Algebra and Trigonometry, Complete Solutions Manual [4 ed.]
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CONTENTS ¥
CHAPTER
PROLOGUE: Principles of Problem Solving 1
P PREREQUISITES P.1
Modeling the Real World with Algebra 3
P.2 P.3
Real Numbers 4 Integer Exponents and Scientific Notation 9
P.4
Rational Exponents and Radicals 14
P.5
Algebraic Expressions 18
P.6
Factoring 22
P.7
Rational Expressions 27
P.8
Solving Basic Equations 34
P.9
Modeling with Equations 39
3
Chapter P Review 45 Chapter P Test 51 ¥
CHAPTER
FOCUS ON MODELING: Making the Best Decisions 54
1 EQUATIONS AND GRAPHS 1.1 1.2
The Coordinate Plane 57 Graphs of Equations in Two Variables; Circles 65
1.3 1.4
Lines 79 Solving Quadratic Equations 90
1.5
Complex Numbers 98
1.6
Solving Other Types of Equations 101
1.7
Solving Inequalities 110
1.8
Solving Absolute Value Equations and Inequalities 129
1.9
Solving Equations and Inequalities Graphically 131
1.10
57
Modeling Variation 139 Chapter 1 Review 143 Chapter 1 Test 161 iii
iv
Contents
¥
CHAPTER
FOCUS ON MODELING: Fitting Lines to Data 165
2 FUNCTIONS 2.1 2.2
Functions 169 Graphs of Functions 178
2.3
Getting Information from the Graph of a Function 190
2.4
Average Rate of Change of a Function 201
2.5 2.6 2.7
Linear Functions and Models 206 Transformations of Functions 212 Combining Functions 226
2.8
One-to-One Functions and Their Inverses 234 Chapter 2 Review 243
169
Chapter 2 Test 255 ¥
CHAPTER
FOCUS ON MODELING: Modeling with Functions 259
3 POLYNOMIAL AND RATIONAL FUNCTIONS 3.1
Quadratic Functions and Models 267
3.2
Polynomial Functions and Their Graphs 276
3.3
Dividing Polynomials 291
3.4
Real Zeros of Polynomials 301
3.5
Complex Zeros and the Fundamental Theorem of Algebra 334
3.6
Rational Functions 344 Chapter 3 Review 377
267
Chapter 3 Test 395 ¥
CHAPTER
FOCUS ON MODELING: Fitting Polynomial Curves to Data 398
4 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 4.1
Exponential Functions 401
4.2
The Natural Exponential Function 409
4.3
Logarithmic Functions 414
4.4
Laws of Logarithms 422
4.5
Exponential and Logarithmic Equations 426
401
Contents
4.6
Modeling with Exponential Functions 433
4.7
Logarithmic Scales 438 Chapter 4 Review 440 Chapter 4 Test 448
¥
CHAPTER
FOCUS ON MODELING: Fitting Exponential and Power Curves to Data 450
5 TRIGONOMETRIC FUNCTIONS: RIGHT TRIANGLE APPROACH 5.1
Angle Measure 455
5.2
Trigonometry of Right Triangles 459
5.3
Trigonometric Functions of Angles 464
5.4
Inverse Trigonometric Functions and Right Triangles 468
5.5 5.6
The Law of Sines 471 The Law of Cosines 476 Chapter 5 Review 481
455
Chapter 5 Test 486 ¥
CHAPTER
FOCUS ON MODELING: Surveying 536
6 TRIGONOMETRIC FUNCTIONS: UNIT CIRCLE APPROACH 6.1 6.2
The Unit Circle 491 Trigonometric Functions of Real Numbers 495
6.3
Trigonometric Graphs 500
6.4
More Trigonometric Graphs 511
6.5
Inverse Trigonometric Functions and Their Graphs 519
6.6
Modeling Harmonic Motion 521
491
Chapter 6 Review 527 Chapter 6 Test 534 ¥
CHAPTER
FOCUS ON MODELING: Fitting Sinusoidal Curves to Data 487
7 ANALYTIC TRIGONOMETRY 7.1
Trigonometric Identities 541
7.2 7.3
Addition and Subtraction Formulas 549 Double-Angle, Half-Angle, and Product-Sum Formulas 556
541
v
vi
Contents
7.4
Basic Trigonometric Equations 567
7.5
More Trigonometric Equations 571 Chapter 7 Review 578 Chapter 7 Test 584
¥
CHAPTER
FOCUS ON MODELING: Traveling and Standing Waves 586
8 POLAR COORDINATES AND PARAMETRIC EQUATIONS 8.1 8.2
Polar Coordinates 589 Graphs of Polar Equations 593
8.3
Polar Form of Complex Numbers; De Moivre’s Theorem 600
8.4
Plane Curves and Parametric Equations 612
589
Chapter 8 Review 623 Chapter 8 Test 630 ¥
CHAPTER
FOCUS ON MODELING: The Path of a Projectile 631
9 VECTORS IN TWO AND THREE DIMENSIONS 9.1 9.2 9.3
Vectors in Two Dimensions 635 The Dot Product 641 Three-Dimensional Coordinate Geometry 644
9.4 9.5 9.6
Vectors in Three Dimensions 646 The Cross Product 649 Equations of Lines and Planes 652
635
Chapter 9 Review 654 Chapter 9 Test 658 ¥
CHAPTER
FOCUS ON MODELING: Vector Fields 659
10 SYSTEMS OF EQUATIONS AND INEQUALITIES 10.1
Systems of Linear Equations in Two Variables 663
10.2
Systems of Linear Equations in Several Variables 670
10.3 10.4
Partial Fractions 678 Systems of Nonlinear Equations 689
10.5
Systems of Inequalities 696
663
Contents
vii
Chapter 10 Review 709 Chapter 10 Test 717 ¥
CHAPTER
FOCUS ON MODELING: Linear Programming 720
11 MATRICES AND DETERMINANTS 11.1
Matrices and Systems of Linear Equations 729
11.2
The Algebra of Matrices 740
11.3
Inverses of Matrices and Matrix Equations 748
11.4
Determinants and Cramer’s Rule 758 Chapter 11 Review 772
729
Chapter 11 Test 782 ¥
CHAPTER
FOCUS ON MODELING: Computer Graphics 785
12 CONIC SECTIONS 12.1 12.2
Parabolas 789 Ellipses 794
12.3
Hyperbolas 803
12.4 12.5 12.6
Shifted Conics 810 Rotation of Axes 822 Polar Equations of Conics 834
789
Chapter 12 Review 842 Chapter 12 Test 856 ¥
CHAPTER
FOCUS ON MODELING: Conics in Architecture 858
13 SEQUENCES AND SERIES 13.1
Sequences and Summation Notation 861
13.2
Arithmetic Sequences 866
13.3
Geometric Sequences 871
13.4 13.5 13.6
Mathematics of Finance 879 Mathematical Induction 883 The Binomial Theorem 892 Chapter 13 Review 896
861
viii
Contents
Chapter 13 Test 903 ¥
CHAPTER
FOCUS ON MODELING: Modeling with Recursive Sequences 904
14 COUNTING AND PROBABILITY 14.1
Counting 907
14.2
Probability 914
14.3
Binomial Probability 922
14.4
Expected Value 927
907
Chapter 14 Review 929 Chapter 14 Test 935 ¥
FOCUS ON MODELING: The Monte Carlo Method 936
APPENDIXES A
Geometry Review 939
B
Calculations and Significant Figures 940
C
Graphing with a Graphing Calculator 941
939
PROLOGUE: Principles of Problem Solving 1 1 distance ; the ascent takes h, the descent takes h, and the rate 15 r 1 1 1 1 1 2 h. Thus we have 0, which is impossible. So the car cannot go total trip should take 30 15 15 r 15 r fast enough to average 30 mi/h for the 2-mile trip.
1. Let r be the rate of the descent. We use the formula time
2. Let us start with a given price P. After a discount of 40%, the price decreases to 06P. After a discount of 20%, the price decreases to 08P, and after another 20% discount, it becomes 08 08P 064P. Since 06P 064P, a 40% discount is better. 3. We continue the pattern. Three parallel cuts produce 10 pieces. Thus, each new cut produces an additional 3 pieces. Since the first cut produces 4 pieces, we get the formula f n 4 3 n 1, n 1. Since f 142 4 3 141 427, we see that 142 parallel cuts produce 427 pieces. 4. By placing two amoebas into the vessel, we skip the first simple division which took 3 minutes. Thus when we place two amoebas into the vessel, it will take 60 3 57 minutes for the vessel to be full of amoebas. 5. The statement is false. Here is one particular counterexample:
First half Second half Entire season
Player A
Player B
1 1 hit in 99 at-bats: average 99 1 hit in 1 at-bat: average 11
0 hit in 1 at-bat: average 01
2 2 hits in 100 at-bats: average 100
98 hits in 99 at-bats: average 98 99
99 99 hits in 100 at-bats: average 100
6. Method 1: After the exchanges, the volume of liquid in the pitcher and in the cup is the same as it was to begin with. Thus, any coffee in the pitcher of cream must be replacing an equal amount of cream that has ended up in the coffee cup. Method 2: Alternatively, look at the drawing of the spoonful of coffee and cream
cream
mixture being returned to the pitcher of cream. Suppose it is possible to separate the cream and the coffee, as shown. Then you can see that the coffee going into the
coffee
cream occupies the same volume as the cream that was left in the coffee. Method 3 (an algebraic approach): Suppose the cup of coffee has y spoonfuls of coffee. When one spoonful of cream 1 coffee y cream and . is added to the coffee cup, the resulting mixture has the following ratios: mixture y1 mixture y1 1 So, when we remove a spoonful of the mixture and put it into the pitcher of cream, we are really removing of a y1 y spoonful of coffee. Thus the amount of cream left in the mixture (cream in the coffee) is spoonful of cream and y 1 1 y 1 of a spoonful. This is the same as the amount of coffee we added to the cream. y 1 y1 7. Let r be the radius of the earth in feet. Then the circumference (length of the ribbon) is 2r. When we increase the radius by 1 foot, the new radius is r 1, so the new circumference is 2 r 1. Thus you need 2 r 1 2r 2 extra feet of ribbon. 1
2
Principles of Problem Solving
8. The north pole is such a point. And there are others: Consider a point a1 near the south pole such that the parallel passing through a1 forms a circle C1 with circumference exactly one mile. Any point P1 exactly one mile north of the circle C1 along a meridian is a point satisfying the conditions in the problem: starting at P1 she walks one mile south to the point a1 on the circle C1 , then one mile east along C1 returning to the point a1 , then north for one mile to P1 . That’s not all. If a point a2 (or a3 , a4 , a5 , ) is chosen near the south pole so that the parallel passing through it forms a circle C2 (C3 , C4 , C5 , ) with a circumference of exactly 12 mile ( 13 mi, 14 mi, 15 mi, ), then the point P2 (P3 , P4 , P5 , ) one mile north of a2 (a3 , a4 , a5 , ) along a meridian satisfies the conditions of the problem: she walks one mile south from P2 (P3 , P4 , P5 , ) arriving at a2 ( a3 , a4 , a5 , ) along the circle C2 (C3 , C4 , C5 , ), walks east along the circle for one mile thus traversing the circle twice (three times, four times, five times, ) returning to a2 (a3 , a4 , a5 , ), and then walks north one mile to P2 ( P3 , P4 , P5 , ).
P
PREREQUISITES
P.1
MODELING THE REAL WORLD WITH ALGEBRA
1. Using this model, we find that if S 12, L 4S 4 12 48. Thus, 12 sheep have 48 legs. 2. If each gallon of gas costs $350, then x gallons of gas costs $35x. Thus, C 35x. 3. If x $120 and T 006x, then T 006 120 72. The sales tax is $720.
4. If x 62,000 and T 0005x, then T 0005 62,000 310. The wage tax is $310.
5. If 70, t 35, and d t, then d 70 35 245. The car has traveled 245 miles. 6. V r 2 h 32 5 45 1414 in3 240 N 30 miles/gallon G 8 175 175 G 7 gallons (b) 25 G 25 9. (a) V 95S 95 4 km3 38 km3
8. (a) T 70 0003h 70 0003 1500 655 F
7. (a) M
(b) 64 70 0003h 0003h 6 h 2000 ft
10. (a) P 006s 3 006 123 1037 hp
(b) 19 km3 95S S 2 km3
11. (a) Depth (ft)
(b) 75 006s 3 s 3 125 so s 5 knots
Pressure (lb/in2 ) 045 0 147 147
0
045 10 147 192
10
045 20 147 237
20
(b) We know that P 30 and we want to find d, so we solve the equation 30 147 045d 153 045d
153 340. Thus, if the pressure is 30 lb/in2 , the depth 045 is 34 ft. d
045 30 147 282
30
045 40 147 327
40
045 50 147 372
50
045 60 147 417
60 12. (a) Population
(b) We solve the equation 40x 120,000
Water use (gal)
0
0
1000
40 1000 40,000
2000 3000 4000 5000
x
120,000 3000. Thus, the population is about 3000. 40
40 2000 80,000
40 3000 120,000 40 4000 160,000 40 5000 200,000
13. The number N of cents in q quarters is N 25q. ab 14. The average A of two numbers, a and b, is A . 2 15. The cost C of purchasing x gallons of gas at $350 a gallon is C 35x.
16. The amount T of a 15% tip on a restaurant bill of x dollars is T 015x. 17. The distance d in miles that a car travels in t hours at 60 mi/h is d 60t.
3
4
CHAPTER P Prerequisites
18. The speed r of a boat that travels d miles in 3 hours is r
d . 3
19. (a) $12 3 $1 $12 $3 $15
(b) The cost C, in dollars, of a pizza with n toppings is C 12 n.
(c) Using the model C 12 n with C 16, we get 16 12 n n 4. So the pizza has four toppings.
20. (a) 3 30 280 010 90 28 $118 daily days cost miles (b) The cost is , so C 30n 01m. rental rented per mile driven (c) We have C 140 and n 3. Substituting, we get 140 30 3 01m 140 90 01m 50 01m m 500. So the rental was driven 500 miles. 21. (a) (i) For an all-electric car, the energy cost of driving x miles is Ce 004x.
(ii) For an average gasoline powered car, the energy cost of driving x miles is C g 012x.
(b) (i) The cost of driving 10,000 miles with an all-electric car is Ce 004 10,000 $400.
(ii) The cost of driving 10,000 miles with a gasoline powered car is C g 012 10,000 $1200.
22. (a) If the width is 20, then the length is 40, so the volume is 20 20 40 16,000 in3 . (b) In terms of width, V x x 2x 2x 3 .
4a 3b 2c d 4a 3b 2c 1d 0 f . abcd f abcd f (b) Using a 2 3 6, b 4, c 3 3 9, and d f 0 in the formula from part (a), we find the GPA to be 463429 54 284. 649 19
23. (a) The GPA is
P.2
THE REAL NUMBERS
1. (a) The natural numbers are 1 2 3 .
(b) The numbers 3 2 1 0 are integers but not natural numbers. p 5 , 1729 . (c) Any irreducible fraction with q 1 is rational but is not an integer. Examples: 32 , 12 23 q p (d) Any number which cannot be expressed as a ratio of two integers is irrational. Examples are 2, 3, , and e. q
2. (a) ab ba; Commutative Property of Multiplication
(b) a b c a b c; Associative Property of Addition (c) a b c ab ac; Distributive Property
3. The set of numbers between but not including 2 and 7 can be written as (a) x 2 x 7 in interval notation, or (b) 2 7 in interval notation. 4. The symbol x stands for the absolute value of the number x. If x is not 0, then the sign of x is always positive.
5. The distance between a and b on the real line is d a b b a. So the distance between 5 and 2 is 2 5 7. a c ad bc . b d bd (b) No, the sum of two irrational numbers can be irrational ( 2) or rational ( 0).
6. (a) Yes, the sum of two rational numbers is rational:
7. (a) No: a b b a b a in general.
(b) No; by the Distributive Property, 2 a 5 2a 2 5 2a 10 2a 10.
8. (a) Yes, absolute values (such as the distance between two different numbers) are always positive. (b) Yes, b a a b.
SECTION P.2 The Real Numbers
10. (a) Natural number: 16 4 (b) Integers: 500, 16, 20 5 4
9. (a) Natural number: 100 (b) Integers: 0, 100, 8 (c) Rational numbers: 15, 0, 52 , 271, 314, 100, 8 (d) Irrational numbers: 7,
(c) Rational numbers: 13, 13333 , 534, 500, 1 23 , 20 16, 246 579 , 5 (d) Irrational number: 5
11. Commutative Property of addition
12. Commutative Property of multiplication
13. Associative Property of addition
14. Distributive Property
15. Distributive Property
16. Distributive Property
17. Commutative Property of multiplication
18. Distributive Property
19. x 3 3 x
20. 7 3x 7 3 x
21. 4 A B 4A 4B
22. 5x 5y 5 x y
23. 3 x y 3x 3y
24. a b 8 8a 8b 26. 43 6y 43 6 y 8y
25. 4 2m 4 2 m 8m 27. 52 2x 4y 52 2x 52 4y 5x 10y 3 4 9 8 17 29. (a) 10 15 30 30 30 5 4 9 (b) 14 15 20 20 20
28. 3a b c 2d 3ab 3ac 6ad
9 1 30. (a) 23 35 10 15 15 15 15 4 35 (b) 1 58 16 24 24 24 24 24
31. (a) 23 6 32 23 6 23 32 4 1 3 1 5 4 13 1 13 (b) 3 14 1 45 12 4 4 5 5 4 5 20
2 32. (a) 2 3 2 32 23 12 3 13 93 13 83 2
33. (a) 2 3 6 and 2 72 7, so 3 72
34. (a) 3 23 2 and 3 067 201, so 23 067
2
3 2 1 2 1 2 1 45 9 (b) 15 23 51 21 51 21 10 10 12 3 3 10 15 10 5 10 5
(b) 6 7
(b) 23 067
(c) 35 72
(c) 067 067
35. (a) False
36. (a) False:
(b) True
(b) False
3 173205 17325.
37. (a) True
(b) False
38. (a) True
(b) True
39. (a) x 0
(b) t 4
40. (a) y 0
(b) z 1
(c) a
(d) 5 x 13
(e) p 3 5 41. (a) A B 1 2 3 4 5 6 7 8 (b) A B 2 4 6
5
(c) b 8
(d) 0 17
(e) y 2 42. (a) B C 2 4 6 7 8 9 10 (b) B C 8
6
CHAPTER P Prerequisites
43. (a) A C 1 2 3 4 5 6 7 8 9 10
44. (a) A B C 1 2 3 4 5 6 7 8 9 10 (b) A B C ∅
(b) A C 7
46. (a) A C x 1 x 5
45. (a) B C x x 5 (b) B C x 1 x 4
(b) A B x 2 x 4 48. 2 8] x 2 x 8
47. 3 0 x 3 x 0 _3
0
8
_6
2
1
53. x 1 x 1]
54. 1 x 2 x [1 2] 1
1
55. 2 x 1 x 2 1]
_5
1
58. 5 x 2 x 5 2
_1
_5
(b) 3 5]
1
63. [4 6] [0 8 [0 6] 0
(b) 2 0]
60. (a) [0 2
_1
0
64. [4 6] [0 8 [4 8 6
65. 4 4 _4
2
62. 2 0 1 1 0
61. 2 0 1 1 2 1 _2
2
56. x 5 x [5
57. x 1 x 1
59. (a) [3 5]
1
_ _2
52. 1 x x 1
51. [2 x x 2
_2
8
50. 6 12 x 6 x 12
49. [2 8 x 2 x 8 2
2
_4
8
66. 6] 2 10 2 6] 4
2
6
SECTION P.2 The Real Numbers
67. (a) 100 100 (b) 73 73 69. (a) 6 4 6 4 2 2
68. (a) 5 5 5 5 5 5, since 5 5. (b) 10 10 , since 10 .
70. (a) 2 12 2 12 10 10
1 1 1 (b) 1 1
71. (a) 2 6 12 12 (b) 13 15 5 5
73. 2 3 5 5 75. (a) 17 2 15
(b) 21 3 21 3 24 24 3 12 55 67 67 (c) 10 11 8 40 40 40 40
(b) 1 1 1 1 1 1 1 0 1 1 1 72. (a) 6 24 4 4 5 (b) 712 127 5 1 1 74. 25 15 4 4
7 1 49 5 54 18 18 76. (a) 15 21 105 105 105 35 35 (b) 38 57 38 57 19 19.
(c) 26 18 26 18 08 08.
77. (a) Let x 0777 . So 10x 77777 x 07777 9x 7. Thus, x 79 .
13 (b) Let x 02888 . So 100x 288888 10x 28888 90x 26. Thus, x 26 90 45 . 19 (c) Let x 0575757 . So 100x 575757 x 05757 99x 57. Thus, x 57 99 33 .
78. (a) Let x 52323 . So 100x 5232323 1x 52323 99x 518. Thus, x 518 99 .
62 (b) Let x 13777 . So 100x 1377777 10x 137777 90x 124. Thus, x 124 90 45 .
1057 (c) Let x 213535 . So 1000x 21353535 10x 213535 990x 2114. Thus, x 2114 990 495 .
2 1, so 1 2 2 1.
79. 3, so 3 3.
80.
81. a b, so a b a b b a.
82. a b a b a b b a 2b
83. (a) a is negative because a is positive.
(b) bc is positive because the product of two negative numbers is positive. (c) a ba b is positive because it is the sum of two positive numbers.
(d) ab ac is negative: each summand is the product of a positive number and a negative number, and the sum of two negative numbers is negative. 84. (a) b is positive because b is negative.
(b) a bc is positive because it is the sum of two positive numbers.
(c) c a c a is negative because c and a are both negative. (d) ab2 is positive because both a and b2 are positive.
85. Distributive Property
7
8
CHAPTER P Prerequisites
86. Day
TO
TG
TO TG
TO TG
Sunday
68
77
9
9
Monday
72
75
3
3
Tuesday
74
74
0
0
Wednesday
80
75
5
5
Thursday
77
69
8
8
Friday
71
70
1
1
Saturday
70
71
1
1
TO TG gives more information because it tells us which city had the higher temperature. 87. (a) When L 60, x 8, and y 6, we have L 2 x y 60 2 8 6 60 28 88. Because 88 108 the post office will accept this package. When L 48, x 24, and y 24, we have L 2 x y 48 2 24 24 48 96 144, and since 144 108, the post office will not accept this package. (b) If x y 9, then L 2 9 9 108 L 36 108 L 72. So the length can be as long as 72 in. 6 ft.
m2 m1 m m1n2 m2n1 m1 and y be rational numbers. Then x y 2 , n1 n2 n1 n2 n1 n2 m m n m 2 n1 m m m m m , and x y 1 2 1 2 . This shows that the sum, difference, and product xy 1 2 1 2 n1 n2 n1 n2 n1 n2 n1n2 of two rational numbers are again rational numbers. However the product of two irrational numbers is not necessarily irrational; for example, 2 2 2, which is rational. Also, the sum of two irrational numbers is not necessarily irrational; for example, 2 2 0 which is rational.
88. Let x
89. 12 2 is irrational. If it were rational, then by Exercise 6(a), the sum 12 2 12 2 would be rational, but
this is not the case. Similarly, 12 2 is irrational. (a) Following the hint, suppose that r t q, a rational number. Then by Exercise 6(a), the sum of the two rational numbers r t and r is rational. But r t r t, which we know to be irrational. This is a contradiction, and hence our original premise—that r t is rational—was false. a (b) r is rational, so r for some integers a and b. Let us assume that rt q, a rational number. Then by definition, b c a c bc q for some integers c and d. But then rt q t , whence t , implying that t is rational. Once again d b d ad we have arrived at a contradiction, and we conclude that the product of a rational number and an irrational number is irrational.
90. x
1
2
10
100
1000
1 x
1
1 2
1 10
1 100
1 1000
As x gets large, the fraction 1x gets small. Mathematically, we say that 1x goes to zero. x 1 x
1
05
01
001
0001
1
1 05 2
1 01 10
1 001 100
1 0001 1000
As x gets small, the fraction 1x gets large. Mathematically, we say that 1x goes to infinity.
SECTION P.3 Integer Exponents and Scientific Notation
91. (a) Construct the number
2 on the number line by transferring
Ï2
the length of the hypotenuse of a right triangle with legs of length 1 and 1.
_1
0
1
(b) Construct a right triangle with legs of length 1 and 2. By the Pythagorean Theorem, the length of the hypotenuse is 12 22 5. Then transfer the length of the hypotenuse to the number line.
(c) Construct a right triangle with legs of length 2 and 2 [construct 2 as in part (a)]. By the Pythagorean Theorem, 2 the length of the hypotenuse is 2 22 6. Then transfer the length of the hypotenuse to the number line.
1 Ï2
Ï5
_1
0
2
3
2 Ï5
3
2
3
1
1 Ï6 Ï2
_1
0
Ï2
1 1
Ï2
Ï6
92. (a) Subtraction is not commutative. For example, 5 1 1 5. (b) Division is not commutative. For example, 5 1 1 5.
(c) Putting on your socks and putting on your shoes are not commutative. If you put on your socks first, then your shoes, the result is not the same as if you proceed the other way around. (d) Putting on your hat and putting on your coat are commutative. They can be done in either order, with the same result. (e) Washing laundry and drying it are not commutative. (f) Answers will vary. (g) Answers will vary. 93. Answers will vary. 94. (a) If x 2 and y 3, then x y 2 3 5 5 and x y 2 3 5. If x 2 and y 3, then x y 5 5 and x y 5. If x 2 and y 3, then x y 2 3 1 and x y 5. In each case, x y x y and the Triangle Inequality is satisfied. (b) Case 0: If either x or y is 0, the result is equality, trivially. xy Case 1: If x and y have the same sign, then x y x y
if x and y are positive x y. if x and y are negative
Case 2: If x and y have opposite signs, then suppose without loss of generality that x 0 and y 0. Then x y x y x y.
P.3
INTEGER EXPONENTS AND SCIENTIFIC NOTATION
1. Using exponential notation we can write the product 5 5 5 5 5 5 as 56 .
2. Yes, there is a difference: 54 5 5 5 5 625, while 54 5 5 5 5 625. 3. In the expression 34 , the number 3 is called the base and the number 4 is called the exponent.
4. When we multiply two powers with the same base, we add the exponents. So 34 35 39 . 35 5. When we divide two powers with the same base, we subtract the exponents. So 2 33 . 3 2 6. When we raise a power to a new power, we multiply the exponents. So 34 38 .
9
10
CHAPTER P Prerequisites
1 7. (a) 21 2
(b) 23
1 8
(c)
1 1 2 2
1 (d) 3 23 8 2
8. Scientists express very large or very small numbers using scientific notation. In scientific notation, 8,300,000 is 83 106
and 00000327 is 327 105 . 2 2 9 3 2 . 9. (a) No, 3 2 4 3 10. (a) No, x 2 x 23 x 6 .
(b) Yes, 54 625 and 54 54 625. 3 3 (b) No, 2x 4 23 x 4 8x 12 .
11. (a) 26 64
(b) 26 64
12. (a) 53 125
(b) 53 125
13. (a)
0 1 5 21 3 2
(b)
23 1 1 3 8 30 2
1 1 14. (a) 23 20 3 8 2
(b) 23 20 23 8
15. (a) 53 5 54 625
(b) 32 30 32 9
16. (a) 38 35 313 1,594,323
(b) 60 6 6
17. (a) 54 52 52 25
(b)
1 1 18. (a) 33 31 34 4 81 3
(b)
19. (a) x 2 x 3 x 23 x 5 20. (a) y 5 y 2 y 52 y 7
107 103 1000 104
54 53 125 5 3 (b) x 2 13 x 23 x 6
2 12 33 27 1 33 5 25 52 2 2 2 2 5 2 (c) 52 4 5 52 2 1 (c) 42 16 4 33 27 2 3 (c) 3 3 8 2 3 (c) 22 26 64 (c)
2 (c) 54 58 390,625 32 1 1 (c) 4 2 9 3 3
72 1 1 (c) 5 3 343 7 7 (c) t 3 t 5 t 35 t 2
(b) 8x2 82 x 2 64x 2
(c) x 4 x 3 x 43 x
1 21. (a) x 5 x 3 x 53 x 2 2 x y 10 y 0 (c) y 1007 y 3 y7
(b) 2 4 5 245 1
1
1 22. (a) y 2 y 5 y 25 y 3 3 y
1 x6 1 (b) z 5 z 3 z 4 z 534 z 2 2 (c) 10 x 610 4 z x x 3 3 3 (b) a 2 a 4 a 24 a 6 a 63 a 18
a 9 a 2 a 921 a 6 a (c) 2x2 5x 6 22 x 2 5x 6 20x 26 20x 8
23. (a)
4 4 4 z2 z4 z 24 z6 31 2 z 62 z 4 (b) 2a 3 a 2 2a 32 2a 5 24 a 54 16a 20 3 1 z z z z 3 2z 3 33 z 23 2z 3 54z 63 54z 9 (c) 3z 2 25. (a) 3x 2 y 2x 3 3 2x 23 y 6x 5 y (b) 2a 2 b1 3a 2 b2 2 3a 22 b12 6b 24. (a)
SECTION P.3 Integer Exponents and Scientific Notation
(c)
2 4y 2 x 4 y 4y 2 x 42 y 2 4x 8 y 22 4x 8 y 4
4x 3 y 2 7y 5 4 7x 3 y 25 28x 3 y 7 (b) 9y 2 z 2 3y 3 z 9 3y 23 z 21 27yz 3
26. (a)
(c)
27. (a)
2 8x 7 y 2 22 8x 7 y 2 32x 7 y 2 8x 7 y 2 12 x 3 y 6 2 32x 76 y 22 32x 2 32 2 x y x y 1 x3 y 2 2 2x 2 y 3 3y 22 x 22 y 32 3y 12x 4 y 7
x 2 y 1 x7 x 25 y 1 x 7 y 1 5 y x 3 x 23 y 3 x 6 y3 x2 y (c) 3 3 27 3
(b)
28. (a)
2 5x 4 y 3 8x 3 5x 4 y 3 82 x 32 5 82 x 46 y 3 320x 2 y 3
y 2 z 3 y 1 2 3 3 1 y y z yz 2 3 2 6 4 a b a6 a b (c) b3 b6 b10 (b)
1
29. (a)
x 3 y3
30. (a)
x 2 y 4
1 3 3 x y 3 2 b6 a3 (b) a 2 b2 a 23 b23 a 32 a 6 b6 a 6 12 a 2 3 2 3 8 2y x (c) x 22 y 22 23 y 33 x 23 x 4 y 4 8y 9 x 6 8x 46 y 49 10 13 y 2 x2 x y 3
y4 x2
3
x2 y4
3
x6 12 y
1 3 x9 2x 3 y 4 y 2 23 x 33 y 43 14 8y 3 2 1 1 b 2a 1 (c) 23 a 13 b23 b12 22 a 22 2 2 b 2a 32ab8 (b)
y2
3x 2 y 5 x y3 3 9x 3 y 2 2 2 y 32 y6 2x 3 2x 3 y 1 (b) y2 y3 22 x 32 4x 6 1 2 x 4 y5 3x 3 y 1 (c) y 11 x 21 32 x 32 y 22 2 2 9 x y
31. (a)
1 a 3 b4 a2 32. (a) 2 5 1 12 21 a 35 b41 14 a 2 b3 3 2a b 4b
11
12
CHAPTER P Prerequisites
x2 y 5x 4
2
5x 4 x2 y
2
5x 2 y
2
25x 4 y2 1 1 y3 y 2 y2 2y 1 z 2 (c) 2 2 4 yz z 3z 9z 18z 3 (b)
b3 3a 1 31 a 1 b31 3 3a b 1 r 5 sq 8 s3 q 1r 1 s 2 1 1 2 q 81 r 51 s 12 7 4 (b) 5 8 r sq q r s q r
33. (a)
34. (a)
s 2 t 4 5s 1 t
(b)
x y 2 z 3 x 2 y 3 z 4
2
s 2212 t 4212 52
3
25t 10 s6
x 323 y 2333 z 3343
35. (a) 69,300,000 693 107
x 3 y 15 z3
36. (a) 129,540,000 12954 108
(b) 7,200,000,000,000 72 1012
(b) 7,259,000,000 7259 109
(c) 0000028536 28536 105
(c) 00000000014 14 109
(d) 00001213 1213 104
(d) 00007029 7029 104
37. (a) 319 105 319,000
38. (a) 71 1014 710,000,000,000,000
(b) 2721 108 272,100,000
(b) 6 1012 6,000,000,000,000
(c) 2670 108 000000002670
(c) 855 103 000855
(d) 9999 109 0000000009999
(d) 6257 1010 00000000006257
39. (a) 5,900,000,000,000 mi 59 1012 mi
(b) 00000000000004 cm 4 1013 cm
(c) 33 billion billion molecules 33 109 109 33 1019 molecules
40. (a) 93,000,000 mi 93 107 mi
(b) 0000000000000000000000053 g 53 1023 g
(c) 5,970,000,000,000,000,000,000,000 kg 597 1024 kg 41. 72 109 1806 1012 72 1806 109 1012 130 1021 13 1020 42. 1062 1024 861 1019 1062 861 1024 1019 914 1043
1295643 1295643 109 109176 01429 1019 1429 1019 43. 3610 2511 3610 1017 2511 106 731 10 16341 1028 731 16341 1028 731 16341 101289 63 1038 44. 00000000019 19 19 109 162 105 1582 102 162 1582 00000162 001582 105283 0074 1012 45. 594621 58 594621000 00058 594621 108 58 103 74 1014
SECTION P.3 Integer Exponents and Scientific Notation
9 3542 106 8774796 35429 1054 105448 319 104 10102 319 10106 46. 12 12 1048 27510376710 4 505 505 10 47. 1050 1010 1050 , whereas 10101 10100 10100 10 1 9 10100 1050 . So 1010 is closer to 1050 than 10100 is to 10101 .
48. (a) b5 is negative since a negative number raised to an odd power is negative.
(b) b10 is positive since a negative number raised to an even power is positive. (c) ab2 c3 we have positive negative2 negative3 positive positive negative which is negative. (d) Since b a is negative, b a3 negative3 which is negative.
(e) Since b a is negative, b a4 negative4 which is positive. (f)
a 3 c3 negative positive3 negative3 positive negative which is negative. 6 6 positive positive positive b c negative6 negative6
49. Since one light year is 59 1012 miles, Centauri is about 43 59 1012 254 1013 miles away or 25,400,000,000,000 miles away. 93 107 mi t st s 500 s 8 13 min. s 186 000 103 liters 3 14 2 133 1021 liters 51. Volume average depth area 37 10 m 36 10 m m3
50. 93 107 mi 186 000
52. Each person’s share is equal to
1674 1013 national debt $52,900. population 3164 108
53. The number of molecules is equal to liters molecules 602 1023 3 5 10 3 10 403 1027 volume 224 liters 224 m3 54. (a) BMI 703
W H2
Person
Weight
Height
Result
Brian
295 lb
5 ft 10 in. 70 in.
4232
obese
Linda
105 lb
5 ft 6 in. 66 in.
1695
underweight
Larry
220 lb
6 ft 4 in. 76 in.
2678
overweight
Helen
110 lb
5 ft 2 in. 62 in.
2012
normal
(b) Answers will vary. 55. Year
Total interest
1
$15208
2
30879
3
47026
4
63664
5
80808
56. Since 106 103 103 it would take 1000 days 274 years to spend the million dollars.
Since 109 103 106 it would take 106 1,000,000 days 273972 years to spend the billion dollars.
13
14
CHAPTER P Prerequisites
57. (a)
185 95
18 5 25 32 9
(b) 206 056 20 056 106 1,000,000 am 58. (a) We wish to prove that n a mn for positive integers m n. By definition, a m factors
a a a ak a a a . Thus, . Because m n, m n 0, so we can write an a a a am
k factors
n factors
n factors
mn factors
mn factors
am a a a a a a a a a a mn . an a a a 1 n factors
(b) We wish to prove that
a n b
an for positive integers m n. By definition, bn
n factors
a n b
a a a an a a a n. b b b b b b b n factors
n factors bn
a n
n . By definition, and using 59. (a) We wish to prove that b a a n n 1 1 b a n n n . a b a b bn 1 n a n bm a n a (b) We wish to prove that m n . By definition, m 1 b a b bm
P.4
the result from Exercise 58(b),
bm 1 bm n. n a 1 a
RATIONAL EXPONENTS AND RADICALS
1. Using exponential notation we can write 3 5 as 513 . 2. Using radicals we can write 512 as 5. 2 12 2 5212 5 and 5 512 5122 5. 3. No. 52 52 3 12 4. 412 23 8; 43 6412 8 5. Because the denominator is of the form
a, we multiply numerator and denominator by a: 1 1 3 33 . 3
3
3
6. 513 523 51 5 7. No. If a is negative, then 4a 2 2a. 8. No. For example, if a 2, then a 2 4 8 2 2, but a 2 0. 1 3 9. 312 10. 72 723 3 1 1 1 3 11. 423 42 3 16 12. 1032 1032 103 103 1 1 5 3 13. 5 535 14. 215 232 8 23
SECTION P.4 Rational Exponents and Radicals
15. a 25 17.
3
y 4 y 43
19. (a) (b) (c) 21. (a) (b) (c) 23. (a) (b) (c) 25. (a) (b) (c)
27. 29. 31. 33. 35. 37. 39. 40. 41. 42. 43. 44. 45. 46.
4
5 2 a
16 42 4 4 4 16 24 2 1 1 4 4 4 1 16 2 2 3 3 3 16 3 2 23 6 3 2 18 18 2 2 81 9 3 81 2 3 3 27 33 4 2 22 7 28 7 28 196 14 48 48 16 4 3 3 4 24 4 54 4 24 54 4 1296 6 216 216 36 6 6 6 3 2 3 32 3 64 4 1 1 1 4 1 4 1 4 4 4 64 256 4 256
1 1 16. 52 x 52 x x5 1 1 18. y 53 53 3 5 y y 2 20. (a) 64 8 8 3 (b) 3 64 43 4 5 (c) 5 32 25 2
3 22. (a) 2 3 81 2 3 33 6 3 3 2 3 12 3 22 (b) 5 5 25 3 2 18 2 32 (c) 2 49 7 7 24. (a) 12 24 12 24 288 2 122 12 2 54 54 (b) 93 6 6 (c) 3 15 3 75 3 15 75 3 1125 3 125 9 5 3 9 1 1 1 1 26. (a) 5 5 5 8 4 32 2 1 (b) 6 6 128 6 64 2 2 3 4 1 1 3 4 3 1 (c) 3 3 108 27 3 108 27 15 5 28. x 10 x 10 x2 3 3 3 30. 8a 5 23 a 3 a 2 2a a 2 13 32. 3 x 3 y 6 x 3 y 6 x y2 12 34. x 4 y 4 x 4 y 4 x 2 y2
x 4 x 5 32y 6 5 25 y 6 2 5 y 6 2y 5 y 4 4 16x 8 24 x 8 2x 2 13 3 3 x y x3 y 13 x 3 y 2 4 4 4 36r 2 t 4 6rt 2 6 r t 2 36. 48a 7 b4 24 a 4 b4 3a 3 2 ab 3a 3 13 3 4 64x 6 8 x 3 2 x 38. 4 x 4 y 2 z 2 x 4 4 y 2 z 2 x 4 y 2 z 2 32 18 16 2 9 2 42 2 32 2 4 2 3 2 7 2 75 48 25 3 16 3 52 3 42 3 5 3 4 3 9 3 125 45 25 5 9 5 52 5 32 5 5 5 3 5 2 5 3 3 3 54 3 16 2 33 23 2 3 3 2 2 3 2 3 2 9a 3 a 32 a 2 a a 3a a a 3a 1 a 2 16x x 5 42 x x 2 x 4 x x 2 x x 2 4 x 3 4 3 3 x 3 8x x 3 x 23 x x 3 x 2 3 x x 2 3 x 3 2y 4 3 2y 3 2y y 3 3 2y 3 2y y 3 3 2y y 1 3 2y
15
16
CHAPTER P Prerequisites
81x 2 81 81 x 2 1 81 x 2 1 9 x 2 1 48. 36x 2 36y 2 36 x 2 y 2 36 x 2 y 2 6 x 2 y 2
47.
49. (a) 1614 2
(b) 12513 5
50. (a) 2713 3
(b) 813 2
2 51. (a) 3225 3215 22 4
12 12 4 9 3 9 4 2 32 3 125 25 5 (b) 64 8 512
52. (a) 12523 52 25 53. (a) 523 513 52313 51 5
1 1 (c) 912 12 3 9 13 1 1 (c) 8 2 34 3 16 2 8 (c) 81 3 27
(b)
(c) 2743 34
335 (b) 25 33525 5 3 3
(c)
1 81
3 3 4 4133 4
10 723 1 1 54. (a) 327 3127 327127 32 9 (b) 53 72353 (c) 5 6 61510 7 36 7 55. When x 3, y 4, z 1 we have x 2 y 2 32 42 9 16 25 5. 4 56. When x 3, y 4, z 1 we have 4 x 3 14y 2z 4 33 14 4 2 1 4 27 56 2 4 81 34 3. 57. When x 3, y 4, z 1 we have
23 23 23 113 9x23 2y23 z 23 9 323 2 423 123 33 32 22 1 9 4 1 14.
1 . 58. When x 3, y 4, z 1 we have x y2z 3 42 1 122 144
59. (a) x 34 x 54 x 3454 x 2
(b) y 23 y 43 y 2343 y 2
60. (a) r 16r 56 r 1656 r
(b) a 35 a 310 a 35310 a 910
43 23 61. (a) 432313 53 13
(b)
x 34 x 74 62. (a) x 347454 x 54 x 54 23 823 a 623 b3223 4a 4 b 63. (a) 8a 6 b32 64. (a)
23 64a 6 b3 6423 a 623 b323 16a 4 b2
(b)
3 a 54 2a 34 a 14 2
2y 43
23 a 5234314 8a 134
y 23
y 73
4 4y 832373 4y 13 3 y
(b) 4a 6 b8 32 432 a 632 b832 8a 9 b12
34 (b) 168 z 32 1634 834 z 3234 86 z 98
23 13 1 1 8y 3 823 y 323 2 (b) u 4 6 u 413 613 43 2 4y u 35 x3 66. (a) x 5 y 13 x 535 y 1335 15 y 12 15 32t 54 (b) 4r 8 t 12 412r 812 t 1212 3215 t 5415 2r 4 t 14 12 t 14 r 4 t 0 r 4 65. (a)
67. (a)
x 23 y 12
x 2 y 3
16
x 23216 y 12316
1 x
SECTION P.4 Rational Exponents and Radicals
(b)
x 12 y 2 2y 14
4
813 y 3413313 z 613
(b)
8y 34 y3 z6
9
73.
x 2 y 8 24 y 1 2x 1 y 2 y 1 241 x 21 y 8121
13
71.
x 124 y 24 214 y 144 412 x 212 y 412 y 212
1614 x 814 y 4144314
68. (a)
x 8 y 4 16y 43
12
14
69.
4x 2 y 4 y2
x 3 x 32 x 5 x 59
6
y5
3
y 2 y 56 y 23 y 5623 y 32
75. 5 3 x 2 4 x 5 2x 1314 10x 712 4 7 x 4 77. x4 x 4 3 x 16u 3 16u 2 4u 79. 2 u 5 4 xy x 14 y 14 81. 1614 x 1214 y 1214 4 2 16x y 13 83. 3 y y y 112 y 3213 y 12 1 6 6 1 85. (a) 6 6 6 6 3 3 2 6 (b) 2 2 2 2 9 234 948 9 14 32 (c) 4 2 2 2 2 1 5x 5x 1 87. (a) 5x 5x 5x 5x x x 5 5x (b) 5 5 5 5 25 1 x x 25 1 (c) 5 3 35 25 x x x x 3 2 3 2 1 1 x x 89. (a) 3 3 x x x 3 x2 6 6 1 1 x x (b) 6 5 6 5 6 x x x x 7 4 7 4 1 x x 1 (c) 7 3 7 3 7 4 x x x x
70.
x y4 8
2y 43 x2 y 34 z 2 2
x 5 x 52
1 1 72. 35 x 35 5 3 x x 4 74. b3 b b3412 b54 3 2 76. 2 a a 2a 1223 2a 76 3 8x 2 78. 2x 23 x 12 2x 16 2 6 x x 2 4 3 3y 3 27y 3 54x y 80. x 2x 5 y x3 a3b 82. a 3234 b1224 a 34 4 3 2 a b 12 84. s s 3 s 132 s 54 12 3 12 12 3 86. (a) 4 3 3 3 3 3 12 12 5 60 2 15 (b) 5 5 5 5 5 8 513 835 8 23 13 (c) 3 2 5 5 5 5 s s 3t 3st 88. (a) 3t 3t 3t 3t a a b23 ab23 (b) 6 2 b b13 b23 b 25 25 1 1 c c (c) 35 35 25 c c c c 3 3 1 1 x x 90. (a) 3 2 3 2 3 x x x x 4 4 1 1 x x (b) 4 3 4 3 4 x x x x
3 2 x 1 1 1 1 (c) 3 4 3 3 3 2 3 3 x x x x x x x x 3 2 3 2 x x 2 3 x x x3
17
18
CHAPTER P Prerequisites
91. (a) Since 12 13 , 212 213 . 12 13 12 13 (b) 12 212 and 12 213 . Since 12 13 , we have 12 12 . 112 112 92. (a) We find a common root: 714 7312 73 343112 ; 413 4412 44 256112 . So 714 413 . 16 16 2516 ; 3 312 336 33 2716 . So (b) We find a common root: 3 5 513 526 52 3 5 3. 1 mile 0215 mi. Thus the distance you can see is given 93. First convert 1135 feet to miles. This gives 1135 ft 1135 5280 feet by D 2r h h 2 2 3960 0215 02152 17028 413 miles. 94. (a) Using f 04 and substituting d 65, we obtain s 30 f d 30 04 65 28 mi/h. (b) Using f 05 and substituting s 50, we find d. This gives s 30 f d 50 30 05 d 50 15d 2500 15d d 500 3 167 feet.
95. (a) Substituting, we get 030 60038 340012 3 65013 18038 58313 866 1822162598 1418. Since this value is less than 16, the sailboat qualifies for the race. (b) Solve for A when L 65and V 600. Substituting, we get 030 65 038A12 3 60013 16
195 038A12 2530 16 038A12 580 16 038A12 2180 A12 5738 A 32920. Thus, the largest possible sail is 3292 ft2 .
7523 005012 17707 ft/s. 24123 0040 (b) Since the volume of the flow is V A, the canal discharge is 17707 75 13280 ft3 s.
96. (a) Substituting the given values we get V 1486 97. (a) n
1
2
5
10
100
21n
211 2
212 1414
215 1149
2110 1072
21100 1007
So when n gets large, 21n decreases toward 1. (b) n 1n 1 2
1 11 1 2
2 05
12 1 2
0707
1n So when n gets large, 12 increases toward 1.
P.5
5 15 1 2
0871
10 110 1 0933 2
100 1100 1 0993 2
ALGEBRAIC EXPRESSIONS
1. (a) 2x 3 12 x 3 is a polynomial. (The constant term is not an integer, but all exponents are integers.) (b) x 2 12 3 x x 2 12 3x 12 is not a polynomial because the exponent 12 is not an integer. (c)
1 is not a polynomial. (It is the reciprocal of the polynomial x 2 4x 7.) x 2 4x 7
(d) x 5 7x 2 x 100 is a polynomial. 3 (e) 8x 6 5x 3 7x 3 is not a polynomial. (It is the cube root of the polynomial 8x 6 5x 3 7x 3.) (f) 3x 4 5x 2 15x is a polynomial. (Some coefficients are not integers, but all exponents are integers.)
SECTION P.5 Algebraic Expressions
2. To add polynomials we add like terms. So 3x 2 2x 4 8x 2 x 1 3 8 x 2 2 1 x 4 1 11x 2 x 5.
19
3. To subtract polynomials we subtract like terms. So 2x 3 9x 2 x 10 x 3 x 2 6x 8 2 1 x 3 9 1 x 2 1 6 x 10 8 x 3 8x 2 5x 2.
4. We use FOIL to multiply two polynomials:x 2 x 3 x x x 3 2 x 2 3 x 2 5x 6. 5. The Special Product Formula for the “square of a sum” is A B2 A2 2AB B 2 . So 2x 32 2x2 2 2x 3 32 4x 2 12x 9.
6. The Special Product Formula for the “product of the sum and difference of terms” is A B A B A2 B 2 . So 5 x 5 x 52 x 2 25 x 2 .
7. (a) No, x 52 x 2 10x 25 x 2 25.
(b) Yes, if a 0, then x a2 x 2 2ax a 2 .
8. (a) Yes, x 5 x 5 x 2 5x 5x 25 x 2 25.
(b) Yes, if a 0, then x a x a x 2 ax ax a 2 x 2 a 2 .
9. Binomial, terms 5x 3 and 6, degree 3 11. Monomial, term 8, degree 0 13. Four terms, terms x, x 2 , x 3 , and x 4 , degree 4
10. Trinomial, terms 2x 2 , 5x, and 3, degree 2 12. Monomial, term 12 x 7 , degree 7 14. Binomial, terms 2x and 3, degree 1
15. 6x 3 3x 7 6x 3x 3 7 9x 4
16. 3 7x 11 4x 7x 4x 3 11 11x 8 17. 2x 2 5x x 2 8x 3 2x 2 x 2 [5x 8x] 3 x 2 3x 3 18. 2x 2 3x 1 3x 2 5x 4 2x 2 3x 2 3x 5x 1 4 x 2 2x 3
19. 3 x 1 4 x 2 3x 3 4x 8 7x 5
20. 8 2x 5 7 x 9 16x 40 7x 63 9x 103 21. 5x 3 4x 2 3x x 2 7x 2 5x 3 4x 2 x 2 3x 7x 2 5x 3 3x 2 10x 2 22. 4 x 2 3x 5 3 x 2 2x 1 4x 2 12x 20 3x 2 6x 3 x 2 6x 17 23. 2x x 1 2x 2 2x
25. x 2 x 3 x 3 3x 2 27. 2 2 5t t t 10 4 10t t 2 10t t 2 4 29. r r 2 9 3r 2 2r 1 r 3 9r 6r 3 3r 2 7r 3 3r 2 9r
31. x 2 2x 2 x 1 2x 4 x 3 x 2
33. x 3 x 5 x 2 5x 3x 15 x 2 2x 15
24. 3y 2y 5 6y 2 15y 26. y y 2 2 y 3 2y
28. 5 3t 4 2t t 3 2t 2 21t 20 30. 3 9 2 2 2 2 4 5 3 4 2
32. 3x 3 x 4 4x 2 5 3x 7 12x 5 15x 3
34. 4 x 2 x 8 4x 2x x 2 x 2 6x 8
35. s 6 2s 3 2s 2 3s 12s 18 2s 2 15s 18 36. 2t 3 t 1 2t 2 2t 3t 3 2t 2 t 3 37. 3t 2 7t 4 21t 2 12t 14t 8 21t 2 26t 8 38. 4s 1 2s 5 8s 2 18s 5 39. 3x 5 2x 1 6x 2 10x 3x 5 6x 2 7x 5 40. 7y 3 2y 1 14y 2 13y 3
20
CHAPTER P Prerequisites
41. x 3y 2x y 2x 2 5x y 3y 2
42. 4x 5y 3x y 12x 2 19x y 5y 2
43. 2r 5s 3r 2s 6r 2 19rs 10s 2
44. 6u 5 u 2 6u 2 7u 10 2
45. 5x 12 25x 2 10x 1
46. 2 7y2 49y 2 28y 4
47. 3y 12 3y2 2 3y 1 12 9y 2 6y 1
48. 2y 52 2y2 2 2y 5 52 4y 2 20y 25
49. 2u 2 4u 2 4u 2
50. x 3y2 x 2 6x y 9y 2
51. 2x 3y2 4x 2 12x y 9y 2 2 53. x 2 1 x 4 2x 2 1
52. r 2s2 r 2 4rs 4s 2 2 54. 2 y 3 y 6 4y 3 4
57. 3x 4 3x 4 3x2 42 9x 2 16
58. 2y 5 2y 5 4y 2 25
55. x 6 x 6 x 2 36
56. 5 y 5 y 25 y 2
60. 2u 2u 4u 2 2 59. x 3y x 3y x 2 3y2 x 2 9y 2 61. x 2 x 2 x 4 62. y 2 y 2 y 2 63. y 23 y 3 3y 2 2 3y 22 23 y 3 6y 2 12y 8
64. x 33 x 3 3x 2 3 3x 32 33 x 3 9x 2 27x 27 65. 1 2r3 13 3 12 2r 3 1 2r2 2r3 8r 3 12r 2 6r 1 66. 3 2y3 33 3 32 2y 3 3 2y2 2y3 8y 3 36y 2 54y 27 67. x 2 x 2 2x 3 x 3 2x 2 3x 2x 2 4x 6 x 3 4x 2 7x 6 68. x 1 2x 2 x 1 2x 3 x 2 x 2x 2 x 1 2x 3 x 2 1 69. 2x 5 x 2 x 1 2x 3 2x 2 2x 5x 2 5x 5 2x 3 7x 2 7x 5 70. 1 2x x 2 3x 1 x 2 3x 1 2x 3 6x 2 2x 2x 3 5x 2 x 1 2 x x x x x x x xx 73. y 13 y 23 y 53 y 1323 y 1353 y 2 y 71.
75.
x 2 y2
2
x 1 x x 2 x 74. x 14 2x 34 x 14 2x x
72. x 32
2 2 x 2 y 2 2x 2 y 2 x 4 y 4 2x 2 y 2
1 2 1 c 2 c2 2 c c 77. x 2 a 2 x 2 a 2 x 4 a 4
76.
ab a b a b2 81. 1 x 23 1 x 23 1 x 43
79.
78. x 12 y 12 x 12 y 12 x y 80. h2 1 1 h2 1 1 h2 82. 1 b2 1 b2 b4 2b2 1
2 83. x 1 x 2 x 1 x 2 x 12 x 2 x 2 2x 1 x 4 x 4 x 2 2x 1 x 2 x 2 x 4 3x 2 4 84. x 2 x 2 85. 2x y 3 2x y 3 2x y2 32 4x 2 4x y y 2 9
SECTION P.5 Algebraic Expressions
21
86. x y z x y z x 2 y 2 z 2 2yz 87. (a) RHS 12 a b2 a 2 b2 12 a 2 b2 2ab a 2 b2 12 2ab ab LHS 2 2 2 2 2 2 (b) LHS a 2 b2 a 2 b2 a 2 b2 2a 2 b2 a 2 b2 2a 2 b2 4a 2 b2 RHS
88. LHS a 2 b2 c2 d 2 a 2 c2 a 2 d 2 b2 c2 b2 d 2 a 2 c2 b2 d 2 2abcd a 2 d 2 b2 c2 2abcd ac bd2 ad bc2 RHS
89. (a) The height of the box is x, its width is 6 2x, and its length is 10 2x. Since Volume height width length, we have V x 6 2x 10 2x. (b) V x 60 32x 4x 2 60x 32x 2 4x 3 , degree 3. (c) When x 1, the volume is V 60 1 32 12 4 13 32, and when x 2, the volume is V 60 2 32 22 4 23 24. 90. (a) The width is the width of the lot minus the setbacks of 10 feet each. Thus width x 20 and length y 20. Since Area width length, we get A x 20 y 20. (b) A x 20 y 20 x y 20x 20y 400
(c) For the 100 400 lot, the building envelope has A 100 20 400 20 80 380 30,400. For the 200 200, lot the building envelope has A 200 20 200 20 180 180 32,400. The 200 200 lot has a larger building envelope. 91. (a) A 2000 1 r3 2000 1 3r 3r 2 r 3 2000 6000r 6000r 2 2000r 3 , degree 3. (b) Remember that % means divide by 100, so 2% 002. Interest rate r
2%
3%
45%
6%
10%
Amount A $212242 $218545 $228233 $238203 $266200 92. (a) P R C 50x 005x 2 50 30x 01x 2 50x 005x 2 50 30x 01x 2 005x 2 20x 50. (b) The profit on 10 calculators is P 005 102 20 10 50 $155. The profit on 20 calculators is P 005 202 20 20 50 $370 . 93. (a) When x 1, x 52 1 52 36 and x 2 25 12 25 26. (b) x 52 x 2 10x 25
94. (a) The degree of the product is the sum of the degrees of the original polynomials. (b) The degree of the sum could be lower than either of the degrees of the original polynomials, but is at most the largest of the degrees of the original polynomials. (c) Product: 2x 3 x 3 2x 3 x 7 4x 6 2x 4 14x 3 2x 4 x 2 7x 6x 3 3x 21 4x 6 4x 4 20x 3 x 2 10x 21 Sum: 2x 3 x 3 2x 3 x 7 4.
22
CHAPTER P Prerequisites
P.6
FACTORING
1. The polynomial 2x 5 6x 4 4x 3 has three terms: 2x 5 , 6x 4 , and 4x 3 . 2. The factor 2x 3 is common to each term, so 2x 5 6x 4 4x 3 2x 3 x 2 3x 2 . [In fact, the polynomial can be factored further as 2x 3 x 2 x 1.]
3. To factor the trinomial x 2 7x 10 we look for two integers whose product is 10 and whose sum is 7. These integers are 5 and 2, so the trinomial factors as x 5 x 2. 4. The greatest common factor in the expression 4 x 12 x x 12 is x 12 , and the expression factors as 4 x 12 x x 12 x 12 4 x.
5. The Special Factoring Formula for the “difference of squares” is A2 B 2 A B A B. So 4x 2 25 2x 5 2x 5.
6. The Special Factoring Formula for a “perfect square” is A2 2AB B 2 A B2 . So x 2 10x 25 x 52 . 7. 5a 20 5 a 4 9. 2x 3 x x 2x 2 1
11. 2x 2 y 6x y 2 3x y x y 2x 6y 3 13. y y 6 9 y 6 y 6 y 9
8. 3b 12 3 b 4 3 b 4 10. 3x 4 6x 3 x 2 x 2 3x 2 6x 1 12. 7x 4 y 2 14x y 3 21x y 4 7x y 2 x 3 2y 3y 2
14. z 22 5 z 2 z 2 [z 2 5] z 2 z 3 15. x 2 8x 7 x 7 x 1
16. x 2 4x 5 x 5 x 1
17. x 2 2x 15 x 5 x 3
18. 2x 2 5x 7 x 1 2x 7
19. 3x 2 16x 5 3x 1 x 5
20. 5x 2 7x 6 5x 3 x 2
21. 3x 22 8 3x 2 12 [3x 2 2] [3x 2 6] 3x 4 3x 8
22. 2 a b2 5 a b 3 [a b 3] [2 a b 1] a b 3 2a 2b 1 23. x 2 25 x 5 x 5
24. 9 y 2 3 y 3 y
25. 49 4z 2 7 2z 7 2z
26. 9a 2 16 3a 4 3a 4
28. a 2 36b2 a 6b a 6b 27. 16y 2 z 2 4y z 4y z 29. x 32 y 2 x 3 y x 3 y x y 3 x y 3 30. x 2 y 52 x y 5 x y 5 x y 5 x y 5 31. x 2 10x 25 x 52
32. 9 6y y 2 3 y2
33. z 2 12z 36 z 62
34. 2 16 64 82
35. 4t 2 20t 25 2t 52
36. 16a 2 24a 9 4a 32
37. 9u 2 6u 2 3u 2 39. x 3 27 x 3 x 2 3x 9
38. x 2 10x y 25y 2 x 5y2 40. y 3 64 y 4 y 2 4y 16
SECTION P.6 Factoring
41. 8a 3 1 2a 1 4a 2 2a 1 43. 27x 3 y 3 3x y 9x 2 3x y y 2
3 45. u 3 6 u 3 2 u 2 u 2 u 2 4 47. 48. 49. 50. 51. 52. 53. 54.
23
42. 8 273 2 3 4 6 92 44. 1 1000y 3 1 10y 1 10y 100y 2
46. 8r 3 64t 6 2r 4t 2 4r 2 8r t 2 16t 4 x 3 4x 2 x 4 x 2 x 4 1 x 4 x 4 x 2 1 3x 3 x 2 6x 2 x 2 3x 1 2 3x 1 3x 1 x 2 2 5x 3 x 2 5x 1 x 2 5x 1 5x 1 x 2 1 5x 1 18x 3 9x 2 2x 1 9x 2 2x 1 2x 1 9x 2 1 2x 1 x 3 x 2 x 1 x 2 x 1 1 x 1 x 1 x 2 1 x 5 x 4 x 1 x 4 x 1 1 x 1 x 1 x 4 1 x 52 x 12 x 12 x 2 1 x x 1 x 1 1 12 12 32 12 2 3x 3 4x x 4x x x 3 x 1 x x
55. Start by factoring out the power of x with the smallest exponent, that is, x 32 . So 1 x2 . x 32 2x 12 x 12 x 32 1 2x x 2 x 32 56. x 172 x 132 x 132 x 12 1 x 132 [x 1 1] [x 1 1] x 132 x 2 x
12 57. Start by factoring out the power of x 2 1 with the smallest exponent, that is, x 2 1 . So
12 12 12 x2 3 x2 1 x2 1 2 2 x2 1 x2 1 . x2 1
2x 1 58. x 12 x 112 x 12 x 112 x 12 x 112 [x 1 x] x x 1 59. 2x 13 x 223 5x 43 x 213 x 13 x 213 [2 x 2 5x] x 13 x 213 2x 4 5x 3x 4 3 x x 13 x 213 3x 4 3 x 2 54 14 14 60. 3x 12 x 2 1 3 x 2 1 x 2 1 x 32 x 2 1 x 12 x 2 1 4 2 14 14 x 1 2x 2 3 3x 2 3 x 2 x 12 x 2 1 2x 2 3 x 12 x 2 1 x 61. 12x 3 18x 6x 2x 2 3 62. 30x 3 15x 4 15x 3 2 x
63. 6y 4 15y 3 3y 3 2y 5
64. 5ab 8abc ab 5 8c
65. x 2 2x 8 x 4 x 2
66. x 2 14x 48 x 8 x 6
24
CHAPTER P Prerequisites
67. y 2 8y 15 y 3 y 5
68. z 2 6z 16 z 2 z 8
69. 2x 2 5x 3 2x 3 x 1 71. 9x 2 36x 45 9 x 2 4x 5 9 x 5 x 1
70. 2x 2 7x 4 2x 1 x 4
73. 6x 2 5x 6 3x 2 2x 3
74. 6 5t 6t 2 3 2t 2 3t
75. x 2 36 x 6 x 6
76. 4x 2 25 2x 5 2x 5
77. 49 4y 2 7 2y 7 2y
78. 4t 2 9s 2 2t 3s 2t 3s
79. t 2 6t 9 t 32
80. x 2 10x 25 x 52
81. 4x 2 4x y y 2 2x y2 83. t 3 1 t 1 t 2 t 1
82. r 2 6rs 9s 2 r 3s2
72. 8x 2 10x 3 4x 3 2x 1
84. x 3 27 x 3 33 x 3 x 2 3x 9 85. 8x 3 125 2x3 53 2x 5 2x2 2x 5 52 2x 5 4x 2 10x 25 86. 125 27y 3 53 3y3 5 3y 52 5 3y 3y2 3y 5 9y 2 15y 25 87. x 3 2x 2 x x x 2 2x 1 x x 12 88. 3x 3 27x 3x x 2 9 3x x 3 x 3
89. x 4 2x 3 3x 2 x 2 x 2 2x 3 x 2 x 1 x 3
90. 35 54 23 3 32 5 2 3 3 1 2
91. x 4 y 3 x 2 y 5 x 2 y 3 x 2 y 2 x 2 y 3 x y x y
92. 18y 3 x 2 2x y 4 2x y 3 9x y 3 2 x 2 x 2 2y 2y2 x 2 2y x 4 2x 2 y 4y 2 93. x 6 8y 3 x 2 2y3 x 2 2y
3 2 3a b2 9a 2 3ab2 b4 94. 27a 3 b6 3a3 b2 3a b2 3a2 3a b2 b2 95.
y 3 3y 2 4y 12
y 3 3y 2 4y 12 y 2 y 3 4 y 3 y 3 y 2 4
y 3 y 2 y 2 (factor by grouping) 96. y 3 y 2 y 1 y 2 y 1 1 y 1 y 2 1 y 1
97. 3x 3 x 2 12x 4 3x 3 12x x 2 4 3x x 2 4 x 2 4 3x 1 x 2 4 3x 1 x 2 x 2 (factor by grouping)
98. 9x 3 18x 2 x 2 9x 2 x 2 x 2 9x 2 1 x 2 3x 1 3x 1 x 2
99. a b2 a b2 [a b a b] [a b a b] 2b 2a 4ab
SECTION P.6 Factoring
100.
25
1 1 1 1 2 1 2 1 1 1 1 1 1 1 x x x x x x 1 1 1 2 4 1 1 1 1 1 2 x x x x x x
101. x 2 x 2 1 9 x 2 1 x 2 1 x 2 9 x 1 x 1 x 3 x 3 102. a 2 1 b2 4 a 2 1 a 2 1 b2 4 a 1 a 1 b 2 b 2
103. x 1 x 22 x 12 x 2 x 1 x 2 [x 2 x 1] 3 x 1 x 2
104. x 13 x 2 x 12 x 2 x 3 x 1 x x 1 x 12 2 x 1 x x 2 x x 1 [x 1 x]2 x x 1 12 x x 1
105. y 4 y 23 y 5 y 24 y 4 y 23 1 y y 2 y 4 y 23 y 2 2y 1 y 4 y 23 y 12 106. n x y n 1 y x n x y n 1 x y x y [n n 1] x y 107. Start by factoring y 2 7y 10, and then substitute a 2 1 for y. This gives
2 a 2 1 7 a 2 1 10 a 2 1 2 a 2 1 5 a 2 1 a 2 4 a 1 a 1 a 2 a 2
2 108. a 2 2a 2 a 2 2a 3 a 2 2a 3 a 2 2a 1 a 2 2a 3 a 2 2a 1 a 1 a 3 a 12
109. 3x 2 4x 122 x 3 2 4x 12 4 x 2 4x 12 [3 4x 12 x 2 4] 4x 2 x 3 12x 36 8x 4x 2 x 3 20x 36 16x 2 x 3 5x 9 4 5 4 110. 5 x 2 4 2x x 24 x 2 4 4 x 23 2 x 2 4 x 23 5 x x 2 x 2 4 2
4 4 2 x 2 4 x 23 5x 2 10x 2x 2 8 2 x 2 4 x 23 7x 2 10x 8
111. 3 2x 12 2 x 312 2x 13 12 x 312 2x 12 x 312 6 x 3 2x 1 12 2x 12 x 312 6x 18 x 12 2x 12 x 312 7x 35 2 112. 13 x 623 2x 32 x 613 2 2x 3 2 13 x 623 2x 3 [2x 3 3 x 6 4] 13 x 623 2x 3 [2x 3 12x 72] 13 x 623 2x 3 14x 69 113.
1 2 13 43 43 43 1 x2 3 3 x 3 x 2 3 23 x 2 x 2 3 23 x 2 x 2 3 x2 3 x2 3 43 3 x2 3
114. 12 x 12 3x 412 32 x 12 3x 412 12 x 12 3x 412 [3x 4 3x] 12 x 12 3x 412 6x 4 x 12 3x 412 3x 2
26
CHAPTER P Prerequisites
115. The volume of the shell is the difference between the volumes of the outside cylinder (with radius R) and the inside cylinder R r h R r. The (with radius r). Thus V R 2 h r 2 h R 2 r 2 h R r R r h 2 2 R r R r average radius is and 2 is the average circumference (length of the rectangular box), h is the height, and 2 2 R r h R r 2 average radius R r is the thickness of the rectangular box. Thus V R 2 h r 2 h 2 2 height thickness R h
length rl
h thickness
116. (a) Mowed portion field habitat
(b) Using the difference of squares, we get b2 b 2x2 [b b 2x] [b b x] 2x 2b 2x 4x b x.
117. (a) 5282 5272 528 527 528 527 1 1055 1055 (b) 1222 1202 122 120 122 120 2 242 484
(c) 10202 10102 1020 1010 1020 1010 10 2030 20,300
118. (a) 501 499 500 1 500 1 5002 1 250,000 1 249,999 (b) 79 61 70 9 70 9 702 92 4900 81 4819
(c) 2007 1993 2000 7 2000 7 20002 72 4,000,000 49 3,999,951
119. (a) A4 B 4 A2 B 2 A2 B 2 A B A B A2 B 2 A6 B 6 A3 B 3 A3 B 3 (difference of squares) A B A2 AB B 2 A B A2 AB B 2 (difference and sum of cubes) (b) 124 74 20,736 2,401 18,335; 126 76 2,985,984 117,649 2,868,335 (c) 18,335 124 74 12 7 12 7 122 72 5 19 144 49 5 19 193 2,868,335 126 76 12 7 12 7 122 12 7 72 122 12 7 72 5 19 144 84 49 144 84 49 5 19 277 109
120. (a) A 1 A 1 A2 A A 1 A2 1 A 1 A2 A 1 A3 A2 A A2 A 1 A3 1 A 1 A3 A2 A 1 A4 A3 A2 A A3 A2 A 1 (b) We conjecture that A5 1 A 1 A4 A3 A2 A 1 . Expanding the right-hand side, we have A 1 A4 A3 A2 A 1 A5 A4 A3 A2 A A4 A3 A2 A 1 A5 1, verifying our conjecture. Generally, An 1 A 1 An1 An2 A 1 for any positive integer n.
SECTION P.7 Rational Expressions
121. (a)
A2 A 1
A 1
A 1
A 1
A2 A A2
A3 A2 A 1
A 1
A2 A 1
A3 A2 A
A3
1
1
27
A 1
A3 A2 A 1
A 4 A3 A2 A A4
1 (b) Based on the pattern in part (a), we suspect that A5 1 A 1 A4 A3 A2 A 1 . Check:
A 4 A 3 A2 A 1
A 1
A4 A3 A2 A 1
A 5 A 4 A 3 A2 A
A5 1 The general pattern is An 1 A 1 An1 An2 A2 A 1 , where n is a positive integer.
P.7
RATIONAL EXPRESSIONS
3x is a rational expression. x2 1 x 1 (b) is not a rational expression. A rational expression must be a polynomial divided by a polynomial, and the 2x 3 numerator of the expression is x 1, which is not a polynomial.
1. (a)
x3 x xx 2 1 is a rational expression. x 3 x 3 2. To simplify a rational expression we cancel factors that are common to the numerator and denominator. So, the expression x 1x 2 x 1 simplifies to . x 3x 2 x 3 3. To multiply two rational expressions we multiply their numerators together and multiply their denominators together. So 2 x 2x 2x is the same as . 2 x 1 x 3 x 1 x 3 x 4x 3 1 2 x 4. (a) has three terms. x x 1 x 12 (c)
(b) The least common denominator of all the terms is x x 12 . (c)
2 x 2x x 1 x x 1 x 12 x 12 2x x 1 x 2 x x 1 x 12 x 1 x x 12 x x 12 x 12
x 2 2x 1 2x 2 2x x 2
x x 12 x x 1 x . 5. (a) Yes. Cancelling x 1, we have x 1 x 12
2x 2 1
x x 12
(b) No; x 52 x 2 10x 25 x 2 25, so x 5 x 2 10x 25 x 2 25. 3a 3 a a 6. (a) Yes, 1 . 3 3 3 3 (b) No. We cannot “separate” the denominator in this way; only the numerator, as in part (a). (See also Exercise 101.)
7. The domain of 4x 2 10x 3 is all real numbers.
8. The domain of x 4 x 3 9x is all real numbers.
28
CHAPTER P Prerequisites
9. Since x 3 0 we have x 3. Domain: x x 3 11. Since x 3 0, x 3. Domain; x x 3
10. Since 3t 6 0 we have t 2. Domain: t t 2 12. Since x 1 0, x 1. Domain; x x 1
13. x 2 x 2 x 1 x 2 0 x 1 or 2, so the domain is x x 1 2.
14. 2x 0 and x 1 0 x 0 and x 1, so the domain is x x 0. 4 x2 1 2x 1 4 x 1 x 1 x 1 5 x 3 2x 1 5 x 3 2x 1 16. 15. 2 5 x 3 2 x 3 2 x 3 12 x 2 x 1 12 x 2 x 1 3 x 2 10 x 3
17. 19.
1 x 2 x 2 2 x 2 x 2 x 2 x 4
x 2 x 2 5x 6 x 2 x 3 2 x 5 5 3 x x x 8x 15
18.
x2 x 2 x 2 x 2 x 1 2 x 1 x 1 x 1 x 1
x 2 x 12 x 4 x 4 x 3 20. 2 x 2 2 3 x x x 5x 6
y2 y y 2 3y 18 y y 1 y y6 y 6 y 3 22. y1 2y 1 y 1 y 1 2y 1 y 3 y2 1 2y 2 7y 3 x 2x 2 x 6 x 2x 3 x 2 x 2x 3 2x 3 x 2 6x 23. 2 2x 3 3 2 3 2 2x x 2x x 2x 7x 6
21.
1 x2 x 1 1 x x 1 1 x 1 x x3 1 x2 x 1 x 1 x 2 x 1 x 1 x 2 x 1 4x 1 x 2 x 2 4x 25. 2 4 x 2 x 2 x 2 16x x 4 16x 24.
26. 27.
x 2 25 x 4 x 5 x 5 x 5 x 4 2 x 4 x 4 x 4 x 5 x 16 x 5
x 2 2x 15 x 5 x 3 x 5 x 3 x 5 x 2 x 2 x 5 x 5 x 2 x 2 25
x 2 2x x 2 2x t 3 29. 2 t 9 28.
30. 31. 32. 33. 34.
3 3x x 1 x 1 1x x 3 x 1 x 3 x 3 x 1 x 1 1x x 3 x 1 3 3x t 3 1 t 3 t 3 2 2 t 9 t 3 t 3 t2 9 t 9
x2 x 6 x 2 x 1 x3 x2 x 3 x 2 2 x 2 x x 2 x 3 x 1 x 2x x 2x 3
x 4 x 2 7x 12 x 2 5x 6 x 3 x 4 x 2 x 3 x 1 x 1 x 2 x 3 x 3 x 2 3x 2 x 2 6x 9
x 2 2x y y 2 2x 2 x y y 2 2x y x y x y x y 2x y 2 x 2y x y x y x 2y x y x 2 y2 x x y 2y 2
x 3 x 2 7x 12 x 3 2x 2 7x 15 x 3 x 5 x 5 2x 3 2 2 2 2x 3 2x 3 x 3 x 4 2x 3 x 4 4x 9 2x 7x 15 4x 9 x 2 7x 12 2x 1
2x 2 x 15 x3
2x 1 6x 2 x 2 x 3 1 x 3 x 3 2x 5 2x 1 3x 2 2x 5 3x 2
x3 x 2 2x 1 x 3 x 1 x 1 x 1 x 2 x 1 x x 1 x x 1 x x 2 2x 1 2x 2 3x 2 x 2 x2 x 2 x 2 2x 1 x 1 x 2 2x 2 3x 2 x2 1 36. x 1 x 1 x 1 x 2 2x 1 x2 1 2x 2 5x 2 2x 2 5x 2 x2 x 2 35.
SECTION P.7 Rational Expressions
29
x 1 x xy z y z yz x y x z xz 38. x yz z 1 y y x 3 1 x 4 1 39. 1 x 3 x 3 x 3 x 3 3x 2 2 x 1 3x 2 2x 2 x 4 3x 2 2 40. x 1 x 1 x 1 x 1 x 1 2 x 3 1 2 x 5 x 3 2x 10 3x 7 41. x 5 x 3 x 5 x 3 x 5 x 3 x 5 x 3 x 5 x 3 1 x 1 1 x 1 x 1x 1 2x 42. x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 1 3 1 3 x 2 x 1 3x 6 x 1 2x 5 43. x 1 x 2 x 1 x 2 x 1 x 2 x 1 x 2 x 1 x 2 37.
3 x x 6 3 x 4 x 2 6x 3x 12 x 2 3x 12 x x 4 x 6 x 4 x 6 x 4 x 6 x 4 x 6 x 4 x 6 5 3 5 2x 3 3 10x 15 3 10x 18 2 5x 9 45. 2x 3 2x 32 2x 32 2x 32 2x 32 2x 32 2x 32 x x 2 2 x 1 3x 2 x 2x 2 46. x 1 x 1 x 1 x 12 x 12 x 12 x 12 44.
47. u 1
u u 2 2u 1 u u 2 3u 1 u u 1 u 1 u1 u1 u1 u1 u1
4 3 2b2 3ab 4a 2 2b2 3ab 4a 2 2 2 2 2 2 2 2 2 2 ab b a a b a b a b a 2 b2 x 1 1 1 1 1 x 2x 1 2 2 49. 2 2 2 2 x x 1 x x x x x x 1 x x 1 x x 1 48.
1 1 x2 1 2 3 3 x x x x 1 2 51. x 3 x 2 7x 12
x 1 x2 x 1 3 3 x x x3 1 2 2 x 4 1 x 3 x 3 x 4 x 3 x 4 x 3 x 4 2x 8 1 2x 7 x 3 x 4 x 3 x 4 x x 1 x 1 x 2 52. 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 4 x 2 2x 2 2 x 1 x 2 x 2 x 2 x 2 1 1 1 1 x 3 1 x 2 53. x 3 x2 9 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 2 x 2 54. 2 x 1 x 2 x 1 x 4 x x 2 x 2 5x 4
50.
2 x 2 x 2 4x 2x 4 x 2 6x 4 x x 4 x 1 x 2 x 4 x 1 x 2 x 4 x 1 x 2 x 4 x 1 x 2 x 4 3 4 3 4 2 x 1 3x 4 2x 2 3x 4 5x 6 2 2 55. x x 1 x2 x x x 1 x x 1 x x 1 x x 1 x x 1 x x 1 x x 1 x 2 x 2 1 1 56. 2 x 2 x 3 x 2 x 3 3 2 x x x x 6 1 x 3 2 x 2 x x 3 2x 4 2x 1 x x 3 x 2 x 3 x 2 x 3 x 2 x 3 x 2 x 3 x 2
30
CHAPTER P Prerequisites
1 1 1 2 x 2 x 1 x 3 x 1 x 2x 3 x 3 x 2 x 3x 2 5 x 3 x 2 x 1 x 3 x 2 x 1 x 3 x 2 x 1 x 3 x 2 x 1 2 2 1 3 3 1 58. 2 x 1 x 12 x 1 x 12 x 1 x 1 x 1 2 x 1 3 x 1 x 1 x 1 2 2 x 1 x 1 x 1 x 1 x 1 x 12
57.
1
x 2 3x 2
x2 1
2x 2 3x 3 x 2 1 2x 2 3x 3 x2 x 4 x 1 x 12 x 1 x 12 x 1 x 12 x 1 x 12 x 1 x 12 x 1 1x 1 1 x 1 59. 1 x 1 2 1 2x 2 x x x y 1 2y 1 2y y 2 60. 3 3 3y 1 y 1
y
61.
62.
63.
64.
65.
66.
y
1 1 2 1 x 1 x 3 x 2 1 x 2 x 2 1 1 x 1 x 2 1 1 x 2 1 x 2 x 2 1 1 c1 c11 c 1 c11 c2 1 c1 1 1 1 1 1 3 x x 2 x 1 x 3 x 1 x 1 x 3 x 1 x 3 x 1 x 1 x 3 x 1 x 1 x 1 x 3 x 1 x 1 x 3 2 x 1 x 3 x 3 x 2 x 2 2x 3 x 2 2x 8 5 x 4 x 1 x 3 x 1 x 2 x 4 x 3 x 4 x 3 x 1 x 4 x 3 x 1 x 4 x 3 x 1 x x xy x x x 2 y 1 x2 y x2 y y y y x y 2 y 2 y 2 x 1 y xy y x x y y 2 xy x x 2 2 y y x x x x yy x x x y2 x 2 x x y2 y x y y y y
y x x 2 y2 x 2 y2 x 2 y2 xy y x xy 67. 2 x y. An alternative method is to multiply the 2 2 1 1 xy 1 y x2 y x x2 y2 x 2 y2
numerator and denominator by the common denominator of both the numerator and denominator, in this case x 2 y 2 : y x y x x y x 2 y2 x 2 y2 x 3 y x y3 y x y x 2 x y. 1 1 1 1 x 2 y2 y x2 y2 x 2 x2 y2 x2 y2
SECTION P.7 Rational Expressions
x x 2 y2 x y2 y xy y x y2 x 3 x y2 x y2 x3 x x 68. x x 2 2 y y xy x 2 2 2 2 2 2 2 x y x y x y x y x y2 y x y x 1 x2 1 y2 2 2 2 2 2 2 y2 x 2 xy yx y x y x x y x y x y x y x y 1 2 y2 2 y 2 y x 1 1 y x xy y x x xy xy x y 1 1 yx y2 x 2 x 2 y 2 x 2 y2 y x y x x2 y2 . Alternatively, 1 1 1 x y y x xy x y 1 x 2 y2 x y2 x 2 y x y
x 2 69. 1 x
y 2
1 1 1 1 y x y x x y x y x y x y x y 70. 1 1 1 x y y xy x x y xy xy x 1 y 1
71. 1
72. 1
1 1 1 x
x y y2 x 2 x y x 2 2x y y 2 x y2 xy xy xy
1
1 1 1 1x
x 1x 1 x x 1 x 1 1x
1
x 2x 1 2x 3 x 1 1x 1 x 2 x 2 x 2 1 x 1
1 1 1 1 x h 1 x 1 x 1 x h 73. h h 1 x 1 x h 1 x 1 x h 74. In calculus it is necessary to eliminate the h in the denominator, and we do this by rationalizing the numerator: 1 1 x x h x x h x x h 1 x x h . h h x x h x x h h x x h x x h x x h x x h
1 1 2 x 2 2xh h 2 2 x 2 2 2 x x h 2x h x h x 75. h hx 2 x h2 hx 2 x h2 x 2 x h2 x h3 7 x h x 3 7x x 3 3x 2 h 3xh 2 h 3 7x 7h x 3 7x 3x 2 h 3xh 2 h 3 7h 76. h h h
h 3x 2 3xh h 2 7 h
3x 2 3xh h 2 7
2 x2 x2 1 1 x2 1 x 77. 1 1 2 2 2 2 1x 1x 1x 1x 1 x2 1 x2 1 2 2x 3 1 1 1 1 3 6 6 78. 1 x 3 1x 3 1x 2 x 6 12 6 6 6 4x 4x 16x 16x 16x 2 1 1 x3 3 x 3 3 4x 4x
31
32
79.
CHAPTER P Prerequisites
3 x 22 x 32 x 23 2 x 3 x 34
x 22 x 3 [3 x 3 x 2 2] x 34 x 22 3x 9 2x 4
x 22 x 13
x 33 x 33 x 63 2x x 6 4x 2 2x x 64 x 2 4 x 63 2x 2 12x 4x 2 12x 2x 2 2x 6 x 80. 8 8 5 x 6 x 6 x 6 x 65 x 65 x 2 2 1 x12 x 1 x12 1 x12 [2 1 x x] 1x 1x 1 x32 12 12 12 1 x2 1 x2 1 x2 x2 x2 1 x2 1 82. 32 2 2 1x 1x 1 x2 81.
83.
84. 85.
86.
87. 88.
89. 90. 91.
92.
3 1 x13 x 1 x23
1 x23 [3 1 x x]
2x 3 1 x23 1 x43 7 3x12 7 3x 32 x 7 32 x 7 3x12 32 x 7 3x12 7 3x 7 3x 7 3x32 5 3 1 1 5 3 5 3 25 3 22 5 3 5 3 5 3 2 5 3 3 63 5 4 5 6 3 5 2 5 2 5 2 5 2 2 2 2 7 2 7 7 2 2 2 2 7 27 5 5 2 7 2 7 2 7 1 1 x 1 x 1 x 1 x 1 x 1 x 1 3 y y y 3y y 3 y y y 3 y 3y 3 y 3 y 3 y 2 x y x y x y 2 x y 2 x y 2 x y 2 x 2 y xy x y x y x y 1 5 1 5 1 5 15 4 3 3 1 5 3 1 5 3 1 5 3 5 3 5 3 5 35 2 1 2 2 3 5 3 5 2 3 5 2 3 5 1 x23
r 2 r 2 r 2 r 2 93. 5 5 r 2 5 r 2 x x h x x h x x h x x h 94. h x x h h x x h x x h h x x h x x h h 1 h x x h x x h x x h x x h x2 1 x x2 1 x x2 1 x2 1 95. x 2 1 x 2 2 2 1 x 1x x 1x x 1x
SECTION P.7 Rational Expressions
96.
x 1
97. (a) R
x
x 1 1
33
x x 1 x x 1x 1 x 1 x x 1 x x 1 x
1 R R R1 R2 1 1 2 1 1 1 1 R1 R2 R2 R1 R1 R2 R1 R2
(b) Substituting R1 10 ohms and R2 20 ohms gives R 98. (a) The average cost A
200 10 20 67 ohms. 30 20 10
500 6x 001x 2 Cost . number of shirts x
(b) x
10
20
50
100
200
500
1000
Average cost
$5610
$3120
$1650
$1200
$1050
$1200
$1650
99. x x2 9 x 3
280
290
295
299
2999
3
3001
301
305
310
320
580
590
595
599
5999
?
6001
601
605
610
620
From the table, we see that the expression
x2 9 approaches 6 as x approaches 3. We simplify the expression: x 3
x2 9 x 3 x 3 x 3, x 3. Clearly as x approaches 3, x 3 approaches 6. This explains the result in the x 3 x 3 table. 2 2 100. No, squaring changes its value by a factor of . x x 101. Answers will vary. Algebraic Error 1 1 1 a b ab
a b2 a 2 b2 a 2 b2 a b ab b a 1 a ab b am mn a an 102. (a)
Counterexample 1 1 1 2 2 22
1 32 12 32 52 122 5 12 26 6 2 1 1 11 35 352 32
5 a a 5a 1 , so the statement is true. 5 5 5 5
(b) This statement is false. For example, take x 5 and y 2. Then LHS RHS
x 5 5 , and 2 . y 2 2
(c) This statement is false. For example, take x 0 and y 1. Then LHS RHS
1 1 1 1 , and 0 . 1y 11 2 2
x 1 51 6 2, while y1 21 3 0 x 0, while xy 01
34
CHAPTER P Prerequisites
(d) This statement is false. For example, take x 1 and y 1. Then LHS 2
a
2
1 2, while 1
b 2a 2 RHS 1, and 2 1. 2b 2 a 1 1 a a a 1 a 1 . (e) This statement is true: b b b b b 2 2 2 1 (f) This statement is false. For example, take x 2. Then LHS , while 4x 42 6 3 1 2 1 2 3 1 3 RHS , and . 2 x 2 2 2 3 2 103. (a) 1 9 99 999 9999 x 1 3 2 10 100 1000 10,000 1 x 2 3333 25 2011 20001 2000001 200000001 x 1 It appears that the smallest possible value of x is 2. x 1 1 2x (b) Because x 0, we can multiply both sides by x and preserve the inequality: x 2 x x x x
x 2 1 2x x 2 2x 1 0 x 12 0. The last statement is true for all x 0, and because each step is 1 reversible, we have shown that x 2 for all x 0. x
P.8
SOLVING BASIC EQUATIONS
1. Substituting x 3 in the equation 4x 2 10 makes the equation true, so the number 3 is a solution of the equation. 2. Subtracting 4 from both sides of the given equation, 3x 4 10, we obtain 3x 4 4 10 4 3x 6. Multiplying
by 13 , we have 13 3x 13 6 x 2, so the solution is x 2. x 2x 10 is equivalent to 52 x 10 0, so it is a linear equation. 3. (a) 2 2 2 (b) 2x 1 is not linear because it contains the term , a multiple of the reciprocal of the variable. x x (c) x 7 5 3x 4x 2 0, so it is linear. 4. (a) x x 1 6 x 2 x 6 is not linear because it contains the square of the variable. (b) x 2 x is not linear because it contains the square root of x 2.
(c) 3x 2 2x 1 0 is not linear because it contains a multiple of the square of the variable.
5. (a) This is true: If a b, then a x b x.
(b) This is false, because the number could be zero. However, it is true that multiplying each side of an equation by a nonzero number always gives an equivalent equation.
(c) This is false. For example, 5 5 is false, but 52 52 is true.
6. To solve the equation x 3 125 we take the cube root of each side. So the solution is x
3 125 5.
7. (a) When x 2, LHS 4 2 7 8 7 1 and RHS 9 2 3 18 3 21. Since LHS RHS, x 2 is not a solution.
(b) When x 2, LHS 4 2 7 8 7 15 and RHS 9 2 3 18 3 15. Since LHS RHS, x 2 is a solution.
8. (a) When x 1, LHS 2 5 1 2 5 7 and RHS 8 1 7. Since LHS RHS, x 1 is a solution. (b) When x 1, LHS 2 5 1 2 5 3 and RHS 8 1 9. Since LHS RHS, x 1 is not a solution.
SECTION P.8 Solving Basic Equations
35
9. (a) When x 2, LHS 1 [2 3 2] 1 [2 1] 1 1 0 and RHS 4 2 6 2 8 8 0. Since LHS RHS, x 2 is a solution. (b) When x 4 LHS 1 [2 3 4] 1 [2 1] 1 3 2 and RHS 4 4 6 4 16 10 6. Since LHS RHS, x 4 is not a solution.
1 1 12 12 12 1 and RHS 1. Since LHS RHS, x 2 is a solution. 24 2 1 is not defined, so x 4 is not a solution. (b) When x 4 the expression 44
10. (a) When x 2, LHS 12
11. (a) When x 1, LHS 2 113 3 2 1 3 2 3 5. Since LHS 1, x 1 is not a solution. (b) When x 8 LHS 2 813 3 2 2 3 4 3 1 RHS. So x 8 is a solution.
432 23 8 4 and RHS 4 8 4. Since LHS RHS, x 4 is a solution. 46 2 2 32 23 832 292 When x 8, LHS 272 and RHS 8 8 0. Since LHS RHS, x 8 is not a 86 2 2 solution. 0a a a When x 0, LHS RHS. So x 0 is a solution. 0b b b ba ba When x b, LHS is not defined, so x b is not a solution. bb 0 2 b b2 b2 b b2 b b When x , LHS 14 b2 0 RHS. So x is a solution. b 2 2 2 4 2 4 2 2 1 1 1 1 b2 1 14 b2 2 1 , so x is not a solution. b When x , LHS b b b 4 b b
12. (a) When x 4, LHS (b)
13. (a) (b) 14. (a) (b)
15. 5x 6 14 5x 20 x 4
16. 3x 4 7 3x 3 x 1
17. 7 2x 15 2x 8 x 4
18. 4x 95 1 4x 96 x 24
19. 12 x 7 3 12 x 4 x 8
20. 2 13 x 4 13 x 6 x 18
21. 3x 3 5x 3 0 8x x 0
22. 2x 3 5 2x 4x 2 x 12
23. 7x 1 4 2x 9x 3 x 13
24. 1 x x 4 3 2x x 32
25. x 3 4x 3 5x x 35
26. 2x 3 7 3x 5x 4 x 45
27. x3 1 53 x 7 x 3 5x 21 4x 24 x 6
3 x 3 4x 10 3x 30 x 40 28. 25 x 1 10
29. 2 1 x 3 1 2x 5 2 2x 3 6x 5 2 2x 8 6x 6 8x x 34
30. 5 x 3 9 2 x 2 1 5x 15 9 2x 4 1 5x 24 2x 3 7x 21 x 3 31. 4 y 12 y 6 5 y 4y 2 y 30 6y 3y 2 30 6y 9y 32 y 32 9
32. r 2 [1 3 2r 4] 61 r 2 1 6r 12 61 r 2 6r 11 61 r 12r 22 61 13r 39 r 3 33. x 13 x 12 x 5 0 6x 2x 3x 30 0 (multiply both sides by 6) x 30 y1 34. 23 y 12 y 3 8y 6 y 3 3 y 1 8y 6y 18 3y 3 14y 18 3y 3 11y 21 4 y 21 11
36
CHAPTER P Prerequisites
x 1 x 1 6x 8x 2x x 1 24x 7x 1 24x 1 17x x 17 2 4 x 1 1 5x 18x 15x 2 x 1 1 3x 2x 1 x 1 36. 3x 2 3 6 35. 2x
37. x 1 x 2 x 2 x 3 x 2 x 2 x 2 5x 6 x 2 5x 6 6x 8 x 43 38. x x 1 x 32 x 2 x x 2 6x 9 x 6x 9 5x 9 x 95
39. x 1 4x 5 2x 32 4x 2 x 5 4x 2 12x 9 x 5 12x 9 13x 14 x 14 13 40. t 42 t 42 32 t 2 8t 16 t 2 8t 16 32 16t 32 t 2 4 1 1 3 4 3x (multiply both sides by the LCD, 3x) 1 3x x 13 41. x 3x 6 2 42. 5 4 2 5x 6 4x 4 9x 49 x x x 2x 1 4 43. 5 2x 1 4 x 2 10x 5 4x 8 6x 13 x 13 6 x 2 5 2x 7 23 2x 7 3 2 2x 4 (cross multiply) 6x 21 4x 8 2x 29 x 29 44. 2 2x 4 3 2 2 t 1 3 t 6 [multiply both sides by the LCD, t 1 t 6] 2t 2 3t 18 20 t 45. t 6 t 1 6 5 46. 6 x 4 5 x 3 6x 24 5x 15 x 39 x 3 x 4 3 1 47. 1 3 6 3x 3 2 [multiply both sides by 6 x 1] 18 3x 3 2 3x 15 2 x 1 2 3x 3 3x 13 x 13 3 48.
49.
50.
51.
52. 53.
54.
5 12x 5 2 12x 5 x 2x 6x 3 5 6x 3 12x 2 5x 12x 2 6x 30x 15 6x 3 x 12x 2 5x 12x 2 24x 15 19x 15 x 15 19
1 1 10 1 10 z 1 5 z 1 2 z 1 10 10z [multiply both sides by 10z z 1] z 2z 5z z1 3 z 3 z 1 100z 3z 3 100z 3 97z 97
4 15 1 0 3 t 4 3 t 15 0 3 t 12 4t 15 0 3t 30 0 3t 30 3 t 3 t 9 t2 t 10 x 1 2 x 2 2x 4 2 [multiply both sides by 2 x 2] x 4x 8 2 3x 6 x 2. 2x 4 x 2 But substituting x 2 into the original equation does not work, since we cannot divide by 0. Thus there is no solution. 1 5 2 x 3 5 2 x 3 x 2 2x 6 x 4 x 3 x2 9 x 3 1 6x 12 3 3 x x 4 6x 12 (multiply both sides by x x 4] 3x 7x 16 4x 16 2 x 4 x x 4x x 4. But substituting x 4 into the original equation does not work, since we cannot divide by 0. Thus, there is no solution. 1 2 1 2 2x 1 2 x 1 1 1. This is an identity for x 0 and x 12 , so the solutions are x 2x 1 2x x all real numbers except 0 and 12 .
55. x 2 25 x 5
56. 3x 2 48 x 2 16 x 4 57. 5x 2 15 x 2 3 x 3
SECTION P.8 Solving Basic Equations
37
58. x 2 1000 x 1000 10 10
59. 8x 2 64 0 x 2 8 0 x 2 8 x 8 2 2 60. 5x 2 125 0 5 x 2 25 0 x 2 25 x 5 61. x 2 16 0 x 2 16 which has no real solution.
62. 6x 2 100 0 6x 2 100 x 2 50 3 , which has no real solution. 2 63. x 3 5 x 3 5 x 3 5 4 7 64. 3x 42 7 3x 4 7 3x 4 7 x 3 65. x 3 27 x 2713 3
66. x 5 32 0 x 5 32 x 3215 2 67. 0 x 4 16 x 2 4 x 2 4 x 2 4 x 2 x 2 x 2 4 0 has no real solution. If x 2 0, then x 2. If x 2 0, then x 2. The solutions are 2. 16 27 2716 3 27 6 6 68. 64x 27 x x 16 64 64 2 64
69. x 4 64 0 x 4 64 which has no real solution.
70. x 13 8 0 x 13 8 x 1 813 2 x 1. 14 8114 x 2 3. So x 2 3, then x 1. If 71. x 24 81 0 x 24 81 x 24 x 2 3, then x 5. The solutions are 5 and 1.
72. x 14 16 0 x 14 16, which has no real solution.
73. 3 x 33 375 x 33 125 x 3 12513 5 x 3 5 8 74. 4 x 25 1 x 25 14 x 2 5 14 x 2 5 14 75. 3 x 5 x 53 125 3 3 14 76. x 43 16 0 x 43 16 24 x 43 24 212 x 4 212 x 212 23 8 15 23 8 77. 2x 53 64 0 2x 53 64 x 53 32 x 3235 25 32 32 78. 6x 23 216 0 6x 23 216 x 23 36 62 x 23 62 x 63 216 944 313 302 161 80. 836 095x 997 095x 161 x 169 095 582 506 81. 215x 463 x 119 115x 582 x 119 195 059 82. 395 x 232x 200 195 332x x 332
79. 302x 148 1092 302x 944 x
4497 4366 103 84. 214 x 406 227 011x 214x 86684 227 011x 225x 109584 x 48704 487 026x 194 85. 176 026x 194 176 303 244x 026x 194 533 429x 455x 727 303 244x 727 160 x 455 173x 320 86. 151 173x 151 212 x 173x 320 151x 022x 320 x 1455 212 x 022 83. 316 x 463 419 x 724 316x 1463 419x 3034 4497 103x x
38
CHAPTER P Prerequisites
87. r
89. P V n RT R
88. d r T H T
PV nT
91. P 2l 2 2 P 2l 92.
93. 94. 95. 96. 97. 98.
d rH mM Fr 2 90. F G 2 m GM r
12 12 M M r
P 2l 2
1 1 1 R1 R2 R R2 R R1 (multiply both sides by the LCD, R R1 R2 ). Thus R1 R2 R R1 R R2 R R1 R2 R R2 . R1 R2 R R R2 R1 R2 R 3V 3V r V 13 r 2 h r 2 h h mM mM mM 2 r G F G 2 r G F F r 3V 3V 3 r V 43 r 3 r 3 4 4 2 2 2 2 2 2 a b c b c a b c2 a 2 i 2 i i 2 A i A A A 1 1 i 100 100 A P 1 1 100 P 100 100 P 100 P P a 2 x a 1 a 1 x a 2 x a 1 x a 1 a 2 a 1 x a 1 a 2 a 1 x a 1
a 1 x 2 a a1 2d b ax b 2 ax b 2 cx d ax b 2cx 2d ax 2cx 2d b a 2c x 2d b x 99. cx d a 2c a1 a1 b1 100. a a 1 a a 1 b b 1 a 2 a a 2 a b2 b 2a b2 b b b a a 12 b2 b
8 25 0032 250 25 000055. So the beam shrinks 10,000 10,000 000055 12025 0007 m, so when it dries it will be 12025 0007 12018 m long. 0032 25 (b) Substituting S 000050 we get 000050 5 0032 25 75 0032 10,000 75 234375. So the water content should be 234375 kg/m3 . 0032 3150 102. Substituting C 3600 we get 3600 450 375x 3150 375x x 840. So the toy manufacturer can 375 manufacture 840 toy trucks. 101. (a) The shrinkage factor when 250 is S
103. (a) Solving for when P 10,000 we get 10,000 156 3 3 64102 86 km/h.
(b) Solving for when P 50,000 we get 50,000 156 3 3 320513 147 km/h.
104. Substituting F 300 we get 300 03x 34 1000 103 x 34 x 14 10 x 104 10,000 lb. 105. (a) 3 0 k 5 k 0 k 1 k 5 k 1 2k 6 k 3
(b) 3 1 k 5 k 1 k 1 3 k 5 k k 1 k 2 1 k 3
(c) 3 2 k 5 k 2 k 1 6 k 5 2k k 1 k 1 k 1. x 2 is a solution for every value of k. That is, x 2 is a solution to every member of this family of equations.
106. When we multiplied by x, we introduced x 0 as a solution. When we divided by x 1, we are really dividing by 0, since x 1 x 1 0.
SECTION P.9 Modeling with Equations
P.9
39
MODELING WITH EQUATIONS
1. An equation modeling a real-world situation can be used to help us understand a real-world problem using mathematical methods. We translate real-world ideas into the language of algebra to construct our model, and translate our mathematical results back into real-world ideas in order to interpret our findings. 2. In the formula I Pr t for simple interest, P stands for principal, r for interest rate, and t for time (in years). 3. (a) A square of side x has area A x 2 .
