

615

616

CHAPTER 8  Polar Coordinates; Vectors

8.1 Polar Coordinates PREPARING FOR THIS SECTION  Before getting started, review the following: r
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Now Work the ‘Are You Prepared?’ problems on page 623. OBJECTIVES 1 2 3 4

Figure 1 y

Polar axis O Pole

x

Plot Points Using Polar Coordinates (p. 616) Convert from Polar Coordinates to Rectangular Coordinates (p. 618) Convert from Rectangular Coordinates to Polar Coordinates (p. 620) Transform Equations between Polar and Rectangular Forms (p. 622)

So far, we have always used a system of rectangular coordinates to plot points in the plane. Now we are ready to describe another system, called polar coordinates. In many instances, polar coordinates offer certain advantages over rectangular coordinates. In a rectangular coordinate system, you will recall, a point in the plane is represented by an ordered pair of numbers 1x, y2, where x and y equal the signed distances of the point from the y-axis and from the x-axis, respectively. In a polar coordinate system, we select a point, called the pole, and then a ray with vertex at the pole, called the polar axis. See Figure 1. Comparing the rectangular and polar coordinate systems, notice that the origin in rectangular coordinates coincides with the pole in polar coordinates, and the positive x-axis in rectangular coordinates coincides with the polar axis in polar coordinates.

1 Plot Points Using Polar Coordinates A point P in a polar coordinate system is represented by an ordered pair of numbers 1r, u2. If r 7 0, then r is the distance of the point from the pole; u is an angle (in degrees or radians) formed by the polar axis and a ray from the pole through the point. We call the ordered pair 1r, u2 the polar coordinates of the point. See Figure 2. p As an example, suppose that a point P has polar coordinates a2, b . Locate P 4 p by first drawing an angle of radian, placing its vertex at the pole and its initial side 4 along the polar axis. Then go out a distance of 2 units along the terminal side of the angle to reach the point P. See Figure 3. Figure 2

Figure 3 P  (r, )

r  O Pole

2 Polar axis

 – 4

O Pole

P  (2, –4 ) Polar axis

In using polar coordinates 1r, u2, it is possible for r to be negative. When this happens, instead of the point being on the terminal side of u, it is on the ray from the pole extending in the direction opposite the terminal side of u at a distance 0 r 0 units from the pole. See Figure 4 for an illustration. 2p For example, to plot the point a - 3, b , use the ray in the opposite direction 3 2p of and go out 0 - 3 0 = 3 units along that ray. See Figure 5. 3

SECTION 8.1   Polar Coordinates

Figure 4

617

Figure 5

2 –– 3

 O

O

–– ) (3, 2 3

⏐r⏐

P  (r, ), r  0

EXAM PL E 1

Plotting Points Using Polar Coordinates Plot the points with the following polar coordinates: (a) a3,

Solution

5p b 3

(b) a2, -

p b 4

(c) 13, 02

(d) a - 2,

p b 4

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 – 4

O

 –

O

( 2,

( 3, 5––3 ) (a)

 –)

O

4

(2, –4 )

4

(c)

(b)

Now Work

O

(3, 0)

PROBLEMS

9, 17,

AND

(d)

r

27

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EXAM PL E 2

Finding Several Polar Coordinates of a Single Point

Figure 7

p Consider again the point P with polar coordinates a2, b , as shown in Figure 7(a). 4 7p p 9p Because , , and all have the same terminal side, this point P could also 4 4 4 9p 7p b or a2, b , as shown have been located by using the polar coordinates a2, 4 4 p in Figures 7(b) and (c). The point a2, b can also be represented by the polar 4 5p b . See Figure 7(d). coordinates a - 2, 4

2

 –

P  (2, –4 )

2

4

O

9––  4

 P  (2, 9––– 4 )

O

2   7––

 P  ( 2,  7––– 4 )

2

5 –– 4

O

 –

) P  (2, 5–– 4

4

O

4

(a)

(b)

(c)

(d)

r

618

CHAPTER 8  Polar Coordinates; Vectors

Finding Other Polar Coordinates of a Given Point

EX A MPL E 3

Plot the point P with polar coordinates a3,

p b , and find other polar coordinates 

(a) r 7 0, 2p … u 6 4p (c) r 7 0, - 2p … u 6 0

(b) r 6 0, 0 … u 6 2p

1r, u2 of this same point for which:

Solution P  (3,

6

p b is plotted in Figure 8. 

p p (a) Add 1 revolution 12p radians2 to the angle to get P = a3, + 2pb =   13p a3, b . See Figure 9. 

Figure 8  –

The point a3,

)

 – 6

O

1 p revolution 1p radians2 to the angle and replace 3 by - 3 to get 2  7p p b . See Figure 10. P = a - 3, + pb = a - 3,  

(b) Add

p p 11p (c) Subtract 2p from the angle to get P = a3, - 2pb = a3, b . See    Figure 11. Figure 10

Figure 9

7––  6

) ––– P  (3, 13 6 13  ––– 6

O

Figure 11 P  (3,

 7–– 6

)

11  ––– 6

 ––– P  (3, 11 6 )

O

O

r

These examples show a major difference between rectangular coordinates and polar coordinates. A point has exactly one pair of rectangular coordinates; however, a point has infinitely many pairs of polar coordinates.

SUMMARY A point with polar coordinates 1r, u2 , u in radians, can also be represented by either of the following: 1r, u + 2pk2

or

1 - r, u + p + 2pk2

k any integer

The polar coordinates of the pole are 10, u2, where u can be any angle.

Now Work

PROBLEM

31

2 Convert from Polar Coordinates to Rectangular Coordinates Sometimes it is necessary to convert coordinates or equations in rectangular form to polar form, and vice versa. To do this, recall that the origin in rectangular coordinates is the pole in polar coordinates and that the positive x-axis in rectangular coordinates is the polar axis in polar coordinates.

THEOREM

Conversion from Polar Coordinates to Rectangular Coordinates If P is a point with polar coordinates 1r, u2 u2,, the rectangular coordinates 1x, y2 of P are given by x = r cos u

y = r sin u

(1)

SECTION 8.1   Polar Coordinates

Figure 12 y

P r

y

 x

O

x

619

Proof Suppose that P has the polar coordinates 1r, u2. We seek the rectangular coordinates 1x, y2 of P
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 If r = 0, then, regardless of u, the point P is the pole, for which the rectangular coordinates are 10, 02. Formula (1) is valid for r = 0. If r 7 0, the point P is on the terminal side of u, and r = d 1O, P2 = 2x2 + y2 . Since y x cos u = sin u = r r we have x = r cos u

y = r sin u

If r 6 0 and u is in radians, then the point P = 1r, u2 can be represented as 1 - r, p + u2, where - r 7 0. Since cos 1p + u2 = - cos u =

sin 1p + u2 = - sin u =

x -r

y -r

we have x = r cos u

EXAM PL E 4

y = r sin u

■

Converting from Polar Coordinates to Rectangular Coordinates Find the rectangular coordinates of the points with the following polar coordinates: (a) a,

Solution

p b 

(b) a - 4, -

p b 4

Use formula (1): x = r cos u and y = r sin u. p p b plotted. Notice that a, b lies in quadrant I of the   rectangular coordinate system. So the x-coordinate and the y-coordinate should p both be positive. With r =  and u = , we have 

(a) Figure 13(a) shows a, Figure 13 y 3

6

(6, –6 )

 –

p 23 = # = 323  2 p 1 y = r sin u =  sin =  # = 3  2

6

3 3

x = r cos u =  cos

x

(a) y

(4, –4 )

p The rectangular coordinates of the point a, b are 1 323, 3 2 , which lies in  quadrant I, as expected.

2 2

–  4

2 2

x

4

(b)

COMMENT Most calculators have the capability of converting from polar coordinates to rectangular coordinates. Consult your owner’s manual for the proper keystrokes. Since in most cases this procedure is tedious, you will find that using formula (1) is faster. ■

p p b plotted. Notice that a - 4, - b lies in quadrant II 4 4 p of the rectangular coordinate system. With r = - 4 and u = - , we have 4

(b) Figure 13(b) shows a - 4, -

x = r cos u = - 4 cos a y = r sin u = - 4 sin a -

p b = -4 4

#

22 = - 222 2

p 22 b = - 4a b = 222 4 2

p The rectangular coordinates of the point a - 4, - b are 1 - 222, 222 2 , which 4 lies in quadrant II, as expected.

r

Now Work

PROBLEMS

39

AND

51

620

CHAPTER 8  Polar Coordinates; Vectors

3 Convert from Rectangular Coordinates to Polar Coordinates

Converting from rectangular coordinates 1x, y2 to polar coordinates 1r, u2 is a little more complicated. Notice that each solution begins by plotting the given rectangular coordinates.

How to Convert from Rectangular Coordinates to Polar Coordinates with the Point on a Coordinate Axis

EX A MPL E 5

Find polar coordinates of a point whose rectangular coordinates are 10, 32 .

Step-by-Step Solution

Figure 14

y

Plot the point 10, 32 in a rectangular coordinate system. See Figure 14. The point lies on the positive y-axis.

(x, y)  (0, 3)

Step 1 Plot the point (x, y) and note the quadrant the point lies in or the coordinate axis the point lies on.

3

 – 2

x

The point (0, 3) lies on the y-axis a distance of 3 units from the origin (pole), so r = 3.

Step 2 Determine the distance r from the origin to the point.

p with the polar axis. 2 p Polar coordinates for this point can be given by a3, b . Other possible 2 p 5p representations include a - 3, - b and a3, b. 2 2

Step 3 Determine u.

A ray with vertex at the pole through (0, 3) forms an angle u =

COMMENT Most graphing calculators have the capability of converting from rectangular coordinates to polar coordinates. Consult your owner’s manual for the proper keystrokes. ■

r

Figure 15 shows polar coordinates of points that lie on either the x-axis or the y-axis. In each illustration, a 7 0.

Figure 15 y

a

(x, y)  (a, 0) (r, θ)  (a, 0) a

 –

(x, y)  (a, 0) (r, )  (a, )

2

x

y

y

y (x, y)  (0, a ) –) (r, )  (a,  2

x

a



3 ––– 2

x

x (x, y)  (0, a) (r, )  (a, 3–– 2 )

(a) (x, y)  (a, 0), a  0

(b) (x, y)  (0, a), a  0

Now Work

EX A MPL E 6

PROBLEM

(c) (x, y)  ( a, 0), a  0

a

(d) (x, y)  (0, a), a  0

55

How to Convert from Rectangular Coordinates to Polar Coordinates with the Point in a Quadrant

Find the polar coordinates of a point whose rectangular coordinates are 12, - 22 .

Step-by-Step Solution Step 1 Plot the point (x, y) and note the quadrant the point lies in or the coordinate axis the point lies on.

Plot the point 12, - 22 in a rectangular DPPSEJOBUF
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The point lies in quadrant IV.

Figure 16

y 1 1 1 2



2

x

r (x, y)  (2, 2)

SECTION 8.1   Polar Coordinates

621

r = 2x2 + y2 = 2 122 2 + 1 - 22 2 = 28 = 222

Step 2 Determine the distance r from the origin to the point using r = 2x 2 + y 2.

y y p p 6 u 6 . Since 12, - 22 lies , so u = tan - 1 , x x 2 2 p in quadrant IV, we know that 6 u 6 0. As a result, 2

Step 3 Determine u.

Find u by recalling that tan u =

u = tan-1

y -2 p b = tan-1 1 - 12 = = tan-1 a x 2 4

A set of polar coordinates for the point 12, - 22 is a222, representations include a222,

EXAM PL E 7

p b . Other possible 4

7p 3p b and a - 222, b. 4 4

r

Converting from Rectangular Coordinates to Polar Coordinates Find polar coordinates of a point whose rectangular coordinates are

Solution

1 - 1, - 23 2 .

STEP 1: See Figure 17. The point lies in quadrant III. STEP 2: The distance r from the origin to the point 1 - 1, - 23 2 is

Figure 17

r = 3 1 - 12 2 +

y u

1 - 23 2 2

= 24 = 2

y - 23 p p = tan - 1 23, 6 a 6 . = tan - 1 x -1 2 2 Since the point 1 - 1, - 23 2 lies in quadrant III, and the inverse tangent function gives an angle in quadrant I, add p to the result to obtain an angle in quadrant III.

STEP 3: To find u, use a = tan - 1

x r (x, y) 5 (21, 2 3)

u = p + tan-1 a

- 23 p 4p b = p + tan-1 23 = p + = -1 3 3

4p A set of polar coordinates for this point is a2, b . Other possible 3 p 2p representations include a - 2, b and a2, b. 3 3

r

Figure 18 shows how to find polar coordinates of a point that lies in a quadrant when its rectangular coordinates 1x, y2 are given.

Figure 18 y

y (x, y )

(x, y ) r

r

y

y





 x

x



x r

r

(x, y )

(a) r  x 2 y 2   tan1 y x

(b) r  x 2 y 2    tan1 y x

(x, y)

(c) r  x 2 y 2    tan1 y x

(d) r  x 2 y 2   tan1 y x

Based on the preceding discussion, we have the formulas r 2 = x2 + y 2

tan u =

y x

if x ≠ 0

(2)

x

622

CHAPTER 8  Polar Coordinates; Vectors

To use formula (2) effectively, follow these steps:

Steps for Converting from Rectangular to Polar Coordinates STEP 1: Always plot the point 1x, y2
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GJSTU BT TIPXO JO &YBNQMFT   BOE  Note the quadrant the point lies in or the coordinate axis the point lies on. STEP 2: If x = 0 or y = 0, use your illustration to find r. If x ≠ 0 and y ≠ 0, then r = 2x 2 2 + y 2. STEP 3: Find u. If x = 0 or y = 0, use your illustration to find u. If x ≠ 0 and y ≠ 0, note the quadrant in which the point lies. u = tan-1

Quadrant I or IV:

y x

Quadrant II or III: u = p + tan-1

Now Work

PROBLEM

y x

59

4 Transform Equations between Polar and Rectangular Forms Formulas (1) and (2) may also be used to transform equations from polar form to rectangular form, and vice versa. Two common techniques for transforming an equation from polar form to rectangular form are 1. Multiplying both sides of the equation by r 2. Squaring both sides of the equation

EX A MPL E 8

Transforming an Equation from Polar to Rectangular Form Transform the equation r =  cos u from polar coordinates to rectangular coordinates, and identify the graph.

Solution

Multiplying each side by r makes it easier to apply formulas (1) and (2). r =  cos u r 2 = r cos u x + y = x 2

2

Multiply each side by r. r 2 = x 2 + y 2; x = r cos u

This is the equation of a circle. Proceed to complete the square to obtain the standard form of the equation. x2 + y2 = x

1x2 - x2 + y2 = 0

1x2 - x + 92 + y2 = 9 1x - 32 + y = 9 2

2

General form Complete the square in x. Factor.

This is the standard form of the equation of a circle with center 13, 02 and radius 3.

Now Work

EX A MPL E 9

PROBLEM

r

75

Transforming an Equation from Rectangular to Polar Form Transform the equation 4xy = 9 from rectangular coordinates to polar coordinates.

Solution

Use formula (1): x = r cos u and y = r sin u. 4xy = 9 41r cos u2 1r sin u2 = 9 x = r cos u, y = r sin u 4r 2 cos u sin u = 9

SECTION 8.1   Polar Coordinates

623

This is the polar form of the equation. It can be simplified as follows: 2r 2 12 sin u cos u2 = 9 Factor out 2r 2.

2r 2 sin 12u2 = 9 Double-angle Formula

Now Work

PROBLEM

r

69

8.1 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. Plot the point whose rectangular coordinates are 13, - 12. What quadrant does the point lie in? (pp. 34–35)

3. If P = 1a, b2 is a point on the terminal side of the angle u at a distance r from the origin, then tan u = . (p. 433) b p a -1 . (pp. 502–504) 4. tan 1 - 12 = 4

2. To complete the square of x2 + x, add 9 . (pp. A38–A39)

Concepts and Vocabulary 7. True or False The polar coordinates of a point are unique.

5. The origin in rectangular coordinates coincides with the pole in polar coordinates; the positive x-axis in rectangular coordinates coincides with the polar axis in polar coordinates.

8. If P is a point with polar coordinates (r, u), the rectangular coordinates (x, y) of P are given by x = r cos u and y = r sin u.

6. True or False In the polar coordinates (r, u), r can be negative. True

7. False

Skill Building In Problems 9–16, match each point in polar coordinates with either A, B, C, or D on the graph. 9. a2, 13. a2,

11p b 

5p b 

10. a - 2, -

A

14. a - 2,

B

p b 

5p b 

B D

11. a - 2,

p b 

15. a - 2,

7p b 

C A

12. a2,

7p b 

16. a2,

11p b 

B

C

A

2

D

π 6

D

C

In Problems 17–30, plot each point given in polar coordinates. p b 

*17. 13, 90°2

*18. 14, 270°2

*19. 1 - 2, 02

*20. 1 - 3, p2

*21. a,

*22. a5,

*23. 1 - 2, 135°2

*24. 1 - 3, 120°2

*25. a4, -

*26. a2, -

*28. a - 3, -

*29. 1 - 2, - p2

*30. a - 3, -

5p b 3

*27. a - 1, -

p b 3

3p b 4

2p b 3

5p b 4

p b 2

In Problems 31–38, plot each point given in polar coordinates, and find other polar coordinates 1r, u2 of the point for which: (a) r 7 0,

- 2p … u 6 0

2p b *31. a5, 3 *35. a1,

p b 2

(b) r 6 0, 0 … u 6 2p

(c) r 7 0, 2p … u 6 4p

3p *32. a4, b 4

*33. 1 - 2, 3p2

*36. 12, p2

*37. a - 3, -

*34. 1 - 3, 4p2

p b 4

*38. a - 2, -

2p b 3

In Problems 39–54, the polar coordinates of a point are given. Find the rectangular coordinates of each point. 39. a3,

p b 2

43. 1, 150°2

(0, 3)

( - 323, 3)

40. a4,

3p b 2

44. 15, 300°2

(0, - 4) 5 523 a ,b 2 2

41. 1 - 2, 02 45. a - 2,

3p b 4

( - 2, 0)

1 22, - 22 2

42. 1 - 3, p2 46. a - 2,

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

2p b 3

(3, 0)

1 1, - 23 2

624

CHAPTER 8  Polar Coordinates; Vectors

47. a - 1, -

p b 3

51. 17.5, 110°2

1 23 a- , b 2 2

48. a - 3, -

3p b 4

( - 2.57, 7.05) 52. 1 - 3.1, 182°2

a

322 322 , b 2 2

(3.10, 0.11)

49. 1 - 2, - 180°2 53. 1.3, 3.82

50. 1 - 3, - 90°2

(2, 0)

54. 18.1, 5.22

( - 4.98, - 3.85)

(0, 3) (3.79, - 7.1)

In Problems 55–66, the rectangular coordinates of a point are given. Find polar coordinates for each point. p p 55. 13, 02 (3, 0) 56. 10, 22 a2, b 58. 10, - 22 a2, - b 57. 1 - 1, 02 (1, p) 2 2 p 59. 11, - 12 a 22, - p b 62. 1 - 2, - 223 2 a4, - 2p b 60. 1 - 3, 32 a322, 3p b 61. 1 23, 1 2 a2, b  4 3 4 65. 18.3, 4.22 (9.30, 0.47) 66. 1 - 2.3, 0.22 (2.31, 3.05) 63. 11.3, - 2.12 (2.47, - 1.02) 64. 1 - 0.8, - 2.12 (2.25, 4.35) In Problems 67–74, the letters x and y represent rectangular coordinates. Write each equation using polar coordinates 1r, u2. *67. 2x2 + 2y2 = 3 71. 2xy = 1

68. x2 + y2 = x

r 2 sin 12u2 = 1 *72. 4x2 y = 1

r = cos u

*69. x2 = 4y 73. x = 4

*70. y2 = 2x r cos u = 4

74. y = - 3

r sin u = - 3

In Problems 75–82, the letters r and u represent polar coordinates. Write each equation using rectangular coordinates 1x, y2. *75. r = cos u 79. r = 2

*76. r = sin u + 1

x2 + y 2 = 4

80. r = 4

x2 + y2 = 1

*77. r 2 = cos u 81. r =

*78. r = sin u - cos u

4 1 - cos u

3 3 - cos u

*82. r =

Applications and Extensions 83. Chicago In Chicago, the road system is set up like a Cartesian plane, where streets are indicated by the number of blocks they are from Madison Street and State Street. 'PS
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Madison Street and State Street as the origin of a coordinate system, with east being the positive x-axis. (a) Write the location of Wrigley Field using rectangular coordinates. ( - 10, 3) *(b) Write the location of Wrigley Field using polar coordinates. Use the east direction for the polar axis. Express u in degrees.

City of Chicago, Illinois

1 mile 1 km N Wrigley Field 1060 West Addison Madison Street

d =

2r 21

+

r 22

- 2r1 r2 cos 1u2 - u1 2

State Street

(c) U.S. Cellular Field, home of the White Sox, is located at 35th and Princeton, which is 3 blocks west of State Street and 35 blocks south of Madison. Write the location of U.S. Cellular Field using rectangular coordinates. ( - 3, - 35) *(d) Write the location of U.S. Cellular Field using polar coordinates. Use the east direction for the polar axis. Express u in degrees. *84. Show that the formula for the distance d between two points P1 = 1r1 , u1 2 and P2 = 1r2 , u2 2 is

Addison Street

Addison Street

U.S. Cellular Field 35th and Princeton 35th Street

35th Street

Explaining Concepts: Discussion and Writing 85. In converting from polar coordinates to rectangular coordinates, what formulas will you use? 86. Explain how you proceed to convert from rectangular coordinates to polar coordinates.

87. Is the street system in your town based on a rectangular coordinate system, a polar coordinate system, or some other system? Explain.

SECTION 8.2   Polar Equations and Graphs

625

Retain Your Knowledge Problems 88–91 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 19 88. Solve: log 4 1x + 32 - log 4 1x - 12 = 2 e 15 f 89.
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real zeros for the function f 1x2 = - 2x3 + x2 - 7x - 8. 2 or 0 positive real zeros; 1 negative real zero 1 5 9 90. Find the midpoint of the line segment connecting the points 1 - 3, 72 and a , 2b. a - , b 2 4 2 91. Given that the point (3, 8) is on the graph of y = f 1x2 , what is the corresponding point on the graph of y = - 2f 1x + 32 + 5? (0, - 11)

‘Are You Prepared?’ Answers y

1.

;quadrant IV

2. 9

3.

2 2 2

b a

4. -

p 4

2 4 x (3, 1)

8.2 Polar Equations and Graphs PREPARING FOR THIS SECTION  Before getting started, review the following: r
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(Section 5.3, pp. 448–449) Now Work the ‘Are You Prepared?’ problems on page 637.

OBJECTIVES 1 Identify and Graph Polar Equations by Converting to Rectangular Equations (p. 626) 2 Test Polar Equations for Symmetry (p. 629) 3 Graph Polar Equations by Plotting Points (p. 630)

Just as a rectangular grid may be used to plot points given by rectangular coordinates, as in Figure 19(a), a grid consisting of concentric circles (with centers at the pole) and rays (with vertices at the pole) can be used to plot points given by polar coordinates, as shown in Figure 19(b). We shall use such polar grids to graph polar equations. Figure 19

y 4



3 2 1 B  (3, 1) 4 3 2 1 O

–   2

3  –– 4

( )

A  (1, 2) 1

2

3 4

P  2, –4 x



2 3 4

(

O

r1

5

)

Q  4, –– 4

5

  –– 4

 (a) Rectangular grid

   –4

r3 r5

3  –– 2

(b) Polar grid

7   –– 4

0

626

CHAPTER 8  Polar Coordinates; Vectors

DEFINITION

An equation whose variables are polar coordinates is called a polar equation. The graph of a polar equation consists of all points whose polar coordinates satisfy the equation.

1 Identify and Graph Polar Equations by Converting to Rectangular Equations One method that can be used to graph a polar equation is to convert the equation to rectangular coordinates. In the following discussion, 1x, y2 represents the rectangular coordinates of a point P, and 1r, u2 represents the polar coordinates of the point P.

EX A MPL E 1

Identifying and Graphing a Polar Equation (Circle) Identify and graph the equation: r = 3

Solution

Convert the polar equation to a rectangular equation. r = 3 r2 = 9

Square both sides.

x + y = 9 r2 = x2 + y 2 2

2

The graph of r = 3 is a circle, with center at the pole and radius 3. See Figure 20. Figure 20

y

r = 3 or x2 + y2 = 9



  –2

   3–– 4



O

5   –– 4

Now Work

EX A MPL E 2

PROBLEM

Solution



Convert the polar equation to a rectangular equation.

x 4 O 1 2 3 4 5 0

p 4

tan u = tan

–  4

y = 1 x y = x

 –



p 4

u =

–   2

r

Identifying and Graphing a Polar Equation (Line)

y

  3–– 4

   7–– 4

13

Figure 21

p or y = x 4

x 1 2 3 4 5 0

3   –– 2

Identify and graph the equation: u = u =

   –4

p 4

Take the tangent of both sides.

tan u =

y p ; tan = 1 x 4

p p is a line passing through the pole making an angle of with the 4 4 polar axis. See Figure 21. The graph of u =

   5–– 4 

  3–– 2

   7–– 4

r

Now Work

PROBLEM

15

SECTION 8.2   Polar Equations and Graphs

EXAM PL E 3

627

Identifying and Graphing a Polar Equation (Horizontal Line) Identify and graph the equation: r sin u = 2

Solution

Since y = r sin u, we can write the equation as

Figure 22 r sin u = 2 or y = 2 y

y = 2

–   2

   3–– 4

Thus, the graph of r sin u = 2 is a horizontal line 2 units above the pole. See Figure 22.

x 1 2 3 4 5 O 0



COMMENT A graphing utility can be used to graph polar equations. 3FBE
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EXAM PL E 4

–  4

   5–– 4

   3–– 2

   7–– 4

r

Identifying and Graphing a Polar Equation (Vertical Line) Identify and graph the equation: r cos u = - 3

Solution

Since x = r cos u, we can write the equation as

Figure 23 r cos u = - 3 or x = - 3 y

x = -3 

Thus, the graph of r cos u = - 3 is a vertical line 3 units to the left of the pole. See Figure 23.

 ⫽ 3–– 4

⫽

– ⫽ 2

O

– ⫽  4

x 1 2 3 4 5 ⫽0

  ⫽ 7–– 4

  ⫽ 5–– 4   ⫽ 3–– 2

r Based on Examples 3 and 4, we are led to the following results. (The proofs are left as exercises. See Problems 81 and 82.)

THEOREM

Let a be a nonzero real number. Then the graph of the equation r sin u = a is a horizontal line a units above the pole if a 7 0 and 0 a 0 units below the pole if a 6 0. The graph of the equation r cos u = a is a vertical line a units to the right of the pole if a 7 0 and 0 a 0 units to the left of the pole if a 6 0.

Now Work

PROBLEM

19

628

CHAPTER 8  Polar Coordinates; Vectors

Identifying and Graphing a Polar Equation (Circle)

EX A MPL E 5

Identify and graph the equation: r = 4 sin u

Solution

To transform the equation to rectangular coordinates, multiply each side by r.

Figure 24 r = 4 sin u or x2 + (y - 2)2 = 4

r 2 = 4r sin u

y

Now use the facts that r 2 = x2 + y2 and y = r sin u. Then

– ⫽ 2

  ⫽ 3–– 4

– ⫽  4

x2 + y2 = 4y

x2 + 1y2 - 4y2 = 0

x2 + 1y2 - 4y + 42 = 4

x

⫽

O



x2 + 1y - 22 2 = 4

1 2 3 4 5 ⫽0

 ⫽ 5–– 4



Complete the square in y. Factor.

This is the standard equation of a circle with center at 10, 22 in rectangular coordinates and radius 2. See Figure 24.

 ⫽ 7–– 4

r

  ⫽ 3–– 2

Identifying and Graphing a Polar Equation (Circle)

EX A MPL E 6

Identify and graph the equation: r = - 2 cos u

Solution Figure 25 2 r = - 2 cos u or (x + 1) + y2 = 1

Proceed as in Example 5. r 2 = - 2r cos u Multiply both sides by r.

y   ⫽ 3–– 4

– ⫽ 2

x2 + y2 = - 2x – ⫽  4

x2 + 2x + y2 = 0

1x2 + 2x + 12 + y2 = 1 1x + 12 + y = 1 2

x

⫽

O

r 2 = x 2 + y 2; x = r cos u

2

Complete the square in x Factor.

1 2 3 4 5 ⫽0

This is the standard equation of a circle with center at 1 - 1, 02 in rectangular coordinates and radius 1. See Figure 25.

  ⫽ 7–– 4

  ⫽ 5–– 4   ⫽ 3–– 2

r

Exploration

Using a square screen, graph r1 = sin u, r2 = 2 sin u, and r3 = 3 sin u. Do you see the pattern? Clear the screen and graph r1 = - sin u, r2 = - 2 sin u, and r3 = - 3 sin u. Do you see the pattern? Clear the screen and graph r1 = cos u, r2 = 2 cos u, and r3 = 3 cos u. Do you see the pattern? Clear the screen and graph r1 = - cos u, r2 = - 2 cos u, and r3 = - 3 cos u. Do you see the pattern?

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THEOREM

Let a be a positive real number. Then Equation (a) r (b) r (c) r (d) r

= = = =

2a sin u - 2a sin u 2a cos u - 2a cos u

Description

Circle: radius a; center at 10, a2 in rectangular coordinates Circle: radius a; center at 10, - a2 in rectangular coordinates Circle: radius a; center at 1a, 02 in rectangular coordinates Circle: radius a; center at 1 - a, 02 in rectangular coordinates

Each circle passes through the pole.

Now Work

PROBLEM

21

SECTION 8.2   Polar Equations and Graphs

629

The method of converting a polar equation to an identifiable rectangular equation to obtain the graph is not always helpful, nor is it always necessary. Usually, a table is created that lists several points on the graph. By checking for symmetry, it may be possible to reduce the number of points needed to draw the graph.

2 Test Polar Equations for Symmetry

In polar coordinates, the points 1r, u2 and 1r, - u2 are symmetric with respect to the polar axis (the xBYJT 
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Figure 26 y 2

   3–– 4

–  4

   3–– 4

(r, ) 



O

   5–– 4 

y

y

–  

x

1 2 3 4 5  (r, )

  3–– 2

0



 O

   3–– 4

–  4

(r,   )

(r, ) 

1 2 3 4 5

–   2

x 0



O

   5–– 4

   3–– 2

   7–– 4

(b) Points symmetric with  respect to the line   –– 2

   5–– 4

–  4

(r, )



(r, )

   7–– 4

(a) Points symmetric with respect to the polar axis



–   2



x

1 2 3 4 5

0

 (r,   )

   3–– 2

   7–– 4

(c) Points symmetric with respect to the pole

The following tests are a consequence of these observations.

THEOREM

Tests T Te sts for Symmetry Symmetry with Respect to the Polar Axis ((x (x-Axis) -Axis) In a polar equation, replace u by - u. If an equivalent equation results, the graph is symmetric with respect to the polar axis. P Symmetry with Respect to the Line U = ((y-Axis) (y -Axis) 2 In a polar equation, replace u by p - u. If an equivalent equation results, p the graph is symmetric with respect to the line u = . 2 Symmetry with Respect to the Pole (Origin) In a polar equation, replace r by - r. If an equivalent equation results, the graph is symmetric with respect to the pole.

The three tests for symmetry given here are sufficient conditions for symmetry, but they are not necessary conditions. That is, an equation may fail these tests and p still have a graph that is symmetric with respect to the polar axis, the line u = , 2 or the pole. For example, the graph of r = sin 12u2 turns out to be symmetric with p respect to the polar axis, the line u = , and the pole, but all three tests given here 2 fail. See also Problems 87–89.

630

CHAPTER 8  Polar Coordinates; Vectors

3 Graph Polar Equations by Plotting Points EX A MPL E 7

Graphing a Polar Equation (Cardioid) Graph the equation: r = 1 - sin u

Solution Table 1 U p 2 p 3

-

p 6

Check for symmetry first. Polar Axis:
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u by - u. The result is r = 1 - sin 1 - u2 = 1 + sin u

r = 1 − sin U 1 - ( - 1) = 2 23 b ≈ 1.87 1 - a2 1 3 1 - a- b = 2 2

0 p 6

1 - 0 = 1 1 1 1 = 2 2

p 3

1 -

p 2

1 - 1 = 0

sin (- u) = - sin u

The test fails, so the graph may or may not be symmetric with respect to the polar axis. The Line U =

P :
3FQMBDF
u by p - u. The result is 2

r = 1 - sin 1p - u2 = 1 - 1sin p cos u - cos p sin u2

= 1 - 3 0 # cos u - 1 - 12 sin u4 = 1 - sin u

23 ≈ 0.13 2

The test is satisfied, so the graph is symmetric with respect to the line u =

p . 2

The Pole:
3FQMBDF
r by - r. Then the result is - r = 1 - sin u, so r = - 1 + sin u. The test fails, so the graph may or may not be symmetric with respect to the pole. Next, identify points on the graph by assigning values to the angle u and calculating the corresponding values of r. Due to the symmetry with respect to the line p p p u = , just assign values to u from - to , as given in Table 1. 2 2 2 Now plot the points 1r, u2 from Table 1 and trace out the graph, beginning at p p the point a2, - b and ending at the point a0, b . Then reflect this portion of the 2 2 p graph about the line u = (the y-axis) to obtain the complete graph. See Figure 27. 2 Figure 27 r = 1 - sin u

y   ⫽ 3–– 4

–  ⫽ 2

(0.13, –3 ) ⫽

– ⫽ 4

 ( ––12 , –– 6 ) (1, 0) 1 2

O

3

x 

⫽0

( ––2 , ⫺––6 )

Exploration Graph r1 = 1 + sin u. Clear the screen and graph r1 = 1 - cos u. Clear the screen and graph r1 = 1 + cos u. Do you see a pattern?

DEFINITION

  ⫽ 5–– 4

 –

(2, ⫺ 2 )   ⫽ 3––

(1.87, ⫺ –3 )   ⫽ 7–– 4

r

2

The curve in Figure 27 is an example of a cardioid (a heart-shaped curve). Cardioids are characterized by equations of the form r = a11 + cos u2

r = a11 + sin u2

r = a11 - cos u2

r = a11 - sin u2

where a 7 0. The graph of a cardioid passes through the pole.

Now Work

PROBLEM

37

SECTION 8.2   Polar Equations and Graphs

EXAM PL E 8

631

Graphing a Polar Equation (Limaçon without an Inner Loop) Graph the equation: r = 3 + 2 cos u

Solution

Check for symmetry first. Polar Axis: 3FQMBDF
u by - u. The result is r = 3 + 2 cos 1 - u2 = 3 + 2 cos u

cos (- u) = cos u

The test is satisfied, so the graph is symmetric with respect to the polar axis. P : 3FQMBDF
u by p - u. The result is 2

The Line U =

Table 2 U

r = 3 + 2 cos U

0

3 + 2(1) = 5

p 6

23 b ≈ 4.73 3 + 2a 2

p 3

1 3 + 2a b = 4 2

p 2

3 + 2(0) = 3

2p 3

1 3 + 2a - b = 2 2

5p 6

3 + 2a -

p

3 + 2( - 1) = 1

23 b ≈ 1.27 2

r = 3 + 2 cos 1p - u2 = 3 + 21cos p cos u + sin p sin u2 = 3 - 2 cos u The test fails, so the graph may or may not be symmetric with respect to the p line u = . 2 The Pole: 3FQMBDF
r by - r. The test fails, so the graph may or may not be symmetric with respect to the pole. Next, identify points on the graph by assigning values to the angle u and calculating the corresponding values of r. Due to the symmetry with respect to the polar axis, just assign values to u from 0 to p, as given in Table 2. Now plot the points 1r, u2 from Table 2 and trace out the graph, beginning at the point 15, 02 and ending at the point 11, p2. Then reflect this portion of the graph about the polar axis (the x-axis) to obtain the complete graph. See Figure 28.

Figure 28 r = 3 + 2 cos u

y   ⫽ 3–– 4

–  ⫽ 2

(4, –3 )

(3, –2 ) (2, 2––3)

⫽

(1.27, 5––6 ) (1, ) O

  ⫽ 5–– 4

Exploration Graph r1 = 3 - 2 cos u. Clear the screen and graph r1 = 3 + 2 sin u. Clear the screen and graph r1 = 3 - 2 sin u. Do you see a pattern?

DEFINITION

– ⫽ 4

(4.73, –6 )

(5, 0) x 1 2 3 4 5 ⫽0

  ⫽ 3–– 2

  ⫽ 7–– 4

r

The curve in Figure 28 is an example of a limaçon (a French word for snail) without an inner loop. Limaçons without an inner loop are characterized by equations of the form r = a + b cos u

r = a + b sin u

r = a - b cos u

r = a - b sin u

where a 7 0, b 7 0, and a 7 b. The graph of a limaÇon without an inner loop does not pass through the pole.

Now Work

PROBLEM

43

632

CHAPTER 8  Polar Coordinates; Vectors

EX A MPL E 9

Graphing a Polar Equation (Limaçon with an Inner Loop) Graph the equation: r = 1 + 2 cos u

Solution

First, check for symmetry. Polar Axis:
3FQMBDF
u by - u. The result is r = 1 + 2 cos 1 - u2 = 1 + 2 cos u

Table 3 U

r = 1 + 2 cos U

0

1 + 2(1) = 3

p 6

1 + 2a

p 3

1 1 + 2a b = 2 2

p 2

1 + 2(0) = 1

2p 3

1 1 + 2a - b = 0 2

5p 6

1 + 2a -

p

1 + 2( - 1) = - 1

23 b ≈ 2.73 2

The test is satisfied, so the graph is symmetric with respect to the polar axis. The Line U =

P :
3FQMBDF
u by p - u. The result is 2

r = 1 + 2 cos 1p - u2 = 1 + 21cos p cos u + sin p sin u2 = 1 - 2 cos u The test fails, so the graph may or may not be symmetric with respect to the line p u = . 2

23 b ≈ - 0.73 2

The Pole: 3FQMBDF
r by - r. The test fails, so the graph may or may not be symmetric with respect to the pole. Next, identify points on the graph of r = 1 + 2 cos u by assigning values to the angle u and calculating the corresponding values of r. Due to the symmetry with respect to the polar axis just assign values to u from 0 to p, as given in Table 3. Now plot the points 1r, u2 from Table 3, beginning at 13, 02 and ending at 1 - 1, p2. See Figure 29(a). Finally, reflect this portion of the graph about the polar axis (the x-axis) to obtain the complete graph. See Figure 29(b).

Figure 29

y



y

–  ⫽ 2

 ⫽3–– 4

(2,  –

 – 3

(1, 2 ) (0, 2––3 )

⫽

5 ⫺0.73, –– 6

( 

Exploration Graph r1 = 1 - 2 cos u. Clear the screen and graph r1 = 1 + 2 sin u. Clear the screen and graph r1 = 1 - 2 sin u. Do you see a pattern?

DEFINITION

 ⫽5–– 4

–  ⫽ 4



–  ⫽ 2

 ⫽3–– 4

(2, –3 ) (1, –2 ) (0, 2––3 )

) ( 2.73, – ) 6 2

)

(3, 0) 4

x ⫽0

⫽

(⫺1, )

  ⫽3–– 2



5 ⫺0.73, –– 6

(

 ⫽ 7–– 4



 ⫽5–– 4

–  ⫽ 4

( 2.73, –6 ) 2

(3, 0) 4

x ⫽0

) (⫺1, )

  ⫽3–– 2

  ⫽ 7–– 4

(b) r ⫽ 1 ⫹ 2 cos 

(a)

The curve in Figure 29(b) is an example of a limaçon with an inner loop.

Limaçons with an inner loop are characterized by equations of the form r = a + b cos u

r = a + b sin u

r = a - b cos u

r = a - b sin u

where a 7 0, b 7 0, and a 6 b. The graph of a limaçon with an inner loop will pass through the pole twice.

Now Work

PROBLEM

45

r

SECTION 8.2   Polar Equations and Graphs

EX AM PL E 10

Solution

633

Graphing a Polar Equation (Rose) Graph the equation: r = 2 cos 12u2 Check for symmetry. Polar Axis: 3FQMBDF
u by - u. The result is r = 2 cos 3 21 - u2 4 = 2 cos 12u2 The test is satisfied, so the graph is symmetric with respect to the polar axis. P :
3FQMBDF
u by p - u. The result is 2

The Line U =

r = 2 cos 3 21p - u2 4 = 2 cos 12p - 2u2 = 2 cos 12u2 p . 2 The Pole: Since the graph is symmetric with respect to both the polar axis and the p line u = , it must be symmetric with respect to the pole. 2 The test is satisfied, so the graph is symmetric with respect to the line u =

Table 4 U

r = 2 cos(2U)

0

2(1) = 2

p 6

1 2a b = 1 2

p 4

2(0) = 0

p 3

1 2a - b = - 1 2

p 2

2( - 1) = - 2

Next, construct Table 4. Due to the symmetry with respect to the polar axis, the p p line u = , and the pole, just consider values of u from 0 to . 2 2 Plot and connect these points as shown in Figure 30(a). Finally, because of symmetry, reflect this portion of the graph first about the polar axis (the x-axis) and p then about the line u = (the y-axis) to obtain the complete graph. See Figure 30(b). 2

Figure 30

y 2

  ⫽3–– 4

Exploration

Graph r1 = 2 cos 14u2 ; clear the screen and graph r1 = 2 cos 16u2 . How many petals did each of these graphs have? Clear the screen and graph, in order, each on a clear screen, r1 = 2 cos 13u2 , r1 = 2 cos 15u2 , and r1 = 2 cos 17u2 . What do you notice about the number of petals?

(0,

⫽

y

–  ⫽

 ⫺1, –

(

3

(1,  – 4

–  ⫽ 4

 – 6

)

)

(⫺2,   ⫽5–– 4 

 – 2

x ⫽0

)

 ⫽3–– 2

–  ⫽ 4

(1, –6 )

(2, 0)

1 2 3 4 5

)



–  ⫽ 2

 ⫽3–– 4

(2, 0) x 2 3 4 5 ⫽0

⫽

(⫺1, –3 )

(⫺2, –2 )

  ⫽5–– 4

  ⫽ 7–– 4



 ⫽3–– 2

  ⫽ 7–– 4

(b) r ⫽ 2 cos (2)

(a)

The curve in Figure 30(b) is called a rose with four petals.

DEFINITION

Rose curves are characterized by equations of the form r = a cos 1nu2,

r = a sin 1nu2,

a ≠ 0

and have graphs that are rose shaped. If n ≠ 0 is even, the rose has 2n petals; if n ≠ {1 is odd, the rose has n petals.

Now Work

PROBLEM

49

r

634

CHAPTER 8  Polar Coordinates; Vectors

EX AM PL E 11

Solution Table 5 U

r2 = 4 sin(2U)

r

0

4(0) = 0

0

p 6

4a

p 4

4(1) = 4

{2

p 3

23 b = 223 4a 2

{ 1.9

p 2

4(0) = 0

23 b = 223 2

{ 1.9

0

Graphing a Polar Equation (Lemniscate) Graph the equation: r 2 = 4 sin 12u2

We leave it to you to verify that the graph is symmetric with respect to the pole. Because of the symmetry with respect to the pole, consider only those values of u between u = 0 and u = p. Note that there are no points on the graph for p 6 u 6 p (quadrant II), since sin 12u2 6 0 for such values. Table 5 lists points on 2 p the graph for values of u = 0 through u = . The points from Table 5 where r Ú 0 2 are plotted in Figure 31(a). The remaining points on the graph may be obtained by using symmetry. Figure 31(b) shows the final graph drawn.

Figure 31 y



 ⫽3–– 4

⫽

  ⫽5–– 4

y

–  ⫽ 2

(1.9,–3 )

(0, 0)

  ⫽3–– 2

1

–  ⫽ 4  –



(2, 4 ) (1.9,–6 ) 2

x ⫽0

⫽

(1.9,–3 )

(⫺ 1.9, –6 ) (⫺ 2, –4 )   ⫽5–– 4

  ⫽ 7–– 4

–  ⫽ 2

 ⫽3–– 4

–  ⫽ 4

(2, –4 ) (1.9,–6 )

(0, 0)1

(⫺ 1.9, –3 )   ⫽3–– 2

2

  ⫽ 7–– 4

(b) r 2 ⫽ 4 sin (2)

(a)

x ⫽0

r

The curve in Figure 31(b) is an example of a lemniscate (from the Greek word ribbon).

DEFINITION

Lemniscates are characterized by equations of the form r 2 = a2 sin 12u2

r 2 = a2 cos 12u2

where a ≠ 0, and have graphs that are propeller shaped.

Now Work

EX AM PL E 12

PROBLEM

53

Graphing a Polar Equation (Spiral) Graph the equation: r = e u>5

Solution

p 2 fail. Furthermore, there is no number u for which r = 0, so the graph does not pass through the pole. Observe that r is positive for all u, r increases as u increases, r S 0 as u S - q , and r S q as u S q . With the help of a calculator, the values JO
5BCMF

DBO
CF
PCUBJOFE
4FF
'JHVSF


The tests for symmetry with respect to the pole, the polar axis, and the line u =

635

SECTION 8.2   Polar Equations and Graphs

Figure 32 r = eu/5

Table 6

-

U>5

U

r = e

3p 2

0.39

-p p 2 p 4

y



0.53

–  ⫽ 2

 ⫽3–– 4

(1.37, –2 ) (1.17, –4 )

0.73

(1.87, )

⫽

(1, 0) (3.51, 2) x 2 4 ⫽0

0.85

0 p 4 p 2

1

p

1.87

3p 2

2.57

2p

3.51

–  ⫽ 4



1.17

 ⫽5–– 4

(2.57, 3––2 )   ⫽3–– 2

  ⫽ 7–– 4

r

1.37

The curve in Figure 32 is called a logarithmic spiral, since its equation may be written as u = 5 ln r and it spirals infinitely both toward the pole and away from it.

Classification of Polar Equations The equations of some lines and circles in polar coordinates and their corresponding equations in rectangular coordinates are given in Table 7. Also included are the names and graphs of a few of the more frequently encountered polar equations.

Table 7 Lines Description

Line passing through the pole making an angle a with the polar axis

Rectangular equation

y = (tan a)x

Polar equation

u = a

Typical graph

Vertical line

y

Horizontal line

x = a

y = b

r cos u = a

r sin u = b

y

y

 x

x

x

Circles Description

Center at the pole, radius a

Passing through the pole, p tangent to the line u = , 2 center on the polar axis, radius a

Passing through the pole, tangent to the polar axis, p center on the line u = , 2 radius a

Rectangular equation

x2 + y2 = a2, a 7 0

x2 + y2 = { 2ax, a 7 0

x2 + y2 = { 2ay, a 7 0

Polar equation

r = a, a 7 0

r = { 2a cos u, a 7 0

r = { 2a sin u, a 7 0

Typical graph

y

y

y

a a

a x

x

x

636

CHAPTER 8  Polar Coordinates; Vectors

Table 7 (Continued) Other Equations Name

Cardioid

Limaçon without inner loop

Limaçon with inner loop

Polar equations

r = a { a cos u, a 7 0

r = a { b cos u, 0 6 b 6 a

r = a { b cos u, 0 6 a 6 b

r = a { a sin u, a 7 0

r = a { b sin u, 0 6 b 6 a

r = a { b sin u, 0 6 a 6 b

Typical graph

y

y

x

y

x

x

Name

Lemniscate

Rose with three petals

Rose with four petals

Polar equations

r2 = a2 cos(2u), a ≠ 0

r = a sin(3u), a 7 0

r = a sin(2u), a 7 0

r2 = a2 sin(2u), a ≠ 0

r = a cos(3u), a 7 0

r = a cos(2u), a 7 0

Typical graph

y

y

x

y

x

x

Sketching Quickly If a polar equation involves only a sine (or cosine) function, you can quickly obtain a sketch of its graph by making use of Table 7, periodicity, and a short table.

EX AM PL E 13

Sketching the Graph of a Polar Equation Quickly Graph the equation: r = 2 + 2 sin u

Solution

You should recognize the polar equation: Its graph is a cardioid. The period of sin u is 2p, so form a table using 0 … u … 2p, compute r, plot the points 1r, u2 , and sketch the graph of a cardioid as u varies from 0 to 2p. See Table 8 and Figure 33.

Table 8 U

r = 2 + 2 sin U

0 p 2

2 + 2(0) = 2 2 + 2(1) = 4

p

2 + 2(0) = 2

3p 2

2 + 2( - 1) = 0

2p

2 + 2(0) = 2

Figure 33 r = 2 + 2 sin u y

   3–– 4



(2, )

   5–– 4

–   2

(4, ––2 )  – 4

(2, 0) 1 2 3 4 5

x 0

(0, 3––2)    3–– 2

   7–– 4

r

SECTION 8.2   Polar Equations and Graphs

637

Calculus Comment For those of you who are planning to study calculus, a comment about one important role of polar equations is in order. In rectangular coordinates, the equation x2 + y2 = 1, whose graph is the unit circle, is not the graph of a function. In fact, it requires two functions to obtain the graph of the unit circle: y1 = 21 - x2

Upper semicircle

y2 = - 21 - x2

Lower semicircle

In polar coordinates, the equation r = 1, whose graph is also the unit circle, does define a function. For each choice of u, there is only one corresponding value of r—that is, r = 1. Since many problems in calculus require the use of functions, the opportunity to express relations that are nonfunctions in rectangular coordinates as functions in polar coordinates becomes extremely useful. Note also that the vertical-line test for functions is valid only for equations in rectangular coordinates.

Historical Feature

P

olar coordinates seem to have been invented by Jakob Bernoulli (1654–1705) in about 1691, although, as with most such ideas, earlier traces of the notion exist. Early users of calculus remained committed to rectangular coordinates, and polar coordinates did not become widely used until the early

1800s. Even then, it was mostly geometers who used them for describing odd curves. Finally, about the mid-1800s, applied mathematicians realized the tremendous simplification that polar coordinates make possible in the description of objects with circular or cylindrical symmetry. From then on, their use became widespread.

Jakob Bernoulli (1654–1705)

8.2 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. If the rectangular coordinates of a point are 14, - 2, the point symmetric to it with respect to the origin is 1 - 4, 2 . QQ
m

2. The difference formula for cosine is cos 1A - B2 = ______. (p. 531) cos A cos B + sin A sin B

3. The standard equation of a circle with center at 1 - 2, 52 and radius 3 is ______.
QQ
m 1x + 22 2 + 1y - 52 2 = 9

4. Is the sine function even, odd, or neither? Q


5. sin

5p = ______. (pp. 430–432) 4

-

22 2

6. cos

2p = ______. (pp. 430–432) 3

-

1 2

Odd

Concepts and Vocabulary 7. An equation whose variables are polar coordinates is called a(n) polar equation. 8. True or False The tests for symmetry in polar coordinates are necessary but not sufficient. False 9. To test whether the graph of a polar equation may be symmetric with respect to the polar axis, replace u by - u .

10. To test whether the graph of a polar equation may be symmetric p with respect to the line u = , replace u by p - u . 2 11. True or False A cardioid passes through the pole. True 12.
3PTF
DVSWFT
BSF
DIBSBDUFSJ[FE
CZ
FRVBUJPOT
PG
UIF
GPSN
r = a cos (n u) or r = a sin (n u), a ≠ 0. If n ≠ 0 is even, the rose has 2n petals; if n ≠ {1 is odd, the rose has n petals.

638

CHAPTER 8  Polar Coordinates; Vectors

Skill Building In Problems 13–28, transform each polar equation to an equation in rectangular coordinates. Then identify and graph the equation. *13. r = 4

*14. r = 2

*15. u =

p 3

*16. u = -

p 4

*17. r sin u = 4

*18. r cos u = 4

*19. r cos u = - 2

*20. r sin u = - 2

*21. r = 2 cos u

*22. r = 2 sin u

*23. r = - 4 sin u

*24. r = - 4 cos u

*25. r sec u = 4

*26. r csc u = 8

*27. r csc u = - 2

*28. r sec u = - 4

In Problems 29–36, match each of the graphs (A) through (H) to one of the following polar equations. 29. r = 2

30. u =

E

33. r = 1 + cos u

   3–– 4



   5–– 4 

   3–– 4



   5–– 4

–   4

x 0

2

O

  3–– 2

31. r = 2 cos u

A

34. r = 2 sin u

H

y –   2

p 4

   3–– 4

G

y –   2



O

   5–– 4

   7–– 4

3p 35. u = 4



1

  3–– 2

–   4

D

   3–– 4

x 0

3

   7–– 4

F

y –   2



O

   5–– 4 

32. r cos u = 2

B

36. r sin u = 2

C

   3–– 4

–  4

x 0

2



O

   5–– 4

   7–– 4

  3–– 2

y –   2



2

  3–– 2

(A)

(B)

(C)

(D)

y –   2

y –   2

y –   2

y –   2

O

1

   3–– 2

–  4

3

x 0

   7–– 4

   3–– 4



   5–– 4

(E)

O

1

   3–– 2

   3–– 4

–  4

3

x 0

   7–– 4



   5–– 4

O

   3–– 2

–   4

2

4

   3–– 4

x 0

   7–– 4



   5–– 4

O

   3–– 2

(G)

(F)

–  4

x 0

4

   7–– 4

–   4

x 2 0

   7–– 4

(H)

In Problems 37–60, identify and graph each polar equation. *37. r = 2 + 2 cos u

*38. r = 1 + sin u

*39. r = 3 - 3 sin u

*40. r = 2 - 2 cos u

*41. r = 2 + sin u

*42. r = 2 - cos u

*43. r = 4 - 2 cos u

*44. r = 4 + 2 sin u

*45. r = 1 + 2 sin u

*46. r = 1 - 2 sin u

*47. r = 2 - 3 cos u

*48. r = 2 + 4 cos u

*53. r = 9 cos 12u2

*54. r = sin12u2

*55. r = 2

*58. r = 3 + cos u

*59. r = 1 - 3 cos u

*49. r = 3 cos 12u2 2

*57. r = 1 - cos u

*50. r = 2 sin 13u2 2

*51. r = 4 sin 15u2 u

*52. r = 3 cos 14u2 *56. r = 3u

*60. r = 4 cos 13u2

Mixed Practice In Problems 61–66, graph each pair of polar equations on the same polar grid. Find the polar coordinates of the point(s) of intersection. *61. r = 8 cos u; r = 2 sec u

*62. r = 8 sin u; r = 4 csc u

*63. r = sin u; r = 1 + cos u

*64. r = 3; r = 2 + 2 cos u

*65. r = 1 + sin u; r = 1 + cos u

*66. r = 1 + cos u; r = 3 cos u

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

SECTION 8.2   Polar Equations and Graphs

639

Applications and Extensions In Problems 67–70, the polar equation for each graph is either r = a + b cos u or r = a + b sin u, a 7 0. Select the correct equation and find the values of a and b. 67.

r = 3 + 3 cos u

y

68.

2

   3–– 4

(

 3, – 2

)

– 

0 2 4 6 8 10

   5–– 4

2

   3–– 4

–4   x

(6, 0)

 

 

 0

   7–– 4

(3, –2 )

(

2

   3–– 4

)

   7–– 4

0 1 2 3 4 5

   5–– 4

70.    3–– 4

x

 0

*77. r = csc u - 2, 0 6 u 6 p (conchoid) *79. r = tan u,

-

p p 6 u 6 (kappa curve) 2 2

–2 

   5–– 4

   3–– 2

*75. r = u, u Ú 0 (spiral of Archimedes)

(5, –2 )

–4  

(1, 0) 0 1 2 3 4 5

 

   7–– 4

In Problems 71–80, graph each polar equation. 2 *71. r = (parabola) 1 - cos u 1 *73. r = (ellipse) 3 - 2 cos u

r = 1 + 4 sin u

y

–4 

(4, 0)

 

 0

   3–– 2

r = 4 + sin u

–2  

x 0 2 4 6 8 10

   5–– 4

y

 5, –

–4  

(6, )

   3–– 2

69.

r = 3 - 3 cos u

y

– 

x

 0

   7–– 4    3–– 2

2 (hyperbola) 1 - 2 cos u 1 *74. r = (parabola) 1 - cos u 3 *76. r = (reciprocal spiral) u *72. r =

*78. r = sin u tan u (cissoid) *80. r = cos

u 2

*81. Show that the graph of the equation r sin u = a is a horizontal line a units above the pole if a Ú 0 and 0 a 0 units below the pole if a 6 0.

*82. Show that the graph of the equation r cos u = a is a vertical line a units to the right of the pole if a Ú 0 and 0 a 0 units to the left of the pole if a 6 0.

*83. Show that the graph of the equation r = 2a sin u, a 7 0, is a circle of radius a with center at 10, a2 in rectangular coordinates.

*84. Show that the graph of the equation r = - 2a sin u, a 7 0, is a circle of radius a with center at 10, - a2 in rectangular coordinates.

*85. Show that the graph of the equation r = 2a cos u, a 7 0, is a circle of radius a with center at 1a, 02 in rectangular coordinates.

*86. Show that the graph of the equation r = - 2a cos u, a 7 0, is a circle of radius a with center at 1 - a, 02 in rectangular coordinates.

Discussion and Writing 87.
&YQMBJO
XIZ
UIF
GPMMPXJOH
UFTU
GPS
TZNNFUSZ
JT
WBMJE
3FQMBDF
r by - r and u by - u in a polar equation. If an equivalent equation results, the graph is symmetric with respect to the p line u = (y-axis). 2 * B

4IPX
UIBU
UIF
UFTU
PO
QBHF

GBJMT
GPS
r 2 = cos u, yet this new test works. * C

4IPX
UIBU
UIF
UFTU
PO
QBHF

XPSLT
GPS
r 2 = sin u, yet this new test fails.

88. Write down two different tests for symmetry with respect to the polar axis. Find examples in which one test works and the other fails. Which test do you prefer to use? Justify your answer. 89.
5IF
UFTUT
GPS
TZNNFUSZ
HJWFO
PO
QBHF

BSF
TVGGJDJFOU
CVU
not necessary. Explain what this means. 90. Explain why the vertical-line test used to identify functions in rectangular coordinates does not work for equations expressed in polar coordinates.

640

CHAPTER 8  Polar Coordinates; Vectors

Retain Your Knowledge Problems 91–94 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 91. Using the given graph of y = f 1x2 , find the absolute maximum and the absolute minimum, if it exists. Absolute maximum: f(4) = 1; y Absolute minimum; none 4

(4, 1) 26

(1, 21)

O

6 x

(25, 22)

7p radians to degrees. 420° 3 93. Determine the amplitude and period of y = -2 sin (5x) without graphing. 94. Find any asymptotes for the graph of x + 3 . R 1x2 = 2 x - x - 12 Horizontal asymptote: y = 0; Vertical asymptote: x = 4 92. Convert

93. Amplitude = 2; Period =

2p 5

(4, 25) (23, 26) 28

‘Are You Prepared?’ Answers 1. 1 - 4, 2

2. cos A cos B + sin A sin B

3. 1x + 22 2 + 1y - 52 2 = 9

4. Odd

5. -

22 2

6. -

1 2

8.3 The Complex Plane; De Moivre’s Theorem PREPARING FOR THIS SECTION  Before getting started, review the following: r
$PNQMFY
/VNCFST
"QQFOEJY
"
4FDUJPO
"
pp. A89–A94) r
7BMVF
PG
UIF
4JOF
BOE
$PTJOF
'VODUJPOT
BU
$FSUBJO
Angles (Section 5.2, pp. 425–432)

r
4VN
BOE
%JGGFSFODF
'PSNVMBT
GPS
4JOF
BOE
$PTJOF
4FDUJPO

QQ

BOE


Now Work the ‘Are You Prepared?’ problems on page 646. OBJECTIVES 1 Plot Points in the Complex Plane (p. 640) 2 Convert a Complex Number between Rectangular Form and Polar Form (p. 641) 3 Find Products and Quotients of Complex Numbers in Polar Form (p. 642) 4 Use De Moivre’s Theorem (p. 643) 5 Find Complex Roots (p. 644)

1 Plot Points in the Complex Plane Figure 34 Imaginary axis

z  x  yi

y

O

x

Real axis

Complex numbers are discussed in Appendix A, Section A.11. In that discussion, we were not prepared to give a geometric interpretation of a complex number. Now we are ready. A complex number z = x + yi can be interpreted geometrically as the point 1x, y2 in the xy-plane. Each point in the plane corresponds to a complex number, and conversely, each complex number corresponds to a point in the plane. The collection of such points is referred to as the complex plane. The x-axis is referred to as the real axis, because any point that lies on the real axis is of the form z = x + 0i = x, a real number. The y-axis is called the imaginary axis, because any point that lies on it is of the form z = 0 + yi = yi, a pure imaginary number. See Figure 34.

SECTION 8.3  The Complex Plane; De Moivre’s Theorem

641

Plotting a Point in the Complex Plane

EXAM PL E 1

Plot the point corresponding to z = 13 - i in the complex plane.

Solution

The point corresponding to z = 23 - i has the rectangular coordinates 1 13, - 12 . This point, located in quadrant IV, is plotted in Figure 35.

Figure 35 Imaginary axis 2 2

O 2

0z0 = 2 2x2 + y2

Figure 36 Imaginary axis 2

 y z  x  yi x  y ⏐z ⏐ Real axis x O

z 3i

r

Let z = x + yi be a complex number. The magnitude or modulus of z, denoted by 0 z 0 , is defined as the distance from the origin to the point 1x, y2. That is,

DEFINITION

2

Real axis

2

(1)

4FF
'JHVSF

GPS
BO
JMMVTUSBUJPO This definition for 0 z 0 is consistent with the definition for the absolute value of a real number: If z = x + yi is real, then z = x + 0i and

0 z 0 = 2x2 + 02 = 2x2 = 0 x 0

For this reason, the magnitude of z is sometimes called the absolute value of z. 3FDBMM
UIBU
JG
z = x + yi, then its conjugate, denoted by z, is z = x - yi. Since zz = x2 + y2, which is a nonnegative real number, it follows from equation (1) that the magnitude of z can be written as

0 z 0 = 2zz

(2)

2 Convert a Complex Number between Rectangular Form and Polar Form When a complex number is written in the standard form z = x + yi, we say that it is in rectangular, or Cartesian, form, because 1x, y2 are the rectangular coordinates of the corresponding point in the complex plane. Suppose that 1r, u2 are the polar coordinates of this point. Then x = r cos u,

z = x + yi = 1rr cos u2 + 1rr sin u2i = r 1cos u + i sin u2

Figure 37 Imaginary axis

z r x



y Real axis

z  x  yi  r (cos   i sin ), r ≥ 0, 0 ≤   2

(3)

If r Ú 0 and 0 … u 6 2p, the complex number z = x + yi may be written in polar form as

DEFINITION

O

y = r sin u

(4)

See Figure 37. If z = r 1cos u + i sin u2 is the polar form of a complex number,* the angle u, 0 … u 6 2p, is called the argument of z. Also, because r Ú 0, we have r = 2x2 + y2 . From equation (1), it follows that the magnitude of z = r 1cos u + i sin u2 is

0z0 = r Some texts abbreviate the polar form using z = r(cos u + i sin u) = r cis u.

*

642

CHAPTER 8  Polar Coordinates; Vectors

EX A MPL E 2

Writing a Complex Number in Polar Form Write an expression for z = 23 - i in polar form.

Solution

The point, located in quadrant IV, is plotted in Figure 35. Because x = 23 and y = - 1, it follows that r = 2x2 + y2 = 3 1 23 2 + 1 - 12 2 = 24 = 2 2

so sin u =

y -1 = , r 2

cos u =

23 x = , r 2

0 … u 6 2p

The angle u, 0 … u 6 2p, that satisfies both equations is u = and r = 2, the polar form of z = 23 - i is z = r 1cos u + i sin u2 = 2acos

Now Work

EX A MPL E 3

PROBLEM

11p 11p . With u =  

11p 11p + i sin b  

r

11

Plotting a Point in the Complex Plane and Converting from Polar to Rectangular Form Plot the point corresponding to z = 21cos 30° + i sin 30°2 in the complex plane, and write an expression for z in rectangular form.

Solution Figure 38

z = 21cos 30° + i sin 30°2 = 2a

Imaginary axis 2

z  2(cos 30°  i sin 30°) 2

O

To plot the complex number z = 21cos 30° + i sin 30°2, plot the point whose polar coordinates are 1r, u2 = 12, 30°2, as shown in Figure 38. In rectangular form,

30° 2

Real axis

2

Now Work

PROBLEM

23 1 + ib = 23 + i 2 2

r

23

3 Find Products and Quotients of Complex Numbers in Polar Form The polar form of a complex number provides an alternative method for finding products and quotients of complex numbers.

THEOREM In Words The magnitude of a complex number z is r and its argument is u, so when z = r (cos u + i sin u) the magnitude of the product (quotient) of two complex numbers equals the product (quotient) of their magnitudes; the argument of the product (quotient) of two complex numbers is determined by the sum (difference) of their arguments.

Let z1 = r1 1cos u1 + i sin u1 2 and z2 = r2 1cos u2 + i sin u2 2 be two complex numbers. Then z1 z2 = r1 r2 3 cos 1u1 + u2 2 + i sin 1u1 + u2 2 4

(5)

If z2 ≠ 0, then z1 r1 = 3 cos 1u1 - u2 2 + i sin 1u1 - u2 2 4 z2 r2

(6)

Proof 8F
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  z1 z2 = 3 r1 1cos u1 + i sin u1 2 4 3 r2 1cos u2 + i sin u2 2 4 = r1 r2 3 1cos u1 + i sin u1 2 1cos u2 + i sin u2 2 4

= r1 r2 3 1cos u1 cos u2 - sin u1 sin u2 2 + i1sin u1 cos u2 + cos u1 sin u2 2 4

= r1 r2 3 cos 1u1 + u2 2 + i sin 1u1 + u2 2 4

Let’s look at an example of how this theorem can be used.

■

SECTION 8.3  The Complex Plane; De Moivre’s Theorem

EXAM PL E 4

643

Finding Products and Quotients of Complex Numbers in Polar Form If z = 31cos 20° + i sin 20°2 and w = 51cos 100° + i sin 100°2, find the following (leave your answers in polar form): (a) zw

Solution

z w

(b)

(a) zw = 3 31cos 20° + i sin 20°2 4 3 51cos 100° + i sin 100°2 4 = 13 # 52 3 cos 120° + 100°2 + i sin 120° + 100°2 4

Apply equation (5).

= 151cos 120° + i sin 120°2 (b)

31cos 20° + i sin 20°2 z = w 51cos 100° + i sin 100°2 3 3 cos 120° - 100°2 + i sin 120° - 100°2 4 5 3 = 3 cos 1 - 80°2 + i sin 1 - 80°2 4 5 3 = 1cos 280° + i sin 280°2 5 =

Now Work

PROBLEM

Apply equation (6).

The argument must lie between 0° and 360°.

r

33

4 Use De Moivre’s Theorem %F
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already known to many people by 1710, is important for the following reason: The fundamental processes of algebra are the four operations of addition, subtraction, multiplication, and division, together with powers and the extraction of roots. De Moivre’s Theorem allows these latter fundamental algebraic operations to be applied to complex numbers. De Moivre’s Theorem, in its most basic form, is a formula for raising a complex number z to the power n, where n Ú 1 is a positive integer. Let’s try to conjecture the form of the result. Let z = r 1cos u + i sin u2 be a complex number. Then, based on equation (5), we have n = 2: z2 = r 2 3 cos 12u2 + i sin 12u2 4

n = 3: z3 = z2 # z

Equation (5)

= 5 r 2 3 cos 12u2 + i sin 12u2 4 6 3 r 1cos u + i sin u2 4 = r 3 3 cos 13u2 + i sin 13u2 4

n = 4: z = z 4

3

#z

Equation (5)

= 5 r 3 3 cos 13u2 + i sin 13u2 4 6 3 r 1cos u + i sin u2 4 = r 4 3 cos 14u2 + i sin 14u2 4

Equation (5)

Do you see the pattern?

THEOREM

De Moivre’s Theorem If z = r 1cos u + i sin u2 is a complex number, then zn = r n 3 cos 1nu2 + i sin 1nu2 4 where n Ú 1 is a positive integer.

(7)

644

CHAPTER 8  Polar Coordinates; Vectors

The proof of De Moivre’s Theorem requires mathematical induction (which is not discussed until Section 11.4), so it is omitted here. The theorem is actually true for all integers, n
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EX A MPL E 5

Solution

Using De Moivre’s Theorem

Write 3 21cos 20° + i sin 20°2 4 3 in the standard form a + bi. 3 21cos 20° + i sin 20°2 4 3 = 23 3 cos 13 # 20°2 + i sin 13 # 20°2 4 Apply De Moivre’s Theorem. = 81cos 0° + i sin 0°2 = 8a

Now Work

EX A MPL E 6

Solution

PROBLEM

1 23 + ib = 4 + 423i 2 2

r

41

Using De Moivre’s Theorem

Write 11 + i2 5 in the standard form a + bi. To apply De Moivre’s Theorem, first write the complex number in polar form. Since the magnitude of 1 + i is 212 + 12 = 22, begin by writing 1 + i = 22 a

1 22

+

1 22

ib = 22 acos

p p + i sin b 4 4

Now 11 + i2 5 = c 22 acos =

NOTE In the solution of Example 6, the approach used in Example 2 could also be used to write 1 + i in polar form. j

p p 5 + i sin b d 4 4

1 22 2 5 c cos a5 #

p p b + i sin a5 # b d 4 4

= 422 acos

5p 5p + i sin b 4 4

= 422 c -

1 22

+ a-

1 22

b i d = - 4 - 4i

r

5 Find Complex Roots Let w be a given complex number, and let n Ú 2 denote a positive integer. Any complex number z that satisfies the equation zn = w is called a complex nth root of w. In keeping with previous usage, if n = 2, the solutions of the equation z2 = w are called complex square roots of w, and if n = 3, the solutions of the equation z3 = w are called complex cube roots of w.

THEOREM

Finding Complex Roots Let w = r 1cos u0 + i sin u0 2 be a complex number, and let n Ú 2 be an integer. If w ≠ 0, there are n distinct complex nth roots of w, given by the formula n

zk = 1r Jcos ¢

u0 u0 2kp 2kp + ≤ + i sin ¢ + ≤R n n n n

where k = 0, 1, 2, c , n - 1.

(8)

SECTION 8.3  The Complex Plane; De Moivre’s Theorem

645

Proof (Outline) We will not prove this result in its entirety. Instead, we shall show only that each zk in equation (8) satisfies the equation znk = w, proving that each zk is a complex nth root of w. znk

n u0 u0 2kp 2kp = b 1r Jcos ¢ + ≤ + i sin ¢ + ≤R r n n n n n

= 1 1r2 n b cos Jn ¢ n

u0 u0 2kp 2kp + ≤ R + i sin Jn ¢ + ≤ R r Apply De Moivre’s Theorem. n n n n

= r 3 cos 1u0 + 2kp2 + i sin 1u0 + 2kp2 4

Simplify.

= r 1cos u0 + i sin u0 2 = w

The Periodic Property

Thus each zk , k = 0, 1, c , n - 1, is a complex nth root of w. To complete the proof, we would need to show that each zk , k = 0, 1, c , n - 1, is, in fact, distinct and that there are no complex nth roots of w other than those given by equation (8). ■

EXAM PL E 7

Finding Complex Cube Roots Find the complex cube roots of - 1 + 23i. Leave your answers in polar form, with the argument in degrees.

Solution

First, express - 1 + 23i in polar form using degrees. - 1 + 23i = 2 a -

WARNING Most graphing utilities will provide only the answer z0 to the calculation 1 - 1 + 23i 2 ^ 11>32. The paragraph following Example 7 explains how to obtain z1 and z2 j from z0.

1 23 + ib = 21cos 120° + i sin 120°2 2 2

The three complex cube roots of - 1 + 23i = 21cos 120° + i sin 120°2 are 3 zk = 2 2 c cos a

120° 30°k 120° 30°k + b + i sin a + bd 3 3 3 3

3 = 2 23 cos 140° + 120°k2 + i sin 140° + 120°k2 4

k = 0, 1, 2

Therefore, z0 = 22 3 cos 140° + 120° # 02 + i sin 140° + 120° # 02 4 = 221cos 40° + i sin 40°2 3

3

3 3 z1 = 2 2 3 cos 140° + 120° # 12 + i sin 140° + 120° # 12 4 = 2 21cos 10° + i sin 10°2 3 3 z2 = 2 2 3 cos 140° + 120° # 22 + i sin 140° + 120° # 22 4 = 2 21cos 280° + i sin 280°2

r Notice that each of the three complex roots of - 1 + 23i has the same 3 magnitude, 2 2. This means that the points corresponding to each cube root lie the same distance from the origin; that is, the three points lie on a circle with center at 3 the origin and radius 2 2. Furthermore, the arguments of these cube roots are 40°, 30° ž
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646

CHAPTER 8  Polar Coordinates; Vectors

Figure 39

Imaginary axis 2 2

2

3

2

x  y  ( 2) z1 

3

2(cos 160°  i sin 160°)

1

z0 

120° 2

1

1

2(cos 40°  i sin 40°)

40°

O 120°

3

1

2

Real axis

120° z2 

3

2(cos 280°  i sin 280°)

2

Now Work

PROBLEM

53

Historical Feature

T

John Wallis

he Babylonians, Greeks, and Arabs considered square roots of negative quantities to be impossible and equations with complex solutions to be unsolvable. The first hint that there was some connection between real solutions of equations and complex numbers came when Girolamo Cardano (1501–1576) and Tartaglia (1499–1557) found real roots of cubic equations by taking cube roots of complex quantities. For centuries

thereafter, mathematicians worked with complex numbers without much belief in their actual existence. In 1673, John Wallis appears to have been the first to suggest the graphical representation of complex numbers, a truly significant idea that was not pursued further until about 1800. Several people, including Karl Friedrich Gauss (1777–1855), then rediscovered the idea, and the graphical representation helped to establish complex numbers as equal members of the number family. In practical applications, complex numbers have found their greatest uses in the study of alternating current, where they are a commonplace tool, and in the field of subatomic physics.

Historical Problems 1. The quadratic formula works perfectly well if the coefficients of a quadratic equation are complex numbers. Solve the following. [Hint: The answers are “nice.”] (a) z2 - (2 + 5i )z - 3 + 5i = 0

1 + 4i, 1 + i

(b) z2 - (1 + i )z - 2 - i = 0

- 1, 2 + i

8.3 Assess Your Understanding ‘Are You Prepared?’  Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. The conjugate of - 4 - 3i is - 4 + 3i . (p. A91) 2. The sum formula for the sine function is sin1A + B2 = _______ . (p. 534) sin A cos B + cos A sin B

Concepts and Vocabulary

3. The sum formula for the cosine function is cos 1A + B2 = _______. (p. 531) cos A cos B - sin A sin B 4. sin 120° = _______; cos 240° = _______ . (pp. 430–432) 1 23 2 2

5. In the complex plane, the x-axis is referred to as the real axis and the y-axis is called the imaginary axis.

8. If z = r(cos u + i sin u) is a complex number, then zn = r n 3 cos 1 nu 2 + i sin 1 nu 2 4 .

6. When a complex number z is written in the polar form z = r 1cos u + i sin u2 , the nonnegative number r is the magnitude or modulus of z, and the angle u, 0 … u 6 2p, is the argument of z.

9. Every nonzero complex number has exactly three distinct cube roots.

7. Let z1 = r1 1cos u1 + i sin u1 2 and z2 = r2 1cos u2 + i sin u2 2 be two complex numbers. Then z1z2 = r1r2 3 cos 1u1 + u2 2 + i sin 1u1 + u2 2 4 .

10. True or False The polar form of a nonzero complex number is unique. True

Skill Building In Problems 11–22, plot each complex number in the complex plane and write it in polar form. Express the argument in degrees. *11. 1 + i

*12. - 1 + i

*13. 23 - i

*14. 1 - 23i

*15. - 3i

*16. - 2

*17. 4 - 4i

*18. 923 + 9i

*19. 3 - 4i

*20. 2 + 23i

*21. - 2 + 3i

*22. 25 - i

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

647

SECTION 8.3  The Complex Plane; De Moivre’s Theorem

In Problems 23–32, write each complex number in rectangular form. 23. 21cos 120° + i sin 120°2

7p 7p + i sin b 222 - 222i 4 4 3p p p 3p 27. 3acos + i sin b - 3i 28. 4acos + i sin b 4i 2 2 2 2 30. 0.41cos 200° + i sin 200°2 - 0.37 - 0.137i

- 1 + 23i 24. 31cos 210° + i sin 210°2

5p 5p + i sin b - 23 + i   *29. 0.21cos 100° + i sin 100°2 26. 2acos

31. 2acos

p p + i sin b 18 18

1.970 + 0.347i 32. 3acos

p p + i sin b 10 10

-

323 3 - i 2 2

25. 4acos

2.853 + 0.927i

z . Leave your answers in polar form. w *34. z = cos 120° + i sin 120° *33. z = 21cos 40° + i sin 40°2 w = cos 100° + i sin 100° w = 41cos 20° + i sin 20°2

In Problems 33–40, find zw and

*35. z = 31cos 130° + i sin 130°2 w = 41cos 270° + i sin 270°2

p p + i sin b 8 8 p p + i sin b w = 2acos 10 10

*36. z = 21cos 80° + i sin 80°2 w = 1cos 200° + i sin 200°2

*37. z = 2acos

*39. z = 2 + 2i w = 23 - i

*40. z = 1 - i w = 1 - 23i

3p 3p + i sin b 8 8 9p 9p + i sin b w = 2acos 1 1

*38. z = 4acos

In Problems 41–52, write each expression in the standard form a + bi. 42. 331cos 80° + i sin 80°2 4 3

*41. 341cos 40° + i sin 40°2 4 3

3 23 1cos 10°

*44. c 22 acos

5p 5p 4 + i sin bd 1 1

*45.

*47. c 25 acos

3p 3p 4 + i sin bd 1 1

*48. c 23 acos

50.

1 23

- i2



- 4

51.

1 22

- i2

+ i sin 10°2 4

-

27 2723 i 2 2

p 5 p + i sin b d 10 10

32i

5 1 46. c 1cos 72° + i sin 72°2 d 2



5p 5p  + i sin bd 18 18 

43. c 2acos

49. 11 - i2 5

- 23 + 14.142i

52.

11

1 32

- 4 + 4i

- 25i 2

8

- 124 - 28.217i

In Problems 53–60, find all the complex roots. Leave your answers in polar form with the argument in degrees. *53. The complex cube roots of 1 + i

*54. The complex fourth roots of 23 - i

*55. The complex fourth roots of 4 - 423i

*56. The complex cube roots of - 8 - 8i

*57. The complex fourth roots of - 1i

*58. The complex cube roots of - 8

*59. The complex fifth roots of i

*60. The complex fifth roots of - i

Applications and Extensions *61. Find the four complex fourth roots of unity (1) and plot them. *62. Find the six complex sixth roots of unity (1) and plot them. *63. Show that each complex nth root of a nonzero complex number w has the same magnitude. *64.
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nth roots of a nonzero complex number w are equally spaced on the circle. *66.

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  *67. Prove that De Moivre’s Theorem is true for all integers, n, by assuming it is true for integers n Ú 1 and then showing it is true for 0 and for negative integers. Hint: Multiply the numerator and the denominator by the conjugate of the denominator and use even-odd properties. *68. Mandelbrot Sets (a) Consider the expression an = 1an - 1 2 2 + z, where z is some complex number (called the seed) and a0 = z. Compute a1 1=a20 + z2, a2 1=a21 + z2, a3 1=a22 + z2, a4 , a5, and a for the following seeds: z1 = 0.1 - 0.4i, z2 = 0.5 + 0.8i, z3 = - 0.9 + 0.7i, z4 = - 1.1 + 0.1i, z5 = 0 - 1.3i, and z = 1 + 1i.

(b) The dark portion of the graph represents the set of all values z = x + yi that are in the Mandelbrot set. Determine which complex numbers in part (a) are in this set by plotting them on the graph. Do the complex numbers that are not in the Mandelbrot set have any common characteristics regarding the values of a found in part (a)? (c) Compute 0 z 0 = 2x2 + y2 for each of the complex numbers in part (a). Now compute 0 a 0 for each of the complex numbers in part (a). For which complex numbers is 0 a 0 … 0 z 0 and 0 z 0 … 2? Conclude that the criterion for a complex number to be in the Mandelbrot set is that 0 an 0 … 0 z 0 and 0 z 0 … 2. Imaginary axis y 1

Real axis 1 x

–2

–1

648

CHAPTER 8  Polar Coordinates; Vectors

Retain Your Knowledge Problems 69–72 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 69. Find the area of the triangle with a = 8, b = 11, and C = 113°. ≈40.50 70. Determine whether f(x) = 5x2 - 12x + 4 has a maximum value or a minimum value, and then find the value. 4 71. Convert 240° to radians. Express your answer as a multiple of p. p 3 3 3 1  72. Simplify: 224x2y5 2y23x2y2 70. Minimum; f a b = 5 5

‘Are You Prepared?’ Answers 1. - 4 + 3i

2. sin A cos B + cos A sin B

3. cos A cos B - sin A sin B

4.

1 13 ;2 2

8.4 Vectors OBJECTIVES 1 2 3 4 5 6 7

Graph Vectors (p. 650) Find a Position Vector (p. 651) Add and Subtract Vectors Algebraically (p. 652) Find a Scalar Multiple and the Magnitude of a Vector (p. 653) Find a Unit Vector (p. 654) Find a Vector from Its Direction and Magnitude (p. 654) Model with Vectors (p. 655)

In simple terms, a vector (derived from the Latin vehere, meaning “to carry”) is a quantity that has both magnitude and direction. It is customary to represent a vector by using an arrow. The length of the arrow represents the magnitude of the vector, and the arrowhead indicates the direction of the vector. Many quantities in physics can be represented by vectors. For example, the velocity of an aircraft can be represented by an arrow that points in the direction of movement; the length of the arrow represents speed. If the aircraft speeds up, we lengthen the arrow; if the aircraft changes direction, we introduce an arrow in the new direction. See Figure 40. Based on this representation, it is not surprising that vectors and directed line segments are somehow related.

Figure 40

Geometric Vectors If P and Q are two distinct points in the xy-plane, there is exactly one line containing both P and Q [Figure 41(a)]. The points on that part of the line that joins P to Q, including P and Q, form what is called the line segment PQ [Figure 41(b)]. Ordering the points so that they proceed from P to Q, we have a> directed line segment from P to > Q, or a geometric vector, which we denote by PQ . In a directed line segment PQ , P is called the initial point and Q the terminal point, as indicated in Figure 41(c). Figure 41

Q

Q

Q

Terminal point

P (a) Line containing P and Q

P (b) Line segment PQ

Initial point

P

(c) Directed line segment PQ

SECTION 8.4   Vectors

649

>

Figure 42 U Q S T P R

Figure 43

The magnitude of the directed line segment PQ is the distance from the point > P to the point Q; that is, it is the length of the line segment. The direction of PQ is from P to Q. If a vector v> * has the same magnitude and the same direction as the directed line segment PQ , we write > v = PQ The vector v whose magnitude is 0 is called the zero vector, 0. The zero vector is assigned no direction. Two vectors v and w are equal, written v = w if they have the same magnitude and the same direction. For example, the three vectors shown in Figure 42 have the same magnitude and the same direction, so they are equal, even though they have different initial points and different terminal points. As a result, it is useful to think of a vector simply as an arrow, keeping in mind that two arrows (vectors) are equal if they have the same direction and the same magnitude (length).

Adding Vectors Geometrically Terminal point of w vw

w

v Initial point of v

The sum v + w of two vectors is defined as follows: Position the vectors v and w so that the terminal point of v coincides with the initial point of w, as shown in Figure 43. The vector v + w is then the unique vector whose initial point coincides with the initial point of v and whose terminal point coincides with the terminal point of w. Vector addition is commutative. That is, if v and w are any two vectors, then v + w = w + v

Figure 44 v v w w w  v

w

Figure 44 illustrates this fact. (The commutative property is another way of saying that opposite sides of a parallelogram are equal and parallel.) Vector addition is also associative. That is, if u, v, and w are vectors, then u + 1v + w2 = 1u + v2 + w

v

Figure 45 illustrates the associative property for vectors. Figure 45

1u + v2 + w = u + 1v + w2 vw

u

The zero vector 0 has the property that

w

v

v + 0 = 0 + v = v

uv

for any vector v. If v is a vector, then - v is the vector that has the same magnitude as v but whose direction is opposite to v
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Figure 46 v

v

v + 1 - v2 = 0 If v and w are two vectors, then the difference v - w is defined as v - w = v + 1 - w2

Figure 47

Figure 47 illustrates the relationships among v, w, v + w, and v - w.

w

v

v w

w

v

w v

Multiplying Vectors by Numbers Geometrically When dealing with vectors, real numbers are referred to as scalars. Scalars are quantities that have only magnitude. Examples of scalar quantities from physics are temperature, speed, and time. We now define how to multiply a vector by a scalar. *

Boldface letters are used to denote vectors, to distinguish them from numbers. For handwritten work, S an arrow is placed over the letter to signify a vector. For example, a vector is written by hand as v .

650

CHAPTER 8  Polar Coordinates; Vectors

DEFINITION

If a is a scalar and v is a vector, the scalar multiple av is defined as follows: 1.. If a 7 0, av is the vector whose magnitude is a times the magnitude of v and whose direction is the same as v. 2.. If a 6 0, av is the vector whose magnitude is 0 a 0 times the magnitude of v and whose direction is opposite that of v. 3. If a = 0 or if v = 0, then av = 0.

Figure 48

2v v

1v

See Figure 48 for some illustrations. For example, if a is the acceleration of an object of mass m due to a force F being exerted on it, then, by Newton’s second law of motion, F = ma. Here, ma is the product of the scalar m and the vector a. Scalar multiples have the following properties: 0v = 0

1v = v

1a + b2v = av + bv

- 1v = - v a1v + w2 = av + aw

a 1bv2 = 1ab2v

1 Graph Vectors EX A MPL E 1

Graphing Vectors Use the vectors illustrated in Figure 49 to graph each of the following vectors: (a) v - w

Solution

(c) 2v - w + u

Figure 50 shows each graph. Figure 50

Figure 49

v

(b) 2v + 3w

w

u

u

2v  w  u

w

2v

w

3w

vw v

2v 2v  3w (b) 2v  3w

(a) v  w

Now Work

PROBLEMS

9

AND

(c) 2v  w  u

11

r

Magnitude of Vectors

The symbol 7 v 7 is used to represent the magnitude of a vector v. Since 7 v 7 equals the length of a directed line segment, it follows that 7 v 7 has the following properties:

THEOREM

Properties of 7 v 7 If v is a vector and if a is a scalar, then

(a) 7 v 7 Ú 0 (c) 7 - v 7 = 7 v 7

(b) 7 v 7 = 0 if and only if v = 0 (d) 7 av 7 = 0 a 0 7 v 7

Property (a) is a consequence of the fact that distance is a nonnegative number. > Property (b) follows because the length of the directed line segment PQ is

SECTION 8.4   Vectors

651

positive unless P and Q are the same point, in which case the length is 0. Property (c) follows because the length of the line segment PQ equals the length of the line segment QP. Property (d) is a direct consequence of the definition of a scalar multiple.

DEFINITION

A vector u for which 7 u 7 = 1 is called a unit vector.

2 Find a Position Vector To compute the magnitude and direction of a vector, an algebraic way of representing vectors is needed.

DEFINITION

An algebraic vector v is represented as v = 8a, b b99

Figure 51 y

,b

v

a =


where a and b are real numbers (scalars) called the components of the vector v.

P  (a, b)

x

THEOREM

A rectangular coordinate system is used to represent algebraic vectors in the plane. If v = 8a, b9 is an algebraic vector whose initial point is at the origin, then v is called a position vector. See Figure 51. Notice that the terminal point of the position vector v = 8a, b9 is P = 1a, b2. The next result states that any vector whose initial point is not at the origin is equal to a unique position vector. Suppose that v is a vector with initial point P1 = 1x1 ,> y1 2, not necessarily the origin, and terminal point P2 = 1x2 , y2 2. If v = P1P2 , then v is equal to the position vector v = 8x2 - x1 , y2 - y1 9

In Words An algebraic vector represents “driving directions” to get from the initial point to the terminal point of a vector. So if v = 85, 4 9 , travel 5 units right and 4 units up from the initial point to arrive at the terminal point.

(1)

To see why this is true, look at Figure 52. Figure 52 v = 8a, b 9 = 8x2 - x1, y2 - y1 9 y P2  (x2, y2) P  (a, b)

b

b

v

A O

a

a

v P1  (x1, y1)

y2  y1 Q

x2  x1

x

Triangle OPA and triangle P1P2Q are congruent. [Do you see why? The line segments have the same magnitude, so d 1O, P2 = d 1P1 , P2 2; and they have the same direction, so ∠POA = ∠P2P1Q. Since the triangles are right triangles, we have angle–side–angle.] It follows that corresponding sides are equal. As a result, x2 - x1 = a and y2 - y1 = b, so v may be written as v = 8a, b9 = 8x2 - x1 , y2 - y1 9

Because of this result, any algebraic vector can be replaced by a unique position vector, and vice versa. This flexibility is one of the main reasons for the wide use of vectors.

652

CHAPTER 8  Polar Coordinates; Vectors

Finding a Position Vector

EX A MPL E 2

>

Find the position vector of the vector v = P1P2 , if P1 = 1 - 1, 22 and P2 = 14, 2 .

Solution

By equation (1), the position vector equal to v is

v = 84 - 1 - 12,  - 29 = 85, 49

r

See Figure 53.

Figure 53 y

Two position vectors v and w are equal if and only if the terminal point of v is the same as the terminal point of w. This leads to the following result:

P2  (4, 6)

5

(5, 4)

THEOREM

v 

P1  (1, 2)

5

O

x

Equality of Vectors Two vectors v and w are equal if and only if their corresponding components are equal. That is, If v = 8a1 , b1 9

v = w if and only if a1 = a2 and b1 = b2 .

then

Figure 54

and w = 8a2 , b2 9

We now present an alternative representation of a vector in the plane that is common in the physical sciences. Let i denote the unit vector whose direction is along the positive x-axis; let j denote the unit vector whose direction is along the positive y-axis. Then i = 81, 09 and j = 80, 19, as shown in Figure 54. Any vector v = 8a, b9 can be written using the unit vectors i and j as follows:

y (0, 1) j

v = 8a, b 9 = a81, 09 + b 80, 19 = ai + bj i

(1, 0)

x

The quantities a and b are called the horizontal and vertical components of v, respectively. For example, if v = 85, 49 = 5i + 4j, then 5 is the horizontal component and 4 is the vertical component.

Now Work

PROBLEM

29

3 Add and Subtract Vectors Algebraically The sum, difference, scalar multiple, and magnitude of algebraic vectors are defined in terms of their components.

DEFINITION

Let v = a1i + b1j = 8a1 , b1 9 and w = a2 i + b2 j = 8a2 , b2 9 be two vectors, and let a be a scalar. Then v + w = 1a1 + a2 2i + 1b1 + b2 2j = 8a1 + a2 , b1 + b2 9 v - w = 1a1 - a2 2i + 1b1 - b2 2j = 8a1 - a2 , b1 - b2 9 av = 1aa1 2i + 1ab1 2j = 8aa1 , ab1 9

In Words To add two vectors, add corresponding components. To subtract two vectors, subtract corresponding components.

7v7 = 2 2a21 + b21

(2) (3) (4) (5)

These definitions are compatible with the geometric definitions given earlier in this section. See Figure 55.

653

SECTION 8.4   Vectors

Figure 55 y (a2, b2)

w

(a1, b1)

w

b2

v

b2

y

y

(a1  a2, b1  b2)

v

b1 O

a1 a2

v b1

(a1, b1) x

a2

a1

x

a1

(b) Illustration of property (4),  0

O

P1  (a1, b1) v a1

b1 x

(c) Illustration of property (5): || v ||  Distance from O to P1 || v ||  a 21  b 21

Adding and Subtracting Vectors

If v = 2i + 3j = 82, 39 and w = 3i - 4j = 83, - 49, find: (a) v + w

Solution

v O

(a) Illustration of property (2)

EXAM PL E 3

b1

b1

(a1, b1)

(b) v - w

(a) v + w = 12i + 3j2 + 13i - 4j2 = 12 + 32 i + 13 - 42j = 5i - j or

v + w = 82, 39 + 83, - 49 = 82 + 3, 3 + 1 - 42 9 = 85, - 19

(b) v - w = 12i + 3j2 - 13i - 4j2 = 12 - 32i + 3 3 - 1 - 42 4 j = - i + 7j or

v - w = 82, 39 - 83, - 49 = 82 - 3, 3 - 1 - 42 9 = 8 - 1, 79

r

4 Find a Scalar Multiple and the Magnitude of a Vector EXAM PL E 4

Finding Scalar Multiples and Magnitudes of Vectors If v = 2i + 3j = 82, 39 and w = 3i - 4j = 83, - 49, find: (a) 3v

Solution

(b) 2v - 3w

(c) 7 v 7

(a) 3v = 312i + 3j2 = i + 9j or

3v = 382, 39 = 8, 99

(b) 2v - 3w = 212i + 3j2 - 313i - 4j2 = 4i + j - 9i + 12j = - 5i + 18j or 2v - 3w = 282, 39 - 383, - 49 = 84, 9 - 89, - 129 = 84 - 9,  - 1 - 122 9 = 8 - 5, 189

(c) 7 v 7 = 7 2i + 3j 7 = 222 + 32 = 213

Now Work

PROBLEMS

35

AND

r 41

For the remainder of the section, we will express a vector v in the form ai + bj.

654

CHAPTER 8  Polar Coordinates; Vectors

5 Find a Unit Vector

3FDBMM
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u is a vector for which 7 u 7 = 1. In many applications, it is useful to be able to find a unit vector u that has the same direction as a given vector v.

THEOREM

Unit Vector in the Direction of v For any nonzero vector v, the vector u =

v 7v7

(6)

is a unit vector that has the same direction as v.

Proof Let v = ai + bj. Then 7 v 7 = 2a2 + b2 and u =

ai + bj v a b = = i + j 2 2 2 2 2 7v7 2a + b 2a + b 2a + b2

The vector u is in the same direction as v, since 7 v 7 7 0. Furthermore,

7u7 =

a2 b2 a 2 + b2 + = = 1 B a 2 + b2 B a 2 + b2 a 2 + b2

That is, u is a unit vector in the direction of v.

■

As a consequence of this theorem, if u is a unit vector in the same direction as a vector v, then v may be expressed as v = 7v7u

(7)

This way of expressing a vector is useful in many applications.

EX A MPL E 5

Finding a Unit Vector Find a unit vector in the same direction as v = 4i - 3j.

Solution

Find 7 v 7 first.

7 v 7 = 7 4i - 3j 7 = 21 + 9 = 5

Now multiply v by the scalar

1 1 = . A unit vector in the same direction as v is 5 7v7 v

7v7

=

4i - 3j 4 3 = i - j 5 5 5

Check: This vector is indeed a unit vector, because 4 2 3 2 1 9 25 a b + a- b = + = = 1 5 5 25 25 25

Now Work

PROBLEM

r

51

6 Find a Vector from Its Direction and Magnitude If a vector represents the speed and direction of an object, it is called a velocity vector. If a vector represents the direction and amount of a force acting on an object, it is called a force vector. In many applications, a vector is described in terms of its

SECTION 8.4   Vectors

Figure 56 y 1 v j

u

(cos , sin )



x

1

i

655

magnitude and direction, rather than in terms of its components. For example, a ball thrown with an initial speed of 25 miles per hour at an angle of 30° to the horizontal is a velocity vector. Suppose that we are given the magnitude 7 v 7 of a nonzero vector v and the direction angle a, 0° … a 6 30°, between v and i. To express v in terms of 7 v 7 and a, first find the unit vector u having the same direction as v. -PPL
BU
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u are 1cos a, sin a2. Then u = cos a i + sin a j and, from (7), v = 7 v 7 1cos ai + sin aj2

(8)

where a is the direction angle between v and i.

Writing a Vector When Its Magnitude and Direction Are Given

EXAM PL E 6

A ball is thrown with an initial speed of 25 miles per hour in a direction that makes an angle of 30° with the positive x-axis. Express the velocity vector v in terms of i and j. What is the initial speed in the horizontal direction? What is the initial speed in the vertical direction?

Solution

The magnitude of v is 7 v 7 = 25 miles per hour, and the angle between the direction of v and i, the positive x-axis, is a = 30°. By equation (8),

v = 7 v 7 1cos ai + sin aj2 = 251cos 30°i + sin 30°j2 = 25a

Figure 57 y 12.5

v = 25(cos 30°i + sin 30°j)

12.5 j 25

x

21.65

The initial speed of the ball in the horizontal direction is the horizontal

2523 ≈ 21.5 miles per hour. The initial speed in the vertical 2 25 direction is the vertical component of v, = 12.5 miles per hour. See Figure 57. 2 component of v,

21.65 i

30°

23 1 2523 25 i + jb = i + j 2 2 2 2

r

Now Work

EXAM PL E 7

PROBLEM

57

Finding the Direction Angle of a Vector Find the direction angle a of v = 4i - 4j.

Solution

See Figure 58. The direction angle a of v = 4i - 4j can be found by solving

Figure 58

tan a =

 4

x

-4 = -1 4

Because 0° … a 6 30°, the direction angle is a = 315°.

v  4i  4j

Now Work

PROBLEM

r

63

(4, 4)

4

7 Model with Vectors Figure 59 Resultant F1 + F2

F1

F2

Because forces can be represented by vectors, two forces “combine” the way that vectors “add.” If F1 and F2 are two forces simultaneously acting on an object, the vector sum F1 + F2 is the resultant force. The resultant force produces the same effect on the object as that obtained when the two forces F1 and F2 act on the object. See Figure 59.

656

CHAPTER 8  Polar Coordinates; Vectors

EX A MPL E 8

A Boeing 737 aircraft maintains a constant airspeed of 500 miles per hour headed due south. The jet stream is 80 miles per hour in the northeasterly direction.

N W

E S

Orlando

Wind

Naples

Finding the Actual Speed and Direction of an Aircraft

Miami

Solution Figure 60 N

(a) Express the velocity va of the 737 relative to the air and the velocity vw of the jet stream in terms of i and j. (b) Find the velocity of the 737 relative to the ground. (c) Find the actual speed and direction of the 737 relative to the ground. (a) Set up a coordinate system in which north (N) is along the positive y-axis. See 'JHVSF

5IF
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UIF

SFMBUJWF
UP
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va = - 500j. The velocity of the jet stream vw has magnitude 80 and direction NE (northeast), so the angle between vw and i is 45°. Express vw in terms of i and j as

y vw W

500 x

E

vw = 801cos 45° i + sin 45° j2 = 80a

22 22 i + jb = 4022 1i + j2 2 2

(b) The velocity of the 737 relative to the ground vg is va  500j

vg = va + vw = - 500j + 4022 1i + j2 = 4022i +

vg

500

1 4022

- 500 2 j

(c) The actual speed of the 737 is

S

7 vg 7 = 3 1 4022 2 2 + 1 4022 - 500 2 2 ≈ 447 miles per hour To find the actual direction of the 737 relative to the ground. Determine the direction angle of vg. The direction angle is found by solving tan a =

4022 - 500 4022

Then a ≈ 277.3°. The 737 is traveling S7.3°E.

Now Work

EX A MPL E 9

PROBLEM

r

77

Finding the Weight of a Piano Two movers require a magnitude of force of 300 pounds to push a piano up a ramp inclined at an angle 20° from the horizontal. How much does the piano weigh?

Solution Figure 61

Let F1 represent the force of gravity, F2 represent the force required to move the piano up the ramp, and F3 represent the force of the piano against the ramp. See 'JHVSF

5IF
BOHMF
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BT
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F1 and F3 because triangles ABC and BDE are similar, so ∠BAC = ∠DBE = 20°. To find the magnitude of F1, calculate sin 20° =

7 F1 7 =

B A

7 F2 7 300 = 7 F1 7 7 F1 7 300 lb ≈ 877 lb sin 20°

F3

20° C F1 D

20° E

F2

The piano weighs approximately 877 pounds.

r

An object is said to be in static equilibrium if (1) the object is at rest and (2) the sum of all forces acting on the object is zero—that is, if the resultant force is 0.

SECTION 8.4   Vectors

EX AM PL E 10

657

An Object in Static Equilibrium A box of supplies that weighs 1200 pounds is suspended by two cables attached to UIF
DFJMJOH
BT
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8IBU
BSF
UIF
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JO
UIF
UXP
DBCMFT

Solution

Figure 62

30°

F1 = 7 F1 7 1cos 150°i + sin 150°j2 = 7 F1 7 a -

45° 30°

%SBX
B
GPSDF
EJBHSBN
VTJOH
UIF
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BT
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JO
'JHVSF

5IF
UFOTJPOT
JO
UIF
cables are the magnitudes 7 F1 7 and 7 F2 7 of the force vectors F1 and F2 . The magnitude of the force vector F3 equals 1200 pounds, the weight of the box. Now write each force vector in terms of the unit vectors i and j. For F1 and F2 , use equation (8). 3FNFNCFS
UIBU
a is the angle between the vector and the positive x-axis.

45°

F2 = 7 F2 7 1cos 45°i + sin 45°j2 = 7 F2 7 a

1200 pounds

23 1 23 1 7 F1 7 i + 7 F1 7 j i + jb = 2 2 2 2

22 22 22 22 7 F2 7 i + 7 F2 7 j i + jb = 2 2 2 2

F3 = - 1200j

Figure 63

For static equilibrium, the sum of the force vectors must equal zero. y

F1

F2 150°

30°

F1 + F2 + F3 = -

23 1 22 22 7 F1 7 i + 7 F1 7 j + 7 F2 7 i + 7 F2 7 j - 1200j = 0 2 2 2 2

45° x

The i component and j component must each equal zero. This results in the two equations

F3

23 22 7 F1 7 + 7 F2 7 = 0 2 2

(9)

1 22 7 F1 7 + 7 F2 7 - 1200 = 0 2 2

(10)

-

Solve equation (9) for 7 F2 7 to obtain

7 F2 7 =

23 22

7 F1 7

(11)

Substituting into equation (10) and solving for 7 F1 7 yields 1 22 23 7 F1 7 + 7 F1 7 b - 1200 = 0 a 2 2 22 1 23 7F 7 + 7 F1 7 - 1200 = 0 2 1 2 1 + 23 7 F1 7 = 1200 2 2400 7 F1 7 = ≈ 878.5 pounds 1 + 23 Substituting this value into equation (11) gives 7 F2 7 .

7 F2 7 =

23 22

7 F1 7 =

23 #

2400

22 1 + 23

≈ 1075.9 pounds

The left cable has tension of approximately 878.5 pounds, and the right cable has tension of approximately 1075.9 pounds.

Now Work

r

PROBLEM

85

658

CHAPTER 8  Polar Coordinates; Vectors

Historical Feature

T

he history of vectors is surprisingly complicated for such a natural concept. In the xy-plane, complex numbers do a good job of imitating vectors. About 1840, mathematicians became interested in finding a system that would do for three dimensions what the complex numbers do for two dimensions. Hermann Grassmann (1809–1877), in Germany, Josiah Gibbs and William Rowan Hamilton (1805–1865), in (1839 –1903) Ireland, both attempted to find solutions. Hamilton’s system was the quaternions, which are best thought of as a real number plus a vector, and do for four dimensions what complex numbers do for two dimensions. In this system the order of multiplication matters; that is, ab ≠ ba. Also, two

products of vectors emerged, the scalar product (or dot product) and the vector product (or cross product). Grassmann’s abstract style, although easily read today, was almost impenetrable during the nineteenth century, and only a few of his ideas were appreciated. Among those few were the same scalar and vector products that Hamilton had found. About 1880, the American physicist Josiah Willard Gibbs (1839–1903) worked out an algebra involving only the simplest concepts: the vectors and the two products. He then added some calculus, and the resulting system was simple, flexible, and well adapted to expressing a large number of physical laws. This system remains in use essentially unchanged. Hamilton’s and Grassmann’s more extensive systems each gave birth to much interesting mathematics, but little of this mathematics is seen at elementary levels.

8.4 Assess Your Understanding Concepts and Vocabulary 1. A vector is a quantity that has both magnitude and direction.

5. If v = ai + bj, then a is called the horizontal component of v and b is called the vertical component of v.

0 . 3. A vector u for which 7 u 7 = 1 is called a(n) unit vector.

6. If F1 and F2 are two forces simultaneously acting on an object, the vector sum F1 + F2 is called the resultant force.

4. If v = 6a, b7 is an algebraic vector whose initial point is the origin, then v is called a(n) position vector.

7. True or False Force is an example of a vector. True

2. If v is a vector, then v + 1 - v2 =

8. True or False Mass is an example of a vector. False

Skill Building In Problems 9–16, use the vectors in the figure at the right to graph each of the following vectors. *9. v + w

*10. u + v

*11. 3v

*12. 4w

*13. v - w

*14. u - v

*15. 3v + u - 2w

*16. 2u - 3v + w

w u

In Problems 17–24, use the figure at the right. Determine whether each statement given is true or false. 17. A + B = F True 19. C = D - E + F 21. E + D = G + H

18. K + G = F

20. G + H + E = D True

False

22. H - C = G - F

False

23. A + B + K + G = 0 True 25. If 7 v 7 = 4, what is 7 3v 7 ?

v

False False

B

24. A + B + C + H + G = 0 True 26. If 7 v 7 = 2, what is 7 - 4v 7 ?

12

A

8

F G

27. P = 10, 02; Q = 13, 42

v = 3i + 4j v = 2i + 4j

31. P = 1 - 2, - 12; Q = 1, - 22 33. P = 11, 02; Q = 10, 12

v = 8i - j

v = -i + j

E

28. P = 10, 02; Q = 1 - 3, - 52 30. P = 1 - 3, 22; Q = 1, 52

32. P = 1 - 1, 42; Q = 1, 22 34. P = 11, 12; Q = 12, 22

v = - 3i - 5j v = 9i + 3j v = 7i - 2j

v = i + j

In Problems 35–40, find 7 v 7 . 35. v = 3i - 4j

5

36. v = - 5i + 12j

13

38. v = - i - j

22

39. v = - 2i + 3j

213

H D

In Problems 27–34, the vector v has initial point P and terminal point Q. Write v in the form ai + bj; that is, find its position vector. 29. P = 13, 22; Q = 15, 2

C

K

37. v = i - j 40. v = i + 2j

22 2210

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

SECTION 8.4   Vectors

In Problems 41–46, find each quantity if v = 3i - 5j and w = - 2i + 3j. 41. 2v + 3w

-j

44. 7 v + w 7

42. 3v - 2w

45. 7 v 7 - 7 w 7

25

13i - 21j 234 - 213

43. 7 v - w 7

46. 7 v 7 + 7 w 7

659

289 234 + 213

In Problems 47–52, find the unit vector in the same direction as v. 47. v = 5i

48. v = - 3j

i

50. v = - 5i + 12j

-

5 12 i + j 13 13

-j 22 22 i j 2 2

51. v = i - j

4 3 i - j 5 5 225 25 i j 5 5

49. v = 3i - 4j

*53. Find a vector v whose magnitude is 4 and whose component in the i direction is twice the component in the j direction.

52. v = 2i - j

*54. Find a vector v whose magnitude is 3 and whose component in the i direction is equal to the component in the j direction. 56. If P = 1 - 3, 12 and Q = 1x, 42, > find all numbers x such that the vector represented by PQ has length 5. { - 7, 1}

55. If v = 2i - j and w = xi + 3j, find all numbers x for which 7 v + w 7 = 5. { - 2 + 221, - 2 - 221}

In Problems 57–62, write the vector v in the form ai + bj, given its magnitude 7 v 7 and the angle a it makes with the positive x-axis. *57. 7 v 7 = 5, a = 0°

*58. 7 v 7 = 8,

*60. 7 v 7 = 3, a = 240°

a = 45°

*61. 7 v 7 = 25, a = 330°

*59. 7 v 7 = 14, a = 120° *62. 7 v 7 = 15, a = 315°

In Problems 63–70, find the direction angle of v for each vector. 63. v = 3i + 3j

45°

64. v = i + 23 j

67. v = 4i - 2j

333.4°

68. v = i - 4j

ž ž

65. v = - 323i + 3j 69. v = - i - 5j

150°

258.7°

66. v = - 5i - 5j

225°

70. v = - i + 3j

108.4°

Applications and Extensions 71. Computer Graphics The field of computer graphics utilizes vectors to compute translations of points. For example, if the point 1 - 3, 22 is to be translated by v = 85, 29 , then the new location will be u′ = u + v = 8 - 3, 29 + 85, 29 = 82, 49. As illustrated in the figure, the point 1 - 3, 22 is translated to 12, 42 by v. Source: Phil Dadd. Vectors and Matrices: A Primer.

www.gamedev.net/reference/articles/article1832.asp

(a) Determine the new coordinates of 13, - 12 if it is translated by v = 8 - 4, 59. 1 - 1, 42 *(b) Illustrate this translation graphically.

*73. Force Vectors A child pulls a wagon with a force of 40 pounds. The handle of the wagon makes an angle of 30° with the ground. Express the force vector F in terms of i and j. 74. Force Vectors A man pushes a wheelbarrow up an incline of 20° with a force of 100 pounds. Express the force vector F in terms of i and j. F = 1001cos 20°i + sin 20°j2 *75. Resultant Force Two forces of magnitude 40 newtons (N) BOE

/
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30° and - 45° with the positive x-axis, as shown in the figure. Find the direction and magnitude of the resultant force; that is, find F1 + F2 . y

y

F1

 40 N

5

(2, 4)

30°

v

45°

(5, 2) (3, 2)

x

u' u

v

5

5

F2

 60 N

x

5

72. Computer Graphics 3FGFS
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1 - 3, 02, 1 - 1, - 22, 13, 12 , and 11, 32 are the vertices of a parallelogram ABCD. *(a) Find the new vertices of a parallelogram A′B′C′D′ if it is translated by v = 83, - 29. *(b) Find the new vertices of a parallelogram A′B′C′D′ if it 1 is translated by - v. 2

*76. Resultant Force Two forces of magnitude 30 newtons (N) and 70 N act on an object at angles of 45° and 120° with the positive x-axis, as shown in the figure. Find the direction and magnitude of the resultant force; that is, find F1 + F2 .

F2

 70 N y

F1

 30 N 120° 45° x

660

CHAPTER 8  Polar Coordinates; Vectors

77. Finding the Actual Speed and Direction of an Aircraft A Boeing 747 jumbo jet maintains a constant airspeed of 550 miles per hour (mph) headed due north. The jet stream is 100 mph in the northeasterly direction. *(a) Express the velocity va of the 747 relative to the air and the velocity vw of the jet stream in terms of i and j. *(b) Find the velocity of the 747 relative to the ground. (c) Find the actual speed and direction of the 747 relative to the ground. 
NQI
/ž& 78. Finding the Actual Speed and Direction of an Aircraft An Airbus A320 jet maintains a constant airspeed of 500 mph headed due west. The jet stream is 100 mph in the southeasterly direction. *(a) Express the velocity va of the A320 relative to the air and the velocity vw of the jet stream in terms of i and j. *(b) Find the velocity of the A320 relative to the ground. (c) Find the actual speed and direction of the A320 relative to the ground. 
NQI
4ž8

85. Static Equilibrium A weight of 1000 pounds is suspended from two cables as shown in the figure. What are the tensions in the two cables? 3JHIU

MC
-FGU

MC

25°

1000 pounds

86. Static Equilibrium A weight of 800 pounds is suspended from two cables, as shown in the figure. What are the tensions in the two cables? 3JHIU

MC
-FGU

MC

*79. Ground Speed and Direction of an Airplane An airplane has an airspeed of 500 kilometers per hour (km/h) bearing /ž&
5IF
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WFMPDJUZ
JT

LNI
JO
UIF
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/ž8
Find the resultant vector representing the path of the plane relative to the ground. What is the ground speed of the plane? What is its direction? *80. Ground Speed and Direction of an Airplane An airplane IBT
BO
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is 40 km/h in the direction S45°E. Find the resultant vector representing the path of the plane relative to the ground. What is the ground speed of the plane? What is its direction? 81. Weight of a Boat A magnitude of 700 pounds of force is required to hold a boat and its trailer in place on a ramp whose incline is 10° to the horizontal. What is the combined weight of the boat and its trailer? ≈4031 lb 82. Weight of a Car A magnitude of 1200 pounds of force is required to prevent a car from rolling down a hill whose incline is 15° to the horizontal. What is the weight of the car? ≈43 lb *83. Correct Direction for Crossing a River A river has a constant current of 3 km/h. At what angle to a boat dock should a motorboat capable of maintaining a constant speed of 20 km/h be headed in order to reach a point directly 1 opposite the dock? If the river is kilometer wide, how long 2 will it take to cross?

40°

35°

50°

800 pounds

87. Static Equilibrium A tightrope walker located at a certain point deflects the rope as indicated in the figure. If the weight of the tightrope walker is 150 pounds, how much tension is in each part of the rope? 3JHIU

MC
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MC

4.2°

3.7°

150 pounds

88. Static Equilibrium 3FQFBU
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JG
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left is 3.8°, the angle on the right is 2.°, and the weight of the tightrope walker is 135 pounds. 3JHIU

MC
MFGU

MC 89. Truck Pull At a county fair truck pull, two pickup trucks are attached to the back end of a monster truck as illustrated in the figure. One of the pickups pulls with a force of 2000 pounds, and the other pulls with a force of 3000 pounds. There is an angle of 45° between them. With how much force must the monster truck pull in order to remain unmoved?

Current

Boat Direction of boat due to current

84. Finding the Correct Compass Heading The pilot of an aircraft wishes to head directly east but is faced with a wind speed of 40 mph from the northwest. If the pilot maintains an airspeed of 250 mph, what compass heading should be maintained to head directly east? What is the actual speed of the aircraft? /ž&

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[Hint: Find the resultant force of the two trucks.]

0 lb 200 45˚ 300 0 lb


MC

SECTION 8.4   Vectors

90. Removing a Stump A farmer wishes to remove a stump from a field by pulling it out with his tractor. Having removed many stumps before, he estimates that he will OFFE

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However, his tractor is only capable of pulling with a force of 7000 pounds, so he asks his neighbor to help. His neighbor’s tractor can pull with a force of 5500 pounds. They attach the two tractors to the stump with a 40° angle between the forces as shown in the figure. * B

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is correct, will the farmer be successful in removing the stump? Explain. No *(b) Had the farmer arranged the tractors with a 25° angle between the forces, would he have been successful in removing the stump? Explain. Yes

661

94. Inclined Ramp A 2-pound weight is attached to a 3-pound weight by a rope that passes over an ideal pulley. The smaller weight hangs vertically, while the larger weight sits on a frictionless inclined ramp with angle u. The rope exerts a tension force T on both weights along the direction of the rope. Find the angle measure for u that is needed to keep UIF
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answer to the nearest tenth of a degree. 41.8°

T

T

3 pounds

2 pounds

u

b

l 00

55 40˚

7000 lb

91. Charting a Course A helicopter pilot needs to travel to a regional airport 25 miles away. She flies at an actual heading PG
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blowing directly east at 20 mph. (a) Determine the compass heading that the pilot needs to reach her destination. N 7.05°E (b) How long will it take her to reach her destination? 3PVOE
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12 min 92. Crossing a River A captain needs to pilot a boat across a river that is 2 km wide. The current in the river is 2 km/h and the speed of the boat in still water is 10 km/h. The desired landing point on the other side is 1 km upstream. (a) Determine the direction in which the captain should aim the boat. "CPVU
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TIPSF (b) How long will the trip take? 0.25 h, or 15 min 93. Static Friction A 20-pound box sits at rest on a horizontal surface, and there is friction between the box and the surface. One side of the surface is raised slowly to create a ramp. The friction force f opposes the direction of motion and is proportional to the normal force FN exerted by the surface on the box. The proportionality constant is called the coefficient of friction, m. When the angle of the ramp, u, reaches 20°, the box begins to slide. Find the value of m to two decimal places. m = 0.3

95. Inclined Ramp A box sitting on a horizontal surface is attached to a second box sitting on an inclined ramp by a rope that passes over an ideal pulley. The rope exerts a tension force T on both weights along the direction of the rope, and the coefficient of friction between the surface BOE
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the right weighs 100 pounds and the angle of the ramp is 35°, how much must the box on the left weigh for the TZTUFN
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two decimal places. 
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96. Muscle Force Two muscles exert force on a bone at the same point. The first muscle exerts a force of 800 N at a 10° angle with the bone. The second muscle exerts a force of 710 N at a 35° angle with the bone. What are the direction and magnitude of the resulting force on the bone? 1474.3 N at an angle of about 21.7° *97. Static Equilibrium Show on the following graph the force needed for the object at P to be in static equilibrium.

F2 P F3 F4

f

20 pounds

u

35°

W2

F1

FN

FN 2

662

CHAPTER 8  Polar Coordinates; Vectors

Explaining Concepts: Discussion and Writing 98. Explain in your own words what a vector is. Give an example of a vector.

100. Explain the difference between an algebraic vector and a position vector.

99. Write a brief paragraph comparing the algebra of complex numbers and the algebra of vectors.

Retain Your Knowledge Problems 101–104 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 3 101. Solve: 2 x - 2 = 3.

5296

102. Factor - 3x + 12x + 3x completely. 3

2

- 3x(x + 2)(x - )

*103. Find the amplitude, period, and phase shift of y = 1 104. Find the exact value of tan c cos - 1 a b d . 2

23

3 cos (x + 3p). Graph the function, showing at least two periods. 2 p p 3 Amplitude = ; Period = ; Phase shift = 2 3 2

8.5 The Dot Product PREPARING FOR THIS SECTION  Before getting started, review the following: r
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Now Work the ‘Are You Prepared?’ problem on page 668. OBJECTIVES 1 2 3 4 5 6

Find the Dot Product of Two Vectors (p. 662) Find the Angle between Two Vectors (p. 663) Determine Whether Two Vectors Are Parallel (p. 664) Determine Whether Two Vectors Are Orthogonal (p. 664) Decompose a Vector into Two Orthogonal Vectors (p. 665) Compute Work (p. 667)

1 Find the Dot Product of Two Vectors The definition for a product of two vectors is somewhat unexpected. However, such a product has meaning in many geometric and physical applications.

DEFINITION

If v = a1 i + b1 j and w = a2 i + b2 j are two vectors, the dot product v # w is defined as v # w = a1 a2 + b1 b2

EX A MPL E 1

Finding Dot Products If v = 2i - 3j and w = 5i + 3j, find: (a) v # w

Solution

(1)

(b) w # v

(c) v # v

(a) v # w = 2152 + 1 - 323 = 1 (c) v # v = 2122 + 1 - 32 1 - 32 = 13 (e) 7 v 7 = 222 + 1 - 32 2 = 213

(d) w # w

(e) 7 v 7

(f) 7 w 7

(b) w # v = 5122 + 31 - 32 = 1 (d) w # w = 5152 + 3132 = 34 (f) 7 w 7 = 252 + 32 = 234

r

663

SECTION 8.5   The Dot Product

COMMENT A scalar multiple av is a vector. A dot product u # v is a scalar (real number). ■

THEOREM

Since the dot product v # w of two vectors v and w is a real number (scalar), it is sometimes referred to as the scalar product. The results obtained in Example 1 suggest some general properties.

Properties of the Dot Product If u, v, and w are vectors, then Commutative Property u#v = v#u

(2)

u # 1v + w2 = u # v + u # w

(3)

v # v = 7v72

(4)

Distributive Property

0#v = 0

(5)

Proof We prove properties (2) and (4) here and leave properties (3) and (5) as FYFSDJTFT
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  To prove property (2), let u = a1 i + b1 j and v = a2 i + b2 j. Then u # v = a1 a2 + b1 b2 = a2 a1 + b2 b1 = v # u To prove property (4), let v = ai + bj. Then

v # v = a 2 + b2 = 7 v 7 2

■

2 Find the Angle between Two Vectors

Figure 64 uv

u  A

v

One use of the dot product is to calculate the angle between two vectors. Let u and v be two vectors with the same initial point A. Then the vectors u, v, and u - v form a triangle. The angle u at vertex A of the triangle is the angle between the vectors u and v
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the angle u. The sides of the triangle have lengths 7 v 7 , 7 u 7 , and 7 u - v 7 , and u is the included angle between the sides of length 7 v 7 and 7 u 7 . The Law of Cosines (Section 7.3) can be used to find the cosine of the included angle.

7 u - v 7 2 = 7 u 7 2 + 7 v 7 2 - 2 7 u 7 7 v 7 cos u

Now use property (4) to rewrite this equation in terms of dot products. 1u - v2 # 1u - v2 = u # u + v # v - 2 7 u 7 7 v 7 cos u

(6)

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= u#u - u#v - v#u + v#v = u#u + v#v - 2u#v

c

Property (2)

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7 u 7 7 v 7 cos u = u # v cos u =

u#v 7u7 7v7

(7)

664

CHAPTER 8  Polar Coordinates; Vectors

THEOREM

Angle between Vectors If u and v are two nonzero vectors, the angle u, 0 … u … p, between u and v is determined by the formula cos u =

EX A MPL E 2

u#v 7u7 7v7

(8)

Finding the Angle between Two Vectors Find the angle u between u = 4i - 3j and v = 2i + 5j.

Solution

Find u # v, 7 u 7 , and 7 v 7 .

Figure 65

u # v = 4122 + 1 - 32 152 = - 7

7 u 7 = 242 + 1 - 32 2 = 5

y

7 v 7 = 222 + 52 = 229

v  2i  5j

By formula (8), if u is the angle between u and v, then

 u  4i  3j

x

cos u =

u#v -7 = ≈ - 0.2 7u7 7v7 5229

Thus, u ≈ cos-1 1 - 0.22 ≈ 105°.
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Now Work

PROBLEMS

7(a)

AND

r

(b)

3 Determine Whether Two Vectors Are Parallel Two vectors v and w are said to be parallel if there is a nonzero scalar a so that v = aw. In this case, the angle u between v and w is 0 or p.

EX A MPL E 3

Determining Whether Vectors Are Parallel The vectors v = 3i - j and w = i - 2j are parallel, since v = since cos u =

1 w. Furthermore, 2

v#w 18 + 2 20 = = = 1 7v7 7w7 210 240 2400

the angle u between v and w is 0.

r

4 Determine Whether Two Vectors Are Orthogonal Figure 66 v is orthogonal to w.

w

v

p If the angle u between two nonzero vectors v and w is , the vectors v and w are 2 * called orthogonal.
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 p Since cos = 0, it follows from formula (8) that if v and w are orthogonal, then 2 v # w = 0. On the other hand, if v # w = 0, then v = 0 or w = 0 or cos u = 0. If cos u = 0, p then u = , and v and w are orthogonal. If v or w is the zero vector, then, since the 2 zero vector has no specific direction, we adopt the convention that the zero vector is orthogonal to every vector. *

Orthogonal, perpendicular, and normal are all terms that mean “meet at a right angle.” It is customary to refer to two vectors as being orthogonal, to two lines as being perpendicular, and to a line and a plane or a vector and a plane as being normal.

SECTION 8.5   The Dot Product

THEOREM

665

Two vectors v and w are orthogonal if and only if v#w = 0

EXAM PL E 4 Figure 67

Determining Whether Two Vectors Are Orthogonal The vectors

y

v = 2i - j and w = 3i + j are orthogonal, since

w  3i  6j

v#w =  -  = 0

x

r

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v  2i  j

Now Work

PROBLEM

7(c)

5 Decompose a Vector into Two Orthogonal Vectors Figure 68 F1

F

F2

Figure 69 v2

v

P

v1

w (a)

v

v2

v1 P

w (b)

In the last section, we discussed how to add two vectors to find the resultant vector. Now, we discuss the reverse problem, decomposing a vector into the sum of two components. In many physical applications, it is necessary to find “how much” of a vector JT
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F due to gravity is pulling straight down (toward the center of Earth) on the block. To study the effect of gravity on the block, it is necessary to determine how much of F is actually pushing the block down the incline 1F1 2 and how much is pressing the block against the incline 1F2 2, at a right angle to the incline. Knowing the decomposition of F often enables us to determine when friction (the force holding the block in place on the incline) is overcome and the block slides down the incline. Suppose that v and w are two nonzero vectors with the same initial point P. We seek to decompose v into two vectors: v1 , which is parallel to w, and v2 , which is orthogonal to w
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v1 is called the vector projection of v onto w. The vector v1 is obtained as follows: From the terminal point of v, drop a perpendicular to the line containing w. The vector v1 is the vector from P to the foot of this perpendicular. The vector v2 is given by v2 = v - v1 . Note that v = v1 + v2 , the vector v1 is parallel to w, and the vector v2 is orthogonal to w. This is the decomposition of v that was sought. Now we seek a formula for v1 that is based on a knowledge of the vectors v and w. Since v = v1 + v2 , we have v # w = 1v1 + v2 2 # w = v1 # w + v2 # w

(9)

Since v2 is orthogonal to w, we have v2 # w = 0. Since v1 is parallel to w, we have v1 = aw for some scalar a. Equation (9) can be written as v # w = aw # w = a 7 w 7 2 v1 = aw ; v2 # w = 0 a =

v#w 7w72

Then v1 = aw =

v#w w 7w72

666

CHAPTER 8  Polar Coordinates; Vectors

THEOREM

If v and w are two nonzero vectors, the vector projection of v onto w is v1 =

v#w w 7w72

(10)

The decomposition of v into v1 and v2 , where v1 is parallel to w and v2 is orthogonal to w, is v#w w 7w72

v1 =

EX A MPL E 5

v2 = v - v1

(11)

Decomposing a Vector into Two Orthogonal Vectors Find the vector projection of v = i + 3j onto w = i + j. Decompose v into two vectors, v1 and v2 , where v1 is parallel to w and v2 is orthogonal to w.

Solution

Use formulas (10) and (11).

Figure 70

v1 =

y v  i  3j

v#w 1 + 3 w = w = 2w = 21i + j2 2 7w7 1 22 2 2

v2 = v - v1 = 1i + 3j2 - 21i + j2 = - i + j

v2  i  j

r

See Figure 70.

v1  2(i  j) wij x

EX A MPL E 6

Now Work

PROBLEM

19

Finding the Force Required to Hold a Wagon on a Hill A wagon with two small children as occupants that weighs 100 pounds is on a hill with a grade of 20°. What is the magnitude of the force that is required to keep the wagon from rolling down the hill?

Solution Figure 71

See Figure 71. We wish to find the magnitude of the force v that is acting to cause the wagon to roll down the hill. A force with the same magnitude in the opposite direction of v will keep the wagon from rolling down the hill. The force of gravity is orthogonal to the level ground, so the force of the wagon due to gravity can be represented by the vector Fg = - 100j

v

w 20°

Fg

Determine the vector projection of Fg onto w, which is the force parallel to the hill. The vector w is given by w = cos 20°i + sin 20°j The vector projection of Fg onto w is v = =

Fg # w

7w72

w - 100(sin 20°)

1 2cos2 20°

+ sin2 20° 2 2

(cos 20°i + sin 20°j)

≈ - 34.2(cos 20°i + sin 20°j) The magnitude of v is 34.2 pounds, so the magnitude of the force required to keep the wagon from rolling down the hill is 34.2 pounds.

r

SECTION 8.5   The Dot Product

Figure 72

667

6 Compute Work A

F θ

B

A

In elementary physics, the work W done by a constant force F in moving an object from a point A to a point B is defined as

>

W = 1magnitude of force2 1distance2 = 7 F 7 7 AB 7

Work is commonly measured in foot-pounds or in newton-meters (joules). In this definition, it is assumed that the force F is applied along the line of motion. If the constant force F is not along the line of motion, but instead is at an angle u to the direction of the motion, as illustrated in Figure 72, then the work W done by F in moving an object from A to B is defined as W = F # AB

>

(12)

This definition is compatible with the force-times-distance definition, since

>

W = 1amount of force in the direction of AB 2 1distance2

>

= 7 projection of F on AB 7 7 AB 7 =

F # AB

7 AB

>

>

72

> > > 7 AB 7 7 AB 7 = F # AB

c Use formula 1102.

EXAM PL E 7

Computing Work Figure 73(a) shows a girl pulling a wagon with a force of 50 pounds. How much work is done in moving the wagon 100 feet if the handle makes an angle of 30° with the ground?

Figure 73

y

F 50(sin 30°)j

F

 50 30° 30° (0, 0)

(100, 0) x

(b)

(a)

Solution

50(cos 30°)i

Position the vectors in a coordinate system in such a way that the wagon is moved from> 10, 02 to 1100, 02. The motion is from A = 10, 02 to B = 1100, 02, so AB = 100i. The force vector F, as shown in Figure 73(b), is F = 501cos 30°i + sin 30°j2 = 50a

23 1 i + jb = 25 1 23i + j 2 2 2

By formula (12), the work done is

W = F # AB = 25 1 23i + j 2 # 100i = 250023 foot@pounds

>

Now Work

PROBLEM

r

25

Historical Feature

W

e stated in an earlier Historical Feature that complex numbers were used as vectors in the plane before the general notion of a vector was clarified. Suppose that we make the correspondence Vector 4 Complex number ai + bj 4 a + bi ci + dj 4 c + di

Show that (ai + bj) # (ci + dj) = real part [(a + bi)(c + di)] This is how the dot product was found originally. The imaginary part is also interesting. It is called a determinant (see Section 10.3) and represents the area of the parallelogram whose edges are the vectors. This is close to some of Hermann Grassmann’s ideas and is also connected with the scalar triple product of three-dimensional vectors.*

668

CHAPTER 8  Polar Coordinates; Vectors

8.5 Assess Your Understanding ‘Are You Prepared?’  The answer is given at the end of these exercises. If you get the wrong answer, read the page listed in red. 1. In a triangle with sides a, b, c and angles A, B, C, the Law of Cosines states that ________. (p. 587) c 2 = a2 + b2 - 2ab cos C

Concepts and Vocabulary 2. If v = a1i + b1j and w = a2i + b2j are two vectors, then the dot product is defined as v # w = a1a2 + b1b2. 3. If v # w = 0, then the two vectors v and w are orthogonal .

5. True or False Given two nonzero vectors v and w, it is always possible to decompose v into two vectors, one parallel to w and the other perpendicular to w. True

4. If v = 3w, then the two vectors v and w are parallel .

6. True or False Work is a physical example of a vector. False

Skill Building In Problems 7–16, (a) find the dot product v # w; (b) find the angle between v and w; (c) state whether the vectors are parallel, orthogonal, or neither. *7. v = i - j, w = i + j

*8. v = i + j, w = - i + j

*9. v = 2i + j, w = i - 2j

*10. v = 2i + 2j, w = i + 2j

*11. v = 23i - j, w = i + j

*12. v = i + 23j, w = i - j

*13. v = 3i + 4j, w = - i - 8j

*14. v = 3i - 4j, w = 9i - 12j

*15. v = 4i, w = j

*16. v = i, w = - 3j 17. Find a so that the vectors v = i - aj 18. Find b so that the vectors v = i + j and w = 2i + 3j are orthogonal. 2 and w = i + bj are orthogonal. - 1 3 In Problems 19–24, decompose v into two vectors v1 and v2 , where v1 is parallel to w and v2 is orthogonal to w. *19. v = 2i - 3j, w = i - j

*20. v = - 3i + 2j, w = 2i + j

*21. v = i - j, w = - i - 2j

*22. v = 2i - j, w = i - 2j

*23. v = 3i + j, w = - 2i - j

*24. v = i - 3j, w = 4i - j

Applications and Extensions 25. Computing Work Find the work done by a force of 3 pounds acting in the direction 0° to the horizontal in moving an PCKFDU

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10, 02 to 1, 02. 9 ft@lb 26. Computing Work A wagon is pulled horizontally by exerting a force of 20 pounds on the handle at an angle of 30° with the horizontal. How much work is done in moving the wagon 100 feet? 100023 ft@lb ≈ 1732 ft@lb 27. Solar Energy The amount of energy collected by a solar panel depends on the intensity of the sun’s rays and the area of the panel. Let the vector I represent the intensity, in watts per square centimeter, having the direction of the sun’s rays. Let the vector A represent the area, in square centimeters, whose direction is the orientation of a solar panel. See the figure. The total number of watts collected by the panel is given by W = 0 I # A 0 .

*(c) If the solar panel is to collect the maximum number of watts, what must be true about I and A? 28. Rainfall Measurement Let the vector R represent the amount of rainfall, in inches, whose direction is the inclination of the rain to a rain gauge. Let the vector A represent the area, in square inches, whose direction is the orientation of the opening of the rain gauge. See the figure. The volume of rain collected in the gauge, in cubic inches, is given by V = 0 R # A 0 , even when the rain falls in a slanted direction or the gauge is not perfectly vertical. R A 9 8 7 6 5 4 3 2 1

I A

Suppose that I = 8 - 0.02, - 0.019 and A = 8300, 4009 . *(a) Find 7 I 7 and 7 A 7 and interpret the meaning of each. *(b) Compute W and interpret its meaning.

Suppose that R = 80.75, - 1.759 and A = 80.3, 19. *(a) Find 7 R 7 and 7 A 7 and interpret the meaning of each. *(b) Compute V and interpret its meaning. *(c) If the gauge is to collect the maximum volume of rain, what must be true about R and A?

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

SECTION 8.5   The Dot Product

*29. Braking Load A Toyota Sienna with a gross weight of 5300 pounds is parked on a street with an 8° grade. See the figure. Find the magnitude of the force required to keep the Sienna from rolling down the hill. What is the magnitude of the force perpendicular to the hill? 
MCT

MCT

669

34. Given vectors u = xi + 2j and v = 7i − 3j, find x so that the angle between the vectors is 75°. State the answer correct to three decimal places. 1.574 35. Find the acute angle that a constant unit force vector makes with the positive x-axis if the work done by the force in moving a particle from 10, 02 to 14, 02 equals 2. ž *36. Prove the distributive property:

u # 1v + w2 = u # v + u # w

Weight  5300 pounds

30. Braking Load A Chevrolet Silverado with a gross weight of 4500 pounds is parked on a street with a 10° grade. Find the magnitude of the force required to keep the Silverado from rolling down the hill. What is the magnitude of the force perpendicular to the hill? 
MCT

MCT 31. Ramp Angle Billy and Timmy are using a ramp to load furniture into a truck. While rolling a 250-pound piano up the ramp, they discover that the truck is too full of other furniture for the piano to fit. Timmy holds the piano in place on the ramp while Billy repositions other items to make room for it in the truck. If the angle of inclination of the ramp is 20°, how many pounds of force must Timmy exert to hold the piano in position? 85.5 lb

*37. Prove property (5), 0 # v = 0.

*38. If v is a unit vector and the angle between v and i is a, show that v = cos ai + sin aj. *39. Suppose that v and w are unit vectors. If the angle between v and i is a and that between w and i is b, use the idea of the dot product v # w to prove that cos 1a - b2 = cos a cos b + sin a sin b

*40. Show that the projection of v onto i is 1v # i2i. Then show that we can always write a vector v as v = 1v # i2i + 1v # j2j

*41. (a) If u and v have the same magnitude, show that u + v and u - v are orthogonal. (b) Use this to prove that an angle inscribed in a semicircle is a right angle (see the figure).

u v

20° 250 lb

32. Incline Angle A bulldozer exerts 1000 pounds of force to prevent a 5000-pound boulder from rolling down a hill. Determine the angle of inclination of the hill. 11.5° 33. Given vectors u = i + 5j and v = 4i + yj, find y so that the angle between the vectors is 114°. State the answer correct to three decimal places.+ −2.833

v

*42. Let v and w denote two nonzero vectors. Show that the v#w vector v - aw is orthogonal to w if a = . 7w72 *43. Let v and w denote two nonzero vectors. Show that the vectors 7 w 7 v + 7 v 7 w and 7 w 7 v - 7 v 7 w are orthogonal. *44. In the definition of work given in this > section, what is the work done if F is orthogonal to AB ? *45. Prove the polarization identity,

7 u + v 7 2 - 7 u - v 7 2 = 41u # v2

Discussion and Writing 46. Create an application (different from any found in the text) that requires a dot product.

Retain Your Knowledge Problems 47–50 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 47. Find the average rate of change of f(x) = x3 - 5x2 + 27 from −3 to 2. 12 49. (1 - sin 2u)(1 + tan 2u) = ( cos 2u)( sec 2u) 1 p 9 = cos 2u # 48. Find the exact value of 5 cos 0° + 2 tan . Do not use a calculator. cos 2u 4 2 = 1 49. Establish the identity: (1 - sin2 u)(1 + tan2 u) = 1 50. Volume of a Box An open-top box is made from a sheet of metal by cutting squares from each corner and folding up the sides. The sheet has a length of 19 inches and a width of 13 inches. If x is the length of one side of the square to be cut out, write a function, V(x), for the volume of the box in terms of x. V(x) = x(19 - 2x)(13 - 2x) or V(x) = 4x3 - 4x2 + 247x

‘Are You Prepared?’ Answer 1. c 2 = a2 + b2 - 2ab cos C +

Courtesy of the Joliet Junior College Mathematics Department

670

CHAPTER 8  Polar Coordinates; Vectors

8.6 Vectors in Space PREPARING FOR THIS SECTION  Before getting started, review the following: r
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'PVOEBUJPOT
4FDUJPO

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Now Work the ‘Are You Prepared?’ problem on page 677.

OBJECTIVES 1 2 3 4 5 6

Figure 74

Rectangular Coordinates in Space z 4 2 2

2 2 4 x

Find the Distance between Two Points in Space (p. 671) Find Position Vectors in Space (p. 671) Perform Operations on Vectors (p. 672) Find the Dot Product (p. 673) Find the Angle between Two Vectors (p. 674) Find the Direction Angles of a Vector (p. 675)

O 2

2

4

y

In the plane, each point is associated with an ordered pair of real numbers. In space, each point is associated with an ordered triple of real numbers. Through a fixed point, called the origin O, draw three mutually perpendicular lines: the x-axis, the y-axis, and the z-axis. On each of these axes, select an appropriate scale and the positive direction. See Figure 74. The direction chosen for the positive z-axis in Figure 74 makes the system right-handed. This conforms to the right-hand rule, which states that if the index finger of the right hand points in the direction of the positive x-axis and the middle finger points in the direction of the positive y-axis, then the thumb will point in the direction of the positive z-axis. See Figure 75. Associate with each point P an ordered triple 1x, y, z2 of real numbers, the coordinates of P. For example, the point 12, 3, 42 is located by starting at the origin and moving 2 units along the positive x-axis, 3 units in the direction of the positive y-axis, and 4 units in the direction of the positive zBYJT
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 Figure 75

Figure 76 z 8

z

6 4

O

2

y

(0, 0, 4)

(2, 3, 4)

x

(0, 3, 0) 2 4

(2, 0, 0) 2

4

y

(2, 3, 0)

x

'JHVSF

BMTP
TIPXT
UIF
MPDBUJPO
PG
UIF
QPJOUT
12, 0, 02, 10, 3, 02, 10, 0, 42, and 12, 3, 02 . Points of the form 1x, 0, 02 lie on the x-axis, and points of the forms 10, y, 02 and 10, 0, z2 lie on the y-axis and z-axis, respectively. Points of the form 1x, y, 02 lie in a plane called the xy-plane. Its equation is z = 0. Similarly, points of the form 1x, 0, z2 lie in the xz-plane (equation y = 0), and points of the form 10, y, z2 lie in the yz-plane (equation x = 0). See Figure 77(a). By extension of these ideas, all points obeying the equation z = 3 will lie in a plane parallel to and 3 units above the xy-plane. The equation y = 4 represents a plane parallel to the xz-plane and 4 units to the right of the plane y = 0. See Figure 77(b).

SECTION 8.6  Vectors in Space

Figure 77

671

z z

Plane z  3

3 y0 xz-plane

Plane y  4

x0 yz-plane z0 xy-plane

4

y

y

x

x (a)

Now Work

(b) PROBLEM

9

1 Find the Distance between Two Points in Space The formula for the distance between two points in space is an extension of the Distance Formula for points in the plane given in Foundations, Section 1.

THEOREM

Distance Formula in Space If P1 = 1x1 , y1 , z1 2 and P2 = 1x2 , y2 , z2 2 are two points in space, the distance d from P1 to P2 is d = 2 1x2 - x1 2 2 + 1y2 - y1 2 2 + 1z2 - z1 2 2

(1)

The proof, which we omit, utilizes a double application of the Pythagorean Theorem.

EXAM PL E 1

Solution

Using the Distance Formula

Find the distance from P1 = 1 - 1, 3, 22 to P2 = 14, - 2, 52. d = 2 3 4 - 1 - 12 4 2 + 3 - 2 - 34 2 + 3 5 - 24 2 = 225 + 25 + 9 = 259

Now Work

15

2 Find Position Vectors in Space

Figure 78

To represent vectors in space, we introduce the unit vectors i, j, and k whose directions are along the positive x-axis, the positive y-axis, and the positive z-axis, respectively. If v is a vector with initial point at the origin O and terminal point at P = 1a, b, c2, then we can represent v in terms of the vectors i, j, and k as

z

P  (a, b, c) k O i

PROBLEM

r

j

v = ai + bj + ck

v  ai  bj  ck y

x

THEOREM

See Figure 78. The scalars a, b, and c are called the components of the vector v = ai + bj + ck, with a being the component in the direction i, b the component in the direction j, and c the component in the direction k. A vector whose initial point is at the origin is called a position vector. The next result states that any vector whose initial point is not at the origin is equal to a unique position vector. Suppose that v is a vector with initial point P1 = 1x1 , y1 , z1 2, not necessarily > the origin, and terminal point P2 = 1x2 , y2 , z2 2. If v = P1P2 , then v is equal to the position vector v = 1x2 - x1 2i + 1y2 - y1 2j + 1z2 - z1 2k

(2)

672

CHAPTER 8  Polar Coordinates; Vectors

Figure 79 illustrates this result. Figure 79

z P1  (x1, y1, z1)

O

P2  (x2, y2, z2)

v  P1P2  (x2  x1)i  (y2  y1)j  (z2  z1)k y

x

EX A MPL E 2

Solution

Finding a Position Vector

>

Find the position vector of the vector v = P1 P2 if P1 = 1 - 1, 2, 32 and P2 = 14, , 22. By equation (2), the position vector equal to v is v = 3 4 - 1 - 12 4 i + 1 - 22j + 12 - 32k = 5i + 4j - k

Now Work

PROBLEM

r

29

3 Perform Operations on Vectors Equality, addition, subtraction, scalar product, and magnitude can be defined in terms of the components of a vector.

DEFINITION

Let v = a1 i + b1 j + c1 k and w = a2 i + b2 j + c2 k be two vectors, and let a be a scalar. Then v = w if and only iff

a1 = a2 , b1 = b2 , and c1 = c2

v + w = 1a1 + a2 2i + 1b1 + b2 2j + 1c1 + c2 2k

v - w = 1a1 - a2 2i + 1b1 - b2 2j + 1c1 - c2 2k av = 1aa1 2i + 1ab1 2j + 1ac1 2k

‘v‘ = 2 2a21 + b21 + c 21

These definitions are compatible with the geometric definitions given in Section 8.4 for vectors in a plane.

EX A MPL E 3

Adding and Subtracting Vectors If v = 2i + 3j - 2k and w = 3i - 4j + 5k, find: (a) v + w

Solution

(b) v - w

(a) v + w = 12i + 3j - 2k2 + 13i - 4j + 5k2

= 12 + 32i + 13 - 42j + 1 - 2 + 52k = 5i - j + 3k

(b) v - w = 12i + 3j - 2k2 - 13i - 4j + 5k2

= 12 - 32i + 3 3 - 1 - 42 4 j + 3 - 2 - 54 k = - i + 7j - 7k

r

SECTION 8.6  Vectors in Space

EXAM PL E 4

673

Finding Scalar Products and Magnitudes If v = 2i + 3j - 2k and w = 3i - 4j + 5k, find: (b) 2v - 3w

(a) 3v

Solution

(c) ‘v‘

(a) 3v = 312i + 3j - 2k2 = i + 9j - k (b) 2v - 3w = 212i + 3j - 2k2 - 313i - 4j + 5k2 = 4i + j - 4k - 9i + 12j - 15k = - 5i + 18j - 19k

(c) ‘v‘ = ‘2i + 3j - 2k ‘ = 222 + 32 + 1 - 22 2 = 217

Now Work

PROBLEMS

33

AND

r

39

3FDBMM
UIBU
B
VOJU
WFDUPS
u is one for which ‘u ‘ = 1. In many applications, it is useful to be able to find a unit vector u that has the same direction as a given vector v.

THEOREM

Unit Vector in the Direction of v For any nonzero vector v, the vector u =

v ‘v‘

is a unit vector that has the same direction as v.

As a consequence of this theorem, if u is a unit vector in the same direction as a vector v, then v may be expressed as v = ‘v‘u

EXAM PL E 5

Finding a Unit Vector Find the unit vector in the same direction as v = 2i - 3j - k.

Solution

Find ‘v‘ first. ‘v‘ = ‘2i - 3j - k ‘ = 24 + 9 + 3 = 249 = 7 Now multiply v by the scalar u =

Now Work

1 1 = . The result is the unit vector 7 ‘v‘

2i - 3j - k v 2 3  = i - j - k = 7 7 7 7 ‘v‘

PROBLEM

r

47

4 Find the Dot Product The definition of dot product in space is an extension of the definition given for vectors in a plane.

DEFINITION

If v = a1 i + b1 j + c1 k and w = a2 i + b2 j + c2 k are two vectors, the dot product v ~ w is defined as v # w = a1 a2 + b1 b2 + c1 c2

(3)

674

CHAPTER 8  Polar Coordinates; Vectors

EX A MPL E 6

Finding Dot Products If v = 2i - 3j + k and w = 5i + 3j - k, find: (a) v ~ w (d) w ~ w

Solution

(a) (b) (c) (d)

(b) w ~ v (e) ‘v‘

v ~ w = 2152 + 1 - 323 + 1 - 12 = - 5 w ~ v = 5122 + 31 - 32 + 1 - 12 12 = - 5 v ~ v = 2122 + 1 - 32 1 - 32 + 12 = 49 w ~ w = 5152 + 3132 + 1 - 12 1 - 12 = 35

(c) v ~ v (f) ‘ w ‘

(e) ‘v‘ = 222 + 1 - 32 2 + 2 = 249 = 7

(f) ‘w ‘ = 252 + 32 + 1 - 12 2 = 235

The dot product in space has the same properties as the dot product in a plane.

THEOREM

r

Properties of the Dot Product If u, v, and w are vectors, then

Commutative Property u~v = v~u

Distributive Property u ~ 1v + w2 = u ~ v + u ~ w v ~ v = ‘v‘ 2 0~v = 0

5 Find the Angle between Two Vectors The angle u between two vectors in space follows the same formula as for two vectors in a plane.

THEOREM

Angle between Vectors If u and v are two nonzero vectors, the angle u, 0 … u … p, between u and v is determined by the formula cos u =

EX A MPL E 7

u~v ‘u ‘ ‘v‘

Finding the Angle between Two Vectors Find the angle u between u = 2i - 3j + k and v = 2i + 5j - k.

Solution

Compute the quantities u ~ v, ‘u ‘, and ‘v‘. u ~ v = 2122 + 1 - 32 152 + 1 - 12 = - 17 ‘u ‘ = 222 + 1 - 32 2 + 2 = 249 = 7 ‘v‘ = 222 + 52 + 1 - 12 2 = 230

(4)

SECTION 8.6  Vectors in Space

675

By formula (4), if u is the angle between u and v, then cos u =

u~v - 17 ≈ - 0.443 = ‘u ‘ ‘v‘ 7230

Thus, u ≈ cos -1 1 - 0.4432 ≈ 11.3°.

Now Work

PROBLEM

r

51

6 Find the Direction Angles of a Vector A nonzero vector v in space can be described by specifying its magnitude and its three direction angles a, b, and g. These direction angles are defined as a = the angle between v and i, the positive x@axis, 0 … a … p b = the angle between v and j, the positive y@axis, 0 … b … p g = the angle between v and k, the positive z@axis, 0 … g … p See Figure 80. Figure 80

z C 5 (0, 0, c)

␥ P 5 (a, b, c)

v ␤ ␣

B 5 (0, b, 0)

A 5 (a, 0, 0)

y

x 0 … ␣ … ␲, 0 … ␤ … ␲, 0 … ␥ … ␲

Our first goal is to find expressions for a, b, and g in terms of the components of a vector. Let v = ai + bj + ck denote a nonzero vector. The angle a between v and i, the positive x-axis, obeys cos a =

v~i a = ‘v‘ ‘i‘ ‘v‘

Similarly, cos b =

b ‘v‘

and cos g =

c ‘v‘

Since ‘ v‘ = 2a2 + b2 + c 2 , the following result is obtained.

THEOREM

Direction Angles If v = ai + bj b + ck is a nonzero vector in space, the direction angles a, b, and g obey cos a =

a 2a + b + c 2 c 2

2

2

=

a ‘v‘

c cos g = = ‘v‘ 2a2 + b2 + c 2 2

cos b =

b 2a + b + c 2 2

2

2

=

b ‘v‘ (5)

676

CHAPTER 8  Polar Coordinates; Vectors

The numbers cos a, cos b, and cos g are called the direction cosines of the vector v.

EX A MPL E 8

Finding the Direction Angles of a Vector Find the direction angles of v = - 3i + 2j - k.

Solution

‘v‘ = 2 1 - 32 2 + 22 + 1 - 2 2 = 249 = 7 Using the formulas in equation (5), we have -3 7 a ≈ 115.4°

2 7 b ≈ 73.4°

cos a =

THEOREM

cos b =

- 7 g ≈ 149.0°

cos g =

r

Property of the Direction Cosines If a, b, and g are the direction angles of a nonzero vector v in space, then cos2 a + cos2 b + cos2 g = 1

(6)

The proof is a direct consequence of the equations in (5). #BTFE
PO
FRVBUJPO



XIFO
UXP
EJSFDUJPO
DPTJOFT
BSF
LOPXO
UIF
UIJSE
JT
determined up to its sign. Knowing two direction cosines is not sufficient to uniquely determine the direction of a vector in space.

EX A MPL E 9

Finding a Direction Angle of a Vector

p p with the positive x-axis, an angle of b = 3 3 with the positive y-axis, and an acute angle g with the positive z-axis. Find g. The vector v makes an angle of a =

Solution

#Z
FRVBUJPO



XF
IBWF p p cos2 a b + cos2 a b + cos2 g = 1 3 3

0 6 g 6

p 2

1 2 1 2 a b + a b + cos2 g = 1 2 2 cos2 g = cos g =

22 2 g =

Since g must be acute, g =

1 2

or cos g = p 4

or g =

22 2

3p 4

p . 4

r

The direction cosines of a vector give information about only the direction of the vector; they provide no information about its magnitude. For example, any p vector that is parallel to the xy-plane and makes an angle of radian with the 4 positive x-axis and y-axis has direction cosines cos a =

22 2

cos b =

22 2

cos g = 0

However, if the direction angles and the magnitude of a vector are known, the vector is uniquely determined.

SECTION 8.6  Vectors in Space

677

Writing a Vector in Terms of Its Magnitude and Direction Cosines

EX AM PL E 10

Show that any nonzero vector v in space can be written in terms of its magnitude and direction cosines as v = ‘v‘ 3 1cos a2i + 1cos b2j + 1cos g2k 4

(7)

Let v = ai + bj + ck. From the equations in (5), note that

Solution

a = ‘v‘ cos a

b = ‘v‘ cos b

c = ‘v‘ cos g

Substituting gives

v = ai + bj + ck = ‘v‘ 1cos a2i + ‘v‘ 1cos b2j + ‘v‘ 1cos g2k = ‘v‘ 3 1cos a2i + 1cos b2j + 1cos g2k 4

Now Work

PROBLEM

59

r

Example 10 shows that the direction cosines of a vector v are also the components of the unit vector in the direction of v.

8.6 Assess Your Understanding ‘Are You Prepared?’  The answer is given at the end of these exercises. If you get the wrong answer, read the page listed in red. 1. The distance d from P1 = 1x1 , y1 2 to P2 = 1x2 , y2 2 is d = _______. (p. 35)

21x2 - x1 2 2 + 1y2 - y1 2 2

Concepts and Vocabulary 2. In space, points of the form 1x, y, 02 lie in a plane called the xy@plane.

5. True or False In space, the dot product of two vectors is a positive number. False

3. If v = ai + bj + ck is a vector in space, the scalars a, b, c are called the components of v.

6. True or False A vector in space may be described by specifying its magnitude and its direction angles. True

4. The squares of the direction cosines of a vector in space add up to 1 .

Skill Building In Problems 7–14, describe the set of points 1x, y, z2 defined by the equation(s). *7. y = 0

*8. x = 0

*11. x = - 4

*12. z = - 3

*9. z = 2 *13. x = 1 and y = 2

In Problems 15–20, find the distance from P1 to P2 . 15. P1 = 10, 0, 02 and P2 = 14, 1, 22

221

17. P1 = 1 - 1, 2, - 32 and P2 = 10, - 2, 12

19. P1 = 14, - 2, - 22 and P2 = 13, 2, 12

233 22

*10. y = 3 *14. x = 3 and z = 1

16. P1 = 10, 0, 02 and P2 = 11, - 2, 32

18. P1 = 1 - 2, 2, 32 and P2 = 14, 0, - 32

214

20. P1 = 12, - 3, - 32 and P2 = 14, 1, - 12

2219 22

In Problems 21–26, opposite vertices of a rectangular box whose edges are parallel to the coordinate axes are given. List the coordinates of the other six vertices of the box. *21. 10, 0, 02;

*24. 15, , 12;

12, 1, 32

*22. 10, 0, 02;

13, 8, 22

*25. 1 - 1, 0, 22;

14, 2, 22

14, 2, 52

*23. 11, 2, 32;

13, 4, 52

*26. 1 - 2, - 3, 02;

1 - , 7, 12

In Problems 27–32, the vector v has initial point P and terminal point Q. Write v in the form ai + bj + ck; that is, find its position vector. 27. P = 10, 0, 02; Q = 13, 4, - 12

29. P = 13, 2, - 12; Q = 15, , 02

v = 3i + 4j - k v = 2i + 4j + k

31. P = 1 - 2, - 1, 42; Q = 1, - 2, 42

v = 8i - j

28. P = 10, 0, 02; Q = 1 - 3, - 5, 42

30. P = 1 - 3, 2, 02; Q = 1, 5, - 12

32. P = 1 - 1, 4, - 22; Q = 1, 2, 22

v = - 3i - 5j + 4k v = 9i + 3j - k v = 7i - 2j + 4k

678

CHAPTER 8  Polar Coordinates; Vectors

In Problems 33–38, find ‘v‘. 33. v = 3i - j - 2k 36. v = - i - j + k

7

34. v = - i + 12j + 4k

14

23

37. v = - 2i + 3j - 3k

222

35. v = i - j + k

23

38. v = i + 2j - 2k

2211

In Problems 39–44, find each quantity if v = 3i - 5j + 2k and w = - 2i + 3j - 2k. 39. 2v + 3w

- j - 2k

40. 3v - 2w

42. ‘v + w ‘

25

43. ‘v‘ - ‘w ‘

13i - 21j + 10k

41. ‘v - w ‘

238 - 217

44. ‘v‘ + ‘w ‘

46. v = - 3j

i

*48. v = - i + 12j + 4k

-j

49. v = i + j + k

23 23 23 i + j + k 3 3 3

In Problems 51–58, find the dot product v ~ w and the angle between v and w. 51. v = i - j, w = i + j + k

v # w = 0; u = 90°

53. v = 2i + j - 3k, w = i + 2j + 2k 55. v = 3i - j + 2k, w = i + j - k

v#w

52. v = i + j, w = - i + j - k

v # w = 0; u = 90°

= - 2; u ≈ 100.3° 54. v = 2i + 2j - k, w = i + 2j + 3k

v # w = 0; u = 90°

57. v = 3i + 4j + k, w = i + 8j + 2k

238 + 217

 2 3 i - j - k 7 7 7 47. v = 3i - j - 2k 2 2 2 50. v = 2i - j + k i j + k 3  

In Problems 45–50, find the unit vector in the same direction as v. 45. v = 5i

2105

v#w

= 52; u = 0°

56. v = i + 3j + 2k, w = i - j + k

v # w = 3; u ≈ 74.5°

v # w = 0; u = 90°

58. v = 3i - 4j + k, w = i - 8j + 2k

v # w = 52; u = 0°

In Problems 59 – 66, find the direction angles of each vector. Write each vector in the form of equation (7). *59. v = 3i - j - 2k

*60. v = - i + 12j + 4k

*61. v = i + j + k

*62. v = i - j - k

*63. v = i + j

*64. v = j + k

*65. v = 3i - 5j + 2k

*66. v = 2i + 3j - 4k

Applications and Extensions 67. Robotic Arm Consider the double-jointed robotic arm shown in the figure. Let the lower arm be modeled by a = 82, 3, 49, the middle arm be modeled by b = 81, - 1, 39, and the upper arm be modeled by c = 84, - 1, - 29, where units are in feet.

In Problems 69 and 70, find the equation of a sphere with radius r and center P0 .

69. r = 1; P0 = 13, 1, 12 (x - 3)2 + (y - 1)2 + (z - 1)2 = 1

70. r = 2; P0 = 11, 2, 22 (x - 1)2 + (y - 2)2 + (z - 2)2 = 4 In Problems 71–76, find the radius and center of each sphere. [Hint: Complete the square in each variable.]

c b

71. x2 + y2 + z2 + 2x - 2y = 2

r = 2; ( - 1, 1, 0)

72. x + y + z + 2x - 2z = - 1 2

2

r = 1; ( - 1, 0, 1)

2

73. x2 + y2 + z2 - 4x + 4y + 2z = 0

r = 3; (2, - 2, - 1)

74. x + y + z - 4x = 0

r = 2; (2, 0, 0) 322 r = ; 12, 0, - 12 2 2 2 2 76. 3x + 3y + 3z + x - y = 3 r = 23; ( - 1, 1, 0) 2

a

2

2

75. 2x2 + 2y2 + 2z2 - 8x + 4z = - 1

*(a) Find a vector d that represents the position of the hand. *(b) Determine the distance of the hand from the origin. *68. The Sphere In space, the collection of all points that are the same distance from some fixed point is called a sphere. See the illustration. The constant distance is called the radius, and the fixed point is the center of the sphere. Show that the equation of a sphere with center at 1x0 , y0 , z0 2 and radius r is 1x - x0 2 2 + 1y - y0 2 2 + 1z - z0 2 2 = r 2

[Hint: Use the Distance Formula (1).]

The work W done by a constant force F in moving an object from a point A > in space to a point B in space is defined as W = F ~ AB . Use this definition in Problems 77–79. 77. Work Find the work done by a force of 3 newtons acting in the direction 2i + j + 2k in moving an object 2 meters from 10, 0, 02 to 10, 2, 02. 2 newton@meters = 2 joules 78. Work Find the work done by a force of 1 newton acting in the direction 2i + 2j + k in moving an object 3 meters from 10, 0, 02 to 11, 2, 22. 79. Work Find the work done in moving an object along a vector u = 3i + 2j - 5k if the applied force is F = 2i - j - k. Use meters for distance and newtons for force.

z

9 newton@meters = 9 joules P ⫽ (x, y, z ) r

78.

P0 ⫽ (x0, y0, z0)

8 8 newton@meters = joules 3 3

y x

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

SECTION 8.7  The Cross Product

679

Retain Your Knowledge Problems 80–83 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 80. Solve:

e x 2 6 x …

3 Ú 5 x - 2

13 f 5

or

a2,

13 d 5

81. Given f(x) = 2x - 3 and g(x) = x2 + x - 1, find (f ∘ g)(x). 2x2 + 2x - 5 1 82. Find the exact value of sin 80° cos 50° − cos 80° sin 50°. 2 83. Solve the triangle. c = 325 ≈ .71; A ≈ 2.°; B ≈ 3.4° B c

3

A 6

‘Are You Prepared?’ Answer 1. 21x2 - x1 2 2 + 1y2 - y1 2 2

8.7 The Cross Product OBJECTIVES 1 2 3 4 5

Find the Cross Product of Two Vectors (p. 679) Know Algebraic Properties of the Cross Product (p. 681) Know Geometric Properties of the Cross Product (p. 682) Find a Vector Orthogonal to Two Given Vectors (p. 682) Find the Area of a Parallelogram (p. 683)

1 Find the Cross Product of Two Vectors For vectors in space, and only for vectors in space, a second product of two vectors is defined, called the cross product. The cross product of two vectors in space is also a vector that has applications in both geometry and physics.

DEFINITION

If v = a1 i + b1 j + c1 k and w = a2 i + b2 j + c2 k are two vectors in space, the cross product v * w is defined as the vector v * w = 1b1 c2 - b2 c1 2i - 1a1 c2 - a2 c1 22jj + 1a1 b2 - a2 b1 2k

(1)

Notice that the cross product v * w of two vectors is a vector. Because of this, it is sometimes referred to as the vector product.

EXAM PL E 1

Finding a Cross Product Using Equation (1) If v = 2i + 3j + 5k and w = i + 2j + 3k, find v * w.

Solution

v * w = 13 # 3 - 2 # 52i - 12 # 3 - 1 # 52j + 12 # 2 - 1 # 32k Equation (1) = 19 - 102i - 1 - 52j + 14 - 32k

= -i - j + k

r

680

CHAPTER 8  Polar Coordinates; Vectors

Determinants* may be used as an aid in computing cross products. A 2 by 2 determinant, symbolized by `

a1 a2

b1 ` b2

has the value a1 b2 - a2 b1; that is, `

a1 a2

b1 ` = a1 b2 - a2 b1 b2

A 3 by 3 determinant has the value A 3 a1

B b1 b2

a2

EX A MPL E 2

C b c1 3 = ` 1 b2 c2

c1 a `A - ` 1 c2 a2

c1 a `B + ` 1 c2 a2

b1 `C b2

Evaluating Determinants 2 3 ` = 2#2 1 2 A B C (b) 3 2 3 5 3 = 1 2 3 = (a) `

- 1#3 = 4 - 3 = 1 `

3 5 2 5 2 3 `A - ` `B + ` `C 2 3 1 3 1 2

19 - 102A - 1 - 52B + 14 - 32 C

r

= -A - B + C

Now Work

PROBLEM

7

The cross product of the vectors v = a1 i + b1 j + c1 k and w = a2 i + b2 j + c2 k, that is, v * w = 1b1 c2 - b2 c1 2i - 1a1 c2 - a2 c1 2j + 1a1 b2 - a2 b1 2k

may be written symbolically using determinants as i v * w = 3 a1 a2

EX A MPL E 3

j b1 b2

k b c1 3 = ` 1 b2 c2

c1 a `i - ` 1 c2 a2

c1 a `j + ` 1 c2 a2

b1 `k b2

Using Determinants to Find Cross Products If v = 2i + 3j + 5k and w = i + 2j + 3k, find: (a) v * w

Solution

(b) w * v

(c) v * v

(d) w * w

i j k 3 5 2 5 2 3 (a) v * w = 3 2 3 5 3 = ` `i - ` `j + ` ` k = -i - j + k 2 3 1 3 1 2 1 2 3 i j k 2 3 1 3 1 2 3 (b) w * v = 1 2 3 3 = ` `i - ` `j + ` `k = i + j - k 3 5 2 5 2 3 2 3 5 i j k 3 5 2 5 2 3 (c) v * v = 3 2 3 5 3 = ` `i - ` `j + ` ` k = 0i - 0j + 0k = 0 3 5 2 5 2 3 2 3 5 *

Determinants are discussed in detail in Section 10.3.

681

SECTION 8.7  The Cross Product

i j k 3 (d) w * w = 1 2 3 3 1 2 3 = `

2 3 1 3 1 2 `i - ` `j + ` ` k = 0i - 0j + 0k = 0 2 3 1 3 1 2

Now Work

PROBLEM

r

15

2 Know Algebraic Properties of the Cross Product Notice in Examples 3(a) and (b) that v * w and w * v are negatives of one another. From Examples 3(c) and (d), one might conjecture that the cross product of a vector with itself is the zero vector. These and other algebraic properties of the cross product are given next.

THEOREM

Algebraic Properties of the Cross Product If u, v, and w are vectors in space and if a is a scalar, then u * u = 0

(2)

u * v = - 1v * u2

(3)

a1u a 1u * v2 = 1au2 * v = u * 1av2

(4)

u * 1v + w2 = 1u * v2 + 1u * w2

(5)

Proof We will prove properties (2) and (4) here and leave properties (3) and (5) as FYFSDJTFT
TFF
1SPCMFNT

BOE
  To prove property (2), let u = a1 i + b1 j + c1 k. Then i u * u = 3 a1 a1

j b1 b1

k b c1 3 = ` 1 b1 c1

c1 a `i - ` 1 c1 a1

c1 a `j + ` 1 c1 a1

b1 `k b1

= 0i - 0j + 0k = 0 To prove property (4), let u = a1 i + b1 j + c1 k and v = a2 i + b2 j + c2 k. Then a1u * v2 = a3 1b1 c2 - b2 c1 2i - 1a1 c2 - a2 c1 2j + 1a1 b2 - a2 b1 2k 4 c

Apply (1).

= a1b1 c2 - b2 c1 2i - a1a1 c2 - a2 c1 2j + a1a1 b2 - a2 b1 2k

(6)

Since au = aa1 i + ab1 j + ac1 k, we have 1au2 * v = 1ab1 c2 - b2ac1 2i - 1aa1 c2 - a2ac1 2j + 1aa1 b2 - a2ab1 2k = a1b1 c2 - b2 c1 2i - a1a1 c2 - a2 c1 2j + a1a1 b2 - a2 b1 2k

(7)

#BTFE
PO
FRVBUJPOT

BOE



UIF
GJSTU
QBSU
PG
QSPQFSUZ

GPMMPXT
5IF
TFDPOE
QBSU
can be proved in like fashion. ■

Now Work

PROBLEM

17

682

CHAPTER 8  Polar Coordinates; Vectors

3 Know Geometric Properties of the Cross Product THEOREM

Geometric Properties of the Cross Product Let u and v be vectors in space. u * v is orthogonal to both u and v.

(8)

7 u * v 7 = 7 u 7 7 v 7 sin u,

(9)

where u is the angle between u and v.

7 u * v 7 is the area of the parallelogram

having u ≠ 0 and v ≠ 0 as adjacent sides.

(10)

u * v = 0 if and only if u and v are parallel.

(11)

Proof of Property (8) Let u = a1 i + b1 j + c1 k and v = a2 i + b2 j + c2 k. Then

Figure 81

u * v = 1b1 c2 - b2 c1 2i - 1a1 c2 - a2 c1 2j + 1a1 b2 - a2 b1 2k Now compute the dot product u ~ 1u * v2.

u

u ~ 1u * v2 = 1a1 i + b1 j + c1 k2 # 3 1b1 c2 - b2 c1 2i - 1a1 c2 - a2 c1 2j + 1a1 b2 - a2 b1 2k 4 = a1 1b1 c2 - b2 c1 2 - b1 1a1 c2 - a2 c1 2 + c1 1a1 b2 - a2 b1 2 = 0

v

Since two vectors are orthogonal if their dot product is zero, it follows that u and u * v are orthogonal. Similarly, v ~ 1u * v2 = 0, so v and u * v are orthogonal. ■ Figure 82

4 Find a Vector Orthogonal to Two Given Vectors As long as the vectors u and v are not parallel, they will form a plane in space. See Figure 81. Based on property (8), the vector u * v is normal to this plane. As Figure 81 illustrates, there are essentially (without regard to magnitude) two vectors normal to the plane containing u and v. It can be shown that the vector u * v is the one determined by the thumb of the right hand when the other fingers of the right hand are cupped so that they point in a direction from u to v. See Figure 82.*

u

v

Finding a Vector Orthogonal to Two Given Vectors

EX A MPL E 4

Find a vector that is orthogonal to u = 3i - 2j + k and v = - i + 3j - k.

Solution

Based on property (8), such a vector is u * v.

u * v = 3

i 3 -1

j -2 3

k 1 3 = 12 - 32 i - 3 - 3 - 1 - 12 4 j + 19 - 22k = - i + 2j + 7k -1

The vector - i + 2j + 7k is orthogonal to both u and v. Check: Two vectors are orthogonal if their dot product is zero.

u ~ 1 - i + 2j + 7k2 = 13i - 2j + k2 ~ 1 - i + 2j + 7k2 = - 3 - 4 + 7 = 0 v ~ 1 - i + 2j + 7k2 = 1 - i + 3j - k2 ~ 1 - i + 2j + 7k2 = 1 +  - 7 = 0

Now Work

PROBLEM

41

5IF
QSPPG
PG
QSPQFSUZ

JT
MFGU
BT
BO
FYFSDJTF
4FF
1SPCMFN
 *

This is a consequence of using a “right-handed” coordinate system.

r

SECTION 8.7  The Cross Product

Figure 83

683

5 Find the Area of a Parallelogram Proof of Property (10) Suppose that u and v are adjacent sides of a parallelogram. See Figure 83. Then the lengths of these sides are 7 u 7 and 7 v 7 . If u is the angle between u and v, then the height of the parallelogram is 7 v 7 sin u and its area is

v

Area of parallelogram = Base * Height = 7 u 7 3 7 v 7 sin u4 = 7 u * v 7

θ

c

u

Property (9)

j

Finding the Area of a Parallelogram

EXAM PL E 5

Find the area of the parallelogram whose vertices are P1 = 10, 0, 02, P2 = 13, - 2, 12, P3 = 1 - 1, 3, - 12, and P4 = 12, 1, 02.

Solution

Two adjacent sides of this parallelogram are

>

>

u = P1 P2 = 3i - 2j + k and v = P1 P3 = - i + 3j - k

WARNING Not all pairs of vertices> give rise to a side. For example, P1P4 is a diagonal > of the > parallelogram > since> P1P3 + P>3P4 = P1P4 . Also, P1P3 and P2P4 are not adjacent sides; they are parallel sides. ■

Since u * v = - i + 2j + 7k (Example 4), the area of the parallelogram is

Area of parallelogram = 7 u * v 7 = 21 + 4 + 49 = 254 = 32 square units

Now Work

PROBLEM

r

49

Proof of Property (11) The proof requires two parts. If u and v are parallel, then there is a scalar a such that u = av. Then u * v = 1av2 * v = a 1v * v2 = 0 c

c

Property (4)

Property (2)

If u * v = 0, then, by property (9), we have

7 u * v 7 = 7 u 7 7 v 7 sin u = 0

Since u ≠ 0 and v ≠ 0, we must have sin u = 0, so u = 0 or u = p. In either case, since u is the angle between u and v, then u and v are parallel. j

8.7 Assess Your Understanding Concepts and Vocabulary 1. True or False If u and v are parallel vectors, then u * v = 0. True 2. True or False For any vector v, v * v = 0. True

5. True or False 7 u * v 7 = 7 u 7 7 v 7 cos u, where u is the angle between u and v. False

3. True or False If u and v are vectors, then u * v + v * u = 0. True 4. True or False u * v is a vector that is parallel to both u and v. False

6. True or False The area of the parallelogram having u and v as adjacent sides is the magnitude of the cross product of u and v. True

Skill Building In Problems 7–14, find the value of each determinant. 7. `

3 4 ` 1 2

2

A B 11. 3 2 1 1 3

C 4 3 1

- 11A + 2B + 5C

8. `

-2 2

5 ` -3

A B 12. 3 0 2 3 1

-4

C 4 3 3

2A + 12B - C

9. `

 -2

5 ` -1

A B 13. 3 - 1 3 5 0

4 C 53 -2

- A + 23B - 15C

10. `

-4 0 ` 5 3

A 14. 3 1 0

B -2 2

- 12 C -3 3 -2

10A + 2B + 2C

684

CHAPTER 8  Polar Coordinates; Vectors

In Problems 15–22, find (a) v * w, (b) w * v, (c) w * w, and (d) v * v. *15. v = 2i - 3j + k w = 3i - 2j - k

*16. v = - i + 3j + 2k w = 3i - 2j - k

*17. v = i + j w = 2i + j + k

*18. v = i - 4j + 2k w = 3i + 2j + k

*19. v = 2i - j + 2k w = j - k

*20. v = 3i + j + 3k w = i - k

*21. v = i - j - k w = 4i - 3k

*22. v = 2i - 3j w = 3j - 2k

In Problems 23–44, use the given vectors u, v, and w to find each expression. u = 2i - 3j + k 23. u * v

- 9i - 7j - 3k

26. w * v

- 7i - 11j + k

29. 13u2 * v

32. 1 - 3v2 * w

35. u ~ 1v * w2

38. 1v * u2 ~ w

- 27i - 21j - 9k - 21i - 33j + 18k - 25

v = - 3i + 3j + 2k

24. v * w

7i + 11j - k

27. v * v

0

30. v * 14w2

33. u ~ 1u * v2

36. 1u * v2 ~ w

25. v * u 28. w * w

28i + 44j - 24k

39. u * 1v * v2

25

w = i + j + 3k

0

9i + 7j + 3k 0

31. u * 12v2

34. v ~ 1v * w2

- 18i - 14j - k 0

37. v ~ 1u * w2

- 25

40. 1w * w2 * v

0

25 0

*41. Find a vector orthogonal to both u and v.

*42. Find a vector orthogonal to both u and w.

*43. Find a vector orthogonal to both u and i + j.

*44. Find a vector orthogonal to both u and j + k.

>

>

In Problems 45–48, find the area of the parallelogram with one corner at P1 and adjacent sides P1 P2 and P1 P3 . 45. P1 = 10, 0, 02, P2 = 11, 2, 32, P3 = 1 - 2, 3, 02

47. P1 = 11, 2, 02, P2 = 1 - 2, 3, 42, P3 = 10, - 2, 32

21 2555

46. P1 = 10, 0, 02, P2 = 12, 3, 12, P3 = 1 - 2, 4, 12

2213

48. P1 = 1 - 2, 0, 22, P2 = 12, 1, - 12, P3 = 12, - 1, 22 2217

In Problems 49–52, find the area of the parallelogram with vertices P1 , P2 , P3 , and P4 . 49. P1 = 11, 1, 22, P2 = 11, 2, 32, P3 = 1 - 2, 3, 02, P4 = 1 - 2, 4, 12 234

51. P1 = 11, 2, - 12, P2 = 14, 2, - 32, P3 = 1, - 5, 22, P4 = 19, - 5, 02 2998

50. P1 = 12, 1, 12, P2 = 12, 3, 12, P3 = 1 - 2, 4, 12, P4 = 1 - 2, , 12 8

52. P1 = 1 - 1, 1, 12, P2 = 1 - 1, 2, 22, P3 = 1 - 3, 4, - 52, P4 = 1 - 3, 5, - 42 289

Applications and Extensions *53. Find a unit vector normal to the plane containing v = i + 3j - 2k and w = - 2i + j + 3k. *54. Find a unit vector normal to the plane containing v = 2i + 3j - k and w = - 2i - 4j - 3k. 55. Volume of a Parallelepiped A parallelepiped is a prism whose faces are all parallelograms. Let A, B, and C be the vectors that define the parallelepiped shown in the figure. The volume V of the parallelepiped is given by the formula V = 0 (A * B) # C 0 .

C

*56. Volume of a Parallelepiped 3FGFS
UP
1SPCMFN

'JOE
the volume of a parallelepiped whose defining vectors are A = 81, 0, 9, B = 82, 3, - 89, and C = 88, - 5, 9. *57. Prove for vectors u and v that

7 u * v 7 2 = 7 u 7 2 7 v 7 2 - 1u ~ v2 2

[Hint: Proceed as in the proof of property (4), computing first the left side and then the right side.] *58. Show that if u and v are orthogonal, then

7u * v7 = 7u7 7v7

*59. Show that if u and v are orthogonal unit vectors, then u * v is also a unit vector.

B

*60. Prove property (3). *61. Prove property (5). A

Find the volume of a parallelepiped if the defining vectors are A = 3i - 2j + 4k, B = 2i + j - 2k, and C = 3i - j - 2k. 48 cubic units

*62. Prove property (9). [Hint: Use the result of Problem 57 and the fact that if u is the angle between u and v, then u ~ v = 7 u 7 7 v 7 cos u.]

Discussion and Writing 63. If u ~ v = 0 and u * v = 0, what, if anything, can you conclude about u and v? *Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

Chapter Review

685

Retain Your Knowledge Problems 64–67 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 64. Find the exact value of cos - 1 a

1 22

b.

p 4

65. Find two pairs of polar coordinates (r, u), one with r 7 0 and the other with r 6 0, for the point with rectangular coordinates ( - 8, - 15). Express u in radians. 117, 4.222, 1 - 17, 1.082 66. Tattoo Cost For custom tattoos, a tattoo shop charges $150 for the first hour and $90 per hour for each additional hour of work. If the total cost for a tattoo is $780, how many hours of custom work were required? 8 hours 67. Use properties of logarithms to write log 4

2x 1 as a sum or difference of logarithms. Express powers as factors. log 4 x - 3 log 4 z 3 2 z

Chapter Review Things to Know Polar Coordinates (pp. 616–623) x = r cos u, y = r sin u

3FMBUJPOTIJQ
CFUXFFO
QPMBS
coordinates 1r, u2 and rectangular coordinates 1x, y2
QQ

BOE


r 2 = x2 + y2, tan u =

1PMBS
GPSN
PG
B
DPNQMFY
OVNCFS
Q


If z = x + yi, then z = r 1cos u + i sin u2,

y , x ≠ 0 x

where r = 0 z 0 = 2x2 + y2 , sin u =

y x , cos u = , 0 … u 6 2p. r r

%F
.PJWSFT
5IFPSFN
Q


If z = r 1cos u + i sin u2, then zn = r n 3cos 1nu2 + i sin1nu2 4, where n Ú 1 is a positive integer.

nth root of a complex number w = r 1cos u0 + i sin u0 2
Q


u0 u0 2kp 2kp + b + i sin a + b d , k = 0, c , n - 1, n n n n where n Ú 2 is an integer zk = 1r c cos a n

Vectors (pp. 648–657)

Quantity having magnitude and direction; equivalent to a directed line ¡ segment PQ

1PTJUJPO
WFDUPS
QQ

BOE


Vector whose initial point is at the origin

6OJU
WFDUPS
QQ

BOE


Vector whose magnitude is 1

%PU
QSPEVDU
QQ

BOE


If v = a1 i + b1 j and w = a2 i + b2 j, then v # w = a1a2 + b1b2. If v = a1 i + b1 j + c1 k and w = a2 i + b2 j + c2 k, then v # w = a1a2 + b1b2 + c1c2.

Angle u between two nonzero vectors u and v
QQ

BOE


Direction angles of a WFDUPS
JO
TQBDF
Q


cos u =

u#v

7u7 7v 7

If v = ai + bj + ck, then v = 7 v 7 3 1cos a2i + 1cos b2j + 1cos g2k4, where cos a =

a

7v7

, cos b =

b

7v7

, and cos g =

c . 7v7

$SPTT
QSPEVDU
Q


If v = a1 i + b1 j + c1 k and w = a2 i + b2 j + c2 k,

"SFB
PG
B
QBSBMMFMPHSBN
Q


7 u * v 7 = 7 u 7 7 v 7 sin u, where u is the angle between the two adjacent sides u and v.

then v * w = 3b1 c2 - b2 c1 4i - 3a1 c2 - a2 c1 4j + 3a1 b2 - a2 b1 4k.

686

CHAPTER 8  Polar Coordinates; Vectors

Objectives Section

You should be able to…

8.1       8.2

1

    8.3  

2

 

3

    8.4             8.5           8.6           8.7        

4

2 3 4 1

3 1 2

5 1 2 3 4 5 6 7 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5

Example(s)

Plot points using polar coordinates (p. 616) Convert from polar coordinates to rectangular coordinates (p. 618) Convert from rectangular coordinates to polar coordinates (p. 620) Transform equations between polar and rectangular forms (p. 622) Identify and graph polar equations by converting to rectangular equations (p. 626) Test polar equations for symmetry (p. 629) Graph polar equations by plotting points (p. 630) Plot points in the complex plane (p. 640) Convert a complex number between rectangular form and polar form (p. 641) Find products and quotients of complex numbers in polar form (p. 642) Use De Moivre’s Theorem (p. 643) Find complex roots (p. 644) Graph vectors (p. 650) Find a position vector (p. 651) Add and subtract vectors algebraically (p. 652) Find a scalar multiple and the magnitude of a vector (p. 653) Find a unit vector (p. 654) Find a vector from its direction and magnitude (p. 654) Model with vectors (p. 655) Find the dot product of two vectors (p. 662) Find the angle between two vectors (p. 663) Determine whether two vectors are parallel (p. 664) Determine whether two vectors are orthogonal (p. 664) Decompose a vector into two orthogonal vectors (p. 665) Compute work (p. 667) Find the distance between two points in space (p. 671) Find position vectors in space (p. 671) Perform operations on vectors (p. 672) Find the dot product (p. 673) Find the angle between two vectors (p. 674) Find the direction angles of a vector (p. 675) Find the cross product of two vectors (p. 679) Know algebraic properties of the cross product (p. 681) Know geometric properties of the cross product (p. 682) Find a vector orthogonal to two given vectors (p. 682) Find the area of a parallelogram (p. 683)

Review Exercises

1–3 4 5–7 8, 9 1– 6

1–3 1–3 4–6 7(a)–10(a) 7(b)–10(b)

7–10 7–13 1 2, 3

11–13 11–13 16–18 14–18

4

19–21

5, 6 7 1 2 3 4 5 6, 7 8–10 1 2 3 4 5, 6 7 1 2 3–5 6 7 8–10 1–3 1 3 4 5

22–25 26 27, 28 29, 30 31 29, 30, 32–34 35 36, 37 59, 60 46, 47 46, 47 50–52 50–52 53, 54, 62 61 38 39 40- 42 48, 49 48, 49 55 43, 44 57, 58 56 45 56

Review Exercises In Problems 1–3, plot each point given in polar coordinates, and find its rectangular coordinates. *1. a3,

p b 6

*2. a - 2,

4p b 3

*3. a - 3, -

p b 2

In Problems 4–6, the rectangular coordinates of a point are given. Find two pairs of polar coordinates 1r, u2 for each point, one with r 7 0 and the other with r 6 0. Express u in radians. *4. 1 - 8, - 82

*5. 1 - 13, 12

*6. 1 - 5, 02

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

Chapter Review

687

In Problems 7–10, the variables r and u represent polar coordinates. (a) Write each polar equation as an equation in rectangular coordinates 1x, y2. (b) Identify the equation and graph it. *7. r = 2 sin u p *9. u = 4

*8. r = 5 *10. r 2 + 4r sin u - 8r cos u = 5

In Problems 11–13, sketch the graph of each polar equation. Be sure to test for symmetry. *11. r = 4 cos u

*12. r = 3 - 3 sin u

*13. r = 4 - cos u

In Problems 14 and 15, write each complex number in polar form. Express each argument in degrees. 14. - 6 + 4i

21cos 146.3° + i sin 146.3°2

15. 4 - 3i

51cos 323.1° + i sin 323.1°2

In Problems 16–18, write each complex number in the standard form a + bi, and plot each in the complex plane. *16. 21cos 150° + i sin 150°2

- 23 + i

*18. 0.11cos 350° + i sin 350°2 In Problems 19–21, find zw and

*17. 3acos

2p 2p + i sin b 3 3

-

3 323 i + 2 2

0.10 - 0.02i z . Leave your answers in polar form. w 5p 5p + i sin b 6 6

*20. z = 3acos

*19. z = 21cos 40° + i sin 40°2 w = - 21cos 10° + i sin 10°2

w = 4acos

*21. z = 61cos 325° + i sin 325°2 w = 1cos 35° + i sin 35°2

p p + i sin b 6 6

In Problems 22–25, write each expression in the standard form a + bi. 22. 321cos 10° + i sin 10°2 4 6 24.

11

+ i28

7p 7p 3 b + i sin a b d 12 12 4 25. 18 - 6i2 - 8432 - 5376i

23. c 25 acos a

32 + 3213i

16

5

5 5 - 5 i A2 A2

*26. Find all the complex fifth roots of 32.

*27. u + v

v

u

In Problems 27 and 28, use the figure to the right to graph each of the following: *28. 2u + 3v

¡ In Problems 29 and 30, the vector v is represented by the directed line segment PQ . Write v in the form ai + bj and find 7 v 7 . 29. P = 11, - 22; Q = 13, - 62

30. P = 10, - 22; Q = 1 - 1, 12

v = 2i - 4j; 225

v = - i + 3j; 210

In Problems 31–35, use the vectors v = - 2i + j and w = 4i - 3j to find: 31. v + w 2i - 2j 33. 7 v 7 25

32. 4v - 3w - 20i + 13j

34. 7 v 7 + 7 w 7 25 + 5 ≈ 7.24

*36. Find the vector v in the xy-plane with magnitude 3 if the direction angle of v is 60°.

*35. Find a unit vector in the same direction as v. 37. Find the direction angle between i and v = - i + 23 j.

120° *38. Find the distance from P1 = 11, 3, - 22 to P2 = 14, - 2, 12.

*39. A vector v has initial point P = 11, 3, - 22 and terminal point Q = 14, - 2, 12. Write v in the form v = ai + bj + ck.

In Problems 40–45, use the vectors v = 3i + j - 2k and w = - 3i + 2j - k to find each expression. 40. 4v - 3w 43. v * w

41. 7 v - w 7 238

21i - 2j - 5k

44. v # (v * w)

3i + 9j + 9k

42. 7 v 7 - 7 w 7

In Problems 46–49, find the dot product v ~ w and the angle between v and w. 46. v = - 2i + j, w = 4i - 3j

v # w = - 11; u ≈ 169.7°

48. v = i + j + k, w = i - j + k

v # w = 1; u ≈ 70.5°

0

*45. Find a unit vector orthogonal to both v and w.

0

47. v = i - 3j, w = - i + j

v # w = - 4; u ≈ 153.4°

49. v = 4i - j + 2k, w = i - 2j - 3k

v # w = 0; u ≈ 90°

In Problems 50–52, determine whether v and w are parallel, orthogonal, or neither. *50. v = 2i + 3j; w = - 4i - 6j

*51. v = - 2i + 2j; w = - 3i + 2j

*52. v = 3i - 2j; w = 4i + 6j

688

CHAPTER 8  Polar Coordinates; Vectors

In Problems 53 and 54, decompose v into two vectors, one parallel to w and the other orthogonal to w. *53. v = 2i + j; w = - 4i + 3j

*54. v = 2i + 3j; w = 3i + j

56. Find the area of the parallelogram with vertices P1 = 11, 1, 12, P2 = 12, 3, 42, P3 = 16, 5, 22, and P4 = 17, 7, 52. 2283

*55. Find the direction angles of the vector v = 3i - 4j + 2k.

58. Suppose that u = 3v. What is u * v?

57. If u * v = 2i - 3j + k, what is v * u? - 2i + 3j - k

59. Actual Speed and Direction of a Swimmer A swimmer can maintain a constant speed of 5 miles per hour. If the swimmer heads directly across a river that has a current moving at the rate of 2 miles per hour, what is the actual speed of the swimmer? (See the figure.) If the river is 1 mile wide, how far downstream will the swimmer end up from the point directly across the river from the starting point? 229 ≈ 5.39 mph; 0.4 mi

0

60. Static Equilibrium A weight of 2000 pounds is suspended from two cables, as shown in the figure. What are the tensions in the two cables? Left: 1843.21 lb; right: 1630.41 lb

40°

30°

2000 pounds Current

61. Computing Work Find the work done by a force of 5 pounds acting in the direction 60° to the horizontal in moving an object 20 feet from 10, 02 to 120, 02. 50 ft@lb

Swimmer's direction

*62. Braking Load A moving van with a gross weight of 8000 pounds is parked on a street with a 5° grade. Find the magnitude of the force required to keep the van from rolling down the hill. What is the magnitude of the force perpendicular to the hill? 697.2 lbs; 7969.6 lbs

Direction of swimmer due to current

Chapter Test Prep Videos include step-by-step solutions to all chapter test exercises and can be found on this text’s Channel. (search “SullivanPrecalcUC3e”)

Chapter Test In Problems 1–3, plot each point given in polar coordinates. *1. a2,

3p b 4

*2. a3, -

p b 6

*3. a - 4,

p b 3

4. Convert 1 2, 223 2 from rectangular coordinates to polar coordinates 1r, u2, where r 7 0 and 0 … u 6 2p.

a4,

p b 3

In Problems 5–7, convert the polar equation to a rectangular equation. Graph the equation. *7. r sin2 u + 8 sin u = r p In Problems 8 and 9, test the polar equation for symmetry with respect to the pole, the polar axis, and the line u = . 2 *9. r = 5 sin u cos2 u *8. r 2 cos u = 5 *5. r = 7

x2 + y2 = 49

*6. tan u = 3

y = 3x

8y = x2

In Problems 10–12, perform the given operation, where z = 21cos 85° + i sin 85°2 and w = 31cos 22° + i sin 22°2. Write your answer in polar form. w 3 10. z # w 6(cos 107° + i sin 107°) (cos 297° + i sin 297°) 12. w 5 343(cos 297° + i sin 297°) 11. z 2 *13. Find all the complex cube roots of - 8 + 823i. Then plot them in the complex plane.

1 322, 722 2 and P2 = 1 822, 222 2 . ¡ 14. Find the position vector v equal toP1P2 . 8522, - 5229

In Problems 14–18, P1 = 15. Find 7 v 7 . 10

16. Find the unit vector in the direction of v. h 22, - 22 i 2 2

17. Find the direction angle of v. 315° 18. Decompose v into its vertical components. v = 522i - 522j

and

horizontal

Chapter Projects

In Problems 19–22, v1 = 84, 9, v2 = 8 - 3, - 9, v3 = 8 - 8, 49, and v4 = 810, 159. 19. Find the vector v1 + 2v2 - v3. 20. Which two vectors are parallel?

8, - 109

v1 and v4

21. Which two vectors are orthogonal?

v2 and v3

22. Find the angle between the vectors v1 and v2 .

172.87°

In Problems 23–25, use the vectors u = 2i - 3j + k and v = - i + 3j + 2k. 23. Find u * v.

- 9i - 5j + 3k

689

*24. Find the direction angles for u. 25. Find the area of the parallelogram that has u and v as adjacent sides. 2115 26. A 1200-pound chandelier is to be suspended over a large ballroom; the chandelier will be hung on two cables of FRVBM
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what is the minimum tension each cable must be able to endure? ≈70.82 lb

Cumulative Review 2

1. Find the real solutions, if any, of the equation e x

-9

= 1. { - 3, 3}

*2. Find an equation for the line containing the origin that makes an angle of 30° with the positive x-axis.

*3. Find an equation for the circle with center at the point 10, 12 and radius 3. Graph this circle. *4. What is the domain of the function f 1x2 = ln11 - 2x2?

*5. Test the equation x2 + y3 = 2x4 for symmetry with respect to the x-axis, the y-axis, and the origin. *6. Graph the function y = 0 ln x 0 .

*7. Graph the function y = 0 sin x 0 .

*8. Graph the function y = sin 0 x 0 .

1 p 9. Find the exact value of sin-1 a - b. 2  *10. Graph the equations x = 3 and y = 4 on the same set of rectangular axes. p *11. Graph the equations r = 2 and u = on the same set of 3 polar axes. *12. What are the amplitude and period of y = - 4 cos(px)?

Chapter Projects Lift

Drag

Thrust

Weight

I.

Modeling Aircraft Motion Four aerodynamic forces act on an airplane in flight: lift, weight, thrust, and drag. While an aircraft is in flight, these four forces continuously battle each other. Weight opposes lift, and drag opposes thrust. See the figure. In balanced flight at constant speed, both the lift and weight are equal and the thrust and drag are equal. 1.

What will happen to the aircraft if the lift is held constant while the weight is decreased (say, from burning off fuel)?

2.

What will happen to the aircraft if the lift is decreased while the weight is held constant?

3.

What will happen to the aircraft if the thrust is increased while the drag is held constant?

4.

What will happen to the aircraft if the drag is increased while the thrust is held constant?

Source: www.aeromuseum.org/eduHowtoFly.html

In 1903 the Wright brothers made the first controlled powered flight. The weight of their plane was approximately 700 pounds (lb). Newton’s Second Law of Motion states that force = mass * acceleration 1F = ma2 . If the mass is measured in kilograms (kg) and acceleration in meters per second squared (m/sec2), then the force will be measured in newtons (N). 3Note: 1 N = 1 kg # m/sec2.4 5.

If 1 kg = 2.205 lb, convert the weight of the Wright brothers’ plane to kilograms.

6.

If acceleration due to gravity is a = 9.80 m/sec2, determine the force due to weight on the Wright brothers’ plane.

690

CHAPTER 8  Polar Coordinates; Vectors

7.

What must be true about the lift force of the Wright brothers’ plane for it to get off the ground?

8.

The weight of a fully loaded Cessna 170B is 2200 lb. What lift force is required to get this plane off the ground?

9.

The maximum gross weight of a Boeing 747 is 255,000 lb. What lift force is required to get this jet off the ground?

The following projects are available at the Instructors’ Resource Center (IRC): II. Project at Motorola Signal Fades due to Interference Complex trigonometric functions are used to ensure that a cellphone has optimal reception as the user travels up and down an elevator. III. Compound Interest The effect of continuously compounded interest is analyzed using polar coordinates. IV. Complex Equations Analysis of complex equations illustrates the connections between complex and real equations. At times, using complex equations is more efficient for proving mathematical theorems.

9

Analytic Geometry The Orbit of the Hale–Bopp Comet The orbits of the Hale–Bopp comet and Earth can be modeled using ellipses, the subject of Section 9.3. The Internet-based Project at the end of this chapter explores the possibility of the Hale–Bopp comet colliding with Earth.

—See the Internet-based Chapter Project I—

In this chapter, geometric definitions are given for the conics, and the distance formula and rectangular coordinates are used to obtain their equations. Historically, Apollonius (200 BC) was among the first to study conics and discover some of their interesting properties. Today, conics are still studied because of their many uses. Paraboloids of revolution (parabolas rotated about their axes of symmetry) are used as signal collectors (the satellite dishes used with radar and dish TV, for example), as solar energy collectors, and as reflectors (telescopes, light projection, and so on). The planets circle the Sun in approximately elliptical orbits. Elliptical surfaces can be used to reflect signals such as light and sound from one place to another. A third conic, the hyperbola, can be used to determine the location of lightning strikes. The Greeks used the methods of Euclidean geometry to study conics. However, we shall use the more powerful methods of analytic geometry, which employs both algebra and geometry, for our study of conics.

Conics The Parabola The Ellipse The Hyperbola Rotation of Axes; General Form of a Conic Polar Equations of Conics Plane Curves and Parametric Equations Chapter Review Chapter Test Cumulative Review Chapter Projects

691

692

CHAPTER 9  Analytic Geometry

9.1 Conics OBJECTIVE 1 Know the Names of the Conics (p. 692)

1 Know the Names of the Conics The word conic derives from the word cone, which is a geometric figure that can be constructed in the following way: Let a and g be two distinct lines that intersect at a point V. Keep the line a fixed. Now rotate the line g about a, while maintaining the same angle between a and g. The collection of points swept out (generated) by the line g is called a (right circular) cone. See Figure 1. The fixed line a is called the axis of the cone; the point V is its vertex; the lines that pass through V and make the same angle with a as g are generators of the cone. Each generator is a line that lies entirely on the cone. The cone consists of two parts, called nappes, that intersect at the vertex. Figure 1

Axis, a

Generators Vertex, V g

Conics, an abbreviation for conic sections, are curves that result from the intersection of a right circular cone and a plane. The conics we shall study arise when the plane does not contain the vertex, as shown in Figure 2. These conics are circles when the plane is perpendicular to the axis of the cone and intersects each generator; ellipses when the plane is tilted slightly so that it intersects each generator, but intersects only one nappe of the cone; parabolas when the plane is tilted farther so that it is parallel to one (and only one) generator and intersects only one nappe of the cone; and hyperbolas when the plane intersects both nappes. If the plane does contain the vertex, the intersection of the plane and the cone is a point, a line, or a pair of intersecting lines. These are usually called degenerate conics. Figure 2

Axis

Axis

Axis

Axis

Generator

(a) Circle

(b) Ellipse

(c) Parabola

(d) Hyperbola

Conic sections are used in modeling many different applications. For example, parabolas are used in describing satellite dishes and telescopes (see Figures 14 and 15 on page 698). Ellipses are used to model the orbits of planets and whispering chambers (see pages 707 and 708). And hyperbolas are used to locate lightning strikes and model nuclear cooling towers (see Problems 76 and 77 in Section 9.4).

SECTION 9.2  The Parabola

693

9.2 The Parabola PREPARING FOR THIS SECTION  Before getting started, review the following: r
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Now Work the ‘Are You Prepared?’ problems on page 699.

OBJECTIVES 1 Analyze Parabolas with Vertex at the Origin (p. 693) 2 Analyze Parabolas with Vertex at (h, k) (p. 696) 3 Solve Applied Problems Involving Parabolas (p. 697)

Section 2.4 taught us that the graph of a quadratic function is a parabola. This section gives a geometric definition of a parabola and uses it to obtain an equation.

DEFINITION

A parabola is the collection of all points P in the plane that are the same distance d from a fixed point F as they are from a fixed line D. The point F is called the focus of the parabola, and the line D is its directrix. As a result, a parabola is the set of points P for which d 1F, P2 = d 1P, D2

Figure 3 P d(P, D)

Axis of symmetry d(F, P) F

2a

a

a

V

(1)

Figure 3 shows a parabola (in blue). The line through the focus F and perpendicular to the directrix D is called the axis of symmetry of the parabola. The point of intersection of the parabola with its axis of symmetry is called the vertex V. Because the vertex V lies on the parabola, it must satisfy equation (1): d 1F, V2 = d 1V, D2. The vertex is midway between the focus and the directrix. We shall let a equal the distance d 1F, V2 from F to V. Now we are ready to derive an equation for a parabola. To do this, we use a rectangular system of coordinates positioned so that the vertex V, focus F, and directrix D of the parabola are conveniently located.

Directrix D

1 Analyze Parabolas with Vertex at the Origin

If we choose to locate the vertex V at the origin 10, 02, we can conveniently position the focus F on either the x-axis or the y-axis. First, consider the case where the focus F is on the positive x-axis, as shown in Figure 4. Because the distance from F to V is a, the coordinates of F will be 1a, 02 with a 7 0. Similarly, because the distance from V to the directrix D is also a, and because D must be perpendicular to the x-axis (since the x-axis is the axis of symmetry), the equation of the directrix D must be x = - a. Now, if P = 1x, y2 is any point on the parabola, P must obey equation (1):

Figure 4 D: x ⫽ ⫺a y

d(P, D ) (⫺a, y)

P ⫽ (x, y )

d 1F, P2 = d 1P, D2

d (F, P ) V (0, 0)

F ⫽ (a, 0)

x

2 1x - a2 + 1y - 02 2 = 2(x - ( - a))2 + (y - y)2 2

1x - a2 2 + y2 = 1x + a2 2

x - 2ax + a + y = x + 2ax + a 2

2

2

2

y2 = 4ax

Use the Distance Formula. Square both sides.

2

Square binomials. Simplify.

694

CHAPTER 9  Analytic Geometry

THEOREM

Equation of a Parabola: Vertex at (0, 0), Focus at (a, 0), a + 0

The equation of a parabola with vertex at 10, 02, focus at 1a, 02, and directrix x = - a, a 7 0, is y2 = 4ax

(2)

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EX A MPL E 1

Solution

Figure 5

Finding the Equation of a Parabola and Graphing It

Find an equation of the parabola with vertex at 10, 02 and focus at 13, 02.
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the equation.

The distance from the vertex 10, 02 to the focus 13, 02 is a = 3. Based on equation (2), the equation of this parabola is

y

D : x  3

6

y2 = 4ax

(3, 6)

y2 = 12x Latus rectum V (0, 0)

6

F  (3, 0)

a = 3

To graph this parabola, find the two points that determine the latus rectum by letting x = 3. Then 6x

y2 = 12x = 12132 = 36 y = {6

6

(3, 6)

The points 13, 62 and 13, - 62 determine the latus rectum. These points help in graphing the parabola because they determine the “opening.” See Figure 5.

Now Work COMMENT To graph the parabola y2 = 12x discussed in Example 1, graph the two functions Y1 = 212x and Y2 = - 212x .
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compare what you see with Figure 5. ■

Figure 6

EX A MPL E 2

D: x  2

V

F  (2, 0) x 5

(2, 4) 5

PROBLEM

19

By reversing the steps used to obtain equation (2), it follows that the graph of an equation of the form of equation (2), y2 = 4ax, is a parabola; its vertex is at 10, 02 , its focus is at 1a, 02, its directrix is the line x = - a, and its axis of symmetry is the x-axis. For the remainder of this section, the direction “Analyze the equation” will mean to find the vertex, focus, and directrix of the parabola and graph it.

Analyzing the Equation of a Parabola Solution The equation y2 = 8x is of the form y2 = 4ax, where 4a = 8, so a = 2.

Latus rectum

(0, 0)

r

Analyze the equation: y2 = 8x

y 5 (2, 4)

5

Solve for y.

Consequently, the graph of the equation is a parabola with vertex at 10, 02 and focus on the positive x-axis at (a, 0) = 12, 02. The directrix is the vertical line x = - 2. The two points that determine the latus rectum are obtained by letting x = 2. Then y2 = 16, so y = {4. The points 12, - 42 and 12, 42 determine the latus rectum. See Figure 6 for the graph.

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SECTION 9.2  The Parabola

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x-axis. If the focus is placed on the negative x-axis, positive y-axis, or negative y-axis, a different form of the equation for the parabola results. The four forms of the equation of a parabola with vertex at 10, 02 and focus on a coordinate axis a distance a from 10, 02 are given in Table 1, and their graphs are given in Figure 7. Notice that each graph is symmetric with respect to its axis of symmetry.

Table 1

Equations of a Parabola: Vertex at (0, 0); Focus on an Axis; a + 0 Vertex

Focus

Directrix

Equation

Description

(0, 0)

(a, 0)

x = -a

y = 4ax

Axis of symmetry is the x-axis, opens right

(0, 0)

( - a, 0)

x = a

y = - 4ax

Axis of symmetry is the x-axis, opens left

(0, 0)

(0, a)

y = -a

x = 4ay

Axis of symmetry is the y-axis, opens up

(0, 0)

(0, - a)

y = a

x = - 4ay

Axis of symmetry is the y-axis, opens down

2 2 2 2

Figure 7 D: x  a y

D: x  a

y

y

y F  (0, a)

F  (a, 0)

V

D: y  a

V

V

V x

x

x

x F  (a, 0)

D: y  a F  (0, a)

(b) y 2  4ax

(a) y 2  4ax

Analyze the equation: x2 = - 12y

Figure 8

Solution

y 6

D: y  3 V 6

(0, 0) (6, 3)

F  (0, 3)

Figure 9 y

V (0, 0) 6

PROBLEM

39

Find the equation of the parabola with focus at 10, 42 and directrix the line y = - 4. (SBQI
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Finding the Equation of a Parabola

10

(8, 4)

The equation x2 = - 12y is of the form x2 = - 4ay, with a = 3. Consequently, the graph of the equation is a parabola with vertex at 10, 02, focus at 10, - 32 , and directrix the line y = 3. The parabola opens down, and its axis of symmetry is the y-axis. To obtain the points defining the latus rectum, let y = - 3. Then x2 = 36, so x = {6. The points 1 - 6, - 32 and 16, - 32 determine the latus rectum. See Figure 8 for the graph.

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x (6, 3)

EXAM PL E 4

10

(d) x 2  4ay

Analyzing the Equation of a Parabola

EXAM PL E 3

6

(c) x 2  4ay

A parabola whose focus is at 10, 42 and whose directrix is the horizontal line y = - 4 will have its vertex at 10, 02.
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(8, 4)

c

10 D: y  4

x

a = 4

Letting y = 4 yields x2 = 64, so x = {8. The points (8, 4) and ( - 8, 4) determine the latus rectum. Figure 9 shows the graph of x2 = 16y.

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696

CHAPTER 9  Analytic Geometry

Finding the Equation of a Parabola

EX A MPL E 5

Find the equation of a parabola with vertex at 10, 02 if its axis of symmetry is the 1 x-axis and its graph contains the point a - , 2b . Find its focus and directrix, and 2 graph the equation.

Solution

The vertex is at the origin, the axis of symmetry is the x-axis, and the graph contains a point in the second quadrant, so the parabola opens to the left. From Table 1, note that the form of the equation is y2 = - 4ax 1 1 Because the point a - , 2b is on the parabola, the coordinates x = - , y = 2 must 2 2 1 satisfy y2 = - 4ax. Substituting x = - and y = 2 into this equation leads to 2

Figure 10 y

D: x  2

5

1 4 = - 4aa - b 2 a = 2

(2, 4) 1 –2 ,

(

2)

V 5 F  (2, 0) (0, 0)

5

x

1 y 2 = - 4ax; x = - , y = 2 2

The equation of the parabola is y2 = - 4122x = - 8x

(2, 4)

The focus is at 1 - 2, 02 and the directrix is the line x = 2. Letting x = - 2 gives y2 = 16, so y = {4. The points 1 - 2, 42 and 1 - 2, - 42 determine the latus rectum. See Figure 10.

5

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PROBLEM

27

2 Analyze Parabolas with Vertex at (h, k) If a parabola with vertex at the origin and axis of symmetry along a coordinate axis is shifted horizontally h units and then vertically k units, the result is a parabola with vertex at 1h, k2 and axis of symmetry parallel to a coordinate axis. The equations of such parabolas have the same forms as those in Table 1, but x is replaced by x - h (the horizontal shift) and y is replaced by y - k (the vertical shift). Table 2 gives the forms of the equations of such parabolas. Figures 11(a)–(d) illustrate the graphs for h 7 0, k 7 0. NOTE It is not recommended that Table 2 be memorized. Rather use the ideas of transformations (shift horizontally h units, vertically k units), along with the fact that a represents the distance from the vertex to the focus, to determine the various components of a parabola. It is also helpful to understand that parabolas of the form “x 2 = ” will open up or down, while parabolas of the form “y 2 = ” will open left or right. j

Table 2

Equations of a Parabola: Vertex at (h, k); Axis of Symmetry Parallel to a Coordinate Axis; a + 0 Vertex

Focus

Directrix

Equation

(h, k)

(h + a, k)

x = h - a

( y - k) = 4a(x - h)

Axis of symmetry is parallel to the x-axis, opens right

(h, k)

(h - a, k)

x = h + a

( y - k)2 = - 4a(x - h)

Axis of symmetry is parallel to the x-axis, opens left

(h, k)

(h, k + a)

y = k - a

(x - h)2 = 4a( y - k)

Axis of symmetry is parallel to the y-axis, opens up

(h, k)

(h, k - a)

y = k + a

(x - h)2 = - 4a( y - k)

Axis of symmetry is parallel to the y-axis, opens down

2

Description

SECTION 9.2  The Parabola

Figure 11

Axis of symmetry x5h D: x 5 h 2 a

y

Axis of symmetry y5k

V 5 (h, k) F 5 (h 1 a, k)

y

D: x 5 h 1 a

y

Axis of symmetry x5h

F 5 (h, k 1 a)

Axis of V5 symmetry (h, k) y5k

y D: y 5 k 1 a V 5 (h, k)

V 5 (h, k)

F 5 (h 2 a, k)

x

x

x

x

D: y 5 k 2 a

EXAM PL E 6

Solution Figure 12 D: x  4

F 5 (h, k 2 a) 2 (d) (x 2 h) 5 24a(y 2 k)

Finding the Equation of a Parabola,Vertex Not at the Origin

Find an equation of the parabola with vertex at 1 - 2, 32 and focus at 10, 32 .
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the equation.

The vertex 1 - 2, 32 and focus 10, 32 both lie on the horizontal line y = 3 (the axis of symmetry). The distance a from the vertex 1 - 2, 32 to the focus 10, 32 is a = 2. Also, because the focus lies to the right of the vertex, the parabola opens to the right. Consequently, the form of the equation is 1y - k2 2 = 4a1x - h2

y

where 1h, k2 = 1 - 2, 32 and a = 2. Therefore, the equation is

8 (0, 7) Axis of symmetry y3

V  (2, 3) F  (0, 3) 6

(c) (x 2 h)2 5 4a(y 2 k)

(b) (y 2 k)2 5 24a(x 2 h)

(a) (y 2 k)2 5 4a(x 2 h)

697

6x

(0, 1)

1y - 32 2 = 4 # 2[x - ( - 2)] 1y - 32 2 = 81x + 22

To find the points that define the latus rectum, let x = 0, so that 1y - 32 2 = 16. Then y - 3 = {4, so y = - 1 or y = 7. The points 10, - 12 and 10, 72 determine the latus rectum; the line x = - 4 is the directrix. See Figure 12.

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Now Work

PROBLEM

29

4

Polynomial equations define parabolas whenever they involve two variables that are quadratic in one variable and linear in the other.

EXAM PL E 7

Analyzing the Equation of a Parabola Analyze the equation: x2 + 4x - 4y = 0

Solution

x2 + 4x - 4y x2 + 4x x2 + 4x + 4 1x + 22 2

Figure 13 Axis of symmetry x  2

y 4

(4, 0) F  (2, 0) 4 V  (2, 1) 3

To analyze the equation x2 + 4x - 4y = 0, complete the square involving the variable x.

(0, 0)

4 x

D: y  2

= = = =

0 4y 4y + 4 41y + 12

Isolate the terms involving x on the left side. Complete the square on the left side. Factor.

This equation is of the form 1x - h2 2 = 4a1y - k2, with h = - 2, k = - 1, and a = 1. The graph is a parabola with vertex at 1h, k2 = 1 - 2, - 12 that opens up. The focus is a = 1 unit above the vertex at 1 - 2, 02, and the directrix is the line y = - 2. See Figure 13.

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PROBLEM

47

3 Solve Applied Problems Involving Parabolas P Parabolas find their way into many applications. For example, as discussed in Section 2.6, suspension bridges have cables in the shape of a parabola. Another S property of parabolas that is used in applications is their reflecting property.

698

CHAPTER 9  Analytic Geometry

Suppose that a mirror is shaped like a paraboloid of revolution, a surface formed by rotating a parabola about its axis of symmetry. If a light (or any other emitting source) is placed at the focus of the parabola, all the rays emanating from the light reflect off the mirror in lines parallel to the axis of symmetry. This principle is used in the design of searchlights, flashlights, certain automobile headlights, and other such devices. See Figure 14. Conversely, suppose that rays of light (or other signals) emanate from a distant source so that they are essentially parallel. When these rays strike the surface of a parabolic mirror whose axis of symmetry is parallel to these rays, they are reflected to a single point at the focus. This principle is used in the design of some solar energy devices, satellite dishes, and the mirrors used in some types of telescopes. See Figure 15. Figure 14 Searchlight

Figure 15 Telescope Rays

ht

of lig

Light at focus

EX A MPL E 8

Satellite Dish A satellite dish is shaped like a paraboloid of revolution. The signals that emanate from a satellite strike the surface of the dish and are reflected to a single point, where the receiver is located. If the dish is 8 feet across at its opening and 3 feet deep at its DFOUFS
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rectangular coordinate system so that the vertex of the parabola is at the origin and its focus is on the positive y-axis. See Figure 16(b).

Figure 16 y

8'

8'

(4, 3)

4

3' 2 4

USA Cable

2

(4, 3) F  (0, a)

0

3' 2

4 x

(b)

(a)

The form of the equation of the parabola is x2 = 4ay

and its focus is at 10, a2. Since 14, 32 is a point on the graph, this gives 42 = 4a132 4 a = 3

x 2 = 4ay ; x = 4, y = 3 Solve for a.

1 The receiver should be located 1 feet (1 foot, 4 inches) from the base of the dish, 3 along its axis of symmetry.

Now Work

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PROBLEM

63

SECTION 9.2  The Parabola

699

9.2 Assess Your Understanding ‘Are You Prepared?’

Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.

1. The formula for the distance d from P1 = 1x1 , y1 2 to 2 2 P2 = 1x2 , y2 2 is d = 2(x2 - x1) + (y2 - y1) . (p. 35) 2. To complete the square of x2 - 4x, add

4. The point that is symmetric with respect to the x-axis to the point 1 - 2, 52 is ( - 2, - 5 ). (pp. 44–45)

5. To graph y = 1x - 32 2 + 1, shift the graph of y = x2 to the right 3 units and then up 1 unit. (pp. 121–123)

4 . (pp. A38–A39)

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Concepts and Vocabulary 6. A(n) parabola is the collection of all points in the plane such that the distance from each point to a fixed point equals its distance to a fixed line. Answer Problems 7–10 using the figure. 7. If a (a) (b) (c) (d)

7 0, the equation of the parabola is of the form 1y - k2 2 = 4a1x - h2 1y - k2 2 = - 4a1x - h2 1x - h2 2 = 4a1y - k2 1x - h2 2 = - 4a1y - k2

8. The coordinates of the vertex are

(3, 2)

y

(c)

F

V  (3, 2)

.

9. If a = 4, then the coordinates of the focus are 10. If a = 4, then the equation of the directrix is

(3, 6)

x

.

y = -2 .

D

Skill Building In Problems 11–18, the graph of a parabola is given. Match each graph to its equation. (A) y2 = 4x (C) y2 = - 4x (E) 1y - 12 2 = 41x - 12 2 2 (B) x = 4y %
x = - 4y (F) 1x + 12 2 = 41y + 12 11.

(

12.

B

13.

14.

E

y

y

y

2

3

2

(2, 1)

(
1y - 12 2 = - 41x - 12 (H) 1x + 12 2 = - 41y + 12 % y 2

(1, 1) (1, 1)

2

2

2 x 2

2

15.

16.

H

2 x

1

18.

C

F

y 2

y

y 2

2

2 x

(2, 1) 2

2

17.

A

y

2

2 x

2

(1, 2)

(1, 1) 2

2 x

2

2 x

2

2 (1, 2)

2

2 x 2

3 (1, 1)

1 x 2

In Problems 19–36, find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation.

*19. Focus at 14, 02; vertex at 10, 02

* 21. Focus at 10, - 32; vertex at 10, 02

y2 = 16x x = - 12y 2

* 23. Focus at 1 - 2, 02; directrix the line x = 2 y2 = - 8x 1 * 25. %JSFDUSJY
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* 31. Vertex at 1 - 1, - 22; focus at 10, - 22

1x - 22 2 = - 81y + 32

* 33. Focus at 1 - 3, 42; directrix the line y = 2

* 20. Focus at 10, 22; vertex at 10, 02

* 22. Focus at 1 - 4, 02; vertex at 10, 02

x2 = 8y y2 = - 16x

* 24. Focus at 10, - 12; directrix the line y = 1 x2 = - 4y 1 * 26. %JSFDUSJY
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x = - ; vertex at 10, 02 y2 = 2x 2 *28. Vertex at 10, 02; axis of symmetry the x-axis; containing the point 12, 32 *30. Vertex at 14, - 22; focus at 16, - 22

1y + 22 2 = 41x + 12 *32. Vertex at 13, 02; focus at 13, - 22

1y + 22 2 = 81x - 42

1x - 32 2 = - 8y

1x + 32 2 = 41y - 32 *34. Focus at 12, 42; directrix the line x = - 4 1y - 42 2 = 121x + 12

* 35. Focus at 1 - 3, - 22; directrix the line x = 1

*36. Focus at 1 - 4, 42; directrix the line y = - 2

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700

CHAPTER 9  Analytic Geometry

In Problems 37–54, find the vertex, focus, and directrix of each parabola. Graph the equation. *37. x2 = 4y

*38. y2 = 8x

*41. 1y - 22 2 = 81x + 12

*39. y2 = - 16x

*42. 1x + 42 2 = 161y + 22 *46. 1x - 22 = 41y - 32

*45. 1y + 32 = 81x - 22 2

2

*49. x2 + 8x = 4y - 8

*50. y2 - 2y = 8x - 1

*53. x - 4x = y + 4

*54. y + 12y = - x + 1

2

*40. x2 = - 4y

*43. 1x - 32 2 = - 1y + 12

*44. 1y + 12 2 = - 41x - 22

*47. y - 4y + 4x + 4 = 0

*48. x2 + 6x - 4y + 1 = 0

*51. y2 + 2y - x = 0

*52. x2 - 4x = 2y

2

2

In Problems 55–62, write an equation for each parabola. y

*55.

*56.

2

2

(1, 2)

y

*58.

2

2

(2, 1)

2

x

2

2

(2, 0)

2

2

x

*60.

2

(0, 1)

(0, 1) 2

2 2

2

y

*62.

2 (0, 1) (1, 0)

2

x

x

(0, 1)

2 (0, 1)

2

2 2

y

*61.

y

(2, 2)

2

x

(1, 0) 2

2

y

2

(1, 2)

(2, 1)

(0, 1)

*59.

y

*57.

y

x

(1, 1)

2

(2, 0)

2

x

2

x

2

2

Applications and Extensions *63. Satellite Dish A satellite dish is shaped like a paraboloid of revolution. The signals that emanate from a satellite strike the surface of the dish and are reflected to a single point, where the receiver is located. If the dish is 10 feet across at its opening and 4 feet deep at its center, at what position TIPVME
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68. Suspension Bridge The cables of a suspension bridge are in the shape of a parabola. The towers supporting the cable are

*73. Parabolic Arch Bridge A bridge is built in the shape of a parabolic arch. The bridge has a span of 120 feet and a maximum height of 25 feet. See the illustration. Choose a

80 ft ? 150 ft

SECTION 9.3  The Ellipse

701

(c)
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suitable rectangular coordinate system and find the height of the arch at distances of 10, 30, and 50 feet from the center.

*76. Show that an equation of the form Ax2 + Ey = 0,

74. Parabolic Arch Bridge A bridge is to be built in the shape of a parabolic arch and is to have a span of 100 feet. The height of the arch a distance of 40 feet from the center is to be 10 feet. Find the height of the arch at its center. 27.78 ft 75. Gateway Arch 5IF
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mistaken to be parabolic in shape. In fact, it is a catenary, which has a more complicated formula than a parabola. The Arch is 630 feet high and 630 feet wide at its base. *(a) Find the equation of a parabola with the same dimensions. Let x equal the horizontal distance from the center of the arc. (b) The following table gives the height of the Arch at various widths; find the corresponding heights for the parabola found in (a). 119.7 ft, 267.3 ft, 479.4 ft Width (ft)

A ≠ 0, E ≠ 0

is the equation of a parabola with vertex at 10, 02 and axis of symmetry the y-axis. Find its focus and directrix.

25 ft 120 ft

*77. Show that an equation of the form Cy2 + Dx = 0,

is the equation of a parabola with vertex at 10, 02 and axis of symmetry the x-axis. Find its focus and directrix. *78. Show that the graph of an equation of the form Ax2 + Dx + Ey + F = 0,

100

478

312.5

308

525

A ≠ 0

(a) Is a parabola if E ≠ 0. (b) Is a vertical line if E = 0 and D2 - 4AF = 0. (c) Is two vertical lines if E = 0 and D2 - 4AF 7 0. (d) Contains no points if E = 0 and D2 - 4AF 6 0. *79. Show that the graph of an equation of the form Cy2 + Dx + Ey + F = 0,

Height (ft)

567

C ≠ 0, D ≠ 0

C ≠ 0

(a) Is a parabola if D ≠ 0. (b) Is a horizontal line if D = 0 and E 2 - 4CF = 0. (c) Is two horizontal lines if D = 0 and E 2 - 4CF 7 0. (d) Contains no points if D = 0 and E 2 - 4CF 6 0.

Retain Your Knowledge Problems 80–83 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 3 2210 83. Find the exact value: tan c cos - 1 a - b d *80. For x = 9y2 - 36, list the intercepts and test for symmetry. 7 3 81. Solve: 4x + 1 = 8x - 1 5 56 5 p 6 u 6 p, find the exact value of *82. (JWFO
tan u = - , 8 2 each of the remaining trigonometric functions.

‘Are You Prepared?’ Answers 1. 21x2 - x1 2 2 + 1y2 - y1 2 2

2. 4

3. x + 4 = {3; 5 - 7, - 16

4. 1 - 2, - 52

5. 3; up

9.3 The Ellipse PREPARING FOR THIS SECTION Before getting started, review the following: r
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pp. 121–129)

Now Work the ‘Are You Prepared?’ problems on page 708.

OBJECTIVES 1 Analyze Ellipses with Center at the Origin (p. 702) 2 Analyze Ellipses with Center at (h, k) (p. 706) 3 Solve Applied Problems Involving Ellipses (p. 707)

702

CHAPTER 9  Analytic Geometry

DEFINITION

Figure 17

Minor axis Major axis

P

Center V1

F2

V2

F1

Figure 18

An ellipse is the collection of all points in the plane the sum of whose distances from two fixed points, called the foci, is a constant. The definition contains within it a physical means for drawing an ellipse. Find a piece of string (the length of this string is the constant referred to in the definition). Then take two thumbtacks (the foci) and stick them into a piece of cardboard so that the distance between them is less than the length of the string. Now attach the ends of the string to the thumbtacks and, using the point of a pencil, pull the string taut. See Figure 17. Keeping the string taut, rotate the pencil around the two thumbtacks. The pencil traces out an ellipse, as shown in Figure 17. In Figure 17, the foci are labeled F1 and F2 . The line containing the foci is called the major axis. The midpoint of the line segment joining the foci is the center of the ellipse. The line through the center and perpendicular to the major axis is the minor axis. The two points of intersection of the ellipse and the major axis are the vertices, V1 and V2 , of the ellipse. The distance from one vertex to the other is the length of the major axis. The ellipse is symmetric with respect to its major axis, with respect to its minor axis, and with respect to its center.

y

1 Analyze Ellipses with Center at the Origin

P ⫽ (x, y ) d(F1, P )

d(F2, P )

F1 ⫽ (⫺c, 0)

F2 ⫽ (c, 0)

x

With these ideas in mind, we are ready to find the equation of an ellipse in a rectangular coordinate system. First, place the center of the ellipse at the origin. Second, position the ellipse so that its major axis coincides with a coordinate axis, say the x-axis, as shown in Figure 18. If c is the distance from the center to a focus, one focus is at F1 = 1 - c, 02 and the other at F2 = 1c, 02. As we shall see, it is convenient to let 2a denote the constant distance referred to in the definition. Then, if P = 1x, y2 is any point on the ellipse, d1F1 , P2 + d 1F2 , P2 = 2a

2 1x - ( - c)2 2 + (y - 0)2 + 2 1x - c2 2 + (y - 0)2 = 2a 2 1x + c2 2 + y2 = 2a - 2 1x - c2 2 + y2

The sum of the distances from P to the foci equals a constant, 2a. Use the Distance Formula. Isolate one radical.

1x + c2 2 + y2 = 4a2 - 4a2 1x - c2 2 + y2 Square both sides. + 1x - c2 2 + y2

x2 + 2cx + c 2 + y2 = 4a2 - 4a2 1x - c2 2 + y2 Square binomials. + x2 - 2cx + c 2 + y2

4cx - 4a2 = - 4a2 1x - c2 2 + y2 cx - a = - a2 1x - c2 + y 2

2

Simplify; isolate the radical.

2

Divide each side by 4.

1cx - a 2 = a 3 1x - c2 + y 4 2 2

2

2

2

Square both sides again.

c x - 2a cx + a = a 1x - 2cx + c + y 2 2 2

2

4

2

2

1c 2 - a2 2x2 - a2 y2 = a2 c 2 - a4 1a2 - c 2 2x2 + a2 y2 = a2 1a2 - c 2 2

2

2

Square binomials. Rearrange the terms. Multiply each side by - 1; factor a 2 on the right side.

(1)

To obtain points on the ellipse off the x-axis, it must be that a 7 c. To see why, look again at Figure 18. Then d 1F1 , P2 + d 1F2 , P2 7 d 1F1 , F2 2 2a 7 2c a 7 c

The sum of the lengths of two sides of a triangle is greater than the length of the third side. d(F1 , P ) + d(F2 , P ) = 2a, d(F1 , F2) = 2c

SECTION 9.3  The Ellipse

703

Because a 7 c 7 0, this means a2 7 c 2, so a2 - c 2 7 0. Let b2 = a2 - c 2, b 7 0. Then a 7 b and equation (1) can be written as b2 x2 + a2 y2 = a2 b2 y2 x2 + = 1 a2 b2

Divide each side by a 2 b 2.

As you can verify, the graph of this equation has symmetry with respect to the x-axis, the y-axis, and the origin. Because the major axis is the x-axis, find the vertices of this ellipse by letting x2 y = 0. The vertices satisfy the equation 2 = 1, the solutions of which are x = {a. a Consequently, the vertices of this ellipse are V1 = 1 - a, 02 and V2 = 1a, 02. The y-intercepts of the ellipse, found by letting x = 0, have coordinates 10, - b2 and 10, b2 . These four intercepts, 1a, 02, 1 - a, 02, 10, b2, and 10, - b2, are used to graph the ellipse.

THEOREM

Equation of an Ellipse: Center at (0, 0); Major Axis along the x-Axis An equation of the ellipse with center at 10, 02, foci at 1 - c, 02 and 1c, 02 , and vertices at 1 - a, 02 and 1a, 02 is y2 x2 + = 1, a2 b2

where a 7 b 7 0 and b2 = a2 - c 2

(2)

The major axis is the x-axis. See Figure 19. Figure 19

y (0, b) V1 ⫽ (⫺a, 0)

a V2 ⫽ (a, 0) c x F1 ⫽ (⫺c, 0) F2 ⫽ (c, 0) b

(0, ⫺b)

Notice in Figure 19 the right triangle formed by the points 10, 02 , 1c, 02, and 10, b2. Because b2 = a2 - c 2 (or b2 + c 2 = a2), the distance from the focus at 1c, 02 to the point 10, b2 is a. This can be seen another way. Look at the two right triangles in Figure 19. They BSF
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EXAM PL E 1

Finding an Equation of an Ellipse

y

The ellipse has its center at the origin, and because the given focus and vertex lie on the x-axis, the major axis is the x-axis. The distance from the center, 10, 02 , to one of the foci, 13, 02, is c = 3. The distance from the center, 10, 02, to one of the vertices, 1 - 4, 02 , is a = 4. From equation (2), it follows that

Figure 20

Solution

5 (0, 7 )

F1 ⫽ (⫺3, 0)

F2 ⫽ (3, 0)

⫺5

5 x

V1 ⫽ (⫺4, 0)

(0, ⫺ 7 ) ⫺5

Find an equation of the ellipse with center at the origin, one focus at 13, 02, and a vertex at 1 - 4, 02.
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b2 = a2 - c 2 = 16 - 9 = 7 so an equation of the ellipse is

V2 ⫽ (4, 0)

y2 x2 + = 1 16 7 Figure 20 shows the graph.

r

704

CHAPTER 9  Analytic Geometry

In Figure 20 the intercepts of the equation are used to graph the ellipse. Following this practice will make it easier for you to obtain an accurate graph of an ellipse. COMMENT The intercepts of the ellipse also provide information about how to set the viewing rectangle for graphing an ellipse. To graph the ellipse y2 x2 + = 1 16 7 discussed in Example 1, we set the viewing rectangle using a square screen that includes the intercepts, perhaps - 4.5 … x … 4.5, - 3 … y … 3. Then we proceed to solve the equation for y: y2 x2 + = 1 16 7 y2 x2 = 1 7 16 Figure 21

y2 = 7¢1 -

(

x2 3 Y1  7 1  16

) y = {

4.5

4.5

(

)

x2 ≤ 16

7¢1 -

x2 from each side. 16

Multiply both sides by 7. x2 ≤ 16

Take the square root of each side.

Now graph the two functions Y1 =

3 x2 Y2   7 1  16

B

Subtract

C

7¢1 -

x2 x2 ≤ and Y2 = - 7¢1 ≤ 16 C 16

Figure 21 shows the result.

Now Work

■

PROBLEM

27

An equation of the form of equation (2), with a 7 b, is the equation of an ellipse with center at the origin, foci on the x-axis at 1 - c, 02 and 1c, 02 , where c 2 = a2 - b2, and major axis along the x-axis. For the remainder of this section, the direction “Analyze the equation” will mean to find the center, major axis, foci, and vertices of the ellipse and graph it.

EX A MPL E 2

Analyzing the Equation of an Ellipse Analyze the equation:

Solution

y2 x2 + = 1 25 9

The given equation is of the form of equation (2), with a2 = 25 and b2 = 9. The equation is that of an ellipse with center 10, 02 and major axis along the x-axis. The vertices are at 1 {a, 02 = 1 {5, 02. Because b2 = a2 - c 2, this means c 2 = a2 - b2 = 25 - 9 = 16

The foci are at 1 {c, 02 = 1 {4, 02. Figure 22 shows the graph. Figure 22

y 6 (0, 3) V1  (5, 0) F  (4, 0) 1

F 2  (4, 0) V2  (5, 0) 6 x

6

(0, 3)

Now Work

PROBLEM

17

r

SECTION 9.3  The Ellipse

705

If the major axis of an ellipse with center at 10, 02 lies on the y-axis, the foci are at 10, - c2 and 10, c2. Using the same steps as before, the definition of an ellipse leads to the following result:

THEOREM

y F 2  (0, c) c

(b, 0) x

F 1  (0, c) V 1  (0, a)

Analyzing the Equation of an Ellipse

EXAM PL E 3

Analyze the equation: 9x2 + y2 = 9

Figure 24

Solution y 3 V2  (0, 3)

To put the equation in proper form, divide each side by 9. x2 +

F2  (0, 2 2)

(1, 0)

3

EXAM PL E 4

Solution Figure 25 y 3 V 2  (0, 3)

PROBLEM

21

Finding an Equation of an Ellipse

Find an equation of the ellipse having one focus at 10, 22 and vertices at 10, - 32 and 10, 32.
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( 5 , 0) 3 x

3

F 1  (0, 2)

The larger denominator, 9, is in the y2@term so, based on equation (3), this is the equation of an ellipse with center at the origin and major axis along the y-axis. Also, a2 = 9, b2 = 1, and c 2 = a2 - b2 = 9 - 1 = 8. The vertices are at 10, {a2 = 10, {32, and the foci are at 10, {c2 = 10, {2222. The x-intercepts are at 1 {b, 02 = 1 {1, 02 . Figure 24 shows the graph.

Now Work

3 V  (0, 3) 1

( 5 , 0)

x

y2 = 1 9

r

F1  (0, 2 2)

F 2  (0, 2)

(3)

Figure 23 illustrates the graph of such an ellipse. Again, notice the right triangle formed by the points at 10, 02, 1b, 02, and 10, c2 , so that a2 = b2 + c 2 (or b2 = a2 - c 2). Look closely at equations (2) and (3). Although they may look alike, there is a difference! In equation (2), the larger number, a2, is in the denominator of the x2@term, so the major axis of the ellipse is along the x-axis. In equation (3), the larger number, a2, is in the denominator of the y2@term, so the major axis is along the y-axis.

a b

where a 7 b 7 0 and b2 = a2 - c 2

The major axis is the y-axis.

V 2  (0, a )

3 (1, 0)

An equation of the ellipse with center at 10, 02, foci at 10, - c2 and 10, c2 , and vertices at 10, - a2 and 10, a2 is y2 x2 + = 1, b2 a2

Figure 23

(b, 0)

Equation of an Ellipse: Center at (0, 0); Major Axis along the y-Axis

y2 x2 + = 1 5 9 3 V1  (0, 3)

r

Figure 25 shows the graph.

Now Work

PROBLEM

29

706

CHAPTER 9  Analytic Geometry

The circle may be considered a special kind of ellipse. To see why, let a = b in equation (2) or (3). Then y2 x2 + = 1 a2 a2 x 2 + y 2 = a2 This is the equation of a circle with center at the origin and radius a. The value of c is c 2 = a 2 - b2 = 0 c

a = b

This indicates that the closer the two foci of an ellipse are to the center, the more the ellipse will look like a circle.

2 Analyze Ellipses with Center at (h, k) If an ellipse with center at the origin and major axis coinciding with a coordinate axis is shifted horizontally h units and then vertically k units, the result is an ellipse with center at 1h, k2 and major axis parallel to a coordinate axis. The equations of such ellipses have the same forms as those given in equations (2) and (3), except that x is replaced by x - h (the horizontal shift) and y is replaced by y - k (the vertical shift). Table 3 gives the forms of the equations of such ellipses, and Figure 26 shows their graphs.

Table 3 NOTE It is not recommended that Table 3 be memorized. Rather, use the ideas of transformations (horizontally h units, vertically k units), along with the fact that a represents the distance from the center to the vertices, c represents the distance from the center to the foci, and b 2 = a 2 - c 2 (or c 2 = a 2 - b 2).

Equations of an Ellipse: Center at (h, k); Major Axis Parallel to a Coordinate Axis Center

Major Axis

Foci

Vertices

(h, k)

Parallel to the x-axis

(h + c, k)

(h + a, k)

(h - c, k)

(h - a, k)

(h, k + c)

(h, k + a)

(h, k - c)

(h, k - a)

(h, k)

Parallel to the y-axis

Equation (y - k)2 (x - h)2 + = 1, a2 b2 2 a 7 b 7 0 and b = a2 - c2 (y - k)2 (x - h)2 + = 1, 2 b a2 2 a 7 b 7 0 and b = a2 - c2

j

y

Figure 26 y

Major axis (h, k  a) (h, k  c)

(h  c, k) Major axis (h  a, k)

(h  c, k) (h , k) (h , k)

(h  a, k) x

(x  h)2 (y  k)2 (a) ––––––  ––––––  1 a2 b2

EX A MPL E 5

Solution

x (h, k  a)

(h, k  c)

(x  h)2 (y  k)2 (b) ––––––  ––––––  1 b2 a2

Finding an Equation of an Ellipse, Center Not at the Origin

Find an equation for the ellipse with center at 12, - 32, one focus at 13, - 32 , and one vertex at 15, - 32.
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SECTION 9.3  The Ellipse

707

is parallel to the x-axis. The distance from the center 12, - 32 to a focus 13, - 32 is c = 1; the distance from the center 12, - 32 to a vertex 15, - 32 is a = 3. Then b2 = a2 - c 2 = 9 - 1 = 8. The form of the equation is 1x - h2 2

+

1y - k2 2

= 1, where h = 2, k = - 3, a = 3, b = 2 22 a2 b2 1x - 22 2 1y + 32 2 + = 1 9 8

To graph the equation, use the center 1h, k2 = 12, - 32 to locate the vertices. The major axis is parallel to the x-axis, so the vertices are a = 3 units left and right of the center 12, - 32. Therefore, the vertices are V1 = 12 - 3, - 32 = 1 - 1, - 32

Figure 27 y 2

V1  (1, 3)

Since c = 1 and the major axis is parallel to the x-axis, the foci are 1 unit left and right of the center. Therefore, the foci are

(2, 3  2 2 )

F1 = 12 - 1, - 32 = 11, - 32

6 x

2 F1

F2

V2  (5, 3)

(2, 3  2 2 )

- 222 2

and

1 2, - 3

+ 222 2

r

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Now Work

EXAM PL E 6

and F2 = 12 + 1, - 32 = 13, - 32

Finally, we use the value of b = 222 to find the two points above and below the center.

1 2, - 3

(2, 3)

and V2 = 12 + 3, - 32 = 15, - 32

PROBLEM

55

Analyzing the Equation of an Ellipse Analyze the equation: 4x2 + y2 - 8x + 4y + 4 = 0

Solution

Complete the squares in x and in y. 4x2 + y2 - 8x + 4y + 4 = 0 4x2 - 8x + y2 + 4y = - 4 41x2 - 2x2 + 1y2 + 4y2 = - 4

Figure 28 y (1, 0) x (1, 2  3 ) (0, 2)

4

(1, 2)

(1, 4)

(2, 2) (1, 2  3 )

41x2 - 2x + 12 + 1y2 + 4y + 42 = - 4 + 4 + 4 41x - 12 2 + 1y + 22 2 = 4 1y + 22 2 1x - 12 2 + = 1 4

Group like variables; place the constant on the right side. Factor out 4 from the first two terms. Complete each square. Factor. Divide each side by 4.

This is the equation of an ellipse with center at 11, - 22 and major axis parallel to the y-axis. Since a2 = 4 and b2 = 1, we have c 2 = a2 - b2 = 4 - 1 = 3. The vertices are at 1h, k { a2 = 11, - 2 { 22 , or 11, 02 and 11, - 42. The foci are at 1h, k { c2 = 11, - 2 { 232 , or 11, - 2 - 232 and 11, - 2 + 232. Figure 28 shows the graph.

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PROBLEM

47

3 Solve Applied Problems Involving Ellipses Ellipses are found in many applications in science and engineering. For example, the orbits of the planets around the Sun are elliptical, with the Sun’s position at a focus. See Figure 29 on the following page.

708

CHAPTER 9  Analytic Geometry

Figure 29

Venus

Jupiter

Mars Earth

Asteroids

Stone and concrete bridges are often shaped as semielliptical arches. Elliptical gears are used in machinery when a variable rate of motion is required. Ellipses also have an interesting reflection property. If a source of light (or sound) is placed at one focus, the waves transmitted by the source reflect off the ellipse and concentrate at the other focus. This is the principle behind whispering galleries, which are rooms designed with elliptical ceilings. A person standing at one focus of the ellipse can whisper and be heard by a person standing at the other focus, because all the sound waves that reach the ceiling are reflected to the other person.

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Solution

Set up a rectangular coordinate system so that the center of the ellipse is at the origin and the major axis is along the x-axis. The equation of the ellipse is y2 x2 + = 1 a2 b2 Since the length of the room is 47.3 feet, the distance from the center of the room 47.3 to each vertex (the end of the room) is = 23.65 feet, so a = 23.65 feet. The 2 distance from the center of the room to each focus is c = 20.3 feet. See Figure 30. Because b2 = a2 - c 2, this means b2 = 23.652 - 20.32 = 147.2325. An equation that describes the shape of the room is given by

Figure 30

y2 x2 + = 1 147.2325 23.652

y 15 (0, 12.1) (⫺23.65, 0) ⫺25 (⫺20.3, 0)

(23.65, 0)

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PROBLEM

r

71

9.3 Assess Your Understanding ‘Are You Prepared?’

Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.

1. The distance d from P1 = 12, - 52 to P2 = 14, - 22 is d = 213 . (p. 35) 9 2. To complete the square of x2 - 3x, add 4 . (pp. A38–A39) 3. The intercepts of the equation y2 = 16 - 4x2 are ( - 2, 0), (2, 0), (0, - 4), (0, 4) . (pp. 43–44) 4. The point that is symmetric with respect to the y-axis to the point 1 - 2, 52 is (2, 5) . (pp. 44–45)

5. To graph y = 1x + 12 2 - 4, shift the graph of y = x2 to the (left/right) left 1 unit(s) and then (up/down) down 4 unit(s). (pp. 121–123)

6. The standard equation of a circle with center at 12, - 32 and radius 1 is . (pp. 66–67) (x - 2)2 + (y + 3)2 = 1

709

SECTION 9.3  The Ellipse

Concepts and Vocabulary 7. A(n) ellipse is the collection of all points in the plane the sum of whose distances from two fixed points is a constant. 8. For an ellipse, the foci lie on a line called the major axis. y2 x2 9. For the ellipse + = 1, the vertices are the points 4 25 (0, - 5) and (0, 5) . 10. For the ellipse value of b is

11. If the center of an ellipse is 12, - 32, the major axis is parallel to the x-axis, and the distance from the center of the ellipse to its vertices is a = 4 units, then the coordinates of the vertices are ( - 2, - 3) and (6, - 3) .

12. If the foci of an ellipse are 1 - 4, 42 and 16, 42 , then the coordinates of the center of the ellipse are (1, 4) .

y2 x2 + = 1, the value of a is 5 , the 25 9 3 , and the major axis is the x -axis.

Skill Building In Problems 13–16, the graph of an ellipse is given. Match each graph to its equation. y2 y2 x2 x2 (A) + y2 = 1 (B) x2 + = 1 (C) + = 1
4 4 16 4 13. C

14. %

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15. B

y

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y2 x2 + = 1 4 16

16. A

y 3

y 3

2 ⫺2

2

⫺4

x

⫺3

4 x

⫺2

⫺4

⫺3

3 x

3x ⫺3

⫺3

In Problems 17–26, find the vertices and foci of each ellipse. Graph each equation. y2 y2 y2 x2 x2 x2 + = 1 *18. + = 1 *19. + = 1 *17. 25 4 9 4 9 25 *21. 4x2 + y2 = 16

*22. x2 + 9y2 = 18

*25. x2 + y2 = 16

*26. x2 + y2 = 4

*20. x2 +

*23. 4y2 + x2 = 8

In Problems 27–38, find an equation for each ellipse. Graph the equation. *27. Center at 10, 02; focus at 13, 02; vertex at 15, 02

y2 = 1 16

*24. 4y2 + 9x2 = 36

*28. Center at 10, 02; focus at 1 - 1, 02; vertex at 13, 02

*29. Center at 10, 02; focus at 10, - 42; vertex at 10, 52

*30. Center at 10, 02; focus at 10, 12; vertex at 10, - 22

*33. Focus at 1 - 4, 02; vertices at 1 {5, 02

*34. Focus at 10, - 42; vertices at 10, {82

*37. Center at 10, 02; vertex at 10, 42; b = 1

*38. Vertices at 1 {5, 02; c = 2

*31. Foci at 1 {2, 02; length of the major axis is 6

*32. Foci at 10, {22; length of the major axis is 8

*35. Foci at 10, {32; x-intercepts are {2

*36. Vertices at 1 {4, 02; y-intercepts are {1

In Problems 39–42, write an equation for each ellipse. *39. (⫺1, 1)

*40.

y 3

*41.

y 3

*42.

y 3

y 3 (0, 1)

⫺3

3 x

⫺3

⫺3

(⫺1, ⫺1)

3 x

⫺3

⫺3

(1, 0)

3 x

⫺3

⫺3

3 x

⫺3

In Problems 43–54, analyze each equation; that is, find the center, foci, and vertices of each ellipse. Graph each equation. 1x - 32 2 1y + 12 2 1x + 42 2 1y + 22 2 *43. + = 1 *44. + = 1 *45. 1x + 52 2 + 41y - 42 2 = 16 4 9 9 4 *46. 91x - 32 2 + 1y + 22 2 = 18 *47. x2 + 4x + 4y2 - 8y + 4 = 0 *48. x2 + 3y2 - 12y + 9 = 0 *49. 2x2 + 3y2 - 8x + 6y + 5 = 0

*50. 4x2 + 3y2 + 8x - 6y = 5

*51. 9x2 + 4y2 - 18x + 16y - 11 = 0

*52. x2 + 9y2 + 6x - 18y + 9 = 0

*53. 4x2 + y2 + 4y = 0

*54. 9x2 + y2 - 18x = 0

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710

CHAPTER 9  Analytic Geometry

In Problems 55–64, find an equation for each ellipse. Graph the equation. *55. Center at 12, - 22; vertex at 17, - 22; focus at 14, - 22

*56. Center at 1 - 3, 12; vertex at 1 - 3, 32; focus at 1 - 3, 02

*57. Vertices at 14, 32 and 14, 92; focus at 14, 82

*58. Foci at 11, 22 and 1 - 3, 22; vertex at 1 - 4, 22

*61. Center at 11, 22; focus at 14, 22; contains the point 11, 32

*62. Center at 11, 22; focus at 11, 42; contains the point 12, 22

*59. Foci at 15, 12 and 1 - 1, 12; length of the major axis is 8

*63. Center at 11, 22; vertex at 14, 22; contains the point 11, 52

*60. Vertices at 12, 52 and 12, - 12; c = 2

*64. Center at 11, 22; vertex at 11, 42; contains the point 11 + 23, 32

In Problems 65–68, graph each function. Be sure to label all the intercepts. [Hint: Notice that each function is half an ellipse.] *66. f 1x2 = 29 - 9x2

*65. f 1x2 = 216 - 4x2

*67. f 1x2 = - 264 - 16x2

*68. f 1x2 = - 24 - 4x2

Applications and Extensions 69. Semielliptical Arch Bridge An arch in the shape of the upper half of an ellipse is used to support a bridge that is to span a river 20 meters wide. The center of the arch is 6 meters above the center of the river. See the figure. Write an equation for the ellipse in which the x-axis coincides with the water level and the y-axis passes through the center of the arch. y2 x2 + = 1 100 36

74. Semielliptical Arch Bridge A bridge is to be built in the shape of a semielliptical arch and is to have a span of 100 feet. The height of the arch, at a distance of 40 feet from the center, is to be 10 feet. Find the height of the arch at its center. 16.67 ft 75. Racetrack Design Consult the figure. A racetrack is in the shape of an ellipse 100 feet long and 50 feet wide. What is UIF
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76. Semielliptical Arch Bridge An arch for a bridge over a highway is in the form of half an ellipse. The top of the arch is 20 feet above the ground level (the major axis). The highway has four lanes, each 12 feet wide; a center safety strip 8 feet wide; and two side strips, each 4 feet wide. What should the span of the bridge be (the length of its major axis) if the IFJHIU

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In Problems 79–83, use the fact that the orbit of a planet about the Sun is an ellipse, with the Sun at one focus. The aphelion of a planet is its greatest distance from the Sun, and the perihelion is its shortest distance. The mean distance of a planet from the Sun is the length of the semimajor axis of the elliptical orbit. See the illustration. *79. Earth The mean distance of Earth from the Sun is 93 million miles. If the aphelion of Earth is 94.5 million NJMFT
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SECTION 9.4  The Hyperbola

711

y-coordinate. Determine the exact y-coordinate of the point M on D that is equidistant from points B and C.† 525 - 4

*82. Pluto The perihelion of Pluto is 4551 million miles, and the distance from the center of its elliptical orbit to the Sun is 897.5 million miles. Find the aphelion of Pluto. What is the mean distance of Pluto from the Sun? Write an equation for the orbit of Pluto about the Sun.

*86. Show that an equation of the form Ax2 + Cy2 + F = 0,

A ≠ 0, C ≠ 0, F ≠ 0

where A and C are of the same sign and F is of opposite sign, (a) is the equation of an ellipse with center at 10, 02 if A ≠ C. (b) is the equation of a circle with center 10, 02 if A = C.

83. Elliptical Orbit A planet orbits a star in an elliptical orbit with the star located at one foci. The perihelion of the planet c is 5 million miles. The eccentricity e of a conic section is e = . a If the eccentricity of the orbit is 0.75, find the aphelion of the planet.† 35 million mi

*87. Show that the graph of an equation of the form Ax2 + Cy2 + Dx + Ey + F = 0,

84. A rectangle is inscribed in an ellipse with major axis of length 14 meters and minor axis of length 4 meters. Find the maximum area of a rectangle inscribed in the ellipse. Round your answer to two decimal places.† 28 m2

A ≠ 0, C ≠ 0

where A and C are of the same sign, D2 E2 (a) is an ellipse if + - F is the same sign as A. 4A 4C 2 2 D E (b) is a point if + - F = 0. 4A 4C D2 E2 (c) contains no points if + - F is of opposite sign to A. 4A 4C

85. Let D be the line given by the equation x + 5 = 0. Let E be the conic section given by the equation x2 + 5y2 = 20. Let the point C be the vertex of E with the smaller x-coordinate, and let B be the endpoint of the minor axis of E with the larger

Discussion and Writing c 88. The eccentricity e of an ellipse is defined as the number , where a is the distance of a vertex from the center and c is the distance a of a focus from the center. Because a 7 c, it follows that e 6 1. Write a brief paragraph about the general shape of each of the following ellipses. Be sure to justify your conclusions. (a) Eccentricity close to 0 (b) Eccentricity = 0.5 (c) Eccentricity close to 1

Retain Your knowledge Problems 89–92 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. *89. Find the zeros of the quadratic function f(x) = (x - 5)2 - 12. What are the x-intercepts, if any, of the graph of the function? 2x - 3 *90. Find the domain of the rational function f(x) = . x - 5 Find any horizontal, vertical, or oblique asymptotes.

92. Solve the right triangle shown. b ≈ 10.94, c ≈ 17.77, B = 38° 528 c

91. Find the work done by a force of 80 pounds acting in the direction of 50° to the horizontal in moving an object 12 feet from (0, 0) to (12, 0). Round to one decimal place. 617.1 ft-lb

b

B 14

‘Are You Prepared?’ Answers 1. 213

2.

9 4

3. 1 - 2, 02, 12, 02, 10, - 42, 10, 42

4. 12, 52

5. left; 1; down: 4

6. 1x - 22 2 + 1y + 32 2 = 1

9.4 The Hyperbola PREPARING FOR THIS SECTION Before getting started, review the following: r
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OBJECTIVES 1 2 3 4 †

Analyze Hyperbolas with Center at the Origin (p. 712) Find the Asymptotes of a Hyperbola (p. 716) Analyze Hyperbolas with Center at (h, k) (p. 718) Solve Applied Problems Involving Hyperbolas (p. 719)

Courtesy off the C h Joliet J li JJunior i C College ll M Mathematics h i D Department

712

CHAPTER 9  Analytic Geometry

DEFINITION

A hyperbola is the collection of all points in the plane the difference of whose distances from two fixed points, called the foci, is a constant.

Figure 31

Conjugate axis

Transverse axis V2

V1

F2

Figure 31 illustrates a hyperbola with foci F1 and F2 . The line containing the foci is called the transverse axis. The midpoint of the line segment joining the foci is the center of the hyperbola. The line through the center and perpendicular to the transverse axis is the conjugate axis. The hyperbola consists of two separate curves, called branches, that are symmetric with respect to the transverse axis, conjugate axis, and center. The two points of intersection of the hyperbola and the transverse axis are the vertices, V1 and V2 , of the hyperbola.

Center

F1

1 Analyze Hyperbolas with Center at the Origin

Figure 32

d 1F1, P2 - d 1F2, P2 = {2a y d(F 1, P ) Transverse axis F 1 ⫽ (⫺c, 0)

P ⫽ (x, y ) d (F 2 , P ) F 2 ⫽ (c, 0) x

With these ideas in mind, we are now ready to find the equation of a hyperbola in the rectangular coordinate system. First, place the center at the origin. Next, position the hyperbola so that its transverse axis coincides with a coordinate axis. Suppose that the transverse axis coincides with the x-axis, as shown in Figure 32. If c is the distance from the center to a focus, one focus is at F1 = 1 - c, 02 and the other at F2 = 1c, 02 . Now we let the constant difference of the distances from any point P = 1x, y2 on the hyperbola to the foci F1 and F2 be denoted by {2a. (If P is on the right branch, the + sign is used; if P is on the left branch, the - sign is used.) The coordinates of P must satisfy the equation d 1F1 , P2 - d 1F2 , P2 = {2a 2 1x - 1 - c2 2 2 + y2 - 2 1x - c2 2 + y2 = {2a 2 1x + c2 2 + y2 = {2a + 2 1x - c2 2 + y2 1x + c2 2 + y2 = 4a2 { 4a2 1x - c2 2 + y2

Difference of the distances from P to the foci equals {2a. Use the Distance Formula. Isolate one radical. Square both sides.

+ 1x - c2 2 + y2 Next, remove the parentheses. x2 + 2cx + c 2 + y2 = 4a2 { 4a2 1x - c2 2 + y2 + x2 - 2cx + c 2 + y2 4cx - 4a2 = {4a2 1x - c2 2 + y2 cx - a = {a2 1x - c2 + y 2

2

2

1cx - a2 2 = a2 3 1x - c2 2 + y2 4 2

c 2 x2 - 2ca2 x + a4 = a2 1x2 - 2cx + c 2 + y2 2 c 2 x 2 + a4 = a2 x 2 + a2 c 2 + a2 y 2 1c - a 2x - a y = a c - a 2

2

2

2 2

2 2

4

1c 2 - a2 2x2 - a2 y2 = a2 1c 2 - a2 2

Simplify; isolate the radical. Divide each side by 4. Square both sides. Simplify. Distribute and simplify. Rearrange terms. Factor a 2 on the right side.

(1)

To obtain points on the hyperbola off the x-axis, it must be that a 6 c. To see why, look again at Figure 32. d 1F1 , P2 6 d 1F2 , P2 + d 1F1 , F2 2

d 1F1 , P2 - d 1F2 , P2 6 d 1F1 , F2 2 2a 6 2c a 6 c

Use triangle F1 P F2 .

P is on the right branch, so d(F1 , P) - d(F2 , P) = 2a; d1F1, F2 2 = 2c.

SECTION 9.4  The Hyperbola

713

Because a 6 c, this means a2 6 c 2, so c 2 - a2 7 0. Let b2 = c 2 - a2, b 7 0. Then equation (1) can be written as b2 x2 - a2 y2 = a2 b2 y2 x2 = 1 a2 b2

Divide each side by a 2 b 2.

To find the vertices of the hyperbola defined by this equation, let y = 0. x2 The vertices satisfy the equation 2 = 1, the solutions of which are x = {a. a Consequently, the vertices of the hyperbola are V1 = 1 - a, 02 and V2 = 1a, 02. Notice that the distance from the center 10, 02 to either vertex is a.

THEOREM

Equation of a Hyperbola: Center at (0, 0); Transverse Axis along the x-Axis An equation of the hyperbola with center at 10, 02, foci at 1 - c, 02 and 1c, 02, and vertices at 1 - a, 02 and 1a, 02 is y2 x2 = 1, a2 b2

Figure 33

where b2 = c 2 - a2

(2)

y

The transverse axis is the x-axis. V 1 ⫽ (⫺a, 0) Transverse axis F 1 ⫽ (⫺c, 0)

V 2 ⫽ (a, 0) F 2 ⫽ (c, 0) x

EXAMPL E 1

Solution

See Figure 33. As you can verify, the hyperbola defined by equation (2) is symmetric with respect to the x-axis, y-axis, and origin. To find the y-intercepts, if y2 any, let x = 0 in equation (2). This results in the equation 2 = - 1, which has no b real solution, so the hyperbola defined by equation (2) has no y-intercepts. In fact, y2 x2 x2 since 2 - 1 = 2 Ú 0, it follows that 2 Ú 1. There are no points on the graph for a b a - a 6 x 6 a.

Finding and Graphing an Equation of a Hyperbola

Find an equation of the hyperbola with center at the origin, one focus at 13, 02, and one vertex at 1 - 2, 02.
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x = {3

714

CHAPTER 9  Analytic Geometry

Figure 34

y2 =

y 5

(3, 5–2)

(3, 5–2)

V1  (2, 0)

y = {

V 2  (2, 0)

5 F 1  (3, 0)

25 4

5 x F 2  (3, 0)

(3, 5–2) (3, 5–2)

5 2

5 5 The points above and below the foci are a {3, b and a {3, - b . These points 2 2 determine the “opening” of the hyperbola. See Figure 34.

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5

COMMENT To graph the hyperbola

y2 x2 = 1 discussed in Example 1, the two functions 4 5

x2 x2 - 1 and Y2 = - 25 - 1
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PROBLEM

19

An equation of the form of equation (2) is the equation of a hyperbola with center at the origin; foci on the x-axis at 1 - c, 02 and 1c, 02 , where c 2 = a2 + b2; and transverse axis along the x-axis. For the next two examples, the direction “Analyze the equation” will mean to find the center, transverse axis, vertices, and foci of the hyperbola and graph it.

EX A MPL E 2

Analyzing the Equation of a Hyperbola Analyze the equation:

Solution

y2 x2 = 1 16 4

The given equation is of the form of equation (2), with a2 = 16 and b2 = 4. The graph of the equation is a hyperbola with center at 10, 02 and transverse axis along the x-axis. Also, c 2 = a2 + b2 = 16 + 4 = 20. The vertices are at 1 {a, 02 = 1 {4, 02, and the foci are at 1 {c, 02 = 1 {225, 02. To locate the points on the graph above and below the foci, let x = {225. Then y2 x2 = 1 16 4

1 {225 2 2 16

-

y2 = 1 4

x = {2 25

y2 20 = 1 16 4 y2 5 = 1 4 4

Figure 35 y 4 (– 2 5 , 1) (2 5 , 1) V 1 = (–4, 0) V = (4, 0) 2 –5

F1 = (–2 5 , 0) (–2 5 , –1) –4

5 x F2 = (2 5 , 0) (2 5 , –1)

y2 1 = 4 4 y = {1 The points above and below the foci are 1 {225, 12 and 1 {225, - 12. See Figure 35.

r

SECTION 9.4  The Hyperbola

THEOREM

715

Equation of a Hyperbola: Center at (0, 0); Transverse Axis along the y-Axis An equation of the hyperbola with center at 10, 02, foci at 10, - c2 and 10, c2 , and vertices at 10, - a2 and 10, a2 is y2 a2

-

x2 = 1, b2

where b2 = c 2 - a2

(3)

The transverse axis is the y-axis.

Figure 36 y2 x2 = 1, b 2 = c 2 - a 2 a2 b2 y F 2  (0, c ) V 2  (0, a) x V 1  (0, a )

F 1  (0, c )

EXAM PL E 3

Figure 36 shows the graph of a typical hyperbola defined by equation (3). y2 x2 An equation of the form of equation (2), 2 - 2 = 1, is the equation of a a b hyperbola with center at the origin; foci on the x-axis at 1 - c, 02 and 1c, 02, where b2 = c 2 - a2 (or c 2 = a2 + b2); and transverse axis along the x-axis. y2 x2 An equation of the form of equation (3), 2 - 2 = 1, is the equation of a a b hyperbola with center at the origin; foci on the y-axis at 10, - c2 and 10, c2 , where b2 = c 2 - a2 (or c 2 = a2 + b2); and transverse axis along the y-axis. Notice the difference in the forms of equations (2) and (3). When the y2@term is subtracted from the x2@term, the transverse axis is along the x-axis. When the x2@term is subtracted from the y2@term, the transverse axis is along the y-axis.

Analyzing the Equation of a Hyperbola Analyze the equation: y2 - 4x2 = 4

Solution

To put the equation in proper form, divide each side by 4: y2 - x2 = 1 4 Since the x2@term is subtracted from the y2@term, the equation is that of a hyperbola with center at the origin and transverse axis along the y-axis. Also, comparing the above equation to equation (3), note that a2 = 4, b2 = 1, and c 2 = a2 + b2 = 5. The vertices are at 10, {a2 = 10, {22, and the foci are at 10, {c2 = 10, { 252. To locate other points on the graph, let x = {2. Then

Figure 37 y 5 (2, 2 5 )

(– 2, 2 5 )

F 2 = (0, 5 ) V 2 = (0, 2) 5 x

–5 V 1 = (0, – 2)

F 1 = (0, – 5 ) (–2, –2 5 )

(2, – 2 5 ) –5

y2 - 4x2 = 4 y2 - 41 {22 2 = 4

x = {2

y2 - 16 = 4 y2 = 20 y = {225 Four other points on the graph are 1 {2, 2252 and 1 {2, - 2252. See Figure 37.

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EXAM PL E 4

Solution

Finding an Equation of a Hyperbola

Find an equation of the hyperbola that has one vertex at 10, 22 and foci at 10, - 32 and 10, 32.
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Because the foci are at 10, - 32 and 10, 32, the center of the hyperbola, which is at their midpoint, is the origin. Because the foci lie on the y-axis, the transverse

716

CHAPTER 9  Analytic Geometry

axis is along the y-axis. The given information also reveals that c = 3, a = 2, and b2 = c 2 - a2 = 9 - 4 = 5. The form of the equation of the hyperbola is given by equation (3):

Figure 38 y 5 F = (0, 3) 2

(– 52 , 3)

y2

( 52 , 3)

a

V 2 = (0, 2) V 1 = (0, – 2)

(– 52 , –3)

( 52 , –3) –5

F 1 = (0, – 3)

-

x2 = 1 b2

y2 x2 = 1 4 5

5 x

–5

2

Let y = {3 to obtain points on the graph on either side of each focus. See Figure 38.

Now Work

PROBLEM

r

21

Look at the equations of the hyperbolas in Examples 2 and 4. For the hyperbola in Example 2, a2 = 16 and b2 = 4, so a 7 b; for the hyperbola in Example 4, a2 = 4 and b2 = 5, so a 6 b. This indicates that for hyperbolas, there are no requirements involving the relative sizes of a and b. Contrast this situation to the case of an ellipse, in which the relative sizes of a and b dictate which axis is the major axis. Hyperbolas have another feature to distinguish them from ellipses and parabolas: Hyperbolas have asymptotes.

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with the property that the distance from the line to points on the graph approaches 0 as x S - q or as x S q . The asymptotes provide information about the end behavior of the graph of a hyperbola.

THEOREM

Asymptotes of a Hyperbola The hyperbola

y2 x2 = 1 has the two oblique asymptotes a2 b2 y =

b b x and y = - x a a

(4)

Proof Begin by solving for y in the equation of the hyperbola. y2 x2 = 1 a2 b2 y2 x2 = - 1 b2 a2 y 2 = b2 ¢

x2 - 1≤ a2

Because x ≠ 0, the right side can be rearranged in the form y2 =

b2 x 2 a2 ¢1 - 2 ≤ 2 a x

y = {

bx a2 1 - 2 a B x

a2 approaches 0, so the expression under the x2 bx radical approaches 1. So, as x S - q or as x S q , the value of y approaches { ; a that is, the graph of the hyperbola approaches the lines

Now, as x S - q or as x S q , the term

b b y = - x and y = x a a These lines are oblique asymptotes of the hyperbola.

■

SECTION 9.4  The Hyperbola

717

The asymptotes of a hyperbola are not part of the hyperbola, but they do serve as a guide for graphing a hyperbola. For example, suppose that we want to graph the equation

Figure 39 y2 x2 = 1 a2 b2 y y

y2 x2 = 1 a2 b2

b– x a

(0, b) V2  (a, 0) x (0, b) V1  (a, 0)

y   b–a x

THEOREM

Begin by plotting the vertices 1 - a, 02 and 1a, 02. Then plot the points 10, - b2 and 10, b2 and use these four points to construct a rectangle, as shown in Figure 39. b b The diagonals of this rectangle have slopes and - , and their extensions are the a a b b asymptotes y = x and y = - x of the hyperbola. If we graph the asymptotes, a a we can use them to establish the “opening” of the hyperbola and avoid plotting other points.

Asymptotes of a Hyperbola The hyperbola

y2 a2

-

x2 = 1 has the two oblique asymptotes b2 y =

a a x and y = - x b b

(5)

You are asked to prove this result in Problem 84. For the remainder of this section, the direction “Analyze the equation” will mean to find the center, transverse axis, vertices, foci, and asymptotes of the hyperbola and graph it.

EXAM PL E 5 Figure 40 y 5 22x

Analyzing the Equation of a Hyperbola Analyze the equation:

y 5

y 5 2x

F2 5 (0, 5) V2 5 (0, 2) 25

5 x V1 5 (0, 2 2) F1 5 (0, 2 5) 25

y2 - x2 = 1 4

Solution Because the x2@term is subtracted from the y2@term, the equation is of the form of equation (3) and is a hyperbola with center at the origin and transverse axis along the y-axis. Also, comparing this equation to equation (3), note that a2 = 4, b2 = 1, and c 2 = a2 + b2 = 5. The vertices are at 10, {a2 = 10, {22, and the foci are at 10, {c2 = 10, { 252. Using equation (5) with a = 2 and b = 1, the a a asymptotes are the lines y = x = 2x and y = - x = - 2x. Form the rectangle b b containing the points 10, {a2 = 10, {22 and 1 {b, 02 = 1 {1, 02. The extensions of the diagonals of this rectangle are the asymptotes. Now graph the rectangle, the asymptotes, and the hyperbola. See Figure 40.

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EXAM PL E 6

Analyzing the Equation of a Hyperbola Analyze the equation: 9x2 - 4y2 = 36

Solution

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718

CHAPTER 9  Analytic Geometry

at 1 {a, 02 = 1 {2, 02, the foci are c = 213 units left and right of the center at 1 {c, 02 = 1 { 213, 02, and the asymptotes have the equations

Figure 41 y   3– x

y

y  3– x

2

2

5

y =

(0, 3) V1  (2, 0)

V2  (2, 0)

5

5

x

To graph the hyperbola, form the rectangle containing the points 1 {a, 02 and 10, {b2 , that is, 1 - 2, 02, 12, 02, 10, - 32, and 10, 32. The extensions of the diagonals of this rectangle are the asymptotes. See Figure 41 for the graph.

r

F 2  ( 13, 0)

F 1  ( 13, 0)

3 b 3 b x = x and y = - x = - x a a 2 2

Now Work

(0, 3)

31

PROBLEM

5

3 Analyze Hyperbolas with Center at (h, k) If a hyperbola with center at the origin and transverse axis coinciding with a coordinate axis is shifted horizontally h units and then vertically k units, the result is a hyperbola with center at 1h, k2 and transverse axis parallel to a coordinate axis. The equations of such hyperbolas have the same forms as those given in equations (2) and (3), except that x is replaced by x - h (the horizontal shift) and y is replaced by y - k (the vertical shift). Table 4 gives the forms of the equations of such hyperbolas. See Figure 42 for typical graphs.

Table 4

Equation of a Hyperbola: Center at (h, k); Transverse Axis Parallel to a Coordinate Axis Center

Transverse Axis

Foci

Vertices

Equation

Asymptotes

(h, k)

Parallel to the x-axis

(h { c, k)

(h { a, k)

(y - k) (x - h) = 1, b2 = c2 - a2 a2 b2

b y - k = { (x - h) a

(h, k)

Parallel to the y-axis

(h, k { c)

(h, k { a)

(y - k)2

a y - k = { (x - h) b

2

a

Figure 42 NOTE It is not recommended that Table 4 be memorized. Rather, use the ideas of transformations (shift horizontally h units, vertically k units), along with the fact that a represents the distance from the center to the vertices, c represents the distance from the center to the foci, and j b 2 = c 2 - a 2 (or c 2 = a 2 + b 2).

(x - h)2 = 1, b2 = c2 - a2 b2

y

Transverse F1 V1 axis

Transverse axis F2

(h, k)

V2

V2 F 2

(h, k ) x

V1

x

F1

2

Solution

-

y

2

(y  k) (x  h ) (a) –––––––  –––––– 1 a2 b2

EX A MPL E 7

2

2

(y  k )2 (x  h)2 (b) –––––––  –––––– 1 2 a b2

Finding an Equation of a Hyperbola, Center Not at the Origin

Find an equation for the hyperbola with center at 11, - 22, one focus at 14, - 22, and one vertex at 13, - 22.
(SBQI
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The center is at 1h, k2 = 11, - 22, so h = 1 and k = - 2. Because the center, focus, and vertex all lie on the line y = - 2, the transverse axis is parallel to the x-axis. The distance from the center 11, - 22 to the focus 14, - 22 is c = 3; the distance from

SECTION 9.4  The Hyperbola

Figure 43

the center 11, - 22 to the vertex 13, - 22 is a = 2. Then b2 = c 2 - a2 = 9 - 4 = 5. The equation is

y 6

1x - h2 2 a2

(1, 2  5)

6 x V2  (3, 2) (1, 2) F2  (4, 2)

1

6

-

1y - k2 2 b2

= 1

1x - 12 2 1y + 22 2 = 1 4 5

V1  (1, 2) 6 Transverse axis F  (2, 2)

719

r

See Figure 43.

Now Work

(1, 2  5)

PROBLEM

41

Analyzing the Equation of a Hyperbola

EXAM PL E 8

Analyze the equation: - x2 + 4y2 - 2x - 16y + 11 = 0

Solution

Complete the squares in x and in y. - x2 + 4y2 - 2x - 16y + 11 = 0

- 1x2 + 2x2 + 41y2 - 4y2 = - 11

Figure 44

- 1x + 2x + 12 + 41y - 4y + 42 = - 11 - 1 + 16

Transverse axis y F2 (1, 2  5 ) V2  (1, 3) 5

(3, 2)

2

- 1x + 12 2 + 41y - 22 2 = 4 1y - 22 2 -

(1, 2)

5 5 x F1 (1, 2 

V1  (1, 1)

5)

2

1x + 12 2 = 1 4

Group terms. Complete each square. Factor. Divide each side by 4.

This is the equation of a hyperbola with center at 1 - 1, 22 and transverse axis parallel to the y-axis. Also, a2 = 1 and b2 = 4, so c 2 = a2 + b2 = 5. Because the transverse axis is parallel to the y-axis, the vertices and foci are located a and c units above and below the center, respectively. The vertices are at 1h, k { a2 = 1 - 1, 2 { 12, or 1 - 1, 12 and 1 - 1, 32. The foci are at 1h, k { c2 = ( - 1, 2 { 25). The asymptotes 1 1 are y - 2 = 1x + 12 and y - 2 = - 1x + 12. Figure 44 shows the graph. 2 2

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Now Work

PROBLEM

55

4 Solve Applied Problems Involving Hyperbolas Figure 45

S

O3 O1

O2

EXAM PL E 9

L Look at Figure 45. Suppose that three microphones are located at points O1 , O2 , aand O3 (the foci of the two hyperbolas). In addition, suppose that a gun is fired at S, and the microphone at O1 records the gunshot 1 second after the microphone at O2 . Because sound travels at about 1100 feet per second, we conclude that the microphone at O1 is 1100 feet farther from the gunshot than O2 . We can model this situation by saying that S lies on a branch of a hyperbola with foci at O1 and O2 . %P
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Lightning Strikes Suppose that two people standing 1 mile apart both see a flash of lightning. After a period of time, the person standing at point A hears the thunder. One second later, the person standing at point B hears the thunder. If the person at B is due west of the person at A, and the lightning strike is known to occur due north of the person standing at point A
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720

CHAPTER 9  Analytic Geometry

Solution

See Figure 46, in which the ordered pair 1x, y2 represents the location of the lightning strike. Sound travels at 1100 feet per second, so the person at point A is 1100 feet closer to the lightning strike than the person at point B. Since the difference of the distance from 1x, y2 to B and the distance from 1x, y2 to A is the constant 1100, the point 1x, y2 lies on a hyperbola whose foci are at A and B.

Figure 46

North (x, y)

East B ⫽ (⫺2640, 0) (⫺a, o)

(a, o)

A ⫽ (2640, 0)

1 mile ⫽ 5280 feet

An equation of the hyperbola is y2 x2 = 1 a2 b2 where 2a = 1100, so a = 550. Because the distance between the two people is 1 mile (5280 feet) and each person is at a focus of the hyperbola, this means 2c = 5280 5280 c = = 2640 2 Because b2 = c 2 - a2 = 26402 - 5502 = 6,667,100, the equation of the hyperbola that describes the location of the lightning strike is y2 x2 = 1 6,667,100 5502 3FGFS
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at the point A = 12640, 02, let x = 2640 and solve the resulting equation. y2 26402 = 1 6,667,100 5502 y2 = - 22.04 6,667,100 y2 = 146,942,884 y = 12,122

Subtract

26402 from both sides. 5502

Multiply both sides by - 6,667,100. y 7 0 since the lightning strike occurred in quadrant I.

Check: The difference between the distance from 12640, 12,1222 to the person at the point B = 1 - 2640, 02 and the distance from 12640, 12,1222 to the person at the point A = 12640, 02 should be 1100. From the distance formula, the difference in the distances is 3 3 2640 - 1 - 26402 4 2 + 112,122 - 02 2 - 3 12640 - 26402 2 + 112,122 - 02 2 = 1100 as required. The lightning strike occurred 12,122 feet north of the person standing at point A.

Now Work

PROBLEM

75

r

SECTION 9.4  The Hyperbola

721

9.4 Assess Your Understanding ‘Are You Prepared?’

Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.

1. The distance d from P1 = 13, - 42 to P2 = 1 - 2, 12 is d = 522 . (p. 35) 25 2 2. To complete the square of x + 5x, add 4 . (pp. A38–A39)

5. To graph y = 1x - 52 3 - 4, shift the graph of y = x3 to the (left/right) 5 unit(s) and then (up/down) 4 unit(s). (pp. 121–123) right; down

y2 = 9 + 4x2.

6. Find the vertical asymptotes, if any, and the horizontal or x2 - 9 . (pp. 268–271) oblique asymptote, if any, of y = 2 x - 4 Vertical: x = - 2, x = 2; horizontal: y = 1

3. Find the intercepts of the equation (p. 44) (0, - 3), (0, 3)

4. True or False The equation y2 = 9 + x2 is symmetric with respect to the x-axis, the y-axis, and the origin. (pp. 44–46) True

Concepts and Vocabulary 7. A(n) hyperbola is the collection of points in the plane the difference of whose distances from two fixed points is a constant. 8. For a hyperbola, the foci lie on a line called the transverse axis . Answer Problems 9–11 using the figure. y

Transverse axis

9. The equation of the hyperbola is of the form (b) 1x - h2 2 1y - k2 2 = 1 (a) 2 a b2 1y - k2 2 1x - h2 2 (b) = 1 2 a b2 10. If the center of the hyperbola is (2, 1) and a = 3, then the coordinates of the vertices are (2, 4) and (2, - 2) .

F2

11. If the center of the hyperbola is (2, 1) and c = 5, then the coordinates of the foci are (2, 6) and (2, - 4) .

V2

12. In a hyperbola, if a = 3 and c = 5, then b = 4 . y2 x2 13. For the hyperbola = 1, the value of a is 2 , the 4 9 value of b is 3 , and the transverse axis is the x -axis. y2 x2 14. For the hyperbola = 1, the asymptotes are 16 81 4 4 y = x y = - x 9 . 9 and

(h, k ) V1

x

F1

Skill Building In Problems 15–18, the graph of a hyperbola is given. Match each graph to its equation. y2 y2 x2 x2 (A) - y2 = 1 (B) x2 = 1 (C) - x2 = 1 (D) y2 = 1 4 4 4 4 15. B

16. C

y 3

⫺3

3 x

17. A

y 4

⫺4

⫺3

4 x

18. D

y 4

⫺4

⫺4

4x ⫺4

y 3

⫺3

3x ⫺3

In Problems 19–28, find an equation for the hyperbola described. Graph the equation. *19. Center at 10, 02; focus at 13, 02; vertex at 11, 02

*21. Center at 10, 02; focus at 10, - 62; vertex at 10, 42

*23. Foci at 1 - 5, 02 and 15, 02; vertex at 13, 02

*25. Vertices at 10, - 62 and 10, 62; asymptote the line y = 2x *27. Foci at 1 - 4, 02 and 14, 02; asymptote the line y = - x

*20. Center at 10, 02; focus at 10, 52; vertex at 10, 32

*22. Center at 10, 02; focus at 1 - 3, 02; vertex at 12, 02 *24. Focus at 10, 62; vertices at 10, - 22 and 10, 22

*26. Vertices at 1 - 4, 02 and 14, 02; asymptote the line y = 2x

*28. Foci at 10, - 22 and 10, 22; asymptote the line y = - x

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

722

CHAPTER 9  Analytic Geometry

In Problems 29–36, find the center, transverse axis, vertices, foci, and asymptotes. Graph each equation. *29.

y2 x2 = 1 25 9

*30.

*33. y2 - 9x2 = 9

y2 x2 = 1 16 4

*34. x2 - y2 = 4

*31. 4x2 - y2 = 16

*32. 4y2 - x2 = 16

*35. y2 - x2 = 25

*36. 2x2 - y2 = 4

In Problems 37–40, write an equation for each hyperbola. *37.

y 3

y ⫽ ⫺x

y⫽x

⫺3

*38.

3x

y 3

⫺3

⫺3

⫺3

y⫽x

*39. y ⫽ ⫺2x

3x

⫺5

y 10

y ⫽ 2x

5 x

*40.

y ⫽ ⫺2x y 5

⫺5

⫺10

y ⫽ ⫺x

y ⫽ 2x

5 x

⫺5

In Problems 41–48, find an equation for the hyperbola described. Graph the equation. *41. Center at 14, - 12; focus at 17, - 12; vertex at 16, - 12

*42. Center at 1 - 3, 12; focus at 1 - 3, 62; vertex at 1 - 3, 42

*45. Foci at 13, 72 and 17, 72; vertex at 16, 72

*46. Focus at 1 - 4, 02 ; vertices at 1 - 4, 42 and 1 - 4, 22

*43. Center at 1 - 3, - 42; focus at 1 - 3, - 82; vertex at 1 - 3, - 22 *47. Vertices at 1 - 1, - 12 and 13, - 12; asymptote the line 3 y + 1 = 1x - 12 2

*44. Center at 11, 42; focus at 1 - 2, 42; vertex at 10, 42

*48. Vertices at 11, - 32 and 11, 12; asymptote the line 3 y + 1 = 1x - 12 2

In Problems 49–62, find the center, transverse axis, vertices, foci, and asymptotes. Graph each equation. 1x - 22 2 1y + 32 2 1y + 32 2 1x - 22 2 = 1 *50. = 1 *51. 1y - 22 2 - 41x + 22 2 = 4 *49. 4 9 4 9 *52. 1x + 42 2 - 91y - 32 2 = 9

*53. 1x + 12 2 - 1y + 22 2 = 4

*54. 1y - 32 2 - 1x + 22 2 = 4

*55. x - y - 2x - 2y - 1 = 0

*56. y - x - 4y + 4x - 1 = 0

*57. y2 - 4x2 - 4y - 8x - 4 = 0

*58. 2x2 - y2 + 4x + 4y - 4 = 0

*59. 4x2 - y2 - 24x - 4y + 16 = 0

*60. 2y2 - x2 + 2x + 8y + 3 = 0

*61. y - 4x - 16x - 2y - 19 = 0

*62. x - 3y + 8x - 6y + 4 = 0

2

2

2

2

2

2

2

2

In Problems 63–66, graph each function. Be sure to label any intercepts. [Hint: Notice that each function is half a hyperbola.] *63. f 1x2 = 216 + 4x2

*64. f 1x2 = - 29 + 9x2

*65. f 1x2 = - 2- 25 + x2

*66. f 1x2 = 2- 1 + x2

Mixed Practice In Problems 67–74, analyze each conic. (x - 3)2 y2 = 1 *67. 4 25 *70. y2 = - 12(x + 1)

= 1 16 4 *71. 25x2 + 9y2 - 250x + 400 = 0

*73. x2 - 6x - 8y - 31 = 0

*74. 9x2 - y2 - 18x - 8y - 88 = 0

*68.

(y + 2)2

-

(x - 2)2

*69. x2 = 16(y - 3) *72. x2 + 36y2 - 2x + 288y + 541 = 0

Applications and Extensions 75. Fireworks Display Suppose that two people standing 2 miles apart both see the burst from a fireworks display. After a period of time the first person, standing at point A, hears the burst. One second later the second person, standing at point B, hears the burst. If the person at point B is due west of the person at point A, and if the display is known to occur due north of the person at point A, where EJE
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50,138 ft north of point A 76. Lightning Strikes Suppose that two people standing 1 mile apart both see a flash of lightning. After a period of time

the first person, standing at point A, hears the thunder. Two seconds later the second person, standing at point B, hears the thunder. If the person at point B is due west of the person at point A, and if the lightning strike is known to occur due north of the person standing at point A, where did UIF
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5236 ft north of point A 77. Nuclear Power Plant Some nuclear power plants utilize “natural draft” cooling towers in the shape of a hyperboloid, a solid obtained by rotating a hyperbola about its conjugate axis. Suppose that such a cooling tower has a base diameter

SECTION 9.4  The Hyperbola

of 400 feet and the diameter at its narrowest point, 360 feet above the ground, is 200 feet. If the diameter at the top of UIF
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592.4 ft Source: Bay Area Air Quality Management District 78. An Explosion Two recording devices are set 2400 feet apart, with the device at point A to the west of the device at point B. At a point between the devices, 300 feet from point B, a small amount of explosive is detonated. The recording devices record the time until the sound reaches each. How far directly north of point B should a second explosion be done so that the measured time difference recorded by the EFWJDFT
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particles follow the path of one branch of a hyperbola. y

723

same point. The rays are collected by the parabolic mirror, reflected toward the (common) focus, and thus are reflected by the hyperbolic mirror through the opening to its front focus, where the eyepiece is located. If the equation of the hyperbola y2 x2 is = 1 and the focal length (distance from the 9 16 vertex to the focus) of the parabola is 6, find the equation of the parabola. Source: www.enchantedlearning.com *81. The eccentricity e of a hyperbola is defined as the number c , where a is the distance of a vertex from the center, and a c is the distance of a focus from the center. Because c 7 a, it follows that e 7 1.
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coordinate axes. *84. Prove that the hyperbola y2

x2 = 1 a b2 has the two oblique asymptotes a a y = x and y = - x b b *85. Show that the graph of an equation of the form

45⬚

2

x

(a) Find an equation of the asymptotes under this scenario. (b) If the vertex of the path of the alpha particles is 10 cm from the center of the hyperbola, find a model that describes the path of the particle. *80. Hyperbolic Mirrors Hyperbolas have interesting reflective properties that make them useful for lenses and mirrors. For example, if a ray of light strikes a convex hyperbolic mirror on a line that would (theoretically) pass through its rear focus, it is reflected through the front focus. This property, and that of the parabola, were used to develop the Cassegrain telescope in 1672. The focus of the parabolic mirror and the rear focus of the hyperbolic mirror are the

-

Ax2 + Cy2 + F = 0

A ≠ 0, C ≠ 0, F ≠ 0

where A and C are opposite in sign, is a hyperbola with center at 10, 02. *86. Show that the graph of an equation of the form Ax2 + Cy2 + Dx + Ey + F = 0

A ≠ 0, C ≠ 0

where A and C are opposite in sign, (a) is a hyperbola if

E2 D2 + - F ≠ 0. 4A 4C

(b) is two intersecting lines if

D2 E2 + - F = 0. 4A 4C

Retain Your Knowledge Problems 87–90 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 3 89. Find the rectangular coordinates of the point with the polar *87. For y = cos (6x + 3p), find the amplitude, period, and 2 p phase shift. Then graph the function, showing at least two coordinates a12, - b. (6, - 623) 3 periods. *90. Transform the polar equation r = 6 sin u to an equation 88. Solve the triangle described: a = 7, b = 10, and C = 100°. in rectangular coordinates. Then identify and graph the c ≈ 13.16, A ≈ 31.6°, B ≈ 48.4°, equation.

‘Are You Prepared?’ Answers 1. 522

2.

25 4

3. 10, - 32, 10, 32

4. True

5. right; 5; down; 4

6. Vertical: x = - 2, x = 2; horizontal: y = 1

724

CHAPTER 9  Analytic Geometry

9.5 Rotation of Axes; General Form of a Conic PREPARING FOR THIS SECTION Before getting started, review the following: r
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OBJECTIVES 1 2 3 4

Identify a Conic (p. 724) Use a Rotation of Axes to Transform Equations (p. 725) Analyze an Equation Using a Rotation of Axes (p. 727) Identify Conics without a Rotation of Axes (p. 729)

In this section, we show that the graph of a general second-degree polynomial containing two variables x and y—that is, an equation of the form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

(1)

where A, B, and C are not simultaneously 0—is a conic.We shall not concern ourselves here with the degenerate cases of equation (1), such as x2 + y2 = 0, whose graph is a single point 10, 02; or x2 + 3y2 + 3 = 0, whose graph contains no points; or x2 - 4y2 = 0, whose graph is two lines, x - 2y = 0 and x + 2y = 0. We begin with the case where B = 0. In this case, the term containing xy is not present, so equation (1) has the form Ax2 + Cy2 + Dx + Ey + F = 0 where either A ≠ 0 or C ≠ 0.

1 Identify a Conic We have already discussed the procedure for identifying the graph of this kind of equation; complete the squares of the quadratic expressions in x or y, or both. Once this has been done, the conic can be identified by comparing it to one of the forms studied in Sections 9.2 through 9.4. In fact, though, the conic can be identified directly from the equation without completing the squares.

THEOREM

Identifying Conics without Completing the Squares Excluding degenerate cases, the equation Ax2 + Cy2 + Dx + Ey + F = 0

(2)

where A and C cannot both equal zero: B
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Proof (a) If AC = 0, then either A = 0 or C = 0, but not both, so the form of equation (2) is either Ax2 + Dx + Ey + F = 0,

A ≠ 0

Cy2 + Dx + Ey + F = 0,

C ≠ 0

or

SECTION 9.5  Rotation of Axes; General Form of a Conic

725

Using the results of Problems 78 and 79 in Exercise 9.2, it follows that, except for the degenerate cases, the equation is a parabola. (b) If AC 7 0, then A and C are of the same sign. Using the results of Problem 87 in Exercise 9.3, except for the degenerate cases, the equation is an ellipse. (c) If AC 6 0, then A and C are of opposite signs. Using the results of Problem 86 in Exercise 9.4, except for the degenerate cases, the equation is a hyperbola. ■ We will not be concerned with the degenerate cases of equation (2). However, in practice, you should be alert to the possibility of degeneracy.

EXAM PL E 1

Identifying a Conic without Completing the Squares Identify the graph of each equation without completing the squares. (a) 3x2 + 6y2 + 6x - 12y = 0 (c) y2 - 2x + 4 = 0

Solution

(b) 2x2 - 3y2 + 6y + 4 = 0

(a) Compare the given equation to equation (2) and conclude that A = 3 and C = 6. Since AC = 18 7 0, the equation defines an ellipse. (b) Here A = 2 and C = - 3, so AC = - 6 6 0. The equation defines a hyperbola. (c) Here A = 0 and C = 1, so AC = 0. The equation defines a parabola.

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PROBLEM

11

Although we can now identify the type of conic represented by any equation of the form of equation (2) without completing the squares, we still need to complete the squares if we desire additional information about the conic, such as its graph.

2 Use a Rotation of Axes to Transform Equations

Figure 47 y′

y θ

x′ θ x

O

Now let’s turn our attention to equations of the form of equation (1), where B ≠ 0. To discuss this case, we introduce a new procedure: rotation of axes. In a rotation of axes, the origin remains fixed while the x-axis and y-axis are rotated through an angle u to a new position; the new positions of the x-axis and the y-axis are denoted by x′ and y′, respectively, as shown in Figure 47(a). Now look at Figure 47(b). There the point P has the coordinates 1x, y2 relative to the xy-plane, while the same point P has coordinates 1x′, y′2 relative to the x′y′@plane. We seek relationships that will enable us to express x and y in terms of x′, y′, and u. As Figure 47(b) shows, r denotes the distance from the origin O to the point P, and a denotes the angle between the positive x′@axis and the ray from O through P. Then, using the definitions of sine and cosine, we have x′ = r cos a y′ = r sin a x = r cos 1u + a2 y = r sin 1u + a2

(a)

Now y′

x = r cos 1u + a2

y r α O

x

P ⫽ (x, y) ⫽ (x ′, y ′) y′ x′ x′ θ

y

= r 1cos u cos a - sin u sin a2

Apply the Sum Formula for cosine.

= x′ cos u - y′ sin u

By equation (3)

= 1r cos a2 1cos u2 - 1r sin a2 1sin u2

x

Similarly, (b)

y = r sin 1u + a2

= r 1sin u cos a + cos u sin a2

Apply the Sum Formula for sine.

= x′ sin u + y′ cos u

By equation (3)

(3) (4)

726

CHAPTER 9  Analytic Geometry

THEOREM

Rotation Formulas If the x- and y-axes are rotated through an angle u, the coordinates 1x, y2 of a point P relative to the xy-plane and the coordinates 1x′, y′2 of the same point relative to the new x′@ and y′@axes are related by the formulas x = x′ cos u - y′ sin u

y = x′ sin u + y′ cos u

(5)

Rotating Axes

EX A MPL E 2

Express the equation xy = 1 in terms of new x′y′@coordinates by rotating the axes through a 45°
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Solution

x′

y

( 2 , 0)

1

45° ⫺1

(⫺ 2 , 0)

22 22 22 - y′ = 1x′ - y′2 2 2 2

y = x′ sin 45° + y′ cos 45° = x′

22 22 22 + y′ = 1x′ + y′2 2 2 2

c

2

⫺2

x = x′ cos 45° - y′ sin 45° = x′

Substituting these expressions for x and y in xy = 1 gives

Figure 48 y′

Let u = 45° in equation (5). Then

1 ⫺1 ⫺2

2

x

22 22 1x′ - y′2 d c 1x′ + y′2 d = 1 2 2 1 1x′2 - y′2 2 = 1 2 y′2 x′2 = 1 2 2

This is the equation of a hyperbola with center at 10, 02 and transverse axis along the x′@axis. The vertices are at 1 { 22, 02 on the x′@axis; the asymptotes are y′ = x′ and y′ = - x′ (which correspond to the original x- and y-axes). See Figure 48 for the graph.

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As Example 2 illustrates, a rotation of axes through an appropriate angle can transform a second-degree equation in x and y containing an xy-term into one in x′ and y′ in which no x′ y′@term appears. In fact, we will show that a rotation of axes through an appropriate angle transforms any equation of the form of equation (1) into an equation in x′ and y′ without an x′ y′@term. To find the formula for choosing an appropriate angle u through which to rotate the axes, begin with equation (1), Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

B ≠ 0

Next rotate through an angle u using the rotation formulas (5).

A 1x′ cos u - y′ sin u2 + B 1x′ cos u - y′ sin u2 1x′ sin u + y′ cos u2 2 + C 1x′ sin u + y′ cos u2 + D1x′ cos u - y′ sin u2 + E 1x′ sin u + y′ cos u2 + F = 0 2

Expanding and collecting like terms gives 1A cos2 u + B sin u cos u + C sin2 u2x′2 + + + +

3 B 1cos2 u - sin2 u2 + 21C - A2 1sin u cos u2 4 x′ y′ 1A sin2 u - B sin u cos u + C cos2 u2y′2 1D cos u + E sin u2x′ (6) 1 - D sin u + E cos u2y′ + F = 0

In equation (6), the coefficient of x′ y′ is

B 1cos2 u - sin2 u2 + 21C - A2 1sin u cos u2

SECTION 9.5  Rotation of Axes; General Form of a Conic

727

Because we want to eliminate the x′ y′@term, we select an angle u so that this coefficient is 0. B 1cos2 u - sin2 u2 + 21C - A2 1sin u cos u2 = 0 Double-angle B cos 12u2 + 1C - A2 sin 12u2 = 0 B cos 12u2 = 1A - C2 sin 12u2 Formulas A - C cot12u2 = B ≠ 0 B

THEOREM

To transform the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

B ≠ 0

into an equation in x′ and y′ without an x′ y′@term, rotate the axes through an angle u that satisfies the equation cot12u2 =

A - C B

(7)

WARNING Be careful if you use a calculator to solve equation (7). 1. If cot 12u2 = 0, then 2u = 90° and u = 45°.

2. If cot 12u2 ≠ 0, first find cos 12u2. Then use the inverse cosine function key(s) to obtain 2u, 0° 6 2u 6 180°. Finally, divide by 2 to obtain the j correct acute angle u.

Equation (7) has an infinite number of solutions for u. We shall adopt the convention of choosing the acute angle u that satisfies (7). There are two possibilities: If cot12u2 Ú 0, then 0° 6 2u … 90°, so 0° 6 u … 45°. If cot12u2 6 0, then 90° 6 2u 6 180°, so 45° 6 u 6 90°. Each of these results in a counterclockwise rotation of the axes through an acute angle u.*

3 Analyze an Equation Using a Rotation of Axes For the remainder of this section, the direction “Analyze the equation” will mean to transform the given equation so that it contains no xy-term and to graph the equation.

EXAM PL E 3

Analyzing an Equation Using a Rotation of Axes Analyze the equation: x2 + 23xy + 2y2 - 10 = 0

Solution

Because an xy-term is present, the axes must be rotated. Using A = 1, B = 23, and C = 2 in equation (7), the appropriate acute angle u through which to rotate the axes satisfies the equation cot12u2 =

A - C -1 23 = = B 3 23

0° 6 2u 6 180°

23 , this means 2u = 120°, so u = 60°. Using u = 60° in the 3 rotation formulas (5) yields Because cot12u2 = -

x = x′cos 60° - y′sin 60° =

1 23 1 x′ y′ = 1 x′ - 23y′ 2 2 2 2

y = x′sin 60° + y′cos 60° =

23 1 1 x′ + y′ = 1 23x′ + y′ 2 2 2 2

A - C eliminates the x′ y′@term. However, the B final form of the transformed equation may be different (but equivalent), depending on the angle chosen. *Any rotation through an angle u that satisfies cot 12u2 =

728

CHAPTER 9  Analytic Geometry

Substituting these values into the original equation and simplifying give 1 1 x′ - 23y′ 2 2 4

x2 + 23xy + 2y2 - 10 = 0 1 1 1 + 23 c 1 x′ - 23y′ 2 d c 1 23x′ + y′ 2 d + 2c 1 23x′ + y′ 2 2 d = 10 2 2 4

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Figure 49 y y′

(2, 0)

(0, 2 5 )

60° (⫺2, 0)

x (0, ⫺2 5 )

This is the equation of an ellipse with center at 10, 02 and major axis along the y′@axis. The vertices are at 10, {2252 on the y′@axis. See Figure 49 for the graph.

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PROBLEM

31

In Example 3, the acute angle u through which to rotate the axes was easy to find because of the numbers used in the given equation. In general, the equation A - C cot12u2 = will not have such a “nice” solution. As the next example shows, B we can still find the appropriate rotation formulas without using a calculator approximation by applying Half-angle Formulas.

EX A MPL E 4

Analyzing an Equation Using a Rotation of Axes Analyze the equation: 4x2 - 4xy + y2 + 525x + 5 = 0

Solution

Letting A = 4, B = - 4, and C = 1 in equation (7), the appropriate angle u through which to rotate the axes satisfies A - C 3 3 = = B -4 4 To use the rotation formulas (5), we need to know the values of sin u and cos u. Since we seek an acute angle u, we know that sin u 7 0 and cos u 7 0. Use the Half-angle Formulas in the form cot12u2 =

sin u =

1 - cos 12u2 B 2

cos u =

1 + cos 12u2 B 2

3 Now we need to find the value of cos 12u2 . Since cot12u2 = - , then 4 3 90° 6 2u 6 180°
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cos 12u2 = - . Then 5 1 - cos 12u2 sin u = = B 2 R

3 1 - a- b 5 4 2 225 = = = 2 B5 5 25

1 + cos 12u2 cos u = = B 2 R

3 1 + a- b 5 1 1 25 = = = 2 B5 5 25

With these values, the rotation formulas (5) are x =

25 225 25 x′ y′ = 1x′ - 2y′2 5 5 5

y =

225 25 25 x′ + y′ = 12x′ + y′2 5 5 5

SECTION 9.5  Rotation of Axes; General Form of a Conic

729

Substituting these values in the original equation and simplifying give 4x2 - 4xy + y2 + 525x + 5 = 0 4c

2 25 25 25 1x′ - 2y′2 d - 4c 1x′ - 2y′2 d c 12x′ + y′2 d 5 5 5

+ c

2 25 25 12x′ + y′2 d + 525 c 1x′ - 2y′2 d = - 5 5 5

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x′

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63.4° x

41x′2 - 4x′ y′ + 4y′2 2 - 412x′2 - 3x′ y′ - 2y′2 2 + 4x′2 + 4x′y′ + y′2 + 251x′ - 2y′2 25y′2 - 50y′ + 25x′ y′2 - 2y′ + x′ y′2 - 2y′ + 1 2 1y′ - 12

= = = = =

- 25 - 25 -1 - x′ - x′

Combine like terms. Divide by 25. Complete the square in y′.

This is the equation of a parabola with vertex at 10, 12 in the x′y′@plane. The axis of 225 , and find symmetry is parallel to the x′@axis. Use a calculator to solve sin u = 5 that u ≈ 63.4°. See Figure 50 for the graph.

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PROBLEM

37

4 Identify Conics without a Rotation of Axes Suppose that we are required only to identify (rather than analyze) the graph of an equation of the form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

B ≠ 0

(8)

Applying the rotation formulas (5) to this equation gives an equation of the form A′ x′2 + B′ x′ y′ + C′ y′2 + D′ x′ + E′ y′ + F′ = 0

(9)

where A′, B′, C′, D′, E′, and F′ can be expressed in terms of A, B, C, D, E, F and the angle u of rotation (see Problem 53). It can be shown that the value of B2 - 4AC in equation (8) and the value of B′2 - 4A′ C′ in equation (9) are equal no matter what angle u of rotation is chosen (see Problem 55). In particular, if the angle u of rotation satisfies equation (7), then B′ = 0 in equation (9), and B2 - 4AC = - 4A′ C′. Because equation (9) then has the form of equation (2), A′ x′2 + C′ y′2 + D′ x′ + E′ y′ + F′ = 0 we can identify its graph without completing the squares, as we did in the beginning of this section. In fact, now we can identify the conic described by any equation of the form of equation (8) without a rotation of axes.

THEOREM

Identifying Conics without a Rotation of Axes Except for degenerate cases, the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 B
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730

CHAPTER 9  Analytic Geometry

Identifying a Conic without a Rotation of Axes

EX A MPL E 5

Identify the graph of the equation: 8x2 - 12xy + 17y2 - 425x - 225y - 15 = 0

Solution

Here A = 8, B = - 12, and C = 17, so B2 - 4AC = - 400. Since B2 - 4AC 6 0, the equation defines an ellipse.

Now Work

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PROBLEM

43

9.5 Assess Your Understanding ‘Are You Prepared?’

Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.

1. The sum formula for the sine function is sin1A + B2 = sin A cos B + cos A sin B. (p. 534) 2. 5IF
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sin12u2 = 2 sin u cos u. (p. 544)

3. If u is acute, the Half-angle Formula for the sine function is 1- cos u u . (p. 547) sin = A 2 2 4. If u is acute, the Half-angle Formula for the cosine function 1 + cos u u is cos = A . (p. 547) 2 2

Concepts and Vocabulary *5. To transform the equation Ax + Bxy + Cy + Dx + Ey + F = 0, 2

2

B ≠ 0

into one in x′ and y′ without an x′y′@term, rotate the axes through an acute angle u that satisfies the equation. 6. Except for degenerate cases, the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 defines a(n) parabola if B2 - 4AC = 0.

8. True or False The equation ax2 + 6y2 - 12y = 0 defines an ellipse if a 7 0. True 9. True or False The equation 3x2 + Bxy + 12y2 = 10 defines a parabola if B = - 12. True 10. True or False To eliminate the xy-term from the equation x2 - 2xy + y2 - 2x + 3y + 5 = 0, rotate the axes through an angle u, where cot u = B2 - 4AC. False

7. Except for degenerate cases, the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 defines an ellipse if B2 - 4AC 6 0.

Skill Building In Problems 11–20, identify the graph of each equation without completing the squares. 11. x2 + 4x + y + 3 = 0

Parabola

13. 6x + 3y - 12x + 6y = 0 2

2

15. 3x - 2y + 6x + 4 = 0 2

2

17. 2y2 - x2 - y + x = 0

Parabola

Ellipse

14. 2x2 + y2 - 8x + 4y + 2 = 0

Hyperbola

16. 4x - 3y - 8x + 6y + 1 = 0

Hyperbola

19. x2 + y2 - 8x + 4y = 0

12. 2y2 - 3y + 3x = 0

Circle

2

2

18. y2 - 8x2 - 2x - y = 0

Ellipse Hyperbola

Hyperbola

20. 2x2 + 2y2 - 8x + 8y = 0

Circle

In Problems 21–30, determine the appropriate rotation formulas to use so that the new equation contains no xy-term. *21. x2 + 4xy + y2 - 3 = 0

*22. x2 - 4xy + y2 - 3 = 0

*23. 5x2 + 6xy + 5y2 - 8 = 0

*24. 3x2 - 10xy + 3y2 - 32 = 0

*25. 13x - 623xy + 7y - 16 = 0

*26. 11x2 + 1023xy + y2 - 4 = 0

*27. 4x2 - 4xy + y2 - 825x - 1625y = 0

*28. x2 + 4xy + 4y2 + 525y + 5 = 0

*29. 25x - 36xy + 40y - 12213x - 8213y = 0

*30. 34x2 - 24xy + 41y2 - 25 = 0

2

2

2

2

In Problems 31–42, rotate the axes so that the new equation contains no xy-term. Analyze and graph the new equation. Refer to Problems 21–30 for Problems 31–40. *31. x2 + 4xy + y2 - 3 = 0

*32. x2 - 4xy + y2 - 3 = 0

*33. 5x2 + 6xy + 5y2 - 8 = 0

*34. 3x2 - 10xy + 3y2 - 32 = 0

*35. 13x - 623xy + 7y - 16 = 0

*36. 11x2 + 1023xy + y2 - 4 = 0

*37. 4x2 - 4xy + y2 - 825x - 1625y = 0

*38. x2 + 4xy + 4y2 + 525y + 5 = 0

*39. 25x - 36xy + 40y - 12213x - 8213y = 0

*40. 34x2 - 24xy + 41y2 - 25 = 0

*41. 16x2 + 24xy + 9y2 - 130x + 90y = 0

*42. 16x2 + 24xy + 9y2 - 60x + 80y = 0

2

2

2

2

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SECTION 9.6  Polar Equations of Conics

731

In Problems 43–52, identify the graph of each equation without applying a rotation of axes. 43. x2 + 3xy - 2y2 + 3x + 2y + 5 = 0 45. x2 - 7xy + 3y2 - y - 10 = 0

Hyperbola

47. 9x2 + 12xy + 4y2 - x - y = 0

Parabola

49. 10x2 - 12xy + 4y2 - x - y - 10 = 0 51. 3x2 - 2xy + y2 + 4x + 2y - 1 = 0

44. 2x2 - 3xy + 4y2 + 2x + 3y - 5 = 0

Hyperbola

46. 2x2 - 3xy + 2y2 - 4x - 2 = 0

48. 10x2 + 12xy + 4y2 - x - y + 10 = 0 50. 4x2 + 12xy + 9y2 - x - y = 0

Ellipse

Ellipse

Parabola

52. 3x2 + 2xy + y2 + 4x - 2y + 10 = 0

Ellipse

Ellipse

Ellipse

Ellipse

Applications and Extensions In Problems 53–56, apply the rotation formulas (5) to

*56. Prove that, except for degenerate cases, the equation

Ax + Bxy + Cy + Dx + Ey + F = 0 2

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

2

B
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B
QBSBCPMB
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B2 - 4AC = 0. C
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B
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B2 - 4AC 6 0. D
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B
IZQFSCPMB
JG
B2 - 4AC 7 0.

to obtain the equation A′ x′2 + B′ x′ y′ + C′ y′2 + D′ x′ + E′ y′ + F′ = 0 *53. Express A′, B′, C′, D′, E′, and F′ in terms of A, B, C, D, E, F, and the angle u of rotation. [Hint:
3FGFS
UP
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 > *54. Show that A + C = A′ + C′, and thus show that A + C is invariant; that is, its value does not change under a rotation of axes. *55. 3FGFS
UP
1SPCMFN

4IPX
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B2 - 4AC is invariant.

*57. Use the rotation formulas (5) to show that distance is invariant under a rotation of axes. That is, show that the distance from P1 = 1x1 , y1 2 to P2 = 1x2 , y2 2 in the xy-plane equals the distance from P1 = 1x1= , y1= 2 to P2 = 1x2= , y2= 2 in the x′ y′@plane. *58. Show that the graph of the equation x1>2 + y1>2 = a1>2 is part of the graph of a parabola.

Discussion and Writing 59. Formulate a strategy for discussing and graphing an equation of the form

60. Explain how your strategy presented in Problem 59 changes if the equation is of the following form:

Ax2 + Cy2 + Dx + Ey + F = 0

Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

Retain Your Knowledge Problems 61–64 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 61. Solve the triangle described: a = 7, b = 9, and c = 11. 62. Find the area of the triangle described: a = 14, b = 11, and C = 30°. 38.5 61. A ≈ 39.4°, B ≈ 54.7°, C ≈ 85.9°

63. Transform the equation xy = 1 from rectangular coordinates to polar coordinates. r 2 cos u sin u = 1 64. Write the complex number 2 - 5i in polar form. 229( cos 291.8° + i sin 291.8°)

‘Are You Prepared?’ Answers 1. sin A cos B + cos A sin B

2. 2 sin u cos u

3.

1 - cos u B 2

4.

1 + cos u B 2

9.6 Polar Equations of Conics PREPARING FOR THIS SECTION Before getting started, review the following: r
1PMBS
$PPSEJOBUFT
4FDUJPO

QQ
m

Now Work the ‘Are You Prepared?’ problems on page 736.

OBJECTIVES 1 Analyze and Graph Polar Equations of Conics (p. 731) 2 Convert the Polar Equation of a Conic to a Rectangular Equation (p. 735)

1 Analyze and Graph Polar Equations of Conics Sections 9.2 through 9.4 gave separate definitions for the parabola, ellipse, and hyperbola based on geometric properties and the distance formula. This section presents an alternative definition that simultaneously defines all these conics. As

732

CHAPTER 9  Analytic Geometry

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3FGFS to Section 8.1.)

DEFINITION

Let D denote a fixed line called the directrix; let F denote a fixed point called the focus, which is not on D; and let e be a fixed positive number called the eccentricity. A conic is the set of points P in the plane such that the ratio of the distance from F to P to the distance from D to P equals e. That is, a conic is the collection of points P for which d 1F, P2 = e d 1D, P2

(1)

If e = 1, the conic is a parabola. If e 6 1, the conic is an ellipse. If e 7 1, the conic is a hyperbola. Observe that if e = 1, the definition of a parabola in equation (1) is exactly the same as the definition used earlier in Section 9.2. In the case of an ellipse, the major axis is a line through the focus perpendicular to the directrix. In the case of a hyperbola, the transverse axis is a line through the focus perpendicular to the directrix. For both an ellipse and a hyperbola, the eccentricity e satisfies

e =

Figure 51 Directrix D

P ⫽ (r, θ) d(D, P ) r

p Pole O (Focus F )

θ Q

Polar axis

c a

(2)

where c is the distance from the center to the focus and a is the distance from the center to a vertex. Just as we did earlier using rectangular coordinates, we derive equations for the conics in polar coordinates by choosing a convenient position for the focus F and the directrix D. The focus F is positioned at the pole, and the directrix D is either parallel or perpendicular to the polar axis. Suppose that we start with the directrix D perpendicular to the polar axis at a distance p units to the left of the pole (the focus F). See Figure 51. If P = 1r, u2 is any point on the conic, then, by equation (1), d 1F, P2 = e or d 1F, P2 = e # d 1D, P2 d 1D, P2

(3)

Now use the point Q obtained by dropping the perpendicular from P to the polar axis to calculate d 1D, P2. d 1D, P2 = p + d 1O, Q2 = p + r cos u Using this expression and the fact that d 1F, P2 = d 1O, P2 = r in equation (3) gives e # d 1D, P2 e1p + r cos u2 ep + er cos u ep ep ep r = 1 - e cos u

d 1F, P2 r r r - er cos u r 11 - e cos u2

= = = = =

SECTION 9.6  Polar Equations of Conics

THEOREM

733

Polar Equation of a Conic; Focus at the Pole; Directrix Perpendicular to the Polar Axis a Distance p to the Left of the Pole The polar equation of a conic with focus at the pole and directrix perpendicular to the polar axis at a distance p to the left of the pole is r =

ep 1 - e cos u

(4)

where e is the eccentricity of the conic.

EXAM PL E 1

Analyzing and Graphing the Polar Equation of a Conic 4 2 - cos u The given equation is not quite in the form of equation (4), since the first term in the EFOPNJOBUPS
JT

JOTUFBE
PG

%JWJEF
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OVNFSBUPS
BOE
EFOPNJOBUPS
CZ

UP
PCUBJO ep 2 r = r = 1 1 - e cos u 1 - cos u 2 This equation is in the form of equation (4), with Analyze and graph the equation: r =

Solution

e =

1 2

and

ep = 2

Then

Figure 52 Directrix

4 3 3

( , π) 4– 3

F

4– 3

( , 0)

(4, 0)

Polar axis

1 p = 2, so p = 4 2 1 Hence the conic is an ellipse, because e = 6 1. One focus is at the pole, and 2 the directrix is perpendicular to the polar axis, a distance of p = 4 units to the left of the pole. It follows that the major axis is along the polar axis. To find the vertices, let 4 u = 0 and u = p. The vertices of the ellipse are 14, 02 and a , pb . 3 4 The midpoint of the vertices, a , 0b in polar coordinates, is the center of the ellipse. 3 4 2

xS1

x S -1 x S -3

xS4

xSp

x S -1

xS2

xS0

1 x

xS2

*35. lim f 1x2, f 1x2 = b

x2 if x Ú 0 2x if x 6 0

*36. lim f 1x2, f 1x2 = b

x - 1 if x 6 0 3x - 1 if x Ú 0

*37. lim f 1x2, f 1x2 = b

3x if x … 1 x + 1 if x 7 1

*38. lim f 1x2, f 1x2 = b

x2 if x … 2 2x - 1 if x 7 2

*40. lim f 1x2, f 1x2 = b

1 if x 6 0 - 1 if x 7 0

*42. lim f 1x2, f 1x2 = b

ex if x 7 0 1 - x if x … 0

xS0

xS1

xS0

xS2

x if x 6 0 *39. lim f 1x2, f 1x2 = c 1 if x = 0 xS0 3x if x 7 0 *41. lim f 1x2, f 1x2 = b xS0

xS0

sin x if x … 0 x2 if x 7 0

xS0

1 x2

In Problems 43–48, use a graphing utility to find the indicated limit rounded to two decimal places. x3 - x2 + x - 1 x S 1 x4 - x3 + 2x - 2

0.67

44. lim

x3 - x2 + 3x - 3 xS1 x2 + 3x - 4

0.8

47. lim

43. lim

46. lim

x3 + x2 + 3x + 3 x S -1 x4 + x3 + 2x + 2

4

45. lim

x3 - 2x2 + 4x - 8 xS2 x2 + x - 6

x3 + 2x2 + x x S -1 x4 + x3 + 2x + 2

0

48. lim

x3 - 3x2 + 4x - 12 x S 3 x4 - 3x3 + x - 3

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

1.6 0.46

SECTION 13.2  Algebra Techniques for Finding Limits

929

Retain Your Knowledge Problems 49–52 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 49. Let A12, - 32 and B 16, - 112
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distance between the points and the midpoint of the line segment connecting the points. d = 42; M = (4, - 7)

*50. 'JOE
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FMMJQTF
1x - 22 2 1y + 12 2 + = 1. 9 13

‘Are You Prepared?’ Answers

51. Logan invests $4000 at an annual interest rate of 6%. How much money will she have after 10 years if interest is compounded continuously? $7288.48 52. Assuming r 7 0 and 0 … u 6 2p, find the polar coordinates of the point whose rectangular coordinates are 1 - 2, 2232. 2p a4, b 3

2. f 102 = 0

1. 4FF
'JHVSF

PO
QBHF


13.2 Algebra Techniques for Finding Limits OBJECTIVES 1 2 3 4 5

Find the Limit of a Sum, a Difference, and a Product (p. 930) Find the Limit of a Polynomial (p. 931) Find the Limit of a Power or a Root (p. 932) Find the Limit of a Quotient (p. 933) Find the Limit of an Average Rate of Change (p. 934)

We mentioned in the previous section that algebra can sometimes be used to find the exact value of a limit. This is accomplished by developing two formulas involving limits and several properties of limits.

THEOREM

Two Formulas: lim b and lim x xSc

xSc

Limit of a Constant

In Words

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f1x2 = b,

The limit of a constant is the constant.

lim f1x2 = lim b = b

(1)

xSc

xSc

where c is any number.

Limit of x 'PS
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f1x2 = x,

In Words The limit of x as x approaches c is c.

lim f1x2 = lim x = c

xSc

(2)

xSc

where c is any number. We use graphs to establish formulas (1) and (2). Since the graph of a constant function is a horizontal line, it follows that, no matter how close x is to c, the corresponding value of f1x2 equals b. That is, lim b = b.
4FF
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 xSc 4FF
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c, as x gets closer to c, the corresponding value of f1x2 is x, which is just as close to c. That is, lim x = c. xSc

Figure 4

(0, b)

y

Figure 5

y f(x ) ⫽ x

f(x)  b c

x

c (c, c)

c

x

930

CHAPTER 13  A Preview of Calculus: The Limit, Derivative, and Integral of a Function

EX A MPL E 1

Using Formulas (1) and (2) (a) lim  = 

(b) lim x = 3

xS3

(c) lim ( - 8) = - 8

xS3

Now Work

PROBLEM

(c)

xS0

lim x = -

x S -1>2

7

1 2

r

'PSNVMBT

BOE



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evaluate limits of more complicated functions.

1 Find the Limit of a Sum, a Difference, and a Product In the following properties, we assume that f and g are two functions for which both lim f1x2 and lim g1x2 exist. xSc

THEOREM

xSc

Limit of a Sum lim 3 f 1x2 + g1x2 4 = lim f1x2 + lim g1x2

xSc

In Words

xSc

(3)

xSc

The limit of the sum of two functions equals the sum of their limits.

EX A MPL E 2

Finding the Limit of a Sum 'JOE


Solution

lim 1x + 42

x S -3

The limit we seek is the sum of two functions: f1x2 = x and g1x2 = 4.
'SPN
GPSNVlas (1) and (2), we know that lim f1x2 = lim x = - 3 and x S -3

x S -3

lim g1x2 = lim 4 = 4

x S -3

x S -3

'SPN
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GPMMPXT
UIBU

lim 1x + 42 = lim x + lim 4 = - 3 + 4 = 1 x S -3

x S -3

THEOREM

r

x S -3

Limit of a Difference lim 3 f 1x2 - g1x2 4 = lim f1x2 - lim g1x2 xSc

xSc

In Words

xSc

(4)

The limit of the difference of two functions equals the difference of their limits.

EX A MPL E 3

Finding the Limit of a Difference 'JOE
lim 16 - x2 xS4

Solution

The limit we seek is the difference of two functions: f 1x2 = 6 and g1x2 = x.
'SPN
formulas (1) and (2), we know that lim f1x2 = lim 6 = 6 and

xS4

xS4

lim g1x2 = lim x = 4

xS4

xS4

'SPN
GPSNVMB



JU
GPMMPXT
UIBU

lim 16 - x2 = lim 6 - lim x = 6 - 4 = 2

xS4

THEOREM

xS4

Limit of a Product lim 3 f 1x2 # g1x2 4 = 3 lim f1x2 4 3 lim g1x2 4

xSc

In Words The limit of the product of two functions equals the product of their limits.