(b) A rectangle of length l and width has area A l. (c) A circle of radius r has area A r 2 .
5 16 ounces 4. Balsamic vinegar contains 5% acetic acid, so a 32 ounce bottle of balsamic vinegar contains 32 5% 32 100
of acetic acid.
1 wall 1 . x hours x d rt d d rt d 6. Solving d rt for r, we find r . Solving d rt for t, we find t . t t t r r r 7. If n is the first integer, then n 1 is the middle integer, and n 2 is the third integer. So the sum of the three consecutive integers is n n 1 n 2 3n 3. 5. A painter paints a wall in x hours, so the fraction of the wall she paints in one hour is
8. If n is the middle integer, then n 1 is the first integer, and n 1 is the third integer. So the sum of the three consecutive integers is n 1 n n 1 3n. 9. If n is the first even integer, then n 2 is the second even integer and n 4 is the third. So the sum of three consecutive even integers is n n 2 n 4 3n 6.
10. If n is the first integer, then the next integer is n 1. The sum of their squares is n 2 n 12 n 2 n 2 2n 1 2n 2 2n 1.
11. If s is the third test score, then since the other test scores are 78 and 82, the average of the three test scores is 160 s 78 82 s . 3 3 12. If q is the fourth quiz score, then since the other quiz scores are 8, 8, and 8, the average of the four quiz scores is 24 q 888q . 4 4 13. If x dollars are invested at 2 12 % simple interest, then the first year you will receive 0025x dollars in interest. 14. If n is the number of months the apartment is rented, and each month the rent is $795, then the total rent paid is 795n. 15. Since is the width of the rectangle, the length is four times the width, or 4. Then area length width 4 42 ft2
16. Since is the width of the rectangle, the length is 4. Then perimeter 2 length 2 width 2 4 2 4 8 ft distance d 17. If d is the given distance, in miles, and distance rate time, we have time . rate 55 1h 18. Since distance rate time we have distance s 45 min 34 s mi. 60 min 19. If x is the quantity of pure water added, the mixture will contain 25 oz of salt and 3 x gallons of water. Thus the 25 . concentration is 3x 20. If p is the number of pennies in the purse, then the number of nickels is 2p, the number of dimes is 4 2 p, and the number of quarters is 2 p 4 2 p 4 p 4. Thus the value (in cents) of the change in the purse is 1 p 5 2 p 10 4 2 p 25 4 p 4 p 10 p 40 20 p 100 p 100 131 p 140.
40
CHAPTER P Prerequisites
21. If d is the number of days and m the number of miles, then the cost of a rental is C 65d 020m. In this case, d 3 80 400. Thus, and C 275, so we solve for m: 275 65 3 020m 275 195 02m 02m 80 m 02 Michael drove 400 miles. 22. If m is the number of messages, then a monthly cell phone bill (above $10) is B 10 010 m 1000. In this case, 285 285 B 385 and we solve for m: 385 10 010 m 1000 010 m 1000 285 m 1000 01 m 1285. Thus, Miriam sent 1285 text messages in June. 23. If x is Linh’s score on her final exam, then because the final counts twice as much as each midterm, her average score 228 2x 114 x 114 x 82 75 71 2x . For her to average 80%, we must have 80% 08 is 3 100 200 500 250 250 114 x 250 08 200 x 86. So Linh scored 86% on her final exam. 24. Six students scored 100 and three students scored 60. Let x be the average score of the remaining 25 6 3 16 students. 6 100 3 60 16x 084 780 16x 084 2500 2100 Because the overall average is 84% 084, we have 25 100 16x 1320 x 1320 16 825. Thus, the remaining 16 students’ average score was 825%. 25. Let m be the amount invested at 4 12 %. Then 12,000 m is the amount invested at 4%.
Since the total interest is equal to the interest earned at 4 12 % plus the interest earned at 4%, we have
525 0045m 004 12,000 m 525 0045m 480 004m 45 0005m m
45 9000. Thus 0005
$9000 is invested at 4 12 %, and $12,000 9000 $3000 is invested at 4%. 26. Let m be the amount invested at 5 12 %. Then 4000 m is the total amount invested. Thus
4 12 % of the total investment interest earned at 4% interest earned at 5 12 %
So 0045 4000 m 004 4000 0055m 180 0045m 160 0055m 20 001m m
20 2000. 001
Thus $2,000 needs to be invested at 5 12 %. 2625 0075 or 75%. 27. Using the formula I Prt and solving for r, we get 26250 3500 r 1 r 3500 28. If $1000 is invested at an interest rate a%, then 2000 is invested at a 12 %, so, remembering that a is expressed as a 1
a 2 a 1 2000 1 10a 20a 10 30a 10. Since the total interest 100 100 is $190, we have 190 30a 10 180 30a a 6. Thus, the $1000 is invested at 6% interest.
percentage, the total interest is I 1000
29. Let x be her monthly salary. Since her annual salary 12 monthly salary Christmas bonus we have 97,300 12x 8,500 88,800 12x x 7,400. Her monthly salary is $7,400. 30. Let s be the husband’s annual salary. Then her annual salary is 115s. Since husband’s annual salarywife’s annual salary total annual income, we have s 115s 69,875 215s 69,875 s 32,500. Thus the husband’s annual salary is $32,500. 31. Let x be the overtime hours Helen works. Since gross pay regular salary overtime pay, we obtain the equation 90 8. Thus Helen 35250 750 35 750 15 x 35250 26250 1125x 90 1125x x 1125 worked 8 hours of overtime. 32. Let x be the hours the assistant worked. Then 2x is the hours the plumber worked. Since the labor charge is equal to the plumber’s labor plus the assistant’s labor, we have 4025 45 2x 25x 4025 90x 25x 4025 115x x 4025 115 35. Thus the assistant works for 35 hours, and the plumber works for 2 35 70 hours.
SECTION P.9 Modeling with Equations
41
33. All ages are in terms of the daughter’s age 7 years ago. Let y be age of the daughter 7 years ago. Then 11y is the age of the movie star 7 years ago. Today, the daughter is y 7, and the movie star is 11y 7. But the movie star is also 4 times his daughter’s age today. So 4 y 7 11y 7 4y 28 11y 7 21 7y y 3. Thus the movie star’s age today is 11 3 7 40 years. 34. Let h be number of home runs Babe Ruth hit. Then h 41 is the number of home runs that Hank Aaron hit. So 1469 h h 41 1428 2h h 714. Thus Babe Ruth hit 714 home runs. 35. Let p be the number of pennies. Then p is the number of nickels and p is the number of dimes. So the value of the coins in the purse is the value of the pennies plus the value of the nickels plus the value of the dimes. Thus 144 001 p 005 p 010 p 144 016 p p 144 016 9. So the purse contains 9 pennies, 9 nickels, and 9 dimes. 36. Let q be the number of quarters. Then 2q is the number of dimes, and 2q 5 is the number of nickels. Thus 300 value of the nickels value of the dimes value of the quarters. So 300 005 2q 5 010 2q 025q 300 010q 025 020q 025q 275 055q q 275 055 5.
Thus Mary has 5 quarters, 2 5 10 dimes, and 2 5 5 15 nickels.
37. Let l be the length of the garden. Since area width length, we obtain the equation 1125 25l l 1125 25 45 ft. So the garden is 45 feet long. 38. Let be the width of the pasture. Then the length of the pasture is 2. Since area length width we have 115,200 2 22 2 57,600 240. Thus the width of the pasture is 240 feet. x
39. Let x be the length of a side of the square plot. As shown in the figure, area of the plot area of the building area of the parking lot. Thus,
x 2 60 40 12,000 2,400 12,000 14,400 x 120. So the plot of
x
land measures 120 feet by 120 feet.
60 40
40. Let be the width of the building lot. Then the length of the building lot is 5. Since a half-acre is 12 43,560 21,780 and area is length times width, we have 21,780 5 52 2 4,356 66. Thus the width of the building lot is 66 feet and the length of the building lot is 5 66 330 feet. base1 base2 height. Putting in the known quantities, we have 2 y 2y 120 y 32 y 2 y 2 80 y 80 4 5. Since length is positive, y 4 5 894 inches. 2
41. The figure is a trapezoid, so its area is
x
42. First we write a formula for the area of the figure in terms of x. Region A has dimensions 10 cm and x cm and region B has dimensions 6 cm and x cm. So the shaded region has area 10 x 6 x 16x cm2 . We are given that this is equal to 144 cm2 , so 144 16x x 144 16 9 cm.
10 cm
A
6 cm B
x
43. Let x be the width of the strip. Then the length of the mat is 20 2x, and the width of the mat is 15 2x. Now the perimeter is twice the length plus twice the width, so 102 2 20 2x 2 15 2x 102 40 4x 30 4x 102 70 8x 32 8x x 4. Thus the strip of mat is 4 inches wide. 44. Let x be the width of the strip. Then the width of the poster is 100 2x and its length is 140 2x. The perimeter of the printed area is 2 100 2 140 480, and the perimeter of the poster is 2 100 2x 2 140 2x. Now we use the
fact that the perimeter of the poster is 1 12 times the perimeter of the printed area: 2 100 2x 2 140 2x 32 480 480 8x 720 8x 240 x 30. The blank strip is thus 30 cm wide.
42
CHAPTER P Prerequisites
45. Let x be the length of the man’s shadow, in meters. Using similar triangles,
x 10 x 20 2x 6x 4x 20 6 2
x 5. Thus the man’s shadow is 5 meters long. 46. Let x be the height of the tall tree. Here we use the property that corresponding sides in similar triangles are proportional. The base of the similar triangles starts at x-5
x 5 150 15 25 25 x 5 15 150 25x 125 2250 25x 2375 x 95. Thus the eye level of the woodcutter, 5 feet. Thus we obtain the proportion
5
tree is 95 feet tall.
25
15
125
47. Let x be the amount (in mL) of 60% acid solution to be used. Then 300 x mL of 30% solution would have to be used to yield a total of 300 mL of solution. 60% acid
30% acid
Mixture
x
300 x
300
Rate (% acid)
060
030
050
Value
060x
030 300 x
050 300
mL
60 200. 03 So 200 mL of 60% acid solution must be mixed with 100 mL of 30% solution to get 300 mL of 50% acid solution.
Thus the total amount of pure acid used is 060x 030 300 x 050 300 03x 90 150 x
48. The amount of pure acid in the original solution is 300 50% 150. Let x be the number of mL of pure acid added. Then 150 x the final volume of solution is 300 x. Because its concentration is to be 60%, we must have 60% 06 300 x 30 150 x 06 300 x 150 x 180 06x 04x 30 x 75. Thus, 75 mL of pure acid must be 04 added. 49. Let x be the number of grams of silver added. The weight of the rings is 5 18 g 90 g.
Grams
Pure silver
Mixture
90
x
90 x
090
0
075
090 90
0x
075 90 x
Rate (% gold) Value
5 rings
So 090 90 0x 075 90 x 81 675 075x 075x 135 x 135 075 18. Thus 18 grams of silver
must be added to get the required mixture.
50. Let x be the number of liters of water to be boiled off. The result will contain 6 x liters.
Liters Concentration Amount
Original
Water
Final
6
x
6x
120
0
200
120 6
0
200 6 x
So 120 6 0 200 6 x 720 1200 200x 200x 480 x 24. Thus 24 liters need to be boiled off.
SECTION P.9 Modeling with Equations
43
51. Let x be the number of liters of coolant removed and replaced by water. 60% antifreeze
60% antifreeze (removed)
Water
Mixture
Liters
36
x
x
36
Rate (% antifreeze)
060
060
0
050
060 36
060x
0x
050 36
Value
036 06. Thus 06 liters 06
So 060 36 060x 0x 050 36 216 06x 18 06x 036 x must be removed and replaced by water.
52. Let x be the number of gallons of 2% bleach removed from the tank. This is also the number of gallons of pure bleach added to make the 5% mixture. Original 2%
Pure bleach
5% mixture
100 x
x
100
002
1
005
002 100 x
1x
005 100
Gallons Concentration Bleach
So 002 100 x x 005 100 2 002x x 5 098x 3 x 306. Thus 306 gallons need to removed and replaced with pure bleach. 53. Let c be the concentration of fruit juice in the cheaper brand. The new mixture that Jill makes will consist of 650 mL of the original fruit punch and 100 mL of the cheaper fruit punch. Original Fruit Punch
Cheaper Fruit Punch
Mixture
mL
650
100
750
Concentration
050
c
048
050 650
100c
048 750
Juice
So 050 650 100c 048 750 325 100c 360 100c 35 c 035. Thus the cheaper brand is only 35% fruit juice. 54. Let x be the number of ounces of $300oz tea Then 80 x is the number of ounces of $275oz tea.
Pounds
$300 tea
$275 tea
Mixture
x
80 x
80
Rate (cost per ounce)
300
275
290
Value
300x
275 80 x
290 80
So 300x 275 80 x 290 80 300x 220 275x 232 025x 12 x 48. The mixture uses 48 ounces of $300oz tea and 80 48 32 ounces of $275oz tea.
55. Let t be the time in minutes it would take Candy and Tim if they work together. Candy delivers the papers at a rate of 1 1 70 of the job per minute, while Tim delivers the paper at a rate of 80 of the job per minute. The sum of the fractions of the
job that each can do individually in one minute equals the fraction of the job they can do working together. So we have 1 1 1 560 8t 7t 560 15t t 37 1 minutes. Since 1 of a minute is 20 seconds, it would take them 70 80 3 3 t 37 minutes 20 seconds if they worked together. 56. Let t be the time, in minutes, it takes Hilda to mow the lawn. Since Hilda is twice as fast as Stan, it takes Stan 2t minutes to 1 1 mow the lawn by himself. Thus 40 40 1 40 20 t t 60. So it would take Stan 2 60 120 minutes t 2t to mow the lawn.
44
CHAPTER P Prerequisites
57. Let t be the time, in hours, it takes Karen to paint a house alone. Then working together, Karen and Betty can paint a house 1 1 1 3 in 23 t hours. The sum of their individual rates equals their rate working together, so 16 2 16 t t 2t t 3
6 t 9 t 3. Thus it would take Karen 3 hours to paint a house alone.
58. Let h be the time, in hours, to fill the swimming pool using Jim’s hose alone. Since Bob’s hose takes 20% less time, it uses 1 1 1 18 08 18 08h 144 18 08h 324 08h only 80% of the time, or 08h. Thus 18 18 h 08h h 405. Jim’s hose takes 405 hours, and Bob’s hose takes 324 hours to fill the pool alone.
59. Let t be the time in hours that Wendy spent on the train. Then 11 2 t is the time in hours that Wendy spent on the bus. We construct a table: Rate By train By bus
Time
40
t
60
11 t 2
Distance
40t
60 11 2 t
The total distance traveled is the sum of the distances traveled by bus and by train, so 300 40t 60 11 2 t
300 40t 330 60t 30 20t t 30 20 15 hours. So the time spent on the train is 55 15 4 hours. 60. Let r be the speed of the slower cyclist, in mi/h. Then the speed of the faster cyclist is 2r. Rate
Time
Distance
Slower cyclist
r
2
2r
Faster cyclist
2r
2
4r
When they meet, they will have traveled a total of 90 miles, so 2r 4r 90 6r 90 r 15. The speed of the slower cyclist is 15 mi/h, while the speed of the faster cyclist is 2 15 30 mi/h.
61. Let r be the speed of the plane from Montreal to Los Angeles. Then r 020r 120r is the speed of the plane from Los Angeles to Montreal. Rate Montreal to L.A.
r
L.A. to Montreal
12r
Time 2500 r 2500 12r
Distance 2500 2500
2500 55 2500 2500 2500 r 12r 6 r 12r ,000 55 12r 2500 6 12 2500 6 66r 18,000 15,000 66r 33,000 r 3366 500. Thus the plane flew The total time is the sum of the times each way, so 9 16
at a speed of 500 mi/h on the trip from Montreal to Los Angeles.
ft 22 62. Let x be the speed of the car in mi/h. Since a mile contains 5280 ft and an hour contains 3600 s, 1 mi/h 5280 3600 s 15 ft/s. 220 220 The truck is traveling at 50 22 15 3 ft/s. So in 6 seconds, the truck travels 6 3 440 feet. Thus the back end
of the car must travel the length of the car, the length of the truck, and the 440 feet in 6 seconds, so its speed must be 1430440 242 ft/s. Converting to mi/h, we have that the speed of the car is 242 15 55 mi/h. 6 3 3 22
63. Let x be the distance from the fulcrum to where the mother sits. Then substituting the known values into the formula given, we have 100 8 125x 800 125x x 64. So the mother should sit 64 feet from the fulcrum.
64. Let be the largest weight that can be hung. In this exercise, the edge of the building acts as the fulcrum, so the 240 lb man is sitting 25 feet from the fulcrum. Then substituting the known values into the formula given in Exercise 43, we have 240 25 5 6000 5 1200. Therefore, 1200 pounds is the largest weight that can be hung.
CHAPTER P
l 2 502 l 102 l 2 2500 l 2 20l 100 20l 2400 l 120.
45
l
65. Let l be the length of the lot in feet. Then the length of the diagonal is l 10. We apply the Pythagorean Theorem with the hypotenuse as the diagonal. So
Review
50
l+10
Thus the length of the lot is 120 feet.
66. Let r be the radius of the running track. The running track consists of two semicircles and two straight sections 110 yards long, so we get the equation 2r 220 440 2r 220 r 110 3503. Thus the radius of the semicircle is about 35 yards. 67. Let h be the height in feet of the structure. The structure is composed of a right cylinder with radius 10 and height 23 h and a cone with base radius 10 and height equation 1400 102 23 h
by
1 h. Using the formulas for the volume of a cylinder and that of a cone, we obtain the 3 1 102 1 h 1400 200 h 100 h 126 6h h (multiply both sides 3 3 3 9
9 ) 126 7h h 18. Thus the height of the structure is 18 feet. 100
68. Let h be the height of the break, in feet. Then the portion of the bamboo above the break is 10 h. Applying the Pythagorean Theorem, we obtain
h 2 32 10 h2 h 2 9 100 20h h 2 91 20h h 91 20 455. Thus the break is 455 ft above the ground.
10-h
h
3
69. Pythagoras was born about 569 BC in Samos, Ionia and died about 475 BC. Euclid was born about 325 BC and died about 265 BC in Alexandria, Egypt. Archimedes was born in 287 BC in Syracuse, Sicily and died in 212 BC in Syracuse. 70. Answers will vary.
CHAPTER P REVIEW 1. (a) Since there are initially 250 tablets and she takes 2 tablets per day, the number of tablets T that are left in the bottle after she has been taking the tablets for x days is T 250 2x. (b) After 30 days, there are 250 2 30 190 tablets left.
(c) We set T 0 and solve: T 250 2x 0 250 2x x 125. She will run out after 125 days.
2. (a) The total cost is $2 per calzone plus the $3 delivery charge, so C 2x 3. (b) 4 calzones would be 2 4 3 $11.
(c) We solve C 2x 3 15 2x 12 x 6. You can order six calzones. 3. (a) 16 is rational. It is an integer, and more precisely, a natural number. (b) 16 is rational. It is an integer, but because it is negative, it is not a natural number. (c) 16 4 is rational. It is an integer, and more precisely, a natural number. (d) 2 is irrational. (e) 83 is rational, but is neither a natural number nor an integer. (f) 82 4 is rational. It is an integer, but because it is negative, it is not a natural number. 4. (a) 5 is rational. It is an integer, but not a natural number. (b) 25 6 is rational, but is neither an integer nor a natural number. (c) 25 5 is rational, a natural number, and an integer. (d) 3 is irrational.
46
CHAPTER P Prerequisites 3 (e) 24 16 2 is rational, but is neither a natural number nor an integer. (f) 1020 is rational, a natural number, and an integer.
5. Commutative Property of addition.
6. Commutative Property of multiplication.
7. Distributive Property.
8. Distributive Property.
5 4 9 3 5 2 6 3 6 6 6 2 5 4 1 5 2 (b) 6 3 6 6 6 15 12 33 15 12 11. (a) 8 5 85 21 15 5 55 15 12 (b) 8 5 8 12 84 9. (a)
7 10 7 (b) 10 30 12. (a) 7 30 (b) 7
10. (a)
9 2 25 32
13. x [2 6 2 x 6 _2
11 21 22 1 15 30 30 30 11 21 22 43 15 30 30 30 12 30 35 55 35 7 12 12 12 30 12 6 12 35 7 35 77
14. x 0 10] 0 x 10 6
15. x 4] x 4
0
_2
18. x 3 x 3
17. x 5 x [5 5
_3
20. 0 x 12 x 0 12
19. 1 x 5 x 1 5] 5
0
21. (a) A B 1 0 12 1 2 3 4
22. (a) C D 1 2]
23. (a) A C 1 2 (b) B D 12 1
24. (a) A D 0 1 (b) B C 12 1
1 1 1 3 13 6 216 216 242 242 121 11 31. 2 2
29. 21613
33. (a) 5 3 2 2 (b) 5 3 8 8
1 2
(b) C D 0 1]
(b) A B 1
25. 7 10 3 3 27. 212 812 2 8 16 4
10
16. x [2 2 x 4
_1
25 2 72 49
26. 3 9 3 9 6 6
9 8 1 28. 23 32 18 19 72 72 72 23 30. 6423 43 42 16
32.
2 50 100 10
34. (a) 4 0 4 4 (b) 4 4 8 8
CHAPTER P
3 7 713 5 (b) 74 745
35. (a)
6
x 5 x 56 9 12 9 (b) x x x 92
37. (a)
3 7 5 573 3 3 (b) 4 5 514 534 12 38. (a) y 3 y 3 y 32 2 2 (b) 8 y y 18 y 14 3 2 4 a3 b b3 a 6 a 6 b2 b12 40. a 2 36. (a)
2 3x 1 y 2 4x 6 y 2 3x 1 y 2 4 3x 61 y 22 39. 2x 3 y 12x 5 y 4
x 4 3x2 x 4 9x 2 9x 423 9x 3 3 x x3 2 3 x 3 y y4 3 x 6 y4 y2 3 x 6 y6 x 2 y2 43. 41.
6
a 66 b212 b14
r 12 s 8 r 2 s 43 2 6 r 122 s 86 r 10 s 2 13 r s r s 2 y 2 x y 2 44. x 2 y 4 x 2 42.
52 8r 12 s 3 122 s 34 4r 52 s 7 4r 4r s7 2r 2 s 4 2 4 6 ab2 c3 a 2 b4 c6 2 a 26 b48 c6 4a 4 b12 c6 4a c 46. 2 2a 3 b4 22 a 6 b8 b12
45.
47. 78,250,000,000 7825 1010
48. 208 108 00000000208 293 106 1582 1014 000000293 1582 1014 ab 293 1582 49. 1061412 12 12 c 28064 28064 10 28064 10 165 1032 50. 80
times 60 minutes 24 hours 365 days 90 years 38 109 times minute hour day year
51. 2x 2 y 6x y 2 2x y x 3y
52. 12x 2 y 4 3x y 5 9x 3 y 2 3x y 2 4x y 2 y 3 3x 2
53. x 2 5x 14 x 7 x 2 2 54. x 4 x 2 2 x 2 x 2 2 x 2 2 x 2 1 x 2 2 x 1 x 1 55. 3x 2 2x 1 3x 1 x 1
56. 6x 2 x 12 3x 4 2x 3
57. 4t 2 13t 12 4t 3 t 4 2 58. x 4 2x 2 1 x 2 1 [x 1 x 1]2 x 12 x 12 59. 16 4t 2 4 4 t 2 4 t 2 t 2 60. 2y 6 32y 2 2y 2 y 4 16 2y 2 y 2 4 y 2 4 2y 2 y 2 4 y 2 y 2 61. x 6 1 x 3 1 x 3 1 x 1 x 2 x 1 x 1 x 2 x 1 62. a 4 b2 ab5 ab2 a 3 b3 ab2 a b a 2 ab b2 63. x 3 27 x 3 x 2 3x 9 64. 3y 3 81x 3 3 y 3 27x 3 3 y 3x y 2 3x y 9x 2
Review
47
48
CHAPTER P Prerequisites
65. 4x 3 8x 2 3x 6 4x 2 x 2 3 x 2 4x 2 3 x 2 66. 3x 3 2x 2 18x 12 x 2 3x 2 6 3x 2 3x 2 x 2 6 67. x y2 7 x y 6 x y 6 x y 1 x y 6 x y 1 68. a b2 3 a b 10 a b 5 a b 2
69. 2y 7 2y 7 4y 2 49
70. 1 x 2 x 3 x 3 x 2 x x 2 9 x 2 2 x x 2 9 x 2 7 x 71. x 2 x 2 x x 22 x 3 2x 2 x x 2 4x 4 x 3 2x 2 x 3 4x 2 4x 2x 3 6x 2 4x x x 2 2x 3 x 3 2x 2 3x x 2 2x 3 72. x x x 1 2 x 1 x x 2 x 1 2x x x 2x x 2x x x x 73. x
74. 2x 13 2x3 3 2x2 1 3 2x 12 13 8x 3 12x 2 6x 1
x 2 2x 3 x 3 x 3 x 1 2 2x 3 3 1 2x x 2x 5x 3 t 1 t 2 t 1 t3 1 t2 t 1 76. 2 t 1 t 1 t 1 t 1 75.
77. 78.
x 2 2x 3 3x 12 3 x 3 x 3 x 1 3 x 4 x 4 x 4 x 4 x 1 x 2 8x 16 x 1
x 2 2x 15 x 2 x 12 x 1 x 5 x 3 x 1 x 1 2 2 x 4 5 1 4 3 x x x x x 6x 5 x 1
79. x 80.
81.
82.
x x 1 1 x2 x 1 1 x 1 x 1 x 1 x 1
x x2 1 x x 1 x2 1 x2 x x 1 1 2 2 2 x 1 x 1 x 1 x 1 x 1 x 1 x 1 x 2 1 x 1 x 2 1
2 x 22 x x 2 3x 1 3 2 2 2 2 x x 2 x 2 x x 2 x x 2 x x 22 2 x 2 4x 4 x 2 2x 3x 2x 2 8x 8 x 2 2x 3x 3x 2 7x 8 2 2 x x 2 x x 2 x x 22 1 1 2 1 2 1 x 2 x2 4 x2 x 2 x 2 x 2 x 2 x 2 x 1 x 1 2 x 2 x 2 x 1 x 2 x 1 x 2 x 2 x 1 x 2 x 2 x 1 x 2
1 x 83. x 1 84. x 1 x
x 2 2x 5 x 2 x 2 x 1 2x 4 x 2 x 1 x 2 x 2 x 1 x 2
x 2 2x 1 1 x 2 1 1 2x 2x 2 x 2 2x x 2 2x x 2 2x 1 1 1 1 x 1 x x 1 x x 1 x 1 x 1 1 1 x x 1 2x 1 x 1 x x 1 x x 1 12
CHAPTER P
85.
86. 87. 88. 89. 90.
91.
92. 93. 94. 95.
96.
3 x h2 5 x h 3x 2 5x
Review
49
3x 2 6xh 3h 2 5x 5h 3x 2 5x 6xh 3h 2 5h h h h h 6x 3h 5 6x 3h 5 h x h x x h x x h x x h x h 1 h h x h x h x h x h x h x x h x 1 1 11 11 11 11 11 11 3 6 6 3 3 6 6 2 6 6 6 10 21 10 10 10 2 10 10 2 21 21 21 21 14 3 2 42 14 2 42 14 2 14 62 2 2 7 3 2 3 2 3 2 32 2 x x 2 x 2 x 2x x x x 2 x 4 2 x 2 x 2 x 22 x x 4 x 4 x 2 x 2 x 2 x 4 x 2 x 2 x 2 x 5 is defined whenever x 10 0 x 10, so its domain is x x 10. x 10 2x is defined whenever x 2 9 0 x 2 9 x 3, so its domain is x x 3 and x 3. 2 x 9 x is defined whenever x 0 (so that x is defined) and x 2 3x 4 x 1 x 4 0 x 1 and x 2 3x 4 x 4. Thus, its domain is x x 0 and x 4. x 3 is defined whenever x 3 0 x 3 and x 2 4x 4 x 22 0 x 2. Thus, its domain is x 2 4x 4 x x 3.
97. This statement is false. For example, take x 1 and y 1. Then LHS x y3 1 13 23 8, while RHS x 3 y 3 13 13 1 1 2, and 8 2. 1a 1 a 1 1 a 1 a 98. This statement is true for a 1: . 1a 1a 1 a 1 a 1 a 1 a
99. This statement is true:
12 y 12 y 12 1. y y y y
100. This statement is false. For example, take a 1 and b 1. Then LHS 3 a b 3 1 1 3 2, while RHS 3 a 3 b 3 1 3 1 1 1 2, and 3 2 2. 101. This statement is false. For example, take a 1. Then LHS a 2 12 1 1, which does not equal a 1. The true statement is a 2 a. 1 1 1 1 1 1 5 1 , while RHS , 102. This statement is false. For example, take x 1. Then LHS x 4 14 5 x 4 1 4 4 1 5 and . 5 4 103. 3x 12 24 3x 12 x 4
104. 5x 7 42 5x 49 x 49 5
50
CHAPTER P Prerequisites
105. 7x 6 4x 9 3x 15 x 5
106. 8 2x 14 x 3x 6 x 2
107. 13 x 12 2 2x 3 12 2x 15 x 15 2 6 3 108. 23 x 35 15 2x 10x 9 3 30x 40x 6 x 40 20 109. 2 x 3 4 x 5 8 5x 2x 6 4x 20 8 5x 2x 26 8 5x 3x 18 x 6
x 5 2x 5 5 3 x 5 2 2x 5 5 3x 15 4x 10 5 x 30 x 30 2 3 6 x 1 2x 1 111. x 1 2x 1 2x 1 x 1 2x 2 3x 1 2x 2 3x 1 6x 0 x 0 x 1 2x 1
110.
112. 113.
1 7 x 3 x 3 x 2 1 x 3x 6 1 2x 7 x x 2 x 2 2
3x 3x x x 1 x 1 x 2 x x 1 x 2 x 2 x 2 x 2 0. Since this x 1 3x 6 3 x 2 x 2 last equation is never true, there is no real solution to the original equation.
114. x 22 x 42 x 22 x 42 0 [x 2 x 4] [x 2 x 4] 0 [x 2 x 4] [x 2 x 4] 6 2x 2 0 2x 2 0 x 1. 115. x 2 144 x 12
7 116. 4x 2 49 x 2 49 4 x 2
117. x 3 27 0 x 3 27 x 3.
118. 6x 4 15 0 6x 4 15 x 4 52 . Since x 4 must be nonnegative, there is no real solution.
119. x 13 64 x 1 4 x 1 4 5. 120. x 22 2 0 x 22 2 x 2 2 x 2 2. 121. 3 x 3 x 33 27. 2 122. x 23 4 0 x 13 4 x 13 2 x 8.
123. 4x 34 500 0 4x 34 500 x 34 125 x 12543 54 625. 124. x 215 2 x 2 25 32 x 2 32 34. 125. A
xy 2A x y x 2A y. 2
126. V x y yz xz V y x z xz V xz y x z y
V xz . x z
1 1 1 1 1 11 11 tJ 1 t , J 0. t 2t 3t 2 3 6 6J q q q q q q 128. F k 1 2 2 r 2 k 1 2 r k 1 2 . (In real-world applications, r represents distance, so we would take the F F r positive root.)
127. Multiply through by t: J
129. Let x be the number of pounds of raisins. Then the number of pounds of nuts is 50 x. Pounds Rate (cost per pound)
Raisins
Nuts
Mixture
x
50 x
50
320
240
272
So 320x 240 50 x 272 50 320x 120 240x 136 08x 16 x 20. Thus the mixture uses 20 pounds of raisins and 50 20 30 pounds of nuts.
CHAPTER P
Test
51
130. Let t be the number of hours that Anthony drives. Then Helen drives for t 14 hours. Anthony Helen
Rate
Time
Distance
45
t
40
t 14
45t 40 t 14
When they pass each other, they will have traveled a total of 160 miles. So 45t 40 t 14 160 45t 40t 10 160 85t 170 t 2. Since Anthony leaves at 2:00 P. M . and travels for 2 hours, they pass each other at 4:00 P. M . 131. Let x be the amount invested in the account earning 15% interest. Then the amount invested in the account earning 25% is 7000 x. Amount invested Interest earned
15% Account
25% Account
Total
x
7000 x
7000
0015x
0025 7000 x
12025
From the table, we see that 0015x 0025 7000 x 12025 0015x 175 0025x 12025 5475 001x x 5475. Thus, Luc invested $5475 in the account earning 15% interest and $1525 in the account earning 25% interest. 132. The amount of interest Shania is currently earning is 6000 003 $180. If she wishes to earn a total of $300, she must earn another $120 in interest at a rate of 125% per year. If the additional amount invested is x, we have the equation 00125x 120 x 9600. Thus, Shania must invest an additional $9600 at 125% simple interest to earn a total of $300 interest per year. 133. Let t be the time it would take Abbie to paint a living room if she works alone. It would take Beth 2t hours to paint the 1 living room alone, and it would take 3t hours for Cathie to paint the living room. Thus Abbie does of the job per hour, t 1 1 1 1 1 of the job per hour, and Cathie does of the job per hour. So 1 6 3 2 6t Beth does 2t 3t t 2t 3t 6t 11 t 11 6 . So it would Abbie 1 hour 50 minutes to paint the living room alone. 134. Let be width of the pool. Then the length of the pool is 2, and its volume is 8 2 8464 162 8464 2 529 23. Since 0, we reject the negative value. The pool is 23 feet wide, 2 23 46 feet long, and 8 feet deep.
CHAPTER P TEST 1. (a) The cost is C 9 15x.
(b) There are four extra toppings, so x 4 and C 9 15 4 $15.
2. (a) 5 is rational. It is an integer, and more precisely, a natural number. (b) 5 is irrational. (c) 93 3 is rational, and it is an integer.
(d) 1,000,000 is rational, and it is an integer.
3. (a) A B 0 1 5 (b) A B 2 0 12 1 3 5 7 4. (a) _4
2
[4 2
0
3
[0 3]
52
CHAPTER P Prerequisites
(b) 0
(c) 4 2 6 6
_4
[4 2 [0 3] [0 2
5. (a) 26 64 (e)
2
2 2 4 2 3 2 3 9
(b) 26 64 5 1 32 2 (f) 4 2 16
3
[4 2 [0 3] [4 3] 1 1 (c) 26 6 64 2 8 32 9 4 3 (g) 16 216 24
6. (a) 186,000,000,000 186 1011
710 1 (d) 12 72 49 7 34 1 (h) 8134 34 33 27
(b) 00000003965 3965 107
a 3 b2 a2 b ab3 2 y4 6 (b) 2x 3 y 2 4x 2 (c) 2x 12 y 2 3x 14 y 1 2 32 x 12214 y 221 18x (d) 20 125 4 5 25 5 2 5 5 5 3 5 2 (e) 18x 3 y 4 9 2 x 2 x y 2 3x y 2 2x 2 1 2x 2 y 22 x 2232 y 2122 10 (f) x 3 y 12 4x y
7. (a)
8. (a) 3 x 6 4 2x 5 3x 18 8x 20 11x 2
(b) x 3 4x 5 4x 2 5x 12x 15 4x 2 7x 15 2 2 a b a b a b ab (c) (d) 2x 32 2x2 2 2x 3 32 4x 2 12x 9
(e) x 23 x3 3 x2 2 3 x 22 23 x 3 6x 2 12x 8 (f) x 2 x 3 x 3 x 2 x 2 9 x 4 9x 2
9. (a) 4x 2 25 2x 5 2x 5
(b) 2x 2 5x 12 2x 3 x 4
(c) x 3 3x 2 4x 12 x 2 x 3 4 x 3 x 3 x 2 4 x 3 x 2 x 2 (d) x 4 27x x x 3 27 x x 3 x 2 3x 9
(e) 2x y2 10 2x y 25 2x y2 2 5 2x y 52 2x y 52 (f) x 3 y 4x y x y x 2 4 x y x 2 x 2 10. (a) (b) (c)
x 2 3x 2 x 2 x 1 x 2 x 2 x 1 x 2 x2 x 2
x 1 2x 2 x 1 x 3 2x 1 x 1 x 3 2 2x 1 x 3 x 3 x 3 2x 1 x 9 x2
x2 4
x2 x2 x 1 x 1 x 2 x 1 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 2 2 x x x 2 x 2 1 x 2 x 2 x 2 x 2 x 2
CHAPTER P
Test
53
y y x x y2 x 2 x y y x y x y x x y x y xy y x (d) 1 1 xy 1 1 xy xy xy y x y x 3 6 6 632 6 2 11. (a) 332 3 3 2 3 2 3 2 4 2 2 2 10 5 2 10 10 52 (b) 50 2 10 5 2 2 10 54 52 52 52 1 1 x 1 x 1 (c) 1x 1 x 1 x 1 x
12. (a) 4x 3 2x 7 4x 2x 7 3 2x 10 x 5. 3 (b) 8x 3 125 8x 3 3 125 2x 5 x 52 . 32 6432 x 83 512. (c) x 23 64 0 x 23 64 x 23
x 3 x x 2x 1 x 3 2x 5 2x 2 x 2x 2 5x 6x 15 x x 15 2x 15 2x 5 2x 1 x 15 2 . (e) 3 x 12 18 0 3 x 12 18 x 12 6 x 1 6 x 6 1 E E 2 2 c c . (We take the positive root because c represents the speed of light, which is positive.) 13. E mc m m 14. Let d be the distance in km, between Bedingfield and Portsmouth. (d)
Direction
Distance
Rate
Bedingfield Portsmouth
d
100
Portsmouth Bedingfield
d
75
Time d 100 d 75
distance to fill in the time column of the table. We are given that the sum of the times is 35 hours. rate d d d 1050 d 35 300 300 35 3d 4d 1050 d 150 km. Thus we get the equation 100 75 100 75 7
We have used time
54
FOCUS ON MODELING
FOCUS ON MODELING Making the Best Decisions
1. (a) The total cost is
cost of
maintenance
copier
cost
number of months
copy cost
number of months
. Each month
the copy cost is 8000 003 240. Thus we get C1 5800 25n 240n 5800 265n. rental number copy number . Each month the copy cost is (b) In this case the cost is cost of months cost of months
(c)
8000 006 480. Thus we get C2 95n 480n 575n. Years
n
Purchase
Rental
1
12
8,980
6,900
2
24
12,160
13,800
3
36
15,340
20,700
4
48
18,520
27,600
5
60
21,700
34,500
6
72
24,880
41,400
(d) The cost is the same when C1 C2 are equal. So 5800 265n 575n 5800 310n n 1871 months.
daily
daily
2. (a) The cost of Plan 1 is 3
cost
The cost of Plan 2 is 3
cost
cost per mile
number of miles
3 65 015x 195 015x.
3 90 270.
(b) When x 400, Plan 1 costs 195 015 400 $255 and Plan 2 costs $270, so Plan 1 is cheaper. When x 800, Plan 1 costs 195 015 800 $315 and Plan 2 costs $270, so Plan 2 is cheaper.
(c) The cost is the same when 195 015x 270 015 75x x 500. So both plans cost $270 when the businessman drives 500 miles.
3. (a) The total cost is
(b) The revenue is
setup cost
cost per
number
price per tire
tire
of tires
number of tires
. So C 8000 22x.
. So R 49x.
(c) Profit Revenue Cost. So P R C 49x 8000 22x 27x 8000.
(d) Break even is when profit is zero. Thus 27x 8000 0 27x 8000 x 2963. So they need to sell at least 297 tires to break even.
Making the Best Decisions
55
4. (a) Option 1: In this option the width is constant at 100. Let x be the increase in length. Then the additional area is increase 100x. The cost is the sum of the costs of moving the old fence, and of installing the width in length
new one. The cost of moving is $6 100 $600 and the cost of installation is 2 10 x 20x, so the total cost is C 600 . Substituting in the area C 20x 600. Solving for x, we get C 20x 600 20x C 600 x 20 C 600 we have A1 100 5 C 600 5C 3,000. 20 Option 2: In this option the length is constant at 180. Let y be the increase in the width. Then the additional area is increase 180y. The cost of moving the old fence is 6 180 $1080 and the cost of installing the new length in width
(b)
one is 2 10 y 20x, so the total cost is C 20y 1080. Solving for y, we get C 20y 1080 20y C 1080 C 1080 C 1080 . Substituting in the area we have A2 180 9 C 1080 9C 9,720. y 20 20 Cost, C
Area gain A1 from Option 1
$1100
2,500 ft2
Area gain A2 from Option 2 180 ft2
$1200
3,000 ft2
1,080 ft2
$1500
4,500 ft2
3,780 ft2
$2000
7,000 ft2
8,280 ft2
$2500
9,500 ft2
12,780 ft2
$3000
12,000 ft2
17,280 ft2
(c) If the farmer has only $1200, Option 1 gives him the greatest gain. If the farmer has only $2000, Option 2 gives him the greatest gain. 5. (a) Design 1 is a square and the perimeter of a square is four times the length of a side. 24 4x, so each side is x 6 feet long. Thus the area is 62 36 ft2 .
12 . Thus, the area is Design 2 is a circle with perimeter 2r and area r 2 . Thus we must solve 2r 24 r 2 12 144 458 ft2 . Design 2 gives the largest area.
(b) In Design 1, the cost is $3 times the perimeter p, so 120 3 p and the perimeter is 40 feet. By part (a), each side is then 40 10 feet long. So the area is 102 100 ft2 . 4
In Design 2, the cost is $4 times the perimeter p. Because the perimeter is 2r, we get 120 4 2r so 2 120 15 225 15 r . The area is r 2 716 ft2 . Design 1 gives the largest area. 8
6. (a) Plan 1: Tomatoes every year. Profit acres Revenue cost 100 1600 300 130,000. Then for n years the profit is P1 130,000n. (b) Plan 2: Soybeans followed by tomatoes. The profit for two years is Profit acres soybean tomato 100 1200 1600 280,000. Remember that no fertilizer is revenue revenue needed in this plan. Then for 2k years, the profit is P2 280,000k.
(c) When n 10, P1 130,000 10 1,300,000. Since 2k 10 when k 5, P2 280,000 5 1,400,000. So Plan B is more profitable.
56
FOCUS ON MODELING
7. (a) Data (GB) 1 15 2 25 3 35 4
Plan A
Plan B $25
25 5 200 $35
Plan C $40
40 5 150 $4750
$60 60 5 100 $65
25 10 200 $45
40 10 150 $55
60 10 100 $70
25 20 200 $65
40 20 150 $70
60 20 100 $80
25 15 200 $55 25 25 200 $75 25 30 200 $85
40 15 150 $6250 40 25 150 $7750
40 30 150 $85
60 15 100 $75 60 25 100 $85 60 30 100 $90
(b) For Plan A: CA 25 2 10x 10 20x 5. For Plan B: CB 40 15 10x 10 15x 25. For Plan C: CC 60 1 10x 10 10x 50. Note that these equations are valid only for x 1.
(c) If Gwendolyn uses 22 GB, Plan A costs 25 12 2 $49, Plan B costs 40 12 15 $58, and Plan C costs 60 12 1 $72. If she uses 37 GB, Plan A costs 25 27 2 $79, Plan B costs 40 27 15 $8050, and Plan C costs 60 27 1 $87. If she uses 49 GB, Plan A costs 25 39 2 $103, Plan B costs 40 39 15 $9850, and Plan C costs 60 39 1 $99.
(d) (i) We set CA CB 20x 5 15x 25 5x 20 x 4. Plans A and B cost the same when 4 GB are used. (ii) We set CA CC 20x 5 10x 50 10x 45 x 45. Plans A and C cost the same when 45 GB are used. (iii) We set CB CC 15x 25 10x 50 5x 25 x 5. Plans B and C cost the same when 5 GB are used. 8. (a) In this plan, Company A gets $32 million and Company B gets $32 million. Company A’s investment is $14 million, so they make a profit of 32 14 $18 million. Company B’s investment is $26 million, so they make a profit of 32 26 $06 million. So Company A makes three times the profit that Company B does, which is not fair.
(b) The original investment is 14 26 $4 million. So after giving the original investment back, they then share the profit of $24 million. So each gets an additional $12 million. So Company A gets a total of 14 12 $26 million and Company B gets 26 12 $38 million. So even though Company B invests more, they make the same profit as Company A, which is not fair.
(c) The original investment is $4 million, so Company A gets 14 4 64 $224 million and Company B gets 26 64 $416 million. This seems the fairest. 4
1
EQUATIONS AND GRAPHS
1.1
THE COORDINATE PLANE
1. The point that is 2 units to the left of the y-axis and 4 units above the x-axis has coordinates 2 4. 2. If x is positive and y is negative, then the point x y is in Quadrant IV. 3. The distance between the points a b and c d is c a2 d b2 . So the distance between 1 2 and 7 10 is 7 12 10 22 62 82 36 64 100 10. 4. The point midway between a b and c d is 8 12 1 7 2 10 4 6. 2 2 2 2
ac bd . So the point midway between 1 2 and 7 10 is 2 2
5. A 5 1, B 1 2, C 2 6, D 6 2, E 4 1, F 2 0, G 1 3, H 2 2 6. Points A and B lie in Quadrant 1 and points E and G lie in Quadrant 3. 7. 0 5, 1 0, 1 2, and 12 23
8. 5 0, 2 0, 26 13, and 25 35
y
(0, 5)
(_1, 0) 1 (_1, _2)
9. x y x 2
(1/2, 2/3) 1
x
10. x y y 2
y 5l
0
y 5l
5
x
0
5
x
57
58
CHAPTER 1 Equations and Graphs
11. x y x 4
12. x y y 3
y
y 5l
5l
0
5
0
x
13. x y 3 x 3
14. x y 0 y 2
y
5l
5
x
5
x
y 5l
0
5
x
15. x y x y 0
x y x 0 and y 0 or x 0 and y 0 y
1
0
16. x y x y 0
x y x 0 and y 0 or x 0 and y 0 y
1 1
x
17. x y x 1 and y 3
1
x
18. x y x 2 and y 1 y
y
1l
1 1
x
1
x
SECTION 1.1 The Coordinate Plane
19. x y 1 x 1 and 2 y 2
20. x y 3 x 3 and 1 y 1
y
y
1
1 1
1
x
x
21. The two points are 0 2 and 3 0. (a) d 3 02 0 22 32 22 9 4 13 30 02 32 1 (b) midpoint: 2 2 22. The two points are 2 1 and 2 2. (a) d 2 22 1 22 42 32 16 9 25 5 2 2 1 2 0 12 (b) midpoint: 2 2 23. The two points are 3 3 and 5 3. (a) d 3 52 3 32 82 62 64 36 100 10 3 5 3 3 (b) midpoint: 1 0 2 2 24. The two points are 2 3 and 4 1. (a) d 2 42 3 12 62 22 36 4 50 2 10 2 4 3 1 1 2 (b) midpoint: 2 2 25. (a)
y
y
26. (a) (_2, 5) (6, 16)
1l
(10, 0) 2
(0, 8) 2l 1
(b) d 0 62 8 162 62 82 100 10 0 6 8 16 (c) Midpoint: 3 12 2 2
x
2 102 5 02 122 52 169 13 2 10 5 0 (c) Midpoint: 4 52 2 2
(b) d
x
59
60
CHAPTER 1 Equations and Graphs y
27. (a) (_4,5)
y
28. (a)
5l (_1, 1)
0
5
1
(3,_2)
x
(_6, _3)
3 42 2 52 72 72 49 49 98 7 2 4 3 5 2 12 32 (c) Midpoint: 2 2
(b) d
y
29. (a)
1l
x
1 62 1 32 52 42 41 6 1 3 1 72 1 (c) Midpoint: 2 2
(b) d
30. (a)
y
(_6, 2) 1l
1l 1
(6, _2)
1
x
(5, 0) x
(0, _6)
6 62 2 22 122 42 144 16 160 4 10 6 6 2 2 (c) Midpoint: 0 0 2 2
(b) d
0 52 6 02 52 62 25 36 61 0 5 6 0 (c) Midpoint: 52 3 2 2
(b) d
SECTION 1.1 The Coordinate Plane
31. d A B 1 52 3 32 42 4. d A C 1 12 3 32 62 6. So the area is 4 6 24.
y
32. The area of a parallelogram is its base times its height. Since two sides are parallel to the x-axis, we use the length of one of these as the base. Thus, the base is d A B 1 52 2 22 42 4. The
height is the change in the y coordinates, thus, the height
A
is 6 2 4. So the area of the parallelogram is
B
base height 4 4 16.
1l 1
y
C
x
C
D 1l
A
D
B
1
33. From the graph, the quadrilateral ABC D has a pair of parallel sides, so ABC D is a trapezoid. The area is b1 b2 h. From the graph we see that 2 b1 d A B 1 52 0 02 42 4; b2 d C D 4 22 3 32 22 2; and h is the difference in y-coordinates is 3 0 3. Thus 42 the area of the trapezoid is 3 9. 2
1l A 1
we find the length of one side and square it. This gives d Q R 5 02 1 62 52 52 25 25 50 2 50 50. So the area is
C
y
R B
Thus point A 6 7 is closer to the origin.
Q
P
1l 1
x
6 02 7 02 62 72 36 49 85. d 0 B 5 02 8 02 52 82 25 64 89.
35. d 0 A
x
34. The point S must be located at 0 4. To find the area,
y
D
61
x
62
CHAPTER 1 Equations and Graphs
6 22 3 12 42 22 16 4 20. d E D 3 22 0 12 52 12 25 1 26.
36. d E C
Thus point C is closer to point E.
1 32 1 12 42 22 16 4 20 2 5. d Q R 1 12 1 32 0 42 16 4. Thus point Q 1 3 is closer to point R.
37. d P R
38. (a) The distance from 7 3 to the origin is 7 02 3 02 72 32 49 9 58. The distance from 3 7 to the origin is 3 02 7 02 32 72 9 49 58. So the points are the same distance from the origin.
(b) The distance from a b to the origin is a 02 b 02 a 2 b2 . The distance from b a to the origin is b 02 a 02 b2 a 2 a 2 b2 . So the points are the same distance from the origin.
39. Since we do not know which pair are isosceles, we find the length of all three sides. d A B 3 02 1 22 32 32 9 9 18 3 2. d C B 3 42 1 32 12 42 1 16 17. d A C 0 42 2 32 42 12 16 1 17. So sides AC and C B have the same length. 40. Since the side AB is parallel to the x-axis, we use this as the base in the formula area 12 base height. The height is the change in the y-coordinates. Thus, the base is 2 4 6 and the height is 4 1 3. So the area is 12 6 3 9.
41. (a) Here we have A 2 2, B 3 1, and C 3 3. So d A B 3 22 1 22 12 32 1 9 10; d C B 3 32 1 32 62 22 36 4 40 2 10; d A C 3 22 3 22 52 52 25 25 50 5 2.
Since [d A B]2 [d C B]2 [d A C]2 , we conclude that the triangle is a right triangle. (b) The area of the triangle is 12 d C B d A B 12 10 2 10 10.
52 42 25 16 41; 11 62 3 72 16 25 41; d A C 2 62 2 72 42 52 d B C 2 112 2 32 92 12 81 1 82.
42. d A B
Since [d A B]2 [d A C]2 [d B C]2 , we conclude that the triangle is a right triangle. The area is 1 41 41 41 2 2.
SECTION 1.1 The Coordinate Plane
63
43. We show that all sides are the same length (its a rhombus) and then show that the diagonals are equal. Here we have A 2 9, B 4 6, C 1 0, and D 5 3. So d A B 4 22 6 92 62 32 36 9 45; d B C 1 42 0 62 32 62 9 36 45; d C D 5 12 3 02 62 32 36 9 45; d D A 2 52 9 32 32 62 9 36 45. So the points form a rhombus. Also d A C 1 22 0 92 32 92 9 81 90 3 10, and d B D 5 42 3 62 92 32 81 9 90 3 10. Since the diagonals are equal, the rhombus is a square.
42 82 16 64 80 4 5. 3 12 11 32 5 32 15 112 22 42 4 16 20 2 5. d B C d A C 5 12 15 32 62 122 36 144 180 6 5. So d A B d B C d A C,
44. d A B
and the points are collinear.
45. Let P 0 y be such a point. Setting the distances equal we get 0 52 y 52 0 12 y 12 25 y 2 10y 25 1 y 2 2y 1 y 2 10y 50 y 2 2y 2 12y 48 y 4. Thus, the point is P 0 4. Check: 0 52 4 52 52 12 25 1 26; 0 12 4 12 12 52 25 1 26.
2 3. So the length of the median CC is d C C 18 02 2 2 92 1 . So the length of the median B B 2 8 3 2 37. The midpoint of AC is B 2 2 9 3 2 1 62 109 . The midpoint of BC is A 3 8 6 2 11 4 . So the length is d B B 2 2 2 2 2 2 11 1 4 02 145 . of the median A A is d A A 2 2
46. The midpoint of AB is C
13 06 2 2
47. As indicated by Example 3, we must find a point S x1 y1 such that the midpoints
y
of P R and of QS are the same. Thus 4 1 2 4 x1 1 y1 1 . Setting the x-coordinates equal, 2 2 2 2
x 1 4 1 1 4 1 x1 1 x1 2Setting the 2 2 y 1 2 4 1 2 4 y1 1 y1 3. y-coordinates equal, we get 2 2 Thus S 2 3. we get
Q 1l
P
1
R
x
64
CHAPTER 1 Equations and Graphs
48. We solve the equation 6 8
2x 2x to find the x coordinate of B. This gives 6 12 2 x x 10. Likewise, 2 2
3y 16 3 y y 13. Thus, B 10 13. 2 y
49. (a)
C D B
1l A
1
2 7 1 7 2 2 41 24 of B D is 52 3 . 2 2
(b) The midpoint of AC is
52 3 , the midpoint
(c) Since the they have the same midpoint, we conclude that the x
diagonals bisect each other.
a b a0 b0 . Thus, 2 2 2 2 2 a 2 b b2 a2 a 2 b2 d C M 0 0 ; 2 2 4 4 2 2 a a 2 b 2 2 b a2 a 2 b2 b2 d A M a 0 ; 2 2 2 2 4 4 2 2 a a 2 b 2 2 b a2 a 2 b2 b2 d B M 0 b . 2 2 2 2 4 4 2
50. We have M
51. (a) The point 5 3 is shifted to 5 3 3 2 8 5. (b) The point a b is shifted to a 3 b 2.
(c) Let x y be the point that is shifted to 3 4. Then x 3 y 2 3 4. Setting the x-coordinates equal, we get x 3 3 x 0. Setting the y-coordinates equal, we get y 2 4 y 2So the point is 0 2.
(d) A 5 1, so A 5 3 1 2 2 1; B 3 2, so B 3 3 2 2 0 4; and C 2 1, so C 2 3 1 2 5 3. 52. (a) The point 3 7 is reflected to the point 3 7. (b) The point a b is reflected to the point a b. (c) Since the point a b is the reflection of a b, the point 4 1 is the reflection of 4 1.
(d) A 3 3, so A 3 3; B 6 1, so B 6 1; and C 1 4, so C 1 4. 53. (a) d A B 32 42 25 5. (b) We want the distances from C 4 2 to D 11 26. The walking distance is 4 11 2 26 7 24 31 blocks. Straight-line distance is 4 112 2 262 72 242 625 25 blocks.
(c) The two points are on the same avenue or the same street. 3 27 7 17 15 12, which is at the intersection of 15th Street and 12th Avenue. 54. (a) The midpoint is at 2 2 (b) They each must walk 15 3 12 7 12 5 17 blocks. 55. The midpoint of the line segment is 66 45. The pressure experienced by an ocean diver at a depth of 66 feet is 45 lb/in2 .
SECTION 1.2 Graphs of Equations in Two Variables: Circles
65
2x 2x to find the x coordinate of B: 6 12 2 x x 10. Likewise, for the y 2 2 3y coordinate of B, we have 8 16 3 y y 13. Thus B 10 13. 2
56. We solve the equation 6
y
57. We need to find a point S x1 y1 such that P Q RS is a parallelogram. As indicated by Example 3, this will be the case if the diagonals P R and QS bisect each other. So the midpoints of P R and QS are the same. Thus x1 2 y1 2 0 5 3 3 . Setting the x-coordinates equal, we get 2 2 2 2
Q 1l
x 2 05 1 0 5 x1 2 x1 3. 2 2 y 2 3 3 1 3 3 y1 2 Setting the y-coordinates equal, we get 2 2 y1 2. Thus S 3 2.
1.2
1
R
x
P
GRAPHS OF EQUATIONS IN TWO VARIABLES: CIRCLES
1. If the point 2 3 is on the graph of an equation in x and y, then the equation is satisfied when we replace x by 2 and y by 3. ?
?
We check whether 2 3 2 1 6 3. This is false, so the point 2 3 is not on the graph of the equation 2y x 1.
To complete the table, we express y in terms of x: 2y x 1 y 12 x 1 12 x 12 . x y
x
y
2
12
1
0
0
1 2
1
1
2
3 2
2 12 1 0 0 12
y 1
0
1
x
1 1 2 32
2. To find the x-intercept(s) of the graph of an equation we set y equal to 0 in the equation and solve for x: 2 0 x 1 x 1, so the x-intercept of 2y x 1 is 1. 3. To find the y-intercept(s) of the graph of an equation we set x equal to 0 in the equation and solve for y: 2y 0 1 y 12 , so the y-intercept of 2y x 1 is 12 .
4. The graph of the equation x 12 y 22 9 is a circle with center 1 2 and radius
9 3.
5. (a) If a graph is symmetric with respect to the x-axis and a b is on the graph, then a b is also on the graph. (b) If a graph is symmetric with respect to the y-axis and a b is on the graph, then a b is also on the graph. (c) If a graph is symmetric about the origin and a b is on the graph, then a b is also on the graph. 6. (a) The x-intercepts are 3 and 3 and the y-intercepts are 1 and 2. (b) The graph is symmetric about the y-axis.
7. Yes, this is true. If for every point x y on the graph, x y and x y are also on the graph, then x y must be on the graph as well, and so it is symmetric about the origin. 8. No, this is not necessarily the case. For example, the graph of y x is symmetric about the origin, but not about either axis.
66
CHAPTER 1 Equations and Graphs ?
?
?
9. y 3 4x. For the point 0 3: 3 3 4 0 3 3. Yes. For 4 0: 0 3 4 4 0 13. No. For 1 1: ?
?
1 3 4 1 1 1. Yes. So the points 0 3 and 1 1 are on the graph of this equation. ? ? ? ? 1 x. For the point 2 1: 1 1 2 1 1. No. For 3 2: 2 1 3 2 4. Yes. For 0 1: ? 1 1 0. Yes. So the points 3 2 and 0 1 are on the graph of this equation.
10. y
?
?
?
?
11. x 2y 1 0. For the point 0 0: 0 2 0 1 0 1 0. No. For 1 0: 1 2 0 1 0 1 1 0. Yes. ?
?
For 1 1: 1 2 1 1 0 1 2 1 0. Yes. So the points 1 0 and 1 1 are on the graph of this equation. ? ? ? ? 12. y x 2 1 1. For the point 1 1: 1 12 1 1 1 2 1. No. For 1 12 : 12 12 1 1 12 2 1. ? ? Yes. For 1 12 : 12 12 1 1 12 2 1. Yes. So the points 1 12 and 1 12 are on the graph of this equation. ?
?
?
13. x 2 2x y y 2 1. For the point 0 1: 02 2 0 1 12 1 1 1. Yes. For 2 1: 22 2 2 1 12 1 ?
?
?
?
4 4 1 1 1 1. Yes. For 2 3: 22 2 2 3 32 1 4 12 9 1 1 1. Yes. So the points 0 1, 2 1, and 2 3 are on the graph of this equation. ?
?
14. 0 1: 02 12 1 0 0 1 1 0. Yes. 2 2 ? ? 1 1 : 1 1 1 0 12 12 1 0. Yes. 2 2 2 2 2 2 ? ? 3 1 : 3 12 1 0 34 14 1 0. Yes. 2 2 2 So the points 0 1, 1 1 , and 23 12 are on the graph of this equation. 2
2
16. y 2x
15. y 3x x
y
x
y
3
9
3
6
2
6
2
4
1
3
1
2
0
0
0
0
1
3
1
2
2
6
2
4
3
9
3
6
SECTION 1.2 Graphs of Equations in Two Variables: Circles
17. y 2 x x
67
18. y 2x 3 y
x
y
4
6
4
5
2
4
2
1
0
2
0
3
2
0
2
7
4
2
4
11
19. Solve for y: 2x y 6 y 2x 6.
20. Solve for x: x 4y 8 x 4y 8.
x
y
x
y
2
10
4
0
6
2
52
2
2
0
2
4
2
2
6
6
4 6
21. y 1 x 2
3
32
1
12
8
0
10
1 2
22. y x 2 2 y
y
x
y
x
y
3
8
3
11
2
3
2
6
1
0
1
3
0
1
0
2
1
0
1
3
2
3
2
6
3
8
3
11
1l 1
x
1l 1
x
68
CHAPTER 1 Equations and Graphs
23. y x 2 2 x
24. y x 2 4 y
x
y
3
7
3
5
2
2
2
0
1
1
1
3
0
2
0
4
1
1
1
3
2
2
2
0
3
7
3
5
25. 9y x 2 . To make a table, we rewrite the equation as
26. 4y x 2 . y
y 19 x 2 . y
x
y
x
y
4
4
9
9
2
1
1
0
0
3 0
0
2
1
3
1
4
4
9
9
p
1
1
x
1 x
2
27. x y 2 4.
2 . x
28. x y 2 y y
x
y
12
4
5
3
0
2
3
1
4
0
3
1
0
2
5
3
12
4
y
1l 1
x
x
y
4
12
2
1
1
2
12 14 1 4 1 2
1
4 8 8 4 2
2
1
4
1 2
1l 1
x
SECTION 1.2 Graphs of Equations in Two Variables: Circles
29. y
x.
30. y 2
69
x.
y
y
x
y
x
y
0
0
0
2
1 4
1 2
1
1
2
2
1 2
3 2 2
4
4
4
2
9
5
9
3
16
4
1l 1
x
1 x
1
31. y 9 x 2 . Since the radicand (the expression inside
32. y
9 x 2.
the square root) cannot be negative, we must have
Since the radicand (the expression inside the square root)
9 x 2 0 x 2 9 x 3.
cannot be negative, we must have 9 x 2 0 x 2 9
x
y
3
0 5 2 2
2 1 0 1 2 3
x 3. x
y
3
0 5 2 2
2
3 2 2 5
1 0 2
3 2 2 5
3
0
1
0
33. y x.
34. x y. In the table below, we insert values of y and find y
x
y
6
6
4
4
2
2
0
0
2
the corresponding value of x. y
x
y
3
3
2
2
1
1
2
0
0
4
4
1
1
6
6
2
2
3
3
1 1
x
1l 1
x
70
CHAPTER 1 Equations and Graphs
35. y 4 x.
36. y 4 x. y
y
x
y
x
y
6
2
6
10
4
0
4
8
2
2
2
6
0
4
0
4
2
2
2
2
4
0
4
0
6
2
6
2
8
4
10
6
1l 1
x
37. x y 3 . Since x y 3 is solved for x in terms of y, we
1l 1
38. y x 3 1. y
insert values for y and find the corresponding values of x in the table below. y
x
y
3
28
x
y
2
9
27
3
1
2
8
2
0
1
1
1
1
1
0
0
2
7
1
1
3
26
8
2
27
3
39. y x 4 .
1l 1
x
40. y 16 x 4 .
y
y
x
y
3
81 16
3
65
2 1
1
0
0
1 2 3
2
0
1
15
0
16
1
1
15
16
2
0
81
3
65
1
x
1l 1
x
1
x
y
x
3l
x
3l
SECTION 1.2 Graphs of Equations in Two Variables: Circles
41. y 001x 3 x 2 5; [100 150] by [2000 2000]
71
42. y 003x 2 17x 3; [100 50] by [50 100]
2000
100
1000 50 -100
-50
50
100
150
-1000
-100
-50
50
-2000
43. y
-50
12x 17; [1 10] by [1 20]
44. y
20
4 256 x 2 ; [20 20] by [2 6] 6 4
10
2
-20 0
2
4
6
8
-10
x ; [50 50] by [02 02] 45. y 2 x 25
46. y x 4 4x 3 ; [4 6] by [50 100] 100
0.1
-20
50
20 -0.1 -0.2
20
-2
0.2
-40
10
10
40 -4
-2
2
4
6
-50
47. y x 6. To find x-intercepts, set y 0. This gives 0 x 6 x 6, so the x-intercept is 6. To find y-intercepts, set x 0. This gives y 0 6 y 6, so the y-intercept is 6. 48. 2x 5y 40. To find x-intercepts, set y 0. This gives 2x 5 0 40 2x 40 x 20, and the x-intercept is 20. To find y-intercepts, set x 0. This gives 2 0 5y 40 y 8, so the y-intercept is 8. 49. y x 2 5. To find x-intercepts, set y 0. This gives 0 x 2 5 x 2 5 x 5, so the x-intercepts are 5. To find y-intercepts, set x 0. This gives y 02 5 5, so the y-intercept is 5.
50. y 2 9 x 2 . To find x-intercepts, set y 0. This gives 02 9 x 2 x 2 9 x 3, so the x-intercepts are 3. To find y-intercepts, set x 0. This gives y 2 9 02 9 y 3, so the y-intercepts are 3.
51. y 2x y 2x 1. To find x-intercepts, set y 0. This gives 0 2x 0 2x 1 2x 1 x is 12 . To find y-intercepts, set x 0. This gives y 2 0 y 2 0 1 y 1, so the y-intercept is 1.
1 , so the x-intercept 2
72
CHAPTER 1 Equations and Graphs
52. x 2 x y y 1. To find x-intercepts, set y 0. This gives x 2 x 0 0 1 x 2 1 x 1, so the x-intercepts are 1 and 1. To find y-intercepts, set x 0. This gives y 02 0 y y 1 y 1, so the y-intercept is 1.
x 1. To find x-intercepts, set y 0. This gives 0 x 1 0 x 1 x 1, so the x-intercept is 1. To find y-intercepts, set x 0. This gives y 0 1 y 1, so the y-intercept is 1.
53. y
54. x y 5. To find x-intercepts, set y 0. This gives x 0 5 0 5, which is impossible, so there is no x-intercept. To find y-intercepts, set x 0. This gives 0 y 5 0 5, which is again impossible, so there is no y-intercept. 55. 4x 2 25y 2 100. To find x-intercepts, set y 0. This gives 4x 2 25 02 100 x 2 25 x 5, so the x-intercepts are 5 and 5.
To find y-intercepts, set x 0. This gives 4 02 25y 2 100 y 2 4 y 2, so the y-intercepts are 2 and 2.
56. 25x 2 y 2 100. To find x-intercepts, set y 0. This gives 25x 2 02 100 x 2 4 x 2, so the x-intercepts are 2 and 2. To find y-intercepts, set x 0. This gives 25 02 y 2 100 y 2 100, which has no solution, so there is no y-intercept. 57. y 4x x 2 . To find x-intercepts, set y 0. This gives 0 4x x 2 0 x 4 x 0 x or x 4, so the x-intercepts are 0 and 4. To find y-intercepts, set x 0. This gives y 4 0 02 y 0, so the y-intercept is 0. 58.
y2 x2 02 x2 x2 1. To find x-intercepts, set y 0. This gives 1 1 x 2 9 x 3, so the 9 4 9 4 9 x-intercepts are 3 and 3. To find y-intercepts, set x 0. This gives
02 y 2 y2 1 1 y 2 4 x 2, so the y-intercepts are 2 and 2. 9 4 4
59. x 4 y 2 x y 16. To find x-intercepts, set y 0. This gives x 4 02 x 0 16 x 4 16 x 2. So the x-intercepts are 2 and 2.
To find y-intercepts, set x 0. This gives 04 y 2 0 y 16 y 2 16 y 4. So the y-intercepts are 4 and 4.
60. x 2 y 3 x 2 y 2 64. To find x-intercepts, set y 0. This gives x 2 03 x 2 02 64 x 2 64 x 8. So the x-intercepts are 8 and 8. To find y-intercepts, set x 0. This gives 02 y 3 02 y 2 64 y 3 64 y 4. So the y-intercept is 4.
61. (a) y x 3 x 2 ; [2 2] by [1 1]
(b) From the graph, it appears that the x-intercepts are 0 and 1 and the y-intercept is 0.
1.0
(c) To find x-intercepts, set y 0. This gives
0.5
0 x 3 x 2 x 2 x 1 0 x 0 or 1. So
the x-intercepts are 0 and 1. -2
-1
1 -0.5 -1.0
2
To find y-intercepts, set x 0. This gives y 03 02 0. So the y-intercept is 0.
SECTION 1.2 Graphs of Equations in Two Variables: Circles
62. (a) y x 4 2x 3 ; [2 3] by [3 3]
73
(b) From the graph, it appears that the x-intercepts are 0 and 2 and the y-intercept is 0.
3
(c) To find x-intercepts, set y 0. This gives
2
0 x 4 2x 3 x 3 x 2 0 x 0 or 2. So
1
the x-intercepts are 0 and 2. -2
-1
1
-1
2
3
To find y-intercepts, set x 0. This gives
y 04 2 03 0. So the y-intercept is 0.
-2 -3
2 ; [5 5] by [3 1] 63. (a) y 2 x 1
(b) From the graph, it appears that there is no x-intercept and the y-intercept is 2.
1
-4
(c) To find x-intercepts, set y 0. This gives
-1
2 0 2 , which has no solution. So there is no x 1 x-intercept.
-2
To find y-intercepts, set x 0. This gives
-2
2
4
2 y 2 2. So the y-intercept is 2. 0 1
-3
x 64. (a) y 2 ; [5 5] by [2 2] x 1
(b) From the graph, it appears that the x- and y-intercepts are 0.
2 1
-4
-2
2
4
-1
(c) To find x-intercepts, set y 0. This gives x 0 2 x 0. So the x-intercept is 0. x 1 To find y-intercepts, set x 0. This gives 0 y 2 0. So the y-intercept is 0. 0 1
-2
65. (a) y
3
x; [5 5] by [2 2]
(b) From the graph, it appears that and the x- and y-intercepts are 0.
2
(c) To find x-intercepts, set y 0. This gives 0
1
-4
-2
x 0. So the x-intercept is 0.
2 -1 -2
4
To find y-intercepts, set x 0. This gives y 3 0 0. So the y-intercept is 0.
3
x
74
CHAPTER 1 Equations and Graphs
66. (a) y
3 1 x 2 ; [5 5] by [5 3]
(b) From the graph, it appears that the x-intercepts are 1 and 1 and the y-intercept is 1.
2
-4
-2
2
4
(c) To find x-intercepts, set y 0. This gives 3 0 1 x 2 1 x 2 0 x 1. So the x-intercepts are 1 and 1. To find y-intercepts, set x 0. This gives 3 y 1 02 1. So the y-intercept is 1.
-2 -4
67. x 2 y 2 9 has center 0 0 and radius 3.
68. x 2 y 2 5 has center 0 0 and radius
y
5.
y
1l
1l 1
1
x
69. x 32 y 2 16 has center 3 0 and radius 4. y
x
70. x 2 y 22 4 has center 0 2 and radius 2. y
1l
1l 1
1
x
x
71. x 32 y 42 25 has center 3 4 and radius 5. 72. x 12 y 22 36 has center 1 2 and y
radius 6.
y
1l 1 1l 1
x
x
SECTION 1.2 Graphs of Equations in Two Variables: Circles
75
73. Using h 3, k 2, and r 5, we get x 32 y 22 52 x 32 y 22 25.
74. Using h 1, k 3, and r 3, we get x 12 y 32 32 x 12 y 32 9.
75. The equation of a circle centered at the origin is x 2 y 2 r 2 . Using the point 4 7 we solve for r 2 . This gives 42 72 r 2 16 49 65 r 2 . Thus, the equation of the circle is x 2 y 2 65.
76. Using h 1 and k 5, we get x 12 y 52 r 2 x 12 y 52 r 2 . Next, using the point 4 6, we solve for r 2 . This gives 4 12 6 52 r 2 130 r 2 . Thus, an equation of the circle is
x 12 y 52 130.
1 5 1 9 77. The center is at the midpoint of the line segment, which is 2 5. The radius is one half the diameter, 2 2 so r 12 1 52 1 92 12 36 64 12 100 5. Thus, an equation of the circle is x 22 y 52 52 x 22 y 52 25.
1 7 3 5 78. The center is at the midpoint of the line segment, which is 3 1. The radius is one half the 2 2 diameter, so r 12 1 72 3 52 4 2. Thus, an equation of the circle is x 32 y 12 32.
79. Since the circle is tangent to the x-axis, it must contain the point 7 0, so the radius is the change in the y-coordinates. That is, r 3 0 3. So the equation of the circle is x 72 y 32 32 , which is x 72 y 32 9. 80. Since the circle with r 5 lies in the first quadrant and is tangent to both the x-axis and the y-axis, the center of the circle is at 5 5. Therefore, the equation of the circle is x 52 y 52 25.
81. From the figure, the center of the circle is at 2 2. The radius is the change in the y-coordinates, so r 2 0 2. Thus the equation of the circle is x 22 y 22 22 , which is x 22 y 22 4.
82. From the figure, the center of the circle is at 1 1. The radius is the distance from the center to the point 2 0. Thus r 1 22 1 02 9 1 10, and the equation of the circle is x 12 y 12 10. 2 2 2 2 2 2 4y 4 y 1 42 83. Completing the square gives x 2 y 2 2x 4y 1 0 x 2 2x 2 2 2 2
x 2 2x 1 y 2 4y 4 1 1 4 x 12 y 22 4. Thus, the center is 1 2, and the radius is 2. 2 2 2 2 84. Completing the square gives x 2 y 2 2x 2y 2 x 2 2x 2 y 2 2y 2 2 2 2 2 2 2 2
x 2 2x 1 y 2 2y 1 2 1 1 x 12 y 12 4. Thus, the center is 1 1, and the radius is 2. 2 2 2 2 85. Completing the square gives x 2 y 2 4x 10y 13 0 x 2 4x 4 y 2 10y 10 13 42 10 2 2 2
x 2 4x 4 y 2 10y 25 13 4 25 x 22 y 52 16. Thus, the center is 2 5, and the radius is 4. 2 2 86. Completing the square gives x 2 y 2 6y 2 0 x 2 y 2 6y 62 2 62 x 2 y 2 6y 9 2 9 x 2 y 32 7. Thus, the circle has center 0 3 and radius 7. 2 2 87. Completing the square gives x 2 y 2 x 0 x 2 x 12 y 2 12 x 2 x 14 y 2 14 2 x 12 y 2 14 . Thus, the circle has center 12 0 and radius 12 .
2 2 2 88. Completing the square gives x 2 y 2 2x y 1 0 x 2 2x 22 y 2 y 12 1 1 12 2 x 2 2x 1 y 2 y 14 14 x 12 y 12 14 . Thus, the circle has center 1 12 and radius 12 .
76
CHAPTER 1 Equations and Graphs
2 2 2 2 89. Completing the square gives x 2 y 2 12 x 12 y 18 x 2 12 x 12 y 2 12 y 12 18 12 12 2 2 2 2 2 2 1 y2 1 y 1 1 1 1 2 1 x 1 x 2 12 x 16 y 14 14 . Thus, the circle has center 2 16 8 16 16 8 4 4 1 1 and radius 1 . 4 4 2 1 0 x 2 1 x 12 2 y 2 2y 2 2 1 12 2 2 2 90. Completing the square gives x 2 y 2 12 x 2y 16 2 2 2 16 2 2 2 1 1 2 x 4 y 1 1. Thus, the circle has center 4 1 and radius 1.
91. Completing the square gives x 2 y 2 4x 10y 21 92. First divide by 4, then complete the square. This gives 2 2 2 4x 2 4y 2 2x 0 x 2 y 2 12 x 0 x 2 12 x 4 21 x 2 4x 42 y 2 10y 10 2 2 2 2 __y 2 0 x 2 12 x 12 y 2 12 10 2 x 22 y 52 21 4 25 50. 2 2 2 2 1 . Thus, the circle has center 1 0 x 14 y 2 16 Thus, the circle has center 2 5 and radius 50 5 2. 4 y
and radius 14 .
y 2
2
x
1 1
x
93. Completing the square gives x 2 y 2 6x 12y 45 0 94. x 2 y 2 16x 12y 200 0 2 2 x 32 y 62 45 9 36 0. Thus, the y 2 12y 12 x 2 16x 16 2 2 center is 3 6, and the radius is 0. This is a degenerate 2 2 circle whose graph consists only of the point 3 6. 200 16 12 2 2 y
1
x 82 y 62 200 64 36 100. Since
completing the square gives r 2 100, this is not the equation of a circle. There is no graph. 1
x
95. x-axis symmetry: y x 4 x 2 y x 4 x 2 , which is not the same as y x 4 x 2 , so the graph is not symmetric with respect to the x-axis. y-axis symmetry: y x4 x2 x 4 x 2 , so the graph is symmetric with respect to the y-axis.
Origin symmetry: y x4 x2 y x 4 x 2 , which is not the same as y x 4 x 2 , so the graph is not symmetric with respect to the origin. 96. x-axis symmetry: x y4 y2 y 4 y 2 , so the graph is symmetric with respect to the x-axis.
y-axis symmetry: x y 4 y 2 , which is not the same as x y 4 y 2 , so the graph is not symmetric with respect to the y-axis.
Origin symmetry: x y4 y2 x y 4 y 2 , which is not the same as x y 4 y 2 , so the graph is not symmetric with respect to the origin.
SECTION 1.2 Graphs of Equations in Two Variables: Circles
77
97. x-axis symmetry: y x 3 10x y x 3 10x, which is not the same as y x 3 10x, so the graph is not symmetric with respect to the x-axis. y-axis symmetry: y x3 10 x y x 3 10x, which is not the same as y x 3 10x, so the graph is not symmetric with respect to the y-axis. Origin symmetry: y x3 10 x y x 3 10x y x 3 10x, so the graph is symmetric with respect to the origin. 98. x-axis symmetry: y x 2 x y x 2 x, which is not the same as y x 2 x, so the graph is not symmetric with respect to the x-axis. y-axis symmetry: y x2 x y x 2 x, so the graph is symmetric with respect to the y-axis. Note that x x.
Origin symmetry: y x2 x y x 2 x y x 2 x, which is not the same as y x 2 x, so the graph is not symmetric with respect to the origin. 99. x-axis symmetry: x 4 y4 x 2 y2 1 x 4 y 4 x 2 y 2 1, so the graph is symmetric with respect to the x-axis. y-axis symmetry: x4 y 4 x2 y 2 1 x 4 y 4 x 2 y 2 1, so the graph is symmetric with respect to the y-axis.
Origin symmetry: x4 y4 x2 y2 1 x 4 y 4 x 2 y 2 1, so the graph is symmetric with respect to the origin. 100. x-axis symmetry: x 2 y2 x y 1 x 2 y 2 x y 1, which is not the same as x 2 y 2 x y 1, so the graph is not symmetric with respect to the x-axis. y-axis symmetry: x2 y 2 x y 1 x 2 y 2 x y 1, which is not the same as x 2 y 2 x y 1, so the graph is not symmetric with respect to the y-axis. Origin symmetry: x2 y2 x y 1 x 2 y 2 x y 1, so the graph is symmetric with respect to the origin. 102. Symmetric with respect to the x-axis.
101. Symmetric with respect to the y-axis. y
0
y
x 0
103. Symmetric with respect to the origin.
104. Symmetric with respect to the origin. y
y
0
x
x 0
x
78
CHAPTER 1 Equations and Graphs
105. x y x 2 y 2 1 . This is the set of points inside (and on) the circle x 2 y 2 1.
106. x y x 2 y 2 4 . This is the set of points outside the circle x 2 y 2 4.
y
1l
y
2l
1
x
107. Completing the square gives x 2 y 2 4y 12 0 4 2 4 2 12 x 2 y 2 4y 2 2
2
x
108. This is the top quarter of the circle of radius 3. Thus, the area is 14 9 94 . y
x 2 y 22 16. Thus, the center is 0 2, and the
radius is 4. So the circle x 2 y 2 4, with center 0 0
3l
and radius 2 sits completely inside the larger circle. Thus, the area is 42 22 16 4 12.
3
x
109. (a) The point 5 3 is shifted to 5 3 3 2 8 5. (b) The point a b is shifted to a 3 b 2.
(c) Let x y be the point that is shifted to 3 4. Then x 3 y 2 3 4. Setting the x-coordinates equal, we get x 3 3 x 0. Setting the y-coordinates equal, we get y 2 4 y 2So the point is 0 2.
(d) A 5 1, so A 5 3 1 2 2 1; B 3 2, so B 3 3 2 2 0 4; and C 2 1, so C 2 3 1 2 5 3. (b) Symmetric about the y-axis.
110. (a) Symmetric about the x-axis.
(c) Symmetric about the origin.
y
y
y
1
1
1
1
x
0
1
x
1
x
111. (a) In 1980 inflation was 14%; in 1990, it was 6%; in 1999, it was 2%. (b) Inflation exceeded 6% from 1975 to 1976 and from 1978 to 1982. (c) Between 1980 and 1985 the inflation rate generally decreased. Between 1987 and 1992 the inflation rate generally increased. (d) The highest rate was about 14% in 1980. The lowest was about 1% in 2002. 112. (a) Closest: 2 Mm. Farthest: 8 Mm.
SECTION 1.3 Lines
79
x 32 1 x 32 22 x 32 1 4 1 34 x 32 75 4 . Taking the square 25 16 25 25 5 3 5 3 5 3 5 3 root of both sides we get x 3 75 4 2 x 3 2 . So x 3 2 133 or x 3 2 733. The distance from 133 2 to the center 0 0 is d 133 02 2 02 57689 240. The distance from 733 2 to the center 0 0 is d 733 02 2 02 577307 760. 2 a 2 a 2 b 2 b y 2 by c 113. Completing the square gives x 2 y 2 ax by c 0 x 2 ax 2 2 2 2 a 2 b2 a 2 b 2 a 2 b2 . This equation represents a circle only when c 0. This x y c 2 2 4 4 (b) When y 2 we have
a 2 b2 a 2 b2 0, and this equation represents the empty set when c 0. 4 4 a b a 2 b2 When the equation represents a circle, the center is , and the radius is c 12 a 2 b2 4ac. 2 2 4
equation represents a point when c
114. (a) (i) x 22 y 12 9, the center is at 2 1, and the radius is 3. x 62
y 42 16, the center is at 6 4, and the radius is 4. The distance between centers is 2 62 1 42 42 32 16 9 25 5. Since 5 3 4, these circles intersect.
(ii) x 2 y 22 4, the center is at 0 2, and the radius is 2. x 52 y 142 9, the center is at 5 14, and the radius is 3. The distance between centers is 0 52 2 142 52 122 25 144 169 13. Since 13 2 3, these circles do not intersect.
(iii) x 32 y 12 1, the center is at 3 1, and the radius is 1. x 22 y 22 25, the center is at 2 2, and the radius is 5. The distance between centers is 3 22 1 22 12 32 1 9 10. Since 10 1 5, these circles intersect.
(b) If the distance d between the centers of the circles is greater than the sum r1 r2 of their radii, then the circles do not intersect, as shown in the first diagram. If d r1 r2 , then the circles intersect at a single point, as shown in the second diagram. If d r1 r2 , then the circles intersect at two points, as shown in the third diagram.
rª CÁ rÁ
Cª
d
CÁ rÁ
rª
Cª
d
rÁ
rª
Cª
CÁ
d
Case 1 d r1 r2
1.3
Case 2 d r1 r2
Case 3 d r1 r2
LINES
1. We find the “steepness” or slope of a line passing through two points by dividing the difference in the y-coordinates of these 51 points by the difference in the x-coordinates. So the line passing through the points 0 1 and 2 5 has slope 2. 20 2. (a) The line with equation y 3x 2 has slope 3. (b) Any line parallel to this line has slope 3.
80
CHAPTER 1 Equations and Graphs
(c) Any line perpendicular to this line has slope 13 .
3. The point-slope form of the equation of the line with slope 3 passing through the point 1 2 is y 2 3 x 1.
4. For the linear equation 2x 3y 12 0, the x-intercept is 6 and the y-intercept is 4. The equation in slope-intercept form is y 23 x 4. The slope of the graph of this equation is 23 .
5. The slope of a horizontal line is 0. The equation of the horizontal line passing through 2 3 is y 3.
6. The slope of a vertical line is undefined. The equation of the vertical line passing through 2 3 is x 2. 7. (a) Yes, the graph of y 3 is a horizontal line 3 units below the x-axis.
(b) Yes, the graph of x 3 is a vertical line 3 units to the left of the y-axis.
(c) No, a line perpendicular to a horizontal line is vertical and has undefined slope. (d) Yes, a line perpendicular to a vertical line is horizontal and has slope 0. y
8.
Yes, the graphs of y 3 and x 3 are perpendicular lines.
5l x=_3
0
x
5 y=_3
y y1 9. m 2 x2 x1 y y1 11. m 2 x2 x1 y y1 13. m 2 x2 x1 y y1 15. m 2 x2 x1
y y1 10. m 2 x2 x1 y y1 12. m 2 x2 x1 y y1 14. m 2 x2 x1 y y1 16. m 2 x2 x1
02 2 2 0 1 1 1 1 2 72 5 44 0 05 3 3 5 2 6 10 4 4
1 0 1 1 30 3 3 3 3 2 1 3 5 8 8 4 4 1 3 14 3 3 2 2 0 63 y y1 20 17. For 1 , we find two points, 1 2 and 0 0 that lie on the line. Thus the slope of 1 is m 2 2. x2 x1 1 0 y y1 32 For 2 , we find two points 0 2 and 2 3. Thus, the slope of 2 is m 2 12 . For 3 we find the points x2 x1 20 y y1 1 2 3. For 4 , we find the points 2 1 and 2 2 and 3 1. Thus, the slope of 3 is m 2 x 2 x1 32 1 1 y y1 2 1 2 2. Thus, the slope of 4 is m 2 . x2 x1 2 2 4 4
y
18. (a)
m=2
m=1
(b)
y
m=3
1 m=_ 2
1 m=_ 2
m=0
1 m=_ 3
1l
1l 1
x
1
x 1 m=_ _ 3
m=_1
SECTION 1.3 Lines
81
04 1. Since the y-intercept is 4, 40 the equation of the line is y mx b 1x 4. So y x 4, or x y 4 0. 04 20. We find two points on the graph, 0 4 and 2 0. So the slope is m 2. Since the y-intercept is 4, the 2 0 equation of the line is y mx b 2x 4, so y 2x 4 2x y 4 0. 0 3 21. We choose the two intercepts as points, 0 3 and 2 0. So the slope is m 32 . Since the y-intercept is 3, 20 the equation of the line is y mx b 32 x 3, or 3x 2y 6 0. 19. First we find two points 0 4 and 4 0 that lie on the line. So the slope is m
22. We choose the two intercepts, 0 4 and 3 0. So the slope is m equation of the line is y mx b 43 x 4 4x 3y 12 0.
0 4 43 . Since the y-intercept is 4, the 3 0
23. Using y mx b, we have y 3x 2 or 3x y 2 0.
24. Using y mx b, we have y 25 x 4 2x 5y 20 0. 25. Using the equation y y1 m x x1 , we get y 3 5 x 2 5x y 7 5x y 7 0.
26. Using the equation y y1 m x x1 , we get y 4 1 x 2 y 4 x 2 x y 2 0. 27. Using the equation y y1 m x x1 , we get y 7 23 x 1 3y 21 2x 2 2x 3y 19 2x 3y 19 0.
28. Using the equation y y1 m x x1 , we get y 5 72 x 3 2y 10 7x 21 7x 2y 31 0.
5 61 y y1 5. Substituting into y y1 m x x1 , we get 29. First we find the slope, which is m 2 x2 x1 12 1 y 6 5 x 1 y 6 5x 5 5x y 11 0. 3 2 y y1 55 1. Substituting into y y1 m x x1 , we get 30. First we find the slope, which is m 2 x2 x1 4 1 y 3 1 x 4 y 3 x 4 x y 1 0. 8 3 5 y y1 8. Substituting 31. We are given two points, 2 5 and 1 3. Thus, the slope is m 2 x2 x1 1 2 1 into y y1 m x x1 , we get y 5 8 [x 2] y 8x 11 or 8x y 11 0. 77 y y1 0. Substituting into 32. We are given two points, 1 7 and 4 7. Thus, the slope is m 2 x2 x1 41 y y1 m x x1 , we get y 7 0 x 1 y 7 or y 7 0. 3 3 0 y y1 3. Using the y-intercept, 33. We are given two points, 1 0 and 0 3. Thus, the slope is m 2 x2 x1 01 1 we have y 3x 3 or y 3x 3 or 3x y 3 0. 60 y y1 68 34 . Using the y-intercept 34. We are given two points, 8 0 and 0 6. Thus, the slope is m 2 x2 x1 0 8 we have y 34 x 6 3x 4y 24 0. 35. Since the equation of a line with slope 0 passing through a b is y b, the equation of this line is y 3.
36. Since the equation of a line with undefined slope passing through a b is x a, the equation of this line is x 1.
37. Since the equation of a line with undefined slope passing through a b is x a, the equation of this line is x 2.
38. Since the equation of a line with slope 0 passing through a b is y b, the equation of this line is y 1.
39. Any line parallel to y 3x 5 has slope 3. The desired line passes through 1 2, so substituting into y y1 m x x1 , we get y 2 3 x 1 y 3x 1 or 3x y 1 0. 1 2. The desired line passes through 3 2, so substituting 40. Any line perpendicular to y 12 x 7 has slope 12 into y y1 m x x1 , we get y 2 2 [x 3] y 2x 8 or 2x y 8 0.
82
CHAPTER 1 Equations and Graphs
41. Since the equation of a horizontal line passing through a b is y b, the equation of the horizontal line passing through 4 5 is y 5. 42. Any line parallel to the y-axis has undefined slope and an equation of the form x a. Since the graph of the line passes through the point 4 5, the equation of the line is x 4. 43. Since x 2y 6 2y x 6 y 12 x 3, the slope of this line is 12 . Thus, the line we seek is given by y 6 12 x 1 2y 12 x 1 x 2y 11 0. 44. Since 2x 3y 4 0 3y 2x 4 y 23 x 43 , the slope of this line is m 23 . Substituting m 23 and b 6 into the slope-intercept formula, the line we seek is given by y 23 x 6 2x 3y 18 0. 45. Any line parallel to x 5 has undefined slope and an equation of the form x a. Thus, an equation of the line is x 1. 46. Any line perpendicular to y 1 has undefined slope and an equation of the form x a. Since the graph of the line passes through the point 2 6, an equation of the line is x 2. 47. First find the slope of 2x 5y 8 0. This gives 2x 5y 8 0 5y 2x 8 y 25 x 85 . So the 1 52 . The equation of the line we seek is slope of the line that is perpendicular to 2x 5y 8 0 is m 25 y 2 52 x 1 2y 4 5x 5 5x 2y 1 0.
48. First find the slope of the line 4x 8y 1. This gives 4x 8y 1 8y 4x 1 y 12 x 18 . So the slope of the 1 2. The equation of the line we seek is y 23 2 x 12 line that is perpendicular to 4x 8y 1 is m 12 y 23 2x 1 6x 3y 1 0.
49. First find the slope of the line passing through 2 5 and 2 1. This gives m of the line we seek is y 7 1 x 1 x y 6 0. 50. First find the slope of the line passing through 1 1 and 5 1. This gives m of the line that is perpendicular is m 2x y 7 0.
(_2, 1)
2 1 1 12 , and so the slope 51 4
1 2. Thus the equation of the line we seek is y 11 2 x 2 12
y
51. (a)
4 15 1, and so the equation 2 2 4
y
52. (a)
1l 1
x 1l 1
(b) y 1 32 x 2 2y 2 3 x 2 2y 2 3x 6 3x 2y 8 0.
(4, _1)
x
(b) y 1 2 x 4 y 1 2x 8 2x y 7 0.
SECTION 1.3 Lines
53.
b=_1 b=_3
8
b=_6
4 _4
54.
m=1.5
4
m=0.75 _4
0
_2
2
_2
4
0
2
4
m=0.25 m=0 m=_0.25
_4
_4
b=6
_8
b=3 b=1
m=_0.75
_8
m=_1.5
b=0
y mx 3, m 0, 025, 075, 15. Each of the
y 2x b, b 0, 1, 3, 6. They have the same
lines contains the point 0 3 because the point 0 3
slope, so they are parallel.
satisfies each equation y mx 3. Since 0 3 is on the y-axis, they all have the same y-intercept.
m=1.5
55. 4
_2
_2
m=6 y
56.
m=0.75
2 0
2
4
83
6
8
m=2 m=1
10
m=0.25 m=0
m=0.5
m=_0.25
_12
_8
_4
x m=0
0
4 m=_0.5
m=_0.75
_4
m=_1
m=_1.5
y m x 3, m 0, 025, 075, 15. Each of the
lines contains the point 3 0 because the point 3 0
satisfies each equation y m x 3. Since 3 0 is on the x-axis, we could also say that they all have the same
_10 m=_6
m=_2
y 2 m x 3, m 0, 05, 1, 2, 6. Each of
the lines contains the point 3 2 because the point 3 2 satisfies each equation y 2 m x 3.
x-intercept.
57. y 3 x x 3. So the slope is 1 and the y-intercept is 3.
58. y 23 x 2. So the slope is 23 and the y-intercept is 2. y
y
1 1
1 1
x
x
84
CHAPTER 1 Equations and Graphs
59. 2x y 7 y 2x 7. So the slope is 2 and the y-intercept is 7.
60. 2x 5y 0 5y 2x y 25 x. So the slope is 2 and the y-intercept is 0. 5
y
y
1l 1
x
1 1
x
61. 4x 5y 10 5y 4x 10 y 45 x 2. So
62. 3x 4y 12 4y 3x 12 y 34 x 3. So the slope is 34 and the y-intercept is 3.
the slope is 45 and the y-intercept is 2.
y
y
1l
1l 1
1
x
x
63. y 4 can also be expressed as y 0x 4. So the slope is 64. x 5 cannot be expressed in the form y mx b. So 0 and the y-intercept is 4.
the slope is undefined, and there is no y-intercept. This is a
y
vertical line. y
1l 1
x
1l 1
x
SECTION 1.3 Lines
85
65. x 3 cannot be expressed in the form y mx b. So the 66. y 2 can also be expressed as y 0x 2. So the slope slope is undefined, and there is no y-intercept. This is a
is 0 and the y-intercept is 2. y
vertical line. y
1l 1
1l 1
x
x
67. 5x 2y 10 0. To find x-intercepts, we set y 0 and
solve for x: 5x 2 0 10 0 5x 10 x 2, so
68. 6x 7y 42 0. To find x-intercepts, we set y 0 and
solve for x: 6x 7 0 42 0 6x 42 x 7, so
the x-intercept is 2.
the x-intercept is 7.
To find y-intercepts, we set x 0 and solve for y:
To find y-intercepts, we set x 0 and solve for y:
y-intercept is 5.
y-intercept is 6.
5 0 2y 10 0 2y 10 y 5, so the y
1
6 0 7y 42 0 7y 42 y 6, so the y
2 1
x
2
x
86
CHAPTER 1 Equations and Graphs
69. 12 x 13 y 1 0. To find x-intercepts, we set y 0 and
70. 13 x 15 y 2 0. To find x-intercepts, we set y 0 and
solve for x: 12 x 13 0 1 0 12 x 1 x 2,
solve for x: 13 x 15 0 2 0 13 x 2 x 6, so
so the x-intercept is 2.
the x-intercept is 6.
1 0 1 y 1 0 1 y 1 y 3, so the 2 3 3
1 0 1 y 2 0 1 y 2 y 10, so the 3 5 5
To find y-intercepts, we set x 0 and solve for y:
To find y-intercepts, we set x 0 and solve for y:
y-intercept is 3.
y-intercept is 10.
y
1
y
2 1
2
x
71. y 6x 4. To find x-intercepts, we set y 0 and solve for x: 0 6x 4 6x 4 x 23 , so the x-intercept is 23 .
To find y-intercepts, we set x 0 and solve for y: y 6 0 4 4, so the y-intercept is 4.
x
72. y 4x 10. To find x-intercepts, we set y 0 and
solve for x: 0 4x 10 4x 10 x 52 , so
the x-intercept is 52 . To find y-intercepts, we set x 0 and solve for y:
y 4 0 10 10, so the y-intercept is 10. y
y
2 1
x
1 1
x
73. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation y 2x 3 has slope 2.
The line with equation 2y 4x 5 0 2y 4x 5 y 2x 52 also has slope 2, and so the lines are parallel. 74. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation y 12 x 4 has slope 12 .
1 , and so the lines are neither The line with equation 2x 4y 1 4y 2x 1 y 12 x 14 has slope 12 12
parallel nor perpendicular. 75. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation 3x 4y 4
4y 3x 4 y 34 x 1 has slope 34 . The line with equation 4x 3y 5 3y 4x 5 y 43 x 53 has 1 , and so the lines are perpendicular. slope 43 34
76. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation 2x 3y 10
2 2 7 3y 2x 10 y 23 x 10 3 has slope 3 . The line with equation 3y 2x 7 0 3y 2x 7 y 3 x 3 also has slope 23 , and so the lines are parallel.
SECTION 1.3 Lines
87
77. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation 7x 3y 2
3y 7x 2 y 73 x 23 has slope 73 . The line with equation 9y 21x 1 9y 21x 1 y 73 19 has 1 , and so the lines are neither parallel nor perpendicular. slope 73 73
78. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation 6y 2x 5
6y 2x 5 y 13 x 56 has slope 13 . The line with equation 2y 6x 1 2y 6x 1 y 3x 12 has 1 , and so the lines are perpendicular. slope 3 13
y
79. We first plot the points to find the pairs of points that determine each side. Next we 3 1 41 , and the find the slopes of opposite sides. The slope of AB is 71 6 2 10 7 3 1 slope of DC is . Since these slope are equal, these two sides 5 1 6 2 are parallel. The slope of AD is
C D
6 71 3, and the slope of BC is 1 1 2
B
6 10 4 3. Since these slope are equal, these two sides are parallel. 57 2 Hence ABC D is a parallelogram.
1l
A
x
1
y
80. We first plot the points to determine the perpendicular sides. Next find the slopes of the sides. The slope of AB is
4 2 3 1 , and the slope of AC is 3 3 6 3
C
8 1 9 3 . Since 9 3 6 2
slope of AB slope of AC 23 32 1 the sides are perpendicular, and ABC is a right triangle.
A
81. We first plot the points to find the pairs of points that determine each side. Next we 31 2 1 find the slopes of opposite sides. The slope of AB is and the 11 1 10 5 68 2 1 slope of DC is . Since these slope are equal, these two sides 0 10 10 5 61 5 are parallel. Slope of AD is 5, and the slope of BC is 01 1 5 38 5. Since these slope are equal, these two sides are parallel. 11 10 1
B
1l 1
x
y
C D
1l
B A
1
x
Since slope of AB slope of AD 15 5 1, the first two sides are each perpendicular to the second two sides. So the sides form a rectangle.
82. (a) The slope of the line passing through 1 1 and 3 9 is and 6 21 is
8 91 4. The slope of the line passing through 1 1 31 2
21 1 20 4. Since the slopes are equal, the points are collinear. 61 5
88
CHAPTER 1 Equations and Graphs
(b) The slope of the line passing through 1 3 and 1 7 is 1 3 and 4 15 is
4 73 2. The slope of the line passing through 1 1 2
15 3 12 . Since the slopes are not equal, the points are not collinear. 4 1 5
83. We need the slope and the midpoint of the line AB. The midpoint of AB is
17 42 2 2
4 1, and the slope of
2 4 6 1 1 1. The slope of the perpendicular bisector will have slope 1. Using the 71 6 m 1 point-slope form, the equation of the perpendicular bisector is y 1 1 x 4 or x y 3 0. AB is m
84. We find the intercepts (the length of the sides). When x 0, we have 2y 3 0 6 0 2y 6 y 3, and when y 0, we have 2 0 3x 6 0 3x 6 x 2. Thus, the area of the triangle is 12 3 2 3.