EX A MPL E 4

r

xS4

Finding the Limit of a Product 'JOE


lim 1 - 4x2

x S -

xSc

xSc

(5)

SECTION 13.2  Algebra Techniques for Finding Limits

Solution

931

The limit we seek is the product of two functions: f1x2 = - 4 and g1x2 = x.
'SPN
formulas (1) and (2), we know that lim f1x2 = lim 1 - 42 = - 4 and

x S -

lim g1x2 = lim x = - 

x S -

x S -

x S -

'SPN
GPSNVMB



JU
GPMMPXT
UIBU

lim 1 - 4x2 = 3 lim 1 - 42 4 3 lim x4 = 1 - 42 1 - 2 = 20

x S -

EXAM PL E 5

x S -

Finding Limits Using Algebraic Properties 'JOE
B
lim 13x - 2

(b) lim 1x2 2

x S -2

Solution

r

x S -

xS2

(a) lim 13x - 2 = lim 13x2 - lim  = 3 lim 34 3 lim x4 - lim  x S -2

x S -2

x S -2

x S -2

x S -2

x S -2

= 132 1 - 22 -  = - 6 -  = - 11

(b) lim 1x2 2 = 3 lim 4 3 lim x2 4 =  lim 1x # x2 = 3 lim x4 3 lim x4 xS2

xS2

Now Work

xS2

xS2

xS2

xS2

=  # 2 # 2 = 20

r

15

PROBLEM

Notice in the solution to part (b) that lim 1x2 2 =  # 22. xS2

THEOREM

Limit of a Monomial If n Ú 1 is a positive integer and a is a constant, then lim 1axn 2 = ac n

(6)

xSc

for any number c.

Proof

lim (axn) = [lim a][lim xn] = a[lim (x · x · x · . . . · x)]

x

c

x

x

c

c

x

c

n factors

= a[lim x][lim x][lim x] . . . [lim x] x

c

x

c

x c n factors

x

c

= a · c · c · c · . . . · c = acn n factors

EXAM PL E 6

■

Finding the Limit of a Monomial 'JOE
lim 1 - 4x3 2 xS2

Solution

lim 1 - 4x3 2 = - 4 # 23 = - 4 # 8 = - 32

r

xS2

2 Find the Limit of a Polynomial Since a polynomial is a sum of monomials, we can use formula (6) and the repeated use of formula (3) to obtain the following result:

THEOREM

Limit of a Polynomial If P is a polynomial function, then lim P1x2 = P1c2

xSc

for any number c.

(7)

932

CHAPTER 13  A Preview of Calculus: The Limit, Derivative, and Integral of a Function

Proof If P is a polynomial function—that is, if P1x2 = an xn + an - 1 xn - 1 + g + a1 x + a0 then

In Words

lim P1x2 = lim 3 an xn + an - 1 xn - 1 + g + a1 x + a0 4 S

To find the limit of a polynomial as x approaches c, all we need to do is evaluate the polynomial at c.

xSc

x

c

x

c

= lim 1an xn 2 + lim 1an - 1 xn - 1 2 + g + lim 1a1 x2 + lim a0 S S S S x

= an c + an - 1 c n

c

n-1

x

c

x

c

+ g + a1 c + a0

= P1c2

EX A MPL E 7

■

Finding the Limit of a Polynomial 'JOE
lim 3 x4 - 6x3 + 3x2 + 4x - 24 xS2

Solution

lim 3 x4 - 6x3 + 3x2 + 4x - 24 =  # 24 - 6 # 23 + 3 # 22 + 4 # 2 - 2

xS2

=  # 16 - 6 # 8 + 3 # 4 + 8 - 2

r

= 80 - 48 + 12 + 6 = 0

Now Work

PROBLEM

17

3 Find the Limit of a Power or a Root THEOREM

Limit of a Power or Root If lim f1x2 exists and if n Ú 2 is a positive integer, then xSc

lim 3 f 1x2 4 n = 3 lim f1x2 4 n

(8)

xSc

xSc

and n

n lim f1x2 lim 2f1x2 = 5 xSc

(9)

xSc

n

n lim f1x2 In formula (9), we require that both 2f1x2 and 5 be defined. xSc

Look carefully at equations (8) and (9) and compare each side.

EX A MPL E 8

Finding the Limit of a Power or a Root 'JOE
B
lim 13x - 2 4 xS1

Solution

(b) lim 2x2 + 8 xS0

(c) lim 1x3 - x + 32

4>3

x S -1

(a) lim 13x - 2 4 = 3 lim 13x - 2 4 4 = 1 - 22 4 = 16 xS1

xS1

(x2 + 8) = 28 = 222 (b) lim 2x2 + 8 = 5 lim xS0 xS0 (c)

lim 1x3 - x + 32

x S -1

4>3

3 lim (x3 - x + 3)4 = 5 x S -1 3 3 3 [ lim (x3 - x + 3)]4 = 2 = 5 1 - 12 4 = 2 1 = 1 x S -1

Now Work

PROBLEM

27

r

SECTION 13.2  Algebra Techniques for Finding Limits

933

4 Find the Limit of a Quotient THEOREM

Limit of a Quotient lim c

In Words The limit of the quotient of two functions equals the quotient of their limits, provided that the limit of the denominator is not zero.

EXAM PL E 9

xSc

(10)

xSc

provided that lim g1x2 ≠ 0. xSc

Finding the Limit of a Quotient 'JOE
lim

xS1

Solution

lim f1x2 f 1x2 xSc d = g1x2 lim g1x2

x3 - x + 2 3x + 4

The limit we seek involves the quotient of two functions: f1x2 = x3 - x + 2 and g1x2 = 3x + 4.
'JSTU
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UIF
MJNJU
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g1x2. lim g1x2 = lim 13x + 42 = 7 xS1

xS1

Since the limit of the denominator is not zero, we can proceed to use formula (10). x3 - x + 2 lim = xS1 3x + 4

Now Work

PROBLEM

lim 1x3 - x + 22

xS1

lim 13x + 42

xS1

=

6 7

r

25

When the limit of the denominator is zero, formula (10) cannot be used. In such cases, other strategies need to be used. Let’s look at two examples.

EX AM PL E 10

Finding the Limit of a Quotient 'JOE
(a) lim

xS3

Solution

x2 - x - 6 x2 - 9

(b) lim

xS0

x - sin x x

(a) The limit of the denominator equals zero, so formula (10) cannot be used. Instead, notice that the expression can be factored as 1x - 32 1x + 22 x2 - x - 6 = 2 1x - 32 1x + 32 x - 9 When we compute a limit as x approaches 3, we are interested in the values of the function when x is close to 3 but unequal to 3. Since x ≠ 3, we can divide out the 1x - 32 >s.
'PSNVMB

DBO
UIFO
CF
VTFE lim

xS3

lim 1x + 22 1x - 32 1x + 22 x2 - x - 6  xS3 = = = lim 2 S x 3 1x 32 1x + 32 lim 1x + 32 6 x - 9 xS3

(b) Again, the limit of the denominator is zero. In this situation, perform the indicated operation and divide by x. lim

xS0

x - sin x x x sin x sin x - lim = lim c d = lim =  - 1 = 4 xS0 x xS0 x xS0 x x x c c Limit of a difference

Refer to Example 3, Section 13.1

r

934

CHAPTER 13  A Preview of Calculus: The Limit, Derivative, and Integral of a Function

EX AM PL E 11

Finding Limits Using Algebraic Properties 'JOE
lim

xS2

Solution

x3 - 2x2 + 4x - 8 x4 - 2x3 + x - 2

The limit of the denominator is zero, so formula (10) cannot be used. We factor the expression. x2 1x - 22 + 41x - 22 1x2 + 42 1x - 22 x3 - 2x2 + 4x - 8 = = x3 1x - 22 + 11x - 22 1x3 + 12 1x - 22 x4 - 2x3 + x - 2 c

Factor by grouping

Then lim

xS2

1x2 + 42 1x - 22 x3 - 2x2 + 4x - 8 8 = lim = 3 4 3 x S 2 1x + 12 1x - 22 9 x - 2x + x - 2

r

which is exact.

Compare the exact solution above with the approximate solution found in Example 6 of Section 13.1.

5 Find the Limit of an Average Rate of Change EX AM PL E 12

Finding the Limit of an Average Rate of Change 'JOE
UIF
MJNJU
BT
x approaches 2 of the average rate of change of the function f1x2 = x2 + 3x from 2 to x.

Solution

The average rate of change of f from 2 to x is

f 1x2 - f 122 1x2 + 3x2 - 10 1x + 2 1x - 22 ∆y = = = ∆x x - 2 x - 2 x - 2

The limit of the average rate of change is lim

xS2

f1x2 - f122 1x2 + 3x2 - 10 1x + 2 1x - 22 = lim = lim = 7 xS2 xS2 x - 2 x - 2 x - 2

r

SUMMARY To find exact values for lim f1x2 , try the following: xSc

1. If f is a polynomial function, lim f1x2 = f1c2 . (formula (7)) xSc 2. If f is a polynomial raised to a power or is the root of a polynomial, use formula (8) or (9) with formula (7). 3. If f is a quotient and the limit of the denominator is not zero, use the fact that the limit of a quotient is the quotient of the limits. (formula (10)) 4. If f is a quotient and the limit of the denominator is zero, use other techniques, such as factoring.

13.2 Assess Your Understanding Concepts and Vocabulary 1. The limit of the product of two functions equals the product of their limits. 2. lim b = b . xSc

3. lim x = xSc

c .

4. True or False The limit of a polynomial function as x BQQSPBDIFT

FRVBMT
UIF
WBMVF
PG
UIF
QPMZOPNJBM
BU

True 5. True or False 5IF
MJNJU
PG
B
SBUJPOBM
GVODUJPO
BU

FRVBMT
UIF
WBMVF
PG
UIF
SBUJPOBM
GVODUJPO
BU

'BMTF 6. True or False The limit of a quotient equals the quotient of the limits. 'BMTF

SECTION 13.2  Algebra Techniques for Finding Limits

935

Skill Building In Problems 7–42, find each limit algebraically. 7. lim  xS1

8. lim 1 - 32



-3

xS1

11. lim 1x2

12. lim 1 - 3x2

- 10

x S -2

15. lim 13x + 22

xS4

19. lim 1x4 - 3x2 + 6x - 92

-1

21. lim 1x2 + 12

22. lim 13x - 42 2

3

8

xS1

25. lim

xS0

x2 - 4 x2 + 4

x2 - 4 x S 2 x2 - 2x

29. lim

x3 - 1 xS1 x - 1

33. lim

-1

26. lim

2

30. lim

xS2

3x + 4 x2 + x

x4 - 1 xS1 x - 1

2 3

38. lim

x3 - x2 + 3x - 3 xS1 x2 + 3x - 4

4 

41. lim

40. lim

lim 18x - 7x3 + 8x2 + x - 42

xS0

x2 - x - 12 x S -3 x2 - 9

35.

lim

x S -1

1x + 12 2 x - 1 2

x S -1

x2 + x - 6 x S -3 x2 + 2x - 3

7 6

32. lim

x3 - 8 x S 2 x2 - 4

36. lim

0

x3 + x2 + 3x + 3 x S -1 x4 + x3 + 2x + 2

4

39. lim

x3 + 2x2 + x x S -1 x + x3 + 2x + 2

0

42. lim

1

28. lim 12x + 12 >3

32

lim

28

2

24. lim 21 - 2x

3

31.

4

xS2

x S -1

xS2

- 4

18. lim 18x2 - 42

8

27. lim 13x - 22 >2

4

x3 - x2 + x - 1 x S 1 x4 - x3 + 2x - 2

37. lim

x S -3

x S -1

xS1

1 2

34. lim

14. lim 12x3 2

80

23. lim 2x + 4

4

 3

x2 + x x S -1 x2 - 1

x S -3

lim 13x2 - x2

17. 20.

xS2

3

xS2

- 13

xS3

xS1

13. lim 1x4 2

-3

10. lim x

4

xS4

- 12

16. lim 12 - x2

8

xS2

9. lim x

x3 - 2x2 + 4x - 8 xS2 x2 + x - 6 x3 - 3x2 + 4x - 12 x S 3 x4 - 3x3 + x - 3

-1  4

3

8  13 28

In Problems 43–52, find the limit as x approaches c of the average rate of change of each function from c to x. 43. c = 2; f 1x2 = x - 3

46. c = 3; f 1x2 = x

3

44. c = - 2; f 1x2 = 4 - 3x



47. c = - 1; f 1x2 = x + 2x 2

27

49. c = 0; f 1x2 = 3x3 - 2x2 + 4 51. c = 1; f 1x2 =

1 x

45. c = 3; f 1x2 = x2

-3

48. c = - 1; f 1x2 = 2x - 3x

0

50. c = 0; f 1x2 = 4x3 - x + 8

0

52. c = 1; f 1x2 =

-1

6 2

1 x2

-7

-

-2

In Problems 53–56, use the properties of limits and the facts that lim

xS0

sin x = 1 x

lim

xS0

cos x - 1 = 0 x

lim sin x = 0

xS0

lim cos x = 1

xS0

where x is in radians, to find each limit. 53. lim

xS0

tan x x

3 sin x + cos x - 1 xS0 4x

55. lim

54. lim

1

sin12x2

2 x [Hint: 6TF
B
%PVCMFBOHMF
'PSNVMB> xS0

3 4

56. lim

xS0

sin2 x + sin x1cos x - 12 x2

1

Retain Your Knowledge Problems 57–60 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 3 - x 23 p -1 . 60° or rad 59. Give the exact value of sin-1 *57. Graph the function f 1x2 = x3 + x2 + 1. 58. g (x) = x - 2 2 3 2x + 3 60. Use the Binomial Theorem to expand 1x + 22 4. 58. 'JOE
UIF
JOWFSTF
PG
UIF
GVODUJPO
g1x2 = . x + 1 x4 + 8x3 + 24x2 + 32x + 16

936

CHAPTER 13  A Preview of Calculus: The Limit, Derivative, and Integral of a Function

13.3 One-sided Limits; Continuous Functions PREPARING FOR THIS SECTION Before getting started, review the following: r
r
r
r


1JFDFXJTFEFGJOFE
'VODUJPOT
4FDUJPO

QQ
m

-JCSBSZ
PG
'VODUJPOT
4FDUJPO

QQ
m

1PMZOPNJBM
'VODUJPOT
4FDUJPO

QQ
m

1SPQFSUJFT
PG
3BUJPOBM
'VODUJPOT
4FDUJPO

QQ
m

r
5IF
(SBQI
PG
B
3BUJPOBM
'VODUJPO
4FDUJPO

QQ
m

r
1SPQFSUJFT
PG
UIF
&YQPOFOUJBM
'VODUJPO
4FDUJPO

pp. 331 and 333) r
1SPQFSUJFT
PG
UIF
-PHBSJUINJD
'VODUJPO
4FDUJPO

p. 346) r
1SPQFSUJFT
PG
UIF
5SJHPOPNFUSJD
'VODUJPOT
4FDUJPO

QQ

BOE

BOE
4FDUJPO

QQ
m

Now Work the ‘Are You Prepared?’ problems on page 940.

OBJECTIVES 1 Find the One-sided Limits of a Function (p. 936) 2 Determine Whether a Function Is Continuous (p. 938)

1 Find the One-sided Limits of a Function Earlier we described lim f1x2 = N by saying that as x gets closer to c but remains xSc unequal to c, the corresponding values of f1x2 get closer to N. Whether we use a numerical argument or the graph of the function f, the variable x can get closer to c in only two ways: by approaching c from the left, through numbers less than c, or by approaching c from the right, through numbers greater than c. If we approach c from only one side, we have a one-sided limit. The notation

lim f1x2 = L

x S c-

is called the left limit. It is read “the limit of f1x2 as x approaches c from the left equals L” and may be described by the following statement:

In Words

x S c - means x is approaching the value of c from the left, so x 6 c.

As x gets closer to c but remains less than c, the corresponding value of f1x2 gets closer to L.

The notation x S c - is used to remind us that x is less than c. The notation

lim f1x2 = R

x S c+

is called the right limit. It is read “the limit of f1x2 as x approaches c from the right equals R” and may be described by the following statement:

In Words

x S c + means x is approaching the value of c from the right, so x 7 c.

As x gets closer to c, but remains greater than c, the corresponding value of f1x2 gets closer to R.

The notation x S c + is used to remind us that x is greater than c.

SECTION 13.3  One-sided Limits; Continuous Functions

937

'JHVSF

JMMVTUSBUFT
MFGU
BOE
SJHIU
MJNJUT Figure 6

y

y

L

R

c

x

x

c

x

x

lim

lim

x

x

(b)

(a)

The left and right limits can be used to determine whether lim f1x2 exists. See xSc 'JHVSF
 Figure 7

y R

y LR

L c

c

x

lim f(x)  lim f(x)

lim f(x)  lim f(x)

x c

x

x c

x c

x c

(b)

(a)

"T
'JHVSF
 B
JMMVTUSBUFT
lim f1x2 exists and equals the common value of the xSc left limit and the right limit 1L = R2.
*O
'JHVSF
 C


XF
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lim f1x2 does not xSc exist because L ≠ R. This leads us to the following result.

THEOREM

Suppose that lim- f1x2 = L and lim+ f1x2 = R. Then lim f1x2 exists if and xSc xSc xSc only if L = R.
'VSUIFSNPSF
JG
L = R, then lim f1x2 = L = R. xSc

Collectively, the left and right limits of a function are called one-sided limits of the function.

Finding One-sided Limits of a Function

EXAM PL E 1

'PS
UIF
GVODUJPO 2x - 1 if x 6 2 f1x2 = c 1 if x = 2 x - 2 if x 7 2 find: (a) lim- f1x2 xS2

Solution

Figure 8 y

(b) lim+ f1x2

(c) lim f1x2

xS2

xS2

'JHVSF

TIPXT
UIF
HSBQI
PG
f. (a) To find lim- f (x), look at the values of f when x is close to 2 but less than 2. xS2

Since f1x2 = 2x - 1 for such numbers, we conclude that lim f1x2 = lim- 12x - 12 = 3

2

x S 2-

(2, 1)

xS2

(b) To find lim+ f1x2, look at the values of f when x is close to 2 but greater than 2. xS2

22

2 22

4

x

Since f1x2 = x - 2 for such numbers, we conclude that lim f1x2 = lim+ 1x - 22 = 0

x S 2+

xS2

(c) Since the left and right limits are unequal, lim f1x2 does not exist. xS2

Now Work

PROBLEMS

21

AND

35

r

938

CHAPTER 13  A Preview of Calculus: The Limit, Derivative, and Integral of a Function

2 Determine Whether a Function Is Continuous We have observed that f1c2, the value of the function f at c, plays no role in determining the one-sided limits of f at c. What is the role of the value of a function at c and its one-sided limits at c? Let’s look at some of the possibilities. See 'JHVSF
 Figure 9

y y

y

y 5 f(x) f(c)

c x lim f(x) 5 lim f(x), so lim f(x) exists;

x

c2

x

c1

x

c

x

lim f(x) 5 f(c)

x

c

x

y 5 f(x)

y 5 f(x)

f(c) c x lim f(x) 5 lim f(x), so lim f(x) exists; c2

x

c1

x

c x lim f(x) 5 lim f(x), so lim f(x) exists; c⫺

x

c

lim f(x) fi f(c)

x

c⫹

x

c

(a)

(b)

y

(c)

y

y 5 f(x)

y y 5 f(x)

y 5 f(x) f(c) c x lim f(x) fi lim f(x), so lim f(x) does not exist;

x

2

c

x

1

c

x

c

f(c) is defined

c x lim f(x) fi lim f(x), so lim f(x) does not exist;

x

c

⫺

c

f(c) is not defined

x

c

⫹

x

c

f(c) is not defined

c lim f(x) 5 f(c) fi lim f(x),

x

c

x

c

x

⫹

so lim f(x) does not exist; x

(e)

(d)

⫺

c

f(c) is defined

(f)

Much earlier in this text, we said that a function f was continuous if its graph DPVME
CF
ESBXO
XJUIPVU
MJGUJOH
QFODJM
GSPN
QBQFS
'JHVSF

SFWFBMT
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HSBQI
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IBT
UIJT
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JT
UIF
HSBQI
JO
'JHVSF
 B


GPS
XIJDI
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POF sided limits at c each exist and are equal to the value of f at c. This leads us to the following definition.

DEFINITION

A function f is continuous at c if 1. f is defined at c; that is, c is in the domain of f so that f1c2 equals a number. 2. lim- f1x2 = f 1c2 xSc

3. lim+ f1x2 = f1c2 xSc

In other words, a function f is continuous at c if lim f1x2 = f1c2

xSc

If f is not continuous at c, we say that f is discontinuous at c. Each function XIPTF
HSBQI
BQQFBST
JO
'JHVSFT
 C
UP
 G
JT
EJTDPOUJOVPVT
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c. Look again at formula (7) on page 931. Based on (7), we conclude that a polynomial function is continuous at every number. Look at formula (10) on page 933. We conclude that a rational function is continuous at every number, except any at which it is not defined. At numbers where a rational function is not defined, either a hole appears in the graph or else an asymptote appears.

Now Work

PROBLEM

27

SECTION 13.3  One-sided Limits; Continuous Functions

939

Determining the Numbers at Which a Rational Function Is Continuous

EXAM PL E 2

(a) Determine the numbers at which the rational function R 1x2 =

x - 2 x - 6x + 8 2

is continuous. (b) Use limits to analyze the graph of R near 2 and near 4. (c) Graph R.

Solution

(a) Since R 1x2 =

x - 2 , the domain of R is 5 x x ≠ 2, x ≠ 46 . 1x - 22 1x - 42

R is a rational function, so it is defined at every number except 2 and 4. We conclude that R is discontinuous at both 2 and 4. (Condition 1 of the definition is violated.) (b) To determine the behavior of the graph near 2 and near 4, look at lim R 1x2 and xS2 lim R 1x2. xS4

'PS
lim R 1x2, we have xS2

x - 2 1 1 = lim = xS2 x - 4 1x - 22 1x - 42 2

lim R 1x2 = lim

xS2

xS2

1 As x gets closer to 2, the graph of R gets closer to - . Since R is not defined at 2, 2 1 the graph will have a hole at a2, - b . 2 'PS
lim R 1x2, we have xS4

lim R 1x2 = lim

xS4

x4

x S 4+

Since 0 R 1x2 0 S q for x close to 4, the graph of R will have a vertical asymptote at x = 4. (c) It is easiest to graph R by observing that

2 1

2

 12

x - 2 1 = lim xS4 x - 4 1x - 22 1x - 42

1 If x 6 4 and x is getting closer to 4, the value of is negative and is becomx - 4 ing unbounded; that is, lim- R 1x2 = - q . xS4 1 If x 7 4 and x is getting closer to 4, the value of is positive and is becomx 4 ing unbounded; that is, lim R 1x2 = q .

Figure 10 y

xS4

2

4

6

x

if x ≠ 2, then R 1x2 =

1

1 x - 2 = 1x - 22 1x - 42 x - 4

1 Therefore, the graph of R is the graph of y = shifted to the right 4 units with x 1 a hole at a2, - b .
4FF
'JHVSF
. 2

r

Now Work

PROBLEM

73

The exponential, logarithmic, sine, and cosine functions are continuous at every number in their domain. The tangent, cotangent, secant, and cosecant functions are continuous except at numbers for which they are not defined, where asymptotes occur. The square root function and absolute value function are continuous at every number in their domain. The function f 1x2 = int 1x2 is continuous except for x = an integer, where a jump occurs in the graph. Piecewise-defined functions require special attention.

940

CHAPTER 13  A Preview of Calculus: The Limit, Derivative, and Integral of a Function

Determining Where a Piecewise-defined Function Is Continuous

EX A MPL E 3

Determine the numbers at which the following function is continuous. x2 if x … 0 f1x2 = c x + 1 if 0 6 x 6 2  - x if 2 … x … 

Solution

The “pieces” of f —that is, y = x2, y = x + 1, and y =  - x—are each continuous for every number since they are polynomials. In other words, when we graph the pieces, we will not lift our pencil. When we graph the function f, however, we have to be careful, because the pieces change at x = 0 and at x = 2. So the numbers we need to investigate for f are x = 0 and x = 2. At x = 0:

f102 = 02 = 0 lim f1x2 = lim- x2 = 0

Figure 11

3

x S 0+

xS0

Since lim+ f1x2 ≠ f102, f is not continuous at x = 0. xS0

(2, 3)

At x = 2:

2

2 (0, 0)

xS0

lim f1x2 = lim+ 1x + 12 = 1

y 4

x S 0-

2

4

6

f122 =  - 2 = 3

lim f1x2 = lim- 1x + 12 = 3

x S 2-

xS2

x S 2+

xS2

lim f1x2 = lim+ 1 - x2 = 3

x

So f is continuous at x = 2. The graph of f,
HJWFO
JO
'JHVSF

EFNPOTUSBUFT
UIF
DPODMVTJPOT
ESBXO
BCPWF

2

Now Work

PROBLEMS

53

AND

61

r

SUMMARY Library of Functions: Continuity Properties Function

Domain

Property

Polynomial function

All real numbers

Continuous at every number in the domain

P1x2 3BUJPOBM
GVODUJPO
R 1x2 = , Q1x2 P, Q are polynomials

5 x Q1x2 ≠ 06

Continuous at every number in the domain Hole or vertical asymptote where R is undefined

Exponential function

All real numbers

Continuous at every number in the domain

Logarithmic function

Positive real numbers

Continuous at every number in the domain

Sine and cosine functions

All real numbers

Continuous at every number in the domain

Tangent and secant functions

All real numbers, except p odd integer multiples of 2 All real numbers, except integer multiples of p

Continuous at every number in the domain p Vertical asymptotes at odd integer multiples of 2 Continuous at every number in the domain Vertical asymptotes at integer multiples of p

Cotangent and cosecant functions

13.3 Assess Your Understanding ‘Are You Prepared?’

Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red.

if x … 0 x2 1. 'PS
UIF
GVODUJPO
f 1x2 = c x + 1 if 0 6 x 6 2,  - x if 2 … x …  find f 102 and f 122. QQ
m f(0) = 0; f(2) = 3

2. What are the domain and range of f 1x2 = ln x? (p. 346)

3. True or False The exponential function f 1x2 = e x is increasing on the interval 1 - q , q 2. (p. 331) True

941

SECTION 13.3  One-sided Limits; Continuous Functions

6. True or False Every polynomial function has a graph that can be traced without lifting pencil from paper. QQ
m

True

4. Name the trigonometric functions that have asymptotes.
QQ
m
m Secant, cosecant, tangent, cotangent 5. True or False Some rational functions have holes in their graph. QQ
m True

Concepts and Vocabulary 10. True or False 'PS
BOZ
GVODUJPO
f, lim- f 1x2 = lim+ f 1x2. 'BMTF

7. If we approach c from only one side, then we have a(n) one@sided limit. lim f 1x2 = R is used to describe the fact 8. The notation x S c + that as x gets closer to c but remains greater than c, the value of f 1x2 gets closer to R. 9. If lim f 1x2 = f 1c2, then f is continuous at xSc

xSc

xSc

11. True or False If f is continuous at c, then lim+ f 1x2 = f 1c2. True xSc

12. True or False Every polynomial function is continuous at every real number. True

c .

Skill Building In Problems 13–32, use the accompanying graph of y = f 1x2.

y

*13. What is the domain of f?

4

*14. What is the range of f?

15. 'JOE
UIF
x-intercept(s), if any, of f. - 8, - , - 3 17. 'JOE
f 1 - 82 and f 1 - 42. 0; 2

16. 'JOE
UIF
y-intercept(s), if any, of f.

19. 'JOE
lim - f 1x2.

q

20. 'JOE
lim + f 1x2.

-q

21. 'JOE
lim - f 1x2.

2

22. 'JOE
lim + f 1x2.

-2

x S -6 x S -4

23. 'JOE
lim- f 1x2.

18. 'JOE
f 122 and f 162. x S -6 x S -4

24. 'JOE
lim+ f 1x2.

1

xS2

xS2

3, 2

26. Does lim f 1x2 exist? If it does, what is it? Yes; 3 xS0

xS1

37.

lim + sin x

x S p>2

41. lim x S -1



x2 - 1 x3 + 1

0

35. lim- 12x3 + x2

38. lim- 13 cos x2

-3

39. lim+

xSp

-

2 3

32. Is f
DPOUJOVPVT
BU

Yes

34. lim- 14 - 2x2 xS2

1

29. Is f continuous at 0? Yes

No

31. Is f continuous at 4? No

In Problems 33–44, find the one-sided limit. 33. lim+ 12x + 32

42. lim+ xS0

x3 - x2 x4 + x2

6

-4

28. Is f continuous at - 4?

30. Is f continuous at 2? No

4

2

xS4

No

(6, 2)

2

25. Does lim f 1x2 exist? If it does, what is it? Yes; 0 27. Is f continuous at - 6?

(2, 3)

3

xS1

xS2

-1

43.

x2 - 4 x - 2

lim +

x S -2

36.

7

40. lim-

4

x2 + x - 2 x2 + 2x

lim 13x2 - 82

xS1

3 2

44.

4

x S -2+

x3 - x x - 1

lim -

x S -4

2

x2 + x - 12 x2 + 4x

In Problems 45–60, determine whether f is continuous at c. 46. f 1x2 = 3x2 - 6x +  c = - 3 Yes

45. f 1x2 = x3 - 3x2 + 2x - 6 c = 2 Yes 47. f 1x2 =

x2 +  x - 6

50. f 1x2 =

x - 6 x + 6

c = 3 Yes c = -6

x3 + 3x 53. f 1x2 = c x2 - 3x 1 x3 + 3x 55. f 1x2 = c x2 - 3x -1

No

48. f 1x2 =

x3 - 8 x2 + 4

51. f 1x2 =

x3 + 3x x2 - 3x

if x ≠ 0 if x = 0

c = 0

No

if x ≠ 0 if x = 0

c = 0

Yes

c = 2 Yes c = 0

No

x2 - 6x 54. f 1x2 = c x2 + 6x -2 x2 - 6x 56. f 1x2 = c x2 + 6x -1

49. f 1x2 =

x + 3 x - 3

52. f 1x2 =

x2 - 6x x2 + 6x

c = 3 c = 0

if x ≠ 0 if x = 0

c = 0

No

c = 0

Yes

if x ≠ 0 if x = 0

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

No No

7 4

x

942

CHAPTER 13  A Preview of Calculus: The Limit, Derivative, and Integral of a Function

x3 - 1 x2 - 1 57. f 1x2 = e 2 3 x + 1

x2 - 2x x - 2 58. f 1x2 = e 2 x - 4 x - 1

if x 6 1 if x = 1

c = 1

No

if x 7 1

2e x 2 59. f 1x2 = d 3 x + 2x2 x2

if x 6 0 if x = 0

c = 0

if x 6 2 if x = 2

if x 7 0

No

if x 7 2

3 cos x 3 60. f 1x2 = d 3 x + 3x2 x2

Yes

c = 2

if x 6 0 if x = 0

c = 0

Yes

if x 7 0

In Problems 61–72, find the numbers at which f is continuous. At which numbers is f discontinuous? *61. f 1x2 = 2x + 3

*62. f 1x2 = 4 - 3x

*63. f 1x2 = 3x2 + x

*64. f 1x2 = - 3x3 + 7

*65. f 1x2 = 4 sin x

*66. f 1x2 = - 2 cos x

*67. f 1x2 = 2 tan x

*68. f 1x2 = 4 csc x

*69. f 1x2 =

2x +  x2 - 4

*70. f 1x2 =

x - 4 x2 - 9 2

*71. f 1x2 =

x - 3 ln x

*72. f 1x2 =

ln x x - 3

In Problems 73–76, discuss whether R is continuous at c. Use limits to analyze the graph of R at c. Graph R. *73. R 1x2 =

x - 1 , c = - 1 and c = 1 x2 - 1

*74. R 1x2 =

3x + 6 , c = - 2 and c = 2 x2 - 4

*75. R 1x2 =

x2 + x , c = - 1 and c = 1 x2 - 1

*76. R 1x2 =

x2 + 4x , c = - 4 and c = 4 x2 - 16

In Problems 77–82, determine where each rational function is undefined. Determine whether an asymptote or a hole appears at such numbers. *77. R 1x2 =

x3 - x2 + x - 1 x4 - x3 + 2x - 2

*78. R 1x2 =

x3 + x2 + 3x + 3 x4 + x3 + 2x + 2

*79. R 1x2 =

x3 - 2x2 + 4x - 8 x2 + x - 6

*80. R 1x2 =

x3 - x2 + 3x - 3 x2 + 3x - 4

*81. R 1x2 =

x3 + 2x2 + x x + x3 + 2x + 2

*82. R 1x2 =

x3 - 3x2 + 4x - 12 x4 - 3x3 + x - 3

4

For Problems 83–88, use a graphing utility to graph the functions R given in Problems 77–82. Verify the solutions found above.*

Discussion and Writing 89. Name three functions that are continuous at every real number.

90. $SFBUF
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Retain Your Knowledge Problems 91–94 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 91. 'JOE
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3x - 4 f 1x2 = . Vertical: x = 4; horizontal: y = 3 x - 4 92. Evaluate P(, 3) 60

93. Write ln x + 2ln y - 4ln z as a single natural logarithm. *94. Write the augmented matrix for the following system: 3x + y + 2z = 4

c

93. lna

x + 2z =  y - 3z = - 2

‘Are You Prepared?’ Answers 1. f 102 = 0; f 122 = 3

2. Domain: 5x x 7 06; range 5y - q 6 y 6 q 6

3. True

4. Secant, cosecant, tangent, cotangent

5. True

6. True

xy2 z4

b

SECTION 13.4  The Tangent Problem; The Derivative

943

13.4 The Tangent Problem; The Derivative PREPARING FOR THIS SECTION Before getting started, review the following: r
1PJOUm4MPQF
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B
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4FDUJPO

Q


r
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3BUF
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$IBOHF
4FDUJPO

QQ
m

Now Work the ‘Are You Prepared?’ problems on page 916.

OBJECTIVES 1 2 3 4

Find an Equation of the Tangent Line to the Graph of a Function (p. 943) Find the Derivative of a Function (p. 945) Find Instantaneous Rates of Change (p. 946) Find the Instantaneous Speed of a Particle (p. 946)

The Tangent Problem One question that motivated the development of calculus was a geometry problem, the tangent problem. This problem asks, “What is the slope of the tangent line to the graph of a function y = f1x2 at a point P
PO
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 We first need to define what we mean by a tangent line. In high school geometry, the tangent line to a circle at a point is defined as the line that intersects the circle at FYBDUMZ
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the graph of the circle. Figure 12

Figure 13 Tangent line

y

P y  f(x) Tangent line to f at P

P

x

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L1 and L2 intersect the graph at only one point P, but neither just touches the graph at P. Additionally, the tangent line LT
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f at P but also intersects the graph elsewhere. So how should we define the tangent line to the graph of f at a point P? Figure 14

Figure 15 y

y

L1

L2 LT P

P x

x

1 Find an Equation of the Tangent Line to the Graph of a Function The tangent line LT to the graph of a function y = f1x2 at a point P necessarily contains the point P. To find an equation for LT
VTJOH
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equation of a line, we need to find the slope mtan of the tangent line.

944

CHAPTER 13  A Preview of Calculus: The Limit, Derivative, and Integral of a Function

Figure 16 y Q  (x, f(x)) y  f(x)

Q1  (x1, f(x1)) LT P  (c, f(c)) c

x

Suppose that the coordinates of the point P are 1c, f1c2 2. Locate another point Q = 1x, f1x2 2 on the graph of f. The line containing P and Q is a secant line. 3FGFS
UP
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msec of the secant line is msec =

f 1x2 - f 1c2 x - c

/PX
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 As we move along the graph of f from Q toward P, we obtain a succession of secant lines. The closer we get to P, the closer the secant line is to the tangent line LT . The limiting position of these secant lines is the tangent line LT . Therefore, the limiting value of the slopes of these secant lines equals the slope of the tangent line. Also, as we move from Q toward P, the values of x get closer to c. Therefore, mtan = lim msec = lim xSc

DEFINITION

xSc

f1x2 - f1c2 x - c

The tangent line to the graph of a function y = f1x2 at a point P = 1c, f1c2 2 on its graph is defined as the line containing the point P whose slope is mtan = lim

xSc

f1x2 - f1c2 x - c

(1)

provided that this limit exists.

THEOREM

If mtan exists, an equation of the tangent line is y - f 1c2 = mtan 1x - c2

(2)

Finding an Equation of the Tangent Line

EX A MPL E 1

Solution

x2 1 'JOE
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f(x) = at the point a1, b . 4 4 Graph f and the tangent line. 1 The tangent line contains the point a1, b . The slope of the tangent line to the 4 x2 1 graph of f(x) = at a1, b is 4 4

mtan

x2 1 f1x2 - f112 1x - 12 1x + 12 4 4 = lim = lim = lim xS1 xS1 x - 1 xS1 x - 1 4 1x - 12 = lim

xS1

Figure 17

x + 1 1 = 4 2

An equation of the tangent line is

y y 1 2

x2 4

y -

1 4

( 1, ) y  12 x  14 1 2

1

1 1 = 1x - 12 4 2 y =

x

'JHVSF

TIPXT
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HSBQI
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y =

Now Work

PROBLEM

y - f(c) = mtan(x - c)

1 1 x 2 4 x2 1 and the tangent line at a1, b . 4 4 11

r

SECTION 13.4  The Tangent Problem; The Derivative

945

2 Find the Derivative of a Function The limit in formula (1) has an important generalization: it is called the derivative of f at c.

DEFINITION

Let y = f1x2 denote a function f. If c is a number in the domain of f, then the derivative of f at c, denoted by f′(c) and read “f prime of c,” is defined as f′1c2 = lim

xSc

f1x2 - f1c2 x - c

(3)

provided that this limit exists.

EXAM PL E 2

Finding the Derivative of a Function 'JOE
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f1x2 = 2x2 - x at 2. That is, find f′(2).

Solution

Since f122 = 2122 2 - 122 = - 2, we have f1x2 - f122 12x2 - x2 - 1 - 22 12x - 12 1x - 22 2x2 - x + 2 = = = x - 2 x - 2 x - 2 x - 2 The derivative of f at 2 is f′122 = lim

xS2

Now Work

f1x2 - f122 12x - 12 1x - 22 = lim = 3 xS2 x - 2 x - 2

PROBLEM

r

21

Example 2 provides a way of finding the derivative at 2 analytically. Graphing utilities have built-in procedures to approximate the derivative of a function at any number c. Consult your owner’s manual for the appropriate keystrokes.

EXAM PL E 3

Finding the Derivative of a Function Using a Graphing Utility Use a graphing utility to find the derivative of f1x2 = 2x2 - x at 2. That is, find f′122.

Solution

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Figure 18 graphing calculator. As shown, f′122 = 3.

Now Work

EXAM PL E 4

PROBLEM

33

r

Finding the Derivative of a Function 'JOE
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f1x2 = x2 at c. That is, find f′(c).

Solution

Since f1c2 = c 2, we have

f1x2 - f1c2 1x + c2 1x - c2 x2 - c 2 = = x - c x - c x - c

The derivative of f at c is f′1c2 = lim

xSc

f1x2 - f1c2 1x + c2 1x - c2 = 2c = lim xSc x - c x - c

r

946

CHAPTER 13  A Preview of Calculus: The Limit, Derivative, and Integral of a Function

As Example 4 illustrates, the derivative of f 1x2 = x2 exists and equals 2c for any number c. In other words, the derivative is itself a function, and using x for the independent variable, we can write f′(x) = 2x. The function f′ is called the derivative function of f or the derivative of f . We also say that f is differentiable. The instruction “differentiate f ” means “find the derivative of f.”

3 Find Instantaneous Rates of Change The average rate of change of a function f from c to x is f1x2 - f1c2 ∆y = x - c ∆x The limit as x approaches c of the average rate of change of f, based on formula (3), is the derivative of f at c.

DEFINITION

The derivative of f at c is also called the instantaneous rate of change of f with respect to x at c. That is, ¢

EX A MPL E 5

f1x2 - f1c2 Instantaneous rate of ≤ = f′1c2 = lim change of f with respect to x at c xSc x - c

(4)

Finding the Instantaneous Rate of Change The volume V of a right circular cone of height h = 6 feet and radius r feet is 1 V = V 1r2 = pr 2h = 2pr 2. If r is increasing, find the instantaneous rate of change 3 of the volume V with respect to the radius r at r = 3.

Solution

The instantaneous rate of change of V with respect to r at r = 3 is the derivative V′(3). V′132 = lim

rS3

V 1r2 - V 132 2p1r 2 - 92 2pr 2 - 18p = lim = lim rS3 rS3 r - 3 r - 3 r - 3

= lim 3 2p1r + 32 4 = 12p rS3

At the instant r = 3 feet, the volume of the cone is increasing with respect to r at a rate of 12p ≈ 37.699 cubic feet per 1-foot change in the radius.

r

Now Work

PROBLEM

43

4 Find the Instantaneous Speed of a Particle If s = f1t2 denotes the position of a particle at time t, then the average speed of the particle from c to t is f 1t2 - f1c2 Change in position ∆s = = Change in time ∆t t - c

DEFINITION

(5)

The limit as t approaches c
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instantaneous speed of the particle at c or the velocity of the particle at c. That is, a

f1t2 - f1c2 Instantaneous speed of b = f′1c2 = lim a particle at time c tSc t - c

(6)

SECTION 13.4  The Tangent Problem; The Derivative

EXAM PL E 6

947

Finding the Instantaneous Speed of a Particle In physics it is shown that the height s of a ball thrown straight up with an initial speed of 80 feet per second (ft/sec) from a rooftop 96 feet high is s = s 1t2 = - 16t 2 + 80t + 96

Figure 19

where t is the elapsed time that the ball is in the air. The ball misses the rooftop on JUT
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 Roof top

96 ft

Solution

(a) (b) (c) (d) (e) (f) (g)

When does the ball strike the ground? That is, how long is the ball in the air? At what time t will the ball pass the rooftop on its way down? What is the average speed of the ball from t = 0 to t = 2? What is the instantaneous speed of the ball at time t 0? What is the instantaneous speed of the ball at t = 2? When is the instantaneous speed of the ball equal to zero? What is the instantaneous speed of the ball as it passes the rooftop on the way down? (h) What is the instantaneous speed of the ball when it strikes the ground? (a) The ball strikes the ground when s = s 1t2 = 0. - 16t 2 + 80t + 96 t 2 - t - 6 1t - 62 1t + 12 t = 6 or t

= = = =

0 0 0 -1

Discard the solution t = - 1. The ball strikes the ground after 6 sec. (b) The ball passes the rooftop when s = s 1t2 = 96. - 16t 2 + 80t + 96 t 2 - t t1t - 2 t = 0 or t

= = = =

96 0 0 

Discard the solution t = 0. The ball passes the rooftop on the way down BGUFS

TFD (c) The average speed of the ball from t = 0 to t = 2 is s 122 - s 102 ∆s 192 - 96 = = = 48 ft/sec ∆t 2 - 0 2

(d) The instantaneous speed of the ball at time t 0 is the derivative s′1t 0 2; that is, s 1t2 - s 1t 0 2 t - t0 1 - 16t 2 + 80t + 962 - 1 - 16t 20 + 80t 0 + 962 lim t S t0 t - t0 - 163 t 2 - t 20 - t + t 0 4 lim t S t0 t - t0 - 163 1t + t 0 2 1t - t 0 2 - 1t - t 0 2 4 lim t S t0 t - t0 - 163 1t + t 0 - 2 1t - t 0 2 4 lim = lim 3 - 161t + t 0 - 2 4 t S t0 t S t0 t - t0 = - 1612t 0 - 2 ft/sec

s′1t 0 2 = lim

t S t0

= = = =

3FQMBDF
t0 by t. The instantaneous speed of the ball at time t is s′1t2 = - 1612t - 2 ft/sec

948

CHAPTER 13  A Preview of Calculus: The Limit, Derivative, and Integral of a Function

(e) At t = 2 sec, the instantaneous speed of the ball is

s′ 122 = - 1612 # 2 - 2 = - 161 - 12 = 16 ft/sec

(f) The instantaneous speed of the ball is zero when s′1t2 = 0 - 1612t - 2 = 0  t = = 2. sec 2 (g) The ball passes the rooftop on the way down when t = . The instantaneous speed at t =  is s′12 = - 16110 - 2 = - 80 ft/sec

Exploration Determine the vertex of the quadratic function given in Example 6. What do you conclude about the velocity when s(t) is a maximum?

At t =  sec, the ball is traveling - 80 ft/sec. When the instantaneous rate of change is negative, it means that the direction of the object is downward. The ball is traveling 80 ft/sec in the downward direction when t =  sec. (h) The ball strikes the ground when t = 6. The instantaneous speed when t = 6 is s′162 = - 16112 - 2 = - 112 ft/sec The speed of the ball at t = 6 sec is - 112 ft/sec. Again, the negative value implies that the ball is traveling downward.

r

SUMMARY The derivative of a function y = f1x2 at c is defined as f′1c2 = lim

xSc

f1x2 - f1c2 x - c

In geometry, f′(c) equals the slope of the tangent line to the graph of f at the point 1c, f1c2 2. In physics, f′(c) equals the instantaneous speed (velocity) of a particle at time c, where s = f 1t2 is the position of the particle at time t. In applications, if two variables are related by the function y = f1x2, then f′(c) equals the instantaneous rate of change of f with respect to x at c.

13.4 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. 'JOE
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DPOUBJOJOH
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point 12, - 42. Q

y = x - 14

2. True or False The average rate of change of a function f from a to b is f 1b2 + f 1a2 b + a

QQ
m
'BMTF

Concepts and Vocabulary 3. If lim

f 1x2 - f 1c2

exists, it equals the slope of the x - c tangent line to the graph of f at the point 1c, f 1c2 2. f 1x2 - f 1c2 4. If lim exists, it is called the derivative of f at c. xSc x - c xSc

5. If s = f 1t2 denotes the position of a particle at time t, the derivative f′1c2 is the velocity of the particle at c.

6. True or False The tangent line to a function is the limiting position of a secant line. True 7. True or False The slope of the tangent line to the graph of f at 1c, f 1c2 2 is the derivative of f at c. True 8. True or False The velocity of a particle whose position at time t is s 1t2 is the derivative s′ 1t2. True

949

SECTION 13.4  The Tangent Problem; The Derivative

Skill Building In Problems 9–20, find the slope of the tangent line to the graph of f at the given point. Graph f and the tangent line. *9. f 1x2 = 3x +  at 11, 82

*12. f 1x2 = 3 - x

2

*10. f 1x2 = - 2x + 1 at 1 - 1, 32

at 11, 22

*13. f 1x2 = 3x

*15. f 1x2 = 2x2 + x at 11, 32

*18. f 1x2 = - 2x + x - 3 at 11, - 42 2

2

*11. f 1x2 = x2 + 2 at 1 - 1, 32

at 12, 122

*14. f 1x2 = - 4x2 at 1 - 2, - 162

*16. f 1x2 = 3x2 - x at 10, 02

*17. f 1x2 = x2 - 2x + 3 at 1 - 1, 62

*19. f 1x2 = x + x at 12, 102

*20. f 1x2 = x3 - x2 at 11, 02

3

In Problems 21–32, find the derivative of each function at the given number. 21. f 1x2 = - 4x +  at 3

24. f 1x2 = 2x + 1 at - 1 2

27. f1x2 = x + 4x at - 1 3

22. f 1x2 = - 4 + 3x at 1

-4

25. f 1x2 = 2x + 3x at 1

-4

2

28. f 1x2 = 2x - x 3

7

30. f1x2 = x3 - 2x2 + x at - 1

2

31. f 1x2 = sin x at 0

8

at 2

23. f 1x2 = x2 - 3 at 0

3

26. f 1x2 = 3x2 - 4x at 2

7

0 8

29. f 1x2 = x + x - 2x at 1 3

20

2

32. f 1x2 = cos x at 0

1

3

0

In Problems 33–42, use a graphing utility to find the derivative of each function at the given number. 33. f 1x2 = 3x3 - 6x2 + 2 at - 2

-x + 1 at 8 x2 + x + 7 p *37. f 1x2 = x sin x at 3 p 2 *40. f 1x2 = x sin x at 4 35. f 1x2 =

3

34. f 1x2 = - x4 + 6x2 - 10 at 

60

- x + 9x + 3 x3 + x2 - 6 p *38. f 1x2 = x sin x at 4

- 0.88777696

4

36. f 1x2 =

*41. f 1x2 = e x sin x at 2

at - 3

- 2440 

*39. f 1x2 = x2 sin x at

p 3

*42. f 1x2 = e -x sin x at 2

Applications and Extensions 43. Instantaneous Rate of Change The volume V of a right circular cylinder of height 3 feet and radius r feet is V = V 1r2 = 3pr 2.
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the volume with respect to the radius r at r = 3. 18p ft 3 >ft 44. Instantaneous Rate of Change The surface area S of a sphere of radius r feet is S = S1r2 = 4pr 2.
'JOE
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instantaneous rate of change of the surface area with respect to the radius r at r = 2. 16p ft 2 >ft 45. Instantaneous Rate of Change The volume V of a sphere of 4 radius r feet is V = V 1r2 = pr 3.
'JOE
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JOTUBOUBOFPVT 3 rate of change of the volume with respect to the radius r at r = 2. 16p ft 3 >ft *46. Instantaneous Rate of Change The volume V of a cube of side x meters is V = V 1x2 = x3.
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of change of the volume with respect to the side x at x = 3. 47. Instantaneous Speed of a Ball In physics it is shown that the height s of a ball thrown straight up with an initial speed of 96 ft/sec from ground level is s = s 1t2 = - 16t 2 + 96t where t is the elapsed time that the ball is in the air. (a) When does the ball strike the ground? That is, how long is the ball in the air? 6 seconds (b) What is the average speed of the ball from t = 0 to t = 2? 64 feet per second *(c) What is the instantaneous speed of the ball at time t? *(d) What is the instantaneous speed of the ball at t = 2? (e) When is the instantaneous speed of the ball equal to zero? 3 seconds (f) How high is the ball when its instantaneous speed equals zero? 144 feet (g) What is the instantaneous speed of the ball when it strikes the ground? - 96 feet per second

48. Instantaneous Speed of a Ball In physics it is shown that the height s of a ball thrown straight down with an initial speed of 48 ft/sec from a rooftop 160 feet high is s = s 1t2 = - 16t 2 - 48t + 160 where t is the elapsed time that the ball is in the air. (a) When does the ball strike the ground? That is, how long is the ball in the air? 2 seconds (b) What is the average speed of the ball from t = 0 to t = 1? - 64 feet per second *(c) What is the instantaneous speed of the ball at time t? *(d) What is the instantaneous speed of the ball at t = 1? (e) What is the instantaneous speed of the ball when it strikes the ground? - 112 feet per second 4 . Instantaneous Speed on the Moon An astronaut throws a 49 49. ball down into a crater on the moon. The height s (in feet) of the ball from the bottom of the crater after t seconds is given in the following table:

Time, t (in seconds)

Distance, s (in feet)

0

1000

1

987

2

969

3

945

4

917

5

883

6

843

7

800

8

749

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

950

CHAPTER 13  A Preview of Calculus: The Limit, Derivative, and Integral of a Function

* B
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t = 1 to t = 4 seconds. * C
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t = 1 to t = 3 seconds. * D
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t = 1 to t = 2 seconds. *(d) Using a graphing utility, find the quadratic function of best fit. *(e) Using the function found in part (d), determine the instantaneous speed at t = 1 second. 50. Instantaneous Rate of Change The data to the right represent the total revenue R (in dollars) received from selling x bicycles at Tunney’s Bicycle Shop. B
'JOE
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x = 2 to x = 10 bicycles. $249.28 per bicycle C
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x = 2 to x = 102 bicycles. ≈$329.87 per bicycle D
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x = 2 to x = 60 bicycles. ≈$48.71 per bicycle (d) Using a graphing utility, find the quadratic function of best fit. R(x) = - 1.22x2 + 97.791x + 7944.40

Number of Bicycles, x

Total Revenue, R (in dollars)

0

0

25

28,000

60

45,000

102

53,400

150

59,160

190

62,360

223

64,835

249

66,525

(e) Using the function found in part (d), determine the instantaneous rate of change of revenue at x = 2 bicycles. ≈$21.69 per bicycle

Retain Your Knowledge Problems 51–54 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. 51. 'JOE
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54. 'JOE
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x2 - 2x - 2y + 7 = 0. places. 23.67 sq. units B y = x + 4 52. Solve the system by substitution: e 2 9 x - y = - 2 (2, 6), ( - 1, 3) 6

53. Evaluate: C(, 3)

10

7 51. Vertex: (1, 3); focus: a1, b 2

C A

13

‘Are You Prepared?’ Answers 1. y = x - 14

2. 'BMTF

13.5 The Area Problem; The Integral PREPARING FOR THIS SECTION Before getting started, review the following: r
(FPNFUSZ
'PSNVMBT
"QQFOEJY
"
4FDUJPO
"
QQ
"m"

r
4VNNBUJPO
/PUBUJPO
4FDUJPO

QQ
m

Now Work the ‘Are You Prepared?’ problems on page 955.

OBJECTIVES 1 Approximate the Area under the Graph of a Function (p. 951) 2 Approximate Integrals Using a Graphing Utility (p. 955)

The Area Problem The development of the integral, like that of the derivative, was originally motivated to a large extent by a problem in geometry: the area problem. Figure 20 y

y  f(x)

Area A a

b

Area A  area under the graph of f from a to b

Area Problem Suppose that y = f1x2 is a function whose domain is a closed interval 3 a, b 4 . We assume that f1x2 Ú 0 for all x in 3 a, b4 .
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graph of f, the x-axis, and the vertical lines x = a and x = b.

x

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A shown as the area under the graph of f from a to b.

SECTION 13.5  The Area Problem; The Integral

951

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f 1x2 = k and for a linear function f1x2 = mx + B, we DBO
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 B
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C  Figure 21

y

y

f(x)  mx  B f(x)  k

k

f(b) k

A a

b

A

f(a) x

a

b

x

ba A  area of trapezoid

ba A  area of rectangle  height  base  k (b  a)

1

 2 [f (b)  f (a)](b  a)

(a)

(b)

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1 Approximate the Area under the Graph of a Function We use rectangles to approximate the area under the graph of a function f. We do this by partitioning or dividing the interval 3 a, b4 into subintervals of equal length. On each subinterval, we form a rectangle whose base is the length of the subinterval and whose height is f 1u2 for some number u in the subinterval. Look at 'JHVSF
 Figure 22

y

y y  f(x)

y  f(x)

f(u2)

f (u1)

f(u1) a  u1

b

u2 ba 2

ba 2

2 subintervals (a)

x

a

u1 ba 4

f (u2) u2

ba ba 4 4 4 subintervals (b)

f(u3)

f(u4) b  u4 x

u3

ba 4

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3 a, b 4 is partitioned into two subintervals, each of b - a length , and the number u is chosen as the left endpoint of each subinterval. 2 *O
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3 a, b4 is partitioned into four subintervals, each of b - a length , and the number u is chosen as the right endpoint of each subinterval. 4 The area A under f from a to b is approximated by adding the areas of the rectangles formed by the partition. 6TJOH
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 B

Area A ≈ area of first rectangle + area of second rectangle b - a b - a = f 1u 1 2 + f1u 2 2 2 2

952

CHAPTER 13  A Preview of Calculus: The Limit, Derivative, and Integral of a Function

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Area A ≈ area of first rectangle + area of second rectangle + area of third rectangle + area of fourth rectangle b - a b - a b - a b - a = f 1u 1 2 + f1u 2 2 + f1u 3 2 + f1u 4 2 4 4 4 4 In approximating the area under the graph of a function f from a to b, the choice of the number u
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u as either the left endpoint of each subinterval or the right endpoint. The choice of how many subintervals to use is also arbitrary. In general, the more subintervals used, the better the approximation will be. Let’s look at a specific example.

EX A MPL E 1

Approximating the Area under the Graph of f(x) = 2x from 0 to 1 Approximate the area A under the graph of f1x2 = 2x from 0 to 1 as follows:

Figure 23 y f (x ) ⫽ 2x

2

1

( 2)

f 1 f (0) 0

x

1

1 2

Solution

1 2

1 2

2 subintervals; u’s are left endpoints (a) y f (x ) ⫽ 2x

2

1

f (1)

( 2)

f 1

0 1 2

x

1

1 2

(a) Partition 3 0, 14 into two subintervals of equal length and choose u as the left endpoint. (b) Partition 3 0, 14 into two subintervals of equal length and choose u as the right endpoint. (c) Partition 3 0, 14 into four subintervals of equal length and choose u as the left endpoint. (d) Partition 3 0, 14 into four subintervals of equal length and choose u as the right endpoint. F
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A is approximated as 1 1 1 A ≈ f102 a b + f a b a b 2 2 2 1 1 = 102 a b + 112 a b 2 2 1 = = 0. 2

1 (b) Partition 3 0, 14 into two subintervals, each of length , and choose u as the right 2 FOEQPJOU
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A is approximated as 1 1 1 A ≈ f a b a b + f112 a b 2 2 2

1 2

2 subintervals; u’s are right endpoints

1 1 = 112 a b + 122 a b 2 2

(b) y f (x ) ⫽ 2x

2

1 f (0) 0

1 4

1 2

3 4

1 1 f(0) = 2 # 0 = 0; f a b = 2 # = 1 2 2

1

x

( 4) f ( 12) f ( 34)

f 1

4 subintervals; u’s are left endpoints (c)

=

3 = 1. 2

1 (c) Partition 3 0, 14 into four subintervals, each of length , and choose u as the left 4 FOEQPJOU
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3 = 0.7 4

SECTION 13.5  The Area Problem; The Integral

(d) Partition 3 0, 14 into four subintervals, each of length

Figure 23 (continued) y f (x ) ⫽ 2x

2

1

953

1 , and choose u as the 4 SJHIU
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A is approximated as 1 1 1 1 3 1 1 A ≈ f a b a b + f a b a b + f a b a b + f112 a b 4 4 2 4 4 4 4

f (1)

1 1 1 3 1 1 = a b a b + 112 a b + a b a b + 122 a b 2 4 4 2 4 4 0

1 4

1 2

3 4

x

1

 = 1.2 4 (e) The actual area under the graph of f1x2 = 2x from 0 to 1 is the area of a right triangle whose base is of length 1 and whose height is 2. The actual area A is therefore =

( 4) f ( 12) f ( 34)

f 1

4 subintervals; u’s are right endpoints (d)

A =

1 1 base * height = a b 112 122 = 1 2 2

r

Now look at Table 7, which shows the approximations to the area under the graph of f1x2 = 2x from 0 to 1 for n = 2, 4, 10, and 100 subintervals. Notice that the approximations to the actual area improve as the number of subintervals increases.

Table 7

Using left endpoints:

n Area

2 0.5

4 0.75

10 0.9

100 0.99

Using right endpoints:

n Area

2 1.5

4 1.25

10 1.1

100 1.01

You are asked to confirm the entries in Table 7 in Problem 31. 5IFSF
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f1x2 = 2x is increasing on 3 0, 14 , the choice of u as the left endpoint gives a lower-bound estimate to the actual area, while choosing u as the right endpoint gives an upper-bound estimate. Do you see why?

Now Work

EXAM PL E 2

PROBLEM

9

Approximating the Area under the Graph of f(x) = x 2 Approximate the area under the graph of f1x2 = x2
GSPN

UP

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GPMMPXT (a) Using four subintervals of equal length (b) Using eight subintervals of equal length In each case, choose the number u to be the left endpoint of each subinterval.

Solution Figure 24 y f (x ) ⫽ x 2

25

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3 1, 4 is  - 1 = 1 as follows: partitioned into subintervals of length 4 3 2, 34 3 3, 44 3 4, 4 3 1, 24 Each of these subintervals is of length 1. Choosing u as the left endpoint of each subinterval, the area A under the graph of f1x2 = x2 is approximated by Area A ≈ f112 112 + f122 112 + f132 112 + f142 112 = 1 + 4 + 9 + 16 = 30

15

5 0

1

2

3

4

5

f (1) f (2) f (3) f (4) 4 subintervals; each of length 1

x

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 - 1 interval 3 1, 4 is partitioned into subintervals of length = 0. as follows: 8 3 1, 1.4

3 1., 24

3 2, 2.4

3 2., 34

3 3, 3.4

3 3., 44

3 4, 4.4

3 4., 4

954

CHAPTER 13  A Preview of Calculus: The Limit, Derivative, and Integral of a Function

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u as the left endpoint of each subinterval, the area A under the graph of f 1x2 = x2 is approximated by

Figure 25 y

15

Area A ≈ f112 10.2 + f 11.2 10.2 + f122 10.2 + f 12.2 10.2

f (4.5) f (x ) ⫽ x 2

25

+ f132 10.2 + f 13.2 10.2 + f142 10.2 + f14.2 10.2

f (3.5) f (2.5) f (1.5)

= 3 f112 + f 11.2 + f122 + f12.2 + f 132 + f13.2 + f142 + f 14.2 4 10.2

= 3 1 + 2.2 + 4 + 6.2 + 9 + 12.2 + 16 + 20.24 10.2

5

= 3.

0

1

3

5

x

In general, we approximate the area under the graph of a function y = f1x2 from a to b as follows:

f (1) f (2) f (3) f (4) 8 subintervals; each of length 1/2

1. Partition the interval 3 a, b4 into n subintervals of equal length. The length ∆x of each subinterval is then

Figure 26 y f(u3) f(u2) f(u1)

a u1 u2 u3 Dx Dx Dx

r

f (un)

∆x =

y 5 f(x)

un b x Dx

b - a n

2. In each of these subintervals, pick a number u and evaluate the function f at each u. This results in n numbers u 1 , u 2 , c , u n and n functional values f1u 1 2, f1u 2 2, c , f1u n 2. 3. 'PSN
n rectangles with base equal to ∆x, the length of each subinterval, and with height equal to the functional value f1u i 2, i = 1, 2, c , n.
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 4. Add up the areas of the n rectangles. A1 + A2 + g + An = f 1u 1 2 ∆x + f 1u 2 2 ∆x + g + f1u n 2 ∆x n

= a f1u i 2 ∆x i=1

This number is the approximation to the area under the graph of f from a to b.

Definition of Area We have observed that the larger the number n of subintervals used, the better the approximation to the area. If we let n become unbounded, we obtain the exact area under the graph of f from a to b.

DEFINITION

Area under a Graph Let f denote a function whose domain is a closed interval 3 a, b4 and suppose f1x2 Ú 0 on 3 a, b 4 . Partition 3 a, b4 into n subintervals, each of length b - a ∆x = . In each subinterval, pick a number u i , i = 1, 2, c , n, and evaluate n f1u i 2.
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f1u i 2 ∆x and add them up, obtaining the sum n

a f1u i 2 ∆x

i=1

If the limit of this sum exists as n S q , that is, n

if

lim f1u i 2 ∆x n Sq a i=1

exists

it is defined as the area under the graph of f from a to b. If this limit exists, it is denoted by b

La

f1x2 dx

which is read as “the integral from a to b of f1x2.”

SECTION 13.5  The Area Problem; The Integral

955

2 Approximate Integrals Using a Graphing Utility Using a Graphing Utility to Approximate an Integral

EXAM PL E 3

Use a graphing utility to approximate the area under the graph of f1x2 = x2 from 1 UP

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L1

Solution

Figure 27

x2 dx

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manual for the proper keystrokes. 124 The value of the integral is . 3 In calculus, techniques are given for evaluating integrals to obtain exact answers.

r

13.5 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 4

1. The formula for the area A of a rectangle of length l and width w is A = lw . Q
"

2. a 12k + 12 = 24 . QQ
m

k=1

Concepts and Vocabulary 3. The integral from a to b of f 1x2 is denoted by 1a f(x) dx . b

4. The area under the graph of f from a to b is denoted by b 1a f(x) dx .

Skill Building In Problems 5 and 6, refer to the illustration. The interval 31, 34 is partitioned into two subintervals 31, 24 and 32, 34. y y ⫽ f(x)

f(3) ⫽ 4

f(2) ⫽ 2 f(1) ⫽ 1

A

x 1 2 3 f(1) ⫽ 1, f(2) ⫽ 2, f(3) ⫽ 4

5. Approximate the area A, choosing u as the left endpoint of each subinterval. 3 6. Approximate the area A, choosing u as the right endpoint of each subinterval. 6

In Problems 7 and 8, refer to the illustration. The interval 30, 84 is partitioned into four subintervals 30, 24, 32, 44, 34, 64, and 36, 84. y 10

5

y ⫽ f(x)

7. Approximate the area A, choosing u as the left endpoint of each subinterval.  8. Approximate the area A, choosing u as the right endpoint of each subinterval. 38

*9. The function f 1x2 = 3x is defined on the interval 30, 64. *(a) Graph f.
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*10. 3FQFBU
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f 1x2 = 4x.

11. The function f 1x2 = - 3x + 9 is defined on the interval 30, 34. *(a) Graph f.
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A under f from 0 to 3 as follows: *(b) Partition 30, 34 into three subintervals of equal length and choose u as the left endpoint of each subinterval. *(c) Partition 30, 34 into three subintervals of equal length and choose u as the right endpoint of each subinterval. *(d) Partition 30, 34 into six subintervals of equal length and choose u as the left endpoint of each subinterval.

2 4 6 8 x *Due to space restrictions, answers to these exercises may be found in the Answers f(0) ⫽ 10, f(2) ⫽ 6, f(4) ⫽ 7, f(6) ⫽ 5, f(8) ⫽ 1 in the back of the book.

956

CHAPTER 13  A Preview of Calculus: The Limit, Derivative, and Integral of a Function

*(e) Partition 30, 34 into six subintervals of equal length and choose u as the right endpoint of each subinterval. *(f) What is the actual area A?

*12. 3FQFBU
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f 1x2 = - 2x + 8.

In Problems 13–22, a function f is defined over an interval 3a, b4. (a) Graph f, indicating the area A under f from a to b. (b) Approximate the area A by partitioning 3a, b4 into four subintervals of equal length and choosing u as the left endpoint of each subinterval. (c) Approximate the area A by partitioning 3a, b4 into eight subintervals of equal length and choosing u as the left endpoint of each subinterval. (d) Express the area A as an integral. (e) Use a graphing utility to approximate the integral. *13. f 1x2 = x2 + 2,

30, 44

*15. f 1x2 = x3, 30, 44 1 *17. f 1x2 = , 31, 4 x *19. f 1x2 = e x, 3 - 1, 34 *21. f 1x2 = sin x,

30, p4

*14. f 1x2 = x2 - 4,

*16. f 1x2 = x3,

32, 64

31, 4

*18. f 1x2 = 1x,

30, 44

*20. f 1x2 = ln x,

33, 74

*22. f 1x2 = cos x,

c 0,

In Problems 23–30, an integral is given. (a) What area does the integral represent? (b) Provide a graph that illustrates this area. (c) Use a graphing utility to approximate this area.

p d 2

4

*23.

L0 

*25.

L2

13x + 12 dx

*24.

1x2 - 12 dx

*26.

p>2

*27.

1 - 2x + 72 dx

4

116 - x2 2 dx

L1 L0

p>4

sin x dx

L0

3

*28.

2

L-p>4

cos x dx

2e

e x dx

ln x dx Le L0 *31. Confirm the entries in Table 7. [Hint: 3FWJFX
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sequence.] *29.

*30.

32. Consider the function f 1x2 = 21 - x2 whose domain is the interval 3 - 1, 14. *(a) Graph f. *(b) Approximate the area under the graph of f from - 1 to 1 by dividing 3 - 1, 14 into five subintervals, each of equal length. *(c) Approximate the area under the graph of f from - 1 to 1 by dividing 3 - 1, 14 into ten subintervals, each of equal length. *(d) Express the area as an integral. *(e) Evaluate the integral using a graphing utility. *(f) What is the actual area?

Retain Your Knowledge Problems 33–36 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. f 1x + h2 - f 1x2 2 19 22 2 and *33. Graph the function f 1x2 = log 2 x. d 35. if f 1x2 = 2x + 3x + 1, find 34. c h 43 0 4 simplify. 4x + 2h + 3 1 2  6 0 16 34. If A = c d and B = c d , find AB. 36. 'JOE
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. 3 4 7 8 1 1x - 22 1x + 22 2 36.

1 4 1 x - 2 x + 2 (x + 2)2

‘Are You Prepared?’ Answers 1. A = lw

2. 24

Chapter Review Things to Know Limit (p. 924)

lim f 1x2 = N

xSc

Limit formulas (p. 929)

  As x gets closer to c, x ≠ c, the value of f gets closer to N.  

lim b = b

The limit of a constant is the constant.

lim x = c

The limit of x as x approaches c is c.

xSc xSc

Review Exercises

Limit properties (pp. 930, 932, 933)

 

lim 3f 1x2 + g1x2 4 = lim f 1x2 + lim g1x2

xSc

xSc

xSc

xSc

xSc

xSc

The limit of a sum equals the sum of the limits.

lim 3f 1x2 - g1x2 4 = lim f 1x2 - lim g1x2 lim 3f 1x2 # g1x2 4

xSc

lim c

xSc

f 1x2 g1x2

=

lim f 1x2

d =

The limit of a difference equals the difference of the limits.

lim f 1x2 # lim g1x2

xSc

xSc

lim g1x2

The limit of a product equals the product of the limits.

xSc

1lim g1x2 ≠ 02

The limit of a quotient equals the quotient of the limits, provided that the limit of the denominator is not zero.

xSc

xSc

957

lim 3f 1x2 4 n = [lim f 1x2]n

Provided lim f(x) exists, n Ú 2 an integer.

lim 2f 1x2 = 2 lim f(x)

Provided 2f(x) and 2 lim f(x) are both defined, n Ú 2 an integer.

xSc

xSc

xSc

n

n

n

xSc

xSc

n

xSc

Limit of a polynomial (p. 931)

Derivative of a function (p. 945)

lim P 1x2 = P 1c2, where P is a polynomial

f′ 1c2 = lim

xSc

f 1x2 - f 1c2 x - c

xSc

Continuous function (p. 938)

, provided that the limit exists

Area under a graph (p. 954) b

lim f 1x2 = f 1c2

xSc

La

n

f 1x2 dx = lim a f 1u i 2 ∆x, provided that the limit exists n Sq i=1

Objectives Section

You should be able to N

Example(s)

Review Exercises

13.1

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1, 2

 

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2 Determine whether a function is continuous (p. 938)

2, 3

m
m

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3

41(e), 42(e), 43(c), 44(c)

Review Exercises In Problems 1–11, find the limit. 1. lim 1x3 - 2x2 + x - 42 xS3

5. lim 1x2 + x + 32 3>2 xS1

x2 - 1 x S -1 x3 - 1

9. lim

2. lim 1x3 - x2 + 2 3

8

x S -2

- 343

x + x + 2 1 4 x2 - 9 x3 - 8 10. lim 3 x S 2 x - 2x2 + 4x - 8 2

27

0

3. lim 2x2 + 7 xS3

4. lim 2x2 - 4

4

x S -3

x - 4 4 x - 2 4 3 x - 2x + x - 2 11. lim 3 x S 2 x - 2x2 + 2x - 4 2

6. lim

7. lim

x S -1

8. lim

xS2

3 2

xS3

1

x - x + 6 x2 - 8x + 1 2

-

1 2

3 2

In Problems 12–15, determine whether f is continuous at c. 12. f 1x2 = x - 2x3 + 6 c = 2 x2 - 9 14. f 1x2 = c x - 3 6

Continuous

if x ≠ 3 if x = 3

c = 3

Continuous

13. f 1x2 =

x2 - 4 x + 2

x2 - 9 15. f 1x2 = c x - 3 -6

c = -2

No

if x ≠ 3 if x = 3

c = 3

Not continuous

958

CHAPTER 13  A Preview of Calculus: The Limit, Derivative, and Integral of a Function

In Problems 16–27, use the accompanying graph of y = f 1x2. 16. What is the domain of f?

y

{x - 6 … x 6 2 or 2 6 x 6  or  6 x … 6}

4

17. What is the range of f? All real numbers

2

18. 'JOE
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x-intercept(s), if any, of f. 1, 6 2

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y-intercept(s), if any, of f. 4 20. 'JOE
f 1 - 62 and f 1 - 42. 21. 'JOE
lim - f 1x2.

4

22. 'JOE
lim + f 1x2.

-2

x S -4 x S -4

4

6

x

f( - 6) = 2; f( - 4) = 1 23. 'JOE
lim- f 1x2. xS2

-q

24. 'JOE
lim+ f 1x2. xS2

q

25. Does lim f 1x2 exist? If it does, what is it? Does not exist 26. Is f continuous at 0? No 27. Is f continuous at 4? Yes xS0 x + 4 is continuous at c = - 4 and c = 4. Use limits to analyze the graph of R at c. *28. Discuss whether R 1x2 = 2 x - 16 x3 - 2x2 + 4x - 8 is undefined. Determine whether an asymptote or a hole *29. Determine where the rational function R 1x2 = x2 - 11x + 18 appears at such numbers. In Problems 30–32, find the slope of the tangent line to the graph of f at the given point. Graph f and the tangent line. *30. f 1x2 = 2x2 + 8x at 11, 102

*31. f 1x2 = x2 + 2x - 3 at 1 - 1, - 42

*32. f 1x2 = x3 + x2 at 12, 122

In Problems 33–35, find the derivative of each function at the number indicated. *33. f 1x2 = - 4x2 +  at 3

34. f 1x2 = x2 - 3x at 0

- 24

35. f 1x2 = 2x2 + 3x + 2 at 1

-3

7

In Problems 36 and 37, find the derivative of each function at the number indicated using a graphing utility. p 36. f 1x2 = 4x4 - 3x3 + 6x - 9 at - 2 - 18 37. f 1x2 = x3 tan x at 0.66621763 6

3

39. Instantaneous Rate of Change The following data represent the revenue R (in dollars) received from selling x wristwatches at Wilk’s Watch Shop.

12

where t is the elapsed time that the ball is in the air. The ball misses the rooftop on its way down and eventually strikes the ground. (a) When does the ball strike the ground? That is, how long is the ball in the air? 7 seconds (b) At what time t will the ball pass the rooftop on its way down? 6 seconds (c) What is the average speed of the ball from t = 0 to t = 2? 64 feet per second *(d) What is the instantaneous speed of the ball at time t? *(e) What is the instantaneous speed of the ball at t = 2? (f) When is the instantaneous speed of the ball equal to zero? At t = 3 sec (g) What is the instantaneous speed of the ball as it passes the rooftop on the way down? −96 feet per second (h) What is the instantaneous speed of the ball when it strikes the ground? −128 feet per second

6

s = s 1t2 = - 16t + 96t + 112 2

9

38. Instantaneous Speed of a Ball In physics it is shown that the height s of a ball thrown straight up with an initial speed of 96 ft/sec from a rooftop 112 feet high is

Wristwatches, x

Revenue, R

0

0

25

2340

40

3600

50

4375

90

6975

130

8775

160

9600

200

10,000

220

9900

250

9375

* B
'JOE
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BWFSBHF
SBUF
PG
DIBOHF
PG
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GSPN
x = 2 to x = 130 wristwatches. * C
'JOE
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BWFSBHF
SBUF
PG
DIBOHF
PG
SFWFOVF
GSPN
x = 2 to x = 90 wristwatches. * D
'JOE
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BWFSBHF
SBUF
PG
DIBOHF
PG
SFWFOVF
GSPN
x = 2 to x = 0 wristwatches. *(d) Using a graphing utility, find the quadratic function of best fit. *(e) Using the function found in part (d), determine the instantaneous rate of change of revenue at x = 2 wristwatches.

Chapter Test

40. The function f 1x2 = 2x + 3 is defined on the interval 30, 44. *(a) Graph f. In (b)–(e), approximate the area A under f from x = 0 to x = 4 as follows: *(b) Partition 30, 44 into four subintervals of equal length and choose u as the left endpoint of each subinterval. *(c) Partition 30, 44 into four subintervals of equal length and choose u as the right endpoint of each subinterval. *(d) Partition 30, 44 into eight subintervals of equal length and choose u as the left endpoint of each subinterval. *(e) Partition 30, 44 into eight subintervals of equal length and choose u as the right endpoint of each subinterval. *(f) What is the actual area A? 28

959

In Problems 41 and 42, a function f is defined over an interval 3a, b4. (a) Graph f, indicating the area A under f from a to b. (b) Approximate the area A by partitioning 3a, b4 into three subintervals of equal length and choosing u as the left endpoint of each subinterval. (c) Approximate the area A by partitioning 3a, b4 into six subintervals of equal length and choosing u as the left endpoint of each subinterval. (d) Express the area A as an integral. (e) Use a graphing utility to approximate the integral. *41. f 1x2 = 4 - x2, *42. f 1x2 =

1 , x2

3 - 1, 24

31, 44

In Problems 43 and 44, an integral is given. (a) What area does the integral represent? (b) Provide a graph that illustrates this area. (c) Use a graphing utility to approximate this area. 3

43.

L-1

1

19 - x2 2 dx

44.

L-1

e x dx

Chapter Test Prep Videos include step-by-step solutions to all chapter test exercises and can be found on this text’s Channel (search “SullivanPrecalcUC3e”).

Chapter Test

0x - 20

In Problems 1–6, find each limit. 1. lim

xS3

4. lim

1 - x2

x S -1

+ 3x - 5 2

x2 - 4x - 5 x3 + 1

-5

2.

-2

lim

xS2 +

5. lim

xS5

3x - 6

3 13x2 1x

x2 - 9 if x … 4 f 1x2 = c x + 3 kx + 5 if x 7 4

In Problems 8–12, use the accompanying graph of y = f 1x2. y

4

9. Find lim- f 1x2 xS3

10. Find lim f 1x2 xS - 2

x

-3 5 2

*11. Does lim f 1x2 exist? If so, what is it? If not, explain why not. xS1

135

6.

lim 27 - 3x

5

tan x 1 + cos2 x

2 3

x S -6

limp

xS

4

*12. Determine whether f is continuous at each of the following numbers. If it is not, explain why not. (a) x = - 2 (b) x = 1 (c) x = 3 (d) x = 4 *13. Determine where the rational function R 1x2 =

x3 + 6x2 - 4x - 24 x2 + 5x - 14

is undefined. Determine whether an asymptote or a hole appears at such numbers.

4

xS3

3.

- 22 2 4

7. Determine the value for k that will make the function continuous at c = 4. - 1

8. Find lim+ f 1x2

1 3

14. For the function f 1x2 = 4x2 - 11x - 3: (a) Find the derivative of f at x = 2. 5 (b) Find the equation of the tangent line to the graph of f at the point 12, - 92 . y = 5x - 19 *(c) Graph f and the tangent line.

15. The function f 1x2 = 216 - x2 is defined on the interval 30, 44 . *(a) Graph f. *(b) Partition 30, 44 into eight subintervals of equal length and choose u as the left endpoint of each subinterval. Use the partition to approximate the area under the graph of f and above the x-axis from x = 0 to x = 4. (c) Find the exact area of the region and compare it to the approximation in part (b). 4p ≈ 12.566

960

CHAPTER 13  A Preview of Calculus: The Limit, Derivative, and Integral of a Function

16. Write the integral that represents the shaded area. Do not attempt to evaluate. 4 ( - x2 + x + 3) dx L1 y

17. A particle is moving along a straight line according to some position function s 1t2 . The distance (in feet) of the particle, s, from its starting point after t seconds is given in the table. 'JOE
UIF
BWFSBHF
SBUF
PG
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t = 3 to t = 6 seconds. 1 3 ft/sec 3 t s

8 f (x) = –x 2 + 5x + 3 4

x 4

8

0

0

1

2.5

2

14

3

31

4

49

5

89

6

137

7

173

8

240

Chapter Projects 2. Graph Y1 = f 1t2, where f 1t2 represents the logistic growth function of best fit found in part (a). 3. Determine the instantaneous rate of growth of population in 1960 using the numerical derivative function on your graphing utility. 4. Use the result from part (c) to predict the population in 1961. What was the actual population in 1961? 5. Determine the instantaneous growth of population in 1970, 1980, 1990, 2000, and 2010. What is happening to the instantaneous growth rate as time passes? Is Malthus’ contention of a geometric growth rate accurate?

I.

World Population Thomas Malthus believed that “population, when unchecked, increases in a geometrical progression of such nature as to double itself every twenty-five years.” However, the growth of population is limited because the resources available to us are limited in supply. If Malthus’s conjecture were true, geometric growth of the world’s population would imply that Pt = r + 1, where r is the growth rate Pt - 1 1. Using world population data and a graphing utility, find the logistic growth function of best fit, treating the year as the independent variable. Let t = 0
SFQSFTFOU

t = 1 SFQSFTFOU

BOE
TP
PO
VOUJM
ZPV
IBWF
FOUFSFE
BMM
UIF
ZFBST
and the corresponding populations up to the current year.

6. Using the numerical derivative function on your graphing utility, graph Y2 = f′(t), where f′(t) represents the derivative of f 1t2 with respect to time. Y2 is the growth rate of the population at any time t. 7. Using the MAXIMUM function on your graphing utility, determine the year in which the growth rate of the population is largest. What is happening to the growth rate JO
UIF
ZFBST
GPMMPXJOH
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NBYJNVN
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UIJT
QPJOU
PO
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graph of Y1 = f 1t2.

8. Evaluate lim f 1t2. This limiting value is the carrying t Sq

capacity of Earth. What is the carrying capacity of Earth? 9. What do you think will happen if the population of Earth exceeds the carrying capacity? Do you think that agricultural output will continue to increase at the same rate as population growth? What effect will urban sprawl have on agricultural output?

The following projects are available on the Instructor’s Resource Center (IRC). II. Project at Motorola: Curing Rates Engineers at Motorola use calculus to find the curing rate of a sealant. III. Finding the Profit-maximizing Level of Output A manufacturer uses calculus to maximize profit.

Appendix

A

Review Outline A.1 A.2 A.3 A.4 A.5 A.6

Algebra Essentials Geometry Essentials Polynomials Factoring Polynomials Synthetic Division Rational Expressions

nth Roots; Rational Exponents Solving Equations Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications A.10 Interval Notation; Solving Inequalities A.11 Complex Numbers A.7 A.8 A.9

A.1 Algebra Essentials PREPARING FOR THIS BOOK  Before getting started, read “To the Student” at the beginning of this text on page xxiv. OBJECTIVES 1 2 3 4 5 6 7 8

Work with Sets (p. A1) Graph Inequalities (p. A4) Find Distance on the Real Number Line (p. A5) Evaluate Algebraic Expressions (p. A6) Determine the Domain of a Variable (p. A7) Use the Laws of Exponents (p. A7) Evaluate Square Roots (p. A9) Use a Calculator to Evaluate Exponents (p. A10)

1 Work with Sets A set is a well-defined collection of distinct objects. The objects of a set are called its elements. By well-defined, we mean that there is a rule that enables us to determine whether a given object is an element of the set. If a set has no elements, it is called the empty set, or null set, and is denoted by the symbol ∅. For example, the set of digits consists of the collection of numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. If we use the symbol D to denote the set of digits, then we can write D = 5 0, 1, 2, 3, 4, 5, 6, 7, 8, 96

In this notation, the braces 5 6 are used to enclose the objects, or elements, in the set. This method of denoting a set is called the roster method. A second way to denote a set is to use set-builder notation, where the set D of digits is written as D =

5

x c

 x is a digit6

c

c

c

5

c

Read as “D is the set of all x such that x is a digit”

A1

A2

APPENDIX A  Review

EX A MPL E 1

Using Set-Builder Notation and the Roster Method

(a) E = 5 x 0 x is an even digit6 = 5 0, 2, 4, 6, 86 (b) O = 5 x 0 x is an odd digit6 = 5 1, 3, 5, 7, 96

r

Because the elements of a set are distinct, we never repeat elements. For example, we would never write 5 1, 2, 3, 26 ; the correct listing is 5 1, 2, 36 . Because a set is a collection, the order in which the elements are listed does not matter. Thus, 5 1, 2, 36 , 5 1, 3, 26 , 5 2, 1, 36 , and so on, all represent the same set. If every element of a set A is also an element of a set B, then A is a subset of B, which is denoted A ⊆ B. If two sets A and B have the same elements, then A equals B, which is denoted A = B. For example, 5 1, 2, 36 ⊆ 5 1, 2, 3, 4, 56 and 5 1, 2, 36 = 5 2, 3, 16 . If A and B are sets, the intersection of A with B, denoted A ∩ B, is the set consisting of elements that belong to both A and B. The union of A with B, denoted A ∪ B, is the set consisting of elements that belong to either set A or set B or to both.

DEFINITION

EX A MPL E 2

Finding the Intersection and Union of Sets

Let A = 5 1, 3, 5, 86 , B = 5 3, 5, 76 , and C = 5 2, 4, 6, 86 . Find: (a) A ∩ B

Solution

(b) A ∪ B

(c) B ∩ 1A ∪ C2

(a) A ∩ B = 5 1, 3, 5, 86 ∩ 5 3, 5, 76 = 5 3, 56 (b) A ∪ B = 5 1, 3, 5, 86 ∪ 5 3, 5, 76 = 5 1, 3, 5, 7, 86 (c) B ∩ 1A ∪ C2 = 5 3, 5, 76 ∩ 1 5 1, 3, 5, 86 ∪ 5 2, 4, 6, 86 2 = 5 3, 5, 76 ∩ 5 1, 2, 3, 4, 5, 6, 86 = 5 3, 56

Now Work

PROBLEM

r

13

When working with sets, it is common practice to designate a universal set U, the set consisting of all the elements to be considered. With a universal set designated, elements of the universal set not found in a given set can be considered.

DEFINITION

EX A MPL E 3

If A is a set, then the complement of A, denoted A, is the set consisting of all the elements in the universal set that are not in A.*

Finding the Complement of a Set

If the universal set is U = 5 1, 2, 3, 4, 5, 6, 7, 8, 96 and if A = 5 1, 3, 5, 7, 96 , then A = 5 2, 4, 6, 86 .

r

Figure 1 Universal set

It follows from the definition of complement that A ∪ A = U and A ∩ A = ∅. Do you see why?

Now Work

B A C

PROBLEM

17

It is often helpful to draw pictures of sets. Such pictures, called Venn diagrams, represent sets as circles enclosed in a rectangle. The rectangle represents the universal set. Such diagrams often make it easier to visualize various relationships among sets. See Figure 1. Some books use the notation A′ or A c for the complement of A.

*

SECTION A.1  Algebra Essentials

A3

The Venn diagram in Figure 2(a) illustrates that A ⊆ B. The Venn diagram in Figure 2(b) illustrates that A and B have no elements in common—that is, A ∩ B = ∅. The sets A and B in Figure 2(b) are disjoint. Figure 2

Universal set

Universal set

B A

A

(a) A ⊆ B subset

B

(b) A ∩ B = ∅ disjoint sets

Figures 3(a), 3(b), and 3(c) use Venn diagrams to illustrate the definitions of intersection, union, and complement, respectively.

Figure 3

Universal set

A

Universal set

Universal set

B

A

(a) A B intersection

A

B

A

(c) A complement

(b) A B union

Real Numbers Real numbers are represented by symbols such as

25, 0,

Figure 4 p =

C d

C d

- 3,

1 , 2

5 - , 0.125, 4

22, p,

3 2 - 2, 0.666 c

The set of counting numbers, or natural numbers, contains the numbers in the set 5 1, 2, 3, 4, c 6 . (The three dots, called an ellipsis, indicate that the pattern continues indefinitely.) The set of integers contains the numbers in the set 5 c , - 3, - 2, - 1, 0, 1, 2, 3, c 6 . A rational number is a number that can be a expressed as a quotient of two integers, where the integer b cannot be 0. Examples b 3 5 0 2 a of rational numbers are , , , and - . Since = a for any integer a, every 4 2 4 3 1 integer is also a rational number. Real numbers that are not rational are called irrational. Examples of irrational numbers are 22 and p (the Greek letter pi), which equals the constant ratio of the circumference to the diameter of a circle. See Figure 4. Real numbers can be represented as decimals. Rational real numbers have decimal representations that either terminate or are nonterminating with repeating 3 2 blocks of digits. For example, = 0.75, which terminates; and = 0.666c, 4 3 in which the digit 6 repeats indefinitely. Irrational real numbers have decimal representations that neither repeat nor terminate. For example, 22 = 1.414213c and p = 3.14159. c In practice, the decimal representation of an irrational number is given as an approximation. We use the symbol ≈ (read as “approximately equal to”) to write 22 ≈ 1.4142 and p ≈ 3.1416.

A4

APPENDIX A  Review

Two frequently used properties of real numbers are given next. Suppose that a, b, and c are real numbers.

Distributive Property a # 1b + c2 = ab + ac

Zero-Product Property

In Words If a product equals 0, then one or both of the factors is 0.

If ab = 0, then either a = 0 or b = 0 or both equal 0. The Distributive Property can be used to remove parentheses: 2(x + 3) = 2x + 2 # 3 = 2x + 6

The Zero-Product Property will be used to solve equations (Section A.8). For example, if 2x = 0, then 2 = 0 or x = 0. Since 2 ≠ 0, it follows that x = 0.

The Real Number Line

Figure 5 Real number line 2 units Scale 1 unit O ⫺3

⫺2

⫺1 ⫺1–2 0 1–2 1 2 2

3␲

DEFINITION

The real numbers can be represented by points on a line called the real number line. There is a one-to-one correspondence between real numbers and points on a line. That is, every real number corresponds to a point on the line, and each point on the line has a unique real number associated with it. Pick a point on the line somewhere in the center, and label it O. This point, called the origin, corresponds to the real number 0. See Figure 5. The point 1 unit to the right of O corresponds to the number 1. The distance between 0 and 1 determines the scale of the number line. For example, the point associated with the number 2 is twice as far from O as 1. Notice that an arrowhead on the right end of the line indicates the direction in which the numbers increase. Points to the left of the origin correspond to the real numbers - 1, - 2, and so on. Figure 5 also shows 1 1 the points associated with the rational numbers - and and the points associated 2 2 with the irrational numbers 12 and p. The real number associated with a point P is called the coordinate of P, and the line whose points have been assigned coordinates is called the real number line.

Figure 6 O ⫺3

⫺2 ⫺ 3–2 ⫺1⫺ 1–2 0 1–2 1 3–2 2 Negative real numbers

Zero

Positive real numbers

Figure 7 a

(a) a ⬍ b

3

b

The real number line consists of three classes of real numbers, as shown in Figure 6. 1. The negative real numbers are the coordinates of points to the left of the origin O. 2. The real number zero is the coordinate of the origin O. 3. The positive real numbers are the coordinates of points to the right of the origin O.

Now Work a b (b) a ⫽ b

b

(c) a ⬎ b

PROBLEM

21

2 Graph Inequalities

a

An important property of the real number line follows from the fact that, given two numbers (points) a and b, either a is to the left of b, or a is at the same location as b, or a is to the right of b. See Figure 7. If a is to the left of b, then “a is less than b,” which is denoted a 6 b. If a is to the right of b, then “a is greater than b,” which is denoted a 7 b. If a is at the same location as b, then a = b. If a is either less than or equal to b, then write a … b.

SECTION A.1  Algebra Essentials

A5

Similarly, a Ú b means that a is either greater than or equal to b. Collectively, the symbols 6 , 7 , … , and Ú are called inequality symbols. Note that a 6 b and b 7 a mean the same thing. It does not matter whether we write 2 6 3 or 3 7 2. Furthermore, if a 6 b or if b 7 a, then the difference b - a is positive. Do you see why? An inequality is a statement in which two expressions are related by an inequality symbol. The expressions are referred to as the sides of the inequality. Statements of the form a 6 b or b 7 a are called strict inequalities, whereas statements of the form a … b or b Ú a are called nonstrict inequalities. The following conclusions are based on the discussion so far. a 7 0 is equivalent to a is positive a 6 0 is equivalent to a is negative The inequality a 7 0 is sometimes read as “a is positive.” If a Ú 0, then either a 7 0 or a = 0, and this is read as “a is nonnegative.”

Now Work

PROBLEMS

25

AND

35

Graphing Inequalities

EXAM PL E 4

(a) On the real number line, graph all numbers x for which x 7 4. (b) On the real number line, graph all numbers x for which x … 5.

Solution

Figure 8 –2 –1

0

1

2

3

4

5

6

7

r

Figure 9 ⫺2 ⫺1

(a) See Figure 8. Note that we use a left parenthesis to indicate that the number 4 is not part of the graph. (b) See Figure 9. Note that we use a right bracket to indicate that the number 5 is part of the graph.

Now Work 0

1

2

3

4

5

6

PROBLEM

41

7

3 Find Distance on the Real Number Line Figure 10 4 units

3 units

⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

DEFINITION

The absolute value of a number a is the distance from 0 to a on the number line. For example, - 4 is 4 units from 0, and 3 is 3 units from 0. See Figure 10. Thus the absolute value of - 4 is 4, and the absolute value of 3 is 3. A more formal definition of absolute value is given next. The absolute value of a real number a, denoted by the symbol 0 a 0 , is defined by the rules

0 a 0 = a if a Ú 0

and

0 a 0 = - a if a 6 0

For example, because - 4 6 0, the second rule must be used to get 0 - 4 0 = - 1 - 42 = 4.

EXAM PL E 5

Computing Absolute Value (a) 0 8 0 = 8

(b) 0 0 0 = 0

(c) 0 - 15 0 = - 1 - 152 = 15

r

Look again at Figure 10. The distance from - 4 to 3 is 7 units. This distance is the difference 3 - 1 - 42 , obtained by subtracting the smaller coordinate from

A6

APPENDIX A  Review

the larger. However, because 0 3 - 1 - 42 0 = 0 7 0 = 7 and 0 - 4 - 3 0 = 0 - 7 0 = 7, absolute value can be used to calculate the distance between two points without being concerned about which is smaller.

DEFINITION

If P and Q are two points on a real number line with coordinates a and b, respectively, then the distance between P and Q, denoted by d 1P, Q2, is d 1P, Q2 = 0 b - a 0 Since 0 b - a 0 = 0 a - b 0 , it follows that d 1P, Q2 = d 1Q, P2.

EX A MPL E 6

Finding Distance on a Number Line Let P, Q, and R be points on a real number line with coordinates - 5, 7, and - 3, respectively. Find the distance (a) between P and Q

Solution

(b) between Q and R

See Figure 11. Figure 11 P

R

Q

⫺5 ⫺4 ⫺3 ⫺2 ⫺1

0

1

2

3

4

5

6

7

d (P, Q) ⫽ ⏐7 ⫺ (⫺5)⏐ ⫽ 12 d (Q, R) ⫽ ⏐ ⫺3 ⫺ 7 ⏐ ⫽ 10

(a) d 1P, Q2 = 0 7 - 1 - 52 0 = 0 12 0 = 12 (b) d 1Q, R2 = 0 - 3 - 7 0 = 0 - 10 0 = 10

Now Work

PROBLEM

r

47

4 Evaluate Algebraic Expressions In algebra, letters such as x, y, a, b, and c are used to represent numbers. If the letter used represents any number from a given set of numbers, it is called a variable. A constant is either a fixed number, such as 5 or 13, or a letter that represents a fixed (possibly unspecified) number. Constants and variables are combined using the operations of addition, subtraction, multiplication, and division to form algebraic expressions. Examples of algebraic expressions include x + 3

3 1 - t

7x - 2y

To evaluate an algebraic expression, substitute a numerical value for each variable.

EX A MPL E 7

Evaluating an Algebraic Expression Evaluate each expression if x = 3 and y = - 1. 3y (d) 0 - 4x + y 0 2 - 2x (a) Substitute 3 for x and - 1 for y in the expression x + 3y. (a) x + 3y

Solution

(b) 5xy

(c)

x + 3y = 3 + 31 - 12 = 3 + 1 - 32 = 0

c

x = 3, y = - 1

SECTION A.1  Algebra Essentials

A7

(b) If x = 3 and y = - 1, then 5xy = 5132 1 - 12 = - 15 (c) If x = 3 and y = - 1, then 31 - 12 3y -3 -3 3 = = = = 2 - 2x 2 - 2132 2 - 6 -4 4 (d) If x = 3 and y = - 1, then

0 - 4x + y 0 = 0 - 4132 + 1 - 12 0 = 0 - 12 + 1 - 12 0 = 0 - 13 0 = 13 Now Work

PROBLEMS

49

AND

r

57

5 Determine the Domain of a Variable In working with expressions or formulas involving variables, the variables may be allowed to take on values from only a certain set of numbers. For example, in the formula for the area A of a circle of radius r, A = pr 2, the variable r is restricted to the positive real numbers (since a radius cannot be 0 or negative). In the expression 1 , the variable x cannot take on the value 0, since division by 0 is not defined. x

DEFINITION

EXAM PL E 8

The set of values that a variable may assume is called the domain of the variable.

Finding the Domain of a Variable The domain of the variable x in the expression 5 x - 2 is 5 x 0 x ≠ 26 , since if x = 2, the denominator becomes 0, which is not defined.

EXAM PL E 9

r

Circumference of a Circle In the formula for the circumference C of a circle of radius r, which is C = 2pr the domain of the variable r, representing the radius of the circle, is the set of positive real numbers, {r 0 r 7 0}. The domain of the variable C, representing the circumference of the circle, is also the set of positive real numbers, {C 0 C 7 0}.

r

In describing the domain of a variable, either set notation or words may be used, whichever is more convenient.

Now Work

PROBLEM

67

6 Use the Laws of Exponents Integer exponents provide a shorthand notation for representing repeated multiplications of a real number. For example, 23 = 2 # 2 # 2 = 8

34 = 3 # 3 # 3 # 3 = 81

A8

APPENDIX A  Review

DEFINITION

If a is a real number and n is a positive integer, then the symbol an represents the product of n factors of a. That is, an = a # a # g # a

c

(1)

n factors

WARNING Be careful with negatives and exponents. - 24 = - 1 # 24 = - 16 whereas ( - 2)4 = ( - 2)( - 2)( - 2)( - 2) = 16 䊏

In the definition, it is understood that a1 = a. In addition, a2 = a # a, a3 = a # a # a, and so on. In the expression an, a is called the base and n is called the exponent, or power. an is read as “a raised to the power n” or as “a to the nth power.” Usually, a2 is read as “a squared” and a3 is read as “a cubed.” In working with exponents, the operation of raising to a power is performed before any other operation. Here are some examples: 4 # 32 = 4 # 9 = 36

- 24 = - 16

22 + 32 = 4 + 9 = 13

5 # 32 + 2 # 4 = 5 # 9 + 2 # 4 = 45 + 8 = 53

Parentheses are used to indicate operations to be performed first. For example, 1 - 22 4 = 1 - 22 1 - 22 1 - 22 1 - 22 = 16

DEFINITION

12 + 32 2 = 52 = 25

If a ≠ 0, then a0 = 1

DEFINITION

If a ≠ 0 and if n is a positive integer, then 1 an

a -n =

Whenever you encounter a negative exponent, think “reciprocal.”

EX AM PL E 10

Evaluating Expressions Containing Negative Exponents (a) 2-3 =

1 1 = 3 8 2

(b) x -4 =

Now Work

PROBLEMS

1 -2 (c) a b = 5

1 x4

85

AND

1 1 a b 5

2

=

1 = 25 1 25

105

r

The following properties, called the Laws of Exponents, can be proved using the preceding definitions. In the list, a and b are real numbers, and m and n are integers.

THEOREM

Laws of Exponents aman = am + n am 1 = am - n = n - m an a

1am 2 n = amn if a ≠ 0

1ab2 n = anbn a n an a b = n b b

if b ≠ 0

SECTION A.1  Algebra Essentials

EX AM PL E 11

A9

Using the Laws of Exponents Write each expression so that all exponents are positive. (a)

Solution

(a)

x5 y -2 x3 y x5 y -2 x3 y

(b) ¢

x ≠ 0, y ≠ 0

=

(b) ¢

x -3 -2 ≤ x ≠ 0, y ≠ 0 3y -1

x2 x5 # y -2 5 - 3 # -2 - 1 2 -3 2# 1 y = x y = x = = x x3 y y3 y3

1x -3 2 -2 x -3 -2 x6 x6 9x6 ≤ = = = = 1 2 y2 3y -1 13y -1 2 -2 3-2 1y -1 2 -2 y 9

Now Work

PROBLEMS

87

AND

r

97

7 Evaluate Square Roots In Words

136 means “give me the nonnegative number whose square is 36.”

DEFINITION

A real number is squared when it is raised to the power 2. The inverse of squaring is finding a square root. For example, since 62 = 36 and 1 - 62 2 = 36, the numbers 6 and - 6 are square roots of 36. The symbol 1 , called a radical sign, is used to denote the principal, or nonnegative, square root. For example, 136 = 6. If a is a nonnegative real number, the nonnegative number b, such that b2 = a, is the principal square root of a and is denoted by b = 1a.

The following comments are noteworthy: 1. Negative numbers do not have square roots (in the real number system), because the square of any real number is nonnegative. For example, 1- 4 is not a real number, because there is no real number whose square is - 4. 2. The principal square root of 0 is 0, since 02 = 0. That is, 10 = 0. 3. The principal square root of a positive number is positive. 4. If c Ú 0, then 1 1c2 2 = c. For example, 1 122 2 = 2 and 1 132 2 = 3.

EX AM PL E 12

Evaluating Square Roots (a) 164 = 8

(b)

1 1 = A 16 4

(c)

1 11.4 2 2

r

= 1.4

Examples 12(a) and (b) are examples of square roots of perfect squares, since 1 1 2 64 = 82 and = a b . 16 4 Consider the expression 2a2. Since a2 Ú 0, the principal square root of a2 is defined whether a 7 0 or a 6 0. However, since the principal square root is nonnegative, an absolute value is needed to ensure the nonnegative result. That is, 2a2 = 0 a 0

EX AM PL E 13

where a is any real number

Simplifying Expressions Using Equation (2) (a) 2 12.32 2 = 0 2.3 0 = 2.3

Now Work

(b) 2 1 - 2.32 2 = 0 - 2.3 0 = 2.3

(2)

(c) 2x2 = 0 x 0

r PROBLEM

93

A10

APPENDIX A  Review

Calculators Calculators are finite machines. As a result, they are incapable of displaying decimals that contain a large number of digits. For example, some calculators are capable of displaying only eight digits. When a number requires more than eight digits, the calculator either truncates or rounds. To see how your calculator handles decimals, divide 2 by 3. How many digits do you see? Is the last digit a 6 or a 7? If it is a 6, your calculator truncates; if it is a 7, your calculator rounds. There are different kinds of calculators. An arithmetic calculator can only add, subtract, multiply, and divide numbers; therefore, this type is not adequate for this course. Scientific calculators have all the capabilities of arithmetic calculators and also contain function keys labeled ln, log, sin, cos, tan, xy, inv, and so on. Graphing calculators have all the capabilities of scientific calculators and contain a screen on which graphs can be displayed. In this book, the graphing calculator is optional. The symbol is shown whenever a graphing calculator is used.

8 Use a Calculator to Evaluate Exponents Your calculator has a caret key, ^ , which is used for computations involving exponents.

Exponents on a Graphing Calculator

EX AM PL E 14

Evaluate: 12.32 5

Solution

r

Figure 12 shows the result using a TI-84 graphing calculator. Figure 12

Now Work

PROBLEM

123

A.1 Assess Your Understanding Concepts and Vocabulary 1. A(n) variable is a letter used in algebra to represent any number from a given set of numbers.

5. True or False The product of two negative real numbers is always greater than zero. True

2. On the real number line, the real number zero is the coordinate of the origin .

6. True or False The distance between two distinct points on the real number line is always greater than zero. True

3. An inequality of the form a 7 b is called a(n) strict inequality.

7. True or False The absolute value of a real number is always greater than zero. False

4. In the expression 24, the number 2 is called the base and 4 is called the exponent or power .

8. True or False To multiply two expressions having the same base, retain the base and multiply the exponents. False

Skill Building In Problems 9–20, use U = universal set = 5 0, 1, 2, 3, 4, 5, 6, 7, 8, 96, A = 51, 3, 4, 5, 96 , B = 52, 4, 6, 7, 86 , and C = 51, 3, 4, 66 to find each set. *9. A ∪ B

13. 1A ∪ B2 ∩ C *17. A ∩ B

51, 3, 4, 66

10. A ∪ C

5 1, 3, 4, 5, 6, 9 6

18. B ∪ C

5 0, 5, 96

14. 1A ∩ B2 ∪ C

51, 3, 4, 66

11. A ∩ B 15. A

5 46

50, 2, 6, 7, 86

*19. A ∪ B

3 5 *21. On the real number line, label the points with coordinates 0, 1, - 1, , - 2.5, , and 0.25. 2 4 2 3 1 *22. On the real number line, label the points with coordinates 0, - 2, 2, - 1.5, , , and . 2 3 3 *Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

12. A ∩ C 16. C

51, 3, 46

50, 2, 5, 7, 8, 96

20. B ∩ C

50, 5, 96

A11

SECTION A.1  Algebra Essentials

In Problems 23–32, replace the question mark by 6, 7, or =, whichever is correct. 1 5 23. ? 0 7 24. 5 ? 6 6 25. - 1 ? - 2 7 26. - 3 ? 2 2 28. 22 ? 1.41

7

29.

1 ? 0.5 2

=

30.

1 ? 0.33 3

7

6

2 ? 0.67 3

31.

27. p ? 3.14

6

32.

7

1 ? 0.25 4

=

In Problems 33–38, write each statement as an inequality. 33. x is positive

x 7 0

36. y is greater than - 5

z 6 0

34. z is negative y 7 -5

35. x is less than 2 x … 1

37. x is less than or equal to 1

x 6 2 x Ú 2

38. x is greater than or equal to 2

In Problems 39–42, graph the numbers x on the real number line. *39. x Ú - 2

*40. x 6 4

*41. x 7 - 1

*42. x … 7

In Problems 43–48, use the given real number line to compute each distance. A

43. d1C, D2

1

44. d1C, A2

B

C

D

⫺4 ⫺3 ⫺2 ⫺1

0

1

45. d1D, E2

3

2

E 2

3

4

5

46. d1C, E2

6

47. d1A, E2

3

48. d1D, B2

6

2

In Problems 49–56, evaluate each expression if x = - 2 and y = 3. 49. x + 2y

53.

2x x - y

50. 3x + y

4 4 5

54.

x + y x - y

-3 -

51. 5xy + 2

1 5

55.

In Problems 57–66, find the value of each expression if x = 3 and y = - 2. 57. 0 x + y 0

62.

0y0 y

1

58. 0 x - y 0

5

63. 0 4x - 5y 0

-1

22

3x + 2y 2 + y

- 28

52. - 2x + xy

0

56.

59. 0 x 0 + 0 y 0

5

60. 0 x 0 - 0 y 0

64. 0 3x + 2y 0

5

65. 0 0 4x 0 - 0 5y 0 0

61.

1

2x - 3 y

0x0 x

-

-2 7 3

1

66. 3 0 x 0 + 2 0 y 0

2

13

In Problems 67–74, determine which of the value(s) (a) through (d), if any, must be excluded from the domain of the variable in each expression. (a) x = 3

(b) x = 1

(c) x = 0

(d) x = - 1

67.

x2 - 1 x

x = 0

68.

x2 + 1 x

x = 0

71.

x2 x + 1

None

72.

x3 x - 1

x = 1, x = - 1

2

2

x x - 9

69.

x = 3

2

x2 + 5x - 10 x3 - x

*73.

70.

x x + 9

74.

- 9x2 - x + 1 x3 + x

78.

x - 2 x - 6

None

2

x = 0

In Problems 75–78, determine the domain of the variable x in each expression. 75.

4 x - 5

5x x ≠ 56

76.

-6 x + 4

5 x x ≠ - 46

77.

x x + 4

5x x ≠ - 46

5x x ≠ 66

5 In Problems 79–82, use the formula C = 1F - 322 for converting degrees Fahrenheit into degrees Celsius to find the Celsius measure 9 of each Fahrenheit temperature. 79. F = 32 °

80. F = 212 °

0°C

81. F = 77 °

100°C

82. F = - 4°

25°C

- 20°C

In Problems 83–94, simplify each expression. 83. 1 - 42 2

16

89. 13-2 2 -1

9

84. - 42

- 16

90. 12-1 2 -3

8

85. 4-2 91. 225

1 16 5

86. - 4-2

-

92. 236

6

1 16

87. 3-6 # 34

88. 4-2 # 43

1 9

93. 21 - 42 2

4

4

94. 21 - 32 2

3

A12

APPENDIX A  Review

In Problems 95–104, simplify each expression. Express the answer so that all exponents are positive. Whenever an exponent is 0 or negative, assume that the base is not 0. y3 x2 y 3 x 1 x4 3 2 2 96. 1 - 4x2 2 -1 - 2 97. 1x2 y -1 2 98. 1x -1 y2 99. 95. 18x3 2 64x6 2 3 y 4x x xy4 y 100.

x -2 y xy

2

1 x3y

*101.

1 - 22 3 x4 1yz2 2 2

*102.

3

3 xy z

4x -2 1yz2 -1

103. ¢

3 4

2x y

In Problems 105–116, find the value of each expression if x = 2 and y = - 1. 3 105. 2xy -1 - 4 106. - 3x -1 y 107. x2 + y2 2 109. 1xy2 2

110. 1x + y2 2

4

111. 2x2

1

3x -1 4y

-1

25

114. 2x2 + 2y2

3

16662 4

12222 4

?

16x2 9y2

104. ¢

112.

2

115. xy

216x6 125y6

4

1 2x 2 2

116. yx

2

1

10; 0 8; 44

120. What is the value of 10.12 3 1202 3 ?

81

5x -2 -3 ≤ 6y -2

108. x2 y2

118. Find the value of the expression 4x3 + 3x2 - x + 2 if x = 1. What is the value if x = 2? 119. What is the value of

-2

5

1 2 117. Find the value of the expression 2x3 - 3x2 + 5x - 4 if x = 2. What is the value if x = 1?

113. 2x2 + y2

≤

8

In Problems 121–128, use a calculator to evaluate each expression. Round your answer to three decimal places. 121. 18.22 6 125. 1 - 2.82 6

122. 13.72 5

304,006.671

126. - 12.82 6

481.890

693.440 - 481.890

123. 16.12 -3

124. 12.22 -5

0.004

127. 1 - 8.112 -4

0.019

128. - 18.112 -4

0.000

- 0.000

Applications and Extensions In Problems 129–138, express each statement as an equation involving the indicated variables. 129. Area of a Rectangle  The area A of a rectangle is the product of its length l and its width w. A = lw

133. Area of an Equilateral Triangle    The area A of an 13 equilateral triangle is times the square of the length x of 4 one side.

l

A = A

w x

x

x

130. Perimeter of a Rectangle  The perimeter P of a rectangle is twice the sum of its length l and its width w. P = 21l + w2 131. Circumference of a Circle  The circumference C of a circle is the product of p and its diameter d. C = pd C

13 2 x 4

134. Perimeter of an Equilateral Triangle    The perimeter P of an equilateral triangle is 3 times the length x of one side. P = 3x 4 135. Volume of a Sphere  The volume V of a sphere is times p 3 times the cube of the radius r.

d

V = 132. Area of a Triangle  The area A of a triangle is one-half the product of its base b and its height h. A =

4 3 pr 3

r

1 bh 2

h b

136. Surface Area of a Sphere  The surface area S of a sphere is 4 times p times the square of the radius r. S = 4pr 2

SECTION A.2  Geometry Essentials

137. Volume of a Cube  The volume V of a cube is the cube of the length x of a side. V = x3

(a) Show that a voltage of 108 volts is acceptable. 2 … 5 *(b) Show that a voltage of 104 volts is not acceptable. 144. Foreign Voltage    In some countries, normal household voltage is 220 volts. It is acceptable for the actual voltage x to differ from normal by at most 8 volts. A formula that describes this is

x

0 x - 220 0 … 8

x

x

A13

138. Surface Area of a Cube  The surface area S of a cube is 6 times the square of the length x of a side. S = 6x2 139. Manufacturing Cost    The weekly production cost C of manufacturing x watches is given by the formula C = 4000 + 2x, where the variable C is in dollars. (a) What is the cost of producing 1000 watches? $6000 (b) What is the cost of producing 2000 watches? $8000 140. Balancing a Checkbook    At the beginning of the month, Mike had a balance of $210 in his checking account. During the next month, he deposited $80, wrote a check for $120, made another deposit of $25, and wrote two checks: one for $60 and the other for $32. He was also assessed a monthly service charge of $5. What was his balance at the end of the month? $98 In Problems 141 and 142, write an inequality using an absolute value to describe each statement.

0x - 40 Ú 6 142. x is more than 5 units from 2. 0 x - 2 0 7 5

(a) Show that a voltage of 214 volts is acceptable. 6 … 8 *(b) Show that a voltage of 209 volts is not acceptable. 145. Making Precision Ball Bearings    The FireBall Company manufactures ball bearings for precision equipment. One of its products is a ball bearing with a stated radius of 3 centimeters (cm). Only ball bearings with a radius within 0.01 cm of this stated radius are acceptable. If x is the radius of a ball bearing, a formula describing this situation is

0 x - 3 0 … 0.01

(a) Is a ball bearing of radius x = 2.999 acceptable? Yes (b) Is a ball bearing of radius x = 2.89 acceptable? No 146. Body Temperature    Normal human body temperature is 98.6°F. A temperature x that differs from normal by at least 1.5°F is considered unhealthy. A formula that describes this is

0 x - 98.6 0 Ú 1.5

141. x is at least 6 units from 4.

143. U.S. Voltage  In the United States, normal household voltage is 110 volts. It is acceptable for the actual voltage x to differ from normal by at most 5 volts. A formula that describes this is

0 x - 110 0 … 5

*(a) Show that a temperature of 97°F is unhealthy. *(b) Show that a temperature of 100°F is not unhealthy. 1 *147. Does equal 0.333? If not, which is larger? By how much? 3 *148. Does

2 equal 0.666? If not, which is larger? By how much? 3

Discussion and Writing 149. Is there a positive real number “closest” to 0? No 150. Number Game  I’m thinking of a number! It lies between 1 and 10; its square is rational and lies between 1 and 10. The number is larger than p. Correct to two decimal places (that is, truncated to two decimal places) name the number. Now think of your own number, describe it, and challenge a fellow student to name it. 3.14, 3.15, or 3.16

151. Write a brief paragraph that illustrates the similarities and differences between “less than” 1 6 2 and “less than or equal to” 1 … 2. 152. Give a reason why the statement 5 6 8 is true.

A.2 Geometry Essentials OBJECTIVES 1 Use the Pythagorean Theorem and Its Converse (p. A13) 2 Know Geometry Formulas (p. A15) 3 Understand Congruent Triangles and Similar Triangles (p. A16)

1 Use the Pythagorean Theorem and Its Converse

Figure 13 Hypotenuse c

b Leg 90°

a Leg

The Pythagorean Theorem is a statement about right triangles. A right triangle is one that contains a right angle—that is, an angle of 90°. The side of the triangle opposite the 90° angle is called the hypotenuse; the remaining two sides are called legs. In Figure 13, c represents the length of the hypotenuse, and a and b represent the lengths of the legs. Note the use of the symbol Pythagorean Theorem is stated next.

to show the 90° angle. The

A14

APPENDIX A  Review

PYTHAGOREAN THEOREM

In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. That is, in the right triangle shown in Figure 13, c 2 = a 2 + b2

(1)

A proof of the Pythagorean Theorem is given at the end of this section.

EX A MPL E 1

Finding the Hypotenuse of a Right Triangle In a right triangle, one leg has length 4 and the other has length 3. What is the length of the hypotenuse?

Solution

Since the triangle is a right triangle, use the Pythagorean Theorem with a = 4 and b = 3 to find the length c of the hypotenuse. From equation (1), c 2 = a 2 + b2 c 2 = 42 + 32 = 16 + 9 = 25 c = 125 = 5

Now Work P R O B L E M 13 The converse of the Pythagorean Theorem is also true.

CONVERSE OF THE PYTHAGOREAN THEOREM

r

In a triangle, if the square of the length of one side equals the sum of the squares of the lengths of the other two sides, the triangle is a right triangle. The 90° angle is opposite the longest side. A proof is given at the end of this section.

EX A MPL E 2

Verifying That a Triangle Is a Right Triangle Show that a triangle whose sides are of lengths 5, 12, and 13 is a right triangle. Identify the hypotenuse.

Solution

Figure 14

Square the lengths of the sides. 52 = 25

13 5 90° 12

EX A MPL E 3

122 = 144

132 = 169

Notice that the sum of the first two squares (25 and 144) equals the third square (169). That is, because 52 + 122 = 132, the triangle is a right triangle. The longest side, 13, is the hypotenuse. See Figure 14.

Now Work

PROBLEM

21

r

Applying the Pythagorean Theorem The tallest building in the world is Burj Khalifa in Dubai, United Arab Emirates, at 2717 feet and 163 floors. The observation deck is 1483 feet above ground level. How far can a person standing on the observation deck see (with the aid of a telescope)? Use 3960 miles for the radius of Earth. Source: Council on Tall Buildings and Urban Habitat

Solution From the center of Earth, draw two radii: one through Burj Khalifa and the other to the farthest point a person can see from the observation deck. See Figure 15. Apply the Pythagorean Theorem to the right triangle. 1483 Since 1 mile = 5280 feet, 1483 feet = mile. Thus 5280 1483 2 d 2 + 139602 2 = a3960 + b 5280 1483 2 d 2 = a3960 + b - 139602 2 ≈ 2224.58 5280 d ≈ 47.17 A person can see more than 47 miles from the observation deck.

SECTION A.2  Geometry Essentials

A15

Figure 15 1483 ft

d

3960 mi

r Now Work

PROBLEM

53

2 Know Geometry Formulas Certain formulas from geometry are useful in solving algebra problems. For a rectangle of length l and width w, w

Area = lw

Perimeter = 2l + 2w

l

For a triangle with base b and altitude h, h

Area = b

1 bh 2

For a circle of radius r (diameter d = 2r), d

r

Area = pr 2

Circumference = 2pr = pd

For a closed rectangular box of length l, width w, and height h, h w

l

Volume = lwh

Surface area = 2lh + 2wh + 2lw

For a sphere of radius r, r

Volume =

4 3 pr 3

Surface area = 4pr 2

For a right circular cylinder of height h and radius r,

r

h

Volume = pr 2 h

Now Work

EXAM PL E 4

PROBLEM

Surface area = 2pr 2 + 2prh 29

Using Geometry Formulas A Christmas tree ornament is in the shape of a semicircle on top of a triangle. How many square centimeters (cm) of copper is required to make the ornament if the height of the triangle is 6 cm and the base is 4 cm?

A16

APPENDIX A  Review

Solution Figure 16 4

See Figure 16. The amount of copper required equals the shaded area. This area is the sum of the areas of the triangle and the semicircle. The triangle has height h = 6 and base b = 4. The semicircle has diameter d = 4, so its radius is r = 2. Area = Area of triangle + Area of semicircle =

l6

1 1 1 1 bh + pr 2 = 142 162 + p # 22 2 2 2 2

b = 4; h = 6; r = 2

= 12 + 2p ≈ 18.28 cm2

r

About 18.28 cm2 of copper is required.

Now Work

PROBLEM

47

3 Understand Congruent Triangles and Similar Triangles In Words Two triangles are congruent if they have the same size and shape.

DEFINITION

Throughout the text we will make reference to triangles, beginning here, with a discussion of congruent triangles. According to dictionary.com, the word congruent means “coinciding exactly when superimposed.” For example, two angles are congruent if they have the same measure, and two line segments are congruent if they have the same length.

Two triangles are congruent if each pair of corresponding angles have the same measure and each pair of corresponding sides are the same length. In Figure 17, corresponding angles are equal and the lengths of the corresponding sides are equal: a = d, b = e, and c = f . Hence these triangles are congruent. Figure 17 Congruent Triangles 100⬚

a 30⬚

50⬚

100⬚

d

b

e 50⬚

30⬚

c

f

Actually, it is not necessary to verify that all three angles and all three sides are the same measure to determine whether two triangles are congruent.

Determining Congruent Triangles 1. Angle–Side–Angle Case Two triangles are congruent if the measures of two of the angles are equal and the lengths of the corresponding sides between the two angles are equal. For example, in Figure 18(a), the two triangles are congruent because the measures of two angles and the included side are equal. 2. Side–Side–Side Case Two triangles are congruent if the lengths of the corresponding sides of the triangles are equal. For example, in Figure 18(b), the two triangles are congruent because the lengths of the three corresponding sides are all equal. 3. Side–Angle–Side Case Two triangles are congruent if the lengths of two corresponding sides are equal and the measures of the angles between the two sides are the same. For example, in Figure 18(c), the two triangles are congruent because the lengths of two sides and the measure of the included angle are equal.

A17

SECTION A.2  Geometry Essentials

Figure 18 15

15 20

80⬚

80⬚

20

40⬚

8

40⬚

7

10

10 40⬚

40⬚

8

8

7

8 (b)

(a)

(c)

See the following definition to contrast congruent triangles with similar triangles.

DEFINITION

In Words Two triangles are similar if they have the same shape, but (possibly) different sizes.

Two triangles are similar if the measures of the corresponding angles are equal and the lengths of the corresponding sides are proportional. For example, the triangles in Figure 19 are similar because the corresponding angles are equal. In addition, the lengths of the corresponding sides are proportional because each side in the triangle on the right is twice as long as each corresponding side in the triangle on the left. That is, the ratio of the corresponding sides is a f d e constant: = = = 2. a c b Figure 19

80

d ⫽ 2a a

80

30

e ⫽ 2b

b 30

70

70 f ⫽ 2c

c

It is not necessary to verify that all three angles are equal and all three sides are proportional to determine whether two triangles are congruent.

Determining Similar Triangles 1. Angle–Angle Case Two triangles are similar if two of the corresponding angles have equal measures. For example, in Figure 20(a), the two triangles are similar because two angles have equal measures. 2. Side–Side–Side Case Two triangles are similar if the lengths of all three sides of each triangle are proportional. For example, in Figure 20(b), the two triangles are similar because 10 5 6 1 = = = 30 15 18 3 3. Side–Angle–Side Case Two triangles are similar if two corresponding sides are proportional and the angles between the two sides have equal measure. For example, in Figure 20(c), the two triangles are similar because 4 12 2 = = and the angles between the sides have equal measure. 6 18 3 Figure 20 15 80⬚ 80⬚

30

5 10

18

(a)

120⬚

120⬚

6

35⬚

35⬚

18

12

6

4 (b)

(c)

A18

APPENDIX A  Review

EX A MPL E 5

Using Similar Triangles Given that the triangles in Figure 21 are similar, find the missing length x and the angles A, B, and C. Figure 21 60⬚

6

x

30⬚

B

3

C

5 A

90⬚

Solution

Because the triangles are similar, corresponding angles are equal, so A = 90°, B 3 and C = 30°. Also, the corresponding sides are proportional. That is, = 5 solve this equation for x. 3 5 3 5x # 5 3x x

6 x

=

= 5x #

6 x

= 30 = 10

Multiply both sides by 5x. Simplify. Divide both sides by 3.

r

The missing length is 10 units.

Now Work

= 60°, 6 . We x

PROBLEM

41

Proof of the Pythagorean Theorem    Begin with a square, each side of length a + b. In this square, form four right triangles, each having legs equal in length to a and b. See Figure 22. All these triangles are congruent (two sides and their included angle are equal). As a result, the hypotenuse of each is the same, say c, and the pink shading in Figure 22 indicates a square with an area equal to c 2. Figure 22 b

a

1

Area = –2 ab

1

Area = –2 ab

a

c c

b

Area = c 2 c

b

c

1

Area = –2 ab a

a

1

Area = –2 ab

b

The area of the original square with sides a + b equals the sum of the areas of the 1 four triangles (each of area ab) plus the area of the square with side c. That is, 2

Figure 23 x b

a (a)

1a + b2 2 =

1 1 1 1 ab + ab + ab + ab + c 2 2 2 2 2 a2 + 2ab + b2 = 2ab + c 2 a 2 + b2 = c 2 The proof is complete.

c

b

a 2

2

(b) c = a + b

2

■

Proof of the Converse of the Pythagorean Theorem    Begin with two triangles: one a right triangle with legs a and b and the other a triangle with sides a, b, and c for which c 2 = a2 + b2. See Figure 23. By the Pythagorean Theorem, the length x of the third side of the first triangle is x 2 = a 2 + b2

SECTION A.2  Geometry Essentials

A19

But c 2 = a2 + b2. Hence, x2 = c 2 x = c The two triangles have sides with the same length and are therefore congruent. This means corresponding angles are equal, so the angle opposite side c of the second triangle equals 90°. The proof is complete. ■

A.2 Assess Your Understanding Concepts and Vocabulary 9. True or False The triangles shown are similar. True

1. A(n) right triangle is one that contains an angle of 90 degrees. The longest side is called the hypotenuse . 2. For a triangle with base b and altitude h, a formula for the 1 A = bh area A is . 2

25⬚

25⬚

3. The formula for the circumference C of a circle of radius r is C = 2pr . 100⬚

4. Two triangles are similar if corresponding angles are equal and the lengths of the corresponding sides are proportional. 5. True or False In a right triangle, the square of the length of the longest side equals the sum of the squares of the lengths of the other two sides. True

100⬚

10. True or False The triangles shown are similar. False

6. True or False The triangle with sides of lengths 6, 8, and 10 is a right triangle. True 4 7. True or False The volume of a sphere of radius r is pr 2. 3 False

4

8. True or False The triangles shown are congruent. True

3 120⬚ 120⬚

10

3

2 30 30 29 29 10

Skill Building In Problems 11–16, the lengths of the legs of a right triangle are given. Find the hypotenuse. 11. a = 5, b = 12 14. a = 4, b = 3

13 5

12. a = 6, b = 8 15. a = 7, b = 24

10 25

13. a = 10, b = 24

26

16. a = 14, b = 48

50

In Problems 17–24, the lengths of the sides of a triangle are given. Determine which are right triangles. For those that are, identify the hypotenuse. 17. 3, 4, 5 Yes; 5

18. 6, 8, 10 Yes; 10

19. 4, 5, 6

No

20. 2, 2, 3

No

21. 7, 24, 25 Yes; 25

22. 10, 24, 26 Yes; 26

23. 6, 4, 3

No

24. 5, 4, 7

No

2

25. Find the area A of a rectangle with length 4 inches and width 2 inches. 8 in

26. Find the area A of a rectangle with length 9 centimeters and width 4 centimeters. 36 cm2 27. Find the area A of a triangle with height 4 inches and base 2 inches. 4 in2 28. Find the area A of a triangle with height 9 centimeters and base 4 centimeters. 18 cm2 29. Find the area A and circumference C of a circle of radius 5 meters. A = 25p m2; C = 10p m 30. Find the area A and circumference C of a circle of radius 2 feet. A = 4p ft 2; C = 4p ft

A20

APPENDIX A  Review

*31. Find the volume V and surface area S of a rectangular box with length 8 feet, width 4 feet, and height 7 feet. *32. Find the volume V and surface area S of a rectangular box with length 9 inches, width 4 inches, and height 8 inches. 256 p cm3; S = 64p cm2 33. Find the volume V and surface area S of a sphere of radius 4 centimeters. V = 3 3 34. Find the volume V and surface area S of a sphere of radius 3 feet. V = 36p ft ; S = 36p ft 2 *35. Find the volume V and surface area S of a right circular cylinder with radius 9 inches and height 8 inches. *36. Find the volume V and surface area S of a right circular cylinder with radius 8 inches and height 9 inches. In Problems 37–40, find the area of the shaded region. 37.

38.

2

39.

2

40. 2

2 2

2

p square units

4- p square units

2

2

2p - 4 square units

2p square units

In Problems 41–44, each pair of triangles is similar. Find the missing length x and the missing angles A, B, and C. *41. 60⬚

*42.

2

90⬚

30°

16

*43.

*44. 20 60⬚

75°

4

125⬚

50⬚

95⬚

12

30⬚

10

50 75°

45

25⬚

x

B

5⬚ x A

A

B

8

6

A

C 30

C

A

8 B

B

x C x C

Applications and Extensions 45. How many feet does a wheel with a diameter of 16 inches travel after four revolutions? About 16.8 ft *46. How many revolutions will a circular disk with a diameter of 4 feet have completed after it has rolled 20 feet? 47. In the figure shown, ABCD is a square, with each side of length 6 feet. The width of the border (shaded portion) between the outer square EFGH and ABCD is 2 feet. Find the area of the border. 64 ft2

48. Refer to the figure. Square ABCD has an area of 100 square feet; square BEFG has an area of 16 square feet. What is the area of the triangle CGF? 12 ft2 A G

D E

E F

C

F A

B 6 ft

D H

B

2 ft C G

49. Architecture    A Norman window consists of a rectangle surmounted by a semicircle. Find the area of the Norman window shown in the illustration. How much wood frame is needed to enclose the window? 24 + 2p ≈ 30.28 ft 2; 16 + 2p ≈ 22.28 ft

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

6'

4'

SECTION A.2  Geometry Essentials

50. Construction  A circular swimming pool that is 20 feet in diameter is enclosed by a wooden deck that is 3 feet wide. What is the area of the deck? How much fence is required to enclose the deck?

3' 20'

69 p ≈ 216.77 ft 2; 26 p ≈ 81.68 ft 51. How Tall Is the Great Pyramid?    The ancient Greek philosopher Thales of Miletus is reported on one occasion to have visited Egypt and calculated the height of the Great Pyramid of Cheops by means of shadow reckoning. Thales knew that each side of the base of the pyramid was 252 paces and that his own height was 2 paces. He measured the length of the pyramid’s shadow to be 114 paces and determined the length of his shadow to be 3 paces. See the illustration. Using similar triangles, determine the height of the Great Pyramid in terms of the number of paces. 160 paces*

A21

52. The Bermuda Triangle    Karen is doing research on the Bermuda Triangle, which she defines roughly by Hamilton, Bermuda; San Juan, Puerto Rico; and Fort Lauderdale, Florida. On her atlas Karen measures the straight-line distances from Hamilton to Fort Lauderdale, Fort Lauderdale to San Juan, and San Juan to Hamilton to be approximately 57 millimeters (mm), 58 mm, and 53.5 mm, respectively. If the actual distance from Fort Lauderdale to San Juan is 1046 miles, approximate the actual distances from San Juan to Hamilton and from Hamilton to Fort Lauderdale.

Source: Reprinted with permission from Red River Press, Inc. Winnipeg, Canada. *San Juan to Hamilton: ≈ 965 mi; Hamilton to Fort Lauderdale: ≈ 1028 mi

Source: Diggins, Julia E., illustrations by Corydon Bell, String, Straightedge and Shadow: The Story of Geometry, 2003 Whole Spirit Press, http://wholespiritpress.com.

In Problems 53–55, use the facts that the radius of Earth is 3960 miles and 1 mile = 5280 feet. 53. How Far Can You See?  The conning tower of the U.S.S. Silversides, a World War II submarine now permanently stationed in Muskegon, Michigan, is approximately 20 feet above sea level. How far can you see from the conning tower? About 5.477 mi* 54. How Far Can You See?    A person who is 6 feet tall is standing on the beach in Fort Lauderdale, Florida, and looks out onto the Atlantic Ocean. Suddenly, a ship appears on the horizon. How far is the ship from shore? About 3.0 mi*

Discussion and Writing 57. You have 1000 feet of flexible pool siding and wish to construct a swimming pool. Experiment with rectangular-shaped pools with perimeters of 1000 feet. How do their areas vary? What is the shape of the rectangle with the largest area? Now compute the area enclosed by a circular pool with a perimeter (circumference) of 1000 feet. What would be your choice of shape for the pool? If rectangular, what is your preference for dimensions? Justify your choice. If your only consideration is to have a pool that encloses the most area, what shape should you use?

*55. How Far Can You See?  The deck of a destroyer is 100 feet above sea level. How far can a person see from the deck? How far can a person see from the bridge, which is 150 feet above sea level? *56. Suppose that m and n are positive integers with m 7 n. If a = m2 - n2, b = 2mn, and c = m2 + n2, show that a, b, and c are the lengths of the sides of a right triangle. (This formula can be used to find the sides of a right triangle that are integers, such as 3, 4, 5; 5, 12, 13; and so on. Such triplets of integers are called Pythagorean triples.)

A22

APPENDIX A  Review

58. The Gibb’s Hill Lighthouse, Southampton, Bermuda, in operation since 1846, stands 117 feet high on a hill 245 feet high, so its beam of light is 362 feet above sea level. A brochure states that the light itself can be seen on the horizon about 26 miles from the lighthouse. Verify the accuracy of this information. The brochure further states ships 40 miles away can see the light and that planes flying at 10,000 feet can see it 120 miles away. Verify the accuracy of these statements. What assumption did the brochure make about the height of the ship?

120 miles 40 miles

A.3 Polynomials OBJECTIVES 1 2 3 4 5 6 7

Recognize Monomials (p. A22) Recognize Polynomials (p. A23) Add and Subtract Polynomials (p. A24) Multiply Polynomials (p. A25) Know Formulas for Special Products (p. A26) Divide Polynomials Using Long Division (p. A27) Work with Polynomials in Two Variables (p. A30)

We have described algebra as a generalization of arithmetic in which letters are used to represent real numbers. From now on, we shall use the letters at the end of the alphabet, such as x, y, and z, to represent variables and shall use the letters at the beginning of the alphabet, such as a, b, and c, to represent constants. In the expressions 3x + 5 and ax + b, it is understood that x is a variable and that a and b are constants, even though the constants a and b are unspecified. As you will find out, the context usually makes the intended meaning clear.

1 Recognize Monomials DEFINITION

A monomial in one variable is the product of a constant and a variable raised to a nonnegative integer power. A monomial is of the form

NOTE The nonnegative integers are the integers 0, 1, 2, 3, …. 䊏

axk where a is a constant, x is a variable, and k Ú 0 is an integer. The constant a is called the coefficient of the monomial. If a ≠ 0, then k is called the degree of the monomial.

EX AMPL E 1

Examples of Monomials Monomial

Coefficient

Degree

6

2

(b) - 22x3

- 22

3

(c) 3

3

0

Since 3 = 3 # 1 = 3x 0, x ≠ 0

(d) - 5x

-5

1

Since - 5x = - 5x 1

(e) x4

1

4

(a) 6x

2

Since x 4 = 1 # x 4

r

SECTION A.3  Polynomials

EXAM PL E 2

A23

Examples of Nonmonomial Expressions 1 1 (a) 3x1>2 is not a monomial, since the exponent of the variable x is , and is not a 2 2 nonnegative integer. (b) 4x -3 is not a monomial, since the exponent of the variable x is - 3, and - 3 is not a nonnegative integer.

r

Now Work

PROBLEM

7

2 Recognize Polynomials Two monomials with the same variable raised to the same power are called like terms. For example, 2x4 and - 5x4 are like terms. In contrast, the monomials 2x3 and 2x5 are not like terms. Like terms may be added or subtracted using the Distributive Property. For example, 2x2 + 5x2 = 12 + 52x2 = 7x2 and 8x3 - 5x3 = 18 - 52x3 = 3x3

The sum or difference of two monomials having different degrees is called a binomial. The sum or difference of three monomials with three different degrees is a trinomial. For example, x2 - 2 is a binomial x3 - 3x + 5 is a trinomial 2x2 + 5x2 + 2 = 7x2 + 2 is a binomial

DEFINITION

A polynomial in one variable is an algebraic expression of the form an xn + an - 1 xn - 1 + g + a1 x + a0

(1)

where an , an - 1 , c , a1 , a0 are constants,* called the coefficients of the polynomial, n Ú 0 is an integer, and x is a variable. If an ≠ 0, it is called the leading coefficient, and n is called the degree of the polynomial.

In Words A polynomial is a sum of monomials.

The monomials that make up a polynomial are called its terms. If all the coefficients are 0, the polynomial is called the zero polynomial, which has no degree. Polynomials are usually written in standard form, beginning with the nonzero term of highest degree and continuing with terms in descending order according to degree. If a power of x is missing, it is because its coefficient is zero.

EXAM PL E 3

Examples of Polynomials Polynomial

Coefficients

Degree

- 8x + 4x - 6x + 2

- 8, 4, - 6, 2

3

3, 0, - 5

2

1, - 2, 8

2

5, 22

1

3

0

0

No degree

3

2

3x2 - 5 = 3x2 + 0 # x + 1 - 52

8 - 2x + x = 1 # x + 1 - 22x + 8 2

2

5x + 22 = 5x1 + 22

3 = 3 # 1 = 3 # x0 0

*The notation an is read as “a sub n.” The number n is called a subscript and should not be confused with an exponent. Subscripts are used to distinguish one constant from another when a large or undetermined number of constants are required.

r

A24

APPENDIX A  Review

Although we have been using x to represent the variable, letters such as y or z are also commonly used. 3x4 - x2 + 2 is a polynomial (in x) of degree 4. 9y3 - 2y2 + y - 3 is a polynomial (in y) of degree 3. z5 + p is a polynomial (in z) of degree 5. Algebraic expressions such as 1 x

and

x2 + 1 x + 5

1 are not polynomials. The first is not a polynomial because = x -1 has an exponent x that is not a nonnegative integer. Although the second expression is the quotient of two polynomials, the polynomial in the denominator has degree greater than 0, so the expression cannot be a polynomial.

Now Work

PROBLEM

17

3 Add and Subtract Polynomials Polynomials are added and subtracted by combining like terms.

EX A MPL E 4

Adding Polynomials Find the sum of the polynomials: 8x3 - 2x2 + 6x - 2 and 3x4 - 2x3 + x2 + x

Solution

The sum can be found using a horizontal or vertical format. Horizontal Addition: The idea here is to group the like terms and then combine them. 18x3 - 2x2 + 6x - 22 + 13x4 - 2x3 + x2 + x2 = 3x4 + 18x3 - 2x3 2 + 1 - 2x2 + x2 2 + 16x + x2 - 2 = 3x4 + 6x3 - x2 + 7x - 2

Vertical Addition: The idea here is to vertically line up the like terms in each polynomial and then add the coefficients. x4

x3

+

x2

x1

x0

8x - 2x + 6x - 2 3x4 - 2x3 + x2 + x 3x4 + 6x3 - x2 + 7x - 2 3

2

r

Subtraction of polynomials can be performed horizontally or vertically as well.

EX A MPL E 5

Solution

Subtracting Polynomials

Find the difference: 13x4 - 4x3 + 6x2 - 12 - 12x4 - 8x2 - 6x + 52 Horizontal Subtraction: 13x4 - 4x3 + 6x2 - 12 - 12x4 - 8x2 - 6x + 52 = 3x4 - 4x3 + 6x2 - 1 + 1 - 2x4 + 8x2 + 6x - 52

w Be sure to change the sign of each term in the second polynomial.

= 13x4 - 2x4 2 + 1 - 4x3 2 + 16x2 + 8x2 2 + 6x + 1 - 1 - 52 c Group like terms.

= x4 - 4x3 + 14x2 + 6x - 6

A25

SECTION A.3  Polynomials

Vertical Subtraction: We line up like terms, change the sign of each coefficient of the second polynomial, and add vertically. x4

-

x3

x2

x1

x0

3x - 4x + 6x - 1 = 3 2x4 - 8x2 - 6x + 54 = + 4

3

2

x4

x3

x2

x1

x0

3x - 4x + 6x - 1 - 2x4 + 8x2 + 6x - 5 x4 - 4x3 + 14x2 + 6x - 6 4

3

2

r

The choice of which of these methods to use for adding and subtracting polynomials is left to you. To save space, we most often use the horizontal format.

Now Work

PROBLEM

29

COMMENT Vertical subtraction will be used when we divide polynomials.

■

4 Multiply Polynomials Two monomials may be multiplied using the Laws of Exponents and the Commutative and Associative Properties. For example, 12x3 2 # 15x4 2 = 12 # 52 # 1x3 # x4 2 = 10x3 + 4 = 10x7 Products of polynomials are found by repeated use of the Distributive Property and the Laws of Exponents. Again, you have a choice of a horizontal or vertical format.

EXAM PL E 6

Solution

Multiplying Polynomials

Find the product: 12x + 52 1x2 - x + 22 Horizontal Multiplication: 12x + 52 1x2 - x + 22 = 2x1x2 - x + 22 + 51x2 - x + 22 c Distributive Property

= 12x # x2 - 2x # x + 2x # 22 + 15 # x2 - 5 # x + 5 # 22 c

Distributive Property

= 12x3 - 2x2 + 4x2 + 15x2 - 5x + 102 c

Law of Exponents

= 2x3 + 3x2 - x + 10 c Combine like terms.

Vertical Multiplication: The idea here is very much like multiplying a two-digit number by a three-digit number. x2 -

x 2x 5x2 - 5x 3 1 + 2 2x - 2x2 + 4x 2x3 + 3x2 - x

Now Work

+ 2 + 5 + 10

This line is 5(x 2 - x + 2). This line is 2x(x 2 - x + 2).

+ 10

Sum of the above two lines

r PROBLEM

45

A26

APPENDIX A  Review

5 Know Formulas for Special Products Certain products, called special products, occur frequently in algebra. They can be calculated using the FOIL (First, Outer, Inner, Last) method of multiplying two binomials. Outer First

(ax + b)(cx + d) = ax(cx + d) + b(cx + d) b

Inner

Last b

Outer

= ax # cx + ax # d

b

First

b

Inner Last

+ b # cx + b # d

= acx2 + adx + bcx + bd = acx2 + (ad + bc)x + bd

EX A MPL E 7

Using FOIL F

(a) (b) (c) (d) (e)

O

I

L

1x - 32 1x + 32 = x2 + 3x - 3x - 9 = x2 - 9 1x + 22 2 = 1x + 22 1x + 22 = x2 + 2x + 2x + 4 1x - 32 2 = 1x - 32 1x - 32 = x2 - 3x - 3x + 9 1x + 32 1x + 12 = x2 + x + 3x + 3 = x2 + 4x + 12x + 12 13x + 42 = 6x2 + 8x + 3x + 4 = 6x2 +

Now Work

PROBLEMS

47

AND

= x2 + 4x + 4 = x2 - 6x + 9 3 11x + 4

r

55

Some products have been given special names because of their form. The following special products are based on Examples 7(a), (b), and (c).

Difference of Two Squares 1a - b2 1a + b2 = a2 - b2

(2)

Squares of Binomials, or Perfect Squares 1a + b2 2 = a2 + 2ab + b2

(3a)

1a - b2 2 = a2 - 2ab + b2

EX A MPL E 8

(3b)

Using Special Product Formulas

(a) 1x - 52 1x + 52 = x2 - 52 = x2 - 25 (b) 1x + 72 2 = x2 + 2 # 7 # x + 72 = x2 + 14x + 49 (c) 12x + 12 2 = 12x2 2 + 2 # 1 # 2x + 12 = 4x2 + 4x + 1

Difference of two squares Square of a binomial Note that we used 2x in place of a in formula (3a).

(d) 13x - 42 2 = 13x2 2 - 2 # 4 # 3x + 42 = 9x2 - 24x + 16 Replace a by 3x in formula (3b).

Now Work

PROBLEMS

65, 67,

AND

69

Let’s look at some more examples that lead to general formulas.

r

SECTION A.3  Polynomials

EXAM PL E 9

A27

Cubing a Binomial

(a) 1x + 22 3 = 1x + 22 1x + 22 2 = 1x + 22 1x2 + 4x + 42

Formula (3a)

= x1x + 4x + 42 + 21x + 4x + 42 = 1x3 + 4x2 + 4x2 + 12x2 + 8x + 82 = x3 + 6x2 + 12x + 8 2

2

(b) 1x - 12 3 = 1x - 12 1x - 12 2 = 1x - 12 1x2 - 2x + 12

Formula (3b)

= x1x - 2x + 12 - 11x - 2x + 12 = 1x3 - 2x2 + x2 + 1 - x2 + 2x - 12 = x3 - 3x2 + 3x - 1 2

2

r

Cubes of Binomials, or Perfect Cubes 1a + b2 3 = a3 + 3a2b + 3ab2 + b3 1a - b2 3 = a3 - 3a2b + 3ab2 - b3

Now Work

EX AM PL E 10

EX AM PL E 11

PROBLEM

(4a) (4b)

85

Forming the Difference of Two Cubes

1x - 12 1x2 + x + 12 = x1x2 + x + 12 - 11x2 + x + 12 = x3 + x2 + x - x2 - x - 1 = x3 - 1

r

Forming the Sum of Two Cubes

1x + 22 1x2 - 2x + 42 = x1x2 - 2x + 42 + 21x2 - 2x + 42 = x3 - 2x2 + 4x + 2x2 - 4x + 8 = x3 + 8

r

Examples 10 and 11 lead to two more special products.

Difference of Two Cubes 1a - b2 1a2 + ab + b2 2 = a3 - b3

(5)

1a + b2 1a2 - ab + b2 2 = a3 + b3

(6)

Sum of Two Cubes

6 Divide Polynomials Using Long Division The procedure for dividing two polynomials is similar to the procedure for dividing two integers.

A28

APPENDIX A  Review

EX AM PL E 12

Dividing Two Integers Divide 842 by 15.

Solution Divisor S

Thus,

56 15) 842 75 92 90 2

d Quotient d Dividend

d 5 # 15 (subtract) d 6 # 15 (subtract) d Remainder

842 2 = 56 + . 15 15

r

In the long-division process detailed in Example 12, the number 15 is called the divisor, the number 842 is called the dividend, the number 56 is called the quotient, and the number 2 is called the remainder. To check the answer obtained in a division problem, multiply the quotient by the divisor and add the remainder. The answer should be the dividend. 1Quotient2 1Divisor2 + Remainder = Dividend For example, check the results obtained in Example 12 as follows: 1562 1152 + 2 = 840 + 2 = 842

To divide two polynomials, first write each polynomial in standard form. The process then follows a pattern similar to that of Example 12. The next example illustrates the procedure.

EX AM PL E 13

Dividing Two Polynomials Find the quotient and the remainder when 3x3 + 4x2 + x + 7 is divided by x2 + 1

Solution NOTE Remember, a polynomial is in standard form when the terms are written according to descending degree. 䊏

Each polynomial is in standard form. The dividend is 3x3 + 4x2 + x + 7, and the divisor is x2 + 1. STEP 1: Divide the leading term of the dividend, 3x3, by the leading term of the divisor, x2. Enter the result, 3x, over the term 3x3, as follows: 3x x2

+ 1) 3x3 + 4x2 + x + 7

STEP 2: Multiply 3x by x2 + 1, and enter the result below the dividend.

x2

3x + 1) 3x3 + 4x2 + x + 7 3x3 + 3x

d 3x # (x 2 + 1) = 3x 3 + 3x

c Align the 3x term under the x to make the next step easier.

STEP 3: Subtract and bring down the remaining terms. 3x x2 + 1) 3x3 + 4x2 + x + 7 3x3 + 3x 4x2 - 2x + 7

d Subtract (change the signs and add). d Bring down the 4x 2 and the 7.

SECTION A.3  Polynomials

A29

STEP 4: Repeat Steps 1–3 using 4x2 - 2x + 7 as the dividend.

COMMENT If the degree of the divisor is greater than the degree of the dividend, the process ends. 䊏

3x + 4 x2 + 1) 3x3 + 4x2 + x 3x3 + 3x 4x2 - 2x 4x2 - 2x

d

+ 7 + 7 + 4 + 3

d Divide 4x 2 by x 2 to get 4. d Multiply x 2 + 1 by 4; subtract.

Since x2 does not divide into - 2x evenly (that is, the result is not a monomial), the process ends. The quotient is 3x + 4, and the remainder is - 2x + 3. Check: 1Quotient2 1Divisor2 + Remainder = 13x + 42 1x2 + 12 + 1 - 2x + 32

= 3x3 + 3x + 4x2 + 4 + 1 - 2x + 32

= 3x3 + 4x2 + x + 7 = Dividend Then 3x3 + 4x2 + x + 7 - 2x + 3 = 3x + 4 + 2 2 x + 1 x + 1

r

The next example combines the steps involved in long division.

EX AM PL E 14

Dividing Two Polynomials Find the quotient and the remainder when x4 - 3x3 + 2x - 5 is divided by x2 - x + 1

Solution

In setting up this division problem, it is necessary to leave a space for the missing x2 term in the dividend.

Divisor S Subtract S Subtract S Subtract S

x2 - 2x - 3 x - x + 1) x4 - 3x3 x4 - x3 + x2 - 2x3 - x2 - 2x3 + 2x2 - 3x2 - 3x2 2

+ 2x - 5 + + +

2x 2x 4x 3x x

d Quotient d Dividend

- 5 - 5 - 3 - 2

d Remainder

Check: 1Quotient2 1Divisor2 + Remainder

= 1x2 - 2x - 32 1x2 - x + 12 + 1x - 22 = x4 - x3 + x2 - 2x3 + 2x2 - 2x - 3x2 + 3x - 3 + x - 2 = x4 - 3x3 + 2x - 5 = Dividend

As a result, x4 - 3x3 + 2x - 5 x - 2 = x2 - 2x - 3 + 2 2 x - x + 1 x - x + 1

r

A30

APPENDIX A  Review

The process of dividing two polynomials leads to the following result:

THEOREM

Let Q be a polynomial of positive degree, and let P be a polynomial whose degree is greater than or equal to the degree of Q. The remainder after dividing P by Q is either the zero polynomial or a polynomial whose degree is less than the degree of the divisor Q.

Now Work

PROBLEM

93

7 Work with Polynomials in Two Variables A monomial in two variables x and y has the form axn ym, where a is a constant, x and y are variables, and n and m are nonnegative integers. The degree of a monomial is the sum of the powers of the variables. For example, 2xy3, x2 y2, and x3 y are all monomials that have degree 4. A polynomial in two variables x and y is the sum of one or more monomials in two variables. The degree of a polynomial in two variables is the highest degree of all the monomials with nonzero coefficients.

Examples of Polynomials in Two Variables

EX AM PL E 15

3x2 + 2x3 y + 5 Two variables, degree is 4.

px3 - y2

x4 + 4x3 y - xy3 + y4

Two variables, degree is 3.

Two variables, degree is 4.

r

Multiplying polynomials in two variables is handled in the same way as multiplying polynomials in one variable.

Using a Special Product Formula

EX AM PL E 16

To multiply 12x - y2 2, use the Square of a Binomial formula (3b) with 2x instead of a and with y instead of b. 12x - y2 2 = 12x2 2 - 2 # y # 2x + y2

r

= 4x2 - 4xy + y2

Now Work

PROBLEM

79

A.3 Assess Your Understanding Concepts and Vocabulary 1. The polynomial 3x4 - 2x3 + 13x2 - 5 is of degree The leading coefficient is 3 . 2. 1x2 - 42 1x2 + 42 =

x4 - 16.

3. 1x - 22 1x2 + 2x + 42 =

x3 - 8 .

4

.

4. True or False 4x -2 is a monomial of degree - 2.

False

5. True or False The degree of the product of two nonzero polynomials equals the sum of their degrees. True 6. True or False

1x + a2 1x2 + ax + a2 = x3 + a3.

False

SECTION A.3  Polynomials

A31

Skill Building In Problems 7–16, tell whether the expression is a monomial. If it is, name the variable(s) and the coefficient and give the degree of the monomial. If it is not a monomial, state why not. 8 *7. 2x3 *8. - 4x2 *9. *10. - 2x -3 *11. - 2xy2 x 2x2 8x *12. 5x2 y3 *13. *14. - 3 *15. x2 + y2 *16. 3x2 + 4 y y In Problems 17–26, tell whether the expression is a polynomial. If it is, give its degree. If it is not, state why not. 17. 3x2 - 5 Yes; 2 *22.

3 + 2 x

18. 1 - 4x Yes; 1

19. 5 Yes; 0

20. - p Yes; 0

23. 2y3 - 22 Yes; 3

24. 10z2 + z Yes; 2

*25.

*21. 3x2 -

x2 + 5 x3 - 1

5 x

3x3 + 2x - 1 x2 + x + 1

*26.

In Problems 27–46, add, subtract, or multiply, as indicated. Express your answer as a single polynomial in standard form. 27. (x2 + 4x + 5) + (3x - 3)

x2 + 7x + 2

28. (x3 + 3x2 + 2) + (x2 - 4x + 4)

29. (x3 - 2x2 + 5x + 10) - (2x2 - 4x + 3) 31. (6x + x + x) + (5x - x + 3x ) 5

3

4

3

2

33. (x2 - 3x + 1) + 2(3x2 + x - 4) 2

3

x3 - 4x2 + 9x + 7 30. (x2 - 3x - 4) - (x3 - 3x2 + x + 5)

6x + 5x + 3x + x 5

4

3

3

38. (x2 + 1) - (4x2 + 5) + (x2 + x - 2)

3

40. 8(1 - y ) + 4(1 + y + y + y )

2

3

2

x3 + x2 - 4x

42. 4x2(x3 - x + 2)

43. - 2x (4x + 5)

- 8x - 10x

44. 5x (3x - 4)

3

5

*45. (x + 1)(x2 + 2x - 4)

- 7x2 - 3x

2x2 - 4x + 6

2

41. x(x2 + x - 4) 2

5

36. 8(4x - 3x - 1) - 6(4x + 8x - 2) 3

2

3

x3 + 3x2 - 2x - 4

- x3 + 4x2 - 4x - 9

10x + 3x3 - 10x2 + 6

2

34. - 2(x2 + x + 1) + ( - 5x2 - x + 2)

15y - 27y + 30

2

2

2

37. (x2 - x + 2) + (2x2 - 3x + 5) - (x2 + 1) 39. 9(y - 3y + 4) - 6(1 - y )

5

- 2x + 18x - 18

2

2

32. (10x - 8x ) + (3x - 2x + 6)

2

7x2 - x - 7

35. 6(x + x - 3) - 4(2x - 3x ) 3

x3 + 4x2 - 4x + 6

8x3 - 24x2 - 48x + 4 - 2x2 + x - 6

- 4y + 4y2 + 4y + 12

3

3

4x5 - 4x3 + 8x2

15x - 20x3 4

46. (2x - 3)(x2 + x + 1)

2x3 - x2 - x - 3

In Problems 47–64, multiply the polynomials using the FOIL method. Express your answer as a single polynomial in standard form. 47. (x + 2)(x + 4)

x2 + 6x + 8 6x + 5x + 1

50. (3x + 1)(2x + 1) 53. (x - 3)(x - 2)

2

x2 - 5x + 6 6x - 10x - 4

56. (2x - 4)(3x + 1)

2

59. ( - x - 2)( - 2x - 4)

2x2 + 8x + 8

62. (2x + 3y)(x - y)

2

2x + xy - 3y

48. (x + 3)(x + 5)

x2 + 8x + 15

49. (2x + 5)(x + 2)

51. (x - 4)(x + 2)

x - 2x - 8

52. (x + 4)(x - 2)

54. (x - 5)(x - 1)

x2 - 6x + 5

55. (2x + 3)(x - 2)

2

x2 + 2x - 8 2x2 - x - 6

57. ( - 2x + 3)(x - 4)

- 2x + 11x - 12

58. ( - 3x - 1)(x + 1) - 3x2 - 4x - 1

60. ( - 2x - 3)(3 - x)

2x2 - 3x - 9

61. (x - 2y)(x + y)

2

*63. ( - 2x - 3y)(3x + 2y)

2

2x2 + 9x + 10

x2 - xy - 2y2

*64. (x - 3y)( - 2x + y)

In Problems 65–88, multiply the polynomials using the special product formulas. Express your answer as a single polynomial in standard form. 65. (x - 7)(x + 7)

x2 - 49 9x - 4

68. (3x + 2)(3x - 2) 71. (x - 4)

x - 8x + 16

2

2

25x - 9

74. (5x - 3)(5x + 3) 77. (x + y)(x - y)

x - y 2

83. (x - 2y)2

9x - 16y 2

x + 8x + 16

70. (x + 5)

72. (x - 5)

2

x - 10x + 25

73. (3x + 4)(3x - 4)

2

2

4x - 12x + 9 x + 2xy + y

2

87. (2x + 1)

2

x - 9y 2

2

3

82. (x - y)

2

4x2 + 12xy + 9y2 8x + 12x + 6x + 1 3

2

2

2

9x2 - 16

9x - 24x + 16 2

79. (3x + y)(3x - y)

2

4x2 - 9

x + 10x + 25

2

76. (3x - 4)

2

84. (2x + 3y)2

x + 3x + 3x + 1 3

69. (x + 4)

81. (x + y)

2

67. (2x + 3)(2x - 3)

2

78. (x + 3y)(x - 3y)

2

x2 - 4xy + 4y2

3

x2 - 1

2

75. (2x - 3)

2

80. (3x + 4y)(3x - 4y) 86. (x + 1)

66. (x - 1)(x + 1)

2

9x2 - y2

x - 2xy + y2

2

2

85. (x - 2)3

x3 - 6x2 + 12x - 8

88. (3x - 2) 27x3 - 54x2 + 36x - 8 3

In Problems 89–104, find the quotient and the remainder. Check your work by verifying that (Quotient)(Divisor) + Remainder = Dividend 89. 4x - 3x + x + 1 divided by x + 2 3

2

91. 4x - 3x + x + 1 divided by x 3

2

4x2 - 11x + 23; R: - 45 90. 3x3 - x2 + x - 2 divided by x + 2

4x - 3; R: x + 1

2

93. 5x - 3x + x + 1 divided by x + 2 4

2

5x - 13; R: x + 27

2

2

95. 4x - 3x + x + 1 divided by 2x - 1 5

2

2x ; R: - x + x + 1

3

2

2

*97. 2x4 - 3x3 + x + 1 divided by 2x2 + x + 1 99. - 4x + x - 4 divided by x - 1 3

2

101. 1 - x2 + x4 divided by x2 + x + 1 103. x - a divided by x - a 3

3

- 4x - 3x - 3; R: - 7 2

x2 - x - 1; R: 2x + 2

x + ax + a ; R: 0 2

2

92. 3x - x + x - 2 divided by x 3

2

3x2 - 7x + 15; R: - 32

3x - 1; R: x - 2

2

94. 5x - x + x - 2 divided by x + 2 4

2

2

96. 3x - x + x - 2 divided by 3x - 1 5

2

3

5x2 - 11; R: x + 20 x2; R: x - 2

*98. 3x4 - x3 + x - 2 divided by 3x2 + x + 1 *100. - 3x4 - 2x - 1 divided by x - 1 102. 1 - x2 + x4 divided by x2 - x + 1 104. x - a divided by x - a 5

5

x2 + x - 1; R: - 2x + 2

x + ax + a2x2 + a3x + a4; R: 0

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

4

3

A32

APPENDIX A  Review

Discussion and Writing 105. Explain why the degree of the product of two nonzero polynomials equals the sum of their degrees. 106. Explain why the degree of the sum of two polynomials of different degrees equals the larger of their degrees. 107. Give a careful statement about the degree of the sum of two polynomials of the same degree.

108. Do you prefer adding two polynomials using the horizontal method or the vertical method? Write a brief position paper defending your choice. 109. Do you prefer to memorize the rule for the square of a binomial (x + a)2 or to use FOIL to obtain the product? Write a brief position paper defending your choice.

A.4 Factoring Polynomials OBJECTIVES 1 Factor the Difference of Two Squares and the Sum and Difference of Two Cubes (p. A33) 2 Factor Perfect Squares (p. A34) 3 Factor a Second-Degree Polynomial: x2 + Bx + C (p. A34) 4 Factor by Grouping (p. A36) 5 Factor a Second-Degree Polynomial: Ax2 + Bx + C, A ≠ 1 (p. A37) 6 Complete the Square (p. A38)

Consider the following product:

12x + 32 1x - 42 = 2x2 - 5x - 12

COMMENT Over the real numbers, 4 3x + 4 factors into 3ax + b . It is 3 4 the noninteger that causes 3x + 4 3 to be prime over the integers. ■

EX A MPL E 1

The two polynomials on the left side are called factors of the polynomial on the right side. Expressing a given polynomial as a product of other polynomials—that is, finding the factors of a polynomial—is called factoring. We shall restrict our discussion here to factoring polynomials in one variable into products of polynomials in one variable, where all coefficients are integers. We call this factoring over the integers. Any polynomial can be written as the product of 1 times itself or as - 1 times its additive inverse. If a polynomial cannot be written as the product of two other polynomials (excluding 1 and - 1), then the polynomial is prime. When a polynomial has been written as a product consisting only of prime factors, it is factored completely. Examples of prime polynomials (over the integers) are 2, 3, 5, x, x + 1, x - 1, 3x + 4, x2 + 4 The first factor to look for in a factoring problem is a common monomial factor present in each term of the polynomial. If one is present, use the Distributive Property to factor it out. Continue factoring out monomial factors until none are left.

Identifying Common Monomial Factors

Polynomial

Common Monomial Factor

Remaining Factor

Factored Form

2x + 4

2

x + 2

2x + 4 = 21x + 22

3x - 6

3

x - 2

3x - 6 = 31x - 22

2x2 - 4x + 8

2

x2 - 2x + 4

2x2 - 4x + 8 = 21x2 - 2x + 42

8x - 12

4

2x - 3

8x - 12 = 412x - 32

x + x

x

x + 1

x2 + x = x1x + 12

x3 - 3x2

x2

x - 3

x3 - 3x2 = x2 1x - 32

6x2 + 9x

3x

2x + 3

6x2 + 9x = 3x12x + 32

2

r

SECTION A.4  Factoring Polynomials

A33

Notice that, once all common monomial factors have been removed from a polynomial, the remaining factor is either a prime polynomial of degree 1 or a polynomial of degree 2 or higher. (Do you see why?)

Now Work

PROBLEM

5

1 Factor the Difference of Two Squares and the Sum and Difference of Two Cubes When you factor a polynomial, first check for common monomial factors. Then see whether you can use one of the special formulas discussed in the previous section. Difference of Two Squares

a2 - b2 = 1a - b2 1a + b2

Perfect Squares

a2 + 2ab + b2 = 1a + b2 2 a2 - 2ab + b2 = 1a - b2 2

Sum of Two Cubes

a3 + b3 = 1a + b2 1a2 - ab + b2 2

Difference of Two Cubes

EXAM PL E 2

a3 - b3 = 1a - b2 1a2 + ab + b2 2

Factoring the Difference of Two Squares Factor completely: x2 - 4

Solution

Note that x2 - 4 is the difference of two squares, x2 and 22. x2 - 4 = 1x - 22 1x + 22

r EXAM PL E 3

Factoring the Difference of Two Cubes Factor completely: x3 - 1

Solution

Because x3 - 1 is the difference of two cubes, x3 and 13, x3 - 1 = (x - 1)(x2 + x + 1)

r EXAM PL E 4

Factoring the Sum of Two Cubes Factor completely: x3 + 8

Solution

Because x3 + 8 is the sum of two cubes, x3 and 23, x3 + 8 = 1x + 22 1x2 - 2x + 42

r EXAM PL E 5

Factoring the Difference of Two Squares Factor completely: x4 - 16

Solution

Because x4 - 16 is the difference of two squares, x4 = 1x2 2 and 16 = 42, 2

x4 - 16 = 1x2 - 42 1x2 + 42 But x2 - 4 is also the difference of two squares, so x4 - 16 = 1x2 - 42 1x2 + 42 = 1x - 22 1x + 22 1x2 + 42

Now Work

r PROBLEMS

15 AND 33

A34

APPENDIX A  Review

2 Factor Perfect Squares When the first term and third term of a trinomial are both positive and are perfect squares, such as x2, 9x2, 1, and 4, check to see whether the trinomial is a perfect square.

EX A MPL E 6

Factoring Perfect Squares Factor completely: x2 + 6x + 9

Solution

The first term, x2, and the third term, 9 = 32, are perfect squares. Because the middle term, 6x, is twice the product of x and 3, the trinomial is a perfect square. x2 + 6x + 9 = 1x + 32 2

EX A MPL E 7

r

Factoring Perfect Squares Factor completely: 9x2 - 6x + 1

Solution

The first term, 9x2 = 13x2 2, and the third term, 1 = 12, are perfect squares. Because the middle term, - 6x, is - 2 times the product of 3x and 1, the trinomial is a perfect square. 9x2 - 6x + 1 = 13x - 12 2

EX A MPL E 8

r

Factoring Perfect Squares Factor completely: 25x2 + 30x + 9

Solution

The first term, 25x2 = (5x)2, and the third term, 9 = 32, are perfect squares. Because the middle term, 30x, is twice the product of 5x and 3, the trinomial is a perfect square. 25x2 + 30x + 9 = 15x + 32 2

Now Work

PROBLEMS

25

AND

99

r

If a trinomial is not a perfect square, it may be possible to factor it using the technique discussed next.

3 Factor a Second-Degree Polynomial: x 2 + Bx + C The idea behind factoring a second-degree polynomial like x2 + Bx + C is to see whether it can be made equal to the product of two (possibly equal) first-degree polynomials. For example, consider (x + 3)(x + 4) = x2 + 7x + 12 The factors of x2 + 7x + 12 are x + 3 and x + 4. Note the following: x2 + 7x + 12 = (x + 3)(x + 4) 12 is the product of 3 and 4. 7 is the sum of 3 and 4.

In general, if x2 + Bx + C = 1x + a2 1x + b2 = x2 + (a + b)x + ab, then ab = C and a + b = B. To factor a second-degree polynomial x2 + Bx + C, find integers whose product is C and whose sum is B. That is, if there are numbers a, b, where ab = C and a + b = B, then x2 + Bx + C = 1x + a2 1x + b2

SECTION A.4  Factoring Polynomials

EXAM PL E 9

A35

Factoring Trinomials Factor completely: x2 + 7x + 10

Solution

First determine all integers whose product is 10, and then compute their sums. Integers whose product is 10 Sum

1, 10

- 1, - 10

2, 5

- 2, - 5

11

- 11

7

-7

The integers 2 and 5 have a product of 10 and add up to 7, the coefficient of the middle term. As a result, x2 + 7x + 10 = 1x + 22 1x + 52

EX AM PL E 10

r

Factoring Trinomials Factor completely: x2 - 6x + 8

Solution

First determine all integers whose product is 8, and then compute each sum. Integers whose product is 8 Sum

1, 8

- 1, - 8

2, 4

- 2, - 4

9

-9

6

-6

Since - 6 is the coefficient of the middle term,

x2 - 6x + 8 = 1x - 22 1x - 42

r EX AM PL E 11

Factoring Trinomials Factor completely: x2 - x - 12

Solution

First determine all integers whose product is - 12, and then compute each sum. Integers whose product is − 12 Sum

1, - 12

- 1, 12

2, - 6

- 2, 6

3, - 4

- 3, 4

11

-4

4

-1

1

- 11

Since - 1 is the coefficient of the middle term,

x2 - x - 12 = 1x + 32 1x - 42

r EX AM PL E 12

Factoring Trinomials Factor completely: x2 + 4x - 12

Solution

The integers - 2 and 6 have a product of - 12 and have the sum 4, so x2 + 4x - 12 = 1x - 22 1x + 62

r

To avoid errors in factoring, always check your answer by multiplying it out to see whether the result equals the original expression. When none of the possibilities works, the polynomial is prime.

A36

APPENDIX A  Review

EX AM PL E 13

Identifying Prime Polynomials Show that x2 + 9 is prime.

Solution

First list the integers whose product is 9, and then compute their sums. Integers whose product is 9

1, 9

- 1, - 9

3, 3

- 3, - 3

Sum

10

- 10

6

-6

Since the coefficient of the middle term in x2 + 9 = x2 + 0x + 9 is 0, and none of the sums equals 0, this means that x2 + 9 is prime. Example 13 demonstrates a more general result:

THEOREM

r

Any polynomial of the form x2 + a2, a real, is prime.

Now Work

PROBLEMS

39

AND

83

4 Factor by Grouping Sometimes a common factor does not occur in every term of the polynomial but does occur in each of several groups of terms that together make up the polynomial. When this happens, the common factor can be factored out of each group by means of the Distributive Property. This technique is called factoring by grouping.

EX AM PL E 14

Solution

Factoring by Grouping

Factor completely by grouping: 1x2 + 22x + 1x2 + 22 # 3 Note the common factor x2 + 2. Applying the Distributive Property yields 1x2 + 22x + 1x2 + 22 # 3 = 1x2 + 22 1x + 32 Since x2 + 2 and x + 3 are prime, the factorization is complete.

r

The next example shows a factoring problem that occurs in calculus.

EX AM PL E 15

Solution

Factoring by Grouping

Factor completely by grouping: 31x - 12 2 1x + 22 4 + 41x - 12 3 1x + 22 3

Here, 1x - 12 2 1x + 22 3 is a common factor of both 31x - 12 2 1x + 22 4 and 41x - 12 3 1x + 22 3. As a result,

31x - 12 2 1x + 22 4 + 41x - 12 3 1x + 22 3 = 1x - 12 2 1x + 22 3 3 31x + 22 + 41x - 124 = 1x - 12 2 1x + 22 3 3 3x + 6 + 4x - 44 = 1x - 12 2 1x + 22 3 17x + 22

r EX AM PL E 16

Factoring by Grouping Factor completely by grouping: x3 - 4x2 + 2x - 8

Solution

To see whether factoring by grouping will work, group the first two terms and the last two terms. Then look for a common factor in each group. In this example, factor

SECTION A.4  Factoring Polynomials

A37

x2 from x3 - 4x2 and 2 from 2x - 8. The remaining factor in each case is the same, x - 4. This means that factoring by grouping will work, as follows: x3 - 4x2 + 2x - 8 = 1x3 - 4x2 2 + 12x - 82 = x2 1x - 42 + 21x - 42 = 1x - 42 1x2 + 22

r

Since x2 + 2 and x - 4 are prime, the factorization is complete.

Now Work

PROBLEMS

51

AND

127

5 Factor a Second-Degree Polynomial: Ax 2 + Bx + C, A 3 1 To factor a second-degree polynomial Ax 2 + Bx + C, when A ≠ 1 and A, B, and C have no common factors, follow these steps:

Steps for Factoring Ax 2 + Bx + C, when A 3 1 and A, B, and C Have No Common Factors STEP 1: Find the value of AC. STEP 2: Find integers whose product is AC that add up to B. That is, find a and b such that ab = AC and a + b = B. STEP 3: Write Ax2 + Bx + C = Ax2 + ax + bx + C. STEP 4: Factor this last expression by grouping.

EX AM PL E 17

Factoring Trinomials Factor completely: 2x2 + 5x + 3

Solution

Comparing 2x2 + 5x + 3 to Ax2 + Bx + C, note that A = 2, B = 5, and C = 3.

STEP 1: The value of AC is 2 # 3 = 6. STEP 2: Determine the integers whose product is AC = 6, and compute their sums. Integers whose product is 6 Sum

1, 6

- 1, - 6

2, 3

- 2, - 3

7

-7

5

-5

STEP 3: The integers whose product is 6 that add up to B ⫽ 5 are 2 and 3.

2x2 ⫹ 5x ⫹ 3 ⫽ 2x2 ⫹ 2x ⫹ 3x ⫹ 3 STEP 4: Factor by grouping.

2x2 + 2x + 3x + 3 = 12x2 + 2x2 + 13x + 32 = 2x1x + 12 + 31x + 12 = 1x + 12 12x + 32

As a result,

2x2 + 5x + 3 = 1x + 12 12x + 32

r EX AM PL E 18

Factoring Trinomials Factor completely: 2x2 - x - 6

Solution

Comparing 2x2 - x - 6 to Ax2 + Bx + C, note that A = 2, B = - 1, and C = - 6.

A38

APPENDIX A  Review

STEP 1: The value of AC is 2 # 1 - 62 = - 12. STEP 2: Determine the integers whose product is AC = - 12, and compute their sums. Integers whose product is − 12 Sum

1, - 12

- 1, 12

2, - 6

- 2, 6

3, - 4

- 3, 4

- 11

11

-4

4

-1

1

STEP 3: The integers whose product is ⫺12 that add up to B ⫽ ⫺1 are ⫺4 and 3. 2x2 ⫺ x ⫺ 6 ⫽ 2x2 ⫺ 4x ⫹ 3x ⫺ 6 STEP 4: Factor by grouping. 2x2 - 4x + 3x - 6 = 12x2 - 4x2 + 13x - 62 = 2x1x - 22 + 31x - 22 = 1x - 22 12x + 32 As a result,

Now Work

2x2 - x - 6 = 1x - 22 12x + 32 PROBLEM

57

r

SUMMARY Type of Polynomial Any polynomial

Method Look for common monomial factors. (Always do this first!)

Example

Binomials of degree 2 or higher    

Check for a special product: Difference of two squares, a2 - b2 Difference of two cubes, a3 - b3 Sum of two cubes, a3 + b3

x2 - 16 = 1x - 42 1x + 42 x3 - 64 = 1x - 42 1x2 + 4x + 162 x3 + 27 = 1x + 32 1x2 - 3x + 92

Trinomials of degree 2       Four or more terms

Check for a perfect square, 1a { b2 2   Factoring x2 + Bx + C (p. A34) Factoring Ax2 + Bx + C (p. A37) Grouping

x2 + 8x + 16 = 1x + 42 2 x2 - 10x + 25 = 1x - 52 2 x2 - x - 2 = 1x - 22 1x + 12 6x2 + x - 1 = 12x + 12 13x - 12 2x3 - 3x2 + 4x - 6 = 12x - 32 1x2 + 22

6x2 + 9x = 3x12x + 32  

6 Complete the Square The idea behind completing the square in one variable is to “adjust” an expression of the form x2 + bx to make it a perfect square. Perfect squares are trinomials of the form x2 + 2ax + a2 = 1x + a2 2

or

x2 - 2ax + a2 = 1x - a2 2

For example, x2 + 6x + 9 is a perfect square because x2 + 6x + 9 = 1x + 32 2. And p2 - 12p + 36 is a perfect square because p2 - 12p + 36 = 1p - 62 2. So how do we “adjust” x2 + bx to make it a perfect square? We do it by adding a number. For example, to make x2 + 6x a perfect square, add 9. But how do we know to add 9? Divide the coefficient on the first-degree term, 6, by 2, and then square the result to obtain 9. This approach works in general.

SECTION A.4  Factoring Polynomials

A39

Completing the Square 1 2 and then square the result. That is, determine the value of b in x2 + bx and 1 2 compute a b b . 2 Identify the coefficient of the first-degree term. Multiply this coefficient by

EX AM PL E 19

Completing the Square Determine the number that must be added to each expression to complete the square. Then factor the expression. Start

Add

Result

Factored Form

y2 + 8y

a

1# 2 8b = 16 2

y2 + 8y + 16

1y + 42 2

x2 + 12x

a

2 1# 12b = 36 2

x2 + 12x + 36

1x + 62 2

a2 - 20a

a

2 1# 1 - 202 b = 100 2

a2 - 20a + 100

1a - 102 2

p2 - 5p

a

2 1# 25 1 - 52 b = 2 4

p2 - 5p +

ap -

25 4

5 2 b 2

r Note that the factored form of a perfect square is either

Figure 24

y

y

4

Area = y 2

Area = 4y

Now Work

PROBLEM

or

b 2 b 2 x2 - bx + a b = ax - b 2 2

69

Are you wondering why making an expression a perfect square is called “completing the square”? Look at the square in Figure 24. Its area is 1y + 42 2. The yellow area is y2 and each orange area is 4y (for a total area of 8y). The sum of these areas is y2 + 8y. To complete the square, we need to add the area of the green region: 4 # 4 = 16. As a result, y2 + 8y + 16 = 1y + 42 2.

Area = 4y

4

b 2 b 2 x2 + bx + a b = ax + b 2 2

A.4 Assess Your Understanding Concepts and Vocabulary 1. If factored completely, 3x3 - 12x = 3x(x - 2)(x + 2).

3. True or False The polynomial x2 + 4 is prime. True

2. If a polynomial cannot be written as the product of two other polynomials (excluding 1 and - 1), then the polynomial is said to be prime .

4. True or False 3x3 - 2x2 - 6x + 4 = (3x - 2)(x2 + 2). False

Skill Building In Problems 5–14, factor each polynomial by removing the common monomial factor. 5. 3x + 6

3(x + 2)

*9. x + x + x 3

2

13. 3x2 y - 6xy2 + 12xy

6. 7x - 14

7(x - 2)

10. x - x + x 3

3xy(x - 2y + 4)

2

x(x - x + 1) 2

7. ax2 + a 11. 2x - 2x 2

a(x2 + 1) 2x(x - 1)

14. 60x2 y - 48xy2 + 72x3 y

8. ax - a 12. 3x - 3x 2

a(x - 1) 3x(x - 1)

12xy(5x - 4y + 6x2)

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

A40

APPENDIX A  Review

In Problems 15–22, factor the difference of two squares. 15. x2 - 1

(x + 1)(x - 1)

19. x2 - 16

16. x2 - 4

(x + 4)(x - 4)

(x + 2)(x - 2)

20. x2 - 25

17. 4x2 - 1

(x + 5)(x - 5)

(2x + 1)(2x - 1)

21. 25x2 - 4

18. 9x2 - 1

(5x + 2)(5x - 2)

(3x + 1)(3x - 1)

22. 36x2 - 9 9(2x + 1)(2x - 1)

In Problems 23–32, factor the perfect squares. 23. x2 + 2x + 1

(x + 1)2

24. x2 - 4x + 4

(x - 2)2

27. x2 - 10x + 25

(x - 5)2 28. x2 + 10x + 25

31. 16x + 8x + 1

(4x + 1) 32. 25x + 10x + 1

2

2

25. x2 + 4x + 4

(x + 5)2

29. 4x2 + 4x + 1

(5x + 1)

2

(x + 2)2

26. x2 - 2x + 1

(2x + 1)2

(x - 1)2

30. 9x2 + 6x + 1

(3x + 1)2

2

In Problems 33–38, factor the sum or difference of two cubes. *33. x3 - 27

*34. x3 + 125

*35. x3 + 27

*36. 27 - 8x3

*37. 8x3 + 27

*38. 64 - 27x3

In Problems 39–50, factor each polynomial. 39. x2 + 5x + 6

(x + 2)(x + 3)

40. x2 + 6x + 8

42. x + 9x + 8

(x + 1)(x + 8)

43. x + 7x + 10

2

45. x - 10x + 16

(x - 8)(x - 2)

2

48. x2 - 2x - 8

(x + 2)(x + 4)

2

(x + 5)(x + 2)

46. x - 17x + 16 49. x2 + 7x - 8

(x + 6)(x + 1)

44. x2 + 11x + 10

(x - 16)(x - 1)

2

(x - 4)(x + 2)

41. x2 + 7x + 6

(x + 8)(x - 1)

(x + 1)(x + 10)

47. x - 7x - 8

(x - 8)(x + 1)

50. x2 + 2x - 8

(x - 2)(x + 4)

2

In Problems 51–56, factor by grouping. 51. 2x2 + 4x + 3x + 6 53. 2x2 - 4x + x - 2

(x + 2)(2x + 3)

52. 3x2 - 3x + 2x - 2

(x - 2)(2x + 1)

55. 6x2 + 9x + 4x + 6

54. 3x2 + 6x - x - 2

(2x + 3)(3x + 2)

(x - 1)(3x + 2) (x + 2)(3x - 1)

56. 9x2 - 6x + 3x - 2

(3x - 2)(3x + 1)

In Problems 57–68, factor each polynomial. *57. 3x2 + 4x + 1

*58. 2x2 + 3x + 1

*59. 2z2 + 5z + 3

*60. 6z2 + 5z + 1

*61. 3x + 2x - 8

*62. 3x + 10x + 8

*63. 3x - 2x - 8

*64. 3x2 - 10x + 8

*65. 3x2 + 14x + 8

*66. 3x2 - 14x + 8

*67. 3x2 + 10x - 8

*68. 3x2 - 10x - 8

2

2

2

In Problems 69–74, determine the number that should be added to complete the square of each expression. Then factor each expression. 69. x2 + 10x 72. x2 - 4x

25; (x + 5)2

70. p2 + 14p

4; (x - 2)2

73. x2 -

1 x 2

49; (p + 7)2

71. y2 - 6y

1 1 2 ; ¢x - ≤ 16 4

74. x2 +

1 x 3

9; (y - 3)2 1 1 2 ; ¢x + ≤ 36 6

Mixed Practice In Problems 75–122, factor each polynomial completely. If the polynomial cannot be factored, say it is prime. 75. x2 - 36

(x + 6)(x - 6)

76. x2 - 9

(x + 3)(x - 3)

*77. 2 - 8x2

*78. 3 - 27x2

*81. x - 10x + 21

*82. x2 - 6x + 8

*79. x + 11x + 10

*80. x + 5x + 4

*83. 4x2 - 8x + 32

*84. 3x2 - 12x + 15

*85. x2 + 4x + 16

*86. x2 + 12x + 36

*87. 15 + 2x - x

*88. 14 + 6x - x

*89. 3x - 12x - 36

*90. x3 + 8x2 - 20x

2

2

2

2

2

2

*91. y4 + 11y3 + 30y2

*92. 3y3 - 18y2 - 48y

*93. 4x2 + 12x + 9

*94. 9x2 - 12x + 4

*95. 6x + 8x + 2

*96. 8x + 6x - 2

*97. x - 81

*98. x4 - 1

*99. x6 - 2x3 + 1

*100. x6 + 2x3 + 1

*101. x7 - x5

*103. 16x + 24x + 9

*104. 9x - 24x + 16

*105. 5 + 16x - 16x

*107. 4y2 - 16y + 15

*108. 9y2 + 9y - 4

*109. 1 - 8x2 - 9x4

2

2

2

4

2

*102. x8 - x5 *106. 5 + 11x - 16x2

2

*110. 4 - 14x2 - 8x4

*111. x(x + 3) - 6(x + 3)

*112. 5(3x - 7) + x(3x - 7)

*114. (x - 1) - 2(x - 1)

*115. (3x - 2) - 27

*117. 3(x + 10x + 25) - 4(x + 5)

*118. 7(x - 6x + 9) + 5(x - 3)

*119. x + 2x - x - 2

*120. x - 3x - x + 3

*121. x - x + x - 1

*122. x4 + x3 + x + 1

2

2

3

2

3

2

4

3

*113. (x + 2) - 5(x + 2) 2

116. (5x + 1)3 - 1 3

5x(25x2 + 15x + 3)

2

Applications and Extensions In Problems 123–132, expressions that occur in calculus are given. Factor each expression completely. *123. 2(3x + 4)2 + (2x + 3) # 2(3x + 4) # 3 125. 2x(2x + 5) + x

2

#2

2x(3x + 5)

*127. 2(x + 3)(x - 2)3 + (x + 3)2 # 3(x - 2)2

*124. 5(2x + 1)2 + (5x - 6) # 2(2x + 1) # 2 126. 3x2(8x - 3) + x3 # 8

x2(32x - 9)

*128. 4(x + 5)3(x - 1)2 + (x + 5)4 # 2(x - 1)

SECTION A.5  Synthetic Division

129. (4x - 3)2 + x # 2(4x - 3) # 4

3(4x - 3)(4x - 1)

130. 3x2(3x + 4)2 + x3 # 2(3x + 4) # 3

3x2(3x + 4)(5x + 4)

* 131. 2(3x - 5) # 3(2x + 1)3 + (3x - 5)2 # 3(2x + 1)2 # 2

*132. 3(4x + 5)2 # 4(5x + 1)2 + (4x + 5)3 # 2(5x + 1) # 5

* 133. Show that x + 4 is prime.

*134. Show that x2 + x + 1 is prime.

2

A41

Discussion and Writing 135. Make up a polynomial that factors into a perfect square.

136. Explain to a fellow student what you look for first when presented with a factoring problem. What do you do next?

A.5 Synthetic Division OBJECTIVE 1 Divide Polynomials Using Synthetic Division (p. A41)

1 Divide Polynomials Using Synthetic Division To find the quotient as well as the remainder when a polynomial of degree 1 or higher is divided by x - c, a shortened version of long division, called synthetic division, makes the task simpler. To see how synthetic division works, first consider long division for dividing the polynomial 2x3 - x2 + 3 by x - 3. d Quotient 2x2 + 5x + 15 3 2 x - 3) 2x - x + 3 2x3 - 6x2 5x2 5x2 - 15x 15x + 3 15x - 45 48 d Remainder

Check: 1Divisor2 # 1Quotient2 + Remainder = 1x - 32 12x2 + 5x + 152 + 48 = 2x3 + 5x2 + 15x - 6x2 - 15x - 45 + 48 = 2x3 - x2 + 3 The process of synthetic division arises from rewriting the long division in a more compact form, using simpler notation. For example, in the long division above, the terms in blue are not really necessary because they are identical to the terms directly above them. Removing these terms gives the following display: 2x2 + 5x + 15 x - 3) 2x3 - x2 + 3 2 - 6x 5x2 - 15x 15x - 45 48

APPENDIX A  Review

Most of the x’s that appear in this process can also be removed, provided that care is used in positioning each coefficient. In this regard, use 0 as the coefficient of x in the dividend, because that power of x is missing. This gives 2x2 + 5x + 15 x - 3) 2 - 1 0 - 6 5 - 15 15

3

- 45 48 To make this display more compact, move the lines up until the numbers in blue align horizontally. 2x2 + 5x + 15 x - 3) 2 - 1 0 3 - 6 - 15 - 45 5 15 48 ○

Row 1 Row 2 Row 3 Row 4

Because the leading coefficient of the divisor is always 1, the leading coefficient of the dividend will also be the leading coefficient of the quotient. Place the leading coefficient of the quotient, 2, in the circled position. Now, the first three numbers in row 4 are precisely the coefficients of the quotient, and the last number in row 4 is the remainder. Thus row 1 is not really needed, so the process can be compressed to three rows, with the bottom row containing both the coefficients of the quotient and the remainder. - 1 0 3 - 6 - 15 - 45 5 15 48

x - 3) 2 2

Row 1 Row 2 (subtract) Row 3

Recall that the entries in row 3 are obtained by subtracting the entries in row 2 from those in row 1. Rather than subtracting the entries in row 2, change the sign of each entry and add. With this modification, the display will look like this: x - 3) 2 2

- 1 6 5

0 15 15

3 45 48

Row 1 Row 2 (add) Row 3

Note that the entries in row 2 are three times the prior entries in row 3. As the last modification to the display, replace the x - 3 with 3. The entries in row 3 give the quotient and the remainder, as shown next. 3) 2

0 15 15

3 45 48

d

2

- 1 6 5 e Quotient

Row 1 Row 2 (add) Row 3 Remainder

e

d

A42

2x2 + 5x + 15 48

Let’s go through an example step by step.

EX A MPL E 1

Using Synthetic Division to Find the Quotient and Remainder Use synthetic division to find the quotient and remainder when x3 - 4x2 - 5 is divided by x - 3

SECTION A.5  Synthetic Division

Solution

A43

STEP 1: Write the dividend in descending powers of x. Then copy the coefficients, remembering to insert a 0 for any missing powers of x. -4 0

1

-5

Row 1

STEP 2: Insert the usual division symbol. In synthetic division, the divisor is of the form x - c, and c is the number placed to the left of the division symbol. Here, since the divisor is x - 3, insert 3 to the left of the division symbol. -4 0

3) 1

-5

Row 1

STEP 3: Bring the 1 down two rows, and enter it in row 3. -4 0

3) 1 T 1

-5

Row 1 Row 2 Row 3

STEP 4: Multiply the latest entry in row 3 by 3, and place the result in row 2, one column over to the right. 3) 1

-5

1

Row 1 Row 2 Row 3

c

x3

-4 0 3

STEP 5: Add the entry in row 2 to the entry above it in row 1, and enter the sum in row 3. -4 0 3 x3 1 -1

-5

3) 1

Row 1

c

Row 2 Row 3

STEP 6: Repeat Steps 4 and 5 until no more entries are available in row 1. -5 -9 x3 - 14 c

c

c

3) 1 - 4 0 3 -3 x3 x3 1 -1 -3

Row 1 Row 2 Row 3

STEP 7: The final entry in row 3, the - 14, is the remainder; the other entries in row 3, the 1, - 1, and - 3, are the coefficients (in descending order) of a polynomial whose degree is 1 less than that of the dividend. This is the quotient. Thus, Quotient = x2 - x - 3

Remainder = - 14

Check: 1Divisor2 1Quotient2 + Remainder

= 1x - 32 1x2 - x - 32 + 1 - 142 = 1x3 - x2 - 3x - 3x2 + 3x + 92 + 1 - 142 = x3 - 4x2 - 5 = Dividend

Let’s do an example in which all seven steps are combined.

EXAM PL E 2

r

Using Synthetic Division to Verify a Factor Use synthetic division to show that x + 3 is a factor of 2x5 + 5x4 - 2x3 + 2x2 - 2x + 3

Solution

If the remainder is 0, then the divisor is a factor of the dividend, and the quotient is the other factor. The divisor is x + 3 = x - ( - 3), so place - 3 to the left of the division symbol. Then the row 3 entries will be multiplied by - 3, entered in row 2, and added to row 1. - 3) 2 2

5 -6 -1

-2 3 1

2 -3 -1

-2 3 1

3 -3 0

Row 1 Row 2 Row 3

A44

APPENDIX A  Review

The remainder is 0, so Dividend = 1Divisor2 1Quotient2 + Remainder

2x + 5x - 2x3 + 2x2 - 2x + 3 = 1x + 32 12x4 - x3 + x2 - x + 12 5

4

Hence, x + 3 is a factor of 2x5 + 5x4 - 2x3 + 2x2 - 2x + 3.

r As Example 2 illustrates, the remainder after division gives information about whether the divisor is or is not a factor. More will come about this in Chapter 3.

Now Work

PROBLEMS

7

AND

17

A.5 Assess Your Understanding Concepts and Vocabulary 1. To check division, ensure that the expression being divided—the dividend—equals the product of the quotient and the divisor plus the remainder . 2. To divide 2x3 - 5x + 1 by x + 3 using synthetic division, the first step is to write - 3

)2 0 - 5 1 . 3. True or False In using synthetic division, the divisor is always a polynomial of degree 1, whose leading coefficient is 1. True 4. True or False - 2) 5 5

3 2 - 10 14 - 7 16

1 means - 32 - 31

- 31 5x3 + 3x2 + 2x + 1 = 5x2 - 7x + 16 + . True x + 2 x + 2

Skill Building In Problems 5–16, use synthetic division to find the quotient and remainder when: *5. x3 - x2 + 2x + 4 *7. 3x + 2x - x + 3 3

2

*9. x - 4x + x 5

3

*13. 0.1x + 0.2x *15. x5 - 1

is divided by x - 3

is divided by x + 3

*11. 4x6 - 3x4 + x2 + 5 3

is divided by x - 2

is divided by x + 1.1

is divided by x + 1

*8. - 4x3 + 2x2 - x + 1 *10. x + x + 2 4

is divided by x - 1

is divided by x - 1

*6. x3 + 2x2 - 3x + 1

*12. x5 + 5x3 - 10 *14. 0.1x - 0.2 2

*16. x5 + 1

is divided by x + 2

is divided by x - 2

2

is divided by x + 1

is divided by x + 2.1

is divided by x + 1

In Problems 17–26, use synthetic division to determine whether x - c is a factor of the given polynomial. 17. 4x3 - 3x2 - 8x + 4; x - 2

No

18. - 4x3 + 5x2 + 8; x + 3

No

19. 3x - 6x - 5x + 10; x - 2 Yes

20. 4x - 15x - 4; x - 2 Yes

21. 3x6 + 82x3 + 27; x + 3 Yes

22. 2x6 - 18x4 + x2 - 9; x + 3 Yes

4

3

23. 4x6 - 64x4 + x2 - 15; x + 4 25. 2x4 - x3 + 2x - 1; x -

1 2

No

Yes

4

2

24. x6 - 16x4 + x2 - 16; x + 4 Yes 26. 3x4 + x3 - 3x + 1; x +

1 3

No

Applications and Extensions 27. Find the sum of a, b, c, and d if d x3 - 2x2 + 3x + 5 = ax2 + bx + c + x + 2 x + 2

-9

Discussion and Writing 28. When dividing a polynomial by x - c, do you prefer to use long division or synthetic division? Does the value of c make a difference to you in choosing? Give reasons. *Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

SECTION A.6  Rational Expressions

A45

A.6 Rational Expressions OBJECTIVES 1 2 3 4 5

Reduce a Rational Expression to Lowest Terms (p. A45) Multiply and Divide Rational Expressions (p. A46) Add and Subtract Rational Expressions (p. A47) Use the Least Common Multiple Method (p. A48) Simplify Complex Rational Expressions (p. A50)

1 Reduce a Rational Expression to Lowest Terms The quotient of two polynomials forms a result called a rational expression. Some examples of rational expressions are (a)

x3 + 1 x

(b)

3x2 + x - 2 x2 + 5

(c)

x x2 - 1

(d)

xy2 (x - y)2

Expressions (a), (b), and (c) are rational expressions in one variable, x, whereas (d) is a rational expression in two variables, x and y. Rational expressions are described in the same manner as rational numbers. In expression (a), the polynomial x3 + 1 is the numerator, and x is the denominator. When the numerator and denominator of a rational expression contain no common factors (except 1 and - 1), the rational expression is reduced to lowest terms, or simplified. The polynomial in the denominator of a rational expression cannot be equal x3 + 1 to 0 because division by 0 is not defined. For example, for the expression , x x cannot take on the value 0. The domain of the variable x is 5 x 0 x ≠ 06 . A rational expression is reduced to lowest terms by factoring the numerator and the denominator completely and dividing out any common factors using the Reduction Property: ac a = bc b

EXAM PL E 1

b ≠ 0, c ≠ 0

(1)

Reducing a Rational Expression to Lowest Terms Reduce to lowest terms:

Solution

if

x2 + 4x + 4 x2 + 3x + 2

Begin by factoring the numerator and the denominator. x2 + 4x + 4 = (x + 2)(x + 2) x2 + 3x + 2 = (x + 2)(x + 1)

WARNING Apply the Reduction Property only to rational expressions written in factored form. Be sure to divide out only common factors, not common terms! 䊏

Since a common factor, x + 2, appears, the original expression is not in lowest terms. To reduce it to lowest terms, use the Reduction Property: (x + 2) (x + 2) x2 + 4x + 4 x + 2 = = 2 (x + 2) (x + 1) x + 1 x + 3x + 2

x ≠ - 2, - 1

r EXAM PL E 2

Reducing Rational Expressions to Lowest Terms Reduce each rational expression to lowest terms. (a)

x3 - 8 x3 - 2x2

(b)

8 - 2x x2 - x - 12

A46

APPENDIX A  Review

Solution

(a)

(x - 2) (x2 + 2x + 4) x3 - 8 x2 + 2x + 4 = = x3 - 2x2 x2 (x - 2) x2

(b)

2( - 1) (x - 4) 2(4 - x) 8 - 2x -2 = = = (x - 4)(x + 3) (x - 4) (x + 3) x + 3 x - x - 12

x ≠ 0, 2

2

Now Work

PROBLEM

x ≠ - 3, 4

r

5

2 Multiply and Divide Rational Expressions The rules for multiplying and dividing rational expressions are the same as the rules a c for multiplying and dividing rational numbers. If and , b ≠ 0, d ≠ 0, are two b d rational expressions, then a#c ac = b d bd

if

a b a d ad = # = c c b bc d

if

b ≠ 0, d ≠ 0

b ≠ 0, c ≠ 0, d ≠ 0

(2)

(3)

In using equations (2) and (3) with rational expressions, be sure first to factor each polynomial completely so that common factors can be divided out. Leave your answer in factored form.

EX A MPL E 3

Multiplying and Dividing Rational Expressions Perform the indicated operation and simplify the result. Leave your answer in factored form. x + 3 2 2 x - 2x + 1 # 4x + 4 x2 - 4 (a) (b) x3 + x x2 + x - 2 x2 - x - 12 x3 - 8

Solution

(a)

4(x2 + 1) (x - 1)2 x2 - 2x + 1 # 4x2 + 4 # = x3 + x x2 + x - 2 x(x2 + 1) (x + 2)(x - 1) = =

(x - 1) 2 (4) (x2 + 12 x (x2 + 12 (x + 2) (x - 12 4(x - 1) x(x + 2)

x ≠ - 2, 0, 1

x + 3 x + 3 # x3 - 8 x2 - 4 = 2 (b) 2 x - 4 x2 - x - 12 x - x - 12 3 x - 8 =

2 x + 3 # (x - 2)(x + 2x + 4) (x - 2)(x + 2) (x - 4)(x + 3)

=

(x + 32 (x - 2) (x2 + 2x + 4) (x - 2) (x + 2)(x - 4) (x + 32

=

x2 + 2x + 4 (x + 2)(x - 4)

Now Work

PROBLEMS

17

x ≠ - 3, - 2, 2, 4 AND

25

r

SECTION A.6  Rational Expressions

A47

3 Add and Subtract Rational Expressions In Words To add (or subtract) two rational expressions with the same denominator, keep the common denominator and add (or subtract) the numerators.

The rules for adding and subtracting rational expressions are the same as the rules for adding and subtracting rational numbers. If the denominators of two rational expressions to be added (or subtracted) are equal, then add (or subtract) the numerators and keep the common denominator.

If

a c and are two rational expressions, then b b a c a + c + = b b b

EXAM PL E 4

a c a - c = b b b

if b ≠ 0

(4)

Adding and Subtracting Rational Expressions with Equal Denominators Perform the indicated operation and simplify the result. Leave your answer in factored form.

Solution

(a)

2x2 - 4 x + 3 + 2x + 5 2x + 5

(a)

(2x2 - 4) + (x + 3) 2x2 - 4 x + 3 + = 2x + 5 2x + 5 2x + 5

x ≠ -

= (b)

(b)

x 3x + 2 x - 3 x - 3

x ≠ 3

(2x - 1)(x + 1) 2x2 + x - 1 = 2x + 5 2x + 5

x - (3x + 2) x 3x + 2 x - 3x - 2 = = x - 3 x - 3 x - 3 x - 3 =

EXAM PL E 5

5 2

- 2(x + 1) - 2x - 2 = x - 3 x - 3

r

Adding Rational Expressions Whose Denominators Are Additive Inverses of Each Other Perform the indicated operation and simplify the result. Leave your answer in factored form. 5 2x + x - 3 3 - x

Solution

x ≠ 3

Notice that the denominators of the two rational expressions are different. However, the denominator of the second expression is the additive inverse of the denominator of the first. That is, 3 - x = - x + 3 = - 1 # (x - 3) = - (x - 3) Then 2x 5 2x 5 2x -5 + = + = + x - 3 3 - x x - 3 - (x - 3) x - 3 x - 3

c

c

a -a = -b b

3 - x = - (x - 3)

=

Now Work

PROBLEMS

37

AND

43

2x + ( - 5) 2x - 5 = x - 3 x - 3

r

A48

APPENDIX A  Review

If the denominators of two rational expressions to be added or subtracted are not equal, the general formulas for adding and subtracting rational expressions can be used. a c a#d b#c ad + bc + = # + # = b d b d b d bd a c a#d b#c ad - bc = # - # = b d b d b d bd

EX A MPL E 6

if

b ≠ 0, d ≠ 0

(5a)

if

b ≠ 0, d ≠ 0

(5b)

Adding and Subtracting Rational Expressions with Unequal Denominators Perform the indicated operation and simplify the result. Leave your answer in factored form. x - 3 x x2 1 (a) + x ≠ - 4, 2 (b) 2 x ≠ - 2, 0, 2 x x + 4 x - 2 x - 4

Solution

(a)

x x - 3#x - 2 x + 4# x x - 3 + = + x + 4 x - 2 x + 4 x - 2 x + 4 x - 2 æ (5a)

(b)

=

(x - 3)(x - 2) + (x + 4)(x) (x + 4)(x - 2)

=

x2 - 5x + 6 + x2 + 4x 2x2 - x + 6 = (x + 4)(x - 2) (x + 4)(x - 2)

x2(x) - (x2 - 4)(1) x2 x2 # x x2 - 4 # 1 1 = = x x2 - 4 x2 - 4 x x2 - 4 x (x2 - 4)(x) æ (5b)

=

x3 - x2 + 4 x (x - 2)(x + 2)

Now Work

PROBLEM

r

47

4 Use the Least Common Multiple Method If the denominators of two rational expressions to be added (or subtracted) have common factors, it usually is better not to use the general rules given by equations (5a) and (5b). Instead, just as with fractions, the least common multiple (LCM) method is used. The LCM method uses the polynomial of least degree that has each denominator polynomial as a factor.

The LCM Method for Adding or Subtracting Rational Expressions The least common multiple (LCM) method requires four steps: STEP 1: Factor completely the polynomial in the denominator of each rational expression. STEP 2: The LCM of the denominators is the product of each of these factors raised to a power equal to the greatest number of times that the factor occurs in the polynomials. STEP 3: Write each rational expression using the LCM as the common denominator. STEP 4: Add or subtract the rational expressions using equation (4). Let’s consider an example that requires only Steps 1 and 2.

SECTION A.6  Rational Expressions

EXAM PL E 7

A49

Finding the Least Common Multiple Find the least common multiple of the following pair of polynomials: x(x - 1)2(x + 1) and 4 (x - 1)(x + 1)3

Solution

STEP 1: The polynomials are already factored completely as x(x - 1)2(x + 1) and 4 (x - 1)(x + 1)3 STEP 2: Start by writing the factors of the left-hand polynomial. (Or you could start with the one on the right.) x(x - 1)2(x + 1) Now look at the right-hand polynomial. Its first factor, 4, does not appear in our list, so we insert it. 4x(x - 1)2(x + 1) The next factor, x - 1, is already in our list, so no change is necessary. The final factor is (x + 1)3. Since our list has x + 1 to the first power only, we replace x + 1 in the list by (x + 1)3. The LCM is 4x(x - 1)2(x + 1)3

r

Notice that the LCM in Example 7 is, in fact, the polynomial of least degree that contains x(x - 1)2(x + 1) and 4(x - 1)(x + 1)3 as factors.

Now Work

EXAM PL E 8

PROBLEM

53

Using the Least Common Multiple to Add Rational Expressions Perform the indicated operation and simplify the result. Leave your answer in factored form. x 2x - 3 + 2 x + 3x + 2 x - 1 2

Solution

x ≠ - 2, - 1, 1

STEP 1: Factor completely the polynomials in the denominators. x2 + 3x + 2 = (x + 2)(x + 1) x2 - 1 = (x - 1)(x + 1) STEP 2: The LCM is (x + 2)(x + 1)(x - 1). Do you see why? STEP 3: Write each rational expression using the LCM as the denominator. x(x - 1) x x x #x - 1 = = = (x + 2)(x + 1) (x + 2)(x + 1) x 1 (x + 2)(x + 1)(x - 1) x + 3x + 2 2

æ

Multiply numerator and denominator by x − 1 to get the LCM in the denominator.

2x - 3 2x - 3 2x - 3 # x + 2 = (2x - 3)(x + 2) = = 2 (x 1)(x + 1) (x 1)(x + 1) x + 2 (x - 1)(x + 1)(x + 2) x - 1 æ

Multiply numerator and denominator by x + 2 to get the LCM in the denominator.

A50

APPENDIX A  Review

STEP 4: Now add by using equation (4). (2x - 3)(x + 2) x(x - 1) x 2x - 3 + + 2 = (x + 2)(x + 1)(x - 1) (x + 2)(x + 1)(x - 1) x + 3x + 2 x - 1 2

EX A MPL E 9

=

(x2 - x) + (2x2 + x - 6) (x + 2)(x + 1)(x - 1)

=

3(x2 - 2) 3x2 - 6 = (x + 2)(x + 1)(x - 1) (x + 2)(x + 1)(x - 1)

r

Using the Least Common Multiple to Subtract Rational Expressions Perform the indicated operation and simplify the result. Leave your answer in factored form. 3 x + 4 x ≠ - 1, 0 - 2 2 x + x x + 2x + 1

Solution

STEP 1: Factor completely the polynomials in the denominators. x2 + x = x (x + 1) x2 + 2x + 1 = (x + 1)2 STEP 2: The LCM is x(x + 1)2. STEP 3: Write each rational expression using the LCM as the denominator. 3 3 3 # x + 1 = 3(x + 1)2 = = x(x + 1) x(x + 1) x + 1 x + x x(x + 1) 2

x(x + 4) x + 4 x + 4 #x x + 4 = = = 2 2 x x + 2x + 1 (x + 1) (x + 1) x(x + 1)2 2

STEP 4: Subtract, using equation (4). x(x + 4) 3(x + 1) 3 x + 4 - 2 = 2 2 x + x x + 2x + 1 x(x + 1) x(x + 1)2 3(x + 1) - x(x + 4) = x(x + 1)2

Now Work

PROBLEM

=

3x + 3 - x2 - 4x x(x + 1)2

=

- x2 - x + 3 x(x + 1)2

r

63

5 Simplify Complex Rational Expressions When sums and/or differences of rational expressions appear as the numerator and/ or denominator of a quotient, the quotient is called a complex rational expression.* For example, 1 x 1 1 x

1 +

x2 - 3 x - 4 x - 3 - 1 x + 2 2

and

are complex rational expressions. To simplify a complex rational expression means to write it as a rational expression reduced to lowest terms. This can be accomplished in either of two ways. *Some texts use the term complex fraction.

SECTION A.6  Rational Expressions

A51

Simplifying a Complex Rational Expression METHOD 1: Treat the numerator and denominator of the complex rational expression separately, performing whatever operations are indicated and simplifying the results. Follow this by simplifying the resulting rational expression. METHOD 2: Find the LCM of the denominators of all rational expressions that appear in the complex rational expression. Multiply the numerator and denominator of the complex rational expression by the LCM and simplify the result. Both methods are shown in the next example. By carefully studying each method, you can discover situations in which one method may be easier to use than the other.

EX AM PL E 10

Simplifying a Complex Rational Expression

Simplify:

Solution

1 3 + x 2 x + 3 4

x ≠ - 3, 0

Method 1: First perform the indicated operation in the numerator, and then divide. 1 3 1#x + 2#3 x + 6 + # x 2 2 x 2x x + 6# 4 = = = x + 3 x + 3 x + 3 2x x + 3 æ æ 4 4 4 Rule for adding quotients

=

Rule for dividing quotients

(x + 6) # 4 2(x + 6) 2 # 2 # (x + 6) = = # # # # 2 x (x + 3) 2 x (x + 3) x(x + 3)

æ Rule for multiplying quotients

Method 2: The rational expressions that appear in the complex rational expression are 1 , 2

3 , x

x + 3 4

The LCM of their denominators is 4x. Multiply the numerator and denominator of the complex rational expression by 4x and then simplify. 1 3 1 3 1 3 4x # a + b 4x # + 4x # + x x x 2 2 2 = = # 4x (x + 3) x + 3 x + 3 4x # a b 4 4 4 æ

æ Use the Distributive Multiply the Property in the numerator. numerator and denominator by 4x.

2 # 2x # =

1 3 + 4x # 2(x + 6) x 2 2x + 12 = = # 4 x (x + 3) x(x + 3) x(x + 3) 4 æ Simplify.

æ Factor.

r

A52

APPENDIX A  Review

EX AM PL E 11

Simplifying a Complex Rational Expression

Simplify:

Solution

x2 + 2 x - 4 2x - 2 - 1 x

x ≠ 0, 2, 4

Here Method 1 is used. 2(x - 4) x2 x2 x2 + 2x - 8 + 2 + x - 4 x - 4 x - 4 x - 4 = = 2x - 2 2x - 2 2x - 2 - x x - 1 x x x x (x + 4)(x - 2) (x + 4) (x - 2) x - 4 = = x - 2 x - 4 x (x + 4) # x x(x + 4) = = x - 4 x - 4

Now Work

PROBLEM

#

x x - 2

r

73

Application EX AM PL E 12

Solving an Application in Electricity An electrical circuit contains two resistors connected in parallel, as shown in Figure 25. If their resistances are R1 and R2 ohms, respectively, their combined resistance R is given by the formula

Figure 25 R1

R = R2

1 1 1 + R1 R2

Express R as a rational expression; that is, simplify the right-hand side of this formula. Evaluate the rational expression if R1 = 6 ohms and R2 = 10 ohms.

Solution

1 Using Method 2, consider 1 as the fraction . Then the rational expressions in the 1 complex rational expression are 1 , 1

1 , R1

1 R2

The LCM of the denominators is R1R2 . Multiply the numerator and denominator of the complex rational expression by R1R2 and simplify. 1 1 1 + R1 R2

=

1 # R1R2 a

1 1 # + b R1R2 R1 R2

=

R1R2 1 # 1 # R1R2 + RR R1 R2 1 2

=

R1R2 R2 + R1

Thus, R = If R1 = 6 and R2 = 10, then R =

R1R2 R2 + R1

6 # 10 60 15 = = ohms 10 + 6 16 4

r

SECTION A.6  Rational Expressions

A53

A.6 Assess Your Understanding Concepts and Vocabulary 2x3 - 4x is reduced x - 2

1. When the numerator and denominator of a rational expression contain no common factors (except 1 and - 1), the rational expression is in lowest terms.

3. True or False The rational expression to lowest terms. True

2. LCM is an abbreviation for least common multiple .

4. True or False The LCM of 2x3 + 6x2 and 6x4 + 4x3 is 4x3(x + 1). False

Skill Building In Problems 5–16, reduce each rational expression to lowest terms. 5.

3x + 9 x2 - 9

9.

24x2 12x2 - 6x

13.

3 x - 3

6.

4x 2x - 1

x2 + 4x - 5 x2 - 2x + 1

x + 5 x - 1

4x2 + 8x 12x + 24

x 3

10.

x2 + 4x + 4 x2 - 4

14.

x - x2 x + x - 2 2

7. x + 2 x - 2

11.

x x + 2

15.

-

x2 - 2x 3x - 6

x 3

8.

y2 - 25

y + 5

2y - 8y - 10

2(y + 1)

12.

- (x + 7)

16.

2

x2 + 5x - 14 2 - x

15x2 + 24x 3x2

5x + 8 x

3y2 - y - 2

y - 1

3y + 5y + 2

y + 1

2x2 + 5x - 3 1 - 2x

- (x + 3)

2

In Problems 17–34, perform the indicated operation and simplify the result. Leave your answer in factored form. 17.

3x + 6 # x 5x2 x2 - 4

21.

12 4x - 8 # - 3x 12 - 6x

3 5x(x - 2)

18.

3 # x2 2x 6x + 10

8 3x

22.

8x x - 1 27. 10x x + 1

4 5(x - 1)

x2 + 7x + 12 x2 - 7x + 12 31. 2 x + x - 12 x2 - x - 12

28.

(x + 3)2 (x - 3)2

*19.

6x - 27 # 2 5x 4x - 18

6x x2 - 4 25. 3x - 9 2x + 4

x2 + x - 6 # x2 - 25 *24. 2 x + 4x - 5 x2 + 2x - 15

2

3x 4(3x + 5)

x - 2 4x x - 4x + 4 12x 2

x2 + 7x + 6 x2 + x - 6 32. 2 x + 5x - 6 x2 + 5x + 6

4x2 # x3 - 64 2x x - 16

3 5x

*23.

4x (x - 2)(x - 3)

26.

12 # x3 + 1 x + x 4x - 2

*20.

2

2

x2 - 3x - 10 # x2 + 4x - 21 x2 + 2x - 35 x2 + 9x + 14 12x 5x + 20

3(x - 4)

2

4x x - 16

5x

2

4 - x 4 + x 29. 4x x2 - 16

3 x - 2

-

(x - 4)2

2x2 - x - 28 3x2 - x - 2 33. (x - 2)(x - 1) 4x2 + 16x + 7 3x2 + 11x + 6 (x + 1)(x + 2)

30.

4x

3 + x 3 - x x - 9 9x3 2

-

9x3 (x - 3)2

9x2 + 3x - 2 12x2 + 5x - 2 *34. (x - 1)(2x + 1) 9x2 - 6x + 1 8x2 - 10x - 3 (x - 4)(x + 3)

In Problems 35–52, perform the indicated operation and simplify the result. Leave your answer in factored form. x + 5 2

35.

5 x + 2 2

38.

3x2 9 2x - 1 2x - 1

3(x2 - 3)

41.

2x - 4 3x + 5 2x - 1 2x - 1

x + 9 2x - 1

44.

x 6 x - 1 1 - x

47.

2x - 3 x + x + 1 x - 1

50.

2x + 1 2x - 3 x - 1 x + 1

2x - 1

x + 6 x - 1 3x - 2x - 3 (x + 1)(x - 1)

36.

3 6 x x

39.

x + 1 2x - 3 + x - 3 x - 3

42.

5x - 4 x + 1 3x + 4 3x + 4

45.

4 2 x - 1 x + 2

(x - 1)(x + 2)

48.

3x 2x + x - 4 x + 3

(x - 4)(x + 3)

51.

1 x + x x2 - 4

2

-

2 (x - 1)(x + 1)

-

3 x 3x - 2 x - 3 4x - 5 3x + 4 2(x + 5) x(5x + 1) 2(x2 - 2) x(x - 2)(x + 2)

*37.

x2 4 2x - 3 2x - 3

40.

2x - 5 x + 4 + 3x + 2 3x + 2

43.

4 x + x - 2 2 - x

4 - x x - 2

46.

2 5 x + 5 x - 5

-

49.

x - 3 x + 4 x + 2 x - 2

*52.

x - 1 x + 2 x + 1 x3

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

3x - 1 3x + 2

3x + 35 (x - 5)(x + 5) - (11x + 2)

(x + 2)(x - 2)

A54

APPENDIX A  Review

In Problems 53–60, find the LCM of the given polynomials. *53. x2 - 4, x2 - x - 2

*54. x2 - x - 12, x2 - 8x + 16

*55. x3 - x, x2 - x

*56. 3x2 - 27, 2x2 - x - 15

*57. 4x3 - 4x2 + x, 2x3 - x2, x3

*58. x - 3, x2 + 3x, x3 - 9x

*59. x3 - x, x3 - 2x2 + x, x3 - 1

60. x2 + 4x + 4, x3 + 2x2, (x + 2)3

x2(x + 2)3

In Problems 61–72, perform the indicated operations and simplify the result. Leave your answer in factored form. x x - 2 x2 - 7x + 6 x - 2x - 24 3x 3x2 - 4x + 4 x - 4 64. - 2 x - 1 x - 2x + 1 (x - 1)2

*61.

*67.

2x + 3 x + 4 - 2 x2 - x - 2 x + 2x - 8

*70.

x 2 x + 1 + - 3 x (x - 1)2 x - x2

x x + 1 - 2 x - 3 x + 5x - 24 2 3 + *65. (x - 1)2(x + 1) (x - 1)(x + 1)2

4x 2 - 2 x2 - 4 x + x - 6 6 2 *66. (x + 2)2(x - 1) (x + 2)(x - 1)2

*62.

*68.

71.

*63.

2x - 3 x - 2 x2 + 8x + 7 (x + 1)2 -1 x(x + h)

1 1 1 a - b h x + h x

3 1 2 + 3 - 2 x x + x x - x2

*69.

72.

- (2x + h)

1 1 1 J - 2R h (x + h)2 x

x2(x + h)2

In Problems 73–84, perform the indicated operations and simplify the result. Leave your answer in factored form. 1 x 73. 1 1 x 1 +

1 x2 74. 1 3 - 2 x 4 +

x + 1 x - 1

x - 3 x + 4 x - 2 x + 1 *77. x + 1

81. 1 -

1 1 1 x

x + 1 x 75. x - 1 3 + x + 1 2 -

4x2 + 1 3x2 - 1

x - 2 x x + 1 x - 2 *78. x + 3

-1 x - 1

82. 1 -

1 1 1 1 - x

x x + 1 76. x - 1 2 x 1 -

(x - 1)(x + 1) 2x(2x + 1)

x - 2 x - 1 + x + 2 x + 1 *79. x 2x - 3 x + 1 x 1 x

83.

2(x - 1) - 1 + 3 3(x - 1)

-1

+ 2

*80.

3x - 1 2x + 1

84.

x (x + 1)2

2x + 5 x x x - 3 (x + 1)2 x2 x - 3 x + 3 4(x + 2) - 1 - 3 3(x + 2)

-1

- 1

3x + 2 x - 1

In Problems 85–92, expressions that occur in calculus are given. Reduce each expression to lowest terms. 85.

88.

91.

(2x + 3) # 3 - (3x - 5) # 2

19 (3x - 5)2

86.

x2 + 4 (x + 2)2(x - 2)2

89.

(3x - 5)2 x # 2x - (x2 - 4) # 1 (x2 - 4)

2

(x2 + 1) # 3 - (3x + 4) # 2x (x2 + 1)

2

-

(4x + 1) # 5 - (5x - 2) # 4 (5x - 2)2

13 (5x - 2)2

(3x + 1) # 2x - x2 # 3

x(3x + 2)

(3x + 1)2

(3x + 1)2

(x + 3)(3x - 1) (x2 + 1)2

92.

87.

90.

(x2 + 9) # 2 - (2x - 5) # 2x (x2 + 9)

2

x # 2x - (x2 + 1) # 1 (x + 1) 2

(x + 1)(x - 1)

2

(x2 + 1)2

(2x - 5) # 3x2 - x3 # 2

x2(4x - 15)

(2x - 5)2

(2x - 5)2

-

2(x2 - 5x - 9) (x2 + 9)2

Applications and Extensions 93. The Lensmaker’s Equation The focal length f of a lens with index of refraction n is 1 1 1 = (n - 1) J + R f R1 R2 where R1 and R2 are the radii of curvature of the front and back surfaces of the lens. Express f as a rational expression. Evaluate the rational expression for n = 1.5, R1 = 0.1 meter, and R2 = 0.2 meter. f =

R1 # R2 2 m ; (n - 1)(R1 + R2) 15

94. Electrical Circuits An electrical circuit contains three resistors connected in parallel. If their resistances are R1 , R2 , and R3 ohms, respectively, their combined resistance R is given by the formula 1 1 1 1 + + = R R1 R2 R3 Express R as a rational expression. Evaluate R for R1 = 5 ohms, R2 = 4 ohms, and R3 = 10 ohms. R =

R1R2R3 20 ; ohms R1R2 + R1R3 + R2R3 11

SECTION A.7  nth Roots; Rational Exponents

A55

Discussion and Writing 95. The following expressions are called continued fractions: 1 +

1 , 1 + x

1 1 1 + x

1

, 1 + 1 +

1

, 1 +

1

1 +

1 x

1 +

,...

1 1 +

1 1 +

1 x

Each simplifies to an expression of the form ax + b bx + c Trace the successive values of a, b, and c as you “continue” the fraction. Can you discover the patterns that these values follow? Go to the library and research Fibonacci numbers. Write a report on your findings. 96. Explain to a fellow student when you would use the LCM method to add two rational expressions. Give two examples of adding two rational expressions, one in which you use the LCM and the other in which you do not.

97. Which of the two methods given in the text for simplifying complex rational expressions do you prefer? Write a brief paragraph stating the reasons for your choice.

A.7 nth Roots; Rational Exponents PREPARING FOR THIS SECTION  Before getting started, review the following: r
&YQPOFOUT
4RVBSF
3PPUT
"QQFOEJY
"
4FDUJPO
"
QQ
"m"

Now Work the ‘Are You Prepared?’ problems on page A60.

OBJECTIVES 1 2 3 4

Work with nth Roots (p. A55) Simplify Radicals (p. A56) Rationalize Denominators (p. A57) Simplify Expressions with Rational Exponents (p. A58)

1 Work with nth Roots DEFINITION

The principal nth root of a real number a, n Ú 2 an integer, symbolized by n 2a, is defined as follows: n

1a = b means a = bn

In Words

n

The symbol 1a means “give me the number that, when raised to the power n, equals a.”

where a Ú 0 and b Ú 0 if n is even, and a, b are any real numbers if n is odd.

n

Notice that if a is negative and n is even, then 2a is not defined. When it is defined, the principal nth root of a number is unique. n The symbol 2a for the principal nth root of a is called a radical; the integer n is 2 called the index, and a is called the radicand. If the index of a radical is 2, then 2 a is called the square root of a, and the index 2 is omitted to simply write 2a. If the 3 index is 3, then 2 a is called the cube root of a.

EXAM PL E 1

Simplifying Principal nth Roots 3 3 3 (a) 2 8 = 2 2 = 2

(c)

The number whose cube is 8 is 2.

4 1 = 4 a1 b = 1 A 16 2 B 2 4

3 3 (b) 2 - 64 = 2 1 - 42 3 = - 4 6 (d) 2 1 - 22 6 = 0 - 2 0 = 2

r

A56

APPENDIX A  Review

These are examples of perfect roots, since each simplifies to a rational number. Notice the absolute value in Example 1(d). If n is even, then the principal nth root must be nonnegative. In general, if n Ú 2 is an integer and a is a real number, then n

2a n = a

2a = 0 a 0 n

Now Work

n

PROBLEM

if n Ú 3 is odd

(1a)

if n Ú 2 is even

(1b)

7

Radicals provide a way of representing many irrational real numbers. For example, there is no rational number whose square is 2. In terms of radicals, it can be said that 12 is the positive number whose square is 2.

EX A MPL E 2

Using a Calculator to Approximate Roots 5 Use a calculator to approximate 1 16.

Solution Figure 26

Figure 26 shows the result using a TI-84 Plus graphing calculator.

Now Work

PROBLEM

r

101

2 Simplify Radicals Let n Ú 2 and m Ú 2 denote positive integers, and let a and b represent real numbers. Assuming that all radicals are defined, the following properties are useful.

Properties of Radicals n

n

n

2ab = 2a 2b

(2a)

n

2a n a = n Ab 2b

2am = ( 2aa22 m n

n

(2b) (2c)

When used in reference to radicals, the direction to “simplify” will mean to remove from the radicals any perfect roots that occur as factors.

EX AMPL E 3

Simplifying Radicals

(a) 232 = 216 # 2 = 216 # 22 = 422 c

c

Factor out 16, (2a) a perfect square.

3 3 3 3 3 3# 3 3 (b) 2 16 = 2 8#2 = 2 8# 2 2 = 2 2 22 = 22 2

c

c

Factor out 8, a perfect cube.

(2a)

SECTION A.7  nth Roots; Rational Exponents

A57

3 3 3 (c) 2 - 16x4 = 2 - 8 # 2 # x3 # x = 2 1 - 8x3 2 12x2

c

c

Factor perfect cubes inside radical

Group perfect cubes.

3 3 3 3 = 2 1 - 2x2 3 # 2x = 2 1 - 2x2 3 # 2 2x = - 2x2 2x

c

(2a)

(d)

2x 16x5 24 x4 x = 4 = 4 a b A 81 A 34 B 3 4

Now Work

PROBLEMS

4

#x =

2x 4 # 4 2x 4 2 4 b 2x = ` ` 2x =  x 2 x 3 3 3 B 4

a

11 AND 17

r

Two or more radicals can be added or subtracted, provided that they have the same index and the same radicand. Such radicals are called like radicals.

EXAM PL E 4

Combining Like Radicals

(a) - 8212 + 23 = - 824 # 3 + 23

= - 8 # 2423 + 23 = - 1623 + 123 = - 1523

3 3 3 3 3 3 3 3 3 (b) 2 8x4 + 2 - x + 42 27x = 2 2xx + 2 - 1 # x + 42 3 x

3 3 3 3 3 3# 3 = 2 12x2 3 # 2 x + 2 -1 # 2 x + 42 3 2x 3 3 3 = 2x2 x - 1# 2 x + 122 x 3 = 12x + 112 2 x

Now Work

PROBLEM

r

33

3 Rationalize Denominators When radicals occur in quotients, it is customary to rewrite the quotient so that the new denominator contains no radicals. This process is referred to as rationalizing the denominator. The idea is to multiply by an appropriate expression so that the new denominator contains no radicals. For example: If a Denominator Contains the Factor

Multiply by

To Obtain a Denominator Free of Radicals

13

13

1 23 2 2

13 + 1

13 - 1

12 - 3

12 + 3

15 - 13

15 + 13

1 25 2 2

3 24

3 22

3 3 3 # 22 24 = 28 = 2

1 23 2

2

1 22 2 2

= 3 - 12 = 3 - 1 = 2 - 32 = 2 - 9 = - 7 -

1 23 2 2

= 5 - 3 = 2

In rationalizing the denominator of a quotient, be sure to multiply both the numerator and the denominator by the expression.

EXAM PL E 5

Rationalizing Denominators Rationalize the denominator of each expression: (a)

1 23

(b)

5 3

424

(c)

22 23 - 322

A58

APPENDIX A  Review

Solution

(a) The denominator contains the factor 13, so multiply the numerator and denominator by 13 to obtain 1 23

=

# 23

1

23 23

23

1 23 2

=

2

=

23 3

3 (b) The denominator contains the factor 2 4, so multiply the numerator and 3 denominator by 22 to obtain

5 3 42 4

=

5

# 23 2 3

3 42 4 22

=

3 52 2 3 42 8

=

3 52 2 8

(c) The denominator contains the factor 13 - 312, so multiply the numerator and denominator by 13 + 312 to obtain 22 23 - 322

=

=

Now Work

22

# 23 + 322

23 - 322 23 + 322

2223 + 3 1 22 2

2

=

3 - 18

PROBLEM

=

22 1 23 + 322 2

1 23 2 2

-

1 322 2 2

26 + 6 6 + 26 = - 15 15

r

47

4 Simplify Expressions with Rational Exponents Radicals are used to define rational exponents.

DEFINITION

If a is a real number and n Ú 2 is an integer, then n

a1>n = 2a

(3)

n

provided that 2a exists.

n

Note that if n is even and a 6 0, then 2a and a1>n do not exist.

EX A MPL E 6

Writing Expressions Containing Fractional Exponents as Radicals (a) 41>2 = 24 = 2 3 (c) 1 - 272 1>3 = 2 - 27 = - 3

DEFINITION

(b) 81>2 = 28 = 222 3 3 (d) 161>3 = 2 16 = 22 2

r

If a is a real number and m and n are integers containing no common factors, with n Ú 2, then am>n = 2am = 1 2aa22 m n

n

(4)

n

provided that 2a exists. Two comments about equation (4) should be noted: m must be in lowest terms, and n must be positive. n n n 2. In simplifying the rational expression am>n, either 2am or 1 2a2 m may be used, the choice depending on which is easier to simplify. Generally, taking the root first, n as in 1 2a2 m, is easier. 1. The exponent

SECTION A.7  nth Roots; Rational Exponents

EXAM PL E 7

Using Equation (4)

1 24 2 3

(a) 43>2 =

(c) 1322 -2>5 =

5 12 32 2 -2

Now Work

4 3 12 - 82

(b) 1 - 82 4>3 =

= 23 = 8

PROBLEM

= 1 - 22 4 = 16

1 225 2 3

1 (d) 256>4 = 253>2 = 4

= 2-2 =

A59

= 53 = 125

r

55

It can be shown that the Laws of Exponents (page A8) hold for rational exponents. The next example illustrates using the Law of Exponents to simplify.

EXAM PL E 8

Simplifying Expressions Containing Rational Exponents Simplify each expression. Express your answer in such a way that only positive exponents occur. Assume that the variables are positive.

Solution

(a) 1x2>3 y2 1x -2 y2

1>2

(a) 1x2>3 y21x -2 y2

1>2

(b) ¢

2x1>3 y2>3

≤

= 1x2>3 y231x -2 2

-3

(c) ¢

1>2 1>2

y

4

= x2>3 yx -1 y1>2

(a m)n = a mn

= x -1>3 y3>2

am # an = am + n

=

(c) ¢

2x1>3 y2>3

≤

-3

9x2 y1>3 x1>3 y

≤

= ¢ 1>2

y2>3 2x

= ¢

Now Work

x1>3 y

y3>2

a -n =

x1>3 3

≤ = 1>3 9x2 - 11>32 y1 - 11>32

1y2>3 2

3

12x1>3 2 ≤

1>2

PROBLEM

3

= ¢

=

y2

23 1x1>3 2

9x5>3 y2>3

≤

≤

1>2

(ab)n = a nb n

= 1x2>3 # x -1 2 1y # y1>2 2

(b) ¢

9x2 y1>3

1>2

3

=

=

1 an

y2 8x

91>2 1x5>3 2 1y2>3 2

1>2

1>2

=

3x5>6

r

y1>3

71

The next two examples illustrate some algebra needed for certain calculus problems.

EXAM PL E 9

Writing an Expression as a Single Quotient Write the following expression as a single quotient in which only positive exponents appear. 1x2 + 12

Solution

1x2 + 12

1>2

+ x#

1>2

+ x#

1 2 1x + 12 -1>2 # 2x 2

1 2 x2 - 1>2 # 1>2 1x + 12 2 x = 1x2 + 12 + 1>2 2 1x2 + 12 = = =

Now Work

PROBLEM

77

1x2 + 12

1>2

1x2 + 12

1x2 + 12

1>2

+ x2

1>2

1x2 + 12 + x2 1x2 + 12

1>2

2x2 + 1

1x2 + 12

1>2

r

A60

APPENDIX A  Review

Factoring an Expression Containing Rational Exponents

EX AM PL E 10

4 Factor: x1>3 12x + 12 + 2x4>3 3 Begin by writing 2x4>3 as a fraction with 3 as the denominator.

Solution

4x1>3(2x + 1) 4x1>3(2x + 1) + 6x4>3 4 1>3 6x4>3 x 12x + 12 + 2x4>3 = + = 3 3 3 c 3 Add the two fractions.

2x1>3[2(2x + 1) + 3x] 2x1>3(7x + 2) = = 3 3 c

c

2 and x1>3 are common factors.

Simplify.

Now Work

PROBLEM

r

89

Historical Note

T

he radical sign, 2 , was first used in print by Christoff Rudolff in 1525. It is thought to be the manuscript form of the letter r (for the Latin word radix = root), although this is not quite conclusively confirmed. It took a long time for 2 to become the standard symbol for a square root and much longer to standardize 4 5 3 2, 2, 2 and so on. The indexes of the root were placed in every conceivable position, with

3

4

all being variants for 28. The notation 2 216 was popular for 216. By the 1700s, the index had settled where we now put it. The bar on top of the present radical symbol, as follows,

2a2 + 2ab + b2 is the last survivor of the vinculum, a bar placed atop an expression to indicate what we would now indicate with parentheses. For example,

3

28,

2 3 8, and

ab + c = a(b + c)

28 3

A.7 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages in red. 1. ( - 3)2 =

; - 32 =

9

-9

2. 216 =

(p. A8)

; 2( - 4)2 =

4

4

(p. A9)

Concepts and Vocabulary n

3. In the symbol 2a, the integer n is called the 3 5. We call 2a the

cube

root

index

5 4. True or False 2 - 32 = - 2 True

.

4 6. True or False 2 ( - 3)4 = - 3

of a.

False

Skill Building In Problems 7–42, simplify each expression. Assume that all variables are positive when they appear. 3 7. 227

11. 28

4 8. 2 16

3

4 12 8 15. 2 x y

19. 236x 23.

1 25 29 2

2

35.

20. 29x 3

24.

1523

5

223 *32. + 1

*36.

17.

3x 2x 4

4

x9 y 7

B xy

3

3 10. 2 -1

-2 4

3

x2y

18. 6x2x

1 326 2 1 222 2

30013

25.

225

29. - 218 + 228

1 25 - 2 2 1 25 1 1x + 25 2 2

+ 32

3

3

33. 522 - 2254 3

-1

14. 248x

2

3

4

- 2x2x

21. 23x 212x

2

28. 625 - 425

722

1 23 + 3 2 1 23 - 1 2 1 1x - 1 2 2 x - 22x

13. 2 - 8x

x2y

1 23 210 2 3

3

322

5 10 5 16. 2 x y

62x

3

3

12. 254

x3y2

27. 322 + 422 31.

3

222

3 9. 2 -8

2

3

*37. 216x4 - 22x

3

- 22

3xy2

3

B 81x y

4 2

4 2x1 3x

1 3x

22. 25x 220x3

1223 26. 22

5

10x2

1 528 2 1 - 323 2

30. 2212 - 3227 3

3

34. 9224 - 281 4

4

*38. 232x + 22x5

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

- 3026 - 523

3 152 3

A61

SECTION A.7  nth Roots; Rational Exponents

39. 28x3 - 3250x

(2x - 15) 12x

3 3 3 41. 2 16x4 y - 3x2 2xy + 52 - 2xy4

40. 3x29y + 4225y 3 - (x + 5y) 1 2xy

(9x + 20) 1y

3 42. 8xy - 225x2 y2 + 2 8x3 y3

5xy

In Problems 43–54, rationalize the denominator of each expression. Assume that all variables are positive when they appear. 43.

47.

51.

22 2

1 22

44.

23

(5 + 12) 13

5 - 22

23

3 524 2

5 3

22

48.

52.

2

223 3

23

45.

22

( 27 - 2) 12

27 + 2

3

-2 3

29

-

3 21 3 3

49.

*53.

- 23

215 5

-

25

2 - 25

46.

825 - 19 41

2 + 325

50.

2x + h - 1x

*54.

2x + h + 1x

- 23

-

28

26 4 9 - 523 3

23 - 1 223 + 3

2x + h + 2x - h 2x + h - 2x - h

In Problems 55–66, simplify each expression. 55. 82>3

56. 43>2

4

59. 163>2

60. 253>2

64

9 3>2 63. a b 8

64. a

2722 32

57. ( - 27)1>3

8

27 2>3 b 8

8 -3>2 65. a b 9

9 4

58. 163>4

1 27

61. 9-3>2

125

-3

8 1 64

62. 16 -3>2 66. a

2722 32

8 -2>3 b 27

9 4

In Problems 67–74, simplify each expression. Express your answer in such a way that only positive exponents occur.Assume that the variables are positive. 67. x3>4 x1>3 x -1>2 71.

(x2 y)

1>3

68. x2>3 x1>2 x -1>4

x7>12

(xy2)

2>3

x2>3y

x2>3 y2>3

72.

(xy)1>4(x2 y2)

x11>12 1>2

(x2 y)3>4

69. (x3 y6)

y1>2

73.

x1>4

1>3

xy2

(16x2 y -1>3)

70. (x4y8)

3>4

8x5>4

(xy2)1>4

74.

y3>4

3>4

x3y6

(4x -1 y1>3)

3>2

(xy)3>2

8 x3y

Applications and Extensions In Problems 75–88, expressions that occur in calculus are given. Write each expression as a single quotient in which only positive exponents and/or radicals appear. 75.

x

+ 2(1 + x)1>2

(1 + x)1>2

x 7 -1

3x + 2

1 77. 2x(x2 + 1)1>2 + x2 # (x2 + 1)-1>2 # 2x 2 *79. 24x + 3 #

83.

(x + 4)1>2 - 2x(x + 4)-1>2 x + 4 x2

85.

- (x - 1)

1>2

x 1 + x2

87.

2

(x2 - 1)

22x

2

x 7 5

*80.

x 7 -4

+ x1>2

x 7 0

3 2 8x + 1 3

32(x - 2)

2

+

82.

3>2

4 - x (x + 4)

3x + 1 2x1>2

84.

3>2

x ≠ -1

3 2 x - 2 3

242(8x + 1)

2x2 + 1 - x #

2 + x 2(1 + x)

2x1>2

4x + 3 3(x + 1)2>3

x ≠ 2, x ≠ -

2

1 8

2x

22x2 + 1 x + 1 2

1 (x2 + 1)3>2

(9 - x2)1>2 + x2(9 - x2)-1>2

-3 6 x 6 3

9 - x2

9 (9 - x2)3>2

1>2

x 6 - 1 or x 7 1

2

- 2x2x

(1 + x2)

x 7 -1

1 + x

1 78. (x + 1)1>3 + x # (x + 1)-2>3 3

(x2 + 1)1>2

1

221 + x 1 + x

81.

x(3x2 + 2)

1 1 + 2x - 5 # 21x - 5 514x + 3

21 + x - x #

76.

(1 + x)1>2

x 7 0

1 - 3x2 22x(1 + x2)2

1 x (x - 1) 2

2

1>2

86.

(x2 + 4)1>2 - x2(x2 + 4)-1>2

4

x + 4

(x + 4)3>2

2

2x(1 - x2)1>3 +

88.

2 3 x (1 - x2)-2>3 3

(1 - x2)2>3

2

x ≠ - 1, x ≠ 1

2x(3 - 2x2) 3(1 - x2)4>3

A62

APPENDIX A  Review

In Problems 89–98, expressions that occur in calculus are given. Factor each expression. Express your answer in such a way that only positive exponents occur. 3 4 1 1 3>2 4>3 1>3 89. (x + 1) + x # (x + 1)1>2 x Ú - 1 (5x + 2)(x + 1)1>2 90. (x2 + 4) + x # (x2 + 4) # 2x (x2 + 4)1>3(11x2 + 12) 2 2 3 3 *91. 6x1>2(x2 + x) - 8x3>2 - 8x1>2 93. 3(x2 + 4)

4>3

+ x # 4(x2 + 4)

1>3

92. 6x1>2(2x + 3) + x3>2 # 8 x Ú 0

x Ú 0

# 2x

*95. 4(3x + 5)1>3(2x + 3)3>2 + 3(3x + 5)4>3(2x + 3)1>2 x Ú 97. 3x -1>2 +

3 1>2 x 2

x 7 0

94. 2x(3x + 4)4>3 + x2 # 4(3x + 4)1>3

(x2 + 4)1>3(11x2 + 12) 3 2

2x(3x + 4)1>3(5x + 4)

*96. 6(6x + 1)1>3(4x - 3)3>2 + 6(6x + 1)4>3(4x - 3)1>2 x Ú

3(x + 2) 2x

2x1>2(10x + 9)

98. 8x1>3 - 4x -2>3

1>2

x ≠ 0

3 4

4(2x - 1) x2>3

In Problems 99–106, use a calculator to approximate each radical. Round your answer to two decimal places. 99. 22 103.

100. 27

1.41

2 + 23 3 - 25

4.89

104.

3 101. 2 4

2.65

25 - 2 22 + 4

3 102. 2 -5

1.59

3

105.

0.04

325 - 22 23

2.15

106.

- 1.71

3 223 - 2 4

22

1.33

107. Calculating the Amount of Gasoline in a Tank A Shell station stores its gasoline in underground tanks that are right circular cylinders lying on their sides. See the illustration. The volume V of gasoline in the tank (in gallons) is given by the formula V = 40h2

96 - 0.608 Ah

where h is the height of the gasoline (in inches) as measured on a depth stick. (a) If h = 12 inches, how many gallons of gasoline are in the tank? (b) If h = 1 inch, how many gallons of gasoline are in the tank?

15,660.4 gal

390.7 gal

108. Inclined Planes The final velocity y of an object in feet per second (ft>sec) after it slides down a frictionless inclined plane of height h feet is y = 264h + y20 where y0 is the initial velocity (in ft/sec) of the object. v0

(a) What is the final velocity y of an object that slides down a frictionless inclined plane of height 4 feet? Assume that the initial velocity is 0. 16 ft/sec (b) What is the final velocity y of an object that slides down a frictionless inclined plane of height 16 feet? Assume that the initial velocity is 0. 32 ft/sec (c) What is the final velocity y of an object that slides down a frictionless inclined plane of height 2 feet with an initial velocity of 4 ft/sec ? 12 ft/sec

h v

Problems 109–112 require the following information. Period of a Pendulum The period T, in seconds, of a pendulum of length l, in feet, may be approximated using the formula T = 2p

l A 32

In Problems 109–112, express your answer both as a square root and as a decimal. 109. Find the period T of a pendulum whose length is 64 feet. 222p ≈ 8.89 sec 110. Find the period T of a pendulum whose length is 16 feet. p22 ≈ 4.44 sec p23 111. Find the period T of a pendulum whose length is 8 inches. ≈ 0.91 sec 6 112. Find the period T of a pendulum whose length is 4 inches. p26 ≈ 0.64 sec 12

Discussion and Writing 113. Give an example to show that 2a2 is not equal to a. Use it to explain why 2a2 = 0 a 0 .

‘Are You Prepared?’ Answers 1. 9; - 9

2. 4; 4

SECTION A.8   Solving Equations

A63

A.8 Solving Equations PREPARING FOR THIS SECTION  Before getting started, review the following: r
4RVBSF
3PPUT
"QQFOEJY
"
4FDUJPO
"
QQ
"m"

r
nth Roots; Rational Exponents (Appendix A, Section A.7, pp. A55–A60)

r
'BDUPSJOH
1PMZOPNJBMT
"QQFOEJY
"
4FDUJPO
"
pp. A32–A38) r
;FSP1SPEVDU
1SPQFSUZ
"QQFOEJY
"
4FDUJPO
"
p. A4) r
3BUJPOBM
&YQSFTTJPOT
"QQFOEJY
"
4FDUJPO
"
pp. A45–A50) Now Work the ‘Are You Prepared?’ problems on page A70.

OBJECTIVES 1 2 3 4

Solve Linear Equations (p. A64) Solve Rational Equations (p. A66) Solve Equations by Factoring (p. A67) Solve Radical Equations (p. A68)

An equation in one variable is a statement in which two expressions, at least one containing the variable, are set equal. The expressions are called the sides of the equation. Since an equation is a statement, it may be true or false, depending on the value of the variable. Unless otherwise restricted, the admissible values of the variable are those in the domain of the variable. Those admissible values of the variable, if any, that result in a true statement are called solutions, or roots, of the equation. To solve an equation means to find all the solutions of the equation. For example, the following are all equations in one variable, x: x + 5 = 9

x2 + 5x = 2x - 2

x2 - 4 = 0 x + 1

2x2 + 9 = 5

The first of these statements, x + 5 = 9, is true when x = 4 and false for any other choice of x. Thus 4 is a solution of the equation x + 5 = 9. Also, 4 satisfies the equation x + 5 = 9, because, when 4 is substituted for x, a true statement results. Sometimes an equation will have more than one solution. For example, the equation x2 - 4 = 0 x + 1 has x = - 2 and x = 2 as solutions. Usually, set notation is used to write the solution of an equation. This set is called the solution set of the equation. For example, the solution set of the equation x2 - 9 = 0 is 5 - 3, 36 . Some equations have no real solution. For example, x2 + 9 = 5 has no real solution, because there is no real number whose square, when added to 9, equals 5. An equation that is satisfied for every choice of the variable for which both sides are defined is called an identity. For example, the equation 3x + 5 = x + 3 + 2x + 2 is an identity, because this statement is true for any real number x. One method for solving equations requires that a series of equivalent equations be developed from the original equation until an obvious solution results. For example, all the following equations are equivalent, because each has only the solution x = 5. 2x + 3 = 13 2x = 10 x = 5

A64

APPENDIX A  Review

The question, though, is “How do I obtain an equivalent equation?” In general, there are five ways to do so.

Procedures That Result in Equivalent Equations 1. Interchange the two sides of the equation: 3 = x is equivalent to x = 3 2. Simplify the sides of the equation by combining like terms, eliminating parentheses, and so on: x + 2 + 6 = 2x + 31x + 12 is equivalent to x + 8 = 5x + 3 3. Add or subtract the same expression on both sides of the equation: 3x - 5 = 4 is equivalent to 13x - 52 + 5 = 4 + 5 4. Multiply or divide both sides of the equation by the same nonzero expression: 3x 6 = x ≠ 1 x - 1 x - 1 3x # 6 # is equivalent to 1x - 12 = 1x - 12 x - 1 x - 1 5. If one side of the equation is 0 and the other side can be factored, use the Zero-Product Property and set each factor equal to 0. x1x - 32 = 0 is equivalent to x = 0 or x - 3 = 0

WARNING Squaring both sides of an equation does not necessarily lead to an equivalent equation. 䊏

Whenever it is possible to solve an equation in your head, do so. For example: The solution of 2x = 8 is x = 4. The solution of 3x - 15 = 0 is x = 5.

Now Work

PROBLEM

9

Some specific types of equations that can be solved algebraically to obtain exact solutions are now introduced, starting with linear equations.

1 Solve Linear Equations Linear equations are equations such as 3x + 12 = 0

3 1 x - = 0 4 5

0.62x - 0.3 = 0

A linear equation in one variable is equivalent to an equation of the form ax + b = 0 where a and b are real numbers and a ≠ 0. Sometimes a linear equation is called a first-degree equation, because the left side is a polynomial in x of degree 1.

SECTION A.8   Solving Equations

EXAM PL E 1

A65

Solving a Linear Equation Solve the equation: 31x - 22 = 51x - 12 31x - 22 = 51x - 12

Solution

3x - 6 = 5x - 5 3x - 6 - 5x = 5x - 5 - 5x - 2x - 6 = - 5

Subtract 5x from each side. Simplify.

- 2x - 6 + 6 = - 5 + 6 - 2x = 1 -2x 1 = -2 -2 x = -

Use the Distributive Property.

Add 6 to each side. Simplify. Divide each side by - 2.

1 2

Simplify.

1 in the expression on the left side of the equation and simplify. 2 1 Let x = - in the expression on the right side of the equation and simplify. 2 If the two expressions are equal, the solution checks.

Check: Let x = -

31x - 22 = 3a -

1 5 15 - 2b = 3a - b = 2 2 2

51x - 12 = 5a -

1 3 15 - 1b = 5a - b = 2 2 2

1 Since the two expressions are equal, the solution x = - checks. The solution 2 1 set is e - f. 2

r

Now Work

PROBLEM

15

The next example illustrates the solution of an equation that does not appear to be linear but leads to a linear equation upon simplification.

EXAM PL E 2

Solving an Equation That Leads to a Linear Equation Solve the equation: 12x - 12 1x - 12 = 1x - 52 12x - 52

Solution

12x - 12 1x - 12 = 1x - 52 12x - 52 2x2 - 3x + 1 = 2x2 - 15x + 25 2x - 3x + 1 - 2x - 3x + 1 - 3x + 1 - 1 - 3x - 3x + 15x 12x 2

2

= = = = = =

2x - 15x - 15x - 15x - 15x 24 2

12x 24 = 12 12 x = 2

15x + 25 - 2x + 25 + 25 - 1 + 24 + 24 + 15x

Multiply and combine like terms. 2

Subtract 2x2 from each side. Simplify. Subtract 1 from each side. Simplify. Add 15x to each side. Simplify. Divide each side by 12. Simplify.

A66

APPENDIX A  Review

Check: 12x - 12 1x - 12 = 12 # 2 - 12 12 - 12 = 132 112 = 3

1x - 52 12x - 52 = 12 - 52 12 # 2 - 52 = 1 - 32 1 - 12 = 3

Since the two expressions are equal, the solution checks. The solution set is 5 26 .

Now Work

PROBLEM

r

25

2 Solve Rational Equations A rational equation is an equation that contains a rational expression. Examples of rational equations are 3 2 = + 7 and x + 1 x - 1

x - 5 3 = x - 4 x + 2

To solve a rational equation, multiply both sides of the equation by the least common multiple of the denominators of the rational expressions that make up the rational equation.

Solving a Rational Equation

EX A MPL E 3

Solve the equation:

Solution

3 1 7 = + x - 2 x - 1 1x - 12 1x - 22

First, note that the domain of the variable is 5 x 0 x ≠ 1, x ≠ 26 . Clear the equation of rational expressions by multiplying both sides by the least common multiple of the denominators of the three rational expressions, 1x - 12 1x - 22.

3 1 7 = + x - 2 x - 1 1x - 12 1x - 22 1x - 12 1x - 22

3 1 7 = 1x - 12 1x - 22 c + d x - 2 x - 1 1x - 12 1x - 22 3x - 3 = 1x - 12 1x - 22

1 7 + 1x - 12 1x - 22 x - 1 1x - 12 1x - 22

3x - 3 = 1x - 22 + 7

Use the Distributive Property on each side; divide out common factors on the right. Rewrite the equation.

3x - 3 = x + 5 2x = 8

Combine like terms. Add 3 to each side; subtract x from each side.

x = 4

Check:

Multiply both sides by (x - 1)(x - 2); divide out common factors on the left.

Divide by 2.

3 3 3 = = x - 2 4 - 2 2 1 7 1 7 1 7 2 7 9 3 + = + = + # = + = = x - 1 1x - 12 1x - 22 4 - 1 14 - 12 14 - 22 3 3 2 6 6 6 2 Since the two expressions are equal, x = 4 checks. The solution set is 5 46 .

Now Work

PROBLEM

35

r

SECTION A.8   Solving Equations

A67

Sometimes the process of creating equivalent equations leads to apparent solutions that are not solutions of the original equation. These are called extraneous solutions.

EXAM PL E 4

A Rational Equation with No Solution Solve the equation:

Solution

3x 3 + 2 = x - 1 x - 1

First, note that the domain of the variable is 5 x 0 x ≠ 16 . Since the two rational expressions in the equation have the same denominator, x - 1, multiply both sides by x - 1 to clear the rational expressions from the equation. The resulting equation is equivalent to the original equation for x ≠ 1. 3x 3 + 2 = x - 1 x - 1 a

3x x - 1

#

3x 3 + 2b # 1x - 12 = x - 1 x - 1

#

1x - 12

1x - 12 + 2 # 1x - 12 = 3

Multiply both sides by x - 1; divide out common factors on the right. Use the Distributive Property on the left side; divide out common factors on the left.

3x + 2x - 2 = 3

Simplify.

5x - 2 = 3

Combine like terms.

5x = 5

Add 2 to each side.

x = 1

Divide both sides by 5.

The solution appears to be 1. But recall that x = 1 is not in the domain of the variable, so x = 1 is an extraneous solution. The equation has no solution. The solution set is ∅.

r

Now Work

PROBLEM

37

3 Solve Equations by Factoring Many equations can be solved by factoring and then using the Zero-Product Property.

EXAM PL E 5

Solving Equations by Factoring Solve the equations: (a) x2 = 4x

Solution

(b) x3 - x2 - 4x + 4 = 0

(a) Begin by collecting all terms on one side. This results in 0 on one side and an expression to be factored on the other. x2 = 4x x2 - 4x = 0 x(x - 4) = 0 x = 0 or x - 4 = 0 x = 4

Factor. Apply the Zero-Product Property.

A68

APPENDIX A  Review

x = 0: 02 = 4 # 0 x = 4: 42 = 4 # 4

Check:

So 0 is a solution. So 4 is a solution.

The solution set is {0, 4}. (b) Group the terms of x3 - x2 - 4x + 4 = 0 as follows: 1x3 - x2 2 - 14x - 42 = 0

Factor out x2 from the first grouping and 4 from the second. x2 1x - 12 - 41x - 12 = 0

This reveals the common factor 1x - 12, so 1x2 - 42 1x - 12 = 0

1x - 22 1x + 22 1x - 12 = 0 x - 2 = 0 or

Factor again.

x + 2 = 0 or

x = 2

x - 1 = 0

x = -2

Set each factor equal to 0.

x = 1 Solve.

1 - 22 3 - 1 - 22 2 - 41 - 22 + 4 = - 8 - 4 + 8 + 4 = 0

Check: x = - 2:

−2 is a solution.

x = 1: 1 - 1 - 4112 + 4 = 1 - 1 - 4 + 4 = 0

1 is a solution.

x = 2: 23 - 22 - 4122 + 4 = 8 - 4 - 8 + 4 = 0

2 is a solution.

3

2

The solution set is 5 - 2, 1, 26 .

Now Work

PROBLEM

r

41

4 Solve Radical Equations When the variable in an equation occurs in a square root, cube root, and so on—that is, when it occurs in a radical—the equation is called a radical equation. Sometimes a suitable operation changes a radical equation to one that is linear or quadratic. A commonly used procedure is to isolate the most complicated radical on one side of the equation and then eliminate it by raising each side to a power equal to the index of the radical. Care must be taken, however, because apparent solutions that are not, in fact, solutions of the original equation may result. Recall that these are called extraneous solutions. In radical equations, extraneous solutions may occur when the index of the radical is even. Therefore, it is important to check all answers when working with radical equations.

EX A MPL E 6

Solving a Radical Equation 3

Find the real solutions of the equation: 22x - 4 - 2 = 0

Solution

The equation contains a radical whose index is 3. Isolate it on the left side: 3

22x - 4 - 2 = 0 3 22x - 4 = 2 Because the index of the radical is 3, raise each side to the third power and solve. 3 12 2x

- 42 2x - 4 2x x 3

= = = =

23 8 12 6

Raise each side to the power 3. Simplify. Add 4 to both sides. Divide both sides by 2.

3 3 3 Check: 2 2162 - 4 - 2 = 2 12 - 4 - 2 = 2 8 - 2 = 2 - 2 = 0

The solution set is 5 66 .

Now Work

PROBLEM

r 45

SECTION A.8   Solving Equations

EXAM PL E 7

A69

Solving a Radical Equation Find the real solutions of the equation: 2x - 1 = x - 7

Solution

Square both sides since the index of a square root is 2. 2x - 1 = 2 - 12 = x - 1 = 2 x - 15x + 50 = 1x - 102 1x - 52 = x = 10 or x =

1 2x

Check: x = 10: x = 5:

x - 7 1x - 72 2 Square both sides. 2 x - 14x + 49 Remove parentheses. 0 Put in standard form. 0 Factor. 5 Apply the Zero-Product Property and solve.

2x - 1 = 210 - 1 = 29 = 3 and x - 7 = 10 - 7 = 3 2x - 1 = 25 - 1 = 24 = 2 and x - 7 = 5 - 7 = - 2

The apparent solution x = 5 is extraneous; the only solution of the equation is x = 10. The solution set is 5 106 .

Now Work

r

PROBLEM

51

Sometimes, it is necessary to raise each side to a power more than once in order to solve a radical equation.

EXAM PL E 8

Solving a Radical Equation Find the real solutions of the equation: 22x + 3 - 2x + 2 = 2

Solution

First, isolate the more complicated radical expression (in this case, 22x + 3) on the left side: 22x + 3 = 2x + 2 + 2 Now square both sides (the index of the radical is 2).

1 22x

+ 32 = 2

2x + 3 =

1 2x 1 2x

+ 2 + 222

+ 2 2 + 42x + 2 + 4 Remove parentheses. 2

2x + 3 = x + 2 + 42x + 2 + 4

Simplify.

2x + 3 = x + 6 + 42x + 2

Combine like terms.

The equation still contains a radical. Isolate the remaining radical on the right side and again square both sides. x - 3 1x - 32 2 x2 - 6x + 9 x2 - 22x - 23 1x - 232 1x + 12 x = 23 or x Check: x = 23: x = - 1:

= = = = = =

42x + 2 161x + 22 16x + 32 0 0 -1

Isolate the radical on the right side. Square both sides. Remove parentheses. Put in standard form. Factor. Apply the Zero-Product Property and solve.

221232 + 3 - 223 + 2 = 249 - 225 = 7 - 5 = 2 221 - 12 + 3 - 2 - 1 + 2 = 21 - 21 = 1 - 1 = 0

The apparent solution x = - 1 is extraneous; the only solution is x = 23. The solution set is 5 236 .

Now Work

r

PROBLEM

57

A70

APPENDIX A  Review

A.8 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. Find the least common multiple of the denominators of 5 3 . (p. A49) and 2 2 x - 4 x - 3x + 2 1x - 22 1x + 22 1x - 12

3. Factor: (a) 2x2 - 10x (p. A32)

2x1x - 52

(b) x3 + 3x2 - 4x - 12 (p. A36)

2. The solution set of the equation 1x - 32 13x + 52 = 0 is . (p. A4) 5 e 3, - f 3

3 3 4. (a) 2 x =

(b)

4 12 x

x

- 324 =

. (p. A56) x - 3

1x + 22 1x - 22 1x + 32 . (p. A56)

Concepts and Vocabulary 5. Two equations that have the same solution set are called equivalent equations .

7. When an apparent solution does not satisfy the original equation, it is called a(n) extraneous solution.

6. An equation that is satisfied for every choice of the variable for which both sides are defined is called a(n) identity .

8. True or False Some equations have no solution. True

Skill Building In Problems 9–66, solve each equation. 9. 2x - 3 = 5

10. 3x + 4 = - 8

{4}

13. 6 - x = 2x + 9

14. 3 - 2x = 2 - x

{91}

21. 0.9t = 0.4 + 0.1t

22. 0.9t = 1 + t

{0.5}

25. (x + 7)(x - 1) = (x + 1)2 28. w(4 - w 2) = 8 - w 3 31. t 3 - 9t = 0

{2}

-3 -2 = x + 4 x + 1

37.

x 2 + 3 = x - 2 x - 2

{910} ∅

19.

3 1 x - 4 = x 2 4

23.

4 2 + = 3 y y

26. (x + 2)(x - 3) = (x - 3)2

35.

2 3 10 = + x - 2 x + 5 (x + 5)(x - 2)

38.

2x -6 = - 2 x + 3 x + 3

43. 2x3 + 4 = x2 + 8x

1 e - 2, , 2 f 2

44. 3x3 + 4x2 = 27x + 36 47. 23t + 4 = - 6

3 49. 2 1 - 2x - 3 = 0

52. 212 - x = x

{913}

{3}

{91, 1} 4 e - 3, - , 3 f 3

No real solution

59. 13x + 12 1>2 = 4

64.

5 4 -3 + = 5z - 11 2z - 3 5 - z

62. 12x + 12 1>3 = - 1 {2}

No real solution {3}

54. 2 + 212 - 2x = x

{18}

{2}

48. 25t + 3 = - 2

{8}

65.

{92}

{91}

x + 3 -3 x - 2 = 2 x2 - 1 x - x x + x

{96}

{4}

57. 23x + 1 - 2x - 1 = 2 60. 13x - 52 1>2 = 2

{5}

{94, - 1, 1}

{1}

53. 3 + 23x + 1 = x

58. 23x - 5 - 2x + 7 = 2 61. 15x - 22 1>3 = 2

45. 22t - 1 = 1

51. 215 - 2x = x

56. 23x + 7 + 2x + 2 = 1

{95, 0, 4}

42. x3 + 4x2 - x - 4 = 0

{0}

{91, 3}

{3}

1 1 1 1 e- f + = 3 2x + 3 x - 1 (2x + 3)(x - 1)

3 50. 2 1 - 2x - 1 = 0

55. 22x + 3 - 2x + 1 = 1

3 f 10

{21}

39. x3 + x2 - 20x = 0

∅

41. x3 + x2 - x - 1 = 0

36.

e

{0, 4}

2 3 = 33. 2x - 3 x + 5 {6}

7 e f 5

{98}

4 5 - 5 = y 2y

27. z(z2 + 1) = 3 + z3

{3}

27 f 4

1 x = 5 2

20. 1 24.

{2}

5 - 22, 0, 226

{97, 0, 1}

{0}

{916}

e

2 9 x = 3 2

{ - 18} 16. 3(2 - x) = 2x - 1

30. x3 = 4x2

40. x3 + 6x2 - 7x = 0

46. 23t + 4 = 2

12.

{0, 9}

32. 4z3 - 8z = 0

{93, 0, 3}

34.

{92}

{910}

29. x2 = 9x

{2}

5 e f 4

15. 2(3 + 2x) = 3(x - 4)

{1}

17. 8x - (2x + 1) = 3x - 10 18. 5 - (2x - 1) = 10 { - 3}

5 1 x = 3 12

11.

{94}

{1, 5}

{3} e-

11 f 6

63.

2 3 5 + = y + 3 y - 4 y + 6

66.

x + 1 x + 4 -3 - 2 = 2 x2 + 2x x + x x + 3x + 2

{97}

SECTION A.8   Solving Equations

A71

Problems 67–72, list some formulas that occur in applications. Solve each formula for the indicated variable. R1R2 1 1 A - P 1 + for R R = 68. Finance  A = P 11 + rt2 for r r = 67. Electricity   = R R1 R2 R1 + R2 Pt 69. Mechanics  F =

mv2 for R R

71. Mathematics  S =

R =

a for r 1 - r

mv2 F

r =

S - a S

70. Chemistry  PV = nRT for T

T =

72. Mechanics  v = - gt + v0 for t

t =

PV nR v0 - v g

Applications and Extensions 73. Physics: Using Sound to Measure Distance  The distance to the surface of the water in a well can sometimes be found by dropping an object into the well and measuring the time elapsed until a sound is heard. If t 1 is the time (measured in seconds) that it takes for the object to strike the water, then t 1 will obey the equation s = 16t 21 , where s is the distance 1s (measured in feet). It follows that t 1 = . Suppose that t 2 4 is the time that it takes for the sound of the impact to reach your ears. Because sound travels at a speed of approximately 1100 feet per second, the time t 2 for the sound to travel

s . See the illustration. Now 1100 t 1 + t 2 is the total time that elapses from the moment that the object is dropped to the moment that a sound is heard. We have the equation the distance s will be t 2 =

Total time elapsed =

s 1s + 4 1100

Find the distance to the water’s surface if the total time elapsed from dropping a rock to hearing it hit water is 4 seconds. ≈229.94 ft 74. Crushing Load  A civil engineer relates the thickness T, in inches, and the height H, in feet, of a square wooden pillar LH 2 . B 25 If a square wooden pillar is 4 inches thick and 10 feet high, what is its crushing load? 64 tons

to its crushing load L, in tons, using the model T = Sound waves: s t2 ⫽ –––– 1100

Falling object: s t1 ⫽ –– 4

4

75. Foucault’s Pendulum    The period of a pendulum is the time it takes the pendulum to make one full swing back and forth. The period T, in seconds, is given by the formula l , where l is the length, in feet, of the pendulum. A 32 In 1851, Jean-Bernard-Leon Foucault demonstrated the axial rotation of Earth using a large pendulum that he hung in the Pantheon in Paris. The period of Foucault’s pendulum was approximately 16.5 seconds. What was its length? ≈221 ft T = 2p

Discussion and Writing 76. Make up a radical equation that has no solution. 77. Make up a radical equation that has an extraneous solution. *78. Discuss the step in the solving process for radical equations that leads to the possibility of extraneous solutions. Why is there no such possibility for linear and quadratic equations? *79. What Went Wrong?  On an exam, Jane solved the equation 22x + 3 - x = 0 and wrote that the solution set was { - 1, 3}. Jane received 3 out of 5 points for the problem. Jane asks you why she received 3 out of 5 points. Provide an explanation. 80. Which of the following pairs of equations are equivalent? Explain. (b) (a) x2 = 9; x = 3

(b) x = 29; x = 3

(c) 1x - 12 1x - 22 = 1x - 12 2; x - 2 = x - 1 *81. The equation 5 8 + x + 3 = x + 3 x + 3 has no solution, yet when we go through the process of solving it we obtain x = - 3. Write a brief paragraph to explain what causes this to happen. 82. Make up an equation that has no solution and give it to a fellow student to solve. Ask the student to write a critique of your equation.

‘Are You Prepared?’ Answers 1. (x - 2)(x + 2)(x - 1)

3. (a) 2x(x - 5)

(b) (x + 2)(x - 2)(x + 3)

4. (a) x (b) x - 3 5 2. e - , 3 f 3 *Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

A72

APPENDIX A  Review

A.9 Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications OBJECTIVES 1 2 3 4 5

The icon is a Model It! icon. It indicates that the discussion or problem involves modeling.

Translate Verbal Descriptions into Mathematical Expressions (p. A72) Solve Interest Problems (p. A73) Solve Mixture Problems (p. A74) Solve Uniform Motion Problems (p. A75) Solve Constant Rate Job Problems (p. A77)

Applied (word) problems do not come in the form “Solve the equation. c” Instead, they supply information using words, a verbal description of the real problem. Solving applied problems requires the ability to translate the verbal description into the language of mathematics. This can be done by using variables to represent unknown quantities and then finding relationships (such as equations) that involve these variables. The process of doing all this is called mathematical modeling. The equation or inequality that represents the relationship among the variables is called a model. Any solution to the mathematical problem must be checked against the mathematical problem, the verbal description, and the real problem. See Figure 27 for an illustration of the modeling process.

Figure 27 Real problem

Verbal description

Language of mathematics

Mathematical problem Check

Check

Check Solution

1 Translate Verbal Descriptions into Mathematical Expressions Let’s look at a few examples that will help with translating certain words into mathematical symbols.

EX A MPL E 1

Translating Verbal Descriptions into Mathematical Expressions (a) The (average) speed of an object equals the distance traveled divided by the time required. d Translation: If r is the speed, d the distance, and t the time, then r = . t (b) Let x denote a number. The number 5 times as large as x is 5x. The number 3 less than x is x - 3. The number that exceeds x by 4 is x + 4. The number that, when added to x, gives 5 is 5 - x.

r

Now Work

PROBLEM

7

Always check the units used to measure the variables of an applied problem. In Example 1(a), if r is measured in miles per hour, then the distance d must be expressed in miles, and the time t must be expressed in hours. It is a good practice to check units to be sure that they are consistent and make sense.

SECTION A.9  Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications

A73

Steps for Solving Applied Problems STEP 1: Read the problem carefully, perhaps two or three times. Pay particular attention to the question being asked in order to identify what you are looking for. If you can, determine realistic possibilities for the answer. STEP 2: Assign a letter (variable) to represent what you are looking for, and, if necessary, express any remaining unknown quantities in terms of this variable. STEP 3: Make a list of all the known facts, and translate them into mathematical expressions. These may take the form of an equation or an inequality involving the variable. If possible, draw an appropriately labeled diagram to assist you. Sometimes, creating a table or chart helps. STEP 4: Solve the equation for the variable, and then answer the question. STEP 5: Check the answer with the facts in the problem. If it agrees, congratulations! If it does not agree, try again.

2 Solve Interest Problems Interest is money paid for the use of money. The total amount borrowed (whether by an individual from a bank in the form of a loan or by a bank from an individual in the form of a savings account) is called the principal. The rate of interest, expressed as a percent, is the amount charged for the use of the principal for a given period of time, usually on a yearly (that is, on a per annum) basis.

Simple Interest Formula If a principal of P dollars is borrowed for a period of t years at a per annum interest rate r, expressed as a decimal, the interest I charged is I = Prt

(1)

Interest charged according to formula (1) is called simple interest. When using formula (1), be sure to express r as a decimal. For example, if the rate of interest is 4%, then r = 0.04.

EXAM PL E 2

Finance: Computing Interest on a Loan Suppose that Juanita borrows $500 for 6 months at the simple interest rate of 9% per annum. What is the interest that Juanita will be charged on the loan? How much does Juanita owe after 6 months?

Solution

The rate of interest is given per annum, so the actual time that the money is borrowed must be expressed in years. The interest charged would be the principal, $500, times 1 the rate of interest 19, = 0.092 times the time in years, : 2 1 Interest charged = I = Prt = 15002 10.092 a b = +22.50 2 After 6 months, Juanita will owe what she borrowed plus the interest: +500 + +22.50 = +522.50

EXAM PL E 3

r

Financial Planning Candy has $70,000 to invest and wants an annual return of $2800, which requires an overall rate of return of 4%. She can invest in a safe, government-insured certificate of deposit, but it pays only 2%. To obtain 4%, she agrees to invest some of her

A74

APPENDIX A  Review

money in noninsured corporate bonds paying 7%. How much should be placed in each investment to achieve her goal?

Solution

STEP 1: The question is asking for two dollar amounts: the principal to invest in the corporate bonds and the principal to invest in the certificate of deposit. STEP 2: Let b represent the amount (in dollars) to be invested in the bonds. Then 70,000 - b is the amount that will be invested in the certificate. (Do you see why?) STEP 3: Now set up Table 1:

Table 1

Principal ($)

Rate

Time (yr)

Interest ($)

Bonds

b

7% = 0.07

1

0.07b

Certificate

70,000 - b

2% = 0.02

1

0.02(70,000 - b)

Total

70,000

4% = 0.04

1

0.04(70,000) = 2800

The total interest from the investments is equal to 0.04 170,0002 = 2800, which leads to the equation 0.07b + 0.02170,000 - b2 = 2800 (Note that the units are consistent: the unit is dollars on each side.) STEP 4: 0.07b + 1400 - 0.02b = 2800 0.05b = 1400 Simplify. b = 28,000 Divide both sides by 0.05. Candy should place $28,000 in the bonds and +70,000 - +28,000 = +42,000 in the certificate.

STEP 5: The interest on the bonds after 1 year is 0.07 1+28,0002 = +1960; the interest on the certificate after 1 year is 0.02 1 +42,0002 = +840. The total annual interest is $2800, the required amount.

r

Now Work

PROBLEM

17

3 Solve Mixture Problems Oil refineries sometimes produce gasoline that is a blend of two or more types of fuel; bakeries occasionally blend two or more types of flour for their bread. These problems are referred to as mixture problems because they combine two or more quantities to form a mixture.

EX A MPL E 4

Blending Coffees The manager of a local coffee shop decides to experiment with a new blend of coffee. She will mix some B grade Colombian coffee that sells for $5 per pound with some A grade Arabica coffee that sells for $10 per pound to get 100 pounds of the new blend. The selling price of the new blend is to be $7 per pound, and there is to be no difference in revenue between selling the new blend and selling the other types. How many pounds of the B grade Colombian coffee and how many pounds of the A grade Arabica coffee are required?

Solution

STEP 1: The question is asking how many pounds of Colombian coffee and how many pounds of Arabica coffee are needed to make 100 pounds of the mixture. STEP 2: Let c represent the number of pounds of the B grade Colombian coffee. Then 100 - c equals the number of pounds of the A grade Arabica coffee. STEP 3: See Figure 28.

SECTION A.9  Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications

Figure 28

$5 per pound

$10 per pound

⫹

B Grade Colombian c pounds

⫹

A75

$7 per pound

⫽

A Grade Arabica 100 − c pounds

Blend ⫽

100 pounds

There is to be no difference in revenue between selling the A and B grades separately and the blend. This leads to the following equation: e

Price per pound Pounds of Price per pound Pounds of Price per pound Pounds of fe f + e fe f = e fe f of B grade B grade of A grade A grade of blend blend # # (100 - c) = # c + +10 +7 100 +5 STEP 4: Solve the equation. 5c + 101100 - c2 5c + 1000 - 10c - 5c c

= = = =

700 700 - 300 60

The manager should blend 60 pounds of B grade Colombian coffee with 100 - 60 = 40 pounds of A grade Arabica coffee to get the desired blend.

STEP 5: The 60 pounds of B grade coffee would sell for 1+52 1602 = +300, and the 40 pounds of A grade coffee would sell for 1+102 1402 = +400; the total revenue, $700, equals the revenue obtained from selling the blend, as desired.

r

Now Work

PROBLEM

21

4 Solve Uniform Motion Problems Objects that move at a constant speed are said to be in uniform motion. When the average speed of an object is known, it can be interpreted as that object’s constant speed. For example, a bicyclist traveling at an average speed of 25 miles per hour can be modeled as being in uniform motion with a constant speed of 25 miles per hour.

Uniform Motion Formula If an object moves at an average speed (rate) r, the distance d covered in time t is given by the formula d = rt

(2)

That is, Distance = Rate # Time.

EXAM PL E 5

Physics: Uniform Motion Tanya, who is a long-distance runner, runs at an average speed of 8 miles per hour (mi>hr). Two hours after Tanya leaves your house, you leave in your Honda and

A76

APPENDIX A  Review

follow the same route. If your average speed is 40 mi>hr, how long will it be before you catch up to Tanya? How far will each of you be from your home?

Solution

Refer to Figure 29. Use t to represent the time (in hours) that it takes the Honda to catch up to Tanya. When this occurs, the total time elapsed for Tanya is t + 2 hours.

Figure 29

t⫽0

2 hr

Time t

Time t

t⫽0

Set up Table 2:

Table 2

Rate (mi/hr)

Time (hr)

Distance (mi)

Tanya

8

t + 2

8(t + 2)

Honda

40

t

40t

The distance traveled is the same for both, which leads to the following equation: 81t + 22 = 40t 8t + 16 = 40t 32t = 16 1 t = hour 2 It will take the Honda

1 hour to catch up to Tanya. Each will have gone 20 miles. 2

1 Check: In 2.5 hours, Tanya travels a distance of 12.52 182 = 20 miles. In hour, 2 1 the Honda travels a distance of a b 1402 = 20 miles. 2

r

EX A MPL E 6

Physics: Uniform Motion A motorboat heads upstream a distance of 24 miles on a river whose current is running at 3 miles per hour (mi/hr). The trip up and back takes 6 hours. Assuming that the motorboat maintained a constant speed relative to the water, what was its speed?

Solution

See Figure 30. Use r to represent the constant speed of the motorboat relative to the water. Then the true speed going upstream is r - 3 mi/hr, and the true speed going Distance downstream is r + 3 mi/hr. Since Distance = Rate * Time, then Time = . Rate Set up Table 3:

Figure 30 24 miles r ⫺ 3 mi/hr r ⫹ 3 mi/hr

Rate (mi/hr)

Table 3

Distance (mi)

Distance Time = (hr) Rate

Upstream

r - 3

24

24 r - 3

Downstream

r + 3

24

24 r + 3

SECTION A.9  Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications

A77

The total time up and back is 6 hours, which gives the equation 24 24 + = 6 r - 3 r + 3 241r + 32 + 241r - 32 = 6 1r - 32 1r + 32

Add the quotients on the left.

48r = 6 r2 - 9

Simplify.

48r = 61r 2 - 92 6r 2 - 48r - 54 r 2 - 8r - 9 1r - 92 1r + 12 r = 9 or r

= = = =

0 0 0 -1

Multiply both sides by r 2 − 9. Place in standard form. Divide by 6. Factor. Apply the Zero-Product Property and solve.

Discard the solution r = - 1 mi/hr, and conclude that the speed of the motorboat relative to the water is 9 mi/hr.

Now Work

r

PROBLEM

27

5 Solve Constant Rate Job Problems This section involves jobs that are performed at a constant rate. The assumption is 1 that if a job can be done in t units of time, then of the job is done in 1 unit of time. t Thus, if a job takes 4 hours, then ¼ of the job is done in 1 hour.

EXAM PL E 7

Solution Table 4 Hours to Do Job

Part of Job Done in 1 Hour

Danny

4

1 4

Mike

6

1 6

Together

t

1 t

Working Together to Do a Job At 10 am Danny is asked by his father to weed the garden. From past experience, Danny knows that this will take him 4 hours, working alone. His older brother, Mike, when it is his turn to do this job, requires 6 hours. Since Mike wants to go golfing with Danny and has a reservation for 1 pm, he agrees to help Danny. Assuming no gain or loss of efficiency, when will they finish if they work together? Can they make the golf date? 1 1 Set up Table 4. In 1 hour, Danny does of the job, and in 1 hour, Mike does of the 4 6 job. Let t be the time (in hours) that it takes them to do the job together. In 1 hour, 1 then, of the job is completed. Reason as follows: t a

Part done by Danny Part done by Mike Part done together b + a b = a b in 1 hour in 1 hour in 1 hour

From Table 4, 1 1 + 4 6 3 2 + 12 12 5 12 5t

= = = =

t =

1 t 1 t 1 t 12 12 5

The model LCD = 12 (left side)

Multiply both sides by 12t. Divide both sides by 5.

12 Working together, Danny and Mike can do the job in hours, or 2 hours, 5 24 minutes. They should make the golf date, since they will finish at 12:24 pm.

Now Work

r

PROBLEM

33

A78

APPENDIX A  Review

A.9 Assess Your Understanding Concepts and Vocabulary 1. The process of using variables to represent unknown quantities and then finding relationships that involve these variables is referred to as mathematical modeling .

5. True or False If an object moves at an average speed r, the distance d covered in time t is given by the formula d = rt. True

2. The money paid for the use of money is interest .

6. Suppose that you want to mix two coffees in order to obtain 100 pounds of the blend. If x represents the number of pounds of coffee A, write an algebraic expression that represents the number of pounds of coffee B. 100 - x

3. Objects that move at a constant rate are said to be in uniform motion . 4. True or False The amount charged for the use of principal for a given period of time is called the rate of interest. False

Applications and Extensions In Problems 7–16, translate each sentence into a mathematical equation. Be sure to identify the meaning of all symbols. *7. Geometry  The area of a circle is the product of the number p and the square of the radius. A = pr 2 *8. Geometry  The circumference of a circle is the product of the number p and twice the radius. C = 2pr *9. Geometry  The area of a square is the square of the length of a side. A = s2 *10. Geometry  The perimeter of a square is four times the length of a side. P = 4s *11. Physics  Force equals the product of mass and acceleration. *12. Physics  Pressure is force per unit area. P = F>A *13. Physics  Work equals force times distance. W = Fd *14. Physics  Kinetic energy is one-half the product of the mass and the square of the velocity. *15. Business    The total variable cost C of manufacturing x dishwashers is $150 per dishwasher times the number of dishwashers manufactured. C = 150x *16. Business    The total revenue R derived from selling x dishwashers is $250 per dishwasher times the number of dishwashers sold. R = 250x *17. Financial Planning    Betsy, a recent retiree, requires $6000 per year in extra income. She has $50,000 to invest and can invest in B-rated bonds paying 15% per year or in a certificate of deposit (CD) paying 7% per year. How much money should Betsy invest in each to realize exactly $6000 in interest per year? $31,250 in bonds; $18,750 in CDs 18. Financial Planning    After 2 years, Betsy (see Problem 17) finds that she will now require $7000 per year. Assuming that the remaining information is the same, how should the money be reinvested? $43,750 in bonds; $6250 in CDs

of the new blend. The selling price of the new blend is to be $4.50 per pound, and there is to be no difference in revenue from selling the new blend versus selling the other types. How many pounds of the Earl Grey tea and of the Orange Pekoe tea are required? Earl Grey: 75 lb; Orange Pekoe: 25 lb 22. Business: Blending Coffee  A coffee manufacturer wants to market a new blend of coffee that sells for $3.90 per pound by mixing two coffees that sell for $2.75 and $5 per pound, respectively. What amounts of each coffee should be blended to obtain the desired mixture? [Hint: Assume that the total weight of the desired blend is 100 pounds.] Coffee I: 49 lb; Coffee II: 51 lb 23. Business: Mixing Nuts  A nut store normally sells cashews for $9.00 per pound and almonds for $3.50 per pound. But at the end of the month the almonds had not sold well, so, in order to sell 60 pounds of almonds, the manager decided to mix the 60 pounds of almonds with some cashews and sell the mixture for $7.50 per pound. How many pounds of cashews should be mixed with the almonds to ensure no change in the profit? 160 lb of cashews 24. Business: Mixing Candy  A candy store sells boxes of candy containing caramels and cremes. Each box sells for $12.50 and holds 30 pieces of candy (all pieces are the same size). If the caramels cost $0.25 to produce and the cremes cost $0.45 to produce, how many of each should be in a box to yield a profit of $3? 20 caramels and 10 cremes 25. Physics: Uniform Motion    A motorboat can maintain a constant speed of 16 miles per hour relative to the water. The boat makes a trip upstream to a certain point in 20 minutes; the return trip takes 15 minutes. What is the speed of the current? See the figure. 2.286 mi/hr

19. Banking  A bank loaned out $12,000, part of it at the rate of 8% per year and the rest at the rate of 18% per year. If the interest received totaled $1000, how much was loaned at 8%? $11,600 20. Banking  Wendy, a loan officer at a bank, has $1,000,000 to lend and is required to obtain an average return of 18% per year. If she can lend at the rate of 19% or at the rate of 16%, how much can she lend at the 16% rate and still meet her requirement? $333,333.33 21. Blending Teas    The manager of a store that specializes in selling tea decides to experiment with a new blend. She will mix some Earl Grey tea that sells for $5 per pound with some Orange Pekoe tea that sells for $3 per pound to get 100 pounds

26. Physics: Uniform Motion    A motorboat heads upstream on a river that has a current of 3 miles per hour. The trip upstream takes 5 hours, and the return trip takes 2.5 hours. What is the speed of the motorboat? (Assume that the motorboat maintains a constant speed relative to the water.) 9 mi/hr

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

SECTION A.9  Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications

40 30 TE DB

20

20 10

31. Tennis    A regulation doubles tennis court has an area of 2808 square feet. If it is 6 feet longer than twice its width, determine the dimensions of the court. 78 ft by 36 ft Source: United States Tennis Association

30

30. High-Speed Walkways    Toronto’s Pearson International Airport has a high-speed version of a moving walkway. If Liam walks while riding this moving walkway, he can travel 280 meters in 60 seconds less time than if he stands still on the moving walkway. If Liam walks at a normal rate of 1.5 meters per second, what is the speed of the high-speed walkway? 2 m/sec

[Hint: At time t = 0, the defensive back is 5 yards behind the tight end.]

40

29. Moving Walkways    The speed of a moving walkway is typically about 2.5 feet per second. Walking on such a moving walkway, it takes Karen a total of 40 seconds to travel 50 feet with the movement of the walkway and then back again against the movement of the walkway. What is Karen’s normal walking speed? 4.05 ft/sec Source: Answers.com

37. Football    A tight end can run the 100-yard dash in 12 seconds. A defensive back can do it in 10 seconds. The tight end catches a pass at his own 20-yard line with the defensive back at the 15-yard line. (See the figure.) If no other players are nearby, at what yard line will the defensive back catch up to the tight end? tight end’s 45-yd line

10

*28. Physics: Uniform Motion    Two cars enter the Florida Turnpike at Commercial Boulevard at 8:00 am, each heading for Wildwood. One car’s average speed is 10 miles per hour more than the other’s. The faster car arrives at 1 Wildwood at 11:00 am, hour before the other car. What 2 was the average speed of each car? How far did each travel?

36. Construction   A pond is enclosed by a wooden deck that is 3 feet wide. The fence surrounding the deck is 100 feet long. * (a) If the pond is square, what are its dimensions? * (b) If the pond is rectangular and the length of the pond is to be three times its width, what are its dimensions? (c) If the pond is circular, what is its diameter? 25.83 ft (d) Which pond has the greatest area? the circular one

SOUTH

27. Physics: Uniform Motion    A motorboat maintained a constant speed of 15 miles per hour relative to the water in going 10 miles upstream and then returning. The total time for the trip was 1.5 hours. Use this information to find the speed of the current. 5 mi/hr

A79

*32. Laser Printers    It takes an HP LaserJet 1300 laser printer 10 minutes longer to complete a 600-page print job by itself than it takes an HP LaserJet 2420 to complete the same job by itself. Together the two printers can complete the job in 12 minutes. How long does it take each printer to complete the print job alone? What is the speed of each printer? Source: Hewlett-Packard

38. Computing Business Expense    Therese, an outside salesperson, uses her car for both business and pleasure. Last year, she traveled 30,000 miles, using 900 gallons of gasoline. Her car gets 40 miles per gallon on the highway and 25 in the city. She can deduct all highway travel, but no city travel, on her taxes. How many miles should Therese deduct as a business expense? 20,000 mi

33. Working Together on a Job   Trent can deliver his newspapers in 30 minutes. It takes Lois 20 minutes to do the same route. How long would it take them to deliver the newspapers if they worked together? 12 min

*39. Mixing Water and Antifreeze  How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 60% antifreeze?

34. Working Together on a Job    Patrick, by himself, can paint four rooms in 10 hours. If he hires April to help, they can do the same job together in 6 hours. If he lets April work alone, how long will it take her to paint four rooms? 15 hr 35. Enclosing a Garden  A gardener has 46 feet of fencing to be used to enclose a rectangular garden that has a border 2 feet wide surrounding it. See the figure. (a) If the length of the garden is to be twice its width, what will be the dimensions of the garden? 10 ft by 5 ft (b) What is the area of the garden? 50 sq ft * (c) If the length and width of the garden are to be the same, what will be the dimensions of the garden? * (d) What will be the area of the square garden? 2 ft 2 ft

40. Mixing Water and Antifreeze    The cooling system of a certain foreign-made car has a capacity of 15 liters. If the system is filled with a mixture that is 40% antifreeze, how much of this mixture should be drained and replaced by pure antifreeze so that the system is filled with a solution that is 60% antifreeze? 5 L *41. Chemistry: Salt Solutions    How much water must be evaporated from 32 ounces of a 4% salt solution to make a 6% salt solution? 42. Chemistry: Salt Solutions    How much water must be evaporated from 240 gallons of a 3% salt solution to produce a 5% salt solution? 96 gal *43. Purity of Gold    The purity of gold is measured in karats, with pure gold being 24 karats. Other purities of gold are expressed as proportional parts of pure gold. Thus, 18-karat 18 12 gold is , or 75% pure gold; 12-karat gold is , or 50% 24 24 pure gold; and so on. How much 12-karat gold should be mixed with pure gold to obtain 60 grams of 16-karat gold?

A80

APPENDIX A  Review

44. Chemistry: Sugar Molecules  A sugar molecule has twice as many atoms of hydrogen as of oxygen and one more atom of carbon than of oxygen. If a sugar molecule has a total of 45 atoms, how many are oxygen? How many are hydrogen? 11 oxygen; 22 hydrogen *45. Running a Race Mike can run the mile in 6 minutes, and Dan can run the mile in 9 minutes. If Mike gives Dan a head start of 1 minute, how far from the start will Mike pass Dan? How long does it take? See the figure. Dan

Mike

Start 1– 4

1– 2

mi

mi

3– 4

mi

46. Range of an Airplane An air rescue plane averages 300 miles per hour in still air. It carries enough fuel for 5 hours of flying time. If, upon takeoff, it encounters a head wind of 30 mi/hr, how far can it fly and return safely? (Assume that the wind speed remains constant.) 742.5 mi 47. Emptying Oil Tankers An oil tanker can be emptied by the main pump in 4 hours. An auxiliary pump can empty the tanker in 9 hours. If the main pump is started at 9 am, when is the latest the auxiliary pump can be started so that the tanker is emptied by noon? 9:45 am

faucets closed and the stopper removed, the tub will empty in 20 minutes. How long will it take for the tub to fill if both faucets are open and the stopper is removed? 1 hr 50. Using Two Pumps A 5-horsepower (hp) pump can empty a pool in 5 hours. A smaller, 2-hp pump empties the same pool in 8 hours. The pumps are used together to begin emptying this pool. After two hours, the 2-hp pump breaks down. How long will it take the larger pump to empty the pool? 1hr, 45 min *51. A Biathlon Suppose that you have entered an 87-mile biathlon that consists of a run and a bicycle race. During your run, your average speed is 6 miles per hour, and during your bicycle race, your average speed is 25 miles per hour. You finish the race in 5 hours. What is the distance of the run? What is the distance of the bicycle race? *52. Cyclists Two cyclists leave a city at the same time, one going east and the other going west. The westbound cyclist bikes 5 mph faster than the eastbound cyclist. After 6 hours they are 246 miles apart. How fast is each cyclist riding? 53. Comparing Olympic Heroes In the 2012 Olympics, Usain Bolt of Jamaica won the gold medal in the 100-meter race with a time of 9.69 seconds. In the 1896 Olympics, Thomas Burke, of the United States, won the gold medal in the 100-meter race in 12.0 seconds. If they ran in the same race repeating their respective times, by how many meters would Bolt beat Burke? 19.25 m 54. Constructing a Coffee Can A 39-ounce can of Hills Bros.® coffee requires 188.5 square inches of aluminum. If its height is 7 inches, what is its radius? [Hint: The surface area S of a right cylinder is S = 2pr 2 + 2prh, where r is the radius and h is the height.] 3 in.

48. Cement Mix A 20-pound bag of Economy brand cement mix contains 25% cement and 75% sand. How much pure cement must be added to produce a cement mix that is 40% cement? 5 lb 49. Emptying a Tub A bathroom tub will fill in 15 minutes with both faucets open and the stopper in place. With both

7 in.

39 oz.

Discussion and Writing 55. Critical Thinking   You are the manager of a clothing store and have just purchased 100 dress shirts for $20.00 each. After 1 month of selling the shirts at the regular price, you plan to have a sale giving 40% off the original selling price. However, you still want to make a profit of $4 on each shirt at the sale price. What should you price the shirts at initially to ensure this? If, instead of 40% off at the sale, you give 50% off, by how much is your profit reduced? $ 40; no profit 56. Critical Thinking   Make up a word problem that requires solving a linear equation as part of its solution. Exchange problems with a friend. Write a critique of your friend’s problem. 57. Critical Thinking    Without solving, explain what is wrong with the following mixture problem: How many liters of 25% ethanol should be added to 20 liters of 48% ethanol to obtain a solution of 58% ethanol? Now go through an algebraic solution. What happens?

58. Computing Average Speed    In going from Chicago to Atlanta, a car averages 45 miles per hour, and in going from Atlanta to Miami, it averages 55 miles per hour. If Atlanta is halfway between Chicago and Miami, what is the average speed from Chicago to Miami? Discuss an intuitive solution. Write a paragraph defending your intuitive solution. Then solve the problem algebraically. Is your intuitive solution the same as the algebraic one? If not, find the flaw. 49.5 mi/hr *59. Speed of a Plane    On a recent flight from Phoenix to Kansas City, a distance of 919 nautical miles, the plane arrived 20 minutes early. On leaving the aircraft, I asked the captain, “What was our tail wind?” He replied, “I don’t know, but our ground speed was 550 knots.” How can you determine whether enough information is provided to find the tail wind? If possible, find the tail wind. (1 knot = 1 nautical mile per hour)

SECTION A.10  Interval Notation; Solving Inequalities

A81

A.10 Interval Notation; Solving Inequalities PREPARING FOR THIS SECTION  Before getting started, review the following: r
"MHFCSB
&TTFOUJBMT
"QQFOEJY
"
4FDUJPO
" ƭQQ
"m"

Now Work the ‘Are You Prepared?’ problems on page A86.

OBJECTIVES 1 2 3 4

Use Interval Notation (p. A81) Use Properties of Inequalities (p. A82) Solve Inequalities (p. A84) Solve Combined Inequalities (p. A85)

Suppose that a and b are two real numbers and a 6 b. The notation a 6 x 6 b means that x is a number between a and b. The expression a 6 x 6 b is equivalent to the two inequalities a 6 x and x 6 b. Similarly, the expression a … x … b is equivalent to the two inequalities a … x and x … b. The remaining two possibilities, a … x 6 b and a 6 x … b, are defined similarly. Although it is acceptable to write 3 Ú x Ú 2, it is preferable to reverse the inequality symbols and write instead 2 … x … 3 so that the values go from smaller to larger, reading left to right. A statement such as 2 … x … 1 is false because there is no number x for which 2 … x and x … 1. Finally, never mix inequality symbols, as in 2 … x Ú 3.

1 Use Interval Notation Let a and b represent two real numbers with a 6 b.

DEFINITION

In Words The notation [a, b] represents all real numbers between a and b, inclusive. The notation (a, b) represents all real numbers between a and b, not including either a or b.

A closed interval, denoted by [a, b], consists of all real numbers x for which a … x … b. An open interval, denoted by (a, b), consists of all real numbers x for which a 6 x 6 b. The half-open, or half-closed, intervals are (a, b], consisting of all real numbers x for which a 6 x … b, and [a, b), consisting of all real numbers x for which a … x 6 b.

In each of these definitions, a is called the left endpoint and b the right endpoint of the interval. The symbol q (read as “infinity”) is not a real number, but a notational device used to indicate unboundedness in the positive direction. The symbol - q (read as “negative infinity”) also is not a real number, but a notational device used to indicate unboundedness in the negative direction. The symbols q and - q , are used to define five other kinds of intervals: [a, ˆ ) Consists of all real numbers x for which x Ú (a, ˆ ) Consists of all real numbers x for which x 7 ( − ˆ , a] Consists of all real numbers x for which x ( − ˆ , a) Consists of all real numbers x for which x ( − ˆ , ˆ ) Consists of all real numbers x

a a … a 6 a

Note that q and - q are never included as endpoints, since neither is a real number. Table 5 on the following page summarizes interval notation, corresponding inequality notation, and their graphs.

A82

APPENDIX A  Review

Table 5

EX A MPL E 1

Interval

Inequality

Graph

The open interval (a, b)

a⬍x⬍b

a

b

The closed interval [a, b]

aⱕxⱕb

a

b

The half-open interval [a, b)

aⱕx⬍b

a

b

The half-open interval (a, b]

a⬍xⱕb

a

b

The interval [a, ⬁)

xⱖa

a

The interval (a, ⬁)

x⬎a

a

The interval (⫺⬁, a]

xⱕa

The interval (⫺⬁, a)

x⬍a

The interval (⫺⬁, ⬁)

All real numbers

a a

Writing Inequalities Using Interval Notation Write each inequality using interval notation. (a) 1 … x … 3

Solution

(b) - 4 6 x 6 0

(c) x 7 5

(d) x … 1

(a) 1 … x … 3 describes all real numbers x between 1 and 3, inclusive, which is 3 1, 34 in interval notation. (b) In interval notation, - 4 6 x 6 0 is written 1 - 4, 02. (c) x 7 5 consists of all real numbers x greater than 5, which in interval notation, is 15, q 2 . (d) In interval notation, x … 1 is written 1 - q , 14 .

r

EX A MPL E 2

Writing Intervals Using Inequality Notation Write each interval as an inequality involving x. (a) 3 1, 42

Solution

(a) (b) (c) (d)

(b) 12, q 2

(c) 3 2, 34

(d) 1 - q , - 34

3 1, 42 consists of all real numbers x for which 1 … x 6 4. 12, q 2 consists of all real numbers x for which x 7 2. 3 2, 34 consists of all real numbers x for which 2 … x … 3. 1 - q , - 34 consists of all real numbers x for which x … - 3.

Now Work

PROBLEMS

11, 23,

AND

r

31

2 Use Properties of Inequalities The product of two positive real numbers is positive, the product of two negative real numbers is positive, and the product of 0 and 0 is 0. For any real number a, the value of a2 is 0 or positive; that is, a2 is nonnegative. This is called the nonnegative property.

Nonnegative Property

In Words The square of a real number is never negative.

For any real number a, a2 Ú 0

(1)

SECTION A.10  Interval Notation; Solving Inequalities

A83

If the same number is added to both sides of an inequality, an equivalent inequality results. For example, since 3 6 5, then 3 + 4 6 5 + 4 or 7 6 9. This is called the addition property of inequalities.

Addition Property of Inequalities For real numbers a, b, and c, If a 6 b, then a + c 6 b + c.

(2a)

If a 7 b, then a + c 7 b + c.

(2b)

The addition property states that the sense, or direction, of an inequality remains unchanged when the same number is added to each side. Now let’s see what happens when each side of an inequality is multiplied by a nonzero number. Begin with 3 6 7 and multiply each side by 2. The numbers 6 and 14 that result obey the inequality 6 6 14. Now start with 9 7 2 and multiply each side by - 4. The numbers - 36 and - 8 that result obey the inequality - 36 6 - 8. Note that the effect of multiplying both sides of 9 7 2 by the negative number - 4 is that the direction of the inequality symbol is reversed. This leads to the following general multiplication properties for inequalities:

Multiplication Properties for Inequalities For real numbers a, b, and c:

In Words Multiplying by a negative number reverses the inequality.

If a 6 b and if c 7 0, then ac 6 bc. If a 6 b and if c 6 0, then ac 7 bc.

(3a)

If a 7 b and if c 7 0, then ac 7 bc. If a 7 b and if c 6 0, then ac 6 bc.

(3b)

The multiplication properties state that the sense, or direction, of an inequality remains the same when each side is multiplied by a positive real number, whereas the direction is reversed when each side is multiplied by a negative real number.

EXAM PL E 3

Multiplication Property of Inequalities 1 1 (a) If 2x 6 6, then 12x2 6 162 , or x 6 3. 2 2 x x (b) If 7 12, then - 3a b 6 - 31122 , or x 6 - 36. -3 -3 - 4x -8 7 , or x 7 2. -4 -4 (d) If - x 7 8, then 1 - 12 1 - x2 6 1 - 12 182 , or x 6 - 8.

(c) If - 4x 6 - 8, then

Now Work

PROBLEM

r

45

The reciprocal property states that the reciprocal of a positive real number is positive and that the reciprocal of a negative real number is negative.

A84

APPENDIX A  Review

Reciprocal Property for Inequalities If a 7 0, then

1 7 0. a

If

1 7 0, then a 7 0. a

(4a)

If a 6 0, then

1 6 0. a

If

1 6 0, then a 6 0. a

(4b)

3 Solve Inequalities An inequality in one variable is a statement involving two expressions, at least one containing the variable, separated by one of the inequality symbols 6 , … , 7 , or Ú . To solve an inequality means to find all values of the variable for which the statement is true. These values are solutions of the inequality. For example, the following are all inequalities involving one variable x: x + 1 x + 5 6 8 2x - 3 Ú 4 x2 - 1 … 3 7 0 x - 2 As with equations, one method for solving an inequality is to replace it by a series of equivalent inequalities until an inequality with an obvious solution, such as x 6 3, is obtained. Equivalent inequalities are obtained by applying some of the same properties as those used to find equivalent equations. The addition property and the multiplication properties form the basis for the following procedures.

Procedures That Leave the Inequality Symbol Unchanged 1. Simplify both sides of the inequality by combining like terms and eliminating parentheses: is equivalent to

1x + 22 + 6 7 2x + 51x + 12 x + 8 7 7x + 5

2. Add or subtract the same expression on both sides of the inequality: is equivalent to

3x - 5 6 4 13x - 52 + 5 6 4 + 5

3. Multiply or divide both sides of the inequality by the same positive expression: 4x 7 16 is equivalent to

4x 16 7 4 4

Procedures That Reverse the Sense or Direction of the Inequality Symbol 1. Interchange the two sides of the inequality: 3 6 x is equivalent to x 7 3 2. Multiply or divide both sides of the inequality by the same negative expression: - 2x 7 6 is equivalent to

- 2x 6 6 -2 -2

As the examples that follow illustrate, inequalities can be solved by using many of the same steps that would be used to solve equations. In writing the solution of an inequality, either set notation or interval notation may be used, whichever is more convenient.

SECTION A.10  Interval Notation; Solving Inequalities

EXAM PL E 4

A85

Solving an Inequality Solve the inequality 4x + 7 Ú 2x - 3, and graph the solution set. 4x + 7 Ú 2x - 3

Solution

4x + 7 - 7 Ú 2x - 3 - 7 4x Ú 2x - 10 4x - 2x Ú 2x - 10 - 2x

⫺6

⫺5

⫺4

⫺3

⫺2

⫺1

Simplify. Subtract 2x from both sides.

2x Ú - 10

Simplify.

2x - 10 Ú 2 2

Divide both sides by 2. (The direction of the inequality symbol is unchanged.)

x Ú -5 Figure 31

Subtract 7 from both sides.

Simplify.

The solution set is 5 x 0 x Ú - 56 or, using interval notation, all numbers in the interval 3 - 5, q 2. See Figure 31 for the graph.

r

Now Work

PROBLEM

57

4 Solve Combined Inequalities EXAM PL E 5

Solving Combined Inequalities Solve the inequality - 5 6 3x - 2 6 1, and graph the solution set.

Solution

Recall that the inequality - 5 6 3x - 2 6 1 is equivalent to the two inequalities - 5 6 3x - 2 and 3x - 2 6 1 Solve each of these inequalities separately. - 5 6 3x - 2 - 5 + 2 6 3x - 2 + 2 - 3 6 3x -3 3x 6 3 3 -1 6 x

3x - 2 6 1 Add 2 to both sides.

3x - 2 + 2 6 1 + 2

Simplify. Divide both sides by 3. Simplify.

3x 6 3 3x 3 6 3 3 x 6 1

The solution set of the original pair of inequalities consists of all x for which - 1 6 x and x 6 1 Figure 32 ⫺3

⫺2

⫺1

0

1

2

This may be written more compactly as 5 x - 1 6 x 6 16 . In interval notation, the solution is 1 - 1, 12. See Figure 32 for the graph.

r

A86

APPENDIX A  Review

Observe in the preceding process that solving the two inequalities required exactly the same steps. A shortcut to solving the original inequality algebraically is to deal with the two inequalities at the same time, as follows: -5 -5 + 2 -3 -3 3 -1

Now Work

EX A MPL E 6

Solution

6 3x - 2 6 1 6 3x - 2 + 2 6 1 + 2 6 3x 6 3 3x 3 6 6 3 3 6 x 6 1 PROBLEM

Add 2 to each part. Simplify. Divide each part by 3. Simplify.

73

Using the Reciprocal Property to Solve an Inequality

Solve the inequality 14x - 12 -1 7 0, and graph the solution set. Recall that 14x - 12 -1 = then a 7 0. Hence,

1 1 . The Reciprocal Property states that when 7 0, a 4x - 1

14x - 12 -1 1 4x - 1 4x - 1 4x

7 0 7 0

7 0 Reciprocal Property 7 1 1 x 7 4

The solution set is b x 0 x 7

Figure 33 0

1 – 4

1

1 1 r —that is, all x in the interval a , q b . Figure 33 4 4

r

illustrates the graph.

Now Work

PROBLEM

83

A.10 Assess Your Understanding ‘Are You Prepared?’ Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. Graph the inequality: x Ú - 2. (pp. A4–A5)

2. Graph the inequality: x 6 1. (pp. A4–A5)

Concepts and Vocabulary 3. If each side of an inequality is multiplied by a(n) negative number, then the direction of the inequality symbol is reversed. 4. A(n) closed interval , denoted 3a, b4, consists of all real numbers x for which a … x … b.

5. The Multiplication Property states that the sense, or direction, of an inequality remains the same if each side is multiplied by a positive number, whereas the direction is reversed if each side is multiplied by a negative number.

In Problems 6–9, determine whether the statement is True or False if a 6 b and c 6 0. 6. a + c 6 b + c

True

7. a - c 6 b - c

True

8. ac 7 bc

10. True or False The square of any real number is always nonnegative. True

True

9.

b a 6 c c

False

A87

SECTION A.10  Interval Notation; Solving Inequalities

Skill Building In Problems 11–16, express the graph shown in blue using interval notation. Also express each as an inequality involving x. *11.

*12.

–1

0

1

2

3

–2

–1

0

1

2

*14.

*15.

–2

–1

0

1

2

–1

0

1

2

3

*13.

–1

0

1

2

3

–1

0

1

2

3

*16.

In Problems 17–22, an inequality is given. Write the inequality obtained by: (a) Adding 3 to each side of the given inequality.  (b) Subtracting 5 from each side of the given inequality.  (c) Multiplying each side of the given inequality by 3.  (d) Multiplying each side of the given inequality by - 2. *17. 3 6 5

*18. 2 7 1

*19. 4 7 - 3

*20. - 3 7 - 5

*21. 2x + 1 6 2

*22. 1 - 2x 7 5

In Problems 23–30, write each inequality using interval notation, and illustrate each inequality using the real number line. *23. 0 … x … 4 *27. x Ú 4

*24. - 1 6 x 6 5

[0, 4]

34, q )

*28. x … 5

( - 1, 5)

( - q , 54

*25. 4 … x 6 6

*26. - 2 6 x 6 0

[4, 6)

( - q , - 4)

*29. x 6 - 4

*30. x 7 1

( - 2, 0)

(1, q )

In Problems 31–38, write each interval as an inequality involving x, and illustrate each inequality using the real number line. *31. 32, 54

*35. 34, q 2

*32. 11, 22

2 … x … 5

*36. 1 - q , 24

x Ú 4

*33. 1 - 3, - 22

1 6 x 6 2

*37. 1 - q , - 32

x … 2

*34. 30, 12

-3 6 x 6 -2

0 … x 6 1

*38. 1 - 8, q 2

x 6 -3

x 7 -8

In Problems 39–52, fill in the blank with the correct inequality symbol. 39. If x 6 5, then x - 5

6

0.

40. If x 6 - 4, then x + 4

42. If x 7 6, then x - 6

7

0.

43. If x Ú - 4, then 3x

45. If x 7 6, then - 2x

6

- 12.

Ú

12.

49. If 2x 7 6, then x

1 x … 3, then x 2

Ú

- 6.

52. If -

0.

Ú

46. If x 7 - 2, then - 4x

48. If x … - 4, then - 3x 51. If -

6

- 12. 6

7

1 x 7 1, then x 4

8. 3.

6

41. If x 7 - 4, then x + 4 44. If x … 3, then 2x

7 …

6.

47. If x Ú 5, then - 4x

…

50. If 3x … 12, then x

…

- 4.

In Problems 53–88, solve each inequality. Express your answer using set notation or interval notation. Graph the solution set. *53. x + 1 6 5

*54. x - 6 6 1

*55. 1 - 2x … 3

*56. 2 - 3x … 5

*57. 3x - 7 7 2

*58. 2x + 5 7 1

*59. 3x - 1 Ú 3 + x

*60. 2x - 2 Ú 3 + x

*61. - 21x + 32 6 8

*62. - 311 - x2 6 12

*63. 4 - 311 - x2 … 3

*64. 8 - 412 - x2 … - 2x

1 1x - 42 7 x + 8 2 x x *68. Ú 2 + 3 6

*65.

*71. - 5 … 4 - 3x … 2 *74. 0 6

3x + 2 6 4 2

*66. 3x + 4 7

1 1x - 22 3

*70. 4 … 2x + 2 … 10

*72. - 3 … 3 - 2x … 9

*73. - 3 6

*75. 1 6 1 -

1 x 6 4 2

*80. x19x - 52 … 13x - 12 2

*81.

*83. 14x + 22 -1 6 0

*84. 12x - 12 -1 7 0

1 x + 1 3 … 6 2 3 4

*87. 0 6 12x - 42 -1 6

2x - 1 6 0 4

*76. 0 6 1 -

*78. 1x - 12 1x + 12 7 1x - 32 1x + 42

4 2 6 x 3

x x Ú 1 2 4

*69. 0 … 2x - 6 … 4

*77. 1x + 22 1x - 32 7 1x - 12 1x + 12

*86. 0 6

*67.

1 x + 1 2 6 … 3 2 3

*85. 0 6 1 2

1 x 6 1 3

*79. x14x + 32 … 12x + 12 2 *82.

2 3 6 x 5

*88. 0 6 13x + 62 -1 6

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

1 3

0.

- 20. 4.

A88

APPENDIX A  Review

Applications and Extensions 89. What is the domain of the variable in the expression 2 3x + 6 ? {x x Ú - 2} 90. What is the domain of the variable in the expression 2 8 + 2x ? {x x Ú - 4} 91. A young adult may be defined as someone older than 21 and less than 30 years of age. Express this statement using inequalities. 21 6 age 6 30 92. Middle-aged may be defined as being 40 or more and less than 60. Express this statement using inequalities. 40 … age 6 60 93. Life Expectancy    According to the National Center for Health Statistics, an average 30-year-old male in 2008 could expect to live at least 47.3 more years, and an average 30-year-old female in 2008 could expect to live at least 51.6 more years. (a) To what age could an average 30-year-old male expect to live? Express your answer as an inequality. male Ú 77.3 yr (b) To what age could an average 30-year-old female expect to live? Express your answer as an inequality. female Ú 81.6 yr (c) Who could expect to live longer, a male or a female? By how many years? female; 4.3 yr Source: National Vital Statistics Reports, Vol. 61, No. 3, September 2012.

JAN 2008

AUG 2059

APR 2055

94. General Chemistry    For a certain ideal gas, the volume V (in cubic centimeters) equals 20 times the temperature T (in degrees Celsius). If the temperature varies from 80° to 120°C, inclusive, what is the corresponding range of the volume of the gas? 1600 cm3 to 2400 cm3, inclusive *95. Real Estate  A real estate agent agrees to sell an apartment complex according to the following commission schedule: $45,000 plus 25% of the selling price in excess of $900,000. Assuming that the complex will sell at some price between $900,000 and $1,100,000, inclusive, over what range does the agent’s commission vary? How does the commission vary as a percent of the selling price? *96. Sales Commission    A used-car salesperson is paid a commission of $25 plus 40% of the selling price in excess of owner’s cost. The owner claims that used cars typically sell for at least owner’s cost plus $200 and at most owner’s cost plus $3000. For each sale made, over what range can the salesperson expect the commission to vary? *97. Federal Tax Withholding    The percentage method of withholding for federal income tax (2013) states that a single person whose weekly wages, after subtracting withholding allowances, are over $739, but not over $1732, shall have $95.95 plus 25% of the excess over $739 withheld. Over

what range does the amount withheld vary if the weekly wages vary from $800 to $1000, inclusive? Source: Employer’s Tax Guide. Department of the Treasury, Internal Revenue Service, Publication 2013. *98. Exercising    Sue wants to lose weight. For healthy weight loss, the American College of Sports Medicine (ACSM) recommends 200 to 300 minutes of exercise per week. For the first six days of the week, Sue exercised 40, 45, 0, 50, 25, and 35 minutes. How long should Sue exercise on the seventh day in order to stay within the ACSM guidelines? *99. Electricity Rates  Commonwealth Edison Company’s charge for electricity in January 2013 is 10.62¢ per kilowatt-hour. In addition, each monthly bill contains a customer charge of $13.04. If last year’s bills ranged from a low of $82.07 to a high of $278.54, over what range did usage vary (in kilowatthours)? Source: Commonwealth Edison Co., January 2013. * 100. Water Bills  The Village of Oak Lawn charges homeowners $52.74 per quarter-year, plus $5.37 per 1000 gallons for water usage in excess of 10,000 gallons. In 2013 one homeowner’s quarterly bill ranged from a high of $165.51 to a low of $101.07. Over what range did water usage vary? Source: Village of Oak Lawn, Illinois, January 2013. * 101. Markup of a New Car    The markup over dealer’s cost of a new car ranges from 12% to 18%. If the sticker price is $18,000, over what range will the dealer’s cost vary? * 102. IQ Tests    A standard intelligence test has an average score of 100. According to statistical theory, of the people who take the test, the 2.5% with the highest scores will have scores of more than 1.96s above the average, where s (sigma, a number called the standard deviation) depends on the nature of the test. If s = 12 for this test and there is (in principle) no upper limit to the score possible on the test, write the interval of possible test scores of the people in the top 2.5%. 103. Computing Grades  In your Economics 101 class, you have scores of 68, 82, 87, and 89 on the first four of five tests. To get a grade of B, the average of the first five test scores must be greater than or equal to 80 and less than 90. (a) Solve an inequality to find the range of the score that you need on the last test to get a B. at least a 74 (b) What score do you need if the fifth test counts double? at least a 77 What do I need to get a B?

82 68 89 87

* 104. “Light” Foods  For food products to be labeled “light,” the U.S. Food and Drug Administration requires that the altered product must either contain at least one-third calories of the regular product or contain at least one-half less fat than the regular product. If a serving of Miracle Whip® Light contains 20 calories and 1.5 grams of fat, then what must be true about either the number of calories or the grams of fat in a serving of regular Miracle Whip®?

SECTION A.11  Complex Numbers

* 105. Arithmetic Mean  If a 6 b, show that a 6

a + b 6 b. 2

* 109. Harmonic Mean  For 0 6 a 6 b, let h be defined by 1 1 1 1 = a + b h 2 a b

a + b The number is called the arithmetic mean of a and b. 2 * 106. Refer to Problem 105. Show that the arithmetic mean of a and b is equidistant from a and b. * 107. Geometric Mean  If 0 6 a 6 b, show that a 6 1 ab 6 b. The number 1 ab is called the geometric mean of a and b. * 108. Refer to Problems 105 and 107. Show that the geometric mean of a and b is less than the arithmetic mean of a and b.

A89

Show that a 6 h 6 b. The number h is called the harmonic mean of a and b. * 110. Refer to Problems 105, 107, and 109. Show that the harmonic mean of a and b equals the geometric mean squared divided by the arithmetic mean. * 111. Another Reciprocal Property  Prove that if 0 6 a 6 b, then 1 1 6 . 0 6 b a

Discussion and Writing 112. Make up an inequality that has no solution. Make up one that has exactly one solution. * 113. The inequality x 2 + 1 6 - 5 has no real solution. Explain why. 114. Do you prefer to use inequality notation or interval notation to express the solution to an inequality? Give your reasons. Are there particular circumstances when you prefer one to the other? Cite examples.

115. How would you explain to a fellow student the underlying reason for the multiplication properties for inequalities (page A83)? That is, the sense (direction) of an inequality remains the same if each side is multiplied by a positive real number, whereas the direction is reversed if each side is multiplied by a negative real number.

‘Are You Prepared?’ Answers 1.

⫺4

⫺2

2.

0

–1

0

1

A.11 Complex Numbers PREPARING FOR THIS SECTION  Before getting started, review the following: r
$MBTTJGJDBUJPO
PG
/VNCFST
"QQFOEJY
" Section A.1, p. A3)

r
3BUJPOBMJ[JOH
%FOPNJOBUPST
"QQFOEJY
" Section A.7, pp. A57–A58)

Now Work the ‘Are You Prepared?’ problems on page A94.

OBJECTIVE 1 Add, Subtract, Multiply, and Divide Complex Numbers (p. A90)

Complex Numbers One property of a real number is that its square is nonnegative (greater than or equal to 0). For example, there is no real number x for which x2 = - 1 The introduction of the imaginary unit will remedy this situation.

DEFINITION

The imaginary unit, denoted by i, is the number whose square is - 1; that is, i2 = - 1 This should not be a surprise. If our universe were to consist only of integers, there would be no number x for which 2x = 1. This unfortunate circumstance was 1 2 remedied by introducing numbers such as and , the rational numbers. If our 2 3 universe were to consist only of rational numbers, there would be no x whose square

A90

APPENDIX A  Review

equals 2. That is, there would be no number x for which x2 = 2. To remedy this, 3 mathematicians introduced numbers such as 2 2 and 2 5, the irrational numbers. Recall that the real numbers consist of the rational numbers and the irrational numbers. Now, if our universe were to consist only of real numbers, there would be no number x whose square is - 1. To remedy this, mathematicians introduced a number i, whose square is - 1. In the progression outlined, each time that a situation was encountered that was unsuitable, a new number system was introduced to remedy the situation. And each new number system contained the earlier number system as a subset. The number system that results from introducing the number i is called the complex number system.

DEFINITION

Complex numbers are numbers of the form a + bi, where a and b are real numbers. The real number a is called the real part of the number a + bi; the real number b is called the imaginary part of a + bi. For example, the complex number - 5 + 6i has the real part - 5 and the imaginary part 6. When a complex number is written in the form a + bi, where a and b are real numbers, it is in standard form. However, if the imaginary part of a complex number is negative, such as in the complex number 3 + 1 - 22i, then it is written in the form 3 - 2i instead. Also, the complex number a + 0i is usually written simply as a. This serves to remind us that the real numbers are a subset of the complex numbers. The complex number 0 + bi is usually written as bi. Sometimes the complex number bi is called a pure imaginary number.

1 Add, Subtract, Multiply, and Divide Complex Numbers Equality, addition, subtraction, and multiplication of complex numbers are defined so as to preserve the familiar rules of algebra for real numbers. Thus, two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

Equality of Complex Numbers a + bi = c + di if and only if a = c and b = d

(1)

Two complex numbers are added by forming the complex number whose real part is the sum of the real parts and whose imaginary part is the sum of the imaginary parts.