85. (a) We start with the two points a 0 and 0 b. The slope of the line that contains them is
b b0 . So the equation 0a a
b of the line containing them is y x b (using the slope-intercept form). Dividing by b (since b 0) gives a x x y y 1 1. b a a b y x 1 4x 3y 24 4x 3y 24 0. (b) Setting a 6 and b 8, we get 6 8
86. (a) The line tangent at 3 4 will be perpendicular to the line passing through the points 0 0 and 3 4. The slope of 4 1 3 4 0 . Thus, the slope of the tangent line will be . Then the equation of the tangent this line is 30 3 4 43 line is y 4 34 x 3 4 y 4 3 x 3 3x 4y 25 0.
(b) Since diametrically opposite points on the circle have parallel tangent lines, the other point is 3 4. 87. (a) The slope represents an increase of 002 C every year. The T -intercept is the average surface temperature in 1950, or 15 C. (b) In 2050, t 2050 1950 100, so T 002100 15 17 degrees Celsius. 88. (a) The slope is 00417D 00417 200 834. It represents the increase in dosage for each one-year increase in the child’s age. (b) When a 0, c 834 0 1 834 mg. 89. (a)
y
(b) The slope, 4, represents the decline in number of spaces sold for
each $1 increase in rent. The y-intercept is the number of spaces at the flea market, 200, and the x-intercept is the cost per space when the
200l
manager rents no spaces, $50. 100l
20
40
60
80 100
x
SECTION 1.3 Lines y
90. (a)
89
(b) The slope is the cost per toaster oven, $6. The y-intercept, $3000, is the monthly fixed cost—the cost that is incurred no matter how many toaster ovens are produced.
10,000l
5000l
500
1000
x
91. (a)
(b) Substituting a for both F and C, we have C F
30 22
20 4
10 14
0
10
20
30
32
50
68
86
a 95 a 32 45 a 32 a 40 . Thus both scales agree at
40 .
10 5 t t1 80 70 . So the linear 92. (a) Using n in place of x and t in place of y, we find that the slope is 2 n2 n1 168 120 48 24 5 n 168 t 80 5 n 35 t 5 n 45. equation is t 80 24 24 24
5 150 45 7625 F 76 F. (b) When n 150, the temperature is approximately given by t 24
93. (a) Using t in place of x and V in place of y, we find the slope of the line
y
(b)
using the points 0 4000 and 4 200. Thus, the slope is
4000l
200 4000 3800 m 950. Using the V -intercept, the 40 4 linear equation is V 950t 4000.
3000l 2000l
(c) The slope represents a decrease of $950 each year in the value of the 1000l
computer. The V -intercept represents the cost of the computer. (d) When t 3, the value of the computer is given by
1
V 950 3 4000 1150.
94. (a) We are given
434 change in pressure 0434. Using P for 10 feet change in depth 10
pressure and d for depth, and using the point P 15 when d 0, we have P 15 0434 d 0 P 0434d 15.
(c) The slope represents the increase in pressure per foot of descent. The y-intercept represents the pressure at the surface. (d) When P 100, then 100 0434d 15 0434d 85 d 1959 ft. Thus the pressure is 100 lb/in3 at a depth of
approximately 196 ft.
(b)
2
3
4
x
y 60l 50l 40l 30l 20l 10l 10
20
30
40
50
60 x
95. The temperature is increasing at a constant rate when the slope is positive, decreasing at a constant rate when the slope is negative, and constant when the slope is 0.
90
CHAPTER 1 Equations and Graphs
96. We label the three points A, B, and C. If the slope of the line segment AB is equal to the slope of the line segment BC, then the points A, B, and C are collinear. Using the distance formula, we find the distance between A and B, between B and C, and between A and C. If the sum of the two smaller distances equals the largest distance, the points A, B, and C are collinear. Another method: Find an equation for the line through A and B. Then check if C satisfies the equation. If so, the points are collinear.
1.4
SOLVING QUADRATIC EQUATIONS
1. (a) The Quadratic Formula states that x
b
(b) In the equation 12 x 2 x 4 0, a 1 12 4 12 4 x 2 12
b2 4ac . 2a
1 , b 1, and c 4. So, the solution of the equation is 2
13 2 or 4. 1
2. (a) To solve the equation x 2 4x 5 0 by factoring, we write x 2 4x 5 x 5 x 1 0 and use the Zero-Product Property to get x 5 or x 1. 2 (b) To solve by completing the square, we add 5 to both sides to get x 2 4x 5, and then add 42 to both sides to get x 2 4x 4 5 4 x 22 9 x 2 3 x 5 or x 1.
(c) To solve using the Quadratic Formula, we substitute a 1, b 4, and c 5, obtaining 4 42 4 1 5 4 36 2 3 x 5 or x 1. x 2 1 2 3. For the quadratic equation ax 2 bx c 0 the discriminant is D b2 4ac. If D 0, the equation has two real solutions; if D 0, the equation has one real solution; and if D 0, the equation has no real solution.
4. There are many possibilities. For example, x 2 1 has two solutions, x 2 0 has one solution, and x 2 1 has no solution. 5. x 2 8x 15 0 x 3 x 5 0 x 3 0 or x 5 0. Thus, x 3 or x 5. 6. x 2 5x 6 0 x 3 x 2 0 x 3 0 or x 2 0. Thus, x 3 or x 2.
7. x 2 x 6 x 2 x 6 0 x 2 x 3 0 x 2 0 or x 3 0. Thus, x 2 or x 3.
8. x 2 4x 21 x 2 4x 21 0 x 3 x 7 0 x 3 0 or x 7 0. Thus, x 3 or x 7.
9. 5x 2 9x 2 0 5x 1 x 2 0 5x 1 0 or x 2 0. Thus, x 15 or x 2.
10. 6x 2 x 12 0 3x 4 2x 3 0 3x 4 0 or 2x 3 0. Thus, x 43 or x 32 . 11. 2s 2 5s 3 2s 2 5s 3 0 2s 1 s 3 0 2s 1 0 or s 3 0. Thus, s 12 or s 3.
12. 4y 2 9y 28 4y 2 9y 28 0 4y 7 y 4 0 4y 7 0 or y 4 0. Thus, y 74 or y 4.
13. 12z 2 44z 45 12z 2 44z 45 0 6z 5 2z 9 0 6z 5 0 or 2z 9 0. Thus, z 56 or z 92 . 14. 42 4 3 42 4 3 0 2 1 2 3 0 2 1 0 or 2 3 0. If 2 1 0, then 12 ; if 2 3 0, then 32 .
15. x 2 5 x 100 x 2 5x 500 x 2 5x 500 0 x 25 x 20 0 x 25 0 or x 20 0. Thus, x 25 or x 20. 16. 6x x 1 21 x 6x 2 6x 21 x 6x 2 5x 21 0 2x 3 3x 7 0 2x 3 0 or 3x 7 0. If 2x 3 0, then x 32 ; if 3x 7 0, then x 73 .
SECTION 1.4 Solving Quadratic Equations
91
17. x 2 8x 1 0 x 2 8x 1 x 2 8x 16 1 16 x 42 15 x 4 15 x 4 15. 18. x 2 6x 2 0 x 2 6x 2 x 2 6x 9 2 9 x 32 11 x 3 11 x 3 11. 19. x 2 6x 11 0 x 2 6x 11 x 2 6x 9 11 9 x 32 20 x 3 2 5 x 3 2 5. 2 3 3 20. x 2 3x 74 0 x 2 3x 74 x 2 3x 94 74 94 x 32 16 4 4 x 2 2 x 2 2 x 12 or x 72 .
2 3 1 x 12 1 x 12 1. So 21. x 2 x 34 0 x 2 x x 2 x 14 34 14 x 12 4 x 12 1 32 or x 12 1 12 . 25 x 5 2 21 x 5 21 21 1 22. x 2 5x 1 0 x 2 5x 1 x 2 5x 25 4 4 2 4 2 4 2
x 52 221 .
23. x 2 22x 21 0 x 2 22x 21 x 2 22x 112 21 112 21 121 x 112 100 x 11 10 x 11 10. Thus, x 1 or x 21.
24. x 2 18x 19 x 2 18x 92 19 92 19 81 x 92 100 x 9 10 x 9 10, so x 1 or x 19. 12 25. 5x 2 10x 7 0 x 2 2x 75 0 x 2 2x 75 x 2 2x 1 75 1 x 12 12 x 1 5 5
x 1 2 515
26. 2x 2 16x 5 0 x 2 8x 52 0 x 2 8x 52 x 2 8x 16 52 16 x 42 27 2 3 6 x 4 27 2 x 4 2 . 49 7 2 17 27. 2x 2 7x 4 0 x 2 72 x 2 0 x 2 72 x 2 x 2 72 x 49 16 2 16 x 4 16 17 7 17 7 x 4 16 x 4 4 . 25 x 5 2 153 x 5 153 28. 4x 2 5x 8 0 x 2 54 x 2 0 x 2 54 x 2 x 2 54 x 25 2 64 64 8 64 8 64
x 58 3 817 .
29. x 2 8x 12 0 x 2 x 6 0 x 2 or x 6.
30. x 2 3x 18 0 x 3 x 6 0 x 3 or x 6.
31. x 2 8x 20 0 x 10 x 2 0 x 10 or x 2.
32. 10x 2 9x 7 0 5x 7 2x 1 0 x 75 or x 12 . 33. 2x 2 x 3 0 x 1 2x 3 0 x 1 0 or 2x 3 0. If x 1 0, then x 1; if 2x 3 0, then x 32 .
34. 3x 2 7x 4 0 3x 4 x 1 0 3x 4 0 or x 1 0. Thus, x 43 or x 1.
35. 3x 2 6x 5 0 x 2 2x 53 0 x 2 2x 53 x 2 2x 1 53 1 x 12 83 x 1 83
x 1 2 3 6 .
0 36. x 2 6x 1 6 62 4 1 1 6 36 4 6 32 64 2 b b2 4ac 3 2 2. x 2a 2 1 2 2 2 37. x 2 43 x 49 0 9x 2 12x 4 0 3x 22 0 x 23 .
92
CHAPTER 1 Equations and Graphs
6 b b2 4ac 38. 2x 2 3x 12 0 4x 2 6x 1 0 x 2a
6 52 3 13 62 4 4 1 . 2 4 8 4
39. 4x 2 16x 9 0 2x 1 2x 9 0 2x 1 0 or 2x 9 0. If 2x 1 0, then x 12 ; if 2x 9 0, then x 92 . x 2 4x 1 0 4 42 4 1 1 b b2 4ac 4 16 4 4 12 42 3 x 2 3. 2a 2 1 2 2 2 3 32 4 1 3 3 9 12 3 3 . Since the 41. 2 3 1 2 3 3 0 2 1 2 2 discriminant is less than 0, the equation has no real solution. 2 4ac 5 52 4 1 3 5 25 12 5 13 b b 2 . 42. 3 5z z 0 z 2a 2 1 2 2 40. 0
43. 10y 2 16y 5 0 16 162 4 10 5 b b2 4ac 16 256 200 16 56 8 14 x . 2a 2 10 20 20 10 44. 25x 2 70x 49 0 5x 72 0 5x 7 0 5x 7 x 75 . 2 2 22 4 3 2 2 4 24 2 20 b b 4ac . Since the 45. 3x 2 2x 2 0 x 2a 2 3 6 6 discriminant is less than 0, the equation has no real solution. 2 4ac 7 72 4 5 5 7 49 100 7 51 b b 2 . 46. 5x 7x 5 x 2a 2 5 10 10 Since the discriminant is less than 0, the equation has no real solution. 47. x 2 0011x 0064 0 0011 00112 4 1 0064 0011 0000121 0256 0011 0506 x . 2 1 2 2 0011 0506 0011 0506 Thus, x 0259 or x 0248. 2 2 48. x 2 2450x 1500 0 2450 24502 4 1 1500 2450 60025 6 2450 00025 2450 0050 x . Thus, 2 1 2 2 2 2450 0050 2450 0050 x 1250 or x 1200. 2 2 49. x 2 2450x 1501 0 2450 24502 4 1 1501 2450 60025 6004 2450 00015 x . 2 1 2 2 Thus, there is no real solution. 50. x 2 1800x 0810 0 1800 18002 4 1 0810 1800 324 324 1800 0 x 0900. Thus the only 2 1 2 2 solution is x 0900.
SECTION 1.4 Solving Quadratic Equations
51. h
93
1 gt 2 t 1 gt 2 t h 0 0 2 2
0. Using the Quadratic Formula, 0 0 2 4 12 g h 0 02 2gh t . g 2 12 g
n n 1 2S n 2 n n 2 n 2S 0. 2 1 12 4 1 2S 1 1 8S n . 2 1 2
52. S
Using the Quadratic Formula,
53. A 2x 2 4xh 2x 2 4xh A 0. Using the Quadratic Formula, 4h 4 4h 2 2A 4h 4h2 4 2 A 4h 16h 2 8A 4h 2 4h 2 2A x 2 2 4 4 4 2 2h 4h 2 2A 2h 4h 2 2A 4 2 54. A 2r 2 2r h 2r 2 2r h A 0. Using the Quadratic Formula, 2h 2h2 4 2 A 2h 4 2 h 2 8 A h 2 h 2 2 A . r 2 2 4 2 1 1 1 c s b c s a s a s b cs bc cs ac s 2 as bs ab 55. sa s b c s 2 a b 2c s ab ac bc 0. Using the Quadratic Formula, a b 2c a b 2c2 4 1 ab ac bc s 2 1 a b 2c a 2 b2 4c2 2ab 4ac 4bc 4ab 4ac 4bc 2 a b 2c a 2 b2 4c2 2ab 2 2 4 2 1 4 1 2 r 2 1 r r 2 1 r 2 r 1 r 2r 2 4 1 r r r 2 2r 2 4 4r 56. r 1r r 1r r r 5 52 4 1 4 5 25 16 5 41 2 r 5r 4 0. Using the Quadratic Formula, r . 2 1 2 2 57. D b2 4ac 62 4 1 1 32. Since D is positive, this equation has two real solutions.
58. x 2 6x 9 x 2 6x 9, so D b2 4ac 62 4 1 9 36 36 0. Since D 0, this equation has one real solution. 59. D b2 4ac 2202 4 1 121 484 484 0. Since D 0, this equation has one real solution. 60. D b2 4ac 2212 4 1 121 48841 484 00441. Since D 0, this equation has two real solutions. 61. D b2 4ac 52 4 4 13 8 25 26 1. Since D is negative, this equation has no real solution.
62. D b2 4ac r2 4 1 s r 2 4s. Since D is positive, this equation has two real solutions. 1 63. a 2 x 2 2ax 1 0 ax 12 0 ax 1 0. So ax 1 0 then ax 1 x . a 64. ax 2 2a 1 x a 1 0 [ax a 1] x 1 0 ax a 1 0 or x 1 0. If ax a 1 0, a1 then x ; if x 1 0, then x 1. a
94
CHAPTER 1 Equations and Graphs
65. We want to find the values of k that make the discriminant 0. Thus k 2 4 4 25 0 k 2 400 k 20
66. We want to find the values of k that make the discriminant 0. Thus D 362 4 k k 0 4k 2 362 2k 36 k 18.
67. Let n be one number. Then the other number must be 55 nsince n 55 n 55. Because
the product is 684, we have n 55 n 684 55n n 2 684 n 2 55n 684 0
55 552 41684 5517 72 36 or n 55 30252736 552 289 5517 21 2 2 . So n 2 2 5517 38 n 2 2 19. In either case, the two numbers are 19 and 36.
68. Let n be one even number. Then the next even number is n 2. Thus we get the equation n 2 n 22 1252 n 2 n 2 4n 4 1252 0 2n 2 4n 1248 2 n 2 2n 624 2 n 24 n 26. So n 24 or n 26. Thus the consecutive even integers are 24 and 26 or 26 and 24.
69. Let be the width of the garden in feet. Then the length is 10. Thus 875 10 2 10 875 0 35 25 0. So 35 0 in which case 35 which is not possible, or 25 0 and so 25. Thus the width is 25 feet and the length is 35 feet. 70. Let be the width of the bedroom. Then its length is 7. Since area is length times width, we have
228 7 2 7 2 7 228 0 19 12 0 19 0 or 12 0. Thus 19 or 12. Since the width must be positive, the width is 12 feet.
71. Let be the width of the garden in feet. We use the perimeter to express the length l of the garden in terms of width. Since the perimeter is twice the width plus twice the length, we have 200 2 2l 2l 200 2 l 100 . Using the formula for area, we have 2400 100 100 2 2 100 2400 0 40 60 0. So 40 0 40, or 60 0 60. If 40, then l 100 40 60. And if 60, then l 100 60 40. So the length is 60 feet and the width is 40 feet.
72. First we write a formula for the area of the figure in terms of x. Region A has dimensions 14 in. and x in. and region B has dimensions 13 x in. and x in. So
the area of the figure is 14 x [13 x x] 14x 13x x 2 x 2 27x. We
are given that this is equal to 160 in2 , so 160 x 2 27x x 2 27x 160 0
x A
14 in. 13 in. B
x
x 32 x 5 x 32 or x 5. x must be positive, so x 5 in.
73. The shaded area is the sum of the area of a rectangle and the area of a triangle. So A y 1 12 y y 12 y 2 y. We
are given that the area is 1200 cm2 , so 1200 12 y 2 y y 2 2y 2400 0 y 50 y 48 0. y is positive, so y 48 cm.
1 x 300 x 30x 1 x 2 1 x 2 30x 1250 0. 74. Setting P 1250 and solving for x, we have 1250 10 10 10 1 1250 30 302 4 10 30 900 500 30 20 . Thus Using the Quadratic Formula, x 1 02 02 2 10
30 20 30 20 50 or x 250. Since he must have 0 x 200, he should make 50 ovens per week. 02 02 75. Let x be the length of one side of the cardboard, so we start with a piece of cardboard x by x. When 4 inches are x
removed from each side, the base of the box is x 8 by x 8. Since the volume is 100 in3 , we get 4 x 82 100
x 2 16x 64 25 x 2 16x 39 0 x 3 x 13 0. So x 3 or x 13. But x 3 is not possible, since then the length of the base would be 3 8 5 and all lengths must be positive. Thus x 13, and the piece of cardboard is 13 inches by 13 inches.
SECTION 1.4 Solving Quadratic Equations
95
76. Let r be the radius of the can. Now using the formula V r 2 h with V 40 cm3 and h 10, we solve for r. Thus 40 r 2 10 4 r 2 r 2. Since r represents radius, r 0. Thus r 2, and the diameter is 4 cm.
77. Let be the width of the lot in feet. Then the length is 6. Using the Pythagorean Theorem, we have
2 62 1742 2 2 12 36 30,276 22 12 30240 0 2 6 15120 0 126 120 0. So either 126 0 in which case 126 which is not possible, or 120 0 in which case 120. Thus the width is 120 feet and the length is 126 feet.
78. Let h be the height of the flagpole, in feet. Then the length of each guy wire is h 5. Since the distance between the points where the wires are fixed to the ground is equal to one guy wire, the triangle is equilateral, and the flagpole is the perpendicular bisector of the base. Thus from the Pythagorean Theorem, we get 1 h 5 2 h 2 h 52 h 2 10h 25 4h 2 4h 2 40h 100 h 2 30h 75 0 2
30 302 4175 3 . Since h 3020 3 0, we reject it. Thus 30 900300 30 2 1200 3020 h 21 2 2 2 3 15 103 3232 ft 32 ft 4 in. the height is h 3020 2
79. Let x be the rate, in mi/h, at which the salesman drove between Ajax and Barrington. Direction
Distance
Rate
Ajax Barrington
120
x
Barrington Collins
150
x 10
Time 120 x 150 x 10
distance to fill in the “Time” column of the table. Since the second part of the trip rate 1 hour) more than the first, we can use the time column to get the equation 120 1 150 took 6 minutes (or 10 x 10 x 10 120 10 x 10 x x 10 150 10x 1200x 12,000 x 2 10x 1500x x 2 290x 12,000 0 We have used the equation time
290 2902 4112,000 290 84,10048,000 290 36,100 290190 145 95. Hence, the salesman x 2 2 2 2
drove either 50 mi/h or 240 mi/h between Ajax and Barrington. (The first choice seems more likely!) 80. Let x be the rate, in mi/h, at which Kiran drove from Tortula to Cactus. Direction
Distance
Rate
Tortula Cactus
250
x
Cactus Dry Junction
360
x 10
Time 250 x 360 x 10
distance to fill in the time column of the table. We are given that the sum of rate 360 250 11 250 x 10 360x the times is 11 hours. Thus we get the equation x x 10 11x x 10 250x 2500 360x 11x 2 110x 11x 2 500x 2500 0 500 5002 4 11 2500 500 250,000 110,000 500 360,000 500 600 . Hence, x 2 11 22 22 22 Kiran drove either 454 mi/h (impossible) or 50 mi/h between Tortula and Cactus. We have used time
96
CHAPTER 1 Equations and Graphs
81. Let r be the rowing rate in km/h of the crew in still water. Then their rate upstream was r 3 km/h, and their rate downstream was r 3 km/h. Direction
Distance
Rate
Upstream
6
r 3
Downstream
6
r 3
Time 6 r 3 6 r 3
Since the time to row upstream plus the time to row downstream was 2 hours 40 minutes 83 hour, we get the equation
6 8 6 6 3 r 3 6 3 r 3 8 r 3 r 3 18r 54 18r 54 8r 2 72 r 3 r 3 3 0 8r 2 36r 72 4 2r 2 9r 18 4 2r 3 r 6 2r 3 0 or r 6 0. If 2r 3 0, then r 32 ,
which is impossible because the rowing rate is positive. If r 6 0, then r 6. So the rate of the rowing crew in still water is 6 km/h.
82. Let r be the speed of the southbound boat. Then r 3 is the speed of the eastbound boat. In two hours the southbound boat has traveled 2r miles and the eastbound boat has traveled 2 r 3 2r 6 miles. Since they are traveling is directions with are 90 apart, we can use the Pythagorean Theorem to get 2r 2 2r 62 302 4r 2 4r 2 24r 36 900 8r 2 24r 864 0 8 r 2 3r 108 0 8 r 12 r 9 0. So r 12 or r 9. Since speed is positive, the speed of the southbound boat is 9 mi/h.
83. Using h 0 288, we solve 0 16t 2 288, for t 0. So 0 16t 2 288 16t 2 288 t 2 18 t 18 3 2. Thus it takes 3 2 424 seconds for the ball the hit the ground.
84. (a) Using h 0 96, half the distance is 48, so we solve the equation 48 16t 2 96 48 16t 2 3 t 2 t 3. Since t 0, it takes 3 1732 s. (b) The ball hits the ground when h 0, so we solve the equation 0 16t 2 96 16t 2 96 t 2 6 t 6. Since t 0, it takes 6 2449 s. 85. We are given o 40 ft/s.
(a) Setting h 24, we have 24 16t 2 40t 16t 2 40t 24 0 8 2t 2 5t 3 0 8 2t 3 t 1 0 t 1 or t 1 12 . Therefore, the ball reaches 24 feet in 1 second (ascending) and again after 1 12 seconds (descending).
(b) Setting h 48, we have 48 16t 2 40t 16t 2 40t 48 0 2t 2 5t 6 0 5 25 48 5 23 t . However, since the discriminant D 0, there is no real solution, and hence the ball 4 4 never reaches a height of 48 feet. (c) The greatest height h is reached only once. So h 16t 2 40t 16t 2 40t h 0 has only one solution. Thus D 402 4 16 h 0 1600 64h 0 h 25. So the greatest height reached by the ball is 25 feet.
(d) Setting h 25, we have 25 16t 2 40t 16t 2 40t 25 0 4t 52 0 t 1 14 . Thus the ball reaches the highest point of its path after 1 14 seconds. (e) Setting h 0 (ground level), we have 0 16t 2 40t 2t 2 5t 0 t 2t 5 0 t 0 (start) or t 2 12 . So the ball hits the ground in 2 12 s.
86. If the maximum height is 100 feet, then the discriminant of the equation, 16t 2 o t 100 0, must equal zero. So
0 b2 4ac o 2 4 16 100 o2 6400 o 80. Since o 80 does not make sense, we must have o 80 ft/s.
SECTION 1.4 Solving Quadratic Equations
97
87. (a) The fish population on January 1, 2002 corresponds to t 0, so F 1000 30 17 0 02 30 000. To find
when the population will again reach this value, we set F 30 000, giving 30000 1000 30 17t t 2 30000 17000t 1000t 2 0 17000t 1000t 2 1000t 17 t t 0 or
t 17. Thus the fish population will again be the same 17 years later, that is, on January 1, 2019. (b) Setting F 0, we have 0 1000 30 17t t 2 t 2 17t 30 0 17 289 120 17 409 17 2022 t . Thus t 1612 or t 18612. Since 2 2 2 t 0 is inadmissible, it follows that the fish in the lake will have died out 18612 years after January 1, 2002, that is on August 12, 2020. 88. Let y be the circumference of the circle, so 360 y is the perimeter of the square. Use the circumference to find the 2 radius, r, in terms of y: y 2r r y 2. Thus the area of the circle is y 2 y 2 4. Now if the 2 perimeter of the square is 360 y, the length of each side is 14 360 y and the area of the square is 14 360 y . 2 Setting these areas equal, we obtain y 2 4 14 360 y y 2 14 360 y 2y 360 y 2 y 360 . Therefore, y 360 2 1691. Thus one wire is 1691 in. long and the other is 1909 in. long. 89. Let be the uniform width of the lawn. With cut off each end, the area of the factory is 240 2 180 2. Since
the lawn and the factory are equal in size this area, is 12 240 180. So 21,600 43,200 480 360 42 0 42 840 21,600 4 2 210 5400 4 30 180 30 or 180. Since 180 ft is too wide, the width of the lawn is 30 ft, and the factory is 120 ft by 180 ft.
2 2 90. Let h be the height the ladder reaches (in feet). Using the Pythagorean Theorem we have 7 12 h 2 19 12 2 2 2 2 15 225 1296 h 2 39 h 2 39 15 1521 2 4 4 2 4 4 4 324. So h 324 18 feet. 91. Let t be the time, in hours it takes Irene to wash all the windows. Then it takes Henry t 32 hours to wash all the windows, and the sum of the fraction of the job per hour they can do individually equals the fraction of the 1 1 1 4 9 9 job they can do together. Since 1 hour 48 minutes 1 48 60 1 5 5 , we have t 3 t 2
5
2 1 59 9 2t 3 2 9t 5t 2t 3 18t 27 18t 10t 2 15t 10t 2 21t 27 0 t 2t 3 21 212 4 10 27 21 441 1080 21 39 21 39 9 t . So t 2 10 20 20 20 10 21 39 or t 3. Since t 0 is impossible, all the windows are washed by Irene alone in 3 hours and by Henry alone in 20 3 32 4 12 hours.
92. Let t be the time, in hours, it takes Kay to deliver all the flyers alone. Then it takes Lynn t 1 hours to deliver all the flyers 1 1 1 1 1 14 04t 04t alone, and it takes the group 04t hours to do it together. Thus 14 04t 1 t t 1 04t t t 1 4t t 4 10 t t 1 4 t 1 4t 10 t 1 t 2 t 4t 4 4t 10t 10 t 2 t 6 0 t 1 t 3 t 2 0. So t 3 or t 2. Since t 2 is impossible, it takes Kay 3 hours to deliver all the flyers alone.
98
CHAPTER 1 Equations and Graphs
93. Let x be the distance from the center of the earth to the dead spot (in thousands of miles). Now setting 0012K K 0012K K 2 K 239 x2 0012K x 2 F 0, we have 0 2 x x 239 x2 239 x2
57121 478x x 2 0012x 2 0988x 2 478x 57121 0. Using the Quadratic Formula, we obtain
478 4782 4098857121 2741808 47852362 241903 26499. x 478 228484225742192 4781976 20988 1976 1976
So either x 241903 26499 268 or x 241903 26499 215. Since 268 is greater than the distance from the earth to the moon, we reject it; thus x 215,000 miles.
94. If we have x 2 9x 20 x 4 x 5 0, then x 4 or x 5, so the roots are 4 and 5. The product is 4 5 20, and
the sum is 4 5 9. If we have x 2 2x 8 x 4 x 2 0, then x 4 or x 2, so the roots are 4 and 2. The
product is 42 8, and the sum is 42 2. Lastly, if we have x 2 4x 2 0, then using the Quadratic Formula, 4 42 4 1 2 4 8 4 2 2 we have x 2 2. The roots are 2 2 and 2 2. The 2 1 2 2 product is 2 2 2 2 4 2 2, and the sum is 2 2 2 2 4. In general, if x r1
and x r2 are roots, then x 2 bx c x r1 x r2 x 2 r1 x r2 x r1r2 x 2 r1 r2 x r1r2 . Equating the coefficients, we get c r1r2 and b r1 r2 . 95. Let x equal the original length of the reed in cubits. Then x 1 is the piece that fits 60 times along the length of the field, that is, the length is 60 x 1. The width is 30x. Then converting cubits to ninda, we have 2 2 375 60 x 1 30x 12 25 2 x x 1 30 x x x x 30 0 x 6 x 5 0. So x 6 or 12 x 5. Since x must be positive, the original length of the reed is 6 cubits.
1.5
COMPLEX NUMBERS
1. The imaginary number i has the property that i 2 1. 2. For the complex number 3 4i the real part is 3 and the imaginary part is 4. 3. (a) The complex conjugate of 3 4i is 3 4i 3 4i. (b) 3 4i 3 4i 32 42 25
4. If 3 4i is a solution of a quadratic equation with real coefficients, then 3 4i 3 4i is also a solution of the equation. 5. Yes, every real number a is a complex number of the form a 0i. 6. Yes. For any complex number z, z z a bi a bi a bi a bi 2a, which is a real number. 7. 5 7i: real part 5, imaginary part 7. 2 5i 9. 23 53 i: real part 23 , imaginary part 53 . 3
8. 6 4i: real part 6, imaginary part 4. 10.
4 7i 2 72 i: real part 2, imaginary part 72 . 2
13. 23 i: real part 0, imaginary part 23 . 15. 3 4 3 2i: real part 3, imaginary part 2.
12. 12 : real part 12 , imaginary part 0. 14. i 3: real part 0, imaginary part 3. 16. 2 5 2 i 5: real part 2, imaginary part 5.
17. 3 2i 5i 3 2 5 i 3 7i
18. 3i 2 3i 2 [3 3] i 2 6i
19. 5 3i 4 7i 5 4 3 7 i 1 10i
20. 3 4i2 5i 3 2[4 5] i 59i 22. 3 2i 5 13 i 3 5 2 13 i 2 73 i
11. 3: real part 3, imaginary part 0.
21. 6 6i 9 i 6 9 6 1 i 3 5i 23. 7 12 i 5 32 i 7 5 12 32 i 2 2i
SECTION 1.5 Complex Numbers
24. 4 i 2 5i 4 i 2 5i 4 2 1 5 i 6 6i
25. 12 8i 7 4i 12 8i 7 4i 12 7 8 4 i 19 4i 26. 6i 4 i 6i 4 i 4 6 1 i 4 7i
27. 4 1 2i 4 8i
28. 2 3 4i 6 8i
29. 7 i 4 2i 28 14i 4i 2i 2 28 2 14 4 i 30 10i
30. 5 3i 1 i 5 5i 3i 3i 2 5 3 5 3 i 8 2i
31. 6 5i 2 3i 12 18i 10i 15i 2 12 15 18 10 i 27 8i 32. 2 i 3 7i 6 14i 3i 7i 2 6 7 14 3 i 1 17i
33. 2 5i 2 5i 22 5i2 4 25 1 29
34. 3 7i 3 7i 32 7i2 58
35. 2 5i2 22 5i2 2 2 5i 4 25 20i 21 20i
36. 3 7i2 32 7i2 2 3 7i 40 42i 1 1 i i i 37. 2 i i i i 1 i 1 1i 1i 1i 1 1 1 1i 2 2i 38. 1i 1i 1i 11 2 1 i2 39. 40.
2 3i 1 2i 2 4i 3i 6i 2 8i 2 3i 2 6 4 3 i or 85 15 i 1 2i 1 2i 1 2i 14 5 1 4i 2
5i 5 i 3 4i 15 20i 3i 4i 2 11 23i 15 4 20 3 i 23 11 25 25 i 3 4i 3 4i 3 4i 9 16 25 9 16i 2
10i 1 2i 10i 20i 2 5 4 2i 20 10i 10i 4 2i 1 2i 1 2i 1 2i 14 5 1 4i 2 1 2 3i 2 3i 2 3i 1 2 3i 2 3i 13 42. 2 3i1 13 2 3i 2 3i 2 3i 49 13 4 9i 2
41.
43.
4 6i 3i 12i 18i 2 18 12 18 12i 4 6i i 2 43 i 3i 3i 3i 9 9 9 9i 2
3 5i 15i 45i 75i 2 75 45 75 45i 3 5i i 13 15 i 15i 15i 15i 225 225 225 225i 2 1 1 1i 1 1i 1i 1 i 1i 1i 1 i 45. 1i 1i 1i 1i 1i 1i 2 2 1 i2 1 i2 44.
3 i 6i 2i 2 5 5i 2 i 10 5i 10i 5i 2 10 5 5 10 i 1 2i 3 i 2i 2i 2i 2i 5 4 i2 15 5i 5 15 5 5i 3 i 5 5 48. i 10 i 2 15 1 47. i 3 i 2 i i
46.
2 49. 3i5 35 i 2 i 243 12 i 243i 250 51. i 1000 i 4 1250 1
50. 2i4 24 i 4 16 1 16
250 52. i 1002 i 4 i 2 1i 2 1 9 53. 49 49 1 7i 54. 81 16 4 i 55. 3 12 i 3 2i 3 6i 2 6 56. 13 27 13 3i 3 3i 57. 3 5 1 1 3 i 5 1 i 3 3i i 5 i 2 5 3 5 3 5 i
99
100
CHAPTER 1 Equations and Graphs
58. 3 4 6 8 3 2i 6 2i 2 18 2i 6 2i 6 4i 2 2 3 2 4 2 2 6 2 6 i 2 4i 6 2 1i 2 2 2i 2 2 8 2 59. 1 2 1i 2 1i 2 2 i 2 2i 2 i 2 6i 36 60. i 2 2 1 2i 2 9 i 2 3i i 2 i 2 61. x 2 49 0 x 2 49 x 7i
62. 3x 2 1 0 3x 2 1 x 2 13 x 33 i 2 4ac 1 12 4 1 2 1 7 b b 12 27 i 63. x 2 x 2 0 x 2a 2 1 2 2 22 4 1 2 2 4 8 2 4 2 2i 64. x 2 2x 2 0 x 1 i 2 1 2 2 2 3 19 3 32 4 1 7 32 219 i 65. x 2 3x 7 0 x 2 1 2 6 62 4 1 10 6 36 40 6 4 6 2i 3i 66. x 2 6x 10 0 x 2 1 2 2 2 1 12 4 1 1 1 1 4 1 3 1 i 3 2 67. x x 1 0 x 12 23 i 2 1 2 2 2 3 32 4 1 3 3 9 12 3 3 3i 3 2 68. x 3x 3 0 x 32 23 i 2 1 2 2 2 2 22 4 2 1 2 48 2 4 2 2i 69. 2x 2 2x 1 0 x 12 12 i 2 2 4 4 4
3 32 413 3 3 912 3 3 3i 3 3 3 i 70. t 3 0 t 2 3t 3 0 t 21 2 2 2 2 2 t 2 71. 6x 12x 7 0 12 122 467 144168 6 12 2i 6 1 6 i 12 12 1212 24 122i x 26 12 12 12 6 72. x 2 12 x 1 0
x
2 1 12 4 1 1 2 2 1
12
2
1 4 4
12
15 4
2
12 12 i 15 14 415 i 2
73. z 3 4i 5 2i 3 4i 5 2i 8 2i 74. z 3 4i 5 2i 8 2i 8 2i 75. z z 3 4i 3 4i 32 42 25
76. z 3 4i 5 2i 15 6i 20i 8i 2 23 14i
77. LHS z a bi c di a bi c di a c b d i a c b d i. RHS z a bi c di a c b d i a c b d i. Since LHS RHS, this proves the statement.
78. LHS z a bi c di ac adi bci bdi 2 ac bd ad bc i ac bd ad bc i. RHS z a bi c di a bi c di ac adi bci bdi 2 ac bd ad bc i. Since LHS RHS, this proves the statement.
SECTION 1.6 Solving Other Types of Equations
101
2 79. LHS z2 a bi a bi2 a 2 2abi b2 i 2 a 2 b2 2abi. RHS z 2 a bi2 a 2 2abi b2 i 2 a 2 b2 2abi a 2 b2 2abi. Since LHS RHS, this proves the statement.
80. z a bi a bi a bi z. 81. z z a bi a bi a bi a bi 2a, which is a real number.
82. z z a bi a bi a bi a bi a bi a bi 2bi, which is a pure imaginary number.
83. z z a bi a bi a bi a bi a 2 b2 i 2 a 2 b2 , which is a real number.
84. Suppose z z. Then we have a bi a bi a bi a bi 0 2bi b 0, so z is real. Now if z is real, then z a 0i(where a is real). Since z a 0i, we have z z. b b2 4ac . Since both solutions are imaginary, 85. Using the Quadratic Formula, the solutions to the equation are x 2a b 4ac b2 we have b2 4ac 0 4ac b2 0, so the solutions are x i, where 4ac b2 is a real number. 2a 2a Thus the solutions are complex conjugates of each other. 86. i i, i 5 i 4 i i, i 9 i 8 i i;
i 2 1, i 6 i 4 i 2 1, i 10 i 8 i 2 1;
i 3 i, i 7 i 4 i 3 i, i 11 i 8 i 3 i; i 4 1, i 8 i 4 i 4 1, i 12 i 8 i 4 1. Because i 4 1, we have i n i r , where r is the remainder when n is divided by 4, that is, n 4 k r , where k is an integer and 0 r 4. Since 4446 4 1111 2, we must have i 4446 i 2 1.
1.6
SOLVING OTHER TYPES OF EQUATIONS
Note: In cases where both sides of an equation are squared, the implication symbol is sometimes used loosely. For example, 2 x x 1 “” x x 12 is valid only for positive x. In these cases, inadmissible solutions are identified later in the
solution.
1. (a) To solve the equation x 3 4x 2 0 we factor the left-hand side: x 2 x 4 0, as above. (b) The solutions of the equation x 2 x 4 0 are x 0 and x 4. 2. (a) Isolating the radical in 2x x 0, we obtain 2x x. 2 2x x2 2x x 2 . (b) Now square both sides:
(c) Solving the resulting quadratic equation, we find 2x x 2 x 2 2x x x 2 0, so the solutions are x 0 and x 2. (d) We substitute these possible solutions into the original equation: 2 0 0 0, so x 0 is a solution, but 2 2 2 4 0, so x 2 is not a solution. The only real solution is x 0.
3. The equation x 12 5 x 1 6 0 is of quadratic type. To solve the equation we set W x 1. The resulting quadratic equation is W 2 5W 6 0 W 3 W 2 0 W 2 or W 3 x 1 2 or x 1 3 x 1 or x 2. You can verify that these are both solutions to the original equation.
4. The equation x 6 7x 3 8 0 is of quadratic type. To solve the equation we set W x 3 . The resulting quadratic equation is W 2 7W 8 0. 5. x 2 x 0 x x 1 0 x 0 or x 1 0. Thus, the two real solutions are 0 and 1.
6. 3x 3 6x 2 0 3x 2 x 2 0 x 0 or x 2 0. Thus, the two real solutions are 0 and 2. 7. x 3 25x x 3 25x 0 x x 2 25 0 x x 5 x 5 0 x 0 or x 5 0 or x 5 0. The three real solutions are 5, 0, and 5.
102
CHAPTER 1 Equations and Graphs
8. x 5 5x 3 x 5 5x 3 0 x 3 x 2 5 0 x 0 or x 2 5 0. The solutions are 0 and 5. 9. x 5 3x 2 0 x 2 x 3 3 0 x 0 or x 3 3 0. The solutions are 0 and 3 3. 10. 6x 5 24x 0 6x x 4 4 0 6x x 2 2 x 2 2 0. Thus, x 0, or x 2 2 0 (which has no solution), or x 2 2 0. The solutions are 0 and 2. 11. 0 4z 5 10z 2 2z 2 2z 3 5 . If 2z 2 0, then z 0. If 2z 3 5 0, then 2z 3 5 z 3 52 . The solutions are 0 and 3 52 . 2 3 2 . The solutions are 0 12. 0 125t 10 2t 7 t 7 125t 3 2 . If t 7 0, then t 0. If 125t 3 2 0, then t 3 125 5 3
and 52 .
13. 0 x 5 8x 2 x 2 x 3 8 x 2 x 2 x 2 2x 4 x 2 0, x 2 0, or x 2 2x 4 0. If x 2 0, then
x 0; if x 2 0, then x 2, and x 2 2x 4 0 has no real solution. Thus the solutions are x 0 and x 2. 14. 0 x 4 64x x x 3 64 x 0 or x 3 64 0. If x 3 64 0, then x 3 64 x 4. The solutions are 0 and 4.
15. 0 x 3 5x 2 6x x x 2 5x 6 x x 2 x 3 x 0, x 2 0, or x 3 0. Thus x 0, or x 2, or x 3. The solutions are x 0, x 2, and x 3. 16. 0 x 4 x 3 6x 2 x 2 x 2 x 6 x 2 x 3 x 2. Thus either x 2 0, so x 0,or x 3, or x 2. The solutions are 0, 3, and 2.
17. 0 x 4 4x 3 2x 2 x 2 x 2 4x 2 . So either x 2 0 x 0, or using the Quadratic Formula on x 2 4x 2 0, 42 412 8 42 2 2 2. The solutions are 0, 2 2, and 4 2 168 4 we have x 4 21 2 2 2 2. 18. 0 y 5 8y 4 4y 3 y 3 y 2 8y 4 . If y 3 0, then y 0. If y 2 8y 4 0, then using the Quadratic Formula, we 8 82 4 1 4 8 48 have y 4 2 3. Thus, the three solutions are 0, 4 2 3, and 4 2 3. 2 1 2 19. 3x 54 3x 53 0. Let y 3x 5. The equation becomes y 4 y 3 0 y y 3 1 y y 1 y 2 y 1 0. If y 0, then 3x 5 0 x 53 . If y 1 0, then 3x 5 1 0 x 43 . If y 2 y 1 0, then 3x 52 3x 5 1 0 9x 2 33x 31 0. The discriminant is b2 4ac 332 4 9 31 27 0, so this case gives no real solution. The solutions are x 53 and x 43 .
20. x 54 16 x 52 0. Let y x 5. The equation becomes y 4 16y 2 y 2 y 4 y 4 0. If y 2 0, then x 5 0 and x 5. If y 4 0, then x 5 4 0 and x 1. If y 4 0, then x 5 4 0 and x 9. Thus, the solutions are 9, 5, and 1. 21. 0 x 3 5x 2 2x 10 x 2 x 5 2 x 5 x 5 x 2 2 . If x 5 0, then x 5. If x 2 2 0, then x 2 2 x 2. The solutions are 5 and 2. 22. 0 2x 3 x 2 18x 9 x 2 2x 1 9 2x 1 2x 1 x 2 9 2x 1 x 3 x 3. The solutions are 12 , 3, and 3.
23. x 3 x 2 x 1 x 2 1 0 x 3 2x 2 x 2 x 2 x 2 x 2 x 2 x 2 1 . Since x 2 1 0 has no real solution, the only solution comes from x 2 0 x 2.
SECTION 1.6 Solving Other Types of Equations
103
24. 7x 3 x 1 x 3 3x 2 x 0 6x 3 3x 2 2x 1 3x 2 2x 1 2x 1 2x 1 3x 2 1 2x 1 0 or 3x 2 1 0. If 2x 1 0, then x 12 . If 3x 2 1 0, then 3x 2 1 x 2 13 x 13 . The solutions are 12 and 13 .
4 4 3 z 1 z z 1 3 z 2 z 4 3z 3 z 2 2z 1 0 z 12 0. The z1 z1 solution is z 1. We must check the original equation to make sure this value of z does not result in a zero denominator. 10 10 15 3m m 5 15 m 5 3m 10 15m 75 3m 2 15m 3m 2 85 0 26. m5 m5 m 85 3 . Verifying that neither of these values of m results in a zero denominator in the original equation, we see that 85 . and the solutions are 85 3 3 25. z
27.
1 5 1 1 1 5 4 x 1 x 2 4 x 1 x 2 x 1 x 2 4 x 1 x 2 4
4 x 2 4 x 1 5 x 1 x 2 4x 8 4x 4 5x 2 5x 10 5x 2 3x 14 0 7 5x 7 x 2 0. If 5x 7 0, then x 75 ; if x 2 0, then x 2. The solutions are and 2. 5 10 12 12 10 28. 4 0 x x 3 4 0 x 3 10 12x 4x x 3 0 x x 3 x x 3
10x 30 12x 4x 2 12x 0 4x 2 14x 30 0. Using the Quadratic Formula, we have 14 142 4 4 30 14 196 480 14 676 14 26 . So the solutions are 5 and 32 . x 2 4 8 8 8
29.
30.
x2 50 x 2 50 x 100 50x 5000 x 2 50x 5000 0 x 100 x 50 0 x 100 0 x 100 or x 50 0. Thus x 100 or x 50. The solutions are 100 and 50. 2x x2 1
1 2x x 2 1 x 2 2x 1 x 12 0, so x 1. This is indeed a solution to the original equation.
1 2 1 x 1 x 2 1 2 x 2 x 1 x 2 3x 2 1 2x 4 x 1 x 1 x 2 x 1 x 2 x 2 2 0 x 2. We verify that these are both solutions to the original equation.
31. 1
32.
33.
2 1 x x x 3 2 x 3 1 x 2 3x 2x 6 1 x 2 5x 5 0. Using the Quadratic x 3 x 3 x2 9 5 52 4 1 5 53 5 Formula, x . We verify that both are solutions to the original equation. 2 2 x x 1 1 x x 3 x 1 2x 7 2x 7 x 3 x 2 3x 2x 2 9x 7 2x 2 13x 21 2x 7 x 3 3x 2 19x 28 0 3x 7 x 4 0. Thus either 3x 7 0, so x 73 , or x 4. The solutions are 73 and 4.
34.
2 1 2 x 1 x
0 x 2 2 x 1 0 x 2 2x 2 0 2 22 4 1 2 2 48 2 4 . Since the radicand is negative, there is no real solution. x 2 1 2 2
104
CHAPTER 1 Equations and Graphs
x x2 x x2 2 5x x 2 2 5x 3x 4 x 2 2 15x 2 20x 0 14x 2 20x 2 4 x 3x 4 3 x 20 202 4 14 2 20 400 112 20 512 20 16 2 5 4 2 . The x 2 14 28 28 28 7 5 4 2 . solutions are 7 3 1 3 1x x 4 x 2 4x x 3x 1 2x 2 4x 2x 2 7x 1 0. Using the Quadratic x x 2 x 36. 4 4 2 x 2 x 7 72 4 2 1 7 57 7 57 . Both are admissible, so the solutions are . Formula, we find x 2 2 4 4 2 37. 5 4x 3 52 4x 3 25 4x 3 4x 28 x 7 is a potential solution. Substituting into the original equation, we get 5 4 7 3 5 25, which is true, so the solution is x 7. 2 38. 8x 1 3 8x 1 32 8x 1 9 x 54 . Substituting into the original equation, we get 8 54 1 3 9 3, which is true, so the solution is x 54 . x 2x 35. 5x 3 4x
39.
40.
41.
42.
43.
44.
2 2 2x 1 3x 5 2x 1 3x 5 2x 1 3x 5 x 4. Substituting into the original equation, we get 2 4 1 3 4 5 7 7, which is true, so the solution is x 4. 2 2 3 x x2 1 3x x 2 1 3 x x 2 1 x 2 x 2 0 x 1 x 2 0 x 1 or x 2. Substituting into the original equation, we get 3 1 12 1 2 2, which is true, and 3 2 22 1, which is also true. So the solutions are x 1 and x 2. 2 x 2 x x 2 x 2 x 2 x 2 x 2 x 2 x 1 x 2 0 x 1 or x 2. Substituting into the original equation, we get 1 2 1 1 1, which is false, and 2 2 2 4 2, which is true. So x 2 is the only real solution. 2 4 6x 2x 4 6x 2x2 4 6x 4x 2 2x 2 3x 2 x 2 2x 1 0 x 2 or x 12 . Substituting into the original equation, we get 4 6 2 2 2 16 4, which is false, and 4 6 12 2 12 1 1, which is true. So x 12 is the only real solution. 2x 1 1 x 2x 1 x 1 2x 1 x 12 2x 1 x 2 2x 1 0 x 2 4x x x 4. Potential solutions are x 0 and x 4 x 4. These are only potential solutions since squaring is not a reversible operation. We must check each potential solution in the original equation. Checking x 0: 2 0 1 1 0 1 1 0 is false. Checking x 4: 2 4 1 1 4 9 1 4 3 1 4 is true. The only solution is x 4. x 9 3x 0 x 9 3x x 2 9 3x 0 x 2 3x 9. Using the Quadratic Formula to find the potential 3 32 4 1 9 3 45 3 3 5 solutions, we have x . Substituting each of these solutions into the 2 1 2 2
5 is a solution, but x 33 5 is not. Thus x 33 5 is the only solution. original equation, we see that x 33 2 2 2 2 2 2 x 1 x 6x 9 x 1 x 2 7x 10 0 45. x x 1 3 x 3 x 1 x 3 x 2 x 5 0. Potential solutions are x 2 and x 5. We must check each potential solution in the original equation. Checking x 2: 2 2 1 3, which is false, so x 2 is not a solution. Checking x 5: 5 5 1 3 5 2 3, which is true, so x 5 is the only solution.
SECTION 1.6 Solving Other Types of Equations
46.
105
2 3 x 2 1 x 3 x 1 x 3 x 1 x2 3 x x 2 2x 1 x 2 3x 2 0. Using 3 17 3 32 4 1 2 . Substituting the Quadratic Formula to find the potential solutions, we have x 2 1 2
each of these solutions into the original equation, we see that x 32 17 is a solution, but x 32 17 is not. Thus
x 32 17 is the only solution. 2 2 47. 3x 1 2 x 1 3x 1 2 x 1 3x 1 4 4 x 1 x 1 2x 4 4 x 1 2 x 2 2 x 1 x 22 2 x 1 x 2 4x 4 4 x 1 x 2 8x 0 x x 8 0 x 0 or x 8. Substituting each of these solutions into the original equation, we see that x 0 is not a solution but x 8 is a solution. Thus, x 8 is the only solution. 2 1 x 1 x 22 1 x 1 x 2 1 x 1 x 4 48. 1 x 1 x 2 2 2 1 x 1 x 4 1 x 1 x 1 1 x 1 x 1 1 x 2 1 x 2 0, so x 0. We verify that this is a solution to the original equation. 49. x 4 4x 2 3 0. Let y x 2 . Then the equation becomes y 2 4y 3 0 y 1 y 3 0, so y 1 or y 3. If y 1, then x 2 1 x 1, and if y 3, then x 2 3 x 3. 50. x 4 5x 2 6 0. Let y x 2 . Then the equation becomes y 2 5y 6 0 y 2 y 3 0, so y 2 or y 3. If y 2, then x 2 2 x 2, and if y 3, then x 2 3 x 3.
51. 2x 4 4x 2 1 0. The LHS is the sum of two nonnegative numbers and a positive number, so 2x 4 4x 2 1 1 0. This equation has no real solution. 52. 0 x 6 2x 3 3 x 3 3 x 3 1 . If x 3 3 0, then x 3 3 x 3 3, or if x 3 1 x 1. Thus x 3 3 or x 1. The solutions are 3 3 and 1. 53. 0 x 6 26x 3 27 x 3 27 x 3 1 . If x 3 27 0 x 3 27, so x 3. If x 3 1 0 x 3 1, so x 1. The solutions are 3 and 1.
54. x 8 15x 4 16 0 x 8 15x 4 16 x 4 16 x 4 1 . If x 4 16 0, then x 4 16 which is impossible (for real numbers). If x 4 1 0 x 4 1, so x 1. The solutions are 1 and 1.
55. 0 x 52 3 x 5 10 [x 5 5] [x 5 2] x x 7 x 0 or x 7. The solutions are 0 and 7. x 1 x 1 2 x 1 56. Let . Then 0 3 becomes 0 2 4 3 1 3. Now if 1 0, 4 x x x x 1 x 1 x 1 x 1 then 10 1 x 1 x x 12 , and if 3 0, then 3 0 3 x x x x x 1 3x x 14 . The solutions are 12 and 14 . 2 1 1 1 . Then 8 0 becomes 2 2 8 0 4 2 0. So 4 0 2 57. Let x 1 x 1 x 1 1 4, and 2 0 2. When 4, we have 4 1 4x 4 3 4x x 34 . When x 1 1 2, we have 2 1 2x 2 3 2x x 32 . Solutions are 34 and 32 . x 1 2 x 4x x 58. Let . Then 4 becomes 2 4 4 0 2 4 4 22 . Now if x 2 x 2 x 2 x x 20 2 x 2x 4 x 4. The solution is 4. 2 0, then x 2 x 2
106
CHAPTER 1 Equations and Graphs
59. Let u x 23 . Then 0 x 43 5x 23 6 becomes u 2 5u 6 0 u 3 u 2 0 u 3 0 or u 2 0. If u 3 0, then x 23 3 0 x 23 3 x 332 3 3. If u 2 0, then x 23 2 0 x 23 2 x 232 2 2. The solutions are 3 3 and 2 2. 60. Let u 4 x; then0 x 3 4 x 4 u 2 3u 4 u 4 u 1. So u 4 4 x 4 0 4 x 4 x 44 256, or u 1 4 x 1 0 4 x 1. However, 4 x is the positive fourth root, so this cannot equal 1. The only solution is 256. 61. 4 x 112 5 x 132 x 152 0 x 1 4 5 x 1 x 12 0 x 1 4 5x 5 x 2 2x 1 0 x 1 x 2 3x 0 x 1 x x 3 0 x 1 or x 0 or x 3. The solutions are 1, 0, and 3.
62. Let u x 4; then 0 2 x 473 x 443 x 413 2u 73 u 43 u 13 u 13 2u 1 u 1. So
u x 4 0 x 4, or 2u 1 2 x 4 1 2x 7 0 2x 7 x 72 , or u 1 x 4 1 x 5 0
x 5. The solutions are 4, 72 ,and 5.
63. x 32 10x 12 25x 12 0 x 12 x 2 10x 25 0 x 12 x 52 0. Now x 12 0, so the only solution is x 5.
64. x 12 x 12 6x 32 0 x 32 x 2 x 6 0 x 32 x 2 x 3 0. Now x 12 0, and furthermore the original equation cannot have a negative solution. Thus, the only solution is x 3.
65. Let u x 16 . (We choose the exponent 16 because the LCD of 2, 3, and 6 is 6.) Then x 12 3x 13 3x 16 9 x 36 3x 26 3x 16 9 u 3 3u 2 3u 9 0 u 3 3u 2 3u 9 u 2 u 3 3 u 3 u 3 u 2 3 . So u 3 0 or u 2 3 0. If u 3 0, then x 16 3 0 x 16 3 x 36 729. If u 2 3 0, then
x 13 3 0 x 13 3 x 33 27. The solutions are 729 and 27. 66. Let u x. Then 0 x 5 x 6 becomes u 2 5u 6 u 3 u 2 0. If u 3 0, then x 3 0 x 3 x 9. If u 2 0, then x 2 0 x 2 x 4. The solutions are 9 and 4. 67.
4 4 1 2 0 1 4x 4x 2 0 1 2x2 0 1 2x 0 2x 1 x 12 . The solution is 12 . x x3 x
x4 we get, 0 1 4x 2 x 4 . Substituting u x 2 , we get 0 1 4u u 2 , and 4 4 42 411 4 2164 42 12 422 3 2 3. Substituting using the Quadratic Formula, we get u 21 back, we have x 2 2 3, and since 2 3 and 2 3 are both positive we have x 2 3 or x 2 3. Thus the solutions are 2 3, 2 3, 2 3, and 2 3. x 5 x 5. Squaring both sides, we get x 5 x 25 x 5 25 x. Squaring both sides again, we 69.
68. 0 4x 4 16x 2 4. Multiplying by
get x 5 25 x2 x 5 625 50x x 2 0 x 2 51x 620 x 20 x 31. Potential solutions are x 20 and x 31. We must check each potential solution in the original equation. Checking x 20: 20 5 20 5 25 20 5 5 20 5, which is true, and hence x 20 is a solution. Checking x 31: 36 31 5 37 5, which is false, and hence x 31 is not a 31 5 31 5 solution. The only real solution is x 20. 3 70. 4x 2 4x x 4x 2 4x x 3 0 x 3 4x 2 4x x x 2 4x 4 x x 22 . So x 0 or x 2. The solutions are 0 and 2.
SECTION 1.6 Solving Other Types of Equations
71. x 2 x 3 x 332 0 x 2 x 3 x 332 0 x 3
107
x 2 x 3 0 x 3 x 2 x 3 .
If x 312 0, then x 3 0 x 3. If x 2 x 3 0, then using the Quadratic Formula x 12 13 . The
solutions are 3 and 12 13 . 2 2 72. Let u 11 x 2 . By definition of u we require it to be nonnegative. Now 11 x 2 1 u 1. 2 u 11 x
Multiplying both sides by u we obtain u 2 2 u 0 u 2 u 2 u 2 u 1. So u 2 or u 1. But since u must be nonnegative, we only have u 2 11 x 2 2 11 x 2 4 x 2 7 x 7. The solutions are 7. 73. x x 2 2. Squaring both sides, we get x x 2 4 x 2 4 x. Squaring both sides again, we get
74.
75.
76.
77.
x 2 4 x2 16 8x x 2 0 x 2 9x 14 0 x 7 x 2. If x 7 0, then x 7. If x 2 0, then x 2. So x 2 is a solution but x 7 is not, since it does not satisfy the original equation. 1 x 2x 1 5 x. We square both sides to get 1 x 2x 1 5 x 2 x 2x 1 4 x 16 8 x x 2x 1 16 8 x. Again, squaring both sides, we obtain 2 2x 1 16 8 x 256 256 x 64x 62x 255 256 x. We could continue squaring both sides until we found possible solutions; however, consider the last equation. Since we are working with real numbers, for x to be defined, we must have x 0. Then 62x 255 0 while 256 x 0, so there is no solution. 0 x 4 5ax 2 4a 2 a x 2 4a x 2 . Since a is positive, a x 2 0 x 2 a x a. Again, since a is positive, 4a x 2 0 x 2 4a x 2 a. Thus the four solutions are a and 2 a. b 0 a 3 x 3 b3 ax b a 2 x 2 abx b2 . So ax b 0 ax b x or a ab ab2 4 a 2 b2 ab 3a 2 b2 b , but this gives no real solution. Thus, the solution is x . x a 2 a2 2a 2 x a x a 2 x 6. Squaring both sides, we have x a x a 2 x 6 2x 2 x a x a 2x 12 2 x a x a 12 x a 2 x a x a x a 6. Squaring both sides again we have x a x a 36 x 2 a 2 36 x 2 a 2 36 x a 2 36. Checking these answers, we see that x a 2 36 is not a solution (for example, try substituting a 8), but x a 2 36 is a solution.
78. Let x 16 . Then x 13 2 and x 12 3 , and so
0 3 a2 b ab 2 a b a 2 b a 3 x b 6 x a . So 6 x a 0 a 6 x x a 6 is one solution. Setting the first factor equal to zero, we have 3 x b 0 3 x b x b3 . However, the original equation includes the term b 6 x, and we cannot take the sixth root of a negative number, so this is not a solution. The only solution is x a 6 .
900 . After 5 people 79. Let x be the number of people originally intended to take the trip. Then originally, the cost of the trip is x 900 900 4500 cancel, there are now x 5 people, each paying 2. Thus 900 x 5 2 900 900 2x 10 x x x 4500 0 2x 10 0 2x 2 10x 4500 2x 100 x 45. Thus either 2x 100 0, so x 50, or x x 45 0, x 45. Since the number of people on the trip must be positive, originally 50 people intended to take the trip.
108
CHAPTER 1 Equations and Graphs
120,000 . If one person joins the group, then there would n 120,000 120,000 be n 1 members in the group, and each person would pay 6000. So n 1 6000 120,000 n n n n 120,000 6000 n 1 120,000 20 n n 1 20n n 2 19n 20 20n 6000 n 6000
80. Let n be the number of people in the group, so each person now pays
0 n 2 n 20 n 4 n 5. Thus n 4 or n 5. Since n must be positive, there are now 4 friends in the group.
t and substituting, we have 500 3t 10 t 140 5 1105 500 3u 2 10u 140 0 3u 2 10u 360 u . Since u t, we must have u 0. So 3 5 1105 9414 t 8862. So it will take 89 days for the fish population to reach 500. t u 3
81. We want to solve for t when P 500. Letting u
82. Let d be the distance from the lens to the object. Then the distance from the lens to the image is d 4. So substituting 1 1 1 . Now we multiply by the F 48, x d, and y d 4, and then solving for x, we have 48 d d 4 LCD, 48d d 4, to get d d 4 48 d 4 48d d 2 4d 96d 192 0 d 2 136d 192 136 104 . So d 16 or d 12. Since d 4 must also be positive, the object is 12 cm from the lens. d 2 83. Let x be the height of the pile in feet. Then the diameter is 3x and the radius is 32 x feet. Since the volume of the cone is 3x 2 3x 3 4000 4000 1000 ft3 , we have 1000 x 3 x 3 752 feet. x 1000 3 2 4 3 3 84. Let r be the radius of the tank, in feet. The volume of the spherical tank is 43 r 3 and is also 750 01337 100275. So 4 r 3 100275 r 3 23938 r 288 feet. 3
85. Let r be the radius of the larger sphere, in mm. Equating the volumes, we have 43 r 3 43 23 33 43 r 3 23 33 44 r 3 99 r 3 99 463. Therefore, the radius of the larger sphere is about 463 mm. 86. We have that the volume is 180 ft3 , so x x 4 x 9 180 x 3 5x 2 36x 180 x 3 5x 2 36x 180 0 x 2 x 5 36 x 5 0 x 5 x 2 36 0 x 5 x 6 x 6 0 x 6 is the only positive solution. So the box is 2 feet by 6 feet by 15 feet.
87. Let x be the length, in miles, of the abandoned road to be used. Then the length of the abandoned road not used is 40 x, and the length of the new road is 102 40 x2 miles, by the Pythagorean Theorem. Since the cost of the road is cost per mile number of miles, we have 100,000x 200,000 x 2 80x 1700 6,800,000 2 x 2 80x 1700 68 x. Squaring both sides, we get 4x 2 320x 6800 4624 136x x 2
1 18488 x 136 3x 2 184x 2176 0 x 184 3385626112 6 6 3 or x 16. Since 45 3 is longer than the existing road, 16 miles of the abandoned road should be used. A completely new road would have length 102 402 (let x 0) and would cost 1700 200,000 83 million dollars. So no, it would not be cheaper.
SECTION 1.6 Solving Other Types of Equations
109
88. Let x be the distance, in feet, that he goes on the boardwalk before veering off onto the sand. The distance along the boardwalk from where he started to the point on the boardwalk closest to the umbrella is 7502 2102 720 ft. Thus the distance that he walks on the sand is 720 x2 2102 518,400 1440x x 2 44,100 x 2 1440x 562,500. Along boardwalk Across sand
Distance
Rate
x
4
x 2 1440x 562,500
2
Time x 4
x 2 1440x 562,500 2
Since 4 minutes 45 seconds 285 seconds, we equate the time it takes to walk along the boardwalk and across the sand x 2 1440x 562,500 x 1140 x 2 x 2 1440x 562,500. Squaring both to the total time to get 285 4 2 sides, we get 1140 x2 4 x 2 1440x 562,500 1,299,600 2280x x 2 4x 2 5760x 2,250,000 0 3x 2 3480x 950,400 3 x 2 1160x 316,800 3 x 720 x 440. So x 720 0 x 720, and x 440 0 x 440. Checking x 720, the distance across the sand is
210 210 feet. So 720 4 2 180 105 285 seconds. Checking x 440, the distance across the sand is 350 720 4402 2102 350 feet. So 440 4 2 110 175 285 seconds. Since both solutions are less than or equal
to 720 feet, we have two solutions: he walks 440 feet down the boardwalk and then heads towards his umbrella, or he walks 720 feet down the boardwalk and then heads toward his umbrella.
89. Let x be the length of the hypotenuse of the triangle, in feet. Then one of the other sides has length x 7 feet, and since the perimeter is 392 feet, the remaining side
must have length 392 x x 7 399 2x. From the Pythagorean Theorem,
x-7
x
we get x 72 399 2x2 x 2 4x 2 1610x 159250 0. Using the
Quadratic Formula, we get 2 44159250 x 1610 161024 16108 44100 1610210 , and so x 2275 or x 175. But if x 2275, then the 8 side of length x 7 combined with the hypotenuse already exceeds the perimeter of 392 feet, and so we must have x 175. Thus the other sides have length 175 7 168 and 399 2 175 49. The lot has sides of length 49 feet, 168 feet, and 175 feet. 90. Let h be the height of the screens in inches. The width of the smaller screen is h 7 inches, and the width of the bigger screen is 18h inches. The diagonal measure of the smaller screen is h 2 h 72 , and the diagonal measure of the larger screen is h 2 18h2 424h 2 206h. Thus h 2 h 72 3 206h h 2 h 72 206h 3. Squaring both sides gives h 2 h 2 14h 49 424h 2 1236h 9 0 224h 2 2636h 40. Applying
2636 26362 422440 2636 10532496 2636 3245 . So the Quadratic Formula, we obtain h 2224 448 448
2636 3245 1313. Thus, the screens are approximately 131 inches high. 448 d d 1 2 1 2 1 3 0 . Letting d, we have 3 14 1090 91. Since the total time is 3 s, we have 3 1090 4 4 1090 545 591054 . Since 0, we have d 1151, so d 13256. The well 22 545 6540 0 4 is 1326 ft deep. h
110
CHAPTER 1 Equations and Graphs
92. (a) Method 1: Let u x, so u 2 x. Thus x x 2 0 becomes u 2 u 2 0 u 2 u 1 0. So u 2 or u 1. If u 2, then x 2 x 4. If u 1, then x 1 x 1. So the possible solutions are 4 and 1. Checking x 4 we have 4 4 2 4 2 2 0. Checking x 1 we have 1 1 2 1 1 2 0. The only solution is 4. Method 2: x x 2 0 x 2 x x 2 4x 4 x x 2 5x 4 0 x 4 x 1 0. So the possible solutions are 4 and 1. Checking will result in the same solution. 1 1 12 10 (b) Method 1: Let u , so u 2 1 0 becomes 12u 2 10u 1 0. Using . Thus 2 2 x 3 x 3 x 3 x 3 102 4121 52 102 13 5 13 . If u 5 13 , the Quadratic Formula, we have u 10 212 10 24 24 12 12 12 5 13 1 5 13 5 13 12 then x 3 5 13. So x 2 13. 12 5 13 5 13 x 3 12 12 5 13 5 1 13 13 , then 12 513 x 3 5 13. So If u 5 12 12 5 13 5 13 x 3 12 x 2 13. The solutions are 2 13. 10 12 1 0 x 32 Method 2: Multiplying by the LCD, x 32 , we get x 32 x 3 x 32 12 10 x 3 x 32 0 12 10x 30 x 2 6x 9 0 x 2 4x 9 0. Using the Quadratic 2 13 2 13. The solutions are 2 13. Formula, we have u 4 4 419 42 52 42 2 22
1.7
SOLVING INEQUALITIES
1. (a) If x 5, then x 3 5 3 x 3 2. (b) If x 5, then 3 x 3 5 3x 15.
(c) If x 2, then 3 x 3 2 3x 6.
(d) If x 2, then x 2.
2. To solve the nonlinear inequality
x 1 0 we x 2
Interval
1
1 2
2
first observe that the numbers 1 and 2 are zeros
Sign of x 1
of the numerator and denominator. These numbers divide the real line into the three intervals 1, 1 2, and 2 .
Sign of x 2
The endpoint 1 satisfies the inequality, because
Sign of x 1 x 2
21 1 1 0 0, but 2 fails to satisfy the inequality because is not 1 2 22
defined. Thus, referring to the table, we see that the solution of the inequality is [1 2. 3. (a) No. For example, if x 2, then x x 1 2 1 2 0. (b) No. For example, if x 2, then x x 1 2 3 6.
4. (a) To solve 3x 7, start by dividing both sides of the inequality by 3.
(b) To solve 5x 2 1, start by adding 2 to both sides of the inequality.
SECTION 1.7 Solving Inequalities
5. x
2 3x 13
6. x
1 2x 5x
5 17 13 ; no
11 25; yes
1
5 13 ; no
5 1
3 5; yes
0
2 0; no
2 3 5 6
1 5 3
0
1 2
1 47 7
0
1 ; no 3 1 ; yes 3 1 ; yes 3 1 ; yes 3 1 ; yes 3 1 ; yes 3
13 10 3 ; no 23 25 6 ; no
1 5
5 13 The elements 56 , 1, 5, 3, and 5 satisfy the inequality. 7.
1 0; yes
2 3 5 6
1 5; no 347 1118; no
3
5 15; no
5
9 25; no
The elements 5, 1, and 0 satisfy the inequality. 8.
x 5 1 0 2 3 5 6
1 5 3 5
1 2x 4 7
x
1 6 7; no
5 1 0 2 3 5 6
1 5 3 5
Graph:
2 3 5 6
1 2 7; no
2 73 2; no
2 13 6 2; no 2 2 2; no
1 5
1 047 7; no
1 2 7; yes
2 076 2; yes
2 0 2; yes
3
1 6 7; yes
2 2 2; yes
5 The elements
5, 3, and 5 satisfy the inequality.
10. x 5 1
1 is undefined; no 0 3 1 ; no 2 2 6 1 ; no 5 2 1 12 ; no 045 12 ; yes 1 1 ; yes 3 2 1 1 ; yes 5 2
x2 2 4 27 4; no
3 4; yes
0
2 4; yes
2 3 5 6
22 4; yes 9 97 4; yes 36
1 5
3 4; yes 7 4; no
3
11 4; no
5
27 4; no
The elements 1, 0, 23 , 56 , and 1 satisfy the inequality. 12. 2x 8 x 4. Interval: [4 Graph:
6 5
2 3 2; no
0
1 83 7; no 1 73 7; no
The elements 5, 1, 5, 3, and 5 satisfy the inequality. 11. 5x 6 x 65 . Interval: 65
2 4 2; no
1
1 4 7; no
1 1 x 2 1 5 12 ; yes 1 12 ; yes
2 8 2; no
5
The elements 3 and 5 satisfy the inequality. 9.
2 3 x 2
x
1 14 7; no
4
111
112
CHAPTER 1 Equations and Graphs
13. 2x 5 3 2x 8 x 4 Interval: 4 Graph:
Graph:
4
15. 2 3x 8 3x 2 8 x 2 Interval: 2 Graph:
_2
17. 2x 1 0 2x 1 x 12 Interval: 12 Graph:
1
__ 2
19. 1 4x 5 2x 6x 4 x 23 Interval: 23 Graph:
2 _ 3
21. 12 x 23 2 12 x 83 x 16 3 16 Interval: 3 Graph:
16 __ 3
x 1 Interval: 1]
_1
25. 2 x 5 4 3 x 1 Interval: [3 1 Graph:
_3
_1
27. 6 3x 7 8 1 3x 15 13 x 5 Interval: 13 5 Graph:
1 _ 3
_2
16. 1 5 2x 2x 5 1 x 2 Interval: 2 Graph:
2
18. 0 5 2x 2x 5 x 52 Interval: 52 Graph:
5 _ 2
20. 5 3x 2 9x 6x 3 x 12 Interval: 12 Graph:
1
__ 2
22. 23 12 x 16 x (multiply both sides by 6) 4 3x 1 6x 3 9x 13 x Interval: 13 Graph:
23. 4 3x 1 8x 4 3x 1 8x 5x 5
Graph:
14. 3x 11 5 3x 6 x 2 Interval: 2
5
1 _ 3
24. 2 7x 3 12x 16 14x 6 12x 16 2x 22 x 11 Interval: 11] Graph:
11
26. 5 3x 4 14 9 3x 18 3 x 6 Interval: [3 6] Graph:
3
6
28. 8 5x 4 5 4 5x 9 45 x 95 Interval: 45 95 Graph:
4
_ _5_
9 __ 5
SECTION 1.7 Solving Inequalities
29. 2 8 2x 1 10 2x 9 5 x 92
30. 3 3x 7 12 10 3x 13 2
92 x 5 Interval: 92 5 Graph:
31.
9 _ 2
13 10 3 x 6 13 Interval: 10 3 6
Graph:
5
2x 3 1 2 8 2x 3 2 (multiply each 3 12 6
5 __ 2
10
13
_ _3_
_ _6_
4 3x 1 1 (multiply each expression by 20) 2 5 4 10 4 4 3x 5 10 16 12x 5
32.
5 expression by 12) 11 2x 5 11 2 x 2 Interval: 52 11 2
Graph:
113
11 11 13 26 12x 11 13 6 x 12 12 x 6 13 Interval: 11 12 6
11 __ 2
Graph:
11 __ 12
13 __ 6
33. x 2 x 3 0. The expression on the left of the inequality changes sign where x 2 and where x 3. Thus we must check the intervals in the following table. 2
2 3
3
Sign of x 2
Sign of x 3
Interval
Sign of x 2 x 3
From the table, the solution set is x 2 x 3. Interval: 2 3. Graph:
_2
3
34. x 5 x 4 0. The expression on the left of the inequality changes sign when x 5 and x 4. Thus we must check the intervals in the following table. 4
4 5
5
Sign of x 5
Sign of x 4
Interval
Sign of x 5 x 4
From the table, the solution set is x x 4 or 5 x. Interval: 4] [5 . Graph:
_4
5
35. x 2x 7 0. The expression on the left of the inequality changes sign where x 0 and where x 72 . Thus we must check the intervals in the following table. Interval Sign of x Sign of 2x 7
Sign of x 2x 7
72
72 0
0
From the table, the solution set is x x 72 or 0 x . Interval: 72 [0 . Graph:
7
__ 2
0
114
CHAPTER 1 Equations and Graphs
36. x 2 3x 0. The expression on the left of the inequality changes sign when x 0 and x 23 . Thus we must check the intervals in the following table. 0 23
0
Sign of x
Sign of 2 3x
Sign of x 2 3x
From the table, the solution set is x x 0 or 23 x . Interval: 0] 23 .
2 3
Interval
Graph:
0
2 _ 3
37. x 2 3x 18 0 x 3 x 6 0. The expression on the left of the inequality changes sign where x 6 and where x 3. Thus we must check the intervals in the following table. 3
3 6
6
Sign of x 3
Sign of x 6
Interval
Sign of x 3 x 6
From the table, the solution set is x 3 x 6. Interval: [3 6]. Graph:
_3
6
38. x 2 5x 6 0 x 3 x 2 0. The expression on the left of the inequality changes sign when x 3 and x 2. Thus we must check the intervals in the following table. 3
3 2
2
Sign of x 3
Sign of x 2
Interval
Sign of x 3 x 2
From the table, the solution set is x x 3 or 2 x.
Interval: 3 2 . Graph:
_3
_2
39. 2x 2 x 1 2x 2 x 1 0 x 1 2x 1 0. The expression on the left of the inequality changes sign where x 1 and where x 12 . Thus we must check the intervals in the following table. Interval
1
1 12
Sign of x 1
Sign of 2x 1
Sign of x 1 2x 1
1 2
From the table, the solution set is x x 1 or 12 x . Interval: 1] 12 . Graph:
_1
1 _ 2
40. x 2 x 2 x 2 x 2 0 x 1 x 2 0. The expression on the left of the inequality changes sign when x 1 and x 2. Thus we must check the intervals in the following table. 1
1 2
2
Sign of x 1
Sign of x 2
Interval
Sign of x 1 x 2
From the table, the solution set is x 1 x 2. Interval: 1 2. Graph:
_1
2
SECTION 1.7 Solving Inequalities
115
41. 3x 2 3x 2x 2 4 x 2 3x 4 0 x 1 x 4 0. The expression on the left of the inequality changes sign where x 1 and where x 4. Thus we must check the intervals in the following table. 1
1 4
4
Sign of x 1
Sign of x 4
Interval
Sign of x 1 x 4
From the table, the solution set is x 1 x 4. Interval: 1 4. Graph:
_1
4
42. 5x 2 3x 3x 2 2 2x 2 3x 2 0 2x 1 x 2 0. The expression on the left of the inequality changes sign when x 12 and x 2. Thus we must check the intervals in the following table. Interval
2
2 12
Sign of 2x 1
Sign of x 2
Sign of 2x 1 x 2
1 2
From the table, the solution set is x x 2 or 12 x . Interval: 2] 12 . Graph:
_2
1 _ 2
43. x 2 3 x 6 x 2 3x 18 0 x 3 x 6 0. The expression on the left of the inequality changes sign where x 6 and where x 3. Thus we must check the intervals in the following table. 3
3 6
6
Sign of x 3
Sign of x 6
Interval
Sign of x 3 x 6
From the table, the solution set is x x 3 or 6 x. Interval: 3 6 . Graph:
_3
6
44. x 2 2x 3 x 2 2x 3 0 x 3 x 1 0. The expression on the left of the inequality changes sign when x 3 and x 1. Thus we must check the intervals in the following table. 3
3 1
1
Sign of x 3
Sign of x 1
Interval
Sign of x 3 x 1
From the table, the solution set is x x 3 or 1 x. Interval: 3 1 . Graph:
_3
1
45. x 2 4 x 2 4 0 x 2 x 2 0. The expression on the left of the inequality changes sign where x 2 and where x 2. Thus we must check the intervals in the following table. 2
2 2
2
Sign of x 2
Sign of x 2
Interval
Sign of x 2 x 2
From the table, the solution set is x 2 x 2. Interval: 2 2. Graph:
_2
2
116
CHAPTER 1 Equations and Graphs
46. x 2 9 x 2 9 0 x 3 x 3 0. The expression on the left of the inequality changes sign when x 3 and x 3. Thus we must check the intervals in the following table. 3
3 3
3
Sign of x 3
Sign of x 3
Interval
Sign of x 3 x 3
From the table, the solution set is x x 3 or 3 x. Interval: 3] [3 . Graph:
_3
3
47. x 2 x 1 x 3 0. The expression on the left of the inequality changes sign when x 2, x 1, and x 3. Thus we must check the intervals in the following table. 2
2 1
1 3
3
Sign of x 2
Sign of x 1
Sign of x 3
Interval
Sign of x 2 x 1 x 3
From the table, the solution set is x x 2 or 1 x 3. Interval: 2][1 3]. Graph:
_2
1
3
48. x 5 x 2 x 1 0. The expression on the left of the inequality changes sign when x 5, x 2, and x 1. Thus we must check the intervals in the following table. 1
1 2
2 5
5
Sign of x 5
Sign of x 2
Sign of x 1
Interval
Sign of x 5 x 2 x 1
From the table, the solution set is x 1 x 2 or 5 x. Interval: 1 25 . Graph:
_1
2
5
49. x 4 x 22 0. Note that x 22 0 for all x 2, so the expression on the left of the original inequality changes sign only when x 4. We check the intervals in the following table. Interval Sign of x 4
Sign of x 22
Sign of x 4 x 22
2
2 4
4
From the table, the solution set is x x 2 and x 4. We exclude the
endpoint 2 since the original expression cannot be 0. Interval: 2 2 4. Graph:
_2
4
SECTION 1.7 Solving Inequalities
117
50. x 32 x 1 0. Note that x 32 0 for all x 3, so the expression on the left of the original inequality changes sign only when x 1. We check the intervals in the following table. Interval Sign of x 32
3
3 1
1
Sign of x 1
Sign of x 32 x 1
From the table, the solution set is x x 1. (The endpoint 3 is already excluded.) Interval: 1 . Graph:
_1
51. x 22 x 3 x 1 0. Note that x 22 0 for all x, so the expression on the left of the original inequality changes sign only when x 1 and x 3. We check the intervals in the following table. Interval Sign of x 22
1
1 2
2 3
3
Sign of x 3
Sign of x 1
Sign of x 22 x 3 x 1
From the table, the solution set is x 1 x 3. Interval: [1 3]. Graph:
_1
3
52. x 2 x 2 1 0 x 2 x 1 x 1 0. The expression on the left of the inequality changes sign when x 1 and x 0. Thus we must check the intervals in the following table. Interval Sign of x 2
1
1 0
0 1
1
Sign of x 1
Sign of x 1 Sign of x 2 x 2 1
From the table, the solution set is x x 1, x 0, or 1 x. (The endpoint 0 is included since the original expression is allowed to be 0.) Interval: 1] 0 [1 . Graph:
_1
0
1
53. x 3 4x 0 x x 2 4 0 x x 2 x 2 0. The expression on the left of the inequality changes sign where x 0, x 2 and where x 4. Thus we must check the intervals in the following table. Interval
2
2 0
0 2
2
Sign of x
Sign of x 2
Sign of x 2
Sign of x x 2 x 2
From the table, the solution set is x 2 x 0 or x 2. Interval: 2 02 . Graph:
_2
0
2
118
CHAPTER 1 Equations and Graphs
54. 16x x 3 0 x 3 16x x x 2 16 x x 4 x 4. The expression on the left of the inequality changes sign when x 4, x 0, and x 4. Thus we must check the intervals in the following table. 4
4 0
0 4
4
Sign of x 4
Sign of x
Sign of x 4
Interval
Sign of x x 4 x 4
From the table, the solution set is x 4 x 0 or 4 x. Interval: [4 0] [4 . Graph:
55.
3
3 12
Sign of x 3
Sign of 2x 1
Sign of
x 3 2x 1
1 2
4
From the table, the solution set is x x 3 or x 12 . Since the denominator
cannot equal 0, x 12 . Interval: 3] 12 . Graph:
_3
1 _ 2
4x 0. The expression on the left of the inequality changes sign when x 4 and x 4. Thus we must check the x 4 intervals in the following table. 4
4 4
4
Sign of 4 x
Sign of x 4
Sign of
Interval
57.
0
x 3 0. The expression on the left of the inequality changes sign where x 3 and where x 12 . Thus we must 2x 1 check the intervals in the following table. Interval
56.
_4
4x x 4
From the table, the solution set is x x 4 or x 4. Interval: 4 4 . Graph:
_4
4
4x 0. The expression on the left of the inequality changes sign where x 4. Thus we must check the intervals in x 4 the following table. Interval
4
4 4
4
Sign of 4 x
Sign of x 4
Sign of
4x x 4
From the table, the solution set is x x 4 or x 4. Interval: 4 4 . Graph:
_4
4
SECTION 1.7 Solving Inequalities
119
x 1 x 1 2 x 3 3x 5 x 1 0 20 0 The expression on the left of the inequality x 3 x 3 x 3 x 3 x 3 changes sign when x 53 and x 3. Thus we must check the intervals in the following table.
58. 2
53
Interval Sign of 3x 5
Sign of x 3
3x 5 Sign of x 3
59.
53 3
3
From the table, the solution set is x x 53 or 3 x . Interval: 53 3 . Graph:
5 _ 3
3
2x 1 2x 1 3 x 5 x 16 2x 1 3 3 0 0 0. The expression on the left of the inequality x 5 x 5 x 5 x 5 x 5 changes sign where x 16 and where x 5. Thus we must check the intervals in the following table. 5
5 16
16
Sign of x 16
Sign of x 5
Interval
x 16 Sign of x 5 60.
From the table, the solution set is x x 5 or x 16. Since the denominator cannot equal 0, we must have x 5. Interval: 5 [16 . Graph:
5
16
3x 3x 3x 2x 3x 1 1 0 0 0. The expression on the left of the inequality changes 3x 3x 3x 3x 3x sign when x 0 and x 3. Thus we must check the intervals in the following table. Since the denominator cannot equal 0, we must 0
0 3
3
Sign of 3 x
Sign of 2x
Interval
Sign of
61.
2x 3x
have x 3. The solution set is x 0 x 3. Interval: [0 3. Graph:
0
3
4 4 xx 4 x2 4 2 x 2 x x x 0 0 0 0. The expression on the left of the x x x x x x inequality changes sign where x 0, where x 2, and where x 2. Thus we must check the intervals in the following table. 2
2 0
0 2
2
Sign of 2 x
Sign of x
Sign of 2 x
Sign of
Interval
2 x 2 x x
From the table, the solution set is x 2 x 0 or 2 x. Interval: 2 02 . Graph:
_2
0
2
120
62.
CHAPTER 1 Equations and Graphs
x x 3x x 1 2x 3x 2 x 2 3x x 3x 3x 0 0 0 0. The expression on x 1 x 1 x 1 x 1 x 1 x 1 the left of the inequality changes sign whenx 0, x 23 , and x 1. Thus we must check the intervals in the following table. 1
Interval
63. 1
_1
2
__ 3
23 0
0
Sign of x
Sign of 2 3x
Sign of x 1
Sign of
2 x 2 x x
From the table, the solution set is
Graph:
1 23
x x 1 or 23 x 0 . Interval: 1 23 0 .
0
2 2 2 2 x x 1 2x 2 x 1 x 2 x 2x 2x 2 1 0 0 0 x 1 x x 1 x x x 1 x x 1 x x 1 x x 1
x2 x 2 x 2 x 1 0 0. The expression on the left of the inequality changes sign where x 2, where x x 1 x x 1 x 1, where x 0, and where x 1. Thus we must check the intervals in the following table. 2
2 1
1 0
0 1
1
Sign of x 2
Sign of x 1
Sign of x
Sign of x 1
Interval
x 2 x 1 Sign of x x 1
Since x 1 and x 0 yield undefined expressions, we cannot include them in the solution. From the table, the solution set is x 2 x 1 or 0 x 1. Interval: [2 1 0 1]. Graph:
_2
_1
0
1
SECTION 1.7 Solving Inequalities
64.
121
4 3 4 3x 4 x 1 x x 1 3x 4x 4 x 2 x 3 1 1 0 0 0 x 1 x x 1 x x x 1 x x 1 x x 1 x x 1 4 x2 2 x 2 x 0 0 The expression on the left of the inequality changes sign when x 2, x 2, x x 1 x x 1 x 0, and x 1. Thus we must check the intervals in the following table. 2
2 0
0 1
1 2
2
Sign of 2 x
Sign of 2 x
Sign of x
Sign of x 1
Interval
2 x 2 x Sign of x x 1
Since x 0 and x 1 give undefined expressions, we cannot include them in the solution. From the table, the solution set is x 2 x 0 or 1 x 2. Interval: [2 0 1 2]. Graph:
65.
_2
0
1
2
6 6 6 6x 6 x 1 x x 1 6 1 10 0 x 1 x x 1 x x x 1 x x 1 x x 1 6x 6x 6 x 2 x x 2 x 6 x 3 x 2 0 0 0. The x x 1 x x 1 x x 1 expression on the left of the inequality changes sign where x 3, where x 2, where x 0, and where x 1. Thus we must check the intervals in the following table. 2
2 0
0 1
1 3
3
Sign of x 3
Sign of x 2
Sign of x
Sign of x 1
Interval
x 3 x 2 Sign of x x 1
From the table, the solution set is x 2 x 0 or 1 x 3. The points x 0 and x 1 are excluded from the solution set because they make the denominator zero. Interval: [2 0 1 3]. Graph:
_2
0
1
3
122
66.
CHAPTER 1 Equations and Graphs
5 x 5 x x 1 25 4 2 x 1 x 2 x 10 8x 8 x 4 4 0 0 0 2 x 1 2 x 1 2 x 1 2 x 1 2 x 1 2 x 1 x 2 7x 18 x 9 x 2 0 0. The expression on the left of the inequality changes sign when x 9, x 2, 2 x 1 2 x 1 and x 1. Thus we must check the intervals in the following table. 2
2 1
1 9
9
Sign of x 9
Sign of x 2
Sign of x 1
Interval
x 9 x 2 Sign of 2 x 1
From the table, the solution set is x 2 x 1 or 9 x. The point x 1 is excluded from the solution set because it makes the expression undefined. Interval: [2 1 [9 . Graph:
67.
_1 _2
9
x 1 x 2 x 1 x 2 x 2 x 2 x 1 x 3 0 0 x 3 x 2 x 3 x 2 x 3 x 2 x 2 x 3 x 2 4 x 2 2x 3 2x 1 0 0. The expression on the left of the inequality x 3 x 2 x 3 x 2 changes sign where x 12 , where x 3, and where x 2. Thus we must check the intervals in the following table. 3
Interval
1
__ 2
2
Sign of x 3
Sign of x 2
From the table, the solution set is
_3
12 2
Sign of 2x 1
2x 1 Sign of x 3 x 2
Graph:
3 12
2
x 3 x 12 or 2 x . Interval: 3 12 2 .
SECTION 1.7 Solving Inequalities
68.
123
1 x 2 1 x 1 x 2x 1 2x 3 0 0 0 0. The x 1 x 2 x 1 x 2 x 1 x 2 x 1 x 2 x 1 x 2 expression on the left of the inequality changes sign when x 32 , x 1, and x 2. Thus we must check the intervals in the following table. 2
Interval
2 32
32 1
1
Sign of 2x 3
Sign of x 1
Sign of x 2
Sign of
2x 3 x 1 x 2
From the table, the solution set is x x 2 or 32 x 1 . The points x 2 and x 1 are excluded from the solution because the expression is undefined at those values. Interval: 2 32 1 . Graph:
69.
_2
x 1 x 2 x 22
3
_1
__ 2
0. Note that x 22 0 for all x. The expression on the left of the original inequality changes sign
when x 2 and x 1. We check the intervals in the following table. 2
2 1
1 2
2
Sign of x 1
Sign of x 2
Interval
Sign of x 22 Sign of
x 1 x 2 x 22
From the table, and recalling that the point x 2 is excluded from the solution because the expression is
undefined at those values, the solution set is x x 2 or x 1 and x 2. Interval: 2] [1 2 2 . Graph:
_2
1
2
124
70.
CHAPTER 1 Equations and Graphs
2x 1 x 32 0. Note that x 32 0 for all x 3. The expression on the left of the inequality changes sign x 4
when x 12 and x 4. We check the intervals in the following table. 13 12 Interval 2 Sign of 2x 1
Sign of x 32 Sign of x 4
3 4
4
2x 1 x 32 Sign of x 4 From the table, the solution set is x x 3 and 12 x 4 . We exclude the endpoint 3 because the original expression
cannot be 0. Interval: 12 3 3 4. Graph:
1 2
3
4
71. x 4 x 2 x 4 x 2 0 x 2 x 2 1 0 x 2 x 1 x 1 0. The expression on the left of the inequality changes sign where x 0, where x 1, and where x 1. Thus we must check the intervals in the following table. 1
1 0
0 1
1
Sign of x 1
Sign of x 1
Interval Sign of x 2
Sign of x 2 x 1 x 1
From the table, the solution set is x x 1 or 1 x. Interval: 1 1 . Graph:
_1
1
72. x 5 x 2 x 5 x 2 0 x 2 x 3 1 0 x 2 x 1 x 2 x 1 0. The expression on the left of the inequality 1 12 4 1 1 1 3 2 . changes sign when x 0 and x 1. But the solution of x x 1 0 are x 2 1 2 Since these are not real solutions. The expression x 2 x 1 does not change signs, so we must check the intervals in the
following table.
0
0 1
Sign of x 2
1
Sign of x 1
Interval
Sign of x 2 x 1
Sign of x 2 x 1 x 2 x 1
From the table, the solution set is x 1 x. Interval: 1 . Graph:
1
SECTION 1.7 Solving Inequalities
73. For
125
16 9x 2 to be defined as a real number we must have 16 9x 2 0 4 3x 4 3x 0. The expression in the
inequality changes sign at x 43 and x 43 . Interval Sign of 4 3x
43
43 43
Sign of 4 3x
Sign of 4 3x 4 3x
4 3
Thus 43 x 43 . 74. For 3x 2 5x 2 to be defined as a real number we must have 3x 2 5x 2 0 3x 2 x 1 0. The
expression on the left of the inequality changes sign whenx 23 and x 1. Thus we must check the intervals in the following table. 21 23 Interval 1 3 Sign of 3x 2
Sign of x 1
Sign of 3x 2 x 1
Thus x 23 or 1 x. 12 1 to be defined as a real number we must have x 2 5x 14 0 x 7 x 2 0. The 75. For x 2 5x 14 expression in the inequality changes sign at x 7 and x 2.
2
2 7
7
Sign of x 7
Sign of x 2
Interval
Sign of x 7 x 2
Thus x 2 or 7 x, and the solution set is 2 7 . 1x 1x to be defined as a real number we must have 0. The expression on the left of the inequality changes 76. For 4 2x 2x sign when x 1 and x 2. Thus we must check the intervals in the following table. 2
2 1
1
Sign of 1 x
Sign of 2 x
Sign of
Interval
1x 2x
Thus 2 x 1. Note that x 2 has been excluded from the solution set because the expression is undefined at that
value.
77. a bx c bc (where a, b, c 0) bx c
bc 1 bc bx cx a a b
78. We have a bx c 2a, where a,b, c 0 a c bx 2a c
bc c c c c c x . a a b a b
ac 2a c x . b b
126
CHAPTER 1 Equations and Graphs
79. Inserting the relationship C 59 F 32, we have 20 C 30 20 59 F 32 30 36 F 32 54 68 F 86. 80. Inserting the relationship F 95 C 32, we have 50 F 95 50 95 C 32 95 18 95 C 63 10 C 35. 81. Let x be the average number of miles driven per day. Each day the cost of Plan A is 30 010x, and the cost of Plan B is
50. Plan B saves money when 50 30 010x 20 01x 200 x. So Plan B saves money when you average more than 200 miles a day.
82. Let m be the number of minutes of long-distance calls placed per month. Then under Plan A, the cost will be 25 005m, and under Plan B, the cost will be 5 012m. To determine when Plan B is advantageous, we must solve
25 005m 5 012m 20 007m 2857 m. So Plan B is advantageous if a person places fewer than 286 minutes of long-distance calls during a month.
83. We need to solve 6400 035m 2200 7100 for m. So 6400 035m 2200 7100 4200 035m 4900 12,000 m 14,000. She plans on driving between 12,000 and 14,000 miles. h , where T is the temperature in C, and h is the height in meters. 100 (b) Solving the expression in part (a) for h, we get h 100 20 T . So 0 h 5000 0 100 20 T 5000 0 20 T 50 20 T 30 20 T 30. Thus the range of temperature is from 20 C down to 30 C.
84. (a) T 20
85. (a) Let x be the number of $3 increases. Then the number of seats sold is 120 x. So P 200 3x
3x P 200 x 13 P 200. Substituting for x we have that the number of seats sold is
120 x 120 13 P 200 13 P 560 3 .
(b) 90 13 P 560 3 115 270 360 P 200 345 270 P 560 345 290 P 215 290 P 215. Putting this into standard order, we have 215 P 290. So the ticket prices are between $215 and $290. c . Since the 65 c 303 scale’s accuracy is 003 lb, and the scale shows 3 lb, we have 3 003 x 3 003 297 65 650 297 c 650 303 19305 c 19695. Since the customer paid $1950, he could have been over- or
86. If the customer buys x pounds of coffee at $650 per pound, then his cost c will be 650x. Thus x
undercharged by as much as 195 cents.
87. 00004
4,000,000 001. Since d 2 0 and d 0, we can multiply each expression by d 2 to obtain d2
00004d 2 4,000,000 001d 2 . Solving each pair, we have 00004d 2 4,000,000 d 2 10,000,000,000
d 100,000 (recall that d represents distance, so it is always nonnegative). Solving 4,000,000 001d 2
400,000,000 d 2 20,000 d. Putting these together, we have 20,000 d 100,000.
SECTION 1.7 Solving Inequalities
88.
127
600,000 500 600,000 500 x 2 300 (Note that x 2 300 300 0, so we can multiply both sides by the 2 x 300
denominator and not worry that we might be multiplying both sides by a negative number or by zero.) 1200 x 2 300 0 x 2 900 0 x 30 x 30. The expression in the inequality changes sign at x 30 and x 30. However, since x represents distance, we must have x 0.
0 30
30
Sign of x 30
Sign of x 30
Interval
Sign of x 30 x 30
So x 30 and you must stand at least 30 meters from the center of the fire. 89. 128 16t 16t 2 32 16t 2 16t 96 0 16 t 2 t 6 0 16 t 3 t 2 0. The expression on the left of the inequality changes sign at x 2, at t 3, and at t 2. However, t 0, so the only endpoint is t 3. 0 3
3
Sign of 16
Sign of t 3
Sign of t 2
Interval
Sign of 16 t 3 t 2 So 0 t 3.
90. Solve 30 10 09 001 2 for 10 75. We have 30 10 09 001 2 001 2 09 20 0 01 4 01 5 0. The possible endpoints are 01 4 0 01 4 40 and 01 5 0 01 5 50.
10 40
40 50
50 75
Sign of 01 4
Sign of 01 5
Interval
Sign of 01 4 01 5 Thus he must drive between 40 and 50 mi/h.
2 1 2 240 0 1 3 80 0. The expression in the inequality changes sign at 20 20 20 60 and 80. However, since represents the speed, we must have 0.
91. 240
Interval 1 3 Sign of 20
Sign of 80 1 3 80 Sign of 20
So Kerry must drive between 0 and 60 mi/h.
0 60
60
128
CHAPTER 1 Equations and Graphs
92. Solve 2400 20x 2000 8x 00025x 2 2400 20x 2000 8x 00025x 2 00025x 2 12x 4400 0 00025x 1 x 4400 0. The expression on the left of the inequality changes sign whenx 400 and x 4400.
Since the manufacturer can only sell positive units, we check the intervals in the following table. 0 400
400 4400
4400
Sign of 00025x 1
Sign of x 4400
Interval
Sign of 00025x 1 x 4400
So the manufacturer must sell between 400 and 4400 units to enjoy a profit of at least $2400. 93. Let x be the length of the garden and its width. Using the fact that the perimeter is 120 ft, we must have 2x 2 120 60 x. Now since the area must be at least 800 ft2 , we have 800 x 60 x 800 60x x 2
x 2 60x 800 0 x 20 x 40 0. The expression in the inequality changes sign at x 20 and x 40.
However, since x represents length, we must have x 0.
0 20
20 40
40
Sign of x 20
Sign of x 40
Interval
Sign of x 20 x 40 The length of the garden should be between 20 and 40 feet. 94. Case 1: a b 0 a b0
We have a a a b, since a 0, and b a b b, since b 0. So a 2 a b b2 , that is
a 2 b2 . Continuing, we have a a 2 a b2 , since a 0 and b2 a b2 b, since b2 0. So
a 3 ab2 b3 . Thus a b 0 a 3 b3 . So a b 0 a n bn , if n is even, and a n b, if n is odd. Case 2: 0 a b
We have a a a b, since a 0, and b a b b, since b 0. So a 2 a b b2 . Thus 0
a b a 2 b2 . Likewise, a 2 a a 2 b and b a 2 b b2 , thus a 3 b3 . So 0 a b a n bn , for all positive integers n.
Case 3: a 0 b
If n is odd, then a n bn , because a n is negative and bn is positive. If n is even, then we could have
either a n bn or a n bn . For example, 1 2 and 12 22 , but 3 2 and 32 22 .
95. The rule we want to apply here is “a b ac bc if c 0 and a b ac bc if c 0 ”. Thus we cannot simply
Case 1: x 0
3 , we must consider two cases. x Multiplying both sides by x, we have x 3. Together with our initial condition, we have 0 x 3.
Case 2: x 0
Multiplying both sides by x, we have x 3. But x 0 and x 3 have no elements in common, so this
multiply by x, since we don’t yet know if x is positive or negative, so in solving 1
gives no additional solution. Hence, the only solutions are 0 x 3. 96. a b, so by Rule 1, a c b c. Using Rule 1 again, b c b d, and so by transitivity, a c b d. 97.
a c a c ad ad , so by Rule 3, d d c. Adding a to both sides, we have a c a. Rewriting the left-hand b d b d b b ad ab a b d a ac side as and dividing both sides by b d gives . b b b b bd cb c b d ac c Similarly, a c c , so . d d bd d
SECTION 1.8 Solving Absolute Value Equations and Inequalities
1.8
129
SOLVING ABSOLUTE VALUE EQUATIONS AND INEQUALITIES
1. The equation x 3 has the two solutions 3 and 3. 2. (a) The solution of the inequality x 3 is the interval [3 3].
(b) The solution of the inequality x 3 is a union of two intervals 3] [3 .
3. (a) The set of all points on the real line whose distance from zero is less than 3 can be described by the absolute value inequality x 3.