Sum of Complex Numbers 1a + bi2 + 1c + di2 = 1a + c2 + 1b + d2i

(2)

To subtract two complex numbers, use this rule:

Difference of Complex Numbers 1a + bi2 - 1c + di2 = 1a - c2 + 1b - d2i

EX A MPL E 1

(3)

Adding and Subtracting Complex Numbers

(a) 13 + 5i2 + 1 - 2 + 3i2 = 3 3 + 1 - 22 4 + 15 + 32i = 1 + 8i (b) 16 + 4i2 - 13 + 6i2 = 16 - 32 + 14 - 62 i = 3 + 1 - 22i = 3 - 2i

Now Work

PROBLEM

13

Products of complex numbers are calculated as illustrated in Example 2.

r

SECTION A.11  Complex Numbers

EXAM PL E 2

A91

Multiplying Complex Numbers

15 + 3i2 # 12 + 7i2 = 512 + 7i2 + 3i12 + 7i2 = 10 + 35i + 6i + 21i 2 = 10 + 41i + 211 - 12 = - 11 + 41i

Distributive Property Distributive Property i 2 =−1

r

Based on the procedure of Example 2, the product of two complex numbers is defined by the following formula:

Product of Complex Numbers 1a + bi2 # 1c + di2 = 1ac - bd2 + 1ad + bc2i

(4)

Do not bother to memorize formula (4). Instead, whenever it is necessary to multiply two complex numbers, follow the usual rules for multiplying two binomials, as in Example 2, remembering that i 2 = - 1. For example, 12i2 12i2 = 4i 2 = - 4

12 + i2 11 - i2 = 2 - 2i + i - i 2 = 3 - i

Now Work

PROBLEM

19

Algebraic properties for addition and multiplication, such as the Commutative, Associative, and Distributive Properties, hold for complex numbers. However, the property that every nonzero complex number has a multiplicative inverse, or reciprocal, requires a closer look.

Conjugates DEFINITION

If z = a + bi is a complex number, then its conjugate, denoted by z, is defined as z = a + bi = a - bi For example, 2 + 3i = 2 - 3i and - 6 - 2i = - 6 + 2i.

EXAM PL E 3

Multiplying a Complex Number by Its Conjugate Find the product of the complex number z = 3 + 4i and its conjugate z.

Solution

Since z = 3 - 4i,

zz = 13 + 4i2 13 - 4i2 = 9 - 12i + 12i - 16i 2 = 9 + 16 = 25

r

The result obtained in Example 3 has an important generalization.

THEOREM

The product of a complex number and its conjugate is a nonnegative real number. That is, if z = a + bi, then zz = a2 + b2

(5)

Proof  If z = a + bi, then

zz = 1a + bi2 1a - bi2 = a2 - abi + abi - 1bi2 2 = a2 - b2 i 2 = a2 + b2

■

A92

APPENDIX A  Review

To express the reciprocal of a nonzero complex number z in standard form, 1 multiply the numerator and denominator of by its conjugate z. That is, if z z = a + bi is a nonzero complex number, then 1 1 z a - bi a b 1 z = = # = = = 2 - 2 i z z z a + bi zz c a2 + b2 a + b2 a + b2 Use (5).

EX A MPL E 4

Writing the Reciprocal of a Complex Number in Standard Form 1 in standard form a + bi; that is, find the reciprocal of 3 + 4i. 3 + 4i 1 Multiply the numerator and denominator of by the conjugate of 3 + 4i, the 3 + 4i complex number 3 - 4i. The result is Write

Solution

1 1 # 3 - 4i 3 - 4i 3 4 = = = i 3 + 4i 3 + 4i 3 - 4i 9 + 16 25 25

r

To express the quotient of two complex numbers in standard form, multiply the numerator and denominator of the quotient by the conjugate of the denominator.

EX A MPL E 5

Writing the Quotient of Complex Numbers in Standard Form Write each of the following in standard form. (a)

Solution

(a)

(b)

1 + 4i 5 - 12i

2 - 3i 4 - 3i

1 + 4i 1 + 4i # 5 + 12i 5 + 12i + 20i + 48i 2 = = 5 - 12i 5 - 12i 5 + 12i 25 + 144 - 43 + 32i 43 32 = = + i 169 169 169 2 - 3i 2 - 3i # 4 + 3i 8 + 6i - 12i - 9i 2 17 - 6i 17 6 = = = = i 4 - 3i 4 - 3i 4 + 3i 16 + 9 25 25 25

Now Work

EX A MPL E 6

(b)

PROBLEM

27

r

Writing Other Expressions in Standard Form If z = 2 - 3i and w = 5 + 2i, write each of the following expressions in standard form. z (a) (c) z + z (b) z + w w

Solution

(a)

12 - 3i2 15 - 2i2 z 10 - 4i - 15i + 6i 2 z#w = = # = w w w 15 + 2i2 15 - 2i2 25 + 4 4 - 19i 4 19 = = i 29 29 29

(b) z + w = 12 - 3i2 + 15 + 2i2 = 7 - i = 7 + i (c) z + z = 12 - 3i2 + 12 + 3i2 = 4

r

The conjugate of a complex number has certain general properties that will be useful later. For a real number a = a + 0i, the conjugate is a = a + 0i = a - 0i = a.

SECTION A.11  Complex Numbers

THEOREM

A93

The conjugate of a real number is the real number itself. Other properties that are direct consequences of the definition of the conjugate are given next. In each statement, z and w represent complex numbers.

THEOREM

The conjugate of the conjugate of a complex number is the complex number itself. z = z

(6)

The conjugate of the sum of two complex numbers equals the sum of their conjugates. z + w = z + w

(7)

The conjugate of the product of two complex numbers equals the product of their conjugates. z#w = z#w

(8)

The proofs of equations (6), (7), and (8) are left as exercises. See Problems 60–62.

Powers of i The powers of i follow a pattern that is useful to know.

i5 = i4 # i = 1 # i = i

i1 = i i2 = - 1

i6 = i4 # i2 = - 1

i 4 = i 2 # i 2 = 1 - 12 1 - 12 = 1

i8 = i4 # i4 = 1

i3 = i2 # i = - i

i7 = i4 # i3 = - i

And so on. The powers of i repeat with every fourth power.

EXAM PL E 7

Evaluating Powers of i (a) i 27 = i 24 # i 3 = 1i 4 2

# i3 = 16 # i3 = - i 25 1i 4 2 # i = 125 # i = i 6

(b) i 101 = i 100 # i 1 =

EXAM PL E 8

Solution

r

Writing the Power of a Complex Number in Standard Form Write 12 + i2 3 in standard form.

Use the special product formula for 1a + b2 3. 1a + b2 3 = a3 + 3a2 b + 3ab2 + b3

Using this special product formula,

NOTE: If you did not remember the special product formula for (a ⫹ b)3, you could find (2 ⫹ i)3 by simplifying (2 ⫹ i)2 (2 ⫹ i ). 䊏

12 + i2 3 = 23 + 3 # 22 # i + 3 # 2 # i 2 + i 3 = 8 + 61 - 12 + 12i + 1 - i2 = 2 + 11i

Now Work

PROBLEMS

33

AND

41

r

A94

APPENDIX A  Review

Square Roots of Negative Numbers Because i 2 = - 1, the square root of a negative number can be defined.

DEFINITION

If N is a positive real number, then the principal square root of −N, denoted by 2 −N, −N N , is 2 - N = 2Ni 2Ni 2N where i is the imaginary unit and i 2 = - 1.

EX A MPL E 9

Evaluating the Square Root of a Negative Number (a) 2 - 1 = 21i = i

(b) 2 - 16 = 216i = 4i

r

(c) 2 - 8 = 28i = 222i

Now Work Problem 47

A.11 Assess Your Understanding ‘Are You Prepared?’Answers are given at the end of these exercises. If you get a wrong answer, read the pages listed in red. 1. Name the integers and the rational numbers in the set 6 e - 3, 0, 22, , p f. (p. A3) 5 2. True or False Rational numbers and irrational numbers are in the set of real numbers. (p. A3) True

3. Rationalize the denominator of 312 - 232

3 2 + 23

. (pp. A57–A58)

Concepts and Vocabulary 4. In the complex number 5 + 2i, the number 5 is called the real part; the number 2 is called the imaginary part; the number i is called the imaginary unit .

6. True or False The conjugate of 2 + 5i is - 2 - 5i.

False

7. True or False All real numbers are complex numbers. True 8. True or False The product of a complex number and its conjugate is a nonnegative real number. True

5. i 2 = - 1 ; i 3 = - i ; i 4 = 1

Skill Building In Problems 9–46, write each expression in the standard form a + bi. 9. 12 - 3i2 + 16 + 8i2

12. 13 - 4i2 - 1 - 3 - 4i2 15. 312 - 6i2

27.

6 - i 1 + i

30. a

33. i 23 36. i

-23

37

5 12 + i 13 13 5 7 - i 2 2

1 2 23 - ib 2 2 -i

13. 12 - 5i2 - 18 + 6i2

1 23 i 2 2

- 6 - 11i

- 8 - 32i

11. 1 - 3 + 2i2 - 14 - 4i2

14. 1 - 8 + 4i2 - 12 - 2i2 17. 2i12 - 3i2

20. 15 + 3i2 12 - i2

22. 1 - 3 + i2 13 + i2

- 10

23.

10 3 - 4i

26.

2 - i - 2i

25.

2 + i i

1 - 2i

28.

2 + 3i 1 - i

-

31. 11 + i2 2

6

- 5

1 23 2 + ib 2 2

2i

32. 11 - i2 2

-6 3 - 4i

13 + i

1 + i 2

29. a

40. 4i 3 - 2i 2 + 1

- 10 + 6i

6 8 + i 5 5

1 5 + i 2 2

-1

- 7 + 6i

6 + 4i

10 - 5i

37. i - 10i

- 4 + 7i

19. 13 - 4i2 12 + i2

34. i 14

i

39. 6i 3 - 4i 5

10. 14 + 5i2 + 1 - 8 + 2i2

16. - 412 + 8i2

- 12 - 9i

21. 1 - 6 + i2 1 - 6 - i2 13 5 - 12i

6

6 - 18i

18. 3i1 - 3 + 4i2

24.

8 + 5i

35. i -15

i

38. 4 + i

3

- 2i

4 - i

-

1 23 + i 2 2

SECTION A.11  Complex Numbers

41. 11 + i2 3

44. 2i 11 + i 2 4

2

42. 13i2 4 + 1

- 2 + 2i

45. i

0

6

+ i

4

43. i 7 11 + i 2 2

82

+ i + 1 2

46. i

0

+ i

7

0

+ i + i

5

A95

3

0

In Problems 47–52, perform the indicated operations and express your answer in the standard form a + bi. 47. 2- 4

48. 2- 9

2i

50. 2- 64

51. 213 + 4i2 14i - 32

8i

49. 2- 25

3i

5i

52. 214 + 3i2 13i - 42

5i

5i

Applications and Extensions In Problems 53–56, z = 3 - 4i and w = 8 + 3i. Write each expression in the standard form a + bi. 53. z + z

6

54. w - w

55. zz

6i

57. Electrical Circuits    The impedance Z, in ohms, of a circuit element is defined as the ratio of the phasor voltage V, in volts, across the element to the phasor current I, in amperes, V through the elements. That is, Z = . If the voltage across I a circuit element is 18 + i volts and the current through the element is 3 - 4i amperes, determine the impedance. 2 + 3i *58. Parallel Circuits  In an ac circuit with two parallel pathways, the total impedance Z, in ohms, satisfies the formula 1 1 1 = + , where Z1 is the impedance of the first Z Z1 Z2

56. z - w

25

- 5 + 7i

pathway and Z2 is the impedance of the second pathway. Determine the total impedance if the impedances of the two pathways are Z1 = 2 + i ohms and Z2 = 4 - 3i ohms. *59. Use z = a + bi to show that z + z = 2a and that z - z = 2bi. *60. Use z = a + bi to show that z = z. *61. Use z = a + bi z + w = z + w.

and

w = c + di

to

show

that

*62. Use z = a + bi and w = c + di to show that z # w = z # w.

Discussion and Writing 63. Explain to a friend how you would add two complex numbers and how you would multiply two complex numbers. Explain any differences between the two explanations. 64. Write a brief paragraph that compares the method used to rationalize denominators and the method used to write the quotient of two complex numbers in standard form.

66. Explain how the method of multiplying two complex numbers is related to multiplying two binomials. *67. What Went Wrong?  A student multiplied 2- 9 and 2- 9 as follows:

65. Use an Internet search engine to investigate the origins of complex numbers. Write a paragraph describing what you find, and present it to the class.

2- 9

# 2- 9 =

= 281 = 9 The instructor marked the problem incorrect. Why?

‘Are You Prepared?’ Answers 6 1. Integers: 5 - 3, 06; rational numbers: e - 3, 0, f 5

2. True

2( - 9)( - 9)

3. 3(2 - 13)

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

Appendix

B

Graphing Utilities Outline B.1 The Viewing Rectangle B.2 Using a Graphing Utility to Graph Equations B.3 Using a Graphing Utility to Locate Intercepts

and Check for Symmetry B.4 Using a Graphing Utility to Solve Equations B.5 Square Screens

B.6 Using a Graphing Utility to Graph

B.9 Using a Graphing Utility to Graph Parametric

Inequalities B.7 Using a Graphing Utility to Solve Systems of Linear Equations B.8 Using a Graphing Utility to Graph a Polar Equation

Equations

B.1 The Viewing Rectangle Figure 1 y = 2x

Figure 2

All graphing utilities—that is, all graphing calculators and all computer software graphing packages—graph equations by plotting points on a screen. The screen itself actually consists of small rectangles called pixels. The more pixels the screen has, the better the resolution. Most graphing calculators have 2048 pixels per square inch; most computer screens have 4096 to 8192 pixels per square inch. When a point to be plotted lies inside a pixel, the pixel is turned on (lights up). The graph of an equation is a collection of pixels. Figure 1 shows how the graph of y = 2x looks on a TI-84 Plus graphing calculator. The screen of a graphing utility will display the coordinate axes of a rectangular coordinate system. However, the scale must be set on each axis. The smallest and largest values of x and y to be included in the graph must also be set. This is called setting the viewing rectangle or viewing window. Figure 2 illustrates a typical viewing window. To select the viewing window, values must be given to the following expressions: Xmin: Xmax: Xscl: Ymin: Ymax: Yscl:

the smallest value of x the largest value of x the number of units per tick mark on the x-axis the smallest value of y the largest value of y the number of units per tick mark on the y-axis

Figure 3 illustrates these settings and their relation to the Cartesian coordinate system.

Figure 3

y Ymax Yscl

x

Xmin

Xmax Xscl

Ymin

B1

B2

APPENDIX B  Graphing Utilities

If the scale used on each axis is known, the minimum and maximum values of x and y shown on the screen can be determined by counting the tick marks. Look again at Figure 2 on the previous page. For a scale of 1 on each axis, the minimum and maximum values of x are - 10 and 10, respectively, the minimum and maximum values of y are also - 10 and 10. If the scale is 2 on each axis, then the minimum and maximum values of x are - 20 and 20, respectively, and the minimum and maximum values of y are - 20 and 20, respectively. Conversely, if the minimum and maximum values of x and y are known, then the scales can be found by counting the tick marks displayed. This text follows the practice of showing the minimum and maximum values of x and y in illustrations so that the reader will know how the viewing window was set. See Figure 4. Figure 4

4

⫺3

3

means

Xmin = - 3 Xmax = 3 Xscl = 1

Ymin = - 4 Ymax = 4 Yscl = 2

⫺4

Finding the Coordinates of a Point Shown on a Graphing Utility Screen

EX A MPL E 1 Figure 5

Find the coordinates of the point shown in Figure 5. Assume that the coordinates are integers.

4

Solution First note that the viewing window used in Figure 5 is

⫺3

3

Xmin = - 3 Xmax = 3 Xscl = 1

⫺4

Ymin = - 4 Ymax = 4 Yscl = 2

The point shown is 2 tick units to the left on the horizontal axis 1scale = 12 and 1 tick unit up on the vertical axis 1scale = 22. The coordinates of the point shown are 1 - 2, 22.

r

B.1 Exercises In Problems 1–4, determine the coordinates of the points shown. Tell in which quadrant each point lies. Assume that the coordinates are integers. 1.

2.

10

⫺5

5

⫺10 ( - 1, 4); II

3.

10

⫺5

⫺5

5

4.

5

⫺10

5

⫺10 (3, 4); I

10

10

⫺10 ( - 6, - 4); III

⫺5 (3, 1); I

In Problems 5–10, determine the viewing window used. *5.

*6.

4

⫺6

6

*7.

2

⫺3

3

3 ⫺6

⫺4

⫺2

6 ⫺1

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

SECTION B.2  Using a Graphing Utility to Graph Equations

*8.

*9.

4 ⫺9

*10.

10

B3

8

9

⫺12

3

9

⫺22

⫺10 4

2

In Problems 11–16, select a setting such that each of the given points will lie within the viewing rectangle. *11. 1 - 10, 52, 13, - 22, 14, - 12

*14. 1 - 80, 602, 120, - 302, 1 - 20, - 402

*12. 15, 02, 16, 82, 1 - 2, - 32

*15. 10, 02, 1100, 52, 15, 1502

*13. 140, 202, 1 - 20, - 802, 110, 402 *16. 10, - 12, 1100, 502, 1 - 10, 302

B.2 Using a Graphing Utility to Graph Equations From Examples 2 and 3 in Foundations, Section 2, recall that a graph can be obtained by plotting points in a rectangular coordinate system and connecting them. Graphing utilities perform these same steps when graphing an equation. For example, the TI-84 Plus determines 95 evenly spaced input values,* starting at Xmin and ending at Xmax, uses the equation to determine the output values, plots these points on the screen, and finally (if in the connected mode) draws a line between consecutive points. To graph an equation in two variables x and y using a graphing utility requires that the equation be written in the form y = 5 expression in x6 . If the original equation is not in this form, replace it by equivalent equations until the form y = 5 expression in x6 is obtained.

Steps for Graphing an Equation Using a Graphing Utility STEP 1: Solve the equation for y in terms of x. STEP 2: Get into the graphing mode of the graphing utility. The screen will usually display Y1 = , prompting the user to enter the expression involving x that was found in Step 1. (Consult your manual for the correct way to enter the expression. For example, y = x2 might be entered as x ¿ 2 or as x*x or as x xY 2). STEP 3: Select the viewing window. Without prior knowledge about the behavior of the graph of the equation, it is common to select the standard viewing window** initially. The viewing window is then adjusted based on the graph that appears. In this text the standard viewing window is Xmin = - 10 Xmax = 10 Xscl = 1

Ymin = - 10 Ymax = 10 Yscl = 1

STEP 4: Graph. STEP 5: Adjust the viewing window until a complete graph is obtained.

*These input values depend on the values of Xmin and Xmax. For example, if Xmin = - 10 and Xmax = 10, 10 - 1 - 102 then the first input value will be - 10 and the next input value will be - 10 + = - 9.7872, 94 and so on. **Some graphing utilities have a ZOOM-STANDARD feature that automatically sets the viewing window to the standard viewing window and graphs the equation.

B4

APPENDIX B  Graphing Utilities

Graphing an Equation on a Graphing Utility

EX A MPL E 1

Graph the equation: 6x2 + 3y = 36

Solution

STEP 1: Solve for y in terms of x. 6x2 + 3y = 36 3y = - 6x2 + 36 Subtract 6x2 from both sides of the equation. y = - 2x2 + 12 Divide both sides of the equation by 3 and simplify. STEP 2: STEP 3: STEP 4: STEP 5:

From the Y1 = screen, enter the expression - 2x2 + 12 after the prompt. Set the viewing window to the standard viewing window. Graph. The screen should look like Figure 6. The graph of y = - 2x2 + 12 is not complete. The value of Ymax must be increased so that the top portion of the graph is visible. Increasing the value of Ymax to 12 gives the graph in Figure 7. The graph is now complete. Figure 6

Figure 7 10

⫺10

Figure 8

12

10

⫺10

10

12

⫺10 ⫺4

⫺10

r

4

Look again at Figure 7. Although a complete graph is shown, the graph might be improved by adjusting the values of Xmin and Xmax. Figure 8 shows the graph of y = - 2x2 + 12 using Xmin = - 4 and Xmax = 4. Do you think this is a better choice for the viewing window?

⫺10

EX A MPL E 2

Creating a Table and Graphing an Equation Create a table and graph the equation: y = x3

Solution

Most graphing utilities have the capability of creating a table of values for an equation. (Check your manual to see if your graphing utility has this capability.) Table 1 illustrates a table of values for y = x3 on a TI-84 Plus. See Figure 9 for the graph. Figure 9

Table 1

10

⫺3

3

⫺10

B.2 Exercises In Problems 1–16, graph each equation using the following viewing windows: (a) Xmin Xmax Xscl Ymin Ymax Yscl

= = = = = =

-5 5 1 -4 4 1

(b) Xmin Xmax Xscl Ymin Ymax Yscl

= = = = = =

- 10 10 2 -8 8 2

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

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SECTION B.3  Using a Graphing Utility to Locate Intercepts and Check for Symmetry

*1. y = x + 2

*2. y = x - 2

*3. y = - x + 2

B5

*4. y = - x - 2

*5. y = 2x + 2

*6. y = 2x - 2

*7. y = - 2x + 2

*8. y = - 2x - 2

*9. y = x2 + 2

*10. y = x2 - 2

*11. y = - x2 + 2

*12. y = - x2 - 2

*13. 3x + 2y = 6

*14. 3x - 2y = 6

*15. - 3x + 2y = 6

*16. - 3x - 2y = 6

*17–32. For each of the above equations, create a table, - 3 … x … 3, and list points on the graph.

B.3 Using a Graphing Utility to Locate Intercepts and Check for Symmetry Value and Zero (or Root) Most graphing utilities have an eVALUEate feature that, given a value of x, determines the value of y for an equation. This feature can be used to evaluate an equation at x = 0 to determine the y-intercept. Most graphing utilities also have a ZERO (or ROOT) feature that can be used to determine the x-intercept(s) of an equation.

Finding Intercepts Using a Graphing Utility

EXAM PL E 1

Use a graphing utility to find the intercepts of the equation y = x3 - 8.

Solution

Figure 10(a) shows the graph of y = x3 - 8.

Figure 10

10

−5

5

−20 (a)

(b)

(c)

The eVALUEate feature of a TI-84 Plus graphing calculator accepts as input a value of x and determines the value of y. If we let x = 0, we find that the y-intercept is - 8. See Figure 10(b). The ZERO feature of a TI-84 Plus is used to find the x-intercept(s). See Figure 10(c). The x-intercept is 2.

EXAM PL E 2

1 Graphing the Equation y = x Graph the equation y = and symmetry.

Solution

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1 . Based on the graph, infer information about intercepts x

Figure 11 shows the graph. Infer from the graph that there are no intercepts. Also infer that symmetry with respect to the origin is a possibility. The TABLE feature on a graphing utility can provide further evidence of symmetry with respect to the origin. Using a TABLE, observe that for any ordered pair 1x, y2 , the ordered pair 1 - x, - y2 is also a point on the graph.

Figure 11

Y1 ⫽ 1x

4

⫺3

3

⫺4

r

B6

APPENDIX B  Graphing Utilities

B.3 Exercises In Problems 1–6, use ZERO (or ROOT) to approximate the smaller of the two x-intercepts of each equation. Express the answer rounded to two decimal places. 1. y = x2 + 4x + 2 4. y = 3x + 5x + 1 2

- 3.41

2. y = x2 + 4x - 3

- 1.43

- 4.65

5. y = 2x - 3x - 1 2

- 0.28

3. y = 2x2 + 4x + 1

- 1.71

6. y = 2x - 4x - 1

- 0.22

2

In Problems 7–12, use ZERO (or ROOT) to approximate the positive x-intercepts of each equation. Express each answer rounded to two decimal places. 7. y = x3 + 3.2x2 - 16.83x - 5.31

8. y = x3 + 3.2x2 - 7.25x - 6.3

3.00

9. y = x4 - 1.4x3 - 33.71x2 + 23.94x + 292.41 11. y = x + 19.5x - 1021x + 1000.5 3

2

4.50

1.00; 23.00

2.00

10. y = x4 + 1.2x3 - 7.46x2 - 4.692x + 15.2881 12. y = x + 14.2x - 4.8x - 12.4 3

2

1.70

1.07

B.4 Using a Graphing Utility to Solve Equations For many equations, there are no algebraic techniques that lead to a solution. For such equations, a graphing utility can often be used to investigate possible solutions. When a graphing utility is used to solve an equation, approximate solutions usually are obtained. Unless otherwise stated, this text follows the practice of giving approximate solutions rounded to two decimal places. The ZERO (or ROOT) feature of a graphing utility can be used to find the solutions of an equation when one side of the equation is 0. In using this feature to solve equations, make use of the fact that the x-intercepts (or zeros) of the graph of an equation are found by letting y = 0 and solving the equation for x. Solving an equation for x when one side of the equation is 0 is equivalent to finding where the graph of the corresponding equation crosses or touches the x-axis.

Using ZERO (or ROOT) to Approximate Solutions of an Equation

EX A MPL E 1

Find the solution(s) of the equation x2 - 6x + 7 = 0. Round answers to two decimal places. The solutions of the equation x2 - 6x + 7 = 0 are the same as the x-intercepts of the graph of Y1 = x2 - 6x + 7. Begin by graphing the equation. See Figure 12(a). From the graph there appear to be two x-intercepts (solutions to the equation): one between 1 and 2, the other between 4 and 5.

Solution

Figure 12

8

8

⫺1

8

7 ⫺1 ⫺2

7 ⫺2

(a)

⫺1

7 ⫺2

(b)

(c)

Using the ZERO (or ROOT) feature of the graphing utility, determine that the x-intercepts, and thus the solutions to the equation, are x = 1.59 and x = 4.41, rounded to two decimal places. See Figures 12(b) and (c).

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A second method for solving equations using a graphing utility involves the INTERSECT feature of the graphing utility. This feature is used most effectively when one side of the equation is not 0.

SECTION B.4  Using a Graphing Utility to Solve Equations

EXAM PL E 2

B7

Using INTERSECT to Approximate Solutions of an Equation Find the solution(s) to the equation 31x - 22 = 51x - 12. Round answers to two decimal places.

Solution

Begin by graphing each side of the equation as follows: graph Y1 = 31x - 22 and Y2 = 51x - 12. See Figure 13(a).

Figure 13

5 ⫺4

5 4

⫺15 (a)

⫺4

4

⫺15 (b)

At the point of intersection of the graphs, the values of the y-coordinates are the same. Conclude that the x-coordinate of the point of intersection represents the solution to the equation. Do you see why? The INTERSECT feature on a graphing utility determines the point of intersection of the graphs. Using this feature, find that the graphs intersect at 1 - 0.5, - 7.52. See Figure 13(b). The solution of the equation is therefore x = - 0.5.

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SUMMARY Here are the steps to follow for approximating solutions of equations. Steps for Approximating Solutions of Equations Using ZERO (or ROOT) STEP 1: Write the equation in the form 5 expression in x6 = 0. STEP 2: Graph Y1 = 5 expression in x6 . Be sure that the graph is complete. That is, be sure that all the intercepts are shown on the screen. STEP 3: Use ZERO (or ROOT) to determine each x-intercept of the graph. Steps for Approximating Solutions of Equations Using INTERSECT STEP 1: Graph Y1 = 5 expression in x on the left side of the equation6 . Graph Y2 = 5 expression in x on the right side of the equation6 . STEP 2: Use INTERSECT to determine each x-coordinate of the point(s) of intersection, if any. Be sure that the graphs are complete. That is, be sure that all the points of intersection are shown on the screen.

EXAM PL E 3

Solving a Radical Equation 3 Find the real solutions of the equation 2 2x - 4 - 2 = 0.

Solution

3 Figure 14 shows the graph of the equation Y1 = 2 2x - 4 - 2. From the graph, note that one x-intercept is near 6. Using ZERO (or ROOT), find that the x-intercept is 6. The only solution is x = 6.

Figure 14

1 ⫺1

10

⫺4

r

B8

APPENDIX B  Graphing Utilities

B.5 Square Screens Figure 15 4

(4, 4)

⫺4

4

(⫺4, ⫺4)

Most graphing utilities have a rectangular screen. Because of this, using the same settings for both x and y will result in a distorted view. For example, Figure 15 shows the graph of the line y = x connecting the points 1 - 4, - 42 and 14, 42. The line is expected to bisect the first and third quadrants, but it doesn’t. The selections for Xmin, Xmax, Ymin, and Ymax must be adjusted so that a square screen results. On most graphing utilities, this is accomplished by setting the ratio of x to y at 3∶2.* For example, if Xmin = - 6 Xmax = 6

⫺4

Ymin = - 4 Ymax = 4

then the ratio of x to y is 6 - 1 - 62 Xmax - Xmin 12 3 = = = Ymax - Ymin 4 - 1 - 42 8 2 for a ratio of 3∶2, resulting in a square screen.

Examples of Viewing Rectangles That Result in Square Screens

EX A MPL E 1

Figure 16 4

⫺6

(4, 4)

6

(⫺4, ⫺4)

⫺4

(a) Xmin Xmax Xscl Ymin Ymax Yscl

= -3 = 3 = 1 = -2 = 2 = 1

(b) Xmin Xmax Xscl Ymin Ymax Yscl

= -6 = 6 = 1 = -4 = 4 = 1

(c) Xmin Xmax Xscl Ymin Ymax Yscl

= - 12 = 12 = 1 = -8 = 8 = 2

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Figure 16 shows the graph of the line y = x on a square screen using the viewing rectangle given in part (b). Notice that the line now bisects the first and third quadrants. Compare this illustration to Figure 15.

B.5 Exercises In Problems 1–8, determine which of the given viewing rectangles result in a square screen. 1. Xmin Xmax Xscl Ymin Ymax Yscl

= - 3 Yes = 3 = 2 = -2 = 2 = 2

2. Xmin Xmax Xscl Ymin Ymax Yscl

= - 5 No = 5 = 1 = -4 = 4 = 1

3. Xmin Xmax Xscl Ymin Ymax Yscl

= 0 Yes = 9 = 3 = -2 = 4 = 2

4. Xmin Xmax Xscl Ymin Ymax Yscl

= - 6 Yes = 6 = 1 = -4 = 4 = 2

5. Xmin Xmax Xscl Ymin Ymax Yscl

= - 6 No = 6 = 1 = -2 = 2 = 0.5

6. Xmin Xmax Xscl Ymin Ymax Yscl

= - 6 Yes = 6 = 2 = -4 = 4 = 1

7. Xmin Xmax Xscl Ymin Ymax Yscl

= 0 Yes = 9 = 1 = -2 = 4 = 1

8. Xmin Xmax Xscl Ymin Ymax Yscl

= - 6 Yes = 6 = 2 = -4 = 4 = 2

*9. If Xmin = - 4, Xmax = 8, and Xscl = 1, how should Ymin, Ymax, and Yscl be selected so that the viewing rectangle contains the point 14, 82 and the screen is square? *10. If Xmin = - 6, Xmax = 12, and Xscl = 2, how should Ymin, Ymax, and Yscl be selected so that the viewing rectangle contains the point 14, 82 and the screen is square? *Some graphing utilities have a built-in function that automatically squares the screen. For example, the TI-84 has a ZSquare function that does this. Some graphing utilities require a ratio other than 3 ∶ 2 to square the screen. For example, the HP 48G requires the ratio of x to y to be 2 ∶ 1 for a square screen. Consult your manual.

*Due to space restrictions, answers to these exercises may be found in the Answers in the back of the book.

SECTION B.7  Using a Graphing Utility to Solve Systems of Linear Equations

B9

B.6 Using a Graphing Utility to Graph Inequalities EXAM PL E 1

Graphing an Inequality Using a Graphing Utility Use a graphing utility to graph: 3x + y - 6 … 0

Solution

Begin by graphing the equation 3x + y - 6 = 0 1Y1 = - 3x + 62. See Figure 17. As with graphing by hand, test points selected from each region and determine whether they satisfy the inequality. To test the point 1 - 1, 22, for example, enter 31 - 12 + 2 - 6 … 0. See Figure 18(a). The 1 that appears indicates that the statement entered (the inequality) is true. When the point 15, 52 is tested, a 0 appears, indicating that the statement entered is false. Thus 1 - 1, 22 is a part of the graph of the inequality, and 15, 52 is not. Figure 18(b) shows the graph of the inequality on a TI-84 Plus.*

Figure 17

Figure 18

10

⫺2

10

⫺2

6

6

Y1 ⫽ ⫺3x ⫹ 6 ⫺10

⫺10 (a)

(b)

r

The steps to follow to graph an inequality using a graphing utility are given next.

Steps for Graphing an Inequality Using a Graphing Utility STEP 1: Replace the inequality symbol by an equal sign, solve the equation for y, and graph the equation. STEP 2: In each region, select a test point P and determine whether the coordinates of P satisfy the inequality. (a) If the test point satisfies the inequality, then so do all the points in the region. Indicate this by using the graphing utility to shade the region. (b) If the coordinates of P do not satisfy the inequality, then none of the points in that region do.

B.7 Using a Graphing Utility to Solve Systems of Linear Equations Most graphing utilities have the capability to put the augmented matrix of a system of linear equations in row echelon form. The next example, Example 6 from Section 10.2, demonstrates this feature using a TI-84 Plus graphing calculator.

*Consult your owner’s manual for shading techniques.

B10

APPENDIX B  Graphing Utilities

EX A MPL E 1

Solving a System of Linear Equations Using a Graphing Utility x - y + z = 8 Solve: c 2x + 3y - z = - 2 3x - 2y - 9z = 9

Solution

(1) (2) (3)

The augmented matrix of the system is -1 3 -2

1 C2 3

1 8 3 -1 -2S -9 9

Enter this matrix into the graphing utility and name it A. See Figure 19(a). Use the REF (row echelon form) command on matrix A. The results are shown in Figure 19(b). Figure 19

(a)

(b)

The system of equations represented by the matrix in row echelon form is 1

-

2 3

E0

1

0

0

-3 15 13 1

x -

3 5

-

24 U 13 1

e

2 y 3 y +

3z =

3

15 24 z = 13 13 z = 1

(1) (2) (3)

Using z = 1, back-substitute to get x d

2 y 3

3112 =

x -

3 (1)

15 24 (2) ¡ y + 112 = 13 13 Simplify.

d

2 y = 3 y =

6 - 39 = -3 13

(1) (2)

Solve the second equation for y, to find that y = - 3. Back-substitute y = - 3 into 2 x - y = 6, to find that x = 4. The solution of the system is x = 4, y = - 3, z = 1. 3

r

Figure 20

Notice that the row echelon form of the augmented matrix using the graphing utility differs from the row echelon form in Chapter 10 (p. 746), yet both matrices provide the same solution! This is because the two solutions used different row operations to obtain the row echelon form. In all likelihood, the two solutions parted ways in Step 4 of the algebraic solution, where fractions were avoided by interchanging rows 2 and 3. Most graphing utilities also have the ability to put a matrix in reduced row echelon form. Figure 20 shows the reduced row echelon form of the augmented matrix from Example 1 using the RREF command on a TI-84 Plus graphing calculator. From this command, note that the solution of the system is x = 4, y = - 3, z = 1.

SECTION B.9  Using a Graphing Utility to Graph Parametric Equations

B11

B.8 Using a Graphing Utility to Graph a Polar Equation Most graphing utilities require the following steps in order to obtain the graph of a polar equation. Be sure to be in POLar mode.

Graphing a Polar Equation Using a Graphing Utility STEP 1: Set the mode to POLar. Solve the equation for r in terms of u. STEP 2: Select the viewing rectangle in polar mode. Besides setting Xmin, Xmax, Xscl, and so forth, the viewing rectangle in polar mode requires setting the minimum and maximum values for u and an increment setting for u (ustep). In addition, a square screen and radian measure should be used. STEP 3: Enter the expression involving u that you found in Step 1. (Consult your manual for the correct way to enter the expression.) STEP 4: Graph.

Graphing a Polar Equation Using a Graphing Utility

EXAM PL E 1

Use a graphing utility to graph the polar equation r sin u = 2.

Solution

STEP 1: Solve the equation for r in terms of u. r sin u = 2 r =

2 sin u

STEP 2: From the POLar mode, select the viewing rectangle. umin = 0 umax = 2p p ustep = 24

6

9

STEP 3: Enter the expression STEP 4: Graph.

⫺6

Ymin = - 6 Ymax = 6

Xscl = 1

Yscl = 1

ustep determines the number of points that the graphing utility will plot. For example, p p 2p 3p if ustep is , the graphing utility will evaluate r at u = 01umin2, , , , and 24 24 24 24 so forth, up to 2p1umax2. The smaller ustep is, the more points the graphing utility will plot. The reader is encouraged to experiment with different values for umin, umax, and ustep to see how the graph is affected.

Figure 21

⫺9

Xmin = - 9 Xmax = 9

2 after the prompt r1 = . sin u

The graph is shown in Figure 21.

r

B.9 Using a Graphing Utility to Graph Parametric Equations Most graphing utilities have the capability of graphing parametric equations. The following steps are usually required to obtain the graph of parametric equations. Check your owner’s manual to see how yours works.

B12

APPENDIX B  Graphing Utilities

Graphing Parametric Equations Using a Graphing Utility STEP 1: Set the mode to PARametric. Enter x1t2 and y1t2. STEP 2: Select the viewing window. In addition to setting Xmin, Xmax, Xscl, and so on, the viewing window in parametric mode requires setting minimum and maximum values for the parameter t and an increment setting for t (Tstep). STEP 3: Graph.

EX A MPL E 1

Graphing a Curve Defined by Parametric Equations Using a Graphing Utility Graph the curve defined by the parametric equations x = 3t 2,

Solution

5

0

15

-2 … t … 2

STEP 1: Enter the equations x1t2 = 3t 2, y1t2 = 2t with the graphing utility in PARametric mode. STEP 2: Select the viewing window. The interval is - 2 … t … 2, so select the following square viewing window: Tmin = - 2 Tmax = 2 Tstep = 0.1

Figure 22

y = 2t,

Xmin = 0 Xmax = 15 Xscl = 1

Ymin = - 5 Ymax = 5 Yscl = 1

Choose Tmin = - 2 and Tmax = 2 because - 2 … t … 2. Finally, the choice for Tstep will determine the number of points that the graphing utility will plot. For example, with Tstep at 0.1, the graphing utility will evaluate x and y at t = - 2, - 1.9, - 1.8, and so on. The smaller the Tstep, the more points the graphing utility will plot. The reader is encouraged to experiment with different values of Tstep to see how the graph is affected. STEP 3: Graph. Note the direction in which the graph is drawn. This direction shows the orientation of the curve.

⫺5

The graph shown in Figure 22 is complete.

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Exploration Graph the following parametric equations using a graphing utility with Xmin = 0, Xmax = 15, Ymin = - 5, Ymax = 5, and Tstep = 0.1. 1. x =

3t 2 , y = t, 4

-4 … t … 4

2. x = 3t2 + 12t + 12, y = 2t + 4, 3

3. x = 3t2>3, y = 22t ,

-4 … t … 0

-8 … t … 8

Compare these graphs to the graph in Figure 22. Conclude that parametric equations defining a curve are not unique; that is, different parametric equations can represent the same graph.

Exploration 3y2

4x 4x aY1 = and Y2 = b with Xmin = 0, Xmax = 15, 4 A3 A3 Ymin = - 5, Ymax = 5. Compare this graph with Figure 22. Why do the graphs differ?

In FUNCtion mode, graph x =

Answers CHAPTER F Foundations: A Prelude to Functions F.1 Assess Your Understanding (page 38) 7. x-coordinate or abscissa; y-coordinate or ordinate 8. quadrants 9. 2 1x2 - x1 2 2 + 1y2 - y1 2 2 10. midpoint 29. d 1A, B2 = 213 11. (a) Quadrant II (b) Positive x-axis 13. The points will be on a 15. 25 (c) Quadrant III (d) Quadrant I vertical line that is 2 units 17. 210 d 1B, C2 = 213 (e) Negative y-axis (f) Quadrant IV to the right of the y-axis. 19. 2 217 d 1A, C2 = 226 (2, 4) y 21. 285 y 1 213 2 2 + 1 213 2 2 = 1 226 2 2 5 D  (6, 5) (2, 1) 6 23. 3 25 13 (2, 0) Area = square units A  (3, 2) 25. 26.89 ≈ 2.62 2 B  (6, 0) 2 2 5 x 27. 2a + b y 7 x C  (2, 2) (2, 1) (2, 3)

E  (0, 3)

A  (2, 5) 5

F  (6, 3)

C  (1, 0)

31. d 1A, B2 = 2130 d 1B, C2 = 226 d 1A, C2 = 2 226 1 226 2 2 + 1 2 226 2 2 = Area = 26 square units y 10 A  (5, 3)

33. d 1A, B2 = 4 d 1A, C2 = 5 d 1B, C2 = 241 42 + 52 = 16 + 25 = 1 241 2 2 Area = 10 square units

1 2130 2 2

y 5 C  (5, 5) B  (6, 0) 10 x

B  (0, 3)

C  (4, 2) 5 x A  (4, 3)

B  (1, 3) 5 x

3 7 a b 35. (4, 0) 37. a , 1 b 39. a1, - b 41. (1.05, 0.7) 43. a , b 2 2 2 2 45. (2, 2); 12, - 42 47. (0, 0); (8, 0) 49. 217; 2 25; 229 s s 51. a , b 53. d 1P1 , P2 2 = 6; d 1P2 , P3 2 = 4; d 1P1 , P3 2 = 2 213; 2 2 right triangle 55. d 1P1 , P2 2 = 2 217; d 1P2 , P3 2 = 234; d 1P1 , P3 2 = 234; isosceles right triangle 57. 90 22 ≈ 127.28 ft 59. (a) (90, 0), (90, 90), (0, 90) (b) 5 22161 ≈ 232.43 ft (c) 30 2149 ≈ 366.20 ft 61. d = 50t mi 63. (a) (2.65, 1.6) (b) ≈1.285 units 65. $21,220; a slight underestimate 67. 15,- 22

F.2 Assess Your Understanding (page 48)

3. intercepts 4. zeros; roots 5. y-axis 6. True 7. True 8. False 9. (0, 0) is on the graph. 11. (0, 3) is on the graph. 13. (0, 2) and 1 22 , 22 2 are on the graph. 15. 1 - 2, 02, 10, 22 17. 1 - 4, 02, 10, 82 19. 1 - 1, 02, 11, 02, 10, - 12 21. 1 - 2, 02, 12, 02, 10, 42 23. 13, 02, 10, 22 yx2

y 5

y 10

y 5

y  2x  8 (0, 8)

(2, 0)

(0, 2)

(4, 0) 10 x

5 x

y 5 (0, 4)

y  x2  1 (2, 0)

(1, 0) 5 x

(1, 0)

y 10 (0, 9) (2, 0)

(c)  (3, 4)

y 5 (2, 1) (a)  (2, 1)

5 x

43. (a) ( - 2, 0), (0, 0), (2, 0) (b) Symmetric with respect to the origin

31. (b)  (2, 1) 5 x (c)  (2, 1)

(c)  (5, 2)

y 5

p p , 0 b , 10, 12, a , 0 b 2 2 (b) Symmetric with respect to the y-axis

39. (a) a -

45. (a) x-intercepts: 3 - 2, 1 4 ; y-intercept: 0 (b) No symmetry

33. (a)  (5, 2)

5 (3, 0)

y 5 (a)  (3, 4)

(5, 2)

x

(c)  (3, 4) 5 x

5 x (b)  (5, 2)

(a)  (3, 4)

37. (a) ( - 1, 0), (1, 0) (b) Symmetric with respect to the x-axis, the y-axis, and the origin

y 5 (a)  (0, 3) (c)  (0, 3) (0, 3) (b)  (0, 3)

29.

y 5 (b)  (3, 4) (3, 4) 5 x

(2, 0) 10 x 9x 2  4y  36

35.

27.

2x  3y  6

(0, 2)

(2, 0) 5 x

y  x 2  4

(0, 1)

25. 1 - 2, 02, 12, 02, 10, 92

y 5

(3, 4)

(b)  (3, 4)

41. (a) (0, 0) (b) Symmetric with respect to the x-axis

47. (a) No intercepts (b) Symmetric with respect to the origin

AN1

AN2

ANSWERS  Section F.2

49.

51.

y 4 5 x

(2, 5) (0, 9)

69.

(, 0) (0, 0) 5 x

  , 2 2

71.

y 5 (1, 1) 5 x

(1, 1)

(, 0)

(2, 5)

53. 1 - 4, 02, 10, - 22, 10, 22; symmetric with respect to the x-axis 55. (0, 0); symmetric with respect to the origin 57. 10, - 92, 13, 02, 1 - 3, 02; symmetric with respect to the y-axis 59. 1 - 2, 02, 12, 02, 10, - 32, 10, 32; symmetric with respect to the x-axis, y-axis, and origin 61. 10, - 92, 13, 02, 1 - 3, 02, 1 - 1, 02 ; no symmetry 63. 10, - 42, 14, 02, 1 - 4, 02 ; symmetric with respect to the y-axis 65. (0, 0); symmetric with respect to the origin 67. (0, 0); symmetric with respect to the origin

y 5 , 2 2

(0, 0)

73. b = 13 75. - 4 or 1

y (1, 1) 5 (4, 2) (0, 0)

5 x

77. (a) 10, - 52, 1 - 25, 02, 1 25, 02 (b) Symmetric with respect to the y-axis (c) y 5 (兹5, 0)

79. (a) 1 - 9, 02, 10, - 32, 10, 32 (b) Symmetric with respect to the x-axis (c) y 5

(兹5, 0)

(2, 1)

(0, 3)

5 x (2, 1)

81. (a) 10, 32, 10, - 32, 1 - 3, 02, 13, 02 83. (a) 10, 02, 1 - 2, 02, 12, 02 (b) Symmetric with respect to the x-axis, (b) Symmetric with respect y-axis, and origin to the origin (c) (c) y y 5

5

(0, 3)

(1, 3) (2, 0)

(3, 0) 5 x

(3, 0)

(0, 0)

(0, 3)

(2, 0) 5 x

1 x

(9, 0)

(1, 4) (0, 5) (1, 4)

(0, 3)

85. 1 - 1, - 22 87. 4 89. (a) 10, 02, 12, 02, 10, 12, 10, - 12 (b) Symmetric with respect to the x-axis 91. (a) y = 2x2 and y =  x have the same graph. (b) 2x2 =  x (c) x Ú 0 for y = 1 2x2 2, while x can be any real number for y = x (d) y Ú 0 for y = 2x2

(1, 3)

F.3 Assess Your Understanding (page 61) 1 1. undefined; 0 2. 3; 2 3. y = b; y-intercept 4. True 5. False 6. True 7. m1 = m2; y-intercepts; m1m2 = - 1 8. 2 9. 10. False 2 1 1 (b) If x increases by 2 units, y will increase by 1 unit. 13. (a) Slope = - (b) If x increases by 3 units, y will decrease by 1 unit. 11. (a) Slope = 2 3 21. Slope undefined 3 19. Slope = 0 1 15. Slope = 17. Slope = y 2 y 2 y 8

5

(3, 1) (2,1)

x

(4, 0)

(1, 2)

5

(2, 1)

(2, 3)

(2, 3)

5

5

y 5

x

5 x

(1, 2)

8 x

23.

25.

y 8 (2, 5) 3

P  (1, 2)

27.

y 8 3

P  (2, 4) 4

1 5 x

29.

y 5

y 5 P  (0, 3)

P  (1, 3) 5 x

(6, 1) 7 x

5 x

1 37. x - 2y = 0 or y = x 2 1 5 39. x + y = 2 or y = - x + 2 41. 2x - y = 3 or y = 2x - 3 43. x + 2y = 5 or y = - x + 45. 3x - y = - 9 or y = 3x + 9 2 2 2 1 1 5 1 47. 2x + 3y = - 1 or y = - x 49. x - 2y = - 5 or y = x + 51. 3x + y = 3 or y = - 3x + 3 53. x - 2y = 2 or y = x - 1 3 3 2 2 2 55. x = 2; no slope–intercept form 57. y = 2 59. 2x - y = - 4 or y = 2x + 4 61. 2x - y = 0 or y = 2x 1 3 63. x = 4; no slope–intercept form 65. 2x + y = 0 or y = - 2x 67. x - 2y = - 3 or y = x + 69. y = 4 2 2 1 1 75. Slope = ; y@intercept = 2 77. Slope = - ; y@intercept = 2 71. Slope = 2; y@intercept = 3 73. Slope = 2; y@intercept = - 2 2 2 31. (2, 6); (3, 10); (4, 14)

33. (4, - 7); (6, - 10); (8, - 13)

y 5

y 8 (0, 3)

35. ( - 1, - 5); (0, - 7); (1, - 9)

(1, 5) (0, 2)

(1, 0) 5 x

y 5 (0, 2)

y 5 (0, 2)

(2, 3)

5 5 x

(4, 0)

x

5 x

79. Slope =

2 ; y@intercept = - 2 3

y 5

5 x

83. Slope undefined; no y-intercept y 5

y 2.5 (0, 1)

(3, 0) (0, 2)

81. Slope = - 1; y@intercept = 1

(1, 0) 2.5 x

85. Slope = 0; y@intercept = 5 y 8 (0, 5)

(4, 0) 5 x

5 x

AN3

ANSWERS  Section F.4 87. Slope = 1; y@intercept = 0

89. Slope =

y 5

3 ; y@intercept = 0 2

(0, 0)

(1, 1)

(0, 2)

(0, 0)

5 x

5 x

(10, 0)

5

8 x

x

(3, 0)

97. (a) x-intercept: 2; y-intercept: 3 (b) y

101. y = 0 103. Parallel 105. Neither 107. (b) 109. (d)

99. (a) x-intercept: 5; y-intercept: - 2 (b) y

5

12

10

(0, 8)

(2, 3)

95. (a) x-intercept: 3; 21 y-intercept: 2 (b) y 21 0, 2

y 5

(b)

y 5 (2, 2)

93. (a) x-intercept: - 10; y-intercept: 8 y (b)

91. (a) x-intercept: 3; y-intercept: 2

5

(0, 3)

(2, 0) 5 x

(3, 0) 5 x

(5, 0) 6 x (0, 2)

Cost (dollars)

3 2 111. P1 = ( - 2, 5), P2 = (1, 3), m1 = - ; P2 = (1, 3), P3 = ( - 1, 0), m2 = ; because m1m2 = - 1, the lines are perpendicular and the points 3 2 ( - 2, 5), (1, 3), and ( - 1, 0) are the vertices of a right triangle. 113. P1 = ( - 1, 0), P2 = (2, 3), m = 1; P3 = (1, - 2), P4 = (4, 1), m = 1; P1 = ( - 1, 0), P3 = (1, - 2), m = - 1; P2 = (2, 3), P4 = (4, 1), m = - 1; opposite sides are parallel, and adjacent sides are perpendicular; the points are the vertices of a rectangle. 115. C = 0.60x + 39; +105.00; +177.00 117. C = 0.45x + 6735 2 5 119. (a) C = 0.1062x + 13.04, 0 … x … 800 121. °C = (°F - 32); approximately 21.1°C 123. (a) y = - x + 30 (b) x-intercept: 375; The ramp 9 25 (b) C = 0.1062x + 13.04 meets the floor 375 in. (31.25 ft) from the base of the platform. (c) The ramp does not meet design 0 ≤ x ≤ 800 C (800, 98.00) requirements. It has a run of 31.25 ft. (d) The only slope possible for the ramp to comply with the 100 requirement is for it to drop 1 in. for every 12-in. run. 80 60 (0, 13.04) 1 40 125. (a) A = x + 20,000 (b) +80,000 (c) Each additional box sold requires an additional +0.20 in 5 20 y 2x  y  0 advertising. 127. All have the same slope, 2; the lines are parallel. 0 100 300 500 700 900 x 6 (0, 4) (0, 0)

kW-hr

(c) +34.28 (d) +66.14 (e) Each additional kilowatt-hour used adds +0.1062 to the bill. 129. (b), (c), (e), (g)

131. (c)

5 x (0, 2)

2 x  y  4

137. No; no

2x  y  2

3 139. They are the same line. 141. Yes, if the y-intercept is 0. 143. No, the slope should be negative, - . 2

F.4 Assess Your Understanding (page 70) 4. radius 5. 1x - h2 2 + 1y - k2 2 = r 2 6. unit 9 5 3 5 2 9. Center a , 2 b ; radius = ; ax - b + 1y - 22 2 = 2 2 2 4

3. False

11. x2 + y2 = 4; x2 + y 2 - 4 = 0

13. x2 + 1y - 22 2 = 4; x2 + y2 - 4y = 0

y 5

y 5

7. Center (2, 1); radius = 2; 1x - 22 2 + 1y - 12 2 = 4

1 2 1 17. 1x + 22 2 + 1y - 12 2 = 16; 19. ax - b + y2 = ; 2 4 x2 + y2 + 4x - 2y - 11 = 0 x2 + y 2 - x = 0

15. 1x - 42 2 + 1y + 32 2 = 25; x2 + y2 - 8x + 6y = 0 y 2

(0, 2)

y (2, 1) 6

(0, 0)

y 2.5

9 x

5 x

(4, 3)

5 x

4 x 1, 0 2

21. (a) 1h, k2 = 10, 02; r = 2 (b) y

23. (a) 1h, k2 = 13, 02; r = 2 (b) y 5

5

(c) 1 {2, 02; 10, {22 1 1 29. (a) 1h, k2 = a , - 1 b ; r = 2 2 (b) y

(2, 2) 6

6 x

5 x

5 x (1, 2)

4 x

(c) 1 1 { 25 , 0 2 ; 1 0, 2 { 2 22 2

(c) (1, 0); (5, 0) 31. (a) 1h, k2 = 13, - 22; r = 5 (b) y 3

2.5

33. (a) 1h, k2 = 1 - 2, 02; r = 2 (b) y (2, 0)

8

5

x 2.5 x

(c) 10, - 12 51. x2 + y2 + 2x + 4y - 4168.16 = 0

27. (a) 1h, k2 = 1 - 2, 22; r = 3 (b) y

5

(3, 0)

(0, 0)

1, 1 2

25. (a) 1h, k2 = 11, 22; r = 3 (b) y

(3, 2)

5 x

(c) 1 3 { 221 , 0 2 ; 10, - 62, 10, 22 53. 22x + 4y = 9 22

2.5 x

(c) 10, 02, 1 - 4, 02

55. (1, 0) 57. y = 2

59. (b), (c), (e), (g)

(c)

1 -2

{ 25 , 0 2 ; 1 0, 2 { 25 2

35. x2 + y2 = 13 37. 1x - 22 2 + 1y - 32 2 = 9 39. 1x + 12 2 + 1y - 32 2 = 5 41. 1x + 12 2 + 1y - 32 2 = 1 43. (c) 45. (b) 47. 18 units2 49. x2 + (y - 139)2 = 15,625

AN4

ANSWERS  Section 1.1

CHAPTER 1 Functions and Their Graphs 1.1 Assess Your Understanding (page 85) 5. independent; dependent 6. range 7. [0, 5] 8. ≠; ƒ; g 9. 1g - f 2 1x2 10. F 11. T 12. T 13. F 14. F 15. Function; Domain: 5Elvis, Colleen, Kaleigh, Marissa 6 ; Range: 5January 8, March 15, September 17 6 17. Not a function 19. Not a function 21. Function; Domain: 51, 2, 3, 4 6; Range: 53 6 23. Not a function 25. Function; Domain: 5 - 2, - 1, 0, 1 6; Range: 50, 1, 4 6 27. Function 29. Function 31. Not a function 33. Not a function 35. Function 37. Not a function 39. (a) - 4 (b) 1 (c) - 3 (d) 3x2 - 2x - 4 1 -x 1 (c) (d) 2 (e) - 3x2 - 2x + 4 (f) 3x2 + 8x + 1 (g) 12x2 + 4x - 4 (h) 3x2 + 6xh + 3h2 + 2x + 2h - 4 41. (a) 0 (b) 2 2 x + 1 -x x + 1 2x x + h (e) 2 (f) 2 (g) 2 (h) 2 43. (a) 4 (b) 5 (c) 5 (d)  x + 4 (e) -  x - 4 (f)  x + 1 + 4 x + 1 x + 2x + 2 4x + 1 x + 2xh + h2 + 1 3 1 2x - 1 - 2x - 1 2x + 3 4x + 1 2x + 2h + 1 1 (b) (c) (d) (e) (f) (g) (h) (g) 2 x + 4 (h)  x + h + 4 45. (a) 5 2 8 3x + 5 3x - 5 3x - 2 6x - 5 3x + 3h - 5 47. All real numbers 49. All real numbers 51. 5x x ≠ - 4, x ≠ 4 6 53. 5x x ≠ 0 6 55. 5x x Ú 4 6 57. 5x x 7 9 6 59. 5x x 7 1 6 61. 5t t Ú 4, t ≠ 7 6 63. (a) 1f + g2 1x2 = 5x + 1; All real numbers (b) 1f - g2 1x2 = x + 7; All real numbers (c) 1ƒ # g2 1x2 = 6x2 - x - 12; f 3x + 4 3 All real numbers (d) a b 1x2 = ; e x ` x ≠ f (e) 16 (f) 11 (g) 10 (h) - 7 65. (a) 1f + g2 1x2 = 2x2 + x - 1; All real numbers g 2x - 3 2 f x - 1 (b) 1f - g2 1x2 = - 2x2 + x - 1; All real numbers (c) 1ƒ # g2 1x2 = 2x3 - 2x2; All real numbers (d) a b 1x2 = ; 5x x ≠ 0 6 g 2x2 (e) 20 (f) - 29 (g) 8 (h) 0 67. (a) 1f + g2 1x2 = 1x + 3x - 5; 5x x Ú 0 6 (b) 1 f - g2 1x2 = 1x - 3x + 5; 5x x Ú 0 6 f 1x 5 1 ; e x ` x Ú 0, x ≠ f (e) 23 + 4 (f) - 5 (g) 22 (h) (c) 1ƒ # g2 1x2 = 3x 1x - 5 1x; 5x x Ú 0 6 (d) a b 1x2 = g 3x - 5 3 2 2 1 1 + 2 ; 5x x ≠ 0 6 69. (a) 1f + g2 1x2 = 1 + ; 5x x ≠ 0 6 (b) 1f - g2 1x2 = 1; 5x x ≠ 0 6 (c) 1ƒ # g2 1x2 = x x x f 5 3 6x + 3 2 - 2x + 3 2 (f) 1 (g) (h) 2 71. (a) 1f + g2 1x2 = ; e x ` x ≠ f (b) 1f - g2 1x2 = ; ex ` x ≠ f (d) a b 1x2 = x + 1; 5x x ≠ 0 6 (e) g 3 4 3x - 2 3 3x - 2 3 2 f 8x + 12x 2 7 2x + 3 2 1 7 5 (c) 1ƒ # g2 1x2 = ; e x ` x ≠ f (d) a b 1x2 = ; e x ` x ≠ 0, x ≠ f (e) 3 (f) (g) (h) 73. g 1x2 = 5 - x 3 g 4x 3 2 2 4 2 13x - 22 2

- 12x + h2 1 7 1 85. A = - 4 87. A = 8; undefined at x = 3 89. A1x2 = x2 81. 83. A = 2 2 x2 1x + h2 2 2x + h + 2x 91. G1x2 = 10x 93. (a) P is the dependent variable; a is the independent variable. (b) P1202 = 243.706 million; There were 243.706 million Americans 20 years of age or older in 2011. (c) P102 = 317.946 million; There were 317.946 million Americans in 2011. 95. (a) 15.1 m, 14.071 m, 12.944 m, L1x2 101. H1x2 = P1x2 # I 1x2 11.719 m (b) 1.01 sec, 1.43 sec, 1.75 sec (c) 2.02 sec 97. (a) $222 (b) $225 (c) $220 (d) $230 99. R1x2 = P1x2 3 2 103. (a) P1x2 = - 0.05x + 0.8x + 155x - 500 (b) P1152 = $1836.25 (c) When 15 hundred cellphones are sold, the profit is $1836.25. 75. 4

77. 2x + h - 1

79.

105. (a) D1v2 = 0.05v2 + 2.6v - 15

(b) 321 feet

(c) The car will need 321 feet to stop once the impediment is observed.

(x + 1)(x - 1) x2 - 1 107. No. g(x) = = = x - 1, x ≠ - 1. x + 1 x + 1

109. H1x2 =

3x - x3 your age

1.2 Assess Your Understanding (page 93) 3. vertical 4. 5; - 3 5. a = - 2 6. F 7. F 8. T 9. (a) f 102 = 3; f 1 - 62 = - 3 (b) f 162 = 0; f 1112 = 1 (c) Positive (d) Negative (e) - 3, 6, and 10 (f) - 3 6 x 6 6; 10 6 x … 11 (g) 5x - 6 … x … 11 6 (h) 5y - 3 … y … 4 6 (i) - 3, 6, 10 (j) 3 (k) Three times (l) Once (m) 0, 4 (n) - 5, 8 (o) - 3, 6, 10 11. Not a function 13. Function (a) Domain: 5x - p … x … p 6; Range: 5y - 1 … y … 1 6 p p (b) a - , 0 b , a , 0 b , 10, 12 (c) y-axis 15. Not a function 17. Function (a) Domain: 5x 0 6 x 6 3 6; Range: {y|y 6 2} (b) 11, 02 (c) None 2 2 19. Function (a) Domain: all real numbers; Range: 5y y … 2 6 (b) 1 - 3, 02, 13, 02, 10, 22 (c) y-axis 21. Function (a) Domain: all real 1 1 numbers; Range: 5y y Ú - 3 6 (b) 11, 02, 13, 02, 10, 92 (c) None 23. (a) Yes (b) f 1 - 22 = 9; 1 - 2, 92 (c) 0, ; 10, - 12, a , - 1 b 2 2 1 1 1 (d) All real numbers (e) - , 1 (f) - 1 (g) - , 1 25. (a) No (b) f 142 = - 3; 14, - 32 (c) 14; 114, 22 (d) 5x x ≠ 6 6 (e) - 2 (f) (g) - 2 2 2 3 8 8 27. (a) Yes (b) f 122 = ; a2, b (c) - 1, 1; 1 - 1, 12, 11, 12 (d) All real numbers (e) 0 (f) 0 (g) 0 17 17 29. (a) Approximately 10.4 ft high (b) h 1122 ≈ 9.9 represents the height of the ball, in feet, after it has traveled 12 feet in front of the foul line. (c) h (d) The ball will not go through (8, 10.4) (12, 9.9) 10

(15, 8.4)

5 (0, 6)

(20, 3.6)

the hoop; h 1152 ≈ 8.4 ft . If v = 30 ft/sec, h 1152 = 10 ft .

31. (a) About 81.07 ft (b) About 129.59 ft (c) h 15002 ≈ 26.63 represents the height of the golf ball, in feet, after it has traveled a horizontal distance of 500 feet. (d) About 528.13 ft (e) 150 (f) About 115.07 ft and 413.05 ft (g) 275 ft; maximum height shown in the table is 131.8 ft (h) 264 ft

(22.6, 0) 0

5 10 15 20 25

x

0 0

550

ANSWERS  Section 1.3 (b) 5x x 7 0 6

33. (a) $223; $220 (c) 500

1 3 37. (a) $80; it costs $80 if you use 0 minutes. (b) $80; it costs $80 if you use 1000 minutes. (c) $210; it costs $210 if you use 2000 minutes. (d) {m|0 … m … 14,400}. There are at most 14,400 anytime minutes in a month. 39. The x-intercepts can number anywhere from 0 to infinitely many. There is at most one y-intercept. 41. (a) III (b) IV (c) I (e) 600 mi/hr (d) V (e) II 45. (a) 2 hr elapsed during which Kevin was between 0 and 3 mi from home (b) 0.5 hr elapsed during which Kevin was 3 mi from home (c) 0.3 hr elapsed during which Kevin was between 0 and 3 mi from home (d) 0.2 hr elapsed during which Kevin was 0 mi from home (e) 0.9 hr elapsed during which Kevin was between 0 and 2.8 mi from home (f) 0.3 hr elapsed during which Kevin was 2.8 mi from home (g) 1.1 hr elapsed during which Kevin was between 0 and 2.8 mi from home (h) 3 mi (i) Twice 47. No points whose x-coordinate is 5 or whose y-coordinate is 0 can be on the graph.

1000 0 y Distance (blocks)

43.

(22, 5) 5 4 3 (5, 2) 2 (7, 0) 1 0

(6, 0)

35. (a) 3 (b) - 2

(d)

0

AN5

(29, 0)

x 10 20 Time (minutes)

(c) - 1

(d) 1

(e) 2

(f) -

1.3 Assess Your Understanding (page 106) 6. increasing 7. even; odd 8. T 9. T 10. F 11. Yes 13. No 15. 1 - 8, - 22; 10, 22; 15, q 2 17. Yes; 10 19. - 2, 2; 6, 10 21. (a) 1 - 2, 02, 10, 32, 12, 02 (b) Domain: 5x - 4 … x … 4 6 or [ - 4, 4]; Range: 5y 0 … y … 3 6 or [0, 3] (c) Increasing on 1 - 2, 02 and 12, 42 ; Decreasing on 1 - 4, - 22 and 10, 22 (d) Even 23. (a) 10, 12 (b) Domain: all real numbers; Range: 5y y 7 0 6 or 10, q 2 (c) Increasing on 1 - q , q 2 (d) Neither 25. (a) 1 - p, 02, 10, 02, 1p, 02 (b) Domain: 5x - p … x … p 6 or [ - p, p]; Range: 5y - 1 … y … 1 6 or [ - 1, 1] p p 1 1 5 p p (c) Increasing on a - , b ; Decreasing on a - p, - b and a , p b (d) Odd 27. (a) a0, b , a , 0 b , a , 0 b (b) Domain: 5x - 3 … x … 3 6 or 2 2 2 2 2 3 2 [ - 3, 3]; Range: 5y - 1 … y … 2 6 or [ - 1, 2] (c) Increasing on 12, 32 ; Decreasing on 1 - 1, 12; Constant on 1 - 3, - 12 and 11, 22 (d) Neither p p 29. (a) 0; 3 (b) - 2, 2; 0, 0 31. (a) ; 1 (b) - ; - 1 33. Odd 35. Even 37. Odd 39. Neither 41. Even 43. Odd 45. Absolute maximum: 2 2 f 112 = 4; absolute minimum: f 152 = 1; local maximum: f 132 = 3; local minimum: f 122 = 2 47. Absolute maximum: f 132 = 4; absolute minimum: f 112 = 1; local maximum: f 132 = 4; local minimum: f 112 = 1 49. Absolute maximum: none; absolute minimum: f 102 = 0; local maximum: f 122 = 3; local minima: f 102 = 0 and f 132 = 2 51. Absolute maximum: none; absolute minimum: none; local maximum: none; local minimum: none 55. 0 59. 53. 0.5 4 57. 8 6 2

2

4

2

3

2 0.5

0

Increasing: 1 - 2, - 0.772, 10.77, 22 Decreasing: 1 - 0.77, 0.772 Local maximum: 1 - 0.77, 0.192 Local minimum: 10.77, - 0.192

Increasing: 1 - 2, - 12, 11, 22 Decreasing: 1 - 1, 12 Local maximum: 1 - 1, 42 Local minimum: 11, 02

2

20

0

Increasing: 1 - 3.77, 1.772 Decreasing: 1 - 6, - 3.772, 11.77, 42 Local maximum: 11.77, - 1.912 Local minimum: 1 - 3.77, - 18.892

Increasing: 1 - 1.87, 02, 10.97, 22 Decreasing: 1 - 3, - 1.872, 10, 0.972 Local maximum: 10, 32 Local minima: 1 - 1.87, 0.952, 10.97, 2.652

75. (a)

77. (a), (b)

2500

0

40 0

(b) 10 riding lawn mowers/hr (c) $239/mower

Debt (trillions of dollars)

61. (a) - 4 (b) - 8 (c) - 10 63. (a) 17 (b) - 1 (c) 11 65. (a) 5 (b) y = 5x - 2 67. (a) - 1 (b) y = - x 69. (a) 4 (b) y = 4x - 8 71. (a) Odd (b) Local maximum value: 54 at x = - 3 73. (a) Even (b) Local maximum value: 24 at x = - 2 79. (a) 1 (b) 0.5 (c) 0.1 (d) 0.01 (e) 0.001 (f) y  x2

D 14 12 10 8 6 4 2000

2

(c) 47.4 sq. units

yx y  0.5x y  0.1x

2005 2010 Year

y

The slope represents the average rate of change of the debt from 2001 to 2006. (c) $575.5 billion (d) $759 billion (e) $1252 billion (f) The average rate of change is increasing over time.

3

3

y  0.01x y  0.001x

2

(g) They are getting closer to the tangent line at 10, 02 . (h) They are getting closer to 0.

AN6

ANSWERS  Section 1.3

81. (a) 2 (b) 2; 2; 2; 2 (c) y = 2x + 5 (d)

83. (a) 2x + h + 2 (b) 4.5; 4.1; 4.01; 4 (c) y = 4.01x - 1.01 (d) 10

10

10

1 1x + h2x 2 10 100 (b) - ; - ; ; -1 3 11 101 100 201 (c) y = x + 101 101 (d) 2

87. (a) -

85. (a) 4x + 2h - 3 (b) 2; 1.2; 1.02; 1 (c) y = 1.02x - 1.02 (d) 10

10 2 5

4

3

10

2

2

4

5 2

91. At most one

93. Yes; the function f 1x2 = 0 is both even and odd. 95. Not necessarily. It just means f 152 7 f 122.

1.4 Assess Your Understanding (page 117) 4. 1 - q , 02 y 17.

5. piecewise-defined

6. T 7. F 19.

8. F y 10

10 (4, 4)

(0, 0)

11. E

13. B

15. F 21.

5 x

(1, 1)

(2, 8)

25. (a) 4 (b) 2 (c) 5 27. (a) - 4 29. (a) All real numbers (b) 10, 12 (c) y 2.5 (0, 1)

(b) - 2

(c) 0

(1, 2)

(d) 5y y ≠ 0 6; 1 - q , 02 h 10, q 2 (e) Discontinuous at x = 0

35. (a) All real numbers (b) 1 - 1, 02, 10, 02 (c) y 2.5 (0, 1) (1, 0)

23. 1 2, 2

5

(1, 1) (0, 0) 5 x (1, 1)

33. (a) 5x x Ú - 2 6; 3 - 2, q 2 (b) 10, 32, 12, 02 (c) y 5 (0, 3) (2, 1)

(2, 4)

(0, 3)

(1, 1) 5 x

(d) 5y y Ú 1 6; 3 1, q 2 (e) Continuous

37. (a) 5x x Ú - 2 6; 3 - 2, q 2 (b) (0, 0) (c) y

2.5 x

(ⴚ2, 2)

(1, 5) (1, 4) (1, 1) (2, 0) 5 x

(d) 5y y 6 4 or y = 5 6; 1 - q , 42 h 55 6 (e) Discontinuous at x = 1

39. (a) All real numbers (b) 1x, 02 for 0 … x 6 1 (c) y 5

2.5 (1, 1)

y 5

5 x

(d) 25 31. (a) All real numbers (b) 10, 32 (c) y

2.5 x

(1, 2)

y 5 (1, 1)

(2, 8)

(0, 0)

10 x

(4, 4)

9. C

(1, 1) 5 x

2.5 x

(0, 0)

(d) All real numbers (e) Discontinuous at x = 0

(d) 5y y Ú 0 6; 3 0, q 2 (e) Continuous

(d) Set of even integers (e) Discontinuous at 5x x is an integer 6

0.10x 892.50 + 0.15 1x - 89252 4991.25 + 0.25 1x - 36,2502 51. f 1x2 = g 17,891.25 + 0.28 1x - 87,8502 44,603.25 + 0.33 1x - 183,2502 115,586.25 + 0.35 1x - 398,3502 116,163.75 + 0.396 1x - 400,002

if 0 … x … 8925 if 8925 6 x … 36,250 if 36,250 6 x … 87,850 if 87,850 6 x … 183,250 if 183,250 6 x … 398,350 if 398,350 6 x … 400,000 if x 7 400,000

Charge (dollars)

- x if - 1 … x … 0 -x 41. f 1x2 = e 1 (Other answers are possible.) 43. f 1x2 = e if 0 6 x … 2 -x + 2 2x 45. (a) 2 (b) 3 (c) - 4 47. (a) $39.99 (b) $46.74 (c) $40.44 0.65183x + 22.25 if 0 … x … 50 49. (a) $54.84 (b) $284.46 (c) C1x2 = e (d) 0.51026x + 29.3285 if x 7 50

if x … 0 (Other answers are possible.) if 0 6 x … 2

C 100

50

(100, 80.35) (50, 54.84) (0, 22.25)

0

50 100 x Usage (therms)

ANSWERS  Section 1.5 y 275 (800, 270)

9000 if 7500 if 5250 if 55. (a) C1s2 = g 3000 if 1500 if 750 if

(960, 270)

200 (400, 170) 100 (100, 50)

(0, 0)

500 1000 x Distance (miles)

(b) C1x2 = 10 + 0.4x

s … 660 680 700 720 s Ú

659 … s … s … s … s 740

… … … …

679 699 719 739

(b) $1500

(c) $7500

(c) C1x2 = 70 + 0.25x

57. (a) 10 °C (b) 4 °C (c) - 3°C (d) - 4°C (e) The wind chill is equal to the air temperature. (f) At wind speed greater than 20 m/s, the wind chill factor depends only on the air temperature.

61. Each graph is that of y = x2, but shifted horizontally. If y = 1x - k2 2, k 7 0, the shift is right k units; if y = 1x + k2 2, k 7 0, the shift is left k units. 59. C1x2 = 63. The graph of y = - f 1x2 is the reflection about the x-axis of the graph of y = f 1x2. 3 3 65. Yes. The graph of y = 1x - 12 + 2 is the graph of y = x shifted right 1 unit and up 2 units. 67. They all have the same general shape. All three go through the points 1 - 1, - 12, 10, 02 , and 11, 12 . As the exponent increases, the steepness of the curve increases (except near x = 0).

i

0.92 1.12 1.32 1.52 1.72 1.92 2.12 2.32 2.52 2.72 2.92 3.12 3.32

0 6 x … 1 1 6 x … 2 2 6 x … 3 3 6 x … 4 4 6 x … 5 5 6 x … 6 6 6 x … 7 7 6 x … 8 8 6 x … 9 9 6 x … 10 10 6 x … 11 11 6 x … 12 12 6 x … 13

Postage (dollars)

Cost (dollars)

53. (a)

AN7

C 3.32 2.92 2.52 2.12 1.72 1.32 0.92 0

2 4 6 8 10 12 x Weight (ounces)

1.5 Assess Your Understanding (page 129) 1. horizontal; right 2. y 3. vertical; up 4. T 5. F 6. T 7. B 9. H 11. I 13. L 15. F 17. G 19. y = 1x - 42 3 21. y = x3 + 4 23. y = - x3 25. y = 4x3 27. y = - 1 2 - x + 22 29. y = - 2x + 3 + 2 31. (c) 33. (c) 35. (a) - 7 and 1 (b) - 3 and 5 (c) - 5 and 3 (d) - 3 and 5 37. (a) 1 - 3, 32 (b) 14, 102 (c) Decreasing on 1 - 1, 52 (d) Decreasing on 1 - 5, 12 y y 39. 45. 47. y 43. 41. y y 5

2.5 (1, 0) 2.5 x

(1, 0)

(21, 1)

(1, 2)

(0, 1)

(0, 1)

23 (22, 0)

5 x

(1, 0)

(0, 1)

5

(2, 2) 3

x

9

(2, 3) (1, 2) 5 x

(4, 8) (1, 4) (0, 0) 8 x

Domain: 1 - q, q2; Range: [ - 1, q2

Domain: 1 - q, q2; Range: 1 - q, q2

Domain: [ - 2, q2; Range: [0, q2

Domain: 1 - q, q2; Range: 1 - q, q2

Domain: [0, q2; Range: [0, q2

49.

51.

53.

55.

57.

y 2.5

1, 1 2 1,

1, 

1 2

y (1, 1) 10 (8, 2) 10

1  , 1 2

y 5 (1, 3) (0, 2)

10

x (8, 2) (1, 1)

2.5 x

1 2

y (1, 1) 10 (8, 2) x (8, 2) (1, 1)

(1, 1) 5 x

Domain: 1 - q, q2; Domain: 1 - q, 02 h 10, q2 Range: 1 - q, q2 Range: 1 - q, 02 h 10, q2

Domain: 1 - q, q2; Range: 1 - q, q2

Domain: 1 - q, q2; Range: 1 - q, q2

59.

63.

65.

61.

y 8 (6, 5)

y 5 (4, 0)

(3, 3) (2, 1) 8 x

y 5

(1, 1)

5 x

(1, 5)

5 x (0, 1)

(2, 1) (1, 3)

Domain: 1 - q, q2; Range: [ - 3, q2

67.

y 5 (0, 2)

(2, 0) 5 x (0, 2)

(1, 1)

y 6

(3, 5)

y 5

(2, 2) (1, 0) 5 x

5 x

(0, 2)

Domain: [2, q2; Range: [1, q2

Domain: 1 - q, 0]; Range: [ - 2, q2

Domain: 1 - q, q2; Range: 1 - q, q2

Domain: 1 - q, q2; Range: [0, q2

Domain: 1 - q, q2; Range: 5 y y is an even integer 6

69. (a) F1x2 = f 1x2 + 3

(b) G1x2 = f 1x + 22

(c) P1x2 = - f 1x2

(d) H1x2 = f 1x + 12 - 2

(e) Q 1x2 =

y 7 (0, 5) (4, 1)

(2, 5)

y 5 (2, 2)

(4, 3) 5 x

(6, 2)

y 5 (0, 2) (2, 0) 3 x

(4, 2) (0, 2)

(4, 0) 5 x (2, 2)

y 3 (1, 0) (1, 0) 4 x (3, 2) (5, 4)

1 f 1x2 2

y 5 (0, 1) (4, 1)

(2, 1) (4, 0) 5 x

AN8

ANSWERS  Section 1.5

(f) g 1x2 = f 1 - x2

(g) h 1x2 = f 12x2

y 5 (2, 2) (4, 0)

y 5 (0, 2)

(0, 2)

5 x (4, 2)

y 2.5

2 x

(4, 15)

c  2

y 9

1 2 3 b + 2 4

3 7 , 2 4 

c0

y (2, 3) 5 1 7 , 2 4

1 3 , 2 4

F 320 256 192 128 64

(100, 212)

2

2

x

83. (a)

y 5 (6, 0) (0, 0)

7

x

(3, −3)

(2, 5) (3, 8)

(b) 9 square units

(c) The time at which the temperature adjusts between the daytime and overnight settings is moved to 1 hr sooner. It begins warming up at 5:00 am instead of 6:00 am, and it begins cooling down at 8:00 pm instead of 9:00 pm. T

76 72 68 64 60 5 10 15 20 25 t Time (hours after midnight)

76 72 68 64 60 0

45,000

Y2

5 10 15 20 25 t Time (hours after midnight)

Y1

(373, 212)

(b)

y 2.5 (1, 1) (1, 1)

(0, 0) 5 x

(1, 1)

5 x

0

K

15,000

93. (a)

(2, 0)

y 1

91. (a)

(273, 32) 0 280 310 350

(0, 32) 0 20 40 60 80 100 C

y 5

, 0 2

Temperature (F)

Temperature (F)

F 320 256 192 128 64

73. f 1x2 = 1x + 12 2 - 1

(1, 8)

0

89.

 , 1 2

81. f 1x2 = - 3 1x + 22 2 - 5

(4, 3)

T 5 x

 x

, 1 4

  , 1 4

87. (a) 72°F; 65°F (b) The temperature decreases by 2° to 70°F during the day and 63°F overnight.

c3



 x

(3, 1) 5 x

(0, 1) 2.5 x

y 2.5

 x

79. f 1x2 = 2 1x - 32 2 + 1

y 4

7.5 

y   , 1 2.5 2

(  2, 0) 2 x

(  2, 0)

  ,0 2

 , 1 2

77. f 1x2 = ax +

(8, 1) 9 x

  2, 1 2

(g) h 1x2 = f 12x2

y 2.5



 1  , 2 2

y (0, 1)

  ,1 2

, 1 2 2  x

   1, 3 2

75. f 1x2 = 1x - 42 2 - 15

(f) g 1x2 = f 1 - x2

1 f 1x2 2

(, 0)

(  1, 2)

(c) P1x2 = - f 1x2

   2, 1 2

  1, 1 2

(  1, 2)

(, 3)  x

(, 3)

(2, 2)

(1, 2)

y 2.5 (2, 0)

(1, 2) (2, 0) 5 x

(e) Q 1x2 =

y 1

(b) G1x2 = f 1x + 22

, 4 2

  ,2 y 2 5

(d) H1x2 = f 1x + 12 - 2

85.

71. (a) F1x2 = f 1x2 + 3

y 2.5 (1, 1) (1, 1) 2

2

x

2000

(b) 10% tax (c) Y1 is the graph of p(x) shifted down vertically 10,000 units. Y2 is the graph of p(x) vertically compressed by a factor of 0.9. (d) 10% tax

95. (a) 1 - 4, 22 (b) 11, - 122 (c) 1 - 4, 52 97. The graph of y = f 1x2 - 2 is the graph of y = f 1x2 shifted down 2 units. The graph of y = f 1x - 22 is the graph of y = f 1x2 shifted right 2 units. 99. The range of f 1x2 = x2 is [0, q 2. The graph of g 1x2 = f 1x2 + k is the graph of f shifted up k units if k 7 0 and down 0 k 0 units if k 6 0, so the range of g is [k, q 2 .

ANSWERS  Review Exercises

AN9

1.6 Assess Your Understanding (page 135) 3. (a) d 1x2 = 2x2 - x + 1

1. (a) d 1x2 = 2x4 - 15x2 + 64 (b) d 102 = 8 (c) d 112 = 250 ≈ 7.07 (d)

(b)

1 4 x 2 7. (a) A1x2 = x 116 - x2 2 (b) Domain: 5x 0 6 x 6 4 6 (c) The area is largest when x ≈ 2.31.

5. A1x2 =

2

40

30

0

2 0

10

10

(c) d is smallest when x = 0.5.

5

0

4 0

(e) d is smallest when x ≈ - 2.74 or x ≈ 2.74

25 - 20x + 4x2 p (b) Domain: 5x 0 6 x 6 2.5 6 (c) A is smallest when x ≈ 1.40 m.

(b) p 1x2 = 4x + 4 24 - x2 (d) p is largest when x ≈ 1.41.

9. (a) A1x2 = 4x 24 - x2 (c) A is largest when x ≈ 1.41.

11. (a) A1x2 = x2 +

12

10

8

0

0

2

2 0

0

0

2.5 0

13. (a) C1x2 = x

(b) A1x2 =

x2 4p

15. (a) A1r2 = 2r 2

19. (a) d 1t2 = 22500t 2 - 360t + 13 (b) d is smallest when t ≈ 0.07 hr.

(b) p 1r2 = 6r

21. V1r2 =

17. A1x2 = a

pH1R - r2r 2

p 23 2 bx 3 4

25. (a) V1x2 = x 124 - 2x2 2 (b) 972 in .3 (c) 160 in .3 (d) V is largest when x = 4.

R

12 - x 2x2 + 4 + 5 3 (b) 5x 0 … x … 12 6 (c) 3.09 hr (d) 3.55 hr

23. (a) T1x2 =

5

0

1100

0.2 0 0

12 0

1.7 Assess Your Understanding (page 141) 1 250 1 9d 2 8a3 4p 3 1 x 5. A = px2 7. F = 2 9. z = 1x2 + y2 2 11. M = 13. T 2 = 2 15. V = r 17. A = bh 5 5 3 2 d d 2 2x mM 429 (b) 143 bags 19. F = 6.67 * 10 - 11 a 2 b 21. p = 0.00649B; $941.05 23. 144 ft; 2 s 25. 2.25 27. R = 3.95g; $41.48 29. (a) D = p d 3 3 2 31. 450 cm 33. 124.76 lb 35. V = pr h 37. 54.86 lb 39. 26 ≈ 1.82 in . 41. 2812.5 joules 43. 384 psi 1. y = kx

2. F 3. y =

Review Exercises (page 145) 1. Function; domain 5 - 2, 1, 3 6 , range 52, 5 6 (f) (c)

9x 9x2 - 1 7 - x2 x

2

4. (a) 0 (d)

x2 - 7 x

2

10. 5x 0 x Ú 0, x ≠ 1 6

(c) 2x2 + 3x

(b) 3 12 (e)

- x2 + 2x + 6 x - 2x + 1 2

11. {x 0 x 7 - 8}

(f)

(b) - 9>8

2. Not a function

3. (a) 9>8

(d) - 2x2 - 3x

(e) 2x2 - 5x + 4

7 - 9x2 9x2

6. 5x 0 x ≠ 2, x ≠ - 2 6

12. 1f + g2 1x2 = 2x + 3; Domain: all real numbers 1f - g2 1x2 = - 4x + 1; Domain: all real numbers 1ƒ # g2 1x2 = - 3x2 + 5x + 2; Domain: all real numbers f 2 - x 1 a b 1x2 = ; Domain: e x ` x ≠ - f g 3x + 1 3

(c) -

3x x2 - 1

(f) {3 2x2 - x

3x 3 (e) x + 1 x2 - 1 2 2 5. (a) (b) 9 9

(d) -

7. 5x 0 x … 2 6 8. 5x 0 x ≠ 0 6

9. 5x 0 x ≠ - 3, x ≠ 1 6

13. 1f + g2 1x2 = 3x2 + 4x + 1; Domain: all real numbers 1f - g2 1x2 = 3x2 - 2x + 1; Domain: all real numbers 1ƒ # g2 1x2 = 9x3 + 3x2 + 3x; Domain: all real numbers f 3x2 + x + 1 a b 1x2 = ; Domain: 5x x ≠ 0 6 g 3x

AN10

ANSWERS  Review Exercises

x2 + 1 ; Domain: 5x x ≠ 0, x ≠ 1 6 x 1x - 12

1f - g2 1x2 = 1ƒ # g2 1x2 =

15. - 4x + 1 - 2h 16. (a) Domain: 5x - 4 … x … 3 6; Range: 5y - 3 … y … 3 6 (b) 10, 02 (c) - 1 (d) - 4 (e) 5x 0 6 x … 3 6 (g) (h) (f) y y

x2 + 2x - 1 ; Domain: 5x x ≠ 0, x ≠ 1 6 x 1x - 12

14. 1f + g2 1x2 =

5

x + 1 ; Domain: 5x x ≠ 0, x ≠ 1 6 x 1x - 12

(1, 3)

x 1x + 12 f a b 1x2 = ; Domain: 5x x ≠ 0, x ≠ 1 6 g x - 1

23.

20

3

y 2

33. (0, 0) 5 x

(4, 0)

(2, 4)

5 x (2, 2)

(2, 4)

Intercepts: 1 - 4, 02, 14, 02, 10, - 42 Domain: all real numbers Range: 5y y Ú - 4 6 or 3 - 4, q 2

Intercept: 10, 02 Domain: all real numbers Range: 5y y … 0 6 or 1 - q , 0 4

35.

36.

y (0, 3)

5

(2, 3)

y 30 (3.6, 0)

(1, 2) 5 x

3 (0, 24)

(3, 6) (2, 8)

Intercepts: 10, - 242 , 3 1 - 2- 2 4 , 02 or about 1 - 3 .6 , 02 Domain: all real numbers Range: all real numbers

Intercept: 10, 32 Domain: all real numbers Range: 5y y Ú 2 6 or 32, q 2

39. A = 11

40 40. (a) A1x2 = 2x + x (d) 100

0

(b) 42 ft

7 0

(1, 1)

5 x

(4, 2) (0, 0) 5 x

34.

y 5 (2, 1)

A is smallest when x ≈ 2 .15 ft .

2

(c) 28 ft

2

y 5

(5, 2) (3, 2)

(1, 0)

7 x

(0, 1) (1, 0) 5 x

Intercept: 11, 02 Domain: 5x x Ú 1 6 or 3 1, q 2 Range: 5y y Ú 0 6 or 3 0, q 2

Intercepts: 10, 12, 11, 02 Domain: 5x x … 1 6 or 1 - q , 1 4 Range: 5y y Ú 0 6 or 30, q 2

37. (a) 5x x 7 - 2 6 or 1 - 2, q 2 (b) 10, 02 (c) y (1, 3)

38. (a) 5x x Ú - 4 6 or [ - 4, q 2 (b) 10, 12 (c) y

5

(1, 10)

2

5 (4, 4)

Local maximum value: 1.53 at x = 0.41 Local minima values: 0.54 at x = - 0.34 and - 3.56 at x = 1.80 Increasing: 1 - 0.34, 0.412; 11.80, 32 Decreasing: 1 - 2, - 0.342; 10.41, 1.802

32.

y 5

(2, 2) (0, 4)

26. 44 27. No 30. y

4

Local maximum value: 4.04 at x = - 0.91 Local minimum value: - 2.04 at x = 0.91 Increasing: 1 - 3, - 0.912; 10.91, 32 Decreasing: 1 - 0.91, 0.912

(4, 0)

y 5

(0, 0)

3

20

5 x (3, 3)

(c) 47 25. 2

(4, 4) 2

(2, 1) (0, 0)

10 x

(8, 3)

24. (a) 23 (b) 7 28. Yes 29.

40

3

31.

(0, 0) (4, 1)

5 x (1, 1)

y (4, 3) 5

(6, 3)

(d) Absolute maximum: f 142 = 3 Absolute minimum: none (e) No symmetry (f) Neither (g) x-intercepts: - 3, 0, 3; y-intercept: 0 18. Odd 19. Even 20. Neither 21. Odd

17. (a) Domain: 5x x … 4 6 or 1 - q , 4]; Range: 5y y … 3 6 or 1 - q , 3] (b) Increasing on 1 - q , - 22 and 12, 42 ; Decreasing on 1 - 2, 22 (c) Local maximum value is 1 and occurs at x = - 2. Local minimum value is - 1 and occurs at x = 2. 22.

5

(6, 3) (3, 0)

5

(2, 3) (1, 2) 5 x

(2, 6)

(d) 5y y 7 - 6 6 or 1 - 6, q 2 (e) Discontinuous at x = 1

(0, 1) (4, 4)

(1, 3) 5 x

(d) 5y - 4 … y 6 0 or y 7 0 6 or 3 - 4, 02 h 10, q 2 (e) Discontinuous at x = 0

41. (a) A1x2 = 10x - x3 (b) The largest area that can be enclosed by the rectangle is approximately 12.17 square units. 427 42. P = B; $1083.92 65,000 43. 199.9 lb 44. 189 BTU

ANSWERS  Section 2.1

AN11

Chapter Test (page 147) 1. (a) Function; Domain: 52, 4, 6, 8 6; Range: 55, 6, 7, 8 6 (b) Not a function (c) Not a function (d) Function; Domain: all real numbers; Range: 4 1 5y y Ú 2 6 2. Domain: e x ` x … f ; f 1 - 12 = 3 3. Domain: 5x x ≠ - 2 6; g 1 - 12 = 1 4. Domain: 5x x ≠ - 9, x ≠ 4 6; h 1 - 12 = 5 8 5. (a) Domain: 5x - 5 … x … 5 6; Range: 5y - 3 … y … 3 6 (b) 10, 22 , 1 - 2, 02, and 12, 02 (c) f 112 = 3 (d) x = - 5 and x = 3 (e) 5x - 5 … x 6 - 2 or 2 6 x … 5 6 or [ - 5, - 22 h 12, 5] 6. Local maximum values: f 1 - 0.852 ≈ - 0.86; f 12.352 ≈ 15.55; local minimum value: f 102 = - 2; the function is increasing on the intervals 1 - 5, - 0.852 and 10, 2.352 and decreasing on the intervals 1 - 0.85, 02 and 12.35, 52 . y 10. (a) (b) 8. 19 y 7. (a) (b) 10, - 42, 14, 02 y 3 10 10 9. (a) 1 f - g2 1x2 = 2x2 - 3x + 3 (c) g 1 - 52 = - 9 (2, 5) (0, 1) (b) 1 ƒ # g2 1x2 = 6x3 - 4x2 + 3x - 2 (6, 4) 5 x (1, 3) (2, 4) (d) g 122 = - 2 2.5 x 10 x (c) f 1x + h2 - f 1x2 = 4xh + 2h2 (4, 2) 11. (a) 8.67% occurring in 1997 1x ≈ 52 (b) The model predicts that the interest rate will be - 10.343,. This is not reasonable. 5x px2 x2 + (b) 1297.61 ft 3 13. 14.69 ohms 12. (a) V1x2 = 8 4 64

CHAPTER 2 Linear and Quadratic Functions 2.1 Assess Your Understanding (page 158) 7. slope; y-intercept 8. - 4; 3 13. (a) m = 2; b = 3 (b) y

9. positive

12. False 1 17. (a) m = ; b = - 3 4 (b) y

8

(0, 3) 1

(1, 5) 2

(0, 4) 1

(d) Increasing

(c) - 3 25. (a) 8 (b)

y

3 (1, 1)

(0, 3)

(d) Decreasing

5 (0, 4) 9 (2, 0) 5 x

(8, 0)

(c)

1 4

21. (a) 4 (b) y

5 (0, 4) (4, 2) 1 4

(4, 0)

1

9 x

8

5 x

x

(0, 8)

(c) 0 (d) Constant

(d) Increasing

27. Linear; f 1x2 = - 3x - 2 1 1 1 35. (a) (b) e x ` x 7 f or a , q b 29. Nonlinear 4 4 4 (c) 1 (d) 5x x … 1 6 or 1 - q , 1 4 31. Nonlinear y (e) 33. Linear; f 1x2 = 8

y

16 (0, 10)

19. (a) m = 0; b = 4 (b) y

5

8 x

8 x

23. (a) 2 (b)

11. False

15. (a) m = - 3; b = 4 (b) y

8

(c) 2

10. True

x

4

(1, 3)

y  f (x)

5 x y  g (x)

37. (a) 40 (b) 88 (c) - 40 (d) 5x x 7 40 6 or 140, q 2 (e) 5x x … 88 6 or ( - q , 88] (f) 5x - 40 6 x 6 88 6 or 1 - 40, 882 39. (a) - 4 (b) 5x x 6 - 4 6 or 1 - q , - 42 41. (a) - 6 (b) 5x - 6 … x 6 5 6 or 3 - 6, 52 43. (a) $59 (b) 180 mi (c) 300 mi (d) 5x x Ú 0 6 or 30, q ) (e) The cost of renting the car for a day increases $0.35 for each mile driven, or there is a charge of $0.35 per mile to rent the car in addition to a fixed charge of $45. (f) It costs $45 to rent the car if 0 miles are driven, or there is a fixed charge of $45 to rent the car, in addition to a charge that depends on mileage. 45. (a) $24; 600 T-shirts (b) $0 … p 6 +24 (c) The price will increase.

5,000 4,000 3,000 2,000 1,000

51. (a) V(x) = - 1000x + 3000 (b) 5x 0 … x … 3 6 or 3 0, 3 4 (c) V(x) 3000

(36,250, 4991.25)

2000 1000 0

(8925, 892.50)

0 12,000 30,000 x Adjusted Gross Income ($)

(e) $27,600 (f) For each additional dollar of taxable income between $8925 and $36,250, the tax bill of a single person in 2013 increased by $0.15.

3000 2000 1000 x 0 4 8 12 Number of Bicycles

1 2 3 x Age (years)

(d) $1000 (e) After 1 year 55. (a) C (x) = 0.89x + 31.95 (b) $129.85; $236.65

53. (a) C(x) = 90x + 1800 (b) C(x) Cost (dollars)

49. (a) x = 5000 (b) x 7 5000

Book Value (dollars)

Tax Bill ($)

47. (a) 5x 8925 … x … 36,250 6 or 3 8925, 36,250 4 (b) $2553.75 (c) The independent variable is adjusted gross income, x. The dependent variable is the tax bill, T. (d) T (20,000, 2553.75)

(c) $3060 (d) 22 bicycles

ANSWERS  Section 2.1 n

(b) Since each input (memory) corresponds to a single output (number of songs), we know that number of songs is a function of memory. Also, because the average rate of change is a constant 218.75 songs per gigabyte, the function is linear. (c) n(m) = 218.75 m (d) {m 0 m Ú 0} or [0, q )

16,000 12,000 8,000 4,000 0 16 32 48 64 m Memory (gigabytes)

61. b = 0; yes, f 1x2 = b

59. (d), (e)

63.

y

64. 6

(e)

Number of Songs

57. (a)

Number of Songs

AN12

n 16,000 12,000 8,000 4,000 0 16 32 48 64 m Memory (gigabytes)

(f) If memory increases by 1 GB, then the number of songs increases by 218.75.

65. 7

y

66.

2 10 x

(−2, 4)

5 (4, 3)

(−1, 1) (2, −5)

(1, 2) 5

x

2.2 Assess Your Understanding (page 165) 3. scatter diagram 4. decrease; 0.008 11. (a) 20

0

5. Linear relation, m 7 0

7. Linear relation, m 6 0 (c) 20

10

0

0

10

13. (a)

3

6

3

3

25

25

0

0 90

(b) Answers will vary. Using ( - 20, 100) and ( - 10, 140), y = 4x + 180.

150

25

0 90

38 46 54 62 70 x Weight (grams)

(d)

Number of Calories

(d) y = 3.8613x + 180.2920

(c) Answers will vary. Using the points (39.52, 210) and (66.45, 280), y = 2.599x + 107.288.

y 280 260 240 220 200

(e)

150

90

3

6

(d) y = 2.2x + 1.2 (c)

150

Number of Calories

6

6

9 1 (b) Answers will vary. Using ( - 2, - 4) and (2, 5), y = x + . 4 2 15. (a)

(e)

6

3

3

10 0

(d) y = 2.0357x - 2.3571 (c)

6

20

0

0

(b) Answers will vary. Using (4, 6) and (8, 14), y = 2x - 2.

17. (a)

9. Nonlinear relation (e)

(e) 269 calories (f) If the weight of a candy bar is increased by 1 gram, the number of calories will increase by 2.599, on average.

y 280 260 240 220 200 38 46 54 62 70 x Weight (grams)

(b) Linear with positive slope 19. (a) The independent variable is the number of hours spent playing video games, and the dependent variable is cumulative grade-point average, because we are using number of hours playing video games to predict (or explain) cumulative grade-point average.

(b)

4

1

13 0

(c) G(h) = - 0.0942h + 3.2763 (d) If the number of hours playing video games in a week increases by 1 hour, the cumulative grade-point average decreases 0.09, on average. (e) 2.52 (f) Approximately 9.3 hours

ANSWERS  Section 2.3 (c) D = - 1.3355p + 86.1974 (d) If the price increases $1, the quantity sold per day decreases by about 1.34 pairs of jeans, on average. (e) D1p2 = - 1.3355p + 86.1974 (f) 5p 0 6 p … 64 6 (g) About 49 pairs

D 60 55 50 45 40 18 22 26 30 34 p Price (dollars/pair)

23.

y Incidence Rate (per 1000)

Demand (pairs of jeans sold per day)

21. (a) No (b)

AN13

50 40 30 20 10 x 40 35 Age of Mother

No, the data do not follow a linear pattern. 25. No linear relation 27. 34.8 hours; A student whose GPA is 0 spends 34.8 hours each week playing video games.; G(0) = 3.28; The average GPA of a student who does not play video games is 3.28. 28. 2x + y = 3 or y = - 2x + 3 29. {x x ≠ - 5, x ≠ 5} 30. (g - f )(x) = x2 - 8x + 12 31. y = (x + 3)2 - 4

2.3 Assess Your Understanding (page 177) - b { 2b2 - 4ac 11. Zeros: 0, 9; x-intercepts: 0, 9 2a 1 1 13. Zeros: - 5, 5; x-intercepts: - 5, 5 15. Zeros: - 3, 2; x-intercepts: - 3, 2 17. Zeros: - , 3; x-intercepts: - , 3 19. Zeros: - 4, 4; x-intercepts: - 4, 4 2 2 3 3 21. Zeros: - 6, - 2; x-intercepts: - 6, - 2 23. Zero: ; x-intercept: 25. Zeros: - 2 22, 2 22; x-intercepts: - 2 22, 2 22 2 2 7. repeated; multiplicity 2

8. discriminant; negative 9. 0, 1, 2

27. Zeros: - 1, 3; x-intercepts: - 1, 3

29. Zeros:

10. x =

- 3 - 4 22 - 3 + 4 22 - 3 - 4 22 - 3 + 4 22 , ; x-intercepts: , 2 2 2 2

1 3 1 3 31. Zeros: - 2 - 2 23, - 2 + 2 23; x-intercepts: - 2 - 2 23, - 2 + 2 23 33. Zeros: - , ; x-intercepts: - , 4 4 4 4 - 1 - 27 - 1 + 27 - 1 - 27 - 1 + 27 35. Zeros: , ; x-intercepts: , 37. Zeros: 2 - 22, 2 + 22; x-intercepts: 2 - 22, 2 + 22 6 6 6 6 3 3 39. Zeros: 2 - 25, 2 + 25; x-intercepts: 2 - 25, 2 + 25 41. Zeros: 1, ; x-intercepts: 1, 43. No real zeros; no x-intercepts 2 2 - 1 - 25 - 1 + 25 - 1 - 25 - 1 + 25 - 2 - 210 - 2 + 210 45. Zeros: , ; x-intercepts: , 47. Zeros: , ; 4 4 4 4 2 2 - 2 - 210 - 2 + 210 1 1 1 1 1 , 49. Zero: ; x-intercept: 51. 5 - 6, 0 6; 1 - 6, 32, (0, 3) 53. e - 1, - f ; 1 - 1, - 12, a - , b x-intercepts: 2 2 3 3 2 2 2 55. 5 - 3, 5 6; 1 - 3, 132, (5, 21) 57. Zeros: - 2 22, 2 22; x-intercepts: - 2 22, 2 22 59. Zeros: - 2, - 1, 1, 2; x-intercepts: - 2, - 1, 1, 2 1 1 61. Zeros: - 1, 1; x-intercepts: - 1, 1 63. Zeros: - 2, 1; x-intercepts: - 2, 1 65. Zeros: - 6, - 5; x-intercepts: - 6, - 5 67. Zero: - ; x-intercept: 3 3 3 3 69. Zeros: - , 2; x-intercepts: - , 2 71. Zeros: 0, 16; x-intercepts: 0, 16 73. Zero: 16; x-intercept: 16 75. Zeros: - 5 22, 5 22; 2 2 1 1 3 5 3 5 1 2 1 2 79. Zeros: - , ; x-intercepts: - , 81. Zeros: - , ; x-intercepts: - , x-intercepts: - 5 22, 5 22 77. Zero: ; x-intercept: 4 4 5 2 5 2 2 3 2 3 - 22 + 2 - 22 - 2 - 22 + 2 - 22 - 2 83. Zeros: , ; x-intercepts: , 2 2 2 2 - 1 - 217 - 1 + 217 - 1 - 217 - 1 + 217 85. Zeros: , ; x-intercepts: , 2 2 2 2 87. (a)

(b) - 1, 3

y 5 (1, 0) (0, 3)

(3, 0) 5 x (2, 3) (1, 4)

1 2 1 25 2 10 93. e - , f ; a - , b , a , - b 4 3 4 16 3 9

89. (a)

(b) - 2, - 6

y 40

(6, 0)

91. (a)

y (3, 6) 6 (4, 3) x 3 (3  兹2, 0) (3  兹2, 0)

(b) 3 - 22, 3 + 22

(2, 3)

(2, 0) 4 x

(4, 8)

95. { - 8}; ( - 8, 180)

5 5 45 97. e f ; a , - b 3 3 88

99. (a) 92 - 213, - 2 + 213 (b) 5 (c) - 7, - 4, 1, 2 101. 11 ft by 13 ft 103. 4 ft by 4 ft 105. (a) The ball strikes the ground after 6 sec. (b) The ball passes the top of the building on its way down after 5 sec. 107. 20 consecutive integers - b + 2b2 - 4ac - b - 2b2 - 4ac - 2b b 1 1 + = = 111. k = or k = 2a 2a 2a a 2 2 b { 2 1 - b2 2 - 4ac - b { 2b2 - 4ac b { 2b2 - 4ac - b { 2b2 - 4ac 2 2 113. ax + bx + c = 0, x = ; ax - bx + c = 0, x = = = 2a 2a 2a 2a 115. (b) y 122. Domain: { - 3, - 1, 1, 3}; Range: {2, 4}; 121. Function 6 3 123. a - 4, b 6 2 x 124. (1, 4) 109.

AN14

ANSWERS  Section 2.4

2.4 Assess Your Understanding (page 189) 7. -

5. parabola 6. axis or axis of symmetry 19.

21.

y 5 (4, 2)

29. f 1x2 = 2 1x + 12 2 - 4

23.

(0, 1) 5 x

(1, 1)

31. (a)

y 5

(2, 0)

5 x (0, 2)

39. (a)

(2, 1)

5 x

(2, 1) 5 x (1, 1)

(4, 2) (3, 1) (2, 2)

(2, 0)

5 x

5 x

35. (a)

y (2, 0) 5 x

1

(0, 0) 2 x

(6, 0)

(b) Domain: 1 - q , q 2 Range: 1 - q , 9] (c) Increasing: 1 - q , - 32 Decreasing: 1 - 3, q 2

1 4

1, 5  2 2

y 2

1, 2 2

5 x (0, 3)

1 , 15 4 8

(0, 8) (1, 9) x  1

x  3

41. (a)

y 5

x

(0, 0)

(1, 1)

y 9

(3, 9)

(1, 1)

(0, 2)

y 5

5 x

(0, 2)

x  1

27. f 1x2 = - 1x + 12 2 + 1

(0, 0)

(0, 1)

(1, 0)

17. H y 5

x  1

y 5

15. G

(4, 0)

(b) Domain: 1 - q , q 2 Range: [ - 1, q 2 (c) Decreasing: 1 - q , - 12 Increasing: 1 - 1, q 2

37. (a)

13. F

25. f(x) = (x + 2)2 - 2

33. (a)

y 2

(1, 1) (2, 2) (1, 4)

10. True 11. C y 5

(0, 2)

(3, 1) (2, 2)

5 x (2, 1)

(2, 1) (0, 2)

8. True 9. True

y 5 (4, 2)

(4, 2)

b 2a

(1, 3)

(b) Domain: 1 - q , q 2 Range: [ - 9, q 2 (c) Decreasing: 1 - q , - 12 Increasing: 1 - 1, q 2 43. (a)

3 17  , 4 4

y 5 (0, 2)

(1.78, 0)

2.5 x (0.28, 0)

3 x x

1 2

x

3 4

(b) Domain: 1 - q , q 2 (b) Domain: 1 - q , q 2 (b) Domain: 1 - q , q 2 15 5 17 Range: c , q b Range: a - q , - d d Range: a - q , 8 2 4 1 1 3 (c) Decreasing: a - q , b (c) Increasing: a - q , b (c) Increasing: a - q , - b 4 2 4 1 1 3 Increasing: a , q b Decreasing: a , q b Decreasing: a - , q b 4 2 4 2 2 2 2 2 45. f 1x2 = 1x + 12 - 2 = x + 2x - 1 47. f 1x2 = - 1x + 32 + 5 = - x - 6x - 4 49. f 1x2 = 2 1x - 12 - 3 = 2x2 - 4x - 1 51. Minimum value; - 18 53. Minimum value; - 21 55. Maximum value; 21 57. Maximum value; 13 y y y 65. (a) y 61. (a) 63. (a) 59. (a) (3, 2) (b) Domain: 1 - q , q 2 Range: [0, q 2 (c) Decreasing: 1 - q , - 12 Increasing: ( - 1, q)

2

(3, 0)

4

(5, 0)

(b) Domain: 1 - q , q 2 Range: [ - 16, q 2 (c) Decreasing: 1 - q , 12 Increasing: 11, q 2

67. (a)

(b) Domain: 1 - q , q 2 Range: 1 - q , q 2 (c) Increasing: 1 - q , q 2

69. (a)

y 10

1  ,0 2

(0, 4) 12 (10, 0) x

(b) Domain: 1 - q , q 2 Range: 1 - q , q 2 (c) Decreasing: 1 - q , q 2

(1, 1)

1  ,1 2

(2, 0)

(b) Domain: 1 - q , q 2 Range: 1 - q , 2] (c) Increasing: 1 - q , 32 Decreasing: 13, q 2

71. a = 6, b = 0, c = 2 73. (a), (c), (d)

y 2 5 x

y 6

(0, 1)

(b) Domain: 1 - q , q 2 Range: 1 - q , 0] 1 (c) Increasing: a - q , - b 2 1 Decreasing: a - , q b 2

(0, 1)

(6, 16)

(0, 16)

(0, 5)

8

7 x

5 x (1, 16)

(4, 0)

2

5, 0 2

7 x

5 x 1 7  , 4 8

(b) Domain: 1 - q , q 2 7 Range: c , q b 8 1 (c) Decreasing: a - q , - b 4 1 Increasing: a - , q b 4 75. (a), (c), (d) y (1, 3) 1

(3, 5)

(1, 3)

(b) 5 - 1, 3 6

5 x

5 x (3, 5)

(b) 5 - 1, 3 6

ANSWERS  Section 2.6 79. (a) a = 1: f 1x2 = 1x + 32 1x - 12 = x2 + 2x - 3 a = 2: f 1x2 = 2 1x + 32 1x - 12 = 2x2 + 4x - 6 a = - 2: f 1x2 = - 2 1x + 32 1x - 12 = - 2x2 - 4x + 6 a = 5: f 1x2 = 5 1x + 32 1x - 12 = 5x2 + 10x - 15 (b) The value of a does not affect the x-intercepts, but it changes the y-intercept by a factor of a. (c) The value of a does not affect the axis of symmetry. It is x = - 1 for all values of a. (d) The value of a does not affect the x-coordinate of the vertex. However, the y-coordinate of the vertex is multiplied by a. (e) The mean of the x-intercepts is the x-coordinate of the vertex.

77. (a), (c), (d) y 10

(2, 6) 10 x

(1, 6)

(b) 5 - 1, 2 6

AN15

81. (a) 1 - 2, - 252 (b) - 7, 3 (c) - 4, 0; 1 - 4, - 212, 10, - 212 y (d) 12 (7, 0) (4, 21) (2, 25)

(3, 0) 10 x (0, 21)

83. $500; $1,000,000 85. (a) 70,000 mp3 players (b) $2500 87. (a) 171 ft (b) 49 mph (c) Reaction time 89. (a) 187 or 188 watches; $7031.20 (b) P1x2 = - 0.2x2 + 43x - 1750 (c) 107 or 108 watches; $561.20 91. f(x2 = 2 1x + 42 1x - 22 93. If x is even, then ax2 and bx 2 2 are even and ax + bx is even, which means that ax + bx + c is odd. If x is odd, then ax2 and bx are odd and ax2 + bx is even, which means that ax2 + bx + c is odd. In either case, f(x) is odd. 95. 97. b2 - 4ac 6 0 99. No y 5

y  x 2  2x  1 y  x 2  2x  3 5 x

y  x 2  2x

101. Symmetric with respect to the x-axis, the y-axis, and the origin. 102. {x  x … 4} or ( - q , 4] 103. Center (5, - 2); radius = 3 104. Zeros: 3 - 25, 3 + 25; x-intercepts: 3 - 25, 3 + 25

2.5 Assess Your Understanding (page 194) 3. (a) 5x x 6 - 2 or x 7 2 6; 1 - q , - 22 h 12. q 2 (b) 5x - 2 … x … 2 6; [ - 2, 2] 5. (a) 5x - 2 … x … 1 6; [ - 2, 1] (b) 5x x 6 - 2 or x 7 1 6; 1 - q , - 22 h 11, q 2 7. 5x - 2 … x … 5 6; [ - 2, 5] 9. 5x x 6 0 or x 7 4 6; 1 - q , 02 h 14, q 2 11. 5x x 6 - 4 or x 7 3 6; 1 - q , - 42 h 13, q 2 13. { - 3} 15. e x ` -

1 1 … x … 3 f ; c - , 3 d 17. 5x x 6 - 1 or x 7 8 6; 1 - q , - 12 h 18, q 2 2 2 19. No real solution 21. All real numbers; 1 - q , q 2 23. 5x x … - 4 or x Ú 4 6; 1 - q , - 4] h [4, q ) 25. (a) 5 - 1, 1 6 (b) 5 - 1 6 (c) 5 - 1, 4 6 (d) 5x x 6 - 1 or x 7 1 6; 1 - q , - 12 h 11, q 2 (e) 5x x … - 1 6 ; 1 - q , - 1] (f) 5x x 6 - 1 or x 7 4 6; 1 - q , - 12 h 14, q 2 1 (g) 5 x x … - 22or x Ú 22 6 ; 1 - q , - 22 4 h 3 2 2, q 2 27. (a) 5 - 1, 1 6 (b) e - f (c) 5 - 4, 0 6 (d) 5x - 1 6 x 6 1 6; 1 - 1, 12 4 1 1 (e) e x ` x … - f or a - q , - d (f) 5x - 4 6 x 6 0 6; 1 - 4, 02 (g) 50 6 29. (a) 5 - 2, 2 6 (b) 5 - 2, 2 6 (c) 5 - 2, 2 6 4 4 (d) 5x x 6 - 2 or x 7 2 6; 1 - q , - 22 h 12, q 2 (e) 5x x … - 2 or x Ú 2 6; 1 - q , - 2] h [2, q ) (f) 5x x 6 - 2 or x 7 2 6; 1 - q , - 22 h 12, q 2 (g) 5 x x … - 25or x Ú 25 6 ; 1 - q , - 25 4 h 3 25, q 2 31. (a) 5 - 1, 2 6 (b) 5 - 2, 1 6 (c) 50 6 (d) 5x x 6 - 1 or x 7 2 6; 1 - q , - 12 h 12, q 2 (e) 5x - 2 … x … 1 6; [ - 2, 1] 1 - 213 1 + 213 1 - 213 1 + 213 or x Ú f; a - q, d h c , qb 2 2 2 2 33. (a) 5 sec (b) The ball is more than 96 ft above the ground for time t between 2 and 3 sec, 2 6 t 6 3. 35. (a) $0, $1000 (b) The revenue is more than $800,000 for prices between $276.39 and $723.61, +276.39 6 p 6 +723.61. 41. When the inequality is not strict. 42. {x x … 5} 43. Odd 44. (a) 9 45. (a) 18 y (b) (f) 5x x 6 0 6; 1 - q , 02

(g) e x ` x …

5 (9, 0) 5

x 24

28 16

(0, −6)

(b) C(H) = 0.3734H + 7.3268 (c) If height increases by 1 in., then head circumference increases by about 0.3734 in., on average. (d) Approximately 17.0 in. (e) Approximately 26.98 in.

2.6 Assess Your Understanding (page 202) 1 2 1 x + 100x (b) 5x 0 … x … 600 6 (c) $13,333.33 (d) 300; $15,000 (e) $50 5. (a) R1x2 = - x2 + 20x (b) $255 6 5 (d) $10 (e) Between $8 and $12 7. (a) A1w2 = - w 2 + 200w (b) A is largest when w = 100 yd. (c) 10,000 yd2 9. 2,000,000 m2

3. (a) R1x2 = (c) 50; $500

AN16 11. (a) (d)

ANSWERS  Section 2.6

625 ≈ 39 ft 16

(b)

7025 ≈ 219.5 ft 32

(c) About 170 ft 13. 18.75 m 15. (a) 3 in. (b) Between 2 in. and 4 in. 750 ≈ 238.73 m by 375 m 17. p a 38 248 19. x = 21. 23. 2 3 3

220

(f) When the height is 100 ft, the projectile is about 135.7 ft from the cliff. 0

200 0

25. (a)

50,000

15

75

27. (a)

50,000

(e)

15

0

75 0

The data appear to follow a quadratic relation with a 6 0. (b) I(x) = - 43.335x2 + 4184.883x - 54, 062.439 (c) About 48.3 years of age (d) Approximately $46,972 32. 15i

33. 13

34. (x + 6)2 + y2 = 7

35.

29. (a)

1800

400 1000

1100

125

10

45 0

The data appear to follow a linear relation with positive slope. (b) R(x) = 0.953x + 704.186 (c) $1514

The data appear to follow a quadratic relation with a 6 0. (b) B(a) = - 0.483a2 - 26.356a - 251.342 (c) 79.4 births per 1000 population

- 4 - 231 - 4 + 231 , 5 5

2.7 Assess Your Understanding (page 209) 7. 2 + 3i 8. True 9. - 2i, 2i

11. - 4, 4

y 7

13. 3 - 2i, 3 + 2i y 4

(0, 4)

(4, 0)

(4, 0) 10 x

(0, 13)

y 16

18 (0, 10)

(6, 10)

(3, 4)

5 x

17. 2 - 23, 2 + 23

19. -

y

1 1 1 1 - i, - + i 2 2 2 2

21. -

1 23 1 23 i, - + i 2 2 2 2

y 8

6

(3, 1) 8 x

8 x

(0, 16)

(0.27, 0)

15. 3 - i, 3 + i

y

23.

4 - 3 22 4 + 3 22 , 2 2 y 10

y 8

(3.73, 0) 7 x

(2, 9)

(0, 1) (1, 1)

(2, 3)

(0, 1) 5 x

1 1  , 2 2

25. Two complex solutions that are conjugates of each other

(1, 1)

(0.12, 0)

(0, 1)

1 3  , 2 4

27. Two unequal real solutions

5 x

3

(4, 1) x (4.12, 0)

29. A repeated real solution 31. - 2, 2, - 2i, 2i

1 23 1 23 3x + 2 33. - i, - + i, - 1 - 23i, - 1 + 23i, 1, 2 35. (g - f)(x) = ; Domain: {x x ≠ - 1, x ≠ 0} 2 2 2 2 x(x + 1) 36. (a) Domain: [ - 3, 3]; Range: [ - 2, 2] (b) ( - 3, 0), (0, 0), (3, 0) (c) Symmetric with respect to the origin (d) The relation is a function. 600 25 38. y = 2 37. x −4

4

−25

Local maximum: (0, 0) Local minima: ( - 2.12, - 20.25), (2.12, - 20.25) Increasing: ( - 2.12, 0), (2.12, 4) Decreasing: ( - 4, - 2.12), (0, 2.12)

2.8 Assess Your Understanding (page 213) 7. - a; a 8. - a 6 u 6 a 9. … 10. True 11. False 12. False 13. (a) 5 - 9, 3 6 (b) 5x - 9 … x … 3 6; [ - 9, 3] (c) 5x x 6 - 9 or x 7 3 6 ; 1 - q , - 92 h 13, q 2 15. (a) 5 - 2, 3 6 (b) 5x x … - 2 or x Ú 3 6; 1 - q , - 2] h [3, q 2 (c) 5x - 2 6 x 6 3 6; 1 - 2, 32 17. 5 - 6, 6 6

AN17

ANSWERS  Review Exercises 19. 5 - 4, 1 6

21. e - 1,

3 1 1 f 23. 5 - 4, 4 6 25. e - , f 2 2 2 35. 5 - 1, 3, 1 - 22i, 1 + 22i 6 37. 5 - 2, - 1, 0, 1 6 39. 5x - 6 6 x 6 6}; 1 - 6, 62

27. e -

27 27 , f 2 2

41. 5x x 6 - 4 or x 7 4}; 1 - q , - 42 h (4, q )

29. e -

36 24 , f 5 5

31. No real solution

43. 5x - 4 6 x 6 4}; 1 - 4, 42 4

4

47. 5x 1 6 x 6 3}; (1, 3)

49. e t ` -

4

45. 5x x 6 - 4 or x 7 4}; 1 - q , - 42 h (4, q )

4 4

51. 5x x … 1 or x Ú 5}; 1 - q, 1 4 h [5, q 2

2 2 … t … 2 f; c - , 2 d 3 3

33. 5 - 3, 3 6

4

53. e x ` - 1 6 x 6

3 3 f ; a - 1, b 2 2

3

1



55. 5x x 6 - 1 or x 7 2 6; 1 - q , - 12 h 12, q 2 1

2 3

1

2

59. e x `

57. No real solution; ∅

65. (a) { - 2, 6}

19 17 19 17 6 x 6 f; a , b 6 6 6 6

2

19 6

17 6

3 2

61. {x - 2 6 x 6 4}; ( - 2, 4)

0

2

1 63. (a) e - , 1 f 5

1

5

4

(b) e x ` -

1 1 1 1 6 x 6 1 f ; a - , 1 b (c) e x ` x … - or x Ú 1 f ; a - q , - d h 3 1, q ) 5 5 5 5 (b) {x x … - 2 or x Ú 6}; ( - q , - 2 4 h 3 6, q ) (c) {x - 2 6 x 6 6}; ( - 2, 6) 67.  x - 10 6 2; 5x 8 6 x 6 12 6

69.  2x + 1 7 5; 5x x 6 - 3 or x 7 2 6 71. Between 5.6995 in. and 5.7005 in. 73. IQs less than 71 or greater than 129 75.  5x + 1 + 7 = 5 is equivalent to  5x + 1 = - 2. Because  u Ú 0 for any real number u, the equation has no real solution. 77.  2x - 1 is greater than or equal to 0 for 1 all real numbers x. The only real number x such that  2x - 1 … 0 is the value of x for which 2x - 1 = 0, which is . 2 78. f( - 4) = 15 79. (0, q ) −2 −1 0

1 2

3

4

5

6

80. 17 + 7i 81. (a) (0, 0), (4, 0) (b) Domain: [ - 2, 5]; Range: [ - 2, 4] (c) Increasing: (3, 5); Decreasing: ( - 2, 1); Constant: (1, 3) (d) Neither

7

Review Exercises (page 217) 1. (a) m = 2; b = - 5 (b) 5 y 2

2

2. (a) m = ,0

(b)

5 x (0, 5)

(1, 3)

(0, 4)

(4, 4) 5 x

(5, 2)

5. Linear; y = 5x + 3

13. { - 1, 7}

1 5 1 5 8. Zeros: - , ; x-intercepts: - , 9. Zeros: 1, 5; x-intercepts: 1, 5 3 2 3 2 5 5 5 5 1 1 11. Zeros: 1 + ,1 ; x-intercepts: 1 + ,1 12. Zeros: - , 1; x-intercepts: - , 1 A3 A3 A3 A3 2 2

6. Nonlinear

7. Zeros: - 5, 3; x-intercepts: - 5, 3

y 10

y 20

(2, 7) (7, 16) (4, 5) 8 x

1 x

(c) Domain: 1 - q , q 2 Range: 5y y = 4 6 (d) 0 (e) Constant

(c) Domain and Range: 1 - q , q 2 4 (e) Increasing (d) 5

14. { - 2, 2}

(0, 9)

(7, 0)

(0, 6)

1 1 10. Zero: - ; x-intercept: 3 3

(3, 0)

(3, 4)

8 x

(c) Domain and Range: 1 - q , q 2 (d) 2 (e) Increasing

(1, 16)

y 18 (0, 14)

5

15 ,0 2

y 2

4. Zero: - 7; y-intercept: 14

3. (a) m = 0; b = 4 (b) y

4 ; b = -6 5

(2, 9)

4 x (0, 5)

15. Zeros: - 12, 12; x-intercepts: - 12, 12 16. Zero: - 1; x-intercept: - 1 17. Zeros: - 3, 5; x-intercepts: - 3, 5 1 1 1 1 18. Zeros: - , ; x-intercepts: - , 2 6 2 6 y 19. 8 (1, 3) (3, 3) (2, 2) 8 x

20.

21.

y 5

(2, 6)

(4, 0) (3, 1)

y 9

8 x (5, 1)

(0, 6)

(1, 4) 5 x

AN18

ANSWERS  Review Exercises

22. (a)

23. (a)

y (0, 6)

24. (a)

y 2

8 (ⴚ8, 0)

(4, 6)

y 2.5

(8, 0) 10 x

(0, 0) (2, 2) 8 x

(0, 1) 2.5 x ⴚ2.5 1 3

1 2

xⴝⴚ

(b) Domain: 1 - q , q 2 Range: ( - q , 1] 1 (c) Increasing: a - q , b 2 1 Decreasing: a , q b 2

(b) Domain: 1 - q , q 2 1 Range: c , q b 2 1 (c) Decreasing: a - q , - b 3 1 Increasing: a - , q b 3

14 28. Maximum value; 12 29. Maximum value; 5 5 2 30. f(x) = - 3(x - 2) - 4 = - 3x2 + 12x - 16 31. f(x) = (x + 1)2 + 2 = x2 + 2x + 3 1 1 32. 5x 0 - 5 6 x 6 2 6; 1 - 5, 22 33. e x ` x … - 3 or x Ú - f ; a - q , - 3 d h c - , q ≤ 2 2

(0.22, 0) 5 x (0, ⴚ1)

7 2 ⴚ ,ⴚ ⴚ5 3 3 2 xⴝⴚ 3

34. - 2 22i, 2 22i

(b) Domain: 1 - q , q 2 7 Range: c - , q b 3

y 5 (ⴚ1ⴚÏ5, 0) (0, 8) (ⴚ1, ⴚ5)

5 x

3 9 38. e - , f 5 5

7 3 7 3 6 x 6 - f; a - , - b 2 6 2 6

22 22 i, 1 + i 2 2

37. -

1 1 22 22 i, - + i 2 2 2 2

y 2 (ⴚ1ⴙÏ5, 0) 5 x

y 9 (ⴚ1, 3)

5 x (1, ⴚ1)

(0, ⴚ3)

(0, 3)

1 ⴚ ,2 2

(2, ⴚ3)

(0, ⴚ4)

3 x

39. 51, 5 6

41. {x x … - 2 or x Ú 7}; ( - q , - 2 4 h 3 7, q ) ⴚ2

7 ⴚ 6

36. 1 -

35. - 1 - 25, - 1 + 25

y 18

2 (c) Decreasing: a - q , - b 3 2 Increasing: a - , q b 3

3 ⴚ 2

(1, 0) 2.5 x

27. Minimum value;

y

40. e x ` -

xⴝ

(b) Domain: 1 - q , q 2 Range: [ - 16, q 2 (c) Decreasing: ( - q , 0) Increasing: (0, q )

(b) Domain: 1 - q , q 2 Range: 32, q ) (c) Decreasing: ( - q , 2) Increasing: (2, q )

(ⴚ1.55, 0)

y 1 1 ⴚ , 3 2

(0, ⴚ16)

xⴝ2

26. (a)

25. (a)

1, 1 2

42. e x ` 0 … x …

4 4 f ; c 0, d 3 3

0

7

43. e x ` x 6 - 1 or x 7

7 f; 3

7 ( - q , - 1) h a , q b 3

4 3

ⴚ1

7 3

120

52. (a)

80 40 0

100 200 300 Price (dollars)

p

(b) Since each input (price) corresponds to a (e) single output (quantity demanded), we know that quantity demanded is a function of price. Also, because the average rate of change is a constant - 0.4 24-in. LCD monitor per dollar, the function is linear. (c) q( p) = - 0.4p + 160 (d) 5p 0 … p … 400 6 or 3 0, 400 4 53. (a) Quadratic, a 6 0

Tibia (mm)

y 38 37 36 35 34 24 25 26 27 28 x Humerus (mm)

(b) Yes; the two variables appear to be linearly related. (c) y = 1.3902x + 1.114 (d) 38.0 mm

Quantity Demanded

q

Total Revenue (thousands of dollars)

51. (a)

Quantity Demanded

44. (a) Company A: C1x2 = 0.06x + 7; Company B: C1x2 = 0.08x (b) 350 min (c) 0 … x 6 350 45. (a) R(p) = - 10p2 + 1500p (b) {p 0 6 p … 150} (c) $75 (d) $56,250 (e) 750 units (f) Between $70 and $80 46. The width is 6 in. and the length is 8 in. 47. (a) 63 clubs (b) $151.90 48. 50 by 50 ft 49. 25 square units 50. 3.6 ft q 120 80 40 0

100 200 300 Price (dollars)

p

(f ) If price increases by $1, quantity demanded of 24-in. LCD monitors decreases by 0.4 monitor. (g) q-intercept: When the price is $0, 160 24-in. LCD monitors will be demanded. p-intercept: There will be 0 24-in. LCD monitors demanded when the price is $400. (d)

6500

R 6500 6400 6300 6200 6100 25 30 A 20 Advertising (thousands of dollars)

(b) R(A) = - 7.76 A2 + 411.88A + 942.72; $26.5 thousand (c) $6408 thousand

15 6000

35

AN19

ANSWERS  Cumulative Review

Chapter Test (page 219) 1. (a) Slope: - 4; y-intercept: 3 (b) Decreasing (c) y

2. Linear; y = f(x) = - 5x + 2

4 3. - , 2 3

7.

y 9

y 21

(0, 3)

2 - 26 2 + 26 , 2 2

6. - 3, 0

5. { - 1, 3}

5

4.

(0, 7)

(6, 7)

(3, 18) (4, ⴚ1)

5 x (1, ⴚ1)

(0, 0) 4 x (ⴚ1, ⴚ2)

(ⴚ3, 0)

8. (a) Opens up (b) 12, - 82 (c) x = 2

(e)

(2, ⴚ1)

(f) Domain: All real numbers; ( - q, q) Range: 5y y Ú - 8 6; [ - 8, q2 (g) Decreasing: ( - q, 2) ; Increasing: 12, q2

y 10 (0, 4)

8 x (3, ⴚ2)

(4, 4)

10 x (3.63, 0) (2, ⴚ8) xⴝ2

6 - 2 26 6 + 2 26 , ; 3 3 y-intercept: 4

(0.37, 0)

(d) x-intercepts:

9. (a) Opens down (b) 11, - 32 (c) x = 1 (d) No x-intercepts: y-intercept: - 5

(e)

(f) Domain: All real numbers; ( - q, q) Range: {y y … - 3}; ( - q, - 3] (g) Increasing: ( - q, 1); Decreasing: 11, q2

y 2 5 x (1, ⴚ3) (0, ⴚ5)

(2, ⴚ5)

xⴝ1

10. f 1x2 = 2x2 - 4x - 30

16. {x x … - 5 or x Ú 2}; ( - q , - 5 4 h 3 2, q )

15. {x - 11 6 x 6 5}; ( - 11, 5) ⴚ11

0

26 26 i, - 1 + i 2 2 17. (a) C1m2 = 0.15m + 129.50 (b) $258.50 (c) 562 miles

11. Maximum value; 21 12. {x x … 4 or x Ú 6}; ( - q , 4 4 h 3 6, q )

ⴚ5

5

1 2 x + 1000x 18. (a) R1x2 = 10 10 19. (a) ⴚ3

0

(b) $384,000

13. - 1 -

7 14. e - 3, f 3

2

(c) 5000 units; $2,500,000

(d) $500 (b) y = - 4.234x - 2.362

15

(c) y = 1.993x 2 + 0.289x + 2.503

3

ⴚ3

3

ⴚ15

0

Quadratic that opens up

Linear with negative slope

Cumulative Review (page 220) 1. 5 22; (1.5, 0.5)

2. 1 - 2, - 12 and (2, 3) are on the graph.

3 3 3. e x ` x Ú - f or c - , q b 5 5 ⴚ

3 5

4. y = - 2x + 2 (ⴚ1, 4)

y 5

6. 1x - 22 2 + 1y + 42 2 = 25

1 13 5. y = - x + 2 2

y 10

y 9 5 x (2, ⴚ2)

(2, 1)

(3, 5) (ⴚ3, ⴚ4) 5 x

7. Yes, a function 8. (a) - 3

(b) x2 - 4x - 2

(c) x2 + 4x + 1

(d) - x2 + 4x - 1

10 x (7, ⴚ4) (2, ⴚ4) (2, ⴚ9)

(e) x2 - 3

(f) 2x + h - 4

9. e z ` z ≠

7 f 6

10. Yes, a function 11. (a) No (b) - 1; 1 - 2, - 12 is on the graph. (c) - 8; 1 - 8, 22 is on the graph. 12. Neither 13. Local maximum is 5.30 and occurs at x = - 1.29. Local minimum is - 3.30 and occurs at x = 1.29. Increasing: 1 - 4, - 1.292 or (1.29, 4) Decreasing: 1 - 1.29, 1.292 14. (a) - 4 (b) 5x x 7 - 4 6 or 1 - 4, q 2 15. (a) Domain: 5x - 4 … x … 4 6; Range: 5y - 1 … y … 3 6 (b) 1 - 1, 02, 10, - 12, 11, 02 (d) 1 (e) - 4 and 4 (f ) 5x - 1 6 x 6 1 6 y (g) (h) (i) y y (ⴚ4, 5) (4, 5) 5 (2, 3) (ⴚ2, 3) (1, 2) (ⴚ1, 2) 5 x (0, 1)

(ⴚ4, 3) (ⴚ2, 1) (ⴚ1, 0)

5

(0, ⴚ1)

( j) Even

(k) (0, 4)

(4, 3) (2, 1) 5 x (1, 0)

10 (ⴚ4, 6) (ⴚ2, 2) (ⴚ1, 0) (0, ⴚ2)

(c) y-axis

(4, 6) (2, 2) 5 x (1, 0)

AN20

ANSWERS  Section 3.1

CHAPTER 3: Polynomial and Rational Functions 3.1 Assess Your Understanding (page 239) 7. smooth; continuous 8. touches 9. ( - 1, 1); (0, 0); (1, 1) 10. r is a real zero of f; r is an x-intercept of the graph of f; x - r is a factor of f. 11. turning points 12. y = 3x4 13. q ; - q 14. As x increases in the positive direction, f(x) decreases without bound. 1 1 1 1 15. Yes; degree 3; f(x) = x3 + 4x; leading term: x3; constant term: 0 17. Yes; degree 2; g(x) = - x2 + ; leading term: - x2; constant term: 2 2 2 2 3 1 1 power. 23. Yes; degree 4; F (x) = 5x4 - px3 + ; leading term: 5x4; constant term: . 2 2 2 25. Yes; degree 4; G(x) = 2x4 - 4x3 + 4x2 - 4x + 2; leading term: 2x4; constant term: 2 19. No; x is raised to the - 1 power.

27.

29.

y 5 (0, 1) (ⴚ2, 1) 5 x (ⴚ1, 0)

35.

31.

y 5

5 x (1, ⴚ2)

37. (ⴚ2, 3) (ⴚ1, 1)

(2, 3) (1, 2) 5 x

1,

(0, 3)

y 6 (1, 5)

1 2

(2, 4) (3, 3)

5 x

5 x

y 5 (ⴚ1, 1) (0, 0)

5 x

(0, 0)

39.

y 5

33.

y 5

1 ⴚ1, 2

(0, ⴚ3) (ⴚ1, ⴚ4)

y 5 (0, 1)

21. No; x is raised to the

5 x (1, ⴚ1)

41. f(x) = x3 - 3x2 - x + 3 for a = 1 43. f(x) = x3 - x2 - 12x for a = 1 45. f(x) = x4 - 15x2 + 10x + 24 for a = 1 47. f(x) = x3 - 5x2 + 3x + 9 for a = 1

49. (a) 7, multiplicity 1; - 3, multiplicity 2

(b) Graph touches the x-axis at - 3 and crosses it at 7. (c) 2 (d) y = 3x3 1 5 51. (a) 2, multiplicity 3 (b) Graph crosses the x-axis at 2. (c) 4 (d) y = 4x 53. (a) - , multiplicity 2; - 4, multiplicity 3 2 1 (b) Graph touches the x-axis at - and crosses it at - 4. (c) 4 (d) y = - 2x5 55. (a) 5, multiplicity 3; - 4, multiplicity 2 2 (b) Graph touches the x-axis at - 4 and crosses it at 5. (c) 4 (d) y = x5 57. (a) No real zeros (b) Graph neither crosses nor touches the x-axis. (c) 5

(d) y = 3x6

(c) 3

(d) y = - 2x

59. (a) 0, multiplicity 2; - 22, 22, multiplicity 1 61. Could be; zeros: - 1, 1, 2; Least degree is 3

4

65. f(x) = x(x - 1)(x - 2)

(4, 16) (3, 0) 5 x (2, ⴚ4)

79. Step 1: y = - x3 Step 2: x-intercepts: - 4, 1; y-intercept: 16 Step 3: - 4: multiplicity 2, touches; 1: multiplicity 1, crosses Step 4: At most 2 turning points (−2, 12) y Step 5: 25 (−5, 6) −6

(−4, 0)

(0, 16) (1, 0) 4 x

69. f(x) = 0.2(x + 4)(x + 1)2(x - 3)

75. Step 1: y = x3 Step 2: x-intercepts: - 4, 2; y-intercept: 16 Step 3: - 4: multiplicity 1, crosses; 2: multiplicity 2, touches Step 4: At most 2 turning points Step 5: y (ⴚ2, 32) 40

24

(0, 0) (ⴚ1, ⴚ4)

63. Cannot be the graph of a polynomial; gap at x = - 1

1 67. f(x) = - (x + 1)(x - 1)2 (x - 2) 2

73. Step 1: y = x3 Step 2: x-intercepts: 0, 3; y-intercept: 0 Step 3: 0: multiplicity 2, touches; 3: multiplicity 1, crosses Step 4: At most 2 turning points y Step 5:

(b) Graph touches the x-axis at 0 and crosses it at - 22 and 22.

(ⴚ4, 0)

(4, 32) (0, 16) (2, 0) 5 x (ⴚ5, ⴚ49)

81. Step 1: y = x3 Step 2: x-intercepts: - 4, - 1, 2; y-intercept: - 8 Step 3: - 4, - 1, 2: multiplicity 1, crosses Step 4: At most 2 turning points Step 5: y (0, ⴚ8) 50 (ⴚ2, 8) (ⴚ4, 0) (ⴚ5, ⴚ28)

−25

77. Step 1: y = - 2x4 Step 2: x-intercepts: - 2, 2; y-intercept: 32 Step 3: - 2: multiplicity 1, crosses; 2: multiplicity 3, crosses Step 4: At most 3 turning points Step 5: (ⴚ1, 54) y (0, 32) (ⴚ2, 0)

(ⴚ3, ⴚ250)

(2, 0) 5 x (3, ⴚ10) ⴚ240

83. Step 1: y = x4 Step 2: x-intercepts: - 2, 0, 2; y-intercept: 0 Step 3: - 2, 2: multiplicity 1, crosses; 0, multiplicity 2, touches Step 4: At most 3 turning points Step 5: y 90

(3, 28) 5 x (2, 0) (1, ⴚ10)

(ⴚ1, 0)

71. f(x) = - x(x + 3)2(x - 3)2

(ⴚ3, 45) (1, ⴚ3) (ⴚ2, 0) (ⴚ1, ⴚ3)

(3, 45) (0, 0) (2, 0) 5 x

ANSWERS   Section 3.1 85. Step 1: y = x4 Step 2: x-intercepts: - 1, 2; y-intercept: 4 Step 3: - 1, 2: multiplicity 2, touches Step 4: At most 3 turning points y Step 5: (ⴚ2, 16)

20

(0, 4)

2

(3, 16) (1, 4)

(ⴚ1, 0)

87. Step 1: y = x3 Step 2: x-intercepts: 0, 1, 2; y-intercept: 0 Step 3: 0, 1, 2: multiplicity 1, crosses Step 4: At most 2 turning points y Step 5: Q52 , 15 8R

(1, 0)

89. Step 1: y = x4 Step 2: x-intercepts: - 3, - 1, 0; y-intercept: 0 Step 3: - 1, - 3: multiplicity 1, crosses; 0: multiplicity 2, touches Step 4: At most 3 turning points y Step 5: 4

(2, 0) 3 x

(0, 0) −2

(2, 0) 5 x

AN21

(−3, 0) (−1, 0) −2

Q− 12 , − 15 R 8

2 x (0, 0)

(−2, −4) −6

91. Step 1: y = x4 Step 2: x-intercepts: - 4, - 2, 2; y-intercept: 32 Step 3: - 2, - 4: multiplicity 1, crosses; 2: multiplicity 2, touches Step 4: At most 3 turning points y Step 5:

93. Step 1: y = 5x4 Step 2: x-intercepts: - 3, - 2, 0, 2; y-intercept: 0 Step 3: - 3, - 2, 0, 2: multiplicity 1, crosses Step 4: At most 3 turning points (−1, 30) y Step 5: 40

200

(ⴚ4, 0) (ⴚ3, ⴚ25)

(3, 35) (2, 0) 5 x

97. Step 1: y = x3 Step 2:

−60

99. Step 1: y = x3 Step 2:

2

ⴚ2

2

160

(0, 0)

(−3, 0) −3 (−2, 0)

(0, 32)

(ⴚ2, 0)

95. Step 1: y = x5 Step 2: x-intercepts: 0, 2; y-intercept: 0 Step 3: 2: multiplicity 1, crosses; 0: multiplicity 2, touches Step 4: At most 4 turning points y Step 5:

(2, 0) 2 x (0, 0) (ⴚ1, ⴚ12) (1, −60)

101. Step 1: y = x4 Step 2:

10

ⴚ4

2

3

ⴚ2

2

ⴚ10

ⴚ2

(3, 108) (1, ⴚ4) (2, 0) 5 x

ⴚ2

Step 3: x-intercepts: - 1.26, - 0.20, 1.26 y-intercept: - 0.31752 Step 4:

Step 3: x-intercepts: - 3.56, 0.50 y-intercept: 0.89 Step 4:

Step 3: x-intercepts: - 1.5, - 0.5, 0.5, 1.5 y-intercept: 0.5625 Step 4:

Step 5: ( - 0.80, 0.57); (0.66, - 0.99) y Step 6:

Step 5: ( - 2.21, 9.91); (0.50, 0) y Step 6:

Step 5: ( - 1.12, - 1); (1.12, - 1); (0, 0.56) Step 6: (0, 0.56)

(ⴚ0.5, 0.40) 2.5 (ⴚ0.80, 0.57) (ⴚ1.26, 0) (ⴚ1.5, ⴚ0.86) (0, ⴚ0.32)

(ⴚ0.20, 0) (1.5, 1.13) 2.5 x (1.26, 0) (0.66, ⴚ0.99)

Step 7: Domain: ( - q , q ); Range: ( - q , q ) Step 8: Increasing on ( - q , - 0.80) and (0.66, q ) Decreasing on ( - 0.80, 0.66)

(ⴚ2.21, 9.91)

10 (ⴚ1, 5.76)

(ⴚ3.56, 0) (0, 0.89) (ⴚ3.9, ⴚ6.58)

(0.5, 0)

1 x (1, 1.14)

Step 7: Domain: ( - q , q ); Range: ( - q , q ) Step 8: Increasing on ( - q , - 2.21) and (0.50, q ) Decreasing on ( - 2.21, 0.50)

(ⴚ0.5, 0) (ⴚ1.75, 2.29) (ⴚ1.5, 0)

y 5

(0.5, 0) (1.75, 2.29) (1.5, 0)

2.5 x (ⴚ1.12, ⴚ1) (1.12, ⴚ1) (ⴚ1, ⴚ0.94) (1, ⴚ0.94)

Step 7: Domain: ( - q , q ); Range: [ - 1, q ) Step 8: Increasing on ( - 1.12, 0) and (1.12, q ) Decreasing on ( - q , - 1.12) and (0, 1.12)

AN22

ANSWERS  Section 3.1

103. Step 1: y = 2x4 Step 2:

Step 5: ( - 0.42, - 4.64) Step 6: y

6

(ⴚ1.25, 4.22)

ⴚ2

5

(1.75, 1.83) (1.62, 0) 2 x

(ⴚ1.07, 0)

2

(ⴚ0.42, ⴚ4.64) ⴚ5

(0, ⴚ4)

107. f(x) = x(x + 4)(x - 3) Step 1: y = x3 Step 2: x-intercepts: - 4, 0, 3; y-intercept: 0 Step 3: - 4, 0, 3: multiplicity 1, crosses Step 4: At most 2 turning points y Step 5:

700 (ⴚ7, 630) (ⴚ1, 30)

(4, 32) (3, 0) 5 x (0, 0) (2, ⴚ12)

(ⴚ2, 0) (ⴚ6, 0) (ⴚ4, ⴚ192)

111. f(x) = - x2 (x + 1)2 (x - 1) Step 1: y = - x5 Step 2: x-intercepts: - 1, 0, 1; y-intercept: 0 Step 3: 1: multiplicity 1, crosses; - 1, 0: multiplicity 2, touches Step 4: At most 4 turning points 113. f(x) = 3(x + 3)(x - 1)(x - 4)

(ⴚ0.54, 0.10) (0, 0)

117. (a) - 3, 2

121. (a)

2 4 6 8 10

T 66 60 54 48 42 36

x

0

The relation appears to be cubic. (b) H(x) = 0.3948x3 - 5.9563x2 + 26.1965x - 7.4127 (c) Approximately 24 (d) 35

0

10

(3, 270) (2, 0) 3 x (1, ⴚ42)

(0.74, 0.43) (1, 0) 2.5 x (1.2, ⴚ1.39392)

(b) - 6, - 1 (b) Approximately 4.67ºF per hour. (c) Approximately - 2.33°F per hour

6 12 18 24 x

The relation appears to be cubic. (d) T(x) = - 0.02357x3 + 0.809x2 - 6.2893x + 52.0714; 63.2°F (e)

66

0

0

y 2.5 (ⴚ1.5, 1.40625) (ⴚ1, 0)

115. f(x) = - 2(x + 5)2 (x - 2)(x - 4)

H

(0, 0)

Step 5:

50 40 30 20 10 0

(ⴚ2, 0)

109. f(x) = 2x(x + 6)(x - 2)(x + 2) Step 1: y = 2x4 Step 2: x-intercepts: - 6, - 2, 0, 2; y-intercept: 0 Step 3: - 6, - 2, 0, 2: multiplicity 1, crosses Step 4: At most 3 turning points y Step 5:

50 (ⴚ2, 20) (ⴚ4, 0)

119. (a)

20 (0, 0) (1, 3) (2, 0) 2.5 x (ⴚ1, ⴚ3) (3, ⴚ15)

(ⴚ3, 15)

Step 7: Domain: ( - q , q ); Range: [ - 4.64, q ) Step 8: Increasing on ( - 0.42, q ) Decreasing on ( - q , - 0.42)

Step 3: x-intercepts: - 1.07, 1.62; y-intercept: - 4 Step 4:

(ⴚ5, ⴚ40)

105. f(x) = - x(x + 2)(x - 2) Step 1: y = - x3 Step 2: x-intercepts: - 2, 0, 2; y-intercept: 0 Step 3: - 2, 0, 2: multiplicity 1, crosses Step 4: At most 2 turning points y Step 5:

26 0

(e) Approximately 54; no

(f) The predicted temperature at midnight is 52.1°F.

Account value of deposit 1

Account value of deposit 2

5

c

e

123. (a) T(r) = 500(1 + r)(1 + r) + 500(1 + r) + 500 Deposit 3

= 500(1 + 2r + r 2) + 500 + 500r + 500 = 500 + 1000r + 500r 2 + 500 + 500r + 500 = 500r 2 + 1500r + 1500 (b) $2155.06 2 11 - 2 - 27 - 2 + 27 4 129. (a)–(d) 132. y = - x 133. 5x x ≠ - 5 6 134. , 135. e - , 2 f 5 5 2 2 5

AN23

ANSWERS   Section 3.2

Historical Problems (page 255) b 3 b 2 b b + b ax - b + c ax - b + d = 0 3 3 3

ax -

1. 2

3

2

3

bx b 2b x b bc + bx2 + + cx + d = 0 3 27 3 9 3

x3 - bx2 +

x3 + ac Let p = c -

b2 2b3 bc bx + a + db = 0 3 27 3

(H + K)3 + p(H + K) + q H 3 + 3H 2K + 3HK2 + K3 + pH + pK + q Let 3HK H 3 - pH - pK + K3 + pH + pK + q = 0, H 3 + K3

= = = =

0 0 - p. -q

2b3 bc b2 and q = + d. Then x3 + px + q = 0. 3 27 3

3HK = - p p K = 3H p 3 b = -q H3 + a 3H

3.

2.

H3 -

p3 27H 3

4. H 3 + K3 = - q K3 = - q - H 3 K3 = - q - c K3 =

= -q

27H 6 - p3 = - 27qH 3

K =

27H 6 + 27qH 3 - p3 = 0 H3 = H3 = H3 = H =

2 # 27

-q 2 -q 2 3

{

-q

B 2

B4 +

+

q2 A4

4(27)p

+

B 22(272) q2

x =

3

27 q

{

2

2

3

-q

B 2

2 q2

B4 -

+ q2

A4

q2

+

B4

p3 27

d

p3 27 +

p3 27

3

-q

B 2

+

q2 B4

+

p3 27

+

3

-q

B 2

-

q2 B4

+

p3 27

(Note that if we had used the negative root in 3, the result would have been the same.) 6. x = 3 7. x = 2 8. x = 2

2 (27 )

p3 27 +

2

-

+

5. x = H + K

- 27q { 2(27q)2 - 4(27)( - p3) 2 2

-q

-q

p3 27

Choose the positive root for now.

3.2 Assess Your Understanding (page 255) 7. - 4 8. F 9. 0 10. T 11. R = f(2) = 8; no 13. R = f(2) = 0; yes 15. R = f( - 3) = 0; yes 1 21. 7; 3 or 1 positive; 2 or 0 negative 23. 6; 2 or 0 positive; 2 or 0 negative 17. R = f( - 4) = 1; no 19. R = f a b = 0; yes 2 25. 3; 2 or 0 positive; 1 negative 27. 4; 2 or 0 positive; 2 or 0 negative 29. 5; 0 positive; 3 or 1 negative 31. 6; 1 positive; 1 negative 1 1 1 1 1 1 3 9 33. {1, { 35. {1, {3 37. {1, {2, { , { 39. {1, {3, {9, { , { , { , { , { 3 4 2 2 3 6 2 2 1 3 1 5 1 2 4 5 10 20 1 5 41. {1, {2, {3, {4, {6, {12, { , { 43. {1, {2, {4, {5, {10, {20, { , { , { , { , { , { , { , { , { , { 2 2 2 2 3 3 3 3 3 3 6 6 1 1 2 45. - 3, - 1, 2; f(x) = (x + 3)(x + 1)(x - 2) 47. ; f(x) = 2 ax b (x + 1) 49. 2, 25, - 25; f(x) = 2(x - 2)(x - 25)(x + 25) 2 2 1 1 51. - 1, , 23, - 23; f(x) = 2(x + 1) ax b (x - 23)(x + 23) 53. 1, multiplicity 2; - 2, - 1; f(x) = (x + 2)(x + 1)(x - 1)2 2 2 6. f(c)

5. Remainder; dividend

55. - 1, 65. e -

1 1 ; f(x) = 4(x + 1) ax + b (x2 + 2) 4 4

1 f 3

67. e

1 , 2, 5 f 2

77. LB = - 3; UB = 3 85. 0.21 87. - 4.04 93.

y 30 (ⴚ2, 4) (ⴚ3, 0) (ⴚ1, 0) (ⴚ4, ⴚ18)

69. LB = - 2; UB = 2

79. f(0) = - 1; f(1) = 10 89. 1.15 91. 2.53 95.

(3, 24)

57. { - 1, 2}

97.

5 x 1 ,0 2

1 , 25, - 25 f 3

63. { - 3, - 2}

75. LB = - 1; UB = 1

83. f(1.4) = - 0.17536; f(1.5) = 1.40625

99.

y 2.5 (ⴚ1, 0)

61. e

73. LB = - 2; UB = 2

81. f( - 5) = - 58; f( - 4) = 2

y 5

(0, ⴚ1)

2 , - 1 + 22, - 1 - 22 f 3

71. LB = - 1; UB = 1

(1, 2) (2, 0) 5 x (0, ⴚ6)

59. e

(1, 0) 2.5 x (0, ⴚ2)

y 1 ⴚ ,0 2

2

1 ,0 2 2.5 x (0, ⴚ2)

AN24

ANSWERS  Section 3.2

101.

103.

y 16 (2, 12)

(1, 0)

兹2 ,0 2

y (0, 2) 4

兹2  ,0 2

105. - 8, - 4, -

7 3

107. k = 5

109. - 7

111. If f(x) = xn - c n, then f(c) = c n - c n = 0, so x - c is a factor of f. 113. 5 115. 7 in. 117. All the potential rational zeros are integers, so (1, 3) (1, 9) (1.5, 1.5625) r either is an integer or is not a rational zero (and is therefore irrational). 119. 0.215 1 2 121. No; by the Rational Zeros Theorem, is not a potential rational zero. 123. No; by the Rational Zeros Theorem, is not a potential rational zero. 3 3 2 3 127. ( - 3, 2) and (5, q ) 124. y = x 125. [3, 8) 126. (0, - 2 23), (0, 2 23), (4, 0) 5 5 (0, 2) (1, 0) 5 x

(2, 0)

(2, 0) 5 x

3.3 Assess Your Understanding (page 262) 4. 3 - 4i

3. one

5. T

7. 4 + i

6. F

9. - i, 1 - i

17. f(x) = x - 14x + 77x - 200x + 208; a = 1 4

3

21. f(x) = x4 - 6x3 + 10x2 - 6x + 9; a = 1 31. 1, -

11. - i, - 2i

19. f(x) = x - 4x

2

5

4

+ 7x

1 25. 2i, - 3, 2

23. - 2i, 4

13. - i 3

15. 2 - i, - 3 + i

- 8x + 6x - 4; a = 1 2

27. 3 + 2i, - 2, 5

29. 4i, - 211, 211, -

2 3

1 23 1 23 1 23 1 23 i, - + i; f(x) = (x - 1) ax + + i b ax + ib 2 2 2 2 2 2 2 2

33. 2, 3 - 2i, 3 + 2i; f(x) = (x - 2)(x - 3 + 2i)(x - 3 - 2i)

35. - i, i, - 2i, 2i; f(x) = (x + i)(x - i)(x + 2i)(x - 2i)

37. - 5i, 5i, - 3, 1; f(x) = (x + 5i)(x - 5i)(x + 3)(x - 1)

1 1 39. - 4, , 2 - 3i, 2 + 3i; f(x) = 3(x + 4) ax - b (x - 2 + 3i)(x - 2 - 3i) 3 3

43. (a) f(x) = (x2 - 22x + 1)(x2 + 22x + 1)

22 22 22 22 22 22 22 22 i, + i, i, + i 2 2 2 2 2 2 2 2

(b) -

41. 130

45. Zeros that are complex numbers must occur in conjugate pairs; or a polynomial with real coefficients of odd degree must have at least one real zero. 47. If the remaining zero were a complex number, its conjugate would also be a zero, creating a polynomial of degree 5. 50. { - 22} 49. y 7 51. 6x3 - 13x2 - 13x + 20 52. A = 9p ft 2 ( ≈28.274 ft 2); C = 6p ft ( ≈18.850 ft) −2

10 x

−5

3.4 Assess Your Understanding (page 272) 5. F

6. horizontal asymptote

7. vertical asymptote

13. All real numbers except 3; 5x  x ≠ 3 6

8. proper

9. T

10. F

11. y = 0

15. All real numbers except 2 and - 4; 5x  x ≠ 2, x ≠ - 4 6

12. T

1 1 and 3; e x  x ≠ - , x ≠ 3 f 19. All real numbers except 2; 5x  x ≠ 2 6 21. All real numbers 2 2 23. All real numbers except - 3 and 3; 5x  x ≠ - 3, x ≠ 3 6 25. (a) Domain: 5x  x ≠ 2 6; Range: 5y  y ≠ 1 6 (b) (0, 0) (c) y = 1 (d) x = 2 (e) None 27. (a) Domain: 5x  x ≠ 0 6; Range: all real numbers (b) ( - 1, 0), (1, 0) (c) None (d) x = 0 (e) y = 2x 29. (a) Domain: 5x  x ≠ - 2, x ≠ 2 6; Range: 5y  y … 0, y 7 1 6 (b) (0, 0) (c) y = 1 (d) x = - 2, x = 2 (e) None 17. All real numbers except -

31.

33.

y 8

35.

y 10

(ⴚ2, 2) yⴝ0

(1, 3) yⴝ2 5 x

(ⴚ1, 1)

37.

y 5

yⴝ0

(0, 1)

(1, ⴚ1)

(2, 1)

xⴝ1

(ⴚ3, ⴚ1)

5

39.

y 2 4 x (ⴚ1, ⴚ1)

y 10

yⴝ0

(2, 3) x ⴝ ⴚ2

x ⴝ ⴚ1

y 3 yⴝ1 5 x (2, 0)

(ⴚ2, 0)

x

5 x

41.

xⴝ3

(4, 3)

yⴝ1 9 x

xⴝ0

xⴝ0

43. Vertical asymptote: x = - 4; horizontal asymptote: y = 3

45. Vertical asymptote: x = 3; oblique asymptote: y = x + 5 1 2 47. Vertical asymptotes: x = 1, x = - 1; horizontal asymptote: y = 0 49. Vertical asymptote: x = - ; horizontal asymptote: y = 3 3 51. Vertical asymptote: none; oblique asymptote: y = 2x - 1 53. Vertical asymptote: x = 0; no horizontal or oblique asymptote 55. (a)

57. (a)

x = −3 y 4 (−2, −1) 4 x y = −2

−6 (−4, −3) −6

(b) Domain: {x x ≠ - 3}; Range: {y y ≠ - 2} (c) Vertical asymptote: x = - 3; Horizontal asymptote: y = - 2

59. (a)

y x=1 6

y x=1 7 (0, 5) (2, 5)

y=3 (0, 1) −5

(2, 1) 5 x

−6

(b) Domain: {x x ≠ 1}; Range: {y y 6 3} (c) Vertical asymptote: x = 1; Horizontal asymptote: y = 3

y=1 6x

−6 −3

(b) Domain: {x x ≠ 1}; Range: {y y 7 1} (c) Vertical asymptote: x = 1; Horizontal asymptote: y = 1

ANSWERS   Section 3.5 61. (a) 9.8208 m/sec2 63. (a)

(b) 9.8195 m/sec2

(c) 9.7936 m/sec2

(e) ∅

(d) h-axis

65. (a) R(x) = 2 +

Rtot

(b)

10

y x=1 10 (2, 7)

5 0

5 1 = 5¢ ≤ + 2 x - 1 x - 1

5 10 15 20 25 R2

(b) Horizontal: Rtot = 10; as the resistance of R2 increases without bound, the total resistance approaches 10 ohms, the resistance R1. (c) R1 ≈ 103.5 ohms 71. x = 5

72. e -

4 r 19

y=2 5 x

y=2 −5 (0, −3) −10

(c) Vertical asymptote: x = 1; Horizontal asymptote: y = 2

74. ( - 3, 11), (2, - 4)

73. x-axis symmetry

3.5 Assess Your Understanding (page 287) 3. in lowest terms 4. vertical 5. True 6. True 7. 1. Domain: { x  x ≠ 0, x ≠ - 4} 2. R is in lowest terms 4. R is in lowest terms; vertical asymptotes: x = 0, x = - 4 4 1 6. Interval

(, 4)

(4, 1)

Number Chosen

5

2

Value of R

R(5)

  45

Location of Graph Below x-axis

(

9. 1. R(x) =

)

3(x + 1) 2(x + 2)

R

2,

1

( ) 1 2

2 7

R(1) 

2 5

Below x-axis

Above x-axis

(

(

( )

)

)

2. R is in lowest terms

4. R is in lowest terms; vertical asymptote: x = - 2 2 1 6. Interval

(, 2)

(2, 1)

Number Chosen

3

2

Value of R

R(3)  3

R 2

1 4 4 x

5. Horizontal asymptote: y =

3 ; x-intercept: - 1 4

3 , not intersected 2 7.

x  2 y 5 (3, 3)

(1, )

3

0,

3

3 2

R(0) 

3 4

Location of Graph Above x-axis

Below x-axis

Above x-axis

Point on Graph

(

(0, )

)

3, 3 2 2

3 4

y

0

( )

y0

1 2  , 2 7

(1, 0)

3. y-intercept:

2 5

1,

4 5,  5

2 1, 5

1 2 2,7

; Domain: 5x  x ≠ - 2 6

(3, 3)

x  4 y 2.5

Above x-axis 1 2, 4

4 5,  5

Point on Graph

1

2

R(2) 

0

(0, )

(1, 0)

1 4

3. no y-intercept; x-intercept: - 1 5. Horizontal asymptote: y = 0, intersected at ( - 1, 0) 7. x0

3 2

5 x (1, 0)

3 3  , 2 2

3 4

3 3 ; Domain: {x x ≠ - 2, x ≠ 2} 2. R is in lowest terms 3. y-intercept: - ; no x-intercept (x + 2)(x - 2) 4 4. R is in lowest terms; vertical asymptotes: x = 2, x = - 2 5. Horizontal asymptote: y = 0, not intersected 2 2 6. 7. x  2 y x  2

11. 1. R(x) =

Interval

(, 2)

(2, 2)

(2, )

Number Chosen

3

0

3

Value of R

R(3)  5

R(0)   34

R(3) 

3

Location of Graph Above x-axis

(3, ) 3 5

Point on Graph

13. 1. P(x) =

Below x-axis

Above x-axis

(3, )

3 4

1x2 + x + 12 1x2 - x + 12 (x + 1)(x - 1)

0, 

3 4

2

3,

3 5 y0

3 5

; Domain: {x x ≠ - 1, x ≠ 1}

4. P is in lowest terms; vertical asymptotes: x = - 1, x = 1 1 1 6. Interval

(, 1)

(1, 1)

(1, )

Number Chosen

2

0

2

Value of P

P(2)  7

P(0)  1

P(2)  7

Location of Graph Above x-axis

Below x-axis

Above x-axis

Point on Graph

(0, 1)

(2, 7)

(2, 7)

3 5

5 x

3 5

(0,  )

3,

2. P is in lowest terms

3. y-intercept: - 1; no x-intercept

5. No horizontal or oblique asymptote 7.

y xⴝ1 (ⴚ2, 7)

6

(2, 7)

(0, ⴚ1)

x ⴝ ⴚ1

5 x

AN25

AN26

ANSWERS  Section 3.5 (x - 1)(x2 + x + 1)

15. 1. H(x) =

(x + 3)(x - 3)

1 3. y-intercept: ; x-intercept: 1 9 1 1 5. Oblique asymptote: y = x, intersected at a , b 9 9 7. (4, 9) y

; Domain: {x x ≠ - 3, x ≠ 3}

2. H is in lowest terms

4. H is in lowest terms; vertical asymptotes: x = 3, x = - 3 6.

3

1

3

0,

Interval

(, 3)

(3, 1)

(1, 3)

(3, )

Number Chosen

4

0

2

4

Value of H

H(4) ⬇ 9.3

H(0)  9

H(2)  1.4

H(4)  9

Above x-axis

Below x-axis

Above x-axis

(0, )

(2, 1.4)

(4, 9)

Location of Graph Below x-axis Point on Graph

(4, 9.3)

1

1 9

1 9

8

yx (1, 0) 10 x (2, 1.4)

(4, 9.3)

x  3 x  3

x2 ; Domain: {x ≠ - 3, x ≠ 2} 2. R is in lowest terms 3. y-intercept: 0; x-intercept: 0 (x + 3)(x - 2) 4. R is in lowest terms; vertical asymptotes: x = 2, x = - 3 5. Horizontal asymptote: y = 1, intersected at (6, 1) 3 0 2 6. 7.

17. 1. R(x) =

Interval Number Chosen Value of R

(, 3)

(3, 0)

(0, 2)

(2, )

6

1

1

3

R(6)  1.5

R(1)   6

R(1)  0.25

R(3)  1.5

1

Location of Graph Above x-axis

Below x-axis

Below x-axis

Above x-axis

Point on Graph

(1,  )

(1, 0.25)

(3, 1.5)

(6, 1.5)

1 6

x ; Domain: {x x ≠ - 2, x ≠ 2} (x + 2)(x - 2) 4. G is in lowest terms; vertical asymptotes: x = - 2, x = 2 2 0 6.

19. 1. G(x) =

(2, 0)

(0, 2)

(2, )

Number Chosen

3

1

1

3

G(3)

Location of Graph Below x-axis Point on Graph

(

)

3

3,  5

1

1 3

10 x (0, 0) (1, 0.25) x  3 x  2

3. y-intercept: 0; x-intercept: 0

1, 3

G(1) 

Above x-axis

Below x-axis

Above x-axis

(1, )

(1,  )

(3, )

G(3)  5

3,

1 3 (0, 0)

3 5

2 5 x

3, 

3 5

1 3

(3, 1.5)

y

G(1)  3 1 3

1 6

2

5. Horizontal asymptote: y = 0, intersected at (0, 0) 7.

2

(, 2)   35

1, 

2. G is in lowest terms

Interval

Value of G

y

(6, 1.5) y1

3 5

y0

1,  x  2 x  2

1 3

3 3 ; Domain: {x x ≠ 1, x ≠ - 2, x ≠ 2} 2. R is in lowest terms 3. y-intercept: ; no x-intercept (x - 1)(x + 2)(x - 2) 4 4. R is in lowest terms; vertical asymptotes: x = - 2, x = 1, x = 2 5. Horizontal asymptote: y = 0, not intersected 2 1 2 6. 7. x  2 y

21. 1. R(x) =

Interval

(, 2)

(2, 1)

Number Chosen

3

0

Value of R

3 R(3)   20

R(0) 

Location of Graph Below x-axis Point on Graph

(3,  ) 3 20

3 4

0,

(1, 2)

(2, )

1.5

3

R(1.5)   24 7

R(3)  10

Above x-axis

Below x-axis

Above x-axis

(0, )

(1.5,  )

(3, )

3 4

24 7

3, 

3

3 4

5

3, 5 x

3 20

x1 x2

3 10

3 10

y0 3 24 , 2 7

(x + 1)(x - 1)

1 ; Domain: {x x ≠ - 2, x ≠ 2} 2. H is in lowest terms 3. y-intercept: ; x-intercepts: - 1, 1 16 (x + 4)(x + 2)(x - 2) 4. H is in lowest terms; vertical asymptotes: x = - 2, x = 2 5. Horizontal asymptote: y = 0, intersected at ( - 1, 0) and (1, 0) 1 2 6. 2 1 7. 1

23. 1. H(x) =

2

(2, 1)

(1, 1)

3

1.5

0

H(3) ⬇ 0.12

H(1.5) ⬇ 0.11 H(0)  16

Location of Graph Above x-axis

Below x-axis

Point on Graph

(1.5, 0.11)

Interval

(, 2)

Number Chosen Value of H

25. 1. F(x) =

(3, 0.12)

(1, 2)

(2, )

1.5

3

H(1.5) ⬇ 0.11

H(3) ⬇ 0.12

Above x-axis

Below x-axis

Above x-axis

(0, 161 )

(1.5, 0.11)

(3, 0.12)

1

(x + 1)(x - 4)

; Domain: {x x ≠ - 2} x + 2 4. F is in lowest terms; vertical asymptote: x = - 2 2 1 6.

2. F is in lowest terms

(, 2)

(2, 1)

(1, 4)

(4, )

Number Chosen

3

1.5

0

5

Value of F

F(3)  14

F(1.5)  5.5

F(0)  2

F(5) ⬇ 0.86

Above x-axis

Below x-axis

Above x-axis

(1.5, 5.5)

(0, 2)

(5, 0.86)

Point on Graph

(3, 14)

0,

16

(3, 0.12)

(3, 0.12) y0 5 x (1, 0) (1.5, 0.11) x  2 x  2

(1, 0) (1.5, 0.11)

3. y-intercept: - 2; x-intercepts: - 1, 4

5. Oblique asymptote: y = x - 5, not intersected 4 7.

Interval

Location of Graph Below x-axis

y 0.3

y

12 (1.5, 5.5) (4, 0)

yx5 (5, 0.86) 10 x (0, 2)

(1, 0)

(3, 14) x  2

AN27

ANSWERS  Section 3.5

27. 1. R(x) =

(x + 4)(x - 3)

; Domain: {x x ≠ 4} x - 4 4. R is in lowest terms; vertical asymptote: x = 4 4 3 6. Interval

(, 4)

Number Chosen

5

Value of R

R(5)   9

8

Location of Graph Below x-axis Point on Graph

29. 1. F(x) =

(5,  ) 8 9

2. R is in lowest terms

5. Oblique asymptote: y = x + 5, not intersected 4 7.

(4, 3)

(3, 4)

(4, )

0

3.5

5

R(0)  3

R(3.5)  7.5

R(5)  18

Above x-axis

Below x-axis

Above x-axis

(0, 3)

(3.5, 7.5)

(5, 18)

(x + 4)(x - 3)

; Domain: {x x ≠ - 2} x + 2 4. F is in lowest terms; vertical asymptote: x = - 2 6. 4 2 Interval

(, 4)

Number Chosen

5

Value of F

F (5)  –3

8

Location of Graph Below x-axis Point on Graph

(5,  ) 8– 3

3. y-intercept: 3; x-intercepts: - 4, 3

2. F is in lowest terms

(2, 3)

(3, )

3

0

4

F (3)  6

F (0)  6

F (4)  –3

3. y-intercept: - 6; x-intercepts: - 4, 3

Above x-axis

Below x-axis

Above x-axis

(3, 6)

(0, 6)

(4, )

4

ⴚ5, ⴚ

(0, 1)

(1, )

Number Chosen

4

1

1– 2

2

Value of R

R(4)  100

R (1)  0.5

R

( ) ⬇ 0.003

R(2)  0.016

Location of Graph Above x-axis

Below x-axis

Above x-axis

Above x-axis

Point on Graph

(1, 0.5)

(

(2, 0.016)

Interval

(ⴚⴥ, ⴚ4)

Number Chosen

ⴚ5

Value of R

R(ⴚ5) ⴝ

1– 3

Location of Graph Above x-axis

35. 1. R(x) =

)

0.003

; Domain: {x x ≠ - 2, x ≠ 3}

7 4. Vertical asymptote: x = - 2; hole at a3, b 5 ⴚ4 ⴚ2 6.

Point on Graph

1– 2

(ⴚ5, ) 1– 3

(3x + 1)(2x - 3) (x - 2)(2x - 3)

x ⴝ ⴚ2

0

4

R(ⴚ3) ⴝ ⴚ1

R(0) ⴝ 2

R(4) ⴝ –3

Below x-axis

Above x-axis

Above x-axis

(ⴚ3, ⴚ1)

(0, 2)

(4, )

Number Chosen

1

0

Value of R

R(1)  23

R(0)   2

Location of Graph Above x-axis Point on Graph

37. 1. R(x) =

(1, ) 2 3

2. In lowest terms, R(x) =

(0, 0)

(1, 0)

0.01 (1, 0) 1.25 x

Enlarged view

y

1 3

(ⴚ4, 0)

7 5 3, 5 (0, 2) yⴝ1 10 x

(ⴚ3, ⴚ1)

4,

x ⴝ ⴚ2

3x + 1 x - 2

4 3

1 1 3. y-intercept: - ; x-intercept: 2 3

7.

2 (2, )

1.7

6

R(1.7) ⬇ 20.3

R(6)  4.75

Below x-axis

Below x-axis

Above x-axis

(0,  )

(1.7, 20.3)

(6, 4.75)

1 2

yⴝ1 10 x

5. Horizontal asymptote: y = 3, not intersected

3 2,

1

y

3. y-intercept: 2; x-intercept: - 4

4– 3

( 2)

)

x + 4 x + 2

4

3 , x ≠ 2f 2

3 4. Vertical asymptote: x = 2; hole at a , - 11 b 2 1 3 6. 3 2

(

See enlarged view at right. 10

x ⴝ ⴚ3

ⴚ5,

ⴚ3

(, )

(0, 0)

7. (3, ⴥ)

Interval

7.

3 (ⴚ2, 3)

1 3 3 , 2

10 x (3, 0) (0, ⴚ6)

5. Horizontal asymptote: y = 1, not intersected

(ⴚ4, ⴚ2)

; Domain: e x x ≠

1 3

2. In lowest terms, R(x) =

yⴝxⴚ1 4 4, 3

10

8 3

y

(3, 0)

(x + 2)(x - 3)

y

4– 3

(, 3)

(x + 4)(x - 3)

x4

(ⴚ3, 6) (ⴚ4, 0)

Interval

33. 1. R(x) =

8 9

5, 

(3, 0) 10 x (3.5, 7.5)

5. Oblique asymptote: y = x - 1, not intersected 3 7.

(4, 2)

1– 2,

(5, 18) yx5

(0, 3)

(4, 0)

31. 1. Domain: {x x ≠ - 3} 2. R is in lowest terms 3. y-intercept: 0; x-intercepts: 0, 1 4. Vertical asymptote: x = - 3 5. Horizontal asymptote: y = 1, not intersected 1 3 0 6.

(4, 100)

y 21

ⴚ1,

2 3

y xⴝ2 8 (6, 4.75)

yⴝ3 ⴚ

10 x 3 , ⴚ11 2

1 ,0 3

0, ⴚ

1 2

(x + 3)(x + 2)

; Domain: {x x ≠ - 3} 2. In lowest terms, R(x) = x + 2 3. y-intercept: 2; x-intercept: - 2 x + 3 4. Vertical asymptote: none; hole at ( - 3, - 1) 5. Oblique asymptote: y = x + 2 intersected at all points except x = - 3 3 2 6. 7. y Interval

(, 3)

(3, 2)

Number Chosen

4

2

Value of R

R(4)  2

Location of Graph Below x-axis Point on Graph

(4, 2)

5

R

0

( ) 5 2

1 2

R(0)  2

Below x-axis

Above x-axis

(

(0, 2)

)

5, 1 2 2

5

(2, ) (ⴚ2, 0) (ⴚ3, ⴚ1) (ⴚ4, ⴚ2)

(0, 2) 5 x 5 1 ⴚ ,ⴚ 2 2

AN28

ANSWERS  Section 3.5

39. 1. H(x) =

- 3(x - 2) (x - 2)(x + 2)

; Domain: {x x ≠ - 2, x ≠ 2}

3 4. Vertical asymptote: x = - 2; hole at ¢ 2, - ≤ 4 2 2 6. (2, 2)

(2, )

Number Chosen

3

0

3 3

H(3)   5

Above x-axis

Below x-axis

Below x-axis

Point on Graph

(3, 3)

(0 ,  )

(3 ,  )

3 2

x - 4 ; Domain: {x x ≠ 1} 2. In lowest terms, F(x) = x - 1 (x - 1)2 4. Vertical asymptote: x = 1 5. Horizontal asymptote: y = 1; not intersected 1 4 6. Interval

(, 1)

(1, 4)

(4, )

Number Chosen

0

2

5

x Q0, − 32R

F(5) 

3. y-intercept: 4; x-intercept: 4

7.

1 4

Value of F

F(0)  4

F(2)  2

Location of Graph

Above x-axis

Below x-axis

Above x-axis

(2, 2)

(5 , )

(0, 4)

Q3, − 35R

3 5

(x - 1)(x - 4)

Point on Graph

x = −2 y Q2, − 34R 10 (−3, 3) y=0 −7

3

H(0)   2

Location of Graph

41. 1. F(x) =

3 3. y-intercept: - ; no x-intercept 2

7.

(, 2)

H(3)  3

-3 x + 2

5. Horizontal asymptote: y = 0; not intersected

Interval

Value of H

2. In lowest terms, H(x) =

y x=1 5 Q5, 14R (0, 4) y=1 −5

5 x (4, 0) (2, −2)

−5

1 4

x

; Domain: {x x ≠ - 2} 2. G is in lowest terms 3. y-intercept: 0; x-intercept: 0 (x + 2)2 4. Vertical asymptote: x = - 2 5. Horizontal asymptote: y = 0; intersected at (0, 0) 2 0 6. 7.

43. 1. G(x) =

Interval

(, 2)

(2, 0)

(0, )

Number Chosen

3

1

1 G(1) 

x = −2

y Q1, 19R

3 −7 y=0 1 9

Value of G

G(3)  3

G(1) 1

Location of Graph

Below x-axis

Below x-axis

Above x-axis

Point on Graph

(3, 3)

(1, 1)

(1 , )

3 x (0, 0) (−1, −1) −7

(−3, −3)

1 9

x2 + 1 ; Domain: {x x ≠ 0} 2. f is in lowest terms 3. no y-intercept; no x-intercepts x 4. f is in lowest terms; vertical asymptote: x = 0 5. Oblique asymptote: y = x, not intersected 0 6. 7.

45. 1. f(x) =

Interval

(, 0)

(0, )

Number Chosen

1

1

Value of f

f(1)  2

Location of Graph Below x-axis Point on Graph

(1, 2)

x0 y 5 (1, 2) yx 5 x (1, 2)

f(1)  2 Above x-axis (1, 2)

(x + 1)(x2 - x + 1) x3 + 1 = ; Domain: {x x ≠ 0} 2. f is in lowest terms 3. no y-intercept; x-intercept: - 1 x x 4. f is in lowest terms; vertical asymptote: x = 0 5. No horizontal or oblique asymptote 1 0 6. 7. y 7

47. 1. f(x) =

Interval Number Chosen Value of f

(, 1)

(1, 0)

2

1 2

f(2) 

7 2

Location of Graph Above x-axis Point on Graph

49. 1. f(x) =

(2, ) 7 2

f

(0, )

2,

5

2

1

( ) 1 2

7 4

f (1)  2

Below x-axis

Above x-axis

(

(1, 2)

)

1, 7 2 4

(1, 0) 1 7  , 2 4

x0

x4 + 1

; Domain: {x x ≠ 0} 2. f is in lowest terms 3. no y-intercept; no x-intercepts x3 4. f is in lowest terms; vertical asymptote: x = 0 5. Oblique asymptote: y = x, not intersected 0 6. 7. Interval

(, 0)

(0, )

Number Chosen

1

1

Value of f

f(1)  2

f(1)  2

Location of Graph Below x-axis Point on Graph

(1, 2)

Above x-axis (1, 2)

(1, 2) 5 x

yx

y 5

(1, 2) 5 x (1, 2) x0

ANSWERS  Section 3.6 53. Minimum value: 1.89 at x = 0.79

51. Minimum value: 2.00 at x = 1.00 5

55. Minimum value: 1.75 at x = 1.32

14

0

10

0

5 0

4

0

0

57. One possibility: R(x) =

x2 - 4

59. One possibility: R(x) =

61. (a) t-axis; C(t) S 0 (b) 0.4

4 b 3

(x + 1)2 (x - 2)2

63. (a) C(x) = 16x +

65. (a) S(x) = 2x2 +

5000 + 100 x

(b)

(b) x 7 0 (c) 10,000

0

5 0

(x - 1)(x - 3) ax2 +

x2

12

40,000 x

10,000

0

0

60 0

(c) 0.71 h after injection

0

300

(c) 2784.95 in.2 (d) 21.54 in. * 21.54 in. * 21.54 in. (e) To minimize the cost of materials needed for construction

0

(d) Approximately 17.7 ft by 56.5 ft (longer side parallel to river) 67. (a) C(r) = 12pr 2 +

4000 r

x2 + x - 12 x - x - 6 2

6000

(b)

69. (a) R(x) = e

7 5

0

10 0

The cost is smallest when r = 3.76 cm.

6x2 - 7x - 3

if x ≠ 3 (b) R(x) = e if x = 3

2x2 - 7x + 6 - 11

4. F

5. (a) 5x 0 0 6 x 6 1 or x 7 2 6; (0, 1) h (2, q )

7. (a) 5x 0 - 1 6 x 6 0 or x 7 1 6; ( - 1, 0) h (1, q ) 9. 5x 0 x 6 0 or 0 6 x 6 3 6; ( - q , 0) h (0, 3)

11. 5x 0 x Ú - 4 6; 3 - 4, q )

31. 5x 0 x 6 - 1 or x 7 1 6; ( - q , - 1) h (1, q )

35. 5x 0 x … - 1 or 0 6 x … 1 6; ( - q , - 1] h (0, 1 4

13. 5x 0 x … - 2 or x Ú 2 6; ( - q , - 2 4 h 3 2, q )

17. 5x 0 - 2 6 x … - 1 6; ( - 2, - 1]

23. 5x 0 - 4 6 x 6 0 or x 7 0 6; ( - 4, 0) h (0, q )

27. 5x 0 - 1 6 x 6 0 or x 7 3 6; ( - 1, 0) h (3, q )

39. 5x 0 x 6 2 6; ( - q , 2)

(b) 5x 0 x … 0 or 1 … x … 2 6; ( - q , 0 4 h [1, 2]

(b) 5x 0 x 6 - 1 or 0 … x 6 1 6; ( - q , - 1) h 3 0, 1)

15. 5x 0 - 4 6 x 6 - 1 or x 7 0 6; ( - 4, - 1) h (0, q ) 21. 5x 0 x 7 4 6; (4, q )

19. 5x 0 x 6 - 2 6; ( - q , - 2)

25. 5x 0 x … 1 or 2 … x … 3 6; ( - q , 1] h 3 2, 3 4

29. 5x 0 x 6 - 1 or x 7 1 6; ( - q , - 1) h (1, q )

33. 5x 0 x 6 - 1 or x 7 1 6; ( - q , - 1) h (1, q )

37. 5x 0 x 6 - 1 or x 7 1 6; ( - q , - 1) h (1, q )

41. 5x 0 - 2 6 x … 9 6; ( - 2, 9 4

43. 5x 0 x 6 2 or 3 6 x 6 5 6; ( - q , 2) h (3, 5)

45. 5x 0 x 6 - 5 or - 4 … x … - 3 or x = 0 or x 7 1 6; ( - q , - 5) h 3 - 4, - 3 4 h 50 6 h (1, q ) 47. e x ` -

1 1 6 x 6 1 or x 7 3 f ; a - , 1 b h (3, q ) 2 2

49. 5x 0 - 1 6 x 6 3 or x 7 5 6; ( - 1, 3) h (5, q )

1 1 55. 5x 0 x 6 2 6 ; ( - q , 2) r; (- q, - 44 h c , q b 53. 5x 0 x 6 3 or x Ú 7 6 ; ( - q , 3) h [7, q ) 2 2 2 3 2 3 57. b x ` x 6 - or 0 6 x 6 r ; a - q , - b h a0, b 59. 5x 0 x … - 3 or 0 … x … 3 6 ; ( - q , - 3 4 h 3 0, 3 4 3 2 3 2

51. b x ` x … - 4 or x Ú

if x ≠

3 2

if x =

3 2

71. No. Each function is a quotient of polynomials, but it is not written in lowest terms Each function is undefined for x = 1; each graph has a hole at x = 1. 77. If there is a common factor between the numerator and the denominator, and the factor yields a real zero, then the graph will have a hole. 1 17 79. e - f 80. 81. ≈ - 0.164 78. 4x3 - 5x2 + 2x - 2 10 2

3.6 Assess Your Understanding (page 295) 3. T

AN29

AN30

ANSWERS  Section 3.6 63. (a)

1 61. (a) - 4, - , 3 2 (b) f(x) = (x + 4)2(2x + 1)(x - 3) (c) (1.87, 60.54)

y 10 (6, 0) y1 0 ,

y

75



1 ,0 2 (0, 48)

15 2

65. (a) 9 10 , 4

(1, 0) 10 x

3 2

x2

(3, 0) 5 x

(4, 0)

4,

y y  x 3 10 28 3, 5 (4, 0) (0, 2) 10 x (1, 0) x  2

(b) 3 - 4, - 2) h 3 - 1, 3) h (3, q )

(b) ( - q , - 6] h [1, 2 ) h (2, q )

(1.74, 185.98)

1 (d) a - , 3 b 2 67. ¢ - 3,

-4 4 ≤ h ¢ , 3≤ 3 3

75.

69. 5x 0 x 7 4 6 ; (4, q ) 77.

y f(x)  x4  1 2.5 (1, 0) 2.5 x

(1, 0)

71. 5x 0 x … - 2 or x Ú 2 6 ; ( - q , - 2 4 h 3 2, q )

y g(x)  3x 2 32 (2, 12)

(2, 12)

2.5 x f (x)  x4  4

g(x)  2x2  2

ƒ(x) … g(x) if - 1 … x … 1

ƒ(x) … g(x) if - 2 … x … 2

73. 5x 0 x 6 - 4 or x Ú 2 6 ; ( - q , - 4) h 32, q )

79. Produce at least 250 bicycles 81. (a) The stretch is less than 39 ft. (b) The ledge should be at least 84 ft above the ground for a 150-lb jumper. 4 83. At least 50 students must attend. 88. J , q ≤ 3 2 2 2 4 89. y = 2x 90. x - x - 4 91. 3x y (x + 2y)(2x - 3y) 3

Review Exercises (page 299) 1. Polynomial of degree 7 y 5.

2. Neither 6.

3. Rational y 4 (1, 0)

15 (2, 0)

(0, 8)

(0, 1)

2 x

(4, 8)

18 7 x

(2, 1)

(4, 0) (5, 15)

20

(1, 2) (2, 3) 3 x

(0, 3)

8. 1. y = x3 2. x-intercepts: - 4, - 2, 0; y-intercept: 0 3. - 4, - 2, 0, multiplicity 1, crosses 4. 2 y 5. (2, 0) (3, 3)

4. Polynomial of degree 0 y 7.

(1, 15) (0, 0) 2 x (1, 3)

9. 1. y = x3 2. x-intercepts: - 4, 2; y-intercept: 16 3. - 4 multiplicity 1, crosses; 2, multiplicity 2, touches 4. 2 y (0, 16) 5. (2, 32) (3, 7) (2, 0) 10 x

(4, 0) (5, 49)

10. 1. y = - 2x3 2. f(x) = - 2x2 (x - 2) x-intercepts: 0, 2; y-intercept: 0 3. 0, multiplicity 2, touches: 2, multiplicity 1, crosses 4. 2 y 5. 20 (1, 2) (2, 0) 6 x (0, 0)

(1, 6)

(3, 18)

12. R = 0; g is a factor of f.

60

11. 1. y = x4 2. x-intercepts: - 3, - 1, 1; y-intercept: 3 3. - 3, - 1, multiplicity 1, crosses; 1, multiplicity 2, touches 4. 3 y 5. (4, 75)

80

(2, 15) (1, 0) 5 x (2, 9) (0, 3) (1, 0)

(3, 0)

13. R = - 5x + 10; g is not a factor of f.. 14. f(4) = 47,105 15. 5, 3, or 1 positive; 3 or 1 negative 1 3 1 1 3 1 1 16. 1 positive; 2 or 0 negative 17. {1, {3, { , { , { , { , { , { , { 18. - 2, 1, 4; f(x) = (x + 2)(x - 1)(x - 4) 2 2 3 4 4 6 12 2 1 1 1 19. , multiplicity 2; - 2; f(x) = 4 ax - b (x + 2) 20. 2, multiplicity 2; f(x) = (x - 2)2 (x2 + 5) 21. 5 - 3, 2 6 22. e - 3, - 1, - , 1 f 2 2 2

ANSWERS  Review Exercises 23. LB: - 2, UB; 3

24. LB: - 5, UB: 5

25. f(0) = - 1; f(2) = 5

26. f(0) = - 1; f(1) = 1

28. 6 + i, 2 - 5i; f(x) = x5 - 19x4 + 162x3 - 838x2 + 2561x - 3219 30. - 2,

1 1 (multiplicity 2); f(x) = 4(x + 2) ax - b 2 2

32. - 3, 2, -

2

27. 2 - 3i; f(x) = x3 - 8x2 + 29x - 52

29. - 2, 1, 4; f(x) = (x + 2)(x - 1)(x - 4)

31. 2 (multiplicity 2), - 25i, 25i; f(x) = (x + 25i)(x - 25i)(x - 2)2

22 22 22 22 i, i; f(x) = 2(x + 3)(x - 2) ax + i b ax ib 2 2 2 2

33. Domain: 5x 0 x 34. Domain: 5x x 35. Domain: 5x x 2(x 36. 1. R(x) =

≠ - 3 6 ; horizontal asymptote: y = 0; vertical asymptotes: x = - 3 ≠ 2, x ≠ - 2 6 ; horizontal asymptote: y = 2; vertical asymptote: x = 2, x = - 2 ≠ - 2}; horizontal asymptote: y = 1; vertical asymptote: x = - 2 - 3) ; Domain: {x x ≠ 0} 2. R is in lowest terms 3. no y-intercept; x-intercept: 3 x 4. R is in lowest terms; vertical asymptote: x = 0 5. Horizontal asymptote: y = 2; not intersected 0 3 6. 7. Interval

(, 0)

(0, 3)

(3, )

Number Chosen

2

1

4

R(2)  5

R(1)  4

R(4)

Below x-axis

Above x-axis

Point on Graph

(1, 4)

(4, )

(2, 5)

10 (2, 5)

3. no y-intercept; x-intercept: - 2 5. Horizontal asymptote: y = 0; intersected at ( - 2, 0) x0 2 7. y 1,

Interval

(, 2)

(2, 0)

(0, 2)

(2, )

Number Chosen

3

1

1

3

Value of H

H(3)   –– 15

H(1)  1–3

H(1)  3

H(3)  5–3

Above x-axis

Below x-axis

Above x-axis

(1, )

(1, 3)

(3, )

1

Location of Graph Below x-axis Point on Graph

(3,  ) 1 –– 15

1– 3

1 2

x0

1– 2

37. 1. Domain: {x x ≠ 0, x ≠ 2} 2. H is in lowest terms 4. H is in lowest terms; vertical asymptotes: x = 0, x = 2 2 0 6.

4,

y2 (3, 0) x (1, 4)

10

1  –2

Location of Graph Above x-axis

Value of R

y

1 3

5

3,

5 3

(2, 0) 5 x 1 3,  15

y0

(1, 3) x2

5– 3

(x + 3)(x - 2)

; Domain: {x x ≠ - 2, x ≠ 3} 2. R is in lowest terms 3. y-intercept: 1; x-intercepts: - 3, 2 (x - 3)(x + 2) 4. R is in lowest terms; vertical asymptotes: x = - 2, x = 3 5. Horizontal asymptote: y = 1; intersected at (0, 1) 3 2 2 3 6. 7. y (0, 1)

38. 1. R(x) =

(3, 2)

Interval

(, 3)

Number Chosen

4

Value of R

R(4)  –– 7

5

5– 2

R(0)  1

R

Below x-axis

Above x-axis

Below x-axis

Above x-axis

(

(0, 1)

(

(4, )

5

3 –– 7

4,

0

( )

(4, )

(3, )

(2, 3)

9 R  –2 –– 11

 –2 3

Location of Graph Above x-axis Point on Graph

(2, 2)

5– 2,

9

)

–– 11

4

( )  5– 2

5– 2,

11 –– 9

)

11

 –– 9

4,

5

3 7

y1 5 x (2, 0)

(3, 0)

7

R (4)  –– 3 

7 –– 3

7 3

5 9 5 11 , , 2 11 2 9 x  2 x  3

x3 ; Domain: {x x ≠ - 2, x ≠ 2} 2. F is in lowest terms 3. y-intercept: 0; x-intercept: 0 (x + 2)(x - 2) 4. F is in lowest terms; vertical asymptotes: x = - 2, x = 2 5. Oblique asymptote: y = x; intersected at (0, 0) 2 6. 0 7. 2 x  2

39. 1. F(x) =

Interval

(, 2)

(2, 0)

(0, 2)

(2, )

Number Chosen

3

1

1

3

Value of F

F(3)  –– 5

F(1 )  1–3

F(1)  1–3

–– F(3)  27 5

Above x-axis

Below x-axis

Above x-axis

27

Location of Graph Below x-axis Point on Graph

(3,  ) 27 –– 5

(1, )

(1 ,  )

1– 3

1– 3

40. 1. Domain: {x x ≠ 1} 2. R is in lowest terms 4. R is in lowest terms; vertical asymptote: x = 1 0 1 6. (, 0)

(0, 1)

(1, )

2

1– 2

2

Value of R

R(2)

Location of Graph Above x-axis Point on Graph

(2, ) 32 –– 9

( )

R

1– 2

R(2)  32

Above x-axis

Above x-axis

(

(2, 32)

1– 2

,

1– 2

)

yx

4 10 x 1 3

1, 

(3, ) 27 – 5

27 3,  5

7.

Number Chosen

1– 2

1 3 (0, 0)

1,

27 5

3,

x2

3. y-intercept: 0; x-intercept: 0 5. No oblique or horizontal asymptote

Interval

32  –– 9

y

y 40 (2, 32) 1 1 , 2 2

32 2, 9 (0, 0) x1

5 x

AN31

AN32

ANSWERS  Review Exercises (x + 2)(x - 2)

41. 1. G(x) =

(x + 1)(x - 2)

; Domain: {x x ≠ - 1, x ≠ 2}

4 4. Vertical asymptote: x = - 1; hole at a2, b 3 1 2 6. Interval

(, 2)

Number Chosen

3

Value of G

G(3) 

(2, 1) 3

Location of Graph Above x-axis

(3, ) 1– 2

Point on Graph

3 G (3)  –4

Below x-axis

Above x-axis

Above x-axis

(

(0, 2)

(3, )

3– 2 , 1

(b) ( - 3, 2) ∪ (2, q )

43. (a) y = 0.25

(b) x = - 2, x = 2

)

5– 4

(c) ( - 3, - 2) ∪ ( - 1, 2)

44. {x x 6 - 2 or - 1 6 x 6 2}; ( - q , - 2) ∪ ( - 1, 2)

3

2

2

(d) f(x) = (x - 2)2(x + 3) (d) ( - q , - 3] ∪ ( - 2, - 1] ∪ (2, q )

3

(b) 223.22 cm2 (d) 1000

4x2 - 16

2

1

48. {x x 6 - 4 or 2 6 x 6 4 or x 7 6}; ( - q , - 4) h (2, 4) ∪ (6, q )

3 4

49. (a) A(r) = 2pr 2 +

x2 + 4x + 3

(e) R(x) =

46. {x x 6 1 or x 7 2}; ( - q , 1) ∪ (2, q )

45. {x - 3 6 x … 3}; ( - 3, 3]

47. {x 1 … x … 2 or x 7 3}; [1, 2] ∪ (3, q ) 1

4 5 2, 3 (0, 2) y1 5 x (2, 0) 5 3, 3 4  , 1 2 x  1

5

(c) ( - q , - 3] ∪ 52 6

y 1 2

3,

G (0)  2

42. (a) 5 - 3, 2 6

2 1

7. (2, )

0 3

3. y-intercept: 2; x-intercept: - 2

2

G  –2  1

( )

x + 2 x + 1

5. Horizontal asymptote: y = 1, not intersected

(1, 2)

 –2 1– 2

2. In lowest terms, G(x) =

50. (a)

500 r (c) 257.08 cm2

2

4 6

(c)

17

−1

17

−1

23 10

0

23 10

Data appear to follow a cubic relation. (b) p(t) = 0.0029t 3 - 0.0723t 2 + 0.2990t + 14.0082; 17.9%

8 0

A is smallest when r ≈ 3.41 cm. 53. (a) Even

(b) Positive

(d) The graph touches the x-axis at x = 0 but does not cross it there.

(c) Even

(e) 8

Chapter Test (page 302) 1.

(4, 1) 8 x (3, 2) (2, 1)

p

1 3 5 15 : { , {1, { , { , {3, {5, { , {15 q 2 2 2 2 1 (c) - 5, - , 3; g(x) = (x + 5)(2x + 1)(x - 3) 2 1 (d) y-intercept: - 15; x-intercepts: - 5, - , 3 2 1 (e) Crosses at - 5, - , 3 (f) y = 2x3 2

2. (a) 3

y 7

(b)

(g) (2, 45) y (3, 60) 60

1  ,0 2

(3, 0) 5 x

(5, 0)

(2, 35) (0, 15) (1, 36)

5 - 261 5 + 261 , r 5. Domain: {x x ≠ - 10, x ≠ 4}; asymptotes: x = - 10, y = 2 6 6 6. Domain: {x x ≠ - 1}; asymptotes: x = - 1, y = x + 1 yx1 7. y 3. 4, - 5i, 5i

4. b 1,

5 (3, 0)

(1, 0) 5 x (0, 3)

x  1

8. Answers may vary. One possibility is f(x) = x4 - 4x3 - 2x2 + 20x. 9. Answers may vary. One possibility is r(x) =

2(x - 9)(x - 1) (x - 4)(x - 9)

.

10. f(0) = 8; f(4) = - 36 Since f(0) = 8 7 0 and f(4) = - 36 6 0, the Intermediate Value Theorem guarantees that there is at least one real zero between 0 and 4. 11. 5x x 6 3 or x 7 8}; ( - q , 3) ∪ (8, q )

AN33

ANSWERS  Section 4.1

Cumulative Review (page 302) 1. 226

4. f(x) = - 3x + 1

3. {x  - 1 6 x 6 4}; ( - 1, 4)

2. { x  x … 0 or x Ú 1}; ( - q , 0] or [1, q )

y 1

4

5

(1, 4)

1

0

5 x

5. y = 2x - 1 3 3 7. Not a function; 3 has two images. 8. {0, 2, 4} 9. e x ` x Ú f ; J , q b 0 1 2 3 2 2 11. x-intercepts: - 3, 0, 3; y-intercept: 0; symmetric with respect to the origin 10. Center: ( - 2, 1); radius: 3 2 17 y 12. y = - x + 3 3 (2, 4) 5 13. Not a function; it fails the vertical-line test. (5, 1) (1, 1) 2 x (2, 1) 14. (a) 22 (b) x2 - 5x - 2 (c) - x2 - 5x + 2 (d) 9x2 + 15x - 2 (2, 2)

15. (a) {x  x ≠ 1}

(b) No; (2, 7) is on the graph.

y 6

6.

(3, 5) 5 x

y 10

(1, 1)

(2, 8) (1, 1) 10 x

(2, 8)

(e) 2x + h + 5

(c) 4; (3, 4) is on the graph.

7 7 d) ; a , 9 b is on the graph. 4 4

(e) Rational 16.

17.

y (0, 7)

y

8

6 7 ,0 3

(0, 1)

兹2 1 ,0 2 3 x (1, 1)

兹2 1 ,0 2 x1

8 x

21. (a) Domain: {x  x 7 - 3} or ( - 3, q ) 1 (b) x-intercept: - ; y-intercept: 1 2 (c) y 1  ,0 2

5

18. 6; y = 6x - 1 19. (a) x-intercepts: - 5, - 1, 5; y-intercept: - 3 (b) No symmetry (c) Neither (d) Increasing: ( - q , - 3) and (2, q ); decreasing: ( - 3, 2) (e) Local maximum is 5 and occurs at x = - 3. (e) Local minimum is - 6 and occurs at x = 2. 20. Odd

22.

23. (a) (f + g)(x) = x2 - 9x - 6; domain: all real numbers

y 8

(b) a

(1, 5) (2, 2)

(0, 2) 2 x

(2, 5)

g

b (x) =

24. (a) R(x) = -

(0, 1)

(3, 5)

f

(b) $14,000

5 x (2, 2)

x2 - 5x + 1 7 ; domain: e x ` x ≠ - f - 4x - 7 4

1 2 x + 150x 10 (c) 750; $56,250

(d) $75

(d) Range: {y  y 6 5} or ( - q , 5)

CHAPTER 4 Exponential and Logarithmic Functions 4.1 Assess Your Understanding (page 311) 4. composite function; f 1g 1x2 2

5. False 6. False 7. (a) - 1 (b) - 1 (c) 8 (d) 0 (e) 8 (f) - 7 9. (a) 4 (b) 5 (c) - 1 (d) - 2 11. (a) 98 163 3 1 1 1 (b) 49 (c) 4 (d) 4 13. (a) 97 (b) (c) 1 (d) 15. (a) 2 22 (b) 2 22 (c) 1 (d) 0 17. (a) (b) (c) 1 (d) 2 2 17 5 2 3 6 3 19. (a) (b) 1 (c) (d) 0 21. {x x ≠ 0, x ≠ 2} 23. {x x ≠ - 4, x ≠ 0} 25. e x ` x Ú - f 27. {x x Ú 1} 29. {x x Ú 0, x ≠ 4} 3 5 2 2 4 + 1 31. (a) 1f ∘ g2 1x2 = 6x + 3; all real numbers (b) 1g ∘ f2 1x2 = 6x + 9; all real numbers (c) 1f ∘ f2 1x2 = 4x + 9; all real numbers (d) 1g ∘ g2 1x2 = 9x; all real numbers

37. (a) 1f ∘ g2 1x2 =

3x ; {x x ≠ 0, x ≠ 2} 2 - x

(d) 1g ∘ g2 1x2 = x; {x x ≠ 0} (c) 1f ∘ f2 1x2 = x; {x x ≠ 1}

33. (a) 1f ∘ g2 1x2 = 3x2 + 1; all real numbers (b) 1g ∘ f2 1x2 = 9x2 + 6x + 1; all real numbers (c) 1f ∘ f2 1x2 = 9x + 4; all real numbers (d) 1g ∘ g2 1x2 = x4; all real numbers (b) 1g ∘ f2 1x2 =

39. (a) 1f ∘ g2 1x2 =

2 1x - 12

4 ; {x x ≠ - 4, x ≠ 0} 4 + x

(d) 1g ∘ g2 1x2 = x; {x x ≠ 0}

(b) 1g ∘ f2 1x2 = 2 2x + 3; {x x Ú 0}

3

; {x x ≠ 1}

35. (a) (f ∘ g2(x2 = x4 + 8x2 + 16; all real numbers (b) (g ∘ f2(x2 = x4 + 4; all real numbers (c) 1f ∘ f2 1x2 = x4; all real numbers (d) 1g ∘ g2 1x2 = x4 + 8x2 + 20; all real numbers

(c) 1f ∘ f2 1x2 =

3 1x - 12

; {x x ≠ 1, x ≠ 4} 4 - x - 4 1x - 12 (b) 1g ∘ f2 1x2 = ; {x x ≠ 0, x ≠ 1} x

3 41. (a) 1f ∘ g2 1x2 = 22x + 3; e x ` x Ú - f 2

4 (c) 1f ∘ f2 1x2 = 2 x; {x x Ú 0}

(d) 1g ∘ g2 1x2 = 4x + 9; all real numbers

AN34

ANSWERS  Section 4.1

43. (a) 1f ∘ g2 1x2 = x; {x x Ú 1}

(b) 1g ∘ f2 1x2 =  x  ; all real numbers

(d) 1g ∘ g2 1x2 = 2 1x - 1 - 1; {x x Ú 2}

(c) 1f ∘ f2 1x2 = x4 + 2x2 + 2; all real numbers

4x - 17 1 ; e x ` x ≠ 3; x ≠ f 2x - 1 2 2x + 5 (c) 1f ∘ f2 1x2 = ; {x x ≠ - 1; x ≠ 2} x - 2

45. (a) 1f ∘ g2 1x2 = -

(b) 1g ∘ f2 1x2 = -

3x - 3 ; {x x ≠ - 4; x ≠ - 1} 2x + 8

(d) 1g ∘ g2 1x2 = -

3x - 4 11 ; ex ` x ≠ ; x ≠ 3f 2x - 11 2

1 1 1 47. 1f ∘ g2 1x2 = f 1g 1x2 2 = ƒ a x b = 2 a x b = x; 1g ∘ f2 1x2 = g 1f 1x2 2 = g 12x2 = 12x2 = x 2 2 2

3 3 3 3 49. 1f ∘ g2 1x2 = f 1g 1x2 2 = f 1 1 x2 = 1 1 x2 3 = x; 1g ∘ f2 1x2 = g 1f 1x2 2 = g 1x3 2 = 2 x = x 1 1 51. 1f ∘ g2 1x2 = f 1g 1x2 2 = f a 1x + 62 b = 2 c 1x + 62 d - 6 = x + 6 - 6 = x; 2 2

1g ∘ f2 1x2 = g 1f 1x2 2 = g 12x - 62 =

1 1 12x - 6 + 62 = 12x2 = x 2 2 1 1 1 53. 1f ∘ g2 1x2 = f 1g 1x2 2 = ƒ a 1x - b2 b = a c 1x - b2 d + b = x; 1g ∘ f2 1x2 = g 1f 1x2 2 = g 1ax + b2 = 1ax + b - b2 = x a a a

55. f 1x2 = x4; g 1x2 = 2x + 3 (Other answers are possible.)

57. f 1x2 = 2x; g 1x2 = x2 + 1 (Other answers are possible.)

59. f 1x 2 =  x ; g 1x 2 = 2x + 1 (Other answers are possible.) 65. (a) 1f ∘ g2(x2 = acx + ad + b

61. 1f ∘ g2 1x2 = 11; 1g ∘ f2 1x2 = 2

63. - 3, 3

(b) 1g ∘ f2 1x2 = acx + bc + d (c) The domains of both f ∘ g and g ∘ f are all real numbers. 16 6 pt 69. C1t2 = 15,000 + 800,000t - 40,000t 2 (d) f ∘ g = g ∘ f when ad + b = bc + d 67. S 1t2 = 9 2 2100 - p 71. C1p2 = + 600, 0 … p … 100 73. V1r2 = 2pr 3 25 75. (a) f 1x2 = 0.7643x (b) g 1x2 = 128.4594x (c) g 1f 1x2 2 = g 10.7643x2 = 98.18151942x (d) g 1f 110002 2 = 98,181.51942; On April 18, 2013, one thousand dollars could purchase 98,181.51942 yen. 77. (a) f 1p2 = p - 200 (b) g 1p2 = 0.8p (c) 1f ∘ g2 1p2 = 0.8p - 200; 1g ∘ f2 1p2 = 0.8p - 160; The 20% discount followed by the $200 rebate is the better deal. 79. - 2 22, - 23, 0, 23, 2 22 81. f and g are odd functions, so f 1 - x2 = - f 1x2 and g 1 - x2 = - g 1x2. Then 1f ∘ g2 1 - x2 = f 1g 1 - x2 2 = f 1 - g 1x2 2 = - f 1g 1x2 2 = - 1f ∘ g2 1x2. So f ∘ g is also odd. 83. 15 84. (f + g)(x) = 4x + 3; Domain: all real numbers (f - g)(x) = 2x + 13; Domain: all real numbers

(f # g)(x) = 3x2 - 7x - 40; Domain: all real numbers 3x + 8 ; Domain: {x x ≠ 5} ¢ ≤(x) = g x - 5

86.

87. Domain: {x x ≠ 3} Vertical asymptote: x = 3 Oblique asymptote: y = x + 9

10

−3

3

f

−10

1 85. , 4 4

Local minimum: - 5.08 at x = - 1.15 Local maximum: 1.08 at x = 1.15 Decreasing: ( - 3, - 1.15); (1.15, 3) Increasing: ( - 1.15, 1.15)

4.2 Assess Your Understanding (Page 322) 5. ƒ(x1) ≠ ƒ(x2) 6. one-to-one 7. 3 8. y = x 9. [4, q 2 10. True 11. one-to-one 13. not one-to-one 15. not one-to-one 17. one-to-one 19. one-to-one 21. not one-to-one 23. one-to-one 25.

Annual Rainfall (inches) 49.7 43.8 4.2 61.9 12.8

27. Location Atlanta, Georgia Boston, Massachusetts Las Vegas, Nevada Miami, Florida Los Angeles, California

Monthly Cost of Life Insurance

Age

$10.55 $12.89 $19.29

30 40 45

Domain: {$10.55, $12.89, $19.29} Range: 530, 40, 45 6

Domain: {49.7, 43.8, 4.2, 61.9, 12.8} Range: {Atlanta, Boston, Las Vegas, Miami, Los Angeles} 29. 5 15, - 32, 19, - 22, 12, - 12, 111, 02, 1 - 5, 12 6 Domain: 55, 9, 2, 11, - 5 6 Range: 5 - 3, - 2, - 1, 0, 1 6

31. 5 11, - 2 2, 12, - 32, 10, - 102, 19, 12, 14, 22 6 Domain: 51, 2, 0, 9, 4 6 Range: 5 - 2, - 3, - 10, 1, 2 6

ANSWERS   Section 4.2 1 1 33. f 1g 1x2 2 = f a 1x - 42 b = 3 c 1x - 42 d + 4 = 1x - 42 + 4 = x 3 3 1 1 g 1f 1x2 2 = g 13x + 42 = 3 13x + 42 - 4 4 = 13x2 = x 3 3

35. f 1g 1x 2 2 = f a

37. f 1g 1x2 2 = f 1 2x + 82 = 1 2x + 82 3 - 8 = 1x + 82 - 8 = x

1 39. f 1g 1x2 2 = f a b = x

3

x x + 2 b = 4 c + 2 d - 8 = 1x + 82 - 8 = x 4 4

3 3 3 g 1f 1x2 2 = g 1x3 - 82 = 2 1x3 - 82 + 8 = 2 x = x

1 = x; x ≠ 0 1 a b x

1 g 1f 1x 2 2 = g a b = x

4x - 3 41. f 1g 1x2 2 = ƒ a b = 2 - x

2x + 3 g 1f 1x2 2 = g a b = x + 4 43.

4x - 3 b + 3 2 14x - 32 + 3 12 - x2 2 - x 5x = = x, x ≠ 2 = 4x - 3 4x - 3 + 4 12 - x2 5 + 4 2 - x

2x + 3 b - 3 4 12x + 32 - 3 1x + 42 x + 4 5x = = x, x ≠ - 4 = 2x + 3 2 1x + 42 - 12x + 32 5 2 x + 4

4a

y 2.5 (1, 2)

(2, 1) f 1 2.5 x

(2, 2)

f

47.

yx

y 2.5

yx

2.5 x

(1, 1)

2.5 x

f 1

(1, 0)

1

1 = x, x ≠ 0 1 a b x

2a

45.

yx

y 2.5

4x - 8 + 2 = 1x - 22 + 2 = x 4

g 1f 1x2 2 = g 14x - 82 =

3

AN35

f 1

(0, 1)

49. f - 1 1x2 =

51. f -1 1x2 =

1 x 3

1 1 f 1f - 1 1x2 2 = ƒ a x b = 3 a x b = x 3 3 1 -1 -1 f 1f 1x2 2 = f 13x2 = 13x2 = x 3

3 f 1f -1 1x2 2 = f 1 2 x + 12

f 1f - 1 1x2 2 = f a

x 1 x 1 - b = 4a - b + 2 4 2 4 2 = 1x - 22 + 2 = x 4x + 2 1 -1 f 1f 1x2 2 = f - 1 14x + 22 = 4 2 1 1 = x = ax + b 2 2

yx

y 5

3 53. f -1 1x2 = 2 x + 1

x 1 4 2

f(x)  3x 5 x

f -1 1f 1x2 2 = f -1 1x3 - 12

3 = 2 1x3 - 12 + 1 = x

f

1

3

y 5

yx

(x)  兹x  1

yx

y 5

1 f 1(x)  x 3

3 = 12 x + 12 3 - 1 = x

5 x f (x)  x 3  1

f (x)  4x  2 5 x f 1(x) 

55. f - 1 1x) = 2x - 4, x Ú 4 f 1f

-1

1x)) = f 1 2x - 4) = 1 2x - 4) + 4 = x 2

f - 1 1f 1x)) = f - 1 1x2 + 4) = 2 1x2 + 4) - 4 = 2x2 = x, x Ú 0 y 8

f(x)  x 2  4, xⱖ0

f

57. f - 1 1x2 = ƒ 1f

-1

4 x

y 5

4 1x2 2 = ƒ a b = x

4

= x

4 a b x

4 f - 1 1 ƒ 1x2 2 = f - 1 a b = x

yx 1

1 x  2 4

(x)  兹x  4

4 4 a b x

8 x

2x + 1 x 2x + 1 1 x f 1f -1 1x2 2 = f a b = = = x x 2x + 1 12x + 12 - 2x - 2 x

59. f

f

-1

-1

1x2 =

1f 1x2 2 = f

-1

1 a b = x - 2

2a

1 b + 1 2 + 1x - 22 x - 2 = = x 1 1 x - 2

f 1(x) 

2x  1 x

x0 y x2 yx 5 y2 y0

5 x

f(x) 

1 x2

= x

yx 4 f(x)  f 1(x)  x 5 x

AN36

ANSWERS  Section 4.2

2 - 3x x 2 - 3x f 1f -1 1x2 2 = f a b = x

61. f -1 1x2 =

f

65. f

-1

-1

1f 1x2 2 = f

1x2 =

f 1f

-1

-1

63. f 2 2x 2x = = x = 2 - 3x 3x + 2 - 3x 2 3 + x

2 a b = 3 + x

- 2x x - 3

1x2 =

2 - 3a

2 b 2 13 + x2 - 3 # 2 3 + x 2x = = = x 2 2 2 3 + x

=

- 2x b x - 3 - 2x + 2 x - 3 3a

- 2x f 1f -1 1x2 2 = f a b = x - 3 3 1 - 2x2

- 2x + 2 1x - 32

x 1x2 2 = f a b = 3x - 2

2a 3a

x b 3x - 2

=

x b - 1 3x - 2

- 2 13x2

3x - 3 1x + 22

- 6x = x -6

=

- 2a

3x b = f -1 1f 1x2 2 = f -1 a x + 2

x 3x - 2

=

-1

3x b x + 2

3x - 3 x + 2 =

- 6x = x -6

2x 2x = = x 3x - 13x - 22 2

2x f -1 1f 1x2 2 = f -1 a b = 3x - 1

2x 3x - 1

2x b - 2 3x - 1 2x 2x = = = x 6x - 2 13x - 12 2

67. f

-1

1x2 =

3x + 4 2x - 3

69. f

3x + 4 f 1f -1 1x2 2 = f a b = 2x - 3

=

f

-1

-1

1x2 =

3a

3x + 4 b + 4 2x - 3

2a

3x + 4 b - 3 2x - 3

3 13x + 42 + 4 12x - 32

=

2 13x + 42 - 3 12x - 32

1f 1x2 2 = f

=

71. f

3a

-1

3x + 4 b = a 2x - 3

3a 2a

17x = x 17

=

f

17x = x 17

2 21 - 2x

4 - 4 4 - 4 11 - 2x2 1 2x f 1f -1 1x2 2 = f a b = = 4 2#4 21 - 2x 2# 1 - 2x 2

8x = = x 8 f

-1

1f 1x2 2 = f

-1

a

x2 - 4 2x2

b =

2 B

1 - 2a

= 2x2 = x, since x 7 0

x2 - 4 2x2

- 2x + 3 x - 2

=

3x + 4 b - 3 2x - 3

2 13x + 42 - 3 12x - 32

1x2 =

- 2x + 3 f 1f -1 1x2 2 = f a b = x - 2

3x + 4 b + 4 2x - 3

3 13x + 42 + 4 12x - 32

-1

b

=

2 4 B x2

-1

2a

- 2x + 3 b + 3 x - 2 - 2x + 3 + 2 x - 2

2 1 - 2x + 32 + 3 1x - 22 - 2x + 3 + 2 1x - 22

=

-x = x -1

2x + 3 b + 3 x + 2 1f 1x2 2 = f 2x + 3 - 2 x + 2 - 2 12x + 32 + 3 1x + 22 -x = = = x 2x + 3 - 2 1x + 22 -1 -1

2x + 3 b = a x + 2

- 2a

73. (a) 0 (b) 2 (c) 0 (d) 1 75. 7 77. Domain of f -1: [ - 2, q 2; range of f -1: [5, q 2 79. Domain of g -1: [0, q 2; range of g -1: ( - q , 0] 81. Increasing on the interval 1f 102, f 152 2 1 83. f -1 1x2 = 1x - b2, m ≠ 0 85. Quadrant I m

87. Possible answer: f 1x2 =  x , x Ú 0, is one-to-one; f -1 1x2 = x, x Ú 0 d + 90.39 89. (a) r 1d2 = 6.97 6.97r - 90.39 + 90.39 6.97r (b) r 1d 1r2 2 = = = r 6.97 6.97 d 1r 1d2 2 = 6.97 a (c) 56 miles per hour

d + 90.39 b - 90.39 = d + 90.39 - 90.39 = d 6.97

AN37

ANSWERS   Section 4.3

93. (a) 5g 36,250 … g … 87,850 6 (b) 5T  4991.25 … T … 17,891.25 6 T - 4991.25 (c) g 1T2 = + 36,250; 0.25 Domain: 5T  4991.25 … T … 17,891.25 6; Range: 5g 36,250 … g … 87,850 6 95. (a) t represents time, so t Ú 0. 100 - H H - 100 = (c) 2.02 seconds (b) t 1H2 = A 4.9 A - 4.9 - dx + b 97. f -1 1x2 = ; f = f -1 if a = - d cx - a

91. (a) 77.6 kg W + 88 W - 50 + 60 = (b) h 1W2 = 2.3 2.3 50 + 2.3 1h - 602 + 88 2.3h (c) h 1W1h2 2 = = = h 2.3 2.3 W1h 1W2 2 = 50 + 2.3 a

W + 88 - 60 b 2.3

= 50 + W + 88 - 138 = W (d) 73 inches 101. No 105. 6xh + 3h2 - 7h y 5

106.

- 5 - 213 - 5 + 213 , ; 6 6

107. Zeros:

- 5 - 213 - 5 + 213 , 6 6

x-intercepts:

−4

3 x

3 3 108. Domain: e x ` x ≠ - , x ≠ 2 f ; Vertical asymptote: x = - ; 2 2 Horizontal asymptote: y = 3

−5

4.3 Assess Your Understanding (page 337) 1 10. True 11. a - 1, b ; (0, 1); (1, a) 12. 1 a 15. (a) 11.212 (b) 11.587 (c) 11.664 (d) 11.665 17. (a) 8.815 (b) 8.821 (c) 8.824 (d) 8.825 19. (a) 21.217 (b) 22.217 (c) 22.440 (d) 22.459 21. 3.320 23. 0.427 25. Neither 27. Exponential; H1x2 = 4x 3 x 29. Linear; H1x2 = 2x + 4 31. Exponential; f 1x2 = - a b 33. B 35. D 37. A 39. E 2 6. exponential function; growth factor; initial value 7. a

41.

43.

y 9 1,

3 2

y 5

45. (2, 3)

(1, 3) (0, 2)

0,

2.5 x

49.

(2, 6)

Domain: All real numbers Range: 5y y 7 2 6 or 12, q 2 Horizontal asymptote: y = 2 57. y5

(2, e2)

7 3 (0, 3)

(2, 5)

Domain: All real numbers Range: 5y y 7 2 6 or 12, q 2 Horizontal asymptote: y = 2

(0, 4)

(2, e 2) (1, e)

y 8

Domain: All real numbers Range: 5y y 6 2 6 or 1 - q , 22 Horizontal asymptote: y = 2

(1, e) (0, 1) 5 x

Domain: 1 - q , q 2 Range: [1, q 2 Intercept: (0, 1)

2, 

1, 

1 e2

y 3

1 e (0, 1)

(0, 1) 2.5 x

y  2

1, 

1 e

y0 5 x 1 2,  2 e

Domain: 1 - q , q 2 Range: [ - 1, 02 Intercept: 10, - 12

y0

5 3

Domain: All real numbers Range: 5y y 7 - 2 6 or 1 - 2, q 2 Horizontal asymptote: y = - 2 55. (0, e2)

1 1, e

y 8

(1, e) 5 x

1 3, e

Domain: All real numbers Range: 5y y 7 0 6 or 10, q 2 Horizontal asymptote: y = 0 63. 5 - 4 6

95. (a) 60; 1 - 6, 602

(2, 1) y05 x

Domain: All real numbers Range: 5y y 7 0 6 or 10, q 2 Horizontal asymptote: y = 0

3 67. e f 69. 5 - 22 , 0, 22 6 2 75. 5 - 4, 2 6 77. 5 - 4 6 79. 51, 2 6

65. 52 6

87. f 1x2 = - 6 x

91. (a) 16; (4, 16) (b) - 4; a - 4,

99. (2, e 2)

y0

71. 56 6 73. 5 - 1, 7 6 1 1 81. 83. 85. f 1x2 = 3x 49 4

y2

5 x

97.

(1, 1)

y 8

(1, e) (0, 1) y0

61. 53 6

y 5

5 x

Domain: All real numbers Range: 5y y 6 5 6 or 1 - q , 52 Horizontal asymptote: y = 5

3 2

Domain: All real numbers Range: 5y y 7 0 6 or 10, q 2 Horizontal asymptote: y = 0 53.

(0, 1)

1,

y 10

14. False

1, 

y2 2.5 x

59.

y 8

1 3

y 8 2,

(1, 3) y2 6 x

(0, 3)

13. 4

47.

y 7

2.5 x

Domain: All real numbers Range: 5y y 7 0 6 or 10, q 2 Horizontal asymptote: y = 0 51.

y 8 33 1, 16

(1, 6)

(1, 1) y0 2.5 x

y1

Domain: All real numbers Range: 5y y 7 1 6 or 11, q 2 Horizontal asymptote: y = 1

8. True 9. False

89. f 1x2 = 3x + 2

1 9 9 b 93. (a) ; a - 1, b 16 4 4 (b) - 4; 1 - 4, 122 (c) - 2

(b) 3; (3, 66)

101. (a) 74% (b) 47% (c) Each pane allows only 97% of light to pass through. 103. (a) $16,231 (b) $8626 (c) As each year passes, the sedan is worth 90% of its value the previous year. 105. (a) 30% (b) 9% (c) As each year passes, only 30% of the previous survivors survive again. 107. 3.35 mg; 0.45 mg 109. (a) 0.632 (b) 0.982 (c) 1 (d) y  1  e0.1t 1 (e) About 7 min

0

40 0

AN38

ANSWERS  Section 4.3 117. 36

111. (a) 0.0516 (b) 0.0888 113. (a) 70.95% (b) 72.62% (c) 100% 115. (a) 5.41 amp, 7.59 amp, 10.38 amp (b) 12 amp (d) 3.34 amp, 5.31 amp, 9.44 amp (e) 24 amp (c), (f) I 30

119.

Final Denominator 1 + 1

Value of Expression

Compare Value to e ? 2.718281828 2.5 6 e

2.5

I2 (t)  24(1  e0.5t )

2 + 2

2.8

2.8 7 e

I1 (t)  12(1  e2t )

3 + 3

2.7

2.7 6 e

5 t

121. f 1A + B2 = aA + B = aA # aB = f 1A2 # f 1B2 123. f 1ax2 = aax = 1ax 2 a = 3 f 1x2 4 a 1 -x 1e + e -1 - x2 2 2 1 = 1e - x + e x 2 2 1 x = 1e + e - x 2 2 = f 1x2

125. (a) f 1 - x2 =

(b)

y

6

4 + 4

2.721649485

2.721649485 7 e

5 + 5

2.717770035

2.717770035 6 e

6 + 6

2.718348855

2.718348855 7 e

(c) 1cosh x2 2 - 1sinh x2 2 2 2 1 1 = c 1e x + e - x 2 d - c 1e x - e - x 2 d 2 2 1 = 3 e 2x + 2 + e - 2x - e 2x + 2 - e - 2x 4 4 1 = 142 = 1 4

1 x (e  ex ) 2

6

6 1

1 132. ( - q , - 5] ∪ [ - 2, 2] 133. (2, q ) 134. Linear; y = - x - 5 3 y (b) Domain: ( - q , q ) 5 Range: [ - 4, q ) (c) Decreasing: ( - q , - 1) (1, 0) 5 x Increasing: ( - 1, q )

127. 59 minutes 135. (a) (3, 0) (2, 3)

(0, 3) (1, 4) x  1

4.4 Assess Your Understanding (page 352) 4. 5x x 7 0 6 or 10, q 2 17. 23 = 8

19. a6 = 3

1 5. a , - 1 b , 11, 02, 1a, 12 a

21. 3x = 2

23. e x = 4

25. 0

6. 1

7. False

27. 2

29. - 4

59.

y f(x)  3 x yx (1, 3) 5 (3, 1) 1 1, (0, 1) f 1 (x)  log 3 x 3

61.

5 (1, 2)

5 x (1, 0)

47. 5x x Ú 1 6; [1, q 2

1 2 y

63. B

1 1, 2

65. D

51. 30.099 53. 2.303 67. A

55. - 53.991

69. E

(3, 0) 2 x

x  4

75. (a) Domain: 10, q 2 (b) y 2

5

(0, 3)

x0

8

(5, 2) 8 x

x0

(c) Range: 1 - q , q 2 Vertical asymptote: x = 0 1 (d) f - 1 1x2 = e x + 3 2 (e) Domain of f -1:1 - q , q 2 Range of f -1: 10, q 2 y (f) 8

y0

7 x

y0

(c) Range: 1 - q , q 2 Vertical asymptote: x = 4 (d) f - 1 1x2 = 10x - 2 + 4 (e) Domain of f - 1: 1 - q , q 2 Range of f -1: 14, q 2 y (f) 8 (2, 5)

1 3, 2

(2, 1)

y  4

2.5

8 1, 3 2

(c) Range: 1 - q , q 2 Vertical asymptote: x = 0 (d) f - 1 1x2 = e x - 2 (e) Domain of f -1: 1 - q , q 2 Range of f -1: 10, q 2 (f) y

5 x

77. (a) Domain: 14, q 2 (b) y x  4

x

(1, 2) 5 x

57. 22

5

(c) Range: 1 - q , q 2 Vertical asymptote: x = - 4 73. (a) Domain: 10, q 2 y (b)

15. x = ln 8

(d) f - 1 1x2 = e x - 4 (e) Domain of f - 1: 1 - q , q 2 Range of f -1: 1 - 4, q 2 y (f)

5

x f 1 (x)  log 1/2 x (2, 1)

1 , 1 3

49. 0.511

71. (a) Domain: 1 - 4, q 2 y (b)

1 ,1 yx 2

5

13. x = log 27.2

31.

x

f (x) 

11. 2 = log a 1.6

1 1 33. 4 35. 37. 5x x 7 3 6; 13, q 2 2 2 41. 5x x 7 10 6; 110, q 2 43. 5x x 7 - 1 6; 1 - 1, q 2

39. All real numbers except 0; 5x x ≠ 0 6; ( - q , 0) ∪ (0, q )

45. 5x x 6 - 1 or x 7 0 6; 1 - q , - 12 ∪ 10, q 2

9. 2 = log 3 9

8. True

y4

5 x 3 x

ANSWERS   Section 4.5 81. (a) Domain: 1 - 2, q 2 (b) x  2 y

79. (a) Domain: 10, q 2 (b) y 2.5

5,

1 2

(1, 3)

5 (1, 4) 5 x y  3

(c) Range: 1 - q , q 2 Vertical asymptote: x = - 2 (d) f - 1 1x2 = 3x - 3 - 2 (e) Domain of f - 1: 1 - q , q 2 Range of f -1: 1 - 2, q 2 y (f)

(c) Range: 1 - q , q 2 Vertical asymptote: x = 0 1 # 2x 10 (d) f - 1 1x2 = 2 -1 (e) Domain of f : 1 - q , q 2 Range of f -1: 10, q 2 (f) y

(c) Range: 1 - 3, q 2 Horizontal asymptote: y = - 3 (d) f - 1 1x2 = ln 1x + 32 - 2 (e) Domain of f - 1: 1 - 3, q 2 Range of f -1: 1 - q , q 2 y (f) 5

5 (4, 1)

8

y0

(2, 2)

5 x

x0

0,

83. (a) Domain: 1 - q , q 2 y (b)

5

8 x

1, 0 2

AN39

1, 5 2

1 2

(2, 2)

5 x (3, 1)

y  2

x  3

5 x

85. (a) Domain: 1 - q , q 2 y (b)

87. 59 6

7 89. e f 2

91. 52 6

95. 53 6

97. 52 6

99. e

9 (6, 8)

(0, 5) y4

8 x

(c) Range: 14, q 2 Horizontal asymptote: y = 4 (d) f - 1 1x2 = 3 log 2 1x - 42 (e) Domain of f - 1:14, q 2 Range of f -1: 1 - q , q 2 (f) y 8

5 x

101. e

ln 8 - 5 f 2

105. 5 - 1 6

103.

107. e 5 ln

109. e 2 - log

5

113.

93. 55 6

y 5 (1, 0)

ln 10 f 3

5 x

(1, 0)

- 2 22 , 2 22 6

x0

Domain: 5x x ≠ 0 6 Range: 1 - q , q 2 Intercepts: 1 - 1, 02, (1, 0)

7 f 5

5 f 2

1 1 111. (a) e x ` x 7 - f ; a - , q b 2 2

(8, 6)

(b) 2; (40, 2) (c) 121; (121, 3) (d) 4

(5, 0) 10 x x4 y 2.5

(1, 0) 5 x x0

Domain: 5x x 7 0 6 Range: 5y y Ú 0 6 Intercept: (1, 0)

117. (a) 1 (b) 2 (c) 3 (d) It increases. (e) 0.000316 (f) 3.981 * 10-8 119. (a) 5.97 km (b) 0.90 km 121. (a) 6.93 min (b) 16.09 min 123. h ≈ 2.29, so the time between injections is about 2 h, 17 min.

1 1 1 1 137. Zeros: - 3, - , , 3; x-intercepts: - 3, - , , 3 2 2 2 2

125. 0.2695 s 0.8959 s 2.0 1.6 1.2 0.8 0.4

(0.8959, 1)

(0.2695, 0.5) 0 0.4 1.2 2.0 Seconds

139. f(1) = - 5; f(2) = 17

138. 12

127. 50 decibels (dB) 129. 90 dB 131. 8.1 133. (a) k ≈ 11.216 (b) 6.73 (c) 0.41% (d) 0.14% 135. Because y = log 1 x means 1y = 1 = x, which cannot be true for x ≠ 1

y Amperes

115.

x

140. 3 + i; f(x) = x4 - 7x3 + 14x2 + 2x - 20; a = 1

4.5 Assess Your Understanding (page 363) 5. log a M; log a N 6. log a M; log a N 7. r log a M 8. 6 9. 7 10. False 11. False 12. False 13. 71 15. - 4 17. 7 5 1 19. 1 21. 1 23. 3 25. 27. 4 29. a + b 31. b - a 33. 3a 35. 1a + b2 37. 2 + log 5 x 39. 3 log 2 z 41. 1 + ln x 43. ln x - x 4 5 1 45. 2 log a u + 3 log av 47. 2 ln x + ln 11 - x2 49. 3 log 2 x - log 2 1x - 32 51. log x + log 1x + 22 - 2 log 1x + 32 2 1 1 2 1 1 x - 1 53. ln 1x - 22 + ln 1x + 12 - ln 1x + 42 55. ln 5 + ln x + ln 11 + 3x2 - 3 ln 1x - 42 57. log 5 u3v4 59. log 3 a 5/2 b 61. log 4 c d 3 3 3 2 1x + 12 4 x 1. 0

2. 1

3. M 4. r

63. - 2 ln 1x - 12

65. log 2[x 13x - 22 4]

67. log a a

25x6 22x + 3

b

69. log 2 c

1x + 12 2

1x + 32 1x - 12

d

71. 2.771

73. - 3.880

75. 5.615

77. 0.874

AN40 79. y =

ANSWERS  Section 4.5

log x

81. y =

log 4

log 1x + 22

83. y =

log 2

2

4

2

0

5

2

log 1x - 12

4

3

1

log 1x + 12

5

4

2

85. (a) 1f ∘ g2 1x2 = x; 5x x is any real number 6 or 1 - q , q 2 (b) 1g ∘ f2 1x2 = x; 5x x 7 0 6 or 10, q 2 (c) 5 (d) 1f ∘ h2 1x2 = ln x2; {x x ≠ 0} or ( - q , 0) ∪ (0, q ) (e) 2

89. y = Cx 1x + 12

87. y = Cx 95. y =

3 2 C12x + 12 1>6

1x + 42 1>9

97. 3

91. y = Ce 3x

93. y = Ce -4x + 3

99. 1

101. log a 1x + 2x2 - 12 + log a 1x - 2x2 - 12 = log a 3 1x + 2x2 - 12 1x - 2x2 - 12 4 = log a[x2 - 1x2 - 12 4 = log a 1 = 0

103. ln 11 + e 2x 2 = ln[e 2x 1e -2x + 12 4 = ln e 2x + ln 1e -2x + 12 = 2x + ln 11 + e -2x 2 1 -y 105. y = f 1x2 = log a x; ay = x implies ay = a b = x, so - y = log 1>a x = - f 1x2. a 1 1 107. f 1x2 = log a x; f a b = log a = log a 1 - log a x = - f 1x2 x x M -1 -1 109. log a = log a 1M # N -1 2 = log a M + log a N -1 = log a M - log a N, since aloga N = N -1 implies a -loga N = N; that is, log a N = - log a N -1. N 1 - 5 - 221 - 5 + 221 116. - 2, , , 5 2 2

115. { - 1.78, 1.29, 3.49} 118.

y 10 −10

117. A repeated real solution (double root)

10 x −10

Domain: {x x … 2} or ( - q , 2] Range: {y y Ú 0} or [0, q )

4.6 Assess Your Understanding (page 369)

16 1 21 f 9. 56 6 11. 516 6 13. e f 15. 53 6 17. 55 6 19. e f 21. 5 - 6 6 23. 5 - 2 6 25. 5 - 1 + 21 + e 4 6 ≈ 56.456 6 5 3 8 - 5 + 3 25 1 9 27. e f ≈ {0.854} 29. 52 6 31. e f 33. {7} 35. { - 2 + 4 22} 37. { - 23, 23} 39. e , 729 f 41. 58 6 2 2 3 8 ln ln 10 ln 1.2 8 5 1 43. {log 2 10} = e f ≈ {3.322} 45. { - log 81.2} = e f ≈ { - 0.088} 47. e log 2 f = d t ≈ 50.226 6 ln 2 ln 8 3 5 3 ln 2 5. 516 6

49. e

7. e

ln 3 f ≈ 50.307 6 2 ln 3 + ln 4

51. e

61. 5log 4 1 - 2 + 272 6 ≈ 5 - 0.315 6 71. { - 0.57}

87. {81}

73. { - 0.70}

2 89. e - 1, f 3

97. (a) 55 6 ; (5, 3)

75. {0.57}

91. 50 6

(b) 55 6 ; (5, 4)

93.

ln 7 f ≈ 51.356 6 ln 0.6 + ln 7

63. 5log 5 4 6 ≈ 50.861 6

77. {0.39}

5 ln 1 2

79. {1.32}

99. (a), (b)

(d) 55 6

83. {2, 3}

y 18

95. c e (e) e -

g(x)  2 2.5 x

(c) 5x x 7 0.710 6 or 10.710, q 2

g(x)  10

85. e

57. e

ln 3 f ≈ 51.585 6 ln 2

67. 5log 4 5 6 ≈ 51.161 6

69. {2.79}

26 + 8 210 26 - 8 210 , f ≈ {5.7, 0.078} 9 9

ln 15

s ≈ {1.921}

1 f 11 103. (a), (b), (c) y

f (x)  3x

5 x  2

g(x)  2

x2

(0.710, 6.541)

ln p f ≈ 50.534 6 1 + ln p

ln 5 # ln 3

101. (a), (b), (c)

y f(x)  3x  1 18

55. e

65. No real solution

81. {1.31}

+ 25 2 6 ≈ 51.444 6

(c) 51 6; yes, at (1, 2)

53. 50 6

(log3 10, 10) 3

x

f (x)  2x  1

1 , 2兹2 2 5 x

59. 50 6

ANSWERS  Section 4.9 105. (a)

107. (a) 2048 (b) 2068 109. (a) After 4.2 yr (b) After 6.5 yr (c) After 12.8 yr 1 112. e - 3, , 2 f 113. {x x Ú 1} or [1, q ) 4 x + 5 114. (f ∘ g)(x) = ; {x x ≠ 3, x ≠ 11} 115. One-to-one - x + 11

y 5

f(x)  2x  4

AN41

5 x y  4

(c) 5x x 6 2 6or 1 - q , 22

(b) 2

4.7 Assess Your Understanding (page 378) 3. principal 4. I; Prt; simple interest

5. 4

6. effective rate of interest 7. $108.29 9. $609.50 11. $697.09 13. $1246.08 15. $88.72 1 17. $860.72 19. $554.09 21. $59.71 23. 5.095% 25. 5.127% 27. 6 , compounded annually 29. 9% compounded monthly 31. 25.992% 4 33. 24.573% 35. (a) About 8.69 yr (b) About 8.66 yr 37. 6.823% 39. 5.09 yr; 5.07 yr 41. 15.27 yr or 15 yr, 3 mo 43. $104,335 45. $12,910.62 47. About $30.17 per share or $3017 49. Not quite. Jim will have $1057.60. The second bank gives a better deal, since Jim will have $1060.62 after 1 yr. 51. Will has $11,632.73; Henry has $10,947.89. 53. (a) $60,933 (b) $33,441 55. About $1020 billion; about $233 billion 57. $940.90 59. 2.53% 61. 34.31 yr 63. (a) $1364.62 (b) $1353.35 65. $4631.93 67. (a) 6.12 yr (b) 18.45 yr 69. (a) 2.47% (b) In 2023 71. 22.7 yr 76. R = 0; yes r nt 2x (c) mP = Pa1 + b 77. - 2, 5; f(x) = (x + 2)2(x - 5)(x2 + 1) 78. f -1(x) = n x - 1 r nt m = a1 + b n r nt r ln m = ln a1 + b = nt ln a1 + b n n ln m t = r n ln a1 + b n

79. {6}

4.8 Assess Your Understanding (page 388) 1. (a) 500 insects (b) 0.02 = 2, (c) About 611 insects (d) After about 23.5 days (e) After about 34.7 days 3. (a) - 0.0244 = - 2.44, (b) About 391.7 g (c) After about 9.1 yr (d) 28.4 yr 5. (a) N(t) = N0 e kt (b) 5832 (c) 3.9 days 7. (a) N(t) = N0 e kt (b) 25,198 9. 9.797 g 11. 9953 yr ago 13. (a) About 5:18 pm (b) About 14.3 min (c) The temperature of the pizza approaches 70°F. 15 18.63°C; 25.07°C 17. 1.7 ppm; 7.17 days or 172 h 19. 0.26 mol; 6.58 h or 395 min 21. 26.6 days 23. (a) 1000 g (b) 43.9% (c) 30 g (d) 616.6 g (e) After 9.85 h (f) About 7.9 h 25. (a) 9.23 * 10-3, or about 0 (b) 0.81, or about 1 (c) 5.01, or about 5 (d) 57.91°, 43.99°, 30.07°

27. (a) In 1984, 91.8% of households did not own a personal computer. (b) 100

29. (a)

120

6

0

100 0

0 0

40 0

100 0

(c) 70.6%

3 30. f(x) = - x + 7 2

31. Neither

32. 2 ln x +

1 ln y - ln z 2

(d) During 2011

(b) 0.78 or 78% (c) 50 people (d) As n increases, the probability decreases.

3 33. 2 2 5

4.9 Assess Your Understanding (page 395) 1. (a)

(d)

1

1

7 0

(b) y = 0.0903 11.33842 x (c) N1t2 = 0.0903e 0.2915t

3. (a)

1

1

7 0

(e) 0.69 (f) After about 7.26 hr

(d)

100

0

4 20

(b) y = 118.7226(0.7013)x (c) A(t) = 118.7226e -0.3548t

100

0

4 20

(e) 28.7% (f) k = - 0.3548 = - 35.48% is the exponential decay rate. It represents the rate at which the percentage of patients surviving advanced-stage breast cancer is decreasing.

AN42 5. (a)

ANSWERS  Section 4.9 (c)

210

20 110

20 110

380

(b) y = 330.0549 - 34.5008 ln x

(c)

0

30

13. (a)

286.2055

(d)

110

(d)

225

90

20

(b) Quadratic with a 6 0 because of the “upside down U-shape” of the data (c) y = - 0.0311x2 + 3.4444x + 118.2493

110

90

3 22 2

(e) 201

1 (x + 3)(x + 1)2(x - 2) 3

15. f(x) =

0

30 0

17. 23

30

18.

y 10 −10

0

(b) Exponential (c) y = 115.5779(0.9012)x = 115.5779e -0.1040x

225

175

16.

0

120

(d) 762,176,844 (e) Approximately 315,203,288 (f) 2023

1 + 8.7428e - 0.0162x

175

(d) 286.2 thousand cell sites (e) 281.3 thousand cell sites

1 + 226.8644e -0.2873x

762,176,844.4

320,000,000

10 70,00,000

120

20

30 0

0

(b) y =

(b) y =

11. (a)

300

(c)

320,000,000

10 70,000,000

380

(d) 185 billion pounds (e) The prediction was under by 5 billion pounds.

9. (a) 300

0

7. (a)

210

10 x −10

(e) 5.1%

Review Exercises (page 401) 1. (a) - 4 (b) 1 (c) - 6 (d) - 6 2. (a) - 65 (b) 665 (c) 23 (d) 2 3. (a) 121 4. (a) 115 (b) 3 18 - 5 (c) 12 (d) - 38 5. 1f ∘ g2 1x2 = 1 - 3x, all real numbers; 1g ∘ f2 1x2 = 7 - 3x, all real numbers; 1f ∘ f2 1x2 = x, all real numbers; 1g ∘ g2 1x2 = 9x + 4, all real numbers

(b) 3 (c) 2 (d) 163

6. 1f ∘ g2 1x2 = 2x2 + 3x, 5x 0 x Ú 0 6 h 5x 0 x … - 3 6 1g ∘ f2 1x2 = x + 3 2x - 1, 5x 0 x Ú 1 6

1f ∘ f2 1x2 = 2 1x - 1 - 1, 5x 0 x Ú 2 6 1g ∘ g2 1x2 = x4 + 6x3 + 14x2 + 15x + 5, all real numbers 1 + x x - 1 , 5x x ≠ 0, x ≠ 1 6; 1g ∘ f2 1x2 = , 5x x ≠ - 1, x ≠ 1 6; 1f ∘ f2 1x2 = x, 5x x ≠ 1 6; 1g ∘ g2 1x2 = x, 5x x ≠ 0 6 7. 1f ∘ g2 1x2 = 1 - x x + 1 8. (a) one-to-one (b) 5 12, 32, 110, 52, 15, 22, 13, 72 6 9. The graph passes the y yx 5 horizontal line test. (0, 2) (2, 0)

(3, 3) 5 x

(3, 1)

10. f -1 1x2 =

2x + 3 5x - 2

f 1f -1 1x2 2 =

f

-1

1f 1x2 2 =

2a

11. f

2x + 3 b + 3 5x - 2

2x + 3 5a b - 2 5x - 2 2a

2x + 3 b + 3 5x - 2

2x + 3 5a b - 2 5x - 2

Domain of f = range of f

-1

Range of f = domain of f

-1

=

=

2(2x + 3) + 3(5x - 2) 5(2x + 3) - 2(5x - 2)

2(2x + 3) + 3(5x - 2) 5(2x + 3) - 2(5x - 2)

2 5 2 = all real numbers except 5

= all real numbers except

=

19x = x 19

=

19x = x 19

-1

1x2 =

x + 1 x

f 1f -1 1x2 2 =

1 x = = x x + 1 x + 1 - x - 1 x 1 + 1 x - 1 1 + x - 1 -1 f 1f 1x2 2 = = = x 1 1 x - 1

Domain of f = range of f

-1

= all real numbers except 1

Range of f = domain of f

-1

= all real numbers except 0

ANSWERS  Review Exercises 12. f -1 1x2 = x2 + 2, x Ú 0 f 1f f

-1

-1

13. ƒ - 1 1x2 = 1x - 12 3

f 1f -1 1x2 2 = 1 1x - 12 3 2 1>3 + 1 = x f -1 1f 1x2 2 = 1x1>3 + 1 - 12 3 = x Domain of f = range of f -1 = 1 - q , q 2

1x2 2 = 2x + 2 - 2 =  x = x, x Ú 0 2

1f 1x2 2 = 1 2x - 22 2 + 2 = x

Domain of f = range of f

-1

Range of f = domain of f

-1

= 3 2, q 2

Range of f = domain of f -1 = 1 - q , q 2 1 14. (a) 16 (b) 4 (c) (d) - 3 15. log 3 z = 5 64

= 3 0, q 2

18. 5x x 6 1 or x 7 2 6; 1 - q , 12 ∪ 12, q 2

AN43

19. 3

20. 22

21. 0.3

22. log 5 x + log 5 y - 2 log 5 z

1 1 24. ln 1x2 + 12 + ln y - ln x 25. 2 ln 12x + 32 - 2 ln 1x - 12 - 2 ln 1x - 22 2 2

26. log 2

x

16. 37 = z

(x + 1)

3

2 2 f; a , q b 3 3

23. 8 log 2 a + 2 log 2 b

3 2

2

17. e x ` x 7

27. ln

x8 13(x + 3) 2

4

28. ln c

16 2x2 + 1 2x 1x - 42

d

29. 2.124 30.

31. (a) Domain of f:1 - q , q 2 (b) y

3

(c) Range of f: 10, q 2 Horizontal asymptote: y = 0 (d) f -1 1x2 = 3 + log 2 x (e) Domain of f - 1: 10, q 2 Range of f -1:1 - q , q 2

9 1

8 (4, 2) (3, 1) 9 x

3

32. (a) Domain of f: 1 - q , q 2 y (b) (1, 4)

5 (0, 2)

y0

(d) f -1 1x2 = - log 3 1x - 12 (e) Domain of f - 1: 11, q 2 Range of f -1: 1 - q , q 2 y (f)

y1 5 x

33. (a) Domain of f: 1 - q , q 2 y (b)

x0

34. (a) Domain of ƒ: 1 - 3, q 2 (b) y 5

y0 5 x 1,

x

(2, 0) 5 x

3 e

0, x  3

(c) Range of f: 10, q 2 Horizontal asymptote: y = 0 x (d) f -1 1x2 = 2 + ln a b 3 (e) Domain of f - 1: 10, q 2 Range of f -1: 1 - q , q 2 y (f)

x1

(2, 4) (1, 3) 5 x

(2, 3)

(2, 0) 5 (4, 1)

y 5

5

5

(c) Range of f: 11, q 2 Horizontal asymptote: y = 1

(f)

1 ln 3 2

(c) Range of f: 1 - q , q 2 Vertical asymptote: x = - 3 (d) f -1 1x2 = e 2x - 3 (e) Domain of f - 1: 1 - q , q 2 Range of f - 1: 1 - 3, q 2 y (f) 5

5

1 ln 3, 0 2

(0, 2)

(3, 2)

5 x y  3

5 x 3 ,1 e x0

5 35. e - f 3

36. e

1 41. e , - 3 f 2 46. (a), (e)

- 1 - 23 - 1 + 23 , f ≈ 5 - 1.366, 0.366 6 2 2

42. 51 6

43. 55 6

44. 51 - ln 5 6 ≈ 5 - 0.609 6

y f 1 (x)  2 x1  2 14 yx 5 (3, 6) 0, f (x)  log2 (x  2)  1 2 (6, 3) 10 x 5 ,0 2

1 37. e f 4

38. e

2 ln 3 f ≈ 54.301 6 ln 5 - ln 3

45. 5log 3 1 - 2 + 27 2 6 = c

(b) 3; (6, 3) (c) 10; (10, 4) 5 5 (d) e x ` x 7 f or a , q b 2 2 (e) f -1 1x2 = 2x - 1 + 2

39. 5 - 2, 6 6

ln 1 - 2 + 27 2 ln 3

40. 583 6

s ≈ 5 - 0.398 6

47. (a) 37.3 W (b) 6.9 dB 48. (a) 11.77 (b) 9.56 in. 49. (a) 9.85 yr (b) 4.27 yr 50. $20,398.87; 4.04%; 17.5 yr 51. $41,668.97 52. 24,765 yr ago 53. 55.22 min or 55 min, 13 sec 54. 7,412,094,029 55. $1.316 trillion; ≈ +216 billion

AN44

ANSWERS   Review Exercises

56. (a) 0.3 (b) 0.8 1 (c)

57. (a)

(c)

165.1

58. (a)

165.1

10

0

0 0

0

16 153

25

40

16 10

153

0

(b) y = 165.73(0.9951)x

(d) In 2026 (c)

59. (a)

10

0

50

(d) Approximately 83 s

(b) y = 18.921 - 7.096 ln x

(c)

(e) 2.4 days; during the tenth hour of day 3 (f) 9.5 days

50

40

1

10

1

9 0

(d) ≈ - 3°F

(b) C =

9 0

46.93

(d) About 47 people; 50 people

1 + 21.273e -0.7306t

Chapter Test (page 404) 2x + 7 3 ; domain: e x ` x ≠ - f (b) 1g ∘ f2 1 - 22 = 5 (c) 1f ∘ g2 1 - 22 = - 3 2x + 3 2 2. (a) The function is not one-to-one. (b) The function is one-to-one. 2 + 5x 5 ; domain of f = b x 0 x ≠ r , range of f = 5 y y ≠ 0 6 ; domain of f - 1 = 3. f -1 1x2 = 3x 3 1. (a) f ∘ g =

4. The point 1 - 5, 32 must be on the graph of f -1. ln 21 ≈ 2.771 11. ln 133 ≈ 4.890 10. log 3 21 = ln 3 12. (a) Domain of f: 5x - q 6 x 6 q 6 or 1 - q , q 2 y (b) 16

(1, 1)

(0, 2) 2.5 y  2

15. 591 6

6. {4}

7. {625}

8. e 3 + 2 ≈ 22.086

(e) Domain of f - 1: {x 0 x 7 - 2} or 1 - 2, q 2 Range of f - 1: {y 0 - q 6 y 6 q } or 1 - q , q 2 y (f) 5

x

16. 5 - ln 2 6 ≈ 5 - 0.693 6

17. e

13. (a) Domain of f: 5x x 7 2 6 or (2, q ) (b) y 8 (3, 1)

(e) Domain of f - 1: {x 0 - q 6 x 6 q } or (- q, q) Range of f - 1: 5y y 7 2} or (2, q ) (f) y 8

x

x2

1 - 213 1 + 213 , f ≈ 5 - 1.303, 2.303 6 2 2

(0, 7) (1, 3) y2

(c) Range of f: 5y - q 6 y 6 q 6 or 1 - q , q 2; vertical asymptote: x = 2 (d) f - 1 1x2 = 51 - x + 2

x  2

5 r 3

9. log 20 ≈ 1.301

18 5 x

(1, 1)

≠ 0 6 ; range of f - 1 = b y 0 y ≠

(7, 0)

(2, 0)

(c) Range of f: 5y y 7 - 2 6 or 1 - 2, q 2 ; Horizontal asymptote: y = -2 (d) f - 1 1x2 = log 4 1x + 22 - 1 14. 51 6

5. {5}

5 x x

18. e

8

x

3 ln 7 f ≈ 5 - 6.172 6 1 - ln 7

19. 5 2 26 6 ≈ 54.899 6 20. 2 + 3 log 2 x - log 2 1x - 62 - log 2 1x + 32 21. About 250.39 days 22. (a) $1033.82 (b) $963.42 23. (a) About 83 dB (b) The pain threshold will be exceeded if 31,623 people shout at the same time.

(c) 11.9 yr

Cumulative Review (page 405) 1. (a) Yes; no (b) Polynomial; The graph is smooth and continuous. 2. (a) 10 (b) 2x2 + 3x + 1 1 23 3. a , b is on the graph. 4. { - 26} 2 2 5.

6. (a)

y 10 (8, 0) 10 x (0, 4)

y 10

(0, 3)

(1, 2) 10 x

7. f 1x2 = 2 1x - 42 2 - 8 = 2x2 - 16x + 24 8. Exponential; f 1x2 = 2 # 3x y 9. 10 (0, 1)

(b) 5x - q 6 x 6 q 6

(1, 2)

5 x

(c) 2x2 + 4xh + 2h2 - 3x - 3h + 1

10. (a) f 1g 1x2 2 =

4

1x - 32 2

domain: 5x x ≠ 3 6; 3

+ 2;

(b) f 1g 1x2 2 = log 2 x + 2; domain: 5x x 7 0 6; 2 + log 2 14

AN45

ANSWERS   Section 5.2 12. (a),(c)

1 11. (a) Zeros: - 4, - , 2 4

60

2

(4, 0) (0, 8)

x2

(b) 5x x 7 - 1 6 or 1 - 1, q 2 16. (a) 20

5 x (3, 0) g1(x)

yx

Domain g = range g -1 = 1 - q , q 2 Range g = domain g -1 = 12, q 2 Vertical asymptote: x = 2 (b) g -1 1x2 = log 3 1x - 22

4 70

1  ,0 4

3 13. e - f 14. 52 6 2 15. (a) 5 - 1 6

y 5

g(x) y2

1 (b) x-intercepts: - 4, - , 2; y-intercept: - 8 4 (c) Local maximum value of 60.75 occurs at x = - 2.5. Local minimum value of - 25 occurs at x = 1. (d) 5, 3 y 

(0, 3)

0

(c) 525 6

80 0

(2, 0) 5 x (1, 25)

(b) Logarithmic; y = 49.293 - 10.563 ln x (c) Highest value of  r

CHAPTER 5 Trigonometric Functions 5.1 Assess Your Understanding (page 417) 3. standard position 4. central angle 11. y

13. 30

1 6. ru; r 2u 2

5. radian

15.

y

7. p

s u 8. ; t t

9. T

17.

y

10. F 19. y

y 3 4

x 135

450

x

– 6

x

21.

y 16 3

x

x

p p 3p p 4p 37. 39. 41. p 43. 45. 47. 60° 6 3 3 4 2 49. - 225° 51. 90° 53. 15° 55. - 90° 57. - 30° 59. 0.30 61. - 0.70 63. 2.18 65. 179.91° 67. 114.59° 69. 362.11° 71. 5 m 73. 6 ft p p 75. 0.6 radian 77. ≈ 1.047 in. 79. 25 m2 81. 2 13 ≈ 3.464 ft 83. 0.24 radian 85. ≈ 1.047 in .2 87. s = 2.094 ft; A = 2.094 ft 2 3 3 675p 1075 p ≈ 1060.29 ft 2 97. ≈ 1125.74 in2. 89. s = 14.661 yd; A = 87.965 yd2 91. 3p ≈ 9.42 in; 5p ≈ 15.71 in . 93. 2p ≈ 6.28 m2 95. 2 3 1 1 radian/s; v = cm/s 101. ≈23.2 mph 103. ≈120.6 km>h 105. ≈452.5 rpm 107. ≈359 mi 109. ≈898 mi/h 111. ≈2292 mi/h 99. v = 60 12 3 113. rpm 115. ≈2.86 mi/h 117. ≈31.47 rpm 119. 63 p ft 2 121. ≈1037 mi/h 123. Radius ≈ 3979 mi; circumference ≈ 25,000 mi 4 r1 v2 7 1 = . 134. x = 135. e - 3, f 136. y = -  x + 3  - 4 125. v1 = r1v1, v2 = r2v2, and v1 = v2, so r1v1 = r2v2 and r2 v1 3 5 137. Horizontal asymptote: y = 3; Vertical asymptote: x = 7 23. 40.17°

25. 1.03°

27. 9.15°

29. 40°19′12″

31. 18°15′18″

33. 19°59′24″

35.

5.2 Assess Your Understanding (page 434) 7. cosine

8. (0, 1)

9. a

22 22 , b 2 2

1 23 10. a , b 2 2

y x 11. ; r r

12. F

13. sin t =

1 23 23 2 23 ; cos t = ; tan t = ; csc t = 2; sec t = ; cot t = 23 2 2 3 3

2 221 5 2 221 22 221 5 221 22 ; cos t = - ; tan t = ; csc t = ; sec t = - ; cot t = 17. sin t = ; cos t = ; 5 5 2 21 2 21 2 2 1 2 22 22 3 22 tan t = - 1; csc t = 22; sec t = - 22; cot t = - 1 19. sin t = - ; cos t = ; tan t = ; csc t = - 3; sec t = ; cot t = - 2 22 3 3 4 4 1 1 4 23 21. - 1 23. 0 25. - 1 27. 0 29. - 1 31. 1 22 + 1 2 33. 2 35. 37. 26 39. 4 41. 0 43. 2 22 + 45. 1 2 2 3 15. sin t =

47. sin

2p 23 2p 1 2p 2p 2 23 2p 2p 23 = ; cos = - ; tan = - 23; csc = ; sec = - 2; cot = 3 2 3 2 3 3 3 3 3 3

1 23 23 2 23 ; tan 210° = ; csc 210° = - 2; sec 210° = ; cot 210° = 23 49. sin 210° = - ; cos 210° = 2 2 3 3 51. sin

3p 22 3p 22 3p 3p 3p 3p = ; cos = ; tan = - 1; csc = 22; sec = - 22; cot = -1 4 2 4 2 4 4 4 4

53. sin

8p 23 8p 1 8p 8p 2 23 8p 8p 23 = ; cos = - ; tan = - 23; csc = ; sec = - 2; cot = 3 2 3 2 3 3 3 3 3 3

55. sin 405° = 57. sin a -

22 22 ; cos 405° = ; tan 405° = 1; csc 405° = 22; sec 405° = 22; cot 405° = 1 2 2

p p 23 p p p 2 23 p 1 23 b = - ; cos a - b = ; tan a - b = ; csc a - b = - 2; sec a - b = ; cot a - b = - 23 6 2 6 2 6 3 6 6 3 6

59. sin 1 - 135°2 = -

22 22 ; cos 1 - 135°2 = ; tan 1 - 135°2 = 1; csc 1 - 135°2 = - 22; sec 1 - 135°2 = - 22; cot 1 - 135°2 = 1 2 2

x

AN46 61. sin

ANSWERS  Section 5.2

5p 5p 5p 5p 5p 5p = 1; cos = 0; tan is undefined; csc = 1; sec is undefined; cot = 0 2 2 2 2 2 2 23 14p 23 14p 1 14p 14p 2 23 14p 14p b = ; cos a b = - ; tan a b = 23; csc a b = ; sec a b = - 2; cot a b = 3 2 3 2 3 3 3 3 3 3 4 5 3 4 5 3 67. 1.07 69. 0.32 71. 3.73 73. 0.84 75. 0.02 77. sin u = ; cos u = - ; tan u = - ; csc u = ; sec u = - ; cot u = 5 5 3 4 3 4

63. sin a 65. 0.47

2 213 213 3 213 3 213 2 ; cos u = ; tan u = - ; csc u = ; sec u = ; cot u = 13 13 2 3 2 3

79. sin u = -

22 22 ; cos u = ; tan u = 1; csc u = - 22; sec u = - 22; cot u = 1 2 2 3 4 3 5 5 4 83. sin u = ; cos u = ; tan u = ; csc u = ; sec u = ; cot u = 85. 0 87. 0 89. - 0.1 5 5 4 3 4 3 81. sin u = -

23 22 22 p 22 113. 115. (a) ;a , b 2 4 2 4 2 11p 5p p 7p 13p 117. Answers may vary. One set of possible answers is ,, , , . 3 3 3 3 3 119. 105. -

103. 23

23 2

107.

1 + 23 2

91. 3

109. -

1 2

111.

93. 5

(b) a

95.

23 2

22 p , b 2 4

97.

1 2

99.

3 4

101.

23 2

p (c) a , - 2 b 4

U

0.5

0.4

0.2

0.1

0.01

0.001

0.0001

0.00001

sin U

0.4794

0.3894

0.1987

0.0998

0.0100

0.0010

0.0001

0.00001

sin U U

0.9589

0.9735

0.9933

0.9983

1.0000

1.0000

1.0000

1.0000

sin u approaches 1 as u approaches 0. u 121. R ≈ 310.56 ft; H ≈ 77.64 ft 123. R ≈ 19,541.95 m; H ≈ 2278.14 m 127. (a) 1.9 hr; 0.57 hr (b) 1.69 hr; 0.75 hr 135. (a) 16.56 ft

(b)

(c) 1.63 hr; 0.86 hr

(d) 1.67 hr; tan 90° is undefined

(c) 67.5°

20

45

90 0

2 2 143. e x  x 7 - f or a - , q b 5 5

144. 4 - 3i , - 5

125. (a) 1.20 s (b) 1.12 s

(c) 1.20 s

129. 15.4 ft

131. 71 ft

133. y = -

4 23 7

137. (a) Values estimated to the nearest tenth: sin 1 ≈ 0.8; cos 1 ≈ 0.5; tan 1 ≈ 1.6; csc 1 ≈ 1.3; sec 1 ≈ 2.0; cot 1 ≈ 0.6; actual values to the nearest tenth: sin 1 ≈ 0.8; cos 1 ≈ 0.5; tan 1 ≈ 1.6; csc 1 ≈ 1.2; sec 1 ≈ 1.9; cot 1 ≈ 0.6 (b) Values estimated to the nearest tenth: sin 5.1 ≈ - 0.9; cos 5.1 ≈ 0.4; tan 5.1 ≈ - 2.3; csc 5.1 ≈ - 1.1; sec 5.1 ≈ 2.5; cot 5.1 ≈ - 0.4; actual values to the nearest tenth: sin 5.1 ≈ - 0.9; cos 5.1 ≈ 0.4; tan 5.1 ≈ - 2.4; csc 5.1 ≈ - 1.1; sec 5.1 ≈ 2.6; cot 5.1 ≈ - 0.4

145. R = 134

146. 81p ft 2

5.3 Assess Your Understanding (page 449) 5. 2p; p 23. 22

6. All real numbers except odd multiples of 25.

23 3

27. II

29. IV

31. IV

33. II

p 2

7. [ - 1, 1]

8. T

9. 1

10. F

11.

22 2

22 21. 0 2 1 37. tan u = 2; cot u = ; sec u = 25; 2

13. 1

15. 1

3 4 5 5 35. tan u = - ; cot u = - ; sec u = ; csc u = 4 3 4 3

17. 23

19.

25 23 2 23 22 3 22 39. tan u = ; cot u = 23; sec u = ; csc u = 2 41. tan u = ; cot u = - 2 22; sec u = ; csc u = - 3 2 3 3 4 4 5 12 13 13 5 3 3 5 5 4 43. cos u = - ; tan u = - ; csc u = ; sec u = - ; cot u = 45. sin u = - ; tan u = ; csc u = - ; sec u = - ; cot u = 13 5 12 5 12 5 4 3 4 3 csc u =

47. cos u = -

12 5 13 12 13 ; tan u = - ; csc u = ; sec u = - ; cot u = 13 12 5 12 5

51. cos u = -

25 2 25 3 3 25 25 ; tan u = ; csc u = ; sec u = ; cot u = 3 5 2 5 2

csc u = -

2 23 23 ; cot u = 3 3

csc u = 210; sec u = 81. 0

83. 1

85. - 1

103. All real numbers

87. 0

105.  y Ú 1

22 2 22 3 22 ; tan u = - 2 22; csc u = ; sec u = - 3; cot u = 3 4 4

53. sin u = -

23 1 ; cos u = ; tan u = - 23; 2 2

3 4 5 5 4 55. sin u = - ; cos u = - ;csc u = - ; sec u = - ; cot u = 5 5 3 4 3

210 ; cot u = - 3 3 89. 0.9

49. sin u =

91. 9

59. -

93. 0

23 2

61. -

23 3

95. All real numbers

107. Odd; yes; origin

63. 2 65. - 1

67. - 1

97. Odd multiples of

109. Odd; yes; origin

69. p 2

57. sin u =

22 2

210 3 210 ; cos u = ; 10 10

71. 0 73. - 22

99. Odd multiples of

111. Even; yes; y-axis

113. (a) -

1 3

p 2

75.

2 23 3

77. 1 79. 1

101. - 1 … y … 1

(b) 1

AN47

ANSWERS   Section 5.4

119. ≈ 15.81 min 121. Let a be a real number and P = 1x, y2 be the point on the unit circle that y 1 a = a. Then y = ax. But x2 + y2 = 1, so x2 + a2x2 = 1. So x = { and y = { ; corresponds to t. Consider the equation tan t = 2 x 21 + a2 21 + a that is, for any real number a, there is a point P = 1x, y2 on the unit circle for which tan t = a. In other words, the range of the tangent function 115. (a) - 2

117. (a) - 4

(b) 6

(b) - 12

is the set of all real numbers. 123. Suppose that there is a number p, 0 6 p 6 2p, for which sin 1u + p2 = sin u for all u. If u = 0, then p p p 3p p sin 10 + p2 = sin p = sin 0 = 0, so p = p. If u = , then sin a + p b = sin a b . But p = p. Thus, sin a b = - 1 = sin a b = 1. This is 2 2 2 2 2 1 ; since cos u has period 2p, so impossible. Therefore, the smallest positive number p for which sin 1u + p2 = sin u for all u is 2p. 125. sec u = cos u does sec u. 127. If P = 1a, b2 is the point on the unit circle corresponding to u, then Q = 1 - a, - b2 is the point on the unit circle corresponding -b b to u + p. Thus, tan 1u + p2 = = = tan u. Suppose that there exists a number p, 0 6 p 6 p, for which tan 1u + p2 = tan u for all u. Then, if -a a u = 0, then tan p = tan 0 = 0. But this means that p is a multiple of p. Since no multiple of p exists in the interval 10, p2, this is a contradiction. Therefore, the period of f 1u2 = tan u is p. 1 1 1 1 a 1 1 129. Let P = 1a, b2 be the point on the unit circle corresponding to u. Then csc u = = ; sec u = = ; and cot u = = = . b sin u a cos u b b/a tan u 2 2 2 2 2 2 2 2 2 2 2 2 2 2 131. 1sin u cos f2 + 1sin u sin f2 + cos u = sin u cos f + sin u sin f + cos u = sin u 1cos f + sin f2 + cos u = sin u + cos u = 1 137. All real numbers

138.

139. E ln 6 + 4 F 140. 9

y

6

(3, 5)

(2, 3)

(4, 3)

-6

6x -6

xⴝ3

5.4 Assess Your Understanding (page 463) p 2

p p p 6. T 7. F 8. T 9. (a) 0 (b) 6 x 6 (c) 1 (d) 0, p, 2p 3 2 2 3p p p 3p 5p p 7p 11p , ; f 1x2 = - 1 for x = - , (f) ,- , , (g) 5xƒ x = kp, k an integer 6 (e) f 1x2 = 1 for x = 2 2 2 2 6 6 6 6 3. 1;

4. 3; p

5. 3;

11. Amplitude = 2; Period = 2p 19. Amplitude = 35.

5 ; Period = 3 3

y (22P, 4) 5 (0, 4)

(2P, 4)

13. Amplitude = 4; Period = p 21. F 37.

23. A

25. H

P 2 ,4 2

y (0, 0) 5

3P ,0 2 25P 2 (2P, 24)

5P x 2 (P, 24)

y (23P, 2) 2.5 (P, 2)

22P

P ,0 2

3  , 10 2

45.

10

(3, 4)

2p, 2

0, 2

53. (3, 4)

P 2 ,5 2

P 2 ,8 4 y

(0, 4)

10 x 3 , 2 2

Domain: 1 - q , q 2 Range: 3 - 2, 10 4

P 2 2

P , 21 4

3P 2 ,5 2

x

(0, 5)

(P, 0) 22P

(0, 3) y 6

(0, 0)

Domain: 1 - q , q 2 Range: 3 - 1, 1 4 49.

P ,5 2

55.

3 5  , 4 3

y (2, 2)

(P, 3)

3 (0, 2) (2, 2)

2

(2P, 3) 3P ,1 2P x 2

3  , 3 2

2

P

Domain: 1 - q , q 2 Range: 3 2, 8 4

x

(1, 8)

9

57. 9 5 , 4 3

2.5

Domain: 1 - q , q 2 5 5 Range: c - , d 3 3

3 , 3 2

Domain: 1 - q , q 2 Range: 3 - 8, 2 4

(3, 0) 5 x

3 5 , 4 3

x

1 , 3 2

y (0, 0) 2.5 5

9 5  , 4 3

P ,0 2

P , 21 4

P , 21 4

Domain: 1 - q , q 2 Range: 3 1, 5 4

3P ,8 4

x

2P

24

(P, 5) P ,2 4 22

P 2 ,1 4

3P ,0 8

x

3P ,1 4

y 1.25

22.5

8

2P

P 2

P 2 ,1 2 22P

P ,5 2

(2P, 5) 3P 2 ,2 4

1 4p ; Period = 2 3

41.

Domain: 1 - q , q 2 Range: 3 - 1, 1 4

5P ,0 4

1 2

P ,0 8 y (0, 1) P ,1 2.5 2

47.

2P

1 2

3P ,0 8

P 2 ,0 8

3P ,0 4

(6, 4) 10 9  , 2 10 2

2

Domain: 1 - q , q 2 1 1 Range: c - , d 2 2

9 , 10 2

2 P 2 ,1 2

P , 24 2

y P 1 , 2.5 2 2 P ,0 4 22P

(3P, 22) (0, 0)

y

39.

Domain: 1 - q , q 2 Range: 3 - 4, 4 4

(2P, 0) (4P, 0)

17. Amplitude =

33. B

2P x

3P 2 , 24 2

Domain: 1 - q , q 2 Range: 3 - 2, 2 4

51.

3P ,4 2

31. A

25

5P x (2P, 22)

29. J

(2P, 0)

Domain: 1 - q , q 2 Range: 3 - 4, 4 4

43.

27. C

15. Amplitude = 6; Period = 2

3 ,0 2

2, (4, 2) 1 6, 2

y 2.5

1 2 (4, 2) 6, 1 2 10 x

(8, 1) (8, 1) 2.5 (0, 1)

Domain: 1 - q , q 2 Range: 3 - 1, 2 4

AN48

ANSWERS  Section 5.4

59. y = {3 sin 12x2 71. y = - cos a

4p xb + 1 3

p 73. y = 3 sin a x b 2

81. 1f ∘ g2 1x2 = sin 14x2 y 1.25

P 2 ,0 2

p 63. y = 5 cos a x b 4

61. y = {3 sin 1px2

y 2.25

x

1g ∘ f2 1x2 = 4 sin x

y 1.25

3 sin 12px2 4

67. y =

3 69. y = - sin a x b 2

12 p 1 s 30 Amplitude = 220 amp

y , 1 (, 0) 1.25 2 (2, 1) 2

(0, 0)

(2P, 22)

5 , 兹2 4 2

87. Period =

I 220

x

3 , 0 2 1 15

220 (P, 1)

P, 21 2

x

220

t

(c) I 1t2 = 22 sin 1120pt2 (d) Amplitude = 22 amp 1 s Period = 60

89. (a) Amplitude = 220 V; 1 period = s 60 (b) , (e)

x

3P , 24 2

25

79.

1g ∘ f2 1x2 = cos 1 - 2x2

P, 4 2

2P

85.

(P, 2)

P 22P

2 p

77.

2P x (0, 22)

y 5

75. y = - 4 cos 13x2

83. 1f ∘ g2 1x2 = - 2 cos x

P, 0 2

2P

1 65. y = - 3 cos a x b 2

V

22 22

I

t

1 30

220

[V0 sin 12pft2]2

V 20

R

0

=

R

sin 2 12pft2

V 20

1 and is of the form y = A cos 1vt2 + B, 2f 2 2 2 2 V0 V0 V0 V0 V 20 1 2p then A = and B = . Since = , then v = 4pf . Therefore, P1t2 = cos 14pft2 + = [1 - cos 14pft2]. 2R 2R 2f v 2R 2R 2R 2p p 2p ; emotional potential: v = ; intellectual potential: v = 93. (a) Physical potential: v = 23 14 33 y (b) 100 (c) No (d) Physical potential peaks at 15 days after 20th birthday. 95. 1.25 Emotional potential is 50% at 17 days, with a maximum at 10 days and a minimum at 24 days. Intellectual potential 2P x starts fairly high, drops to a minimum at 13 days, and rises to a maximum at 29 days. 91. (a) P1t2 =

(b) Since the graph of P has amplitude

2R

and period

36 0

5p 1 p 1 p 1 5p 1 3p p 5p 9p , b , a - , b , a , b , a , b 99. Answers may vary. a ,1b, a ,1b, a ,1b, a ,1b 3 2 3 2 3 2 3 2 4 4 4 4 5 7 106. (2 , 5) 107. 10 , 52, a - , 0 b , a - , 0 b 108. E F or ∅ 3 3

97. Answers may vary. a 105. 2x + h - 5

5.5 Assess Your Understanding (page 473) 3. origin; odd multiples of

p 2

4. y-axis; odd multiples of

p 2

5. y = cos x

11. sec x = 1 for x = - 2p, 0, 2p; sec x = - 1 for x = - p, p 17.

y P ,3 4

6

13. -

19.

9. 1 15. -

3p p p 3p ,- , , 2 2 2 2

y P ,4 4

21.

4

x

1 ,1 2 5 x

P 2 , 24 4

25.

2 (0, 2)

x (2P, 21)

Domain: 5x|x ≠ 4kp, k is an integer 6 Range: 1 - q , q 2

Domain: 5x|x does not equal an odd integer 6 Range: 1 - q , q 2 27.

y

(P, 1) 4P

1  , 1 2

Domain: 5x x ≠ kp, k is an integer 6 Range: 1 - q , q 2

kp , k is an odd integer f 2 Range: 1 - q , q 2 y 5

y

2P x

Domain: e x 2 x ≠

23.

7. 0

3p p p 3p ,- , , 2 2 2 2

16

2P P 2 , 23 4

6. T

(2P, 22)

1

(2P, 2) 2P x (P, 22)

Domain: e x 2 x ≠

kp , k is an odd integer f 2 Range: 5y|y … - 2 or y Ú 2 6

P ,3 2

y 15

3P ,3 2 2P

2

3P , 23 2

x P , 23 2

Domain: 5x 0 x ≠ kp, k is an integer 6 Range: 5y 0 y … - 3 or y Ú 3 6

AN49

ANSWERS   Section 5.6 29.

31.

y 16

(0, 4)

(22P, 24)

1  ,2 2

(2P, 24)

2 13 p

43.

6 13 p

y

2

y 2.5

P , 21 16

Domain: 5x 0 x ≠ 2pk, k is an odd integer 6 Range: 1 - q , q 2 1g ∘ f2 1x2 = 4 tan x

P ,1 16 P 4

P 2

, 0 2

(0, 0)

(c) ≈0.83

x (, 1)

y 5

2.5

2P

(b)

P

x

53.

25

y 5

3P 2 P 2

0 0

54. 15p - 2q2 2 125p2 + 10pq2 + 4q4 2 1 55. 2log 3(x + 3) + log 3(x - 2) - log 3(x - 1) 2 56. E - 1 , 3 F

57.

x

y 5

(d) ≈9.86 ft

3 , 兹2 4

 x 2

1g ∘ f2 1x2 = cot ( - 2x)

y

x

51. (a) L1u2 =

9P , 23 2

3P , 23 2

47. 1f ∘ g2 1x2 = - 2 cot x

P, 4 4

P 2 2 P , 24 2 4

x

3P ,1 2 5P x

Domain: 5x 0 x ≠ 3pk, k is an integer 6 Range: 5y|y … - 3 or y Ú 1 6

(0, 22)

y 5

y 10

9P 2 ,1 2 2

4 3 + cos u sin u = 3sec u + 4 cscu

, 1 4



39.

3 2

x

1

49.

P, 2

5 2P, 2 2

45. 1f ∘ g2 1x2 = tan 14x2

5P x

Domain: 5x 0 x ≠ 2pk, k is an odd integer 6 Range: 1 - q , q 2

5P

3 Domain: e x 2 x ≠ k, k is an odd integer f 4 Range: 5y|y … 1 or y Ú 3 6 41.

y 10

3 ,1 2 x

(P, 2)

(0, 1)

(3P, 0)

37.

2.5

(23P, 2) 1 , 2 2

Domain: 5x 0 x does not equal an integer 6 Range: 5y 0 y … - 2 or y Ú 2 6

(0, 3)

3  ,1 2

2.5 x

y 10

3  , 2 2

Domain:5x 0 x ≠ kp, k is an odd integer 6 Range: 5y 0 y … - 4 or y Ú 4 6 y 8

33.

3 ,2 2

6

5P x

35.

y

y 5 tan x

x

3P 2 P y 5 2cot x 1 2

y 6

26 (0, 23) 26

18 x (4, 22)

5.6 Assess Your Understanding (page 483)

P ,0 2

Phase shift = -

p Phase shift = 6

3P ,4 4

y 5

7. Amplitude = 3 Period = p

5. Amplitude = 2 2p Period = 3

1. phase shift 2. F 3. Amplitude = 4 Period = p p Phase shift = 2

P 2 ,2 6

7P ,4 4

P ,2 2

y 2.5

P 2 ,3 2

2P ,0 3

(0, 0) P , 24 4

3P ,0 2

(P, 0) 5P , 24 4

9. Amplitude = 4 Period = 2 Phase shift = 1 2 2 , 21 2 P

12 2

2

2 , 25 P

y 9

x

2

2 , 25 P

3 2 2 , 29 2 P

2

2 ,8 P

21

2 ,8 P

11

21 1

2 ,2 P

(0, 23)

5P , 22 6

13. Amplitude = 3 Period = p p Phase shift = 4

2 2 2 , 25 P

1 2 2 , 29 2 P

P ,0 3

P , 22 6

11. Amplitude = 3 Period = 2 2 Phase shift = p

2 p

y 1

P 2 ,0 4

P x

x

2P

3

2 ,2 P

y 5

2

P ,3 2

3P ,0 4

y 5

2

P ,0 4

3P ,0 4 x

(P, 23) (0, 23)

x (P, 23)

P ,0 4

1 b d or 2 y = 2 sin 12x - 12

P

x

P

3P ,0 4

15. y = 2 sin c 2 ax -

P ,3 2

(2P, 23)

P ,3 2

p 4

P ,0 4

2 1 17. y = 3 sin c ax + b d or 3 3 2 2 y = 3 sin a x + b 3 9

x

AN50 19.

ANSWERS  Section 5.6 3P 2 ,2 16

21.

5P ,2 16

y 5

P 2 ,0 4

P 2

5P 2 , 22 16

P ,0 4 x

7P P x5 x52 8 y 8 5 5P 2 ,3 8

1 s 15

3P x52 8

29. (a)

P (0, 0) x52 4

P 2 , 21 8

x

10 5

10

x

I 120

(c)

(1, 1) 1

x 1 , 1 2

x

1 4

x

3 4

(d) y = 9.46 sin (1.247x + 2.096) + 24.088 (e)

40

H 0

30 2 t 15

3 1 x 4 y 4

1  , 1 2

3P , 21 8

2p 11 y = 8.5 sin c ax b d + 24.5 5 4 or 2p 11p y = 8.5 sin a x b + 24.5 5 10

20

x

3

(b)

H

25.

P ,1 8

(1, 1) P 2

5P x5 8

30

Amplitude = 120 amp 1 Phase shift = s 90

3P 2 ,1 8

7P , 23 8

(0, 0)

27. Period =

3P ,3 8

P y x5 4 2.5

P x

P 2 , 23 8

3P , 22 16

23.

10 10

20 10 5

31. (a)

(b) y = 23.65 sin c

y 80

(c) 50 20

10

x

p p 2p 1x - 42 d + 51.75 or y = 23.65 sin a x b + 51.75 6 6 3

(d) y = 24.25 sin 10.493x - 1.9272 + 51.61 86 (e)

y 80 50

5

10

x 20 5

10

x 0

13 14

33. (a) 11:55 pm (b) y = 3.105 sin c

24p 24p 1x - 8.39582 d + 2.735 or y = 3.105 sin c x - 4.2485 d + 2.735 149 149

2p x - 1.39 b + 12.135 35. (a) y = 1.615 sin a 365 (b) 12.42 h (c) y

2p 37. (a) y = 6.96 sin a x - 1.39 b + 12.41 365 (b) 13.63 h (c) y

20

20

10

10 x 150 280 420

x 150 280 420

(d) The actual hours of sunlight on April 1, 2013, were 12.45 hours. This is close to the predicted amount of 12.42 hours. 41. f -1 1x2 =

2x - 9 4

7 42. e f 3

(c) 2.12 ft

43. 64x2 + 240xy + 225y2

(d) The actual hours of sunlight on April 1, 2013, were 13.47 hours. This is close to the predicted amount of 13.63 hours.

44. 2 213

Review Exercises (page 490) p 1 5 3. 315° 4. - 450° 5. 6. + 2 13 7. - 3 22 - 2 23 8. 3 9. 0 10. 0 11. 1 12. 1 13. 1 10 2 2 12 5 13 3 4 5 5 3 16. cos u = ; tan u = ; csc u = ; sec u = ; cot u = 17. sin u = - ; cos u = - ; csc u = - ; sec u = 5 3 4 3 4 13 13 12 3 4 3 5 4 12 12 13 18. sin u = ; cos u = - ; tan u = - ; csc u = ; cot u = 19. sin u = ; tan u = - ; csc u = ; sec u = 5 5 4 3 3 13 5 12 1.

5p 6

2.

20. cos u =

12 5 13 13 12 ; tan u = - ; csc u = - ; sec u = ; cot u = 13 12 5 12 5

22. sin u = 23. sin u =

21. sin u = -

2 22 1 3 22 22 ; cos u = ; tan u = - 2 22; csc u = ; cot u = 3 3 4 4

25 2 25 1 25 ; cos u = ; tan u = - ; csc u = 25; sec u = 5 5 2 2

14. - 1

15. 1

13 5 ; cot u = 5 12 13 5 - ; cot u = 5 12

210 3 210 210 ; cos u = ; csc u = - 210; sec u = ; cot u = 3 10 10 3

ANSWERS   Review Exercises 24.

25.

P, 2 8

y 2.5

P

y 5

26.

P, 3 2 P

x

y 5 P x 2

x

(P, 23) 3P , 22 8

Domain: ( - q , q ) Range: 3 - 3, 3 4

Domain: ( - q , q ) Range: 3 - 2, 2 4 27.

x5

y

P 2

28.

P 2 6

Range: 19 q , q 2 30.

kp , k is an odd integer f 2 Range: 1 - q , q 2

29.

p p + k # , k is an integer f 6 3

Domain: e x 2 x ≠ Range: 1 - q , q 2 31.

y

y 5P 4

P x

x

Domain: e x 2 x ≠

Domain: e x 2 x ≠

3P y x5 4 5

8

AN51

23

p + kp, k is an integer f 4

Domain: e x 2 x ≠

kp , k is an odd integer f 4 Range: {y y … - 4 or y Ú 4} 32.

y

5

x P , 24 2

y 25

5 x

4P x

5P x 7

p Domain: e x 2 x ≠ + kp, k is an integer f 4 Range: {y y … - 1 or y Ú 1} 33. Amplitude = 3; Period = 1

P, 4 6

36. Amplitude = 1 Period = 4p Phase shift = - p y 2.5

P , 24 2

39. y = 5 cos

x 4

p 40. y = - 7sin a x b 4

2 3 37. Amplitude = Period =

1 2

38. Amplitude = Period = 2

4p 3

Phase shift =

2p Phase shift = 3 5P x

P x

3p + k # 3p, k is an integerf 4 Range: 1 - q , q 2

Domain: ex 2 x ≠

Range: [ - 6, 2]

34. Amplitude = 2; Period =

35. Amplitude = 4 2p Period = 3 Phase shift = 0 y 5

Domain: ( - q , q )

y 2 3

y 1.25

12 P

2 3 6 p

x

2P x

41. 0.38

42. 1.02

43. Sine, tangent, cosecant, and cotangent: negative; cosine and secant: positive

23 1 2 1 , cos t = , tan t = - 13, csc t = , sec t = 2; cot t = 2 2 13 13 4 3 4 46. sin t = - , cos t = , tan t = 47. Domain: 5x 0 x ≠ np, n is an integer 6; range: 1 - q , q 2 ; period = p 5 5 3 48. (a) 32.34° (b) 63°10′48″ 3p 49. p ≈ 3.14 ft; ≈ 4.71 ft 2 50. 9p ≈ 28.26 in.; 2p ≈ 6.28 in. 51. Approximately 114.59 revolutions/hr 2 p 52. 0.1 revolution/sec = radian/sec 5 54. (a) y (c) y 1 53. (a) 90 90 15 (b) 220 70 70 1 (c) 50 50 180 x x 5 10 5 10 I (d) p 220 (d) y = 19.81 sin(0.543x - 2.296) + 75.66 (b) y = 20 sin c 1x - 42 d + 75 or 6 100 (e) 2p p 2 t b + 75 y = 20 sin a x 15 6 3 44. III

45. sin t = -

220

0

13 40

AN52 55.

ANSWERS  Review Exercises

p 5 2 1 , Ï3 2 2 p 5 (0, 1) 2 2 p 5 1 , Ï3 p 5 2Ï , Ï y P P 2 2 2P 2 2 2 3 P 3 2 2 3P p 5 Ï ,Ï 4 2 2 4 P 5P 6 6 3 1 Ï , 3 1 p5 p 5 2Ï , 2 2 2 2 p 5 (1, 0) 0 P p 5 (21, 0) 1 x Ï3 3 1 p 5 2Ï , 2 , 21 p5 2 2 2 2 11P 7P 6 6 2 2 5P 7P p 5 2Ï , 2Ï 2 2 4 4P 5P 4 3P 2 2 3 2 3 p 5 Ï , 2Ï p 5 2 1 , 2Ï3 2 2 p 5 (0, 21) 2 2 p 5 1 , 2Ï3 2 2

Chapter Test (page 492) 1.

13p 9

2. -

15. - 1.524

20p 9

3.

13p 180

16. 2.747

4. - 22.5°

17.

1 2

9. -

10. -

13 3

6. 135°

7.

sin U

cos U

tan U

sec U

csc U

cot U

+

+

+

+

+

+

U in QI

8. 0

1 2

5. 810°

11. 2

U in QII

+

-

-

-

+

-

U in QIII

-

-

+

-

-

+

U in QIV

-

+

-

+

-

-

12.

3 11 - 122

13. 0.292

2

18. -

14. 0.309

3 5

2 16 5 16 7 7 16 2 16 15 15 3 15 ; tan u = ; csc u = ; sec u = ; cot u = 20. sin u = ; tan u = ; csc u = ; 7 12 5 12 5 3 2 5 3 2 15 5 13 5 5 1146 1 12 13 7 153 sec u = ; cot u = 21. sin u = ; cos u = - ; csc u = ; sec u = - ; cot u = 22. 23. 24. 2 5 13 13 12 5 12 53 146 2

19. cos u = -

25. 11P , 0 2 2

26.

y 5 (24P , 2) (2P, 2)

7P , 0 2

27. y = - 3 sin a3x +

(2P, 3) y (0, 3) 4

28. 78.93 ft 2 29. 143.5 rpm

5P x 2

5P , 0 2

(5P, 22)

P

3p b 4

x

P (2P , 22) 2 , 0

Cumulative Review (page 492) 1 1. e - 1, f 2

2. y - 5 = - 3 1x + 22 or y = - 3x - 1

3. x2 + 1y + 22 2 = 16

2 4. A line; slope ; intercepts (6, 0) and 10, - 42 3

5. A circle; center 11, - 22; radius 3

y 5

6.

y 2

y 8 (2, 3)

(6, 0)

(4, 3) (3, 2) 6 x

5 x 10 x (1, 2)

(0, 4)

7. (a)

(b)

y 4.5

(c)

y 2.5

(1, 1) 2.5 x

2.5 x

(1, 1)

(e)

y 3 (e, 1)

(1, 1)

(0, 0) (1, 1) (0, 0)

(d)

y 4.5 1 1, e

(1, e)

(1, 0) 3

(0, 1) 2.5 x

1 e , 1

8. f -1 1x2 =

1 1x + 22 3

9. - 2

10.

3p 2 ,3 4

p 2 ,0 2 p 2 , 23 4

y p, 3 p ,0 4 5 2 P P x 2 3p, 23 4 (0, 0)

11. 3 -

3 13 2

12. y = 2 13x 2

x

y 3  ,1 2 (0, 0)    , 1 2

p 13. y = 3 cos a x b 6

(f)

x

y 2.5

p 2 , 21 4

p, 1 4 (0, 0) P x

ANSWERS   Section 6.3 14. (a) f 1x2 = - 3x - 3;

(b) f 1x2 = 1x - 12 2 - 6; 10, - 52 ,

m = - 3; 1 - 1, 02, 10, - 32

(2, 3)

(1, 0)

y 4

(2, 3)

(3.45, 0) 5 x

(1.45, 0)

5 x (0, 3) (1, 6)

15. (a) f 1x2 = y 10

(1, 6)

1 1x + 22 1x - 32 1x - 52 6

(b) R1x2 = y 10 (0, 5) (2, 0)

(0, 5)

(2, 0) (3, 0)

(c) We have that y = 3 when x = - 2 and y = - 6 when x = 1. Both points satisfy y = ae x. Therefore, for 1 - 2, 32 we have 3 = ae -2, which implies that a = 3e 2. But for 11, - 62 we have - 6 = ae 1, which implies that a = - 6e -1. Therefore, there is no exponential function y = ae x that contains 1 - 2, 32 and 11, - 62 .

1 - 16 + 1, 02, 1 16 + 1, 02

y 4

AN53

(5, 0) 6 x

1x + 22 1x - 32 1x - 52 3 1x - 22

(5, 0) x 4 (3, 0)

CHAPTER 6 Analytic Trigonometry 6.1 Assess Your Understanding (page 505) 9. - q 6 x 6 q

7. x = sin y 8. 0 … x … p 29. 0.51

31. - 0.38

33. - 0.12

35. 1.08

4p 37. 5

x- 2 5 Range of f = Domain of f - 1 = 3 - 3, 7 4 p p Range of f -1 = c - , d 2 2

53. f -1 1x2 = sin-1

59. f -1 1x2 =

1 x c sin - 1 a b - 1 d 2 3 Range of f = Domain of f - 1 = 3 - 3, 3 4 1 p 1 p Range of f -1 = c - - , - + d 2 4 2 4

77. (a)

p square units 3

(b)

5p square units 12

10. False

11. True

12. True 13. 0

15. -

p 2

17. 0

19.

p 4

21.

p 3

23.

5p 6

25. 0.10

27. 1.37

p p 1 41. 43. 45. 47. 4 49. Not defined 51. p 8 5 4 1 x 55. f -1 1x2 = cos - 1 a - b 3 2 57. f -1 1x2 = - tan - 1 1x + 32 - 1 Range of f = Domain of f - 1 = 3 - 2, 2 4 Range of f = Domain of f - 1 = 1 - q , q 2 p p p Range of f -1 = c 0, d - 1b Range of f -1 = a - 1 - , 3 2 2 22 1 61. e f 63. e - f 65. 5 23 6 67. 5 - 1 6 2 4 69. (a) 13.92 h, or 13 h, 55 min (b) 12 h (c) 13.85 h, or 13 h, 51 min 71. (a) 13.3 h, or 13 h, 18 min (b) 12 h (c) 13.26 h, or 13 h, 15 min 73. (a) 12 h (b) 12 h (c) 12 h (d) It is 12 h. 75. 3.35 min 3p 39. 8

2 81. c - , 2 d 3

79. 4250 mi

82. The graph passes the horizontal-line test. 83. f -1 1x2 = log 2 1x - 12

1,

1 2

y 5 (0, 1)

1

y

x

3

84. (2x + 1)- 2 (x2 + 3)- 2 (3x2 + 2)

(1, 2) 5 x

6.2 Assess Your Understanding (page 512) 23 22 2 23 3p p 22 13. 2 15. 22 17. 19. 21. 23. 25. 3 2 3 4 3 4 25 214 3 210 p p p p 2p 27. 29. 31. 33. 25 35. 37. 39. 41. 43. 45. 1.32 47. 0.46 49. - 0.34 51. 2.72 53. - 0.73 2 2 10 4 6 2 6 3 5 3p 5 p 1 u 2u2 - 1 2u2 - 1 1 3 67. 69. 71. 73. 75. 77. - 215 55. 2.55 57. 59. 61. 63. 65. u 13 4 4 13 6 0u0 0u0 21 + u2 21 - u2 79. (a) u = 31.89° (b) 54.64 ft in diameter (c) 37.96 ft high 81. (a) u = 22.3° (b) y0 = 2940.23 ft/s 4. x = sec y; Ú 1; 0; p

5. cosine

P

83.

25

6. False

7. True

87. - 5i, 5i, - 2, 2

8. True

9.

22 2

88. Neither

11. -

89.

7p 4

90.

5p in. ≈7.85 in. 2

5 0

6.3 Assess Your Understanding (page 519) p 5p p 5p 7p 11p p 2p 4p 5p p 3p 5p 7p ; 8. e u ` u = + 2pk, u = + 2pk, k is any integer f 9. False 10. False 11. e , f 13. e , , , f 15. e , , , f 6 6 6 6 6 6 3 3 3 3 4 4 4 4 p 7p 11p p 2p 4p 5p 4p 8p 16p 7p 11p 3p 7p 2p 4p 3p 5p 17. e , , f 19. e , , , f 21. e , , f 23. e , f 25. e , f 27. e , f 29. e , f 2 6 6 3 3 3 3 9 9 9 6 6 4 4 3 3 4 4 7.

AN54

ANSWERS  Section 6.3

3p 7p 11p p 5p p 5p 13p 17p 25p 29p , f 33. e f 35. e u ` u = + 2kp, u = + 2kp f ; , , , , , 4 4 6 6 6 6 6 6 6 6 6 5p 11p 17p 23p 29p 35p p 3p p 3p 5p 7p 9p 11p 5p + kp f ; , , , , , 39. e u ` u = + 2kp, u = + 2kp f ; , , , , , 37. e u ` u = 6 6 6 6 6 6 6 2 2 2 2 2 2 2 2 p 2p p 2p 4p 5p 7p 8p 8p 10p 8p 10p 20p 22p 32p 34p 41. e u ` u = + kp, u = + kp f ; , , , , , 43. e u ` u = + 4kp, u = + 4kp f ; , , , , , 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 p 2p 4p 3p 45. 50.41, 2.73 6 47. 51.37, 4.51 6 49. 52.69, 3.59 6 51. 51.82, 4.46 6 53. 52.08, 5.22 6 55. 50.73, 2.41 6 57. e , , , f 2 3 3 2 p 7p 11p p 5p p 2p 4p 3p p 5p p 5p p 5p 3p p 59. e , , f 61. e 0, , f 63. e , , , f 65. 5p 6 67. e , f 69. e 0, , p, f 71. e , , f 73. e f 75. 50 6 2 6 6 4 4 2 3 3 2 4 4 3 3 6 6 2 2 p 5p p 2p 4p 5p 77. e , f 79. No real solution 81. { - 1.31, 1.98, 3.84} 83. {0.52} 85. {1.26} 87. { - 1.02, 1.02} 89. {0, 2.15} 91. {0.76, 1.35} 93. , , , 3 3 3 3 3 3 31. e

95. (a) - 2p, - p, 0, p, 2p, 3p, 4p

(b)

5P 3 , 6 2 2

11P 3 , 6 2

11p 7p p 5p 13p 17p ,, , , , f 6 6 6 6 6 6 11p 7p p 5p 13p 17p (d) e x ` 6 x 6 or 6 x 6 or 6 x 6 f 6 6 6 6 6 6 (c) e -

13P 3 , 6 2 17P 3 , 6 2

y 3.75

3P x 7P 3 2 , 6 2 P, 3 6 2

97. (a) e x ` x = -

p + kp, k is any integer f 4

99. (a), (d) P 7 , 12 2

y 7

5P 7 , 12 2 g(x) 5

7 2

(b) -

p p p p 6 x 6 - or a - , - b 2 4 2 4

(b) e

p 5p , f 12 12

(c) e x `

(b) 123.6 m

111. 28.90° 113. Yes; it varies from 1.25 to 1.34 115. 1.47 117. If u is the original angle of incidence and f is the angle sin u = n2 . The angle of incidence of the of refraction, then sin f 1 emerging beam is also f, and the index of refraction is . n2 Thus, u is the angle of refraction of the emerging beam.

90

124. Amplitude: 2 Period: π p Phase shift: 2

9 - 217 9 + 217 , 4 4 1 210 123. tan u = - ; csc u = - 210; sec u = ; cot u = - 3 3 3

y 2.5 3P x

6.4 Assess Your Understanding (page 528) 3. identity; conditional 4. - 1 19. csc u # cos u =

5. 0

6. True 7. False

8. True

9.

1 cos u

11.

1 + sin u cos u

13.

1 sin u cos u

15. 2

17.

3 sin u + 1 sin u + 1

1 # cos u cos u = = cot u 21. 1 + tan2 1 - u2 = 1 + 1 - tan u2 2 = 1 + tan2 u = sec2 u sin u sin u sin u cos u sin2 u + cos2 u 1 1 23. cos u 1tan u + cot u2 = cos ua + b = cos u a b = cos u a b = = csc u cos u sin u cos u sin u cos u sin u sin u 1 25. tan u cot u - cos2 u = tan u # - cos2 u = 1 - cos2 u = sin2 u 27. 1sec u - 12 1sec u + 12 = sec2 u - 1 = tan2 u tan u 1 = 1 29. 1sec u + tan u2 1sec u - tan u2 = sec2 u - tan2 u = 1 31. cos2 u 11 + tan2 u2 = cos2 u sec2 u = cos2 u # cos2 u 2 2 2 2 2 2 33. 1sin u + cos u2 + 1sin u - cos u2 = sin u + 2 sin u cos u + cos u + sin u - 2 sin u cos u + cos u = sin2 u + cos2 u + sin2 u + cos2 u = 1 + 1 = 2 35. sec4 u - sec2 u = sec2 u 1sec2 u - 12 = 11 + tan2 u2 tan2 u = tan4 u + tan2 u 37. cos3x = cos x # cos2 x = cos x 11 - sin2 x2 = cos x - sin2 x cos x

2p 4p , f 3 3

2p 4p 2p 4p 6 x 6 f or a , b 3 3 3 3

0

122.

(b) e

f(x) 5 24 cos x

(c) e x `

0

4P ,2 3 2P x

P x f(x) 5 3 sin(2x) 1 2

109. (a) 30°, 60° 103. (a) 0 s, 0.43 s, 0.86 s (b) 0.21 s (c) (c) 30, 0.03 4 ∪ 30.39, 0.43 4 h 3 0.86, 0.89 4 105. (a) 150 mi (b) 6.06, 8.44, 15.72, 18.11 min 130 (c) Before 6.06 min, between 8.44 and 15.72 min, and after 18.11 min (d) No 107. 2.03, 4.91

g(x) 5 2 cos x 1 3

2P ,2 3

p 5p p 5p 6 x 6 f or a , b 12 12 12 12

121. x = log 6 y

y 5

101. (a), (d)

ANSWERS   Section 6.4

39. sec u - tan u =

AN55

1 sin u 1 - sin u # 1 + sin u 1 - sin2 u cos2 u cos u = = = = cos u cos u cos u 1 + sin u cos u 11 + sin u2 cos u 11 + sin u2 1 + sin u

41. 3 sin2 u + 4 cos2 u = 3 sin2 u + 3 cos2 u + cos2 u = 3 1sin2 u + cos2 u2 + cos2 u = 3 + cos2 u 11 + sin u2 11 - sin u2 cos2 u 1 - sin2 u 43. 1 = 1 = 1 = 1 - 11 - sin u2 = sin u 1 + sin u 1 + sin u 1 + sin u 1 cot y + 1 1 1 + sec u sin u 1 + tan y cot y cot y cot y + 1 cos u sin u = 47. + = + tan u = tan u + tan u = 2 tan u 45. = = + tan u = 1 - tan y 1 cot y - 1 cot y - 1 csc u cos u 1 cos u 1 cot y cot y sin u csc u + 1 1 1 + 1 + sin u csc u csc u csc u + 1 49. = = = 1 1 - sin u csc u - 1 csc u - 1 1 csc u csc u 51. 53.

11 - sin y2 2 + cos2 y 2 11 - sin y2 cos y 1 - 2 sin y + sin2 y + cos2 y 2 - 2 sin y 2 1 - sin y + = = = = = = 2 sec y cos y 1 - sin y cos y11 - sin y2 cos y11 - sin y2 cos y11 - sin y2 cos y11 - sin y2 cos y sin u 1 1 1 = = = cos u sin u - cos u sin u - cos u 1 - cot u 1 sin u sin u

55. 1sec u - tan u2 2 = sec2 u - 2 sec u tan u + tan2 u = 11 - sin u2 2

1 - sin u = = 11 - sin u2 11 + sin u2 1 + sin u 57.

cos u 2

-

2 sin u cos u 2

+

sin2 u cos u 2

=

1 - 2 sin u + sin2 u cos u 2

=

11 - sin u2 2 1 - sin2 u

cos u sin u + = 1 - tan u 1 - cot u

cos u sin2 u sin u cos u sin u cos2 u + + = + = sin u cos u cos u - sin u sin u - cos u cos u - sin u sin u - cos u 1 1 cos u sin u cos u sin u 1cos u - sin u2 1cos u + sin u2 cos2 u - sin2 u = = = sin u + cos u cos u - sin u cos u - sin u

59. tan u + 61.

1

sin u11 + sin u2 + cos2 u cos u sin u cos u sin u + sin2 u + cos2 u sin u + 1 1 = + = = = = = sec u 1 + sin u cos u 1 + sin u cos u 11 + sin u2 cos u 11 + sin u2 cos u 11 + sin u2 cos u

tan u + 1sec u - 12 tan u + 1sec u - 12 tan2 u + 2 tan u1sec u - 12 + sec2 u - 2 sec u + 1 tan u + sec u - 1 # = = tan u - sec u + 1 tan u - 1sec u - 12 tan u + 1sec u - 12 tan2 u - 1sec2 u - 2 sec u + 12 =

sec2 u - 1 + 2 tan u1sec u - 12 + sec2 u - 2 sec u + 1

=

2 sec2 u - 2 sec u + 2 tan u1sec u - 12

- 2 + 2 sec u sec2 u - 1 - sec2 u + 2 sec u - 1 2 sec u1sec u - 12 + 2 tan u1sec u - 12 2 1sec u - 12 1sec u + tan u2 = = = tan u + sec u 2 1sec u - 12 2 1sec u - 12

sin2 u - cos2 u sin u cos u sin2 u - cos2 u tan u - cot u cos u sin u cos u sin u = 63. = = = sin2 u - cos2 u tan u + cot u sin u cos u 1 sin2 u + cos2 u + cos u sin u cos u sin u sin u cos u sin2 u - cos2 u tan u - cot u cos u sin u cos u sin u 65. + 1 = + 1 = + 1 = sin2 u - cos2 u + 1 = sin2 u + 11 - cos2 u2 = 2 sin2 u tan u + cot u sin u cos u sin2 u + cos2 u + cos u sin u cos u sin u 1 sin u 1 + sin u + sec u + tan u cos u cos u cos u 1 + sin u # sin u sin u # 1 67. = = = = = tan u sec u cot u + cos u cos u cos u + cos u sin u cos u cos u 11 + sin u2 cos u cos u + cos u sin u sin u 1 - tan2 u

1 - tan2 u + 1 + tan2 u

2 2 = = = 2 cos2 u 1 + tan u 1 + tan2 u 1 + tan2 u sec2 u sec u - csc u sec u csc u 1 1 71. = = = sin u - cos u sec u csc u sec u csc u sec u csc u csc u sec u 2 2 1 1 - cos u sin u sin u 73. sec u - cos u = - cos u = = = sin u # = sin u tan u cos u cos u cos u cos u 1 1 1 + sin u + 1 - sin u 2 2 75. + = = = = 2 sec2 u 1 - sin u 1 + sin u 11 + sin u2 11 - sin u2 1 - sin2 u cos2 u sec u 11 + sin u2 sec u11 + sin u2 1 + sin u sec u sec u # 1 + sin u = = 77. = = 2 1 - sin u 1 - sin u 1 + sin u cos3 u 1 - sin u cos2 u 69.

2

+ 1 =

AN56

79.

ANSWERS  Section 6.4

1sec y - tan y2 2 + 1 csc y1sec y - tan y2

=

sec2 y - 2 sec y tan y + tan2 y + 1

1 1 sin y a b sin y cos y cos y 2 11 - sin y2 # sin y = 2 sin y = 2 tan y = cos y 1 - sin y cos y

=

2 sec2 y - 2 sec y tan y 1 1 - sin y a b sin y cos y

2 sin y 2 2 - 2 sin y # sin y cos y cos2 y cos2 y = = 1 - sin y 1 - sin y cos2 y sin y cos y

sin u + cos u sin u - cos u sin u cos u sin2 u + cos2 u 1 = + 1 - 1 + = = = sec u csc u cos u sin u cos u sin u cos u sin u cos u sin u 1sin u + cos u2 1sin2 u - sin u cos u + cos2 u2 sin3 u + cos3 u = = sin2 u + cos2 u - sin u cos u = 1 - sin u cos u 83. sin u + cos u sin u + cos u cos2 u - sin2 u cos2 u - sin2 u cos2 u - sin2 u = = = cos2 u 85. 2 1 - tan2 u sin u cos2 u - sin2 u 1 cos2 u cos2 u 2 2 2 2 [2 cos u - 1sin u + cos2 u2]2 12 cos u - 12 1cos2 u - sin2 u2 2 87. = = = cos2 u - sin2 u = 11 - sin2 u2 - sin2 u = 1 - 2 sin2 u 4 4 2 2 2 2 cos u - sin u 1cos u - sin u2 1cos u + sin u2 cos2 u - sin2 u 81.

89.

11 + sin u2 + cos u 11 + sin u2 + cos u 1 + 2 sin u + sin2 u + 2 11 + sin u2 cos u + cos2 u 1 + sin u + cos u # = = 1 + sin u - cos u 11 + sin u2 - cos u 11 + sin u2 + cos u 1 + 2 sin u + sin2 u - cos2 u = =

1 + 2 sin u + sin2 u + 2 11 + sin u2 1cos u2 + 11 - sin2 u2 1 + 2 sin u + sin u - 11 - sin u2 2

2

2 11 + sin u2 + 2 11 + sin u2 1cos u2 2 sin u11 + sin u2

=

=

2 + 2 sin u + 2 11 + sin u2 1cos u2

2 11 + sin u2 11 + cos u2 2 sin u 11 + sin u2

2 sin u + 2 sin2 u =

1 + cos u sin u

91. 1a sin u + b cos u2 2 + 1a cos u - b sin u2 2 = a2 sin2 u + 2ab sin u cos u + b2 cos2 u + a2 cos2 u - 2ab sin u cos u + b2 sin2 u

= a2 1sin2 u + cos2 u2 + b2 1cos2 u + sin2 u2 = a2 + b2 tan a + tan b tan a + tan b tan a tan b tan a + tan b = = = 1tan a + tan b2 # = tan a tan b 93. tan b + tan a cot a + cot b 1 1 tan a + tan b + tan a tan b tan a tan b

95. 1sin a + cos b2 2 + 1cos b + sin a2 1cos b - sin a2 = 1sin2 a + 2 sin a cos b + cos2 b2 + 1cos2 b - sin2 a2 = 2 cos2 b + 2 sin a cos b = 2 cos b 1cos b + sin a2 = 2 cos b 1sin a + cos b2 97. ln 0 sec u 0 = ln 0 cos u 0 -1 = - ln 0 cos u 0 99. ln 0 1 + cos u 0 + ln 0 1 - cos u 0 = ln 1 0 1 + cos u 0 0 1 - cos u 0 2 = ln 0 1 - cos2 u 0 = ln 0 sin2 u 0 = 2 ln 0 sin u 0 1 1 cos2 x 1 - cos2 x sin2 x sin x - cos x = = = = sin x # = sin x # tan x = f 1x2 101. g 1x2 = sec x - cos x = cos x cos x cos x cos x cos x cos x 1 - sin u cos u 1 - sin u # 1 + sin u cos u # cos u 1 - sin2 u cos2 u = = 103. f 1u2 = cos u 1 + sin u cos u 1 + sin u 1 + sin u cos u cos u 11 + sin u2 cos u 11 + sin u2 cos2 u cos2 u = 0 = g 1u2 = cos u11 + sin u2 cos u 11 + sin u2 p p 105. 216 + 16 tan2 u = 216 21 + tan2 u = 4 2sec2 u = 4 sec u, since sec u 7 0 for - 6 u 6 2 2 1 2 1 2 cos2 u 1 2 - cos2 u 2 107. 1200 sec u (2 sec u - 1) = 1200 a 2 - 1 b = 1200 a 2 b = 1200 ¢ ≤ 2 cos u cos u cos u cos u cos u cos u cos2 u =

1200 (1 + 1 - cos2 u) cos u 3

113. Maximum, 1250 x - 1 114. 1f ∘ g2 1x2 = x - 2

=

1200 (1 + sin2 u)

115. sin u =

cos3 u 5 12 5 13 13 12 ; cos u = - ; tan u = - ; csc u = ; sec u = - ; cot u = 13 13 12 5 12 5

116. -

2 p

6.5 Assess Your Understanding (page 540) 1 1 1 22 - 262 15. - 1 22 + 262 17. 2 - 23 4 4 1 1 1 2 25 11 25 2 25 19. - 1 26 + 222 21. 26 - 22 23. 25. 0 27. 1 29. - 1 31. 33. (a) (b) (c) (d) 2 4 2 2 25 25 5 5 + 12 23 4 - 3 23 - 3 - 4 23 4 + 3 23 25 23 + 48 12 - 5 23 - 5 + 12 23 - 240 + 169 23 35. (a) (b) (c) (d) 37. (a) (b) (c) (d) 10 10 10 39 26 26 26 69 2 22 - 2 22 + 23 - 2 22 + 23 9 - 4 22 1 - 2 26 23 - 2 22 8 22 - 9 23 39. (a) (b) (c) (d) 41. 43. 45. 3 6 6 7 6 6 5 p p p 47. sin a + ub = sin cos u + cos sin u = 1 # cos u + 0 # sin u = cos u 2 2 2 49. sin 1p - u2 = sin p cos u - cos p sin u = 0 # cos u - 1 - 12 sin u = sin u 5. -

6. -

7. False

8. False

9. False

10. True

11.

1 1 26 + 222 4

13.

ANSWERS   Section 6.5

AN57

51. sin 1p + u2 = sin p cos u + cos p sin u = 0 # cos u + 1 - 12 sin u = - sin u 0 - tan u 3p tan p - tan u 3p 3p cos u + cos sin u = 1 - 12 cos u + 0 # sin u = - cos u = = - tan u 55. sin a + u b = sin 53. tan 1p - u2 = 1 + tan p tan u 1 + 0 # tan u 2 2 2 57. sin 1a + b2 + sin 1a - b2 = sin a cos b + cos a sin b + sin a cos b - cos a sin b = 2 sin a cos b sin 1a + b2 sin a cos b + cos a sin b sin a cos b cos a sin b 59. = = + = 1 + cot a tan b sin a cos b sin a cos b sin a cos b sin a cos b 61.

cos 1a + b2

=

cos a cos b

cos a cos b - sin a sin b cos a cos b

=

cos a cos b

-

sin a sin b

cos a cos b cos a cos b sin a cos b + cos a sin b

= 1 - tan a tan b

sin a cos b cos a sin b + cos a cos b cos a cos b cos a cos b tan a + tan b 63. = = = = sin a cos b - cos a sin b sin a cos b cos a sin b sin 1a - b2 sin a cos b - cos a sin b tan a - tan b cos a cos b cos a cos b cos a cos b sin 1a + b2

sin a cos b + cos a sin b

cos a cos b - sin a sin b

cos a cos b sin a sin b sin a sin b sin a sin b sin a sin b cot a cot b - 1 = = 65. cot 1a + b2 = = = sin a cos b + cos a sin b sin a cos b cos a sin b sin 1a + b2 sin a cos b + cos a sin b cot b + cot a + sin a sin b sin a sin b sin a sin b cos 1a + b2

cos a cos b - sin a sin b

1 # 1 1 sin a sin b sin a sin b csc a csc b 1 1 67. sec 1a + b2 = = = = = sin a sin b cos a cos b - sin a sin b cos a cos b cos 1a + b2 cos a cos b - sin a sin b cot a cot b - 1 sin a sin b sin a sin b sin a sin b 69. sin 1a - b2 sin 1a + b2 = 1sin a cos b - cos a sin b2 1sin a cos b + cos a sin b2 = sin2 a cos2 b - cos2 a sin2 b = 1sin2 a2 11 - sin2 b2 - 11 - sin2 a2 1sin2 b2 = sin2 a - sin2 b 71. sin 1u + kp2 = sin u cos kp + cos u sin kp = 1sin u2 1 - 12 k + 1cos u2 102 = 1 - 12 k sin u, k any integer 73. 87.

23 2

24 25

75. -

u 21 - y2 - y 21 + u

2

77. -

33 65

79.

63 65

81.

48 + 25 23 39

: - q 6 u 6 q; - 1 … y … 1

89.

83.

4 3

85. u 21 - v 2 - v 21 - u2: - 1 … u … 1; - 1 … v … 1

uy - 21 - u2 21 - y2 y21 - u + u 21 - y 2

2

: - 1 … u … 1; - 1 … y … 1

p 7p f 91. e , 2 6

93. e

p f 4

95. e

11p f 6

p p - ab , cos a - ab = cos b. 2 2 p p p p p p If y Ú 0, then 0 … a … , so a - ab and b both lie on c 0, d . If y 6 0, then … a 6 0, so a - ab and b both lie on a , p d . 2 2 2 2 2 2 p p p - a = b, or a + b = . Either way, cos a - ab = cos b implies 2 2 2 -1

97. Let a = sin

-1

y and b = cos

y.

Then sin a = cos b = y, and since sin a = cos a

1 1 1 p = = cot b, and since tan a = cot a and b = tan-1 y. Because y ≠ 0, a, b ≠ 0. Then tan a = y y tan b 2 p p p p p Because y 7 0, 0 6 a 6 , and so a - ab and b both lie on a0, b . Then cot a - ab = cot b implies - a = 2 2 2 2 2 101. sin 1sin-1 y + cos-1 y2 = sin 1sin-1 y2 cos 1cos-1 y2 + cos 1sin-1 y2 sin 1cos-1 y2 = 1y2 1y2 + 21 - y2 21 - y2 = sin 1x + h2 - sin x cos x sin h - sin x 11 - cos h2 sin x cos h + cos x sin h - sin x sin h 1 103. = = = cos x # - sin x # h h h h 99. Let a = tan-1

105. (a) tan(tan - 1 1 + tan - 1 2 + tan - 1 3) = tan((tan - 1 1 + tan - 1 2) + tan - 1 3) = tan(tan - 1 1) + tan(tan - 1 2)

1 + 2 + 3 1 - 1#2 = = -1 -1 1 + 2 tan(tan 1) + tan(tan 2) #3 1 - 1 - 1#2#3 1 -1 -1 1 - tan(tan 1)tan(tan 2) 1 - tan(tan - 1 1)tan(tan - 1 2)

+ 3

p - ab = cot b. 2 p b, or a = - b. 2 y2 + 1 - y2 = 1 - cos h h - ab , cot a

tan(tan - 1 1 + tan - 1 2) + tan(tan - 1 3) 1 - tan(tan - 1 1 + tan - 1 2) tan(tan - 1 3)

3 + 3 -1 -3 + 3 0 = = = = 0 3 # 1 + 9 10 1 3 -1

p p p (b) From the definition of the inverse tangent function, 0 6 tan - 1 1 6 , 0 6 tan - 1 2 6 , and 0 6 tan - 1 3 6 , 2 2 2 3p so 0 6 tan - 11 + tan - 12 + tan - 13 6 . 2 3p On the interval a0, b , tan u = 0 if and only if u = p. Therefore, from part (a), tan - 11 + tan - 12 + tan - 13 = p. 2 tan u2 - tan u1 m2 - m1 107. tan u = tan 1u2 - u1 2 = = 1 + tan u1 tan u2 1 + m1m2 1 + 1x + 12 1x - 12 1 + tan a tan b 2 1 + x2 - 1 2x2 = 2a b = 2a b = 2a b = = x2 109. 2 cot 1a - b2 = tan 1a - b2 tan a - tan b 1x + 12 - 1x - 12 x + 1 - x + 1 2

AN58

ANSWERS  Section 6.5

1 5 112. a - , - b , ( - 5, 1) 3 9

113. 510°

p - ub 2

sin a

p p 111. tan is not defined; tan a - ub = 2 2

p cos a - ub 2

cos u = cot u. sin u

=

9p cm2 ≈14.14 cm2 2

114.