(b) The set of all points on the real line whose distance from zero is greater than 3 can be described by the absolute value inequality x 3.
4. (a) 2x 1 5 is equivalent to the two equations 2x 1 5 and 2x 1 5. (b) 3x 2 8 is equivalent to 8 3x 2 8.
5. 5x 20 5x 20 x 4.
6. 3x 10 3x 10 x 10 3.
7. 5 x 3 28 5 x 25 x 5 x 5.
8. 12 x 7 2 12 x 9 x 18 x 18.
9. x 3 2 is equivalent to x 3 2 x 3 2 x 1 or x 5.
10. 2x 3 7 is equivalent to either 2x 3 7 2x 10 x 5; or 2x 3 7 2x 4 x 2. The two solutions are x 5 and x 2. 11. x 4 05 is equivalent to x 4 05 x 4 05 x 45 or x 35.
12. x 4 3. Since the absolute value is always nonnegative, there is no solution.
13. 2x 3 11 is equivalent to either 2x 3 11 2x 14 x 7; or 2x 3 11 2x 8 x 4. The two solutions are x 7 and x 4.
14. 2 x 11 is equivalent to either 2 x 11 x 9; or 2 x 11 x 13. The two solutions are x 9 and x 13.
15. 4 3x 6 1 3x 6 3 3x 6 3, which is equivalent to either 3x 6 3 3x 3 x 1; or 3x 6 3 3x 9 x 3. The two solutions are x 1 and x 3.
16. 5 2x 6 14 5 2x 8 which is equivalent to either 5 2x 8 2x 3 x 32 ; or 5 2x 8
3 13 2x 13 x 13 2 . The two solutions are x 2 and x 2 . 17. 3 x 5 6 15 3 x 5 9 x 5 3, which is equivalent to either x 5 3 x 2; or x 5 3 x 8. The two solutions are x 2 and x 8.
18. 20 2x 4 15 2x 4 5. Since the absolute value is always nonnegative, there is no solution. 19. 8 5 13 x 56 33 5 13 x 56 25 13 x 56 5, which is equivalent to either 13 x 56 5 13 x 35 6
1 5 1 25 25 25 35 x 35 2 ; or 3 x 6 5 3 x 6 x 2 . The two solutions are x 2 and x 2 . 3 9 20. 35 x 2 12 4 35 x 2 92 which is equivalent to either 35 x 2 92 35 x 52 x 25 6 ; or 5 x 2 2 3 x 13 x 65 . The two solutions are x 25 and x 65 . 5 2 6 6 6
21. x 1 3x 2, which is equivalent to either x 1 3x 2 2x 3 x 32 ; or x 1 3x 2 x 1 3x 2 4x 1 x 14 . The two solutions are x 32 and x 14 .
22. x 3 2x 1 is equivalent to either x 3 2x 1 x 2 x 2; or x 3 2x 1 x 3 2x 1 3x 4 x 43 . The two solutions are x 2 and x 43 .
23. x 5 5 x 5. Interval: [5 5].
130
CHAPTER 1 Equations and Graphs
24. 2x 20 20 2x 20 10 x 10. Interval: [10 10].
25. 2x 7 is equivalent to 2x 7 x 72 ; or 2x 7 x 72 . Interval: 72 72 .
26. 12 x 1 x 2 is equivalent to x 2 or x 2. Interval: 2] [2 .
27. x 4 10 is equivalent to 10 x 4 10 6 x 14. Interval: [6 14].
28. x 3 9 is equivalent to x 3 9 x 6; or x 3 9 x 12. Interval: 6 12 . 29. x 1 1 is equivalent to x 1 1 x 0; or x 1 1 x 2. Interval: 2] [0 .
30. x 4 0 is equivalent to x 4 0 x 4 0 x 4. The only solution is x 4.
31. 2x 1 3 is equivalent to 2x 1 3 2x 4 x 2; or 2x 1 3 2x 2 x 1. Interval: 2] [1 .
32. 3x 2 7 is equivalent to 3x 2 7 3x 5 x 53 ; or 3x 2 7 3x 9 x 3. Interval: 53 3 .
33. 2x 3 04 04 2x 3 04 26 2x 34 13 x 17. Interval: [13 17]. 34. 5x 2 6 6 5x 2 6 4 5x 8 45 x 85 . Interval: 45 85 . x 2 2 2 x 2 2 6 x 2 6 4 x 8. Interval: 4 8. 35. 3 3 x 1 4 1 x 1 4 1 x 1 4 x 1 8 which is equivalent to either x 1 8 x 7; or 36. 2 2 2 x 1 8 x 9. Interval: 9] [7 .
37. x 6 0001 0001 x 6 0001 6001 x 5999. Interval: 6001 5999. 38. x a d d x a d a d x a d. Interval: a d a d.
39. 4 x 2 3 13 4 x 2 16 x 2 4 4 x 2 4 6 x 2. Interval: 6 2.
40. 3 2x 4 1 2x 4 2 2x 4 2 which is equivalent to either 2x 4 2 2x 2 x 1; or 2x 4 2 2x 6 x 3. Interval: 3] [1 .
41. 8 2x 1 6 2x 1 2 2x 1 2 2 2x 1 2 1 2x 3 12 x 32 . Interval: 12 32 .
42. 7 x 2 5 4 7 x 2 1 x 2 17 . Since the absolute value is always nonnegative, the inequality is true for all real numbers. In interval notation, we have . 1 43. 12 4x 13 56 4x 13 53 , which is equivalent to either 4x 13 53 4x 43 x 13 ; or 4x 53 3 4x 2 x 12 . Interval: 12 13 . 44. 2 12 x 3 3 51 2 12 x 3 48 12 x 3 24 24 12 x 3 24 27 12 x 21 54 x 42. Interval: [54 42].
45. 1 x 4. If x 0, then this is equivalent to 1 x 4. If x 0, then this is equivalent to 1 x 4 1 x 4 4 x 1. Interval: [4 1] [1 4].
46. 0 x 5 12 . For x 5, this is equivalent to 12 x 5 12 92 x 11 2 . Since x 5 is excluded, the solution is 9 5 5 11 . 2 2
47.
1 13 2 1 2 x 7 (x 7) x 7 12 12 x 7 12 15 2 x 2 and x 7. x 7 13 Interval: 15 2 7 7 2 .
SECTION 1.9 Solving Equations and Inequalities Graphically
48.
131
1 5 15 2x 3, since 2x 3 0, provided 2x 3 0 x 32 . Now for x 32 , we have 2x 3
1 2x 3 is equivalent to either 1 2x 3 16 2x 8 x; or 2x 3 1 2x 14 x 7 . 5 5 5 5 5 5 5 7 8 Interval: 5 5 .
49. x 3
50. x 2
51. x 7 5
52. x 2 4
53. x 2
54. x 1
55. x 3
56. x 4
57. (a) Let x be the thickness of the laminate. Then x 0020 0003.
(b) x 0020 0003 0003 x 0020 0003 0017 x 0023. h 682 2 2 h 682 2 58 h 682 58 624 h 740. Thus 95% of the adult males are 58. 29 29 between 624 in and 740 in.
59. x 1 is the distance between x and 1; x 3 is the distance between x and 3. So x 1 x 3 represents those points closer to 1 than to 3, and the solution is x 2, since 2 is the point halfway between 1 and 3. If a b, then the ab . solution to x a x b is x 2
1.9
SOLVING EQUATIONS AND INEQUALITIES GRAPHICALLY
1. The solutions of the equation x 2 2x 3 0 are the x-intercepts of the graph of y x 2 2x 3. 2. The solutions of the inequality x 2 2x 3 0 are the x-coordinates of the points on the graph of y x 2 2x 3 that lie above the x-axis. 3. (a) From the graph, it appears that the graph of y x 4 3x 3 x 2 3x has x-intercepts 1, 0, 1, and 3, so the solutions to the equation x 4 3x 3 x 2 3x 0 are x 1, x 0, x 1, and x 3.
(b) From the graph, we see that where 1 x 0 or 1 x 3, the graph lies below the x-axis. Thus, the inequality x 4 3x 3 x 2 3x 0 is satisfied for x 1 x 0 or 1 x 3 [1 0] [1 3].
4. (a) The graphs of y 5x x 2 and y 4 intersect at x 1 and at x 4, so the equation 5x x 2 4 has solutions x 1 and x 4. (b) The graph of y 5x x 2 lies strictly above the graph of y 4 when 1 x 4, so the inequality 5x x 2 4 is satisfied for those values of x, that is, for x 1 x 4 1 4.
5. Algebraically: x 4 5x 12 16 4x x 4. Graphically: We graph the two equations y1 x 4 and y2 5x 12 in the viewing rectangle [6 4] by
[10 2]. Zooming in, we see that the solution is x 4.
6. Algebraically: 12 x 3 6 2x 9 32 x x 6. Graphically: We graph the two equations y1 12 x 3 and y2 6 2x in the viewing rectangle [10 5] by
[10 5]. Zooming in, we see that the solution is x 6. 5
-6
-4
-2
2 -5
4 -10
-5
5 -5
-10
-10
132
CHAPTER 1 Equations and Graphs
7. Algebraically:
1 2 7 2x x 2x
1 2 x 2x
2x 7
5. 4 1 14x x 14
1 2 x 2x and y2 7 in the viewing rectangle [2 2] by [2 8].
Graphically: We graph the two equations y1
Zooming in, we see that the solution is x 036.
5
6 5 4 8. Algebraically: x 2 2x 2x 4 6 4 5 2x x 2 2x x 2 x 2 2x 2x 4 2x 4 x 2 6 x 5 8x 6x 12 5x 12 3x 4 x. Graphically: We graph the two equations 6 4 5 and y2 in the viewing y1 x 2 2x 2x 4 rectangle [5 5] by [10 10]. Zooming in, we see that there is only one solution at x 4. 10
-2
-1
1
2 -4
-2
2
4
-10
9. Algebraically: x 2 32 0 x 2 32 x 32 4 2. Graphically: We graph the equation y1 x 2 32 and determine where this curve intersects the x-axis. We use the viewing rectangle [10 10] by [5 5]. Zooming in, we see that solutions are x 566 and x 566.
10. Algebraically: x 3 16 0 x 3 16 x 2 3 2. Graphically: We graph the equation y x 3 16 and
determine where this curve intersects the x-axis. We use the viewing rectangle [5 5] by [5 5]. Zooming in, we see that the solution is x 252. 4
4
2
2 -10
-5
-2
5
10
-4
-2
11. Algebraically: x 2 9 0 x 2 9, which has no real solution.
Graphically: We graph the equation y x 2 9 and see that this curve does not intersect the x-axis. We use the viewing rectangle [5 5] by [5 30]. 30 20
12. Algebraically: x 2 3 2x x 2 2x 3 0 2 22 4 1 3 2 8 . x 2 1 2 1 Because the discriminant is negative, there is no real solution. Graphically: We graph the two equations y1 x 2 3 and y2 2x in the viewing rectangle [4 6] by [6 12], and see that the two curves do not intersect.
10 -2
4
-4
-4
-4
2
-2
10 2
5
4 -4
-2
2 -5
4
6
SECTION 1.9 Solving Equations and Inequalities Graphically
13. Algebraically: 16x 4 625 x 4 625 16 x 52 25. Graphically: We graph the two equations y1 16x 4 and
y2 625 in the viewing rectangle [5 5] by [610 640].
Zooming in, we see that solutions are x 25. 640
14. Algebraically: 2x 5 243 0 2x 5 243 x 5 243 2 5 5 243 3 x 2 2 16. Graphically: We graph the equation y 2x 5 243 and
determine where this curve intersects the x-axis. We use the viewing rectangle [5 10] by [5 5]. Zooming in, we see that the solution is x 261. 4
630
2
620 -4
-2
133
0
2
-5
4
5
-2
10
-4
15. Algebraically: x 54 80 0 x 54 80 x 5 4 80 2 4 5 x 5 2 4 5.
Graphically: We graph the equation y1 x 54 80
and determine where this curve intersects the x-axis. We use the viewing rectangle [1 9] by [5 5]. Zooming in, we see that solutions are x 201 and x 799.
32 16. Algebraically: 6 x 25 64 x 25 64 6 3 2 5 2 5 x 2 5 32 3 3 81 x 2 3 81.
Graphically: We graph the two equations y1 6 x 25
and y2 64 in the viewing rectangle [5 5] by [50 70]. Zooming in, we see that the solution is x 039 70
4 2
60 -2
2
4
6
8
-4
-4
-2
0
2
4
17. We graph y x 2 7x 12 in the viewing rectangle [0 6] 18. We graph y x 2 075x 0125 in the viewing by [01 01]. The solutions appear to be exactly x 3
and x 4. [In fact x 2 7x 12 x 3 x 4.]
rectangle [2 2] by [01 01]. The solutions are x 025 and x 050. 0.1
0.1
0.0 2 -0.1
4
6
-2
-1
1 -0.1
2
134
CHAPTER 1 Equations and Graphs
19. We graph y x 3 6x 2 11x 6 in the viewing
20. Since 16x 3 16x 2 x 1 16x 3 16x 2 x 1 0, we graph y 16x 3 16x 2 x 1 in the viewing
rectangle [1 4] by [01 01]. The solutions are x 100, x 200,and x 300.
rectangle [2 2] by [01 01]. The solutions are: x 100, x 025, and x 025.
0.1
0.1 -1
1
2
3
4 -2
-0.1
-1
1
2
-0.1
21. We first graph y x
x 1 in the viewing rectangle [1 5] by [01 01] and
find that the solution is near 16. Zooming in, we see that solutions is x 162.
0.1
-1
1
2
3
4
5
-0.1
1 x2 1 x 1 x 2 0Since x is only defined
22. 1
x
1
0.01
for x 0, we start with the viewing rectangle [1 5] by [1 1]. In this rectangle, there
appears to be an exact solution at x 0 and
another solution between x 2 and x 25. We then use the viewing rectangle [23 235] by
0.00 -1
1
2
3
4
5
-1
2.32
2.34
-0.01
[001 001], and isolate the second solution as x 2314. Thus the solutions are x 0 and x 231. 23. We graph y x 13 x in the viewing rectangle [3 3] by [1 1]. The solutions
1
are x 1, x 0, and x 1, as can be verified by substitution.
-3
-2
-1
1
2
3
-1
24. Since x 12 is defined only for x 0, we start by
1
graphing y x 12 x 13 x in the viewing
0.01
rectangle [1 5] by [1 1] We see a solution at x 0 and another one between x 3 and x 35. We then use the viewing rectangle
[33 34] by [001 001], and isolate the second solution as x 331. Thus, the solutions are x 0 and x 331.
0.00 -1 -1
1
2
3
4
5
3.35 -0.01
3.40
SECTION 1.9 Solving Equations and Inequalities Graphically
2x 1 1 and y x in the viewing rectangle [3 6] by [0 6] and see that the only solution to the equation 2x 1 1 x is x 4, which can
25. We graph y
135
6 4
be verified by substitution.
2 -2
0
2
3 x 2 and y 1 x in the viewing rectangle [7 4] by [2 8] and see that the only solution to the equation 3 x 2 1 x is
4
6
26. We graph y
5
x 356, which can be verified by substitution.
-6
-4
-2
27. We graph y 2x 4 4x 2 1 in the viewing rectangle [2 2] by [5 40] and see
2
4
40
that the equation 2x 4 4x 2 1 0 has no solution.
30 20 10 -2
-1
1
2
1
2
28. We graph y x 6 2x 3 3 in the viewing rectangle [2 2] by [5 15] and see
that the equation x 6 2x 3 3 0 has solutions x 1 and x 144, which can
10
be verified by substitution.
-2
29. x 3 2x 2 x 1 0, so we start by graphing
-1
100
the function y x 3 2x 2 x 1 in the viewing
1
rectangle [10 10] by [100 100]. There
appear to be two solutions, one near x 0 and
-10
another one between x 2 and x 3. We then use the viewing rectangle [1 5] by [1 1] and
-5
5
10
-100
-1
1
2
3
4
5
-1
zoom in on the only solution, x 255. 30. x 4 8x 2 2 0. We start by graphing the
10
function y x 4 8x 2 2 in the viewing
1
rectangle [10 10] by [10 10]. There appear to be four solutions between x 3 and x 3. We then use the viewing rectangle [5 5] by [1 1], and zoom to find the four solutions x 278, x 051, x 051, and x 278.
-10
-5
5 -10
10
-4
-2
2 -1
4
136
CHAPTER 1 Equations and Graphs
31. x x 1 x 2 16 x
10
x x 1 x 2 16 x 0. We start by graphing the function y x x 1 x 2 16 x in the
-4
viewing rectangle [5 5] by [10 10]. There
1
-2
appear to be three solutions. We then use the
2
4
-2
-1
-10
1
2
-1
viewing rectangle [25 25] by [1 1] and zoom into the solutions at x 205, x 000, and x 105. 32. x 4 16 x 3 . We start by graphing the functions y1 x 4 and y2 16 x 3 in the viewing rectangle [10 10] by [5 40]. There appears to be two solutions, one near x 2 and another one near x 2. We then use the viewing rectangle [24 22] by [27 29], and zoom in to find the solution at x 231. We then use the viewing rectangle [17 18] by [95 105], and zoom in to find the solution at x 179. 40
29
30 28
20 10 -10
-5
5
10
-2.4
-2.3
27 -2.2
10.4 10.2 10.0 9.8 9.6 1.70
1.75
33. We graph y x 2 and y 3x 10 in the viewing rectangle [4 7] by [5 30].
1.80
30
The solution to the inequality is [2 5].
20 10 -4
-2
2
34. Since 05x 2 0875x 025 05x 2 0875x 025 0, we graph
4
6
4
y 05x 2 0875x 025 in the viewing rectangle [3 1] by [5 5]. Thus the
2
solution to the inequality is [2 025].
-3
-2
-1
1
-2 -4
35. Since x 3 11x 6x 2 6 x 3 6x 2 11x 6 0, we graph
y x 3 6x 2 11x 6 in the viewing rectangle [0 5] by [5 5]. The solution
set is 10] [20 30].
4 2 0 -2 -4
2
4
SECTION 1.9 Solving Equations and Inequalities Graphically
36. Since 16x 3 24x 2 9x 1
0.01
4
16x 3 24x 2 9x 1 0, we graph
2
y 16x 3 24x 2 9x 1 in the viewing
rectangle [3 1] by [5 5]. From this rectangle,
137
0.00 -3
-2
-1
we see that x 1 is an x-intercept, but it is
1
-2
-1.0
-0.5
-4
unclear what is occurring between x 05 and
-0.01
x 0. We then use the viewing rectangle [1 0] by [001 001]. It shows y 0 at x 025. Thus in interval notation, the solution is 1 025 025 . 37. Since x 13 x x 13 x 0, we graph y x 13 x
05x 2 1 2 x 05x 2 1 2 x 0, we graph y 05x 2 1 2 x in the viewing rectangle
38. Since
in the viewing rectangle [3 3] by [1 1]. From this, we find that the solution set is 1 0 1 .
[1 1] by [1 1]. We locate the x-intercepts at
x 0535. Thus in interval notation, the solution is approximately 0535] [0535 .
1
1 -3
-2
-1
1
2
3
-1 -1.0
-0.5
0.5
1.0
-1
39. Since x 12 x 12 x 12 x 12 0, we graph y x 12 x 12 in the viewing
40. Since x 12 x 3 x 12 x 3 0, we graph
rectangle [2 2] by [5 5]. The solution set is 0.
y x 12 x 3 in the viewing rectangle [4 4] by
[1 1]. The x-intercept is close to x 2. Using a trace function, we obtain x 2148. Thus the solution is [2148 .
4 2
1 -2
-1
-2
1
2
-4 -4
-2
2
4
-1
41. We graph the equations y 3x 2 3x and y 2x 2 4 in the viewing rectangle
[2 6] by [5 50]. We see that the two curves intersect at x 1 and at x 4,
40
the inequality 3x 2 3x 2x 2 4 has the solution set 1 4.
20
and that the first curve is lower than the second for 1 x 4. Thus, we see that
-2
2
4
6
138
CHAPTER 1 Equations and Graphs
42. We graph the equations y 5x 2 3x and y 3x 2 2 in the viewing rectangle
20
[3 2] by [5 20]. We see that the two curves intersect at x 2 and at x 12 , which can be verified by substitution. The first curve is larger than the second for x 2 and for x 12 , so the solution set of the inequality 5x 2 3x 3x 2 2 is 2] 12 .
10
-3
-2
-1
1
2
43. We graph the equation y x 22 x 3 x 1 in the viewing rectangle
[2 4] by [15 5] and see that the inequality x 22 x 3 x 1 0 has the solution set [1 3].
-2
-1
1
2
3
4
-10
44. We graph the equation y x 2 x 2 1 in the viewing rectangle [2 2] by [1 1] and see that the inequality x 2 x 2 1 0 has the solution set 1] 0 [1 .
1
-2
-1
1
2
2
3
-1
45. To solve 5 3x 8x 20 by drawing the graph of a single equation, we isolate all terms on the left-hand side: 5 3x 8x 20 5 3x 8x 20 8x 20 8x 20 11x 25 0 or 11x 25 0. We graph y 11x 25, and see that the solution is x 227, as in Example 2.
1
-1
1 -1
46. Graphing y x 3 6x 2 9x and y
x in the viewing rectangle [001 002] 0.2
by [005 02], we see that x 0 and x 001 are solutions of the equation x 3 6x 2 9x x.
0.1
-0.01
0.01
0.02
SECTION 1.10 Modeling Variation
139
(c) We graph the equations y 15,000 and
47. (a) We graph the equation y 10x 05x 2 0001x 3 5000 in the viewing
y 10x 05x 2 0001x 3 5000 in the viewing
rectangle [0 600] by [30000 20000].
rectangle [250 450] by [11000 17000]. We use a zoom or trace function on a graphing calculator, and find that
20000
the company’s profits are greater than $15,000 for 279 x 400.
0 200
400
600 16000
-20000
14000
(b) From the graph it appears that
12000
0 10x 005x 2 0001x 3 5000 for
300
400
100 x 500, and so 101 cooktops must be produced
to begin to make a profit. 48. (a)
(b) Using a zoom or trace function, we find that y 10 for x 667. We x 2 000036. So for could estimate this since if x 100, then 5280 x 2 15x. Solving 15x 10 we x 100 we have 15x 5280
15 10 5 0 0
50
100
get 15 100 or x 100 15 667 mi.
49. Answers will vary. 50. Calculators perform operations in the following order: exponents are applied before division and division is applied before addition. Therefore, Y_1=x^1/3 is interpreted as y interpreted as y
x x1 , which is the equation of a line. Likewise, Y_2=x/x+4 is 3 3
x 4 1 4 5. Instead, enter the following: Y_1=x^(1/3), Y_2=x/(x+4). x
1.10 MODELING VARIATION 1. If the quantities x and y are related by the equation y 3x then we say that y is directly proportional to x, and the constant of proportionality is 3. 3 2. If the quantities x and y are related by the equation y then we say that y is inversely proportional to x, and the constant x of proportionality is 3. x 3. If the quantities x, y, and z are related by the equation z 3 then we say that z is directly proportional to x and inversely y proportional to y. 4. Because z is jointly proportional to x and y, we must have z kx y. Substituting the given values, we get 10 k 4 5 20k k 12 . Thus, x, y, and z are related by the equation z 12 x y. 5. (a) In the equation y 3x, y is directly proportional to x. (b) In the equation y 3x 1, y is not proportional to x. 3 , y is not proportional to x. 6. (a) In the equation y x 1 3 (b) In the equation y , y is inversely proportional to x. x
140
CHAPTER 1 Equations and Graphs
7. T kx, where k is constant.
8. P k, where k is constant.
k , where k is constant. z ks 11. y , where k is constant. t 9.
10. kmn, where k is constant.
13. z k y, where k is constant.
k , where k is constant. T kx 2 14. A 3 , where k is constant. t
15. V klh, where k is constant.
16. S kr 2 2 , where k is constant.
12. P
k P2t 2 , where k is constant. 18. A k x y, where k is constant. 3 b 19. Since y is directly proportional to x, y kx. Since y 42 when x 6, we have 42 k 6 k 7. So y 7x. k k 24 20. is inversely proportional to t, so . Since 3 when t 8, we have 3 k 24, so . t 8 t k 21 k . 21. A varies inversely as r, so A . Since A 7 when r 3, we have 7 k 21. So A r 3 r 17. R
1 . So P 1 T . 22. P is directly proportional to T , so P kT . Since P 20 when T 300, we have 20 k 300 k 15 15 kx . Since A 42 when x 7 and t 3, we 23. Since A is directly proportional to x and inversely proportional to t, A t 18x k 7 k 18. Therefore, A . have 42 3 t 24. S kpq. Since S 180 when p 4 and q 5, we have 180 k 4 5 180 20k k 9. So S 9 pq. k k k 360. 25. Since W is inversely proportional to the square of r, W 2 . Since W 10 when r 6, we have 10 r 62 360 So W 2 . r xy xy 2 3 26. t k . Since t 25 when x 2, y 3, and r 12, we have 25 k k 50. So t 50 . r 12 r 27. Since C is jointly proportional to l, , and h, we have C klh. Since C 128 when l h 2, we have 128 k 2 2 2 128 8k k 16. Therefore, C 16lh. 2 28. H kl 2 2 . Since H 36 when l 2 and 13 , we have 36 k 22 13 36 49 k k 81. So H 81l 2 2 .
k k k 275 29. R . Since R 25 when x 121, 25 k 275. Thus, R . 11 x x 121 a 2 2 abc abc . Since M 128 when a d and b c 2, we have 128 k 4k k 32. So M 32 . 30. M k d a d
x3 31. (a) z k 2 y
x3 27 k 2 , so z changes by a factor of 27 (b) If we replace x with 3x and y with 2y, then z k 4. 4 y 2y2 3x3
x2 32. (a) z k 4 y
x2 9 9. k 4 , so z changes by a factor of 16 (b) If we replace x with 3x and y with 2y, then z k 16 y 2y4 3x2
33. (a) z kx 3 y 5
(b) If we replace x with 3x and y with 2y, then z k 3x3 2y5 864kx 3 y 5 , so z changes by a factor of 864.
SECTION 1.10 Modeling Variation
141
k 34. (a) z 2 3 x y (b) If we replace x with 3x and y with 2y, then z
k 3x2 2y3
1 k 1 . , so z changes by a factor of 72 72 x 2 y 3
35. (a) The force F needed is F kx.
(b) Since F 30 N when x 9 cm and the spring’s natural length is 5 cm, we have 30 k 9 5 k 75. (c) From part (b), we have F 75x. Substituting x 11 5 6 into F 75x gives F 75 6 45 N.
36. (a) C kpm
(b) Since C 60,000 when p 120 and m 4000, we get 60,000 k 120 4000 k 18 . So C 18 pm.
(c) Substituting p 92 and m 5000, we get C 18 92 5000 $57,500. 37. (a) P ks 3 .
(b) Since P 96 when s 20, we get 96 k 203 k 0012. So P 0012s 3 .
(c) Substituting x 30, we get P 0012 303 324 watts.
38. (a) The power P is directly proportional to the cube of the speed s, so P ks 3 .
80 2 008. (b) Because P 80 when s 10, we have 80 k 103 k 1000 25 2 and s 15, we have P 2 153 270 hp. (c) Substituting k 25 25
39. D ks 2 . Since D 150 when s 40, we have 150 k 402 , so k 009375. Thus, D 009375s 2 . If D 200, then 200 009375s 2 s 2 21333, so s 46 mi/h (for safety reasons we round down). 40. L ks 2 A. Since L 1700 when s 50 and A 500, we have 1700 k 502 500 k 000136. Thus L 000136s 2 A. When A 600 and s 40 we get the lift is L 000136 402 600 13056 lb.
41. F k As 2 . Since F 220 when A 40 and s 5. Solving for k we have 220 k 40 52 220 1000k k 022. Now when A 28 and F 175 we get 175 0220 28 s 2 284090 s 2 so s 284090 533 mi/h. 42. (a) T 2 kd 3
3 (b) Substituting T 365 and d 93 106 , we get 3652 k 93 106 k 166 1019 . 3 (c) T 2 166 1019 279 109 360 109 T 600 104 . Hence the period of Neptune is 6.00104 days 164 years.
43. (a) P
kT . V
(b) Substituting P 332, T 400, and V 100, we get 332 P
k 400 k 83. Thus k 83 and the equation is 100
83T . V
(c) Substituting T 500 and V 80, we have P
83 500 51875 kPa. Hence the pressure of the sample of gas is 80
about 519 kPa. s 2 r (b) For the first car we have 1 1600 and s1 60 and for the second car we have 2 2500. Since the forces are equal
44. (a) F k
we have k
2500 s22 16 602 1600 602 k s22 , so s2 48 mi/h. r r 25
142
CHAPTER 1 Equations and Graphs
k 45. (a) The loudness L is inversely proportional to the square of the distance d, so L 2 . d k (b) Substituting d 10 and L 70, we have 70 2 k 7000. 10 1 k k , so the loudness is changed by a factor of 14 . (c) Substituting 2d for d, we have L 4 d2 2d2 k k , so the loudness is changed by a factor of 4. (d) Substituting 12 d for d, we have L 2 4 d2 1d 2
46. (a) The power P is jointly proportional to the area A and the cube of the velocity , so P k A 3 . (b) Substituting 2 for and 12 A for A, we have P k 12 A 23 4k A 3 , so the power is changed by a factor of 4. 3 (c) Substituting 12 for and 3A for A, we have P k 3A 12 38 Ak 3 , so the power is changed by a factor of 38 .
47. (a) R
kL d2
k 12 7 0002916. k 2400 00052 7 3 4375 (c) Substituting L 3 and d 0008, we have R 137 . 2400 00082 32 (b) Since R 140 when L 12 and d 0005, we get 140
k 3L
3 kL , so the resistance is changed by a factor of 34 . 4 d2 2d2 48. Let S be the final size of the cabbage, in pounds, let N be the amount of nutrients it receives, in ounces, and let c be the N number of other cabbages around it. Then S k . When N 20 and c 12, we have S 30, so substituting, we have c N 20 30 k 12 k 18. Thus S 18 . When N 10 and c 5, the final size is S 18 10 5 36 lb. c 4 E k60004 49. (a) For the sun, E S k60004 and for earth E E k3004 . Thus S 6000 204 160,000. So the sun 300 4 EE k300 produces 160,000 times the radiation energy per unit area than the Earth. (d) If we substitute 2d for d and 3L for L, then R
(b) The surface area of the sun is 4 435,0002 and the surface area of the Earth is 4 3,9602 . So the sun has 4 435,0002 435,000 2 times the surface area of the Earth. Thus the total radiation emitted by the sun is 3,960 4 3,9602 435,000 2 160,000 1,930,670,340 times the total radiation emitted by the Earth. 3,960 50. Let V be the value of a building lot on Galiano Island, A the area of the lot, and q the quantity of the water produced. Since V is jointly proportional to the area and water quantity, we have V k Aq. When A 200 300 60,000 and q 10, we have V $48 000, so 48,000 k 60,000 10 k 008. Thus V 008Aq. Now when A 400 400 160,000 and q 4, the value is V 008 160,000 4 $51,200. 51. (a) Let T and l be the period and the length of the pendulum, respectively. Then T k l.
T2 T2 2T 2 (b) T k l T 2 k 2 l l 2 . If the period is doubled, the new length is 4 2 4l. So we would 2 k k k quadruple the length l to double the period T . 52. Let H be the heat experienced by a hiker at a campfire, let A be the amount of wood, and let d be the distance from A A campfire. So H k 3 . When the hiker is 20 feet from the fire, the heat experienced is H k 3 , and when the amount d 20 2A A 2A k 3 d 3 16 000 d 20 3 2 252 feet. of wood is doubled, the heat experienced is H k 3 . So k 8 000 d d
CHAPTER 1
53. (a) Since f is inversely proportional to L, we have f
Review
143
k , where k is a positive constant. L
k k 12 12 f . So the frequency of the vibration is cut in half. 2L L 54. (a) Since r is jointly proportional to x and P x, we have r kx P x, where k is a positive constant. (b) If we replace L by 2L we have
(b) When 10 people are infected the rate is r k10 5000 10 49,900k. When 1000 people are infected the rate is r k 1000 5000 1000 4,000,000k. So the rate is much higher when 1000 people are infected. Comparing 1000 people infected 4,000,000k these rates, we find that 80. So the infection rate when 1000 people are infected 10 people infected 49,900k is about 80 times as large as when 10 people are infected. (c) When the entire population is infected the rate is r k 5000 5000 5000 0. This makes sense since there are no more people who can be infected.
L 25 1026 14 . 55. Using B k 2 with k 0080, L 25 1026 , and d 24 1019 , we have B 0080 2 347 10 d 24 1019
The star’s apparent brightness is about 347 1014 Wm2 . L L L 2 56. First, we solve B k 2 for d: d k d k because d is positive. Substituting k 0080, L 58 1030 , and B B d 58 1030 16 B 82 10 , we find d 0080 238 1022 , so the star is approximately 238 1022 m from earth. 82 1016 57. Examples include radioactive decay and exponential growth in biology.
CHAPTER 1 REVIEW 1. (a)
y
Q
(b) The distance from P to Q is d P Q 5 22 12 02 49 144 193 3 5 2 12 0 6 . (c) The midpoint is 2 2 2
1 1
(d) The line has slope m
P
x
12 0 12 7 , and has 5 2
12 24 equation y 0 12 7 x 2 y 7 x 7 12x 7y 24 0. Q
(e) The radius of this circle was found in part (b). It is r d P Q 193. So an equation is 2 193 x 22 y 2 193. x 22 y 02 Q
y
y
2
P 2
x
P
1 1
P
x
144
CHAPTER 1 Equations and Graphs
2. (a)
y 1 1
x
P
(b) The distance from P to Q is d P Q 2 72 11 12 25 100 125 5 5 9 2 7 11 1 6 . (c) The midpoint is 2 2 2
Q
10 11 1 2, and 27 5 its equation is y 11 2 x 2
(d) The line has slope m
y 11 2x 4 y 2x 15.
(e) The radius of this circle was found in part (b). It is r d P, Q 5 5. So an equation is 2 x 72 y 12 5 5 x 72 y 12 125.
y
1
y
1
x
P
2 2
x
P
Q Q
y
3. (a)
P
4
x
4 Q
(d) The line has slope m
(c) The midpoint is
16 8 2 14 6 4 10 5
and equation y 2 85 x 6
8 38 y 2 85 x 48 5 y 5x 5 . y
P
(b) The distance from P to Q is d P Q 6 42 [2 14]2 100 256 356 2 89
6 4 2 14 2 2
x
4 Q
1 6.
(e) The radius of this circle was found in part (b). It is r d P Q 2 89. So an equation is 2 [x 6]2 y 22 2 89 x 62 y 22 356.
4
P
y
4 4 Q
x
CHAPTER 1 y
4. (a)
2 2
Q
P
145
(b) The distance from P to Q is d P Q [5 3]2 [2 6]2 64 16 80 4 5.
x
(c) The midpoint is 2 6 48 12 , and 5 3
(d) The line has slope m
Review
has equation y 2 12 x 5 y 2 12 x 52 y 12 x 92 . y
5 3 2 6 2 2
1 4.
(e) The radius of this circle was found in part (b). It is r d P Q 4 5. So an equation is 2 2 x 52 y 2 4 5 x 52 y 22 80. y
2 Q
2
P
x
2 Q
y
5.
6. x y x 4 or y 2
2
x
P
y
1 1
x
1 1
x
7. d A C 74 and 4 12 4 32 4 12 4 32 d B C 5 12 3 32 5 12 3 32 72. Therefore, B is closer to C. 8. The circle with center at 2 5 and radius
2 2 has equation x 22 y 52 2 x 22 y 52 2.
9. The center is C 5 1, and the point P 0 0 is on the circle. The radius of the circle is r d P C 0 52 0 12 = 0 52 0 12 26. Thus, the equation of the circle is x 52 y 12 26.
1 11 21 38 , and the radius is 12 of the distance from P to Q, or 10. The midpoint of segment P Q is 2 2 2 2 r 12 d P, Q 12 2 12 3 82 12 2 12 3 82 r 12 34. Thus the equation is 2 2 x 12 y 11 17 2 2 .
146
CHAPTER 1 Equations and Graphs
11. (a) x 2 y 2 2x 6y 9 0 x 2 2x y 2 6y 9 x 2 2x 1 y 2 6y 9 9 1 9
(b) The circle has center 1 3 and radius 1. y
x 12 y 32 1, an equation of a circle.
1l 1
12. (a) 2x 2 2y 2 2x 8y 12 x 2 x y 2 4y 14 x 2 x 14 y 2 4y 4 14 14 4
2 x 12 y 22 92 , an equation of a circle.
x
(b) The circle has center 12 2
and radius 3 2 2 . y 1l
1
x
13. (a) x 2 y 2 72 12x x 2 12x y 2 72 x 2 12x 36 y 2 72 36 x 62 y 2 36. Since the left side of this equation must be greater than or equal to zero, this equation has no graph.
14. (a) x 2 y 2 6x 10y 34 0 x 2 6x y 2 10y 34 x 2 6x 9 y 2 10y 25 34 9 25 x 32 y 52 0, an equation of a point.
(b) This is the equation of the point 3 5. y
1l 1
x
CHAPTER 1
15. y 2 3x
y
x
y
x
y
2
8
2
3
0
2
0
1
12
0
2 3
17.
16. 2x y 1 0 y 2x 1
1
0
1
x
y x 1 y 72 x 7 2 7
18. y
x
2
14
4
5
0
7
0
0
2
0
4
5
2 x
20. 8x y 2 0 y 2 8x
y
y
x
y
3
7
8
8
1
15
2
4
0
16
0
0
1
15
3
7
1
x
1
x
1
x
1
y
1
x
22. y 1 x 2
y
x
y
x
0
0
1
1
1
2
4 9
1
2
y
3
1
y
x
y
y
y
2
21. x
147
y x 0 5x 4y 0 4 5
x
19. y 16 x 2
Review
1 2
1 1
x
y
y 0
23
0
1
1
0
1
1
x
148
CHAPTER 1 Equations and Graphs
23. y 9 x 2 (a) x-axis symmetry: replacing y by y gives y 9 x 2 , which is not the same as the original equation, so the graph is not symmetric about the x-axis. y-axis symmetry: replacing x by x gives y 9 x2 9 x 2 , which is the same as the original equation, so the graph is symmetric about the y-axis. Origin symmetry: replacing x by x and y by y gives y 9 x2 y 9 x 2 , which is not the same as the original equation, so the graph is not symmetric about the origin. (b) To find x-intercepts, we set y 0 and solve for x: 0 9 x 2 x 2 9 x 3, so the x-intercepts are 3 and 3. To find y-intercepts, we set x 0 and solve for y: y 9 02 9, so the y-intercept is 9.
24. 6x y 2 36 (a) x-axis symmetry: replacing y by y gives 6x y2 36 6x y 2 36, which is the same as the original equation, so the graph is symmetric about the x-axis. y-axis symmetry: replacing x by x gives 6 x y 2 36 6x y 2 36, which is not the same as the original equation, so the graph is not symmetric about the y-axis. Origin symmetry: replacing x by x and y by y gives 6 x y2 36 6x y 2 36, which is not the same as the original equation, so the graph is not symmetric about the origin. (b) To find x-intercepts, we set y 0 and solve for x: 6x 02 36 x 6, so the x-intercept is 6.
To find y-intercepts, we set x 0 and solve for y: 6 0 y 2 36 y 6, so the y-intercepts are 6 and 6.
25. x 2 y 12 1
2 (a) x-axis symmetry: replacing y by y gives x 2 y 1 1 x 2 y 12 1, so the graph is not symmetric about the x-axis. y-axis symmetry: replacing x by x gives x2 y 12 1 x 2 y 12 1, so the graph is symmetric about the y-axis. 2 Origin symmetry: replacing x by x and y by y gives x2 y 1 1 x 2 y 12 1, so the graph is not symmetric about the origin.
(b) To find x-intercepts, we set y 0 and solve for x: x 2 0 12 1 x 2 0, so the x-intercept is 0.
To find y-intercepts, we set x 0 and solve for y: 02 y 12 1 y 1 1 y 0 or 2, so the y-intercepts are 0 and 2.
26. x 4 16 y (a) x-axis symmetry: replacing y by y gives x 4 16 y x 4 16 y, so the graph is not symmetric about the x-axis. y-axis symmetry: replacing x by x gives x4 16 y x 4 16 y, so the graph is symmetric about the y-axis. Origin symmetry: replacing x by x and y by y gives x4 16 y x 4 16 y, so the graph is not symmetric about the origin.
(b) To find x-intercepts, we set y 0 and solve for x: x 4 16 0 x 4 16 x 2, so the x-intercepts are 2 and 2. To find y-intercepts, we set x 0 and solve for y: 04 16 y y 16, so the y-intercept is 16.
27. 9x 2 16y 2 144 (a) x-axis symmetry: replacing y by y gives 9x 2 16 y2 144 9x 2 16y 2 144, so the graph is symmetric about the x-axis. y-axis symmetry: replacing x by x gives 9 x2 16y 2 144 9x 2 16y 2 144, so the graph is symmetric about the y-axis. Origin symmetry: replacing x by x and y by y gives 9 x2 16 y2 144 9x 2 16y 2 144, so the graph is symmetric about the origin.
CHAPTER 1
Review
149
(b) To find x-intercepts, we set y 0 and solve for x: 9x 2 16 02 144 9x 2 144 x 4, so the x-intercepts are 4 and 4. To find y-intercepts, we set x 0 and solve for y: 9 02 16y 2 144 16y 2 144, so there is no y-intercept.
4 x 4 (a) x-axis symmetry: replacing y by y gives y , which is different from the original equation, so the graph is not x symmetric about the x-axis. 4 y-axis symmetry: replacing x by x gives y , which is different from the original equation, so the graph is not x symmetric about the y-axis. 4 4 y , so the graph is symmetric about the Origin symmetry: replacing x by x and y by y gives y x x origin. 4 (b) To find x-intercepts, we set y 0 and solve for x: 0 has no solution, so there is no x-intercept. x To find y-intercepts, we set x 0 and solve for y. But we cannot substitute x 0, so there is no y-intercept.
28. y
29. x 2 4x y y 2 1 (a) x-axis symmetry: replacing y by y gives x 2 4x y y2 1, which is different from the original equation, so the graph is not symmetric about the x-axis. y-axis symmetry: replacing x by x gives x2 4 x y y 2 1, which is different from the original equation, so the graph is not symmetric about the y-axis. Origin symmetry: replacing x by x and y by y gives x2 4 x y y2 1 x 2 4x y y 2 1, so the graph is symmetric about the origin. (b) To find x-intercepts, we set y 0 and solve for x: x 2 4x 0 02 1 x 2 1 x 1, so the x-intercepts are 1 and 1. To find y-intercepts, we set x 0 and solve for y: 02 4 0 y y 2 1 y 2 1 y 1, so the y-intercepts are 1 and 1.
30. x 3 x y 2 5 (a) x-axis symmetry: replacing y by y gives x 3 x y2 5 x 3 x y 2 5, so the graph is symmetric about the x-axis. y-axis symmetry: replacing x by x gives x3 x y 2 5, which is different from the original equation, so the graph is not symmetric about the y-axis. Origin symmetry: replacing x by x and y by y gives x3 x y2 5, which is different from the original equation, so the graph is not symmetric about the origin. (b) To find x-intercepts, we set y 0 and solve for x: x 3 x 02 5 x 3 5 x 3 5, so the x-intercept is 3 5. To find y-intercepts, we set x 0 and solve for y: 03 0y 2 5 has no solution, so there is no y-intercept.
150
CHAPTER 1 Equations and Graphs
31. (a) We graph y x 2 6x in the viewing rectangle [10 10] by [10 10].
32. (a) We graph y
5 x in the viewing rectangle
[10 6] by [1 5].
10
4 2
-10
-5
5
10 -10
-10
(b) From the graph, we see that the x-intercepts are 0
-5
5
(b) From the graph, we see that the x-intercept is 5 and
and 6 and the y-intercept is 0.
the y-intercept is approximately 224.
33. (a) We graph y x 3 4x 2 5x in the viewing rectangle [4 8] by [30 20]. 20
x2 x2 y2 1 y2 1 34. (a) We graph 4 4 x2 in the viewing rectangle [3 3] by y 1 4 [2 2].
-4
-2
2
4
6
8
2 1
-20 -3
(b) From the graph, we see that the x-intercepts are 1,
-2
-1 -1
1
2
3
-2
0, and 5 and the y-intercept is 0.
(b) From the graph, we see that the x-intercepts are 2 and 2 and the y-intercepts are 1 and 1.
35. (a) The line that has slope 2 and y-intercept 6 has the slope-intercept equation
(c)
y
y 2x 6. (b) An equation of the line in general form is 2x y 6 0.
1 1
x
CHAPTER 1
36. (a) The line that has slope 12 and passes through the point 6 3 has
Review
151
y
(c)
equation y 3 12 x 6 y 3 12 x 6 y 12 x. (b) 12 x 3 y 3 x 6 2y 6 x 2y 0. 1 1
37. (a) The line that passes through the points 1 6 and 2 4 has slope
(c)
4 6 2 m , so y 6 23 [x 1] y 6 23 x 23 2 1 3
x
y 1 1
y 23 x 16 3 .
x
(b) y 23 x 16 3 3y 2x 16 2x 3y 16 0.
38. (a) The line that has x-intercept 4 and y-intercept 12 passes through the points
(c)
y
12 0 3 and the equation is 04 y 0 3 x 4 y 3x 12.
4 0 and 0 12, so m
(b) y 3x 12 3x y 12 0. 2
39. (a) The vertical line that passes through the point 3 2 has equation x 3.
(c)
1
x
1
x
y
(b) x 3 x 3 0. 1
152
CHAPTER 1 Equations and Graphs
40. (a) The horizontal line with y-intercept 5 has equation y 5.
(c)
y
(b) y 5 y 5 0. 1 1
41. (a) 2x 5y 10 5y 2x 10 y 25 x 2, so the given line has slope (c) m 25 . Thus, an equation of the line passing through 1 1 parallel to this line is y 1 25 x 1 y 25 x 35 .
y
1
(b) y 25 x 35 5y 2x 3 2x 5y 3 0.
42. (a) The line containing 2 4 and 4 4 has slope
x
(c)
1
x
1
x
y
8 4 4 4, and the line passing through the origin with 42 2 this slope has equation y 4x. m
1
(b) y 4x 4x y 0.
43. (a) The line y 12 x 10 has slope 12 , so a line perpendicular to this one has
(c)
y
1 2. In particular, the line passing through the origin slope 12
perpendicular to the given line has equation y 2x. (b) y 2x 2x y 0.
1 1
x
CHAPTER 1
44. (a) x 3y 16 0 3y x 16 y 13 x 16 3 , so the given line has
(c)
Review
153
y
slope 13 . The line passing through 1 7 perpendicular to the given line has equation y 7
1 x 1 y 7 3 x 1 y 3x 10. 13
(b) y 3x 10 3x y 10 0.
x 1 1
45. The line with equation y 13 x 1 has slope 13 . The line with equation 9y 3x 3 0 9y 3x 3 y 13 x 13 also has slope 13 , so the lines are parallel.
46. The line with equation 5x 8y 3 8y 5x 3 y 58 x 38 has slope 58 . The line with equation 10y 16x 1 1 has slope 8 1 , so the lines are perpendicular. 10y 16x 1 y 85 x 10 5 58
47. (a) The slope represents a stretch of 03 inches for each one-pound increase in weight. The s-intercept represents the length of the unstretched spring. (b) When 5, s 03 5 25 15 25 40 inches.
48. (a) We use the information to find two points, 0 60000 and 3 70500. Then the slope is 10,500 70,500 60,000 3,500. So S 3,500t 60,000. m 30 3 (b) The slope represents an annual salary increase of $3500, and the S-intercept represents her initial salary. (c) When t 12, her salary will be S 3500 12 60,000 42,000 60,000 $102,000. 49. x 2 9x 14 0 x 7 x 2 0 x 7 or x 2.
50. x 2 24x 144 0 x 122 0 x 12 0 x 12.
51. 2x 2 x 1 2x 2 x 1 0 2x 1 x 1 0. So either 2x 1 0 2x 1 x x 1. 52. 3x 2 5x 2 0 3x 1 x 2 0 x 13 or x 2. 53. 0 4x 3 25x x 4x 2 25 x 2x 5 2x 5 0. So either x 0 or 2x 5 0 2x
1 ; or x 1 0 2
5 x 52 ; or
2x 5 0 2x 5 x 52 .
54. x 3 2x 2 5x 10 0 x 2 x 2 5 x 2 0 x 2 x 2 5 0 x 2 or x 5.
55. 3x 2 4x 1 0 2 2 7 4 42 431 b b2 4ac 4 1612 42 7 2 7 . 4 28 x 2a 23 6 6 6 6 3 2 419 3 3 2 56. x 2 3x 9 0 x b 2ab 4ac 3 2936 3 227 , which are not real numbers. 21 There is no real solution. 2 1 3 x 1 2 x 3 x x 1 x 1 2x 3x 2 3x 0 3x 2 6x 1 57. x x 1 2 3 6 6 62 431 b b2 4ac 6 3612 6 24 62 6 3 6 . x 2a 6 6 6 6 3 23 58.
1 8 x 2 x x 2 x 2 8 x 2 2x x 2 8 x 2 3x 10 0 x 2 x 5 0 x 2 x 2 x 4 x 2 or x 5. However, since x 2 makes the expression undefined, we reject this solution. Hence the only solution is x 5.
154
CHAPTER 1 Equations and Graphs
59. x 4 8x 2 9 0 x 2 9 x 2 1 0 x 3 x 3 x 2 1 0 x 3 0 x 3, or x 3 0 x 3, however x 2 1 0 has no real solution. The solutions are x 3.
60. x 4 x 32. Let u x. Then u 2 4u 32 u 2 4u 32 0 u 8 u 4 0 So either u 8 0 or u 4 0. If u 8 0, then u 8 x 8 x 64. If u 4 0, then u 4 x 4, which has no real solution. So the only solution is x 64. 61. x 12 2x 12 x 32 0 x 12 1 2x x 2 0 x 12 1 x2 0. Since x 12 1 x is never 0, the only solution comes from 1 x2 0 1 x 0 x 1.
2 62. 1 x 2 1 x 15 0. Let u 1 x, then the equation becomes u 2 2u 15 0 u 5 u 3 0 u 5 0 or u 3 0. If u 5 0, then u 5 1 x 5 x 4 x 16. If u 3 0, then u 3 1 x 3 x 4, which has no real solution. So the only solution is x 16. 63. x 7 4 x 7 4 x 7 4, so x 11 or x 3. 64. 2x 5 9 is equivalent to 2x 5 9 2x 5 9 x
59 . So x 2 or x 7. 2
65. (a) 2 3i 1 4i 2 1 3 4 i 3 i
(b) 2 i 3 2i 6 4i 3i 2i 2 6 i 2 8 i
66. (a) 3 6i 6 4i 3 6i 6 4i 3 6 6 4 i 3 2i (b) 4i 2 12 i 8i 2i 2 8i 2 2 8i 8 8i 2 4 2i 4 2i 2 i 8 8i 2i 2 6 8i 65 85 i 2i 2i 2i 41 5 4 i2 (b) 1 1 1 1 1 i 1 i 1 i i i 2 1 1 2
67. (a)
8 3i 4 3i 32 12i 9i 2 41 12i 8 3i 32 12i 9 12 41 25 25 i 2 4 3i 4 3i 4 3i 16 9 25 16 9i (b) 10 40 i 10 2i 10 20i 2 20
68. (a)
69. x 2 16 0 x 2 16 x 4i 70. x 2 12 x 12 2 3i b 71. x 2 6x 10 0 x
72. 2x 2 3x 2 0 x
6 62 4 1 10 6 36 40 b2 4ac 3 i 2a 2 1 2
3
32 4 2 2 2 2
3
7 7 3 i 4 4 4
73. x 4 256 0 x 2 16 x 2 16 0 x 4 or x 4i 74. x 3 2x 2 4x 8 0 x 2 x 2 4 x 2 or x 2i
CHAPTER 1
Review
155
75. Let r be the rate the woman runs in mi/h. Then she cycles at r 8 mi/h. Rate Cycle
r 8
Run
r
Time
Distance
4 r 8 25 r
4 25
25 4 1. Multiplying by 2r r 8, we r 8 r get 4 2r 25 2 r 8 2r r 8 8r 5r 40 2r 2 16r 0 2r 2 3r 40 Since the total time of the workout is 1 hour, we have
3 32 4240 3 9320 3 329 . Since r 0, we reject the negative value. She runs at r 22 4 4 3 329 378 mi/h. r 4 x2 1500 20x x 2 x 2 20x 1500 0 x 30 x 50 0. So 76. Substituting 75 for d, we have 75 x
20 x 30 or x 50. The speed of the car was 30 mi/h.
77. Let x be the length of one side in cm. Then 28 x is the length of the other side. Using the Pythagorean Theorem, we have x 2 28 x2 202 x 2 784 56x x 2 400 2x 2 56x 384 0 2 x 2 28x 192 0
2 x 12 x 16 0. So x 12 or x 16. If x 12, then the other side is 28 12 16. Similarly, if x 16, then the other side is 12. The sides are 12 cm and 16 cm. 80 and the total amount of fencing material is 78. Let l be length of each garden plot. The width of each plot is then l 480 80 4 l 6 88. Thus 4l 88 4l 2 480 88l 4l 2 88l 480 0 4 l 2 22l 120 0 l l 4 l 10 l 12 0. So l 10 or l 12. If l 10 ft, then the width of each plot is 80 10 8 ft. If l 12 ft, then the
width of each plot is 80 12 667 ft. Both solutions are possible.
80. 12 x 7x 12 8x 32 x. Interval: 32
79. 3x 2 11 3x 9 x 3. Interval: 3 . Graph:
-3
Graph:
81. 3 x 2x 7 10 3x 10 3 x Interval: 10 3 Graph:
3 2
82. 1 2x 5 3 6 2x 2 3 x 1 Interval: 3 1]. Graph:
10 3
_3
_1
83. x 2 4x 12 0 x 2 x 6 0. The expression on the left of the inequality changes sign where x 2 and where x 6. Thus we must check the intervals in the following table. 6
6 2
2
Sign of x 2
Sign of x 6
Interval
Sign of x 2 x 6
Interval: 6 2 Graph:
_6
2
156
CHAPTER 1 Equations and Graphs
84. x 2 1 x 2 1 0 x 1 x 1 0. The expression on the left of the inequality changes sign when x 1 and x 1. Thus we must check the intervals in the following table. Interval: [1 1] 1
1 1
1
Sign of x 1
Sign of x 1
Interval
Sign of x 1 x 1
85.
Graph:
_1
1
2x 5 2x 5 x 1 x 4 2x 5 1 1 0 0 0. The expression on the left of the inequality x 1 x 1 x 1 x 1 x 1 changes sign where x 1 and where x 4. Thus we must check the intervals in the following table. We exclude x 1, since the expression is not
4
4 1
1
Sign of x 4
defined at this value. Thus the solution is [4 1.
Sign of x 1
Graph:
Sign of
Interval
x 4 x 1
_4
_1
86. 2x 2 x 3 2x 2 x 3 0 2x 3 x 1 0. The expression on the left of the inequality changes sign when 1 and 32 . Thus we must check the intervals in the following table. 1
Interval
3 2
Sign of 2x 3
Sign of x 1
Sign of 2x 3 x 1
87.
1 32
Interval: 1] 32 Graph:
3 2
_1
x 4 x 4 0 0. The expression on the left of the inequality changes sign where x 2, where x 2, 2 2 x 2 x x 4 and where x 4. Thus we must check the intervals in the following table. 2
2 2
2 4
4
Sign of x 4
Sign of x 2
Interval
Sign of x 2 x 4 Sign of x 2 x 2 Since the expression is not defined when x 2we exclude these values and the solution is 2 2 4]. Graph:
_2
2
4
CHAPTER 1
88.
Review
157
5 5 5 0 0 2 0. The 0 2 x 1 x 2 x 2 x x 1 4 x 1 x 1 x 4 expression on the left of the inequality changes sign when 2 1and 2. Thus we must check the intervals in the following table. 5
x 3 x 2 4x 4
2
2 1
1 2
2
Sign of x 1
Sign of x 2
Sign of x 2
Interval
5 Sign of x 1 x 2 x 2 Interval: 2 1 2 Graph:
_2
1
2
90. x 4 002 002 x 4 002
89. x 5 3 3 x 5 3 2 x 8. Interval: [2 8] Graph:
2
398 x 402 Interval: 398 402
8
Graph:
3.98
4.02
91. 2x 1 1 is equivalent to 2x 1 1 or 2x 1 1. Case 1: 2x 1 1 2x 0 x 0. Case 2: 2x 1 1 2x 2 x 1. Interval: 1] [0 . Graph:
_1
0
92. x 1 is the distance between x and 1 on the number line, and x 3 is the distance between x and 3. We want those points that are closer to 1 than to 3. Since 2 is midway between 1 and 3, we get x 2 as the solution. Graph: 2
93. (a) For
24 x 3x 2 to define a real number, we must have 24 x 3x 2 0 8 3x 3 x 0. The expression
on the left of the inequality changes sign where 8 3x 0 3x 8 x 83 ; or where x 3. Thus we must check the intervals in the following table. Interval: 3 83 . 8 3 83 Interval 3 3 Graph: Sign of 8 3x Sign of 3 x
Sign of 8 3x 3 x
_3
8 3
158
CHAPTER 1 Equations and Graphs
(b) For 4
1 x x4
to define a real number we must have x x 4 0 x 1 x 3 0 x 1 x 1 x x 2 0.
The expression on the left of the inequality changes sign where x 0; or where x 1; or where 1 x x 2 0 12 411 x 1 21 1 214 which is imaginary. We check the intervals in the following table. Interval: 0 1. Interval
0
0 1
1
Sign of x
Sign of 1 x
Sign of x 1 x 1 x x 2
Sign of 1 x x 2
6 9 r3 94. We have 8 43 r 3 12
3
6 r
3
9 . Thus r
Graph: 0
3
6
1
3 9 .
95. From the graph, we see that the graphs of y x 2 4x and y x 6 intersect at x 1 and x 6, so these are the solutions of the equation x 2 4x x 6.
96. From the graph, we see that the graph of y x 2 4x crosses the x-axis at x 0 and x 4, so these are the solutions of the equation x 2 4x 0.
97. From the graph, we see that the graph of y x 2 4x lies below the graph of y x 6 for 1 x 6, so the inequality x 2 4x x 6 is satisfied on the interval [1 6].
98. From the graph, we see that the graph of y x 2 4x lies above the graph of y x 6 for x 1 and 6 x , so the inequality x 2 4x x 6 is satisfied on the intervals 1] and [6 .
99. From the graph, we see that the graph of y x 2 4x lies above the x-axis for x 0 and for x 4, so the inequality x 2 4x 0 is satisfied on the intervals 0] and [4 .
100. From the graph, we see that the graph of y x 2 4x lies below the x-axis for 0 x 4, so the inequality x 2 4x 0 is satisfied on the interval [0 4]. 101. x 2 4x 2x 7. We graph the equations y1 x 2 4x 102. x 4 x 2 5. We graph the equations y1 x 4 and y2 2x 7 in the viewing rectangle [10 10] by
[5 25]. Using a zoom or trace function, we get the solutions x 1 and x 7.
-10
-5
and y2 x 2 5 in the viewing rectangle [4 5] by [0 10]. Using a zoom or trace function, we get the solutions x 250 and x 276.
20
10
10
5
5
10
-4
-2
0
2
4
CHAPTER 1
Review
159
103. x 4 9x 2 x 9. We graph the equations y1 x 4 9x 2 104. x 3 5 2. We graph the equations and y2 x 9 in the viewing rectangle [5 5] by
y1 x 3 5 and y2 2 in the viewing rectangle
[25 10]. Using a zoom or trace function, we get the
[20 20] by [0 10]. Using Zoom and/or Trace, we get the
solutions x 272, x 115, x 100, and x 287.
solutions x 10, x 6, x 0, and x 4.
10 -4
10
-2 -10
2
5
4
-20
-20
105. 4x 3 x 2 . We graph the equations y1 4x 3 and
y2 x 2 in the viewing rectangle [5 5] by [0 15]. Using
-10
0
10
20
106. x 3 4x 2 5x 2. We graph the equations
y1 x 3 4x 2 5x and y2 2 in the viewing rectangle
a zoom or trace function, we find the points of intersection
[10 10] by [5 5]. We find that the point of intersection
are at x 1 and x 3. Since we want 4x 3 x 2 , the
is at x 507. Since we want x 3 4x 2 5x 2, the solution is the interval 507 .
solution is the interval [1 3]. 15
4
10
2
5 -4
-2
-10 0
2
-5
-2
5
10
-4
4
108. x 2 16 10 0. We graph the equation y1 x 4 4x 2 and y2 12 x 1 in the viewing rectangle y x 2 16 10 in the viewing rectangle [10 10] by [5 5] by [5 5]. We find the points of intersection are [10 10]. Using a zoom or trace function, we find that the at x 185, x 060, x 045, and x 200. Since x-intercepts are x 510 and x 245. Since we we want x 4 4x 2 12 x 1, the solution is want x 2 16 10 0, the solution is approximately 185 060 045 200. 510] [245 245] [510 .
107. x 4 4x 2 12 x 1. We graph the equations
4
10
2 -4
-2
-2
2
4 -10
-5
5
-4 -10
10
160
CHAPTER 1 Equations and Graphs
109. Here the center is at 0 0, and the circle passes through the point 5 12, so the radius is r 5 02 12 02 25 144 169 13. The equation of the circle is x 2 y 2 132
x 2 y 2 169. The line shown is the tangent that passes through the point 5 12, so it is perpendicular to the line 12 12 0 . The slope of the line we seek is through the points 0 0 and 5 12. This line has slope m 1 5 0 5 5 1 1 5 x 5 y 12 5 x 25 m2 . Thus, an equation of the tangent line is y 12 12 12 12 m1 125 12
5 x 169 5x 12y 169 0. y 12 12 110. Because the circle is tangent to the x-axis at the point 5 0 and tangent to the y-axis at the point 0 5, the center is at
5 5 and the radius is 5. Thus an equation is x 52 y 52 52 x 52 y 52 25. The slope of 4 4 51 , so an equation of the line we seek is the line passing through the points 8 1 and 5 5 is m 58 3 3 y 1 43 x 8 4x 3y 35 0.
111. Since M varies directly as z we have M kz. Substituting M 120 when z 15, we find 120 k 15 k 8. Therefore, M 8z. k k k 192. 112. Since z is inversely proportional to y, we have z . Substituting z 12 when y 16, we find 12 y 16 192 Therefore z . y k 113. (a) The intensity I varies inversely as the square of the distance d, so I 2 . d k (b) Substituting I 1000 when d 8, we get 1000 k 64,000. 82 64,000 64,000 (c) From parts (a) and (b), we have I . Substituting d 20, we get I 160 candles. d2 202 114. Let f be the frequency of the string and l be the length of the string. Since the frequency is inversely proportional to the k k 5280 length, we have f . Substituting l 12 when k 440, we find 440 k 5280. Therefore f . For l 12 l 5280 l 5280 f 660, we must have 660 660 8. So the string needs to be shortened to 8 inches. l 115. Let be the terminal velocity of the parachutist in mi/h and be his weight in pounds. Since the terminal velocity is directly proportional to the square root of the weight, we have k . Substituting 9 when 160, we solve 9 for k. This gives 9 k 160 k 0712. Thus 0712 . When 240, the terminal velocity is 160 0712 240 11 mi/h. 116. Let r be the maximum range of the baseball and be the velocity of the baseball. Since the maximum range is directly
proportional to the square of the velocity, we have r l 2 . Substituting 60 and r 242, we find 242 k 602 k 00672. If 70, then we have a maximum range of r 00672 702 3294 feet.
CHAPTER 1
Test
161
CHAPTER 1 TEST
y
1. (a)
There are several ways to determine the coordinates of S. The diagonals of a
S
P
square have equal length and are perpendicular. The diagonal P R is horizontal R
and has length is 6 units, so the diagonal QS is vertical and also has length 6. Thus, the coordinates of S are 3 6. (b) The length of P Q is 0 32 3 02 18 3 2. So the area of 2 P Q RS is 3 2 18.
1 1
Q
x
(b) The x-intercept occurs when y 0, so 0 x 2 4 x 2 4 x 2. The
y
2. (a)
y-intercept occurs when x 0, so y 4.
(c) x-axis symmetry: y x 2 4 y x 2 4, which is not the same as the original equation, so the graph is not symmetric with respect to the x-axis.
1 1
x
y-axis symmetry: y x2 4 y x 2 4, which is the same as the
original equation, so the graph is symmetric with respect to the y-axis.
Origin symmetry: y x2 4 y x 2 4, which is not the same
as the original equation, so the graph is not symmetric with respect to the origin.
y
3. (a)
Q
P
1 1
x
(b) The distance between P and Q is d P Q 3 52 1 62 64 25 89. 3 5 1 6 (c) The midpoint is 1 72 . 2 2 16 5 5 (d) The slope of the line is . 3 5 8 8 (e) The perpendicular bisector of P Q contains the midpoint, 1 72 , and it slope is
1 8 . Hence the equation the negative reciprocal of 58 . Thus the slope is 58 5
is y 72 85 x 1 y 85 x 85 72 85 x 51 10 . That is, y 85 x 51 10 . (f) The center of the circle is the midpoint, 1 72 , and the length of the radius is 12 89 . Thus the equation of the circle 2 2 2 whose diameter is P Q is x 12 y 72 12 89 x 12 y 72 89 4 .
162
CHAPTER 1 Equations and Graphs
4. (a) x 2 y 2 25 52 has center 0 0 (b) x 22 y 12 9 32 has center 2 1 and radius 3.
and radius 5.
y
y
(c) x 2 6x y 2 2y 6 0
x 2 6x 9 y 2 2y 1 4
x 32 y 12 4 22 has
center 3 1 and radius 2. y 1 1
x
1 1
x 1 1
5. (a) x 4 y 2 . To test for symmetry about the x-axis, we replace y with y:
x
y
x 4 y2 x 4 y 2 , so the graph is symmetric about the x-axis.
To test for symmetry about the y-axis, we replace x with x:
x 4 y 2 is different from the original equation, so the graph is not
1
symmetric about the y-axis.
1
x
1
x
For symmetry about the origin, we replace x with x and y with y:
x 4 y2 x 4 y 2 , which is different from the original
equation, so the graph is not symmetric about the origin.
To find x-intercepts, we set y 0 and solve for x: x 4 02 4, so the x-intercept is 4.
To find y-intercepts, we set x 0 and solve for y:: 0 4 y 2 y 2 4 y 2, so the y-intercepts are 2 and 2.
(b) y x 2. To test for symmetry about the x-axis, we replace y with y:
y
y x 2 is different from the original equation, so the graph is not symmetric about the x-axis.
To test for symmetry about the y-axis, we replace x with x:
y x 2 x 2 is different from the original equation, so the
graph is not symmetric about the y-axis.
To test for symmetry about the origin, we replace x with x and y with
y: y x 2 y x 2, which is different from the original
equation, so the graph is not symmetric about the origin.
To find x-intercepts, we set y 0 and solve for x: 0 x 2 x 2 0 x 2, so the x-intercept is 2.
To find y-intercepts, we set x 0 and solve for y: y 0 2 2 2, so the y-intercept is 2.
1
CHAPTER 1
6. (a) To find the x-intercept, we set y 0 and solve for x: 3x 5 0 15
Test
163
y
(b)
3x 15 x 5, so the x-intercept is 5.
To find the y-intercept, we set x 0 and solve for y: 3 0 5y 15 5y 15 y 3, so the y-intercept is 3.
1
(c) 3x 5y 15 5y 3x 15 y 35 x 3.
x
1
(d) From part (c), the slope is 35 . (e) The slope of any line perpendicular to the given line is the negative 1 5. reciprocal of its slope, that is, 35 3
7. (a) 3x y 10 0 y 3x 10, so the slope of the line we seek is 3. Using the point-slope, y 6 3 x 3 y 6 3x 9 3x y 3 0. x y (b) Using the intercept form we get 1 2x 3y 12 2x 3y 12 0. 6 4 8. (a) When x 100 we have T 008 100 4 8 4 4, so the
(b)
T
temperature at one meter is 4 C.
(c) The slope represents an increase of 008 C for each one-centimeter
5
increase in depth, the x-intercept is the depth at which the temperature is 0 C, and the T -intercept is the temperature at ground level. 20
40
60
80
100 120 x
_5
9. (a) x 2 x 12 0 x 4 x 3 0. So x 4 or x 3. 4 16 8 4 8 4 2 2 2 2 4 42 4 2 1 2 . (b) 2x 4x 1 0 x 2 2 4 4 4 2 2 (c) 3 x 3 x 3 x x 3 3 x2 3 x x 2 6x 9 3 x
x 2 5x 6 x 2 x 3 0. Thus, x 2 and x 3 are potential solutions. Checking in the original equation, we see that only x 3 is valid.
(d) x 12 3x 14 2 0. Let u x 14 , then we have u 2 3u 2 0 u 2 u 1 0. So either u 2 0 or
u 1 0. If u 2 0, then u 2 x 14 2 x 24 16. If u 1 0, then u 1 x 14 1 x 1. So x 1 or x 16. (e) x 4 3x 2 2 0 x 2 1 x 2 2 0. So x 2 1 0 x 1 or x 2 2 0 x 2. Thus the solutions are x 1, x 1, x 2, and x 2.
10 10 10 2 (f) 3 x 4 10 0 3 x 4 10 x 4 10 3 x 4 3 x 4 3 . So x 4 3 3 or 22 2 22 x 4 10 3 3 . Thus the solutions are x 3 and x 3 .
10. (a) 3 2i 4 3i 3 4 2i 3i 7 i
(b) 3 2i 4 3i 3 4 2i 3i 1 5i
(c) 3 2i 4 3i 3 4 3 3i 2i 4 2i 3i 12 9i 8i 6i 2 12 i 6 1 18 i
3 2i 4 3i 12 17i 6i 2 6 17 12 17i 6 3 2i i 4 3i 4 3i 4 3i 16 9 25 25 16 9i 2 24 (e) i 48 i 2 124 1 (d)
164
CHAPTER 1 Equations and Graphs
(f)
2 2 2 8 2 2 8 2 2 2 8 2 4 2i 4i 2 6 2i
4 11. Using the Quadratic Formula, 2x 2 4x 3 0 x
42 4 2 3 4 8 1 22 i. 2 2 4
12. Let be the width of the parcel of land. Then 70 is the length of the parcel of land. Then 2 702 1302
2 2 140 4900 16,900 22 140 12,000 0 2 70 6000 0 50 120 0. So 50 or 120. Since 0, the width is 50 ft and the length is 70 120 ft.
13. (a) 4 5 3x 17 9 3x 12 3 x 4. Expressing in standard form we have: 4 x 3. Interval: [4 3. Graph:
_4
3
(b) x x 1 x 2 0. The expression on the left of the inequality changes sign when x 0, x 1, and x 2. Thus we must check the intervals in the following table. 2
2 0
0 1
1
Sign of x
Sign of x 1
Sign of x 2
Interval
Sign of x x 1 x 2
From the table, the solution set is x 2 x 0 or 1 x. Interval: 2 0 1 . Graph:
_2
0
1
(c) x 4 3 is equivalent to 3 x 4 3 1 x 7. Interval: 1 7. Graph: (d)
1
7
2x 3 2x 3 2x 3 x 1 x 4 1 1 0 0 0. The expression on the left of the inequality x 1 x 1 x 1 x 1 x 1 changes sign where x 4 and where x 1. Thus we must check the intervals in the following table. Interval
1
1 4
4
Sign of x 4
Sign of x 1 x 4 Sign of x 1 Since x 1 makes the expression in the inequality undefined, we exclude this value. Interval: 1 4]. Graph:
_1
4
14. 5 59 F 32 10 9 F 32 18 41 F 50. Thus the medicine is to be stored at a temperature between 41 F and 50 F.
Fitting Lines to Data
165
15. For 6x x 2 to be defined as a real number 6x x 2 0 x 6 x 0. The expression on the left of the inequality changes sign when x 0 and x 6. Thus we must check the intervals in the following table. 0
0 6
6
Sign of x
Sign of 6 x
Interval
Sign of x 6 x From the table, we see that 6x x 2 is defined when 0 x 6.
16. (a) x 3 9x 1 0. We graph the equation
y x 3 9x 1 in the viewing rectangle [5 5]
(b) x 2 1 x 1. We graph the equations
y1 x 2 1 and y2 x 1 in the viewing
by [10 10]. We find that the points of
rectangle [5 5] by [5 10]. We find that the
intersection occur at x 294, 011, 305.
points of intersection occur at x 1 and x 2. Since we want x 2 1 x 1, the solution is
10
the interval [1 2].
10 -4
-2
2
4 5
-10 -4
-2
2
4
-5
17. (a) M k
h 2 L
4 62
h 2 (b) Substituting 4, h 6, L 12, and M 4800, we have 4800 k k 400. Thus M 400 . 12 L 3 102 12,000. So the beam can support 12,000 pounds. (c) Now if L 10, 3, and h 10, then M 400 10
FOCUS ON MODELING Fitting Lines to Data 1. (a)
y
(b) Using a graphing calculator, we obtain the regression line y 18807x 8265.
180
(c) Using x 58 in the equation y 18807x 8265, we get y 18807 58 8265 1917 cm.
160
140 40
50
Femur length (cm)
x
166
FOCUS ON MODELING
2. (a)
y 800
(b) Using a graphing calculator, we obtain the regression line y 164163x 62183. (c) Using x 95 in the equation
600
y 164163x 62183, we get
y 164163 95 62183 938 cans. 400
50
60
70
90 x
80
High temperature (°F)
3. (a)
y 100
(b) Using a graphing calculator, we obtain the regression
80
(c) Using x 18 in the equation y 6451x 01523,
line y 6451x 01523. we get y 6451 18 01523 116 years.
60 40 20 0
2
4
6
8
10
12
14
16
18
20 x
Diameter (in.)
4. (a)
y 400
(b) Letting x 0 correspond to 1990, we obtain the regression line y 18446x 3522.
390
(c) Using x 21 in the equation y 18446x 3522,
380
we get y 18446 21 3522 3909 ppm CO2 ,
370
slightly lower than the measured value.
360 350 1990
1995
2000 2005 Year
2010
x
Fitting Lines to Data
5. (a)
y
167
(b) Using a graphing calculator, we obtain the regression line y 4857x 22097.
200
(c) Using x 100 F in the equation
y 4857x 22097, we get y 265 chirps per
minute. 100
0
50
60
70
80
90
x
Temperature (°F) y
6. (a)
(b) Using a graphing calculator, we obtain the regression line y 01275x 7929.
8
(c) Using x 30 in the regression line equation, we get
6
y 01275 30 7929 410 million km2 .
4 2 0
10
x
20
Years since 1986
7. (a)
y
(b) Using a graphing calculator, we obtain the regression line y 0168x 1989.
20
(c) Using the regression line equation y 0168x 1989, we get y 813% when
x 70%.
10
0
20
40
60
80
100 x
Flow rate (%) 8. (a)
y
(b) Using a graphing calculator, we obtain y 39018x 4197.
100
(c) The correlation coefficient is r 098, so linear model is appropriate for x between 80 dB and 104 dB.
50
(d) Substituting x 94 into the regression equation, we get y 39018 94 4197 53. So the intelligibility is about 53%.
0
80
90
100
Noise level (dB)
110 x
168
FOCUS ON MODELING
9. (a)
y 80
(b) Using a graphing calculator, we obtain y 027083x 4629. (c) We substitute x 2006 in the model y 027083x 4629 to get y 804, that is, a life
70
expectancy of 804 years.
(d) The life expectancy of a child born in the US in 2006
60
was 777 years, considerably less than our estimate in part (b). 1920
1940
1960
1980
2000
x
Year 10. (a)
y
(c) Year
x
Height (m)
1972
0
564
1976
4
564
1980
8
578
1984
12
575
1988
16
590
1992
20
587
1996
24
592
2000
28
590
2004
32
595
2008
36
596
(b) Using a graphing calculator, we obtain the regression line y 5664 000929x.
6.0
5.9 5.8 5.7 5.6 0
20
30
x
Years since 1972
The regression line provides a good model. (d) The regression line predicts the winning pole vault height in 2012 to be y 000929 2012 1972 5664 604 meters.
11. Students should find a fairly strong correlation between shoe size and height. 12. Results will depend on student surveys in each class.
10
2
FUNCTIONS
2.1
FUNCTIONS
1. If f x x 3 1, then (a) the value of f at x 1 is f 1 13 1 0. (b) the value of f at x 2 is f 2 23 1 9.
(c) the net change in the value of f between x 1 and x 2 is f 2 f 1 9 0 9.
2. For a function f , the set of all possible inputs is called the domain of f , and the set of all possible outputs is called the range of f . x 5 3. (a) f x x 2 3x and g x have 5 in their domain because they are defined when x 5. However, x h x x 10 is undefined when x 5 because 5 10 5, so 5 is not in the domain of h. 0 55 (b) f 5 52 3 5 25 15 10 and g 5 0. 5 5 4. (a) Verbal: “Subtract 4, then square and add 3.” (b) Numerical: x
f x
0
19
2
7
4
3
6
7
5. A function f is a rule that assigns to each element x in a set A exactly one element called f x in a set B. Table (i) defines y as a function of x, but table (ii) does not, because f 1 is not uniquely defined. 6. (a) Yes, it is possible that f 1 f 2 5. [For instance, let f x 5 for all x.]
(b) No, it is not possible to have f 1 5 and f 1 6. A function assigns each value of x in its domain exactly one value of f x.
7. Multiplying x by 3 gives 3x, then subtracting 5 gives f x 3x 5. 8. Squaring x gives x 2 , then adding two gives f x x 2 2. 9. Subtracting 1 gives x 1, then squaring gives f x x 12 . 10. Adding 1 gives x 1, taking the square root gives
x 1, then dividing by 6 gives f x
x 1 . 6
11. f x 2x 3: Multiply by 2, then add 3.
12. g x
x 2 : Add 2, then divide by 3. 3
13. h x 5 x 1: Add 1, then multiply by 5.
14. k x
x2 4 : Square, then subtract 4, then divide by 3. 3 169
170
CHAPTER 2 Functions
15. Machine diagram for f x
16. Machine diagram for f x
x 1.
1
subtract 1, then take square root
0
2
subtract 1, then take square root
1
5
subtract 1, then take square root
2
17. f x 2 x 12
3 . x 2
3
subtract 2, take reciprocal, multiply by 3
3
_1
subtract 2, take reciprocal, multiply by 3
_1
1
subtract 2, take reciprocal, multiply by 3
_3
18. g x 2x 3
x
f x
x
g x
1
2 1 12 8
3
2 3 3 3
0 1 2 3
2 12 2
2
2 1 12 0
0
2 2 12 2
1
2 3 12 8
3
2 2 3 1
2 0 3 3 2 1 3 5 2 3 3 9
19. f x x 2 6; f 3 32 6 9 6 3; f 3 32 6 9 6 3; f 0 02 6 6; 2 f 12 12 6 14 6 23 4. 20. f x x 3 2x; f 2 23 2 2 8 4 12; f 1 13 2 1 1 2 3; 3 f 0 03 2 0 0; f 12 12 2 12 18 1 98 . 1 2 2 1 2 2 5 1 2x ; f 2 1; f 2 ; f 3 3 3 3 1 2a 1 2 a 1 3 2a 1 2 a ; f a 1 . f a 3 3 3 3
21. f x
1 2
1 2 12 3
0; f a
1 2a ; 3
22 4 8 8 a2 4 x2 4 x2 4 22 4 x2 4 ; h 2 ; h 2 ; h a ; h x ; 5 5 5 5 5 5 5 5 2 x 4 a 2 4a 8 x 4 a 22 4 ;h x . h a 2 5 5 5 5
22. h x
23. f x x 2 2x; f 0 02 2 0 0; f 3 32 2 3 9 6 15; f 3 32 2 3 9 6 3; 2 1 1 2 1 1 2 f a a 2 2 a a 2 2a; f x x2 2 x x 2 2x; f 2 . a a a a a 1 1 1 1 2; h 2 2 1 5 ; h 1 1 1 1 2 5 ; ; h 1 1 1 2 2 2 1 x 2 2 2 2 1 1 1 1 1 ;h h x 1 x 1 x. 1 x 1 x x x x
24. h x x
SECTION 2.1 Functions
171
1 1 1 1 2 1 1 1 1 1x 1 1 2 2 ; ; g 2 ; g 1 , which is undefined; g 3 1 1x 1 2 3 3 1 1 2 3 1 2 2 1 x2 1 2 x2 1 a 1a 1 a 1 1a1 2a 2 g a ; g a 1 ;g x 1 . 1 a 1a 1 a 1 1a1 a 1 x2 1 x2
25. g x
2 2 22 02 a2 t 2 ; g 2 0; g 2 , which is undefined; g 0 1; g a ; t 2 2 2 22 02 a2 a2 2 2 a3 a2 a12 . 2 ; g a 1 g a2 2 2 a12 a1 a 22 a 4
26. g t
27. k x x 2 2x 3; k 0 02 2 0 3 3; k 2 22 2 2 3 5; k 2 22 2 2 3 3; 2 2 2 2 2 3 1 2 2; k a 2 a 22 2 a 2 3 a 2 6a 5; k 2 k x x2 2 x 3 x 2 2x 3; k x 2 x 2 2 x 2 3 x 4 2x 2 3. 28. k x 2x 3 3x 2 ; k 0 2 03 3 02 0; k 3 2 33 3 32 27; k 3 2 33 3 32 81; 3 2 3 2 a 3 3a 2 ; k x 2 x3 3 x2 2x 3 3x 2 ; k 12 2 12 3 12 12 ; k a2 2 a2 3 a2 4 3 2 k x 3 2 x 3 3 x 3 2x 9 3x 6 .
29. f x 2 x 1; f 2 2 2 1 2 3 6; f 0 2 0 1 2 1 2; f 12 2 12 1 2 12 1; f 2 2 2 1 2 1 2; f x 1 2 x 1 1 2 x; f x 2 2 2 x 2 2 1 2 x 2 1 2x 2 2 (since x 2 1 0 ).
2 1 x 2 1 ; f 2 1; f 1 1; f x is not defined at x 0; x 2 2 1 1 x 2 1x 5 5 x x2 1 . 1; f x 2 2 2 1 since x 2 0, x 0; f f 5 x 5 5 x 1x x x
30. f x
31. Since 2 0, we have f 2 22 4. Since 1 0, we have f 1 12 1. Since 0 0, we have f 0 0 1 1. Since 1 0, we have f 1 1 1 2. Since 2 0, we have f 2 2 1 3.
32. Since 3 2, we have f 3 5. Since 0 2, we have f 0 5. Since 2 2, we have f 2 5. Since 3 2, we have f 3 2 3 3 3. Since 5 2, we have f 5 2 5 3 7.
33. Since 4 1, we have f 4 42 2 4 16 8 8. Since 32 1, we have 2 f 32 32 2 32 94 3 34 . Since 1 1, we have f 1 12 2 1 1 2 1. Since 1 0 1, we have f 0 0. Since 25 1, we have f 25 1.
34. Since 5 0, we have f 5 3 5 15. Since 0 0 2, we have f 0 0 1 1. Since 0 1 2, we have f 1 1 1 2. Since 0 2 2, we have f 2 2 1 3. Since 5 2, we have f 5 5 22 9.
35. f x 2 x 22 1 x 2 4x 4 1 x 2 4x 5; f x f 2 x 2 1 22 1 x 2 1 4 1 x 2 6.
36. f 2x 3 2x 1 6x 1; 2 f x 2 3x 1 6x 2. 2 37. f x 2 x 2 4; f x [x 4]2 x 2 8x 16. x x f x 6x 18 3 2x 6 38. f 6 18 2x 18; 2x 6 3 3 3 3 3 39. f x 3x 2, so f 1 3 1 2 1 and f 5 3 5 2 13. Thus, the net change is f 5 f 1 13 1 12. 40. f x 4 5x, so f 3 4 5 3 11 and f 5 4 5 5 21. Thus, the net change is f 5 f 3 21 11 10.
172
CHAPTER 2 Functions
41. g t 1 t 2 , so g 2 1 22 1 4 3 and g 5 1 52 24. Thus, the net change is g 5 g 2 24 3 21.
42. h t t 2 5, so h 3 32 5 14 and h 6 62 5 41. Thus, the net change is h 6h 3 4114 27.
43. f a 5 2a; f a h 5 2 a h 5 2a 2h; 5 2a 2h 5 2a 5 2a 2h 5 2a 2h f a h f a 2. h h h h 44. f a 3a 2 2; f a h 3 a h2 2 3a 2 6ah 3h 2 2; 2 6ah 3h 2 2 3a 2 2 3a f a h f a 6ah 3h 2 6a 3h h h h f a h f a 55 45. f a 5; f a h 5; 0. h h 1 1 46. f a ; f a h ; a1 ah1 a1 ah1 1 1 f a h f a 1 h 1 1 a h 1 a a a ah1 a1 h h h h 1 a 1 a h 1 . h a 1 a h 1 a ah 47. f a ; f a h ; a1 ah1 a a h 1 a h a 1 a ah f a h f a a h 1 a 1 a h 1 a 1 a h 1 a 1 h h h a h a 1 a a h 1 2 a ah h a 2 ah a a a h 1 a 1 h h a h 1 a 1 1 a h 1 a 1 2 a h 2a ; f a h ; 48. f a a1 ah1 2a a h 1 2a 2h a 1 2a 2 a h f a h f a a h 1 a 1 a h 1 a 1 a h 1 a 1 h h h 2 a h a 1 2a a h 1 2a 2 2ah 2a 2h 2a 2 2ah 2a a h 1 a 1 h h a h 1 a 1 2h 2 h a h 1 a 1 a h 1 a 1 49. f a 3 5a 4a 2 ;
f a h 3 5 a h 4 a h2 3 5a 5h 4 a 2 2ah h 2
3 5a 5h 4a 2 8ah 4h 2 ; 2 8ah 4h 2 3 5a 4a 2 3 5a 5h 4a f a h f a h h 3 5a 5h 4a 2 8ah 4h 2 3 5a 4a 2 5h 8ah 4h 2 h h h 5 8a 4h 5 8a 4h. h
SECTION 2.1 Functions
173
50. f a a 3 ; f a h a h3 a 3 3a 2 h 3ah 2 h 3 ; a 3 3a 2 h 3ah 2 h 3 a 3 f a h f a 3a 2 h 3ah 2 h 3 h h h 2 2 h 3a 3ah h 3a 2 3ah h 2 . h 51. f x 3x. Since there is no restriction, the domain is all real numbers, . Since every real number y is three times the real number 13 y, the range is all real numbers .
52. f x 5x 2 4. Since there is no restriction, the domain is all real numbers, . Since 5x 2 0 for all x, 5x 2 4 4 for all x, so the range is [4 .
53. f x 3x, 2 x 6. The domain is [2 6], f 2 3 2 6, and f 6 3 6 18, so the range is [6 18]. 54. f x 5x 2 4, 0 x 2. The domain is [0 2], f 0 5 02 4 4, and f 2 5 22 4 24, so the range is [4 24]. 1 . Since the denominator cannot equal 0 we have x 3 0 x 3. Thus the domain is x x 3. In x 3 interval notation, the domain is 3 3 .
55. f x
1 . Since the denominator cannot equal 0, we have 3x 6 0 3x 6 x 2. In interval notation, the 3x 6 domain is 2 2 .
56. f x
x 2 57. f x 2 . Since the denominator cannot equal 0 we have x 2 1 0 x 2 1 x 1. Thus the domain is x 1 x x 1. In interval notation, the domain is 1 1 1 1 .
x4 . Since the denominator cannot equal 0, x 2 x 6 0 x 3 x 2 0 x 3 or x 2. 58. f x 2 x x 6 In interval notation, the domain is 3 3 2 2 . 59. f x x 1. We must have x 1 0 x 1. Thus, the domain is [1 . 60. g x x 2 9. The argument of the square root is positive for all x, so the domain is . 61. f t 3 t 1. Since the odd root is defined for all real numbers, the domain is the set of real numbers, . 62. g x 7 3x. For the square root to be defined, we must have 7 3x 0 7 3x 73 x. Thus the domain is 73 . 63. f x 1 2x. Since the square root is defined as a real number only for nonnegative numbers, we require that 1 2x 0 x 12 . So the domain is x x 12 . In interval notation, the domain is 12 . 64. g x x 2 4. We must have x 2 4 0 x 2 x 2 0. We make a table: 2
2 2
2
Sign of x 2
Sign of x 2
Sign of x 2 x 2
Thus the domain is 2] [2 . 2x . We require 2 x 0, and the denominator cannot equal 0. Now 2 x 0 x 2, and 3 x 0 65. g x 3x x 3. Thus the domain is x x 2 and x 3, which can be expressed in interval notation as [2 3 3 .
174
CHAPTER 2 Functions
x . We must have x 0 for the numerator and 2x 2 x 1 0 for the denominator. So 2x 2 x 1 0 2x 2 x 1 2x 1 x 1 0 2x 1 0 or x 1 0 x 12 or x 1. Thus the domain is 0 12 12 .
66. g x
67. g x a table:
4
x 2 6x. Since the input to an even root must be nonnegative, we have x 2 6x 0 x x 6 0. We make 0
0 6
6
Sign of x
Sign of x 6
Sign of x x 6
Thus the domain is 0] [6 . 68. g x
x 2 2x 8. We must have x 2 2x 8 0 x 4 x 2 0. We make a table: 2
2 4
4
Sign of x 4
Sign of x 2
Sign of x 4 x 2
Thus the domain is 2] [4 .
3 . Since the input to an even root must be nonnegative and the denominator cannot equal 0, we have x 4 x 4 0 x 4. Thus the domain is 4 .
69. f x
x2 70. f x . Since the input to an even root must be nonnegative and the denominator cannot equal 0, we have 6x 6 x 0 6 x. Thus the domain is 6. x 12 71. f x . Since the input to an even root must be nonnegative and the denominator cannot equal 0, we have 2x 1 2x 1 0 x 12 . Thus the domain is 12 . x 72. f x . Since the input to an even root must be nonnegative and the denominator cannot equal 0, we have 4 9 x2 9 x 2 0 3 x 3 x 0. We make a table:
3
3 3
3
Sign of 3 x
Sign of 3 x
Interval
Thus the domain is 3 3.
Sign of x 4 x 2
SECTION 2.1 Functions
73. To evaluate f x, divide the input by 3 and add 23 to the result. 2 x (a) f x 3 3
175
y
(c)
(b) x
f x
2
4 3
4
2
6
8 3 10 3
8
1l 1
x
1
x
74. To evaluate g x, subtract 4 from the input and multiply the result by 34 . (a) g x x 4 34 34 x 4
y
(c)
(b) x
g x
2
32
4
0
6
3 2
8
3
1l
75. Let T x be the amount of sales tax charged in Lemon County on a purchase of x dollars. To find the tax, take 8% of the purchase price. (a) T x 008x
(c)
y
(b) x
T x
2
016
4
032
6
048
8
064
1l
1
x
76. Let V d be the volume of a sphere of diameter d. To find the volume, take the cube of the diameter, then multiply by and divide by 6. 3 (a) V d d 3 6 6d
(c)
y
(b) x
f x
2
4 42 3 32 335 3
4 6 8
36 113
256 268 3
10l 1
x
176
CHAPTER 2 Functions
1 if x is rational 77. f x 5 if x is irrational
The domain of f is all real numbers, since every real number is either rational or
irrational; and the range of f is 1 5. 1 if x is rational 78. f x The domain of f is all real numbers, since every real number is either rational or 5x if x is irrational
irrational. If x is irrational, then 5x is also irrational, and so the range of f is x x 1 or x is irrational. 0 2 50 and V 20 50 1 20 2 0. (c) 79. (a) V 0 50 1 20 20 x V x (b) V 0 50 represents the volume of the full tank at time t 0, and 0 50 V 20 0 represents the volume of the empty tank twenty minutes 5 28125 later. 10 125 (d) The net change in V as t changes from 0 minutes to 20 minutes is 15 3125 V 20 V 0 0 50 50 gallons. 20 0 80. (a) S 2 4 22 16 5027, S 3 4 32 36 11310.
(b) S 2 represents the surface area of a sphere of radius 2, and S 3 represents the surface area of a sphere of radius 3. 05c2 075c2 81. (a) L 05c 10 1 866 m, L 10 1 661 m, and 075c c2 c2 09c2 L 09c 10 1 436 m. c2 (b) It will appear to get shorter. 13 7 104 20 2 mm, 82. (a) R 1 (b) 5 1 4 104 x R x 04 1 2 13 7 10 R 10 166 mm, and 04 1 4 10 10 166 100 148 13 7 10004 148 mm. R 100 04 200 144 1 4 100 (c) The net change in R as x changes from 10 to 100 is R 100 R 10 148 166 018 mm. 83. (a) 01 18500 025 012 4440, 04 18500 025 042 1665.
(b) They tell us that the blood flows much faster (about 275 times faster) 01 cm from the center than 01 cm from the edge.
(d) The net change in V as r changes from 01 cm to 05 cm is V 05 V 01 0 4440 4440 cms. 84. (a) D 01 2 3960 01 012 79201 281 miles D 02 2 3960 02 022 158404 398 miles
500
141
1000
139
(c) r
r
0
4625
01
4440
02
3885
03
2960
04
1665
05
0
SECTION 2.1 Functions
(b) 1135 feet 1135 miles 0215 miles. D 2 3960 0215 02152 1702846 413 miles 0215 5280 (c) D 7 2 3960 7 72 55489 2356 miles
177
(d) The net change in D as h changes from 1135 ft (or 0215 mi) to 7 mi is D 7 D 0215 2356413 1943 miles.
85. (a) Since 0 5,000 10,000 we have T 5,000 0. Since 10,000 12,000 20,000 we have T 12,000 008 12,000 960. Since 20,000 25,000 we have T 25,000 1600 015 25,000 5350. (b) There is no tax on $5000, a tax of $960 on $12,000 income, and a tax of $5350 on $25,000.
86. (a) C 75 75 15 $90; C 90 90 15 $105; C 100 $100; and C 105 $105. (b) The total price of the books purchased, including shipping.
75x if 0 x 2 87. (a) T x 150 50 x 2 if x 2
(b) T 2 75 2 150; T 3 150 50 3 2 200; and T 5 150 50 5 2 300. (c) The total cost of the lodgings.
15 40 x if 0 x 40 88. (a) F x 0 if 40 x 65 15 x 65 if x 65
(b) F 30 15 40 10 15 10 $150; F 50 $0; and F 75 15 75 65 15 10 $150. (c) The fines for violating the speed limits on the freeway.
89. We assume the grass grows linearly. h
0
90.
T
0
91.
t
T 60
50 0
5
10
t
W
W
W
W
t
178
CHAPTER 2 Functions P
92.
1000 900 Population (×1000) 800 700 600 0
1980
1990
2000
2010
t
Year
93. Answers will vary. 94. Answers will vary. 95. Answers will vary.
2.2
GRAPHS OF FUNCTIONS
1. To graph the function f we plot the points x f x in a coordinate plane. To graph f x x 2 2, we plot the points x x 2 2 . So, the point 3 32 2 3 7 is
on the graph of f . The height of the graph of f above the
x
f x
x y
2
2
2 2
1
1
0
2
1
1
2
2
x-axis when x 3 is 7.
y 1
1 1
0
0 2
1
x
1 1 2 2
2. If f 4 10 then the point 4 10 is on the graph of f . 3. If the point 3 7 is on the graph of f , then f 3 7. 4. (a) f x x 2 is a power function with an even exponent. It has graph IV. (b) f x x 3 is a power function with an odd exponent. It has graph II. (c) f x x is a root function. It has graph I. (d) f x x is an absolute value function. It has graph III.
5.
6.
y
x
f x x 2
6
4
2
8
4
2
1
6
2
0
0
4
0
2
1
2
2
4
2
0
4
6
3
2
6
8
4
4
x
1 1
x
f x 4 2x
y
1 1
x
SECTION 2.2 Graphs of Functions
7. x
f x x 3, 3 x 3
3
6
2
5
0
3
2
1
3
0
x
f x x 2
4
16
3
9
2
4
1
1
0
0
11.
x
1
2
x
1
g x x 12
5
16
3
4
2 1 0
0
15
1
1
2
05
3
0
4
05
5
1
10.
y
1
x
x 3 , 2 0x 5
f x
x
1
1
9.
8.
y
y
f x x 2 4
5
21
4
12
3
5
2
0
1
3
0
4
y
16
3
4
1
2
1
0
1
0
1
0
1
1
4
1
4
3
16
3
16
13. x
r x 3x 4
3
_5
y
x
r x 1 x 4
243
3
80
2
48
2
15
1
3
1
0
0
0
0
1
1
3
1
0
2
48
2
15
3
243
3
80
0
1
x
10
_1 0
x
1
y
14.
100
x
1
5
x
1
y
1
g x x 2 2x 1
0
x
1
x
_1
y
1
x
12.
5
179
10 _1
0 _10
1
x
180
CHAPTER 2 Functions
15. x
g x x 3 8
3
16.
y
x
g x x 13
35
2
27
2
16
1
8
1
9
0
1
0
8
1
0
1
7
2
1
2
0
3
8
3
19
4
27
5
17. k x
x
y
3 x
27
3
8
2
1
1
0
18.
3
8
2
1
1
0
0
0
1
1
1
1
8
2
8
2
27
3
27
3
f x 1
1 0 _1
x
x
10
20.
y
f x
x
10 0
1 0 _1
y
1
2
0
1
2
3
1
4
3
6
2
1
9
4
11
3
0 2
16
5
18
4
27
5
38
6
25
10
6
21. x
1 C t 2 t
2
1 4
1
12 14
0
x
20
y
22. x 3
1 4 16
2
1 _1 0
1
t
C t 1 2
1
32
2
1
12
2
0
1
1 4 1 2
16 4
1
1
1
2
2
1 4
12 13
x
10
x 2
0
1
x
1
y
k x 3 x
x 27
19. x
x
1
y
1 t 1
x
10
y
2 0
2
t
SECTION 2.2 Graphs of Functions
23. x
H x 2x
5
10
4
8
3
6
2
4
1 0
24.
y
x
H x x 1
5
4
4
3
3
2
2
1
2
1
0
0
0
1
1
2
1 x
1
25. x
G x x x
5
0
2
0
0
0
1 1
x
G x x x
5
10
2
4
1
2
0
0
2
2
4
1
0
5
10
3
0
x
f x
x
f x 2x 2
5
12
2
8
0
2
1
0
2
2
5
8
28.
y
1 1
x
3
1
2
1
1
1
0
29. f x 8x x 2
(a) [5 5] by [5 5]
1
x
1
27.
y
26.
y
1
x
1
x
y
1
y
x x
1 x
1
undefined
1
1
2
1
3
1
(b) [10 10] by [10 10] 10
4 2 -4
-2
-2 -4
2
4
181
-10
-5
5 -10
10
182
CHAPTER 2 Functions
(c) [2 10] by [5 20]
(d) [10 10] by [100 100]
20
100
10 -10 -2
2
4
6
8
-5
10
5
10
5
10
5
10
-100
The viewing rectangle in part (c) produces the most appropriate graph of the equation. 30. g x x 2 x 20
(b) [10 10] by [10 10]
(a) [2 2] by [5 5]
10
4 2 -2
-1
1
-2
2
-10
-5
-4
-10
(c) [7 7] by [25 20]
(d) [10 10] by [100 100] 20
100
10 -6 -4 -2 -10
2
4
6
-10
-5
-20
-100
The viewing rectangle in part (c) produces the most appropriate graph of the equation. 31. h x x 3 5x 4
(b) [3 3] by [10 10]
(a) [2 2] by [2 2] 2
10
1 -2
-1
1
-1
2
-3
-2
-1
-2
3
(d) [10 10] by [10 10] 5
-2
2
-10
(c) [3 3] by [10 5]
-3
1
-1 -5
10
1
2
3 -10
-10
The viewing rectangle in part (c) produces the most appropriate graph of the equation.