115. sin u = -

2 25 25 25 1 ; cos u = ; csc u = ; sec u = 25; cot u = 5 5 2 2

6.6 Assess Your Understanding (page 550) 1. sin2 u; 2 cos2 u; 2 sin2 u 24 25

9. (a)

11. (a) 13. (a)

7 25

(b) -

2 22 3

4 22 9

2 25 5

(c)

(b)

1 3

(c)

(b) -

7 9

(c)

3 5

(c)

15. (a) -

4 5

(b)

17. (a) -

3 5

(b) -

19.

2. 1 - cos u

32 - 22 2

4 5

C

(c)

C

3. sin u

(d) -

25 5

3 + 26 6

(d)

23 3

(d)

5 + 2 25 10

26 3

310 15 - 252

23. -

(e) -

C

(d)

C

24 7

5. False

(e) - 2 22

4 22 7

5 - 2 25 10

(f)

32 + 23 2

7. (a)

24 25

(b)

7 25

(c)

210 10

(d)

3 210 10

(e)

4 3

(e) -

3 + 26

C 3 - 26

5 + 2 25

(f) (e)

2

25.

(f)

4 5

31.

10

32 + 22

C 5 - 2 25 3 4

(f) -

or 49 + 4 25 or 2 + 25

10 - 210

C 10 + 210

or -

= 12 - 222 32 + 22

45. sin 15u2 = 16 sin5 u - 20 sin3 u + 5 sin u

1 1 - tan2 u = = 49. cot 12u2 = tan 12u2 2 tan u

1

cot u - 1

cot 2 u = 1 b 2a cot u

cot 2 u 2 cot u

1 -

1 1 = = cos 12u2 2 cos2 u - 1

411 - 2 210 1 - 210 or 3 3

27. -

32 - 22 2

= - 1

1cos u cos2 u - sin2 u = 1 + sin 12u2 1 + 2 sin u cos u sin2 u sin u cos u - sin u cos u sin u sin u sin u cot u = = = cos u + sin u cos u sin u cot u + sin u sin u sin u

57. sec2

u = 2

y 59. cot 2 = 2

=

1 u cos2 a b 2

=

2 - sec u 2

sin u

=

sec2 u 2 - sec2 u

sec u 2

- sin u2 1cos u + sin u2

+ cos2 u + 2 sin u cos u

=

1cos u - sin u2 1cos u + sin u2 1sin u + cos u2 1sin u + cos u2

=

cos u - sin u cos u + sin u

- 1 + 1

1 2 = 1 + cos u 1 + cos u 2

1 sec y + 1 1 + 1 1 + cos y sec y sec y sec y + 1 # sec y sec y + 1 = = = = = = 1 cos y 1 cos y 1 sec y 1 sec y sec y 1 sec y - 1 y 1 tan2 a b 1 + cos y sec y sec y 2

1 1 61. = 1 u 1 + 1 + tan2 a b 1 2 sin 13u2

cot 2 u - 1 # cot u cot 2 u - 1 = 2 2 2 cot u cot u

1

u 1 - tan2 a b 2

63.

=

1

53. cos2 12u2 - sin2 12u2 = cos[2 12u2] = cos 14u2 cos 12u2

47. cos4 u - sin4 u = 1cos2 u + sin2 u2 1cos2 u - sin2 u2 = cos 12u2

2

1 2 sec2 u

55.

-

1 -

cos 13u2 cos u

=

+ +

1 3

33.

43. cos 13u2 = 4 cos3 u - 3 cos u

51. sec 12u2 =

(f)

or 45 + 2 26 or 22 + 23

4 7 210 215 35. 37. 39. 3 8 4 3 1 - cos 12u2 2 1 1 1 1 4 2 2 b = [1 - 2 cos 12u2 + cos2 12u2] = - cos 12u2 + cos2 12u2 41. sin u = 1sin u2 = a 2 4 4 2 4 1 1 1 1 + cos 14u2 1 1 1 1 3 1 1 - cos 12u2 + a b = - cos 12u2 + + cos 14u2 = - cos 12u2 + cos 14u2 = 4 2 4 2 4 2 8 8 8 2 8

29. -

24 7

22 2

1 10 + 210 2C 5

(d) -

6. False

(f) - 2

3 - 26 6

(e) -

1 10 - 210 2C 5

21. 1 - 22

4. True

1 + cos u - 11 - cos u2 cos u cos u 1 + cos u 2 cos u # 1 + cos u = = = cos u cos u 1 + cos u + 1 - cos u 1 + cos u 2 cos u 1 + cos u

sin 13u2 cos u - cos 13u2sin u sin u cos u

=

sin 13u - u2

1 12 sin u cos u2 2

=

2 sin 12u2 sin 12u2

= 2

ANSWERS   Section 6.7

65. tan 13u2 = tan 1u + 2u2 =

67.

4p 3p 5p p p 2p , , , p, , , f 3 2 3 3 2 3

p 3p , 2 2

= 1 -

0 1 - cos 12u2 0

b

2

1/2

75. No real solution

95. (a) W = 2D(csc u - cot u) = 2Da

(b)

y20

1 - tan2 u = ln 0 sin2 u 0 1/2 = ln 0 sin u 0 77. e 0,

p 5p , p, f 3 3

79.

p 2p 4p 5p 69. e , , , f 3 3 3 3

23 2

81.

7 25

83.

1 cos u 1 - cos u u b = 2D = 2D tan sin u sin u sin u 2

y20 22

cos u 1sin u - cos u2 16 22 = 12 cos u sin u - 2 cos2 u2 32 y20 22 = [sin 12u2 - cos 12u2 - 1] 32

97. (a) R =

2 tan u tan u - tan3 u + 2 tan u 3 tan u - tan3 u 1 - tan2 u = = 2 2 tan u 12 tan u2 1 - tan u - 2 tan u 1 - 3 tan2 u

tan u +

1 - tan u tan 12u2

1 1ln 0 1 - cos 12u2 0 - ln 22 = ln a 2

73. e 0, 93.

tan u + tan 12u2

3p or 67.5° 8

(d)

24 7

85.

71. e 0, 24 25

87.

2p 4p , f 3 3 1 5

89.

25 7

91. 0,

p 5p , p, 3 3

(b) u = 24.45°

u = 67.5° 3p a radians b makes R 8 largest. R = 18.75 ft

20

(c) 32 12 - 222 ≈ 18.75 ft

45

90 0

99. A =

1 1 u u 1 h 1base2 = h a base b = s cos # s sin = s2 sin u 2 2 2 2 2

101. sin 12u2 =

4x 4 + x2

103. -

1 4

a 2 sin a b 2 a 2 tan a b 2

a cos a b 2 a a a 105. = = = = 2 sin a b cos a b = sin a2 # b = sin a 1 2 2 2 1 + z2 2 a 2 a 1 + tan a b sec a b a 2 2 cos2 a b 2 2z

a 2 tan a b 2

107.

y 1.25

2P x

p 22 = 44 - 36 - 22 24 4 p 22 = 44 + 36 + 22 cos 24 4 3 3 111. sin u + sin 1u + 120°2 + sin3 1u + 240°2 = sin3 u + 1sin u cos 120° + cos u sin 120°2 3 + 1sin u cos 240° + cos u sin 240°2 3 109. sin

= sin3 u + a -

3 3 1 23 1 23 sin u + cos u b + a - sin u cos u b 2 2 2 2

1 1 13 23 cos3 u - 9 cos2 u sin u + 3 23 cos u sin2 u - sin3 u2 - 1sin3 u + 3 23 sin2 u cos u + 9 sin u cos2 u + 3 23 cos3 u2 8 8 3 3 9 3 3 3 1 2 3 2 3 = sin u - cos u sin u = 3 sin u - 3 sin u11 - sin u2 4 = 14 sin u - 3 sin u2 = - sin 13u2 (from Example 2) 113. 4 4 4 4 4 2 = sin3 u +

115. y =

1 x - 4 2

y

116.

(3, 16)

117.

23 + 1 2

y (22, 2) (2, 2) 2

118.

(0, 7) (21, 0)

2422

(7, 0)

2 4

x

22 (24, 22) (0, 22)(4, 22)

x

6.7 Assess Your Understanding (page 556) 1 23 1 23 22 1 1 1 a - 1 b 3. - a + 1 b 5. 7. 3 cos 12u2 - cos 16u2 4 9. 3 sin 16u2 + sin 12u2 4 11. 3 cos 12u2 + cos 18u2 4 2 2 2 2 2 2 2 2 1 1 u 13. 3cos u - cos 13u2 4 15. 3 sin 12u2 + sin u4 17. 2 sin u cos 13u2 19. 2 cos 13u2 cos u 21. 2 sin 12u2 cos u 23. 2 sin u sin 2 2 2 2 sin 12u2 cos u sin 14u2 + sin 12u2 2 sin 13u2 cos u sin 13u2 sin u + sin 13u2 = = cos u 27. = = = tan 13u2 25. 2 sin 12u2 2 sin 12u2 cos 14u2 + cos 12u2 2 cos 13u2 cos u cos 13u2 1.

29.

cos u - cos 13u2 sin u + sin 13u2

=

2 sin 12u2 sin u

2 sin 12u2 cos u

=

sin u = tan u cos u

31. sin u 3sin u + sin 13u2 4 = sin u3 2 sin 12u2 cos u4 = cos u 3 2 sin 12u2 sin u 4 = cos u c 2 # 33.

sin 14u2 + sin 18u2

cos 14u2 + cos 18u2

=

2 sin 16u2 cos 12u2

2 cos 16u2 cos 12u2

=

AN59

sin 16u2

cos 16u2

= tan 16u2

1 3 cos u - cos 13u2 4 = cos u 3 cos u - cos 13u2 4 d 2

AN60

ANSWERS  Section 6.7

sin 14u2 + sin 18u2

39.

sin 14u2 - sin 18u2

2 sin 16u2 cos 1 - 2u2

=

=

sin 16u2

#

cos 12u2

= tan 16u2 3 - cot 12u2 4 = -

2 sin 1 - 2u2 cos 16u2 cos 16u2 - sin 12u2 a + b a + b a - b a - b 2 sin cos sin cos sin a + sin b a + b a - b 2 2 2 2 # = = = tan 37. cot a - b a + b a + b a - b sin a - sin b 2 2 2 sin cos cos sin 2 2 2 2 35.

sin a + sin b cos a + cos b

2 sin = 2 cos

a + b 2 a + b 2

cos cos

a - b 2 = a - b 2

sin

tan 16u2 tan 12u2

a + b

a + b 2 = tan a + b 2 cos 2

41. 1 + cos 12u2 + cos 14u2 + cos 16u2 = 3 1 + cos 16u2 4 + 3 cos 12u2 + cos 14u2 4 = 2 cos2 13u2 + 2 cos 13u2 cos 1 - u2 = 2 cos 13u2 3cos 13u2 + cos u4 = 2 cos 13u2 3 2 cos 12u2 cos u4 = 4 cos u cos 12u2 cos 13u2 p p 2p 4p 3p 5p p 2p 3p 4p 6p 7p 8p 9p 43. e 0, , , , p, , , f 45. e 0, , , , , p, , , , f 3 2 3 3 2 3 5 5 5 5 5 5 5 5 47. (a) y = 2 sin 11906 pt2 cos 1512 pt2 (b) ymax = 2 (c)

49. Iu = Ix cos2 u + Iy sin2 u - 2Ixy sin u cos u = Ixcos2 u + Iy sin2 u - Ixy sin 2u

2

= 0

0.01

= Iv =

2

cos 2u + 1 1 - cos 2u b + Iy a b - Ixy sin 2u 2 2 Iy Iy Ix Ix cos 2u + + - cos 2u - Ixy sin 2u 2 2 2 2 Ix - Iy Ix + Iy + cos 2u - Ixy sin 2u 2 2 1 - cos 2u cos 2u + 1 Ix sin2 u + Iy cos2 u + 2Ixy sin u cos u = Ix a b + Iy a b + Ixy sin 2u 2 2 Iy Iy Ix Ix - cos 2u + cos 2u + + Ixy sin 2u 2 2 2 2 Ix + Iy Ix - Iy cos 2u + Ixy sin 2u 2 2

= Ix a

= =

51. sin 12a2 + sin 12b2 + sin 12g2 = 2 sin 1a + b2 cos 1a - b2 + sin 12g2 = 2 sin 1a + b2 cos 1a - b2 + 2 sin g cos g = 2 sin 1p - g2 cos 1a - b2 + 2 sin g cos g = 2 sin g cos 1a - b2 + 2 sin g cos g = 2 sin g[cos 1a - b2 + cos g] a - b + g a - b - g p - 2b 2a - p p p cos b = 4 sin g cos cos = 4 sin g cos a - b b cos aa - b = 2 sin ga2 cos 2 2 2 2 2 2

= 4 sin g sin b sin a = 4 sin a sin b sin g

sin 1a - b2 = sin a cos b - cos a sin b

53.

sin 1a + b2 = sin a cos b + cos a sin b

sin 1a - b2 + sin 1a + b2 = 2 sin a cos b 1 sin a cos b = 3 sin 1a + b2 + sin 1a - b2 4 2 a + b a - b a + b a - b a + b a - b 2b 1 2a cos = 2 # c cos a + b + cos a b d = cos + cos = cos a + cos b 55. 2 cos 2 2 2 2 2 2 2 2 2 p p 2 26 57. {7} 58. Amplitude: 5; Period: ; Phase shift: 59. 2 4 7 x + 5 p p 60. f -1 1x2 = sin-1 a b ; Range of f = Domain of f -1 = 3 - 8, - 2 4 ; Range of f -1 = c - , d 3 2 2

Review Exercises (page 559) p 5p p p 3p 3p 3p p p 6. 7. 8. 9. 10. 11. 12. 13. 14. 0.9 15. 0.6 16. 5 17. Not defined 6 3 4 4 8 4 3 7 9 p 2 23 4 4 1 x p p 18. 19. p 20. - 23 21. 22. 23. 24. f -1(x) = sin - 1 a b ; Range of f = Domain of f -1 = [ - 2, 2]; Range of f -1 = c - , d 6 3 5 3 3 2 6 6 0 u 0 25. f -1 1x2 = cos-1 13 - x2; Range of f = Domain of f -1 = 3 2, 4 4 ; Range of f -1 = 3 0, p 4 26. 21 - u2 27. u 2u2 - 1 28. tan u cot u - sin2 u = 1 - sin2 u = cos2 u 29. cos2 u11 + tan2 u2 = cos2 u sec2 u = 1 30. 5 cos2 u + 3 sin2 u = 2 cos2 u + 3 1cos2 u + sin2 u2 = 3 + 2 cos2 u 11 - cos u2 2 + sin2 u 2 11 - cos u2 sin u 1 - 2 cos u + cos2 u + sin2 u 1 - cos u + = = = = 2 csc u 31. sin u 1 - cos u sin u11 - cos u2 sin u 11 - cos u2 sin u 11 - cos u2 cos u cos u cos u 1 1 32. = = = cos u - sin u cos u - sin u sin u 1 - tan u 1 cos u cos u 1.

p 2

2.

p 2

3.

p 4

4. -

p 6

5.

ANSWERS  Chapter Test 1 sin u 1 1 # 1 - sin u = = = 1 1 + sin u 1 + sin u 1 - sin u 1 + sin u 1 - cos2 u sin2 u 1 - cos u = = = sin u # 34. sec u - cos u = cos u cos u cos u csc u 33. = 1 + csc u

35.

38.

1 - sin u

=

cos2 u

sin u = sin u tan u cos u

sin u11 - cos2 u2 1 + cos u sin3 u 1 - cos u = sin u 11 + cos u2 # = = csc u 1 - cos u 1 - cos u 1 - cos u cos u sin u cos2 u - sin2 u 1 - 2 sin2 u = = sin u cos u sin u cos u sin u cos u

36. cot u - tan u = 37.

1 - sin u 1 - sin2 u

AN61

sin 1a - b2 cos a cos b

cos 1a - b2 cos a cos b

sin a cos b - cos a sin b

=

cos a cos b

=

cos a cos b + sin a sin b

=

sin a cos b cos a cos b

=

cos a cos b

cos a cos b cos a cos b

-

cos a sin b cos a cos b

= tan a - tan b

sin a sin b

+

= 1 + tan a tan b

cos a cos b

u sin u = sin u 39. 11 + cos u2 atan b = 11 + cos u2 # 2 1 + cos u 2 cos u1cos2 u - sin2 u2 cos u cos 2u cos2 u - sin2 u ba b = = = cot 2 u - 1 2 sin u sin 2u 2 sin u cos u sin2 u sin 12u2 sin 13u2 cos u - sin u cos 13u2 = = 1 41. 1 - 8 sin2 u cos2 u = 1 - 2 12 sin u cos u2 2 = 1 - 2 sin2 12u2 = cos 14u2 42. sin 12u2 sin 12u2 sin 12u2 + sin 14u2 2 sin 13u2 cos 1 - u2 43. = = tan 13u2 cos 12u2 + cos 14u2 2 cos 13u2 cos 1 - u2 40. 2 cot u cot 2u = 2 a

44. 46.

cos 12u2 - cos 14u2

cos 12u2 + cos 14u2 1

47.

23 + 1

48. 1

49. 1

23 - 2 22 6

55. (a) -

63 65

(b)

16 65

- 23 - 2 22 6

(b) (c)

2 cos 13u2 cos 1 - u2

50. 0

22 2 22 12 - 3 27 9 + 4 27 53. (a) (b) 20 20 54. (a)

- 2 sin 13u2 sin 1 - u2

- tan u tan 13u2 =

51.

23 + 1

52.

- tan u tan 13u2 = tan 13u2 tan u - tan u tan 13u2 = 0

45.

1 + 2 26 6 33 65

63 16

24 25

(f) -

119 169

- 23 + 2 22 6

(g)

2 213 13

(h) -

8 22 + 9 23 23

210 10 23 2

(f) -

7 9

p 5p , f 3 3

74. e

p p 3p 3p , , , f 4 2 4 2

23 2 27(5 + 3 217) 1 4 25 1 230 26 33 - 25 13 (d) Not defined (e) (f) (g) (h) 58. 57. (a) 1 (b) 0 (c) 59. 9 9 9 6 6 2493 7 217 - 15 48 + 25 23 22 24 7 p 5p 2p 5p 3p 7p p 3p p 2p 4p 5p 61. 62. 63. 64. e , f 65. e , f 66. e , f 67. e 0, , p, f 68. e , , , f 60. 39 10 25 25 3 3 3 3 4 4 2 2 3 3 3 3 56. (a)

69. 50.25, 2.89 6 77. - 1.11

(b)

70. e 0,

1 - 2 26 6

(e)

2 12

42 - 22 2

23 - 1 3 27 27 12 + 3 27 12 - 3 27 7 2 (c) (d) (e) (f) (g) (h) 20 8 25 9 + 4 27 25 2 22 23 + 2 22 23 - 2 22 4 22 22 - 13 23 + 2 22 7 (c) (d) (g) (h) (e) (f) 6 2 2 C 6 9 1 + 2 26

(d) -

13 - 1

(c)

2p 4p , p, f 3 3

71. e 0,

81. 51.11 6

78. 1.77 79. 1.23 80. 2.90

1 - cos 30° 86. sin 15° = = B 2 T

p 5p , f 6 6

13 2

1 -

(d)

72. e

p p 5p , , f 6 2 6

82. 50.87 6

(e) -

73. e

83. 52.22 6

84. e -

23 f 2

(g)

23 3

(h)

75. e

p , pf 2

85. {0}

2 - 23 22 - 13 = ; 4 2 22 # 23 22 # 1 26 22 26 - 22 sin 15° = sin(45° - 30°) = sin 45° cos 30° - cos 45° sin 30° = = = ; 2 2 2 2 4 4 4 2 2 4 12 232 42 - 23 2 - 23 8 - 4 23 6 - 2 212 + 2 26 - 22 J R = = = = = ¢ ≤ 2 4 4 # 4 16 16 4 2

=

C

Chapter Test (page 561) p p p 7 4 2. 3. 4. 5. 3 6. 7. 0.39 8. 0.78 9. 1.25 10. 0.20 6 4 5 3 3 csc u + cot u csc u + cot u # csc u - cot u csc2 u - cot 2 u 1 11. = = = sec u + tan u sec u + tan u csc u - cot u 1sec u + tan u2 1csc u - cot u2 1sec u + tan u2 1csc u - cot u2 sec u - tan u 1 sec u - tan u # sec u - tan u = = = 1sec u + tan u2 1csc u - cot u2 sec u - tan u csc u - cot u 1sec2 u - tan2 u2 1csc u - cot u2 1.

12. sin u tan u + cos u = sin u #

sin u sin2 u cos2 u sin2 u + cos2 u 1 + cos u = + = = = sec u cos u cos u cos u cos u cos u

76. 0.78

AN62

ANSWERS  Chapter Test sin u cos u sin2 u cos2 u sin2 u + cos2 u 1 2 2 + = + = = = = = 2 csc 12u2 cos u sin u sin u cos u sin u cos u sin u cos u sin u cos u 2 sin u cos u sin 12u2

13. tan u + cot u = 14.

sin 1a + b2 tan a + tan b

=

sin a cos b + cos a sin b sin a cos b + cos a sin b sin a cos b + cos a sin b = = sin a cos b sin a cos b + cos a sin b sin b cos a sin b sin a + + cos a cos b cos a cos b cos a cos b cos a cos b

=

sin a cos b + cos a sin b 1

#

cos a cos b sin a cos b + cos a sin b

= cos a cos b

15. sin 13u2 = sin 1u + 2u2 = sin u cos 12u2 + cos u sin 12u2 = sin u # 1cos2 u - sin2 u2 + cos u # 2 sin u cos u = sin u cos2 u - sin3 u + 2 sin u cos2 u = 3 sin u cos2 u - sin3 u = 3 sin u11 - sin2 u2 - sin3 u = 3 sin u - 3 sin3 u - sin3 u = 3 sin u - 4 sin3 u sin u cos u sin2 u - cos2 u 11 - cos2 u2 - cos2 u sin2 u - cos2 u tan u - cot u cos u sin u sin u cos u = = 16. = = 1 - 2 cos2 u = 2 2 2 2 tan u + cot u sin u cos u 1 sin u + cos u sin u + cos u + cos u sin u sin u cos u

2 213 1 25 - 32 25 12 285 2 + 23 20. 21. 22. 5 49 39 4 3p 7p 11p 15p 27. e , , , f 28. 50.285, 3.427 6 29. 50.253, 2.889 6 8 8 8 8 18. 2 + 23

19.

23.

26 2

24.

22 2

25. e

17.

p 2p 4p 5p , , , f 3 3 3 3

1 1 26 + 222 4 26. 50, 1.911, p, 4.373 6

Cumulative Review (page 561) 1. e

- 1 - 213 - 1 + 213 , f 6 6

4.

y 8

2. y + 1 = - 1 1x - 42 or x + y = 3; 6 22; y 5

5. (0, 5)

(6, 5)

6.

3. x-axis symmetry; 10, - 32, 10, 32, 13, 02

y 0.5 2P

(0, 1)

x

6 x y  2

(3, 2) 8 x

y 2.5 y  x3

7. (a)

11, 22

(0, 0) (1, 1)

(b)

y 5

1 1, e

(1, 1) 2.5 x y  ex

(c)

(1, e) (0, 1)

1,

y 1.25

5 x f 1(x)  ln x

P 2 , 21 2

(d)

P 2

P ,1 2

(21, P)

y 3

y 5 sin x

(0, 0)

P

f 21(x) 5 cos21 x

(0, 1) P x y 5 cos x (P, 21)

x

P ,0 2

3

f 1(x)  兹x f 21(x) 5 sin21 x

8. (a) -

2 22 3

(b)

22 4

(c)

4 22 9

(d)

7 9

(e)

C

3 + 2 22 6

11. (a) f 1x2 = 12x - 12 1x - 12 2 1x + 12 2; 1 multiplicity 1; 1 and - 1 multiplicity 2 2 1 (b) 10, - 12; a , 0 b ; 1 - 1, 02; (1, 0) (c) y = 2x5 2 (d) 2

(f) -

C

3 - 2 22 6

25 5

10. (a) -

2 22 3

(b) -

(e) Local minimum value - 1.33 at x = - 0.29 Local minimum value 0 at x = 1 Local maximum value 0 at x = - 1 Local maximum value 0.10 at x = 0.69 (f) 1, 0 y 1.25

2

(1, 0) 2

9.

P 21, 2 2

2 22 3

(c)

7 9

(d)

4 22 9

1 f 2 (b) 5 - 1, 1 6

1 (c) 1 - q , - 12 ∪ a - , q b 2 (d) 1 - q , - 1 4 ∪ 31, q 2

(1, 0)

(g) Increasing: 1 - q , - 12, 1 - 0.29, 0.692, 11, q 2 Decreasing: 1 - 1, - 0.292, 10.69, 12

2

CHAPTER 7 Applications of Trigonometric Functions 7.1 Assess Your Understanding (page 571) 4. False

5. True

6. angle of elevation 7. True

8. False

9. sin u =

5 12 5 12 13 13 ; cos u = ; tan u = ; cot u = ; sec u = ; csc u = 13 13 12 5 12 5

11. sin u =

2 213 3 213 2 3 213 213 ; cos u = ; tan u = ; cot u = ; sec u = ; csc u = 13 13 3 2 3 2

13. sin u =

23 1 23 2 23 ; cos u = ; tan u = 23; cot u = ; sec u = 2; csc u = 2 2 3 3

26 3

12. (a) e - 1, -

1.25 x (0.69, 0.10) (0, 1) (0.29, 1.33)

2

(e)

ANSWERS  Section 7.3

15. sin u =

AN63

26 23 22 26 ; cos u = ; tan u = 22; cot u = ; sec u = 23; csc u = 3 3 2 2

25 2 25 1 25 ; cos u = ; tan u = ; cot u = 2; sec u = ; csc u = 25 5 5 2 2 19. 0 21. 1 23. 0 25. 0 27. 1 29. a ≈ 13.74, c ≈ 14.62, A = 70° 31. b ≈ 5.03, c ≈ 7.83, A = 50° 33. a ≈ 0.71, c ≈ 4.06, B = 80° 35. b ≈ 10.72, c ≈ 11.83, B = 65° 37. b ≈ 3.08, a ≈ 8.46, A = 70° 39. c ≈ 5.83, A ≈ 59.0°, B ≈ 31.0° 41. b ≈ 4.58, A ≈ 23.6°, B ≈ 66.4° 43. 23.6° and 66.4° 45. 4.59 in.; 6.55 in. 47. (a) 5.52 in. or 11.83 in. 49. 70.02 ft 51. 985.91 ft 53. 137.37 m 55. 58.8° 57. (a) 111.96 ft/sec or 76.3 mi/hr (b) 82.42 ft/sec or 56.2 mi/hr (c) Under 18.8° 59. (a) 2.4898 * 1013 miles (b) 0.000214° 61. 554.52 ft 63. S76.6°E 65. The embankment is 30.5 m high. 67. 3.83 mi 69. 1978.09 ft 71. 60.27 ft 73. The buildings are 7979 ft apart. 75. 69.0° 77. 38.9° 79. The white ball should hit the top cushion 4.125 ft from the upper left corner. 84. Yes 17. sin u =

85.

26 - 22 32 - 23 or 4 2

87. e

86. 0.236, 0.243, 0.248

p 7p 11p , , f 2 6 6

7.2 Assess Your Understanding (page 583) sin A sin B sin C = = 6. False 7. False 8. ambiguous case 9. a ≈ 3.23, b ≈ 3.55, A = 40° 11. a ≈ 3.25, c ≈ 4.23, B = 45° a b c 13. C = 95°, c ≈ 9.86, a ≈ 6.36 15. A = 40°, a = 2, c ≈ 3.06 17. C = 120°, b ≈ 1.06, c ≈ 2.69 19. A = 100°, a ≈ 5.24, c ≈ 0.92 21. B = 40°, a ≈ 5.64, b ≈ 3.86 23. C = 100°, a ≈ 1.31, b ≈ 1.31 25. One triangle; B ≈ 30.7°, C ≈ 99.3°, c ≈ 3.86 27. One triangle; C ≈ 36.2°, A ≈ 43.8°, a ≈ 3.51 29. No triangle 31. Two triangles; C1 ≈ 30.9°, A1 ≈ 129.1°, a1 ≈ 9.07 or C2 ≈ 149.1°, A2 ≈ 10.9°, a2 ≈ 2.20 33. No triangle 35. Two triangles; A1 ≈ 57.7°, B1 ≈ 97.3°, b1 ≈ 2.35 or A2 ≈ 122.3°, B2 ≈ 32.7°, b2 ≈ 1.28 37. (a) Station Able is about 143.33 mi from the ship. Station Baker is about 135.58 mi from the ship. (b) Approximately 41 min 39. 1490.48 ft 41. 381.69 ft 43. The tree is 39.39 ft high. 45. Adam receives 100.6 more frequent flyer miles. 47. 84.7°; 183.72 ft 49. 2.64 mi 51. 38.5 in. 53. 449.36 ft 55. 187,600,000 km or 101,440,000 km 57. The diameter is 252 ft. 4. oblique 5.

a - b a b sin A sin B sin A - sin B = = = = 59. c c c sin C sin C sin C

2 sin a

A - B A - B A - B A + B p C sin a sin a b cos a b b cos a - b b 2 2 2 2 2 2 = = C C C C C 2 sin cos sin cos cos 2 2 2 2 2

1 sin c 1A - B2 d 2 C cos 2

1 1 1 a - b tan c 1A - B2 d tan c 1A - B2 d tan c 1A - B2 d 2 2 2 a - b c 61. = = = = = a + b a + b C 1 p 1 C cot cos c 1A - B2 d tan a - b tan c 1A + B2 d c 2 2 2 2 2 C sin 2 69.

4 66. e - 3 , - , 3 f 3

(23P, 4) (24P, 0)

67. 3 25 ≈ 6.71 68. -

215 7

y 5 (P, 4) (2P, 0) (4P, 0) 5P x

(22P, 0) (2P, 24)

(3P, 24) (0, 0)

7.3 Assess Your Understanding (page 590) 3. Cosines 4. Sines 5. Cosines 6. False 7. False 8. True 9. b ≈ 2.95, A ≈ 28.7°, C ≈ 106.3° 11. c ≈ 3.75, A ≈ 32.1°, B ≈ 52.9° 13. A ≈ 48.5°, B ≈ 38.6°, C ≈ 92.9° 15. A ≈ 127.2°, B ≈ 32.1°, C ≈ 20.7° 17. c ≈ 2.57, A ≈ 48.6°, B ≈ 91.4° 19. a ≈ 2.99, B ≈ 19.2°, C ≈ 80.8° 21. b ≈ 4.14, A ≈ 43.0°, C ≈ 27.0° 23. c ≈ 1.69, A = 65.0°, B = 65.0° 25. A ≈ 67.4°, B = 90°, C ≈ 22.6° 27. A = 60°, B = 60°, C = 60° 29. A ≈ 33.6°, B ≈ 62.2°, C ≈ 84.3° 31. A ≈ 97.9°, B ≈ 52.4°, C ≈ 29.7° 33. A = 85°, a = 14.56, c = 14.12 35. A = 40.8°, B = 60.6°, C = 78.6° 37. A = 80°, b = 8.74, c = 13.80 39. Two triangles: B1 = 35.4°, C1 = 134.6°, c1 = 12.29; B2 = 144.6°, C2 = 25.4°, c2 = 7.40 41. B = 24.5°, C = 95.5°, a = 10.44 43. 165 yd 45. (a) 26.4° (b) 30.8 h 47. (a) 63.72 ft (b) 66.78 ft (c) 92.8° 49. (a) 492.58 ft (b) 269.26 ft 51. 342.33 ft 53. The footings should be 7.65 ft apart. 1 - cos u 1 - cos u u b 1 d = 2r = 2r sin . 2 B 2 2 u u Since, for any angle in 10, p2, d is strictly less than the length of the arc subtended by u—that is, d 6 ru—then 2r sin 6 ru, or 2 sin 6 u. 2 2 u u u u Since cos 6 1, then, for 0 6 u 6 p, sin u = 2 sin cos 6 2 sin 6 u. If u Ú p, then, since sin u … 1, sin u 6 u. Thus sin u 6 u for all u 7 0. 2 2 2 2

55. Suppose 0 6 u 6 p. Then, by the Law of Cosines, d 2 = r 2 + r 2 - 2r 2 cos u = 4r 2 a

57. sin

C 1 - cos C = = 2 2 R B =

1 -

12s - 2b2 12s - 2a2

B

4ab

a 2 + b2 - c 2 c 2 - 1a - b2 2 1c + a - b2 1c + b - a2 2ab 2ab - a2 - b2 + c 2 = = = 2 4ab 4ab 4ab B B B =

1s - a2 1s - b2

B

ab

AN64 64.

ANSWERS  Section 7.3

1 2 ,0 2

y

65. e

(4, 9)

8

5,

y=2 22 1 0, 2 3

11 2

5 x

x=3

ln 3 f ≈ E 3.819 F ln 4 - ln 3

66. 2 26 7 26 7 5 26 sin u = ; csc u = ; sec u = - ; cot u = 7 12 5 12 67. y = - 3 sin 14x2

7.4 Assess Your Understanding (page 596) 1 2. ab sin C 2

1 3. 2s(s - a)(s - b)(s - c); (a + b + c) 4. True 5. 2.83 7. 2.99 9. 14.98 11. 9.56 13. 3.86 15. 1.48 17. 2.82 19. 30 2 a2 sin B sin C 1 1 a sin B b = 27. 0.92 29. 2.27 31. 5.44 33. 9.03 sq ft 35. $5446.38 21. 1.73 23. 19.90 25. K = ab sin C = a sin Ca 2 2 sin A 2 sin A 1 37. The area of home plate is about 216.5 in.2 39. K = r 2 1u + sin u2 41. The ground area is 7517.4 ft2. 2 1  OC  #  AC  1 1  OC   AC  = # = sin a cos a 2 2 1 1 2  BC   OC  1 1 1 2 # =  OB 2 sin b cos b (b) Area ∆OCB =  BC   OC  =  OB 2 2 2  OB  OB

43. (a) Area ∆OAC =

(c) Area ∆OAB =

 BD 1 1 1  BD  OA =  OB =  OB sin 1a + b2 2 2 2  OB

 OC 

(e) Area ∆OAB = Area ∆OAC + Area ∆OCB 1 1 1  OB  sin(a + b) = sin a cos a +  OB  2 sin b cos b 2 2 2 1 sin(a + b) = sin a cos a +  OB  sin b cos b  OB  sin(a + b) =

cos a 1 = (d) =  OB cos b  OC 

cos b cos a

sin a cos a +

cos a sin b cos b cos b

sin(a + b) = sin a cos b + cos a sin b

 OB 45. 31,145 ft 2 47. (a) The perimeter and area are both 36. (b) The perimeter and area are both 60. 1 1 a sin B sin C 49. K = ah = ab sin C 1 h = b sin C = 2 2 sin A A B 1 C C C C + b = 180° - (180° - C) = 90° + , and sin a90° + b = cos a- b = cos , since cosine is an even function. 2 2 2 2 2 2 2 A B B A c sin sin c sin sin 2 2 2 2 Therefore, r = . = C C cos sin a90° + b 2 2 3s - (a + b + c) A B C s - a s - b s - c 3s - 2s s + cot + cot = + + = = = 53. cot 2 2 2 r r r r r r

51. ∠POQ = 180° - a

22 27 214 3 22 3 27 214 ; cos t = ; tan t = ; csc t = ; sec t = ; cot t = 3 3 7 2 7 2 1 1 - sin2 u cos2 u cos u - sin u = = = cos u # = cos u cot u 61. csc u - sin u = sin u sin u sin u sin u 58. Maximum value; 17 59. 1 - q , - 32 h [ - 1 , 32

60. sin t =

7.5 Assess Your Understanding (page 606) 2. simple harmonic; amplitude 3. simple harmonic; damped 4. True 5. d = - 5 cos 1pt2 7. d = - 6 cos 12t2 9. d = - 5 sin 1pt2 2p 3 11. d = - 6 sin 12t2 13. (a) Simple harmonic (b) 5 m (c) sec (d) oscillation/sec 15. (a) Simple harmonic (b) 6 m (c) 2 sec 3 2p 1 1 (d) oscillation/sec 17. (a) Simple harmonic (b) 3 m (c) 4p sec (d) oscillation/sec 2 4p 19. (a) Simple harmonic (b) 2 m (c) 1 sec (d) 1 oscillation/sec

21.

25.

27.

y 3P

y5x y 5 cos x 2P x

23.

y 1.25

P

t

2P

29.

y 3P

y5x y 5 2sin x 3P

x

y 1.25

t

y y 5 cos x 1.25 y 5 sin x

3P

x

ANSWERS   Review Exercises 31.

33. (a) f 1x2 =

y 2.5

(b)

3P y 5 sin x

x y 5 sin(2x)

1 3 cos x - cos (3x) 4 2

1 3 cos(6x) + cos(2x) 4 2

35. (a) G(x) = (b)

y 1

1 y 5 cos x 2

1 cos(2x) 2

y5

y 1

2P x 1 y 5 2 cos(3x) 2

2P x

y5

37. (a) H(x) = sin(4x) + sin(2x)

39. (a) d = - 10e -0.7t/50 cos a

(b) y 2

y 5 sin(2x) y 5 sin(4x)

(b)

4p2 0.49 tb B 25 2500

AN65

1 cos(6x) 2

41. (a) d = - 18e -0.6t/60 cos a (b)

10

p2 0.36 tb B 4 3600

18

2P x 0

0

25

ⴚ18

10

43. (a) d = - 5e -0.8t/20 cos a (b)

45. (a) The motion is damped. The bob has mass m = 20 kg with a damping factor of 0.7 kg/s. (b) 20 m leftward (c) 20

4p2 0.64 tb B 9 400

5

0

20

47. (a) The motion is damped. The bob has mass m = 40 kg with a damping factor of 0.6 kg/s. (b) 30 m leftward (c) 30

15 0

0

25

35

5 30

20

(d) 18.33 m leftward

(e) d S 0

51. v = 1040p; d = 0.80 cos(1040pt) 53. v = 880p; d = 0.01 sin(880pt) 55. (a) V

49. (a) The motion is damped. The bob has mass m = 15 kg with a damping factor of 0.9 kg/s. (b) 15 m leftward (c) 15

(d) 28.47 m leftward (e) d S 0 57.

1.25

2.5

0

2

3 t 2.5

0

30

(b) At t = 0, 2; at t = 1, t = 3 (c) During the approximate intervals 0.35 6 t 6 0.67, 1.29 6 t 6 1.75, and 2.19 6 t … 3

15

(d) 12.53 m leftward (e) d S 0 61.

63. y =

1

1 sin x x

y =

1

0

0.1

0 0

1 x3

sin x

0.05

4x - 3 1x2 = x - 1

66. log 7 a

xy3 x + y

b

20.06

67. {4}

3 210 68. (a) 10

7P 0

5P

20.3

65. f

y =

sin x

5P

20.3

-1

1 x2

210 (b) 10

3P

20.015

1 (c) 3

Review Exercises (page 610) 23 4 3 4 3 5 5 1 23 2 23 ; cos u = ; tan u = ; cot u = ; sec u = ; csc u = 2.sin u = ; cos u = ; tan u = 23; cot u = ; sec u = 2; csc u = 5 5 3 4 3 4 2 2 3 3 3. 0 4. 1 5. 1 6. A = 70°, b ≈ 3.42, a ≈ 9.40 7. a ≈ 4.58, A ≈ 66.4°, B ≈ 23.6° 8. C = 75°, b ≈ 2.6, c ≈ 5 9. B ≈ 56.8°, C ≈ 23.2°, b ≈ 4.25 10. No triangle 11. b ≈ 3.32, A ≈ 62.8°, C ≈ 17.2° 12. No triangle 13. No triangle 14. A ≈ 46.6°, B ≈ 104.5°, C ≈ 28.9° 15. c ≈ 2.32, A ≈ 16.1°, B ≈ 123.9° 16. A ≈ 35°, B ≈ 105°, c ≈ 3.3 17. A ≈ 39.6°, B ≈ 18.6°, C ≈ 121.9° 18. Two triangles: B1 ≈ 13.4°, C1 ≈ 156.6°, c1 ≈ 6.86 or B2 ≈ 166.6°, C2 ≈ 3.4°, c2 ≈ 1.02 19. b ≈ 11.52, c ≈ 10.13, C ≈ 60° 20. a ≈ 5.23, B ≈ 46.0°, C ≈ 64.0° 21. 1.93 22. 30.31 23. 6 24. 3.80 25. 10.57 26. 0.76 in2 1. sin u =

AN66

ANSWERS  Review Exercises

27. 34.85° and 55.15°

28. 23.32 ft

29. 2.15 mi

30. 132.55 ft/min

31. 12.7°

32. 29.97 ft

33. 6.22 mi

34. (a) 131.8 mi

(b) 23.1°

(c) 0.21 hr

p 37. 76.94 in. 38. 79.69 in. 39. d = - 3 cos a t b 2 1 40. (a) Simple harmonic (b) 6 ft (c) p s (d) oscillation/s p 1 41. (a) Simple harmonic (b) 2 ft (c) 2 s (d) oscillation/s 2

35. 8798.67 sq ft

36. S4.0°E

42. (a) d = - 15e -0.75t>80 cos a (b)

0.5625 4p2 tb 6400 B 25

43. (a) The motion is damped. The bob has mass m = 20 kg with a damping factor of 0.6 kg/s. (b) 15 m leftward (c) 15 (d) 13.92 m leftward (e) d S 0

15

0

25

0

15

44.

y 2

y 5 2 sin x

x 2P y 5 cos(2 x)

25

15

Chapter Test (page 612) 25 2 25 1 25 ; cos u = ; tan u = ; csc u = 25; sec u = ; cot u = 2 2. 0 3. a = 15.88, B ≈ 57.5°, C ≈ 70.5° 5 5 2 2 4. b ≈ 6.85, C = 117°, c ≈ 16.30 5. A ≈ 52.4°, B ≈ 29.7°, C ≈ 97.9° 6. b ≈ 4.72, c ≈ 1.67, B = 105° 7. No triangle 8. c ≈ 7.62, A ≈ 80.5°, B ≈ 29.5° 9. 15.04 square units 10. 19.81 square units 11. 61.0° 12. 1.3° 13. The area of the shaded region is 9.26 cm2. 14. 54.15 square units 15. Madison will have to swim about 2.23 miles. 16. 12.63 square units 17. The lengths of the sides are 15, 18, and 21. pt pt 18. d = 5(sin 42°) sin a b or d ≈ 3.346 sin a b 3 3 1. sin u =

Cumulative Review (page 613) 1 1. e , 1 f 3

2. 1x + 52 2 + 1y - 12 2 = 9

3. 5x x … - 1 or x Ú 4 6

4.

y 5

(5, 1)

5.

y 3

y 2.5 P

2.5 x

x

1 x

6. (a) 7. (a)

2 25 5

(b)

25 5

(c) -

4 5

(d) -

3 5

(e)

(b)

60

B

5 - 15 10

(f) -

B

5 + 15 10 (c)

1.5

(d)

10 0

0

0

4

8. (a)

(b)

y 5

(d)

y 5

5 x

(h)

y 1.25

2P

x

4 0

(e)

y 5

(f)

y 9

5 x 5 x

(g)

0

50

(c)

y 10

4

4

1.5

0

8

5 x

(i)

y 1.25

y 2

2P

x

P

x

y 5

8 x 5 x

9. Two triangles: A1 ≈ 59.0°, B1 ≈ 81.0°, b1 ≈ 23.05 or A2 ≈ 121.0°, B2 ≈ 19.0°, b2 ≈ 7.59 1 10. e - 2i, 2i, , 1, 2 f 3

ANSWERS   Section 8.2 12x + 12 1x - 42

11. R1x2 =

1x + 52 1x - 32

; Domain: 5x x ≠ - 5, x ≠ 3 6

1 4 Intercepts: a - , 0 b , (4, 0), a0, b 2 15 No symmetry Vertical asymptotes: x = - 5, x = 3 Horizontal asymptote: y = 2 Intersects: a 12. 52.26 6

y 10

8

AN67

26 ,2 11

10 x 10

8

(4, 0) 1  ,0 2

4

0,

4 15

26 , 2b 11

13. 51 6

5 f 4

14. (a) e -

(e) {x| - 8 … x … 3} or [ - 8, 3]

(b) 52 6

(f)

y 6

(c) e

- 1 - 3 213 - 1 + 3 213 , f 2 2 (g)

(d) e x ` x 7 -

5 5 f or a - , q b 4 4

y 15

(0, 5)

(8, 0)

(3, 0) 10 x

(1.25, 0) 5 x

(0, 24) (2.5, 30.25)

CHAPTER 8 Polar Coordinates; Vectors 8.1 Assess Your Understanding (page 623) 8. r cos u; r sin u

5. pole; polar axis 6. True 7. False 19.

17.

(3, 90)

(2, 0)

9. A

11. C

21.

O

90

13. B

27.

O

25. 135 O

P 6

O

15. A

23.

P 6, 6

4, 2

(2, 135)

2P 3

2

2P 3

O

O

29.

O (22, 2P) 2P

31.

(a) a5, -

5, 2P 3

4p b 3

(b) a - 5,

5p b 3

(c) a5,

8p b 3

O

2P 3

P 3

2

P 3

(a) 12, - 2p2 (b) 1 - 2, p2 (c) 12, 2p2

3P

33.

21, 2

(22, 3P)

O

35.

37.

1, P 2

P 2

23, 2

O

P 4

O

3p b 2 3p (b) a - 1, b 2 5p b (c) a1, 2

2

(a) a1, -

41. 1 - 2, 02

5p b 4 7p b (b) a - 3, 4 11p (c) a3, b 4

1 23 b 47. a - , 2 2

45. 1 22, - 222

49. 12, 02

53. 1 - 4.98, - 3.852

18 b b ≈ 137.36, 105.5°2 (c) 1 - 3, - 352 5 5 9 89. 2 or 0 positive real zeros; 1 negative real zero 90. a - , b 91. (0, - 11) 4 2

19 f 15

(b) a2 2349, 180° + tan

43. 1 - 3 23, 32

p p 55. 13, 02 57. 11, p2 59. a 22, - b 61. a2, b 4 6 3 26 69. r 2 cos2 u - 4r sin u = 0 63. 12.47, - 1.022 65. 19.30, 0.472 67. r 2 = or r = 2 2 1 2 1 71. r 2 sin (2u) = 1 73. r cos u = 4 75. x2 + y2 - x = 0 or ax - b + y2 = 2 4 77. 1x2 + y2 2 3>2 - x = 0 79. x2 + y2 = 4 81. y2 = 8 1x + 22

51. 1 - 2.57, 7.052

P 4

(a) a3, -

83. (a) 1 - 10, 362 88. e

39. 10, 32

-1

a-

(d) a 21234, 180° + tan

-1

a

35 b b ≈ 135.13, 265.1°2 3

8.2 Assess Your Understanding (page 637) 7. polar equation 8. False

9. - u

13. x + y = 16; circle, radius 4, center at pole 2

2

y U5 U5

3P 4

5P 4

P U5 4

U5 3P U5 2

11. True

12. 2n; n

15. y = 23x; line through pole, making an p angle of with polar axis 3

17. y = 4; horizontal line 4 units above the pole y

y

x 1 2 3 4 5 U50

U5P

U5

P 2

10. p - u

7P 4

3P U5 4

P U5 2

x 1 2 3 4 5 U50

U5P

U5

P U5 3 P U5 4

U5

5P 4 U5

3P 2

7P 4

U5

3P 4

U5

U5

P 4

x 1 2 3 4 5 U50

U5P

U5

P 2

U5

5P 4 U5

3P 2

7P 4

AN68

ANSWERS  Section 8.2

19. x = - 2; vertical line 2 units to the left of the pole

21. 1x - 12 2 + y2 = 1; circle, radius 1, center 11, 02 in rectangular coordinates

y U5 U5

3P 4

P 4

x 1 2 3 4 5 U50

5P 4

U5

U5

7P 4

3P U5 2

U5

3P 4

U5

U5

U5

P 4

x 1 2 3 4 5 U50

7P U5 4

5P 4 U5

3P U5 4

P 2

4

P 4

x 6 8 10 U 5 0

U5 U5

7P 4

3P U5 4

U5P

U5

3P U5 4

P 2

U5

U5

P 4

7P 4

3P 2

3P U5 4

U5

P U5 4

7P 4

3P U5 4

x 1 2 3 4 5 U50

U5P

U5

5P 4 U5

3P 2

7P 4

U5

3P 4

P 4

U5

5P 4

U5

7P 4

3P U5 2

U5

3P 4

U5

P 2

P 4

x 4 6 8 10 U 5 0

U5P

U5

U5

U5

5P 4

7P 4

3P 2

y P U5 4

7P 4

3P U5 4

U5

U5

P 4

x 1 2 3 4 5 U50

U5P

U5

P 2

5P 4

3P 2

U5

U5

7P 4

U5

P 4

3P 2

55. Spiral U5

y P 2

U5

P 4

x 1 2 3 4 5 U50

U5P

U5

U5

x 1 2 3 4 5 U50

y P U5 4

P 2

49. Rose

53. Lemniscate P 2

U5

U5

U5 U5

35. D

43. Limaçon without inner loop

x 1 2 3 4 5 U50

5P 4

33. H

U5P

3P 2

P U5 2

U5P

3P 2

y P 2

U5

5P 4

y U5

3P 4

3P 2

x 1 2 3 4 5 U50

U5P

7P 4

y

y

51. Rose

U5

x 2 3 4 5 U50

47. Limaçon with inner loop

U5 U5

U5

U5

x 1 2 3 4 5 U50

5P 4

29. E 31. F 37. Cardioid

x 2 3 4 5 U50

U5

5P 4

P 4

P 4

7P U5 4

5P U5 4

y

U5P

1

U5 U5

1

3P 2

45. Limaçon with inner loop U5

U5

P 2 U5

U5P

y

2

5P 4

3P 4

U5

P U5 2

U5

U5

U5P

U5

3P 2

41. Limaçon without inner loop U5

3P 4

7P 4

27. x2 + 1y + 12 2 = 1, y ≠ 0; circle, radius 1, center at 10, - 12 in rectangular coordinates, hole at (0, 0)

y

U5

U5

5P 4

3P 2

39. Cardioid

U5

P U5 4

y

P U5 2

U5P

U5

x 1 2 3 4 5 U50

U5P

y 3P 4

P U5 2

U5

25. 1x - 22 2 + y2 = 4, x ≠ 0; circle, radius 2, center at (2, 0) in rectangular coordinates, hole at (0, 0)

U5

y

y

P 2 U5

U5P

23. x2 + 1y + 22 2 = 4; circle, radius 2, center at 10, - 22 in rectangular coordinates

U5

5P 4 U5

3P 2

7P 4

U5 3P U5 4

U5P

U5

P 2

15 30 45 60

U5

5P 4 U5

3P 2

x U50

7P 4

ANSWERS  Historical Problems 57. Cardioid

59. Limaçon with inner loop y

U5

P 2

U5

3P 4

U5

U5P

U5

1

63. 3P 4

U5

P 2

U5

U5P

U5

1

P 4

U5

x U50

2

3P 4

U5

U5P

1

5P 4

U5

U5

3P 4

U5

U5

U5P

U5

1 2 3 3

5P 4

x U50

1

U5 U5

P 4

3P 2

7P 4

U5 U5

3P 4

U5

3P 4

U5

U5

U5P

U5

U5

3P 2

5P 4

U5

1. (a) 1 + 4i, 1 + i

(b) - 1, 2 + i

3P 2

3P U5 4

3P 2

7P 4

U5

U5

5P 4

2p 5

7P 4

3P 2

r = 2a cos u r 2 = 2ar cos u x2 + y2 = 2ax x2 - 2ax + y2 = 0 1x - a2 2 + y2 = a2 Circle, radius a, center at 1a, 02 in rectangular coordinates 85.

(b) r 2 = sin u; r 2 = sin 1p - u2 r 2 = sin u Test works. 1 - r2 2 = sin 1 - u2 r 2 = - sin u Not equivalent; new test fails.

93. Amplitude = 2; Period =

P 4

x 1 2 3 4 5 U50

U5

r = 2a sin u r 2 = 2ar sin u x2 + y2 = 2ay x2 + y2 - 2ay = 0 x2 + 1y - a2 2 = a2 Circle, radius a, center at 10, a2 in rectangular coordinates

P 2 U5

U5P

83.

7P 4

y

P 4

87. (a) r 2 = cos u; r 2 = cos 1p - u2 r 2 = - cos u Not equivalent; test fails. 1 - r2 2 = cos 1 - u2 r 2 = cos u New test works.

91. Absolute maximum: f 142 = 1; absolute minimum: none 92. 420° 94. Horizontal asymptote: y = 0; vertical asymptote: x = 4

Historical Problems (page 646)

U5

5P 4

U5

x 5P U 5 0

P 4

U5

x 1 2 3 4 5 U50

77.

3P

P 4, 2 3

P 2

7P 4 U5

P 2

P

P 4

7P 4

U5

x U50

3P 2

U5

x 1 2 3 4 5 U50

5P 4

3P 4

U5

81. r sin u = a y = a

P 2 U5

U5P

U5

U5

U5

y

79.

U5

y

75.

P 2

3P 2

7P 4

67. r = 3 + 3 cos u 69. r = 4 + sin u y 71.

P 4

2

U5

5P 4

2 2 Ï2 , 5P 2 4

3P U5 2

y

73.

U5

P 4

U5

U5P

7P 4

U5

5P 4 U5

2 1 Ï2 , P 2 4

P 2

(0, P) U5

U5

3P 2

y

65.

P U5 2

P 3

x 2 4 6 8 10 U 5 0

U5P

7P U5 4

5P 4

3P 4

4,

P 2

U5

P 4

x 1 2 3 4 5 U50

U5

y

U5 U5

U5P

3P U5 2

1, U5

3P U5 4

x U50

2

P 2

U5

P 4

7P U5 4

5P 4

y

61.

y

AN69

AN70

ANSWERS  Section 8.3

8.3 Assess Your Understanding (page 646) 5. real; imaginary 6. magnitude; modulus; argument 7. r1 r2; u1 + u2; u1 + u2 13. 11. Imaginary Imaginary axis

8. r n; nu; nu

9. three

axis

10. True 15.

Imaginary axis 3

1

1 Real axis

1

1

2

Real axis

3

1

17.

19.

Imaginary axis

3 1cos 270° + i sin 270°2 21.

Imaginary axis

4

Imaginary axis

1

2 2

4

3

2

Real axis

Real axis

3

2 1cos 330° + i sin 330°2

22 1cos 45° + i sin 45°2

3

1

3

Real axis 1

2

2

2

4

4

5 1cos 306.9° + i sin 306.9°2

4 22 1cos 315° + i sin 315°2

2

Real axis

1

213 1cos 123.7° + i sin 123.7°2

23. - 1 + 23i

25. 2 22 - 2 22i 27. - 3i 29. - 0.035 + 0.197i 31. 1.970 + 0.347i z z 1 3 33. zw = 8 1cos 60° + i sin 60°2; = 1cos 20° + i sin 20°2 35. zw = 12 1cos 40° + i sin 40°2; = 1cos 220° + i sin 220°2 w 2 w 4 9p 9p z p p z + i sin b; = cos + i sin 39. zw = 4 22 1cos 15° + i sin 15°2; = 22 1cos 75° + i sin 75°2 37. zw = 4 acos 40 40 w 40 40 w 27 25 22 27 23 25 22 + i 47. + i 49. - 4 + 4i 41. - 32 + 32 23i 43. 32i 45. 2 2 2 2 51. - 23 + 14.142i 6 6 6 53. 2 2 1cos 15° + i sin 15°2, 2 2 1cos 135° + i sin 135°2, 2 2 1cos 255° + i sin 255°2 4 4 4 4 55. 28 1cos 75° + i sin 75°2, 28 1cos 165° + i sin 165°2,28 1cos 255° + i sin 255°2,2 8 1cos 345° + i sin 345°2 57. 2 1cos 67.5° + i sin 67.5°2, 2 1cos 157.5° + i sin 157.5°2, 2 1cos 247.5° + i sin 247.5°2, 2 1cos 337.5° + i sin 337.5°2 59. cos 18° + i sin 18°, cos 90° + i sin 90°, cos 162° + i sin 162°, cos 234° + i sin 234°, cos 306° + i sin 306° 61. 1, i, - 1, - i

n

63. Look at formula (8);  zk = 1r for all k.

Imaginary i axis

1

1

65. Look at formula (8). The zk are spaced apart by an angle of

2p . n

Real axis

i

67. Assume the theorem is true for n Ú 1. For n = 0: z0 = r 0 3cos(0 # u) + i sin(0 # u) 4 1 = 1 # 3cos(0) + i sin(0) 4 1 = 1 # 31 + 0 4 1 = 1 True

For negative integers: z-n = (zn)-1 = (r n[cos(nu) + i sin(nu)])-1 with n Ú 1 1 = n r [cos(nu) + i sin(nu)] = = =

1 # cos(nu) - i sin(nu) r n[cos(nu) + i sin(nu)] cos(nu) - i sin(nu) cos(nu) - i sin(nu) r (cos2(nu) + sin2(nu)) cos(nu) - i sin(nu) n

n

sin2 u + cos2 u = 1

r = r -n[cos(nu) - i sin(nu)]

= r -n[cos( - nu) + i sin( - nu)] Thus, De Moivre’s Theorem is true for all integers. 69. ≈40.50

6 16 70. Minimum; f a b = 5 5

4 71. p 3

3 72. 2y 2 3x2y2

cos(- u) = cos(u); sin(- u) = - sin(u)

AN71

ANSWERS  Section 8.5

8.4 Assess Your Understanding (page 658) 1. vector 2. 0

3. unit

4. position

9.

5. horizontal; vertical 6. resultant 11.

7. True 8. False 13.

3v

15.

v

u

3v

vw vw

w

w

3v  u  2w

2w

v

17. T

19. F 21. F

23. T 25. 12

43. 289

45. 234 - 213

41. - j 55.

5-2

+ 221, - 2 - 221 6

71. (a) 1 - 1, 4 2 (b) (1, 4) y

u (4, 5)

5 v

v u 5 x (3, 1)

107.

y 2.5

27. v = 3i + 4j 47. i

57. v =

29. v = 2i + 4j

3 4 49. i - j 5 5

5 5 23 i + j 2 2

51.

31. v = 8i - j

22 22 i j 2 2

59. v = - 7i + 7 23j

53. v = 61. v =

33. v = - i + j 8 25 4 25 i + j 5 5 25 23 25 i j 2 2

35. 5

37. 22

or v = 63. 45°

73. F = 20 23i + 20 j 75. F = 1 20 23 + 30 22 2 i + 1 20 - 30 22 2 j 77. (a) va = 550j; vw = 50 22i + 50 22j (b) vg = 50 22i + 1550 + 50 222j (c) 7 vg 7 = 624.7 mph; N6.5°E 79. v = 1250 22 - 302i + 1250 22 + 30 232j; 518.8 km/h; N38 .6°E 81. Approximately 4031 lb 83. 8.6° left of direct heading across the river; 1.52 min 85. Tension in right cable: 1000 lb; tension in left cable: 845.2 lb 87. Tension in right part: 1088.4 lb; tension in left part: 1089.1 lb 89. The truck must pull with a force of 4635.2 lb. 91. (a) N7.05°E (b) 12 min 93. m = 0.36 95. 13.68 lb

39. 213

8 25 4 25 i j 5 5

65. 150°

67. 333.4°

69. 258.7°

99. F2 P F1

F3

F4

105. {29} 106. - 3x(x + 2)(x - 6)

108. 23 P 3 x P 3 ,2 3 2

3 p ; Period = 2 3 p Phase shift = 2 Amplitude =

Historical Problem (page 667)

1ai + bj2 # 1ci + dj2 = ac + bd Real part [ 1a + bi2 1c + di2 4 = Real part[ 1a - bi2 1c + di2 4 = Real part[ac + adi - bci - bdi 2] = ac + bd

8.5 Assess Your Understanding (page 668) 9. (a) 0 (b) 90° (c) orthogonal 11. (a) 13 - 1 2 5 5 1 1 19. v1 = i - j, v2 = - i - j (b) 75° (c) neither 13. (a) - 50 (b) 180° (c) parallel 15. (a) 0 (b) 90° (c) orthogonal 17. 3 2 2 2 2 1 2 6 3 14 7 1 2 i + j, v2 = i - j 25. 9 ft-lb 27. (a) 7 I 7 ≈ 0.022; the intensity of the sun’s rays is 21. v1 = - i - j, v2 = i - j 23. v1 = 5 5 5 5 5 5 5 5 approximately 0 .022 W/cm2. 7 A 7 = 500; the area of the solar panel is 500 cm2. (b) W = 10; ten watts of energy is collected. (c) Vectors I and A should be parallel with the solar panels facing the sun. 29. Force required to keep Sienna from rolling down the hill: 737.6 lb; force perpendicular to the hill: 5248.4 lb 31. Timmy must exert 85.5 lb. 33. - 2.833 35. 60° 37. Let v = ai + bj. Then 0 # v = 0a + 0b = 0. 39. v = cos ai + sin aj, 0 … a … p; w = cos bi + sin bj, 0 … b … p. If u is the angle between v and w, then v # w = cos u, since 7 v 7 = 1 and 7 w 7 = 1. Now u = a - b or u = b - a. Since the cosine function is even, v # w = cos(a - b). Also, v # w = cos a cos b + sin a sin b. So cos(a - b) = cos a cos b + sin a sin b. 41. (a) If u = a1i + b1j and v = a2i + b2 j, then, since 7 u 7 = 7 v 7 , a21 + b21 = 7 u 7 2 = 7 v 7 2 = a22 + b22, 1u + v 2 # 1u - v 2 = 1a1 + a2 2 1a1 - a2 2 + 1b1 + b2 2 1b1 - b2 2 = 1a21 + b21 2 - 1a22 + b22 2 = 0. (b) The legs of the angle can be made to correspond to vectors u + v and u - v. 43. 1 7 w 7 v + 7 v 7 w2 # 1 7 w 7 v - 7 v 7 w2 = 7 w 7 2v # v - 7 w 7 7 v 7 v # w + 7 v 7 7 w 7 w # v - 7 v 7 2w # w = 7 w 7 2v # v - 7 v 7 2w # w = 7 w 7 2 7 v 7 2 - 7 v 7 2 7 w 7 2 = 0 2. dot product 3. orthogonal

4. parallel

5. T 6. F

7. (a) 0 (b) 90°

(c) orthogonal

45. 7 u + v 7 2 - 7 u - v 7 2 = 1u + v2 # 1u + v2 - 1u - v2 # 1u - v2 = 1u # u + u # v + v # u + v # v2 - 1u # u - u # v - v # u + v # v2 9 = 2 1u # v2 + 2 1v # u2 = 4 1u # v2 47. 12 48. 2 50. V(x) = x(19 - 2x)(13 - 2x) or V(x) = 4x3 - 64x2 + 247x 49. 11 - sin2u2 11 + tan2u2 = 1cos2u2 1sec2u2 1 = cos2u # cos2u = 1

AN72

ANSWERS  Section 8.6

8.6 Assess Your Understanding (page 677) 2. xy-plane 3. components 4. 1 5. F 6. T 7. All points of the form (x, 0, z) 9. All points of the form (x, y, 2) 11. All points of the form ( - 4, y, z) 13. All points of the form (1, 2, z) 15. 221 17. 233 19. 226 21. (2, 0, 0); (2, 1, 0); (0, 1, 0); (2, 0, 3); (0, 1, 3); (0, 0, 3) 23. (1, 4, 3); (3, 2, 3); (3, 4, 3); (3, 2, 5); (1, 4, 5); (1, 2, 5) 25. ( - 1, 2, 2); (4, 0, 2); (4, 2, 2); ( - 1, 2, 5); (4, 0, 5); ( - 1, 0, 5) 27. v = 3i + 4j - k 3 6 2 29. v = 2i + 4j + k 31. v = 8i - j 33. 7 35. 23 37. 222 39. - j - 2k 41. 2105 43. 238 - 217 45. i 47. i - j - k 7 7 7 23 23 23 # # # # 49. i + j + k 51. v w = 0; u = 90° 53. v w = - 2, u ≈ 100.3° 55. v w = 0; u = 90° 57. v w = 52; u = 0° 3 3 3 59. a ≈ 64.6°; b ≈ 149.0°; g ≈ 106.6°; v = 7(cos 64.6°i + cos 149.0°j + cos 106.6°k) 61. a = b = g ≈ 54.7°; v = 23(cos 54.7°i + cos 54.7°j + cos 54.7°k) 63. a = b = 45°; g = 90°; v = 22(cos 45°i + cos 45°j + cos 90°k) 65. a ≈ 60.9°; b ≈ 144.2°; g ≈ 71.1°; v = 238(cos 60.9°i + cos 144.2°j + cos 71.1°k) 67. (a) d = a + b + c = 6 7, 1, 5 7 (b) 8.66 ft 3 22 69. (x - 3)2 + (y - 1)2 + (z - 1)2 = 1 71. Radius = 2, center ( - 1, 1, 0) 73. Radius = 3, center (2, - 2, - 1) 75. Radius = , Center (2, 0, - 1) 2 13 13 1 f or a2, d 81. 2x2 + 2x - 5 82. 77. 2 newton-meters = 2 joules 79. 9 newton-meters = 9 joules 80. e x ` 2 6 x … 5 5 2 83. c = 3 25 ≈ 6.71; A ≈ 26.6°; B ≈ 63.4°

8.7 Assess Your Understanding (page 683) 1. T 2. T 3. T 4. F 5. F 6. T 7. 2 9. 4 11. - 11A + 2B + 5C 13. - 6A + 23B - 15C 15. (a) 5i + 5j + 5k (b) - 5i - 5j - 5k (c) 0 (d) 0 17. (a) i - j - k (b) - i + j + k (c) 0 (d) 0 19. (a) - i + 2j + 2k (b) i - 2j - 2k (c) 0 (d) 0 21. (a) 3i - j + 4k (b) - 3i + j - 4k (c) 0 (d) 0 23. - 9i - 7j - 3k 25. 9i + 7j + 3k 27. 0 29. - 27i - 21j - 9k 31. - 18i - 14j - 6k 33. 0 35. - 25 37. 25 39. 0 41. Any vector of the form c( - 9i - 7j - 3k), where c is a nonzero scalar 43. Any vector of the form c( - i + j + 5k), where c is a nonzero scalar 45. 2166

47. 2555

i 55. 98 cubic units 57. u * v = 3 a1 a2

j b1 b2

49. 234

51. 2998

53.

11 219 219 7 219 i + j + k 57 57 57

or -

11 219 219 7 219 i j k 57 57 57

k c1 3 = (b1c2 - b2c1)i - (a1c2 - a2c1)j + (a1b2 - a2b1)k c2

7 u * v 7 2 = 2(b1c2 - b2c1)2 + (a1c2 - a2c1)2 + (a1b2 - a2b1)2 2 2 7u

72

7u

= b21c 22 - 2b1b2c1c2 + b22c 21 + a21c 22 - 2a1a2c1c2 + a22c 21 + a21b22 - 2a1a2b1b2 + a22b21 =

727

v

a21

72

+ b21 + c 21, 7 v 7 2 = a22 + b22 + c 22

= (a21 + b21 + c 21)(a22 + b22 + c 22) = a21a22 + a21b22 + a21c 22 + b21a22 + b21b22 + b21c 22 + a22c 21 + b22c 21 + c 21c 22

(u # v)2 = (a1a2 + b1b2 + c1c2)2 = (a1a2 + b1b2 + c1c2)(a1a2 + b1b2 + c1c2)

= a21a22 + a1a2b1b2 + a1a2c1c2 + b1b2c1c2 + b1b2a1a2 + b21b22 + b1b2c1c2 + a1a2c1c2 + c 21c 22

7u

= a21a22 + b21b22 + c 21c 22 + 2a1a2b1b2 + 2b1b2c1c2 + 2a1a2c1c2

727

v 7 2 - (u # v)2 = a21b22 + a21c 22 + b21a22 + a22c 21 + b22c 21 + b21c 22 - 2a1a2b1b2 - 2b1b2c1c2 - 2a1a2c1c2, which equals 7 u * v 7 2. 59. By Problem 58, since u and v are orthogonal, 7 u * v 7 = 7 u 7 7 v 7 . If, in addition, u and v are unit vectors, 7 u * v 7 = 1 # 1 = 1. 61. Assume that u = ai + bj + ck, v = di + ej + fk, and w = li + mj + nk. Then u * v = (bf - ec)i - (af - dc)j + (ae - db)k, u * w = 1bn - mc2i - 1an - lc2j + 1am - lb2k, and v + w = 1d + l2i + 1e + m2j + 1f + n2k. Therefore, (u * v) + (u * w) = (bf - ec + bn - mc)i - (af - dc + an - lc)j + (ae - db + am - lb)k and u * 1v + w2 = [b(f + n) - (e + m)c]i - [a(f + n) - (d + l)c]j + [a(e + m) - (d + l)b]k = 1bf - ec + bn - mc2i - 1af - dc + an - lc2j + 1ae - db + am - lb)k, which equals 1u * v2 + 1u * w2. p 1 64. 65. (17, 4.22), ( - 17, 1.08) 66. 8 hours 67. log 4 x - 3log 4 z 4 2

Review Exercises (page 686) 1. a

3 23 3 , b 2 2 3,

P 6

P 6

2. 1 1, 23 2

3. (0, 3) 22,

4P 3

23, 2

4P 3

O

O 2

P 2

P 2

O

4. a8,

5p p b , a - 8, b 4 4

5. a2,

5p p b , a - 2, - b 6 6

6. (5, p), 1 - 5, 02

AN73

ANSWERS  Review Exercises 7. (a) x2 + 1y - 12 2 = 1 (b) circle, radius 1, center (0, 1) in rectangular coordinates

8. (a) x2 + y2 = 25 at pole

y U5 U5

3P 4

1

P 4

2

x U50

U5

3P 4

7P 4

U5

P 4

U5

U5P

4

U5

5P 4 U5

U5P

7P 4

U5

13. Limaçon without inner loop; symmetric with respect to the polar axis

U5

U5

3P 4

P 2

P 2 U5

U5

(5, P)

5P 4

U5

0,

U5

P 4

U5

x 5 U50

P 2

3P 2

3P 4

y U5

P 2

U5

P 4

(3, 0) x 2 4 6 8 10 U 5 0

U5P

7P 4

U5

3P 2

U5

5P 4 U5

14. 2 1cos 146.3° + i sin 146.3°2 16. - 23 + i

P 4

3P 2

P 6, 2 2

17. -

18. 0.10 - 0.02i

3 3 23 + i 2 2

Imaginary axis 0.06

Imaginary axis 3

2

7P 4

15. 5 1cos 323.1° + i sin 323.1°2

2

x 4 5 U50

U5 3P U5 2

7P 4

12. Cardioid; symmetric with respect to the line p u = 2

0.10

Real axis

0.02

(3, 0) 1 2

U5

5P 4

P 3

Imaginary axis

2

2

2

Real axis

Real axis

0.06

1

7P 4

4, 3P 2

19. zw = - 1cos 50° + i sin 50°2; 21. zw = 6;

2,

U5

5P 4

2 U5P

x 1 2 3 4 5 U50

U5

1 2 3

U5

4,

7P 4

P 4

P 2

3P 2

y

3P 4

P 2

U5

(4, 0)

x U50

8

U5

P 2

U5P

3P U5 2

y P 2

3P U5 4

11. Circle; radius 2, center at (2, 0) in rectangular coordinates; symmetric with respect to the polar axis

0,

U5

P U5 4

U5

5P 4

y 3P 4

U5

x 1 2 3 4 5 U50

U5

10. (a) 1x - 42 2 + 1y + 22 2 = 25 (b) circle, radius 5, center 14, - 22 in rectangular coordinates

U5

P 2

U5P

3P U5 2

U5

y

U5

U5

U5

5P 4

(b) line through pole, making p an angle of with polar axis 4

y

P 2

U5P

U5

9. (a) x - y = 0

(b) circle, radius 5, center

z = - 1cos 30° + i sin 30°2 w

z = 6 1cos 70° - i sin 70°2 w

22. 32 + 32 23i

20. zw = - 12;

23. 5

5 5 - 5 i A2 A2

z 3 2p 2p = acos + i sin b w 4 3 3 24. 16

25. - 8432 - 5376i

26. 2, 2 1cos 72° + i sin 72°2, 2 1cos 144° + i sin 144°2, 2 1cos 216° + i sin 216°2, 2 1cos 288° + i sin 288°2 or 2, 27.

25 - 1 25 + 1 25 + 1 25 + 15 25 - 15 25 - 15 25 - 1 25 + 15 + i, + i, i, i 4 4 4 4 2 12 2 12 2 12 2 12 v

28.

u uv

2u

3v

29. v = 2i - 4j; 7 v 7 = 2 25 30. v = - i + 3j; 7 v 7 = 210 31. 2i - 2j 32. - 20i + 13j 33. 25 2 25 25 3 3 23 34. 25 + 5 ≈ 7.24 35. i + j 36. v = i + j 37. 120° 38. 243 ≈ 6.56 5 5 2 2 39. v = 3i - 5j + 3k 40. 21i - 2j - 5k 41. 238 42. 0 43. 3i + 9j + 9k 44. 0

219 3 219 3 219 219 3 219 3 219 i + j + k or i j k 19 19 19 19 19 19 # # # 46. v w = - 11; u ≈ 169.7° 47. v w = - 4; u ≈ 153.4° 48. v w = 1; u ≈ 70.5° 49. v # w = 0; u = 90° 4 3 6 8 9 7 21 50. Parallel 51. Neither 52. Orthogonal 53. v1 = i - j; v2 = i + j 54. v1 = 13i + j2; v2 = - i + j 5 5 5 5 10 10 10 55. a ≈ 56.1°; b ≈ 138°; g ≈ 68.2° 56. 2 283 57. - 2i + 3j - k 58. 0 59. 229 ≈ 5.39 mi/h 0.4 mi 60. Left cable: 1843.21 lb; right cable: 1630.41 lb 61 50 ft-lb 62. A force of 697.2 lb is needed to keep the van from rolling down the hill. The magnitude of the force perpendicular to the hill is 7969.6 lb. 2u  3v

45.

AN74

ANSWERS  Chapter Test

Chapter Test (page 688) 1–3.

P 2

2, 3P 4

P 3

P 4

p b 3

4. a4,

5. x2 + y2 = 49

6.

y U5

P 6

3P U5 4

P 2 U5

y x

= 3 or y = 3x y U5

P 4

3P U5 4

P 2 U5

P 4

0 P 3, 2 6 24,

x U50

U5P

P 3 U5

U5

5P 4 U5

7. 8y = x2 3P 4

P 2 U5

P 4

x U50

U5P

7P 4

U5

3P 2

U5

5P 4 U5

3P 2

3

z1  2兹2(cos 160  i sin 160) U5

U5

5P 4 U5

Imaginary axis

7P 4

2

3P 2

3

z0  2兹2(cos 40  i sin 40) 40 Real axis

2

3

z2  2兹2(cos 280  i sin 280)

15. 7 v 7 = 10

14. v = 8 5 22, - 5 22 9

19. v1 + 2v2 - v3 = 8 6, - 10 9

22 22 ,i 17. 315° off the positive x-axis 18. v = 5 22i - 5 22j 2 2 20. Vectors v1 and v4 are parallel. 21. Vectors v2 and v3 are orthogonal. 22. 172.87° 23. - 9i - 5j + 3k

24. a ≈ 57.7°, b ≈ 143.3°, g ≈ 74.5°

v

16. u =

7v7

25. 2115

= h

26. The cable must be able to endure a tension of approximately 670.82 lb.

Cumulative Review (page 689) 1. 5 - 3, 3 6

2. y =

23 x 3

3. x2 + 1y - 12 2 = 9

y (0, 1) 5 (0, 4) (3, 1)

6.

y 2

7. 1, 1 e

(1, 0)

10.

y 4.5

8.

y 1.25

4. e x ` x 6

1 1 f or a - q , b 2 2

5. Symmetry with respect to the y-axis

(3, 1) 5 x (0, 2)

y 1.25

9. -

p 6

2P

(e, 1)

2P

x

x

4 x

y

11.

x3 y4

7P 4

p 8. r 2 cos u = 5 is symmetric about the pole, the polar axis, and the line u = . p 2 9. r = 5 sin u cos2 u is symmetric about the line u = . The tests for symmetry about the pole and the polar 2 2 axis fail, so the graph of r = 5 sin u cos u may or may not be symmetric about the pole or the polar axis. w 3 10. z # w = 6 1cos 107° + i sin 107°2 11. = 1cos 297° + i sin 297°2 z 2 5 12. w = 243 1cos 110° + i sin 110°2 3 3 3 13. z0 = 2 22 1cos 40° + i sin 40°2, z1 = 2 22 1cos 160° + i sin 160°2, z2 = 2 2 2 1cos 280° + i sin 280°2

y U5

U5

x U50

U5P

3P U5 4

U5

P 3 P U5 4 r52

4 x U5P

U5

1

5P 4

x U50

2

U5

7P 4

12. Amplitude: 4; period: 2

ANSWERS  Section 9.2

AN75

CHAPTER 9 Analytic Geometry 9.2 Assess Your Understanding (page 699) 8. 13, 2)

6. parabola 7. (c)

9. 13, 6)

19. y = 16x 20

V  (0, 0)

13. E

15. H

17. C

23. y2 = - 8x

y

(4, 8) F  (4, 0) 20 x

(6, 3)

(2, 4) F  (2, 0)

10 x (6, 3)

(2, 3) 2, 1 3 3 2.5 x 1 1 D: y   F  0, 3 3 V  (0, 0)

V  (2, 3) 10 x (6, 5)

(2, 3)

D: y  1

2 1  , 3 3

(2, 5)

V  (1, 2)

37. Vertex: (0, 0); Focus: (0, 1); Directrix: y = - 1

D: x  1

F  (0, 1)

2 x

(2, 1)

V  (0, 0)

43. Vertex: 13, - 12; Focus: 5 3 a3, - b ; Directrix: y = 4 4

2.5 x

45. Vertex: 12, - 32; Focus: 14, - 32; Directrix: x = 0 y 2

y 1 V  (3, 1) x F  3, 

5 x F  (0, 2) (0, 4)

5 4

D: x  0

(4, 8) F  (4, 0)

D: x  4

V  (0, 0) (4, 8)

10 x

47. Vertex: (0, 2); Focus: 1 - 1, 22; Directrix: x = 1 y (1, 4) 8

(4, 1) 8 x V  (2, 3)

F  (1, 2)

F  (4, 3) (4, 7)

y

V  (1, 1)

y 2.5

1 8 x 2.5 x

F   3 , 1 4

3 3  , 2 4

77. Cy2 + Dx = 0, C ≠ 0, D ≠ 0 Cy2 = - Dx D y2 = - x C

V  (2, 8)

F  2,  D: y  

33 4

31 4

D: x  3 V  (1, 2)

8 (1, 6) F  (1, 2) 5 (1, 2) x

49. Vertex: 1 - 4, - 22; Focus: 1 - 4, - 12; Directrix: y = - 3 F  (4, 1) y 5

V  (0, 2) (6, 1) 2 x

3 31 51. Vertex: 1 - 1, - 12; Focus: a - , - 1 b ; 53. Vertex: 12, - 82; Focus: a2, - b ; 4 4 5 33 Directrix: x = Directrix: y = 4 4

41. Vertex: 1 - 1, 22; Focus: (1, 2); Directrix: x = - 3

D: x  1

(1, 0)

5 D: x   4 3, 1   2 4

8 (5, 4) (1, 4) V  (3, 3) D: y  2 2 x

y

y 10

2.5

D: y  1

3

33. 1x + 3)2 = 4 1y - 3) F  (3, 4) y

5 (0, 0)

39. Vertex: (0, 0); Focus: 1 - 4, 02; Directrix: x = 4

y

(2, 1) V  (1, 2)

3 4

1, 1 2 x 1 D: y   2 2.5

F  (2, 5)

35. 1y + 22 2 = - 8 1x + 12 y 2

D: x  2

(2, 2)

2.5

(2, 2) 1 1, 2 V  (0, 0)

y 10

4

D: y  

5 x

31. 1y + 2)2 = 4 1x + 1)

y y

(3, 6)

D: x  2

V  (0, 0) (2, 4)

29. 1x - 2)2 = - 8 1y + 3)

F  (3, 2)

1 F  0, 2 y

5

F  (0, 3)

4 y 3

(3, 2)

25. x2 = 2y

y

10 V  (0, 0) D: y  3

(4, 8)

27. x2 =

11. B

2

y D: x  4

10. y = - 2

21. x = - 12y

2

D: y  3

1 x (2, 1)

V  (4, 2)

55. 1y - 12 2 = x

57. 1y - 12 2 = - 1x - 22 1 59. x2 = 4 1y - 12 61. y2 = 1x + 22 2 63. 1.5625 ft from the base of the dish, along the axis of symmetry 65. 1 in. from the vertex, along the axis of symmetry 67. 20 ft 69. 0.78125 ft 71. 4.17 ft from the base, along the axis of symmetry 73. 24.31 ft, 18.75 ft, 7.64 ft 2 2 75. (a) y = x + 630 315 (b) 567 ft: 119.7 ft; 478 ft: 267.3 ft; 308 ft: 479.4 ft (c) No

This is the equation of a parabola with vertex at 10, 02 and axis of symmetry the x-axis. D D The focus is a , 0 b ; the directrix is the line x = . The parabola opens to the right if 4C 4C D D 7 0 and to the left if 6 0. C C

AN76

ANSWERS  Section 9.2 (a) If D ≠ 0, then the equation may be written as E 2 - 4CF E 2 D b = - ax b. ay + 2C C 4CD E 2 - 4CF E This is the equation of a parabola with vertex at a ,b 4CD 2C and axis of symmetry parallel to the x-axis. (b)–(d) If D = 0, the graph of the equation contains no points if E 2 - 4CF 6 0, is a single horizontal line if E 2 - 4CF = 0, and is two horizontal lines if E 2 - 4CF 7 0.

79. Cy2 + Dx + Ey + F = 0, C ≠ 0 Cy2 + Ey = - Dx - F D E F y = - x C C C E 2 D F E2 ay + b = - x + 2C C C 4C 2 E 2 D E 2 - 4CF ay + b = - x + 2C C 4C 2 y2 +

80. (0, 2), (0, −2), ( - 36, 0); symmetric with respect to the x-axis

81. {5}

5 289 8 289 289 289 8 ; cos u = ; csc u = ; sec u = ; cot u = 89 89 5 8 5

82. sin u =

83. -

2 210 3

9.3 Assess Your Understanding (page 708) 8. major 9. 10, - 5); 10, 5)

7. ellipse

17. Vertices: 1 - 5, 02, (5, 0)

Foci: 1 - 221, 0), 1 221, 0)

(0, 5)

(5, 0) (兹21, 0)

5 (0, 4) (3, 0) 5 x (0, 4)

(3, 0)

13. C

(0, 2 兹3)

(0, 5)

(3, 0) (4, 0) 5 x

(4, 0)

y

5 (0, 4)

y 5 (0, 4)

(3, 0)

(4, 0)

y 5

35.

3

(5, 0) (0, 3)

y2 x2 + = 1 4 13

(5, 0) x

(2, 0)

(4, 0)

(0, 兹13)

43. Center: 13, - 12; Vertices: 13, - 42, (3, 2); Foci: 1 3, - 1 - 25 2 , 1 3, - 1 + 25 2

45.

(3, 1  兹5 )

(0, 4)

(5, 6)

y 7 (5  2 兹3, 4)

(9, 4)

(1, 4) (5, 4) 1 x

(5  2 兹3, 4)

1x - 22 2

+

1y + 12 2

= 1

3 2 Center: 12, - 12; Vertices: 1 2 - 23 , - 1 2 ; 1 2 + 23 , - 1 2 ; Foci: 11, - 12, 13, - 12 (2, 1)

1x - 12 2

+

(2  兹3, 1) 5 x (3, 1) (2, 1  兹2)

1y + 22 2

= 1

4 9 Center: 11, - 22; Vertices: 11, - 52, 11, 12 ; Foci: 1 1, - 2 - 25 2 , 1 1, - 2 + 25 2 y 5

y 5 (2, 1  兹2)

(2  兹3, 1) (1, 1)

51.

(1, 2  兹5 ) (1, 2) (1, 2) (1, 5)

5 x (3, 2) (1, 2  兹5 )

4

+ 1y - 12 2 = 1

41. 1x - 12 2 +

47.

y2 4

= 1

1x + 22 2

+ 1y - 12 2 = 1 4 Center: 1 - 2, 12; Vertices: 1 - 4, 12, (0, 1);

Foci: 1 - 2 - 23 , 1 2 , 1 - 2 + 23 , 1 2 y 5 (2, 2) (4, 1) 5 (2  兹3 , 1) (2, 1)

53. x2 +

(2  兹3 , 1) (0, 1) x (2, 0)

1y + 22 2

= 1 4 Center: 10, - 22; Vertices:10, - 42, 10, 02 ; Foci: 1 0, - 2 - 23 2 , 1 0, - 2 + 23 2 y

(1, 1)

1x + 12 2

(0, 兹15 )

1y - 42 2 + = 1 16 4 Center: 1 - 5, 42; Vertices: 1 - 9, 42, 1 - 1, 42;

(3, 2)

(3, 4)

39.

(0, 兹5) (3, 0) 5 x (2, 0)

(0, 兹5)

= 1

1x + 52 2

(5, 2)

49.

y2 16

Foci: 1 - 5 - 2 23 , 4 2 , 1 - 5 + 2 23 , 4 2

x (5, 1)

(3, 0)

(3, 0) 5 x

y 5 (0, 兹15 ) (0, 4) (1, 0) (1, 0) 5 x

(2, 0) 5 x (0, 3)

y (3, 1  兹5 ) 5 (1, 1) 5 (3, 1)

37. x2 +

y2 x2 + = 1 9 5 y (2, 0) 5

(0, 5)

y 5 (0, 3) (0, 兹13)

(0, 3)

31.

(0, 4)

(0, 4)

y2 x2 + = 1 25 9

(0, 兹2)

(0, 5)

2

(2 兹2, 0) 5 x ( 兹6, 0)

(兹6, 0)

(0, 4)

(0, 4)

33.

(2 兹2 , 0)

(0, 4)

(3, 0) (5, 0) 4 x

(5, 0)

y 5 (0, 兹2)

y2 x2 + = 1 9 25

29.

y x2 + = 1 8 2 Vertices: 1 - 2 22 , 0 2 , 1 2 22 , 0 2 Foci: 1 - 26 , 0 2 , 1 26 , 0 2

(2, 0) 5 x

(2, 0)

y2 y2 x2 x2 + = 1; Vertices: 1 - 4, 02, 27. + = 1 16 16 25 16 (4, 0), 10, - 42, 10, 4); Focus: (0, 0) y

2

23.

y 5 (0, 4)

(0, 2 兹3)

25.

15. B

2

y x + = 1 4 16 Vertices: 10, - 4), 10, 4) Foci: 10, - 2 23), 10, 2 23)

21.

y

5

(兹21, 0) (5, 0) 5 x (0, 2)

12. 11, 4) 2

19. Vertices: 10, - 52, (0, 5) Foci: 10, - 4), 10, 4)

y (0, 2)

11. 1 - 2, - 3); 16, - 3)

10. 5; 3; x

(0, 0)

(0, 2  兹3)

3 x

(1, 2)

(1, 2) (0, 2)

(0, 2  兹3)

5

(0, 4)

ANSWERS  Section 9.4

55.

1x - 22 2 25

+

1y + 22 2 21

y (0, 2) 3

10

x (7, 2) (4, 2) (2, 2)

= 1

9

y 9

59.

(4, 9) (4, 8)

1x - 22 2 y 5

(x - 1)2 9 (1, 2)

(1  兹10, 2) 5 x (1, 2) (1, 1)

+

1y - 2)2 9

= 1

65.

= 1

y (0, 4)

y 6 (1, 5)

6 (2, 0) 2.5 x

(2, 0)

(4, 2)

(2, 2)

7

(2, 1  兹7 )

7 x

63.

1y - 12 2

(2, 1  兹7 ) (5, 1) (6, 1) 7 x (2, 1)

(4, 6)

(4, 3)

+

16 (1, 1) (2, 1)

(4  兹5 , 6)

(4, 4)

+ 1y - 22 2 = 1

(1  兹10, 2)

1y - 62 2

+

5 (4  兹5 , 6)

y (2, 2) 5 (1, 3) (4, 2)

67.

1x - 42 2

6

(2, 2  兹21)

1x - 12 2

57.

(2, 2  兹21)

(3, 2)

61.

= 1

AN77

5 x (1, 1)

y2 x2 + = 1 71. 43.3 ft 73. 24.65 ft, 21.65 ft, 13.82 ft 75. 30 ft 100 36 77. The elliptical hole will have a major axis of length 2 241 in. and a minor axis of length 8 in. y2 x2 = 1 79. 91.5 million mi; + 8646.75 1932 2 y2 x2 81. Perihelion: 460.6 million mi; mean distance: 483.8 million mi; + = 1 233,524.2 1483.82 2

y

69.

(2, 0) 2 (2, 0) 2.5 x

(0, 8)

83. 35 million mi 85. 5 25 - 4 87. Ax2 + Cy2 + Dx + Ey + F = 0 A ≠ 0. C ≠ 0 Ax2 + Dx + Cy2 + Ey = - F D E b = -F Aax2 + x b + Cay2 + A Cy 2 D E 2 D2 E2 Aax + b + Cay + b = -F + + 2A 2C 4A 4C D2 E2 D E (a) If + - F is of the same sign as A (and C), this is the equation of an ellipse with center at a ,b. 4A 4C 2A 2C 2 2 D E D E + - F, the graph is the single point a ,b. (b) If 4A 4C 2A 2C D2 E2 (c) If + - F is of the opposite sign to A (and C), the graph contains no points since, in this case, the left side has a sign opposite that of the 4A 4C right side. 89. Zeros: 5 - 2 23, 5 + 2 23; x-intercepts: 5 - 2 23, 5 + 2 23 90. Domain: {x 0 x ≠ 5}; Horizontal asymptote: y = 2; Vertical asymptote: x = 5 91. 617.1 ft-lb 92. b ≈ 10.94, c ≈ 17.77, B = 38°

9.4 Assess Your Understanding (page 721) 7. hyperbola 8. transverse axis 9. b 10. 12, 42; 12, - 22 11. 12, 6); 12, - 4) 4 4 12. 4 13. 2; 3; x 14. y = - x; y = x 9 9 15. B 17. A

19. x2 -

y2 8

= 1

21.

y (0, 2兹2) 5 V1  (1, 0) V2  (1, 0) F2  (3, 0) F1  (3, 0) 5 x (0, 2兹2) y  2兹2 x

y2 16

-

x2 = 1 20

2 兹5 x 5 V2  (0, 4)

y

(2 兹5, 0)

y F2  (0, 6) 2 兹5 10 y x 5 (2 兹5, 0) 10 x

y  2兹2 x V1  (0, 4)

y2 x2 23. = 1 9 16 y

4 x 3

y2

(0, 4) y 10

V1  (3, 0) F1  (5, 0) (0, 4)

y

4 x 3

V2  (3, 0) 10 x F2  (5, 0)

F2  (0, 3 兹5 ) y  2 x y y  2x 10 V2  (0, 6) (3, 0) (3, 0) 10 x V1  (0, 6) F1  (0, 3 兹5 )

F1  (0, 6)

y2 x2 27. = 1 8 8

x2 25. = 1 36 9

y  x V1  (2 兹2, 0) F1  (4, 0)

y 5

(0, 2 兹2 ) yx V2  (2 兹2, 0) 5 x F2  (4, 0) (0, 2 兹2 )

AN78

ANSWERS  Section 9.4 2

29.

y x2 = 1 25 9 Center: 10, 02 Transverse axis: x-axis Vertices: 1 - 5, 02, 15, 02

2

31.

Foci: 1 - 234 , 0 2 , 1 234 , 0 2 3 Asymptotes: y = { x 5

y x2 = 1 4 16 Center: 10, 02 Transverse axis: x-axis Vertices: 1 - 2, 02, 12, 02 Foci: 1 - 2 25 , 0 2 , 1 2 25 , 0 2 Asymptotes: y = {2x y  2 x y 5

y 10 (0, 3)

3 y x 5 F2  ( 兹34, 0) 10 x

3 y x 5 F1  (兹34, 0) V1  (5, 0)

(0, 3)

- x2 = 1 9 Center: 10, 02 Transverse axis: y-axis Vertices: 10, - 32, 10, 32 Foci: 1 0, - 210 2 , 1 0, 210 2 Asymptotes: y = {3x y  3x

y 5

F2  (0, 兹10) V2  (0, 3)

(1, 0)

V2  (2, 0) 5 x F2  (2 兹5, 0)

F1  (2 兹5, 0)

y2

y  3 x

(0, 4) y  2x

V1  (2, 0)

V2  (5, 0)

33.

5 x (1, 0)

V1  (0, 3)

F1  (0, 兹10)

(0, 4)

35.

41.

y2

x2 = 1 25 25 Center: 10, 02 Transverse axis: y-axis Vertices: 10, - 52, 10, 52 Foci: 1 0, - 5 22 2 , 1 0, 5 22 2 Asymptotes: y = {x -

1x - 42 2 4

-

1y + 12 2 5

= 1

V2  (0, 5) y  x

43.

47.

1x - 12 2 4

-

1y + 12 2

3 y  1   (x1) 2 (1, 1)

9

= 1

1y - 22 2

49.

Foci: 1 - 2, 2 - 25 2 , 1 - 2, 2 + 25 2 Asymptotes: y - 2 = {2 1x + 22 y  2  2(x  2) y y  2  2(x  2) F2  (2, 2 兹5 ) V2  (2, 4) (2, 2) 4 (1, 2) (3, 2) V1  (2, 0)

2 x F1  (2, 2 兹5 )

y4

兹3 (x  3) 3

8 x (3  2 兹3, 4) 兹3 y4 (x  3) 3

1x - 22 2

-

53.

-

y 10 V1  (4, 7)

1y + 22 2

= 1 4 4 Center: 1 - 1, - 22 Transverse axis: parallel to x-axis Vertices: 1 - 3, - 22, 11, - 22

Foci: 1 - 1 - 2 22 , - 2 2 , 1 - 1 + 2 22 , - 2 2 Asymptotes: y + 2 = { 1x + 12 (1, 0) y 4

y2x1

4 x F2  (1  2 兹2, 2) V1  (3, 2) V2  (1, 2) y  2  (x  1) (1, 4) (1, 2)

F1  (1  2 兹2, 2)

= 1

y  7  兹3 (x  5)

V2  (6, 7) F2  (7, 7)

F1  (3, 7)

(5, 7)

1y + 32 2

= 1 4 9 Center: 12, - 32 Transverse axis: parallel to x-axis Vertices: 10, - 32, 14, - 32

1x + 12 2

3

(5, 7 兹3 )

8

(5, 7 兹3 )

Foci: 1 2 - 213 , - 3 2 , 1 2 + 213 , - 3 2 3 Asymptotes: y + 3 = { 1x - 22 2

V2  (3, 1)

4 Center: 1 - 2, 22 Transverse axis: parallel to y-axis Vertices: 1 - 2, 02, 1 - 2, 42

y 6

1y - 72 2

45. 1x - 52 2 -

= 1

F1  (3, 8)

x F2  (1  兹13 , 1)

- 1x + 22 2 = 1

12

V1  (3, 6)

(1, 4)

51.

1x + 32 2

(3  2 兹3, 4)

5

V1  (1, 1)

4

-

V2  (3, 2)

3 y (1, 2) y  1  2 (x1) 4

F1  (1  兹13 , 1)

F1  (0, 5 兹2 )

F2  (3, 0)

兹5 y1 (x  4) 2

(4, 1  兹5 )

1y + 42 2

37. x2 - y2 = 1 y2 x2 39. = 1 36 9

(5, 0) 10 x

V1  (0, 5)

(4, 1) 9 x F2  (7, 1) V2  (6, 1)

V1  (2, 1)

F2  (0, 5 兹2 ) yx

(5, 0)

y (4, 1  兹5 ) 兹5 y1 (x  4) 2 4 F1  (1, 1)

y 10

y (2, 3)

x y  7  兹3 (x  5)

(2, 0)

y3

2

3 (x2) 2

6 x F2  (2  兹13 , 3) V2  (4, 3)

V1  (0, 3) F1  (2  兹13 , 3) (2, 6)

3 y  3   (x2) 2

55. 1x - 12 2 - 1y + 12 2 = 1 Center: 11, - 12 Transverse axis: parallel to x-axis Vertices: 10, - 12, 12, - 12

Foci: 1 1 - 22 , - 1 2 , 1 1 + 22 , - 1 2 Asymptotes: y + 1 = { 1x - 12 y  1  (x  1)

y 2 (1, 0)

F1  (1  兹2 , 1) V1  (0, 1)

y1x1 4 x F2  (1  兹2 , 1)

V2  (2, 1) (1, 2)

ANSWERS  Section 9.4

57.

1y - 22 2

- 1x + 12 2 = 1

4 Center: 1 - 1, 22 Transverse axis: parallel to y-axis Vertices: 1 - 1, 02, 1 - 1, 42

59.

Foci: 1 - 1, 2 - 25 2 , 1 - 1, 2 + 25 2 Asymptotes: y - 2 = {2 1x + 12

V1  (1, 0)

F2  (1, 2  兹5 ) V2  (1, 4) (0, 2) 5 x

63.

(3, 6)

yx 10 x (5, 0)

(5, 0) 5 x

1y - 12 2

- 1x + 22 2 = 1 4 Center: 1 - 2, 12 Transverse axis: parallel to y-axis Vertices: 1 - 2, - 12, 1 - 2, 32 Foci: 1 - 2, 1 - 25 2 , 1 - 2, 1 + 25 2 Asymptotes: y - 1 = {2 1x + 22 F2  (2, 1  兹5) y y  1  2(x  2) 6 V2  (2, 3) (2, 1) (1, 1) (3, 1)

y  1  2(x  2)

4 x

V1  (2, 1)

(3, 2)

10 y  x

y  2x

(0, 4)

61.

V2  (5, 2)

y

8 y  2x

= 1

V1  (1, 2)

65.

y

1y + 22 2

4 16 Center: 13, - 22 Transverse axis: parallel to x-axis Vertices: 11, - 22, 15, - 22 Foci: 1 3 - 2 25 , - 2 2 , 1 3 + 2 25 , - 2 2 Asymptotes: y + 2 = {2 1x - 32

F1  (1, 2  兹5 )

5

-

y  2  2(x  3) y  2  2(x  3) y (3, 2) 2 x 8 F1  (3  2兹5, 2) F2  (3  2兹5, 2)

y  2  2(x  1)

y  2  2(x  1) y (1, 2) (2, 2)

1x - 32 2

AN79

F1  (2, 1  兹5)

67. Center: 13, 02 Transverse axis: parallel to x-axis Vertices: 11, 0), 15, 0) Foci: 13 - 229, 0), 13 + 229, 0) 5 Asymptotes: y = { 1x - 3) 2 (3, 5)

y 10 (3, 0)

F1  (3  兹29, 0) 10 x

F2  (3  兹29, 0) V2  (1, 0)

69. Vertex: 10, 3); Focus: 10, 7); Directrix: y = - 1

71.

F  (0, 7) y

1x - 5)

2

2

y + = 1 9 25 Center: 15, 02 ; Vertices: 15, 5), 15, - 5); Foci: 15, - 4), 15, 4)

73. 1x - 3)2 = 8 1y + 5) Vertex: 13, - 5); Focus: 13, - 3); Directrix: y = - 7 y

8 y V  (0, 3) D: y  1

x

2

V1  (5, 5) 5 F1  (5, 4)

10

10 x

F  (3, 3)

(5, 0) (8, 0) 9 x F2  (5, 4)

(2, 0)

V1  (5, 0) (3, 5)

V  (3, 5) D: y  7

V2  (5, 5)

y2 x2 75. The fireworks display is 50,138 ft north of the person at point A. 77. The tower is 592.4 ft tall. 79. (a) y = {x (b) = 1, x Ú 0 100 100 81. If the eccentricity is close to 1, the “opening” of the hyperbola is very small. As e increases, the opening gets bigger. x2 1 x2 - y2 = 1; asymptotes y = { x 83. y2  1 y 4 4 2 2.5 1 2 (0, 1) y  x x 1 2 y2 = 1; asymptotes y = { x (2, 0) (2, 0) 4 2 1.5

(0, 1)

x

x2  y2  1 4

1 y x 2

P, 3 2 2 3

If A and C are opposite in sign and F ≠ 0, this equation may be written as

x2

y2

= 1, F F b a- b A C F F where - and - are opposite in sign. This is the equation of a hyperbola with center 10, 02 . The A C F F transverse axis is the x-axis if 7 0; the transverse axis is the y-axis if 6 0. A A 3 p p y 88. c ≈ 13.16, A ≈ 31.6°, B ≈ 48.4° 87. Amplitude = ; Period = ; Phase shift = P 2 3 2 U5 89. 16, - 6 232 2 y 3P 2 2 P U5 90. x + (y 3) = 9; circle, U5 4 2.5 4 radius 3, center at (0, 3) in P 3 rectangular coordinates x

85. Ax2 + Cy2 + F = 0 Ax2 + Cy2 = - F

a-

+

x 1 2 3 4 5 6 U50

U5P

U5

U5

5P 4 U5

3P 2

7P 4

AN80

ANSWERS  Section 9.5

9.5 Assess Your Understanding (page 730) 5. cot 12u2 =

A - C B

17. Hyperbola 19. Circle 25. x =

7. B2 - 4AC 6 0

6. parabola 21. x =

y 2.5 x (1, 0) 2.5 x

y

23. x =

13. Ellipse

15. Hyperbola

22 22 1x′ - y′2, y = 1x′ + y′2 2 2

25 25 1x′ - 2y′2, y = 12x′ + y′2 5 5

y

29. x =

213 213 13x′ - 2y′2, y = 12x′ + 3y′2 13 13

35. u = 60° (see Problem 25) x′2 + y′2 = 1 4 Ellipse Center at 10, 02 Major axis is the x′@axis. Vertices at 1 {2, 02

y

y 2.5 x

2.5 x

(0, 2) 2.5

(2, 0) 5 x

y

2.5 x

(2, 0)

39. u ≈ 34° (see Problem 29) 1x′ - 22 2 + y′2 = 1 4 Ellipse Center at 12, 02 Major axis is the x′@axis. Vertices at 14, 02 and 10, 02

x

(2, 0)

y

x (0, 2)

37. u ≈ 63°(see Problem 27) y′2 = 8x′ Parabola Vertex at 10, 02 Focus at 12, 02 y 5

27. x =

10. False 11. Parabola

33. u = 45° (see Problem 23) y′2 = 1 x′2 + 4 Ellipse Center at 10, 02 Major axis is the y′@axis. Vertices at 10, {22

31. u = 45° (see Problem 21) y′2 = 1 x′2 3 Hyperbola Center at origin Transverse axis is the x′@axis. Vertices at 1 {1, 02

(1, 0)

9. True

22 22 1x′ - y′2, y = 1x′ + y′2 2 2

1 1 1 x′ - 23y′ 2 , y = 1 23x′ + y′ 2 2 2

y

8. True

y 6 (2, 1) x (4, 0) 5 x (2, 1)

41. cot 12u2 =

7 ; 24 3 u = sin-1 a b ≈ 37° 5 1x′ - 12 2 = - 6 ay′ -

1 b 6

Parabola

1 Vertex at a1, b 6 4 Focus at a1, - b 3 y y

5

1,

1 6 x 5 x 1, 

4 3

43. Hyperbola 45. Hyperbola 47. Parabola 49. Ellipse 51. Ellipse 53. Refer to equation (6): A′ = A cos2 u + B sin u cos u + C sin2 u B′ = B1cos2 u - sin2 u2 + 2 1C - A2 1sin u cos u2 C′ = A sin2 u - B sin u cos u + C cos2 u D′ = D cos u + E sin u E′ = - D sin u + E cos u F′ = F 55. Use Problem 53 to find B′2 - 4A′C′. After much cancellation, B′2 - 4A′C′ = B2 - 4AC. 57. The distance between P1 and P2 in the x′y′@plane equals 2 1x2 = - x1 = 2 2 + 1y2 = - y1 = 2 2 . Assuming that x′ = x cos u - y sin u and y′ = x sin u + y cos u, then 1x2 = - x1 = 2 2 = 1x2 cos u - y2 sin u - x1 cos u + y1 sin u2 2 = cos2 u1x2 - x1 2 2 - 2 sin u cos u1x2 - x1 2 1y2 - y1 2 + sin2 u 1y2 - y1 2 2, and = = 2 1y2 - y1 2 = 1x2 sin u + y2 cos u - x1 sin u - y1 cos u2 2 = sin2 u 1x2 - x1 2 2 + 2 sin u cos u1x2 - x1 2 1y2 - y1 2 + cos2 u 1y2 - y1 2 2. Therefore, 1x2 = - x1 = 2 2 + 1y2 = - y1 = 2 2 = cos2 u1x2 - x1 2 2 + sin2 u 1x2 - x1 2 2 + sin2 u 1y2 - y1 2 2 + cos2 u 1y2 - y1 2 2 = 1x2 - x1 2 2 1cos2 u + sin2 u2 + 1y2 - y1 2 2 1sin2 u + cos2 u2 = 1x2 - x1 2 2 + 1y2 - y1 2 2. 61. A ≈ 39.4°, B ≈ 54.7°, C ≈ 85.9°

62. 38.5

63. r 2cos u sin u = 1

64. 229(cos 291.8° + i sin 291.8°)

9.6 Assess Your Understanding (page 736) 3. conic; focus; directrix 4. 1; 6 1; 7 1 5. True 6. True 7. Parabola; directrix is perpendicular to the polar axis 1 unit to the right of the pole. 4 9. Hyperbola; directrix is parallel to the polar axis units below the pole. 3 3 11. Ellipse; directrix is perpendicular to the polar axis units to the left of the pole. 2

ANSWERS  Section 9.7 15. Ellipse; directrix is parallel to the polar 8 axis units above the pole; vertices are at 3 8 p 3p a , b and a8, b. 7 2 2

13. Parabola; directrix is perpendicular to the polar axis 1 unit to the right of the pole; 1 vertex is at a , 0 b . 2 y Directrix P 2

1,

1 ,0 2

2

2

3P 2

1,

8,

(4, 0)

Polar x axis 4 8 3P (4, P) , Directrix 3 2

(2, P)

Directrix y P 5 3, 2 (23, 0) (1, P) Polar 5 x axis 3P 3, 2

23. Ellipse; directrix is perpendicular to the polar axis 6 units to the left of the pole; vertices are at (6, 0) and 12, p2. Directrix y 5 (2, P)

3,

P 2

(6, 0) 4

Polar x axis

(2, 0)

Polar 5 x axis

3,

3P 2

Directrix

6 3P , 5 2

25. y2 + 2x - 1 = 0

P 6, 2

y 7

4

17. Hyperbola; directrix is perpendicular to the 3 polar axis units to the left of the pole; 2 vertices are at 1 - 3, 02 and 11, p2.

3P 2

y P 2

(2, 0) Polar 5 x axis

21. Ellipse; directrix is parallel to the polar axis 3 units below the pole; vertices are at p 6 3p a6, b and a , b. 2 5 2

19. Ellipse; directrix is parallel to the polar axis 8 units below the pole; vertices are at 8 3p p b. a8, b and a , 2 3 2

8,

8 P , 7 2

Directrix y (2, P) 2

Polar x axis

27. 16x2 + 7y2 + 48y - 64 = 0

29. 3x2 - y2 + 12x + 9 = 0 31. 4x2 + 3y2 - 16y - 64 = 0 1 12 12 39. r = 41. r = 33. 9x2 + 5y2 - 24y - 36 = 0 35. 3x2 + 4y2 - 12x - 36 = 0 37. r = 1 + sin u 5 - 4 cos u 1 - 6 sin u 43. Use d 1D, P2 = p - r cos u in the derivation of equation (a) in Table 5. 45. Use d 1D, P2 = p + r sin u in the derivation of equation (a) in Table 5. p 5p 47. 27.81 48. Amplitude = 4; Period = 10p 49. e , p, f 50. 26 3 3

9.7 Assess Your Understanding (page 746) 2. plane curve; parameter 3. ellipse 7.

4. cycloid 9.

y

5. False

6. True 11.

y 5

13.

y 10

y 5

6 x

3

17.

y 5

x = 3 1y - 12 2

y = x - 8

y = 2x - 2

x - 3y + 1 = 0 15.

10 x

x

5 x

12

19.

y 3

21.

y 5

3 x

y 5

5 x

5 x

5 x

23.

25.

y

y 1

2 (兹2 , 1)

27. x y 29. x y 31. x y 33. x y

back and forth twice 1 x

x

2

x - y = 1 2

y2 x2 + = 1 4 9

y = x3

2y = 2 + x

x + y = 1

2

= = = = = = = =

t or 4t - 1 t or t2 + 1 t or t3 t or t 2/3, t Ú 0

35. x = t + 2, y = t, 0 … t … 5 37. x = 3 cos t, y = 2 sin t, 0 … t … 2p 39. x = 2 cos 1pt2, y = - 3 sin 1pt2, 0 … t … 2 41. x = 2 sin 12pt2, y = 3 cos 12pt2, 0 … t … 1 y y y 43. y (4, 16) 20

(4, 16)

(1, 1) (0, 0) C1

(1, 1) 5 x

16

20

(1, 1)

(1, 1) C2

2.5 x

AN81

(4, 16)

C3

(1, 1) 5 x

20

(4, 16)

C4

(1, 1) 5 x

y2 x2 + = 1 4 9 t + 1 x = 4 y = t x = t3 y = t6 + 1 3 x = 2 t y = t x = t3 y = t 2, t Ú 0

AN82

ANSWERS  Section 9.7

45.

47.

7

4

6 5

6

9

6

5

49. (a) x = 3 y = - 16t 2 + 50t + 6 (b) 3.24 s (c) 1.56 s; 45.06 ft (d) 50

53. (a) x = 1145 cos 20°2t y = - 16t 2 + 1145 sin 20°2t + 5 (b) 3.20 s (c) 435.65 ft (d) 1.55 s; 43.43 ft (e) 170

51. (a) Train: x1 = t 2, y1 = 1; Bill: x2 = 5 1t - 52, y2 = 3 (b) Bill won't catch the train. (c) 5

Bill Train 0 0

100

0

440

0

5 0

120

55. (a) x = 140 cos 45°2t y = - 4.9t 2 + 140 sin 45°2t + 300 (b) 11.23 s (c) 317.52 m (d) 2.89 s; 340.82 m (e) 400

57. (a) Camry: x = 40t - 5, y = 0; Impala: x = 0, y = 30t - 4 (d) 0.2 mi; 7.68 min (b) d = 2 140t - 52 2 + 130t - 42 2 (e) Turn axes off to see the graph: (c) 7 4

6 0

6

0.2 0 4

160

320 0

22 22 y t, y = - 16t 2 + y t + 3 (b) Maximum height is 139.1 ft. (c) The ball is 272.25 ft from home plate. 2 0 2 0 (d) Yes; the ball will clear the wall by about 99.5 ft. 61. The orientation is from 1x1 , y1 2 to 1x2 , y2 2.

59. (a) x = 65.

66.

y 6

y 3

y 5 sin

x 2

4 2 2P 26 24 22 22

2

4

x

6 x y 5 2 cos(2x)

24

67. Approximately 2733 miles 68. (a) Simple harmonic (b) 2 ft p (c) s 2 2 (d) oscillation/s p

26

Review Exercises (page 751) 1. Parabola; vertex (0, 0), focus (0, 3), directrix y = - 3 2. Hyperbola; center (0, 0), vertices (3, 0) and ( - 3, 0), foci (5, 0) and ( - 5, 0), asymptotes 4 4 y = x and y = - x 3. Ellipse; center (0, 0), vertices (6, 0) and ( - 6, 0), foci 13 23, 02 and 1 - 3 23, 02 4. Parabola; vertex (3, 1), 3 3 focus (5, 1), directrix x = 1

5. Hyperbola; center (0, 0), vertices 1 22 , 0 2 and 1 - 22 , 0 2 , foci 1 210 , 0 2 and 1 - 210 , 0 2 ,

3 5 7 3 asymptotes y = 2x and y = - 2x 6. Parabola; vertex a - , 3 b , focus a - , b , directrix y = 2 2 2 2 y - 1 x + 1 7. Hyperbola; center ( - 1, 1), vertices (3, 1) and ( - 5, 1), foci ( - 6, 1) and (4, 1), asymptotes = { 4 3

8. Ellipse; center (1, - 3), vertices 11, 2 23 - 32 and 11, - 2 23 - 32 , foci 11,22 - 32 and 111, - 22 - 322 1 1 1 9. Parabola; vertex a , - 1 b , focus a , - 1 b directrix x = 2 2 2

10. Ellipse; center 11, - 12, vertices (1, 2) and 11, - 42, foci 1 1, - 1 + 25 2 and 1 1, - 1 - 25 2

ANSWERS  Review Exercises

11. y2 = - 8x

12. y 5

5 x

V  (0, 0) (2, 4)

y 2

1x + 12 2

-

(4, 4)

9

1y - 22 2 7

18.

V2  (2, 2) F2  (3, 2) 8 x

(1, 2) y  2  兹7 (x  1) (1, 2 兹7 ) 3

y′2

y2 x2 + = 1 16 7

= 1

y (0, 兹7) 5 V1  (4, 0)

1x - 32 2

-

9

5 x

16.

1x + 42 2

9 Hyperbola Center at the origin Transverse axis is the x′@axis. Vertices at 1 {1, 02 y 5

y

16

1y - 12 2 4

x 9

= 1

= 1

(0, 5) F1  (4, 2) V1  (4, 0)

19. Parabola 20. Ellipse 21. Parabola 22. Hyperbola 23. Ellipse

4 213 x′ 13

26. y′2 = -

Parabola Vertex at the origin Focus on the x′@axis at a -

x

(0, 2) 2

25

(8, 5)

213 , 0b 13

y 5

y x

1y - 52 2

V2  (1, 3)

y′ x′2 + = 1 2 4 Ellipse Center at origin Major axis is the y′@axis. Vertices at 10, {22

(1, 0) 5 x

(1, 0)

+

y V2  (4, 10) 10 F2  (4, 8) (4, 5)

2 x F2  (0, 3)

y (3, 1) y  1   2 (x  3) 5 (3, 3) 3 V2  (6, 1) V1  (0, 1) F2  (3  兹13, 1) F1  (3  兹13, 1) x 5 2 y1 (x  3) (3, 1) 3

25.

F2  (3, 0)

(0, 兹7 )

2

= 1

V2  (4, 0)

F1  (3, 0)

(2, 3 兹3 )

y  2  兹7 (x  1) y (1, 2 兹7 ) 3 12

24. x′2 -

3

(2, 3)

= 1

V1  (4, 2) F1  (5, 2)

1y + 32 2

V1  (3, 3) F1  (4, 3)

F  (2, 4)

17.

13.

y  3  兹3 (x  2) y y  3  兹3 (x  2) 2 (2, 3 兹3 )

8 x V  (2, 3)

(0, 4)

x2 = 1 12

15. 1x + 22 2 -

14. 1x - 22 2 = - 4 1y + 32

D: y  2

4

-

V2  (0, 2) y 5 F2  (0, 4) 兹3 y x 兹3 y 3 x 3 5 x (2兹3, 0) (2兹3, 0) V1  (0, 2) F1  (0, 4)

D: x  2

(2, 4) F  (2, 0)

y2

AN83

y

x

y

5 x

x (0, 2)

27. Parabola; directrix is perpendicular to the polar axis 4 units to the left of the pole; vertex is 12, p2. Directrix y 5

28. Ellipse; directrix is parallel to the polar axis 6 units below the pole; p 3p vertices are a6, b and a2, b. 2 2

P 2 (2, P) Polar 5 x axis 3P 4, 2

y 5

4,

6,

(3, P) 2,

29. Hyperbola; directrix is perpendicular to the polar axis 1 unit to the right of the pole; 2 vertices are a , 0 b and 1 - 2, p2. 3

P 2

2,

2 ,0 3

(3, 0) Polar 5 x axis

3P 2

P 2

2,

y

Directrix (22, P) Polar 3 x axis

1 3P 2

Directrix

30. x = 4 1y + 12 2

32.

31. 3x - y - 8x + 4 = 0 2

2

33.

y (2, 1) 2 (2, 0) 2

x

34.

y (0, 6) 7 (3, 2)

2 (0, 2) (3, 2) 5 x (0, 2)

x + 4y = 2

y

1y - 22 2 x2 + = 1 9 16

(2, 1) 2 (1, 0)

x

1 + y = x

p p t - 4 , y = t, - q 6 t 6 q 36. x = 4 cos a t b , y = 3 sin a t b , 0 … t … 4 35. x = t, y = - 2t + 4, - q 6 t 6 q ; x = -2 2 2 2 y x 1 38. The ellipse + = 1 39. ft, or 3 in. 40. 19.72 ft, 18.86 ft, 14.91 ft 41. 450 ft 16 7 4

37.

y2 x2 = 1 5 4

AN84

ANSWERS  Review Exercises 43. (a) x = 180 cos 35°2t y = - 16t 2 + 180 sin 35°2t + 6 (b) 2.9932 s (c) 1.4339 s; 38.9 ft (d) 196.15 ft (e) 50

3 2 t Mary: x2 = 6 1t - 22 2 y1 = 1 y2 = 3 (b) Mary won't catch the train. (c) 5

42. (a) Train: x1 =

0 0

250

100 t8

0

50

Chapter Test (page 752) 1. Hyperbola; Center: 1 - 1, 02; Vertices: 1 - 3, 02 and 11, 02; Foci: 1 - 1 - 213 , 0 2 and 1 - 1 + 213 , 0 2 ; 3 3 Asymptotes: y = - 1x + 12 and y = 1x + 12 2 2 3 1 5 2. Parabola; Vertex: a1, - b ; Focus: a1, b ; Directrix: y = 2 2 2 3. Ellipse; Center: 1 - 1, 12; Foci: 1 - 1 - 23 , 1 2 and

1 - 1 + 23 , 1 2 ; Vertices: 1 - 4, 12 and 12, 12 y2 x2 + = 1 5. 7 16

4. 1x + 12 2 = 6 1y - 32 F  (1, 4.5)

y

V  (1, 3)

1y - 22 2 4

y 5 V1  (0, 4)

9

(4, 4.5)

6.

(兹7, 0) F1

(2, 4.5) D: y  1.5

F2

5 x

y 8

(兹7, 0) 5 x

1x - 22 2 8 F1

(2  2兹2, 2)

= 1

V1  (2, 4) (2  兹10, 5) (2, 2) (2  2兹2, 2) 8 x

V2  (0, 4)

7. Hyperbola 8. Ellipse

-

F2

V2  (2, 0)

9. Parabola

10. x′2 + 2y′2 = 1. This is the equation of an ellipse with center at 10, 02 in the x′y′@plane. The vertices are at 1 - 1, 02 and 11, 02 in the x′y′@plane.

y 5 y

x

11. Hyperbola; 1x + 22 2 2

y2 3

= 1

1 5 x

12. y = 1 -

B

x + 2 3

y 5 (2, 1)

(1, 0) (10, 1)

13. The microphone should be located

45 x (25, 2)

2 ft (or 8 in.) from the base of the reflector, along its axis of symmetry. 3

Cumulative Review (page 752) 1 2. e - 5, - , 2 f 3. {x  - 3 … x … 2} or [ - 3, 2] 4. (a) Domain: 1 - q , q 2; Range: 12, q 2 3 y2 x2 (b) y = log 3 1x - 22; Domain: 12, q 2; Range: 1 - q , q 2 5. (a) 518 6 (b) (2, 18] 6. (a) y = 2x - 2 (b) (x - 2)2 + y2 = 4 (c) + = 1 9 4 2 x p 5p p (d) y = 2(x - 1)2 (e) y2 = 1 (f) y = 4x 7. u = { pk, k is any integer; u = { pk, k is any integer 8. u = 3 12 12 6 3p x2 y 9. r = 8 sin u 10. e x ` x ≠ { pk, k is an integer f 11. 522.5° 6 12. y = + 5 10 4 5 1. - 6x + 5 - 3h

10 x

ANSWERS  Section 10.2

AN85

CHAPTER 10 Systems of Equations and Inequalities 10.1 Assess Your Understanding (page 765) 3. inconsistent 4. consistent; independent 5. 13, - 22 1 3 122 - 4 a b = 4 2 122 - 1 - 12 = 5 2 7. e 9. d 5 122 + 2 1 - 12 = 8 1 1 1 122 - 3 a b = 2 2 2 17. x = 6, y = 2; 16, 22 27. x =

3 3 , y = 3; a , 3 b 2 2

33. x =

3 3 , y = 1; a , 1 b 2 2

6. consistent; dependent

3 112 + 3 1 - 12 + 2 122 = 4 3 122 + 3 1 - 22 + 2 122 = 4 4 - 1 = 3 13. c 1 - 1 - 12 2 = 0 15. c 2 - 3 1 - 22 + 2 11. c 1 = 10 142 + 1 = 3 2 1 - 12 - 3 122 = - 8 5 122 - 2 1 - 22 - 3 122 = 8 2

1 1 1 1 , y = - ; a , - b 25. Inconsistent 3 6 3 6 4 - x 29. 5 1x, y2 x = 4 - 2y, y is any real number 6 or b 1x, y2 ` y = , x is any real number r 31. x = 1, y = 1; 11, 12 2

19. x = 3, y = - 6; 13, - 62

35. x = 4, y = 3; 14, 32

43. x = 2, y = - 1, z = 1; 12, - 1, 12

21. x = 8, y = - 4; 18, - 42

37. x =

4 1 4 1 ,y = ; a , b 3 5 3 5

23. x =

39. x =

1 1 1 1 ,y = ; a , b 5 3 5 3

41. x = 8, y = 2, z = 0; 18, 2, 02

47. 5 1x, y, z2 x = 5z - 2, y = 4z - 3; z is any real number 6 49. Inconsistent 1 1 51. x = 1, y = 3, z = - 2; 11, 3, - 22 53. x = - 3, y = , z = 1; a - 3, , 1 b 55. Length 30 ft; width 15 ft 57. There were 20 commercial launches 2 2 and 58 noncommercial launches in 2012. 59. 22.5 lb 61. Smartphone: $325; tablet: $640 63. Average wind speed 25 mph; average airspeed 175 mph 65. 80 $25 sets and 120 $45 sets 67. $9.96 69. Mix 50 mg of first compound with 75 mg of second. 10 65 55 4 5 ,I = ,I = 71. a = , b = - , c = 1 73. Y = 9000, r = 0.06 75. I1 = 3 3 71 2 71 3 71 77. 100 orchestra, 210 main, and 190 balcony seats 79. 1.5 chicken, 1 corn, 2 milk 45. Inconsistent

81. If x = price of hamburgers, y = price of fries, and z = price of colas, then 41 1 x = 2.75 - z, y = + z, +0.60 … z … +0.90. 60 3 There is not sufficient information; see table.

x

$2.13

$2.01

$1.86

y

$0.89

$0.93

$0.98

z

$0.62

$0.74

$0.89

83. It will take Beth 30 hr, Bill 24 hr, and Edie 40 hr. y 87. 6 y=2

(1, 1) 26

6 x

(0, 21)

26

88. (a) 2(2x - 3)3(x3 + 5)(10x3 - 9x2 + 20) 3

1

(b) 7(3x - 5)- 2 (x + 3)- 2 p 90. 2(cos 150° + i sin 150°) 89. 9

10.2 Assess Your Understanding (page 781) 1. matrix 1 5. c 4

2. augmented

-5 2 5 d 3 6

2 7. c 4

3. third; fifth 3 2 6 d -6 -2

112 ; 122

17. e

x - 3y = - 2 2x - 5y = 5

21. c

x - 3y + 2z = - 6 2x - 5y + 3z = - 4 - 3x - 6y + 4z = 6

c

1 0

112 122 ; 132

0.01 9. c 0.13 -3 -2 ` d 1 9 1 C0 0

x = 5 y = -1 Consistent; x = 5, y = - 1 or 15, - 12

25. e

4. True

-3 1 - 15

- 0.03 2 0.06 d 0.10 0.20

19. W

1 11. C 3 1

-1 3 1

x - 3y + 4z = 3 3x - 5y + 6z = 6 - 5x + 3y + 4z = 6

2 -6 -1 3 8S 10 - 12

23. c

1 10 0 3 5S 2 2 112 122 ; 132

1 C0 0

1 13. C 3 5 -3 4 - 12

5x - 3y + z = - 2 2x - 5y + 6z = - 2 - 4x + y + 4z = 6

x = 1 27. c y = 2 0 = 3 Inconsistent

1 -2 3

-1 2 0 3 2S -1 1

1 2 15. D -3 4

4 3 -6 3 -3S 24 21

112 122 ; 132

1 C2 0

7 -5 -9

- 11 2 6 3 -2S 16 2

-1 1 4 -5

- 1 10 2 4 -1 T 0 5 1 0

AN86

ANSWERS  Section 10.2 x1 + 4x4 = 2 x1 = 1 x1 + x4 = - 2 31. c x2 + x4 = 2 33. c x2 + x3 + 3x4 = 3 x2 + 2x4 = 2 35. d x3 + 2x4 = 3 0 = 0 x3 - x4 = 0 Consistent: Consistent: 0 = 0 x1 = 2 - 4x4 x1 = 1, x2 = 2 - x4 Consistent: c x3 = 3 - 2x4 c x2 = 3 - x3 - 3x4 x1 = - 2 - x4 x4 is any real number or x3, x4 are any real numbers or x = 2 - 2x4 μ 2 5 1x1, x2, x3, x4 2  x1 = 1, 5 1x1, x2, x3, x4 2  x1 = 2 - 4x4, x3 = x4 x2 = 2 - x4, x3 = 3 - 2x4, x4 is x2 = 3 - x3 - 3x4, x3, x4 are any x4 is any real number or any real number 6 real numbers 6 5 1x1, x2, x3, x4 2  x1 = - 2 - x4, x2 = 2 - 2x4, x3 = x4, x4 is any real number 6

x + 2z = - 1 29. c y - 4z = - 2 0 = 0 Consistent: x = - 1 - 2z c y = - 2 + 4z z is any real number or 5 1x, y, z2  x = - 1 - 2z, y = - 2 + 4z, z is any real number 6

3 1 3 1 , y = ; a , b 41. x = 4 - 2y, y is any real number; 5 1x, y2  x = 4 - 2y, y is any real number 6 2 4 2 4 3 3 4 1 4 1 , y = 1; a , 1 b 45. x = , y = ; a , b 47. x = 8, y = 2, z = 0; 18, 2, 02 49. x = 2, y = - 1, z = 1; 12, - 1, 12 51. Inconsistent 2 2 3 5 3 5 5z - 2, y = 4z - 3, where z is any real number; 5 1x, y, z2  x = 5z - 2, y = 4z - 3, z is any real number 6 55. Inconsistent 1 1 1 2 1 2 1, y = 3, z = - 2; 11, 3, - 22 59. x = - 3, y = , z = 1; a - 3, , 1 b 61. x = , y = , z = 1; a , , 1 b 2 2 3 3 3 3 1, y = 2, z = 0, w = 1; 11, 2, 0, 12 65. y = 0, z = 1 - x, x is any real number; 5 1x, y, z2  y = 0, z = 1 - x, x is any real number 6

37. x = 6, y = 2; 16, 22 43. x = 53. x = 57. x = 63. x =

39. x =

67. x = 2, y = z - 3, z is any real number; 5 1x, y, z2  x = 2, y = z - 3, z is any real number 6

13 7 19 13 7 19 ,y = ,z = ;a , , b 9 18 18 9 18 18 2 8 13 7 8 7 3 7 3 2 7 13 71. x = - z - w, y = - + z + w, where z and w are any real numbers; e 1x, y, z, w2 ` x = - z - w, y = - + z + w, 5 5 5 5 5 5 5 5 5 5 5 5 z and w are any real numbers r

69. x =

75. f 1x2 = 3x3 - 4x2 + 5

73. y = - 2x2 + x + 3

77. 1.5 salmon steak, 2 baked eggs, 1 acorn squash 44 16 28 , I = 2, I3 = ,I = 79. $4000 in Treasury bills, $4000 in Treasury bonds, $2000 in corporate bonds 81. 8 Deltas, 5 Betas, 10 Sigmas 83. I1 = 23 2 23 4 23 85. (a) (b) (c) All the money invested at 7% provides $2100, Amount Invested At Amount Invested At more than what is required. 7% 9% 11% 7% 9% 11%

87.

0

10,000

10,000

12,500

12,500

0

1000

8000

11,000

14,500

8500

2000

2000

6000

12,000

16,500

4500

4000

3000

4000

13,000

18,750

0

6250

4000

2000

14,000

5000

0

15,000

92. {x  x is any real nunber} or ( - q , q ) 93. {x 0 - 1 6 x 6 6} or ( - 1, 6) 94. y 9

First Supplement

Second Supplement

Third Supplement

50 mg

75 mg

0 mg

36 mg

76 mg

8 mg

22 mg

77 mg

16 mg

8 mg

78 mg

24 mg

12 35 , 2 2

(23, 5)

13, 78 2 y=2

26

(1, 0)

12 12 , 0 2

6 x

(0, 21) x = 21

95. 2.42

10.3 Assess Your Understanding (page 792) 1. ad - bc

2. 2

5 -3

3. False 4. False 5. False 6. False 7. 22 9. - 2 11. 10 13. - 26 15. x = 6, y = 2; 16, 22

32 -4

19. x = 8, y = - 4; 18, - 42 29. x =

3 3 , y = 1; a , 1 b 2 2

21. x = 4, y = - 2; 14, - 22

31. x =

4 1 4 1 ,y = ;a , b 3 5 3 5

23. Not applicable

= = = =

1 3 1 3 ,y = ;a , b 2 4 2 4

33. x = 1, y = 3, z = - 2; 11, 3, - 22

37. Not applicable 39. x = 0, y = 0, z = 0; (0, 0, 0) 41. Not applicable 57. 1y1 - y2 2x - 1x1 - x2 2y + 1x1y2 - x2y1 2 1y1 - y2 2x + 1x2 - x1 2y 1x2 - x1 2y - 1x2 - x1 2y1 1x2 - x1 2 1y - y1 2

25. x =

43. - 4

45. 12

0 x2y1 - x1y2 1y2 - y1 2x + x2y1 - x1y2 - 1x2 - x1 2y1 1y2 - y1 2x - 1y2 - y1 2x1 y2 - y1 y - y1 = 1x - x1 2 x2 - x1

27. x =

35. x = - 3, y = 47. 8

17. x = 3, y = 2; 13, 22

1 2 1 2 ,y = ;a , b 10 5 10 5

1 1 , z = 1; a - 3, , 1 b 2 2

13 55. 0 or - 9 11 59. The triangle has an area of 5 square units. 49. 8

51. - 5

53.

AN87

ANSWERS  Section 10.4 61. 50.5 square units 63. (x - 3)2 + (y + 2)2 = 25 65. If a = 0, we have If b = 0, we have by = s ax = s cx + dy = t cx + dy = t s Since D = ad ≠ 0, then Thus, y = and b a ≠ 0 and d ≠ 0. t - dy tb - ds s x = = Thus, x = and c bc a t - cx ta - cs Using Cramer’s Rule, we get y = = d ad sd - tb tb - sd x = = Using Cramer’s Rule, we get - bc bc sd s - sc s x = = y = = ad a - bc b ta - cs y = ad a11 67. 3 ka21 a31

a12 ka22 a32

If c = 0, we have ax + by = s dy = t Since D = ad ≠ 0, then a ≠ 0 and d ≠ 0. t Thus, y = and d s - by sd - bt x = = a ad Using Cramer’s Rule, we get sd - bt x = ad at t y = = ad d

If d = 0, we have ax + by = s cx = t Since D = - bc ≠ 0, then b ≠ 0 and c ≠ 0. t Thus, x = and c s - ax sc - at y = = b bc Using Cramer’s Rule, we get - tb t x = = - bc c at - sc sc - at y = = - bc bc

a13 ka23 3 = - ka21 1a12a33 - a32a13 2 + ka22 1a11a33 - a31a13 2 - ka23 1a11a32 - a31a12 2 a33 a11 = k[ - a21 1a12a33 - a32a13 2 + a22 1a11a33 - a31a13 2 - a23 1a11a32 - a31a12 2] = k 3 a21 a31

69. 3

a11 + ka21 a21 a31

a12 + ka22 a22 a32

a12 a22 a32

a13 a23 3 a33

a13 + ka23 3 = 1a11 + ka21 2 1a22a33 - a32a23 2 - 1a12 + ka22 2 1a21a33 - a31a23 2 + 1a13 + ka23 2 1a21a32 - a31a22 2 a23 a33 = a11a22a33 - a11a32a23 + ka21a22a33 - ka21a32a23 - a12a21a33 + a12a31a23 - ka22a21a33 + ka22a31a23 + a13a21a32 - a13a31a22 + ka23a21a32 - ka23a31a22 = a11a22a33 - a11a32a23 - a12a21a33 + a12a31a23 + a13a21a32 - a13a31a22 = a11 1a22a33 - a32a23 2 - a12 1a21a33 - a31a23 2 + a13 1a21a32 - a31a22 2 a11 a12 a13 = 3 a21 a22 a23 3 a31 a32 a33 72.

70. v = 9i - 4j; 297 1 5 71. { , { , {1, {2, {5, {10 2 2

73. 0

y 5

5 x (22, 23) (0, 23) (21, 24)

Historical Problem (page 808) 1. (a) 2 - 5i · c

-5 1 d , 1 + 3i · c 2 -3

2 5

3 d 1

(b) c

-5 1 dc 2 -3

2 5

3 17 d = c 1 -1

1 d 17

(c) 17 + i

a b a -b a 2 + b2 0 dc d = c d ; the product is a real number. -b a b a 0 b2 + a 2 ka + lc 3. (a) x = k 1ar + bs2 + l 1cr + ds2 = r 1ka + lc2 + s 1kb + ld2 (b) A = c ma + nc y = m1ar + bs2 + n 1cr + ds2 = r 1ma + nc2 + s 1mb + nd2

(d) 17 + i

2. c

kb + ld d mb + nd

10.4 Assess Your Understanding (page 808) 1. square 9. c

4 -1

- 13 23. C - 18 17

2. True 4 5

-5 d 4 7 10 -7

3. columns; rows 4. False 11. c

- 12 - 14 S 34

0 4

12 8

-2 25. c 2

- 20 d 24

4 1

13. c

2 4

8 d 6

5. inverse -8 7

7 0

5 27. c 9

- 15 d 22

14 d 16

7. True 8. A- 1B 1 14 28 - 9 15. c d 17. C 2 22 4 23 3 0

6. singular

9 29. C 34 47

2 13 S 20

1 31. c -1

-1 d 2

- 14 - 18 S 28 33. £

15 19. C 22 - 11 1 -1

-

5 2§ 3

21 34 7

- 16 - 22 S 22 1

35. D -1

21. c

-

1 a T 2 a

25 4

-9 d 20

AN88

ANSWERS  Section 10.4

-3 2 5

3 37. C - 2 -4

1 -1S -2

39. F

1 7 1 7 2 7

5 7 9 7 3 7

3 7 4 - V 7 1 7

41. x = 3, y = 2; 13, 22

43. x = - 5, y = 10; 1 - 5, 102

45. x = 2, y = - 1; 12, - 12

1 1 2 3 2 3 , y = 2; a , 2 b 49. x = - 2, y = 1; 1 - 2, 12 51. x = , y = ; a , b 53. x = - 2, y = 3, z = 5; 1 - 2, 3, 52 2 2 a a a a 1 1 1 1 34 85 12 34 85 12 1 2 1 2 55. x = , y = - , z = 1; a , - , 1 b 57. x = - , y = ,z = ; a - , , b 59. x = , y = 1, z = ; a , 1, b 2 2 2 2 7 7 7 7 7 7 3 3 3 3

47. x =

4 61. c 2 -3 65. C 1 1

-1 1 -7 3 0 5 0

1 -4 2 1

2

5

SF 0

1

2 6 0

0

1

2

0.01 67. C 0.01 - 0.02

1 1 2 3 4 1 0

1 0 dSC 1 2

2 2 1 1 0

0.05 - 0.02 0.01

0 1 0

0 -

1 7

0 1

0

0 1 0S S C 1 1 -3

2 -4 1 1

0

1

1 6

1 V S F 0 6 3 0 7

0 - 0.01 0.01 S 0.03

0.02 - 0.02 69. D 0.02 - 0.02

1 4 1 2

1 2 4

1 S SD

0

0 T 1

5 0 -7 3 0 -1 1

2

5

1

2 6 0

0

0

0 1 0

0 -

1 7

- 0.04 0.05 0.01 0.06

15 63. c 10

3 2 1 2 0

1 1 0S S C0 0 0

0

1

1 6 1 6

1 V 6 11 42

- 0.01 0.03 - 0.04 0.07

0.01 - 0.03 T 0.00 0.06

2 -6 7

1 1 5 3 15 2 0

1 0 d S£ 1 10 5 0 - 12 3 0 14 1

0 1 0

0

§ SD

1

1 0

1 1 5 4 15 2 0 3

0 T 1

1 -1S 3

71. x = 4.57, y = - 6.44, z = - 24.07 or 14.57, - 6.44, - 24.072

73. x = - 1.19, y = 2.46, z = 8.27 or 1 - 1.19, 2.46, 8.272 5 5 75. x = - 5, y = 7; 1 - 5, 72 77. x = - 4, y = 2, z = ; a - 4, 2, b 79. Inconsistent; ∅ 2 2 1 1 1 6 1 1 1 6 81. x = - z + , y = z - , where z is any real number; e 1x, y, z2 2 x = - z + , y = z - , z is any real number f 5 5 5 5 5 5 5 5 6 9 100.00 3591.42 83. (a) A = c d; B = c d (b) AB = c d ; Nikki’s total tuition and fees are $3591.42, and Joe’s total tuition and fees are $4288.56. 332.38 4288.56 3 12 85. (a) c

500 700

350 500

1 87. (a) K -1 = £ - 1 0

500 400 d ; C 350 850 400 0 1 -1

-1 1§ 1

700 500 S 850

15 (b) C 8 S 3

13 (b) M = £ 8 6

(c) c 1 9 21

11,500 d 17,050

20 19 § 14

(d) [0.75

0.80]

(e) $22,265

(c) Math is fun.

89. If D = ad - bc ≠ 0, then a ≠ 0 and d ≠ 0, or b ≠ 0 and c ≠ 0. Assuming the former, a c c

b 2 1 d 0

0 1 d S£ 1 c R1 =

b 1 a 3 a d 0

1 r a 1

2 5 91. (a) B3 = A + A2 + A3 = E 4 2 1

0

§ SD

1 0

1

b 1 a a 4 D c a a

R2 = - cr1 + r2 4 3 2 2 3

5 2 2 3 2

2 5 4 2 1

0

1

b a 4

0

1

T SD 1 R2 =

a r D 2

1 a c D

0 a D

1 T SD 0

d D 4 c 1 D 0

-

b D T a D

b R1 = - r2 + r1 a

3 4 2 U ; Yes, all pages can reach every other page within 3 clicks. (b) Page 3 3 2

567.75 - 2.902 - 1.983 2.335 - 1.75 - 1.167 0.661 0.361 0.565 0.005 - 2.517 - 0.427 3.768 1.307 0.405 1.050 - 0.310 0.050 - 2.123 1.053 - 1.738 - 0.496 0.487 1.164 93. H 0 0.929 - 0.462 0.473 - 3.5 0.565 - 1.657 - 1.744 2.063 - 2.470 - 0.162 - 0.034 0.604 - 1.747 - 0.875 - 1.810 1.211 1.147 - 0.431 1.204 0.652 - 0.292 0.168 0.242

1.092 - 0.433 0.875 - 0.356 0.191 - 0.664 - 0.655 0.851

0.715 - 0.662 - 0.568 - 0.493 X - 1.046 - 0.727 0.985 0.169

ANSWERS  Section 10.6 23 1 2 2 95. (a) 13 - 2 23, 3 23 + 22 (b) R - 1 = E 23 1 2 2 0 0 coordinates back to the original coordinates. 101. x6 - 4x5 - 3x4 + 18x3

0 0

U ; This is the rotation matrix needed to get the translated

1

102. v # w = - 5; u = 180°

103. 50, 3 6

2u2 - 1 u

104.

10.5 Assess Your Understanding (page 819) 7. Improper; 1 +

5. Proper

9 x - 4

5 -4 -5 25. + + x + 2 x + 1 1x + 12 2 35.

1 x2 + 4

2x - 1

+

1x2 + 42 2

37.

2 1 1 1 9 3 6 18 45. + 2 + + x x - 3 x + 3 x

x - 4

1 1 - 1x + 42 4 1 4 27. + 2 + x x x2 + 4

-1 2 -1 + + x x - 3 x + 1

-

58. a

22 22 , b; 2 2

3 4

15.

1 -x + 2 x x + 1

3 1 4 4 33. + x + 3 x - 1 8 4 x - 16x 7 7 41. 2 + 43. + 2x + 1 x - 3 1x + 162 2 1x2 + 162 3

51 51 1 1 5 5 5 5 47. x - 2 + 2 ; + ;x - 2 + x - 1 x + 4 x - 1 x + 3x - 4 x + 4 -

-4 4 + x x - 1

1 2 7 7 + 31. 3x - 2 2x + 1

10x - 11

- 11x - 32 - 11 - 10 + ; ; x2 - 4x + 7 x2 + 4x + 4 x + 2 1x + 22 2 1 1 2x3 + x2 - x + 1 1 4 ; + + + 53. x + 1 + x + 1 x - 1 x4 - 2x2 + 1 1x + 12 2 1x

13.

1 1 1 1 4 4 4 4 + + + 23. x - 1 x + 1 1x - 12 2 1x + 12 2

4 -3 -1 + + x - 2 x - 1 1x - 12 2

39.

- 2 1x - 62

1x + 42 1x - 32

1 2 1x + 12 3 3 + 2 29. x + 1 x + 2x + 4

51. x2 - 4x + 7 +

55. 3.85 years 56. - 2 57. 1

11. Improper; 1 +

2

1 1 - 1x + 42 12 12 21. + 2 x - 2 x + 2x + 4

1 3 1 4 4 2 19. + + x + 1 x - 1 1x - 12 2

2 -1 + 17. x - 1 x - 2

22x - 1

9. Improper; 5x +

2

49. x -

x x2 + 1

11 10 x + 2 1x + 22 2

1 3 1 1 4 4 ;x + 1 + + + + x + 1 x - 1 - 12 2 1x + 12 2 1x - 12 2

121, 5P 4 2

5P 4 o

Historical Problem (page 826) x = 6 units, y = 8 units

10.6 Assess Your Understanding (page 826) 5.

7.

y y  x2  1 yx1

y (4  兹2, 4  兹2 ) (4  兹2, 4  兹2 ) 10

5 y  兹36  x

(1, 2) (0, 1)

13. x 2  2x  y2  0 (2, 0)

9.

y 5

2

10

2.5 x

x 2  y2  4

2

(2, 1) 5 x (1, 2)

5 x

21. No points of intersection yx 9 y 10 2

x2  y2  4 10 x

y  3x  5

x y 5 2

23.

y  x2  4

y 20

y  6 x  13

(3, 5) 10 x

x  2y (8, 4) 9 x x  y2  2y

5 x y2x

17.

y 5

15.

y  兹x

y 5 (0, 0)

(1, 1)

x y8x

y 5

11.

19.

y (1, 兹3) 5 (0, 2) x 2  y2  4 5 x

(1, 兹3) 2 (0, 2) y  x  4

y 5 xy  4 x y 8 (2, 2) 2

2

5 x (2, 2)

AN89

AN90

ANSWERS  Section 10.6

25. x = 1, y = 4; x = - 1, y = - 4; x = 2 22 , y = 22; x = - 2 22 , y = - 22 or 11, 42, 1 - 1, - 42, 12 22, 222, 1 - 2 22, - 222 1 2 1 2 27. x = 0, y = 1; x = - , y = - or 10, 12, a - , - b 3 3 3 3

29. x = 0, y = - 1; x =

7 5 7 5 , y = - or 10, - 12, a , - b 2 2 2 2

1 1 4 1 1 4 ; x = , y = or a2, b , a , b 3 2 3 3 2 3 33. x = 3, y = 2; x = 3, y = - 2; x = - 3, y = 2; x = - 3, y = - 2 or 13, 22, 13, - 22, 1 - 3, 22, 1 - 3, - 22

31. x = 2, y =

35. x =

3 1 3 1 3 1 3 1 3 1 3 1 3 1 3 1 , y = ; x = , y = - ; x = - , y = ; x = - , y = - or a , b , a , - b , a - , b , a - , - b 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

37. x = 22 , y = 2 22; x = - 22 , y = - 2 22 or 1 22, 2 222, 1 - 22, - 2 222 41. x =

39. No real solution exists.

2 210 8 2 210 8 2 210 8 2 210 8 2 210 8 2 210 8 2 210 8 8 2 210 ,y = ;x = - ,y = ;x = ,y = ;x = - ,y = or a , b, a - , b, a , b , a - ,b 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

43. x = 1, y =

1 1 1 1 1 1 1 1 ; x = - 1, y = ; x = 1, y = - ; x = - 1, y = - or a1, b , a - 1, b , a1, - b , a - 1, - b 2 2 2 2 2 2 2 2

45. No real solution exists.

47. x = 23 , y = 23; x = - 23 , y = - 23; x = 2, y = 1; x = - 2, y = - 1 or 1 23, 232 , 1 - 23,- 232, 12, 12, 1 - 2, - 12

49. x = 0, y = - 2; x = 0, y = 1; x = 2, y = - 1 or 10, - 22, 10, 12, 12, - 12 55.

51. x = 2, y = 8 or 12, 82

53. x = 81, y = 3 or 181, 32

57. x = 0.48, y = 0.62 59. x = - 1.65, y = - 0.89 61. x = 0.58, y = 1.86; x = 1.81, y = 1.05; x = 0.58, y = - 1.86; x = 1.81, y = - 1.05 63. x = 2.35, y = 0.85

y x 2  x  y 2  3y  2  0 3 (1, 1) 2 x x1

65.



y2  y x 0

67.

y 5 (x  1)2  (y  1)2  5

6 3 , 5 5

(0, 兹 3  2) 5 x x  2y  0 (2, 1)

69.

y 5 (1, 0)

y 5

y

(5, 2) 5 x (x  1)2  (y  2)2  4

y 2  4y  x  1  0 (0, 兹 3  2)

4 x3

8 x

(1, 4)

x 2  6x  y 2  1  0 (1, 2)

1 1 1 and 77. 5 79. 5 in. by 3 in. 81. 2 cm and 4 cm 83. Tortoise: 7 m/hr, hare: 7 m/hr 2 3 2 P + 2P 2 - 16A P - 2P 2 - 16A ;w = 91. y = 4x - 4 93. y = 2x + 1 85. 12 cm by 18 cm 87. x = 60 ft; y = 30 ft 89. l = 4 4 1 7 - b + 2b2 - 4ac - b - 2b2 - 4ac 95. y = - x + 97. y = 2x - 3 99. r1 = ; r2 = 101. (a) 4.274 ft by 4.274 ft or 0.093 ft by 0.093 ft 3 3 2a 2a 71. 3 and 1; - 1 and - 3

102. e

73. 2 and 2; - 2 and - 2

- 3 - 265 - 3 + 265 , f 7 7

75.

2 103. y = - x - 3 5

104. sin u = -

7 24 7 25 25 ; cos u = - ; tan u = ; csc u = - ; sec u = 25 25 24 7 24

105. ≈15.8°

10.7 Assess Your Understanding (page 835) 7. dashes; solid 8. half-planes y 11.

9. False 13.

10. unbounded 15.

y 5

5 xⱖ0 5 x

8

2x  y ⱖ 6

xⱖ4

17.

y

y x y ⬎1 2

2

5

5 x

5 x 8 x

19.

21.

y 5

23.

y 5

xⴙyⴝ2

xy ⱖ 4 5 x

25.

y

y 5

5 2x ⴙ y ⴝ 4

5 x

3x  2y  6

2x  y  4 5 x

5 x

y ⱕ x2  1

27.

29.

y 5

y 5

31.

33. No solution

y

2x  4y  0

5

y 5

2x ⴙ y ⴝ ⴚ2

2x  3y  0

5 x 3x  2y  6

6 x x  2y  6

5 x 2x ⴙ y ⴝ 2

2x ⴙ 3y ⴝ 6 5 x 2x ⴙ 3y ⴝ 0

AN91

ANSWERS  Section 10.8 35.

37.

y 5

x2 ⴙ y2 ⴝ 9

yⴝx ⴚ4 2

y 5

39.

yⴝxⴚ2

5 x

y 5

5 x

45. Unbounded

y

xⴙyⴝ2 (0, 4)

(0, 0) (3, 0)

x x ⴙ 2y ⴝ 6

x x + y 53. d x y

(1, 0)

x + y x 57. (a) d y y

… Ú … Ú

(b)

(10, 0) 16 x

… … Ú Ú

y

3x ⴙ y ⴝ 12

8

(2, 6) xⴙyⴝ8

(0, 2)

x (4, 0) 2x ⴙ 3y ⴝ 12

9 x (2, 0)

… Ú … … Ú

x y 55. e x + y x - y x

4 6 0 0

2x ⴙ y ⴝ 10

(0, 8) 5 xⴙyⴝ2

24 12 , 7 7

(2, 0)

2x ⴙ y ⴝ 4

x ⴙ 2y ⴝ 10 (0, 5)

49. Bounded y 8

(0, 2)

(2, 0)

y 16

1 2

(0, 4) xⴙyⴝ2 5 x

51. Bounded

0,

5

(2, 2) 8

xy ⴝ 4

47. Bounded

y 8

y ⴝ x2 ⴙ 1

5 x

x 2 ⴙ y 2 ⴝ 16

43. Bounded

(0, 3)

y 5

5 x

xⴙyⴝ3

2x ⴙ y ⴝ 6

41.

y ⴝ x2 ⴚ 4

(5, 0)

20 15 50 0 0

x ⴙ 2y ⴝ 1

50,000 35,000 10,000 0 y

x y 59. (a) d x + 2y 3x + 2y

Ú Ú … …

(b)

x ⴝ 35,000

3x + 2y 2x + 3y 61. (a) d x y

0 0 300 480

(b)

y

80 (0, 50)

40,000 (35,000, 10,000) ⴚ20,000 (35,000, 0)

(40,000, 10,000)

(90, 105) 400

(0, 150)

y ⴝ 10,000 x (50,000, 0) x ⴙ y ⴝ 50,000

U5 3P 4

P 2 U5

U5P

U5

1 6

1 3

U5

5P 4 U5

80

2x ⴙ 3y ⴝ 150 x

160, 0 3

64. f 1 - 12 = - 5; f 122 = 28 2p 4p , f 65. e 0, 3 3

y

U5

(36, 26)

(0, 0)

x ⴙ 2y ⴝ 300

(160, 0)

62. 5 - 1 - 2i, - 1 + 2i 6 1 2 1 ; 63. x2 + ay - b = 6 36 1 1 circle with radius and center a0, b ; 6 6

3x ⴙ 2y ⴝ 160

x

(0, 0)

160 150 0 0

y

400

3x ⴙ 2y ⴝ 480

… … Ú Ú

P 4

x U50

7P 4

3P 2

10.8 Assess Your Understanding (page 842) 1. objective function 2. True 3. Maximum value is 11; minimum value is 3. 5. Maximum value is 65; minimum value is 4. 7. Maximum value is 67; minimum value is 20. 9. The maximum value of z is 12, and it occurs at the point (6, 0). 11. The minimum value of z is 4, and it occurs at the point (2, 0). 13. The maximum value of z is 20, and it occurs at the point (0, 4). 15. The minimum value of z is 8, and it occurs at the point (0, 2). 17. The maximum value of z is 50, and it occurs at the point (10, 0). 19. Produce 8 downhill and 24 cross-country; $1760; $1920, which is the profit when producing 16 downhill and 16 cross-country. 21. Rent 15 rectangular tables and 16 round tables for a minimum cost of $1252. 23. (a) $10,000 in the junk bond and $10,000 in Treasury bills (b) $12,000 in the junk bond and $8000 in Treasury bills 25. Manufacture 10 racing skates and 15 figure skates. 27. 100 lb of ground beef should be mixed with 50 lb of pork. 29. Order 2 metal samples and 4 plastic samples; $34 31. (a) Configure with 10 first-class seats and 120 coach seats. (b) Configure with 15 first-class seats and 120 coach seats. 1 33. e - , 1 f 32 34.

y 3 P

2P x

Domain: 5x 0 x ≠ kp, k is an integer 6 ; range: 1 - q , q 2

35. 89.1 years

36. y = 3x + 7

AN92

ANSWERS  Review Exercises

Review Exercises (page 846) 1. x =

3 1 3 1 ,y = or a , - b 4 20 4 20

6. x = 2, y = 3 or 12, 32

2. x = 2, y =

1 1 or a2, b 2 2

3. x = 2, y = - 1 or 12, - 12

4. x = 1, y = 1 or 11, 12

5. Inconsistent

7. Inconsistent

8. x = - 1, y = 2, z = - 3 or 1 - 1, 2, - 32 7 39 9 69 7 39 9 69 9. x = z + ,y = z + , where z is any real number or b 1x, y, z2 ` x = z + ,y = z + , z is any real number r 4 4 8 8 4 4 8 8 11. e

10. Inconsistent

3x + 2y = 8 x + 4y = - 1 -

1 2 17. D 1 6

-1 2 3

T

18. F

5 7 1 7 3 7

9 7 1 7 4 7

12. W

3 7 2 - V 7 1 7

3x + 2y + 5z = - 1 2x + 7z = 3 - 4x + y - 3z = 2

19. Singular

0 13. £ 8 2

1 2§ 8

22. Inconsistent

45. 3x ⴙ 4y ⱕ 12

2, y = 3 or 12, 32 1 9 x + 10 10 x2 + 9

41. x = 0, y = 0; x = - 3, y = 3; x = 3, y = 3 or

47. Bounded y 5

y 8

xⴙyⴝ2 (0, 2)

5 x

5 x

ⴚ2x ⴙ y ⴝ 2

(0, 2)

xⴙyⴝ4 8 x

(0, 0) (3, 0)

49.

50.

y 5

x 2 ⴙ y 2 ⴝ 16

9 (0, 8)

5 x

2x ⴙ y ⴝ 8

(0, 1) x ⴙ 2y ⴝ 2

xⴙyⴝ2

(4, 0)

13 d -4

43. x = 1, y = - 1 or 11, - 12

y ⱕ x2

y

15 -7

4 2 4 2 22, y = - 22; x = - 22, y = 22 or 3 3 3 3

5

48. Bounded

16. c

- 1; x = - 1, y = 2 or 1 - 2, - 12 1 - 1, 22

46. Unbounded

y 5

x

-1 -3§ 0

2 12 4

+ 1, z = - 1, y is any real number 6 29. x = 2, y = - 1 or 12, - 12 30. x = 3 1 2 -3 3 4 10 35. + + 2 36. + - 4 x - 1 x x + 1 x

42. x = 22, y = - 22; x = - 22, y = 22; x =

4 2 4 2 1 22, - 222, 1 - 22, 222, a 22, - 22 b , a - 22, 22 b 3 3 3 3 y 5

7 15. £ 9 -8

1 2 1 13 13 13 13 2 ,y = or a , b 21. x = 9, y = ,z = or a9, , b 5 10 5 10 3 3 3 3 1 7 6 1 7 6 23. x = - , y = - , z = - or a - , - , - b 5 5 5 5 5 5

40. x = 2 22 , y = 22; x = - 2 22 , y = - 22 or 12 22, 222, 1 - 2 22, - 222

44.