-5
5 -10
10
SECTION 2.2 Graphs of Functions 1 x4 x2 2 32. k x 32
(b) [2 2] by [2 2]
(a) [1 1] by [1 1]
1
2 1
-1.0
-0.5
0.5
1.0
-2
-1
-1
-1
1
2
5
10
-2
(c) [5 5] by [5 5]
(d) [10 10] by [10 10] 10
4 2 -4
-2
2
-2
4
-10
-5
-4
-10
The viewing rectangle in part (d) produces the most appropriate graph of the equation. 0 if x 2 33. f x 1 if x 2
1 if x 1 34. f x x 1 if x 1
y
y
1 2 1
x
3 if x 2 35. f x x 1 if x 2
1
x
1
x
1 x if x 2 36. f x 5 if x 2
y
y
1
1 1
x
183
184
CHAPTER 2 Functions
x if x 0 37. f x x 1 if x 0
2x 3 if x 1 38. f x 3 x if x 1
y
y
1
1 x
1
1 if x 1 39. f x 1 if 1 x 1 1 if x 1
1
1 if x 1 40. f x x if 1 x 1 1 if x 1
y
y
2 2
2 if x 1 41. f x x 2 if x 1
1 x
1
x
1
x
1
x
1 x 2 if x 2 42. f x x if x 2 y
y
1
1 1
0 if x 2 43. f x 3 if x 2
x
x
x 2 if x 1 44. f x 1 if x 1
y
1
y
1 1
x
SECTION 2.2 Graphs of Functions
if x 2 4 45. f x x 2 if 2 x 2 x 6 if x 2
185
if x 0 x 46. f x 9 x 2 if 0 x 3 x 3 if x 3
y
y
1 x
1
1 x
1
x 2 if x 1 47. f x x2 if x 1
5
-6 -4 -2
2
4
6
-5
2x x 2 if x 1 48. f x x 13 if x 1
The first graph shows the output of a typical graphing device. However, the actual graph
of this function is also shown, and its difference from the graphing device’s version should be noted. y 1
2
-3 -2 -1 -2
2 if x 2 49. f x x if 2 x 2 2 if x 2
1
1
2
x
3
if x 1 1 50. f x 1 x if 1 x 2 2 if x 2
51. The curves in parts (a) and (c) are graphs of a function of x, by the Vertical Line Test. 52. The curves in parts (b) and (c) are graphs of functions of x, by the Vertical Line Test.
53. The given curve is the graph of a function of x, by the Vertical Line Test. Domain: [3 2]. Range: [2 2]. 54. No, the given curve is not the graph of a function of x, by the Vertical Line Test. 55. No, the given curve is not the graph of a function of x, by the Vertical Line Test. 56. The given curve is the graph of a function of x, by the Vertical Line Test. Domain: [3 2]. Range: 2 0 3].
57. Solving for y in terms of x gives 3x 5y 7 y 35 x 75 . This defines y as a function of x.
58. Solving for y in terms of x gives 3x 2 y 5 y 3x 2 5. This defines y as a function of x. 59. Solving for y in terms of x gives x y 2 y x. The last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x.
186
CHAPTER 2 Functions
60. Solving for y in terms of x gives x 2 y 12 4 y 12 4 x 2 y 1 4 x 2 y 1 4 x 2 . The last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x. 61. Solving for y in terms of x gives 2x 4y 2 3 4y 2 2x 3 y 12 2x 3. The last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x. 62. Solving for y in terms of x gives 2x 2 4y 2 3 4y 2 2x 2 3 y 12 2x 2 3. The last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x. 63. Solving for y in terms of x using the Quadratic Formula gives 2x y 5y 2 4 5y 2 2x y 4 0 2x 2x2 4 5 4 2x 4x 2 80 x x 2 20 . The last equation gives two values of y for a y 2 5 10 5 given value of x. Thus, this equation does not define y as a function of x. 64. Solving for y in terms of x gives y 5 x y x 52 . This defines y as a function of x. 65. Solving for y in terms of x gives 2 x y 0 y 2 x. This defines y as a function of x.
66. Solving for y in terms of x gives 2x y 0 y 2x. Since a a, the last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x. 67. Solving for y in terms of x gives x y 3 y 3 x. This defines y as a function of x. 68. Solving for y in terms of x gives x y 4 y 4 x. The last equation gives two values of y for any positive value of x. Thus, this equation does not define y as a function of x.
69. (a) f x x 2 c, for c 0, 2, 4, and 6. c=4
10 8
c=2
(b) f x x 2 c, for c 0, 2, 4 , and 6. c=_2
c=6
6
4 2
c=_6
4
c=0
0 -2
-2
c=0
8
6
-4
c=_4
10
2
2
-4
4
0 -2
-2
-4
-4
-6
-6
-8
-8
-10
-10
2
4
(c) The graphs in part (a) are obtained by shifting the graph of f x x 2 upward c units, c 0. The graphs in part (b) are obtained by shifting the graph of f x x 2 downward c units.
70. (a) f x x c2 , for c 0, 1, 2, and 3.
(b) f x x c2 , for c 0, 1, 2, and 3.
c=0 c=1
10 8
c=_1 c=0
c=2
c=_2
c=3
c=_3
8
6
6
4 2 -4
-2
0 -2
10
4 2
2
4
-4
-2
0 -2
-4
-4
-6
-6
-8
-8
-10
-10
2
4
(c) The graphs in part (a) are obtained by shifting the graph of y x 2 to the right 1, 2, and 3 units, while the graphs in part (b) are obtained by shifting the graph of y x 2 to the left 1, 2, and 3 units.
SECTION 2.2 Graphs of Functions
71. (a) f x x c3 , for c 0, 2, 4, and 6. c=2
10
-8
-4
(b) f x x c3 , for c 0, 2, 4, and 6.
c=6
c=_4
8
8
6
6
4
4
2
2
-2
0 -2
4
8
-8
-4
-4
-4
-6
-6
-8
c=0
187
c=0
4
8
-8
c=4
c=_6
-10
c=_2
(c) The graphs in part (a) are obtained by shifting the graph of f x x 3 to the right c units, c 0. The graphs in part (b) are obtained by shifting the graph of f x x 3 to the left c units, c 0.
72. (a) f x cx 2 , for c 1, 12 , 2, and 4. c=1
c=4
(b) f x cx 2 , for c 1, 1, 12 , and 2.
c=2
10
10
1 c=_ 2
6 4
8
c=1
8
6 4 2
2 -4
-4
0 -2
-2
2
0 -2
-2
4
2
4 1
-4
-4
c=__ 2
-6
-6
-8
-8
-10
c=_4
-10
c=_2
(c) As c increases, the graph of f x cx 2 is stretched vertically. As c decreases, the graph of f is flattened. When c 0, the graph is reflected about the x-axis. 73. (a) f x x c , for c 12 , 14 , and 16 .
(b) f x x c , for c 1, 13 , and 15 .
3
2
1
1
1
c= _ 5
1 6
1
c= _
0
1
2
3
-1
4
-1
-2
(c) Graphs of even roots are similar to y x, graphs of odd roots are similar to y y c x becomes steeper near x 0 and flatter when x 1. 1 74. (a) f x n , for n 1 and 3. x
3
x. As c increases, the graph of
1 (b) f x n , for n 2 and 4. x 3
3 2
n=3
-2
-1
0
2
n=1
1
-3
1
c= _ 3 c= _ 4
-1
c=1
2 1 c= _ 2
1
2
n=4
1
3
-3
-2
-1
0
-1
-1
-2
-2
-3
-3
n=2 1
2
3
188
CHAPTER 2 Functions
(c) As n increases, the graphs of y 1x n go to zero faster for x large. Also, as n increases and x goes to 0, the graphs of y 1x n go to infinity faster. The graphs of y 1x n for n odd are similar to each other. Likewise, the graphs for n even are similar to each other.
75. The slope of the line segment joining the points 2 1 and 4 6 is m
6 1 76 . Using the point-slope form, 4 2
we have y 1 76 x 2 y 76 x 73 1 y 76 x 43 . Thus the function is f x 76 x 43 for 2 x 4. 76. The slope of the line containing the points 3 2 and 6 3 is m
2 3 5 59 . Using the point-slope equation 3 6 9
5 1 5 1 of the line, we have y 3 59 x 6 y 59 x 10 3 3 9 x 3 . Thus the function is f x 9 x 3 , for 3 x 6.
77. First solve the circle for y: x 2 y 2 9 y 2 9 x 2 y 9 x 2 . Since we seek the top half of the circle, we choose y 9 x 2 . So the function is f x 9 x 2 , 3 x 3.
78. First solve the circle for y: x 2 y 2 9 y 2 9 x 2 y 9 x 2 . Since we seek the bottom half of the circle, we choose y 9 x 2 . So the function is f x 9 x 2 , 3 x 3. 05 for 10 r 100. As the balloon r2 is inflated, the skin gets thinner, as we would expect.
79. We graph T r
80. We graph P 141 3 for 1 10. As wind speed increases, so does power output, as expected. P
T
20,000
0.004 0.003 10,000
0.002 0.001 0
50
r
600 010x if 0 x 300 81. (a) E x 3600 006 x 300 if 300 x
0
(b)
5
10
E 60 50 40 30 20 10 0
100
200
300
400
500
600
x
SECTION 2.2 Graphs of Functions
200 220 240 82. C x 400
189
C
4.00
if 0 x 1
if 1 x 11
if 11 x 12
3.00 2.00
if 19 x 2 0
1
2
x
P
049 070 83. P x 091 112
1.00
if 0 x 1
0.80
if 1 x 2 if 2 x 3
0.60
if 3 x 35
0.40 0.20 0
1
2
3
4
x
84. The graph of x y 2 is not the graph of a function because both 1 1 and 1 1 satisfy the equation x y 2 . The graph of x y 3 is the graph of a function because x y 3 x 13 y. If n is even, then both 1 1 and 1 1 satisfies the equation x y n , so the graph of x y n is not the graph of a function. When n is odd, y x 1n is defined for all real
numbers, and since y x 1n x y n , the graph of x y n is the graph of a function.
85. Answers will vary. Some examples are almost anything we purchase based on weight, volume, length, or time, for example gasoline. Although the amount delivered by the pump is continuous, the amount we pay is rounded to the penny. An example involving time would be the cost of a telephone call.
86.
y
y
1
y
1 1
f x [[x]]
x
1 1
x
g x [[2x]] 1 The graph of k x [[nx]] is a step function whose steps are each wide. n
1
h x [[3x]]
x
190
CHAPTER 2 Functions
87. (a) The graphs of f x x 2 x 6 and g x x 2 x 6 are shown in the viewing rectangle [10 10] by [10 10].
10
-10
10
-5
5
10
-10
-5
5
-10
10
-10
For those values of x where f x 0 , the graphs of f and g coincide, and for those values of x where f x 0, the graph of g is obtained from that of f by reflecting the part below the x-axis about the x-axis. (b) The graphs of f x x 4 6x 2 and g x x 4 6x 2 are shown in the viewing rectangle [5 5] by [10 15]. 10
-4
-2
10
2
4
-4
-10
-2
2
4
-10
For those values of x where f x 0 , the graphs of f and g coincide, and for those values of x where f x 0, the graph of g is obtained from that of f by reflecting the part below the x-axis above the x-axis. (c) In general, if g x f x, then for those values of x where f x 0, the graphs of f and g coincide, and for those values of x where f x 0, the graph of g is obtained from that of f by reflecting the part below the x-axis above the x-axis. y
y
x
y f x
2.3
x
y g x
GETTING INFORMATION FROM THE GRAPH OF A FUNCTION
1. To find a function value f a from the graph of f we find the height of the graph above the x-axis at x a. From the graph of f we see that f 3 4 and f 1 0. The net change in f between x 1 and x 3 is f 3 f 1 4 0 4. 2. The domain of the function f is all the x-values of the points on the graph, and the range is all the corresponding y-values. From the graph of f we see that the domain of f is the interval and the range of f is the interval 7]. 3. (a) If f is increasing on an interval, then the y-values of the points on the graph rise as the x-values increase. From the graph of f we see that f is increasing on the intervals 2 and 4 5. (b) If f is decreasing on an interval, then y-values of the points on the graph fall as the x-values increase. From the graph of f we see that f is decreasing on the intervals 2 4 and 5 .
SECTION 2.3 Getting Information from the Graph of a Function
191
4. (a) A function value f a is a local maximum value of f if f a is the largest value of f on some interval containing a. From the graph of f we see that there are two local maximum values of f : one maximum is 7, and it occurs when x 2; the other maximum is 6, and it occurs when x 5.
(b) A function value f a is a local minimum value of f if f a is the smallest value of f on some interval containing a. From the graph of f we see that there is one local minimum value of f . The minimum value is 2, and it occurs when x 4.
5. The solutions of the equation f x 0 are the x-intercepts of the graph of f . The solution of the inequality f x 0 is the set of x-values at which the graph of f is on or above the x-axis. From the graph of f we find that the solutions of the equation f x 0 are x 1 and x 7, and the solution of the inequality f x 0 is the interval [1 7]. 6. (a) To solve the equation 2x 1 x 4 graphically we graph the
y
functions f x 2x 1 and g x x 4 on the same set of axes and determine the values of x at which the graphs of f and g intersect. From the graph, we see that the solution is x 1.
1 0
1
x
(b) To solve the inequality 2x 1 x 4 graphically we graph the functions f x 2x 1 and g x x 4 on the same set of axes and find the values of x at which the graph of g is higher than the graph of f . From the graphs in part (a) we see that the solution of the inequality is 1. 7. (a) h 2 1, h 0 1, h 2 3, and h 3 4. (b) Domain: [3 4]. Range: [1 4].
(c) h 3 3, h 2 3, and h 4 3, so h x 3 when x 3, x 2, or x 4.
(d) The graph of h lies below or on the horizontal line y 3 when 3 x 2 or x 4, so h x 3 for those values of x. (e) The net change in h between x 3 and x 3 is h 3 h 3 4 3 1.
8. (a) g 4 3, g 2 2, g 0 2, g 2 1, and g 4 0. (b) Domain: [4 4]. Range: [2 3].
(c) g 4 3. [Note that g 2 1 not 3.]
(d) It appears that g x 0 for 1 x 18 and for x 4; that is, for x 1 x 18 4. (e) g 1 0 and g 2 1, so the net change between x 1 and x 2 is 1 0 1.
9. (a) f 0 3 12 g 0. So f 0 is larger.
(b) f 3 1 25 g 3. So g 3 is larger. (c) f x g x for x 2 and x 2.
(d) f x g x for 4 x 2 and 2 x 3; that is, on the intervals [4 2] and [2 3]. (e) f x g x for 2 x 2; that is, on the interval 2 2.
10. (a) The graph of g is higher than the graph of f at x 6, so g 6 is larger.
(b) The graph of f is higher than the graph of g at x 3, so f 3 is larger.
(c) The graphs of f and g intersect at x 2, x 5, and x 7, so f x g x for these values of x. (d) f x g x for 1 x 2 and approximately 5 x 7; that is, on [1 2] and [5 7].
(e) f x g x for 2 x 5 and approximately 7 x 8; that is, on [2 5 and 7 8].
192
CHAPTER 2 Functions y
11. (a)
y
12. (a)
2 0
x
1
2 0
x
1
(b) Domain: ; Range:
13. (a)
y
(b) Domain: ; Range
14. (a)
y
1 0
1 0
1
x
1
x
(b) Domain: 1 4; Range 4 2 (b) Domain: [2 5]; Range [4 3]
15. (a)
y
16. (a)
y
2 0
1
x
2 0
x
1
(b) Domain: [3 3]; Range: [1 8] (b) Domain: [3 3]; Range [6 3]
18. (a)
17. (a) 10
-2
5
2
4
-5 -6
-4
-2
2
(b) Domain: ; Range: [1
-10
(b) Domain: ; Range: 2]
SECTION 2.3 Getting Information from the Graph of a Function
19. (a)
193
20. (a) 2
2
1
1 2
4
6
-3 -2 -1 -1
-1
(b) Domain: [1 ; Range: [0
1
2
3
(b) Domain: [2 ; Range: [0
21. (a)
22. (a) 5
-8 -6 -4 -2
5
2 4 6 8
-8 -6 -4 -2
-5
-5
(b) Domain: [4 4]; Range: [0 4]
(b) Domain: [5 5]; Range: [5 0]
y
23.
y
24. y=4-x
_2x+3
y=x-2
1l
1l 1
2 4 6 8
3x-7 1
x
x
(a) From the graph, we see that x 2 4 x when
(a) From the graph, we see that 2x 3 3x 7 when
(b) From the graph, we see x 2 4 x when x 3.
(b) From the graph, we see that 2x 3 3x 7 when
x 3.
x 2. x 2.
194
CHAPTER 2 Functions y
25.
y
26.
1l
1
x y=_x@l
y=x@l y=3-4x 1l
y=2-xl 1
x
(a) From the graph, we see that x 2 2 x when
(a) From the graph, we see that x 2 3 4x when
(b) From the graph, we see that x 2 2 x when
(b) From the graph, we see that x 2 3 4x when
x 2 or x 1.
x 1 or x 3.
2 x 1.
1 x 3.
28.
27.
20
10
10 -4
-2
-10
2
-20
-10
-30
-20
(a) We graph y x 3 3x 2 (black) and
2
4
6
(a) We graph y 5x 2 x 3 (black) and
y x 2 3x 7 (gray). From the graph, we see
y x 2 3x 4 (gray). From the graph, we see
that the graphs intersect at x 432, x 112,
that the graphs intersect at x 058, x 129, and
and x 144.
x 529.
(b) From the graph, we see that
(b) From the graph, we see that
x 3 3x 2 x 2 3x 7 on approximately
5x 2 x 3 x 2 3x 4 on approximately
[432 112] and [144 . 29.
[058 129] and [529 .
30. 4 2
3 2
-2
-1
1
1 -2
(a) We graph y 16x 3 16x 2 (black) and y x 1 (gray). From the graph, we see that the graphs intersect at x 1, x 14 , and x 14 . (b) From the graph, we see that 16x 3 16x 2 x 1 on 1 14 and 14 .
-1
0
(a) We graph y 1
1
2
3
4
5
x (black) and y
x2 1
(gray). From the graph, we see that the solutions are x 0 and x 231.
(b) From the graph, we see that 1 approximately 0 231.
x
x 2 1 on
SECTION 2.3 Getting Information from the Graph of a Function
195
31. (a) The domain is [1 4] and the range is [1 3]. (b) The function is increasing on 1 1 and 2 4 and decreasing on 1 2.
32. (a) The domain is [2 3] and the range is [2 3]. (b) The function is increasing on 0 1 and decreasing on 2 0 and 1 3.
33. (a) The domain is [3 3] and the range is [2 2]. (b) The function is increasing on 2 1 and 1 2 and decreasing on 3 2, 1 1, and 2 3. 34. (a) The domain is [2 2] and the range is [2 2]. (b) The function is increasing on 1 1 and decreasing on 2 1 and 1 2. 35. (a) f x x 2 5x is graphed in the viewing rectangle [2 7] by [10 10].
36. (a) f x x 3 4x is graphed in the viewing rectangle [10 10] by [10 10].
10
10
-2
2
4
6
-10
-5
-10
5
10
-10
(b) The domain is and the range is [625 .
(c) The function is increasing on 25 . It is decreasing on 25.
37. (a) f x 2x 3 3x 2 12x is graphed in the viewing rectangle [3 5] by [25 20].
(b) The domain and range are . (c) The function is increasing on 115 and 115 . It is decreasing on 115 115.
38. (a) f x x 4 16x 2 is graphed in the viewing rectangle [10 10] by [70 10].
20 -10
10 -2
-10
-5
5
10
-20 2
4
-40 -60
-20
(b) The domain and range are .
(b) The domain is and the range is [64 .
(c) The function is increasing on 1 and 2 .
(c) The function is increasing on 283 0 and
It is decreasing on 1 2.
283 . It is decreasing on 283 and 0 283.
196
CHAPTER 2 Functions
39. (a) f x x 3 2x 2 x 2 is graphed in the viewing rectangle [5 5] by [3 3].
40. (a) f x x 4 4x 3 2x 2 4x 3 is graphed in the viewing rectangle [3 5] by [5 5]. 4
2
2 -4
-2
2
4
-2
-2
2
-2
4
-4
(b) The domain and range are .
(b) The domain is and the range is [4 .
(c) The function is increasing on 155 and
(c) The function is increasing on 04 1 and 24 .
022 . It is decreasing on 155 022.
41. (a) f x x 25 is graphed in the viewing rectangle [10 10] by [5 5].
It is decreasing on 04 and 1 24.
42. (a) f x 4 x 23 is graphed in the viewing rectangle [10 10] by [10 10].
10
4 2 -10
-5
-2
5
10
-10
-5
-4
5
10
-10
(b) The domain is and the range is [0 .
(b) The domain is and the range is 4].
(c) The function is increasing on 0 . It is decreasing
(c) The function is increasing on 0. It is
on 0.
decreasing on 0 .
43. (a) Local maximum: 2 at x 0. Local minimum: 1 at x 2 and 0 at x 2.
(b) The function is increasing on 2 0 and 2 and decreasing on 2 and 0 2.
44. (a) Local maximum: 2 at x 2 and 1 at x 2. Local minimum: 1 at x 0.
(b) The function is increasing on 2 and 0 2 and decreasing on 2 0 and 2 .
45. (a) Local maximum: 0 at x 0 and 1 at x 3. Local minimum: 2 at x 2 and 1 at x 1.
(b) The function is increasing on 2 0 and 1 3 and decreasing on 2, 0 1, and 3 .
46. (a) Local maximum: 3 at x 2 and 2 at x 1. Local minimum: 0 at x 1 and 1 at x 2.
(b) The function is increasing on 2, 1 1, and 2 and decreasing on 2 1 and 1 2.
47. (a) In the first graph, we see that f x x 3 x has a local minimum and a local maximum. Smaller x- and y-ranges show that f x has a local maximum of about 038 when x 058 and a local minimum of about 038 when x 058. 5
0.5
0.50 -0.3
0.4 -5
-0.4
5 -5
-0.60
-0.55
0.3 -0.50
-0.5
(b) The function is increasing on 058 and 058 and decreasing on 058 058.
0.55
0.60
SECTION 2.3 Getting Information from the Graph of a Function
197
48. (a) In the first graph, we see that f x 3 x x 2 x 3 has a local minimum and a local maximum. Smaller x- and y-ranges show that f x has a local maximum of about 400 when x 100 and a local minimum of about 281 when x 033. 4 2
-2
0
2
-0.40
-0.35
2.9
4.1
2.8
4.0
2.7 -0.30
3.9 0.9
1.0
1.1
(b) The function is increasing on 033 100 and decreasing on 033 and 100 .
49. (a) In the first graph, we see that g x x 4 2x 3 11x 2 has two local minimums and a local maximum. The local maximum is g x 0 when x 0. Smaller x- and y-ranges show that local minima are g x 1361 when x 171 and g x 7332 when x 321. -5
5
-1.75
-1.70
-1.65 -13.4
-73.0
3.1
-50
-13.6
-73.5
-100
-13.8
-74.0
3.2
3.3
(b) The function is increasing on 171 0 and 321 and decreasing on 171 and 0 321.
50. (a) In the first graph, we see that g x x 5 8x 3 20x has two local minimums and two local maximums. The local maximums are g x 787 when x 193 and g x 1302 when x 104. Smaller x- and y-ranges show that local minimums are g x 1302 when x 104 and g x 787 when x 193. Notice that since g x is odd, the local maxima and minima are related. 20
-2.0
-1.8 -7.8
13.1 13.0
-5
-7.9
5
12.9 -20
1.0
-8.0 -1.2
-1.0
1.2
7.90 -12.8 7.85 -13.0 -13.2
7.80 1.90
1.95
2.00
(b) The function is increasing on 193, 104 104, and 193 and decreasing on 193 104 and 104 193.
198
CHAPTER 2 Functions
51. (a) In the first graph, we see that U x x 6 x has only a local maximum. Smaller x- and y-ranges show that U x has a local maximum of about 566 when x 400. 10
5.70
5
5.65 5.60 3.9
5
4.0
4.1
(b) The function is increasing on 400 and decreasing on 400 6. 52. (a) In the first viewing rectangle below, we see that U x x x x 2 has only a local maximum. Smaller x- and y-ranges show that U x has a local maximum of about 032 when x 075. 1.0
0.40
0.5
0.35 0.30
0.0 0.0
0.5
1.0
0.7
0.8
0.9
(b) The function is increasing on 0 075 and decreasing on 075 1. 1 x2 has a local minimum and a local maximum. Smaller x- and y-ranges x3 show that V x has a local maximum of about 038 when x 173 and a local minimum of about 038 when x 173.
53. (a) In the first graph, we see that V x
2
0.40
1.6
1.7
1.8
-0.30 0.35 -5
-0.35
5 -1.8
-2
-1.7
0.30 -1.6
-0.40
(b) The function is increasing on 173 and 173 and decreasing on 173 0 and 0 173. 1 has only a local maximum. Smaller x- and x2 x 1 y-ranges show that V x has a local maximum of about 133 when x 050.
54. (a) In the first viewing rectangle below, we see that V x
2
1.40 1.35
-5
5 -0.6
-0.5
1.30 -0.4
(b) The function is increasing on 050 and decreasing on 050 . 55. (a) At 6 A . M . the graph shows that the power consumption is about 500 megawatts. Since t 18 represents 6 P. M ., the graph shows that the power consumption at 6 P. M . is about 725 megawatts. (b) The power consumption is lowest between 3 A . M . and 4 A . M ..
SECTION 2.3 Getting Information from the Graph of a Function
199
(c) The power consumption is highest just before 12 noon. (d) The net change in power consumption from 9 A . M . to 7 P. M . is P 19 P 9 690 790 100 megawatts. 56. (a) The first noticeable movements occurred at time t 5 seconds. (b) It seemed to end at time t 30 seconds.
(c) Maximum intensity was reached at t 17 seconds.
57. (a) This person appears to be gaining weight steadily until the age of 21 when this person’s weight gain slows down. The person continues to gain weight until the age of 30, at which point this person experiences a sudden weight loss. Weight gain resumes around the age of 32, and the person dies at about age 68. Thus, the person’s weight W is increasing on 0 30 and 32 68 and decreasing on 30 32. (b) The sudden weight loss could be due to a number of reasons, among them major illness, a weight loss program, etc. (c) The net change in the person’s weight from age 10 to age 20 is W 20 W 10 150 50 100 lb. 58. (a) Measuring in hours since midnight, the salesman’s distance from home D is increasing on 8 9, 10 12, and 15 17, constant on 9 10, 12 13, and 17 18, and decreasing on 13 15 and 18 19. (b) The salesman travels away from home and stops to make a sales call between 9 A . M . and 10 A . M ., and then travels further from home for a sales call between 12 noon and 1 P. M . Next he travels along a route that takes him closer to home before taking him further away from home. He then makes a final sales call between 5 P. M . and 6 P. M . and then returns home. (c) The net change in the distance D from noon to 1 P. M . is D 1 P. M . D noon 0. 59. (a) The function W is increasing on 0 150 and 300 and decreasing on 150 300. (b) W has a local maximum at x 150 and a local minimum at x 300.
(c) The net change in the depth W from 100 days to 300 days is W 300 W 100 25 75 50 ft.
60. (a) The function P is increasing on 0 25 and decreasing on 25 50. (b) The maximum population was 50,000, and it was attained at x 25 years, which represents the year 1975. (c) The net change in the population P from 1970 to 1990 is P 40 P 20 40 40 0.
61. Runner A won the race. All runners finished the race. Runner B fell, but got up and finished the race. 62. (a)
F
63. (a)
80
E 400
70 60
300
50 40
200
30
100
20 10 0
1
2
3
4
5
6
7
8
9 10
(b) As the distance x increases, the gravitational attraction F decreases. The rate of decrease is rapid at first, and slows as the distance increases.
x
0
50
100 150 200 250 300
(b) As the temperature T increases, the energy E increases. The rate of increase gets larger as the temperature increases.
T
200
CHAPTER 2 Functions
64. In the first graph, we see the general location of the minimum of V 99987 006426T 00085043T 2 00000679T 3 is around T 4. In the second graph, we isolate the minimum, and from this graph, we see that the minimum volume of 1 kg of water occurs at T 396 C. 1005
999.76
1000
999.75
995
999.74 0
20
3.5
4.0
4.5
10 . In the second graph, we isolate the 5 minimum, and from this graph, we see that energy is minimized when 75 mi/h.
65. In the first graph, we see the general location of the minimum of E 273 3
10000
4700 4650
5000 4600 6
8
10
7.4
7.5
7.6
66. In the first graph, we see the general location of the maximum of r 32 1 r r 2 is around r 07 cm. In the second graph, we isolate the maximum, and from this graph we see that at the maximum velocity is approximately 047 when r 067 cm. 1.0
0.50 0.48
0.5
0.46 0.0 0.6
0.8
1.0
67. (a) f x is always increasing, and f x 0 for all x. y
0
0.60
0.65
0.70
(b) f x is always decreasing, and f x 0 for all x. y
x
0
x
SECTION 2.4 Average Rate of Change of a Function
(c) f x is always increasing, and f x 0 for all x.
201
(d) f x is always decreasing, and f x 0 for all x.
y
y
x 0
0
x
68. Numerous answers are possible. 69. (a) If x a is a local maximum of f x then
f a f x 0 for all x around x a. So 2 2 g a g x and thus g a g x.
(c) Let f x x 4 x 2 6x 9. From the graph, we
Similarly, if x b is a local minimum of f x, then f x f b 0 for all x around x b. So 2 2 g x g b and thus g x g b.
see that f x has a minimum at x 1. Thus g x also has a minimum at x 1 and this minimum value is g 1 14 12 6 1 9 5. 10 5
(b) Using the distance formula, 2 g x x 32 x 2 0 x 4 x 2 6x 9
2.4
0 0
2
AVERAGE RATE OF CHANGE OF A FUNCTION
100 miles 50 mi/h. 2 hours f b f a 2. The average rate of change of a function f between x a and x b is average rate of change . ba
1. If you travel 100 miles in two hours then your average speed for the trip is average speed
3. The average rate of change of the function f x x 2 between x 1 and x 5 is
25 1 24 52 12 f 1 average rate of change f 5 6. 51 4 4 4 4. (a) The average rate of change of a function f between x a and x b is the slope of the secant line between a f a and b f b. (b) The average rate of change of the linear function f x 3x 5 between any two points is 3.
5. (a) Yes, the average rate of change of a function between x a and x b is the slope of the secant line through a f a f b f a . and b f b; that is, ba (b) Yes, the average rate of change of a linear function y mx b is the same (namely m) for all intervals.
6. (a) No, the average rate of change of an increasing function is positive over any interval.
(b) No, just because the average rate of change of a function between x a and x b is negative, it does not follow
that the function is decreasing on that interval. For example, f x x 2 has negative average rate of change between x 2 and x 1, but f is increasing for 0 x 1.
7. (a) The net change is f 4 f 1 5 3 2.
(b) We use the points 1 3 and 4 5, so the average rate of change is
2 53 . 41 3
202
CHAPTER 2 Functions
8. (a) The net change is f 5 f 1 2 4 2. (b) We use the points 1 4 and 5 2, so the average rate of change is 9. (a) The net change is f 5 f 0 2 6 4. (b) We use the points 0 6 and 5 2, so the average rate of change is 10. (a) The net change is f 5 f 1 4 0 4.
24 2 1 . 51 4 2 26 4 . 50 5
(b) We use the points 1 0 and 5 4, so the average rate of change is
4 2 40 . 5 1 6 3
11. (a) The net change is f 3 f 2 [3 3 2] [3 2 2] 7 4 3. 3 f 3 f 2 3. (b) The average rate of change is 32 1 1 12. (a) The net change is r 6 r 3 3 3 6 3 13 3 1 2 1.
1 r 6 r 3 . 63 3 3 13. (a) The net change is h 1 h 4 1 2 4 32 12 11 2 5. 5 h 1 h 4 1. (b) The average rate of change is 1 4 5 14. (a) The net change is g 2 g 3 2 23 2 2 23 3 23 4 10 3 . (b) The average rate of change is
10 g 2 g 3 3 2. 2 3 5 3 15. (a) The net change is h 6 h 3 2 62 6 2 32 3 66 15 51. (b) The average rate of change is
h 6 h 3 51 17. 63 3 16. (a) The net change is f 0 f 2 1 3 02 1 3 22 1 11 12. (b) The average rate of change is
f 0 f 2 12 6. 0 2 2 17. (a) The net change is f 10 f 0 103 4 102 03 4 02 600 0 600. (b) The average rate of change is
f 10 f 0 600 60. 10 0 10 18. (a) The net change is g 2 g 2 24 23 22 24 23 22 12 28 16. (b) The average rate of change is
g 2 g 2 16 4. 2 2 4 19. (a) The net change is f 3 h f 3 5 3 h2 5 32 45 30h 5h 2 45 5h 2 30h. (b) The average rate of change is
f 3 h f 3 5h 2 30h 5h 30. h 3 h 3 20. (a) The net change is f 2 h f 2 1 3 2 h2 1 3 22 3h 2 12h 11 11 3h 2 12h. (b) The average rate of change is
(b) The average rate of change is
3h 2 12h f 2 h f 2 3h 12. h 2 h 2
SECTION 2.4 Average Rate of Change of a Function
203
1a 1 1 . a 1 a 1a 1a 1 g a g 1 a . (b) The average rate of change is a1 a1 a a 1 a
21. (a) The net change is g a g 1
2 2h 2 . h1 01 h1 2h g h g 0 h 1 2h 2 . (b) The average rate of change is h0 h h h 1 h1
22. (a) The net change is g h g 0
23. (a) The net change is f a h f a
2 2 2h . ah a a a h
(b) The average rate of change is 2h 2h 2 f a h f a a a h . h ah a h a a h a h a 24. (a) The net change is f a h f a
a h a.
(b) The average rate of change is f a h f a ah a ah a h 1 a h a . h a h a ah a h ah a h ah a ah a
25. (a) The average rate of change is 1 a h 3 1 a 3 1a 1h 3 1a 3 1h 1 f a h f a 2 2 2 2 2 2 . h h h 2 a h a (b) The slope of the line f x 12 x 3 is 12 , which is also the average rate of change.
26. (a) The average rate of change is [4 a h 2] [4a 2] g a h g a 4a 4h 2 4a 2 4h 4. h h h a h a (b) The slope of the line g x 4x 2 is 4, which is also the average rate of change. 27. The function f has a greater average rate of change between x 0 and x 1. The function g has a greater average rate of change between x 1 and x 2. The functions f and g have the same average rate of change between x 0 and x 15. 28. The average rate of change of f is constant, that of g increases, and that of h decreases. 29. The average rate of change is
50 75 25 1 W 200 W 100 ft/day. 200 100 200 100 100 4
P 40 P 20 40 40 0 0. 40 20 40 20 20 (b) The population increased and decreased the same amount during the 20 years.
30. (a) The average rate of change is
735 1,591 856 245 persons/yr. 2001 1998 3 657 826 1 483 3285 persons/yr. (b) The average rate of change of population is 2004 2002 2 (c) The population was increasing from 1997 to 2001.
31. (a) The average rate of change of population is
(d) The population was decreasing from 2001 to 2006.
204
CHAPTER 2 Functions
400 100 800 400 476 m/s. 152 68 84 21 400 1,600 1,200 268 m/s. (b) The average speed is 412 263 149 (c) Lap Length of time to run lap
32. (a) The average speed is
Average speed of lap.
1
32
625 m/s
2
36
556 m/s
3
40
500 m/s
4
44
455 m/s
5
51
392 m/s
6
60
333 m/s
7
72
278 m/s
8
77
260 m/s
The man is slowing down throughout the run.
635 495 140 14 players/yr. 2013 2003 10 18 513 495 18 players/yr. (b) The average rate of change of sales is 2004 2003 1 103 410 513 103 players/yr. (c) The average rate of change of sales is 2005 2004 1 (d) Year DVD players sold Change in sales from previous year
33. (a) The average rate of change of sales is
2003
495
—
2004
513
18
2005
410
103
2006
402
8
2007
520
118
2008
580
60
2009
631
51
2010
719
88
2011
624
95
2012
582
42
2013
635
53
Sales increased most quickly between 2006 and 2007, and decreased most quickly between 2004 and 2005.
SECTION 2.4 Average Rate of Change of a Function
205
34. Year
Number of books
1980
420
1981
460
1982
500
1985
620
1990
820
1992
900
1995
1020
1997
1100
1998
1140
1999
1180
2000
1220
35. The average rate of change of the temperature of the soup over the first 20 minutes is 119 200 81 T 20 T 0 405 F/min. Over the next 20 minutes, it is 20 0 20 0 20 T 40 T 20 89 119 30 15 F/min. The first 20 minutes had a higher average rate of change of 40 20 40 20 20 temperature (in absolute value).
36. (a) (i) Between 1860 and 1890, the average rate of change was about 84 farms per year. (ii) Between 1950 and 1970, the average rate of change was about 131 farms per year.
4570 2040 y 1890 y 1860 84, a gain of 1890 1860 30 2780 5390 y 1970 y 1950 131, a loss of 1970 1950 20
(b) From the graph, it appears that the steepest rate of decline was during the period from 1950 to 1960.
d 10 d 0 100 10. 10 0 10 (b) Skier A gets a great start, but slows at the end of the race. Skier B maintains a steady pace. Runner C is slow at the beginning, but accelerates down the hill.
37. (a) For all three skiers, the average rate of change is
38. (a) Skater B won the race, because he travels 500 meters before Skater A. A 10 A 0 200 0 (b) Skater A’s average speed during the first 10 seconds is 20 ms. 10 0 10 B 10 B 0 100 0 Skater B’s average speed during the first 10 seconds is 10 ms. 10 0 10 500 395 A 40 A 25 7 ms. (c) Skater A’s average speed during his last 15 seconds is 40 25 15 B 35 B 20 500 200 Skater B’s average speed during his last 15 seconds is 20 ms. 35 20 15
206
CHAPTER 2 Functions
39. t a
t b
3
35
3
31
3
301
3
3001
3
30001
Average Speed
f b f a ba
16 352 16 32 104 35 3
16 312 16 32 976 31 3
16 3012 16 32 9616 301 3
16 30012 16 32 96016 3001 3
16 300012 16 32 960016 30001 3
From the table it appears that the average speed approaches 96 fts as the time intervals get smaller and smaller. It seems reasonable to say that the speed of the object is 96 fts at the instant t 3.
2.5
LINEAR FUNCTIONS AND MODELS
1. If f is a function with constant rate of change, then (a) f is a linear function of the form f x ax b. (b) The graph of f is a line.
2. If f x 5x 7, then (a) The rate of change of f is 5.
(b) The graph of f is a line with slope 5 and y-intercept 7.
3. From the graph, we see that y 2 50 and y 0 20, so the slope of the graph is 50 20 y 2 y 0 15 galmin. m 20 2
4. From Exercise 3, we see that the pool is being filled at the rate of 15 gallons per minute. 5. If a linear function has positive rate of change, its graph slopes upward.
6. f x 3 is a linear function because it is of the form f x ax b, with a 0 and b 3. Its slope (and hence its rate of change) is 0. 7. f x 3 13 x 13 x 3 is linear with a 13 and b 3. 8. f x 2 4x 4x 2 is linear with a 4 and b 2. 9. f x x 4 x 4x x 2 is not of the form f x ax b for constants a and b, so it is not linear. 10. f x x 1 is not linear. 11. f x
x 1 15 x 15 is linear with a 15 and b 15 . 5
12. f x
3 2x 3 2 is not linear. x x
13. f x x 12 x 2 2x 1 is not of the form f x ax b for constants a and b, so it is not linear. 14. f x 12 3x 1 32 x 12 is linear with a 32 and b 12 .
SECTION 2.5 Linear Functions and Models y
15. x
f x 2x 5
1
7
0
5
1
3
2
1
3
1
4
3
2 0
x
1
The slope of the graph of f x 2x 5 is 2. y
16. x
g x 4 2x
1
6
0
4
1
2
2
0
3
2
4
4
2 0
1
x
The slope of the graph of g x 4 2x 2x 4 is 2. 17.
r t 23 t 2
t 1
y
267
0
2
2
1
133
0
2
067
3
0
4
067
2
t
The slope of the graph of r t 23 t 2 is 23 . 18. t 2 1
h t 12 34 t 2 125
0
05
1
025
2
1
3
175
The slope of the graph of h t 12 34 t is 34 .
y
1 0
1
t
207
208
CHAPTER 2 Functions
19. (a)
f
g
20. (a)
1
0
0
_1
1
1
z
x
1
_10
(b) The graph of f x 2x 6 has slope 2.
(b) The graph of g z 3z 9 has slope 3.
(c) f x 2x 6 has rate of change 2.
(c) g z 3z 9 has rate of change 3.
h
21. (a)
s
22. (a)
1
1 _1
_10 0
0
w
t
1
_1
10
(b) The graph of h t 05t 2 has slope 05.
(b) The graph of s 02 6 has slope 02.
(c) h t 05t 2 has rate of change 05.
(c) s 02 6 has rate of change 02.
23. (a)
v
A
24. (a)
10 _2 0 _10
1 2
t
10 (b) The graph of t 10 3 t 20 has slope 3 . 10 (c) t 10 3 t 20 has rate of change 3 .
_1
0 _1
1
r
(b) The graph of A r 23 r 1 has slope 23 . (c) A r 23 r 1 has rate of change 23 .
SECTION 2.5 Linear Functions and Models f
25. (a)
26. (a)
209
g 1 0
x
1
1 _1
0 _1
t
1
(b) The graph of f t 32 t 2 has slope 32 .
(b) The graph of g x 54 x 10 has slope 54 .
(c) f t 32 t 2 has rate of change 32 .
(c) g x 54 x 10 has rate of change 54 .
27. The linear function f with rate of change 3 and initial value 1 has equation f x 3x 1.
28. The linear function g with rate of change 12 and initial value 100 has equation g x 12x 100. 29. The linear function h with slope 12 and y-intercept 3 has equation h x 12 x 3.
30. The linear function k with slope 45 and y-intercept 2 has equation k x 45 x 2. 31. (a) From the table, we see that for every increase of 2 in the value of x, f x increases by 3. Thus, the rate of change of f is 32 . (b) When x 0, f x 7, so b 7. From part (a), a 32 , and so f x 32 x 7. 32. (a) From the table, we see that f 3 11 and f 0 2. Thus, when x increases by 3, f x decreases by 9, and so the rate of change of f is 3. (b) When x 0, f x 2, so b 2. From part (a), a 3, and so f x 3x 2.
33. (a) From the graph, we see that f 0 3 and f 1 4, so the rate of change of f is (b) From part (a), a 1, and f 0 b 3, so f x x 3. 34. (a) From the graph, we see that f 0 4 and f 2 0, so the rate of change of f is (b) From part (a), a 2, and f 0 b 4, so f x 2x 4. 35. (a) From the graph, we see that f 0 2 and f 4 0, so the rate of change of f is (b) From part (a), a 12 , and f 0 b 2, so f x 12 x 2.
43 1. 10 04 2. 20 02 1 . 40 2
36. (a) From the graph, we see that f 0 1 and f 2 0, so the rate of change of f is (b) From part (a), a 12 , and f 0 b 1, so f x 12 x 1. 37.
f
Increasing the value of a makes the graph of f steeper. In other words, it
a=2
increases the rate of change of f .
a=1 a= 21
1 01
0 1 1 . 20 2
t
210
CHAPTER 2 Functions f
38.
Increasing the value of b moves the graph of f upward, but does not affect
b=2
the rate of change of f . b=1 b= 21
1
0
39. (a)
t
1
T 38,000
(b) The slope of T x 150x 32,000 is the value of a, 150. (c) The amount of trash is changing at a rate equal to the slope of the graph, 150 thousand tons per year.
36,000 34,000 32,000
0
40. (a)
10
20
30
x
f 900
(b) The slope of the graph of f x 200 32x is 32.
700
(c) Ore is being produced at a rate equal to the slope of the graph, 32 thousand tons per year.
800 600 500 400 300 200 100
0
5
10
15
20
25 x
41. (a) Let V t at b represent the volume of hydrogen. The balloon is being filled at the rate of 05 ft3 s, so a 05, and initially it contains 2 ft3 , so b 2. Thus, V t 05t 2.
(b) We solve V t 15 05t 2 15 05t 13 t 26. Thus, it takes 26 seconds to fill the balloon. 42. (a) Let V t at b represent the volume of water. The pool is being filled at the rate of 10 galmin, so a 10, and initially it contains 300 gal, so b 300. Thus, V t 10t 300. (b) We solve V t 1300 10t 300 1300 10t 1000 t 100. Thus, it takes 100 minutes to fill the pool.
1 . The ramp 43. (a) Let H x ax b represent the height of the ramp. The maximum rise is 1 inch per 12 inches, so a 12 1 x. starts on the ground, so b 0. Thus, H x 12 1 150 125. Thus, the ramp reaches a height of 125 inches. (b) We find H 150 12
1200 0075, or 75%. 15,000 500 Brianna descends 500 vertical feet over 10,000 feet, so the grade of her road is 005, or 5%. 10,000
44. Meilin descends 1200 vertical feet over 15,000 feet, so the grade of her road is
45. (a) From the graph, we see that the slope of Jari’s trip is steeper than that of Jade. Thus, Jari is traveling faster.
SECTION 2.5 Linear Functions and Models
211
7 70 7 miles per minute or 60 70 mih. (b) The points 0 0 and 6 7 are on Jari’s graph, so her speed is 60 6 6 16 10 The points 0 10 and 6 16 are on Jade’s graph, so her speed is 60 60 mih. 60 1 hmin 1 mi/min and Jari’s speed is (c) t is measured in minutes, so Jade’s speed is 60 mih 60 1 hmin 7 mi/min. Thus, Jade’s distance is modeled by f t 1 t 0 10 t 10 and Jari’s 70 mih 60 6
distance is modeled by g t 76 t 0 0 76 t.
46. (a) Let d t represent the distance traveled. When t 0, d 0, and when
(b)
40 0 t 50, d 40. Thus, the slope of the graph is 08. The 50 0 y-intercept is 0, so d t 08t.
(c) Jacqueline’s speed is equal to the slope of the graph of d, that is, 08 mimin or 08 60 48 mih.
d 110 100 90 80 70 60 50 40 30 20 10 0
20
40
60
80 100 120 t
6 , so if we take 0 0 as the starting point, the 47. Let x be the horizontal distance and y the elevation. The slope is 100
6 x. We have descended 1000 ft, so we substitute y 1000 and solve for x: 1000 6 x elevation is y 100 100 1 16,667 316 mi. x 16,667 ft. Converting to miles, the horizontal distance is 5280
48. (a)
D
(b) The slope of the graph of D x 20 024x is 024.
30
(c) The rate of sedimentation is equal to the slope of the graph, 024 cmyr or 24 mmyr.
20 10
0
10
20
30
40
50
x
480 miles at a cost of $380, and in June she drove 800 miles at a cost of
C 600
$460. Thus, the points 480 380 and 800 460 are on the graph, so the
500
49. (a) Let C x ax b be the cost of driving x miles. In May Lynn drove
slope is a
1 460 380 . We use the point 480 380 to find the 800 480 4
(b)
400 300
value of b: 380 14 480 b b 260. Thus, C x 14 x 260.
200
(c) The rate at which her cost increases is equal to the slope of the line, that is
100
1 . So her cost increases by $025 for every additional mile she drives. 4
0
200 400 600 800 1000 1200 1400 x
The slope of the graph of C x 14 x 260 is the value of a, 14 .
212
CHAPTER 2 Functions
50. (a) Let C x ax b be the cost of producing x chairs in one day. The first (b) day, it cost $2200 to produce 100 chairs, and the other day it cost $4800 to produce 300 chairs.. Thus, the points 100 2200 and 300 4800 are on 4800 2200 13. We use the point 300 100 100 2200 to find the value of b: 2200 13 100 b b 900. Thus,
the graph, so the slope is a C x 13x 900.
(c) The rate at which the factory’s cost increases is equal to the slope of the line, that is $13chair.
C 10,000 9000 8000 7000 6000 5000 4000 3000 2000 1000 0
100 200 300 400 500 600 x
The slope of the graph of C x 13x 900 is the value of a, 13. f x2 f x1 ax ax1 ax2 b ax1 b 2 . x2 x1 x2 x1 x 2 x1 a x2 x1 ax ax1 a. (b) Factoring the numerator and cancelling, the average rate of change is 2 x2 x1 x2 x1
51. (a) By definition, the average rate of change between x1 and x2 is
f x f a c. x a (b) Multiplying the equation in part (a) by x a, we obtain f x f a c x a. Rearranging and adding f a to both sides, we have f x cx f a ca, as desired. Because this equation is of the form f x Ax B with constants A c and B f a ca, it represents a linear function with slope c and y-intercept f a ca.
52. (a) The rate of change between any two points is c. In particular, between a and x, the rate of change is
2.6
TRANSFORMATIONS OF FUNCTIONS
1. (a) The graph of y f x 3 is obtained from the graph of y f x by shifting upward 3 units. (b) The graph of y f x 3 is obtained from the graph of y f x by shifting left 3 units.
2. (a) The graph of y f x 3 is obtained from the graph of y f x by shifting downward 3 units. (b) The graph of y f x 3 is obtained from the graph of y f x by shifting right 3 units.
3. (a) The graph of y f x is obtained from the graph of y f x by reflecting in the x-axis. (b) The graph of y f x is obtained from the graph of y f xby reflecting in the y-axis.
4. (a) The graph of f x 2 is obtained from that of y f x by shifting upward 2 units, so it has graph II.
(b) The graph of f x 3 is obtained from that of y f x by shifting to the left 3 units, so it has graph I.
(c) The graph of f x 2 is obtained from that of y f x by shifting to the right 2 units, so it has graph III. (d) The graph of f x 4 is obtained from that of y f x by shifting downward 4 units, so it has graph IV.
5. If f is an even function, then f x f x and the graph of f is symmetric about the y-axis. 6. If f is an odd function, then f x f x and the graph of f is symmetric about the origin. 7. (a) The graph of y f x 1 can be obtained by shifting the graph of y f x downward 1 unit.
(b) The graph of y f x 2 can be obtained by shifting the graph of y f x to the right 2 units.
8. (a) The graph of y f x 4 can be obtained by shifting the graph of y f x to the left 5 units. (b) The graph of y f x 4 can be obtained by shifting the graph of y f x upward 4 units.
9. (a) The graph of y f x can be obtained by reflecting the graph of y f x in the y-axis.
(b) The graph of y 3 f x can be obtained by stretching the graph of y f x vertically by a factor of 3.
SECTION 2.6 Transformations of Functions
213
10. (a) The graph of y f x can be obtained by reflecting the graph of y f x about the x-axis.
(b) The graph of y 13 f x can be obtained by shrinking the graph of y f x vertically by a factor of 13 .
11. (a) The graph of y f x 5 2 can be obtained by shifting the graph of y f x to the right 5 units and upward 2 units. (b) The graph of y f x 1 1 can be obtained by shifting the graph of y f x to the left 1 unit and downward 1 unit. 12. (a) The graph of y f x 3 2 can be obtained by shifting the graph of y f x to the left 3 units and upward 2 units. (b) The graph of y f x 7 3 can be obtained by shifting the graph of y f x to the right 7 units and downward 3 units. 13. (a) The graph of y f x 5 can be obtained by reflecting the graph of y f x in the x-axis, then shifting the resulting graph upward 5 units. (b) The graph of y 3 f x 5 can be obtained by stretching the graph of y f x vertically by a factor of 3, then shifting the resulting graph downward 5 units. 14. (a) The graph of y 1 f x can be obtained by reflect the graph of y f x about the x-axis, then reflecting about the y-axis, then shifting upward 1 unit. (b) The graph of y 2 15 f x can be obtained by shrinking the graph of y f x vertically by a factor of 15 , then reflecting about the x-axis, then shifting upward 2 units. 15. (a) The graph of y 2 f x 5 1 can be obtained by shifting the graph of y f x to the left 5 units, stretching vertically by a factor of 2, then shifting downward 1 unit. (b) The graph of y 14 f x 3 5 can be obtained by shifting the graph of y f x to the right 3 units, shrinking vertically by a factor of 14 , then shifting upward 5 units. 16. (a) The graph of y 13 f x 2 5 can be obtained by shifting the graph of y f x to the right 2 units, shrinking vertically by a factor of 13 , then shifting upward 5 units. (b) The graph of y 4 f x 1 3 can be obtained by shifting the graph of y f x to the left 1 unit, stretching vertically by a factor of 4, then shifting upward 3 units. 17. (a) The graph of y f 4x can be obtained by shrinking the graph of y f x horizontally by a factor of 14 . (b) The graph of y f 14 x can be obtained by stretching the graph of y f x horizontally by a factor of 4.
18. (a) The graph of y f 2x 1 can be obtained by shrinking the graph of y f x horizontally by a factor of 12 , then shifting it downward 1 unit. (b) The graph of y 2 f 12 x can be obtained by stretching the graph of y f x horizontally by a factor of 2 and stretching it vertically by a factor of 2.
19. (a) The graph of g x x 22 is obtained by shifting the graph of f x to the left 2 units. (b) The graph of g x x 2 2 is obtained by shifting the graph of f x upward 2 units.
20. (a) The graph of g x x 43 is obtained by shifting the graph of f x to the right 4 units. (b) The graph of g x x 3 4 is obtained by shifting the graph of f x downward 4 units.
21. (a) The graph of g x x 2 2 is obtained by shifting the graph of f x to the left 2 units and downward 2 units.
(b) The graph of g x g x x 2 2 is obtained from by shifting the graph of f x to the right 2 units and upward 2 units.
214
CHAPTER 2 Functions
22. (a) The graph of g x x 1 is obtained by reflecting the graph of f x in the x-axis, then shifting the resulting graph upward 1 unit. (b) The graph of g x x 1 is obtained by reflecting the graph of f x in the y-axis, then shifting the resulting graph upward 1 unit. 23. (a)
y
y=x@+1
y
(b)
y=x@
y=x@ 1 1
(c)
y=(x-1)@
1 1
x
y
x
y
(d)
y=(x-1)@+3
y=x@
y=x@
1
1 1
1
x
x
y=_x@
24. (a)
y
(b)
y
y=Ïx+1
y=Ïx y=Ïx-2
1 1
(c)
y
1
x
(d)
y=Ïx+2+2
y=Ïx
1 1
y=Ïx
1
y
y=Ïx
1 x
x
1
x y=_Ïx+1
25. The graph of y x 1 is obtained from that of y x by shifting to the left 1 unit, so it has graph II. 26. y x 1 is obtained from that of y x by shifting to the right 1 unit, so it has graph IV. 27. The graph of y x 1 is obtained from that of y x by shifting downward 1 unit, so it has graph I.
SECTION 2.6 Transformations of Functions
215
28. The graph of y x is obtained from that of y x by reflecting in the x-axis, so it has graph III.
29. f x x 2 3. Shift the graph of y x 2 upward 3 units. 30. f x x 2 4. Shift the graph of y x 2 downward y
4 units.
y y=x@+3 y=x@ y=x@ 1 0
1 0
y=x@-4
x
1
31. f x x 1. Shift the graph of y x downward 1 unit.
32. f x
x 1. Shift the graph of y y
y=| x |
x upward 1 unit.
y=Ïx
1 y=| x |-1
1
y=Ïx+1
y
1
x
1
1
x
x
33. f x x 52 . Shift the graph of y x 2 to the right 5 units.
34. f x x 12 . Shift the graph of y x 2 to the left 1 unit.
y
y y=x@ y=(x-5)@ y=(x+1)@
5 1
x
y=x@ 1 1
x
216
CHAPTER 2 Functions
35. f x x 2. Shift the graph of y x to the left 2 units.
y
36. f x 4 units.
x 4. Shift the graph of y
y=Ïx
y=| x |
y=Ïx-4
1
1
1
x
37. f x x 3 . Reflect the graph of y x 3 in the x-axis. y
1
x
38. f x x. Reflect the graph of y x in the x-axis. y
y=| x |
y=x#
2
1 1
x y=_x#
39. y
x to the right
y
y=| x+2 |
1
x y=_| x |
4 x. Reflect the graph of y 4 x in the y-axis. y
40. y
3 x. Reflect the graph of y 3 x in the y-axis. y
2 y=Îx 4 y=Ï_x
4 y=Ïx
2
1
x y=Î_x
10
x
SECTION 2.6 Transformations of Functions
41. y 14 x 2 . Shrink the graph of y x 2 vertically by a factor of 14 .
42. y 5 x. Stretch the graph of y x vertically by a factor of 5, then reflect it in the x-axis. y
y y=x@
y=Ïx
2 1
x
y=41 x@ y=_5Ïx
1 1
x
43. y 3 x. Stretch the graph of y x vertically by a factor of 3.
44. y 12 x. Shrink the graph of y x vertically by a factor of 12 .
y
y y=3 | x | y=| x | 1 y=| x |
1 1
y=21 | x | 1
3 units and upward 5 units.
46. y
x 4 3. Shift the graph of y
y=x@
x to the left
y=Ïx
1 1 y=Ïx+4-3
y=(x-3)@+5 1
4 units and downward 3 units. y
y
5
x
x
45. y x 32 5. Shift the graph of y x 2 to the right
x
217
x
218
CHAPTER 2 Functions
47. y 3 12 x 12 . Shift the graph of y x 2 to the right one unit, shrink vertically by a factor of 12 , reflect in the x-axis, then shift upward 3 units.
48. y 2
x 1. Shift the graph of y
x to the left
1 unit, reflect the result in the x-axis, then shift upward
2 units. y
y
y=Ïx
y=x@ 1
2 1
y=3-21 (x-1)@
x
1
x y=2-Ïx+1
49. y x 2 2. Shift the graph of y x to the left 2 units and upward 2 units.
y
50. y 2 x. Reflect the graph of y x in the x-axis, then shift upward 2 units.
y
y=| x+2 |+2
y=| x | 1 y=| x |
1
1
x
y=2-| x | 1
x
51. y 12 x 4 3. Shrink the graph of y x vertically by a factor of 12 , then shift the result to the left 4 units and downward 3 units.
52. y 3 2 x 12 . Stretch the graph of y x 2 vertically by a factor of 2, reflect the result in the x-axis, then shift
the result to the right 1 unit and upward 3 units. y
y
y=x@ y=Ïx 1 1
1 4
x
x
y=3-2(x-1)@
1
y=2Ïx+4-3
53. y f x 3. When f x x 2 , y x 2 3. 55. y f x 2. When f x
x, y
x 2.
57. y f x 2 5. When f x x, y x 2 5.
54. y f x 5. When f x x 3 , y x 3 5. 56. y f x 1. When f x
3
x, y
58. y f x 4 3. When f x x, y x 4 3.
3
x 1.
SECTION 2.6 Transformations of Functions
59. y f x 1. When f x
4
x, y
4 x 1.
219
60. y f x 2. When f x x 2 , y x 22 . 62. y 12 f x 1 3. When f x x,
61. y 2 f x 3 2. When f x x 2 , y 2 x 32 2.
y 12 x 1 3.
63. g x f x 2 x 22 x 2 4x 4
64. g x f x 3 x 3 3
65. g x f x 1 2 x 1 2 67. g x f x 2 x 2
66. g x 2 f x 2 x 68. g x f x 2 1 x 22 1 x 2 4x 3
69. (a) y f x 4 is graph #3.
70. (a) y 13 f x is graph #2.
(b) y f x 3 is graph #1.
(b) y f x 4 is graph #3.
(c) y 2 f x 6 is graph #2.
(c) y f x 5 3 is graph #1.
(d) y f 2x is graph #4.
(d) y f x is graph #4.
71. (a) y f x 2
(b) y f x 2
y
1
y
y
1 1
1 1
x
(d) y f x 3
(e) y f x
y
x
x
y
1
x
(c) y g x 2
y
1 1
x
1
x
1
x
1
1
(b) y g x
x
y
1 1
1
(f) y 12 f x 1
y
1
72. (a) y g x 1
(c) y 2 f x
y
1 1
x
220
CHAPTER 2 Functions
(d) y g x 2
(e) y g x
y
1
(f) y 2g x
y
1 1
1 1
x
y
1
1
x
74. (a) y h 3x
1
(b) y h 13 x
y y=h(3x)
76. y 14 x
x
y
1 1
x
2
x
For part (b), shift the graph in (a) to the left 5 units; for part (c), shift the graph
8 (d)
4
2
4
6
in (a) to the left 5 units, and stretch it vertically by a factor of 2; for part (d), shift
(c)
the graph in (a) to the left 5 units, stretch it vertically by a factor of 2, and then shift
(b)
it upward 4 units.
(a) -4
1 y=h( _ 3 x)
3
1
-2 0
y
x
y
75. y [[2x]]
x
y=h(x)
y=h(x)
1
-6
x
y
1
-8
1
x
(b) y g 12 x
73. (a) y g 2x
77.
y
8
SECTION 2.6 Transformations of Functions
78. (a)
-8
-6
-4
6
For (b), reflect the graph in (a) in the x-axis; for (c), stretch the graph in (a)
4
vertically by a factor of 3 and reflect in the x-axis; for (d), shift the graph in (a) to
2
the right 5 units, stretch it vertically by a factor of 3, and reflect it in the x-axis. The
-2
2
4
6
8
(c)
-4 -6
79.
order in which each operation is applied to the graph in (a) is not important to obtain the graphs in part (c) and (d).
-2 (b)
221
(c)
(d)
For part (b), shrink the graph in (a) vertically by a factor of 13 ; for part (c), shrink
4 (a) (b)
the graph in (a) vertically by a factor of 13 , and reflect it in the x-axis; for part (d),
2
shift the graph in (a) to the right 4 units, shrink vertically by a factor of 13 , and then -4
-2
2 -2
4
(c)
6 (d)
reflect it in the x-axis.
-4
80. (b)
4
For (b), shift the graph in (a) to the left 3 units; for (c), shift the graph in (a) to the
2 (a)
left 3 units and shrink it vertically by a factor of 12 ; for (d), shift the graph in (a) to the left 3 units, shrink it vertically by a factor of 12 , and then shift it downward
(c) -6
-4
2
4
-2
6
3 units. The order in which each operation is applied to the graph in (a) is not important to sketch (c), while it is important in (d).
(d) -4
(b) y f 2x 2 2x 2x2 4x 4x 2
81. (a) y f x 2x x 2
-4
-2
1x 2
2 2 12 x 12 x
(c) y f x 14 x 2
4
4
4
2
2
2
-2
2
4
-4
-2
-4
-2
2
4
-4
-2
-4
-2
2
4
-4
The graph in part (b) is obtained by horizontally shrinking the graph in part (a) by a factor of 12 (so the graph is half as wide). The graph in part (c) is obtained by horizontally stretching the graph in part (a) by a factor of 2 (so the graph is twice as wide). 82. (a) y f x 2x x 2 (b) y f x 2 x x2 (c) y f x 2 x x2 2x x 2 2x x 2
-4
-2
4
4
4
2
2
2
-2 -4
2
4
-4
-2
-2 -4
2
4
-4
-2
-2 -4
2
4
222
CHAPTER 2 Functions
2 1 (e) y f 2 x 2 12 x 12 x x 14 x 2
(d) y f 2x 2 2x 2x2 2x x 2 4x 4x 2
-4
-2
4
4
2
2 2
-2
4
-4
-4
-2
2
-2
4
-4
The graph in part (b) is obtained by reflecting the graph in part (a) in the y-axis. The graph in part (c) is obtained by rotating the graph in part (a) through 180 about the origin [or by reflecting the graph in part (a) first in the x-axis and then in the y-axis]. The graph in part (d) is obtained by reflecting the graph in part (a) in the y-axis and then horizontally shrinking the graph by a factor of 12 (so the graph is half as wide). The graph in part (e) is obtained by reflecting the graph in part (a) in the y-axis and then horizontally stretching the graph by a factor of 2 (so the graph is twice as wide).
83. f x x 4 . f x x4 x 4 f x. Thus f x is even.
84. f x x 3 . f x x3 x 3 f x. Thus f x is odd.
y
y
2 1
x
2 1
x
85. f x x 2 x. f x x2 x x 2 x. Thus 86. f x x 4 4x 2 . f x f x. Also, f x f x, so f x is
neither odd nor even.
f x x4 4 x2 x 4 4x 2 f x. Thus f x is even.
y
y
1 1
x
1 1
x
SECTION 2.6 Transformations of Functions
87. f x x 3 x.
88. f x 3x 3 2x 2 1.
f x 3 x3 2 x2 1 3x 3 2x 2 1.
f x x3 x x 3 x x 3 x f x.
Thus f x f x. Also f x f x, so f x is neither odd nor even.
Thus f x is odd.
y
y
1
1
3
x. f x 1
3 x 1 3 x. Thus
f x f x. Also f x f x, so f x is
neither odd nor even.
x
1
x
1
89. f x 1
223
90. f x x 1x. f x x 1 x x 1x x 1x f x.
y
Thus f x is odd. y
1 x
1
1 x
1
91. (a) Even
(b) Odd y
y
1
1 1
x
1
x
224
CHAPTER 2 Functions
92. (a) Even
(b) Odd y
y
1
1 1
1
x
93. Since f x x 2 4 0, for 2 x 2, the graph of
y g x is found by sketching the graph of y f x for
x 2 and x 2, then reflecting in the x-axis the part of
94. g x x 4 4x 2
x
y g
4
the graph of y f x for 2 x 2.
2 _3
95. (a) f x 4x x 2
(b) f x 4x x 2
y
1 1
96. (a) f x x 3
_1
(b) g x x 3
1 1
x
y
1
x
y
3 x
1
1
x
1
x
y
1
SECTION 2.6 Transformations of Functions
97. (a) Luisa drops to a height of 200 feet, bounces up and down, then settles at
(b)
350 feet.
225
y (ft) 500
(c) To obtain the graph of H from that of h, we shift downward 100 feet. Thus, H t h t 100.
0
98. (a) Miyuki swims two and a half laps, slowing down with each successive lap.
(b)
In the first 30 seconds she swims 50 meters, so her average speed is
4
t (s)
80 60
50 167 ms. 30
40 20
(c) Here Miyuki swims 60 meters in 30 seconds, so her average speed is 60 2 ms. 30
0
100
200
t
This graph is obtained by stretching the original graph vertically by a factor of 12. 99. (a) The trip to the park corresponds to the first piece of the graph. The class travels 800 feet in 10 minutes, so their average speed is 800 10 80 ftmin. The second (horizontal) piece of the graph stretches from t 10 to t 30, so the class
spends 20 minutes at the park. The park is 800 feet from the school. y 600
(b)
(c)
800
400
600 400
200 0
y 1000
200 20
40
60 t
0
20
40
60 t
The new graph is obtained by shrinking the original
This graph is obtained by shifting the original graph
graph vertically by a factor of 050. The new average
to the right 10 minutes. The class leaves ten minutes
speed is 40 ftmin, and the new park is 400 ft from
later than it did in the original scenario.
the school. 100. To obtain the graph of g x x 22 5 from that of f x x 22 , we shift to the right 4 units and upward 5 units.
101. To obtain the graph of g x from that of f x, we reflect the graph about the y-axis, then reflect about the x-axis, then shift upward 6 units. 102. f even implies f x f x g even implies g x g x; f odd implies f x f x and g odd implies g x g x If f and g are both even, then f g x f x g x f x g x f g x and f g is even. If f and g are both odd, then f g x f x g x f x g x f g x and f g is odd. If f odd and g even, then f g x f x g x f x g x, which is neither odd nor even.
103. f even implies f x f x; g even implies g x g x; f odd implies f x f x; and g odd implies g x g x. If f and g are both even, then f g x f x g x f x g x f g x. Thus f g is even. If f and g are both odd, then f g x f x g x f x g x f x g x f g x. Thus f g is even If f if odd and g is even, then f g x f x g x f x g x f x g x f g x. Thus f g is odd.
226
CHAPTER 2 Functions
104. f x x n is even when n is an even integer and f x x n is odd when n is an odd integer. These names were chosen because polynomials with only terms with odd powers are odd functions, and polynomials with only terms with even powers are even functions.
2.7
COMBINING FUNCTIONS
1. From the graphs of f and g in the figure, we find f g 2 f 2 g 2 3 5 8, 3 f 2 f . f g 2 f 2 g 2 3 5 2, f g 2 f 2 g 2 3 5 15, and 2 g g 2 5
2. By definition, f g x f g x. So, if g 2 5 and f 5 12, then f g 2 f g 2 f 5 12. 3. If the rule of the function f is “add one” and the rule of the function g is “multiply by 2” then the rule of f g is “multiply by 2, then add one” and the rule of g f is “add one, then multiply by 2.” 4. We can express the functions in Exercise 3 algebraically as f x x 1, g x 2x, f g x 2x 1, and g f x 2 x 1.
5. (a) The function f g x is defined for all values of x that are in the domains of both f and g. (b) The function f g x is defined for all values of x that are in the domains of both f and g.
(c) The function f g x is defined for all values of x that are in the domains of both f and g, and g x is not equal to 0. 6. The composition f g x is defined for all values of x for which x is in the domain of g and g x is in the domain of f .
7. f x x has domain . g x 2x has domain . The intersection of the domains of f and g is . f g x x 2x 3x, and the domain is . f g x x 2x x, and the domain is . 1 x f , and the domain is 0 0 . f g x x 2x 2x 2 , and the domain is . x g 2x 2 8. f x x has domain . g x x has domain [0 . The intersection of the domains of f and g is [0 . f g x x x, and the domain is [0 . f g x x x, and the domain is [0 . x f f g x x x x 32 , and the domain is [0 . x x, and the domain is 0 . g x 9. f x x 2 x and g x x 2 each have domain . The intersection of the domains of f and g is .
f g x 2x 2 x, and the domain is . f g x x, and the domain is . x2 x f 1 1 , and the domain is 0 0 . x f g x x 4 x 3 , and the domain is . 2 g x x
10. f x 3 x 2 and g x x 2 4 each have domain . The intersection of the domains of f and g is .
f g x 1, and the domain is . f g x 2x 2 7, and the domain is . 3 x2 3 x2 f , f g x 3 x 2 x 2 4 x 4 7x 2 12, and the domain is . x 2 g x 2 x 2 x 4 and the domain is 2 2 2 2 .
11. f x 5 x and g x x 2 3x each have domain . The intersection of the domains of f and g is . f g x 5 x x 2 3x x 2 4x 5, and the domain is . f g x 5 x x 2 3x x 2 2x 5, and the domain is . f g x 5 x x 2 3x x 3 8x 2 15x, and the domain is . f 5x 5x , and the domain is 0 0 3 3 . x 2 g x x 3 x 3x
SECTION 2.7 Combining Functions
227
12. f x x 2 2x has domain . g x 3x 2 1 has domain . The intersection of the domains of f and g is . f g x x 2 2x 3x 2 1 4x 2 2x 1, and the domain is . f g x x 2 2x 3x 2 1 2x 2 2x 1, and the domain is . f g x x 2 2x 3x 2 1 3x 4 6x 3 x 2 2x, and the domain is . x 2 2x f , 3x 2 1 0 x 33 , and the domain is x x 33 . x 2 g 3x 1
13. f x 25 x 2 , has domain [5 5]. g x x 3, has domain [3 . The intersection of the domains of f and g is [3 5]. f g x 25 x 2 x 3 , and the domain is [3 5]. f g x 25 x 2 x 3, and the domain is [3 5]. f g x 25 x 2 x 3, and the domain is [3 5]. 25 x 2 f , and the domain is 3 5]. x g x 3
14. f x 16 x 2 has domain [4 4]. g x x 2 1 has domain 1] [1 . The intersection of the domains of f and g is [4 1] [1 4]. f g x 16 x 2 x 2 1, and the domain is [4 1] [1 4]. f g x 16 x 2 x 2 1, and the domain is [4 1] [1 4]. f g x 16 x 2 x 2 1 , and the domain is [4 1] [1 4]. f 16 x 2 , and the domain is [4 1 1 4]. x g x2 1
2 4 has domain x 0. g x , has domain x 4. The intersection of the domains of f and g is x x 4 x x 0 4; in interval notation, this is 4 4 0 0 . 2 4 2 4 2 3x 4 , and the domain is 4 4 0 0 . f g x x x 4 x x 4 x x 4 4 2 x 4 2 , and the domain is 4 4 0 0 . f g x x x 4 x x 4 4 8 2 , and the domain is 4 4 0 0 . f g x x x 4 x x 4 2 f x 4 x , and the domain is 4 4 0 0 . x 4 g 2x x 4
15. f x
228
CHAPTER 2 Functions
x 2 has domain x 1. g x has domain x 1 The intersection of the domains of f and g is x 1 x 1 x x 1; in interval notation, this is 1 1 . 2 x x 2 , and the domain is 1 1 . f g x x 1 x 1 x 1 2 x 2x , and the domain is 1 1 . f g x x 1 x 1 x 1 2 x 2x , and the domain is 1 1 . f g x x 1 x 1 x 12 2 2 f 1 x x x x , so x 0 as well. Thus the domain is 1 1 0 0 . g x 1 17. f x x 3 x. The domain of x is [0 , and the domain of 3 x is 3]. Thus, the domain of f is 3] [0 [0 3]. 1x 1x 18. f x x 4 . The domain of x 4 is [4 , and the domain of is 0 0 1]. Thus, the x x domain of f is [4 0 0 1] [4 0 0 1].
16. f x
1 . Since 14 is an even root and the denominator can not equal 0, x 3 0 x 3 x 314 So the domain is 3 . 1 x 3 20. k x . The domain of x 3 is [3 , and the domain of is x 1. Since x 1 is 1 1 , x 1 x 1 the domain is [3 1 1 [3 1 1 . 19. h x x 314
y
21.
y
22.
f
f+g
0
2
f
x
-4
24. 5
25.
f+g
40
f+g
g
-2
26.
f+g
g
0
2
4
3
f
20
f+g 2
3
f
-4
2
-2
2
-40
1
4
f
1
-20
g
1 0
g
x
f
4
4
f+g
g
0
23.
2
2 27. f x 2x 3 and g x 4 x .
(a) f g 0 f 4 02 f 4 2 4 3 5 (b) g f 0 g 2 0 3 g 3 4 32 5
28. (a) f f 2 f 2 2 3 f 1 2 1 3 1 (b) g g 3 g 4 32 g 5 4 52 21
g -4
-2
00
2
4
SECTION 2.7 Combining Functions
29. (a) f g 2 f g 2 f 4 22 f 0 2 0 3 3
(b) g f 2 g f 2 g 2 2 3 g 7 4 72 45
30. (a) f f 1 f f 1 f 2 1 3 f 5 2 5 3 13 (b) g g 1 g g 1 g 4 12 g 3 4 32 5
31. (a) f g x f g x f 4 x 2 2 4 x 2 3 8 2x 2 3 5 2x 2 (b) g f x g f x g 2x 3 4 2x 32 4 4x 2 12x 9 4x 2 12x 5
32. (a) f f x f f x f 3x 5 3 3x 5 5 9x 15 5 9x 20 2 (b) g g x g g x g 2 x 2 2 2 x 2 2 4 4x 2 x 4 x 4 4x 2 2 33. f g 2 f 5 4
34. f 0 0, so g f 0 g 0 3.
35. g f 4 g f 4 g 2 5
36. g 0 3, so f g 0 f 3 0.
37. g g 2 g g 2 g 1 4
38. f 4 2, so f f 4 f 2 2.
39. From the table, g 2 5 and f 5 6, so f g 2 6. 40. From the table, f 2 3 and g 3 6, so g f 2 6. 41. From the table, f 1 2 and f 2 3, so f f 1 3. 42. From the table, g 2 5 and g 5 1, so g g 2 1. 43. From the table, g 6 4 and f 4 1, so f g 6 1. 44. From the table, f 2 3 and g 3 6, so g f 2 6. 45. From the table, f 5 6 and f 6 3, so f f 5 3. 46. From the table, g 2 5 and g 5 1, so g g 5 1. 47. f x 2x 3, has domain ; g x 4x 1, has domain . f g x f 4x 1 2 4x 1 3 8x 1, and the domain is . g f x g 2x 3 4 2x 3 1 8x 11, and the domain is . f f x f 2x 3 2 2x 3 3 4x 9, and the domain is . g g x g 4x 1 4 4x 1 1 16x 5, and the domain is . x 48. f x 6x 5 has domain . g x has domain . 2 x x 6 5 3x 5, and the domain is . f g x f 2 2 6x 5 3x 52 , and the domain is . g f x g 6x 5 2 f f x f 6x 5 6 6x 5 5 36x 35, and the domain is . x x x 2 , and the domain is . g g x g 2 2 4 49. f x x 2 , has domain ; g x x 1, has domain .
f g x f x 1 x 12 x 2 2x 1, and the domain is . g f x g x 2 x 2 1 x 2 1, and the domain is . 2 f f x f x 2 x 2 x 4 , and the domain is .
g g x g x 1 x 1 1 x 2, and the domain is .
229
230
CHAPTER 2 Functions
50. f x x 3 2 has domain . g x 3 x has domain . 3 f g x f 3 x 3 x 2 x 2, and the domain is . 3 g f x g x 3 2 x 3 2 and the domain is . 3 f f x f x 3 2 x 3 2 2 x 9 6x 6 12x 3 8 2 x 9 6x 6 12x 3 10, and the domain is .
g g x g
13 3 x 3 3 x x 13 x 19 , and the domain is .
1 , has domain x x 0; g x 2x 4, has domain . x 1 . f g x is defined for 2x 4 0 x 2. So the domain is f g x f 2x 4 2x 4 x x 2 2 2 . 1 2 1 2 4 4, the domain is x x 0 0 0 . g f x g x x x 1 1 x. f f x is defined whenever both f x and f f x are defined; that is, f f x f 1 x x whenever x x 0 0 0 . g g x g 2x 4 2 2x 4 4 4x 8 4 4x 12, and the domain is .
51. f x
52. f x x 2 has domain . g x x 3 has domain [3 . 2 x 3 x 3 x 3, and the domain is [3 . f g x f g f x g x 2 x 2 3. For the domain we must have x 2 3 x 3 or x 3. Thus the domain is 3 3 . 2 f f x f x 2 x 2 x 4 , and the domain is . x 3 3. For the domain we must have x 3 3 x 3 9 x 12, so the g g x g x 3 domain is [12 . 53. f x x, has domain ; g x 2x 3, has domain f g x f 2x 4 2x 3, and the domain is . g f x g x 2 x 3, and the domain is . f f x f x x x, and the domain is . g g x g 2x 3 2 2x 3 3 4x 6 3 4x 9. Domain is . 54. f x x 4 has domain . g x x 4 has domain . f g x f x 4 x 4 4, and the domain is . g f x g x 4 x 4 4 x, and the domain is . f f x f x 4 x 4 4 x 8, and the domain is . g g x g x 4 x 4 4 x 4 4 (x 4 4 is always positive). The domain is .
SECTION 2.7 Combining Functions
231
x , has domain x x 1; g x 2x 1, has domain x 1 2x 1 2x 1 , and the domain is x x 0 0 0 . f g x f 2x 1 2x 2x 1 1 2x x x 2 1 1, and the domain is x x 1 1 1 g f x g x 1 x 1 x 1 x x x x 1 x x x 1 . f f x is defined whenever both f x and f f x f x 1 x 1 x x 1 2x 1 1 x 1 f f x are defined; that is, whenever x 1 and 2x 1 0 x 12 , which is 1 1 12 12 .
55. f x
g g x g 2x 1 2 2x 1 1 4x 2 1 4x 3, and the domain is . 1 56. f x has domain x x 0 g x x 2 4x has domain . x 1 . f g x is defined whenever 0 x 2 4x x x 4. The product of two f g x f x 2 4x 2 x 4x numbers is positive either when both numbers are negative or when both numbers are positive. So the domain of f g is x x 0 and x 4 x x 0 and x 4 which is 0 4 . 1 2 1 1 4 1 . g f x is defined whenever both f x and g f x are 4 g f x g x x x x x defined, that is, whenever x 0. So the domain of g f is 0 . 1 1 x 14 . f f x is defined whenever both f x and f f x are defined, that is, f f x f x 1 x whenever x 0. So the domain of f f is 0 . 2 g g x g x 2 4x x 2 4x 4 x 2 4x x 4 8x 3 16x 2 4x 2 16x x 4 8x 3 12x 2 16x,
and the domain is .
1 x , has domain x x 1; g x has domain x x 0. x 1 x 1 1 1 1 1 x . f g x is defined whenever both g x and f g x are f g x f 1 1 x x 1 1 x x
57. f x
x
defined, so the domain is x x 1 0. 1 x 1 x x . g f x is defined whenever both f x and g f x are defined, so the g f x g x 1 x x1
domain is x x 1 0. x x x x xx1 . f f x is defined whenever both f x and f f x f x x 1 1 2x 1 x 1 x1 1 x1 f f x are defined, so the domain is x x 1 12 . 1 1 1 x. g g x is defined whenever both g x and g g x are defined, so the domain is g g x g x x
x x 0.
232
CHAPTER 2 Functions
x 2 has domain x x 0; g x has domain x x 2. x x 2 x 2x 4 2 x . f g x is defined whenever both g x and f g x are defined; that f g x f x 2 x x 2 is, whenever x 0 and x 2. So the domain is x x 0 2. 2 1 2 2 x . g f x is defined whenever both f x and g f x are defined; g f x g 2 x 2 2x 1x 2 x that is, whenever x 0 and x 1. So the domain is x x 0 1. 2 2 x. f f x is defined whenever both f x and f f x are defined; that is, whenever f f x f 2 x x x 0. So the domain is x x 0. x x x x x x 2 . g g x is defined whenever both g x and g g x g x 2 x 2 2 3x 4 x 2 x 2 g g x are defined; that is whenever x 2 and x 43 . So the domain is x x 2 43 . x 1 x 11 f g h x f g h x f g x 1 f 3 g h x g x 2 2 x 2 2 x 6 6x 4 12x 2 8. 1 . f g h x f x 6 6x 4 12x 2 8 6 4 x 6x 12x 2 8 4 x 5 x 5 1 f g h x f g h x f g x f 3 3 3 x x x 3 . f g h x f . x g h x g 3 3 3 x 1 x 1 x 1
58. f x
59. 60.
61. 62.
For Exercises 63–72, many answers are possible. 63. F x x 95 . Let f x x 5 and g x x 9, then F x f g x. 64. F x x 1. If f x x 1 and g x x, then F x f g x.
x2 x 65. G x 2 and g x x 2 , then G x f g x. . Let f x x 4 x 4 1 1 . If f x and g x x 3, then G x f g x. 66. G x x 3 x 67. H x 1 x 3 . Let f x x and g x 1 x 3 , then H x f g x. 68. H x 1 x. If f x 1 x and g x x, then H x f g x.
1 1 69. F x 2 . Let f x , g x x 1, and h x x 2 , then F x f g h x. x x 1 70. F x 3 x 1. If g x x 1 and h x x, then g h x x 1, and if f x 3 x, then F x f g h x. 9 71. G x 4 3 x . Let f x x 9 , g x 4 x, and h x 3 x, then G x f g h x.
2 2 72. G x 2 . If g x 3 x and h x x, then g h x 3 x, and if f x x 2 , then 3 x G x f g h x.
SECTION 2.7 Combining Functions
233
73. Yes. If f x m 1 x b1 and g x m 2 x b2 , then f g x f m 2 x b2 m 1 m 2 x b2 b1 m 1 m 2 x m 1 b2 b1, which is a linear function, because it is of the form y mx b. The slope is m 1 m 2 . 74. g x 2x 1 and h x 4x 2 4x 7.
Method 1: Notice that 2x 12 4x 2 4x 1. We see that adding 6 to this quantity gives
2x 12 6 4x 2 4x 1 6 4x 2 4x 7, which is h x. So let f x x 2 6, and we have
f g x 2x 12 6 h x. Method 2: Since g x is linear and h x is a second degree polynomial, f x must be a second degree polynomial, that is, f x ax 2 bx c for some a, b, and c. Thus f g x f 2x 1 a 2x 12 b 2x 1 c
4ax 2 4ax a 2bx b c 4ax 2 4a 2b x a b c 4x 2 4x 7. Comparing this with f g x, we
have 4a 4 (the x 2 coefficients), 4a 2b 4 (the x coefficients), and a b c 7 (the constant terms) a 1 and
2a b 2 and a b c 7 a 1, b 0 c 6. Thus f x x 2 6.
f x 3x 5 and h x 3x 2 3x 2. Note since f x is linear and h x is quadratic, g x must also be quadratic. We can then use trial and error to find g x.
Another method is the following: We wish to find g so that f g x h x. Thus f g x 3x 2 3x 2 3 g x 5 3x 2 3x 2 3 g x 3x 2 3x 3 g x x 2 x 1.
75. The price per sticker is 015 0000002x and the number sold is x, so the revenue is R x 015 0000002x x 015x 0000002x 2 .
76. As found in Exercise 75, the revenue is R x 015x 0000002x 2 , and the cost is 0095x 00000005x 2 , so the profit is P x 015x 0000002x 2 0095x 00000005x 2 0055x 00000015x 2 .
77. (a) Because the ripple travels at a speed of 60 cm/s, the distance traveled in t seconds is the radius, so g t 60t. (b) The area of a circle is r 2 , so f r r 2 .
(c) f g g t2 60t2 3600t 2 cm2 . This function represents the area of the ripple as a function of time.
78. (a) Let f t be the radius of the spherical balloon in centimeters. Since the radius is increasing at a rate of 1 cm/s, the radius is f t t after t seconds. (b) The volume of the balloon can be written as g r 43 r 3 .
(c) g f 43 t3 43 t 3 . g f represents the volume as a function of time. 79. Let r be the radius of the spherical balloon in centimeters. Since the radius is increasing at a rate of 2 cm/s, the radius is r 2t after t seconds. Therefore, the surface area of the balloon can be written as S 4r 2 4 2t2 4 4t 2 16t 2 .
80. (a) f x 080x
(b) g x x 50
(c) f g x f x 50 080 x 50 080x 40. f g represents applying the $50 coupon, then the 20% discount. g f x g 080x 080x 50. g f represents applying the 20% discount, then the $50 coupon. So applying the 20% discount, then the $50 coupon gives the lower price.
81. (a) f x 090x
(b) g x x 100
(c) f g x f x 100 090 x 100 090x 90. f g represents applying the $100 coupon, then the 10% discount. g f x g 090x 090x 100. g f represents applying the 10% discount, then the $100 coupon. So applying the 10% discount, then the $100 coupon gives the lower price. 82. Let t be the time since the plane flew over the radar station. (a) Let s be the distance in miles between the plane and the radar station, and let d be the horizontal distance that the plane has flown. Using the Pythagorean theorem, s f d 1 d 2 .
234
CHAPTER 2 Functions
(b) Since distance rate time, we have d g t 350t. (c) s t f g t f 350t 1 350t2 1 122,500t 2 .
83. A x 105x A A x A A x A 105x 105 105x 1052 x. A A A x A A A x A 1052 x 105 1052 x 1053 x. A A A A x A A A A x A 1053 x 105 1053 x 1054 x. A represents the amount in
the account after 1 year; A A represents the amount in the account after 2 years; A A A represents the amount in the account after 3 years; and A A A A represents the amount in the account after 4 years. We can see that if we compose n copies of A, we get 105n x.
84. If g x is even, then h x f g x f g x h x. So yes, h is always an even function.
If g x is odd, then h is not necessarily an odd function. For example, if we let f x x 1 and g x x 3 , g is an odd function, but h x f g x f x 3 x 3 1 is not an odd function.
If g x is odd and f is also odd, then h x f g x f g x f g x f g x f g x h x. So in this case, h is also an odd function. If g x is odd and f is even, then h x f g x f g x f g x f g x f g x h x, so in this case, h is an even function.
2.8
ONE-TO-ONE FUNCTIONS AND THEIR INVERSES
1. A function f is one-to-one if different inputs produce different outputs. You can tell from the graph that a function is one-to-one by using the Horizontal Line Test. 2. (a) For a function to have an inverse, it must be one-to-one. f x x 2 is not one-to-one, so it does not have an inverse. However g x x 3 is one-to-one, so it has an inverse. (b) The inverse of g x x 3 is g1 x 3 x.
3. (a) Proceeding backward through the description of f , we can describe f 1 as follows: “Take the third root, subtract 5, then divide by 3.” 3 x 5 (b) f x 3x 53 and f 1 x . 3 4. Yes, the graph of f is one-to-one, so f has an inverse. Because f 4 1, f 1 1 4, and because f 5 3, f 1 3 5.
5. If the point 3 4 is on the graph of f , then the point 4 3 is on the graph of f 1 . [This is another way of saying that f 3 4 f 1 4 3.] 6. (a) False. For instance, if f x x, then f 1 x x, but
1 1 f 1 x. f x x
(b) This is true, by definition. 7. By the Horizontal Line Test, f is not one-to-one. 9. By the Horizontal Line Test, f is one-to-one. 11. By the Horizontal Line Test, f is not one-to-one.
8. By the Horizontal Line Test, f is one-to-one. 10. By the Horizontal Line Test, f is not one-to-one. 12. By the Horizontal Line Test, f is one-to-one.
13. f x 2x 4. If x1 x2 , then 2x1 2x2 and 2x1 4 2x2 4. So f is a one-to-one function. 14. f x 3x 2. If x1 x2 , then 3x1 3x2 and 3x1 2 3x2 2. So f is a one-to-one function.
SECTION 2.8 One-to-One Functions and Their Inverses
235
15. g x x. If x1 x2 , then x1 x2 because two different numbers cannot have the same square root. Therefore, g is a one-to-one function. 16. g x x. Because every number and its negative have the same absolute value (for example, 1 1 1), g is not a one-to-one function. 17. h x x 2 2x. Because h 0 0 and h 2 2 2 2 0 we have h 0 h 2. So f is not a one-to-one function. 18. h x x 3 8. If x1 x2 , then x13 x23 and x13 8 x23 8. So f is a one-to-one function.
19. f x x 4 5. Every nonzero number and its negative have the same fourth power. For example, 14 1 14 , so f 1 f 1. Thus f is not a one-to-one function.
20. f x x 4 5, 0 x 2. If x1 x2 , then x14 x24 because two different positive numbers cannot have the same fourth power. Thus, x14 5 x24 5. So f is a one-to-one function.
21. r t t 6 3, 0 t 5. If t1 t2 , then t16 t26 because two different positive numbers cannot have the same sixth power. Thus, t16 3 t26 3. So r is a one-to-one function.
22. r t t 4 1. Every nonzero number and its negative have the same fourth power. For example, 14 1 14 , so r 1 r 1. Thus r is not a one-to-one function. 1 1 1 23. f x 2 . Every nonzero number and its negative have the same square. For example, 1 , so 2 x 1 12 f 1 f 1. Thus f is not a one-to-one function. 1 1 1 24. f x . If x1 x2 , then . So f is a one-to-one function. x x1 x2 25. (a) f 2 7. Since f is one-to-one, f 1 7 2.
(b) f 1 3 1. Since f is one-to-one, f 1 3.
26. (a) f 5 18. Since f is one-to-one, f 1 18 5. (b) f 1 4 2. Since f is one-to-one, f 2 4.
27. f x 5 2x. Since f is one-to-one and f 1 5 2 1 3, then f 1 3 1. (Find 1 by solving the equation 5 2x 3.)
28. To find g1 5, we find the x value such that g x 5; that is, we solve the equation g x x 2 4x 5. Now
x 2 4x 5 x 2 4x 5 0 x 1 x 5 0 x 1 or x 5. Since the domain of g is [2 , x 1 is
the only value where g x 5. Therefore, g1 5 1. 29. (a) Because f 6 2, f 1 2 6.
(b) Because f 2 5, f 1 5 2.
(c) Because f 0 6, f 1 6 0.
30. (a) Because g 4 2, g1 2 4.
(b) Because g 7 5, g1 5 7.
(c) Because g 8 6, g1 6 8.
31. From the table, f 4 5, so f 1 5 4.
32. From the table, f 5 0, so f 1 0 5. 34. f f 1 6 6 33. f 1 f 1 1 35. From the table, f 6 1, so f 1 1 6. Also, f 2 6, so f 1 6 1. Thus, f 1 f 1 1 f 1 6 1. 36. From the table, f 5 0, so f 1 0 5. Also, f 4 5, so f 1 5 4. Thus, f 1 f 1 0 f 1 5 4. 37. f g x f x 6 x 6 6 x for all x. g f x g x 6 x 6 6 x for all x. Thus f and g are inverses of each other. x x 3 x for all x. 38. f g x f 3 3 3x x for all x. Thus f and g are inverses of each other. g f x g 3x 3
236
CHAPTER 2 Functions
x 4 4 x 4 4 x for all x. 3 3x 4 4 g f x g 3x 4 x for all x. Thus f and g are inverses of each other. 3 2x 2x 40. f g x f 25 2 2 x x for all x. 5 5 2 2 5x 5x g f x g 2 5x x for all x. Thus f and g are inverses of each other. 5 5 1 1 x for all x 0. Since f x g x, we also have g f x x for all x 0. Thus f and 41. f g x f x 1x g are inverses of each other. 5 42. f g x f 5 x 5 x x for all x. 5 g f x g x 5 x 5 x for all x. Thus f and g are inverses of each other.
39. f g x f
x 4 3
3
2 x 9 x 9 9 x 9 9 x for all x 9. g f x g x 2 9 x 2 9 9 x 2 x for all x 0. Thus f and g are inverses of each other. 3 44. f g x f x 113 x 113 1 x 1 1 x for all x. 3 g f x g x 3 1 x 113 1 x 1 1 x for all x. Thus f and g are inverses of each other. 1 1 x for all x 0. 1 45. f g x f 1 x 1 1 x 1 1 1 x 1 1 x for all x 1. Thus f and g are inverses of each other. g f x g 1 x 1 x 1 2 4 x2 4 4 x 2 4 4 x 2 x 2 x, for all 0 x 2. (Note that the last 46. f g x f 43. f g x f
equality is possible since x 0.) 2 2 g f x g 4x 4 4 x 2 4 4 x 2 x 2 x, for all 0 x 2. (Again, the last equality
is possible since x 0.) Thus f and g are inverses of each other. 2x2 2 2x 2 2 x 1 4x 2x 2 x1 2x2 x for all x 1. 47. f g x f x 1 2x 2 2 x 1 4 2 x1 2 x2 2 4x 2 x 2 2 x 2 x 2 x2 x2 x for all x 2. Thus f and g are inverses of g f x g x 2 x 2 1 x 2 4 1 x2
each other.
54x 5 5 4x 5 1 3x 19x 13x x for all x 13 . 54x 3 5 4x 4 1 3x 19 3 13x 4 5 4 x5 19x 5 3x 4 4 x 5 x 5 3x4 g f x g for all x 43 . Thus f and g are inverses x5 3x 4 3x 4 3 5 19 x 1 3 3x4
48. f g x f
5 4x 1 3x
of each other.
49. f x 3x 5. y 3x 5 3x y 5 x 13 y 5 13 y 53 . So f 1 x 13 x 53 .
50. f x 7 5x. y 7 5x 5x 7 y x 15 7 y 15 y 75 . So f 1 x 15 x 75 .
SECTION 2.8 One-to-One Functions and Their Inverses
51. f x 5 4x 3 . y 5 4x 3 4x 3 5 y x 3 14 5 y x 3 14 5 y. So f 1 x 3 14 5 x. 52. f x 3x 3 8. y 3x 3 8 3x 3 y 8 x 3 13 y 83 x 3 13 y 83 . So f 1 x 3 13 x 8.
1 1 1 1 1 .y x 2 x 2. So f 1 x 2. x 2 x 2 y y x x 2 x 2 .y y x 2 x 2 x y 2y x 2 x y x 2 2y x y 1 2 y 1 f x x 2 x 2 2 y 1 2 x 1 x . So f 1 x . y1 x 1 x 4y x . y y x 4 x x y 4y x x x y 4y x 1 y 4y x . So f x x 4 x 4 1y 4x f 1 x . 1x 3x 2y 3x .y y x 2 3x x y 2y 3x x y 3x 2y x y 3 2y x . So f x x 2 x 2 y3 2x f 1 x . x 3 2x 5 2x 5 .y y x 7 2x 5 x y 7y 2x 5 x y 2x 7y 5 x y 2 7y 5 f x x 7 x 7 7y 5 7x 5 x . So f 1 x . y2 x 2 4x 2 4x 2 .y y 3x 1 4x 2 3x y y 4x 2 4x 3x y y 2 x 4 3y y 2 f x 3x 1 3x 1 y2 x 2 x . So f 1 x . 4 3y 4 3x 2x 3 2x 3 f x .y y 1 5x 2x 3 y 5x y 2x 3 2x 5x y y 3 x 2 5y y 3 1 5x 1 5x y3 x 3 x . So f 1 x . 5y 2 5x 2 3 4x y3 3 4x .y y 8x 1 3 4x 8x y y 3 4x 4x 2y 1 y 3 x . f x 8x 1 8x 1 4 2y 1 x 3 So f 1 x . 4 2x 1 f x 4 x 2 , x 0. y 4 x 2 x 2 4 y x 4 y. So f 1 x 4 x, x 4. [Note that x 0 f x 4.] 2 2 2 1 y 14 f x x 2 x x 2 x 14 14 x 12 14 , x . y x 12 14 x 12 2 x 12 y 14 x y 14 12 , y 14 . So f 1 x x 14 12 , x 14 . (Note that x 12 , so that 2 2 x 12 0, and hence x 12 y 14 x 12 y 14 . Also, since x 12 , y x 12 14 14 so that y 14 0, and hence y 14 is defined.) f x x 6 , x 0. y x 6 x 6 y for x 0. The range of f is y y 0, so f 1 x 6 x, x 0. 1 1 1 1 1 f x 2 , x 0. y 2 x 2 x . The range of f is y y 0, so f 1 x , x 0. y y x x x
53. f x
54.
55.
56.
57.
58.
59.
60.
61.
62.
63. 64.
237
2 x3 2 x3 .y 5y 2 x 3 x 3 2 5y x 3 2 5y. Thus, f 1 x 3 2 5x. 5 5 7 7 66. f x x 5 6 . y x 5 6 7 y x 5 6 x 5 7 y 6 x 5 7 y 6. Thus, f 1 x 5 7 x 6. 65. f x
238
CHAPTER 2 Functions
67. f x
5 8x. Note that the range of f (and thus the domain of f 1 ) is [0 . y 5 8x y 2 5 8x
y2 5 x2 5 . Thus, f 1 x , x 0. 8x y 2 5 x 8 8
68. f x 2
3 x. The range of f is [2 . y 2 3 x y 2 3 x y 22 3 x
x y 22 3. Thus, f 1 x x 22 3, x 2.
69. f x 2
3
x. y 2
3
x y2
3
x x y 23 . Thus, f 1 x x 23 .
4 x 2 , 0 x 2. The range of f is [0 2]. y 4 x 2 y 2 4 x 2 x 2 4 y 2 x 4 y 2 . Thus, f 1 x 4 x 2 , 0 x 2.
70. f x
71. (a), (b) f x 3x 6
72. (a), (b) f x 16 x 2 , x 0
y
y
f f Ð!
1
f
x
1
f Ð!
2 2
(c) f x 3x 6. y 3x 6 3x y 6 x 13 y 6. So f 1 x 13 x 6.
x
(c) f x 16 x 2 , x 0. y 16 x 2 x 2 16 y x 16 y. So f 1 x 16 x, x 16. (Note: x 0 f x 16 x 2 16.)
73. (a), (b) f x
y
74. (a), (b) f x x 3 1
x 1
y
f f Ð!
f Ð!
1 1
f
x
1 x
1
(c) f x
x 1, x 1. y
x 1, y 0
y 2 x 1 x y 2 1 and y 0. So f 1 x x 2 1, x 0.
(c) f x x 3 1 y x 3 1 x 3 y 1 x 3 y 1. So f 1 x 3 x 1.
SECTION 2.8 One-to-One Functions and Their Inverses
75. f x x 3 x. Using a graphing device and the
Horizontal Line Test, we see that f is not a one-to-one function. For example, f 0 0 f 1.
-2
76. f x x 3 x. Using a graphing device and the
Horizontal Line Test, we see that f is a one-to-one function.
-2
2
x 12 . Using a graphing device and the x 6 Horizontal Line Test, we see that f is a one-to-one
77. f x
function.
239
78. f x
2
x 3 4x 1. Using a graphing device and the
Horizontal Line Test, we see that f is not a one-to-one function. For example, f 0 1 f 2.
10 2
-20
20 -2
-10
79. f x x x 6. Using a graphing device and the Horizontal Line Test, we see that f is not a one-to-one function. For example f 0 6 f 2.
0
2
80. f x x x. Using a graphing device and the
Horizontal Line Test, we see that f is a one-to-one function. 20
10
-10
-5
10
5 -20
-10
81. (a) y f x 2 x x y 2. So
82. (a) y f x 2 12 x 12 x 2 y x 4 2y.
f 1 x x 2.
So f 1 x 4 2x.
(b)
(b) 10 5
-5
5
-10
10
-5 -10
240
CHAPTER 2 Functions
83. (a) y g x
x 3, y 0 x 3 y 2 , y 0
x y 2 3, y 0. So g1 x x 2 3, x 0.
84. (a) y g x x 2 1, x 0 x 2 y 1, x 0 x y 1. So g1 x x 1.
(b)
(b) 10
10
5
5
5
10 5
85. If we restrict the domain of f x to [0 , then y 4 x 2 x 2 4 y x positive square root). So f 1 x 4 x.
10
4 y (since x 0, we take the
If we restrict the domain of f x to 0], then y 4 x 2 x 2 4 y x 4 y (since x 0, we take the negative square root). So f 1 x 4 x.
86. If we restrict the domain of g x to [1 , then y x 12 x 1 root) x 1 y. So g 1 x 1 x.
87. If we restrict the domain of h x to [2 , then y x 22 x 2 root) x 2 y. So h 1 x 2 x.
y (since x 1 we take the positive square
If we restrict the domain of g x to 1], then y x 12 x 1 y (since x 1 we take the negative square root) x 1 y. So g 1 x 1 x.
y (since x 2, we take the positive square
If we restrict the domain of h x to 2], then y x 22 x 2 y (since x 2, we take the negative square root) x 2 y. So h 1 x 2 x.
x 3 if x 3 0 x 3 88. k x x 3 x 3 if x 3 0 x 3
If we restrict the domain of k x to [3 , then y x 3 x 3 y. So k 1 x 3 x.
If we restrict the domain of k x to 3], then y x 3 y x 3 x 3 y. So k 1 x 3 x.
89.
y f Ð!
90.
y
1
f
1
f
x 1
f Ð! 1
x
SECTION 2.8 One-to-One Functions and Their Inverses y
91. (a)
241
y
92. (a)
1 0
1
(b) Yes, the graph is unchanged upon reflection about the line y x. (c) y
1 1 1 x , so f 1 x . x y x
1
x
0
1
x
(b) Yes, the graph is unchanged upon reflection about the line y x.
x 3 y x 1 x 3 x 1 y3 x y 1 y 3 x . Thus, y1
(c) y
f 1 x
x 3 . x 1
93. (a) The price of a pizza with no toppings (corresponding to the y-intercept) is $16, and the cost of each additional topping (the rate of change of cost with respect to number of toppings) is $150. Thus, f n 16 15n. (b) p f n 16 15n p 16 15n n 23 p 16. Thus, n f 1 p 23 p 16. This function represents the number of toppings on a pizza that costs x dollars.
(c) f 1 25 23 25 16 23 9 6. Thus, a $25 pizza has 6 toppings. 94. (a) f x 500 80x.
p 500 1 p 500 . So x f 1 p . f 80 80 represents the number of hours the investigator spends on a case for x dollars. 720 1220 500 (c) f 1 1220 9. If the investigator charges $1220, he spent 9 hours investigating the case. 80 80 (b) p f x 500 80x. p 500 80x 80x p 500 x
t 2 V t 2 t 2 V t 95. (a) V f t 100 1 1 , 0 t 40. V 100 1 1 40 40 100 40 40 100 t V 1 t 40 4 V . Since t 40, we must have t f 1 V 40 4 V . f 1 represents time that 40 10 has elapsed since the tank started to leak. (b) f 1 15 40 4 15 245 minutes. In 245 minutes the tank has drained to just 15 gallons of water. 96. (a) g r 18,500 025 r 2 . 18,500 025 r 2 4625 18,500r 2 18,500r 2 4625 4625 4625 4625 1 2 1 r r . Since r represents a distance, r 0, so g . g 18,500 18,500 18,500 represents the radial distance from the center of the vein at which the blood has velocity . 4625 30 (b) g 1 30 0498 cm. The velocity is 30 cms at a distance of 0498 cm from the center of the artery 18,500 or vein.
242
CHAPTER 2 Functions
97. (a) D f p 3 p 150. D 3 p 150 3 p 150 D p 50 13 D. So f 1 D 50 13 D. f 1 D represents the price that is associated with demand D. (b) f 1 30 50 13 30 40. So when the demand is 30 units, the price per unit is $40. 98. (a) F g C 95 C 32. F 95 C 32 95 C F 32 C 59 F 32. So g1 F 59 F 32. g1 F represents the Celsius temperature that corresponds to the Fahrenheit temperature of F. (b) F 1 86 59 86 32 59 54 30. So 86 Fahrenheit is the same as 30 Celsius. 99. (a) f 1 U 102396U .
(b) U f x 09766x. U 09766x x 10240U . So f 1 U 10240U . f 1 U represents the value of U US dollars in Canadian dollars.
(c) f 1 12,250 10240 12,250 12,54352. So $12,250 in US currency is worth $12,54352 in Canadian currency.
01x, if 0 x 20,000 100. (a) f x 2000 02 x 20,000 if x 20,000 (b) We will find the inverse of each piece of the function f .
f 1 x 01x. T 01x x 10T . So f 11 T 10T . f 2 x 2000 02 x 20,000 02x 2000. T 02x 2000 02x T 2000 x 5T 10,000. So f 21 T 5T 10,000.
10T , if 0 T 2000 Since f 0 0 and f 20,000 2000 we have f 1 T 5T 10,000 if T 2000
This represents the
taxpayer’s income.
(c) f 1 10,000 5 10,000 10,000 60,000. The required income is 60,000. 101. (a) f x 085x.
(b) g x x 1000.
(c) H x f g x f x 1000 085 x 1000 085x 850.
(d) P H x 085x 850. P 085x 850 085x P 850 x 1176P 1000. So
H 1 P 1176P 1000. The function H 1 represents the original sticker price for a given discounted price P.
(e) H 1 13,000 1176 13,000 1000 16,288. So the original price of the car is $16,288 when the discounted price ($1000 rebate, then 15% off) is $13,000. 102. f x mx b. Notice that f x1 f x2 mx1 b mx2 b mx1 mx2 . We can conclude that x1 x2 if and only if m 0. Therefore f is one-to-one if and only if m 0. If m 0, f x mx b y mx b mx y b x
yb x b . So, f 1 x . m m
2x 1 is “multiply by 2, add 1, and then divide by 5 ”. So the reverse is “multiply by 5, subtract 1, and then 5 5x 1 2 1 5x 1 1 5x 5x 1 5x 1 2 . Check: f f 1 x f x divide by 2 ” or f 1 x 2 2 5 5 5 2x 1 5 1 2x 1 2x 1 1 2x 5 and f 1 f x f 1 x. 5 2 2 2
103. (a) f x
CHAPTER 2
Review
243
1 1 3 is “take the negative reciprocal and add 3 ”. Since the reverse of “take the negative x x reciprocal” is “take the negative reciprocal ”, f 1 x is “subtract 3 and take the negative reciprocal ”, that is, 1 x 3 1 1 1 1 . Check: f f x f 3 3 x 3 x and f x 3 1 1 x 3 x 3 1 x 3 1 1 1 x x. f 1 f x f 1 3 1 1 1 x 1 3 3 x x (c) f x x 3 2 is “cube, add 2, and then take the square root”. So the reverse is “square, subtract 2, then take 3 the cube root ” or f 1 x x 2 2. Domain for f x is 3 2 ; domain for f 1 x is [0 . Check: 3 3 2 3 2 1 x 2 x 2 2 x 2 2 2 x 2 x (on the appropriate domain) and f f x f 2 3 3 3 1 1 3 f f x f x 2 x 3 2 2 x 3 2 2 x 3 x (on the appropriate domain). (b) f x 3
(d) f x 2x 53 is “double, subtract 5, and then cube”. So the reverse is “take the cube root, add 3 x 5 Domain for both f x and f 1 x is . Check: 5, and divide by 2” or f 1 x 2 3 3 3 3 3 x 5 x 5 3 f f 1 x f 2 5 3 x 5 5 3 x x 3 x and 2 2 x 2x 53 5 2x 2x 5 5 1 1 3 f x. f x f 2x 5 2 2 2 In a function like f x 3x 2, the variable occurs only once and it easy to see how to reverse the operations step by step. But in f x x 3 2x 6, you apply two different operations to the variable x (cubing and multiplying by 2) and then add 6, so it is not possible to reverse the operations step by step.
104. f I x f x; therefore f I f . I f x f x; therefore I f f . By definition, f f 1 x x I x; therefore f f 1 I . Similarly, f 1 f x x I x; therefore f 1 f I . 105. (a) We find g 1 x: y 2x 1 2x y 1 x 12 y 1. So g1 x 12 x 1. Thus 2 f x h g1 x h 12 x 1 4 12 x 1 4 12 x 1 7 x 2 2x 1 2x 2 7 x 2 6.
(b) f g h f 1 f g f 1 h I g f 1 h g f 1 h. Note that we compose with f 1 on the left on each side of the equation. We find f 1 : y 3x 5 3x y 5 x 13 y 5. So f 1 x 13 x 5. Thus g x f 1 h x f 1 3x 2 3x 2 13 3x 2 3x 2 5 13 3x 2 3x 3 x 2 x 1.
CHAPTER 2 REVIEW 1. “Square, then subtract 5” can be represented by the function f x x 2 5. 2. “Divide by 2, then add 9” can be represented by the function g x
x 9. 2
3. f x 3 x 10: “Add 10, then multiply by 3.” 4. f x 6x 10: “Multiply by 6, then subtract 10, then take the square root.”
244
CHAPTER 2 Functions
5. g x x 2 4x
6. h x 3x 2 2x 5 x
g x
x
h x
5
2
3
0
0
1
4
1
3
0
5
2
4
1
0
3
3
2
11
1
7. C x 5000 30x 0001x 2 (a) C 1000 5000 30 1000 0001 10002 $34,000 and
C 10,000 5000 30 10,000 0001 10,0002 $205,000.
(b) From part (a), we see that the total cost of printing 1000 copies of the book is $34,000 and the total cost of printing 10,000 copies is $205,000. (c) C 0 5000 30 0 0001 02 $5000. This represents the fixed costs associated with getting the print run ready.
(d) The net change in C as x changes from 1000 to 10,000 is C 10,000 C 1000 205,000 34,000 $171,000, and 171,000 C 10,000 C 1000 $19copy. the average rate of change is 10,000 1000 9000 8. E x 400 003x (a) E 2000 400 003 2000 $460 and E 15 000 400 003 15,000 $850. (b) From part (a), we see that if Reynalda sells $2000 worth of goods, she makes $460, and if she sells $15,000 worth of goods, she makes $850. (c) E 0 400 003 0 $400 is Reynalda’s base weekly salary.
(d) The net change in E as x changes from 2000 to 15,000 is E 15,000 E 2000 850 460 $390, and the average 390 E 15,000 E 2000 $003 per dollar. rate of change is 15,000 2000 13,000 (e) Because the value of goods sold x is multiplied by 003 or 3%, we see that Reynalda earns a percentage of 3% on the goods that she sells. 9. f x x 2 4x 6; f 0 02 4 0 6 6; f 2 22 4 2 6 2;
f 2 22 4 2 6 18; f a a2 4 a 6 a 2 4a 6; f a a2 4 a 6 a 2 4a 6;
f x 1 x 12 4 x 16 x 2 2x 14x 46 x 2 2x 3; f 2x 2x2 4 2x6 4x 2 8x 6. 10. f x 4 3x 6; f 5 4 15 6 1; f 9 4 27 6 4 21; f a 2 4 3a 6 6 4 3a; f x 4 3 x 6 4 3x 6; f x 2 4 3x 2 6.
11. By the Vertical Line Test, figures (b) and (c) are graphs of functions. By the Horizontal Line Test, figure (c) is the graph of a one-to-one function. 12. (a) f 2 1 and f 2 2.
(b) The net change in f from 2 to 2 is f 2 f 2 2 1 3, and the average rate of change is f 2 f 2 3 . 2 2 4 (c) The domain of f is [4 5] and the range of f is [4 4]. (d) f is increasing on 4 2 and 1 4; f is decreasing on 2 1 and 4 5. (e) f has local maximum values of 1 (at x 2) and 4 (at x 4).
(f) f is not a one-to-one, for example, f 2 1 f 0. There are many more examples. 13. Domain: We must have x 3 0 x 3. In interval notation, the domain is [3 . Range: For x in the domain of f , we have x 3 x 3 0 x 3 0 f x 0. So the range is [0 .
CHAPTER 2
Review
245
14. F t t 2 2t 5 t 2 2t 1 5 1 t 12 4. Therefore F t 4 for all t. Since there are no restrictions on t, the domain of F is , and the range is [4 .
15. f x 7x 15. The domain is all real numbers, . 16. f x 17. f x
2x 1 . Then 2x 1 0 x 12 . So the domain of f is x x 12 . 2x 1
x 4. We require x 4 0 x 4. Thus the domain is [4 .
18. f x 3x
19. f x
2 . The domain of f is the set of x where x 1 0 x 1. So the domain is 1 . x 1
1 1 1 . The denominators cannot equal 0, therefore the domain is x x 0 1 2. x x 1 x 2
2x 2 5x 3 2x 2 5x 3 . The domain of g is the set of all x where the denominator is not 0. So the 2 2x 1 x 3 2x 5x 3 domain is x 2x 1 0 and x 3 0 x x 12 and x 3 .
20. g x
21. h x
4 x x 2 1. We require the expression inside the radicals be nonnegative. So 4 x 0 4 x; also
x 2 1 0 x 1 x 1 0. We make a table:
1
1 1
1
Sign of x 1
Sign of x 1
Interval
Sign of x 1 x 1
Thus the domain is 4] 1] [1 1] [1 4].
3 2x 1 . Since we have an odd root, the domain is the set of all x where the denominator is not 0. Now 22. f x 3 2x 2 3 2x 2 0 3 2x 2 2x 8 x 4. Thus the domain of f is x x 4. 23. f x 1 2x
24. f x 13 x 5, 2 x 8
y
y
1
1 1
x
1
x
246
CHAPTER 2 Functions
25. f x 3x 2
26. f x 14 x 2
y
y
1
1
0
x
5 0
27. f x 2x 2 1
1
x
28. f x x 14
y
y 5 0
1
x
1 0
29. f x 1
1
x
30. f x 1
x y
x 2 y
1 0
2
x
1 0
1
x
31. f x 12 x 3
32. f x y
y
1
2 0
3 x
1
x
0
2
x
CHAPTER 2
33. f x x
34. f x x 1
y
Review
247
y
1 0
1
x
1 0
1 35. f x 2 x
36. f x y
1
x
1 x 13
y
1 0
1
x 5 0
1
x
x if x 0 38. f x x 2 if 0 x 2 1 if x 2
1 x if x 0 37. f x 1 if x 0
y
y
1 1
x
1 1
x
39. x y 2 14 y 2 14 x y 14 x, so the original equation does not define y as a function of x. 40. 3x
y 8
y 3x 8 y 3x 82 , so the original equation defines y as a function of x.
13 41. x 3 y 3 27 y 3 x 3 27 y x 3 27 , so the original equation defines y as a function of x (since the cube root function is one-to-one).
42. 2x y 4 16 y 4 2x 16 y 4 2x 16, so the original equation does not define y as a function of x.
248
CHAPTER 2 Functions
43. f x 6x 3 15x 2 4x 1
(ii) [8 8] by [8 8]
(i) [2 2] by [2 2]
2 5
-2
-5
2
5 -5
-2
(iii) [4 4] by [12 12]
(iv) [100 100] by [100 100] 100
10
-4
-2
2
4
-100
-10
100 -100
From the graphs, we see that the viewing rectangle in (iii) produces the most appropriate graph.
44. f x
100 x 3
(i) [4 4] by [4 4]
(ii) [10 10] by [10 10] 10
-4
-2
2
4
-10
10 -10
(iii) [10 10] by [10 40]
(iv) [100 100] by [100 100]
40
100
20 -100 -10
10
100 -100
From the graphs, we see that the viewing rectangle in (iii) produces the most appropriate graph of f .
CHAPTER 2
45. (a) We graph f x
9 x 2 in the viewing rectangle
[4 4] by [1 4].
Review
46. (a) We graph f x x 2 3 in the viewing rectangle [5 5] by [6 1].
4 -5
5
2
-4
-2
2
-5
4
(b) From the graph, the domain of f is [3 3] and the
(b) From the graph, the domain of f is 173] [173 and the range of f is
range of f is [0 3].
0].
47. (a) We graph f x
x 3 4x 1 in the viewing
rectangle [5 5] by [1 5].
48. (a) We graph f x x 4 x 3 x 2 3x 6 in the viewing rectangle [3 4] by [20 100]. 100
4
50
2
-5
5
(b) From the graph, the domain of f is approximately [211 025] [186 and the range of f is
[0 .
49. f x x 3 4x 2 is graphed in the viewing rectangle
[5 5] by [20 10]. f x is increasing on 0 and 267 . It is decreasing on 0 267.
-2
2
4
(b) From the graph, the domain of f is and the range of f is approximately [710 .
50. f x x 4 16 is graphed in the viewing rectangle
[5 5] by [5 20]. f x is increasing on 2 0 and 2 . It is decreasing on 2 and 0 2. 20
-5
5
-20
10
-5
5
f 8 f 4 4 1. 84 4
51. The net change is f 8 f 4 8 12 4 and the average rate of change is
52. The net change is g 30 g 10 30 5 35 and the average rate of change is 53. The net change is f 2 f 1 6 2 4 and the average rate of change is
35 7 g 30 g 10 . 30 10 20 4
4 f 2 f 1 . 2 1 3
54. The net change is f 3 f 1 1 5 6 and the average rate of change is
6 f 3 f 1 3. 31 2
249
250
CHAPTER 2 Functions
55. The net change is f 4 f 1 42 2 4 12 2 1 8 1 9 and the average rate of change is 9 f 4 f 1 3. 41 3
56. The net change is g a h g a a h 12 a 12 2ah 2h h 2 and the average rate of change is g a h g a 2ah 2h h 2 2a 2 h. aha h
57. f x 2 3x2 9x 2 12x 4 is not linear. It cannot be expressed in the form f x ax b with constant a and b. 58. g x
x 3 15 x 35 is linear with a 15 and b 35 . 5 y
59. (a)
y
60. (a)
1 0
1
1
x
0
(b) The slope of the graph is the value of a in the
1
x
(b) The slope of the graph is the value of a in the
equation f x ax b 3x 2; that is, 3.
equation f x ax b 12 x 3; that is, 12 .
(c) The rate of change is the slope of the graph, 3.
(c) The rate of change is the slope of the graph, 12 .
61. The linear function with rate of change 2 and initial value 3 has a 2 and b 3, so f x 2x 3. 62. The linear function whose graph has slope 12 and y-intercept 1 has a 12 and b 1, so f x 12 x 1. 63. Between x 0 and x 1, the rate of change is f x 2x 3. 64. Between x 0 and x 2, the rate of change is is f x 14 x 6.
53 f 1 f 0 2. At x 0, f x 3. Thus, an equation is 10 1 f 2 f 0 55 6 14 . At x 0, f x 6. Thus, an equation 20 2
65. The points 0 4 and 8 0 lie on the graph, so the rate of change is y 12 x 4.
04 1 . At x 0, y 4. Thus, an equation is 80 2
66. The points 0 4 and 2 0 lie on the graph, so the rate of change is is y 2x 4.
0 4 2. At x 0, y 4. Thus, an equation 20
67. P t 3000 200t 01t 2 (a) P 10 3000 200 10 01 102 5010 represents the population in its 10th year (that is, in 1995), and
P 20 3000 200 20 01 202 7040 represents its population in its 20th year (in 2005). P 20 P 10 7040 5010 2030 (b) The average rate of change is 203 peopleyear. This represents the 20 10 10 10 average yearly change in population between 1995 and 2005.
CHAPTER 2
Review
251
68. D t 3500 15t 2 (a) D 0 3500 15 02 $3500 represents the amount deposited in 1995 and D 15 3500 15 152 $6875 represents the amount deposited in 2010. 13500 (b) Solving the equation D t 17,000, we get 17,000 3500 15t 2 15t 2 13,500 t 2 900 15 t 30, so thirty years after 1995 (that is, in the year 2025) she will deposit $17,000. 6875 3500 D 15 D 0 $225year. This represents the average annual (c) The average rate of change is 15 0 15 increase in contributions between 1995 and 2010. 69. f x 12 x 6 (a) The average rate of change of f between x 0 and x 2 is 1 2 6 1 0 6 5 6 1 f 2 f 0 2 2 , and the average rate of change of f between x 15 20 2 2 2 and x 50 is 1 50 6 1 15 6 19 32 f 50 f 15 1 2 2 . 50 15 35 35 2 (b) The rates of change are the same. (c) Yes, f is a linear function with rate of change 12 . 70. f x 8 3x
[8 3 2] [8 3 0] 28 f 2 f 0 3, 20 2 2 and the average rate of change of f between x 15 and x 50 is [8 3 50] [8 3 15] f 50 f 15 142 37 3. 50 15 35 35 (b) The rates of change are the same. (a) The average rate of change of f between x 0 and x 2 is
(c) Yes, f is a linear function with rate of change 3. 71. (a) y f x 8. Shift the graph of f x upward 8 units.
(b) y f x 8. Shift the graph of f x to the left 8 units.
(c) y 1 2 f x. Stretch the graph of f x vertically by a factor of 2, then shift it upward 1 unit. (d) y f x 2 2. Shift the graph of f x to the right 2 units, then downward 2 units. (e) y f x. Reflect the graph of f x about the y-axis.
(f) y f x. Reflect the graph of f x first about the y-axis, then reflect about the x-axis.
(g) y f x. Reflect the graph of f x about the x-axis.
(h) y f 1 x. Reflect the graph of f x about the line y x.
72. (a) y f x 2
(b) y f x
y
1
(c) y 3 f x
y
1 1
x
y
1 1
x
1
x
252
CHAPTER 2 Functions
(d) y 12 f x 1
(e) y f 1 x
y
(f) y f x
y
1
1 1
1 1
x
y
1
x
x
73. (a) f x 2x 5 3x 2 2. f x 2 x5 3 x2 2 2x 5 3x 2 2. Since f x f x, f is not even. f x 2x 5 3x 2 2. Since f x f x, f is not odd. (b) f x x 3 x 7 . f x x3 x7 x 3 x 7 f x, hence f is odd.
1 x2 1 x2 1 x2 . f f x. Since f x f x, f is even. x 1 x2 1 x2 1 x2 1 1 1 1 . f x . f x . Since f x f x , f is not even, and since (d) f x x 2 2x x 2 x 2 f x f x, f is not odd. (c) f x
74. (a) This function is odd. (b) This function is neither even nor odd. (c) This function is even. (d) This function is neither even nor odd. 75. g x 2x 2 4x 5 2 x 2 2x 5 2 x 2 2x 1 5 2 2 x 12 7. So the local minimum value 7 when x 1.
2 76. f x 1 x x 2 x 2 x 1 x 2 x 14 1 14 x 12 54 . So the local maximum value is 54 when x 12 .
77. f x 33 16x 25x 3 . In the first viewing rectangle, [2 2] by [4 8], we see that f x has a local maximum and a local minimum. In the next viewing rectangle, [04 05] by [378 380], we isolate the local maximum value as approximately 379 when x 046. In the last viewing rectangle, [05 04] by [280 282], we isolate the local minimum value as 281 when x 046. 3.80
2.82
3.79
2.81
5
-2
2
3.78 0.40
0.45
0.50
-0.50
-0.45
2.80 -0.40
CHAPTER 2
Review
253
78. f x x 23 6 x13 . In the first viewing rectangle, [10 10] by [10 10], we see that f x has a local maximum and a local minimum. The local minimum is 0 at x 0 (and is easily verified). In the next viewing rectangle, [395 405] by [316 318], we isolate the local maximum value as approximately 3175 when x 400. 10
3.18 3.17
-10
10 3.16 3.95
-10
4.00
4.05
79. h t 16t 2 48t 32 16 t 2 3t 32 16 t 2 3t 94 32 36 2 16 t 2 3t 94 68 16 t 32 68 The stone reaches a maximum height of 68 feet.
1500 12x 00004x 2 00004 x 2 30,000x 1500 00004 x 2 30,000x 225,000,000 1500 90,000 00004 x 15,0002 88,500
80. P x
The maximum profit occurs when 15,000 units are sold, and the maximum profit is $88,500. 81. f x x 2, g x x 2
82. f x x 2 1, g x 3 x 2 5
5
-4
-2
2
-2
2
83. f x x 2 3x 2 and g x 4 3x. (a) f g x x 2 3x 2 4 3x x 2 6x 6 (b) f g x x 2 3x 2 4 3x x 2 2 (c) f g x x 2 3x 2 4 3x 4x 2 12x 8 3x 3 9x 2 6x 3x 3 13x 2 18x 8 x 2 3x 2 f , x 43 (d) x g 4 3x
(e) f g x f 4 3x 4 3x2 3 4 3x 2 16 24x 9x 2 12 9x 2 9x 2 15x 6 (f) g f x g x 2 3x 2 4 3 x 2 3x 2 3x 2 9x 2 84. f x 1 x 2 and g x x 1. (Remember that the proper domains must apply.) 2 (a) f g x f x 1 1 x 1 1x 1 x (b) g f x g 1 x 2 1 x 2 1 x 2 x (c) f g 2 f g 2 f 2 1 f 1 1 12 2. (d) f f 2 f f 2 f 1 22 f 5 1 52 26. (e) f g f x f g f x f x 1 x2 1 x 2 . Note that g f x x by part (b). (f) g f g x g f g x g x x 1. Note that f g x x by part (a).
254
CHAPTER 2 Functions
85. f x 3x 1 and g x 2x x 2 . f g x f 2x x 2 3 2x x 2 1 3x 2 6x 1, and the domain is .
g f x g 3x 1 2 3x 1 3x 12 6x 2 9x 2 6x 1 9x 2 12x 3 , and the domain is f f x f 3x 1 3 3x 1 1 9x 4, and the domain is . 2 g g x g 2x x 2 2 2x x 2 2x x 2 4x 2x 2 4x 2 4x 3 x 4 x 4 4x 3 6x 2 4x, and domain is .
2 , has domain x x 4. x, has domain x x 0. g x x 4 2 2 . f g x is defined whenever both g x and f g x are defined; that is, f g x f x 4 x 4 2 2 whenever x 4 and 0. Now 0 x 4 0 x 4. So the domain of f g is 4 . x 4 x 4 2 . g f x is defined whenever both f x and g f x are defined; that is, whenever g f x g x x 4 x 0 and x 4 0. Now x 4 0 x 16. So the domain of g f is [0 16 16 . x x x 14 . f f x is defined whenever both f x and f f x are defined; that is, f f x f whenever x 0. So the domain of f f is [0 . 2 x 4 2 x 4 2 g g x is defined whenever both g x and g g x g 2 x 4 2 4 x 4 9 2x 4 x 4 g g x are defined; that is, whenever x 4 and 9 2x 0. Now 9 2x 0 2x 9 x 92 . So the domain of g g is x x 92 4 .
86. f x
1 x, g x 1 x 2 and h x 1 x. 2 f 1 12 x x f g h x f g h x f g 1 x f 1 1 x 2 f x 2 x 1 x 2 x 1 2 x x 1 x 1 x
87. f x
88. If h x
x and g x 1 x , then g h x g
1 f g h x f 1 x T x. 1 x
1 x 1 x. If f x , then x
89. f x 3 x 3 . If x1 x2 , then x13 x23 (unequal numbers have unequal cubes), and therefore 3 x13 3 x23 . Thus f is a one-to-one function. 90. g x 2 2x x 2 x 2 2x 1 1 x 12 1. Since g 0 2 g 2 , as is true for all pairs of numbers equidistant from 1, g is not a one-to-one function.
1 . Since the fourth powers of a number and its negative are equal, h is not one-to-one. For example, x4 1 1 1 and h 1 1, so h 1 h 1. h 1 4 1 14 92. r x 2 x 3. If x1 x2 , then x1 3 x2 3, so x1 3 x2 3 and 2 x1 3 2 x2 3. Thus r is one-to-one. 91. h x
CHAPTER 2
93. p x 33 16x 25x 3 . Using a graphing device and the Horizontal Line Test, we see that p is not a one-to-one
Test
255
94. q x 33 16x 25x 3 . Using a graphing device and the Horizontal Line Test, we see that q is a one-to-one
function.
function. 10
10
-5
5
-5
5
-10
-10
95. f x 3x 2 y 3x 2 3x y 2 x 13 y 2. So f 1 x 13 x 2. 2x 1 2x 1 y 2x 1 3y 2x 3y 1 x 12 3y 1 So f 1 x 12 3x 1. 96. f x 3 3 97. f x x 13 y x 13 x 1 3 y x 3 y 1. So f 1 x 3 x 1. 98. f x 1 5 x 2. y 1 5 x 2 y 1 5 x 2 x 2 y 15 x 2 y 15 . So f 1 x 2 x 15 .
99. The graph passes the Horizontal Line Test, so f has an inverse. Because f 1 0, f 1 0 1, and because f 3 4, f 1 4 3.
100. The graph fails the Horizontal Line Test, so f does not have an inverse. 101. (a), (b) f x x 2 4, x 0
102.
y
f
(a) If x1 x2 , then 4 x1 4 x2 , and so 1 4 x1 1 4 x2 . Therefore, f is a one-to-one function.
f Ð!
y
(b), (c)
1 1
f Ð!
x
f 1
(c) f x x 2 4, x 0 y x 2 4, y 4 x 2 y 4 x y 4. So f 1 x x 4, x 4.
x
1
(d) f x 1
4
x. y 1
4
x
4
x y 1
x y 14 . So f 1 x x 14 ,
x 1. Note that the domain of f is [0 , so y 1 4 x 1. Hence, the domain of f 1 is
[1 .
CHAPTER 2 TEST 1. By the Vertical Line Test, figures (a) and (b) are graphs of functions. By the Horizontal Line Test, only figure (a) is the graph of a one-to-one function.
256
CHAPTER 2 Functions
0 2 2 a2 a2 0; f 2 ; f a 2 . 2. (a) f 0 01 21 3 a21 a3 x (b) f x . Our restrictions are that the input to the radical is nonnegative and that the denominator must not be 0. x 1 Thus, x 0 and x 1 0 x 1. (The second restriction is made irrelevant by the first.) In interval notation, the domain is [0 . 10 2 f 10 f 2 3 10 11 2 10 1 2 1 (c) The average rate of change is . 10 2 10 2 264
3. (a) “Subtract 2, then cube the result” can be expressed
y
(c)
algebraically as f x x 23 . (b) x
f x
1
27
0
8
1
1
2
0
3
1
2
1 x
4 8 (d) We know that f has an inverse because it passes the Horizontal Line Test. A verbal description for f 1 is, “Take the cube root, then add 2.” (e) y x 23 3 y x 2 x 3 y 2. Thus, a formula for f 1 is f 1 x 3 x 2. 4. (a) f has a local minimum value of 4 at x 1 and local maximum values of 1 at x 4 and 4 at x 3. (b) f is increasing on 4 and 1 3 and decreasing on 4 1 and 3 .
5. R x 500x 2 3000x
(a) R 2 500 22 3000 2 $4000 represents their total sales revenue when their price is $2 per bar and
R 4 500 42 3000 4 $4000 represents their total
sales revenue when their price is $4 per bar
(c) The maximum revenue is $4500, and it is achieved at a price of x $3.
(b)
R 5000 4000 3000 2000 1000 0
1
2
3
4
5
x
6. The net change is f 2 h f 2 2 h2 2 2 h 22 2 2 4 h 2 4h 4 2h 0 2h h 2 and the average rate of change is
2h h 2 f 2 h f 2 2 h. 2h2 h
CHAPTER 2
7. (a) f x x 52 x 2 10x 25 is not linear because it cannot be
Test
257
y
(b)
expressed in the form f x ax b for constants a and b. g x 1 5x is linear.
y=g(x)
y=f(x)
(c) g x has rate of change 5. 10 0
8. (a) f x x 3
1
x
(b) g x x 13 2. To obtain the graph of g,
y
shift the graph of f to the right 1 unit and downward 2 units. y
1 1
x 1 x
1
9. (a) y f x 3 2. Shift the graph of f x to the right 3 units, then shift the graph upward 2 units. (b) y f x. Reflect the graph of f x about the y-axis.
10. (a) f 2 1 2 1 2 3 (since 2 1 ).
(b)
y
f 1 1 1 0 (since 1 1 ).
1 1
x
11. f x x 2 x 1; g x x 3. (a) f g x f x g x x 2 x 1 x 3 x 2 2x 2 (b) f g x f x g x x 2 x 1 x 3 x 2 4
(c) f g x f g x f x 3 x 32 x 3 1 x 2 6x 9 x 3 1 x 2 5x 7 (d) g f x g f x g x 2 x 1 x 2 x 1 3 x 2 x 2
(e) f g 2 f 1 12 1 1 1. [We have used the fact that g 2 2 3 1.] (f) g f 2 g 7 7 3 4. [We have used the fact that f 2 22 2 1 7.]
(g) g g g x g g g x g g x 3 g x 6 x 6 3 x 9. [We have used the fact that g x 3 x 3 3 x 6.]
258
CHAPTER 2 Functions
12. (a) f x x 3 1 is one-to-one because each real number has a unique cube.
(b) g x x 1 is not one-to-one because, for example, g 2 g 0 1.
1 1 2 2 x the Inverse Function Property,
13. f g x
1 1 x for all x 0, and g f x 2 x 2 2 x for all x 2. Thus, by 1 1 x x 2 f and g are inverse functions.
x 3 x 3 5y 3 . y 2x 5 y x 3 x 2y 1 5x 3 x . Thus, 2x 5 2x 5 2y 1 5x 3 f 1 x . 2x 1 15. (a) f x 3 x, x 3 y 3 x (b) f x 3 x, x 3 and f 1 x 3 x 2 ,
14. f x
y 2 3 x x 3 y 2 . Thus
x 0
y
f 1 x 3 x 2 , x 0.
f 1 x
1
f Ð!
16. The domain of f is [0 6], and the range of f is [1 7]. 17. The graph passes through the points 0 1 and 4 3, so f 0 1 and f 4 3. 18. The graph of f x 2 can be obtained by shifting the graph of f x to the right
y
2 units. The graph of f x 2 can be obtained by shifting the graph of f x
y=f(x)+2
upward 2 units.
y=f(x-2) 1
f 1
19. The net change of f between x 2 and x 6 is f 6 f 2 7 2 5 and the average rate of change is f 6 f 2 5 . 62 4 20. Because f 0 1, f 1 1 0. Because f 4 3, f 1 3 4. 21.
y f fÐ! 1
1
x
x
Modeling with Functions
259
22. (a) f x 3x 4 14x 2 5x 3. The graph is shown in the viewing rectangle [10 10] by [30 10].
-10
10 -20
(b) No, by the Horizontal Line Test. (c) The local maximum is approximately 255 when x 018, as shown in the first viewing rectangle [015 025] by [26 25]. One local minimum is approximately 2718 when x 161, as shown in the second viewing rectangle [165 155] by [275 27]. The other local minimum is approximately 1193 when x 143, as shown is the viewing rectangle [14 15] by [12 119]. 0.15 -2.50
0.20
0.25
-1.65
-1.60
-2.55
-1.55 -27.0 -27.2
1.40 -11.90
1.45
1.50
-11.95
-27.4 -2.60
-12.00
(d) Using the graph in part (a) and the local minimum, 2718, found in part (c), we see that the range is [2718 .
(e) Using the information from part (c) and the graph in part (a), f x is increasing on the intervals 161 018 and 143 and decreasing on the intervals 161 and 018 143.
FOCUS ON MODELING Modeling with Functions 1. Let be the width of the building lot. Then the length of the lot is 3. So the area of the building lot is A 32 , 0. 2. Let be the width of the poster. Then the length of the poster is 10. So the area of the poster is A 10 2 10.
3. Let be the width of the base of the rectangle. Then the height of the rectangle is 12 . Thus the volume of the box is given by the function V 12 3 , 0. 4. Let r be the radius of the cylinder. Then the height of the cylinder is 4r. Since for a cylinder V r 2 h, the volume of the cylinder is given by the function V r r 2 4r 4r 3 .
5. Let P be the perimeter of the rectangle and y be the length of the other side. Since P 2x 2y and the perimeter is 20, we have 2x 2y 20 x y 10 y 10 x. Since area is A x y, substituting gives A x x 10 x 10x x 2 , and since A must be positive, the domain is 0 x 10.
6. Let A be the area and y be the length of the other side. Then A x y 16 y P 2x 2
32 16 2x , where x 0. x x
16 . Substituting into P 2x 2y gives x
260
FOCUS ON MODELING
7.
Let h be the height of an altitude of the equilateral triangle whose side has length x, x
x
h 1 _x 2
as shown in the diagram. Thus the area is given by A 12 xh. By the Pythagorean 2 Theorem, h 2 12 x x 2 h 2 14 x 2 x 2 h 2 34 x 2 h 23 x.
Substituting into the area of a triangle, we get A x 12 xh 12 x 23 x 43 x 2 , x 0.
8. Let d represent the length of any side of a cube. Then the surface area is S 6d 2 , and the volume is V d 3 d 2 Substituting for d gives S V 6 3 V 6V 23 , V 0. A r 9. We solve for r in the formula for the area of a circle. This gives A r 2 r 2 r A
3
V.
A , so the model is
A , A 0.
C . Substituting for r 10. Let r be the radius of a circle. Then the area is A r 2 , and the circumference is C 2r r 2 2 C C2 , C 0. gives A C 2 4 60 11. Let h be the height of the box in feet. The volume of the box is V 60. Then x 2 h 60 h 2 . x The surface area, S, of the box is the sum of the area of the 4 sides and the area of the base and top. Thus 240 240 60 S 4xh 2x 2 4x 2x 2 2x 2 , so the model is S x 2x 2 , x 0. x x x2 5 12 5d 5 L d 12L 5d 7L L . The model is L d 57 d. L L d 7
12. By similar triangles,
dª
13.
Let d1 be the distance traveled south by the first ship and d2 be the distance traveled east by the second ship. The first ship travels south for t hours at 5 mi/h, so
dÁ
d1 15t and, similarly, d2 20t. Since the ships are traveling at right angles to
D
each other, we can apply the Pythagorean Theorem to get D t d12 d22 15t2 20t2 225t 2 400t 2 25t.
14. Let n be one of the numbers. Then the other number is 60 n, so the product is given by the function P n n 60 n 60n n 2 .
Let b be the length of the base, l be the length of the equal sides, and h be the
15. l
h
b
l
height in centimeters. Since the perimeter is 8, 2l b 8 2l 8 b 2 l 12 8 b. By the Pythagorean Theorem, h 2 12 b l 2 h l 2 14 b2 . Therefore the area of the triangle is b1 A 12 b h 12 b l 2 14 b2 8 b2 14 b2 2 4 b b b 64 16b b2 b2 64 16b 4 4 b b 4 b 4 4 4 so the model is A b b 4 b, 0 b 4.
Modeling with Functions
261
16. Let x be the length of the shorter leg of the right triangle. Then the length of the other triangle is 2x. Since it is a right triangle, the length of the hypotenuse is x 2 2x2 5x 2 5 x (since x 0 ). Thus the perimeter of the triangle is P x x 2x 5 x 3 5 x.
2 2 17. Let be the length of the rectangle. By the Pythagorean Theorem, 12 h 2 102 h 2 102 4 2 4 100 h 2 2 100 h 2 (since 0 ). Therefore, the area of the rectangle is A h 2h 100 h 2 , so the model is A h 2h 100 h 2 , 0 h 10.
18. Using the formula for the volume of a cone, V 13 r 2 h, we substitute V 100 and solve for h. Thus 100 13 r 2 h h r
300 . r 2
19. (a) We complete the table. First number
Second number
Product
1 2 3 4 5 6 7 8 9 10 11
18 17 16 15 14 13 12 11 10 9 8
18 34 48 60 70 78 84 88 90 90 88
(b) Let x be one number: then 19 x is the other number, and so the product, p, is p x x 19 x 19x x 2 . (c) p x 19x x 2 x 2 19x 2 2 19 x 2 19x 19 2 2 x 952 9025
So the product is maximized when the numbers are both 95.
From the table we conclude that the numbers is still increasing, the numbers whose product is a maximum should both be 95.
20. Let the positive numbers be x and y. Since their sum is 100, we have x y 100 y 100 x. We wish to minimize
the sum of squares, which is S x 2 y 2 x 2 100 x2 . So S x x 2 100 x2 x 2 10,000 200x x 2 2x 2 200x 10,000 2 x 2 100x 10,000 2 x 2 100x 2500 10,000 5000 2 x 502 5000.
Thus the minimum sum of squares occurs when x 50. Then y 100 50 50. Therefore both numbers are 50.
262
FOCUS ON MODELING
21. (a) Let x be the width of the field (in feet) and l be the length of the field (in feet). Since the farmer has 2400 ft of fencing we must have 2x l 2400. 2000
200
400
200
Area=2000(200)=400,000 1000
1000
700
1000
700 Area=400(1000)=400,000
Area=1000(700)=700,000
Width
Length
Area
200
2000
400,000
300
1800
540,000
400
1600
640,000
500
1400
700,000
600
1200
720,000
700
1000
700,000
800
800
640,000
It appears that the field of largest area is about 600 ft 1200 ft.
(b) Let x be the width of the field (in feet) and l be the length of the field (in feet). Since the farmer has 2400 ft of fencing we must have 2x l 2400 l 2400 2x. The area of the fenced-in field is given by A x l x 2400 2x x 2x 2 2400x 2 x 2 1200x . (c) The area is A x 2 x 2 1200x 6002 2 6002 2 x 6002 720,000. So the maximum area occurs when x 600 feet and l 2400 2 600 1200 feet.
22. (a) Let be the width of the rectangular area (in feet) and l be the length of the field (in feet). Since the farmer has 750 feet of fencing, we must have 5 2l 750 2l 750 5 l 52 150 . Thus the total area of the four pens is A l 52 150 52 2 150 . (b) We complete the square to get A 52 2 150 52 2 150 752 52 752 52 752 140625. Therefore, the largest possible total area of the four pens is 14,0625 square feet.
23. (a) Let x be the length of the fence along the road. If the area is 1200, we have 1200 x width, so the width of the garden 1200 7200 1200 . Then the cost of the fence is given by the function C x 5 x 3 x 2 8x . is x x x (b) We graph the function y C x in the viewing
(c) We graph the function y C x and y 600 in
cost is minimized when x 30 ft. Then the
From this we get that the cost is at most $600
rectangle [0 75] [0 800]. From this we get the width is 1200 30 40 ft. So the length is 30 ft and
the width is 40 ft.
the viewing rectangle [10 65] [450 650].
when 15 x 60. So the range of lengths he can fence along the road is 15 feet to 60 feet.
600 500 500 0 0
50
20
40
60
Modeling with Functions
24. (a) Let x be the length of wire in cm that is bent into a square. So 10 x is the length of wire in 10 x x and , and the area cm that is bent into the second square. The width of each square is 4 4 2 x 2 x2 100 20x x 2 10 x and . Thus the sum of the areas is of each square is 4 16 4 16
x2 100 20x x 2 100 20x 2x 2 18 x 2 54 x 25 4 . 16 16 16 1 x 2 10x 25 1 x 2 10x 25 25 (b) We complete the square. A x 18 x 2 54 x 25 4 8 4 8 4 25 1 x 52 25 So the minimum area is 25 cm2 when each piece is 5 cm long. 8 8 8 8 A x
25. (a) Let h be the height in feet of the straight portion of the window. The circumference of the semicircle is C 12 x. Since the perimeter of the window is 30 feet, we have x 2h 12 x 30. Solving for h, we get 2h 30 x 12 x h 15 12 x 14 x. The area of the window is 2 A x xh 12 12 x x 15 12 x 14 x 18 x 2 15x 12 x 2 18 x 2 . 120 (b) A x 15x 18 4 x 2 18 4 x 2 x 4 2 2 450 120 450 60 60 x 18 4 x 18 4 x 2 4 4 4 4 4 The area is maximized when x
60 840, and hence h 15 12 840 14 840 420. 4
26. (a) The height of the box is x, the width of the box is 12 2x, and the length of the
box is 20 2x. Therefore, the volume of the box is
(b) We graph the function y V x in the viewing rectangle [0 6] [200 270].
250
V x x 12 2x 20 2x 4x 3 64x 2 240x, 0 x 6
200 0
(c) From the graph, the volume of the box with the largest volume is 262682 in3 when x 2427.
5
From the calculator we get that the volume of the box is greater than 200 in3 for 1174 x 3898 (accurate to 3 decimal places).
27. (a) Let x be the length of one side of the base and let h be the height of the box in feet. Since the volume of 12 the box is V x 2 h 12, we have x 2 h 12 h 2 . The surface area, A, of the box is sum of the x area of the four sides and the area of the base. Thus the surface area of the box is given by the formula 12 48 A x 4xh x 2 4x x2 x 2 , x 0. 2 x x
263
264
FOCUS ON MODELING
(b) The function y A x is shown in the first viewing rectangle below. In the second viewing rectangle, we isolate the minimum, and we see that the amount of material is minimized when x (the length and width) is 288 ft. Then the 12 height is h 2 144 ft. x 26
50
25 24
0 0
2
4
3.0
28. Let A, B, C, and D be the vertices of a rectangle with base AB on the x-axis and its other two vertices C and D above the x-axis and lying on the parabola y 8 x 2 . Let C have the coordinates x y, x 0. By symmetry, the coordinates of D must be x y. So the width of the rectangle is 2x, and the length is y 8 x 2 . Thus the area of the rectangle is A x length width 2x 8 x 2 16x 2x 3 . The graphs of A x below show that the area is maximized when
x 163. Hence the maximum area occurs when the width is 326 and the length is 533. y
y=8-x@
D
A
C
x
18
10
17 16
0
x
B
20
0
2
4
1.5
2.0
29. (a) Let be the width of the pen and l be the length in meters. We use the area to establish a relationship between 100 and l. Since the area is 100 m2 , we have l 100 l . So the amount of fencing used is 200 22 100 2 . F 2l 2 2 (b) Using a graphing device, we first graph F in the viewing rectangle [0 40] by [0 100], and locate the approximate location of the minimum value. In the second viewing rectangle, [8 12] by [39 41], we see that the minimum value of F occurs when 10. Therefore the pen should be a square with side 10 m. 100
41
50
40 39
0 0
20
40
8
10
12
Modeling with Functions
30. (a) Let t1 represent the time, in hours, spent walking, and let t2 represent the time spent rowing. Since the distance walked is x and the walking
265
(b) We graph y T x. Using the zoom
function, we see that T is minimized when x 613. He should land at a point
speed is 5 mi/h, the time spent walking is t1 15 x. By the Pythagorean Theorem, the distance rowed is d 22 7 x2 x 2 14x 53, and so the time spent rowing is t2 12 x 2 14x 53. Thus the total time is T x 12 x 2 14x 53 15 x.
613 miles from point B. 4 2 0 0
31. (a) Let x be the distance from point B to C, in miles. Then the distance from A to C is flying from A to C then C to D is f x 14 x 2 25 10 12 x.
2
4
6
x 2 25, and the energy used in
(b) By using a graphing device, the energy expenditure is minimized when the distance from B to C is about 51 miles. 200
169.1
100
169.0 168.9
0 0
5
10
5.0
5.1
5.2
32. (a) Using the Pythagorean Theorem, we have that the height of the upper triangles is 25 x 2 and the height of the lower triangles is 144 x 2 . So the area of the each of the upper triangles is 12 x 25 x 2 , and the area of the each of the lower triangles is 12 x 144 x 2 . Since there are two upper triangles and two lower triangles, we get that the total area 25 x 2 144 x 2 . is A x 2 12 x 25 x 2 2 12 x 144 x 2 x (b) The function y A x x 25 x 2 144 x 2 is shown in the first viewing rectangle below. In the second viewing rectangle, we isolate the maximum, and we see that the area of the kite is maximized when x 4615. So the length of the horizontal crosspiece must be 2 4615 923. The length of the vertical crosspiece is 52 46152 122 46152 1300. 100
60.1
50
60.0
0 0
2
4
59.9 4.60
4.62
4.64
3
POLYNOMIAL AND RATIONAL FUNCTIONS
3.1
QUADRATIC FUNCTIONS AND MODELS
1. To put the quadratic function f x ax 2 bx c in standard form we complete the square. 2. The quadratic function f x a x h2 k is in standard form. (a) The graph of f is a parabola with vertex h k. (b) If a 0 the graph of f opens upward. In this case f h k is the minimum value of f .
(c) If a 0 the graph of f opens downward. In this case f h k is the maximum value of f .
3. The graph of f x 3 x 22 6 is a parabola that opens upward, with its vertex at 2 6, and f 2 6 is the minimum value of f . 4. The graph of f x 3 x 22 6 is a parabola that opens downward, with its vertex at 2 6, and f 2 6 is the maximum value of f . 5. (a) The vertex is 3 4, the x-intercepts are 1 and 5, and it appears that the y-intercept is approximately 5.
6. (a) The vertex is 2 8, the x-intercepts are 6 and 2, and the y-intercept is 6.
(b) Maximum value of f : 4
(b) Maximum value of f : 8
(c) Domain , range: 4]
(c) Domain: , range: 8]
7. (a) The vertex is 1 3, the x-intercepts are
approximately 02 and 22, and the y-intercept is
8. (a) The vertex is 1 4, the x-intercepts are
1.
approximately 22 and 02, and the y-intercept is 1.
(b) Minimum value of f : 3
(b) Minimum value of f : 4
(c) Domain: , range: [3
(c) Domain: , range: [4
9. (a) f x x 2 2x 3 x 12 1 3 x 12 2
(c)
y
(b) The vertex is at 1 2. x-intercepts: y 0 0 x 12 2 x 12 2. This has
no real solution, so there is no x-intercept.
y-intercept: x 0 y 0 12 2 3. The y-intercept is 3. (d) Domain: , range: [2
1 1
x
267
268
CHAPTER 3 Polynomial and Rational Functions
10. (a) f x x 2 4x 1 x 22 4 1 x 22 5
y
(c)
(b) The vertex is at 2 5.
x-intercepts: y 0 0 x 22 5 x 22 5 x 2 5 x 2 5. So x 2 5 or x 2 5. The x-intercepts are 2 5 and 2 5.
1 1
y-intercept: x 0 y 0 22 5 1. The y-intercept is 1.
x
(d) Domain: , range: [5
11. (a) f x x 2 6x x 2 6x x 2 6x 9 9 x 32 9
(c)
(b) The vertex is at 3 9.
y
1
x-intercepts: y 0 0 x 2 6x x x 6. So x 0 or x 6.
x
1
The x-intercepts are 0 and 6.
y-intercept: x 0 y 0. The y-intercept is 0. (d) Domain: , range: [9
12. (a) f x x 2 8x x 42 16
y
(c)
(b) The vertex is at 4 16.
x-intercepts: y 0 0 x 2 8x x x 8. So x 0 or
x 8. The x-intercepts are 0 and 8.
3
y-intercept: x 0 y 0. The y-intercept is 0.
1
x
1
x
(d) Domain: , range: [16
13. (a) f x 3x 2 6x 3 x 2 2x 3 x 12 3
(c)
y
(b) The vertex is at 1 3.
x-intercepts: y 0 0 3 x 12 3 x 12 1 x 2 or 0. The x-intercepts are 2 and 0.
y-intercept: x 0 y 3 02 6 0 0. The y-intercept is 0.
(d) Domain: , range: [3
1
SECTION 3.1 Quadratic Functions and Models
14. (a) f x x 2 10x x 2 10x x 52 25 (b) The vertex is at 5 25.
(c)
x-intercepts: y 0 0 x 52 25 x 52 25
269
y
10
x 0 or 10. The x-intercepts are 0 and 10.
y-intercept: x 0 y 02 10 0 0. The y-intercept is 0.
x
1
(d) Domain: , range: 25]
15. (a) f x x 2 4x 3 x 22 1
y
(c)
(b) The vertex is at 2 1.
x-intercepts: y 0 0 x 2 4x 3 x 1 x 3. So x 1 or x 3. The x-intercepts are 1 and 3.
1
y-intercept: x 0 y 3. The y-intercept is 3.
1
x
1
x
(d) Domain: , range: [1
16. (a) f x x 2 2x 2 x 12 1
y
(c)
(b) The vertex is at 1 1. x-intercepts: y 0 x 12 1 0 x 12 1. Since
this last equation has no real solution, there is no x-intercept.
1
y-intercept: x 0 y 2. The y-intercept is 2. (d) Domain: , range: [1
17. (a) f x x 2 6x 4 x 32 13
(c)
y
(b) The vertex is at 3 13. x-intercepts: y 0 0 x 32 13 x 32 13 x 3 13 x 3 13. The x-intercepts are 3 13 and 3 13. y-intercept: x 0 y 4. The y-intercept is 4. (d) Domain: , range: 13]
2 1
x
270
CHAPTER 3 Polynomial and Rational Functions
18. (a) f x x 2 4x 4 x 22 8
y
(c)
(b) The vertex is at 2 8. x-intercepts: y 0 0 x 2 4x 4 0 x 2 4x 4.
Using the Quadratic Formula,
2
2 22 2 4 42 414 4 32 2 2 2. x 21 2 2
1
x
The x-intercepts are 2 2 2 and 2 2 2.
y-intercept: x 0 y 4. The y-intercept is 4.
(d) Domain: , range: 8]
19. (a) f x 2x 2 4x 3 2 x 12 1
y
(c)
(b) The vertex is at 1 1. x-intercepts: y 0 0 2x 2 4x 3 2 x 12 1
2 x 12 1. Since this last equation has no real solution, there is
no x-intercept.
y-intercept: x 0 y 3. The y-intercept is 3.
1
(d) Domain: , range: [1
20. (a) f x 3x 2 6x 2 3 x 12 1
1
x
y
(c)
(b) The vertex is at 1 1. x-intercepts: y 0 0 3 x 12 1 0 x 12 13 x 1 13 x 1 13 . The x-intercepts are 1 13 and 1 13 .
2 1
x
y-intercept: x 0 y 2. The y-intercept is 2.
(d) Domain: , range: 1]
21. (a) f x 2x 2 20x 57 2 x 52 7
(c)
y
(b) The vertex is at 5 7. x-intercepts: y 0 0 2x 2 20x 57 2 x 52 7
2 x 52 7. Since this last equation has no real solution, there is
no x-intercept.
y-intercept: x 0 y 57. The y-intercept is 57. (d) Domain: , range: [7
2 1
x
SECTION 3.1 Quadratic Functions and Models
22. (a) f x 2x 2 12x 10 2 x 2 6x 9 18 10 2 x 32 8
y
(c)
(b) The vertex is at 3 8. The x-intercepts are 5 and 1 and the
2
y-intercept is 10.
0
(d) Domain: , range: [8
23. (a) f x 4x 2 12x 1 4 x 2 3 1 2 2 4 x 32 9 1 4 x 32 10 (b) The vertex is at 32 10 .
1
x
y
(c)
10
2 2 x-intercepts: y 0 0 4 x 32 10 x 32 52 x 32 52 x 32 52 32 210 . The x-intercepts are
32
271
5
1
10 10 3 2 and 2 2 .
_3
_2
x
_1
y-intercept: x 0 y 4 02 12 0 1 1. The y-intercept
is 1.
(d) Domain: , range: 10]
24. (a) f x 3x 2 2x 2 3 x 2 23 x 2 2 2 3 x 13 13 2 3 x 13 73 (b) The vertex is at 13 73 .
x-intercepts: y 0 0 3 x 13
2
73 x 13
(c)
2
10
79
x 13 37 x 13 37 . The x-intercepts are 13 37 and
13 37 .
y-intercept: x 0 y 3 02 2 0 2 2. The y-intercept
is 2.
(d) Domain: , range: 73
y 20
1
x
272
CHAPTER 3 Polynomial and Rational Functions
25. (a) f x x 2 2x 1 x 2 2x 1 x 2 2x 1 1 1 x 12 2 y
(b)
26. (a) f x x 2 8x 8 x 2 8x 16 8 16 x 42 8
(b)
2
1 1
3 x 12 2 y
x
1
x
(c) The minimum value is f 1 2. 27. (a) f x 3x 2 6x 1 3 x 2 2x 1 3 x 2 2x 1 1 3 (b)
y
(c) The minimum value is f 4 8. 28. (a) f x 5x 2 30x 4 5 x 2 6x 4 5 x 2 6x 9 4 45 5 x 32 41
(b)
y 10 1
x
1 1
x
(c) The minimum value is f 1 2. 29. (a) f x x 2 3x 3 x 2 3x 3 x 2 3x 94 3 94 2 x 32 21 4 (b)
y
(c) The minimum value is f 3 41.
30. (a) f x 1 6x x 2 x 2 6x 1 x 2 6x 9 1 9 x 32 10
(b)
y
2
1 1
(c) The maximum value is f 32 21 4.
x
1
(c) The maximum value is f 3 10.
x
SECTION 3.1 Quadratic Functions and Models
31. (a) g x 3x 2 12x 13 3 x 2 4x 13 3 x 2 4x 4 13 12 3 x 22 1
(b)
y
32. (a) g x 2x 2 8x 11 2 x 2 4x 11 2 x 2 4x 4 11 8 2 x 22 3
y
(b)
2 x
1
2 1
(c) The minimum value is g 2 1.
(c) The minimum value is g 2 3.
34. (a) h x 3 4x 4x 2 4 x 2 x 3 4 x 2 x 14 3 1 2 4 x 12 4
33. (a) h x 1 x x 2 x 2 x 1 x 2 x 14 1 14 2 x 12 54 (b)
x
y
(b)
y
1
1 1
x
1
(c) The maximum value is h 12 54 . (c) The maximum value is h 12 4. 35. f x 2x 2 4x 1 2x 2 4x 1 2 x 2 2x 1 2 x 2 2x 1 2 1 2 x 12 3.
Therefore, the minimum value is f 1 3. 36. f x 3 4x x 2 x 2 4x 3 x 2 4x 3 x 2 4x 4 4 3 x 22 7. Therefore, the maximum value is f 2 7.
37. f t 3 80t 20t 2 20t 2 80t 3 20 t 2 4t 4 80 3 20 t 22 77. Therefore, the maximum value is f 2 77. 38. f x 6x 2 24x 100 6 x 2 4x 4 24 100 6 x 22 124.
Therefore, the minimum value is f 2 124. 39. f s s 2 12s 16 s 2 12s 16 s 2 12s 036 16 036 s 062 1564. Therefore, the minimum value is f 06 1564.
x
273
274
CHAPTER 3 Polynomial and Rational Functions
2 5625 10 x 15 40. g x 100x 2 1500x 100 x 2 15x 100 x 2 15x 225 5625. 4 2 Therefore, the minimum value is g 15 2 5625. 41. h x 12 x 2 2x 6 12 x 2 4x 6 12 x 2 4x 4 6 2 12 x 22 8. Therefore, the minimum value is h 2 8.
x2 2x 7 13 x 2 6x 7 13 x 2 6x 9 7 3 13 x 32 10. 3 Therefore, the maximum value is f 3 10. 43. f x 3 x 12 x 2 12 x 2 2x 3 12 x 2 2x 1 3 12 12 x 1 72 . Therefore, the maximum
42. f x
value is f 1 72 .
44. g x 2x x 4 7 2x 2 8x 7 2 x 2 4x 7 2 x 2 4x 4 7 8 2 x 22 1. Therefore, the minimum value is g 2 1.
45. (a) The graph of f x x 2 179x 321 is shown. The minimum value is f x 401. -1.0
-0.9
(b) f x x 2 179x 321 2 2 321 179 x 2 179x 179 2 2 x 08952 4011025
-0.8 -3.9
Therefore, the exact minimum of f x is 4011025.
-4.0 -4.1
46. (a) The graph of f x 1 x
2 2x is
shown. The maximum value is f x 118. 1.180
Therefore, the exact maximum of f x is 88 2 .
1.175 1.170 0.30
(b) f x 1 x 2 x 2 2 x 2 22 x 1 2 1 82 2 x 2 22 x 42 2 2 x 42 88 2
0.35
0.40
47. The vertex is 2 3, so the parabola has equation y a x 22 3. Substituting the point 3 1, we have 1 a 3 22 3 a 4, so f x 4 x 22 3.
48. The vertex is 1 5, so the parabola has equation y a x 12 5. Substituting the point 3 7, we have
7 a 3 12 5 a 3, so f x 3 x 12 5. 49. Substituting t x 2 , we have f t 3 4t t 2 t 2 4t 4 4 3 t 22 7. Thus, the maximum value is 7, when t 2 (or x 2). 50. Substituting t x 3 , we have f t 2 16t 4t 2 4 t 2 4t 4 16 2 4 t 22 14. Thus, the minimum value is 14, when t 2 (or x 3 2).
SECTION 3.1 Quadratic Functions and Models
275
2 2 2 16 54 16 t 54 25. Thus the maximum 51. y f t 40t 16t 2 16 t 2 52 16 t 2 52 t 54 height attained by the ball is f 54 25 feet. 32 x 2 x 5 2 x 2 25 x 5 2 x 2 25 x 25 2 2 25 2 5 52. (a) We complete the square: y 400 25 2 25 2 4 25 4 2 x 25 2 65 , so the maximum height of the ball is 65 8125 ft. y 25 4 8 8
2 x 2 x 5. Using the Quadratic (b) The ball hits the ground when its vertical displacement y is 0, that is, when 0 25 2 5 1 1 4 25 25 5 65 . Taking the positive root, we find that x 163 ft. Formula, we find x 4 4 25 53. R x 80x 04x 2 04 x 2 200x 04 x 2 200x 10,000 4,000 04 x 1002 4,000. So
revenue is maximized at $4,000 when 100 units are sold. 54. P x 0001x 2 3x 1800 0001 x 2 3000x 1800 0001 x 2 3000x 2 250 000 1800
2250 0001 x 15002 450. The vendor’s maximum profit occurs when he sells 1500 cans and the profit is $450. 1 n 2 1 n 2 60n 1 n 2 60n 900 10 1 n 302 10. Since the maximum of 55. E n 23 n 90 90 90 90 the function occurs when n 30, the viewer should watch the commercial 30 times for maximum effectiveness. 56. C t 006t 00002t 2 00002 t 2 300t 00002 t 2 300t 22,500 45 00002 t 1502 45. The maximum concentration of 45 mg/L occurs after 150 minutes.
57. A n n 900 9n 9n 2 900n is a quadratic function with a 9 and b 900, so by the formula, the maximum 900 b 50 trees, and because a 0, this gives a maximum value. or minimum value occurs at n 2a 2 9 58. A n 700 n 10 001n 001n 2 10n 001 700 n 7000 001n 2 3n 7000. This is a quadratic b 3 function with a 001 and b 3, so the maximum (a 0) occurs at x 150. Since n 150 2a 2 001 is the number of additional vines that should be planted, the total number of vines that maximizes grape production is 700 150 850 vines. 59. The area of the fenced-in field is given by A x 2400 2x x 2x 2 2400x. Thus, by the formula in this section, 2400 b 600. The maximum area occurs when x 600 feet the maximum or minimum value occurs at x 2a 2 2 and l 2400 2 600 1200 feet. 60. The total area of the four pens is A 52 150 52 2 375. Thus, by the formula, the maximum or 375 b 75. Therefore, the largest possible total area of the four pens is minimum value occurs at 2a 2 52 A 75 52 752 375 75 14,0625 square feet.
15 b 84 ft 61. A x 15x 18 4 x 2 , so by the formula, the maximum area occurs when x 2a 2 18 4 and h 15 12 840 14 840 42 ft.
54 b 5. The 62. A x 18 x 2 54 x 25 , so by the formula, the maximum or minimum area occurs where x 4 2a 2 18 25 2 minimum area is 18 52 54 5 25 4 8 cm when each piece is 5 cm long.
276
CHAPTER 3 Polynomial and Rational Functions
63. (a) The area of the corral is A x x 1200 x 1200x x 2 x 2 1200x.
(b) A is a quadratic function with a 1 and b 1200, so by the formula, it has a maximum or minimum at 1200 b 600, and because a 0, this gives a maximum value. The desired dimensions are 600 ft by x 2a 2 1 600 ft.
64. (a) The dimensions of the gutter are x inches and 30 x x 30 2x inches, so the cross sectional area is A x 30 2x 30x 2x 2 .
(b) Since A is a quadratic function with a 2 and b 30, the maximum occurs at x
b 30 75 inches. 2a 2 2
(c) The maximum cross section is A 75 2 752 30 75 1125 in2 .
65. (a) To model the revenue, we need to find the total attendance. Let x be the ticket price. Then the amount by which the ticket price is lowered is 10 x, and we are given that for every dollar it is lowered, the attendance increases by 3000; that is, the increase in attendance is 3000 10 x. Thus, the attendance is 27,000 3000 10 x, and since each spectator pays $x, the revenue is R x x [27,000 3000 10 x] 3000x 2 57,000x.
(b) Since R is a quadratic function with a 3000 and b 57,000, the maximum occurs at 57,000 b 95; that is, when admission is $950. x 2a 2 3000 (c) We solve R x 0 for x: 3000x 2 57,000x 0 3000x x 19 0 x 0 or x 19. Thus, if admission is $19, nobody will attend and no revenue will be generated. 66. (a) Let x be the price per feeder. Then the amount by which the price is increased is x 10, and we are given that for every dollar increase, sales decrease by 2; that is, the change in sales is 2 x 10, so the total number sold is 20 2 x 10 40 2x. The profit per feeder is equal to the sale price minus the cost, that is, x 6. Multiplying the number of feeders sold by the profit per feeder sold, we find the profit to be P x 40 2x x 6 2x 2 52x 240.
(b) Using the formula, profit is maximized when x
b 52 13; that is, when the society charges $13 per 2a 2 2
feeder. The maximum weekly profit is P 13 2 132 52 13 240 $98.
y
67. Because f x x m x n 0 when x m or x n, those are its x-intercepts. By symmetry, we expect that the vertex is halfway between mn these values; that is, at x . We obtain the graph shown at right. 2
Expanding, we see that f x x 2 m n x mn, a quadratic
function with a 1 and b m n. Because a 0, the minimum value occurs at x
3.2
y=(x-a)(x-b)
a
0
a+b 2
b
x
b m n , the x-value of the vertex, as expected. 2a 2
POLYNOMIAL FUNCTIONS AND THEIR GRAPHS
1. Graph I cannot be that of a polynomial because it is not smooth (it has a cusp.) Graph II could be that of a polynomial function, because it is smooth and continuous. Graph III could not be that of a polynomial function because it has a break. Graph IV could not be that of a polynomial function because it is not smooth. 2. (a) y x 3 8x 2 2x 15 has odd degree and a positive leading coefficient, so y as x and y as x .
SECTION 3.2 Polynomial Functions and Their Graphs
277
(b) y 2x 4 12x 100 has even degree and a negative leading coefficient, so y as x and y as x . 3. (a) If c is a zero of the polynomial P, then P c 0.
(b) If c is a zero of the polynomial P, then x c is a factor of P x.
(c) If c is a zero of the polynomial P, then c is an x-intercept of the graph of P.
4. (a) This is impossible. If P has degree n 3, it has at most n 1 2 local extrema.
(b) This is possible. For example, y x 3 has degree 3 and no local maxima or minima.
(c) This is possible. For example, P x x 4 has one local maximum and no local minima.
5. (a) P x x 2 4
(b) Q x x 42
y
y
1 1
(_2, 0)
x
(2, 0)
4
(0, _4)
(c) P x 2x 2 3
(0, 16)
1
(d) P x x 22
y
x
(4, 0)
(_2, 0)
y 1 0
1
x
(0, _4)
1 0
6. (a) P x x 4 16
(0, 3) x
1
(b) P x x 54
y
y 100 0
4 (_2, 0)
1
(2, 0)
x
(0, _625) (0, _16)
1
x
278
CHAPTER 3 Polynomial and Rational Functions
(c) P x 5x 4 5
(d) P x x 54
y
y
(0, 5)
1
(_1, 0)
(1, 0)
0
1
(0, 625)
x
100 0
7. (a) P x x 3 8
(5, 0) x
1
(b) Q x x 3 27
y
y
1 1
(2, 0)
x (0, 27)
(0, _8)
4
(3, 0)
x
1
(c) R x x 23
(_2, 0)
(d) S x 12 x 13 4
y
y
1 x
1
(0, _72) (_1, 0)
(0, _8)
8. (a) P x x 35
1
(b) Q x 2 x 35 64
y
y (0, 422)
(0, 243)
(_3, 0)
x
1
(_1, 0)
50 1x
100 1x
SECTION 3.2 Polynomial Functions and Their Graphs
(c) R x 12 x 25
279
(d) S x 12 x 25 16
y
y (0, 16) (0, 32)
2
(2, 0) 1
x
4
(4, 0)
1
x
9. (a) P x x x 2 4 x 3 4x has odd degree and a positive leading coefficient, so y as x and y as x .
(b) This corresponds to graph III. 10. (a) Q x x 2 x 2 4 x 4 4x 2 has even degree and a negative leading coefficient, so y as x and y as x .
(b) This corresponds to graph I. 11. (a) R x x 5 5x 3 4x has odd degree and a negative leading coefficient, so y as x and y as x . (b) This corresponds to graph V.
12. (a) S x 12 x 6 2x 4 has even degree and a positive leading coefficient, so y as x and y as x . (b) This corresponds to graph II.
13. (a) T x x 4 2x 3 has even degree and a positive leading coefficient, so y as x and y as x . (b) This corresponds to graph VI.
14. (a) U x x 3 2x 2 has odd degree and a negative leading coefficient, so y as x and y as x . (b) This corresponds to graph IV.
15. P x x 1 x 2
16. P x 2 x x 5
y
1 (_2, 0)
1
(1, 0)
(0, _2)
y 20
10 (0, 10) x (_5, 0)
1
(2, 0)
x
280
CHAPTER 3 Polynomial and Rational Functions
17. P x x x 3 x 2
18. P x x x 3 x 2
y
y
1
(_2, 0) (_2, 0) 1
1
1
(3, 0)
(0, 0)
x
(3, 0)
x
20. P x x 3 x 2 3x 2
y
y
(0, 12)
(0, 3) 1
(_1, 0)
(3, 0)
x
19. P x 2x 1 x 1 x 3
(_3, 0)
(0, 0)
(_2, 0)
1
4
1
(1/2, 0) x
21. P x x 2 x 1 x 2 x 3
( _23 , 0)
22. P x x x 1 x 1 2 x y
y
(_1, 0)
1 0
(0, 0) 1 (1, 0)
(2, 0)
x
(0, 12) 5
(2, 0)
0
(_2, 0)
(3, 0) x
1
(_1, 0)
24. P x 15 x x 52
23. P x 2x x 22 y
y
1 (0, 0)
(2, 0) 1
2 (0, 0) x
1
(5, 0)
x
SECTION 3.2 Polynomial Functions and Their Graphs
25. P x x 1 x 12 2x 3
26. P x x 12 x 13 x 2
y
y
1
(_1, 0) (_1, 0) 2 0 (_2, 0)
(1, 0)
0
(2, 0)
1
x
(3/2, 0) 1
x
(0, _2)
(0, _6)
1 x 22 x 32 27. P x 12
28. P x x 12 x 23
y
y
(0, 8) 2
1
(0, 3)
x
(1, 0)
(_2, 0)
1 1
(_2, 0)
x
(3, 0)
29. P x x 3 x 2 x 32
30. P x x 32 x 12
y
10
y
1 (0, 0)
(_2, 0)
(3, 0)
x
(0, 9) 2 1
(_1, 0)
31. P x x 3 x 2 6x x x 2 x 3
2 (0, 0) 1
y
4
(_4, 0) (3, 0)
x
32. P x x 3 2x 2 8x x x 2 x 4
y
(_2, 0)
(3, 0)
x
(0, 0)
1
(2, 0)
x
281
282
CHAPTER 3 Polynomial and Rational Functions
33. P x x 3 x 2 12x x x 3 x 4 y
34. P x 2x 3 x 2 x x 2x 2 x 1 x 2x 1 x 1 y
4 (0, 0) 1
(_3, 0)
(4, 0)
1 x
( _21 , 0)
(0, 0)
1
(_1, 0)
35. P x x 4 3x 3 2x 2 x 2 x 1 x 2 y
1
36. P x x 5 9x 3 x 3 x 3 x 3 y
10
(_3, 0)
(3, 0)
(0, 0) (1, 0) 1
(0, 0)
(2, 0)
y
y
(_2, 0)
(1, 0) 1
(0, _1)
x
38. P x x 3 3x 2 4x 12 x 3 x 2 x 2
(_3, 0) (_1, 0)
1
x
37. P x x 3 x 2 x 1 x 1 x 12
1
x
2
(2, 0) 1
x
(0, _12)
x
SECTION 3.2 Polynomial Functions and Their Graphs
2 40. P x 18 2x 4 3x 3 16x 24 2 18 x 22 2x 32 x 2 2x 4
39. P x 2x 3 x 2 18x 9 x 3 2x 1 x 3 y
y
40 (0, 9)
(_3, 0)
1
( _21 , 0)
(3, 0)
x (0, 72) 20 1
(_ _32 , 0)
41. P x x 4 2x 3 8x 16 x 22 x 2 2x 4
x
42. P x x 4 2x 3 8x 16 x 2 x 2 x 2 2x 4
y
y
(_2, 0)
10
(2, 0)
10 (2, 0)
x
1 (0, _16)
(0, 16) x
1 (2, 0)
43. P x x 4 3x 2 4 x 2 x 2 x 2 1 y
44. P x x 6 2x 3 2 2 1 x 3 1 x 12 x 2 x 1 y
(_2, 0)
1
(2, 0) 1
x
(0, _4) 1
(0, 1)
1 (1, 0)
x
283
284
CHAPTER 3 Polynomial and Rational Functions
45. P x 3x 3 x 2 5x 1; Q x 3x 3 . Since P has odd degree and positive leading coefficient, it has the following end behavior: y as x and y as x . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, we see that the graphs of P and Q have different intercepts. 100
10
50 -3
-2
-1
P 1
5
Q
-50
2
3
-1.5 -1
-0.5 -5
-100
-10
P
Q
0.5
1
1.5
46. P x 18 x 3 14 x 2 12x; Q x 18 x 3 . Since P has odd degree and negative leading coefficient, it has the following end behavior: y as x and y as x . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, the graphs of P and Q look very different and seem (wrongly) to have different end behavior. 60 40
1000 P
Q
Q
500 -6
-30 -20 -10 -500
10
20
P
20
-4
-2
30
0 -20
2
4
6
-40 -60
-1000
-80
47. P x x 4 7x 2 5x 5; Q x x 4 . Since P has even degree and positive leading coefficient, it has the following end behavior: y as x and y as x . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, the graphs of P and Q look very different and we see that they have different intercepts. 600 10
500 Q
400
P
300
P
Q
-3
200
-2
-1
-4
-2 0 -100
1
2
3
-10
100 -6
0
2
4
6
-20
48. P x x 5 2x 2 x; Q x x 5 . Since P has odd degree and negative leading coefficient, it has the following end behavior: y as x and y as x . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, the graphs of P and Q look very different. 3
200 Q
P
-3
-2
Q
100 -1 0 -100 -200
1
2
3
P
-1
2 1 0 -1 -2 -3
1
SECTION 3.2 Polynomial Functions and Their Graphs
285
49. P x x 11 9x 9 ; Q x x 11 . Since P has odd degree and positive leading coefficient, it has the following end behavior: y as x and y as x . On a large viewing rectangle, the graphs of P and Q look like they have the same end behavior. On a small viewing rectangle, the graphs of P and Q look very different and seem (wrongly) to have different end behavior. 600000 400000
Q
P
100
Q
-6
-4
-2 0 -200000
P
50
200000 2
4
6
0
-1 P
-400000
1
-50
Q
-100
-600000
50. P x 2x 2 x 12 ; Q x x 12 . Since P has even degree and negative leading coefficient, it has the following end behavior: y as x and y as x . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, the graphs of P and Q look very different. 200 -3
-2
-1 0 -200
1
2
2
3
-400
P Q
-800
0
-1
-600 P
Q
51. (a) x-intercepts at 0 and 4, y-intercept at 0. (b) Local maximum at 2 4, no local minimum.
53. (a) x-intercepts at 2 and 1, y-intercept at 1. (b) Local maximum at 1 0, local minimum at 1 2.
55. y x 2 8x, [4 12] by [50 30]
No local minimum. Local maximum at 4 16. Domain: , range: 16].
-4
52. (a) x-intercepts at 0 and 45, y-intercept at 0. (b) Local maximum at 0 0, local minimum at 3 3.
54. (a) x-intercepts at 0 and 4, y-intercept at 0. (b) No local maximum, local minimum at 3 3.
56. y x 3 3x 2 , [2 5] by [10 10]
Local minimum at 2 4. Local maximum at 0 0. Domain: , range: . 10
20
-40
-2
-1000 -1200
-4 -2 -20
1
2 4 6 8 10 12 -2
2 -10
4
286
CHAPTER 3 Polynomial and Rational Functions
57. y x 3 12x 9, [5 5] by [30 30]
Local maximum at 2 25. Local minimum at 2 7. Domain: , range: .
58. y 2x 3 3x 2 12x 32, [5 5] by [60 30] Local minimum at 2 52. Local maximum at 1 25.
Domain: , range: .
20 20 -4
-2
2
4
-4
-2 -20
-20
2
4
-40 -60
59. y x 4 4x 3 , [5 5] by [30 30]
Local minimum at 3 27. No local maximum.
Domain: , range: [27 .
60. y x 4 18x 2 32, [5 5] by [100 100]
Local minima at 3 49 and 3 49. Local maximum at 0 32.
Domain: , range: [49 .
20
-4
-2
100 2
4 -4
-20
-2
2
4
-100
61. y 3x 5 5x 3 3, [3 3] by [5 10]
Local maximum at 1 5. Local minimum at 1 1. Domain: , range: .
62. y x 5 5x 2 6, [3 3] by [5 10]
Local minimum at 126 124. Local maximum at 0 6. Domain: , range: .
10
10
5
-3
-2
-1
1
2
3
-3
-2
-1
1
2
3
-5
63. y 2x 2 3x 5 has one local maximum at 075 613.
64. y x 3 12x has no local maximum or minimum. 20
10
10
5
-4
-2
-4 2
-5
4
-2 -10 -20
2
4
SECTION 3.2 Polynomial Functions and Their Graphs
287
65. y x 3 x 2 x has one local maximum at 033 019 66. y 6x 3 3x 1 has no local maximum or minimum. and one local minimum at 100 100.
20 10
2 -3 -3
-2
-1
1
2
-2
3
-1 -10
67. y x 4 5x 2 4 has one local maximum at 0 4 and two local minima at 158 225 and 158 225. 10
3
68. y 12x 5 375x 4 7x 3 15x 2 18x has two local
maxima at 050 465 and 297 1210 and two local minima at 140 2744 and 140 254. 20
5
-2
2
-20
-2
-3
1
-1
1
2
3
-4
-2
2
-5
-20
69. y x 25 32 has no maximum or minimum.
3 70. y x 2 2 has one local minimum at 0 8.
60
10
40 20 -3 -2
2
4
-2
6
-1
1
2
3
-10
71. y x 8 3x 4 x has one local maximum at 044 033 and two local minima at 109 115 and 112 336.
72. y 13 x 7 17x 2 7 has one local maximum at 0 7 and one local minimum at 171 2846. 20
4 2 -3
-2
-1 -2 -4
-3 1
2
3
-2
-1 -20
1
2
3
288
CHAPTER 3 Polynomial and Rational Functions
73. y cx 3 ; c 1, 2, 5, 12 . Increasing the value of c stretches 74. P x x c4 ; c 1, 0, 1, 2. Increasing the value of c shifts the graph to the right.
the graph vertically. c=5
100
c=0
c=1
c=_1
-4
-2
2
c=2
1
4
0
-2
1 c=_ 2
2
-1
-100
c=2
c=1
2
75. P x x 4 c; c 1, 0, 1, and 2. Increasing the value of c moves the graph up.
76. P x x 3 cx; c 2, 0, 2, 4. Increasing the value of c makes the “bumps” in the graph flatter.
c=2 c=1 c=0 c=_1
6 4
c=0 10
c=_2
c=2
c=_4
2 0
-2
0
-1
2
1
-2 -10
77. P x x 4 cx; c 0, 1, 8, and 27. Increasing the value 78. P x x c ; c 1, 3, 5, 7. The larger c gets, the flatter the of c causes a deeper dip in the graph, in the fourth
graph is near the origin, and the steeper it is away from the
quadrant, and moves the positive x-intercept to the right.
origin.
c=1
10
c=5
c=3
c=27
20
c=0
c=7
c=1 -4
-2
0
2 c=8
4
-2
-1
0
1
2
-20
-10 -40
79. (a)
y
y=x#-2x@-x+2 y=_x@+5x+2
5 1
x
(b) The two graphs appear to intersect at 3 points. (c) x 3 2x 2 x 2 x 2 5x 2 x 3 x 2 6x 0 x x 2 x 6 0 x x 3 x 2 0. Then either x 0, x 3, or x 2. If x 0, then y 2; if x 3 then
y 8 if x 2, then y 12. Hence the points where the
two graphs intersect are 0 2, 3 8, and 2 12.
80. Graph 1 belongs to y x 4 . Graph 2 belongs to y x 2 . Graph 3 belongs to y x 6 . Graph 4 belongs to y x 3 . Graph 5 belongs to y x 5 .
SECTION 3.2 Polynomial Functions and Their Graphs
289
81. (a) Let P x be a polynomial containing only odd powers of x. Then each term of P x can be written as C x 2n1 , for some constant C and integer n. Since C x2n1 C x 2n1 , each term of P x is an odd function. Thus by part (a), P x is an odd function. (b) Let P x be a polynomial containing only even powers of x. Then each term of P x can be written as C x 2n , for some constant C and integer n. Since C x2n C x 2n , each term of P x is an even function. Thus by part (b), P x is an even function. (c) Since P x contains both even and odd powers of x, we can write it in the form P x R x Q x, where R x contains all the even-powered terms in P x and Q x contains all the odd-powered terms. By part (d), Q x is an odd function, and by part (e), R x is an even function. Thus, since neither Q x nor R x are constantly 0 (by assumption), by part (c), P x R x Q x is neither even nor odd. (d) P x x 5 6x 3 x 2 2x 5 x 5 6x 3 2x x 2 5 PO x PE x where PO x x 5 6x 3 2x and PE x x 2 5. Since PO x contains only odd powers of x, it is an odd function, and since PE x contains only even powers of x, it is an even function.
82. (a) From the graph, P x x 3 4x x x 2 x 2 has three x-intercepts, one local maximum, and one local minimum.
(b) From the graph, Q x x 3 4x x x 2 4 has one x-intercept and no local maximum or minimum.
10
-5
10
5
-5
-10
5 -10
(c) For the x-intercepts of P x x 3 ax, we solve x 3 ax 0. Then we have x x 2 a 0 x 0 or x 2 a. If x 2 a, then x a. So P has 3 x-intercepts. Since P x x x 2 a x x a x a , by part (c) of problem 67, P has 2 local extrema. For the x-intercepts of Q x x 3 ax, we solve x 3 ax 0. Then we have x x 2 a 0 x 0 or x 2 a. The equation x 2 a has no real solutions because a 0. So Q has 1
x-intercept. We now show that Q is always increasing and hence has no extrema. Ifx1 x2 , then ax1 ax2 (because
a 0) and x13 x23 . So we have x13 ax1 x23 ax2 , and hence Q x1 Q x2 . Thus Q is increasing, that is, its graph always rises, and so it has no local extrema. 83. (a) P x x 1 x 3 x 4. Local maximum at 18 21.
Local minimum at 36 06.
(b) Since Q x P x 5, each point on the graph of Q has y-coordinate 5 units more than the corresponding point on the graph of P. Thus Q has a local maximum at 18 71 and a local minimum at 35 44.
10 10 5 -10
5
290
CHAPTER 3 Polynomial and Rational Functions
84. (a) P x x 2 x 4 x 5 has one local maximum and one local minimum.
(b) Since P a P b 0, and P x 0 for a x b (see the table
below), the graph of P must first rise and then fall on the interval a b, and so P must have at least one local maximum between a and b. Using similar reasoning, the fact that P b P c 0 and P x 0 for
10
b x c shows that P must have at least one local minimum between b and c. Thus P has at least two local extrema.
5
a a b b c c
Interval
-10
Sign of x a
Sign of x b
Sign of x c
Sign of x a x b x c
85. Since the polynomial shown has five zeros, it has at least five factors, and so the degree of the polynomial is greater than or equal to 5. 86. No, it is not possible. Clearly a polynomial must have a local minimum between any two local maxima. 87. P x 8x 03x 2 00013x 3 372 4000
(a) For the firm to break even, P x 0. From the graph, we see that P x 0 when x 252. Of course, the firm cannot produce
fractions of a blender, so the manufacturer must produce at least 26 blenders a year.
2000
(b) No, the profit does not increase indefinitely. The largest profit is approximately $ 327622, which occurs when the firm produces
0 0
200
166 blenders per year.
88. P t 120t 04t 4 1000
(a) A maximum population of approximately 1380 is attained after 422 months. (b) The rabbit population disappears after approximately 842 months.
1000 500 0 0
5
10
89. (a) The length of the bottom is 40 2x, the width of the bottom is 20 2x, and the height is x, so the volume of the box is V x 20 2x 40 2x 4x 3 120x 2 800x.
(c) Using the domain from part (b), we graph V in the viewing rectangle [0 10] by [0 1600]. The maximum volume is V 15396 when x 423.
(b) Since the height and width must be positive, we must have x 0 and 20 2x 0, and so the domain of V is 0 x 10.
1000
0 0
5
10
SECTION 3.3 Dividing Polynomials
90. (a) Let h be the height of the box. Then the total length of all 12 edges is 8x 4h 144 in. Thus, 8x 4h 144
2x h 36 h 36 2x. The volume of the box is equal to area of baseheight x 2 36 2x 2x 3 36x 2 .
291
(c) Using the domain from part (b), we graph V in the viewing rectangle [0 18] by [0 2000]. The maximum volume is V 1728 in3 when x 12 in. 2000
Therefore, the volume of the box is
V x 2x 3 36x 2 2x 2 18 x. (b) Since the length of the base is x, we must have x 0. Likewise, the height must be positive so 36 2x 0
1000 0 0
10
x 18. Putting these together, we get that the domain of V
is 0 x 18.
The graph of y x 100 is close to the x-axis for x 1, but passes through the
y
91. (_1, 1) y=x$ y=x%
points 1 1 and 1 1. The graph of y x 101 behaves similarly except that the
(1, 1)
1
y-values are negative for negative values of x, and it passes through 1 1
y=x@
instead of 1 1. 1
x
y=x#
(1, _1)
92. No, it is impossible. The end behavior of a third degree polynomial is the same as that of y kx 3 , and for this function, the values of y go off in opposite directions as x and x . But for a function with just one extremum, the values of y would head off in the same direction (either both up or both down) on either side of the extremum. An nth-degree polynomial can have n 1 extrema or n 3 extrema or n 5 extrema, and so on (decreasing by 2). A polynomial that has six local extrema must be of degree 7 or higher. For example, P x x 1 x 2 x 3 x 4 x 5 x 6 x 7 has six local extrema.
3.3
DIVIDING POLYNOMIALS
1. If we divide the polynomial P by the factor x c, and we obtain the equation P x x c Q x R x, then we say that x c is the divisor, Q x is the quotient, and R x is the remainder. 2. (a) If we divide the polynomial P x by the factor x c, and we obtain a remainder of 0, then we know that c is a factor of P. (b) If we divide the polynomial P x by the factor x c, and we obtain a remainder of k, then we know that P c k.
292
CHAPTER 3 Polynomial and Rational Functions
3.
2
2
5
7
4
2
2
1
9
4.
P x D x
x 2
5. 2x 1
2x 1
3 3
Thus, the quotient is 2x 1 and the remainder is 9, and 2x 2 5x 7
4
5
1
12
12
28
3
7
29
Thus, the quotient is 3x 2 3x 7 and the remainder is 29, and
9 . x 2
2x 12 4x 2 3x 7
9
3x 3 9x 2 5x 1 P x D x x 4 29 3x 2 3x 7 . x 4
2x 2 3x
6. 3x 4
4x 2 2x
6x 3 x 2 12x 5
6x 3 8x 2
9x 2 12x
x 7
x 12 15 2
9x 2 12x 5
Thus, the quotient is 2x 12 and the remainder is 15 2 , 15 P x 4x 2 3x 7 2 . and 2x 12 D x 2x 1 2x 1
7. x2 4
2x 2 x
1
x 2 3x 1
8x 2
x 3 x 2 x 3
2x 3 6x 2 17x
2x 5 0x 4
3x 5
x 3 2x 2
6x 4 18x 3 6x 2
4x
17x 3 8x 2
4
43x 2 14x 5
43x 2 129x 43
Thus, the quotient is 2x 2 x 1 and the remainder is 2x 4 x 3 9x 2 P x 2 x 1 4x 4 . 2x D x x2 4 x2 4
3x
17x 3 51x 2 17x
4x 4
4x 4, and
43
x 3 2x 2
2x 5 6x 4 2x 3 6x 4
x 2 4x x2
P x 6x 3 x 2 12x 5 2 5 2x 3x . D x 3x 4 3x 4
8.
2x 4 x 3 9x 2 2x 4
Thus, the quotient is 2x 2 3x and the remainder is 5, and
115x 48
Thus, the quotient is 2x 3 6x 2 17x 43 and the remainder is 115x 48 and
2x 5 x 3 2x 2 3x 5 P x D x x 2 3x 1 115x 48 3 2x 6x 2 17x 43 2 x 3x 1
SECTION 3.3 Dividing Polynomials
x 2
9. x 1
x 3
293
10. Long division:
x 3 0x 2 2x 6
x 3 5x 2 15x 35
x 3 x 2
x 3
x 2 2x
x2 x
x 4 2x 3 0x 2 10x
0
x 3 3x 3 5x 3
3x 6
5x 3 15x 2
3x 3
10x
15x 2 10x
9
15x 2 45x
Thus, the quotient is x 2 x 3 and the remainder is 9, so P x x 3 2x 6 x 1 x 2 x 3 9.
35x 35x 105 105 P x x 4 2x 3 10x x 3 x 3 5x 2 15x 35 105
Synthetic division:
3
1 1
2
0
10
0
3
15
45
105
5
15
35
105
Thus, the quotient is x 3 5x 2 15x 35 and the
remainder is 105, as above.
x2
11. 2x 3
1
2x 2
12.
2x 3 3x 2 2x
2x 1
2x 3 3x 2
4x 3
x 4
4x 3 2x 2
7x 9
2x 2 7x
2x
2x 2 x
2x 3
8x 9
3 Thus, the quotient is x 2 1 and the remainder is 3, and P x 2x 3 3x 2 2x 2x 3 x 2 1 3.
8x 4 5 Thus
P x 4x 3 7x 9 2x 1 2x 2 x 4 5.
294
CHAPTER 3 Polynomial and Rational Functions
13.
4x 2 2x 1
2x 2 1
8x 4 4x 3 6x 2 8x 4
9x 3 6x 2
14. 3x 2 3x 1
4x 2
4x 3 2x 2 4x 3
3x 2
27x 5 9x 4 0x 3 3x 2 0x 3 27x 5 27x 4 9x 3
18x 4 9x 3 3x 2
2x 2
18x 4 18x 3 6x 2
2x
9x 3 3x 2
2x
2x 2
9x 3 9x 2 3x
1
6x 2 3x 3
2x 1
6x 2 6x 2
Thus, the quotient is 4x 2 2x 1 and the remainder is 2x 1, and
P x 27x 5 9x 4 3x 2 3 3x 2 3x 1 9x 3 6x 2 3x 2 3x 5
2x 2 1 4x 2 2x 1 2x 1
15.
x 1
x2
16.
x 2 3x 7
x 2
3x 5
Thus
P x 8x 4 4x 3 6x 2
x 3
x 2 2x
x 2
x 3 2x 2 x 3 3x 2
x 1
x 2 x
x 7
x 2 3x
x 2 5
2x 1 2x 6
Thus, the quotient is x 1 and the remainder is 5.
5 Thus, the quotient is x 2 x 2 and the remainder is 5.
2x 2
17. 2x 1
18.
1
4x 3 2x 2 2x 3
4x 3 2x 2
2x 3 2x 1 2 Thus, the quotient is 2x 2 1 and the remainder is 2.
3x 6
1 x2 1 x 2 3 3 3 x 3 3x 2 4x 3
x 3 2x 2
x 2 4x x 2 2x 2x 3 2x 4 1
Thus, the quotient is 13 x 2 13 x 23 and the remainder is 1.
SECTION 3.3 Dividing Polynomials
19. x2 x 3
x
1
20.
x 3 0x 2 2x 1
x 2 5x 1
x 3 x 2 3x
x2 x 1 x2
x 2 2x
9
x 4 3x 3
0x 2
2x 3
x2
x 4 5x 3
x 2
x2
x
2x 3 10x 2
x 3
2x
9x 2
2
295
x 2
9x 2 45x 9
Thus, the quotient is x 1 and the remainder is 2.
44x 11 Thus, the quotient is x 2 2x 9 and the remainder is 44x 11.
21. 2x 2 0x 5
3x
1
22.
6x 3 2x 2 22x 0 6x 3
3 3x 2 7x
15x
2x 2
2x 2
9x 2
x 5
9x 2 21x
7x 0
20x 5
5
Thus, the quotient is 3 and the remainder is 20x 5.
7x 5 Thus, the quotient is 3x 1 and the remainder is 7x 5. x4
23. x2 1
24.
1
x 6 0x 5 x 4 0x 3 x 2 0x 1
x6
x4
x2
0
x2
4x 2 6x 8
1 x3 x2 5 x 7 2 2 4 2x 5 7x 4 0x 3 0x 2 0x 13
2x 5 3x 4 4x 3
4x 4 4x 3 0x 2
1
4x 4 6x 3 8x 2
1
10x 3 8x 2 0x
0
10x 3 15x 2 20x
Thus, the quotient is x 4 1 and the remainder is 0.
7x 2 20x 13
7x 2 21 2 x 14
19 x 1 2
Thus, the quotient is 12 x 3 x 2 52 x 74 and the remainder is 19 2 x 1. 25. The synthetic division table for this problem takes the following form.
26. The synthetic division table for this problem takes the following form.
3
2 2
5
3
6
3
1
6
Thus, the quotient is 2x 1 and the remainder is 6.
1
1 1
1
4
1
2
2
6
Thus, the quotient is x 2 and the remainder is 6.
296
CHAPTER 3 Polynomial and Rational Functions
27. The synthetic division table for this problem takes the following form.
28. The synthetic division table for this problem takes the following form.
3
1
3
1
0
3
2
2
2
2
4
Thus, the quotient is 3x 2 and the remainder is 2. 29. Since x 2 x 2, the synthetic division table for this problem takes the following form. 2
1
2
2
1
2
0
4
0
2
3
1
0
8
2
3
9
3
3
1
1
Thus, the quotient is x 2 3x 1 and the remainder is 1. 33. Since x 5 3x 3 6 x 5 0x 4 3x 3 0x 2 0x 6, the synthetic division table for this problem takes the
3
1
12
9
1
15
15
30
3
6
31
following form. 2
1
1 2
2
6
10
1
1
3
5
12
0
3
0
0
6
1
1
4
4
4
1
4
4
4
2
Thus, the quotient is x 4 x 3 4x 2 4x 4 and the
1
2
1
Thus, the quotient is x 3 x 2 3x 5 and the remainder is 12.
34. The synthetic division table for this problem takes the following form. 3
1
13
Thus, the quotient is 3x 2 3x 6 and the remainder is 31.
following form. 1
16
8
following form.
table for this problem takes the following form.
1
8
32. The synthetic division table for this problem takes the
x 3 8x 2 x 3 0x 2 8x 2, the synthetic division 1
3
30. The synthetic division table for this problem takes the
3
31. Since x 3 x 3 and
0
Thus, the quotient is 4x 8 and the remainder is 13.
5
Thus, the quotient is x 2 2 and the remainder is 3.
3
4
1
9
27
27
3
18
27
1
6
9
0
Thus, the quotient is x 2 6x 9 and the remainder is 0.
remainder is 2.
35. The synthetic division table for this problem takes the following form.
36. The synthetic division table for this problem takes the following form.
1 2
2 2
3
2
1
1
2
0
4
0
1
Thus, the quotient is 2x 2 4x and the remainder is 1.
23
6 6
10
5
1
1
4
4
6
1
23 1 3
29 7 9
Thus, the quotient is 6x 3 6x 2 x 13 and the remainder is 79 .
SECTION 3.3 Dividing Polynomials
37. Since x 3 27 x 3 0x 2 0x 27, the synthetic
division table for this problem takes the following form. 3
1 1
0
0
27
3
9
27
3
9
0
4
5
4
8
8
3
1
3
7
6
1
2
10
6
5
3
12
30
2
0
7
2
0
0
0
0
7
5
40
0
2
9
1
1
5
10
6
Therefore, by the Remainder Theorem, P 12 6. 1
1
1
5
1
2
3
2
3
2
Therefore, by the Remainder Theorem, P 1 2.
2
21
9
200
22
11
220
1
20
20
36
14
35
35
35
497
5
5
71
483
Therefore, by the Remainder Theorem, P 11 20. 46. P x 6x 5 10x 3 x 1, c 2 2
6 6
0
10
0
1
1
12
24
68
136
274
12
34
68
137
273
Therefore, by the Remainder Theorem, P 2 273. 48. P x 2x 6 7x 5 40x 4 7x 2 10x 112, c 3
x 7 0x 6 0x 5 0x 4 0x 3 3x 2 0x 1 c3
1
16
8
2
47. P x x 7 3x 2 1
1
8
4
is 0.
11
Therefore, by the Remainder Theorem, P 7 483.
3
4
2
44. P x 2x 3 21x 2 9x 200, c 11
45. P x 5x 4 30x 3 40x 2 36x 14, c 7 5
2
1
Therefore, by the Remainder Theorem, P 2 7.
7
16
Thus, the quotient is x 3 2x 2 4x 8 and the remainder
1
43. P x x 3 2x 2 7, c 2 1
0
42. P x x 3 x 2 x 5, c 1
Therefore, by the Remainder Theorem, P 2 12.
2
0
2
41. P x x 3 3x 2 7x 6, c 2 1
0
1 2
Therefore, by the Remainder Theorem, P 1 3.
2
1
40. P x 2x 2 9x 1, c 12
12
4
following form.
1
39. P x 4x 2 12x 5 c 1 1
38. The synthetic division table for this problem takes the
2
Thus, the quotient is x 2 3x 9 and the remainder is 0.
297
0
0
0
0
3
0
1
3
9
27
81
243
720
2160
3
9
27
81
240
720
2159
Therefore by the Remainder Theorem, P 3 2159.
3
2 2
7
40
0
7
10
112
6
39
3
9
6
12
13
1
3
2
4
100
Therefore, by the Remainder Theorem, P 3 100.
298
CHAPTER 3 Polynomial and Rational Functions
49. P x 3x 3 4x 2 2x 1, c 23 2 3
3 3
4
2
2
4
6
2
50. P x x 3 x 1, c 14 1 4
1 4 3 7 3
1
0
1
1 4 1 4
1
1 16 15 16
1 15 64 49 64
Therefore, by the Remainder Theorem, P 14 49 64 .
Therefore, by the Remainder Theorem, P 23 73 . 51. P x x 3 2x 2 3x 8, c 01 01
1 1
2
3
8
01
021
0279
21
279
8279
Therefore, by the Remainder Theorem, P 01 8279.
52. (a) P x 6x 7 40x 6 16x 5 200x 4 60x 3 69x 2 13x 139, c 7 7
6 6
40
16
200
60
69
13
139
42
14
210
70
70
7
140
2
30
10
10
1
20
1
Therefore, by the Remainder Theorem, P 7 1. (b) P 7 6 77 40 76 16 75 200 74 60 73 69 72 13 7 139 6 823,543 40 117,649 16 16,807 200 2401 60 343 69 49 13 7 139 1
which agrees with the value obtained by synthetic division, but requires more work. 53. P x x 3 3x 2 3x 1, c 1 1
1 1
54. P x x 3 2x 2 3x 10, c 2
3
3
1
1
2
1
2
1
0
Since the remainder is 0, x 1 is a factor. 55. P x 2x 3 7x 2 6x 5, c 12 1 2
2 2
2
1 1
2
3
10
2
8
10
4
5
0
Since the remainder is 0, x 2 is a factor. 56. P x x 4 3x 3 16x 2 27x 63, c 3, 3
7
6
5
1
4
5
8
10
0
Since the remainder is 0, x 12 is a factor.