- 12 0§ 24

20. x =

24. z = - 1, x = y + 1, where y is any real number or 5 1x, y, z2 0 x = y 25. x = 4, y = 2, z = 3, t = - 2 or 14, 2, 3, - 22 26. 5 27. 67 28. - 100 3 2 + 31. x = - 1, y = 2, z = - 3 or 1 - 1, 2, - 32 32. 15 33. - 8 34. x x 1 1 1 2 4 4 x - 4x + 37. 2 + 38. 2 + 39. x = - 2, y = x - 1 x + 1 x + 4 1x2 + 42 2 x + 1 10, 02, 1 - 3, 32, 13, 32

6 14. £ 18 0

y

2x ⴙ 3y ⴝ 6

xy ⴝ 24

3 3

x

x2 ⴙ y2 ⴝ 25

9 x (2, 0)

51. The maximum value is 32 when x = 0 and y = 8. 52. The minimum value is 3 when x = 1 and y = 0. 53. 9 54. A is any real number, A ≠ 9. 2 1 55. y = - x2 - x + 1 56. Mix 70 lb of $6.00 coffee and 30 lb of $9.00 coffee. 57. Buy 1 small, 5 medium, and 2 large. 3 3 y

x y 58. (a) d 4x + 3y 2x + 3y

Ú Ú … …

0 0 960 576

400 4x ⴙ3y ⴝ 960

(b)

(0, 192)

(192, 64) 400 x 2x ⴙ3y ⴝ 576 (240, 0)

59. Speedboat: 36.67 km/hr; Aguarico River: 3.33 km/hr 60. Bruce: 4 hr; Bryce: 2 hr; Marty: 8 hr 61. Produce 35 gasoline engines and 15 diesel engines; the factory is producing an excess of 15 gasoline engines and 0 diesel engines.

AN93

ANSWERS  Cumulative Review

Chapter Test (page 848) 1. x = 3, y = - 1 or 13, - 12 b 1x, y, z2 ` x = - z +

6. c

3. x = - z +

2. Inconsistent

18 17 ,y = z , where z is any real number or 7 7

18 17 ,y = z , z is any real number r 7 7

4. x =

3x + 2y + 4z = - 6 3x + 2y + 4z = - 6 6 1x + 0y + 8z = 2 or c x + 8z = 2 7. C 1 - 2x + 1y + 3z = - 11 - 2x + y + 3z = - 11 5

4 - 11 - 11 S 8. C - 3 12 6

- 19 4 5 S 9. C 1 - 22 -1

-5 -1 5

4 5. C - 2 1

1 1 , y = - 2, z = 0 or a , - 2, 0 b 3 3

10 - 11 26

1 0 0 3 - 25 S -5 10

26 16 2 S 10. c 3 3

17 d - 10

11. D -

2

-1

5 2

3 2

T

3 3 -4 1 1 1 1 12. C - 2 - 2 3 S 13. x = , y = 3 or a , 3 b 14. x = - y + 7, where y is any real number or b 1x, y2 ` x = - y + 7, y is any real number r 2 2 4 4 -4 -5 7 15. x = 1, y = - 2, z = 0 or 11,- 2, 02 16. Inconsistent 17. - 29 18. - 12 19. x = - 2, y = - 5 or 1 - 2, - 52 20. x = 1, y = - 1, z = 4 or 11, - 1, 42 21. 11, - 32 and (1, 3) 22. (3, 4) and (1, 2) 23.

y 15

24. x 2 ⴙ y 2 ⴝ 100

3 -2 + x + 3 1x + 32 2

26. Unbounded y 9

2x ⴚ 3y ⴝ 2

1 1 x 5x 3 3 + + 25. 2 x 1x + 32 1x2 + 32 2

15 x

4x ⴚ 3y ⴝ 0

(4, 2) 9 (8, 0)

x x ⴙ 2y ⴝ 8

27. The maximum value of z is 64, and it occurs at the point (0, 8). 28. Flare jeans cost $24.50, camisoles cost $8.50, and T-shirts cost $6.00.

Cumulative Review (page 849) 1. e 0,

1 f 2

2. 55 6

1 3. e - 1, - , 3 f 2

4. 5 - 2 6

8. Center: 11, - 22; radius = 4

5 5. e f 2

1 f ln 3

7. Odd; symmetric with respect to the origin 10. f -1 1x2 =

9. Domain: all real numbers Range: 5y 0 y 7 1 6 Horizontal asymptote: y = 1

y 3

(2, 2)

yⴝ1 5 x

11. (a)

(b)

y 8

(c)

y

y

(0, 2) (0, 6)

(ⴚ2, 0)

1 1

(ⴚ2, 0) 2 x

(e)

y

(d)

2 (2, 0) x

5 - 2 x

Domain of f: 5x 0 x ≠ - 2 6 Range of f: 5y 0 y ≠ 0 6 Domain of f -1: 5x 0 x ≠ 0 6 Range of f -1: 5y 0 y ≠ - 2 6

y 5

x

3

6. e

2

(1, 1)

(0, 0)

2

(1, 1)

yⴝ0

x

2

(ⴚ1, ⴚ1)

y

2

(0, 0)

x

(1, 1) 2 x

(ⴚ1, ⴚ1)

(0, ⴚ2)

xⴝ0

(f)

y 2 yⴝ0

(g)

x

(1, 0)

4 x

xⴝ0

12. (a)

zero: - 2.28

10

ⴚ10

10

ⴚ10

y 兹2 ,0 ⴚ 2

2

(0, 1) 2

(h)

y

1

0,

y (ⴚ2, 1)

1 兹5 0, ⴚ 5

兹5 5

(i)

x

兹2 ,0 2

(ⴚ1, 0) (ⴚ2, ⴚ1)

1

yⴝ

兹3 x 3

(j)

y 3 (3, 1)

(2, 1) (1, 0) 2 x (2, ⴚ1) 兹3 yⴝⴚ x 3

(b) Local maximum of 7 at x = - 1; local minimum of 3 at x = 1 (c) 1 - q , - 12, 11, q 2

2,

(ⴚ1, 1) 0,

1 4

(1, 0)

1 4

7 x

AN94

ANSWERS  Section 11.1

CHAPTER 11 Sequences; Induction; the Binomial Theorem 11.1 Assess Your Understanding (page 858)

5. n(n - 1) # # # 3 # 2 # 1 6. recursive 7. summation 8. True 9. 3,628,800 11. 504 13. 1260 15. s1 = 1, s2 = 2, s3 = 3, 1 1 3 2 5 1 2 2 8 8 19. c1 = 1, c2 = - 4, c3 = 9, c4 = - 16, c5 = 25 21. s1 = , s2 = , s3 = , s4 = , s5 = s4 = 4, s5 = 5 17. a1 = , a2 = , a3 = , a4 = , a5 = 3 2 5 3 7 2 5 7 41 61 1 1 1 1 1 1 2 3 4 5 n 1 , t = - , t4 = ,t = 25. b1 = , b2 = 2 , b3 = 3 , b4 = 4 , b5 = 5 27. an = 29. an = n - 1 23. t 1 = - , t 2 = 6 12 3 20 30 5 42 e n + 1 e e e e 2 31. an = ( - 1)n + 1 33. an = ( - 1)n + 1n 35. a1 = 2, a2 = 5, a3 = 8, a4 = 11, a5 = 14 37. a1 = - 2, a2 = 0, a3 = 3, a4 = 7, a5 = 12 3 1 1 1 43. a1 = 1, a2 = 2, a3 = 2, a4 = 4, a5 = 8 39. a1 = 5, a2 = 10, a3 = 20, a4 = 40, a5 = 80 41. a1 = 3, a2 = , a3 = , a4 = , a5 = 2 2 8 40 45. a1 = A, a2 = A + d, a3 = A + 2d, a4 = A + 3d, a5 = A + 4d 3. sequence 4. True

47. a1 = 22 , a2 = 32 + 12 , a3 = 32 + 22 + 12 , a4 = 42 + 32 + 22 + 12 ,

49. 3 + 4 + g + (n + 2) 55.

a5 = 52 + 42 + 32 + 22 + 12

1 1 1 + + g + n 3 9 3

51.

9 n2 1 + 2 + + g + 2 2 2

53. 1 +

57. ln 2 - ln 3 + ln 4 - g + ( - 1)n ln n

1 1 1 + + g + n 3 9 3 20

59. a k k=1

13 6 n n n+1 k 1 3k 61. a 63. a ( - 1)k a k b 65. a 67. a (a + kd) or a [a + (k - 1)d] 69. 200 71. 820 73. 1110 75. 1560 77. 3570 3 k=1 k + 1 k=0 k=1 k k=0 k=1 79. 44,000 81. $2930 83. $18,058.03 85. 21 pairs 87. Fibonacci sequence 89. (a) 3.630170833 (b) 3.669060828 (c) 3.669296668 (d) 12 91. (a) a1 = 0.4; a2 = 0.7; a3 = 1; a4 = 1.6; a5 = 2.8; a6 = 5.2; a7 = 10; a8 = 19.6 (b) Except for term 5, which has no match, Bode’s formula provides excellent approximations for the mean distances of the planets from the Sun. (c) The mean distance of Ceres from the Sun is approximated by a5 = 2.8, and that of Uranus is a8 = 19.6. (d) a9 = 38.8; a10 = 77.2 (e) Pluto’s distance is approximated by a9, but no term approximates Neptune’s mean distance from the Sun. (f) According to Bode’s Law, the mean orbital distance of Eris from the Sun will be 154 AU. 93. a0 = 2; a5 = 2.236067977; 2.236067977 95. a0 = 4; a5 = 4.582575695; 4.582575695

98. $2654.39

99. 22(cos 225° + i sin 225°)

100. - 5i + 5j + 5k

101. (y - 4)2 = 16(x + 3)

11.2 Assess Your Understanding (page 866) 1. arithmetic 2. False 3. 17 4. True 5. sn - sn - 1 = (n + 4) - [(n - 1) + 4] = n + 4 - (n + 3) = n + 4 - n - 3 = 1, a constant; d = 1; s1 = 5, s2 = 6, s3 = 7, s4 = 8 7. an - an - 1 = (2n - 5) - [2(n - 1) - 5] = 2n - 5 - (2n - 2 - 5) = 2n - 5 - (2n - 7) = 2n - 5 - 2n + 7 = 2, a constant; d = 2; a1 = - 3, a2 = - 1, a3 = 1, a4 = 3 9. cn - cn - 1 = (6 - 2n) - [6 - 2(n - 1)] = 6 - 2n - (6 - 2n + 2) = 6 - 2n - (8 - 2n) = 6 - 2n - 8 + 2n = - 2, a constant; d = - 2; c1 = 4, c2 = 2, c3 = 0, c4 = - 2 1 1 1 1 1 5 1 5 1 1 1 1 1 1 1 1 1 1 - n - a - n + b = - n - a - nb = - n + n = - , a constant; 11. t n - t n - 1 = a - n b - c - (n - 1) d = 2 3 2 3 2 3 2 3 3 2 3 6 3 2 3 6 3 3 1 1 1 1 5 d = - ; t1 = , t2 = - , t3 = - , t4 = 3 6 6 2 6 13. sn - sn - 1 = ln 3n - ln 3n - 1 = n ln 3 - (n - 1) ln 3 = n ln 3 - (n ln 3 - ln 3) = n ln 3 - n ln 3 + ln 3 = ln 3, a constant; d = ln 3; s1 = ln 3, s2 = 2 ln 3, s3 = 3 ln 3, s4 = 4 ln 3 1 83 15. an = 3n - 1; a51 = 152 17. an = 8 - 3n; a51 = - 145 19. an = (n - 1); a51 = 25 21. an = 22n; a51 = 51 22 23. 200 25. - 266 27. 2 2 29. a1 = - 13; d = 3; an = an - 1 + 3; an = - 16 + 3n 31. a1 = - 53; d = 6; an = an - 1 + 6; an = - 59 + 6n n 33. a1 = 28; d = - 2; an = an - 1 - 2; an = 30 - 2n 35. a1 = 25; d = - 2; an = an - 1 - 2; an = 27 - 2n 37. n2 39. (9 + 5n) 41. 1260 2 3 43. 324 45. 30,919 47. 10,036 49. 6080 51. - 1925 53. 15,960 55. 57. 24 terms 59. 1185 seats 2 61. 210 beige and 190 blue 63. {Tn} = { - 5.5n + 67}; T5 = 39.5°F 65. The amphitheater has 1647 seats. 67. 8 yr 70. 16.42% 71. v = 4i - 6j 72. Center: (0, 0); Vertices: (0, - 5), (0, 5); Foci: 10, - 2212, 10, 2212 y 6 (0, 5) (0, Ï21) 2 (22, 0) 26 24

(0, 2Ï21)

22

(2, 0) 4 6 x

26 (0, 25)

1

73. c 23 2

0 d -1

ANSWERS  Section 11.4

AN95

Historical Problems (page 875) 1. 1

2 5 1 1 loaves, 10 loaves, 20 loaves, 29 loaves, 38 loaves 3 6 6 3

2. (a) 1 person (b) 2401 kittens

(c) 2800

11.3 Assess Your Understanding (page 876) a 1 - r 1 3 11. r = ; a1 = - , a2 2 2 3 1 17. r = ; t 1 = , t 2 = 2 2 3. geometric 4.

5. divergent series

6. True 7. False

8. True 9. r = 3; s1 = 3, s2 = 9, s3 = 27, s4 = 81

3 3 3 1 1 = - , a3 = - , a4 = 13. r = 2; c1 = , c2 = , c3 = 1, c4 = 2 15. r = 21/3; e1 = 21/3 , e2 = 22/3 , e3 = 2, e4 = 24/3 4 8 16 4 2 3 9 27 n-1 # , t = , t4 = 19. a5 = 162; an = 2 3 21. a5 = 5; an = 5 # ( - 1)n - 1 23. a5 = 0; an = 0 4 3 8 16 1 1 n-1 1 n-2 29. a9 = 1 31. a8 = 0.00000004 33. an = 7 # 2n - 1 35. an = - 3 # a - b 25. a5 = 4 22; an = ( 22)n 27. a7 = = a- b 64 3 3 n 7 1 2 n-1 n-1 n 2 n n 39. an = (15) = 7 # 15 41. - (1 - 2 ) 43. 2 c 1 - a b d 45. 1 - 2 37. an = - ( - 3) 15 4 3 51. 47. 49.

53. Converges;

3 2

55. Converges; 16

69. Arithmetic; d = 1; 1375

57. Converges;

71. Neither

8 5

59. Diverges 61. Converges;

2 73. Arithmetic; d = - ; - 700 3

1 79. Geometric; r = - 2; - 3 1 - 1 - 22 50 4 3 23 1/2 81. Geometric; r = 3 ; 11 + 232 11 - 325 2 2

83. - 4

75. Neither

20 3

63. Diverges

65. Converges;

77. Geometric; r =

18 5

67. Converges; 6

50

2 2 ; 2c 1 - a b d 3 3

85. $39,392.81

87. (a) 0.775 ft (b) 8th (c) 15.88 ft (d) 20 ft 89. $349,496.41 91. $96,885.98 93. $305.10 95. 1.845 * 1019 97. 10 99. $72.67 per share 101. December 20, 2014; $9999.92 103. Option A results in a higher salary in 5 years ($25,250 versus $24,761); option B results in a higher 5-year total ($116,801 versus $112,742). 105. Option 2 results in the most: $16,038,304; option 1 results in the least: $14,700,000. 107. Yes. A constant sequence is both arithmetic and geometric. For example, 3, 3, 3, . . . is an arithmetic sequence with a1 = 3 and d = 0 and is a y2 8 15 x2 = 1 114. 54 geometric sequence with a = 3 and r = 1. 111. 2.121 112. i j 113. 17 17 4 12

11.4 Assess Your Understanding (page 882) 1. (I) n = 1: 2(1) = 2 and 1(1 + 1) = 2 (II) If 2 + 4 + 6 + g + 2k = k(k + 1), then 2 + 4 + 6 + g + 2k + 2(k + 1) = (2 + 4 + 6 + g + 2k) + 2(k + 1) = k(k + 1) + 2(k + 1) = k 2 + 3k + 2 = (k + 1)(k + 2) = (k + 1)[(k + 1) + 1]. 1 1 3. (I) n = 1: 1 + 2 = 3 and (1)(1 + 5) = (6) = 3 2 2 1 (II) If 3 + 4 + 5 + g + (k + 2) = k(k + 5), then 3 + 4 + 5 + g + (k + 2) + [(k + 1) + 2] 2 1 1 1 1 = [3 + 4 + 5 + g + (k + 2)] + (k + 3) = k(k + 5) + k + 3 = (k 2 + 7k + 6) = (k + 1)(k + 6) = (k + 1)[(k + 1) + 5]. 2 2 2 2 1 1 5. (I) n = 1: 3(1) - 1 = 2 and (1)[3(1) + 1] = (4) = 2 2 2 1 (II) If 2 + 5 + 8 + g + (3k - 1) = k(3k + 1), then 2 + 5 + 8 + g + (3k - 1) + [3(k + 1) - 1] 2 1 1 1 = [2 + 5 + 8 + g + (3k - 1)] + (3k + 2) = k(3k + 1) + (3k + 2) = (3k 2 + 7k + 4) = (k + 1)(3k + 4) 2 2 2 1 = (k + 1)[3(k + 1) + 1]. 2 7. (I) n = 1: 21 - 1 = 1 and 21 - 1 = 1 (II) If 1 + 2 + 22 + g + 2k - 1 = 2k - 1, then 1 + 2 + 22 + g + 2k - 1 + 2(k + 1) - 1 = (1 + 2 + 22 + g + 2k - 1) + 2k = 2k - 1 + 2k = 2(2k) - 1 = 2k + 1 - 1. 1 1 9. (I) n = 1: 41 - 1 = 1 and (41 - 1) = (3) = 1 3 3 1 k 2 k-1 = (4 - 1), then 1 + 4 + 42 + g + 4k - 1 + 4(k + 1) - 1 = (1 + 4 + 42 + g + 4k - 1) + 4k (II) If 1 + 4 + 4 + g + 4 3 1 k 1 k 1 1 k = (4 - 1) + 4 = [4 - 1 + 3(4k)] = [4(4k) - 1] = (4k + 1 - 1). 3 3 3 3

AN96

ANSWERS  Section 11.4

1 1 1 1 = and = 1#2 2 1 + 1 2 1 1 1 k 1 1 1 1 1 1 = , then # + # + # + g + + (II) If # + # + # + g + 1 2 2 3 3 4 k(k + 1) k + 1 1 2 2 3 3 4 k(k + 1) (k + 1)[(k + 1) + 1] k(k + 2) + 1 1 1 1 1 k 1 1 = c # + # + # + g + d + = + = 1 2 2 3 3 4 k(k + 1) (k + 1)(k + 2) k + 1 (k + 1)(k + 2) (k + 1)(k + 2) (k + 1)2 k + 1 k + 1 k 2 + 2k + 1 = = = . = (k + 1)(k + 2) (k + 1)(k + 2) k + 2 (k + 1) + 1

11. (I) n = 1:

1# # # 1 2 3 = 1 6 1 2 2 2 (II) If 1 + 2 + 3 + g + k 2 = k(k + 1)(2k + 1), then 12 + 22 + 32 + g + k 2 + (k + 1)2 6 1 1 = (12 + 22 + 32 + g + k 2) + (k + 1)2 = k(k + 1)(2k + 1) + (k + 1)2 = (2k 3 + 9k 2 + 13k + 6) 6 6 1 1 = (k + 1)(k + 2)(2k + 3) = (k + 1)[(k + 1) + 1][2(k + 1) + 1]. 6 6

13. (I) n = 1: 12 = 1 and

1 1 15. (I) n = 1: 5 - 1 = 4 and (1)(9 - 1) = # 8 = 4 2 2 1 (II) If 4 + 3 + 2 + g + (5 - k) = k(9 - k), then 4 + 3 + 2 + g + (5 - k) + [5 - (k + 1)] 2 1 1 1 = [4 + 3 + 2 + g + (5 - k)] + 4 - k = k(9 - k) + 4 - k = (9k - k 2 + 8 - 2k) = ( - k 2 + 7k + 8) 2 2 2 1 1 = (k + 1)(8 - k) = (k + 1)[9 - (k + 1)]. 2 2 1 17. (I) n = 1: 1 # (1 + 1) = 2 and # 1 # 2 # 3 = 2 3 1 (II) If 1 # 2 + 2 # 3 + 3 # 4 + g + k(k + 1) = k(k + 1)(k + 2), then 1 # 2 + 2 # 3 + 3 # 4 + g + k(k + 1) 3 + (k + 1)[(k + 1) + 1] = [1 # 2 + 2 # 3 + 3 # 4 + g + k(k + 1)] + (k + 1)(k + 2) 1 1 1 1 = k(k + 1)(k + 2) + # 3(k + 1)(k + 2) = (k + 1)(k + 2)(k + 3) = (k + 1)[(k + 1) + 1][(k + 1) + 2]. 3 3 3 3 19. (I) n = 1: 12 + 1 = 2, which is divisible by 2. (II) If k 2 + k is divisible by 2, then (k + 1)2 + (k + 1) = k 2 + 2k + 1 + k + 1 = (k 2 + k) + 2k + 2. Since k 2 + k is divisible by 2, and 2k + 2 is divisible by 2, then (k + 1)2 + (k + 1) is divisible by 2. 21. (I) n = 1: 12 - 1 + 2 = 2, which is divisible by 2. (II) If k 2 - k + 2 is divisible by 2, then (k + 1)2 - (k + 1) + 2 = k 2 + 2k + 1 - k - 1 + 2 = (k 2 - k + 2) + 2k. Since k 2 - k + 2 is divisible by 2, and 2k is divisible by 2, then (k + 1)2 - (k + 1) + 2 is divisible by 2. 23. (I) n = 1: If x 7 1, then x1 = x 7 1. (II) Assume, for an arbitrary natural number k, that if x 7 1 then xk 7 1. Multiply both sides of the inequality xk 7 1 by x. If x 7 1, then xk + 1 7 x 7 1. 25. (I) n = 1: a - b is a factor of a1 - b1 = a - b. (II) If a - b is a factor of ak - bk, then ak + 1 - bk + 1 = a(ak - bk) + bk(a - b). Since a - b is a factor of ak - bk and a - b is a factor of a - b, then a - b is a factor of ak + 1 - bk + 1.

27. (I) n = 1: (1 + a)1 = 1 + a Ú 1 + 1 # a (II) Assume that there is an integer k for which the inequality holds. So (1 + a)k Ú 1 + ka. We need to show that (1 + a)k + 1 Ú 1 + (k + 1)a. (1 + a)k + 1 = (1 + a)k (1 + a) Ú (1 + ka)(1 + a) = 1 + ka2 + a + ka = 1 + (k + 1)a + ka2 Ú 1 + (k + 1)a.

6 + g + 2k = k 2 + k + 2, then 2 + 4 + 6 + g + 2k + 2(k + 1) + 4 + 6 + g + 2k) + 2k + 2 = k 2 + k + 2 + 2k + 2 = k 2 + 3k + 4 = (k 2 + 2k + 1) + (k + 1) + 2 + 1)2 + (k + 1) + 2. 2 and 12 + 1 + 2 = 4. The fact is that 2 + 4 + 6 + g + 2n = n2 + n, not n2 + n + 2 (Problem 1). 1 # (1 - 1) 31. (I) n = 1: [a + (1 - 1)d] = a and 1 # a + d = a. 2 k(k - 1) (II) If a + (a + d) + (a + 2d) + g + [a + (k - 1)d] = ka + d , then 2 k(k - 1) a + (a + d) + (a + 2d) + g + [a + (k - 1)d] + [a + ((k + 1) - 1)d] = ka + d + a + kd 2 k(k - 1) + 2k (k + 1)(k) (k + 1)[(k + 1) - 1] = (k + 1)a + d = (k + 1)a + d = (k + 1)a + d . 2 2 2 35. {251} 36. Left: 448.3 kg; right: 366.0 kg 33. (I) n = 3: The sum of the angles of a triangle is (3 - 2) # 180° = 180°. 1 1 (II) Assume for some k Ú 3 that the sum of the angles of a 37. x = , y = - 3; a , - 3 b 2 2 convex polygon of k sides is (k - 2) # 180°. A convex polygon k sides 7 -3 of k + 1 sides consists of a convex polygon of k sides plus 38. c d -7 8 a triangle (see the illustration). The sum of the angles is k ⴙ 1 sides 29. If 2 + 4 + = (2 = (k But 2 # 1 =

(k - 2) # 180° + 180° = (k - 1) # 180° = [(k + 1) - 2] # 180°.

ANSWERS  Chapter Test

AN97

11.5 Assess Your Understanding (page 888) 1. Pascal triangle 2. 1; n 3. False 4. Binomial Theorem 5. 10 7. 21 9. 50 11. 1 13. ≈1.8664 * 1015 15. ≈1.4834 * 1013 17. x5 + 5x4 + 10x3 + 10x2 + 5x + 1 19. x6 - 12x5 + 60x4 - 160x3 + 240x2 - 192x + 64 21. 81x4 + 108x3 + 54x2 + 12x + 1 23. x10 + 5x8y2 + 10x6y4 + 10x4y6 + 5x2y8 + y10 25. x3 + 6 22x5/2 + 30x2 + 40 22x3/2 + 60x + 24 22x1/2 + 8 27. a5x5 + 5a4bx4y + 10a3b2x3y2 + 10a2b3x2y3 + 5ab4xy4 + b5y5 29. 17,010 31. - 101,376 33. 41,472 35. 2835x3 37. 314,928x7 39. 495 41. 3360 43. 1.00501 n # (n - 1)! n n! n! n n! n! n! 45. a b = = = = n; a b = = = = 1 n - 1 (n - 1)! [n - (n - 1)]! (n - 1)! 1! (n - 1)! n n! (n - n)! n!0! n! n n n n n n 47. 2n = (1 + 1)n = a b 1n + a b (1)n - 1(1) + g + a b 1n = a b + a b + g + a b 0 1 n 0 1 n 5 51. e f ≈ 58.827 6 52. (a) 0 (b) 90° (c) Orthogonal 2n 6 - 2n 5 53. x = 1, y = 3, z = - 2; (1, 3, - 2) y 54. Bounded

49. 1

(0, 6) (4, 2) (0, 0) (5, 0)

x

Review Exercises (page 891) 4 5 6 7 8 1. a1 = - , a2 = , a3 = - , a4 = , a5 = 3 4 5 6 7

4 8 16 , a = , a5 = 3 4 9 27 13 n 1 4. a1 = 8, a2 = 19, a3 = 41, a4 = 85, a5 = 173 5. 6 + 10 + 14 + 18 = 48 6. a 1 - 12 k + 1 7. Arithmetic; d = 1; Sn = 1n + 112 8. Neither k 2 k=1 8 n 1 1 n 9. Geometric; r = 8; Sn = 18 - 12 10. Arithmetic; d = 4; Sn = 2n 1n - 12 11. Geometric; r = ; Sn = 6 c 1 - a b d 12. Neither 13. 3825 7 2 2 1 1 1 9 14. 9515 15. - 1320 16. a1 - 10 b ≈ 0.49999 17. 35 18. 12 19. 9 22 20. 5an 6 = 53n + 1 6 21. 5an 6 = 5n - 10 6 22. Converges; 2 2 5 3 4 23. Converges; 24. Diverges 25. Converges; 8 3 3#1 26. (I) n = 1: 3 # 1 = 3 and (1 + 12 = 3 2 3k 1k + 12, then 3 + 6 + 9 + g + 3k + 3 1k + 12 = 13 + 6 + 9 + g + 3k2 + 13k + 32 (II) If 3 + 6 + 9 + g + 3k = 2 3 1k + 12 3k 3k 2 3 3k 6k 6 3 = 1k + 12 + 13k + 32 = + + + = 1k 2 + 3k + 22 = 1k + 12 1k + 22 = 3 1k + 12 + 1 4 . 2 2 2 2 2 2 2 2 2. c1 = 2, c2 = 1, c3 =

8 32 , c = 1, c5 = 9 4 25

3. a1 = 3, a2 = 2, a3 =

27. (I) n = 1: 2 # 31 - 1 = 2 and 31 - 1 = 2 (II) If 2 + 6 + 18 + g + 2 # 3k - 1 = 3k - 1, then 2 + 6 + 18 + g + 2 # 3k - 1 + 2 # 31k + 12 - 1 = 12 + 6 + 18 + g + 2 # 3k - 1 2 + 2 # 3k = 3k - 1 + 2 # 3k = 3 # 3k - 1 = 3k + 1 - 1. 1 28. (I) n = 1: 13 # 1 - 22 2 = 1 and # 1 # 3 6 112 2 - 3 112 - 1 4 = 1 2 1 2 2 2 (II) If 1 + 4 + 7 + g + 13k - 22 2 = k 16k 2 - 3k - 12, then 12 + 42 + 72 + g + 13k - 22 2 + 3 3 1k + 12 - 2 4 2 2 1 1 = 312 + 42 + 72 + g + 13k - 22 2 4 + 13k + 12 2 = k 16k 2 - 3k - 12 + 13k + 12 2 = 16k 3 - 3k 2 - k2 + 19k 2 + 6k + 12 2 2 1 1 1 3 2 2 2 = 16k + 15k + 11k + 22 = 1k + 12 16k + 9k + 22 = 1k + 12 3 6 1k + 12 - 3 1k + 12 - 1 4 . 2 2 2

29. 10 30. 32x5 + 240x4 + 720x3 + 1080x2 + 810x + 243 31. 81x4 - 432x3 + 864x2 - 768x + 256 32. 144 33. 6048 34. (a) 8 bricks (b) 1100 bricks 3 3 135 3 n 35. 360 tiles 36. (a) 20 a b = ft (b) 20 a b ft (c) 13 times (d) 140 ft 37. $151,873.77 4 16 4 38. $23,397.17

Chapter Test (page 892) 1. 0,

3 8 5 24 , , , 10 11 4 13

2. 4, 14, 44, 134, 404

7. Geometric; r = 4; Sn =

2 (1 - 4n) 3

3. 2 -

3 4 61 + = 4 9 36

4. -

10 k + 1 5. a ( - 1)k a b 6. Neither k + 4 k=1 1 n 9. Arithmetic; d = - ; Sn = (27 - n) 2 4

1 14 73 308 680 = 3 9 27 81 81

8. Arithmetic: d = - 8; Sn = n(2 - 4n)

AN98

ANSWERS  Chapter Test

10. Geometric; r =

2 125 2 n ;S = c1 - a b d 5 n 3 5

11. Neither

14. First show that the statement holds for n = 1. a1 +

12. Converges;

1024 5

13. 243m5 + 810m4 + 1080m3 + 720m2 + 240m + 32

1 b = 1 + 1 = 2. The equality is true for n = 1, so Condition I holds. Next, assume that 1

a1 +

1 1 1 1 b a1 + b a1 + b g a1 + b = n + 1 is true for some k, and determine whether the formula then holds for k + 1. Assume that 1 2 3 n 1 1 1 1 1 1 1 1 1 a1 + b a1 + b a1 + b g a1 + b = k + 1. Now show that a1 + b a1 + b a1 + b g a1 + b a1 + b 1 2 3 k 1 2 3 k k + 1 = (k + 1) + 1 = k + 2. Do this as follows: a1 +

1 1 1 1 1 1 1 1 1 1 b a1 + b a1 + b g a1 + b a1 + b = c a1 + b a1 + b a1 + b g a1 + b d a1 + b 1 2 3 k k + 1 1 2 3 k k + 1 1 1 b (induction assumption) = (k + 1) # 1 + (k + 1) # = k + 1 + 1 = k + 2 = (k + 1) a1 + k + 1 k + 1 Condition II also holds. Thus, the formula holds true for all natural numbers. 15. After 10 years, the car will be worth $6103.11. 16. The weightlifter will have lifted a total of 8000 pounds after 5 sets.

Cumulative Review (page 892) 1. { - 3, 3, - 3i, 3i}

2. (a)

(b) e a

y

B

- 1 + 13601 - 1 + 23601 - 1 + 13601 - 1 + 23601 , b, a , bf 18 6 B 18 6

(c) The circle and the parabola intersect at ⴚ4

6

x

a

B

- 1 + 13601 - 1 + 23601 - 1 + 13601 - 1 + 23601 b, a b. , , 18 6 B 18 6

5 6x + 3 1 (d) e x ` x ≠ f 3. e ln a b f 4. y = 5x - 10 5. 1x + 12 2 + 1y - 22 2 = 25 6. (a) 5 (b) 13 (c) 2 2x - 1 2 y2 1 2x x2 -1 -1 2 (g) g (x) = (x - 1); all reals (h) f (x) = ; {x x ≠ 3} 7. + = 1 8. (x + 1) = 4(y - 2) 2 x - 3 7 16

(e)

7x - 2 x - 2

(f) {x x ≠ 2}

CHAPTER 12 Counting and Probability 12.1 Assess Your Understanding (page 899) 5. subset; ⊆ 6. finite 7. n 1A2 + n 1B2 - n 1AxB2 8. True 9. ∅, {a}, {b}, {c}, {d}, {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {c, d}, {a, b, c}, {b, c, d}, {a, c, d}, {a, b, d}, {a, b, c, d} 11. 25 13. 40 15. 25 17. 37 19. 18 21. 5 23. 15 different arrangements 25. 9000 numbers 27. 175; 125 29. (a) 15 (b) 15 (c) 15 (d) 25 (e) 40 31. (a) 13.6 million (b) 78.9 million 33. 480 portfolios y 37. 2, 5, - 2 38. 163.9 ft 39. - 14i + 11j + 13k 36. 6

(x22)2 1 (y11)2 2 9

26

6 x

26

12.2 Assess Your Understanding (page 906) n! n! 6. 7. 30 9. 24 11. 1 13. 1680 15. 28 17. 35 19. 1 21. 10,400,600 1n - r2! 1n - r2!r! 23. 5abc, abd, abe, acb, acd, ace, adb, adc, ade, aeb, aec, aed, bac, bad, bae, bca, bcd, bce, bda, bdc, bde, bea, bec, bed, cab, cad, cae, cba, cbd, cbe, cda, cdb, cde, cea, ceb, ced, dab, dac, dae, dba, dbc, dbe, dca, dcb, dce, dea, deb, dec, eab, eac, ead, eba, ebc, ebd, eca, ecb, ecd, eda, edb, edc 6; 60 25. 5123, 124, 132, 134, 142, 143, 213, 214, 231, 234, 241, 243, 312, 314, 321, 324, 341, 342, 412, 413, 421, 423, 431, 432 6; 24 27. 5abc, abd, abe, acd, ace,ade, bcd, bce, bde, cde 6; 10 29. 5123, 124, 134, 234 6; 4 31. 16 33. 8 35. 24 37. 60 39. 18,278 41. 35 43. 1024 45. 120 47. 132,860 49. 336 51. 90,720 53. (a) 63 (b) 35 (c) 1 55. 1.157 * 1076 57. 362,880 59. 660 61. 15 63. (a) 125,000; 117,600 (b) A better name for a combination lock would be a permutation lock, because the order of the numbers matters. 67. 10 sq. units 3. permutation

4. combination

68. (g ∘ f )(x) = 4x2 - 2x - 2

5.

69. sin 75° =

Historical Problem (page 916)

22 + 26 22 + 26 1 ; cos 15° = 22 + 13 or cos 15° = 4 2 4

70. a5 = 80

1. (a) 5AAAA, AAAB, AABA, AABB, ABAA, ABAB, ABBA, ABBB, BAAA, BAAB, BABA, BABB, BBAA, BBAB, BBBA, BBBB6; 11 5 P(A wins) = ; P(B wins) = 16 16 C14, 22 + C14, 32 + C14, 42 C14, 32 + C14, 42 6 + 4 + 1 11 4 + 1 5 (b) P1A wins2 = = = ; P1B wins2 = = = ; the results are the same. 16 16 16 16 24 24

ANSWERS  Cumulative Review

AN99

12.3 Assess Your Understanding (page 916) 1. equally likely 2. complement

3. F 4. T 5. 0, 0.01, 0.35, 1 7. Probability model 9. Not a probability model 1 1 1 1 11. S = 5HH, HT, TH, TT 6; P1HH2 = , P1HT2 = , P1TH2 = , P1TT2 = 4 4 4 4 13. S = 5HH1, HH2, HH3, HH4, HH5, HH6, HT1, HT2, HT3, HT4, HT5, HT6, TH1, TH2, TH3, TH4, TH5, TH6, TT1, TT2, TT3, TT4, TT5, TT6 6; 1 each outcome has the probability of . 24 1 15. S = 5HHH, HHT, HTH, HTT, THH, THT, TTH, TTT6; each outcome has the probability of . 8 17. S = 5 1 Yellow, 1 Red, 1 Green, 2 Yellow, 2 Red, 2 Green, 3 Yellow, 3 Red, 3 Green, 4 Yellow, 4 Red, 4 Green 6; each outcome has the probability 1 1 1 1 of ; thus, P12 Red2 + P14 Red2 = + = . 12 12 12 6 19. S = 5 1 Yellow Forward, 1 Yellow Backward, 1 Red Forward, 1 Red Backward, 1 Green Forward, 1 Green Backward, 2 Yellow Forward, 2 Yellow Backward, 2 Red Forward, 2 Red Backward, 2 Green Forward, 2 Green Backward, 3 Yellow Forward, 3 Yellow Backward, 3 Red Forward, 3 Red Backward, 3 Green Forward, 3 Green Backward, 4 Yellow Forward, 4 Yellow Backward, 4 Red Forward, 4 Red Backward, 4 Green Forward, 4 1 1 1 1 Green Backward 6 ; each outcome has the probability of ; thus, P11 Red Backward2 + P11 Green Backward2 = + = . 24 24 24 12 21. S = 511 Red, 11 Yellow, 11 Green, 12 Red, 12 Yellow, 12 Green, 13 Red, 13 Yellow, 13 Green, 14 Red, 14 Yellow, 14 Green, 21 Red, 21 Yellow, 21 Green, 22 Red, 22 Yellow, 22 Green, 23 Red, 23 Yellow, 23 Green, 24 Red, 24 Yellow, 24 Green, 31 Red, 31 Yellow, 31 Green, 32 Red, 32 Yellow, 32 Green, 33 Red, 33 Yellow, 33 Green, 34 Red, 34 Yellow, 34 Green, 41 Red, 41 Yellow, 41 Green, 42 Red, 42 Yellow, 42 Green, 43 Red, 43 Yellow, 43 Green, n 1E2 4 1 1 = = . 44 Red, 44 Yellow, 44 Green6; each outcome has the probability of ; thus, E = 522 Red, 22 Green, 24 Red, 24 Green 6; P1E2 = 48 n 1S2 48 12 4 1 1 2 3 1 1 1 23. A, B, C, F 25. B 27. P1H2 = ; P1T2 = 29. P112 = P132 = P152 = ; P122 = P142 = P162 = 31. 33. 35. 37. 5 5 9 9 10 2 6 8 1 1 1 17 11 1 3 2 41. 43. 45. 0.55 47. 0.70 49. 0.30 51. 0.88 53. 0.67 55. 0.936 57. 59. 61. 63. 65. 39. 4 6 18 20 20 2 10 5 25 25 67. (a) 0.57 (b) 0.95 (c) 0.83 (d) 0.38 (e) 0.29 (f) 0.05 (g) 0.78 (h) 0.71 69. (a) (b) 71. 0.167 73. 0.000033069 33 33 74. 2; left; 3; down 75. ( - 3, 3 23) 76. {22} 77. (2, - 3, - 1)

Review Exercises (page 920) 1. ∅, 5Dave 6, 5Joanne 6, 5Erica 6, 5Dave, Joanne 6, 5Dave, Erica 6, 5Joanne, Erica 6, 5Dave, Joanne, Erica 6 2. 17 3. 24 4. 29 5. 34 6. 7 7. 45 8. 25 9. 7 10. 336 11. 56 12. 48 13. 128 14. 3024 15. 1680 16. 1820 17. 1,600,000 18. 216,000 19. 256 (allowing numbers with initial zeros, such as 011) 20. 2522520 21. (a) 381,024 (b) 1260 22. (a) 8.634628387 * 1045 (b) 0.6531 (c) 0.3469 23. (a) 0.081 (b) 0.919 24. 0.13 25. 0.5; 0.24 26. (a) 0.68 (b) 0.58 (c) 0.32

Chapter Test (page 921) 1. 22 2. 3 3. 8 4. 45 5. 5040 6. 151,200 7. 462 8. There are 54,264 ways to choose 6 different colors from the 21 available colors. 9. There are 840 distinct arrangements of the letters in the word REDEEMED. 10. There are 56 different exacta bets for an 8-horse race. 11. There are 155,480,000 possible license plates using the new format. 12. (a) 0.95 (b) 0.30 13. (a) 0.25 (b) 0.55 1 1250 14. 0.19 15. P1win on $1 play2 = ≈ 0.0000000083 16. P1exactly 2 fours2 = ≈ 0.1608 120,526,770 7776

Cumulative Review (page 922) 1. e

1 1 22 22 i, + if 3 3 3 3

3.

y 10

2. (ⴚ5, 0)

(1, 0) 10 x (0, ⴚ5)

(ⴚ2, ⴚ9) x ⴝ ⴚ2

5. b -

1 27 1 27 1 + i, - i, - , 3 r 2 2 2 2 5

6.

y 10 (1, 6)

yⴝ5

5 x

Domain: all real numbers Range: 5y y 7 5 6 Horizontal asymptote: y = 5

4. 5x 3.99 … x … 4.01 6 or 33.99, 4.01 4

y 5

(ⴚ2, ⴚ2)

5 x (0, ⴚ2)

(ⴚ1, ⴚ4)

7. 2

8 8. e f 3

11.

y

9. x = 2, y = - 5, z = 3

3 P x

10. 125; 700

12. a ≈ 6.09, B ≈ 31.9°, C ≈ 108.1°; area ≈ 14.46 square units 1 13. ≈ 0.00000000571; 175,223,510 The new format has decreased the probability of winning the jackpot.

AN100

ANSWERS  Section 13.1

CHAPTER 13 A Preview of Calculus: The Limit, Derivative, and Integral of a Function 13.1 Assess Your Understanding (page 927) 3. lim f 1x2 xSc

23.

4. does not exist

5. True

6. False

25.

y 15

7. 32

9. 1

11. 4

13. 2

27.

y 5

15. 0

17. 3

19. 4 29.

y 8

21. Does not exist y 1.25

31. P

5 x

y 2.5

x

2.5 x

5 x 5 x

lim f 1x2 = 13

lim f 1x2 = - 3

xS4

33.

35.

y 5

37.

lim f 1x2 = - 1

lim f 1x2 = 0

xS0

41.

y 5

y 5

5 x

5 x

lim f 1x2 does not exist.

xS1

47. 0 49. d = 4 25; M = (4, - 7)

45. 1.6

39.

5 x

xS0

lim f 1x2 = 1

x S p/2

y 5

5 x

x S -1

lim f 1x2 = 1

x S -3

y 5

5 x

43. 0.67

lim f 1x2 = 6

xS2

lim f 1x2 = 0

lim f 1x2 = 0

xS0

xS0

50. Center: (2, - 1); foci: (2, - 3), (2, 1); vertices: (2, 213 - 1), (2, - 213 - 1)

2p 52. ¢ 4, ≤ 3

51. $7288.48

13.2 Assess Your Understanding (page 934) 3. c

1. product 2. b 33. 3

35. 0

37.

57.

39.

8 5

41. 0

43. 5

58. g - 1(x) =

y

12 3

2 3

4. True 5. False

2

f (x) 5 x 1x 11

6. False 45. 6

7. 5

47. 0

3 - x x - 2

11. - 10

9. 4

51. - 1

49. 0

59. 60° or

p red 3

13. 80

53. 1

15. 8

55.

17. 8

19. - 1

21. 8

23. 3

25. - 1

27. 32

29. 2

31.

7 6

3 4

60. x4 + 8x3 + 24x2 + 32x + 16

3x

23 212

13.3 Assess Your Understanding (page 940) 7. one-sided 8. lim+ f 1x2 = R xSc

9. continuous; c 10. False

13. 5x - 8 … x 6 - 6 or - 6 6 x 6 4 or 4 6 x … 6 6

11. True 12. True

2 15. - 8, - 5, - 3 17. f 1 - 82 = 0; f 1 - 42 = 2 19. q 21. 2 23. 1 25. Limit exists; 0 27. No 29. Yes 31. No 33. 5 35. 7 37. 1 39. 4 41. 3 3 43. 45. Continuous 47. Continuous 49. Not continuous 51. Not continuous 53. Not continuous 55. Continuous 57. Not continuous 2 59. Continuous 61. Continuous for all real numbers 63. Continuous for all real numbers 65. Continuous for all real numbers kp 67. Continuous for all real numbers except x = , where k is an odd integer 69. Continuous for all real numbers except x = - 2 and x = 2 2 71. Continuous for all positive real numbers except x = 1 73. Discontinuous at x = - 1 and x = 1; 1 1 lim R1x2 = : hole at a1, b xS1 2 2 lim -R1x2 = - q ; lim + R1x2 = q ;

75. Discontinuous at x = - 1 and x = 1; 1 1 lim R1x2 = : hole at a - 1, b x S -1 2 2 lim-R1x2 = - q ; lim+ R1x2 = q ;

y 5

5 x

y0

x S -1

x S -1

vertical asymptote at x = - 1

79. x = - 3: asymptote; x = 2: hole 85. 20

−20

5

20

−3

91. Vertical: x = 4; horizontal: y = 3

−20

92. 60

y1 5 x

xS1

vertical asymptote at x = 1

x  1

3 77. x = - 2 2 : asymptote; x = 1: hole 3 83.

−3

xS1

y 5

93. ln ¢

x 5y 2 z4

≤

3 94. C 1 0

x1

3 81. x = - 2 2 : asymptote; x = - 1: hole 87. 3

−3

1

−2

1 0 1

2 4 23 5S -3 -2

AN101

ANSWERS  Section 13.5

13.4 Assess Your Understanding (page 948) 3. tangent line

4. derivative

5. velocity

9. mtan = 3

6. True 7. True 8. True

11. mtan = - 2

13. mtan = 12

y 10 (1, 8)

f (x)  3x 2 f (x)  x 2  2

(1, 3)

(2, 12) f (x)  2x 2  x

17. mtan = - 4

y 18

10

y  13x  16 (1, 6)

(2, 10) f(x)  x 2  2x  3 5 x

f (x)  x 3  x

5 x

52. (2, 6), ( - 1, 3)

53. 10

5 x y  5x  2

5 x

19. mtan = 13

y

(1, 3)

y  12x  12

5 x

5 x

7 51. Vertex: (1, 3); focus: ¢ 1, ≤ 2

y 8

y 15

y  2x  1

f(x)  3x  5

y  4x  2

15. mtan = 5

y 10

21. - 4 23. 0 25. 7 27. 7 29. 3 31. 1 33. 60 35. - 0.8587776956 37. 1.389623659 39. 2.362110222 41. 3.643914112 43. 18p ft 3/ft 45. 16p ft 3/ft 47. (a) 6 sec (b) 64 ft/sec (c) 1 - 32t + 962 ft/sec (d) 32 ft/sec (e) 3 sec (f) 144 ft (g) - 96 ft/sec 1 49. (a) - 23 ft/sec (b) - 21 ft/sec (c) - 18 ft/sec 3 (d) s 1t2 = - 2.631t 2 - 10.269t + 999.933 (e) Approximately - 15.531 ft/sec

54. 23.66 sq. units

13.5 Assess Your Understanding (page 955) b

3.

La

9. (a)

f 1x2 dx

b

4.

La

f 1x2 dx

5. 3

7. 56 11. (a)

y 24

13. (a)

y 10

15. (a)

y 20

y 90

8 x

(b) 36 (d) 45

17. (a)

(c) 72 (e) 63

(f) 54

(b) 18 (c) 9 63 45 (d) (e) 4 4 19. (a)

y 1.5

5 x

5 x

4 x

51 (b) 22 (c) 2

27 (f) 2

4

(d) 21. (a)

y 20

L0

(b) 36

1x2 + 22 dx

(e)

88 3

y 1.2

(d)

L0

x3 dx

(e) 64

23. (a) Area under the graph of f 1x2 = 3x + 1 from 0 to 4 (b) y P x

5 x

(c) 49 4

16

4 x

(b)

25 12

(c)

(b) 11.475

4609 2520

(c) 15.197

(b) 1.896 p

3

(c) 1.974

5 x

x

(d) e dx (e) 19.718 (d) sin x dx (e) 2 (c) 28 L-1 L0 1 dx (e) 1.609 (d) x L1 27. (a) Area under the graph of f 1x2 = sin x 29. (a) Area under the graph of f 1x2 = e x from 25. (a) Area under the graph of f 1x2 = x2 - 1 p 0 to 2 from 0 to from 2 to 5 2 (b) y (b) y (b) y 8 5

30

1.5

P x

4 x

5 x

(c) 36

(c) 1

31. Using left endpoints: n = 2: 0 + 0.5 = 0.5; n = 4: 0 + 0.125 + 0.25 + 0.375 = 0.75; 10 10 + 0.182 = 0.9; 2 n = 100: 0 + 0.0002 + 0.0004 + 0.0006 + g + 0.0198 100 = 10 + 0.01982 = 0.99; 2 n = 10: 0 + 0.02 + 0.04 + 0.06 + g + 0.18 =

(c) 6.389 Using right endpoints: n = 2: 0.5 + 1 = 1.5; n = 4: 0.125 + 0.25 + 0.375 + 0.5 = 1.25; 10 10.02 + 0.202 = 1.1; n = 10: 0.02 + 0.04 + 0.06 + g + 0.20 = 2 n = 100: 0.0002 + 0.0004 + 0.0006 + g + 0.02 100 10.0002 + 0.022 = 1.01 = 2

AN102 y

33.

4

ANSWERS  Section 13.5

34. J

f(x) 5 log2x

22

19 43

22 50

2 R 4

35. 4x + 2h + 3

36.

1 1 4 x - 2 x + 2 (x + 2)2

x

8

Review Exercises (page 957) 1 1 3 3 7. 4 8. 9. 0 10. 11. 12. Continuous 13. Not continuous 14. Continuous 15. Not continuous 4 2 2 2 16. 5x - 6 … x 6 2 or 2 6 x 6 5 or 5 6 x … 6 6 17. All real numbers 18. 1, 6 19. 4 20. f 1 - 62 = 2; f 1 - 42 = 1 21. 4 22. - 2 23. - q 24. q 25. Does not exist 26. No 27. Yes 1. 8 2. - 343 3. 4 4. 25 5. 27 6. -

28. R is discontinuous at x = - 4 and x = 4. 1 1 lim R1x2 = - : hole at a - 4, - b x S -4 8 8 lim-R1x2 = - q ; lim+ R1x2 = q :

29. Undefined at x = 2 and x = 9; R has a hole at x = 2 and a vertical asymptote at x = 9.

y 5

5 x

y0

xS4

xS4

The graph of R has a vertical asymptote at x = 4.

x4

31. mtan = 0

30. mtan = 12 y 21 f (x)  2x 2  8x

y 5

32. mtan = 16 y 21

f (x)  x 2  2x  3

(2, 12)

(1, 10) 5 x 10 x (1, 4)

y  12x  2

5 x y  16x  20

f (x)  x 3  x 2

y  4

33. - 24 34. - 3 35. 7 36. - 158 37. 0.6662517653 38. (a) 7 sec (b) 6 sec (c) 64 ft/sec (d) 1 - 32t + 962 ft/sec (e) 32 ft/sec (f) At t = 3 sec (g) - 96 ft/sec (h) - 128 ft/sec 39. (a) $61.29/watch (b) $71.31/watch (c) $81.40/watch (d) R1x2 = - 0.25x2 + 100.01x - 1.24 (e) Approximately $87.51/watch 40. (a)

41. (a)

y 14

42. (a)

y 5

43. (a) Area under the graph of f 1x2 = 9 - x2 from - 1 to 3 (b) y

y 2

10 5 x

4 x 3 x

(b) 24 (d) 26

(c) 32 (e) 30

(f) 28

77 (c) 8

(b) 10 2

(d)

L-1

49 ≈ 1.36 (c) 1.02 36 4 1 (d) dx (e) 0.75 L1 x2 (b)

14 - x2 2 dx

44. (a) Area under the graph of f 1x2 = e x from - 1 to 1

(b)

(e) 9

4 x

(c)

(c) 2.35

y 4

2.5 x

Chapter Test (page 959) 1 2 3. 5 4. - 2 5. 135 6. 7. - 1 8. - 3 9. 5 10. 2 11. Limit exists; 2 3 3 12. (a) Yes (b) No; lim f 1x2 ≠ f 112 (c) No; lim- f 1x2 ≠ f 132 (d) Yes 13. x = - 7: asymptote; x = 2: hole 1. - 5

2.

xS1

xS3

4

14. (a) 5 (b) y = 5x - 19 (c) f(x)  4x 2  11x  3 y

15. (a)

16.

y 5

5 x

30 y  5x  19 5 x (2, 9)

(b) 13.359 (c) 4p ≈ 12.566

L1

1 - x2 + 5x + 32 dx

1 17. 35 ft/sec 3

80 3

AN103

ANSWERS  Section A.4

APPENDIX A Review A.1 Assess Your Understanding (page A10) 1. variable 2. origin 3. strict 4. base; exponent or power 5. True 6. True 7. False 8. False 9. 51, 2, 3, 4, 5, 6, 7, 8, 9 6 11. 54 6 13. 51, 3, 4, 6 6 15. 50, 2, 6, 7, 8 6

17. 50, 1, 2, 3, 5, 6, 7, 8, 9 6

19. 50, 1, 2, 3, 5, 6, 7, 8, 9 6

21.

– 2.5

–1

31 4

0

23. 7

5 2

25. 7

0.25

27. 7 51. - 28

29. = 53.

31. 6

4 5

55. 0

77. 5x x ≠ - 4 6 105. - 4

57. 1

79. 0°C

107. 5

131. C = pd

33. x 7 0

109. 4

59. 5

35. x 6 2 61. 1

81. 25°C 111. 2

37. x … 1

63. 22

83. 16

113. 15

115.

39.

41.

67. x = 0

65. 2

85.

1 16

1 2

117. 10; 0

87.

1 9

69. x = 3

89. 9

91. 5

119. 81

47. 6

49. 4

95. 64x6

121. 304,006.671

75. 5x x ≠ 5 6

73. x = 0, x = 1, x = - 1

71. None

93. 4

23 2 4 x 135. V = pr 3 137. V = x3 139. (a) $6000 4 3 1 147. No; is larger; 0.000333 c 149. No 3

(b) No

45. 2

–1

133. A =

145. (a) Yes

43. 1

–2

97.

123. 0.004

x

4

y

2

99.

x y

101. -

125. 481.890

141.  x - 4 Ú 6

(b) $8000

8x3z 9y

127. 0.000

103.

16x2 9y2

129. A = lw

143. (a) 2 … 5

(b) 6 7 5

A.2 Assess Your Understanding (page A19) 1 bh 3. C = 2pr 4. similar 5. True 6. True 7. False 8. True 9. True 10. False 11. 13 13. 26 15. 25 2 17. Right triangle; 5 19. Not a right triangle 21. Right triangle; 25 23. Not a right triangle 25. 8 in2 27. 4 in2 29. A = 25p m2; C = 10p m 256 31. V = 224 ft 3; S = 232 ft 2 33. V = p cm3; S = 64p cm2 35. V = 648p in3; S = 306p in2 37. p square units 39. 2p square units 3 41. x = 4 units; A = 90°; B = 60°; C = 30° 43. x = 67.5 units; A = 60°; B = 95°; C = 25° 45. About 16.8 ft 47. 64 ft 2 1. right; hypotenuse

2. A =

49. 24 + 2p ≈ 30.28 ft 2; 16 + 2p ≈ 22.28 ft

51. 160 paces

53. About 5.477 mi

55. From 100 ft: 12.2 mi; From 150 ft: 15.0 mi

A.3 Assess Your Understanding (page A30) 1. 4; 3 2. x4 - 16 3. x3 - 8 4. False 5. True 6. False 7. Monomial; variable: x; coefficient: 2; degree: 3 9. Not a monomial; the exponent of the variable is not a nonnegative integer 11. Monomial; variables: x, y; coefficient: - 2; degree: 3 13. Not a monomial; the exponent of one of the variables is not a nonnegative integer 15. Not a monomial; it has more than one term 17. Yes; 2 19. Yes; 0 21. No; the exponent of the variable of one of the terms is not a nonnegative integer 23. Yes; 3 25. No; the polynomial of the denominator has a degree greater than 0 27. x2 + 7x + 2 3 2 5 4 2 29. x - 4x + 9x + 7 31. 6x + 5x + 3x + x 33. 7x2 - x - 7 35. - 2x3 + 18x2 - 18 37. 2x2 - 4x + 6 39. 15y2 - 27y + 30 3 2 5 2 3 2 2 2 2 41. x + x - 4x 43. - 8x - 10x 45. x + 3x - 2x - 4 47. x + 6x + 8 49. 2x + 9x + 10 51. x - 2x - 8 53. x2 - 5x + 6 55. 2x2 - x - 6 57. - 2x2 + 11x - 12 59. 2x2 + 8x + 8 61. x2 - xy - 2y2 63. - 6x2 - 13xy - 6y2 65. x2 - 49 67. 4x2 - 9 69. x2 + 8x + 16 71. x2 - 8x + 16 73. 9x2 - 16 75. 4x2 - 12x + 9 77. x2 - y2 79. 9x2 - y2 81. x2 + 2xy + y2 83. x2 - 4xy + 4y2 85. x3 - 6x2 + 12x - 8 87. 8x3 + 12x2 + 6x + 1 89. 4x2 - 11x + 23; remainder - 45 91. 4x - 3; remainder x + 1 1 5 1 2 2 2 2 99. - 4x2 - 3x - 3; remainder - 7 93. 5x - 13; remainder x + 27 95. 2x ; remainder - x + x + 1 97. x - 2x + ; remainder x + 2 2 2 101. x2 - x - 1; remainder 2x + 2 103. x2 + ax + a2; remainder 0

A.4 Assess Your Understanding (page A39) 1. 3x(x - 2)(x + 2) 15. (x + 1)(x - 1) 29. (2x + 1)

2

2. prime 2

79. (x + 1)(x + 10)

81. (x - 7)(x - 3)

95. 2(3x + 1)(x + 1)

123. 2(3x + 4)(9x + 13)

35. (x + 3)(x - 3x + 9)

69. 25; (x + 5)2 83. 4(x2 - 2x + 8)

47. (x - 8)(x + 1)

71. 9; ( y - 3)2 85. Prime

37. (2x + 3)(4x - 6x + 9) 2

49. (x + 8)(x - 1)

27. (x - 5)2 39. (x + 2)(x + 3)

51. (x + 2)(2x + 3)

61. (x + 2)(3x - 4)

63. (x - 2)(3x + 4)

1 1 ; ax - b 75. (x + 6)(x - 6) 77. 2(1 + 2x)(1 - 2x) 16 4 87. - (x - 5)(x + 3) 89. 3(x + 2)(x - 6) 91. y2( y + 5)( y + 6) 73.

99. (x - 1)2(x2 + x + 1)2

119. (x - 1)(x + 1)(x + 2)

127. 5(x + 3)(x - 2)2(x + 1)

25. (x + 2)

23. (x + 1)

13. 3xy(x - 2y + 4)

2

2

109. - (3x - 1)(3x + 1)(x2 + 1)

117. (x + 5)(3x + 11)

11. 2x(x - 1) 2

59. (z + 1)(2z + 3)

97. (x - 3)(x + 3)(x + 9) 2

125. 2x(3x + 5)

9. x(x2 + x + 1)

21. (5x + 2)(5x - 2)

57. (3x + 1)(x + 1)

107. (2y - 5)(2y - 3)

115. 13x - 52 19x - 3x + 72 2

7. a(x2 + 1) 2

45. (x - 8)(x - 2)

55. (2x + 3)(3x + 2) 67. (x + 4)(3x - 2)

105. - (4x - 5)(4x + 1)

19. (x + 4)(x - 4)

33. (x - 3)(x + 3x + 9)

65. (x + 4)(3x + 2)

93. (2x + 3)

5. 3(x + 2)

2

43. (x + 5)(x + 2)

53. (x - 2)(2x + 1)

2

4. False

17. (2x + 1)(2x - 1)

31. (4x + 1)

41. (x + 6)(x + 1)

3. True

101. x5(x - 1)(x + 1)

111. (x + 3)(x - 6)

113. (x + 2)(x - 3)

121. (x - 1)(x + 1)(x - x + 1)

129. 3(4x - 3)(4x - 1)

2

131. 6(3x - 5)(2x + 1)2(5x - 4)

133. The possibilities are (x { 1)(x { 4) = x { 5x + 4 or (x { 2)(x { 2) = x { 4x + 4, none of which equals x2 + 4. 2

2

103. (4x + 3)2

AN104

ANSWERS  Section A.5

A.5 Assess Your Understanding (page A44) 3. True 4. True 5. x2 + x + 4; remainder 12 11. 4x5 + 4x4 + x3 + x2 + 2x + 2; remainder 7 19. Yes 21. Yes 23. No 25. Yes 27. - 9

1. quotient; divisor; remainder 2. - 3) 2 0 - 5 1 9. x4 - 3x3 + 5x2 - 15x + 46; remainder - 138 15. x4 + x3 + x2 + x + 1; remainder 0 17. No

A.6 Assess Your Understanding (page A53) 1. lowest terms 2. least common multiple 3. True 4. False 19.

2x(x2 + 4x + 16) x + 4

21.

8 3x

23.

x - 3 x + 7

25.

5.

3 x - 3

4x (x - 2)(x - 3)

7.

(x - 2)(x + 2) x + 5 3x - 2 x + 9 4 - x 37. 39. 41. 43. 2 2x - 3 x - 3 2x - 1 x - 2 2(x2 - 2) 51. 53. (x - 2)(x + 2)(x + 1) 55. x(x - 1)(x + 1) x(x - 2)(x + 2) 5x (x - 6)(x - 1)(x + 4)

71.

-1 x(x + h)

87.

73.

x + 1 x - 1

(x + 1)(x - 1) (x + 1) 2

63.

2

89.

75.

2(2x2 + 5x - 2)

65.

(x - 2)(x + 2)(x + 3)

(x - 1)(x + 1)

77.

2x(2x + 1)

x(3x + 2) (3x + 1)

(x - 1) (x + 1) 79.

2

93. f =

2

y + 5

4x 47.

(x - 1)(x + 2)

67.

2

13.

2( y + 1)

(x - 4)2

57. x3(2x - 1)2

2

(x - 2)(x + 1)

(x + 1)

29. -

11.

2(x + 5)

45.

2(5x - 1)

2

4x 2x - 1

5x + 1

(x + 3)(3x - 1)

91. -

2

9.

4 5(x - 1)

27.

35.

61.

x 3

7. 3x2 + 11x + 32; remainder 99 13. 0.1x2 - 0.11x + 0.321; remainder - 0.3531

31.

x + 5 x - 1

15. - (x + 7)

(x + 3)2

33.

(x - 3)2

3x2 - 2x - 3 (x + 1)(x - 1)

17.

3 5x(x - 2)

(x - 4)(x + 3) (x - 1)(2x + 1)

49.

- (11x + 2) (x + 2)(x - 2)

59. x(x - 1)2(x + 1)(x2 + x + 1)

- x2 + 3x + 13 (x - 2)(x + 1)(x + 4)

- 2x(x2 - 2)

81.

(x + 2)(x2 - x - 3)

R1 # R2 2 ; m (n - 1)(R1 + R2) 15

69.

-1 x - 1

x3 - 2x2 + 4x + 3 x2(x + 1)(x - 1)

83.

3x - 1 2x + 1

85.

19 (3x - 5)2

A.7 Assess Your Understanding (page A60) 3. index

4. True

25. 12 23 43.

22 2

65.

27 22 32

81.

27. 7 22 45. -

67. x7>12

2 + x 2(1 + x)

215 5

3>2

83.

5. cube root 29. 22 47.

15

93. (x2 + 4)1>3(11x2 + 12) 107. (a) 15,660.4 gal

49.

23 71. x2>3y

4 - x 3>2

85.

7. 3

73.

8x5>4 y

x (x - 1) 2

1>2

87.

15. x3y2

17. x2y

19. 6 1x

3 52 4 2

2x + h - 2 2x2 + xh h

55. 4

x(3x2 + 2)

22x + 5

51.

53.

3x + 2 (1 + x)

77.

1>2

1 - 3x2 2 1x(1 + x2)2

95. (3x + 5)1>3(2x + 3)1>2(17x + 27)

(b) 390.7 gal

3 13. - 2x 1 x

3 37. (2x - 1) 2 2x

75.

3>4

11. 2 22

35. x - 2 1x + 1

8 25 - 19 41

1 2

9. - 2

3 33. - 2 2

31. 2 23

+ 22 2 23

69. xy2

(x + 4)

6. False

109. 2 22p ≈ 8.89 sec

111.

97.

(x2 + 1)1>2

79.

39. (2x - 15) 22x

2x1>2

99. 1.41

3 41. - (x + 5y) 2 2xy

59. 64

61.

1 27

63.

27 22 32

10 2 1x - 52 14x + 32

1 89. (5x + 2)(x + 1)1>2 2 3(x + 2)

57. - 3

3 23. 15 2 3

21. 6x 1x

91. 2x1>2(3x - 4)(x + 1)

101. 1.59

103. 4.89

105. 2.15

p 23 ≈ 0.91 sec 6

A.8 Assess Your Understanding (page A70) 5 5. equivalent equations 6. identity 7. extraneous 8. True 9. {4} 11. e f 13. 5 - 1 6 15. 5 - 18 6 17. 5 - 3 6 19. 5 - 16 6 21. 50.5 6 23. {2} 4 1 25. {2} 27. {3} 29. {0, 9} 31. 5 - 3, 0, 3 6 33. {21} 35. {6} 37. ∅ or { } 39. 5 - 5, 0, 4 6 41. 5 - 1, 1 6 43. e - 2, , 2 f 45. {1} 47. No real solution 2 49. 5 - 13 6

51. {3}

53. {8}

55. 5 - 1, 3 6

57. {1, 5}

59. {5}

61. {2}

63. e -

11 f 6

65. 5 - 6 6

67. R =

R1R2 R1 + R2

69. R =

mv2 F

71. r =

S - a S

73. The distance is approximately 229.94 ft. 75. Approx. 221 ft 79. The apparent solution - 1 is extraneous. 81. When multiplying both sides by x + 3, we are actually multiplying both sides by 0 when x = - 3. This violates the Multiplication Property of Equality.

A.9 Assess Your Understanding (page A78) 1. mathematical modeling 2. interest 3. uniform motion 4. False 5. True 6. 100 - x 7. A = pr 2; r = radius, A = area 9. A = s2; A = area, s = length of a side 11. F = ma; F = force, m = mass, a = acceleration 13. W = Fd; W = work, F = force, d = distance 15. C = 150x; C = total variable cost, x = number of dishwashers 17. Invest $31,250 in bonds and $18,750 in CDs. 19. $11,600 was loaned out at 8%. 21. Mix 75 lb of Earl Grey tea with 25 lb of Orange Pekoe tea. 23. Mix 160 lb of cashews with the almonds. 25. The speed of the current is 2.286 mi/hr. 27. The speed of the current is 5 mi/hr. 29. Karen walked at 4.05 ft/sec. 31. A doubles tennis court is 78 feet long and 36 feet wide. 33. Working together, it takes 12 min. 35. (a) The dimensions are 10 ft by 5 ft. (b) The area is 50 sq ft. (c) The dimensions will be 7.5 ft by 7.5 ft. (d) The area will be 56.25 sq ft. 2 2 37. The defensive back catches up to the tight end at the tight end’s 45-yd line. 39. Add gal of water. 41. Evaporate 10 oz of water. 3 1 3 43. 40 g of 12-karat gold should be mixed with 20 g of pure gold. 45. Mike passes Dan mile from the start, 2 min from the time Mike started to race. 3 47. The latest the auxiliary pump can be started is 9:45 am. 49. The tub will fill in 1 hr. 51. Run: 12 miles; bicycle: 75 miles 53. Bolt would beat Burke by 19.75 m. 55. Set the original price at $40. At 50% off, there will be no profit.

59. The tail wind was 91.47 knots.

AN105

ANSWERS  Section A.11

A.10 Assess Your Understanding (page A86) 3. negative

4. closed interval

13. 32, q 2; x Ú 2 (c) 12 7 - 9

15. 3 0, 32; 0 … x 6 3

(d) - 8 6 6

17. (a) 6 6 8

21. (a) 2x + 4 6 5

23. [0, 4]

6. True

(b) 2x - 4 6 - 3

4

4

31. 2 … x … 5

3

5

41. 7

43. Ú

45. 6

53. {x x 6 4} or 1 - q , 42

47. …

49. 7

5

77. 5x x 6 - 5 6 or 1 - q ,- 52

10 10 r or a , q b 3 3

37. x 6 - 3 3

4

57. {x x 7 3} or 13, q 2

2 3

59. {x x Ú 2} or [2, q 2 2

65. {x x 6 - 20} or ( - q , - 20)

67. b x ` x Ú

4 4 r or c , q b 3 3

20

2 2 … x … 3 r or c , 3 d 3 3 3

79. 5x x Ú - 1 6 or 3 - 1, q 2 1

73. b x ` 

81. b x `

4 3

11 1 11 1 6 x 6 r or a - , b 2 2 2 2

11 2

75. 5x - 6 6 x 6 0 6 or 1 - 6, 02 6

1 2

1 5 1 5 … x 6 r or c , b 2 4 2 4 1 2

87. 5x x 7 3 6 or 13, q 2

10 3

4

3

2 2 f or a - q , d 3 3

5

29. 1 - q , - 42

4

2 3

71. b x `

(b) - 1 7 - 8

51. Ú

7

69. 5x 3 … x … 5 6 or 3 3, 5 4

19. (a) 7 7 0

(d) - 4x - 2 7 - 4

27. 3 4, q 2

11. [0, 2]; 0 … x … 2

10. True

35. x Ú 4

55. {x x Ú - 1} or [ - 1, q 2

63. e x x …

9. False

(d) - 6 7 - 10

(c) 6x + 3 6 6

−1

61. {x x 7 - 7} or 1 - 7, q 2

85. b x ` x 7

(c) 9 6 15

2

4

3

8. True

6

33. - 3 6 x 6 - 2

2

7. True

(b) - 2 6 0

25. [4, 6) 0

39. 6

5. Multiplication Property

1 1 83. b x ` x 6 - r or a - q , - b 2 2

5 4

89. 5x x Ú - 2 6

0



1 2

91. 21 6 Age 6 30

93. (a) Male Ú 77.3 years (b) Female Ú 81.6 years (c) A female can expect to live 4.3 years longer.

3

95. The agent’s commission ranges from $45,000 to $95,000, inclusive. As a percent of selling price, the commission ranges from 5% to 8.6%, inclusive. 97. The amount withheld varies from $111.20 to $161.20, inclusive. 99. The usage varies from 650 kW # hr to 2500 kW # hr, inclusive. 101. The dealer’s cost varies from $15,254.24 to $16,071.43, inclusive. 103. (a) You need at least a 74 on the fifth test. (b) You need at least a 77 on the fifth test. a + b a + b - 2a b - a a + b a + b 2b - a - b b - a a + b - a = = 7 0; therefore, a 6 .b = = 7 0; therefore, b 7 . 105. 2 2 2 2 2 2 2 2 2 2 2 2 2 107. 1 1ab2 - a = ab - a = a 1b - a2 7 0; thus 1 1ab2 7 a and 1ab 7 a. b2 - 1 2ab2 2 = b2 - ab = b 1b - a2 7 0; thus b2 7 1 2ab2 2 and b 7 2ab.

a 1b - a2 b 1b - a2 ab - a2 2ab b2 - ab 2ab - a = = 7 0; thus h 7 a. b - h = b = = 7 0; thus h 6 b. a + b a + b a + b a + b a + b a + b a - b a b 1 1 1 1 1 111. Since 0 6 a 6 b, then a - b 6 0 and 6 0. So 6 0, or 6 0. Therefore, 6 . And 0 6 because b 7 0. ab ab ab b a b a b 113. x2 + 1 Ú 1 for all real numbers x.

109. h - a =

A.11 Assess Your Understanding (page A94) 4. real; imaginary; imaginary unit 5. - 1; - i; 1

6. False

7. True 8. True 9. 8 + 5i

11. - 7 + 6i

13. - 6 - 11i

15. 6 - 18i

6 8 5 7 1 23 19. 10 - 5i 21. 37 23. + i 25. 1 - 2i 27. - i 29. - + i 31. 2i 33. - i 35. i 37. - 6 39. - 10i 41. - 2 + 2i 5 5 2 2 2 2 47. 2i 49. 5i 51. 5i 53. 6 55. 25 57. 2 + 3i 59. z + z = 1a + bi2 + 1a - bi2 = 2a; z - z = 1a + bi2 - 1a - bi2 = 2bi 61. z + w = 1a + bi2 + 1c + di2 = 1a + c2 + 1b + d2i = 1a + c2 - 1b + d2i = 1a - bi2 + 1c - di2 = z + w

67. 2a # 2b = 2ab only when 2a and 2b are real numbers.

17. 6 + 4i 43. 0

45. 0

AN106

ANSWERS  Section B.1

APPENDIX B Graphing Utilities B.1 Exercises (page B2) 1. ( - 1, 4); II 3. (3, 1); I 5. Xmin = - 6, Xmax = 6, Xscl = 2, Ymin = - 4, Ymax = 4, Yscl = 2 7. Xmin = - 6, Xmax = 6, Xscl = 2, Ymin = - 1, Ymax = 3, Yscl = 1 9. Xmin = 3, Xmax = 9, Xscl = 1, Ymin = 2, Ymax = 10, Yscl = 2 11. Xmin = - 11, Xmax = 5, Xscl = 1, Ymin = - 3, Ymax = 6, Yscl = 1 13. Xmin = - 30, Xmax = 50, Xscl = 10, Ymin = - 90, Ymax = 50, Yscl = 10 15. Xmin = - 10, Xmax = 110, Xscl = 10, Ymin = - 10, Ymax = 160, Yscl = 10

B.2 Exercises (page B4) 1. (a)

(b)

4

5

10

5

5

4

5

8

10

5

4

5

8

10

5

10

5

8

8

10

5

4

19.

21.

23.

25.

27.

29.

31.

B.3 Exercises (page B6) 3. - 1.71

5. - 0.28

7. 3.00

9. 4.50

11. 1.00, 23.00

B.5 Exercises (page B8) 1. Yes

3. Yes

5. No

7. Yes

10

8

17.

1. - 3.41

10

8

(b)

4

5

10

4

8

4

15. (a)

10

8

(b)

4

8

(b)

10

5

5

10

4

13. (a)

8

4

11. (a)

10

8

(b)

4

8

(b)

10

5

5

10

4

9. (a)

7. (a)

8

10

5

8

4

8

(b)

4

(b)

4

5

10

4

5. (a)

3. (a)

8

9. Answers may vary. A possible answer is Ymin = 4, Ymax = 12, and Yscl = 1.

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C1

Subject Index Δ (change in), 52, 104 ± (plus or minus), 67 Abel, Niels, 255, 888 Abscissa, 34 Absolute maximum and absolute minimum of function, 102–3 Absolute value, 641, A5 Absolute value equations, 210–11 Absolute value function, 111–12, 114 Absolute value inequalities, 211–13 Acute angles, 564–66 Addition. See also Sum of complex numbers, A90 of polynomials, A24–A25 of rational expressions, A47–A48 least common multiple (LCM) method for, A48–A50 triangular, 888 of vectors, 649, 652–53 in space, 672 Addition principle of counting, 896–97 Addition property of inequalities, A83 Additive inverse, A47 Adjacency matrix, 811 Ahmes (Egyptian scribe), 875 Airfoil, 615 Airplane wings, 615 Algebra essentials, A1–A13 distance on the real number line, A5–A6 domain of variable, A7 evaluating algebraic expressions, A6–A7 evaluating exponents, A10 graphing inequalities, A4–A5 Laws of Exponents, A7–A9 real number line, A4 sets, A1–A4 to simplify trigonometric expressions, 525 square roots, A9–A10 Algebraic functions, 305 Algebraic vector, 651–52 Algorithm, 245n Alpha particles, 723 Altitude of triangle, A15 Ambiguous case, 578–79 Amount of annuity, 874–75 Amplitude of simple harmonic motion, 600 of sinusoidal functions, 456–58, 460, 476–79 Analytic trigonometry, 495–62 algebra to simplify trigonometric expressions, 525 Double-angle Formulas, 543–53 to establish identities, 544–47 to find exact values, 544 Half-angle Formulas, 547–50 to find exact values, 547–50 for tangent, 549–50 Product-to-Sum Formulas, 553–55 Sum and Difference Formulas, 531–43 for cosines, 531–32 defined, 531 to establish identities, 533–36 to find exact values, 532, 534–35 involving inverse trigonometric function, 537 for sines, 534 for tangents, 536 Sum-to-Product Formulas, 555 trigonometric equations, 514–23 calculator for solving, 517 graphing utility to solve, 519 identities to solve, 518–19 involving single trigonometric function, 514–17 linear, 515–16

linear in sine and cosine, 538–39 quadratic in from, 517–18 solutions of, defined, 514 trigonometric identities, 523–30 basic, 524 establishing, 525–28, 533–36, 544–47 Even-Odd, 524 Pythagorean, 524 Quotient, 524 Reciprocal, 524 Angle(s), 408–21 acute, 564–66 central, 411 converting between decimals and degrees, minutes, seconds forms for, 410–11 of depression, 568, 569 direction, 655, 675–77 drawing, 409–10 of elevation, 568 elongation, 586 of incidence, 522 of inclination, 438 initial side of, 408, 409 negative, 408 optical (scanning), 552 positive, 408 quadrantal, 409, 425–27 of refraction, 522 of repose, 513 right, 409, A13 in standard position, 408 straight, 409 terminal side of, 408, 409 trigonometric functions of, 424–27 between vectors, 663–64, 674–75 viewing, 507 Angle–angle case of similar triangle, A16, A17 Angle–side–angle case of congruent triangle, A16 Angular speed, 416 Annuity(ies), 874–75 amount of, 874–75 defined, 874 formula for, 874 ordinary, 874–75 Aphelion, 710, 737, 753 Apollonius of Perga, 691 Applied (word) problems, A72–A80 constant rate job problems, A77 interest problems, A73–A74 mixture problems, A74–A75 steps for solving, A73 translating verbal descriptions into mathematical expressions, A72 uniform motion problems, A75–A77 Approaches infinity, 265–66 Approximate decimals, A3 Approximating area, 951–54 Araybhatta the Elder, 433 Arc length, 411–12 Area definition of, 954 formulas for, A15 under graph of function, 951–54 of parallelogram, 683 of triangle, 593–99, A15 SAS triangles, 594–95 SSS triangles, 595–96 Argument of complex number, 641 of function, 79 Arithmetic calculator, A10 Arithmetic mean, A89 Arithmetic progression. See Arithmetic sequences

Arithmetic sequences, 862–68 common difference in, 862 defined, 862 determining, 862–63 formula for, 863–64 nth term of, 863 recursive formula for, 863–64 sum of, 864–66 Ars Conjectandi (Bernoulli), 916 Ars Magna (Cardano), 255 ASA triangles, 578 Associative property of matrix addition, 797 of matrix multiplication, 802 of vector addition, 649 Asymptote(s), 266–72, 716–18 horizontal (oblique), 267, 269–72, 321 vertical, 267, 268–69, 321 Atomic systems, 808 Augmented matrix, 770–71, 772 in row echelon form, B9–B10 Average cost function, 92 Average rate of change, 52, 329–30 of function, 104–6 defined, 104 finding, 104–5 secant line and, 105–6 limit of, 934 of linear functions, 151–54 Axis/axes of complex plane, 640–41 of cone, 692 coordinate, 34 of ellipse, 702, 732 of hyperbola conjugate, 712 transverse, 712, 732 polar, 616–17 of quadratic function, 183–84 rotation of, 725–27 analyzing equation using, 727–29 formulas for, 726 identifying conic without, 729–30 of symmetry of parabola, 182, 693 of quadratic function, 183–84 Azimuth, 571n Babylonians, ancient, 875 Back substitution, 758 Barry, Rick, 95 Base of exponent, A8 Basic trigonometric identities, 524 Bearing (direction), 571 Bernoulli, Jakob, 637, 916 Bernoulli, Johann, 746 Bessel, Friedrich, 573 Bessel’s Star, 573 Best fit line of, 164–65 polynomial function of, 238–39 quadratic functions of, 201–2 Beta of stock, 150, 221–22 Bezout, Étienne, 826 Binomial(s), A23 cubing, A27 squares of (perfect squares), A26 Binomial coefficient, 886 Binomial Theorem, 883–89 n to evaluate a b , 884–85 j expanding a binomial, 886–88 historical feature on, 888

I1

I2

Subject Index

Binomial Theorem (Continued) proof of, 888 using, 886–88 Bisection Method, 254, 258 Blood alcohol concentration (BAC), 351 Bode Law, 861 Bonds, zero-coupon, 380 Book value, 156–57 Boole, George, 916 Bounded graphs, 835 Bounding curves, 603 Bounds on zeros, 252–53, 254 Box, volume and surface area of, A15 Brachistochrone, 745–46 Brancazio, Peter, 95 Branches of hyperbola, 712 Break-even point, 160 Brewster’s Law, 523 Briggs, Henry, 363 Bürgi, Joost, 363 Calculator(s), A10. See also Graphing utility(ies) approximating roots on, A56 converting from polar coordinates to rectangular coordinates, 619 to evaluate powers of 34, 327 factorial key on, 855 functions on, 80–81 inverse sine on, 499n kinds of, A10 logarithms on, 360–61 nCr key on, 884 trigonometric equations solved using, 517 Calculus angle measurement in, 411 approximating f(x) = ex, 861 area under curve, 133, 507 area under graph, 108 complicated functions in, 84 composite functions in, 310 derivative, 274 difference quotient in, 80, 109, 342, 364, 542 double-angle formulas in, 546 e in, 334 end behavior of graphs, 233 exponential equations in, 368 Extreme Value Theorem, 103 factoring problems occurring in, A36 functions and approximated by polynomial functions, 243 exponential, 327 increasing, decreasing, or constant, 100, 387 local maxima and local minima in, 101 graph of polynomial functions in, 225 independent variable in, 453 integral, 547, 954–55 area under graph, 951–54 graphing utility to approximate, 955 Intermediate Value Theorem, 253–54 limit notation, 872 limits, 233 logarithms and, 359, 368 partial fraction decomposition and, 814 polar equations and, 637 projectile motion, 741 removable discontinuity in, 285 secant line and, 105, 109 simplifying expressions with rational exponents in, A59 Simpson’s Rule, 204 Snell’s Law and, 522 tangent line, 71 trigonometric functions and equations in, 519, 522, 546, 552, 553 trigonometric identities useful in, 547 turning points and, 231, 233 of variations, 746 Carbon dating, 383–84

Cardano, Girolamo, 255, 646, 915 Cardioid, 630, 636 Carlson, Tor, 394 Carrying capacity, 386 Cartesian (rectangular) coordinates, 33, 34–35 Cartesian (rectangular) form of complex number, 641–42 Catenary, 200n, 701 Cayley, Arthur, 754, 808 Ceilometer, 569 Cell division, 381–82, 386 Cell phones, 74 Center of circle, 66 of hyperbolas, 712 of sphere, 678 Central angle, 411 Change-of-Base Formula, 361 Chebyshëv, P.I., 545n Chebyshëv polynomials, 545n Chu Shih-chieh, 888 Circle(s), 66–72, 692 arc length of, 411–12 area of, A15 area of a sector of, 415 center of, 66 circumference of, A15 defined, 66 general form of equation of, 68–69 graphing, 67, 635 inscribed, 599 intercepts of, 67–68 polar equation of, 626, 628 radius of, 66 standard form of equation of, 66–67 unit, 66, 422–38 Circular functions, 423 Circular motion, simple harmonic motion and, 600–1 Circumference, A15 Clark, William, 563 Clock, Huygens’s, 746 Closed interval, A81, A82 Coefficient, 224, A22, A23 binomial, 886 correlation, 165 damping, 603 of friction, 661 leading, 224, 259, A23 Coefficient matrix, 771 Cofactors, 788–89 Cofunctions, 433, 566 names of, 433 Coincident lines, 757 Column index, 770, 795 Column vector, 799–800 Combinations, 903–5 defined, 903 listing, 903–4 of n distinct objects taken r at a time, 904 Combinatorics, 895, 916 Combined variation, 139–40 Common difference, 862 Common logarithms (log), 348, 361, 363 Common ratio, 868 Commutative property of dot products, 663, 674 of vector addition, 649 Commutative property of matrix addition, 797 Complementary angle theorem, 566 Complement of event, 914 Complement of set, A2–A3 Complement Rule, 914–15 Complete graph, 43 Completing the square, 172, A38–A39 identifying conics without, 724–25 Complex fraction, A50n Complex function, 76n

Complex number(s), 658, 667 argument of, 641 conjugates of, 641 De Moivre’s Theorem and, 643–44 geometric interpretation of, 640 magnitude (modulus) of, 641 in polar form converting from rectangular form to, 641–42 converting to rectangular form, 641–42 products and quotients of, 642–43 product of, 642–43 quotient of, 642–43 Complex number system, A90 Complex numbers, 76n, A89–A95 addition, subtraction, and multiplication of, A90–A94 conjugates of, A91–A93 definition of, A90 equality of, A90 imaginary part of, A90 real part of, A90 in standard form, A90 power of, A93 reciprocal of, A92 Complex plane, 640–42 defined, 640 imaginary axis of, 640 plotting points in, 640–41 real axis of, 640 Complex polynomial function, 259 Complex rational expressions, A50–A52 Complex roots, 644–46 Complex variable, 259 Complex zeros, 173n of polynomials, 258–64 Conjugate Pairs Theorem, 259–60 defined, 259 finding, 261–62 polynomial function with specified zeros, 260–61 of quadratic function, 207–9 Components of vectors, 651, 652, 671 Composite functions, 306–13 calculus application of, 310 components of, 310 defined, 306 domain of, 307–10 equal, 309–10 evaluating, 307 finding, 307–10 forming, 306–7 Compound interest, 372–75 computing, 372–73 continuous, 374–75 defined, 372 doubling or tripling time for money, 377 effective rates of return, 375 formula, 373 present value of lump sum of money, 376 Compound probabilities, 912 Compressions, 124–25, 127 Conditional equation, 524 Cone axis of, 692 generators of, 692 right circular, 692 vertex of, 692 Congruent triangles, A16–A19 Conics, 692 defined, 732 degenerate, 692 directrix of, 732 eccentricity of, 732 ellipse, 692, 701–11 with center at (h, k), 706–7 with center at the origin, 702–6 with center not at origin, 706–7 center of, 702 defined, 702, 732

Subject Index eccentricity of, 711, 733, 734 foci of, 702 graphing of, 704 intercepts of, 704 length of major axis, 702 major axis of, 702, 732 minor axis of, 702 solving applied problems involving, 707–8 vertices of, 702 focus of, 732 general form of, 724–31 hyperbolas, 691, 692, 711–23 asymptotes of, 716–18 branches of, 712 with center at (h, k), 718–19 with center at the origin, 712–16 with center not at the origin, 718–19 center of, 712 conjugate, 723 conjugate axis of, 712 defined, 712, 732 eccentricity of, 723, 734 equilateral, 723 foci of, 712 graphing equation of, 713–14 solving applied problems involving, 719–20 transverse axis of, 712, 732 vertices of, 712 identifying, 724–25 without a rotation of axes, 729–30 names of, 692 parabola, 182–88, 692, 693–701 axis of symmetry of, 182, 693 defined, 693, 732 directrix of, 693 focus of, 693 graphing equation of, 694 solving applied problems involving, 697–98 with vertex at (h, k), 696–97 with vertex at the origin, 693–96 vertex of, 182, 693 paraboloids of revolution, 691, 698 parametric equations, 737–49 applications to mechanics, 745–46 for curves defined by rectangular equations, 743–45 cycloid, 745 defined, 737 describing, 740 graphing, using graphing utility, 738 rectangular equation for curve defined parametrically, 738–40 time as parameter in, 740–43 polar equations of, 731–37 analyzing and graphing, 731–35 converting to rectangular equation, 735 focus at pole; eccentricity e, 733–35 rotation of axes to transform equations of, 725–27 analyzing equation using, 727–29 formulas for, 726 Conjugate of complex number, A91–A93 of conjugate of complex number, A93 of product of two complex numbers, A93 of real number, A92–A93 of sum of two complex numbers, A93 Conjugate axis, 712 Conjugate golden ratio, 860 Conjugate hyperbola, 723 Conjugate of complex numbers, 641 Conjugate Pairs Theorem, 259–60 Connected mode, 115 Consistent systems of equations, 756, 757, 761–62 Constant(s), A6, A22 limit of, 929 of proportionality, 138 Constant functions, 100–1, 112 Constant linear functions, 154–55

Constant rate job problems, A77 Constant term, 224 Constraints, 839 Consumer Price Index (CPI), 380 Continued fractions, A55 Continuous compounding, 374–75 Continuous function, 253, 938–40 Continuous graph, 225 Convergent geometric series, 871–73 Cooling, Newton’s Law of, 384–86 Coordinates, 34, 35 of point on number line, A4 rectangular (Cartesian), 33, 34–35 Copernicus, 417 Corner points, 835 Correlation coefficient, 165 Correspondence between two sets, 75 Cosecant, 425, 426, 440 graph of, 471–73 periodic properties of, 442 Cosecant function, 423 continuous, 939, 940 domain of, 440, 441 inverse, 510–11 approximate value of, 511 calculator to evaluate, 511 definition of, 510 exact value of, 510–11 range of, 440, 441 Cosine(s), 425, 426, 439 direction, 676–77 exact value of, 532 Law of, 587–93 in applied problems, 589–90 defined, 587–88 historical feature on, 590 proof of, 588 Pythagorean Theorem as special case of, 588 SAS triangles solved using, 588–89 SSS triangles solved using, 589 periodic properties of, 441, 442 Sum and Difference Formula for, 531–32 trigonometric equations linear in, 538–39 Cosine function, 423 continuous, 940 domain of, 439, 441, 455 graphs of, 452–67 amplitude and period, 456–58 equation for, 462 key points for, 458–61 inverse, 500–2 defined, 500 exact value of, 501–2, 509–10 implicit form of, 500 properties of, 455 range of, 439, 440, 441, 455 Cosine function hyperbolic, 342 Cost(s) fixed, 63 marginal, 191 variable, 63 Cotangent, 425, 426, 440 periodic properties of, 442 Cotangent function, 423 continuous, 939, 940 domain of, 440, 441 graph of, 470–71 inverse, 510–11 approximating the value of, 511 calculator to evaluate, 511 definition of, 510 range of, 440, 441 Counting, 895–900 addition principle of, 896–97 combinations, 903–5 defined, 903 listing, 903–4 of n distinct objects taken r at a time, 904

I3

formula, 896 multiplication principle of, 897–98 permutations, 900–3 computing, 903 defined, 901 distinct objects with repetition, 901 distinct objects without repetition, 901–2 involving n nondistinct objects, 905–6 Counting numbers (natural numbers), 879n, 881–82, A3 Cramer, Gabriel, 754 Cramer’s Rule, 754, 784–91 for three equations containing three variables, 789–91 for two equations containing two variables, 785–87 Cross (vector) product, 658, 679–85 algebraic properties of, 681 area of parallelogram using, 683 defined, 679 geometric properties of, 682 of two vectors, 679–81 vector orthogonal to two given vectors, 682 Cube(s) of binomials (perfect cubes), A27 difference of two, A27, A33 sum of two, A27, A33 Cube function, 79, 113 Cube root, 110–11, 113, A55 complex, 644–46 Cubic function of best fit, 239 Cubic models from data, 238–39 Curve(s) bounding, 603 defined by rectangular equations, 743–45 defined parametrically, 738–40 graphing utility to graph parametrically-defined, B12 of quickest descent, 745–46 sawtooth, 608 Curve fitting, 765 sinusoidal, 479–83 hours of daylight, 482–83 sine function of best fit, 483 temperature data, 479–81 Curvilinear motion, 740 Cycle of sinusoidal graph, 453, 457 Cycloid, 745 Cygni, 573 Damped motion, 600, 603–4 Damping factor (damping coefficient), 603 Dantzig, George, 838n Data arrangement in matrix, 795 fitting exponential functions to, 392–93 sinusoidal models from, 479–83 Day length, 223, 303–4, 407, 493 “Deal or No Deal” (TV show), 894, 922 Decay, Law of, 381, 383–84. See also Exponential growth and decay Decimals, A3 approximate, A3 repeating, 872–73, A3 Declination of the Sun, 506 Decomposition, 665–66 Decreasing functions, 100–2 Decreasing linear functions, 154–55 Deflection, force of, 615 Degenerate conics, 692 Degree(s), 409–11 converting between decimals and, 410–11 converting between radians and, 412–15 Degree of monomial, A22, A30 Degree of polynomial, 224–28, A23, A30 odd, 252, 260 Demand equation, 197 De Moivre, Abraham, 643

I4

Subject Index

De Moivre’s Theorem, 643–44 Denominator, A45 rationalizing the, A57–A58 Dependent systems of equations, 757 containing three variables, 764–65 containing two variables, 761 matrices to solve, 777–78 Dependent variable, 79 Depressed equation, 249–50 Depression, angle of, 568, 569 Derivative, 945–46 Descartes, René, 33, 34n, 74, 522n Descartes’ Rule of Signs, 247–48 Descartes’s Law, 522n Determinants, 667, 680–81, 754, 784–94 cofactors, 788–89 Cramer’s Rule to solve a system of three equations containing three variables, 789–91 Cramer’s Rule to solve a system of two equations containing two variables, 785–87 evaluating, 680 expanding across a row or column, 789 to find cross products, 680–81 minors of, 788–89 properties of, 791–92 3 by 3, 788–89 2 by 2, 785, 791–92 Diagonal entries, 802 Difference(s). See also Subtraction common, 862 first, 475 limits of, 930 of logarithms, 359 of two cubes, A27, A33 of two functions, 83–84 of two matrices, 796–98 of two squares, A26, A33 of vectors, 649 Difference quotient, 80, 109, 342, 364, 542 Diophantus, 875 Directed line segment, 648, 650–51 Direction angles, 655, 675–77 Direction (bearing), 571 Direction cosines, 676–77 Direction of vectors, 648, 654–55 Directrix, 732 of parabola, 693 Direct variation, 138, 139–40 Dirichlet, Lejeune, 74, 76n Discontinuity, 114–15 nonremovable, 285 removable, 285 Discontinuous function, 938–39 Discrete mathematics, 851 Discriminant, 174, 209 Disjoint sets, A3 Distance, mean, 710, 753 Distance formula, 35–37 proof of, 35–36 in space, 671 Distributive Property of dot products, 663, 674 of matrix multiplication, 802 of real numbers, A4 Divergent geometric series, 871–73 Dividend, 245, A28 Division. See also Quotient(s) of complex numbers, A92 of polynomials, A27–A30 algorithm for, 245 synthetic, A41–A44 of rational expressions, A46 of two integers, A28 Divisor, 245, A28 Domain, 76, 81–83 of absolute value function, 114 of composite function, 307–10 of constant function, 112

of cosecant function, 440, 441 of cosine function, 439, 441, 455 of cotangent function, 440, 441 of cube function, 113 of cube root function, 113 of difference function, 83–84 of greatest integer function, 114 of identity function, 112 of inverse function, 317 of logarithmic function, 345 of logistic models, 386 of one-to-one function, 314–15 of product function, 83–84 of quotient function, 84 as range of inverse, 497 of rational function, 265–67 of reciprocal function, 113 of secant function, 439, 440, 441 of sine function, 439, 441, 454 of square function, 113 of square root function, 113 of sum function, 83–84 of tangent function, 439, 441, 469 of trigonometric functions, 439–40 unspecified, 85 of variable, A7 Domain-restricted function, 321–22 Doppler, Christian, 289 Doppler effect, 289 Dot mode, 115 Dot product, 658, 662–69, 673–74 angle between vectors using, 663–64 to compute work, 667 defined, 662, 673 finding, 662–63 historical feature on, 667 orthogonal vectors and, 664–65 parallel vectors and, 664 properties of, 663, 674 of two vectors, 662–63 Double-angle Formulas, 543–53 to establish identities, 544–47 to find exact values, 544 Double root (root of multiplicity 34), 170 Drag, 615 Dry adiabatic lapse rate, 867 e, 333–35, 342 Earthquakes, magnitude of, 355 Eccentricity, 732 of ellipse, 711, 733, 734 of hyperbola, 723, 734 Eddin, Nasir, 417, 590 Effective rates of return, 375 Egyptians, ancient, 875 Elements (Euclid), 590, 875 Elements of sets, 895-658, A1 Elevation, angle of, 568 Elimination, Gauss-Jordan, 777 Elimination method, 754, 759, 761 systems of nonlinear equations solved using, 822–25 Ellipse, 692, 701–11 with center at (h, k), 706–7 with center at the origin, 702–6 major axis along x-axis, 703–5 major axis along y-axis, 705–6 with center not at origin, 706–7 center of, 702 defined, 702, 732 eccentricity of, 711, 733, 734 foci of, 702 graphing of, 704 intercepts of, 704 major axis of, 702, 732 length of, 702 minor axis of, 702

solving applied problems involving, 707–8 vertices of, 702 Ellipsis, A3 Elliptical orbits, 691 Elongation angle, 586 Empty (null) sets, 895, A1 End behavior, 233–35, 267 Engels, Friedrich, 923 Entries of matrix, 770, 795 diagonal, 802 Epicycloid, 749 Equality of complex numbers, A90 of sets, 895, A2 of vectors, 649, 652 Equally likely outcomes, 911–12 Equation(s) conditional, 524 demand, 197 depressed, 249–50 domain of a function defined by, 82 equivalent, A63–A64, B3 even and odd functions identified from, 99–100 exponential, 335–36, 350–51, 367–68 quadratic in form, 368 as function, 78 graphing utility to graph, B3–B5 intercepts from, 44 inverse function defined by, 319–22 linear. See Linear equation(s) polar. See Polar equations quadratic in form, 176–77 solving, 176–77 satisfying the, 41, A63 sides of, 41, A63 solution set of, A63 solving, A63–A71 algebraically, A64 by factoring, A67–A68 with graphing calculator, B6–B7 systems of. See Systems of equations in two variables, graphs of, 41–43 intercepts from, 43 by plotting points, 41–43 symmetry test using, 44–46 x = y2, 47 y = 1 , x, 47–48 y = x3, 46–47 Equilateral hyperbola, 723 Equilateral triangle, 39 Equilibrium, static, 656–57 Equilibrium price, 157 Equilibrium quantity, 157 Equilibrium (rest) position, 600 Equivalent equations, A63–A64, B3 Equivalent systems of equations, 759 Error triangle, 40 Euclid, 590, 875, 888 Euler, Leonhard, 74, 417, 916 Even functions defined, 98 determining from graph, 98–99 identifying from equation, 99–100 Evenness ratio, 354 Even-Odd identity, 448–49, 524 Events, 911 complement of, 914 mutually exclusive, 913–14 probabilities of union of two, 913–15 Explicit form of function, 81 Exponent(s), A8–A9 Laws of, 327, 335, 336, A7–A9 logarithms related to, 343 Exponential equations, 335–36 defined, 335 solving, 335–36, 350–51, 367–68 equations quadratic in form, 368 using graphing utility, 368–69

Subject Index Exponential expressions, changing between logarithmic expressions and, 344 Exponential functions, 326–42, 495, 940 continuous, 940 defined, 328 e, 333–35, 342 evaluating, 326–30 fitting to data, 392–93 graph of, 330–33 using transformations, 333 identifying, 328–30 power function vs., 328 properties of, 331, 333, 337 Exponential growth and decay, 327–30, 381–84 law of decay, 381, 383–84 logistic models, 386–88 defined, 386 domain and range of, 386 graph of, 386 properties of, 386 uninhibited growth, 381–83 Exponential law, 381 Extended Principle of Mathematical Induction, 883 Extraneous solutions, A67, A68 Extreme values of functions, 102–3 Extreme Value Theorem, 103 Factored completely, A32 Factorial symbol, 854–55 Factoring defined, A32 equations solved by, A67–A68 of expression containing rational exponents, A60 over the integers, A32 polynomials, A32–A41 Ax2 + Bx + C, A37–A38 difference of two squares and the sum and the difference of two cubes, A33 by grouping, A36–A37 perfect squares, A34 x2 + Bx + C, A34–A36 quadratic equations, 170 Factors, A32 linear, 814–17 nonrepeated, 814–15 repeated, 816–17 quadratic, 252, 817–19 synthetic division to verify, A43–A44 Factor Theorem, 245–47 Family of lines, 64 of parabolas, 131 Feasible point, 839, 840 Fermat, Pierre de, 33, 342, 915 Ferrari, Lodovico, 255 Ferris, George W., 71, 521 Fertility, population growth and, 851 Fibonacci, 875 Fibonacci numbers, 856 Fibonacci sequences, 856 Financial models, 371–81 effective rates of return, 375 future value of a lump sum of money, 372–75 present value of a lump sum of money, 376 rate of interest or time required to double a lump sum of money, 377 Finck, Thomas, 417, 433 Finite sets, 895 First-degree equation. See Linear equation(s) First differences, 475 Fixed costs, 63 Focus/foci, 732 of ellipse, 702 of hyperbola, 712 of parabola, 693 FOIL method, A26 Foot-pounds, 667

Force, 600–1 of deflection, 615 resultant, 655, 656 Force vector, 654–55 Formulas, geometry, A15–A16 Foucault, Jean-Bernard-Leon, A71 Fractions complex, A50n continued, A55 partial, 814 Frequency, 466 in simple harmonic motion, 601 Friction, coefficient of, 661 Frobenius, Georg, 808 Function(s), 74–149. See also Composite functions; Exponential functions; Inverse functions; Linear functions; Polynomial functions; Trigonometric functions absolute maximum and absolute minimum (extreme values) of, 102–3 absolute value, 111–12, 114 area under graph of, 951–54 argument of, 79 average cost, 92 average rate of change of, 104–6 defined, 104 finding, 104–5 secant line and, 105–6 building and analyzing, 133–35 on calculators, 80–81 circular, 423 complex, 76n constant, 100–1, 112 continuous, 253, 938–40 cube, 79, 113 cube root, 110–11, 113 decreasing, 100–2 defined, 76 derivative of, 945–46 difference of two, 83–84 difference quotient of, 80 discontinuous, 938–39 domain of, 76, 81–83 unspecified, 85 domain-restricted, 321–22 equation as, 78 even and odd determining from graph, 98–99 identifying from equation, 99–100 explicit form of, 81 graph of, 88–97, 121–33 combining procedures, 123–24, 127–29 determining odd and even functions from, 98–99 determining properties from, 100–1 extreme values, 102–3 identifying, 89–90 information from or about, 90–92 using compressions and stretches, 124–25, 127 using reflections about the x-axis or y-axis, 126 using vertical and horizontal shifts, 121–24, 127 greatest integer, 114 identically equal, 524 identity, 112 implicit form of, 81 important facts about, 81 increasing, 100–2 intersection of two, 175–76 library of, 110–20 graphing, 110–17 local maxima and local minima of, 101–2, 231 with no limit at 33, 927 nonlinear, 151, 153 objective, 839–40 one-to-one, 314–16 periodic, 441 piecewise-defined, 115–17, 939–40 power, 225–28 graph of, 226–27

I5

of odd degree, 227–28 properties of, 227–28 product of two, 83–84 properties of, 98–110 quotient of two, 83–84 range of, 76 real zeros of, 176–77 reciprocal, 113, 471, 473. See also Cosecant function; Secant function relation as, 75–78 square, 112–13 square root, 110, 113 step, 115 sum of two, 83–84 graph of, 604–6 value (image) of, 76, 78–81 zeros of, 91–92 Bisection Method for approximating, 258 Function keys, A10 Function notation, 85 Fundamental identities of trigonometric functions, 443–45 quotient, 444 reciprocal, 443–44 Fundamental period, 441 Fundamental Theorem of Algebra, 259 Conjugate Pairs Theorem and, 259–60 proof of, 259 Future value, 372–75 Galois, Evariste, 255 Gauss, Karl Friedrich, 259, 646, 754 Gauss-Jordan method, 777 General addition principle of counting, 897 General form of conics, 724–31 of equation of circle, 68–69 linear equation in, 58–59 General term, 853 Generators of cone, 692 Geometric mean, A89 Geometric progression. See Geometric sequences Geometric sequences, 868–79 common ratio of, 868 defined, 868 determining, 868–69 formula for, 869–70 nth term of, 869–70 sum of, 870–71 Geometric series, 871–73 infinite, 871–73 Geometric vectors, 648–49 Geometry essentials, A13–A22 congruent and similar triangles, A16–A19 formulas, A15–A16 Pythagorean Theorem and its converse, A13–A15 Geometry problems, algebra to solve, 36–37 Gibbs, Josiah, 658 Golden ratio, 860 conjugate, 860 Grade, 65 Graph(s)/graphing area under, 951–54 bounded, 835 of circles, 67–68, 635 complete, 43 of cosecant function, 471–73 using transformations, 472 of the cosine function, 455 of cotangent function, 470–71 of ellipse, 704 of equations in two variables, 41–50 intercepts from, 43 by plotting points, 41–43 symmetry test using, 44–46 x = y2, 47 y = 1 , x, 47–48 y = x3, 46–47

I6

Subject Index

Graph(s)/graphing (Continued) of exponential functions, 330–33 using transformations, 333 of function, 88–97, 121–33 combining procedures, 123–24, 127–29 determining odd and even functions from, 98–99 determining properties from, 100–1 extreme values, 102–3 identifying, 89–90 information from or about, 90–92 using compressions and stretches, 124–25, 127 using reflections about the x-axis or y-axis, 126 using vertical and horizontal shifts, 121–24, 127 functions listed in the library of functions, 110–17 of inequalities, 830–35, A4–A5 linear inequalities, 831–32 steps for, 831 of inverse functions, 318–19 of linear functions, 151 of lines given a point and the slope, 54 using intercepts, 58–59 of logarithmic functions, 345–49 base not 42 or e, 362 inverse, 347–48 of logistic models, 386 of parabola, 694 of parametric equations, 738, B11–B12 of piecewise-defined functions, 115–17 of polar equations, 625–40 cardioid, 630, 636 circles, 635 of conics, 733–35 by converting to rectangular coordinates, 626–29 defined, 626 lemniscate, 634, 636 limaçon with inner loop, 632, 636 limaçon without inner loop, 631, 636 by plotting points, 630–35 polar grids for, 625 rose, 633, 636 sketching, 636–37 spiral, 634–35 using graphing utility, 627, B11 of polynomial functions analyzing, 236–38 smooth and continuous, 225 using bounds on zeros, 254 using transformations, 228 of polynomial inequalities, 291 of quadratic functions properties of, 183 steps for, 187 using its vertex, axis, and intercepts, 184–87 using transformations, 181–83 of rational functions, 275–90 analyzing, 275–85 constructing rational function from, 285–86 using transformations, 266 of rational inequalities, 292–93 of secant function, 471–72 using transformations, 472 of sequences, 852, 853 of sine and cosine functions, 452–67, 476–79, 604–5 amplitude and period, 456–58 equation for, 462 key points for, 458–61 y = sin x, 453 to solve systems of equations, 757 of systems of nonlinear inequalities, 834 of vectors, 650–51 of y = 33/x2, 265–66 Graphing calculator(s), A10 composite functions on, 307 scale setting, 35 TABLE feature, 254

Graphing utility(ies), B1–B12 to analyze graph of polynomial function, 237–38 connected mode, 115 converting between degrees, minutes, seconds, and decimal forms on, 410 coordinates of point shown on, B2 derivative of function using, 945 dot mode, 115 eVALUEate feature, 246, B5 to find limit, 927 to find sum of arithmetic sequence, 864–66 to fit exponential function to data, 391–92 to fit logistic function to data, 393–94 functions on, 103–4 geometric sequences using, 870, 871 to graph a circle, 69 to graph equations, 55, B3–B5 to graph inequalities, B9 to graph parametric equations, B11–B12 to graph polar equations, 627, B11 identity established with, 526 intercepts found using, 44 INTERSECT feature, B6–B7 line of best fit from, 164–65 to locate intercepts and check for symmetry, B5–B6 MAXIMUM and MINIMUM features, 104, 200 PARametric mode, 742 polar equations using, 627 REGression options, 392 row echelon form on, 780 RREF command, B10 sine function of best fit on, 483 to solve equations, B6–B7 to solve logarithmic and exponential equations, 368–69 to solve systems of linear equations, B9–B10 square screens, B8 TABLE feature, 853 tables on, B4 TRACE feature, 853 trigonometric equations solved using, 519 turning points in, 232 viewing rectangle, B1–B3 setting, B1–B2 ZERO (or ROOT) feature, 200, B5, B6 ZOOM-STANDARD feature, B3n ZSQR function on, B8n Grassmann, Hermann, 658, 667 Greatest integer function, 114–15 Grouping, factoring by, A36–A37 Growth, exponential, 327–30 Growth, uninhibited, 381–83 Growth factor, 328 Hale–Bopp comet, 691, 753 Half-angle Formulas, 547–50 to find exact values, 547–50 for tangent, 549–50 Half-life, 383–84 Half-line (ray), 408 Half-open/half-closed intervals, A81, A82 Half-planes, 832 Hamilton, William Rowan, 658 Harmonic mean, A89 Heron of Alexandria, 595, 596, 875 Heron’s Formula, 595–96 historical feature on, 596 proof of, 595–96 Horizontal component of vector, 652 Horizontal compression or stretches, 125 Horizontal lines, 55–56, 627, 635 Horizontal-line test, 315–16 Horizontal (oblique) asymptote, 267, 269–72, 321 Horizontal shifts, 121–24, 127 to the left, 122–23 to the right, 122 HP 48G, B8n

Huygens, Christiaan, 746, 915 Huygens’s clock, 746 Hyperbolas, 691, 692, 711–23 asymptotes of, 716–18 branches of, 712 with center at (h, k), 718–19 with center at the origin, 712–16 transverse axis along x-axis, 713–14 transverse axis along y-axis, 715–16 with center not at the origin, 718–19 center of, 712 conjugate, 723 conjugate axis of, 712 defined, 712, 732 eccentricity of, 723, 734 equilateral, 723 foci of, 712 graphing equation of, 713–14 solving applied problems involving, 719–20 transverse axis of, 712, 732 vertices of, 712 Hyperbolic cosine function, 342 Hyperbolic sine function, 342 Hyperboloid, 722 Hypocycloid, 749 Hypotenuse, A13 i, A93 Identically equal functions, 524 Identity(ies), A63 definition of, 524 polarization, 669 Pythagorean, 445, 524 trigonometric, 523–30 basic, 524 establishing, 525–28, 533–36, 544–47 Even-Odd, 524 Pythagorean, 524 Quotient, 524 Reciprocal, 443–44, 524 trigonometric equations solved using, 518–19 Identity function, 112 Identity matrix, 802–3 Identity Properties, 803 Image (value) of function, 76, 78–81 Imaginary axis of complex plane, 640 Imaginary unit, A89 Implicit form of function, 81 Improper rational expression, 814 Improper rational function, 270 Incidence, angle of, 522 Inclination, angle of, 438 Inconsistent systems of equations, 756, 757, 761–62, 763–64 containing three variables, 763–64 containing two variables, 760 matrices to solve, 778–79 Increasing functions, 100–2 Increasing linear functions, 154–55 Independent systems of equations, 757 Independent variable, 79 in calculus, 453 Index/indices of radical, A55, A60 of refraction, 522 row and column, 770, 795 of sum, 856 Induction, mathematical, 879–83 Extended Principle of, 883 principle of, 880, 882 proving statements using, 879–82 Inequality(ies) absolute value, 211–13 combined, A85–A86 graphing, 830–35 on graphing utility, B9 linear inequalities, 831–32 steps for, 831

Subject Index interval notation to write, A82 involving quadratic functions, 192–96 nonstrict, A5 in one variable, A84 polynomial, 291–92 algebraically and graphically solving, 291–92 multiplicity in, 292 steps for solving, 294 properties of, A82–A84, A86 rational algebraically and graphically solving, 292–94 multiplicity in, 294 steps for solving, 294 satisfying, 830 sides of, A5 solving, A84–A86 strict, A5 systems of, 830–37 graphing, 832–35 in two variables, 830 Inequality symbols, A5 Inertia moment of, 557 product of, 552 Infinite geometric series, 871–73 Infinite limit, 234 Infinite sets, 895 Infinity, A81 approaches, 265–66 limits at, 234 Inflation, 379 Inflection point, 386 Initial point of directed line segment, 648 Initial side of angle, 408, 409 Initial value, 328 Input to relation, 75 Inscribed circle, 599 Instantaneous rate of change, 946 Integers, A3 dividing, A28 factoring over the, A32 positive, 879n Integrals, 547, 955 area under graph, 951–54 graphing utility to approximate, 955 Intercept(s) of circle, 67–68 of the ellipse, 704 from an equation, 44 from a graph, 43 graphing an equation in general form using, 58–59 graphing utility to find, B5–B6 graph of lines using, 58–59 of quadratic function, 184–87 Interest compound computing, 372–73 continuous, 374–75 defined, 372 doubling or tripling time for money, 377 effective rates of return, 375 formula, 373 present value of lump sum of money, 376 problems involving, A73–A74 rate of, 372, A73 simple, 372, A73 Intermediate Value Theorem, 253–54 Intersection of sets, A2 of two functions, 175–76 Intervals, A81–A82 closed, A81, A82 endpoints of, A81 half-open, or half-closed, A81, A82 open, A81, A82 writing, using inequality notation, A82 Invariance, 731

Inverse additive, A47 of matrix, 803–8 finding, 805–6 multiplying matrix by, 804–5 solving system of linear equations using, 807–8 Inverse functions, 316–22, 495, 496–528. See also Logarithmic functions cosine, 500–2 defined, 500 exact value of, 501–2 exact value of expressions involving, 509–10 implicit form of, 500 defined by a map or set of ordered pairs, 316–18 domain of, 317 of domain-restricted function, 321–22 finding, 316–18, 504–5 defined by an equation, 319–22 graph of, 318–19 range of, 317, 322 secant, cosecant, and cotangent, 510–11 approximating the value of, 511 calculator to evaluate, 511 definition of, 510 sine, 496–500 approximate value of, 498–99 defined, 497 exact value of, 497–98, 509–10, 537 implicit form of, 497 properties of, 499–500 solving equations involving, 505 Sum and Difference Formulas involving, 537 tangent, 502–4 defined, 503 exact value of, 503–4 exact value of expressions involving, 509–10 implicit form of, 503 verifying, 318 written algebraically, 511–12 Inverse trigonometric equations, 505 Inverse variation, 139 Irrational numbers, A3, A90 decimal representation of, A3 Irreducible quadratic factor, 252, 817–19 Isosceles triangle, 39 Jiba, 433 Jiva, 433 Joint variation, 139–40 Jordan, Camille, 754 Joules (newton-meters), 667 Karmarkar, Narendra, 838n Kepler, Johannes, 140, 143 Khayyám, Omar, 888 Kirchhoff’s Rules, 768, 783 Koukounas, Teddy, 598n Kôwa, Takakazu Seki, 754 Latus rectum, 694 Law of Cosines, 587–93 in applied problems, 589–90 defined, 587–88 historical feature on, 590 proof of, 588 Pythagorean Theorem as special case of, 588 SAS triangles solved using, 588–89 SSS triangles solved using, 589 Law of Decay, 381, 383–84. See also Exponential growth and decay Law of Sines, 576–77, 588n in applied problems, 581–83 defined, 577 historical feature on, 590 proof of, 582–83 SAA or ASA triangles solved using, 578 SSA triangles solved using, 578–81 Law of Tangents, 587, 590

Laws of Exponents, 327, 335, 336, A7–A9 Leading coefficient, 224, 259, A23 Leading term, 224 Least common multiple (LCM) to add rational expressions, A48–A50 Left endpoint of interval, A81 Left limit, 936 Left stochastic transition matrix, 811 Legs of triangle, A13–A14 Leibniz, Gottfried Wilhelm, 74, 754 Lemniscate, 634, 636 Lensmaker’s equation, A54 Lewis, Meriwether, 563 Lift, 615 Light detector, 569 Light projector, 569 Like radicals, A57 Like terms, A23 Limaçon with inner loop, 632, 636 without inner loop, 631, 636 Limits, 233–34, 924–42 of average rate of change, 934 of constant, 929 of difference, 930 finding, 924–29 algebra techniques for, 929–35 by graphing, 926–27 using a table, 924–26 at infinity, 234 of monomial, 931 one-sided, 936–37 of polynomial, 931–32 of power or root, 932 of product, 930 of quotient, 933–34 of sum, 930 of x, 929 Line(s), 51–65 of best fit, 164–65 coincident, 757 equation of secant, 105–6. See also Linear equation(s); Systems of linear equations equations of perpendicular lines, 664n family of, 64 graphing given a point and the slope, 54 using intercepts, 58–59 horizontal, 55–56, 627, 635 point-slope form of, 55–56 polar equation of, 626–27, 635 slope of, 51–54, 57 containing same point, 52–53 containing two points, 52 from linear equation, 57 tangent, 71 vertical, 51, 627, 635 y-intercept of, 57 Linear algebra, 795 Linear equation(s), A64–A66. See also Line(s); Systems of linear equations defined, 59 in general form, 58–59 given two points, 56 for horizontal line, 55–56 in one variable, A63, A64 for parallel line, 59–60 for perpendicular line, 60–61 slope from, 57 in slope-intercept form, 56–57 solving equations that lead to, A65–A66 for vertical line, 54–55 Linear factors, 814–17 nonrepeated, 814–15 repeated, 816–17 Linear functions, 151–62 average rate of change of, 151–54 defined, 151

I7

I8

Subject Index

Linear functions (Continued) graphing utility to find the line of best fit, 164–65 graph of, 151 identifying, 328–30 increasing, decreasing, or constant, 154–55 models from data, 162–68 models from verbal descriptions, 156–58 scatter diagrams, 162–63 zero of a, 155 Linear programming problems, 754, 838–44 definition of, 839 maximum, 841–42 minimum, 840–41 setting up, 838–39 solution to, 840 location of, 840 solving, 839–42 in two variables, 839 Linear relations, nonlinear relations vs., 163–64 Linear speed, 416–17 Linear trigonometric equation, 515–16 LINE command, 55 Line segment, 648, 650–51 midpoint of, 38 Local maxima and local minima of functions, 101–2, 231 Logarithmic equations, 365–66 defined, 350 solving, 350–51, 365–66 Logarithmic functions, 343–56 changing between logarithmic expressions and exponential expressions, 344 continuous, 940 defined, 343 domain of, 345 evaluating, 344 fitting to data, 393–94 graph of, 345–49 properties of, 346, 352 range of, 345 Logarithmic spiral, 635 Logarithms, 356–64 with bases other than 42 or e evaluating, 360–62 graphing, 362 on calculators, 360–61 common (log), 348, 361, 363 historical feature on, 363 logarithmic expression as single, 359–60 logarithmic expression as sum or difference of, 359 natural (ln), 346, 360–61, 363 properties of, 356–58 establishing, 357 proofs of, 357 summary of, 362 relating to exponents, 343 Logistic models, 386–88 defined, 386 domain and range of, 386 graph of, 386 growth models, 394–95 properties of, 386 Loudness, 355 Lowest terms rational expressions in, A45–A46 rational function in, 265, 268 Magnitude of vectors, 648, 650–51, 652, 654–55, 676–77 in space, 673 Magnitude (modulus), 641, 642 Magnitude of earthquake, 355 Major axis, 732 Malthus, Thomas Robert, 923, 960 Mandel, Howie, 894 Mandelbrot sets, 647 Mapping, 75

Marginal cost, 191 Marginal propensity to consume, 878 Markov chains, 850 Marx, Karl, 923 Mathematical induction, 879–83 Extended Principle of, 883 principle of, 880, 882 proving statements using, 879–82 Mathematical modeling, A72 Matrix/matrices, 754, 770–84, 794–813 adjacency, 811 arranging data in, 795 augmented, 770–71, 772 in row echelon form, B9–B10 coefficient, 771 defined, 770, 795 entries of, 770, 795, 802 equal, 796 examples of, 795–96 historical feature on, 808 identity, 802–3 inverse of, 803–8 finding, 805–6 multiplying matrix by, 804–5 solving system of linear equations using, 807–8 left stochastic transition, 811 m by n, 795 nonsingular, 803, 805–6 product of two, 799–803 in reduced row echelon form, 776–77 row and column indices of, 770, 795 in row echelon form, 773–80 row operations on, 772–73 scalar multiples of, 798–99 singular, 803 to solve system of linear equations, 773–80 square, 795–96 sum and difference of two, 796–98 transition, 850 zero, 798 Maxima of functions, local, 101–2, 231 Mean arithmetic, A89 geometric, A89 harmonic, A89 Mean distance, 710, 753 Mechanics, parametric equations applied to, 745–46 Medians of triangle, 39 Menelaus of Alexandria, 417 Metrica (Heron), 596 Midpoint formula, 37–38 Midpoint of triangle, 39 Mindomo software, 562 Minima of functions, local, 101–2, 231 Minors, 788–89 Minutes, 410–11 Miranda, Kathleen, 598n Mixture problems, A74–A75 Model(s), A72 conic sections in, 692 financial, 371–81 effective rates of return, 375 future value of a lump sum of money, 372–75 present value of a lump sum of money, 376 rate of interest or time required to double a lump sum of money, 377 probability, 909–11 quadratic, 196–206 from data, 201–2 from verbal descriptions, 197–201 using direct variation, 138, 139–40 using inverse variation, 139 using joint variation or combined variation, 139–40 with vectors, 655–57 from verbal descriptions, 156–58 Modulus (magnitude), 641, 642 Mollweide, Karl, 586

Mollweide’s Formula, 586 Moment of inertia, 557 Monomial(s), A22–A23 common factors, A32 degree of, A22, A30 examples of, A22–A23 limit of, 931 recognizing, A22–A23 in two variables, A30 Monter, 65 Monty Hall Game, 922 Motion circular, 600–1 curvilinear, 740 damped, 600, 603–4 Newton’s second law of, 650 projectile, 741–42 simple harmonic, 600–2 uniform, A75–A77 Multiple root, 230n Multiplication. See also Product(s) of complex numbers, A91 of polynomials, A25 of rational expressions, A46 scalar, 798–99 of vectors, by numbers, 649–50 Multiplication principle of counting, 897–98 Multiplication properties for inequalities, A83 Multiplicity of polynomial functions, 229–35 roots of, 170, 230n in solving polynomial inequalities, 292 in solving rational inequalities, 294 vertical asymptote and, 268–69 Multiplier, 878 Mutually exclusive events, 913–14 Napier, John, 363 Nappes, 692 Natural logarithms (ln), 346, 360–61, 363 Natural numbers (counting numbers), 879n, 881–82, A3 Nautical miles, 420 Negative numbers real, A4 square roots of, A94 Newton, Isaac, 384n Newton-meters (joules), 667 Newton’s Law of Cooling, 384–86, 389 Newton’s Law of Heating, 389 Newton’s Law of universal gravitation, 297 Newton’s Method, 274 Newton’s Second Law of Motion, 600, 650 Niccolo of Brescia (Tartaglia), 255, 646 Nonlinear equations, systems of, 821–29 elimination method for solving, 822–25 historical feature on, 826 substitution method for solving, 821–22 Nonlinear functions, 151, 153 Nonlinear inequalities, systems of, 834 Nonlinear relations, 163–64 Nonnegative property of inequalities, A82 Nonremovable discontinuity, 285 Nonsingular matrix, 803, 805–6 Nonstrict inequalities, A5 Normal plane, 664n nth roots, A55–A56 historical feature, A60 rationalizing the denominator, A57–A58 simplifying, A55 simplifying radicals, A56–A57 Null (empty) sets, 895, A1 Numbers Fibonacci, 856 irrational, A3 natural (counting), 879n, 881–82, A3 rational, A3 Numerator, A45

Subject Index Objective function, 839–40 Oblique (horizontal) asymptote, 267, 269–72, 321 Oblique triangle, 577 Odd functions defined, 98 determining from graph, 98–99 identifying from equation, 99–100 One-sided limits, 936–37 One-to-one functions, 314–16 defined, 314 horizontal-line test for, 315–16 inverse of, 320 Open interval, A81, A82 Opens down, 182 Opens up, 182 Optical (scanning) angle, 552 Optimization, quadratic functions and, 197 Orbits elliptical, 691 planetary, 710 Ordered pair(s), 34 inverse function defined by, 316–18 as relations, 75–76 Ordinary annuity, 874–75 Ordinary (statute) miles, 420 Ordinate (y-coordinate), 34 Orientation, 738 Origin, 34 distance from point to, 133–34 of real number line, A4 in rectangular coordinates, 670 symmetry with respect to, 44–45 Orthogonal vectors, 664–67 Outcome of probability, 909 equally likely, 911–12 Output of relation, 75 Parabola, 182–88, 692, 693–701 axis of symmetry of, 182, 693 defined, 693, 732 directrix of, 693 family of, 131 focus of, 693 graphing equation of, 694 solving applied problems involving, 697–98 with vertex at (h, k), 696–97 with vertex at the origin, 693–96 finding equation of, 694–95 focus at (a, 0), a > 0, 694 vertex of, 182, 693 Paraboloids of revolution, 691, 698 Parallax, 573 Parallel lines, 59–60 Parallel vectors, 664 Parallelepiped, 684 Parallelogram, area of, 683 Parameter, 737 time as, 740–43 Parametric equations, 737–49 for curves defined by rectangular equations, 743–45 applications to mechanics, 745–46 cycloid, 745 defined, 737 describing, 740 graphing, 738 using graphing utility, B11 rectangular equation for curve defined parametrically, 738–40 time as parameter in, 740–43 Partial fraction decomposition, 754, 813–20 defined, 814 where denominator has nonrepeated irreducible quadratic factor, 818–19 where denominator has only nonrepeated linear factors, 814–15

where denominator has repeated irreducible quadratic factors, 819 where denominator has repeated linear factors, 816–17 Partial fractions, 814 Participation rate, 88 Partitioning, 951–54 Pascal, Blaise, 746, 885, 915 Pascal triangle, 885, 888 Payment period, 372 Peano, Giuseppe, 916 Pendulum period of, 141, A62 simple, 141 Perfect cubes, A27 Perfect roots, A56 Perfect squares, A26, A34 Perfect triangle, 598 Perihelion, 710, 737, 753 Perimeter, formulas for, A15 Period fundamental, 441 of simple harmonic motion, 600 of sinusoidal functions, 456–58, 460, 476–79 Periodic functions, 441 Periodic properties, 441–42 Period of pendulum, 141, A62 Permutations, 900–3 computing, 903 defined, 901 distinct objects with repetition, 901 distinct objects without repetition, 901–2 involving n nondistinct objects, 905–6 Perpendicular line, 664n Perpendicular lines, equations of, 60–61 Phase shift, 475–86 to graph y = A sin(vx - w), 475–79 Phones, cellular, 74 Physics, vectors in, 648 Piecewise-defined functions, 115–17 continuous, 940 Pitch, 65 Pixels, B1 Plane(s) complex defined, 640 imaginary axis of, 640 plotting points in, 640–41 real axis of, 640 normal, 664n Plane curve, 737 Planets, orbit of, 710 PLOT command, 55 Plotting points, 34, 616–18 graph equations by, 37–39 Point(s) coordinates of, 35 on graphing utility, B2 on number line, A4 corner, 835 distance between two, 35 in space, 671 distance from the origin to, 133–34 feasible, 839, 840 inflection, 386 initial, 648 of intersection of two functions, 175–76 plotting, 34, 616–18 graph equations by, 37–39 polar coordinates of, 617–18 of tangency, 71 terminal, 648 turning, 231–32 Point-slope form of equation of line, 55–56 Polar axis, 616 Polar coordinates, 616–25 conversion from rectangular coordinates to, 620–22

I9

conversion to rectangular coordinates, 618–19 defined, 616 plotting points using, 616–18 of a point, 617–18 polar axis of, 616 pole of, 616 rectangular coordinates vs., 616 Polar equations calculus and, 636 classification of, 635–36 of conics, 731–37 analyzing and graphing, 733–35 converting to rectangular equation, 735 focus at pole; eccentricity e, 733–35 defined, 626 graph of, 625–40 cardioid, 630, 636 circles, 635 by converting to rectangular coordinates, 626–29 defined, 626 lemniscate, 634, 636 limaçon with inner loop, 632, 636 limaçon without inner loop, 631, 636 by plotting points, 630–35 polar grids for, 625 rose, 633, 636 sketching, 636–37 spiral, 634–35 using graphing utility, 627, B11 historical feature on, 637 identifying, 626–28 testing for symmetry, 629 transforming rectangular form to, 622–23 transforming to rectangular form, 622–23 Polar form of complex number, 641–42 Polar grids, 625 Polarization identity, 669 Pole, 616 Polynomial(s), A22–A44 adding, A24–A25 Chebyshëv, 545n degree of, 224–28, A23, A30 odd, 252, 260 second-degree, A34–A36 dividing, 245–47, A27–A30 synthetic division, A41–A44 examples of, A23–A24 factoring, A32–A41 Ax2 + Bx + C, A37–A38 difference of two squares and the sum and the difference of two cubes, A33 by grouping, A36–A37 perfect squares, A34 x2 + Bx + C, A34–A36 limit of, 931–32 monomials, 931 multiplying, A25 prime, A32, A36 recognizing, A23–A24 solving, 251–52 special products formulas, A26–A27 in standard form, A23 subtracting, A24–A25 terms of, A23 in two variables, A30 zero, A23 Polynomial functions, 224–44, 938 of best fit, 238–39 complex, 259 complex zeros of, 258–64 Conjugate Pairs Theorem, 259–60 defined, 259 finding, 261–62 polynomial function with specified zeros, 260–61 continuous, 938 defined, 224 end behavior of, 233–35

I10

Subject Index

Polynomial functions (Continued) graph of analyzing, 236–38 smooth and continuous, 225 using bounds on zeros, 254 using transformations, 228 historical feature on, 255 identifying, 224–28 multiplicity of, 229–35 real zeros (roots) of, 229–35, 244–58 approximating, 254 Descartes’ Rule of Signs for, 247–48 finding, 250–51 Intermediate Value Theorem, 253–54 number of, 247–48 Rational Zeros Theorem, 248–49, 262 Remainder Theorem and Factor Theorem, 245–47 repeated, 230 theorem for bounds on, 252–53 solving, 250–51 in standard form, 224 unbounded in the negative direction, 233–34 writing, 235 Polynomial inequalities, 291–92 algebraically and graphically solving, 291–92 multiplicity in solving, 292 steps for solving, 294 Polynomial in one variable, second-degree, quadratic function as, 169 Population, world, 851 Population increases, 923, 960 Position vector, 651–52, 671–72 Positive integers, 879n Positive real numbers, A4 Power(s). See also Exponent(s) of i, A93 limit of, 932 log of, 358 Power (electrical), 466 Power functions, 225–28 exponential function vs., 328 graph of, 226–27 of odd degree, 227–28 properties of, 227–28 Present value, 373, 376 Price, equilibrium, 157 Prime polynomials, A32, A36 Principal, 372, A73 Principal nth root of real number, A55–A56 Principal square root, A9, A94 Probability(ies), 850, 909–19 Complement Rule to find, 914–15 compound, 912 constructing models, 909–11 defined, 909 of equally likely outcomes, 911–12 of event, 911 mutually exclusive, 913–14 historical feature on, 915–16 outcome of, 909 sample space, 909 of union of two events, 913–15 Product(s). See also Dot product; Multiplication of complex numbers, A91 in polar form, 642–43 of inertia, 552 limits of, 930–31 log of, 358 special, A26–A27, A30 of two functions, 83–84 of two matrices, 799–803 vector (cross), 658 Product function, 83 Product-to-Sum Formulas, 553–55 Projectile motion, 741–42 Projection, vector, 665–66 Projection of P on the x -axis, 601

Projection of P on the y -axis, 601 Prolate spheroid, 710 Proper rational expressions, 814 Proper rational function, 270 Proper subsets, 895 Proportionality, constant of, 138 Ptolemy, 523, 590 Pure imaginary number, A90 Pythagorean Identities, 445, 524 Pythagorean Theorem, A13–A15 applying, A14–A15 converse of, A13–A15, A18–A19 proof of, A18–A19 as special case of Law of Cosines, 588 Pythagorean triples, A21 Quadrantal angles, 409, 425–27 Quadrants, 34 Quadratic equations, 169–76 character of solutions of, 209 completing the square to solve, 172 defined, 169 discriminant of, 174, 209 factoring, 170 quadratic formula for, 173–75 solutions in complex number system, 208 Square Root Method for solving, 170–71 in standard form, 169 Quadratic factors, irreducible, 252, 817–19 Quadratic formula, 173–75 Quadratic functions, 169–80 of best fit to data, 201–2 defined, 169 graph of properties of, 183 steps for, 187 using its vertex, axis, and intercepts, 184–87 using transformations, 181–83 inequalities involving, 192–96 maximum or minimum value of, 188 optimizations and, 197 properties of, 180–92 vertex and axis of symmetry of, 184–87 zeros of completing the square to find, 172 complex, 207–9 factoring to find, 170 quadratic formula to find, 173–75 Square Root Method to find, 170–71 Quadratic models, 196–206 from data, 201–2 from verbal descriptions, 197–201 Quantity, equilibrium, 157 Quantity demanded, 157–58 Quantity supplied, 157–58 Quaternions, 658 Quotient(s), 245, A28. See also Division of complex numbers in polar form, 642–43 difference, 80, 109, 342, 364, 542 limit of, 933–34 log of, 358 synthetic division to find, A42–A43 of two functions, 83–84 Quotient identity(ies), 524 of trigonometric functions, 444 Radians, 411 converting between degrees and, 412–15 Radical equations, A68–A69 defined, A68 graphing utility to solve, B7 solving, A68–A69 Radicals, A55 fractional exponents as, A58–A59 index of, A55, A60 like, A57 properties of, A56

rational exponents defined using, A58 simplifying, A56–A57 Radical sign, A9, A60 Radicand, A55 Radioactive decay, 383–84 Radius, 66 of sphere, 678 Range, 76 of absolute value function, 114 of constant function, 112 of cosecant function, 440, 441 of cosine function, 439, 440, 441, 455 of cotangent function, 440, 441 of cube function, 113 of cube root function, 113 of greatest integer function, 114 of identity function, 112 of inverse function, 317, 322 inverse of, domain as, 497 of logarithmic function, 345 of logistic models, 386 of one-to-one function, 314–15 of projectile, 547 of reciprocal function, 113 of secant function, 440, 441 of sine function, 439, 440, 441, 454 of square function, 113 of square root function, 113 of tangent function, 440, 441, 469 of trigonometric functions, 439–40 Rate of change, average, 52, 329–30 of function, 104–6 defined, 104 finding, 104–5 secant line and, 105–6 limit of, 934 of linear functions, 151–54 Rate of change, instantaneous, 946 Rate of interest, 372, A73 Rates of return, effective, 375 Ratio common, 868 evenness, 354 Rational equations, A66–A67 defined, A66 with no solution, A67 solving, A66–A67 Rational exponents, A58–A60 Rational expressions, A45–A55 adding and subtracting, A47–A48 least common multiple (LCM) method for, A48–A50 application of, A52 complex, A50–A52 decomposing. See Partial fraction decomposition defined, A45 improper, 814 multiplying and dividing, A46 proper, 814 reducing to lowest terms, A45–A46 Rational functions, 223, 264–90 applied problems involving, 286–87 asymptotes of, 266–72 horizontal or oblique, 267 vertical, 267 continuous, 938–39 defined, 264 domain of, 264–67 examples of, 264 graph of, 275–90 analyzing, 275–85 constructing rational function from, 285–86 using transformations, 266 with a hole, 283–85 improper, 270 in lowest terms, 265, 268 proper, 270

Subject Index unbounded in positive direction, 265 zeros of denominator of, 285 Rational inequalities algebraically and graphically solving, 292–94 multiplicity in solving, 294 steps for solving, 294 Rationalizing the denominator, A57–A58 Rational numbers, 264, A3, A89–A90 Rational Zeros Theorem, 248–49, 262 Ray (half-line), 408 Real axis of complex plane, 640 Real number(s), A3–A6, A89–A90 approximate decimals, A3 conjugate of, A92–A93 defined, A3 principal nth root of, A55–A56 Real number line, A4 Real part of complex numbers, A90 Real zeros (roots) of function, 176–77 of polynomial functions, 244–58 approximating, 254 Descartes’ Rule of Signs for, 247–48 finding, 250–51 Intermediate Value Theorem, 253–54 number of, 247–48 Rational Zeros Theorem, 248–49, 262 Remainder Theorem and Factor Theorem, 245–47 repeated, 230 theorem for bounds on, 252–53 Reciprocal function, 113, 471, 473. See also Cosecant function; Secant function Reciprocal identities, 443–44, 524 Reciprocal property for inequalities, A83–A84, A86 Rectangle, area and perimeter of, A15 Rectangular (Cartesian) coordinates converted to polar coordinates, 620–22 polar coordinates converted to, 618–19 polar coordinates vs., 616 polar equations graphed by converting to, 626–29 in space, 670–71 Rectangular (Cartesian) coordinate system, 33, 34–35 Rectangular (Cartesian) form of complex number, 641–42 Rectangular equations for curve defined parametrically, 738–40 polar equations converted to, 622–23, 735 transforming to polar equation, 622–23 Rectangular grid, 625 Recursive formula, 855–56 for arithmetic sequences, 863–64 terms of sequences defined by, 855–56 Reduced row echelon form, 776–77 Reflections about x-axis or y-axis, 126 Refraction, 522 Regiomontanus, 417, 590 Relation(s), 75–78. See also Function(s) defined, 75 as function, 75–78 input to, 75 linear vs. nonlinear, 163–64 ordered pairs as, 75–76 output of, 75 Remainder, 245, A28 synthetic division to find, A42–A43 Remainder Theorem, 245–47 Removable discontinuity, 285 Repeated zeros (solutions), 170, 230 Repeating decimals, 872–73, A3 Repose, angle of, 513 Rest (equilibrium) position, 600–2 Resultant force, 655 Review, A1–A95 of algebra, A1–A13 distance on the real number line, A5–A6 domain of variable, A7 evaluating algebraic expressions, A6–A7 evaluating exponents, A10

graphing inequalities, A4–A5 Laws of Exponents, A7–A9 sets, A1–A4 square roots, A9–A10 complex numbers, A89–A95 of geometry, A13–A22 congruent and similar triangles, A16–A19 formulas, A15–A16 Pythagorean theorem and its converse, A13–A15 inequalities combined, A85–A86 properties of, A82–A84 solving, A84–A86 interval notation, A81–A82 of nth roots, A55–A56 historical feature, A60 rationalizing the denominator, A57–A58 simplifying, A55 simplifying radicals, A56–A57 of polynomials, A22–A44 adding, A24–A25 dividing, A27–A30 factoring, A32–A41 monomials, A22–A23 multiplying, A25 recognizing, A23–A24 special products formulas, A26–A27 subtracting, A24–A25 synthetic division of, A41–A44 in two variables, A30 of rational exponents, A58–A60 of rational expressions, A45–A55 adding and subtracting, A47–A48 application of, A52 complex, A50–A52 multiplying and dividing, A46 reducing to lowest terms, A45–A46 Revolutions, 416 Rhaeticus, 417 Rhind papyrus, 875 Richter scale, 355 Right angle, 409, A13 Right circular cone, 692 Right circular cylinder, volume and surface area of, A15 Right endpoint of interval, A81 Right-hand rule, 670 Right limit, 936–37 Right triangles, 564–76, A13 applications of, 567–71 solving, 566–67 Right triangle trigonometry, 564–76 Rise, 51 Root(s), A63. See also Solution(s); Zeros complex, 644–46 limit of, 932 of multiplicity 34 (double root), 170 of multiplicity m, 230n perfect, A56 Rose, 633, 636 Roster method, A1 Rotation of axes, 725–27 analyzing equation using, 727–29 formulas for, 726 identifying conic without, 729–30 Rounding, A10 Round-off errors, 567 Row echelon form, 773–80 augmented matrix in, B9–B10 reduced, 776–77 Row index, 770, 795 Row operations, 772–73 Row vector, 799–800 Rudolff, Christoff, A60 Ruffini, P., 255 Rule of Signs, Descartes’, 247–48 Run, 51 Rutherford, Ernest, 723

I11

SAA triangles, 578 Sample space, 909 SAS triangles, 577, 588–89, 594–95 Satisfying equations, 41, A63 Satisfying inequalities, 830 Sawtooth curve, 608 Scalar, 649–50, 798 Scalar multiples of matrix, 798–99 Scalar product. See Dot product Scale of number line, A4 Scanning (optical) angle, 552 Scatter diagrams, 162–63, 479–81 Schroeder, E., 916 Scientific calculators, A10 converting between degrees, minutes, seconds, and decimal forms on, 410 Secant, 425, 426, 439 graph of, 471–72 periodic properties of, 442 Secant function, 423 continuous, 939, 940 domain of, 439, 441 inverse, 510–11 approximating the value of, 511 calculator to evaluate, 511 definition of, 510 range of, 440, 441 Secant line, 105–6 Second-degree equation. See Quadratic equations Second-degree polynomials, A34–A36 Seconds, 410–11 Seed, 647 Sequences, 852–61 annuity problems, 874–75 arithmetic, 862–68 common difference in, 862 defined, 862 determining, 862–63 formula for, 863–64 nth term of, 863 recursive formula for, 863–64 sum of, 864–66 defined, 852 factorial symbol, 854–55 Fibonacci, 856 geometric, 868–79 common ratio of, 868 defined, 868 determining, 868–69 formula for, 869–70 nth term of, 869–70 sum of, 870–71 graph of, 852, 853 historical feature on, 875 from a pattern, 854 properties of, 857 summation notation, 856–57 sum of, 857–58 terms of, 852–54, 855–56 alternating, 854 defined by a recursive formula, 855–56 general, 853 Set(s), A1–A4 complement of, A2 correspondence between two, 75 disjoint, A3 elements of, 895–97, A1 empty (null), 895, A1 equal, 895, A2 finite, 895 infinite, 895 intersection of, A2 Mandelbrot, 647 of numbers, A1–A4 subsets of, 895 proper, 895

I12

Subject Index

Set(s) (Continued) union of, A2 universal, 896, A2 as well-defined, A1 Set-builder notation, A1–A2 Shannon’s diversity index, 354 Shifts, graphing functions using vertical and horizontal, 121–24, 127 Side–angle–side case of congruent triangle, A16 Side–angle–side case of similar triangle, A16, A17 Sides of equation, 41, A63 of inequality, A5 Side–side–side case of similar triangle, A16, A17 Similar triangles, 577n, A16–A19 Simple harmonic motion, 600–2 amplitude of, 600 analyzing, 602 circular motion and, 600–1 defined, 600 equilibrium (rest) position, 600 frequency of object in, 601 model for an object in, 600–2 period of, 600 Simple interest, 372, A73 Simple pendulum, 141 Simplex method, 838n Simplifying complex rational expressions, A50–A52 expressions with rational exponents, A58–A60 nth roots, A55 radicals, A56–A57 rational expressions, A45–A46 Simpson’s rule, 204 Sine, 425, 426, 439 historical feature on, 433 Law of, 576–77, 588n in applied problems, 581–83 defined, 577 historical feature on, 590 proof of, 582–83 SAA or ASA triangles solved using, 578 SSA triangles solved using, 578–81 periodic properties of, 441, 442 Sum and Difference Formula for, 534 trigonometric equations linear in, 538–39 Sine function, 423 of best fit, 483 continuous, 940 domain of, 439, 441, 454 graphs of, 452–67 amplitude and period, 456–58 equation for, 462 key points for, 458–61 y = sin x, 453 inverse, 496–500 approximate value of, 498–99 defined, 497 exact value of, 497–98, 509–10 implicit form of, 497 properties of, 499–500 properties of, 454 range of, 439, 440, 441, 454 Sine function, hyperbolic, 342 Singular matrix, 803 Sinusoidal curve fitting, 479–83 hours of daylight, 482–83 sine function of best fit, 483 temperature data, 479–81 Sinusoidal graphs, 452–67, 604–5 amplitude and period, 456–58 defined, 456 equation for, 462 key points for, 458–61 steps for, 479 y = sin x, 453 Six trigonometric functions of u, 425, 447 of t, 423

Slope, 51–54, 57 containing same point, 52–53 containing two points, 52 graphing lines given, 54 from linear equation, 57 of secant line, 105–6 Slope-intercept form of equation of line, 56–57 Smooth graph, 225 Snell, Willebrord, 522 Snell’s Law of Refraction, 522 Solution(s), A63. See also Zeros extraneous, A67, A68 of inequality, A84 of linear programming problems, 840 location of, 840 repeated, 170, 230 of systems of equations, 756, 761 of trigonometric equations, 514 Solution set of equation, A63 Special products, A26–A27, A30 Speed angular, 416 instantaneous, 946–48 linear, 416–17 Sphere, 678 volume and surface area of, A15 Spheroid, prolate, 710 Spiral, 634–35 Square(s) of binomials (perfect squares), A26, A33 difference of two, A26, A33 perfect, A26, A34 Square function, 112–13 Square matrix, 795–96 Square root(s), A9–A10, A55 complex, 644 of negative number, A94 principal, A9, A94 Square root function, 110, 113 Square Root Method, 170–71 SSA triangles, 578–81 SSS triangles, 577, 589, 595–96 Standard deviation, A88 Standard form complex number in, A90 power of, A93 quotient of two, A92 reciprocal of, A92 of equation of circle, 66–67 linear equation in, 58–59 polynomial function in, 224 polynomials in, A23 quadratic equations on, 169 Standard position, angle in, 408 Static equilibrium, 656–57 Statute (ordinary) miles, 420 Step function, 115 Stirling’s formula, 889 Stock risk, beta to measure, 150 Stock valuation, 150, 221–22 Straight angle, 409 Stretches, graphing functions using, 124–25, 127 vertical, 124 Strict inequalities, A5 Subscript, A23n Subscripted letters, 852 Subsets, 895, A2 proper, 895 Substitution method, 754, 757–58 systems of nonlinear equations solved using, 821–22 Subtraction. See also Difference(s) of complex numbers, A90 of polynomials, A24–A25 of rational expressions, A47–A48 least common multiple (LCM) method for, A48–A50

of vectors, 652–53 in space, 672 Sum. See also Addition of arithmetic sequences, 864–66 of geometric sequences, 870–71 index of, 856 of infinite geometric series, 872 limits of, 930 of logarithms, 358 of sequences, 857–58 of two cubes, A27, A33 of two functions, 83–84 graph of, 604–6 of two matrices, 796–98 Sum and Difference Formulas, 531–43 for cosines, 531–32 defined, 531 to establish identities, 533–36 to find exact values, 532, 534–35 involving inverse trigonometric function, 537 for sines, 534 for tangents, 536 Sum function, 83 Sum-to-Product Formulas, 555 Summation notation, 856–57 Sun, declination of, 506 Surface area, formulas for, A15 Sylvester, James J., 808 Symmetry, 44–46 axis of of parabola, 182, 693 of quadratic function, 183–84 graphing utility to check for, B5–B6 of polar equations, 629 with respect to origin, 44–45 p (y-axis), 629 with respect to the line u = 2 with respect to the polar axis (x-axis), 629 with respect to the pole (origin), 629 with respect to the x-axis, 44–45 with respect to the y-axis, 44–45 Synthetic division, A41–A44 Systems of equations consistent, 756, 761–62 dependent, 757 containing three variables, 764–65 containing two variables, 761 equivalent, 759 graphing, 757 inconsistent, 756, 761–62, 763–64 containing three variables, 763–64 containing two variables, 760 independent, 757 solutions of, 756, 761 Systems of inequalities, 830–37 graphing, 832–35 bounded and unbounded graphs, 835 vertices or corner points, 835 Systems of linear equations, 755–94 consistent, 757, 761–62 defined, 755 dependent, 757 containing three variables, 764–65 containing two variables, 761 matrices to solve, 777–78 determinants, 784–94 cofactors, 788–89 Cramer’s Rule to solve a system of three equations containing three variables, 789–91 Cramer’s Rule to solve a system of two equations containing two variables, 785–87 minors of, 788–89 properties of, 791–92 3 by 3, 788–89 2 by 2, 785, 791–92 elimination method of solving, 759, 761 equivalent, 759 examples of, 755–56

Subject Index graphing, 757 inconsistent, 757, 761–62, 763–64 containing three variables, 763–64 containing two variables, 760 matrices to solve, 778–79 independent, 757 matrices. See Matrix/matrices partial fraction decomposition, 813–20 defined, 814 where denominator has a nonrepeated irreducible quadratic factor, 818–19 where denominator has only nonrepeated linear factors, 814–15 where denominator has repeated irreducible quadratic factors, 819 where denominator has repeated linear factors, 816–17 solution of, 756, 761 solving, 756 with graphing utility, B9–B10 substitution method of, 757–58 three equations containing three variables, 761–63 Systems of nonlinear equations, 821–29 elimination method for solving, 822–25 historical feature on, 826 substitution method for solving, 821–22 Systems of nonlinear inequalities, graphing, 834 Tables, on graphing utility, B4 Tangency, point of, 71 Tangent(s), 425, 426, 439 graph of, 468–70 Half-angle Formulas for, 549–50 historical feature on, 433 Law of, 587, 590 periodic properties of, 442 Sum and Difference Formulas for, 536 Tangent function, 423 continuous, 939, 940 domain of, 439, 441, 469 inverse, 502–4 defined, 503 exact value of, 503–4, 509–10 implicit form of, 503 properties of, 469 range of, 440, 441, 469 Tangent line, 71 to the graph of a function, 943–44 Greek method for finding, 71 Tangent problem, 943 Tartaglia (Niccolo of Brescia), 255, 646 Tautochrone, 746 Terminal point of directed line segment, 648 Terminal side of angle, 408, 409 Terminating decimals, A3 Terms constant, 224 leading, 224 like, A23 of polynomial, A23 of sequences, 852–54, 855–56 alternating, 854 defined by a recursive formula, 855–56 general, 853 3 by 3 determinants, 680, 788–89 Thrust, 615 TI-84 Plus, B3, B9, B10 TI-85, B8n Time, as parameter, 740–43 Transcendental functions, 305 Transformations combining, 123–24, 127–29 compressions and stretches, 124–25, 127 cosecant and secant graphs using, 472 of cosine function, 455–56 defined, 121 graphs using of exponential functions, 333

of polynomial functions, 228 of quadratic functions, 181–83 of rational functions, 266 reflections about the x-axis or y-axis, 126 of sine function, 454 vertical and horizontal shifts, 121–24, 127 Transition matrix, 850 Transverse axis, 712, 732 Tree diagram, 898 Triangle(s). See also Law of Sines altitude of, A15 area of, 593–99, A15 ASA, 578 congruent, A16–A19 equilateral, 39 error, 40 isosceles, 39 legs of, A13–A14 medians of, 39 midpoint of, 39 oblique, 576–77 Pascal, 885, 888 perfect, 598 right, 564–76, A13 applied problems involving, 567–71 solving, 566–67 SAA, 578 SAS, 577, 588–89, 594–95 similar, 577n, A16–A19 SSA, 578–81 SSS, 577, 589, 595–96 Triangular addition, 888 Trigonometric equations, 514–23 calculator for solving, 517 graphing utility to solve, 519 identities to solve, 518–19 involving single trigonometric function, 514–17 linear, 515–16 linear in sine and cosine, 538–39 quadratic in from, 517–18 solutions of, defined, 514 Trigonometric expressions, written algebraically, 511–12, 537 Trigonometric functions, 407–94 of acute angles, 564–66 angles and their measure, 408–21 arc length of a circle, 411–12 area of a sector of a circle, 415 degrees, 409–11, 412–15 radians, 411, 412–15 applications of, 563–614 damped motion, 600, 603–4 graphing sum of two functions, 604–6 involving right triangles, 567–71 Law of Cosines, 587–93 Law of Sines, 576–87, 589–90 Law of Tangents, 587, 589–90 simple harmonic motion, 600–2 calculator to approximate values of, 432 circle of radius r to evaluate, 432–33 cosecant and secant graphs, 471–73 exact values, 425–27 given one of the functions and the quadrant of the angle, 445–48 p p = 30°, = 45°, and for integer multiples of 6 4 p = 60°, 428–30 3 p = 45°, 427 4 p p = 30° and = 60°, 428–30 6 3 using a point on the unit circle, 424–25 using periodic properties, 442 fundamental identities of, 443–45 quotient, 444 reciprocal, 443–44 historical feature on, 417 phase shift, 475–86 to graph y = A sin(vx - w), 475–79

I13

properties of, 439–52 domain and range, 439–40 even-odd, 448–49 periodic, 441–42 of quadrantal angles, 425–27 right triangle trigonometry, 564–66 signs of, in a given quadrant, 443 sine and cosine graphs, 452–67 amplitude and period, 456–58, 460, 476–79 equation for, 462 key points for, 458–61 sinusoidal curve fitting, 479–83 of t, 423 tangent and cotangent graphs, 468–71 of u, 425, 447 unit circle approach to finding exact values of, 422–38 Trigonometric identities, 523–30 basic, 524 establishing, 525–28 Double-angle Formulas for, 544–47 Sum and Difference Formulas for, 533–36 Even-Odd, 524 Pythagorean, 524 Quotient, 524 Reciprocal, 524 Trinomials, A23 factoring, A35, A37–A38 Truncation, A10 Turning points, 231–32 2 by 2 determinants, 680, 785 proof for, 791 Umbra versa, 433 Unbounded graphs, 835 Unbounded in positive direction, 265 Unbounded in the negative direction, 233–34 Uniform motion, A75–A77 Uninhibited growth, 381–83 Union of sets, A2 of two events, probabilities of, 913–15 Unit circle, 66, 422–38 Unit vector, 651, 654, 671 in direction of v, 673 Universal sets, 896, A2 Value (image) of function, 76, 78–81 Variable(s), A6, A22 complex, 259 dependent, 79 domain of, A7 independent, 79 in calculus, 453 Variable costs, 63 Variation, 138–43 combined, 139–40 direct, 138, 139–40 inverse, 139 joint, 139–40 Vector(s), 648–62 adding, 649, 652–53 algebraic, 651–52 angle between, 663–64 column, 799–800 components of, 651, 652, 671 horizontal and vertical, 652 cross product of, 679–81 decomposing, 665–66 defined, 648 difference of, 649 direction of, 648, 654–55 dot product of two, 662–63 equality of, 649, 652 finding, 654–55 force, 654–55 geometric, 648–49 graphing, 650–51

I14

Subject Index

Vector(s) (Continued) historical feature on, 658 magnitudes of, 648, 650–51, 652, 654–55 modeling with, 655–57 multiplying by numbers, 649–50 objects in static equilibrium, 656–57 orthogonal, 664–67 parallel, 664 in physics, 648 position, 651–52 row, 799–800 scalar multiples of, 650, 653, 663 in space, 670–79 angle between two vectors, 674–75 direction angles of, 675–77 distance between two points in space, 671 dot product, 673–74 operations on, 672–73 position vectors, 671–72 subtracting, 652–53 unit, 651, 654, 671 in direction of v, 673 velocity, 654–55 writing, 655 zero, 649 Vector product. See Cross (vector) product Vector projection, 665–66 Velocity, instantaneous, 946–48 Velocity vector, 654–55 Venn diagrams, A2 VERT command, 55 Vertex/vertices, 835 of cone, 692 of ellipse, 702 of hyperbola, 712 of parabola, 182, 693 of quadratic function, 183–87 of ray, 408 Vertical asymptote, 267–69, 321 multiplicity and, 268–69

Vertical component of vector, 652 Vertical line, 51, 627, 635 equation of, 54–55 Vertical-line test, 89 Vertically compressed or stretched graphs, 124 Vertical shifts, 121–24, 127 Viète, François, 590 Viewing angle, 507 Viewing rectangle, 35, B1–B3 setting, B1–B2 Vinculum, A60 Volume, formulas for, A15 Wallis, John, 646 Waves, traveling speeds of, 522 Weight, 615 Whispering galleries, 708 Wings, airplane, 615 Work, 678 dot product to compute, 667 World population, 851, 923, 960 x-axis, 34 projection of P on the, 601 reflections about, 126 symmetry with respect to, 44–45 x-coordinate, 34 x-intercept(s), 43 polynomial graphed using, 229, 230–31 of quadratic function, 175, 184. See also Zeros: of quadratic function xy-plane, 34, 670–71 xz-plane, 670–71 Yang Hui, 888 y-axis, 34 projection of P on the, 601 reflections about, 126 symmetry with respect to, 44–45

y-coordinate (ordinate), 34 y-intercept, 43, 57 from linear equation, 57 yz-plane, 670–71 Zero-coupon bonds, 380 Zero-level earthquake, 355 Zero matrix, 798 Zero polynomial, A23 Zero-Product Property, A4 Zeros bounds on, 252–53 complex, of polynomials, 258–64 Conjugate Pairs Theorem, 259–60 defined, 259 finding, 261–62 polynomial function with specified zeros, 260–61 complex, of quadratic function, 207–9 of function, 91–92 real, 176–77 of linear function, 155 of quadratic function completing the square to find, 172 factoring to find, 170 quadratic formula to find, 173–75 Square Root Method to find, 170–71 real, of polynomials, 229–35, 244–58 approximating, 254 Descartes’ Rule of Signs for, 247–48 finding, 250–51 Intermediate Value Theorem, 253–54 number of, 247–48 Rational Zeros Theorem, 248–49, 262 Remainder Theorem and Factor Theorem, 245–47 repeated, 230 theorem for bounds on, 252–53 Zero vector, 649