3
1 1
3
16
27
63
3
18
6
63
6
2
21
0
Since the remainder is 0, x 3 is a factor. We next show that x 3 is also a factor by using synthetic division on the quotient of the above synthetic division, x 3 6x 2 2x 21. 3
1 1
6
2
21
3
9
21
3
7
0
Since the remainder is 0, x 3 is a factor.
SECTION 3.3 Dividing Polynomials
57. P x x 3 2x 2 9x 18, c 2 1
2
58. P x x 3 5x 2 2x 10, c 5
2
9
18
2
0
18
0
9
0
1
5
Hence, the zeros are 3, 2, and 3. .
1
5
2
10
5
0
10
0
2
0
Since the remainder is 0, we know that 5 is a zero and P x x 5 x 2 2 x 5 x 2 x 2 . Hence, the zeros are 5 and 2. 60. P x 3x 4 x 3 21x 2 11x 6, c 2, 13
59. x 3 x 2 11x 15, c 3 1
1 1
Since the remainder is 0, we know that 2 is a zero and P x x 2 x 2 9 x 2 x 3 x 3.
3
1
11
15
3
6
15
2
5
0
3
2
3
Since the remainder is 0, we know that 3 is a zero and P x x 3 x 2 2x 5 . Now x 2 2x 5 0 2 22 4 1 5 1 6. Hence, when x 2 1 the zeros are 3 and 1 6.
1
21
11
6
6
14
14
6
7
7
3
0
Since the remainder is 0, we know that 2 is a zero and P x x 2 3x 3 7x 2 7x 3 . 1 3
3
7
7
3
3
1
2
3
6
9
0
Since the remainder is 0, we know that 13 is a zero and P x x 2 x 13 3x 2 6x 9 3 x 2 x 13 x 1 x 3. Hence, the zeros are 2, 13 , 1, and 3.
61. P x 3x 4 8x 3 14x 2 31x 6, c 2, 3 3
2
3
8
14
31
6
6
28
28
6
14
14
3
0
62. P x 2x 4 13x 3 7x 2 37x 15, c 1, 3 2
1
2
13
7
37
15
2
15
22
15
15
22
15
0
Since the remainder is 0, we know that 2 is a zero and P x x 2 3x 3 14x 2 14x 3 . 14
14
3
Since the remainder is 0, we know that 1 is a zero and P x x 1 2x 3 15x 2 22x 15 . 15
22
15
9
15
3
6
27
15
5
1
0
9
5
0
3
3 3
299
Since the remainder is 0, we know that 3 is a zero and P x x 2 x 3 3x 2 5x 1 . Now
3x 2 5x 1 0 when 5 52 4 3 1 5 37 , and x 2 3 6 5 37 . hence, the zeros are 2, 3, and 6
3
2 2
The remainder is 0, so 3 is a zero and P x x 1 x 3 2x 2 9x 5
x 1 x 3 2x 1 x 5.
Hence, the zeros are 1, 3, 12 , and 5.
300
CHAPTER 3 Polynomial and Rational Functions
63. Since the zeros are x 1, x 1, and x 3, the factors are x 1, x 1, and x 3. Thus P x x 1 x 1 x 3 x 3 3x 2 x 3.
64. Since the zeros are x 2, x 0, x 2, and x 4, the factors are x 2, x, x 2, and x 4.
Thus P x c x 2 x x 2 x 4. If we let c 1, then P x x 4 4x 3 4x 2 16x.
65. Since the zeros are x 1, x 1, x 3, and x 5, the factors are x 1, x 1, x 3, and x 5. Thus P x x 1 x 1 x 3 x 5 x 4 8x 3 14x 2 8x 15.
66. Since the zeros are x 2, x 1, x 0, x 1, and x 2, the factors are x 2, x 1, x, x 1, and x 2. Thus P x c x 2 x 1 x x 1 x 2. If we let c 1, then P x x 5 5x 3 4x.
67. Since the zeros of the polynomial are 2, 0, 1, and 3, it follows that P x C x 2 x x 1 x 3 C x 4 2C x 3
5C x 2 6C x. Since the coefficient of x 3 is to be 4, 2C 4, so C 2. Therefore, P x 2x 4 4x 3 10x 2 12x is the polynomial. 68. Since the zeros of the polynomial are 1, 0, 2, and 12 , it follows that P x C x 1 x x 2 x 12 C x 4
3 C x 3 3 C x 2 C x. Since the coefficient of x 3 is to be 3, 3 C 3, so C 2. Therefore, P x 2x 4 3x 3 3x 2 2x 2 2 2
is the polynomial.
2 and integer coefficients, the fourth zero must be 2, otherwise the constant term would be irrational. Thus, P x C x 1 x 1 x 2 x 2 C x 4 3C x 2 2C. Requiring
69. Since the polynomial degree 4 and zeros 1, 1, and
that the constant term be 6 gives C 3, so P x 3x 4 9x 2 6. 70. Since the polynomial degree 5 and zeros 2, 1, 2, and 5 and integer coefficients, the fourth zero must be 5, otherwise the constant term would be irrational. Thus, P x C x 2 x 1 x 2 x 5 x 5 C x 5 C x 4 9C x 3 9C x 2 20C x 20C. Requiring that the constant term be 40 gives C 2, so P x 2x 5 2x 4 18x 3 18x 2 40x 40.
71. The y-intercept is 2 and the zeros of the polynomial are 1, 1, and 2. It follows that P x C x 1 x 1 x 2 C x 3 2x 2 x 2 . Since P 0 2 we have 2 C 03 2 02 0 2 2 2C C 1 and P x x 1 x 1 x 2 x 3 2x 2 x 2.
72. The y-intercept is 4 and the zeros of the polynomial are 1 and 2 with 2 being degree two. It follows that P x C x 1 x 22 C x 3 3x 2 4 . Since P 0 4 we have 4 C 03 3 02 4 4 4C C 1 and P x x 3 3x 2 4.
73. The y-intercept is 4 and the zeros of the polynomial are 2 and 1 both being degree two. It follows that P x C x 22 x 12 C x 4 2x 3 3x 2 4x 4 . Since P 0 4 we have 4 C 04 2 03 3 02 4 0 4 4 4C C 1. Thus P x x 22 x 12 x 4 2x 3 3x 2 4x 4.
74. The y-intercept is 2 and the zeros of the polynomial are 2, 1, and 1 with 1 being degree two. It follows that P x C x 2 x 1 x 12 C x 4 x 3 3x 2 x 2 . Since P 0 2 we have 4 C 04 03 3 02 0 2 2 2C so C 1 and P x x 4 x 3 3x 2 x 2.
75. A. By the Remainder Theorem, the remainder when P x 6x 1000 17x 562 12x 26 is divided by x 1 is P 1 6 11000 17 1562 12 1 26 6 17 12 26 3.
B. If x 1 is a factor of Q x x 567 3x 400 x 9 2, then Q 1 must equal 0. Q 1 1567 3 1400 19 2 1 3 1 2 1 0, so x 1 is not a factor.
SECTION 3.4 Real Zeros of Polynomials
301
76. R x x 5 2x 4 3x 3 2x 2 3x 4 x 4 2x 3 3x 2 2x 3 x 4 x 3 2x 2 3x 2 x 3 x 4 x 2 2x 3 x 2 x 3 x 4 [x 2 x 3] x 2 x 3 x 4
So to calculate R 3, we start with 3, then subtract 2, multiply by 3, add 3, multiply by 3, subtract 2, multiply by 3, add 3, multiply by 3, and add 4, to get 157.
3.4
REAL ZEROS OF POLYNOMIALS
1. If the polynomial function P x an x n an1 x n1 a1 x a0 has integer coefficients, then the only numbers that p could possibly be rational zeros of P are all of the form , where p is a factor of the constant coefficient a0 and q is a factor q of the leading coefficient an . The possible rational zeros of P x 6x 3 5x 2 19x 10 are 1, 12 , 13 , 16 , 2, 23 , 5, 52 , 53 , 56 , 10, and 10 3.
2. Using Descartes’ Rule of Signs, we can tell that the polynomial P x x 5 3x 4 2x 3 x 2 8x 8 has 1, 3 or 5 positive real zeros and no negative real zero. 3. This is true. If c is a real zero of the polynomial P, then P x x c Q x, and any other zero of P x is also a zero of Q x P x x c. 4. False. Consider the polynomial P x x 1 x 2 x 4 x 3 x 2 10x 8. An upper bound is 3, but 3 is not a lower bound.
5. P x x 3 4x 2 3 has possible rational zeros 1 and 3.
6. Q x x 4 3x 3 6x 8 has possible rational zeros 1, 2, 4, 8.
7. R x 2x 5 3x 3 4x 2 8 has possible rational zeros 1, 2, 4, 8, 12 .
8. S x 6x 4 x 2 2x 12 has possible rational zeros 1, 2, 3, 4, 6, 12, 12 , 32 , 13 , 23 , 43 , 16 .
9. T x 4x 4 2x 2 7 has possible rational zeros 1, 7, 12 , 72 , 14 , 74 .
1 . 10. U x 12x 5 6x 3 2x 8 has possible rational zeros 1, 2, 4, 8 12 , 13 , 23 , 43 , 83 , 14 , 16 , 12
11. (a) P x 5x 3 x 2 5x 1 has possible rational zeros 1, 15 . (b) From the graph, the actual zeros are 1, 15 , and 1.
12. (a) P x 3x 3 4x 2 x 2 has possible rational zeros 1, 2, 13 , 23 . (b) From the graph, the actual zeros are 1 and 23 .
13. (a) P x 2x 4 9x 3 9x 2 x 3 has possible rational zeros 1, 3, 12 , 32 . (b) From the graph, the actual zeros are 12 , 1, and 3.
14. (a) P x 4x 4 x 3 4x 1 has possible rational zeros 1, 12 , 14 . (b) From the graph, the actual zeros are 14 and 1.
15. P x x 3 2x 2 13x 10. The possible rational zeros are 1, 2, 5, 10. P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x x 3 2x 2 13x 10 has 1 variation in sign and hence P has 1 negative real zero. 1
2 2
1
13
10
2
3
10
10 0 x 1 is a zero. P x x 3 2x 2 13x 10 x 1 x 2 3x 10 x 5 x 1 x 2. Therefore, the zeros are 5, 1, and 2.
3
302
CHAPTER 3 Polynomial and Rational Functions
16. P x x 3 4x 2 19x 14. The possible rational zeros are 1, 7, 14. P x has 1 variation in sign and hence 1 positive real zero. P x x 3 4x 2 19x 14 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1
1
1
4
19
14
1
3
22
1
1
4
19
14
1
5
14
x 1 is not a zero. 1 5 14 0 x 1 is a zero. P x x 3 4x 2 19x 14 x 1 x 2 5x 14 x 2 x 1 x 7. Therefore, the zeros are 2, 1, and 3
22
36
7.
17. P x x 3 3x 2 4. The possible rational zeros are 1, 2, 4. P x has 1 variation in sign and hence 1 positive real zero.P x x 3 3x 2 4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1
1
3
0
4
1
4
4
1
4 4 0 x 1 is a zero. P x x 3 3x 2 4 x 1 x 2 4x 4 x 1 x 22 . Therefore, the zeros are 2 and 1.
18. P x x 3 3x 2. The possible rational zeros are 1, 2. P x has 1 variation in sign and hence 1 positive real zero.P x x 3 3x 2 has 2 variations in sign and P has hence 0 or 2 negative real zeros. 1
1 1
0
3
1
1
1
2
1
2
2 2
0
3
2
2
4
2
x 1 is not a zero. 1 2 1 0 x 2 is a zero. P x x 3 3x 2 x 2 x 2 2x 1 x 2 x 12 . Therefore, the zeros are 2 and 1. 4
19. P x x 3 6x 2 12x 8. The possible rational zeros are 1, 2, 4, 8. P x has 3 variations in sign and hence 1 or 3 positive real zeros. P x x 3 6x 2 12x 8 has no variations in sign and hence P has no negative real zero. 1
1 1
6
12
8
1
5
7
7
1
1
2
6
12
8
2
8
8
x 1 is not a zero. 1 4 4 0 x 2 is a zero. P x x 3 6x 2 12x 8 x 2 x 2 4x 4 x 23 . Therefore, the only zero is x 2. 5
20. P x x 3 12x 2 48x 64. The possible rational zeros are 1, 2, 4, 8, 16, 32, 64. P x has 0 variations in sign and hence no positive real zero. P x x 3 12x 2 48x 64 has 3 variations in sign and hence P has 1 or 3 negative real zeros. 1
1 1
12
48
64
1
11
37
11
37
27
2 x 1 is not a zero. 4
1 1
1 1
12
48
64
4
32
64
8
16
12
48
64
2
20
56
10
28
6
x 2 is not a zero.
0 x 4 is a zero. P x x 3 12x 2 48x 64 x 4 x 2 8x 16 x 43 . Therefore, the only zero is x 4.
SECTION 3.4 Real Zeros of Polynomials
303
21. P x x 3 19x 30. The possible rational zeros are 1, 2, 3, 5, 6, 15, 30. P x has 1 variations in sign and hence 1 positive real zero. P x x 3 19x 30 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1
1 1
0
19
30
1
1
18
1
18
12
1
2 x 1 is not a zero. 1
4
1
12
48
64
4
32
64
8
16
1
0
19
30
2
4
30
2
15
0
x 2 is a zero.
0 x 4 is a zero. P x x 3 19x 30 x 2 x 2 2x 15 x 3 x 2 x 5. Therefore, the zeros are 3, 2, and 5.
22. P x x 3 11x 2 8x 20. The possible rational zeros are 1, 2, 4, 5, 10, 20. P x has 1 variation in sign
and hence 1 positive real zeros. P x x 3 11x 2 8x 20 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1
1
11
8
20
1
12
20
1 12 20 0 x 1 is a zero. P x x 3 11x 2 8x 20 x 1 x 2 12x 20 x 10 x 2 x 1. Therefore, the zeros are 10, 2,
and 1.
23. P x x 3 3x 2 x 3. The possible rational zeros are 1, 3. P x has 1 variation in sign and hence 1 positive real zero. P x x 3 3x 2 x 4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1
1
3
1
3
1
2
3
2
3
0
1
x 1 is a zero.
So P x x 3 3x 2 x 3 x 1 x 2 2x 3 x 1 x 3 x 1. Therefore, the zeros are 1, 3, and 1.
24. P x x 3 4x 2 11x 30. The possible rational zeros are 1, 2, 3, 5, 10, 15, 30. P x has 2 variations
in sign and hence 0 or 2 positive real zeros. P x x 3 4x 2 11x 30 has 1 variation in sign and hence P has 1 negative real zero. 1
1 1
4
11
30
1
3
14
2
1
4
11
30
2
4
30
1 2 15 0 x 2 is a zero. So P x x 3 4x 2 11x 30 x 2 x 2 2x 15 x 2 x 5 x 3. Therefore, the zeros are 3, 2, and 5.
3
14
16
304
CHAPTER 3 Polynomial and Rational Functions
25. Method 1: P x x 4 5x 2 4The possible rational zeros are 1, 2, 4. P x has 1 variation in sign and hence 1 positive real zero. P x x 4 5x 2 4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1
1
0
5
0
4
1
1
4
4
1 1 4 4 0 x 1 is a zero. Thus P x x 4 5x 2 4 x 1 x 3 x 2 4x 4 . Continuing with the quotient we have: 1
1
1
4
4
1
0
4
1 0 4 0 x 1 is a zero. P x x 4 5x 2 4 x 1 x 1 x 2 4 x 1 x 1 x 2 x 2. Therefore, the zeros are 1, 2.
Method 2: Substituting u x 2 , the polynomial becomes P u u 2 5u 4, which factors: u 2 5u 4 u 1 u 4 x 2 1 x 2 4 , so either x 2 1 or x 2 4. If x 2 1, then x 1; if x 2 4, then x 2. Therefore, the zeros are 1 and 2.
26. P x x 4 2x 3 3x 2 8x 4. Using synthetic division, we see that x 1 is a factor of P x: 1
1 1
2
3
8
4
1
1
4
4
1
4
4
0
x 1 is a zero.
We continue by factoring the quotient, and we see that x 1 is again a factor: 1
1
1
1
4
4
1
0
4
0
4 0 x 1 is a zero. P x x 4 2x 3 3x 2 8x 4 x 1 x 1 x 2 4 x 12 x 2 x 2
Therefore, the zeros are 1 and 2.
27. P x x 4 6x 3 7x 2 6x 8. The possible rational zeros are 1, 2, 4, 8. P x has 1 variation in sign and hence 1 positive real zero. P x x 4 6x 3 7x 2 6x 8 has 3 variations in sign and hence P has 1 or 3 negative real zeros. 1
1 1
6
7
6
8
1
7
14
8
7
14
8
0
x 1 is a zero
and there are no other positive zeros. Thus P x x 4 6x 3 7x 2 6x 8 x 1 x 3 7x 2 14x 8 . Continuing
by factoring the quotient, we have:
1
1 1
7
14
8
1
6
8
6
8
x 1 is a zero. So P x x 4 6x 3 7x 2 6x 8 x 1 x 1 x 2 6x 8 x 1 x 1 x 2 x 4. Therefore, the zeros are 4, 2, and 1.
0
SECTION 3.4 Real Zeros of Polynomials
305
28. P x x 4 x 3 23x 2 3x 90. The possible rational zeros are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90. Since P x has 2 variations in sign, P has 0 or 2 positive real zeros. Since P x x 4 x 3 23x 2 3x 90 has 2 variations in sign, P has 0 or 2 negative real zeros. 1
1
1
23
3
90
1
0
23
26
0
23
26
64
1 3
1
1
21
45
3
12
27
1
2
1 1
5
1
23
3
90
2
2
42
90
1
21
45
0
1
21
45
5
30
45
x 2 is a zero.
1 6 9 0 x 5 is a zero. 4 9 72 2 2 P x x 2 x 5 x 6x 9 x 2 x 5 x 3 . Therefore, the zeros are 3, 2, and 5. 1
29. P x 4x 4 37x 2 9 has possible rational zeros 1, 3, 9, 12 , 32 , 92 , 14 , 34 , 94 . Since P x has 2 variations in sign, there are 0 or 2 positive real zeros, and since P x 4x 4 37x 2 36 has 2 variations in sign, P has 0 or 2 negative real zeros. 1
4 4
0
37
0
9
4
4
33
33
4
33
33
24 1 2
4
3
4
4 4
12
1
3
2
7
3
14
6
0
Because P is even, we conclude that the zeros are 3 and 12 and
0
37
0
9
12
36
3
9
12
1
3
0
x 3 is a zero.
x 12 is a zero.
P x 4x 4 37x 2 9 x 3 2x 1 2x 1 x 3. Note: Since P x has only even terms, factoring by substitution also works. Let x 2 u; then P u 4u 2 37u 9 4u 1 u 9 4x 2 1 x 2 3 , which gives the same results.
30. P x 6x 4 23x 3 13x 2 32x 16. The possible rational zeros are 1, 2, 4, 8, 16, 12 , 13 , 23 , 43 , 83 , 16 3 , 16 . Since P x has 2 variations in sign, P has 0 or 2 positive real zeros. Since P x 6x 4 23x 3 13x 2 32x 16 has 2 variations in sign, P has 0 or 2 negative real zeros. 1
6 6
23
13
32
16
6
17
30
2
17
30
2
18
4
6
2
6
23
13
32
16
24
4
36
16
6
6
23
13
32
16
12
22
70
76
11
35
38
60
9 4 0 x 4 is a zero P x 6x 4 23x 3 13x 2 32x 16 x 4 6x 3 x 2 9x 4 . We continue: 12
1
6
1
9
4
3
1
4
6 2 8 0 x 12 is a zero Thus, P x x 4 2x 1 3x 2 x 4 x 1 2x 1 3x 4 x 4 and the rational zeros are 1, 12 , 43 ,
and 4.
306
CHAPTER 3 Polynomial and Rational Functions
31. P x 3x 4 10x 3 9x 2 40x 12. The possible rational zeros are 1, 2, 3, 4, 6, 12. P x has 3 variations in sign and hence 1, 3, or 5 positive real zeros. P x 3x 4 10x 3 9x 2 40x 12 has 1 variation in sign and hence P has 1 negative real zero. 1 3 10
9
3 3
40 12
2 3 10
24
6
7 16
40 12
9
12
8 34
24 x 1 is not a zero. 3 4 17 6 0 x 2 is a zero. Thus P x 3x 4 10x 3 9x 2 40x 12 x 2 3x 3 4x 2 17x 6 . Continuing by factoring the quotient, we 7 16
24
have
3
3
4
17
6
9
15
6
3 5 2 0 x 3 is a zero. Thus P x x 3 x 2 3x 2 5x 2 x 3 x 2 3x 1 x 2. Therefore, the zeros are 2, 13 , 2, and 3.
32. P x 2x 3 7x 2 4x 4. The possible rational zeros are 1, 2, 4, 12 . Since P x has 1 variation in sign, P has 1 positive real zero. Since P x 2x 3 7x 2 4x 4 has 2 variations in sign, P has 0 or 2 negative real zeros. 1
2
7
4
4
2
9
13
1 2
2
7
4
4
1
4
4
9 13 9 x 1 is an upper bound. 2 8 8 0 x 12 is a zero. P x x 12 2x 2 8x 8 2 x 12 x 2 4x 4 2 x 12 x 22 . Therefore, the zeros are 2 and 12 . 2
33. Factoring by grouping can be applied to this exercise.
4x 3 4x 2 x 1 4x 2 x 1 x 1 x 1 4x 2 1 x 1 2x 1 2x 1. Therefore, the zeros are 1 and 12 .
34. We use factoring by grouping: P x 2x 3 3x 2 2x 3 2x x 2 1 3 x 2 1 x 2 1 2x 3 x 1 x 1 2x 3. Therefore, the zeros are 32 and 1. 35. P x 4x 3 7x 3. The possible rational zeros are 1, 3, 12 , 32 , 14 , 34 . Since P x has 2 variations in sign, there are 0 or 2 positive zeros. Since P x 4x 3 7x 3 has 1 variation in sign, there is 1 negative zero. 1 2
4
0
7
3
2
1
3
4 2 6 0 x 12 is a zero. P x x 12 4x 2 2x 6 2x 1 2x 2 x 3 2x 1 x 1 2x 3 0. Thus, the zeros are 32 ,
1 , and 1. 2
SECTION 3.4 Real Zeros of Polynomials
307
1 . Since P x has 2 36. P x 12x 3 25x 2 x 2. The possible rational zeros are 1, 2, 12 , 13 , 23 , 14 , 16 , 12
variations in sign, P has 0 or 2 positive real zeros, and since P x 12x 3 25x 2 x 2 has 1 variations in sign, P has 1 negative real zero. 1
12
25
1
2
12
13
12
2
12
25
1
2
24
2
2
12 1 1 0 x 2 is a zero. 10 P x 12x 3 25x 2 x 2 x 2 12x 2 x 1 4x 1 3x 1 x 2. Therefore, the zeros are 14 , 13 , 12
13
12
and 2.
37. P x 24x 3 10x 2 13x 6. The possible rational zeros are 1, 2, 3, 6, 12 , 32 , 13 , 23 , 14 , 34 , 16 , 18 , 1 , 1 . P x has 1 variation in sign and hence 1 positive real zero. P x 24x 3 10x 2 13x 6 has 2 38 , 12 24
variations in sign, so P has 0 or 2 negative real zeros. 1 24
10 13 6 24
14 1
24 14 3 24
2 24
48
1 7 x 1 is not a zero.
10 13
10 13
24 38 6 24
6
72 186 519
6
76 126 63 132 x 2 is not a zero.
10 13
6
144 804 4746
24 62 173 525 x 3 is not a zero. 12
24 24
24 134 791 4752 x 6 is not a zero.
10
13
6
12
1
6
2
12
0
x 12 is a zero.
Thus P x 24x 3 10x 2 13x 6 2x 1 12x 2 x 6 3x 2 2x 1 4x 3 has zeros 23 , 12 , and 3. 4
1 , 3 , 3 . P x has 2 38. P x 12x 3 20x 2 x 3. The possible rational zeros are 1, 2, 3, 12 , 13 , 14 , 16 , 12 2 4
variations in sign and hence 0 or 2 positive real zeros. P x 12x 3 20x 2 x 3 has 1 variations in sign and hence P has 1 negative real zero. 1
12 12
3
12 12
20
1
3
12
8
7
8
7
4
20
1
3
36
48
147
16
49
150
12
2 x 1 is not a zero.
12 1 2
x 3 is not a zero.
12 12
20
1
3
24
8
18
4
9
21
x 2 is not a zero.
20
1
3
6
7
3
14
6
0
x 12 is a zero.
308
CHAPTER 3 Polynomial and Rational Functions
Thus, P x 12x 3 20x 2 x 3 x 12 12x 2 14x 6 . Continuing: 3 2
12 12
14
6
18
6
4
0
x 32 is a zero.
Thus, P x 2x 1 2x 3 3x 1 has zeros 12 , 32 , and 13 .
39. P x 2x 4 7x 3 3x 2 8x 4. The possible rational zeros are 1, 2, 4, 12 . P x has 3 variations in sign and hence 1 or 3 positive real zeros. P x 2x 4 7x 3 3x 2 8x 4 has 1 variation in sign and hence P has 1 negative real zero. 1
2
7
3
8
4
2
5
2
6
2
6
2
1 2
2
7
3
8
4
1
3
0
4
x 1 is not a zero. 2 6 0 8 0 x 12 is a zero. Thus P x 2x 4 7x 3 3x 2 8x 4 x 12 2x 3 6x 2 8 . Continuing by factoring the quotient, we have: 2
5
2
2
6
0
8
4
4
8
2 2 4 0 x 2 is a zero. P x x 12 x 2 2x 2 2x 4 2 x 12 x 2 x 2 x 2 2 x 12 x 22 x 1. Thus, the
zeros are 12 , 2, and 1.
40. P x 6x 4 7x 3 12x 2 3x 2. The possible rational zeros are 1, 2, 12 , 13 , 23 , 16 . Since P x has 2 variations in sign, P has 0 or 2 positive real zeros. Since P x 6x 4 7x 3 12x 2 3x 2 has 2 variations in sign, P
has 0 or 2 negative real zeros.
7
12
3
2
12
10
4
2
6 5 2 P x 6x 4 7x 3 12x 2 3x 2 x 2 6x 3 5x 2 2x 1 .
1
0
1
6 6
7
12
3
2
6
1
13
10
1
13
10
8
2
6
x 2 is a zero.
SECTION 3.4 Real Zeros of Polynomials
309
Continuing by factoring the quotient, we first note that the possible rational zeros are 1, 12 , 13 , 16 . We have: 1 2
6
5
2
1
3
4
1
0 x 12 is a zero. P x x 2 x 12 6x 2 8x 2 2 x 2 x 12 3x 2 4x 1 2 x 2 x 12 x 1 3x 1. 6
8
2
Therefore, the zeros are 1, 13 , 12 , and 2.
41. P x x 5 3x 4 9x 3 31x 2 36. The possible rational zeros are 1, 2, 3, 4, 6, 8, 9, 12, 18. P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x x 5 3x 4 9x 3 31x 2 36 has 3 variations in sign and hence P has 1 or 3 negative real zeros. 1
1 1
3
9
31
0
36
1
4
5
36
36
4
5
36 36 0 x 1 is a zero. So P x x 5 3x 4 9x 3 31x 2 36 x 1 x 4 4x 3 5x 2 36x 36 . Continuing by factoring the quotient, we have:
1
1 1
4
5
36
36
1
5
0
36
1
0
36
72
3
1 1
2
1 1
4
5
36
36
3
21
48
36
7
16
4
5
36
36
2
12
14
44
6
7
22
80
12 0 x 3 is a zero. So P x x 1 x 3 x 3 7x 2 16x 12 . Since we have 2 positive zeros, there are no more positive zeros, so
we continue by factoring the quotient with possible negative zeros. 1
1
7
16
12
1
6
10
6
10
1
2
1
7
16
12
2
10
12
2
1 5 6 0 x 2 is a zero. Then P x x 1 x 3 x 2 x 2 5x 6 x 1 x 3 x 22 x 3. Thus, the zeros are 1, 3, 2,
and 3.
42. P x x 5 4x 4 3x 3 22x 2 4x 24 has possible rational zeros 1, 2, 3, 4, 6, 8, 12, 24. Since P x has 3 variations in sign, there are 1 or 3 positive real zeros. Since P x x 5 4x 4 3x 3 22x 2 4x 24 has 2 variations in sign, P has 0 or 2 negative real zeros. 1
1 1
4
3
22
4
24
1
3
6
16
12
3 6 16 12 12 P x x 2 x 4 2x 3 7x 2 8x 12
2
1 1
4
3
22
4
24
2
4
14
16
24
2
7
8
12
0
x 2 is a zero.
310
CHAPTER 3 Polynomial and Rational Functions
2
1
2
7
8
12
1
2
0
14
12
0
7
6
0
P x x 22 x 3 7x 6 2
1
0
7
6
2
4
6
x 2 is a zero again.
1
3
0
7
6
3
9
6
1 3 2 0 x 3 is a zero. 2 3 12 P x x 22 x 3 x 2 3x 2 x 22 x 3 x 1 x 2 0. Therefore, the zeros are 1, 2, and 3. 1
43. P x 3x 5 14x 4 14x 3 36x 2 43x 10 has possible rational zeros 1, 2, 5, 10, 13 , 23 , 53 , 10 3 . Since P x has 2 variations in sign, there are 0 or 2 positive real zeros. Since P x 3x 5 14x 4 14x 3 36x 2 43x 10 has 3 variations in sign, P has 1 or 3 negative real zeros. 1
3
14
14
36
43
10
3
11
25
11
54
2
3
11 25 11 54 64 P x x 2 3x 4 8x 3 30x 2 24x 5 2
3
8
30
24
5
6
4
68
184
3
2
3 3
5
14
14
36
43
10
6
16
60
48
10
8
30
24
5
0
3
8
30
24
5
15
35
25
5
x 2 is a zero.
34 92 189 3 7 5 1 0 x 5 is a zero. P x x 2 x 5 3x 3 7x 2 5x 1 . Since 3x 3 7x 2 5x 1 has no variation in sign, there are no more
positive zeros.
1
3
7
5
1
3
4
1
3 4 1 0 x 1 is a zero. P x x 2 x 5 x 1 3x 2 4x 1 x 2 x 5 x 1 x 1 3x 1. Therefore, the zeros are 1,
13 , 2, and 5.
SECTION 3.4 Real Zeros of Polynomials
311
44. P x 2x 6 3x 5 13x 4 29x 3 27x 2 32x 12 has possible rational zeros 1, 2, 3, 4,
6, 12, 12 , 32 . Since P x has 5 variations in sign, there are 1 or 3 or 5 positive real zeros. Since P x 2x 6 3x 5 13x 4 29x 3 27x 2 32x 12 has 3 variations in sign, P has 1 or 3 negative real zeros. 1
2
2
3
13
29
2
2
1
1
14
13
29
4
2
1
11
3
2
2
27
32
12
14
15
12
20
15
12
20
8
27
32
12
22
14
26
12
7
13
0 x 2 is a zero. P x x 2 2x 5 x 4 11x 3 7x 2 13x 6 . We continue with the quotient: 2
2
6
1
11
7
13
6
4
10
2
10
6
2 5 1 5 3 0 x 2 is a zero again. P x x 22 2x 4 5x 3 x 2 5x 3 . We continue with the quotient, first noting 2 is no longer a possible rational
solution:
3
2 2
5
1
5
3
6
22
42
94
11
21
47
91
x 3 is an upper bound.
We know that there is at least 1 more positive zero. 1 2
2
5
1
5
3
1
2
1
3
2 6 2 6 0 x 12 is a zero. P x x 22 x 12 2x 3 6x 2 2x 6 . We can factor 2x 3 6x 2 2x 6 by grouping; 2x 3 6x 2 2x 6 2x 3 6x 2 2x 6 2x 6 x 2 1 . So P x 2 x 22 x 12 x 3 x 2 1 .
Since x 2 1 has no real zeros, the zeros of P are 3, 2, and 12 .
45. P x 3x 3 5x 2 2x 4. The possible rational zeros are 1, 2, 4, 13 , 23 , 43 . P x has 1 variation in sign and hence 1 positive real zero. P x 3x 3 5x 2 2x 4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1 3 5 2 4 5 2 4 1 3 3 3
2
0
3
2
4
2 0 4 3 2 4 0 x 1 is a zero. So P x x 1 3x 2 2x 4 . Using the Quadratic Formula on the second factor, we have 2 434 13 13 . Therefore, the zeros of P are 1 and 13 13 . x 2 223
312
CHAPTER 3 Polynomial and Rational Functions
46. P x 3x 4 5x 3 16x 2 7x 15. The possible rational zeros are 1, 3, 5, 15, 13 , 53 . P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x 3x 4 5x 3 16x 2 7x 15 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1
3
5
16
7
15
3
3
5
2 18 11 4 So P x x 3 3x 3 4x 2 4x 5 . Continuing:
3
2
18
11 3
3
1
3
4
4
5
3
1
5
16
7
15
9
12
12
15
4
4
5
0
x 3 is a zero.
3 1 5 0 x 1 is a zero. Thus, P x x 3 3x 3 4x 2 4x 5 x 1 x 3 3x 2 x 5 . Using the Quadratic Formula on the last 2 435 factor, we have x 1 123 16 61 . Therefore, the zeros of P are 1, 3, and 16 61 .
47. P x x 4 6x 3 4x 2 15x 4. The possible rational zeros are 1, 2, 4. P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x x 4 6x 3 4x 2 15x 4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1 1 6 4 15 4 4 15 4 2 1 6 1
1
5
1
14
5
1
14
18
4
1
1
6
4
15
4
4
8
16
4
2
8
8
14
4
4
7
18
2 4 1 0 x 4 is a zero. So P x x 4 x 3 2x 2 4x 1 . Continuing by factoring the quotient, we have: 4
1 1
2
1
4
1
4
8
16
2
4
15
1
x 4 is an upper bound.
1 1
2
4
1
1
3
1
3
1
0
x 1 is a zero.
So P x x 4 x 1 x 2 3x 1 . Using the Quadratic Formula on the third factor, we have: 3 32 411 3 13 . Therefore, the zeros are 4, 1, and 3 13 . x 2 2 21
SECTION 3.4 Real Zeros of Polynomials
313
48. P x x 4 2x 3 2x 2 3x 2. The possible rational zeros are 1, 2. P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x x 4 2x 3 2x 2 3x 2 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1
1 1
2
2
3
2
1
3
1
2
3
1
2
0
x 1 is a zero.
P x x 1 x 3 3x 2 x 2 . Continuing with the quotient: 1 1 3 1 2 1 4
1 1
5
3
1 2
2 1
1
1 2
3
1 2
2 2
2
1 4 5
3 x 1 is an 1 2 1 1 1 1 1 0 x 2 is a zero. upper bound. So P x x 1 x 2 x 2 x 1 . Using the Quadratic Formula on the third factor, we have 1 12 4 1 1 1 5 5. x . Therefore, the zeros are 1, 2, and 1 2 2 1 2
49. P x x 4 7x 3 14x 2 3x 9. The possible rational zeros are 1, 3, 9. P x has 3 variations in sign and hence 1 or 3 positive real zeros. P x x 4 7x 3 14x 2 3x 4 has 1 variation in sign and hence P has 1 negative real zero. 1
1
7
14
3
9
1
6
8
5
1
3
7
14
3
9
3
12
6
9
1
6 8 5 4 1 4 2 3 0 x 3 is a zero. So P x x 3 x 3 4x 2 2x 3 . Since the constant term of the second term is 3, 9 are no longer possible zeros.
Continuing by factoring the quotient, we have:
3
1 1
4
2
3
3
3
3
1
1
0
x 3 is a zero again.
So P x x 32 x 2 x 1 . Using the Quadratic Formula on the second factor, we have: 1 12 411 1 5 . Therefore, the zeros are 3 and 1 5 . x 2 2 21
314
CHAPTER 3 Polynomial and Rational Functions
50. P x x 5 4x 4 x 3 10x 2 2x 4. The possible rational zeros are 1, 2, 4. P x has 3 variations in sign and hence 1 or 3 positive real zeros. P x x 5 4x 4 x 3 10x 2 2x 4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1
1 1
4
1
10
2
4
1
3
4
6
8
2
4 6 8 4 So P x x 2 x 4 2x 3 5x 2 2 . 2
1
3
is 2, 4 are no longer possible zeros.
1
4
1
10
2
4
2
4
10
0
4
2
5
0
2
0
x 2 is a zero.
Since the constant term of the second factor
Continuing by factoring the quotient, we have:
1
2
5
0
2
2
0
10
20
1
0
1
1
2
5
0
2
1
3
2
2
5 10 18 1 3 2 2 0 x 1 is a zero. So P x x 2 x 1 x 3 3x 2 2x 2 . Continuing by factoring the quotient, we have: 1
1
3
2
2
1
4
2
1
4 2 0 x 1 is a zero again. So P x x 2 x 12 x 2 4x 2 . Using the Quadratic Formula on the second factor, we have:
4 42 412 4 8 42 2 2 2. Therefore, the zeros are 1, 2, and 2 2. x 21 2 2
51. P x 4x 3 6x 2 1. The possible rational zeros are 1, 12 , 14 . P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x 4x 3 6x 2 1 has 1 variation in sign and hence P has 1 negative real zero. 1
So P x
4
6
0
1
4
2
2
1 2
4
6
0
1
2
2
1
4 2 2 1 4 4 2 0 x 12 is a zero. x 12 4x 2 4x 2 . Using the Quadratic Formula on the second factor, we have:
4 42 442 4 48 44 3 1 3 . Therefore, the zeros are 1 and 1 3 . x 8 8 2 2 2 24
SECTION 3.4 Real Zeros of Polynomials
315
52. P x 3x 3 5x 2 8x 2. The possible rational zeros are 1, 2, 13 , 23 . P x has 1 variation in sign and hence 1 positive real zero. P x 3x 3 5x 2 8x 2 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1
1 3
3
5
3 3
5
8
2
3
2
10
2
10
12
8
2
43
1 3
2 3
3
46 9
5
8
2
6
2
12
3
1
6
14
3
5
8
2
2
3
10
26 3
28 9
28 3
4
2
3
2
20 3
Thus we have tried all the positive rational zeros, so we try the negative zeros. 1
3 3
5
8
3
8
0
8
0
2
13
2
3
2
5
8
2
1
2
2
3
5
8
2
6
22
28
3
11
14
30
3 6 6 0 x 13 is a zero. So P x x 13 3x 2 6x 6 3 x 13 x 2 2x 2 . Using the Quadratic Formula on the second factor, we 2 22 412 22 12 222 3 1 3. Therefore, the zeros are 13 and 1 3. have: x 21
53. P x 2x 4 15x 3 17x 2 3x 1. The possible rational zeros are 1, 12 . P x has 1 variation in sign and hence 1 positive real zero. P x 2x 4 15x 3 17x 2 3x 1 has 3 variations in sign and hence P has 1 or 3 negative real zeros. 1 2
2 2 12
15
17
3
1
8
16
25
25 2 31 2
2
1 31 4 27 4
x 12 is an upper bound.
15
17
3
1
1
7
5
1
14 10 2 0 x 12 is a zero. So P x x 12 2x 3 14x 2 10x 2 2 x 12 x 3 7x 2 5x 1 . 2
1
1
7
5
1
1
6
1
1 6 1 0 x 1 is a zero. So P x x 12 2x 3 14x 2 10x 2 2 x 12 x 1 x 2 6x 1 Using the Quadratic Formula on the
6 62 411 10 3 10. Therefore, the zeros are 1, 1 , third factor, we have x 62 40 62 21 2 2
and 3
10.
316
CHAPTER 3 Polynomial and Rational Functions
54. P x 4x 5 18x 4 6x 3 91x 2 60x 9. The possible rational zeros are 1, 3, 9, 12 , 32 , 92 , 14 , 34 , 94 .
P x has 4 variations in sign and hence 0 or 2 or 4 positive real zeros. P x 4x 5 18x 4 6x 3 91x 2 60x 9 has 1 variation in sign and hence P has 1 negative real zero. 1
4
18
6
91
60
9
4
14
20
71
1
4
14
3
4
18
6
91
60
9
12
18
72
57
9
20 71 11 10 4 6 24 19 3 0 x 3 is a zero. So P x x 3 4x 4 6x 3 24x 2 19x 3 . Continuing by factoring the quotient, we have: 4
3
6
24
19
3
12
18
18
3
4 6 6 1 0 x 3 is a zero again. So P x x 32 4x 3 6x 2 6x 1 . Continuing by factoring the quotient, we have: 3
4
6
6
1
12
54
144
1 2
4
6
6
1
2
4
1
48 1445 x 3 is an upper bound. 4 8 2 0 x 12 is a zero. So P x x 32 x 12 4x 2 8x 2 2 x 32 x 12 2x 2 4x 1 . Using the Quadratic Formula on 2 421 the second factor, we have x 4 422 44 24 1 26 . Therefore, the zeros are 12 , 3, and 1 26 . 4
18
55. (a) P x x 3 3x 2 4x 12 has possible rational zeros 1, 2, 3,
(b)
y
4, 6, 12. 1
2
1
3
1 1 1
4
12
1
2
6
2
6
6
3
4
12
2
2
12
1
6
0
2 x
1
x 2 is a zero.
So P x x 2 x 2 x 6 x 2 x 2 x 3. The real zeros of P are 2, 2, and 3.
56. (a) P x x 3 2x 2 5x 6 has possible rational zeros 1, 2, 3,
(b)
y
6. 1
1 1
2
1 1
2
5
6
1
3
2
3
2
8
2 1
2
5
6
2
8
6
4
3
0
x
x 2 is a zero.
So P x x 2 x 2 4x 3 x 2 x 2 4x 3 x 2 x 1 x 3. The real zeros of P are 2, 1, and 3.
SECTION 3.4 Real Zeros of Polynomials
57. (a) P x 2x 3 7x 2 4x 4 has possible rational zeros 1, 2, 4,
317
y
(b)
12 .
1 2 7
4
4
2 2 7
2 5 1
4
4
4 6 4
2 5 1 3
2 3 2 0 x 2 is a zero. So P x x 2 2x 2 3x 2 . Continuing: 2
2 2
3
2
4
2
1
0
1 x
1
x 2 is a zero again.
Thus P x x 22 2x 1. The real zeros of P are 2 and 12 . 58. (a) P x 3x 3 17x 2 21x 9 has possible rational zeros 1, 3,
y
(b)
2
9, 13 , 23 . 3
1
1
17
21
9
3
20
41
3
20
41
32
1 3
3
17
3
x
x 1 is an upper bound.
21
9
1
6
9
18
27
0
x 13 is a zero.
So P x x 13 3x 2 18x 27 3 x 13 x 2 6x 9 3 x 13 x 32 . The real zeros of P are 3 and 13 .
59. (a) P x x 4 5x 3 6x 2 4x 8 has possible rational zeros 1,
(b)
y
2, 4, 8. 1
2
1
5
6
4
8
1
4
2
6
1
4
2
6
2
1
5
6
4
8
2
6
0
8
3
0
4
0
1
2 1
x
x 2 is a zero.
So P x x 2 x 3 3x 2 4 and the possible rational zeros are restricted to 1, 2, 4. 2
1
3
0
4
2
2
4
1 1 2 0 x 2 is a zero again. P x x 22 x 2 x 2 x 22 x 2 x 1 x 23 x 1. So the real zeros of P are 1 and 2.
318
CHAPTER 3 Polynomial and Rational Functions
60. (a) P x x 4 10x 2 8x 8 has possible rational zeros 1, 2,
y
(b)
4, 8. 1 1
0 10
8 8
1 1
9 17
1 1 4 1
0
9 17 10
8 8
2 4 12 40
9 8
0 10
2 1
6 20 32
1 2
5
8
1
4 16 24 64 6 16 72 x 4 is an upper bound.
1 4
0
10
8
2
4
12
8
1 9 1 7 1 2 So P x x 2 x 3 2x 2 6x 4 . Continuing, we have:
6
4
0
1
x
1
0
10
8
8
1
1
9
1
1
2
1
2
1
2
6
2
8
4
2
8 x 2 is a zero.
4
4
x 2 is a zero again. P x x 22 x 2 4x 2 . Using the Quadratic Formula on the second factor, we have 2 412 8 42 2 2 2. So the real zeros of P are 2 and 2 2. 4 x 4 421 2 2 1
0
61. (a) P x x 5 x 4 5x 3 x 2 8x 4 has possible rational zeros
(b)
y
1, 2, 4. 1
2
1 1
1
1
5
1
8
4
1
0
5
4
4
1
0
5
4
4
8
1
5
1
8
4
2
2
6
10
4
1 1
1
3 5 2 0 x 2 is a zero. So P x x 2 x 4 x 3 3x 2 5x 2 , and the possible rational zeros are restricted to 1, 2. 2
1
1 2
6
6
2
1
3
3
1
0
1
1
3
3
1
1
2
1
3
5
2
x 2 is a zero again. So P x x 22 x 3 3x 2 3x 1 , and the possible rational zeros are restricted to 1.
1
2 1 0 x 1 is a zero. So P x x 22 x 1 x 2 2x 1 x 22 x 13 ., and the real zeros of P are 1 and 2.
x
SECTION 3.4 Real Zeros of Polynomials
62. (a) P x x 5 x 4 6x 3 14x 2 11x 3 has possible rational zeros
(b)
319
y
1, 3. 1
1
1
6
14
11
3
1
0
6
8
3
1
0
6 8 3 0 So P x x 1 x 4 6x 2 8x 3 : 1
1 1
0
6
8
3
1
1
5
3
x 1 is a zero.
10 1
x
1
5 3 0 x 1 is a zero again. 2 3 2 So P x x 1 x x 5x 3 . 1
1
1
5
3
1
2
3
1 2 3 0 x 1 is a zero again. So P x x 13 x 2 2x 3 x 14 x 3, and the real zeros of P are 1 and 3.
63. P x x 3 x 2 x 3. Since P x has 1 variation in sign, P has 1 positive real zero. Since P x x 3 x 2 x 3 has 2 variations in sign, P has 2 or 0 negative real zeros. Thus, P has 1 or 3 real zeros. 64. P x 2x 3 x 2 4x 7. Since P x has 3 variations in signs, P has 3 or 1 positive real zeros. Since P x 2x 3 x 2 4x 7 has no variation in sign, there is no negative real zero. Thus, P has 1 or 3 real zeros. 65. P x 2x 6 5x 4 x 3 5x 1. Since P x has 1 variation in sign, P has 1 positive real zero. Since P x 2x 6 5x 4 x 3 5x 1 has 1 variation in sign, P has 1 negative real zero. Therefore, P has 2 real zeros. 66. P x x 4 x 3 x 2 x 12. Since P x has no variations in sign, P has no positive real zeros. Since P x x 4 x 3 x 2 x 12 has 4 variations in sign, P has 4, 2, or 0 negative real zeros. Therefore, P x has 0, 2, or 4 real zeros. 67. P x x 5 4x 3 x 2 6x. Since P x has 2 variations in sign, P has 2 or 0 positive real zeros. Since P x x 5 4x 3 x 2 6x has no variation in sign, P has no negative real zero. Therefore, P has a total of 1 or 3 real zeros (since x 0 is a zero, but is neither positive nor negative). 68. P x x 8 x 5 x 4 x 3 x 2 x 1. Since P x has 6 variations in sign, the polynomial has 6, 4, 2, or 0 positive real zeros. Since P x has no variation in sign, the polynomial has no negative real zeros. Therefore, P has 6, 4, 2, or 0 real zeros. 69. P x 2x 3 5x 2 x 2; a 3, b 1 3
2 2
1
2 2
5
1
2
6
3
12
1
4
14
5
1
2
2
7
8
7
8
6
Therefore a 3 and b 1 are lower and upper bounds.
alternating signs lower bound.
all nonnegative upper bound.
320
CHAPTER 3 Polynomial and Rational Functions
70. P x x 4 2x 3 9x 2 2x 8; a 3, b 5 1
3
5
2
9
2
8
3
15
18
48
1
5
6
16
56
1
2
9
2
8
5
15
30
160
3
6
32
168
1
Alternating signs lower bound.
All nonnegative upper bound.
Therefore a 3 and b 5 are lower and upper bounds.
71. P x 8x 3 10x 2 39x 9; a 3, b 2 8
3
8 2
10
39
9
24
42
9
14
3
0
8 8
10
39
16
52
26
26
13
35
alternating signs lower bound.
9 all nonnegative upper bound.
Therefore a 3 and b 2 are lower and upper bounds. Note that x 3 is also a zero.
72. P x 3x 4 17x 3 24x 2 9x 1; a 0, b 6 0
3 3
6
3
17
24
9
1
0
0
0
0
17
24
9
1
Alternating signs lower bound.
17
24
9
1
18
6
180
1026
1
30
171
1027
3
All nonnegative upper bound.
Therefore a 0 and b 6 are lower and upper bounds. Note that because P x alternates in sign, by Descartes’ Rule of Signs, 0 is automatically a lower bound.
73. P x x 4 2x 3 3x 2 5x 1; a 2, b 1 2
1 1
1
1 1
2
3
5
1
2
0
6
2
0
3
1
1
2
3
5
1
1 3
3
6
11
6
11
10
Therefore a 2 and b 1 are lower and upper bounds.
Alternating signs lower bound.
All nonnegative upper bound.
SECTION 3.4 Real Zeros of Polynomials
74. P x x 4 3x 3 4x 2 2x 7; a 4, b 2 1
4
1 2
3
4
2
7
4
4
0
8
1
0
2
1
1 1
3
4
2
7
2
10
12
20
5
6
10
13
Alternating signs lower bound.
All nonnegative upper bound.
Therefore a 4 and b 2 are lower and upper bounds.
75. P x 2x 4 6x 3 x 2 2x 3; a 1, b 3 2
1
2 3
6
1
2 8
2
2
3
8
9
11
9
11
14
6
1
2
3
6
0
3
3
0
1
1
6
2
Alternating signs lower bound.
All nonnegative upper bound.
Therefore a 1 and b 3 are lower and upper bounds.
76. P x 3x 4 5x 3 2x 2 x 1; a 1, b 2 3
1
2
5
2
1
1
3
8
6
5
3
8
6
5
4
3
5
2
1
1
6
2
0
2
1
0
1
1
3
Therefore a 1 and b 2 are lower and upper bounds.
Alternating signs lower bound.
All nonnegative upper bound.
77. P x x 3 3x 2 4 and use the Upper and Lower Bounds Theorem: 1
1
3
0
4
1
4
4
1
4
4
0
1
3
0
4
3
0
0
0
0
4
3
1
alternating signs lower bound.
all nonnegative upper bound.
Therefore 1 is a lower bound (and a zero) and 3 is an upper bound. (There are many possible solutions.)
78. P x 2x 3 3x 2 8x 12 and using the Upper and Lower Bounds Theorem: 2
2 2 3
3
8
12
4
14
12
7
6
0
2
3
8
6
9
3
3
1
15
2 (There are many possible solutions.)
Alternating signs x 2 is a lower bound (and a zero). 12 All nonnegative x 3 is an upper bound.
321
322
CHAPTER 3 Polynomial and Rational Functions
79. P x x 4 2x 3 x 2 9x 2. 1
1
1
2
1
9
2
1
1
0
9
1
0
9
7
3
1
1 1
1 1
2
1
9
2
3
3
12
9
1
4
3
11
1 1
2
2
1
9
2
1
3
4
13
3
4
13
15
2
1
9
2
2
0
2
14
0
1
7
12
all positive upper bound.
alternating signs lower bound.
Therefore 1 is a lower bound and 3 is an upper bound. (There are many possible solutions.)
80. Set P x x 5 x 4 1. 1
1
1
0
0
0
1
1
0
0
0
0
0
0
0
0
1
1 1
1 1
All nonnegative x 1 is an upper bound.
1
0
0
0
1
1
2
2
2
2
2
2
2
2
1
(There are many possible solutions.)
81. P x 2x 4 3x 3 4x 2 3x 2. 1
P x x 1 2x 3 5x 2 x 2
1
2 2
2
Alternating signs x 1 is a lower bound.
3
4
3
2
2
5
1
2
5
1
2
0
5
1
2
2
3
2
x 1 is a zero.
2 3 2 0 x 1 is a zero. P x x 1 x 1 2x 2 3x 2 x 1 x 1 2x 1 x 2. Therefore, the zeros are 2, 12 , 1.
SECTION 3.4 Real Zeros of Polynomials
323
82. P x 2x 4 15x 3 31x 2 20x 4. The possible rational zeros are 1, 2, 4, 12 . Since all of the coefficients are positive, there are no positive zeros. Since P x 2x 4 15x 3 31x 2 20x 4 has 4 variations in sign, there are 0, 2, or 4 negative real zeros. 1 2 15
31
20
4
2 2 15
2 13 18 2 2 13
18
2
31
20
4 22 18 4 2 11
2
9
2
P x x 2 2x 3 11x 2 9x 2 : 2 2 11
9
2
4 2 11
4 14 10 2
7
4
9
12 2 11
2
8 12 12 2
5 12
3
0 x 2 is a zero.
9
2
1 5 2
3 14
2 10
4
0 x 12 is a zero.
P x x 2 2x 1 x 2 5x 2 . Now if x 2 5x 2 0, then x 5 25412 52 17 . Thus, the zeros 2
are 2, 12 , and 52 17 .
83. Method 1: P x 4x 4 21x 2 5 has 2 variations in sign, so by Descartes’ rule of signs there are either 2 or 0 positive zeros. If we replace x with x, the function does not change, so there are either 2 or 0 negative zeros. Possible rational zeros are 1, 12 , 14 , 5, 52 , 54 By inspection, 1 and 5 are not zeros, so we must look for non-integer solutions: 1 2
4
0
4
21
0
5
2
1
10
5
2
20
10
0
x 12 is a zero.
P x x 12 4x 3 2x 2 20x 10 , continuing with the quotient, we have: 12
4 4
P x x 12
2
20
10
2
0
10
0
20
0
x 12 is a zero.
1 2 4x 20 0. If 4x 2 20 0, then x 5. Thus the zeros are 12 5. x 2
Method 2: Substituting u x 2 , the equation becomes 4u 2 21u 5 0, which factors: 4u 2 21u 5 4u 1 u 5 4x 2 1 x 2 5 . Then either we have x 2 5, so that x 5, or we have x 2 14 , so that x 14 12 . Thus the zeros are 12 5.
324
CHAPTER 3 Polynomial and Rational Functions
84. P x 6x 4 7x 3 8x 2 5x x 6x 3 7x 2 8x 5 . So x 0 is a zero. Continuing with the quotient,
1 Q x 6x 3 7x 2 8x 5. The possible rational zeros are 1, 5, 12 , 52 , , 53 , 16 , 56 . Since Q x has 2 3
variations in sign, there are 0 or 2 positive real zeros. Since Q x 6x 4 7x 3 8x 2 5x has 1 variation in sign, there is 1 negative real zero. 6
1
6
7
8
5
6
1
9
1
9
4 1 2
6
7
8
5
30
115
535
6
23
107
540
8
5
5
6
7 3
2
5
6
4
10
0
All positive upper bound.
x 12 is a zero.
P x x 2x 1 3x 2 2x 5 x 2x 1 3x 5 x 1. Therefore, the zeros are 0, 1, 12 and 53 .
85. P x x 5 7x 4 9x 3 23x 2 50x 24. The possible rational zeros are 1, 2, 3, 4, 6, 8, 12, 24. P x has 4 variations in sign and hence 0, 2, or 4 positive real zeros. P x x 5 7x 4 9x 3 23x 2 50x 24 has 1 variation in sign, and hence P has 1 negative real zero. 1
1 1
7
9
23
50
24
1
6
3
26
24
6
3
26
24
0
x 1 is a zero.
P x x 1 x 4 6x 3 3x 2 26x 24 ; continuing with the quotient, we try 1 again. 1
1 1
6
3
26
24
1
5
2
24
5
2
24
0
x 1 is a zero again.
P x x 12 x 3 5x 2 2x 24 ; continuing with the quotient, we start by trying 1 again. 1
1 1
5
2
24
1
4
6
4
6
18
2
1 1
5
2
24
2
6
16
3
8
8
3
1 1
5
2
24
3
6
24
2
8
0
x 3 is a zero.
P x x 12 x 3 x 2 2x 8 x 12 x 3 x 4 x 2. Therefore, the zeros are 2, 1, 3 4.
SECTION 3.4 Real Zeros of Polynomials
325
1 86. P x 8x 5 14x 4 22x 3 57x 2 35x 6. The possible rational zeros are 1, 2, 3, 6, , 2 1 32 , 14 , 34 , , 38 . Since P x has 4 variations in sign, there are 0, 2, or 4 positive real zeros. Since 8 P x 8x 5 14x 4 22x 3 57x 2 35x 6 has 1 variation in sign, there is 1 negative real zero. 1
8
14
22
57
35
6
8 8
22
22
0
57
92
0
57
92
98
2
8
14 16
8
30
57
35
6
60
76
38
6
38
19
3
0
22
x 2 is a zero.
P x x 2 8x 4 30x 3 38x 2 19x 3 . All the other real zeros are positive. 8
1
8
30
38
19
3
8
22
16
3
22
16
3
0
P x x 2 x 1 8x 3 22x 2 16x 3 . 1
8 8
Since f
1 2
22
16
3
8
14
2
14
2
1
x 1 is a zero.
1 2
8 8
22
16
4
9
18
7
3 7 2 1 2
0 f 1, there must be a zero between 12 and 1. We try 34 : 3 4
8 8
22
16
3
6
12
3
16
4
0
x 34 is a zero.
P x x 2 x 1 4x 3 2x 2 4x 1 . Now, 2x 2 4x 1 0 when x 4 16421 22 2 . Thus, the 22
zeros are 1, 34 , 2, and 22 2 .
87. P x x 3 x 2. The only possible rational zeros of P x are 1 and 2. 1
1 1
0
1
1
1
0
1
0
2
2
2
1 1
0
1
2
2
4
6
2
3
4
1
1 1
0
1
1
1
0
1
0
2
2
Since the row that contains 1 alternates between nonnegative and nonpositive, 1 is a lower bound and there is no need to try 2. Therefore, P x does not have any rational zeros.
326
CHAPTER 3 Polynomial and Rational Functions
88. P x 2x 4 x 3 x 2. The only possible rational zeros of P x are 1, 2, 12 . 1 2
2
1
0
1 0
2 2
1
2
1
2
0
0
0
1
1 2 5 2
1
0
1
2
2
3
3
2
3
3
2
4
12
All nonnegative x 12 is an upper bound.
Alternating signs x 1 is a lower bound. 2 2
1
0
1
2
1
1
2
1
12 1 2
14 7 4
Therefore, there is no rational zero.
89. P x 3x 3 x 2 6x 12 has possible rational zeros 1, 2, 3, 4, 6, 12, 13 , 23 , 43 . 3 1
3
1
6
12
2
4
8
2
3
5
4
20
1
3
4
2
14
2
3
7
8
4
1 3 2 3 4 3 13 23 43
all positive x 2 is an upper bound alternating signs x 2 is a lower bound
3
1
6
12
3
0
6
10
3
1
16 3
3
3
2
3
2
16 3
3
3
4
3
5
2 3
76 9 28 3 124 9 44 3 100 9
Therefore, there is no rational zero.
90. P x x 50 5x 25 x 2 1. The only possible rational zeros of P x are 1. Since P 1 150 5 125 12 1 4 and P 1 150 5 125 12 1 6, P x does not have a rational zero.
91. P x x 3 3x 2 4x 12, [4 4] by [15 15]. The
possible rational zeros are 1, 2, 3, 4, 6, 12. By
observing the graph of P, the rational zeros are x 2, 2, 3.
92. P x x 4 5x 2 4, [4 4] by [30 30].
The possible rational solutions are 1, 2, 4.
By observing the graph of the equation, the solutions of the given equation are x 1, 2.
10
-4
-2
20
2 -10
4
-4
-2
2 -20
4
SECTION 3.4 Real Zeros of Polynomials
93. P x 2x 4 5x 3 14x 2 5x 12, [2 5] by
327
94. P x 3x 3 8x 2 5x 2, [3 3] by [10 10]
[40 40]. The possible rational zeros are 1, 2, 3,
The possible rational solutions are 1, 2, 13 , 23 . By
4, 6, 12, 12 , 32 . By observing the graph of P, the
observing the graph of the equation, the only real solution
zeros are 32 , 1, 1, 4.
of the given equation is x 2. 10
40 20 -2
2
-20
-2
4
2 -10
-40
95. x 4 x 4 0. Possible rational solutions are 1, 2, 4. 1
1 1
1
0
0
1
4
1
1
1
0
1
1
0
4
0
2
1
1
0 1
1
1
2
1
1
1
2
2
1
1
4
2
0
0
1
4
2
4
8
14
2
4
7
10
1
4
0
x 2 is an upper bound.
1
0 2
4
8
18
1
2
4
9
14
x 2 is a lower bound.
Therefore, we graph the function P x x 4 x 4 in the viewing rectangle [2 2] by [5 20] and see there are two
solutions. In the viewing rectangle [13 125] by [01 01], we find the solution x 128. In the viewing rectangle
[1516] by [01 01], we find the solution x 153. Thus the solutions are x 128, 153. 20 10 -1.30 -2
-1
1
-1.28
0.1
0.1
0.0
0.0
-1.26
2
1.55 -0.1
1.60
-0.1
96. 2x 3 8x 2 9x 9 0. Possible rational solutions are 1, 3, 9, 12 , 32 , 92 . 1
2 2
8
9
9
2
6
3
6
3
6
3
2 2
8
9
9
6
6
9
2
3
0
x 3 is a zero.
We graph P x 2x 3 8x 2 9x 9 in the viewing rectangle [4 6] by
[40 40]. It appears that the equation has no other real solution. We can factor 2x 3 8x 2 9x 9 x 3 2x 2 2x 3 . Since the quotient is a quadratic expression, we can use the Quadratic Formula to locate the other possible 2 423 , which are not real. So the only solution is x 3. solutions: x 2 222
40 20 -4
-2 -20 -40
2
4
6
328
CHAPTER 3 Polynomial and Rational Functions
97. 400x 4 400x 3 1096x 2 588x 909 0. 1 4 4 1096 588 4
2 4
4 1096 588
8 296 884
8
296 884
025
4 12
4 1096 588
909
4 8 2 4
909
8
8
4 4
296
3 4
592 008 004
24 2608 4040 1304
4 1096 12
901
4
8
909
202 4949 x 2 is an upper bound. 588
909
24 3912
135
1304
45 14409 x 3 is a lower bound.
Therefore, we graph the function P x 400x 4 400x 3 1096x 2 588x 909 in the viewing rectangle [3 2] by [10 40]. There appear to be two solutions. In the viewing rectangle [16 14] by [01 01], we find the solution x 150. In the viewing rectangle [08 12] by [0 1], we see that the graph comes close but does not go through the x-axis. Thus there is no solution here. Therefore, the only solution is x 150. 40
0.1
20
1.0 0.5
0.0 -1.6
-1.5 0.0
-3
-2
-1
1
2
0.8
-0.1
0.9
1.0
1.1
1.2
98. x 5 2x 4 096x 3 5x 2 10x 48 0. Since all the coefficients are positive, there is no positive solution. So x 0 is an upper bound. 2 1
2 096 2
1
5
10
48
3 1
0 192 616 768
0 096
308
384 288
2 096 3
5
10
48
3 1188 2064 9192
1 1 396
688 3064 8712 x 3 is a lower bound.
Therefore, we graph P x x 5 2x 4 096x 3 5x 2 10x 48 in the viewing rectangle [3 0] by [10 5] and see that there are three possible solutions. In the viewing rectangle [175 17] by [01 01], we find the solution x 171. In the viewing rectangle [125 115] by [01 01], we find the solution x 120. In the viewing
rectangle [085 075] by [01 01], we find the solution x 080. So the solutions are x 171, 120, 080. 0.1
0.1
0.1
0 -2
0.0 -1.74 -10
-1.72
0.0 -1.25
-0.1
-1.20
0.0 -0.85
-0.1
-0.80 -0.1
SECTION 3.4 Real Zeros of Polynomials
329
99. Let r be the radius of the silo. The volume of the hemispherical roof is 12 43 r 3 23 r 3 . The volume of the cylindrical section is r 2 30 30r 2 . Because the total volume of the silo is 15,000 ft3 , we get the following equation: 2 r 3 30r 2 15000 2 r 3 30r 2 15000 0 r 3 45r 2 22500 0. Using a graphing device, we 3 3
first graph the polynomial in the viewing rectangle [0 15] by [10000 10000]. The solution, r 1128 ft., is shown in the viewing rectangle [112 114] by [1 1]. 1
10000
0
0 11.3
11.4
10
-1
-10000
100. Given that x is the length of a side of the rectangle, we have that the length of the diagonal is x 10, and the length of the other side of the rectangle is x 102 x 2 . Hence x x 102 x 2 5000 x 2 20x 100 25,000,000 2x 3 10x 2 2,500,000 0 x 3 5x 2 1,250,000 0. The first viewing rectangle, [0 120] by [100 500],
shows there is one solution. The second viewing rectangle, [106 1061] by [01 01], shows the solution is x 10608.
Therefore, the dimensions of the rectangle are 47 ft by 106 ft.
0.1
400 200
0.0 106.05
106.10
0 50
101. h t 1160t 1241t 2 620t 3
158t 4 020t 5 001t 6 is shown in the viewing rectangle
100
-0.1
(a) It started to snow again. (b) No, h t 4. (c) The function h t is shown in the viewing rectangle [6 65] by
[0 10] by [0 6].
[0 05]. The x-intercept of the function is a little less than 65, which means that the snow melted just before midnight on Saturday night.
5
0.4 0.2
0 0
5
10 0.0 6.0
6.2
6.4
330
CHAPTER 3 Polynomial and Rational Functions
102. The volume of the box is V 1500 x 20 2x 40 2x 4x 3 120x 2 800x 4x 3 120x 2 800x 1500 4 x 3 30x 2 200x 375 0. Clearly, we must have 20 2x 0, and so 0 x 10. 5
1
30
200
375
5
125
375
1 25 75 0 x 5 is a zero. x 3 30x 2 200x 375 x 5 x 2 25x 75 0. Using the Quadratic Formula, we find the other zeros:
252 325 2552 13 . Since 2552 13 10, the two answers are: x height 5 cm, x 25 6254175 2
width 20 2 5 10 cm, and length 40 2 5 30 cm; and x height 2552 13 349 cm, width 20 25 5 13 5 13 5 1303 cm, and length 40 25 5 13 15 5 13 3303 cm. 103. Let r be the radius of the cone and cylinder and let h be the height of the cone. r
Since the height and diameter are equal, we get h 2r. So the volume of the
cylinder is V1 r 2 cylinder height 20r 2 , and the volume of the cone is 500 , it follows V2 13 r 2 h 13 r 2 2r 23 r 3 . Since the total volume is 3 500 r 3 30r 2 250 0. By Descartes’ Rule of that 23 r 3 20r 2 3 Signs, there is 1 positive zero. Since r is between 276 and 2765 (see the table),
the radius should be 276 m (correct to two decimals).
r 3 30r 2 250
1
219
2
122
3
47
27
1162
276
233
277
144
2765
144
28
715
104. (a) Let x be the length, in ft, of each side of the base and let h be the height. The volume of the box is V 2 2 hx 2 , and so hx 2 2 2. The length of the diagonal on the base is x 2 x 2 2x 2 , and hence the length of the diagonal between opposite corners is 2x 2 h 2 x 1. Squaring both sides of the equation, we have 2x 2 h 2 x 2 2x 1 x 2 2x 1 x 2 h 2 x 2 2x 1 h x 2 2x 1. Therefore, 2 2 hx 2 x 2 2x 1 x 4 8 x 6 2x 5 x 4 8 0. (b) We graph y x 6 2x 5 x 4 8 in the viewing rectangle [0 5] by [10 10], and we see that there are two solutions. In the second viewing rectangle, [14 15] by [1 1], we see the solution x 145. The third viewing rectangle, [225 235] by [1 1], shows the solution x 231. 1
10
0
0
2 -10
1
0 1.45
4 -1
1.50
2.30
2.35
-1
If x width length 145 ft, then height x 2 2x 1 134 ft, and if x width length 231 ft, then height x 2 2x 1 053 ft.
SECTION 3.4 Real Zeros of Polynomials
331
105. Let b be the width of the base, and let l be the length of the box. Then the length plus girth is l 4b 108, and the volume
is V lb2 2200. Solving the first equation for l and substituting this value into the second equation yields l 108 4b V 108 4b b2 2200 4b3 108b2 2200 0 4 b3 27b2 550 0. Now P b b3 27b2 550 has two variations in sign, so there are 0 or 2 positive real zeros. We also observe that since l 0 b 27, so b 27 is an upper bound. Thus the possible positive rational real zeros are 1 2 3 10 11 22 25. 1
1 1
27
0
550
1
26
26
26
26
524
5
2
1 1
1
27
0
550
5
110
550
1
22
27
0
550
2
50
100
25
50
450
0 b 5 is a zero. 22 924 2230397 . The positive P b b 5 b2 22b 110 . The other zeros are b 22 48441110 2 2 2 110
answer from this factor is b 2620Thus we have two possible solutions, b 5 or b 2620. If b 5, then
l 108 4 5 88; if b 2620, then l 108 4 2620 320. Thus the length of the box is either 88 in. or 320 in.
106. (a) An odd-degree polynomial must have a real zero. The end behavior of such a polynomial requires that the graph of the polynomial heads off in opposite directions as x and x . Thus the graph must cross the x-axis. (b) There are many possibilities one of which is P x x 4 1. (c) P x x x 2 x 2 x 3 2x. (d) P x x 2 x 2 x 3 x 3 x 4 5x 2 6. If a polynomial with integer coefficients has no real zeros, then the polynomial must have even degree.
a for x we have 3 a 3 a 2 a x 3 ax 2 bx c X c a X b X 3 3 3 a2 2a a3 a2 ab 3 2 2 X aX X a X X bX c 3 27 3 9 3
107. (a) Substituting X
a3 a3 ab a2 2a 2 X a X2 X bX c 3 27 3 9 3 2a 2 a3 ab a2 a3 b X c X 3 a a X 2 3 3 27 9 3 ab 4a 3 X 3 b a2 X c 27 3 X3 a X2
(b) x 3 6x 2 9x 4 0. Setting a 6, b 9, and c 4, we have: X 3 9 62 X 32 18 4 X 3 27X 18.
332
CHAPTER 3 Polynomial and Rational Functions
2 3 3 2 3 2 2 22 3 33 3 1 3 1 1 1 2 108. (a) Using the cubic formula, x 2 4 27 2 4 27 2
1
0
3
2
2
4
2
1 2 1 0 So x 2 x 2 2x 1 x 2 x 12 0 x 2, 1. Using the methods from this section, we have
1
1
0
1
3
2
1
1
2
1
2
0 So x 3 3x 2 x 1 x 2 x 2 x 12 x 2 0 x 2, 1.
Since this factors easily, the factoring method was easier.
(b) Using the cubic formula, 2 3 3 54 3 54 54 542 27 273 x 2 4 27 2 4 27 3 54 3 54 3 3 2 2 2 2 2 27 27 2 27 27 27 27 3 3 6 6
1 1
0
27
54
6
36
54
6
9
0 x 3 27x 54 x 6 x 2 6x 9 x 6 x 32 0 x 3, 6.
Using methods from this section, 1
1 1
0
27
54
1
1
26
2
1
0
27
54
2
4
46
3
1 2 23 8 So x 3 27x 54 x 3 x 2 3x 18 x 6 x 32 0 x 3 6. 1
26
28
1 1
Since this factors easily, the factoring method was easier.
(c) Using the cubic formula, 2 3 3 3 4 3 33 4 4 42 x 2 4 27 2 4 27 3 3 3 3 2 4 1 2 4 1 2 5 2 5
0
27
54
3
9
54
3
18
0
SECTION 3.4 Real Zeros of Polynomials
333
Using a graphing calculator, we see that P x x 3 3x 4 has one zero. Using methods from this section, P x has possible rational zeros 1, 2, 4. 1
1 1
0
3
4
1
1
4
1
4
4
1
1
0
3
4
1
1
4
1 is an upper bound. 1 1 4 0 x 1 is a zero. P x x 3 3x 4 x 1 x 2 x 4 . Using the Quadratic Formula we have
1 12 414 1 215 which is not a real number. Since it is not easy to see that x 2
3 3 2 5 2 5 1, we see that the factoring method was much easier.
109. (a) Since z b, we have z b 0. Since all the coefficients of Q x are nonnegative, and since z 0, we have Q z 0 (being a sum of positive terms). Thus, P z z b Q z r 0, since the sum of a positive number and a nonnegative number. (b) In part (a), we showed that if b satisfies the conditions of the first part of the Upper and Lower Bounds Theorem and z b, then P z 0. This means that no real zero of P can be larger than b, so b is an upper bound for the real zeros. (c) Suppose b is a negative lower bound for the real zeros of P x. Then clearly b is an upper bound for P1 x P x. Thus, as in Part (a), we can write P1 x x b Q x r, where r 0 and the coefficients of Q are all nonnegative, and P x P1 x x b Q x r x b [Q x] r . Since the coefficients of Q x are all nonnegative, the coefficients of Q x will be alternately nonpositive and nonnegative, which proves the second part of the Upper and Lower Bounds Theorem. 110. P x x 5 x 4 x 3 5x 2 12x 6 has possible rational zeros 1, 2, 3, 6. Since P x has 1 variation in sign,
there is 1 positive real zero. Since P x x 5 x 4 x 3 5x 2 12x 6 has 4 variations in sign, there are 0, 2, or 4 negative real zeros. 1
1
1
1
5
12
6
1
0
1
6
18
0
1
6
18
24
1
2
1
3 1 1 1 5 12 6 3
1
1
5
12
6
2
2
2
6
36
1
1
3
18
42
1 1 1 1 5 12 6 2 1
6
5 10 18 48 3 is an upper bound. 1 2 1 6 P x x 1 x 4 2x 3 x 2 6x 6 , continuing with the quotient we have
6
1
6 15
1
30 54
1
2
1
1 1
2
1
6
6
1
3
4
10
3
4
10
4
6 0 x 1 is a zero.
1 is a lower bound.
Therefore, there is 1 rational zero, namely 1. Since there are 1, 3 or 5 real zeros, and we found 1 rational zero, there must
be 0, 2 or 4 irrational zeros. However, since 1 zero must be positive, there cannot be 0 irrational zeros. Therefore, there is exactly 1 rational zero and 2 or 4 irrational zeros.
334
CHAPTER 3 Polynomial and Rational Functions
3.5
COMPLEX ZEROS AND THE FUNDAMENTAL THEOREM OF ALGEBRA
1. The polynomial P x 5x 2 x 43 x 7 has degree 6. It has zeros 0, 4, and 7. The zero 0 has multiplicity 2, and the zero 4 has multiplicity 3. 2. (a) If a is a zero of the polynomial P, then x a must be a factor of P x.
(b) If a is a zero of multiplicity m of the polynomial P, then x am must be a factor of P x when we factor P completely.
3. A polynomial of degree n 1 has exactly n zeros, if a zero of multiplicity m is counted m times. 4. If the polynomial function P has real coefficients and if a bi is a zero of P, then a bi is also a zero of P. So if 3 i is a zero of P, then 3 i is also a zero of P. 5. P x x 4 1 (a) True, P has degree 4, so by the Zeros Theorem it has four (not necessarily distinct) complex zeros. (b) True, by the Complete Factorization Theorem, this is true. (c) False, the fourth power of any real number is nonnegative, so P x 1 for all real x and P has no real zeros. 6. P x x 3 x
(a) False. P x x x 2 1 , and x 2 1 has no real zeros. (b) True, P 0 0.
(c) False, P x cannot be factored into linear factors with real coefficients because x 2 1 has no real zeros. 7. (a) x 4 4x 2 0 x 2 x 2 4 0. So x 0 or x 2 4 0. If x 2 4 0 then x 2 4 x 2i. Therefore, the solutions are x 0 and 2i.
(b) To get the complete factorization, we factor the remaining quadratic factor P x x 2 x 4 x 2 x 2i x 2i. 8. (a) x 5 9x 3 0 x 3 x 2 9 0. So x 0 or x 2 9 0. If x 2 9 0 then x 3i. Therefore, the zeros of P are x 0, 3i.
(b) Since 3i and 3i are the zeros from x 2 9 0, x 3i and x 3i are the factors of x 2 9. Thus the complete factorization is P x x 3 x 2 9 x 3 x 3i x 3i.
9. (a) x 3 2x 2 2x 0 x x 2 2x 2 0. So x 0 or x 2 2x 2 0. If x 2 2x 2 0 then 2 22 412 2 2 4 22i x 2 2 1 i. Therefore, the solutions are x 0, 1 i.
(b) Since 1 i and 1 i are zeros, x 1 i x 1 i and x 1 i x 1 i are the factors of x 2 2x 2. Thus the complete factorization is P x x x 2 2x 2 x x 1 i x 1 i.
10. (a) x 3 x 2 x 0 x x 2 x 1 0. So x 0 or x 2 x 1 0. If x 2 x 1 0 then
1 12 411 12 3 12 i 23 . Therefore, the zeros of P are x 0, 12 i 23 . x 21 (b) The zeros of x 2 x 1 0 are 12 i 23 and 12 i 23 , so factoring we get x 2 x 1 x 12 i 23 x 12 i 23 x 12 i 23 x 12 i 23 . Thus the complete factorization is P x x x 2 x 1 x x 12 i 23 x 12 i 23 .
SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra
335
2 11. (a) x 4 2x 2 1 0 x 2 1 0 x 2 1 0 x 2 1 x i. Therefore the zeros of P are x i. (b) Since i and i are zeros, x i and x i are the factors of x 2 1. Thus the complete factorization is 2 P x x 2 1 [x i x i]2 x i2 x i2 .
12. (a) x 4 x 2 2 0 x 2 2 x 2 1 0. So x 2 2 0 or x 2 1 0. If x 2 2 0 then x 2 2 x 2. And if x 2 1 0 then x 2 1 x i. Therefore, the zeros of P are x 2, i. (b) To get the complete factorization, we factor the quadratic factors to get P x x 2 2 x 2 1 x 2 x 2 x i x i.
13. (a) x 4 16 0 0 x 2 4 x 2 4 x 2 x 2 x 2 4 . So x 2 or x 2 4 0. If x 2 4 0 then x 2 4 x 2i. Therefore the zeros of P are x 2, 2i.
(b) Since i and i are zeros, x i and x i are the factors of x 2 1. Thus the complete factorization is P x x 2 x 2 x 2 4 x 2 x 2 x 2i x 2i.
2 14. (a) x 4 6x 2 9 0 x 2 3 0 x 2 3. So x i 3 are the only zeros of P (each of multiplicity 2). (b) To get the complete factorization, we factor the quadratic factor to get 2 2 2 2 x i 3 . P x x 2 3 x i 3 x i 3 x i 3
15. (a) x 3 8 0 x 2 x 2 2x 4 0. So x 2 or x 2 2x 4 0. If x 2 2x 4 0 then
2 22 414 2 12 22i 3 1i 3. Therefore, the zeros of P are x 2, 1 i 3. x 2 2 2
(b) Since 1 i 3 and 1 i 3 are the zeros from the x 2 2x 4 0, x 1 i 3 and x 1 i 3 are the factors of x 2 2x 4. Thus the complete factorization is P x x 2 x 2 2x 4 x 2 x 1 i 3 x 1i 3 x 2 x 1 i 3 x 1 i 3
16. (a) x 3 8 0 x 2 x 2 2x 4 0. So x 2 or x 2 2x 4 0. If x 2 2x 4 0 then
2 22 414 3 1i 3. Therefore, the zeros of P are x 2, 1 i 3. 22 12 22i x 2 2
(b) Since 1 i 3 and 1 i 3 are the zeros from x 2 2x 4 0, x 1 i 3 and x 1 i 3 are the factors of x 2 2x 4. Thus the complete factorization is P x x 2 x 2 2x 4 x 2 x 1 i 3 x 1 i 3 x 2 x 1 i 3 x 1 i 3
17. (a) x 6 1 0 0 x 3 1 x 3 1 x 1 x 2 x 1 x 1 x 2 x 1 . Clearly, x 1 are solutions.
If x 2 x 1 0, then x 1 1411 12 3 12 23 so x 12 i 23 . And if x 2 x 1 0, then 2
1 2 3 12 x 1 1411 2
3 1 i 3 . Therefore, the zeros of P are x 1, 1 i 3 , 1 i 3 . 2 2 2 2 2 2 2
336
CHAPTER 3 Polynomial and Rational Functions 3 and 1 i 3 , so x 2 x 1 factors as 2 2 2 3 1 x 2 i 2 x 12 i 23 . Similarly,since the zeros of x 2 x 1 0 are 12 i 23 and 12 i 23 , so x 2 x 1 factors as 1 x 12 i 23 x 12 i 23 x 12 i 23 . Thus the complete i 23 x 2
(b) The zeros of x 2 x 1 0 are 12 i x 12 i 23 x 12 i 23
factorization is
P x x 1 x 2 x 1 x 1 x 2 x 1
x 1 x 1 x 12 i 23 x 12 i 23 x 12 i 23 x 12 i 23 18. (a) x 6 7x 3 8 0 0 x 3 8 x 3 1 x 2 x 2 2x 4 x 1 x 2 x 1 . Clearly, x 1 and x 2 are solutions. If x 2 2x 4 0, then x 2 4414 22 12 22 2 23 so x 1 i 3. If 2
1 2 3 12 23 12 i 23 . Therefore, the zeros of P are x 1, 2, x 2 x 1 0, then x 1 1411 2 1 i 3, 12 i 23 . (b) From Exercise 10, x 2 2x 4 x 1 i 3 x 1 i 3 and from Exercise 11, x 2 x 1 x 12 i 23 x 12 i 23 . Thus the complete factorization is P x x 2 x 2 2x 4 x 1 x 2 x 1
x 2 x 1 x 1 i 3 x 1 i 3 x 12 i 23 x 12 i 23
19. P x x 2 25 x 5i x 5i. The zeros of P are 5i and 5i, both multiplicity 1.
20. P x 4x 2 9 2x 3i 2x 3i. The zeros of P are 32 i and 32 i, both multiplicity 1.
2 22 412 2 21. Q x x 2x 2. Using the Quadratic Formula x 22 4 22i 1 i. So 21 2
Q x x 1 i x 1 i. The zeros of Q are 1 i (multiplicity 1) and 1 i (multiplicity 1). 22. Q x x 2 8x 17 x 2 8x 16 1 x 42 1 [x 4 i] [x 4 i] x 4 i x 4 i. The zeros of Q are 4 i and 4 i, both multiplicity 1. 23. P x x 3 4x x x 2 4 x x 2i x 2i. The zeros of P are 0, 2i, and 2i (all multiplicity 1). 24. P x x 3 x 2 x x x 2 x 1 . Using the Quadratic Formula, we have
1 12 411 1 2 3 12 i 23 . The zeros of P are 0, 12 i 23 , and 12 i 23 , all of multiplicity 1. And x 21 P x x x 12 i 23 x 12 i 23 .
25. Q x x 4 1 x 2 1 x 2 1 x 1 x 1 x 2 1 x 1 x 1 x i x i. The zeros of Q are
1, 1, i, and i (all of multiplicity 1). 26. Q x x 4 625 x 2 25 x 2 25 x 5 x 5 x 2 25 x 5 x 5 x 5i x 5i. The zeros
of Q are 5, 5, 5i, and 5i, all multiplicity 1. 27. P x 16x 4 81 4x 2 9 4x 2 9 2x 3 2x 3 2x 3i 2x 3i. The zeros of P are 32 , 32 , 32 i, and 32 i (all of multiplicity 1).
SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra
28. P x x 3 64 x 4 x 2 4x 16 .
337
Using the Quadratic Formula, we have
3 2 2i 3. The zeros of P are 4, 2 2i 3, and 2 2i 3, all of x 4 164116 42 48 44i 2 2
multiplicity 1, and P x x 4 x 2 2i 3 x 2 2i 3 . 29. P x x 3 x 2 9x 9 x 2 x 1 9 x 1 x 1 x 2 9 x 1 x 3i x 3i. The zeros of P are 1, 3i, and 3i (all of multiplicity 1). 30. P x x 6 729 x 3 27 x 3 27 x 3 x 2 3x 9 x 3 x 2 3x 9 . Using the
27 2 . Using the 3 9419 3 27 27 3 9 we have x 2 2 32 3 2 3 i 2 2 3 2 3 i, 32 3 2 3 i, 32 3 2 3 i, and 32 3 2 3 i, all multiplicity 1. And 3 2 3 i x 32 3 2 3 i x 32 3 2 3 i x 32 3 2 3 i .
Quadratic Formula on x 2 3x 9 we have x 3 9419 32 27 32 2
Quadratic Formula on x 2 3x The zeros of P are 3, 3, 32 P x x 3 x 3 x 32
2 31. Q x x 4 2x 2 1 x 2 1 x i2 x i2 . The zeros of Q are i and i (both of multiplicity 2).
2 2 2 2 x i 5 . The zeros of Q are i 5 x i 5 32. Q x x 4 10x 2 25 x 2 5 x i 5 x i 5 and i 5, both of multiplicity 2. 33. P x x 4 3x 2 4 x 2 1 x 2 4 x 1 x 1 x 2i x 2i. The zeros of P are 1, 1, 2i, and 2i (all of multiplicity 1).
34. P x x 5 7x 3 x 3 x 2 7 x 3 x i 7 x i 7 . The zeros of P are 0 (multiplicity 3), i 7 and i 7, both of multiplicity 1.
2 2 2 x i 3 . The zeros of P are 0 35. P x x 5 6x 3 9x x x 4 6x 2 9 x x 2 3 x x i 3 (multiplicity 1), i 3 (multiplicity 2), and i 3 (multiplicity 2). 2 2 36. P x x 6 16x 3 64 x 3 8 x 22 x 2 2x 4 . Using the Quadratic Formula, on x 2 2x 4 we have 2 22 414 2 12 22i 3 1 i 3, so P x x 22 x 1 i 3 2 x 1 i 3 2 . x 2 2 2
The zeros of P are 2, 1 i 3, and 1 i 3, all multiplicity 2.
37. Since 1 i and 1 i are conjugates, the factorization of the polynomial must be P x a x [1 i] x [1 i] a x 2 2x 2 . If we let a 1, we get P x x 2 2x 2. 38. Since 1 i 2 and 1 i 2 are conjugates, the factorization of the polynomial must be P x c x 1 i 2 x 1 i 2 c x 2 2x 3 . If we let c 1, we get P x x 2 2x 3.
39. Since 2i and 2i are conjugates, the factorization of the polynomial must be Q x b x 3 x 2i x 2i] b x 3 x 2 4 b x 3 3x 2 4x 12 . If we let b 1, we get Q x x 3 3x 2 4x 12.
40. Since i is a zero, by the Conjugate Roots Theorem, i is also a zero. So the factorization of the polynomial must be Q x b x 0 x i x i bx x 2 1 b x 3 x . If we let b 1, we get Q x x 3 x.
41. Since i is a zero, by the Conjugate Roots Theorem, i is also a zero. So the factorization of the polynomial must be P x a x 2 x i x i a x 3 2x 2 x 2 . If we let a 1, we get P x x 3 2x 2 x 2.
338
CHAPTER 3 Polynomial and Rational Functions
42. Since 1 i is a zero, by the Conjugate Roots Theorem, 1 i is also a zero. So the factorization of the polynomial must be Q x a x 3 x [1 i] x [1 i] a x 3 x 2 2x 2 a x 3 x 2 4x 6 . If we let a 1, we get Q x x 3 x 2 4x 6.
43. Since the zeros are 1 2i and 1 (with multiplicity 2), by the Conjugate Roots Theorem, the other zero is 1 2i. So a factorization is R x c x [1 2i] x [1 2i] x 12 c [x 1] 2i [x 1] 2i x 12 c [x 1]2 [2i]2 x 2 2x 1 c x 2 2x 1 4 x 2 2x 1 c x 2 2x 5 x 2 2x 1 c x 4 2x 3 x 2 2x 3 4x 2 2x 5x 2 10x 5 c x 4 4x 3 10x 2 12x 5 If we let c 1 we get R x x 4 4x 3 10x 2 12x 5.
44. Since S x has zeros 2i and 3i, by the Conjugate Roots Theorem, the other zeros of S x are 2i and 3i. So a factorization of S x is S x C x 2i x 2i x 3i x 3i C x 2 4i 2 x 2 9i 2 C x 2 4 x 2 9 C x 4 13x 2 36 If we let C 1, we get S x x 4 13x 2 36.
45. Since the zeros are i and 1 i, by the Conjugate Roots Theorem, the other zeros are i and 1 i. So a factorization is T x C x i x i x [1 i] x [1 i] C x 2 i 2 [x 1] i [x 1] i C x 2 1 x 2 2x 1 i 2 C x 2 1 x 2 2x 2 C x 4 2x 3 2x 2 x 2 2x 2 C x 4 2x 3 3x 2 2x 2 C x 4 2C x 3 3C x 2 2C x 2C Since the constant coefficient is 12, it follows that 2C 12 C 6, and so T x 6 x 4 2x 3 3x 2 2x 2 6x 4 12x 3 18x 2 12x 12.
46. Since U x has zeros 12 , 1 (with multiplicity two), and i, by the Conjugate Roots Theorem, the other zero is i. So a
factorization of U x is U x c x 12 x 12 x i x i 12 c 2x 1 x 2 2x 1 x 2 1 12 c 2x 5 3x 4 2x 3 2x 2 1 Since the leading coefficient is 4, we have 4 12 c 2 c. Thus we have U x 12 4 2x 5 3x 4 2x 3 2x 2 1 4x 5 6x 4 4x 3 4x 2 2.
47. P x x 3 2x 2 4x 8 x 2 x 2 4 x 2 x 2 x 2 4 x 2 x 2i x 2i. Thus the zeros are 2 and 2i.
48. P x x 3 7x 2 17x 15. We start by trying the possible rational factors of the polynomial: 1
1
7
17
15
1
6
11
3
1
7
17
15
3
12
15
1 4 5 0 x 3 is a zero. 4 So P x x 3 x 2 4x 5 . Using the Quadratic Formula on the second factor, we have 1
6
11
x 4 16415 4 2 4 42i 2 2 2 i. Thus the zeros are 3, 2 i.
SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra
339
49. P x x 3 2x 2 2x 1. By inspection, P 1 1 2 2 1 0, and hence x 1 is a zero. 1
1 1
2
2
1
1
1
1
1
1
0
Thus P x x 1 x 2 x 1 . So x 1 or x 2 x 1 0. 1i2 3 . Hence, the zeros are 1 and 1i2 3 . Using the Quadratic Formula, we have x 1 1411 2 50. P x x 3 7x 2 18x 18 has possible rational zeros 1, 2, 3, 6, 9, 18. Since all of the coefficients are positive, there are no positive real zeros. 1 1
7 18
18
2 1
1 6 12
7
18
18
3 1
2 10 16
7
18
18
3 12 18
1 5 8 2 1 4 6 0 x 3 is a zero. 6 12 6 So P x x 3 x 2 4x 6 . Using the Quadratic Formula on the second factor, we have 2 2 i 2. Thus the zeros are 3, 2 i 2. x 4 16416 42 8 42i 2 2 1
51. P x x 3 3x 2 3x 2. 3
3
2
2
2
2
1 1 1 Thus P x x 2 x 2 x 1 . So x 2 or x 2 x 1 0
0
2
1
Using the Quadratic Formula we have x 1 1411 1i2 3 . Hence, the zeros are 2, and 1i2 3 . 2 52. P x x 3 x 6 has possible zeros 1, 2, 3. 1
1
0
1
6
1
1
0
2
1
0
1
6
2
4
6
1 2 3 0 x 2 is a zero. 0 6 2 1 i 2. Thus the P x x 2 x 2 2x 3 . Now x 2 2x 3 has zeros x 2 4413 22i 2 2 zeros are 2, 1 i 2. 1
1
53. P x 2x 3 7x 2 12x 9 has possible rational zeros 1, 3, 9, 12 , 32 , 92 . Since all coefficients are positive, there are no positive real zeros. 1
2 2
7
12
9
2
5
7
5
7
2
2
2
There is a zero between 1 and 2.
32
2
2
7
12
9
3
6
9
7
12
9
4
6
12
3
6
3
2 4 6 0 x 32 is a zero. P x x 32 2x 2 4x 6 2 x 32 x 2 2x 3 . Now x 2 2x 3 has zeros 22 2 x 2 4431 1 i 2. Hence, the zeros are 32 and 1 i 2. 2 2
340
CHAPTER 3 Polynomial and Rational Functions
54. Using synthetic division, we see that x 3 is a factor of the polynomial: 1
2
8
9
9
2
6
3
2
3
8
9
9
6
6
9
2 2 3 0 x 3 is a zero. So P x 2x 3 8x 2 9x 9 x 3 2x 2 2x 3 . Using the Quadratic Formula, we find the other two solutions: 2 20 2 4 4 3 2 12 25 i. Thus the zeros are 3, 12 25 i. x 2 2 4 2
3
6
6
55. P x x 4 x 3 7x 2 9x 18. Since P x has one change in sign, we are guaranteed a positive zero, and since P x x 4 x 3 7x 2 9x 18, there are 1 or 3 negative zeros. 1
1
1
7
9
18
1
2
9
18
1 2 9 18 0 Therefore, P x x 1 x 3 2x 2 9x 18 . Continuing with the quotient, we try negative zeros.
1
1
2
9
18
1
1
8
2
1
2
9
18
2
0
18
1 0 9 0 1 8 10 P x x 1 x 2 x 2 9 x 1 x 2 x 3i x 3i. Therefore,the zeros are 1, 2, and 3i. 1
56. P x x 4 2x 3 2x 2 2x 3 has possible zeros 1, 3. 1
1
2
2
2
3
1
1
3
5
3
1
2
2
2
3
3
3
3
3
1 1 1 1 0 x 3 is a zero. 5 8 P x x 3 x 3 x 2 x 1 . If we factor the second factor by grouping, we get x 3 x 2 x 1 x 2 x 1 1 x 1 x 1 x 2 1 . So we have P x x 3 x 1 x 2 1 x 3 x 1 x i x i. Thus the zeros are 3, 1, i, and i. 1
1
3
57. We see a pattern and use it to factor by grouping. This gives P x x 5 x 4 7x 3 7x 2 12x 12 x 4 x 1 7x 2 x 1 12 x 1 x 1 x 4 7x 2 12 x 1 x 2 3 x 2 4 x 1 x i 3 x i 3 x 2i x 2i
Therefore,the zeros are 1, i 3, and 2i.
58. P x x 5 x 3 8x 2 8 x 3 x 2 1 8 x 2 1 x 2 1 x 3 8 x 2 1 x 2 x 2 2x 4
(factoring a sum of cubes). So x 2, or x 2 1 0. If x 2 1 0, then x 2 1 x i. If x 2 2x 4 0, then 12 x 2 4414 1 1 i 3. Thus, the zeros are 2, i, 1 i 3. 2 2
SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra
341
59. P x x 4 6x 3 13x 2 24x 36 has possible rational zeros 1, 2, 3, 4, 6, 9, 12, 18. P x has 4 variations in sign and P x has no variation in sign. 1 1 6 13 24 1 5 1 5
36
8 16
8 16
Continuing:
2 1 6 13 24 2 8
20
1 4
36
3 1 6 13 24
10 28
5 14
3 9
8
1 3
36
12 36
4 12
0 x 3 is a zero.
3 1 3 4 12 3 0
12
1 0 4 0 x 3 is a zero. P x x 32 x 2 4 x 32 x 2i x 2i. Therefore,the zeros are 3 (multiplicity 2) and 2i.
60. P x x 4 x 2 2x 2 has possible rational zeros 1, 2. 1 1 0 1 2 2 1
1 1
0 1 2
1 0 2
2
1 1 1 0
1 0 2
1
2
1 2 2
1 1
0 2 4 1 is an upper bound. 1 1 0 2 0 1 2 2 0 P x x 12 x 2 2x 2 . Using the Quadratic Formula on x 2 2x 2, we have x 2 248 22i 2 1 i. Thus, the zeros of P x are 1 (multiplicity 2) and 1 i.
61. P x 4x 4 4x 3 5x 2 4x 1 has possible rational zeros 1, 12 , 14 . Since there is no variation in sign, all real zeros (if there are any) are negative. 1
4
4
5
4
1
4
0
5
1
12
4 0 5 1 2 1 3 P x x 2 4x 2x 2 4x 2 . Continuing:
12
4 4
4 4
2
4
2
2
0
2
0
4
0
4
5
4
1
2
1
2
1
2
4
2
0
x 12 is a zero.
x 12 is a zero again.
2 P x x 12 4x 2 4 . Thus, the zeros of P x are 12 (multiplicity 2) and i.
62. P x 4x 4 2x 3 2x 2 3x 1 has possible rational zeros 1, 12 , 14 . P has one variation in sign, so P has one positive real zero. 4
2
2
3
1
4
6
4
1
4 6 4 P x x 1 4x 3 6x 2 4x 1 . Continuing:
1
0
12
4
1
1
4
6
4
1
4
2
2
1 is a zero.
6
4
1
2
2
1
2 2 3 4 4 2 0 12 is a zero. P x x 1 x 12 4x 2 4x 2 . Using the Quadratic Formula on 4x 2 4x 2, we find 4
x 4 81632 12 12 i. Thus, P has zeros 1, 12 , 12 12 i.
342
CHAPTER 3 Polynomial and Rational Functions
63. P x x 5 3x 4 12x 3 28x 2 27x 9 has possible rational zeros 1, 3, 9. P x has 4 variations in sign and P x has 1 variation in sign.
1
1
1 1
1
3
12
28
27
9
1
2
10
18
9
2
10
18
9
0
2
10
18
9
1
1
9
9
x 1 is a zero. 1
1
1
9
9
1
0
9
1
1 9 9 0 x 1 is a zero. 1 0 9 0 x 1 is a zero. P x x 13 x 2 9 x 13 x 3i x 3i. Therefore,the zeros are 1 (multiplicity 3) and 3i 64. P x x 5 2x 4 2x 3 4x 2 x 2 has possible rational zeros 1, 2. 1
1
2
2
4
1
2
1
1
1
3
2
2
1
2
2
4
1
2
2
0
4
0
2
1
1 1 3 2 4 1 0 2 0 1 0 x 2 is a zero. 2 P x x 2 x 4 2x 2 1 x 2 x 2 1 x 2 x i2 x i2 . Thus, the zeros of P x are 2, i.
65. (a) P x x 3 5x 2 4x 20 x 2 x 5 4 x 5 x 5 x 2 4 (b) P x x 5 x 2i x 2i
66. (a) P x x 3 2x 4 1
1 1
0
2
4
1
1
1
1
1
2
5 P x x 3 2x 4 x 2 x 2 2x 2
1 1
0
2
4
2
4
4
2
2
0
(b) P x x 2 x 1 i x 1 i
67. (a) P x x 4 8x 2 9 x 2 1 x 2 9 x 1 x 1 x 2 9 (b) P x x 1 x 1 x 3i x 3i
2 68. (a) P x x 4 8x 2 16 x 2 4 (b) P x x 2i2 x 2i2
69. (a) P x x 6 64 x 3 8 x 3 8 x 2 x 2 2x 4 x 2 x 2 2x 4 (b) P x x 2 x 2 x 1 i 3 x 1 i 3 x 1 i 3 x 1 i 3 70. (a) P x x 5 16x x x 4 16 x x 2 4 x 2 4 x x 2 x 2 x 2 4 (b) P x x x 2 x 2 x 2i x 2i
SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra
343
71. (a) x 4 2x 3 11x 2 12x x x 3 2x 2 11x 12 0. We first find the bounds for our viewing rectangle.
5
1
2
11
12
5
1
3
4
32
x 5 is an upper bound.
4
1
6
13
50
x 4 is a lower bound.
-20 -40
We graph P x x 4 2x 3 11x 2 12x in the viewing rectangle [4 5] by [50 10] and see that it has 4 real solutions. Since this matches the degree of P x, P x has no nonreal solution. (b) x 4 2x 3 11x 2 12x 5 0. We use the same bounds for our viewing rectangle, [4 5] by [50 10], and see that R x x 4 2x 3 11x 2 12x 5 has
(c) x 4 2x 3 11x 2 12x 40 0. We graph
T x x 4 2x 3 11x 2 12x 40 in the viewing
rectangle [4 5] by [10 50], and see that T has no
2 real solutions. Since the degree of R x is 4, R x
real solution. Since the degree of T is 4, T must have
must have 2 nonreal solutions.
4 nonreal solutions.
5 -20
40 20
-40 5
72. (a) 2x 4i 1 2x 1 4i x 12 2i. (b) x 2 i x 0 x x i 0 x 0, i.
(c) x 2 2i x 1 0 x i2 0 x i.
(d) i x 2 2x i 0. Using the Quadratic Formula, we get 2 22 4ii 2 1 2 1 2 i 1 2 i. x 22i 8 22 i 2i 2i 73. (a) P x x 2 1 i x 2 2i. So P 2i 2i2 1 i 2i 2 2i 4 2i 2 2 2i 0, and P 1 i 1 i2 1 i 1 i 2 2i 1 2i 1 1 1 2 2i 0.
Therefore, 2i and 1 i are solutions of the equation x 2 1 i x 2 2i 0. However, P 2i 2i2 1 i 2i 2 2i 4 2i 2 2 2i 4 4i, and
P 1 i 1 i2 1 i 1 i 2 2i 2 2i. Since, P 2i 0 and P 1 i 0, 2i and 1 i are not solutions. (b) This does not violate the Conjugate Roots Theorem because the coefficients of the polynomial P x are not all real. 74. (a) Because i and 1 i are zeros, i and 1 i are also zeros. Thus,
P x C x i x i x [1 i] x [1 i] C x 2 1 x 2 2x 2 C x 4 2x 3 2x 2 x 2 2x 2 C x 4 2x 3 3x 2 2x 2
Because C 1, the polynomial is P x x 4 2x 3 3x 2 2x 2.
344
CHAPTER 3 Polynomial and Rational Functions
(b) Because i and 1 i are zeros,
P x C x i x [i 1] C x 2 xi x xi 1 i C x 2 1 2i x 1 i
Because C 1, the polynomial is P x x 2 1 2i x 1 i.
75. Because P has real coefficients, the imaginary zeros come in pairs: a bi (by the Conjugate Roots Theorem), where b 0. Thus there must be an even number of nonreal zeros. Since P is of odd degree, it has an odd number of zeros (counting multiplicity). It follows that P has at least one real zero. 76. x 4 1 0 x 2 1 x 2 1 0 x 1 x 1 x i x i 0 x 1, i. So there are four fourth
roots of 1, two that are real and two that are nonreal. Consider P x x n 1, where n is even. P has one change in sign
so P has exactly one real positive zero, namely x 1. Since P x P x, P also has exactly one real negative zero,
namely x 1. Thus P must have n 2 complex roots. As a result, x n 1 has two real n th zeros and n 2 complex roots. 3 . So there is one real cube zero of unity and two nonreal x 3 1 0 x 1 x 2 x 1 0 x 1, 1i 2
roots. Now consider Q x x k 1, where k is odd. Since Q has one change in sign, Q has exactly one real positive zero,
namely x 1. But Q x x k 1 has no change in sign, so there is no negative real zero. As a result, x k 1 has one real kth zero and k 1 nonreal roots.
3.6
RATIONAL FUNCTIONS
1. If the rational function y r x has the vertical asymptote x 2, then as x 2 , either y or y . 2. If the rational function y r x has the horizontal asymptote y 2, then y 2 as x . 3. The function r x
x 1 x 2 has x-intercepts 1 and 2. x 2 x 3
4. The function r has y-intercept 13 . 5. The function r has vertical asymptotes x 2 and x 3. 6. The function r has horizontal asymptote y 1. x2 x x x 1 x for x 1. 2 x 2 x 1 2x 4 x 1 2x 4 (a) False. r does not have vertical asymptote x 1. It has a “hole” at 1 16 .
7. r x
(b) True, r has vertical asymptote x 2.
(c) False, r has a horizontal asymptote y 12 but not a horizontal asymptote y 1.
(d) True, r has horizontal asymptote y 12 .
x2 x crosses its 8. True, the graph of a rational function may cross a horizontal asymptote. For example, r x 2 x x 1 horizontal asymptote y 1 at the point 12 1 . 9. r x
x x 2
SECTION 3.6 Rational Functions
(a) x
r x
x
x
r x
15
3
25
5
10
125
10
0833
19
19
21
21
50
1042
50
0962
199
199
201
201
100
1020
100
0980
1999
1999
2001
2001
1000
1002
1000
0998
x
r x
r x
(b) r x as x 2 and r x as x 2 .
(c) r has horizontal asymptote y 1. 4x 1 10. r x x 2 (a) x
r x
x
r x
x
r x
x
r x
15
14
25
22
10
5125
10
325
19
86
21
94
50
4188
50
3827
199
896
201
904
100
4092
100
3912
1999
8996
2001
9004
1000
4009
1000
3991
(b) r x as x 2 and r x as x 2 .
(c) r has horizontal asymptote y 4. 3x 10 11. r x x 22 (a) x
r x
r x
x
r x
x
r x
x
15
22
25
10
10
03125
10
02778
19
430
21
370
50
00608
50
00592
39,700
100
00302
100
00298
3,997,000
1000
00030
1000
00030
x
r x
x
r x
199 1999
40,300 4,003,000
201 2001
(b) r x as x 2 and r x as x 2 . (c) r has horizontal asymptote y 0.
12. r x (a)
3x 2 1
x 22 x
r x
x
r x
15
31
25
79
10
4703
10
209
19
1183
21
1423
50
3256
50
2774
131,203
100
3124
100
2884
13,012,003
1000
3012
1000
2988
199 1999
128,803 12,988,003
201 2001
(b) r x as x 2 and r x as x 2 .
(c) r has horizontal asymptote y 3. In Exercises 13–20, let f x
1 . x
345
346
CHAPTER 3 Polynomial and Rational Functions
1 f x 1. From this form we see that the graph of r is obtained x 1 from the graph of f by shifting 1 unit to the right. Thus r has vertical asymptote
y
x 1 and horizontal asymptote y 0. The domain of r is 1 1 and
1
13. r x
x
1
its range is 0 0 .
1 f x 4. From this form we see that the graph of r is obtained x 4 from the graph of f by shifting 4 units to the left. Thus r has vertical asymptote
y
x 4 and horizontal asymptote y 0. The domain of r is 4 4
1
14. r x
1
and its range is 0 0 .
15. s x
3 1 3 3 f x 1. From this form we see that the graph x 1 x 1
x
y
of s is obtained from the graph of f by shifting 1 unit to the left and stretching
3
vertically by a factor of 3. Thus s has vertical asymptote x 1 and horizontal
1
asymptote y 0. The domain of s is 1 1 and its range is
x
0 0 .
16. s x
2 1 2 2 f x 2. From this form we see that the x 2 x 2
y
graph of s is obtained from the graph of f by shifting 2 units to the right, stretching 1
vertically by a factor of 2, and then reflecting about the x-axis. Thus s has vertical
x
1
asymptote x 2 and horizontal asymptote y 0. The domain of s is 2 2 and its range is 0 0 .
1 2x 3 2 f x 2 2 (see the long x 2 x 2 division at right). From this form we see that the graph of t is
17. t x
obtained from the graph of f by shifting 2 units to the right and 2 units vertically. Thus t has vertical asymptote x 2 and horizontal asymptote y 2. The domain of t is
2 2 and its range is 2 2 .
y
2 x 2
2x 3 2x 2 1
1 1
x
SECTION 3.6 Rational Functions
9 3x 3 1 3 39 x 2 x 2 x 2 9 f x 2 3
18. t x
y
3 x 2
3x 3 3x 6
From this form we see that the graph of t is obtained from the
347
5
9
graph of f by shifting 2 units to the left, stretching vertically
x
1
by a factor of 9, reflecting about the x-axis, and then shifting 3 units vertically. Thus t has vertical asymptote x 2 and horizontal asymptote y 3. The domain of t is
2 2 and its range is 3 3 .
1 x 2 1 f x 3 1 (see the long x 3 x 3 division at right). From this form we see that the graph of r is
19. r x
y
1 x 3
x 2
obtained from the graph of f by shifting 3 units to the left,
x 3
reflect about the x-axis, and then shifting vertically 1 unit.
1
1 1
Thus r has vertical asymptote x 3 and horizontal
x
asymptote y 1. The domain of r is 3 3 and its range is 1 1 .
1 2x 9 2 2 x 4 x 4 f x 4 2
20. t x
1 x 4
y
2 x 4
2x 9
From this form we see that the graph of t is obtained from the
2x 8
graph of f by shifting 4 units to the right, reflecting about the
1
x-axis, and then shifting 2 units vertically. Thus t has vertical
1 1
x
asymptote x 4 and horizontal asymptote y 2. The domain of r is 4 4 and its range is 2 2 .
x 1 . When x 0, we have r 0 14 , so the y-intercept is 14 . The numerator is 0 when x 1, so the x 4 x-intercept is 1.
21. r x
3x . When x 0, we have s 0 0, so the y-intercept is 0. The numerator is zero when 3x 0 or x 0, so x 5 the x-intercept is 0.
22. s x
23. t x
2 x2 x 2 . When x 0, we have t 0 13 , so the y-intercept is 13 . The numerator is 0 when x 6 6
x 2 x 2 x 2 x 1 0 or when x 2 or x 1, so the x-intercepts are 2 and 1.
2 2 24. r x 2 12 , so the y-intercept is 12 . The numerator is never zero, so . When x 0, we have r 0 4 x 3x 4 there is no x-intercept. 25. r x
x2 9 . Since 0 is not in the domain of r x, there is no y-intercept. The numerator is 0 when x2
x 2 9 x 3 x 3 0 or when x 3, so the x-intercepts are 3.
348
CHAPTER 3 Polynomial and Rational Functions
x3 8 26. r x 2 . When x 0, we have r 0 84 2, so the y-intercept is 2. The x-intercept occurs when x 3 8 0 x 4 x 2 x 2 2x 4 0 x 2 or x 1 i 3, which has only one real solution, so the x-intercept is 2. 27. From the graph, the x-intercept is 3, the y-intercept is 3, the vertical asymptote is x 2, and the horizontal asymptote is y 2.
28. From the graph, the x-intercept is 0, the y-intercept is 0, the horizontal asymptote is y 0, and the vertical asymptotes are x 1 and x 2.
29. From the graph, the x-intercepts are 1 and 1, the y-intercept is about 14 , the vertical asymptotes are x 2 and x 2, and the horizontal asymptote is y 1. 30. From the graph, the x-intercepts are 2, the y-intercept is 6, the horizontal asymptote is y 2, and there are no vertical asymptotes 5 has a vertical asymptote where x 2 0 x 2, and y 0 is a horizontal asymptote because the degree x 2 of the denominator is greater than that of the numerator.
31. r x
2x 3 has are vertical asymptotes where x 2 1 0 x 1 or x 1, and y 0 is a horizontal asymptote 32. r x 2 x 1 because the degree of the denominator is greater than that of the numerator. 3x 1 has no vertical asymptote since 4x 2 1 0 for all x. y 0 is a horizontal asymptote because the degree 4x 2 1 of the denominator is greater than that of the numerator.
33. r x
3x 2 5x has vertical asymptotes where x 4 1 x 2 1 x 2 1 0 x 1, and y 0 is a horizontal 4 x 1 asymptote because the degree of the denominator is greater than that of the numerator.
34. r x
35. s x
6x 2 1 has vertical asymptotes where 2x 2 x 1 0 x 1 2x 1 0 x 1 or x 12 , and 2x 2 x 1
horizontal asymptote y 62 3. 36. s x
8x 2 1
4x 2 2x 6
has vertical asymptotes where 4x 2 2x 6 0 2 2x 3 x 1 0 x 32 or x 1,
and horizontal asymptote y 84 2. 37. r x
x 1 2x 3 has vertical asymptotes where x 2 4x 7 0 x 74 or x 2, and horizontal x 2 4x 7
asymptote y 38. r x
x 3 x 2 has vertical asymptotes where 5x 1 2x 3 0 x 15 or x 32 , and horizontal 5x 1 2x 3
asymptote y 39. r x
1 12 . 14 2
1 11 . 52 10
6x 3 2 6x 3 2 . Because the quadratic in the denominator has no real zero, r has vertical 2x 3 5x 2 6x x 2x 2 5x 6
asymptote x 0 and horizontal asymptote y 62 3.
SECTION 3.6 Rational Functions
349
5x 3 5x 3 5x 2 40. r x 3 . Because the denominator has no real zero, r has no vertical x 2x 2 5x x x 2 2x 5 x 2 2x 5 asymptote. r has horizontal asymptote y 51 5.
x2 2 . A vertical asymptote occurs when x 1 0 x 1. There is no horizontal asymptote because the degree x 1 of the numerator is greater than the degree of the denominator.
41. y
x 2 x 3 x 3 3x 2 . Because the degree of the numerator is greater than the degree of the denominator, 2 x 2 x 2 x 4 the function has no horizontal asymptote. Two vertical asymptotes occur at x 2 and x 2. By using long division, we 4x 12 see that r x x 3 2 so y x 3 is a slant asymptote. x 4
42. r x
4x 4 . When x 0, y 2, so the y-intercept is 2. When y 0, x 2 4x 4 0 x 1, so the x-intercept is 1. Since the degree of the numerator and
y
43. y
denominator are the same, the horizontal asymptote is y 41 4. A vertical 4x 4 , and as asymptote occurs when x 2. As x 2 , y x 2 4x 4 x 2 , y . The domain is x x 2 and the range is x 2 y y 4. 44. r x
4 1
2 x 3 2x 6 . When x 0, we have y 2, so the y-intercept 6x 3 3 2x 1
x
y
is 2. When y 0, we have x 3 0 x 3, so the x-intercept is 3. A
vertical asymptote occurs when 2x 1 0 x 12 . Because the degree of the denominator and the numerator are the same, the horizontal asymptote is 2 1 . The domain is x x 1 and the range is y y 1 . y 6 3 2 3
3 x 2 4x 4 1 3x 2 12x 13 1 45. r x . When x 0, 3 x 2 4x 4 x 2 4x 4 x 22
1 1
x
y
1 13 y 13 is positive 4 , so the y-intercept is 4 . There is no x-intercept since x 22
on its domain. There is a vertical asymptote at x 2. The horizontal asymptote is y 3. The domain is x x 2 and the range is y y 3.
1 1
x
350
CHAPTER 3 Polynomial and Rational Functions
2 x 2 4x 4 1 2x 2 8x 9 1 46. r x . When 2 2 2 x 4x 4 x 4x 4 x 22 x 0, y 94 , so the y-intercept is 94 . There is no x-intercept since
1
x 22
y
is
1 0
positive on its domain. There is a vertical asymptote at x 2. The horizontal
x
1
asymptote is y 2. The domain is x x 2 and the range is y y 2.
x 2 8x 16 2 x 2 8x 18 2 47. r x 2 . When 1 2 x 8x 16 x 8x 16 x 42 x 0, y 98 , and so the y-intercept is 98 . There is no x-intercept since 2
is positive on its domain. There is a vertical asymptote at x 4. The x 42 horizontal asymptote is y 1. The domain is x x 4 and the range is
y
1 0
1
x
y y 1.
1 2x 2 4x 2 2 1 x 2 2x 3 1 2 48. r x 2 . is true. When 2 x 12 2x 4x 2 2x 2 4x 2
y
x 0, we have y 32 , so the y-intercept is 2. There is no x-intercept since 1 is positive on its domain. There is a vertical asymptote at x 1. The x 12
horizontal asymptote is y 12 . The domain is x x 1 and the range is y y 12 .
49. s x
4x 8 8 . When x 0, y 2, so the y-intercept is 2. x 4 x 1 4 1
1 0
1
x
y
When y 0, 4x 8 0 x 2, so the x-intercept is 2. The vertical asymptotes are x 1 and x 4, and because the degree of the numerator is less than the
degree of the denominator, the horizontal asymptote is y 0. The domain is x x 1 4 and the range is R.
1 1
x
SECTION 3.6 Rational Functions
6 6 1, so the y-intercept is 1. . When x 0, y 50. s x 2 6 x 5x 6 Since the numerator is never zero, there is no x-intercept. The vertical asymptotes occur when x 2 5x 6 x 1 x 6 x 1 and x 6, and because the degree of the numerator is less less than the degree of the denominator, the
351
y
1
1
x
horizontal asymptote is y 0. The domain is x x 1 6 and the range is y y 05 or y 0.
2x 4 2 x 2 51. s x 2 . When x 0, y 2, so the y-intercept is 1 x 2 x x x 2 2. When y 0, we have 2x 4 0 x 2, so the x-intercept is 2. A vertical
y
asymptote occurs when x 1 x 2 0 x 1 and x 2. Because the
1
degree of the denominator is greater than the degree of the numerator, the
horizontal asymptote is y 0. The domain is x x 2 1 and the range is
1
x
1
x
y y 02 or y 2.
52. s x
x 2 2 , so the y-intercept is 23 When . When x 0, y 3 x 3 x 1
y
y 0, we have x 2 0 x 2, so the x-intercept is 2. A vertical
asymptote occurs when x 3 x 1 0 x 3 and x 1. Because the
1
degree of the denominator is greater than the degree of the numerator, the
horizontal asymptote is y 0. The domain is x x 3 1 and the range is R.
53. r x
x 1 x 2 . When x 0, y 23 , so the y-intercept is 23 . When x 1 x 3
y
y 0, x 1 x 2 0 x 2, 1, so, the x-intercepts are 2 and 1. The
vertical asymptotes are x 1 and x 3, and because the degree of the
numerator and denominator are the same the horizontal asymptote is y 11 1. The domain is x x 1 3 and the range is R.
2x 2 10x 12 2 x 1 x 6 2 1 6 . When x 0, y 2, x 2 x 3 2 3 x2 x 6 so the y-intercept is 2. When y 0, 2 x 1 x 6 0 x 6, 1, so the
54. r x
x-intercepts are 6 and 1. Vertical asymptotes occur when x 2 x 3 0
x 3 or x 2. Because the degree of the numerator and denominator are the same the horizontal asymptote is y 21 2. The domain is x x 3 2 and the range is R.
1 x
1
y
2 1
x
352
CHAPTER 3 Polynomial and Rational Functions
2x 2 2x 4 2 x 2 x 1 . Vertical asymptotes occur at x 0 x x 1 x2 x and x 1. Since x cannot equal zero, there is no y-intercept. When y 0, we
y
55. r x
have x 2 or 1, so the x-intercepts are 2 and 1. Because the degree of the
5
denominator and numerator are the same, the horizontal asymptote is y 21 2. The domain is x x 1 0 and the range is y y 2 or y 184.
3 x2 2 3x 2 6 56. r x 2 . When x 0, y 2, so the y-intercept x 3 x 1 x 2x 3
1
x
y
is 2. Since the numerator can never equal zero, there is no x-intercept. Vertical
asymptotes occur when x 1, 3. Because the degree of the numerator and
denominator are the same, the horizontal asymptote is.y 31 3. The domain is x x 1 3 and the range is y y 18 or y 25.
57. s x
x 2 2x 1 x 12 . Since x 0 is not in the domain of s x, x 3 3x 2 x 2 x 3
there is no y-intercept. The x-intercept occurs when y 0
4 1
x
1
x
y 1
x 2 2x 1 x 12 0 x 1, so the x-intercept is 1. Vertical asymptotes
occur when x 0, 3. Since the degree of the numerator is less than the degree of
the denominator, the horizontal asymptote is y 0. The domain is x x 0 3 and the range is R.
x2 x 6 x 3 x 2 . The x-intercept occurs when y 0 2 x x 3 x 3x x 3 x 2 0 x 2, 3, so the x-intercepts are 2 and 3. There is no
y
58. y
y-intercept because y is undefined when x 0. The vertical asymptotes are x 0
2
and x 3. Because the degree of the numerator and denominator are the same,
1
the horizontal asymptotes is y 11 1. The domain is x x 3 0 and the range is R.
x 2 2x 1 x 12 59. r x 2 x 2x 1 x 12
x 1 2 . When x 0, y 1, so the x 1
x
y
y-intercept is 1. When y 0, x 1, so the x-intercept is 1. A vertical asymptote
occurs at x 1 0 x 1. Because the degree of the numerator and
denominator are the same the horizontal asymptote is y 11 1. The domain is x x 1 and the range is y y 0.
1 1
x
SECTION 3.6 Rational Functions
4x 2 4x 2 60. r x 2 . When x 0, we have y 0, so the x 3 x 1 x 2x 3 graph passes through the origin. Vertical asymptotes occur at x 1 and x 3.
353
y
Because the degree of the denominator and numerator are the same, the horizontal
asymptote is y 41 4. The domain is x x 1 3 and the range is y y 0 or y 29.
1
5 x2 1 5x 2 5 5 61. r x 2 . When x 0, we have y , so the 2 4 x 4x 4 x 2
2
x
2
x
y
y-intercept is 54 . Since x 2 1 0 for all real x, y never equals zero, and there is
no x-intercept. The vertical asymptote is x 2. Because the degree of the
denominator and numerator are the same, the horizontal asymptote occurs at y 51 5. The domain is x x 2 and the range is y y 10.
x 2 x 1 x3 x2 3 . When x 0, we have y 0, so the 62. r x 3 x 3x 2 x 3x 2
2
y
y-intercept is 0. When y 0, we have x 2 x 1 0, so the x-intercepts are 0
and 1. Vertical asymptotes occur when x 3 3x 2 0. Since x 3 3x 2 0
when x 2, we can factor x 2 x 12 0, so the vertical asymptotes occur
at x 2 and x 1. Because the degree of the denominator and numerator are
the same, the horizontal asymptote is y 11 1. The domain is x x 1 2 and the range is R.
63. r x
x 2 4x 5 x 5 x 5 x 1 for x 1. When x 0, y 52 , 2 x 2 x 2 x 1 x x 2
1 x
1
y
so the y-intercept is 52 . When y 0, x 5, so the x-intercept is 5. Vertical asymptotes occur when x 2 0 x 2. Because the degree of the
denominator and numerator are the same, the horizontal asymptote is y 1. The
domain is x x 2 1 and the range is y y 1 2.
(1, 2) 1 0
1
x
354
CHAPTER 3 Polynomial and Rational Functions
64. r x
x 2 3x 10 x 2 x 2 x 5 for x 1 x 3 x 5 x 1 x 3 x 5 x 1 x 3
y
x 5. When x 0, we have y 23 , so the y-intercept is 23 . When y 0, we have x 2, so the x-intercept is 2. Vertical asymptotes occur when
1
x 1 x 3 0 x 1 or 3, so the vertical asymptotes occur at x 1 and x 3. Because the degree of the denominator is greater than that of the
numerator, the horizontal asymptote is y 0. The domain is x x 5 1 3
0
(_5, _ 327 )
1
x
and the range is R.
x 2 2x 3 x 3 x 1 x 3 for x 1. When x 0, x 1 x 1 y 3, so the y-intercept is 3. When y 0, x 3, so the x-intercept is 3.
y
65. r x
There are no asymptotes. The domain is x x 1 and the range is
1
y y 4.
0
1
x
(_1, _4)
x 3 2x 2 3x x x 1 x 3 x x 1 for x 3. When x 0, x 3 x 3 we have y 0, so the y-intercept is 0. When y 0, x 1 or 0, so the
y
66. r x
(3, 12)
x-intercepts are 1 and 0. There are no asymptotes. The domain is x x 3 and the range is y y 14 .
1 0
x 3 5x 2 3x 9 . We use synthetic division to check whether the x 1 denominator divides the numerator: 1 1 Thus, r x
x 2 6x 9 x 1
5
3
9
1
6
9
6
9
0
x 2 6x 9 x 32 for x 1.
x 1 When x 0, y 9, so the y-intercept is 9. When y 0, x 3, so the x-intercept is 3. There are no asymptotes. The domain is x x 1 and the range is
y y 0.
x
y
67. r x
1
1
(_1, 16)
2 0
1
x
SECTION 3.6 Rational Functions
x 1 x 2 4x 5 x 1 x 5 for x 5. x 0 is 68. r x 3 x x 2 x 5 x x 2 x 7x 2 10x not in the domain of r , so there is no y-intercept. When y 0, x 1, so the
355
y
x-intercept is 1. There are vertical asymptotes at x 2 and x 0. Because the
degree of the denominator is less than the degree of the numerator, y 0 is a horizontal asymptote. The domain is x x 5 2 0 and the range is y y 0134 or y 1866.
69. r x
x2 . When x 0, y 0, so the graph passes through the origin. There x 2
1 0
(_5, _ 327 )
1 x
y
is a vertical asymptote when x 2 0 x 2, with y as x 2 , and y as x 2 . Because the degree of the numerator is greater than the
degree of the denominator, there is no horizontal asymptotes. By using long
2
4 division, we see that y x 2 , so y x 2 is a slant asymptote. x 2 x x 2 x 2 2x . When x 0, we have y 0, so the graph passes x 1 x 1 through the origin. Also, when y 0, we have x 0 or 2, so the x-intercepts are
70. r x
2
x
1
x
y
2 and 0. The vertical asymptote is x 1. There is no horizontal asymptote, and the line y x 3 is a slant asymptote because by long division, we have y x 3
2 . x 1
x 2 2x 8 x 4 x 2 . The vertical asymptote is x 0, thus, x x there is no y-intercept. If y 0, then x 4 x 2 0 x 2, 4, so the
71. r x
x-intercepts are 2 and 4. Because the degree of the numerator is greater than the
degree of the denominator, there are no horizontal asymptotes. By using long
1
y
2 2
8 division, we see that y x 2 , so y x 2 is a slant asymptote. x
x 3 x 3x x 2 . When x 0, we have y 0, so the graph passes 2x 2 2 x 1 through the origin. Also, when y 0, we have x 0 or x 3, so the x-intercepts are 0 and 3. The vertical asymptote is x 1. There is no horizontal asymptote, and
72. r x
the line y 12 x 1 is a slant asymptote because by long division we have 1 y 12 x 1 . x 1
x
y
1 1
x
356
CHAPTER 3 Polynomial and Rational Functions
73. r x
x 2 5x 4 x 4 x 1 . When x 0, y 43 , so the y-intercept x 3 x 3
y
is 43 . When y 0, x 4 x 1 0 x 4, 1, so the two x-intercepts are 4 and 1. A vertical asymptote occurs when x 3, with y as x 3 , and y as x 3 . Using long division, we see that 28 y x 8 , so y x 8 is a slant asymptote. x 3
x3 4 x3 4 . When x 0, we have 2x 1 x 1 2x 2 x 1 04 4, so the y-intercept is 4. Since x 3 4 0 x 3 4, y 001 the x-intercept is x 3 4. There are vertical asymptotes where
74. r x
5 x
2
y
1 1
x
2x 1 x 1 0 x 12 or x 1. Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. By long division, we have y 12 x 14 slant asymptote.
75. r x
3 x 15 4 4 , so the line y 1 x 1 is a 2 4 2x 2 x 1
x3 x2 x 2 x 1 . When x 0, y 0, so the graph passes 2 x 2 x 2 x 4
through the origin. Moreover, when y 0, we have x 2 x 1 0 x 0, 1, so the x-intercepts are 0 and 1. Vertical asymptotes occur when x 2; as
x 2 , y and as x 2 , y . Because the degree of the numerator is greater than the degree of the denominator, there is no horizontal 4x 4 asymptote. Using long division, we see that y x 1 2 , so y x 1 is a x 4 slant asymptote.
2x x 2 1 2x 3 2x 76. r x 2 . When x 0, we have y 0, so the graph x 1 x 1 x 1
passes through the origin. Also, note that x 2 1 0, for all real x, so the only x-intercept is 0. There are two vertical asymptotes at x 1 and x 1. There is no horizontal asymptote, and the line y 2x is a slant asymptote because by long 4x division, we have y 2x 2 . x 1
y
2 1
x
1
x
y
2
SECTION 3.6 Rational Functions
77. f x
357
2x 2 6x 6 , g x 2x. f has vertical asymptote x 3. x 3 50 -10
-5
5 -20
20
-20
78. f x
-50
x 3 6x 2 5 , g x x 4. f has vertical asymptotes x 0 and x 2. x 2 2x 20
20
-5
-20
5
20
-20
79. f x
-20
x 3 2x 2 16 , g x x 2 . f has vertical asymptote x 2. x 2 50
50
-10
10
5 -50
80. f x
x 4 2x 3 2x x 12
-4
, g x 1 x 2 . f has vertical asymptote x 1.
-2
2
4
-4
-2
2
-5
-5
-10
-10
4
2x 2 5x has vertical asymptote x 15, x-intercepts 0 and 25, y-intercept 0, local maximum 39 104, 2x 3 12 . From the graph, we see that the end and local minimum 09 06. Using long division, we get f x x 4 2x 3 behavior of f x is like the end behavior of g x x 4.
81. f x
x 4
2x 3
2x 2 5x
50
20
2x 2 3x 8x
-10
8x 12 12
10
-20
20
-20 -50
358
CHAPTER 3 Polynomial and Rational Functions
x 4 3x 3 x 2 3x 3 has vertical asymptotes are x 0, x 3, x-intercept 082, and no y-intercept. The x 2 3x local minima are 080 263 and 338 1476. The local maximum is 256 488. By using long division, we see that 3 f x x 2 1 2 . From the second graph, we see that the end behavior of f x is the same as the end behavior x 3x 2 of g x x 1.
82. f x
x2
x 2 3x
1
20
x 4 3x 3 x 2 3x 3
50
x 4 3x 3
0x 3 x 2 3x
-5
x 2 3x
5
-5 -50
-20
3
5
x5 83. f x 3 has vertical asymptote x 1, x-intercept 0, y-intercept 0, and local minimum 14 31. x 1 x2 Thus y x 2 3 . From the graph we see that the end behavior of f x is like the end behavior of g x x 2 . x 1 x2 x3 1
10
x5
10
x5 x2 x2
-5
5
-5
5
-10
-10
Graph of f 84. f x
x4
Graph of f and g
has vertical asymptotes x 141, x-intercept 0, and y-intercept 0. The local maximum is 0 0. The
x2 2
4 . From the second graph, local minima are 2 8 and 2 8. By using long division, we see that f x x 2 2 2 x 2 2 we see that the end behavior of f x is the same as the end behavior of g x x 2. x2
x2 2
2
20
x 4 0x 3 0x 2 0x 0
x4
50
2x 2 2x 2 2x 2
4
-5
5 -5
4
5
x 4 3x 3 6 has vertical asymptote x 3, x-intercepts 16 and 27, y-intercept 2, local maxima 04 18 x 3 6 and 24 38, and local minima 06 23 and 34 543. Thus y x 3 . From the graphs, we see that the end x 3 behavior of f x is like the end behavior of g x x 3 .
85. f x
x3
x 3
x 4 3x 3 6
100
100
x 4 3x 3
6
-5
5 -100
-5
5 -100
SECTION 3.6 Rational Functions
359
4 x2 x4 x 4 x 2 4 has vertical asymptotes x 1, x-intercepts 16, and y-intercept 4. The local x 1 x 1 x2 1 maximum is 0 4 and there is no local minimum. 6 Thus y x 2 2 . From the graphs, we see that the end behavior of f x is like the end behavior of g x x 2 . x 1 x 2
86. r x
x2 1
20
x 4 0x 3 x 2 0x 4 x 4
x2
0
20
-5
4
5
-5
5
-20
-20
88. (a)
87. (a)
10
2000 0
0 0
20
40
0
3000 3000t 3000 . So as t , we t 1 t 1 have p t 3000.
(b) p t
5t 89. c t 2 t 1
10
20
30t (b) c t 2 . Since the degree of the denominator t 2 is larger than the degree of the numerator, c t 0 as t .
(a) The highest concentration of drug is 250 mg/L, and it is reached 1 hour after the drug is administered. (b) The concentration of the drug in the bloodstream goes to 0.
2
(c) From the first viewing rectangle, we see that an approximate solution
0 0
10
20
5t is near t 15. Thus we graph y 2 and y 03 in the t 1 viewing rectangle [14 18] by [0 05]. So it takes about 1661 hours for the concentration to drop below 03 mg/L. 0.4 0.2 0.0 14
64 106 2 90. Substituting for R and g, we have h . The vertical 2 98 64 106 2 asymptote is 11,000, and it represents the escape velocity from the earth’s gravitational pull: 11,000 m/s 1900 mi/h.
16
18
2e+7 1e+7 0 0
10000
360
CHAPTER 3 Polynomial and Rational Functions
91. P P0
s0 s0
P 440
332 332
4000
If the speed of the train approaches the speed of sound, the pitch of the whistle becomes very loud. This would be experienced as a “sonic boom”— an effect
2000
seldom heard with trains. 0 0
92. (a)
1 1 1 1 1 1 xF xF 1 y . Using x y F y F x y xF xF
200
F 55, we get y
100
55x . Since y 0, we use the viewing x 55 rectangle [0 1000] by [0 250].
200
400
0
(b) y approaches 55 millimeters.
0
500
1000
(c) y approaches . 2x 1 . Vertical asymptote x 3 and horizontal asymptote y 2: r x . x 3 x 3 x 4 Vertical asymptotes x 1 and x 1, horizontal asymptote 0, and x-intercept 4: q x . Of course, x 1 x 1
93. Vertical asymptote x 3: p x
other answers are possible.
x 6 10 94. r x 4 has no x-intercept since the numerator has no real roots. Likewise, r x has no vertical asymptotes, x 8x 2 15 since the denominator has no real roots. Since the degree of the numerator is two greater than the degree of the denominator, r x has no horizontal or slant asymptotes. 1 1 95. (a) Let f x 2 . Then r x f x 2. From this form we see that x x 22 the graph of r is obtained from the graph of f by shifting 2 units to the right. Thus
y
r has vertical asymptote x 2 and horizontal asymptote y 0.
1 x
1
3 2x 2 4x 5 2 3 f x 1 2. From this form we see that the graph of s is obtained from 2 x 2x 1 x 12 the graph of f by shifting 1 unit to the left, stretching vertically by a factor of 3, and shifting 2 units vertically. Thus r has vertical asymptote x 1 and horizontal asymptote y 2.
(b) s x
y
2 x 2 2x 1
2x 2 2x 2
4x
5
4x
2 3 1 1
x
SECTION 3.7 Polynomial and Rational Inequalities
361
2 3x 2 12x 14 (c) Using long division, we see that p x 2 3 2 which cannot be graphed by transforming x 4x 4 x 4x 4 1 f x 2 . Using long division on q we have: x 3
x 2 4x 4
3x 2 3x 2
0x
2
12x
12
12x
14
3
x 2 4x 4
3x 2 3x 2
12x
0
12x
12 12
12x 3x 2 12 So q x 2 3 12 f x 2 3. From this form we see that the graph of q is obtained x 4x 4 x 22 from the graph of f by shifting 2 units to the right, stretching vertically by a factor of 12, and then shifting 3 units vertically down. Thus the vertical asymptote is x 2 and the horizontal asymptote is y 3. We show y p x just 1 to verify that we cannot obtain p x from y 2 . x
y 1 1
y
1
1
x
y q x
3.7
x
y p x
POLYNOMIAL AND RATIONAL INEQUALITIES
1. To solve a polynomial inequality, we factor the polynomial into irreducible factors and find all the real zeros of the polynomial. Then we find the intervals determined by the real zeros and use test points in each interval to find the sign of the polynomial on that interval. 2
2 0
0 1
1
x
x 2
x 1
Sign of
P x x x 2 x 1
From the table, we see that P x 0 on the intervals [2 0] and [1 .
362
CHAPTER 3 Polynomial and Rational Functions
2. To solve a rational inequality, we factor the numerator and the denominator into irreducible factors. The cut points are the real zeros of the numerator and the real zeros denominator. Then we find the intervals determined by the cut points, and we use test points to find the sign of the rational function on each interval. Sign of
4
4 2
2 1
1 3
3
x 2
x 1
x 3
x 4
x 2 x 1 r x x 3 x 4 From the table, we see that r x 0 on the intervals 4, [2 1], and 3 .
3. The inequality x 3 x 5 2x 5 0 already has all terms on one side and the polynomial is factored. The intervals determined by the zeros 3, 5, and 52 are 5, 5 52 , 52 3 , and 3 . We make a sign diagram: 5 52 52 3 Sign of 5 3 x 3
x 5
2x 5
P x x 3 x 5 2x 5
None of the endpoints satisfies the inequality. The solution is 5 52 3 .
4. The inequality x 1 x 2 x 3 x 4 0 already has all terms on one side and the polynomial is factored. The intervals determined by the zeros 1, 2, 3, and 4 are 4, 4 2, 2 1, 1 3, and 3 . We make a sign diagram: Sign of
4
4 2
2 1
1 3
3
x 1
x 2
x 3
x 4
P x x 1 x 2 x 3 x 4
All of the endpoints satisfy the inequality. The solution is 4] [2 1] [3 .
5. The inequality x 52 x 3 x 1 0 already has all terms on one side and the polynomial is factored. The intervals determined by the zeros 5, 3, and 1 are 5, 5 3, 3 1, and 1 . We make a sign diagram: 5
5 3
3 1
1
x 3
x 1
Sign of x 52
P x x 52 x 3 x 1
None of the endpoints satisfies the inequality. The solution is 5 5 3 1 .
SECTION 3.7 Polynomial and Rational Inequalities
363
6. The inequality 2x 74 x 13 x 1 0 already has all terms on one side and the polynomial is factored. The intervals determined by the zeros 72 , 1, and 1 are 1, 1 1, 1 72 , and 72 . We make a sign diagram: 7 1 72 Sign of 1 1 1 2 2x 74 x 13 x 1
P x 2x 74 x 13 x 1 All of the endpoints satisfy the inequality [note that P 72 0]. The solution is [1 1] 72 .
7. We start by moving all terms to one side and factoring: x 3 4x 2 4x 16 x 3 4x 2 4x 16 x 4 x 2 x 2 0. The intervals determined by the zeros 4, 2, and 2 are 4, 4 2, 2 2, and 2 . We make a sign diagram: 4
4 2
2 2
2
x 4
x 2
x 2
Sign of
P x x 4 x 2 x 2
All of the endpoints satisfy the inequality. The solution is [4 2] [2 .
8. We start by moving all terms to one side and factoring: 2x 3 18x x 2 9 2x 3 x 2 18x 9 x 3 2x 1 x 3 0. The intervals determined by the zeros 3, 12 , and 3 are 3, 3 12 , 12 3 , and 3 . We make a sign diagram:
3 12
13 2
3
Sign of
3
x 3
2x 1
x 3
P x x 3 2x 1 x 3
None of the endpoints satisfies the inequality. The solution is 3
13 . 2
9. We start by moving all terms to one side and factoring: 2x 3 x 2 918x 2x 3 x 2 18x 9 2x 1 x 2 9 0. Note that x 2 9 0 for all x, so the sign of P x 2x 1 x 2 9 is negative where 2x 1 is negative and positive where 2x 1 is positive. The endpoint x 12 does not satisfy the inequality, so the solution is 12 .
364
CHAPTER 3 Polynomial and Rational Functions
10. We start by moving all terms to one side and factoring: x 4 3x 3 x 3 x 4 3x 3 x 3 0. The possible rational zeros of P x x 4 3x 3 x 3 are 1 and 3.
1 1 3 1 3 1
4
3
1 4 3 0 Thus, P x x 1 x 3 4x 2 4x 3 x 1 x 3 x 2 x 1 . The last factor is positive everywhere, so
we test the intervals 3, 3 1, and 1 . Sign of
3
3 1
1
x 3
x 1
P x x 1 x 3 x 2 x 1
Neither of the endpoints satisfies the inequality. The solution is 3 1.
11. All the terms are on the left size. We factor, using the substitution t x 2 : x 4 7x 2 18 0 t 2 7t 18 t 2 t 9 0 x 2 2 x 2 9 0 x 2 2 x 3 x 3 0. The first factor is positive everywhere, so we test 3, 3 3, and 3 : Sign of
3
3 3
3
x 3
x 3
P x x 2 2 x 3 x 3
Neither of the endpoints satisfies the inequality. The solution is 3 3.
12. All the terms are on the left size. We factor, using the substitution t x 2 : 4x 4 25x 2 36 0 4t 2 25t 36 4t 9 t 4 0 4x 2 9 x 2 4 2x 3 2x 3 x 2 x 2 0. The zeros are 32 and 2, so we test 2, 2 32 , 32 32 , 32 2 , and 2 : 32 2 32 32 32 Sign of 2 2 2 x 2
x 32 x 32 x 2 P x
All of the endpoints satisfy the inequality. 2 32 32 2 .
SECTION 3.7 Polynomial and Rational Inequalities
365
13. All the terms are on the left size. To factor, note that the possible rational zeros of P x x 3 x 2 17x 15 are 1, 3, 5, 15. 1 1 1 17 1
15
2 15
1 2 15 0 Thus, P x x 1 x 2 2x 15 x 5 x 1 x 3. The zeros are 5, 1, and 3, so we test 5,
5 1, 1 3, and 3 :
5
5 1
1 3
3
x 5
x 1
x 3
Sign of
P x
All of the endpoints satisfy the inequality P x 0, so the solution is [5 1] [3 .
14. All the terms are on the left size. To factor, note that the possible rational zeros of P x x 4 3x 3 3x 2 3x 4 are 1, 2, 4. 1 1 3 3 3 4 1
4 1
4
1 4
1 4
0
Thus, P x x 1 x 3 4x 2 x 4 x 1 x 4 x 2 1 . The last factor is positive everywhere, so we test 4, 4 1, and 1 :
Sign of
4
4 1
1
x 4
x 1
P x
None of the endpoints satisfies P x 0, so the solution is 4 1.
3 3 3 15. We start by moving all terms to one side and factoring: x 1 x 2 7 1 x 2 x 7 1 x 2 0
P x x 7 1 x3 1 x3 0. The intervals determined by the zeros 1, 1, and 7 are 1, 1 1, 1 7, and 7 . We make a sign diagram: Sign of
1
1 1
1 7
7
x 7
1 x3 1 x3 P x
None of the endpoints satisfies the inequality. The solution is 1 1 7.
366
CHAPTER 3 Polynomial and Rational Functions
16. We start by moving all terms to one side and factoring: x 2 7 6x 1 6x 3 7x 2 1 0. The possible rational zeros of P x 6x 3 7x 2 1 are 1, 12 , 13 , 16 .
1 6
7 0 1 6 1
1
6 1 1 0 Thus, P x x 1 6x 2 x 1 1 x 2x 1 3x 1. The intervals determined by the zeros 13 , 12 , and 1 13 , 13 12 , 12 1 , and 1 . We make a sign diagram: 11 13 13 12 Sign of 1 2 3x 1
2x 1
1x
P x
All of the endpoints satisfy the inequality P x 0. The solution is 13 12 [1 .
x 1 0. Since all nonzero terms are already on one side of the inequality symbol and there is no factoring x 10 needed, we find the intervals determined by the cut points 1 and 10. These are 1, 1 10, and 10 . We make a sign diagram:
17. r x
Sign of
1
1 10
10
x 1
x 10
r x
The cut point 1 does not satisfy the inequality, and the cut point 10 is not in the domain of r. Thus, the solution is 1 10.
3x 7 0. Since all nonzero terms are already on one side of the inequality symbol and there is no factoring x 2 needed, we find the intervals determined by the cut points 73 and 2. These are 2, 2 73 , and 73 . We
18. r x
make a sign diagram:
2 73
2
3x 7
x 2
r x
7 3
Sign of
The cut point 73 satisfies equality, but the cut point 2 is not in the domain of r. Thus, the solution is 2 73 .
SECTION 3.7 Polynomial and Rational Inequalities
19. r x
2x 5
x 2 2x 35 2x 5
0.
367
Since all nonzero terms are already on one side, we factor:
2x 5 . Thus, the cut points are 7, 52 , and 5. The intervals determined by these x 7 x 5 points are 7, 7 52 , 52 5 , and 5 . We make a sign diagram: 7 52 52 5 Sign of 7 5
r x
x 2 2x 35
2x 5
x 7
x 5
r x
The cut point 52 satisfies equality, but the cut points 7 and 5 are not in the domain of r. Thus, the solution is 7 52 5 .
4x 2 25 4x 2 25 2x 5 2x 5 . 0. Since all nonzero terms are already on one side, we factor: r x 2 2 x 3 x 3 x 9 x 9 Thus, the cut points are 52 and 3. The intervals determined by these points are 3, 3 52 , 52 52 , 52 3 ,
20. r x
and 3 . We make a sign diagram:
3 52
52 52
x 3
x 3
Sign of
3
2x 5
2x 5
r x
53 2
3
The cut points 52 satisfy equality, but the cut points 3 are not in the domain of r. Thus, the solution is 3 52 52 3 . x 0. Since all nonzero terms are already on one side but the denominator cannot be factored, we x 2 2x 2 2 22 4 1 2 2 1 3. Thus, the cut points are use the Quadratic Formula: x 2x 2 0 x 2 1 3, 0, and 1 3. The intervals determined by these points are 1 3 , 1 3 0 , 0 3 1 , and 3 1 . We make a sign diagram: 1 3 1 3 0 0 3 1 Sign of 3 1
21. r x
x
x 2 2x 2
r x
The cut point 0 satisfies equality, but the cut points 1 3 are not in the domain of r. Thus, the solution is 1 3 0 3 1 .
368
CHAPTER 3 Polynomial and Rational Functions
x 1
0. Since all nonzero terms are already on one side but the denominator cannot be factored, we 4 42 4 2 1 2 2 use the Quadratic Formula: 2x 2 4x 1 0 x . Thus, the cut points are 2 2 2 1, 1 22 , and 1 22 . The intervals determined by these points are 1, 1 1 22 , 1 22 1 22 , and 1 22 . We make a sign diagram: 1 1 22 1 22 1 22 1 22 Sign of 1
22. r x
2x 2 4x 1
x 1
x 1 22 x 1 22
r x
None of the cut points satisfies the strict inequality, so the solution is 1 1 22 1 22 .
x 2 2x 3 x 3 x 1 0. The intervals determined by the cut points are 3, 3 23 , 23. r x 2 3x 2 x 3 3x 7x 6 2 3 1 , 1 3, and 3 . 3 23 23 1 Sign of 3 1 3 3 x 3
3x 2
x 1
x 3
r x
None of the cut points satisfies the strict inequality, so the solution is 3 23 1 3 .
x 1 x 1 0. The second factor in the denominator is positive for all x, so the intervals 24. r x 3 x 1 x 1 x 2 x 1 determined by the cut points are 1, 1 1, and 1 . Sign of
1
1 1
1
x 1
x 1
Sign of
4
4 3
3
x 4
x 3
r x
The cut point 1 satisfies equality, but 1 is not in the domain of r. Thus, the solution is 1 [1 . x x 2 6x 9 3 x 2 6x 9 x 3 3x 2 9x 27 x 3 x 32 25. r x 0. The second factor in the x 4 x 4 x 4 numerator is positive for all x, so the intervals determined by the cut points are 4, 4 3, and 3 .
r x
The cut point 3 satisfies equality, but 4 is not in the domain of r. Thus, the solution is 4 3].
SECTION 3.7 Polynomial and Rational Inequalities
369
x 2 16 x 4 x 4 x 4 x 4 0. The last factor in the denominator is positive for 26. r x 4 2 x 4 x2 4 x 16 x 2 x 2 x 2 4 all x, so the intervals determined by the cut points are 4, 4 2, 2 2, 2 4, and 4 . Sign of
4
4 2
2 2
2 4
4
x 4
x 2
x 2
x 4
r x
None of the cut points satisfy the strict inequality. Thus, the solution is 4 2 2 4.
x 3 x 3 2x 5 x 3 1 1 0 0 2x 5 2x 5 2x 5 x 8 r x 0. The intervals determined by the cut points are 8, 8 52 , and 52 . 2x 5 8 52 52 Sign of 8
27. We start by moving all terms to one side and simplifying:
x 8
2x 5
r x
The cut point 8 satisfies equality, but 52 is not in the domain of r. Thus, the solution is 8 52 .
1 1 2 1 1 2 0 x x 1 x 2 x x 1 x 2 x 2 3x 2 x 2 2x 2x 2 2x x 1 x 2 x x 2 2x x 1 0 0 x x 1 x 2 x x 1 x 2 3x 2 r x 0. The intervals determined by the cut points are 2, 2 1, 1 23 , 23 0 , x x 1 x 2 and 0 . 1 23 23 0 Sign of 2 2 1 0
28. We start by moving all terms to one side and simplifying:
x 2
x 1
3x 2
x
r x
None of the cut points satisfies the strict inequality. Thus, the solution is 2 1 23 0 .
370
CHAPTER 3 Polynomial and Rational Functions
3 1 3 1 2 0 1x x 1x x 2x 2 6x 3 2x 2 6x 3 2 1 x x x 3 1 x 0 r x 0. The numerator is 0 when x 1 x x 1 x x x 1 6 62 4 2 3 3 3 3 3 x , so the intervals determined by the cut points are 0, 0 , 2 2 2 2 3 3 3 3 3 3 1 , 1 , and . 2 2 2 3 3 3 3 3 3 3 3 0 1 1 Sign of 0 2 2 2 2
29. We start by moving all terms to one side and simplifying: 2
x
x 1
2x 2 6x 3
r x
3 3 satisfy equality, but 0 and 1 are not in the domain of r, so the solution is The cut points 2 3 3 3 3 1 . 0 2 2
1 1 2x 2 x 3 x 2 x x 2 1 1 2x x 2 x 1 x 3 x 1 2x x 3 0 0 x 3 x 2 x 2 x 1 x 2 x 1 x 3 3x 1 0. The intervals determined by the cut points are 2, 2 13 , 13 1 , 1 3, and x 2 x 1 x 3 3 . 2 13 13 1 Sign of 2 1 3 3
30. We start by moving all terms to one side and factoring:
x 2
3x 1
x 1
x 3
r x
The cut point 13 satisfies equality, but 2, 1, and 3 are not in the domain of r, so the solution is 2 13 1 3 .
SECTION 3.7 Polynomial and Rational Inequalities
371
x 12 0. The numerator is nonnegative for all x, but note that x 1 fails to satisfy the strict inequality. x 1 x 2 The intervals determined by the cut points are 2, 2 1, and 1 .
31. r x
2
2 1
1
x 2
x 1
(except at x 1)
Sign of x 12
r x
The cut points 2 and 1 are not in the domain of r, so the solution is 2 1 1 1 . 32. r x 1 .
x 2 2x 1 x 12 0. The intervals determined by the cut points are 1, 1 1, and x 3 3x 2 3x 1 x 13 Sign of x 13
1
1 1
1
x 12
r x
The cut point 1 is not in the domain of r and the cut point 1 satisfies the inequality. Thus, the solution is 1] 1. 6 6 6 6 1 1 0 x 1 x x 1 x 6x 6 x 1 x x 1 x2 x 6 x 2 x 3 0 0 r x 0. The intervals determined by the x x 1 x x 1 x x 1 cut points are 2, 2 0, 0 1, 1 3, and 3 .
33. We start by moving all terms to one side and factoring:
Sign of
2
2 0
0 1
1 3
3
x 2
x
x 1
x 3
r x (note negative sign)
The cut points 0 and 1 are inadmissible in the original inequality, so the solution is [2 0 1 3]. 5 x 5 x 4 4 0 2 x 1 2 x 1 x 2 7x 18 x x 1 5 2 4 2 x 1 x 2 x 9 0 0 0. The intervals determined by the cut 2 x 1 2 x 1 2 x 1 points are 2, 2 1, 1 9, and 9 .
34. We start by moving all terms to one side and factoring:
Sign of
2
2 1
1 9
9
x 2
x 1
x 9
r x
The cut point 1 is inadmissible in the original inequality, so the solution is [2 1 [9 .
372
CHAPTER 3 Polynomial and Rational Functions
x 1 x 2 x 1 x 2 0 35. We start by moving all terms to one side and factoring: x 3 x 2 x 3 x 2 2 x 12 x 2 x 2 x 1 x 3 0 r x 0. The intervals determined by the cut points are x 3 x 2 x 3 x 2 3, 3 12 , 12 2 , and 2 . 3 12 12 2 Sign of 3 2 x 3
x 12 x 2
r x (note negative sign)
The cut points fail to satisfy the strict inequality, so the solution is 3 12 2 .
1 1 1 1 1 1 0 x 1 x 2 x 3 x 1 x 2 x 3 x 2 6x 7 x 2 x 3 x 1 x 3 x 1 x 2 0 r x 0. The numerator is x 1 x 2 x 3 x 1 x 2 x 3 6 62 4 1 7 3 2, so the intervals determined by the cut points are 3 2 , 0 when x 2 1 3 2 3 , 3 2, 2 3 2 , 3 2 1 , and 1 . 3 2 3 2 3 2 3 2 3 2 1 Sign of 3 2 1
36. We start by moving all terms to one side and factoring:
x 2 6x 7
x 3
x 2
x 1
r x
The cut points 3, 2, and 1 are not in the domain, so the solution is 3 2 3 2 3 2 1 .
37. The graph of f lies above that of g where f x g x; that is, where x 2 3x 10 x 2 3x 10 0 x 2 x 5 0. We make a sign diagram: 2
2 5
5
x 2
x 5
Sign of
x 2 x 5
Thus, the graph of f lies above the graph of g on 2 and 5 .
SECTION 3.7 Polynomial and Rational Inequalities
1 1 1 1 x 1 x 0 0 x x 1 x x 1 x x 1
38. The graph of f lies above that of g where f x g x; that is, where
1 . We make a sign diagram: x x 1 0
0 1
1
x 1 1 x x 1 Thus, the graph of f lies above the graph of g on 0 1.
Sign of x
39. The graph of f lies above that of g where f x g x; that is, where 4x r x
2x 1 2x 1 0. We make a sign diagram: x 12 12 0 Sign of
1 4x 2 1 1 4x 0 0 x x x
0 12
x
2x 1
Thus, the graph of f lies above the graph of g on 12 0 and 12 .
1 2
2x 1
r x
373
x3 x2 2 2 2 40. The graph of f lies above that of g where f x g x; that is, where x 2 x x 2 x 0 0 x x x x 1 x 2 2x 2 0. The second factor in the numerator is positive for all x. We make a sign diagram: r x x 0
0 1
1
x
x 1
Sign of
r x
Thus, the graph of f lies above the graph of g on 0 and 1 .
41. f x
6 x x 2 is defined where 6 x x 2 x 2 x 3 0. We make a sign diagram: Sign of
2
2 3
3
x 2
x 3
x 2 x 3 Thus, the domain of f is [2 3].
374
CHAPTER 3 Polynomial and Rational Functions
42. g x
5x 5x is defined where 0 and 5 x 0. We make a sign diagram: 5x 5x 5
5 5
5
5x 5x 5x The cut point 5 is permissible, and so the domain of g is [5 5.
Sign of 5x
43. h x
4
x 4 1 is defined where x 4 1 x 1 x 1 x 2 1 0. The last factor is positive for all x. We make a
sign diagram:
1
1 1
1
x 1
x 1
x 1 x 1 x 2 1
Sign of
x2 1
Thus, the domain of h is 1] [1 .
44. f x
1
x 4 5x 2 4 make a sign diagram:
is defined where x 4 5x 2 4 x 2 4 x 2 1 x 2 x 2 x 1 x 1 0. We 2
2 1
1 1
1 2
2
x 2
x 1
x 1
x 2
Sign of
x 4 5x 2 4
Thus, the domain of h is 2 1 1 2 .
46.
45.
-4
20
20
10
10
-2
2
4
-10
From the graph, we see that x 3 2x 2 5x 6 0 on [2 1] [3 .
-4
-2
-10
2
-20
From the graph, we see that 2x 3 x 2 8x 4 0 on 2] 12 2 .
SECTION 3.7 Polynomial and Rational Inequalities
375
48.
47.
-2
-1
4
4
2
2 1
-2
2
-2
-4
-1 -2
1
2
3
4
-4
From the graph, we see that 2x 3 3x 1 0 on
From the graph, we see that x 4 4x 3 8x 0 on
approximately 137 037 1.
approximately 124 0 2 324 .
49.
50. 20
40
10 20 -2 -1
0
1
1
2
-10
From the graph, we see that 5x 4 8x 3 on 0 85 . 51.
-1
2
From the graph, we see that x 5 x 3 x 2 6x on
approximately [131 0] [151 .
1 x2 x 4 x x 1 1 x2 1 x2 4 x x 1 4 x x 1 0 0 x x x
1 2x x 2 4x 2 4x 3x 2 2x 1 1 x 3x 1 0 0 r x 0. The domain of r is 0 , and x x x both 3x 1 and x are positive there. 1 x 0 for x 1, so the solution is 0 1]. 7x 8 52. 23 x 13 x 212 12 x 23 x 212 0 16 x 212 x 13 [4 x 2 3x] 0 r x 0. 63 x x 2 Note that the domain of r is 2 . We make a sign diagram with cut points 87 and 0: 2 87 87 0 Sign of 0 7x 8 3 x x 2
Neither cut point is a solution. The solution is 87 0 .
r x
53. We want to solve P x x a x b x c x d 0, where a b c d. We make a sign diagram with cut points a, b, c, and d: a
a b
b c
c d
d
x a
x b
x c
x d
Sign of
P x
Each cut points satisfies equality, so the solution is a] [b c] [d .
376
CHAPTER 3 Polynomial and Rational Functions
[x a] x b x 2 a b x ab x a x b 0. Note that x c x c x c 0 a c, so c a 0. We make a sign diagram with cut points c, a, and b:
54. Factoring the numerator, we have r x Sign of
x c x b
x a P x
c
c a
a b
b
The cut points a and b satisfy equality, but c is not in the domain of P. Thus, the solution is c [a b]. 500,000 500,000 55. We want to solve the inequality T x 300 2 300 2 300 0 x 400 x 400 2 500,000 300 x 2 400 300 3800 300x 2 380,000 3 x 0 0 . Because x represents distance, it is x 2 400 x 2 400 x 2 400 3800 356. The positive, and the denominator is positive for all x. Thus, the inequality holds where x 2 3800 3 x 3 temperature is below 300 C at distances greater than 356 meters from the center of the fire.
2 175 0 2 25 4375 0. Using the Quadratic Formula, we find 56. We want to solve d 175 25 25 252 4 1 4375 25 18,125 25 18,125 . Because is positive, we have 548, so 2 1 2 2 Kerry can travel at up to 548 mih. 88x 57. We graph N x x 2 and y 40 in the viewing rectangle 60 17 17 20 40
[0 100] by [0 60], and see that N x 40 for approximately 95 x 423.
20
Thus, cars can travel at between 95 and 423 mih.
0 0
50
100
20000
We graph S x 8x 08x 2 0002x 3 4000 and y 12,000 in the viewing
10000
approximately 183 x 344. Thus, profits are above $12,000 when between 183
58.
rectangle [0 400] by [10000 20000], and see that S x 12,000 for and 344 units are produced.
0 100 200 300 400 -10000
CHAPTER 3
Review
377
CHAPTER 3 REVIEW 1. (a) f x x 2 6x 2 x 2 6x 2 x 2 6x 9 2 9
2. (a) f x 2x 2 8x 4 2 x 2 4x 4 2 x 2 4x 4 4 8 2 x 22 4
x 32 7
y
(b)
y
(b)
1 _1
3. (a) f x 1 10x x 2 x 2 10x 1 x 2 10x 25 1 25 x 52 26
(b)
1
0 x
y
0
1
x
4. (a) f x 2x 2 12x 2 x 2 6x 2 x 2 6x 9 18 2 x 32 18
(b)
y
10 0
1 x
5 0
1
x
2 5. f x x 2 3x 1 x 2 3x 1 x 2 3x 94 1 94 x 32 54 has the maximum value 54 when x 32 .
6. f x 3x 2 18x 5 3 x 2 6x 5 3 x 2 6x 9 5 27 3 x 32 22 has the minimum value 22 when x 3.
7. We write the height function in standard form: h t 16t 2 48t 32 16 t 2 3t 32 16 t 2 3t 94 2 32 36 16 t 32 68. The stone reaches a maximum height of 68 ft. 8. We write the profit function in standard form:
P x 1500 12x 0004x 2 0004 x 2 3000x 1500 0004 x 2 3000x 15002 1500 0004 15002 0004 x 15002 7500
Thus, the maximum profit of $7500 is achieved when 1500 units are sold.
378
CHAPTER 3 Polynomial and Rational Functions
9. P x x 3 64
10. P x 2x 3 16 2 x 2 x 2 2x 4
y
y
5
(0, 64)
(2, 0)
x
1 (0, _16)
10
(4, 0) 1
11. P x 2 x 14 32
x
12. P x 81 x 34 y
y
5
25
(1, 0) x
(_3, 0)
(0, 0)
(6, 0) 1
x
(0, _30)
13. P x 32 x 15
14. P x 3 x 25 96
y
y
(0, 31) 10
(_1, 0)
1
x
20 (0, 0)
1
15. (a) P x x 3 x 1 x 5 x 3 7x 2 7x15 has odd degree and a positive leading coefficient, so
16. (a) P x x 5 x 2 9 x 2
x 4 3x 3 19x 2 27x 90
y as x and y as x . (b)
x
has even degree and a negative leading coefficient, so
y
y as x and y as x . (b)
y 20
10 1
1 x
x
CHAPTER 3
17. (a) P x x 12 x 4 x 22 x 5 2x 4 11x 3 8x 2 20x 16
379
18. (a) P x x 2 x 2 4 x 2 9 x 6 13x 4 36x 2 has even degree and a positive leading coefficient, so y as x and y as x .
has odd degree and a negative leading coefficient, so y as x and y as x .
Review
y
(b)
y
(b)
100
20 x
1 5
_5
x
_100
19. (a) P x x 3 x 22 . The zeros of P are 0 and 2, with multiplicities 3 and 2, respectively.
(b) We sketch the graph using the guidelines on page 295.
20. (a) P x x x 13 x 12 . The zeros of P are 1, 0, and 1, with multiplicities 3, 1, and 2, respectively.
(b) We sketch the graph using the guidelines on page 295.
y
y
1
1
x
0.1 1
21. P x x 3 4x 1. x-intercepts: 21, 03, and 19. y-intercept: 1. Local maximum is 12 41. Local
minimum is 12 21. y as x ; y as x .
22. P x 2x 3 6x 2 2. x-intercepts: 05, 07, and 29. y-intercept: 2. Local maximum is 2 6. Local
minimum is 0 2. y as x and y as x .
4 2 -2
-2 -4
5 2
x
-2
2 -5
4
380
CHAPTER 3 Polynomial and Rational Functions
23. P x 3x 4 4x 3 10x 1. x-intercepts: 01 and 21. y-intercept: 1. Local maximum is 14 145.
24. P x x 5 x 4 7x 3 x 2 6x 3. x-intercepts: 30, 13, and 19. y-intercept: 3. Local maxima are
24 332 and 05 50. Local minima are 06 06
There is no local maximum. y as x .
and 16 16. y as x and y as x .
20 10 -1-10 -20
1
2
40
3
20 -4
25. (a) Use the Pythagorean Theorem and solving for y 2 we have, x 2 y 2 102 y 2 100 x 2 . Substituting we get S 138x 100 x 2 1380x 138x 3 .
-2
2
-20
26. (a) The area of the four sides is 2x 2 2x y 1200
600 x 2 2x y 1200 2x 2 y . Substituting x 600 x 2 600x x 3 . we get V x 2 y x 2 x
(b) Domain is [0 10]. (c)
(b) 5000
4000 2000 0 0
5
0
10
0
(d) The strongest beam has width 58 inches.
27.
x 2 5x 2 x 3 1 1
5
2
3
6
2
4
2x 3 x 2 3x 4 x 5
2
1
3
4
10
55
290
11
58
294
1
5
6
10
5
5
Using synthetic division, we see that Q x 3x 5 and R x 5.
30. 2
3 3
R x 4.
5
3x 2 x 5 x 2 2
Using synthetic division, we see that Q x x 2 and
29.
20
(c) V is maximized when x 1414, y 2828.
28. 3
10
x 3 2x 4 x 7 7
1 1
0
2
4
7
49
329
7
47
325
Using synthetic division, we see that
Using synthetic division, we see that
Q x 2x 2 11x 58 and R x 294.
Q x x 2 7x 47 and R x 325.
CHAPTER 3
31.
x 4 8x 2 2x 7 x 5
32.
1
5
1
33.
0
8
2
7
5
25
85
415
5
17
83
422
Review
2x 4 3x 3 12 x 4 4
2 2
3
0
0
12
8
20
80
320
5
20
80
308
Using synthetic division, we see that
Using synthetic division, we see that
Q x x 3 5x 2 17x 83 and R x 422.
Q x 2x 3 5x 2 20x 80 and R x 308.
2x 3 x 2 8x 15 x 2 2x 1
34. 2x
x 2 2x 1
2x 3
x 4 2x 2 7x x2 x 3
x2
3 x 2 8x 15
2x 3 4x 2 2x
381
x2 x 3
x
4
x 4 0x 3 2x 2 7x 0
x 4 x 3 3x 3
3x 2 6x 15
x 3 5x 2 7x
3x 2 6x 3
x 3 x 2 3x
4x 2 4x 0
12
4x 2 4x 12
Therefore, Q x 2x 3, and R x 12.
12 Therefore, Q x x 2 x 4, and R x 12.
35. P x 2x 3 9x 2 7x 13; find P 5. 5
2 2
9
7
13
10
5
10
1
2
3
36. Q x x 4 4x 3 7x 2 10x 15; find Q 3 3
1 1
Therefore, P 5 3.
4
7
10
15
3
3
12
6
1
4
2
21
By the Remainder Theorem, we have Q 3 21.
37. The remainder when dividing P x x 500 6x 101 x 2 2x 4 by x 1 is P 1 1500 6 1201 12 2 14 8.
38. The remainder when dividing P x x 101 x 4 2 by x 1 is P 1 1101 14 2 0. 39. 12 is a zero of P x 2x 4 x 3 5x 2 10x 4 if P 12 0. 1 2
2 2
Since P
1 2
1
5
10
4
1
1
2
4
2
4
8
0
0, 12 is a zero of the polynomial.
40. x 4 is a factor of
P x x 5 4x 4 7x 3 23x 2 23x 12 if P 4 0. 4
1 1
4
7
23
23
12
4
0
28
20
12
0
7
5
3
0
Since P 4 0, x 4 is a factor of the polynomial.
41. (a) P x x 5 6x 3 x 2 2x 18 has possible rational zeros 1, 2, 3, 6, 9, 18.
(b) Since P x has 2 variations in sign, there are either 0 or 2 positive real zeros. Since P x x 5 6x 3 x 2 2x 18 has 3 variations in sign, there are 1 or 3 negative real zeros.
42. (a) P x 6x 4 3x 3 x 2 3x 4 has possible rational zeros 1, 2, 4, 12 , 13 , 23 , 43 , 16 .
(b) Since P x has no variations in sign, there are no positive real zeros. Since P x 6x 4 3x 3 x 2 3x 4 has 4 variations in sign, there are 0, 2, or 4 negative real zeros.
382
CHAPTER 3 Polynomial and Rational Functions
43. (a) P x 3x 7 x 5 5x 4 x 3 8 has possible rational zeros 1, 2, 4, 8, 13 , 23 , 43 , 83 .
(b) Since P x has 2 variations in sign, there are either 0 or 2 positive real zeros. Since P x 3x 7 x 5 5x 4 x 3 8 has 3 variations in sign, there are 1 or 3 negative real zeros.
44. (a) P x 6x 10 2x 8 5x 3 2x 2 12 has possible rational zeros 1, 2, 3, 4, 6, 12, 12 , 13 , 16 , 23 , 32 , 43 .
(b) Since P x has 2 variations in sign, there are either 0 or 2 positive real zeros. Since P x 6x 10 2x 8 5x 3 2x 2 12 has 2 variations in sign, there are 0 or 2 negative real zeros.
45. (a) P x x 3 16x x x 2 16
46. (a) P x x 3 3x 2 4x
x x 1 x 4
x x 4 x 4
has zeros 0, 1, and 4 (all of multiplicity 1).
has zeros 4, 0, 4 (all of multiplicity 1).
y
(b)
y
(b)
2
5
47. (a) P x x 4 x 3 2x 2 x 2 x 2 x 2 x 2 x 2 x 1
48. (a) P x x 4 5x 2 4 x 2 4 x 2 1 x 2 x 2 x 1 x 1
Thus, the zeros are 1, 1, 2, 2 (all of multiplicity
The zeros are 0 (multiplicity 2), 2 (multiplicity 1),
1).
and 1 (multiplicity 1).
(b)
(b)
y
x
1
x
1
y
5
1 1
x
1
x
CHAPTER 3
Review
383
49. (a) P x x 4 2x 3 7x 2 8x 12. The possible rational zeros are 1, 2, 3, 4, 6, 12. P has 2 variations in sign, so it has either 2 or 0 positive real zeros. 1
1 1
2
7
8
12
1
1
8
0
1
8
0
12
1
2
2
2
7
8
12
2
0
14
12
0
7
6
0
x 2 is a root.
P x x 4 2x 3 7x 2 8x 12 x 2 x 3 7x 6 . Continuing: 2 1 0 6
6
2
4
3
9
6
1 2 2 10
1 3
2
0
4
3 1 0 7 6
(b)
y
so x 3 is a root and
P x x 2 x 3 x 2 3x 2
5
x 2 x 3 x 1 x 2
x
1
Therefore the real roots are 2, 1, 2, and 3 (all of multiplicity 1).
50. (a) P x x 4 2x 3 2x 2 8x 8 x 2 x 2 2x 2 4 x 2 2x 2 x 2 4 x 2 2x 2 x 2 x 2 x 2 2x 2
(b)
y
5 x
1
The quadratic is irreducible, so the real roots are 2 (each of multiplicity 1).
51. (a) P x 2x 4 x 3 2x 2 3x 2. The possible rational roots are 1, 2, 12 . P has one variation in sign, and hence 1 positive real root. P x has 3 variations in sign and hence either 3 or 1 negative real roots. 1
2 2
1
2
3
2
2
3
5
2
5
2
0
3
x 1 is a zero.
P x 2x 4 x 3 2x 2 3x 2 x 1 2x 3 3x 2 5x 2 . Continuing: 1 2
3
5
2
2 1 4 2 12 2
1
4 2
3
5
2
4
3 5
y
2
4 2 14 2 1 7 12
2
1 1 2 2
2 2
(b)
0 x 12 is a zero.
5 1
x
P x x 1 x 12 2x 2 2x 4 . The quadratic is irreducible, so the real zeros are 1 and 12 (each of
multiplicity 1).
384
CHAPTER 3 Polynomial and Rational Functions
52. (a) P x 9x 5 21x 4 10x 3 6x 2 3x 1. The possible rational zeros are 1, 13 , 19 . P has 3 variations in sign, hence 3 or 1 positive real roots. P x has 2 variations in sign, hence 2 or 0 real negative roots. 1 9 21
10
6 3 1
9 12 2
4
1
9 12 2 4 1 0 x 1 is a zero. P x x 1 9x 4 12x 3 2x 2 4x 1 . Continuing: 1 9 12 2
4
1
9 3 5 1 9 3 5 1 P x x 12 9x 3 3x 2 5x 1 .
0 x 1 is a zero again. (b)
y
Continuing:
4
1 9 3 5 1 9
9
6
6
1
1 1
0 x 1 is a zero yet again. P x x 13 9x 2 6x 1
x
x 13 3x 12
So the real zeros of P are 1 (multiplicity 3) and 13 (multiplicity 2).
53. Because it has degree 3 and zeros 12 , 2, and 3, we can write P x C x 12 x 2 x 3
C x 3 92 C x 2 72 C x 3C. In order that the constant coefficient be 12, we must have 3C 12 C 4, so
P x 4x 3 18x 2 14x 12. 54. Because it has degree 4 and zeros 4, 4, and 3i, 3i must also be a zero. Thus, P x C x 42 x 2 9 C x 4 8C x 3 25C x 2 72C x 144C.
The coefficient of x 2 is 25 25C, so C 1 and hence P x x 4 8x 3 25x 2 72x 144.
55. No, there is no polynomial of degree 4 with integer coefficients that has zeros i, 2i, 3i and 4i. Since the imaginary zeros of polynomial equations with real coefficients come in complex conjugate pairs, there would have to be 8 zeros, which is impossible for a polynomial of degree 4. 56. P x 3x 4 5x 2 2 3x 2 2 x 2 1 . Since 3x 2 2 0 and x 2 1 0 have no real zeros, it follows that 3x 4 5x 2 2 has no real zeros.
57. P x x 3 x 2 x 1 has possible rational zeros 1. 1
1
1
1
1
1
0
1
1 0 1 0 x 1 is a zero. So P x x 1 x 2 1 . Therefore, the zeros are 1 and i. 58. P x x 3 8 x 2 x 2 2x 4 0, so 2 is a zero. Using the Quadratic Formula, the other zeros are 2 22 4 1 4 1 12 1 3i. x 2 2
CHAPTER 3
Review
385
59. P x x 3 3x 2 13x 15 has possible rational zeros 1, 3, 5, 15. 1
1 1
3
13
15
1
2
15
2
15
0
x 1 is a zero.
So P x x 3 3x 2 13x 15 x 1 x 2 2x 15 x 1 x 5 x 3. Therefore, the zeros are 3, 1, and 5.
60. P x 2x 3 5x 2 6x 9 has possible rational zeros 1, 3, 9, 12 , 32 , 92 . Since there is one variation in sign, there is a positive real zero. 1 2 5 6 9 2
7
3 2
1
5 6 9
3 2 5 6 9 2
6 33 81
3 12
9
2 11 27 72 x 3 is an upper bound 2 8 6 0 x 32 is a zero. So P x 2x 3 5x 2 6x 9 2x 3 x 2 4x 3 2x 3 x 3 x 1. Therefore, the zeros are 3, 1 2 7
1 8
and 32 .
61. P x x 4 6x 3 17x 2 28x 20 has possible rational zeros 1, 2, 4, 5, 10, 20. Since all of the coefficients are positive, there are no positive real zeros. 1
1
6
17
28
20
1
5
12
16
2
1
17
28
20
8
18
20
9 10 0 x 2 is a zero. P x x 4 6x 3 17x 2 28x 20 x 2 x 3 4x 2 9x 10 . Continuing with the quotient, we have 1
5
12
16
1
6 2
4
2
1
4
1
9
10
2
4
10
2
5
4
0 x 2 is a zero. Thus P x x 4 6x 3 17x 2 28x 20 x 22 x 2 2x 5 . Now x 2 2x 5 0 when
x 2 4451 24i 1 2i. Thus, the zeros are 2 (multiplicity 2) and 1 2i. 2 2
62. P x x 4 7x 3 9x 2 17x 20 has possible rational zeros 1, 2, 4, 5, 10, 20. 1 1 7 1
9 17 20 2 1 7 8
17
0
9 17 20
2 18
54
1 9 27
37
1 1
74
7
9 17 20
1 6
3
20
54 x 2 is an 1 6 3 20 0 x 1 is a zero. upper bound. So P x x 4 7x 3 9x 2 17x 20 x 1 x 3 6x 2 3x 20 . Continuing with the quotient, we have 1 8 17
0 20
1 1
6
3 20
1 5 1
2 1
2
5 2 18
6
3 20
2 8 1
10
4 5 10
4 1
6
3 20
4 8
20
1 2 5 0 x 4 is a zero. So P x x 4 7x 3 9x 2 17x 20 x 1 x 4 x 2 2x 5 . Now using the Quadratic Formula on 2 2 6 2 4 4 1 5 1 6. Thus, the zeros are 4, 1, and 1 6. x 2 2x 5 we have: x 2 2
386
CHAPTER 3 Polynomial and Rational Functions
63. P x x 5 3x 4 x 3 11x 2 12x 4 has possible rational zeros 1, 2, 4. 1 1 3 1 11 12 1 2 3 1 2 3
4
8 4
8 4
0 x 1 is a zero.
P x x 5 3x 4 x 3 11x 2 12x 4 x 1 x 4 2x 3 3x 2 8x 4 . Continuing with the quotient, we have 1 1 2 3
8 4
1 1 4
4
1 1 4 4 0 x 1 is a zero. x 5 3x 4 x 3 11x 2 12x 4 x 12 x 3 x 2 4x 4 x 13 x 2 4 x 13 x 2 x 2
Therefore, the zeros are 1 (multiplicity 3), 2, and 2. 64. P x x 4 81 x 2 9 x 2 9 x 3 x 3 x 2 9 x 3 x 3 x 3i x 3i. Thus, the zeros are 3, 3i.
65. P x x 6 64 x 3 8 x 3 8 x 2 x 2 2x 4 x 2 x 2 2x 4 . Now using the Quadratic
Formula to find the zeros of x 2 2x 4, we have 3 1 i 3, and using the Quadratic Formula to find the zeros of x 2 2x 4, we have x 2 4441 22i 2 2 22i 3 1 i 3. Therefore, the zeros are 2, 2, 1 i 3, and 1 i 3. x 2 4441 2 2
1 . 66. P x 18x 3 3x 2 4x 1 has possible rational zeros 1, 12 , 13 , 16 , 19 , 18
1 18
1 18 2
3 4 1 18 21 17
3 4 1 9
6
1
18 12 2 0 x 12 is a zero. 18 21 17 16 x 1 is an upper bound. So P x 18x 3 3x 2 4x 1 2x 1 9x 2 6x 1 2x 1 3x 12 . Thus the zeros are 12 and 13
(multiplicity 2).
67. P x 6x 4 18x 3 6x 2 30x 36 6 x 4 3x 3 x 2 5x 6 has possible rational zeros 1, 2, 3, 6. 1 6 18
6 30
6 12 6 12
36
6 36
6 36
0 x 1 is a zero. 6 x 1 x 3 2x 2 x 6 .
So P x 6x 4 18x 3 6x 2 30x 36 x 1 6x 3 12x 2 6x 36 Continuing with the quotient we have 1 1 2 1 6
2 1 2 1 6
1 1 2 1 1 2 8
2
0 2
0 1 8
3
3
6
1 2 0 x 3 is a zero. So P x 6x 4 18x 3 6x 2 30x 36 6 x 1 x 3 x 2 x 2 . Now x 2 x 2 0 when
1
3 1 2 1 6 1
7 , and so the zeros are 1, 3, and 1i 7 . x 1 1412 1i 2 2 2 68. P x x 4 15x 2 54 x 2 9 x 2 6 . If x 2 9, then x 3i. If x 2 6, then x i 6. Therefore, the zeros are 3i and i 6.
CHAPTER 3
69. 2x 2 5x 3 2x 2 5x 3 0. The solutions are x 05, 3.
2
Review
387
70. Let P x x 3 x 2 14x 24. The solutions to P x 0 are x 3, 2, and 4.
-5
4
5 -20
-5 -40
71. x 4 3x 3 3x 2 9x 2 0 has solutions x 024, 424.
72. x 5 x 3 x 5 x 3 0. We graph
P x x 5 x 3. The only real solution is 134. 10
5
-2
2
-50 -10
73. P x x 3 2x 4 1
1
0
2
4
1
1
1
2
1
0
2
4
2
4
4
1
1 1 5 1 2 2 0 P x x 3 2x 4 x 2 x 2 2x 2 . Since x 2 2x 2 0 has no real solution, the only real zero of P is
x 2.
74. P x x 4 3x 2 4 x 2 1 x 2 4 . Since x 2 4 0 has no real solution, the only real zeros of P are x 1 and x 1.
3 . The vertical asymptote is x 4. Because the x 4 denominator has higher degree than the numerator, the horizontal
75. (a) r x
asymptote is y 0. When x 0, y 34 , so the y-intercept is 34 . There is no x-intercept because the numerator is never 0. The domain of r is 4 4 and its range is 0 0 . 1 3 1 (b) If f x , then r x 3 3 f x 4, so we x x 4 x 4 obtain the graph of r by shifting the graph of f to the left 4 units and stretching vertically by a factor of 3.
y
1 1
x
388
CHAPTER 3 Polynomial and Rational Functions
76. (a) r x
1 . The vertical asymptote is x 5 and the horizontal x 5
y
asymptote is y 0. When x 0, y 15 , so the y-intercept is 15 .
There is no x-intercept. The domain of r is 5 5 and its range is 0 0 .
1 1 1 f x 5. (b) If f x , then r x x x 5 x 5
1 x
1
Thus, we obtain the graph of r by shifting the graph of f to the right 5 units and reflecting in the x-axis. 3x 4 . The vertical asymptote is x 1. Because the x 1 denominator has the same degree as the numerator, the horizontal
77. (a) r x
y
asymptote is y 31 3. When x 0, y 4, so the y-intercept is 4. When y 0, 3x 4 0 x 43 , so the x-intercept is 43 . The domain of r is 1 1 and its range is 3 3 .
1 , then x 3 x 1 1 1 r x 3 3 f x 1. Thus, we x 1 x 1 obtain the graph of r by shifting the graph of f to the right 1 unit,
(b) If f x
1 x
1
reflecting in the x-axis, and shifting upward 3 units. 78. (a) r x
2x 5 . The vertical asymptote is x 2 and the horizontal x 2
y
asymptote is y 2. When x 0, y 52 , so the y-intercept is 52 .
When y 0, x 52 , so the x-intercept is 52 . The domain of r is 2 2 and its range is 2 2 .
1
1 , then x 2 x 2 1 1 r x 2 2 f x 2. Thus, we x 2 x 2 obtain the graph of r by shifting the graph of f to the left 2 units and
(b) If f x
1x
_1
upward 2 units. 3x 12 12 . When x 0, we have r 0 12, so the y-intercept is x 1 1 12. Since y 0, when 3x 12 0 x 4, the x-intercept is 4. The vertical
79. r x
y
asymptote is x 1. Because the denominator has the same degree as the
numerator, the horizontal asymptote is y 31 3. The domain of r is 1 1 and its range is 3 3 .
5 1
x
CHAPTER 3
80. r x
1 x 22
1 1 . When x 0, we have r 0 2 , so the y-intercept is 14 . 4 2
Review
389
y
Since the numerator is 1, y never equals zero and there is no x-intercept. There is a vertical asymptote at x 2. The horizontal asymptote is y 0 because the
degree of the denominator is greater than the degree of the numerator. The domain of r is 2 2 and its range is 0 . 1
x 2 x 2 1 . When x 0, we have r 0 2 81. r x 2 8 4 , x 2 x 4 x 2x 8
1
x
1
x
y
so the y-intercept is 14 . When y 0, we have x 2 0 x 2, so the
x-intercept is 2. There are vertical asymptotes at x 2 and x 4. The domain
1
of r is 2 2 4 4 and its range is .
82. r x
x 3 27 27 . When x 0, we have r 0 27 4 , so the y-intercept is y 4 . x 4
y
When y 0, we have x 3 27 0 x 3 27 x 3. Thus the x-intercept
is x 3. The vertical asymptote is x 4. Because the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. By long division, we have r x x 2 4x 16
37 . So the end behavior of y is x 4
like the end behavior of g x x 2 4x 16. The domain of r is
20 x
2
4 4 and its range is .
83. r x
x2 9 x 3 x 3 . When x 0, we have r 0 9 1 , so the 2x 2 1 2x 2 1
y-intercept is 9. When y 0, we have x 2 9 0 x 3 so the x-intercepts
are 3 and 3. Since 2x 2 1 0, the denominator is never zero so there are no
vertical asymptotes. The horizontal asymptote is at y 12 because the degree of the denominator and numerator are the same. The domain of r is and its range is 9 12 .
y 1 1
x
390
CHAPTER 3 Polynomial and Rational Functions
84. r x
2x 2 6x 7 7 . When x 0, we have r 0 7 4 4 , so the y-intercept x 4
y
is y 74 . We use the Quadratic Formula to find the x-intercepts:
6 62 427 6 92 3 23 . Thus the x-intercepts are x 22 4 2
x 39 and x 09. The vertical asymptote is x 4. Because the degree of the
numerator is greater than the degree of the denominator, there is no horizontal
2
1 , so the slant asymptote. By long division, we have r x 2x 2 x 4 asymptote is r x 2x 2. The domain of r is 4 4 and its range is
x
1
approximately 717] [1283 .
x 2 5x 14 x 7 x 2 x 7 for x 2. The x-intercept is x 2 x 2 7 and the y-intercept is 7, there are no asymptotes, the domain is x x 2, and
y
85. r x
(2, 9)
the range is y y 9.
1 0
x 3 3x 2 10x
x x 2 3x 10
x x 5 x 2 x x 5 x 2 x 2 x 2 for x 2. The x-intercepts are 0 and 5 and the y-intercept is 0, there are no asymptotes, the domain is x x 2, and the range is y y 25 4 .
86. r x
1
x
y (_2, 14)
2 0
x 2 3x 18 x 6 x 6 x 3 for x 3. The x-intercept is 87. r x 2 x 5 5 3 x x x 8x 15
6 and the y-intercept is 65 , the vertical asymptote is x 5, the horizontal asymptote is y 1, the domain is x x 3 5, and the range is y y 1 92 .
1
x
y
2 2
(3, _ 92 )
x
CHAPTER 3
x 2 2x 15 x 3 x 3 x 5 88. r x 3 for x 1 x 2 x 5 x 1 x 2 x 4x 2 7x 10
1
x 1 and x 2, the horizontal asymptote is y 0, the domain is x x 5 1 2, and the range is y y 19 or y 1 (You can use a
asymptote at x 3 and a horizontal asymptote at
0
(_5, _ 27 )
graphing calculator to find the range.)
x 3 . From the graph we see that the 2x 6 x-intercept is 3, the y-intercept is 05, there is a vertical
391
y
x 5. The x-intercept is 3 and the y-intercept is 32 , the vertical asymptotes are
89. r x
Review
1
x
2x 7 90. r x 2 . From the graph we see that the x 9 x-intercept is 35, the y-intercept is 078, there is a
horizontal asymptote at y 0 and no vertical asymptote,
y 05, and there is no local extremum.
the local minimum is 111 090, and the local maximum is 811 012.
5
-10 -8 -6 -4 -2
2
4
-20
20
-5 -1
x3 8 91. r x 2 . From the graph we see that the x-intercept is 2, the x x 2 y-intercept is 4, there are vertical asymptotes at x 1 and x 2, there is no
10
horizontal asymptote, the local maximum is 0425 3599, and the local minimum is 4216 7175. By using long division, we see that
-5
10 x f x x 1 2 , so f has a slant asymptote of y x 1. x x 2
5 -10
2x 3 x 2 . From the graph we see that the x-intercepts are 0 and 12 , the x 1 y-intercept is 0, there is a vertical asymptote at x 1, the local maximum is
92. r x
50
0 0, and the local minima are 157 1790 and 032 003. Using long division, we see that r x 2x 2 3x 3
3 . So the end x 1
-5
behavior of r is the same as the end behavior of g x 2x 2 3x 3.
5
93. 2x 2 x 3 2x 2 x 3 0 P x x 1 2x 3 0. The cut points occur where x 1 0 and where 2x 3 0; that is, at x 1 and x 32 . We make a sign diagram: 1 32 Sign of 1
x 1
2x 3
P x
Both endpoints satisfy the inequality. The solution is 1] 32 .
3 2
392
CHAPTER 3 Polynomial and Rational Functions
94. x 3 3x 2 4x 12 0 x 2 x 3 4 x 3 0 P x x 2 x 2 x 3 0. The cut points are 2, 2, and 3. We make a sign diagram: 2
2 2
2 3
3
x 2
x 2
x 3
Sign of
P x
All endpoints satisfy the inequality. The solution is 2] [2 3].
95. x 4 7x 2 18 0 x 2 9 x 2 2 0 P x x 3 x 3 x 2 2 0. The last factor is positive for all x. We make a sign diagram:
Sign of
3
3 3
3
x 3
x 3
P x
Neither endpoint satisfies the strict inequality. The solution is 3 3.
96. x 8 17x 4 16 0 x 4 16 x 4 1 0 x 2 4 x 2 4 x 2 1 x 2 1 0 P x x 2 4 x 2 1 x 4 x 4 x 1 x 1 0. The first two factors are positive for all x. We make a sign diagram:
Sign of
4
4 1
1 1
1 4
4
x 4
x 1
x 1
x 4
P x
None of the endpoints satisfies the strict inequality. The solution is 2 1 1 2 .
97.
5 x 3 x 2 4x 4
0
5
0 r x
x x2 4 x2 4
5 0. We make a sign diagram: x 1 x 2 x 2
2
2 1
1 2
2
x 2
x 1
x 2
Sign of
r x
None of the endpoints is in the domain of r . The solution is 2 1 2.
CHAPTER 3
98.
Review
393
2 3x 1 2 3 3x 1 2 x 2 9x 3 2x 4 7x 1 3x 1 0 0 0 r x 0. We x 2 3 x 2 3 3 x 2 3 x 2 3 x 2 make a sign diagram: 1 2 17 Sign of 2 7 x 2
7x 1
r x
The cut point 17 satisfies equality, but 2 is not in the domain of r. The solution is 2 17 . 99.
1 2 3 1 2 3 x x 3 2x x 2 3 x 2 x 3 0 0 x 2 x 3 x x 2 x 3 x x x 2 x 3 4x 18 2 9 2x 0 r x 0. We make a sign diagram: x x 2 x 3 x x 2 x 3 9 2 92 Sign of 3 3 0 0 2 2 x 3
x
x 2
9 2x
r x
The cut point 92 satisfies equality, but the other cut points are not in the domain of r . The solution is 3 0 2 92 . 100.
1 3 4 1 3 4 x x 3 3x x 2 4 x 2 x 3 0 0 x 2 x 3 x x 2 x 3 x x x 2 x 3 7x 24 r x 0. We make a sign diagram: x x 2 x 3 24 24 Sign of 2 0 0 3 3 7 7 2 7x 24
x 2
x
x 3
r x
24 The cut point 24 7 satisfies equality, but the other cut points are not in the domain of r. The solution is 7 2 0 3. 101. f x
24 x 3x 2 x 3 8 3x is defined where r x x 3 8 3x 0. We make a sign diagram: 8 3 83 Sign of 3 3 x 3
8 3x
r x
Both cut points satisfy equality, so the domain of f is 3 83 .
394
CHAPTER 3 Polynomial and Rational Functions
102. g x 4
1 1 is defined where r x x 1 x x 2 x 1 0. 4 4 x x4 x 1 x3 x 1 x x 2 x 1 1
(Equality is excluded because the denominator cannot be 0.) The last factor of r x is positive for all x. We make a sign diagram: 0
0 1
1
x
1x
Sign of
x2 x 1
r x
Neither cut point satisfies the strict inequality, so the domain of g is 0 1. 104.
103. 40
20 20 -2 -3
-2
-1
1
2
3
-1
1
2
3
-20
From the graph, we see that x 4 x 3 5x 2 4x 5 on
From the graph, we see that x 5 4x 4 7x 3 12x 2 0
on approximately 106 017 191 .
approximately [074 195].
105. (a) We use synthetic division to show that 1 is a zero of P x 2x 4 5x 3 x 4: 2
1
2
5
0
1
4
2
3
3
4
3
3
4
0 Thus, 1 is a zero and P x x 1 2x 3 3x 2 3x 4 .
(b) P x has no change of sign, and hence no positive real zeros. But P x x 1 Q x, so Q cannot have a positive real zero either. 106. We want to find the solutions of x 4 x 2 24x 6x 3 20 P x x 4 6x 3 x 2 24x 20 0. The possible rational zeros of P are 1, 2, 4, 5, 10, 20. P x has 3 variations in sign and hence 1 or 3 positive real zeros. P x x 4 6x 3 x 2 24x 20 has 1 variation in sign, and so P has 1 negative real zeros. 1
1
6
1
24
20
1
1
5
4
20
5
4
20
0
2 x 1 is a zero.
1
5
4
20
1
2
6
20
3
10
0
P x x 1 x 2 x 2 3x 10 x 1 x 2 x 2 x 5, and
y
x 2 is a zero.
so the zeros of P (and hence the x-coordinates of the points of intersection) are 2, 1, 2, and 5. From the original equation, the coordinates of the points of intersection are 2 28, 1 26, 2 68, 5 770.
100 1
x
CHAPTER 3
Test
395
CHAPTER 3 TEST y
1. f x x 2 x 6 x2 x 6 x 2 x 14 6 14 2 x 12 25 4
2. g x 2x 2 6x 3 is a quadratic
function with a 2 and b 6, so it has a maximum or minimum value where b 6 3 x . Because 2a 2 2 2
1 1
x
a 0, this gives the minimum value 2 g 32 2 32 6 32 3 32 .
3. (a) We write the function in standard form: h x 10x001x 2 001 x 2 1000x 001 x 2 1000x 5002 001 5002 001 x 5002 2500. Thus, the maximum height reached by the cannonball is 2500 feet. (b) By the symmetry of the parabola, we see that the cannonball’s height will be 0 again (and thus it will splash into the water) when x 1000 ft. y
4. f x x 23 27 has y-intercept y 23 27 19 and x-intercept where x 23 27 x 1.
10
(0, 19) (1, 0)
x
1
5. (a)
2
1 1
0
4
2
5
2
4
0
4
2
0
2
(b) 2x 2 1
9
x 3 2x 2
2x 5 4x 4 x 3 2x 5
Therefore, the quotient is
4x 4
Q x x 3 2x 2 2, and the remainder is
4x 4
x3
1 2 x 2 0x
7
x2 2x 2 x2
R x 9.
x2
7
12 15 2
Therefore, the quotient is Q x x 3 2x 2 12 and the remainder is R x 15 2 . 6. (a) Possible rational zeros are: 1, 3, 12 , 32 . (b)
1
2
5
4
3
2
7
3
2 7 3 0 x 1 is a zero. P x x 1 2x 2 7x 3 x 1 2x 1 x 3 2 x 1 x 12 x 3
(c) The zeros of P are x 1, 3, 12 .
y
2 1
x
396
CHAPTER 3 Polynomial and Rational Functions
7. P x x 3 x 2 4x 6. Possible rational zeros are: 1, 2, 3, 6. 1 1 1 4 1
0
2 1 1 4
6
2
4
3 1 1 4 6
6
2
3
4
6
6
1
0 4 10 1 1 2 10 1 2 2 0 x 3 is a zero. So P x x 3 x 2 2x 2 . Using the Quadratic Formula on the second factor, we have 2 4 2 2 1 2 22 4 1 2 1 i. So zeros of P x are 3, 1 i, and 1 i. x 2 1 2 2 8. P x x 4 2x 3 5x 2 8x 4. The possible rational zeros of P are: 1, 2, and 4. Since there are four changes in sign, P has 4, 2, or 0 positive real zeros. 1
1
2
5
8
4
1
1
4
4
1 1 4 4 0 So P x x 1 x 3 x 2 4x 4 . Factoring the second factor by grouping, we have P x x 1 x 2 x 1 4 x 1 x 1 x 2 4 x 1 x 12 x 2i x 2i.
9. Since 3i is a zero of P x, 3i is also a zero of P x. And since 1 is a zero of multiplicity 2, P x x 12 x 3i x 3i x 2 2x 1 x 2 9 x 4 2x 3 10x 2 18x 9.
10. P x 2x 4 7x 3 x 2 18x 3. (a) Since P x has 4 variations in sign, P x can have 4, 2, or 0 positive real zeros. Since
P x 2x 4 7x 3 x 2 18x 3 has no variations in sign, there are no negative real zeros. (b)
4
2 2
7
1
18
3
8
4
20
8
1
5
2
11
Since the last row contains no negative entry, 4 is an upper bound for the real zeros of P x. 1
2 2
1
18
3
1
9
10
28
9
10
28
31
7
Since the last row alternates in sign, 1 is a lower bound for the real zeros of P x. (c) Using the upper and lower limit from part (b), we graph P x in the viewing rectangle [1 4] by
(d) Local minimum 282 7031.
[1 1]. The two real zeros are 017 and 393.
50
1 -2
2
4
-50 2
4
-1
x2 x 6 2x 1 x 3 27 x 3 9x x 3 6x 2 9x , u x . 11. r x 2 , s x 2 , t x , and x 2 x 2 x 3 x x 2 x 4 x 25
CHAPTER 3
Test
397
(a) r x has the horizontal asymptote y 0 because the degree of the denominator is greater than the degree of the
numerator. u x has the horizontal asymptote y 11 1 because the degrees of the numerator and the denominator are the same.
(b) The degree of the numerator of s x is one more than the degree of the denominator, so s has a slant asymptote. x x 2 6x 9 x x 32 x x 3 (c) The denominator of s x is never 0, so s has no vertical asymptote. x x 3 x 3 for x 3, so has no vertical asymptote.
y
(d) From part (c), has a “hole” at 3 0. 2x 1 2x 1 (e) r x 2 , so r has vertical asymptotes at x 1 x 1 x 2 x x 2 and x 2. y 0 is a horizontal asymptote because the degree of the numerator is less than the degree of the denominator.
1
x2 x 6 x 3 x 2 (f) u x . When x 0, we have 2 x 5 x 5 x 25
2
x
6 6 , so the y-intercept is y 6 . When y 0, we have x 3 or u x 25 25 25 x 2, so the x-intercepts are 3 and 2. The vertical asymptotes are x 5 and
x 5. The horizontal asymptote occurs at y 11 1 because the degree of the denominator and numerator are the same. x 2 2x 5
(g) x 2
x 3 0x 2 9x x 3 2x 2
y
0
2x 2 9x 2x 2 4x 5x 0 5x 10 10
Thus P x x 2 2x 5 and t x
12. x
10 2
x
x 3 9x have the same end behavior. x 2
6x 6 x x 2x 5 2x 2 4x 6 6x x 1 3 x 0 x 0 0 r x. We 2x 5 2x 5 2x 5 2x 5 x5 2
make a sign diagram:
1 52
53 2
3
Sign of
1
x 1
3x
x 52
r x
The cut point 52 is excluded so that the denominator is not 0. The solution is 1] 52 3 .
398
FOCUS ON MODELING
1 13. f x is defined where 4 2x x 2 0. Using the Quadratic Formula to solve x 2 2x 4 0, we 4 2x x 2 2 22 4 1 4 1 5. The radicand is positive between these two roots, so the domain of have x 2 1 f is 1 5 1 5 .
P x x 4 4x 3 8x. From the graph, the x-intercepts are approximately
14. (a) 10
124, 0, 2, and 324, P has local maximum P 1 5, and P has local minima P 073 P 273 4.
-2
2
4
(b) From part (a), P x 0 on approximately 124] [0 2] [324 .
FOCUS ON MODELING Fitting Polynomial Curves to Data
1. (a) Using a graphing calculator, we obtain the quadratic
2. (a) Using a graphing calculator, we obtain the quadratic
polynomial
polynomial
y 0275428x 2 197485x 2735523 (where
y 02783333x 2 184655x 166732 (where
miles are measured in thousands). (b)
y
(b)
100
y 120
80 60
80
40
40
20 0
plants/acre are measured in thousands).
25
30
35
40
45
50 x
Pressure (lb/in@)
(c) Moving the cursor along the path of the polynomial, we find that 3585 lb/in2 gives the longest tire life.
0
10
20
30
40
50
60 x
Density (thousand plants/acre)
(c) Moving the cursor along the path of the polynomial, we find that yield when 37,000 plants are planted per acre is about 135 bushels/acre.
Fitting Polynomial Curves to Data
3. (a) Using a graphing calculator, we obtain the cubic
y 60
4. (a)
polynomial
50
y 000203709x 3 0104522x 2
40
1966206x 145576.
(b)
399
30
y
20 10
20
0
1
10
2
x
3
Time (s) A quadratic model seems appropriate. 35 x
(b) Using a graphing calculator, we obtain the quadratic
(c) Moving the cursor along the path of the polynomial,
(c) Moving the cursor along the path of the polynomial,
0
5
10
15
20
25
30
Seconds
polynomial y 160x 2 518429x 420714.
we find that the subjects could name about
we find that the ball is 20 ft. above the ground 03
43 vegetables in 40 seconds.
seconds and 29 seconds after it is thrown upward.
(d) Moving the cursor along the path of the polynomial,
(d) Again, moving the cursor along the path of the
we find that the subjects could name 5 vegetables in
polynomial, we find that the maximum height is
about 20 seconds.
462 ft.
5. (a) Using a graphing calculator, we obtain the quadratic polynomial y 00120536x 2 0490357x 496571. (c) Moving the cursor along the path of the polynomial, we find that the tank should drain in 190 minutes.
(b)
y 6 5 4 3 2 1 0
10
Time (min)
20 x
4
EXPONENTIAL AND LOGARITHMIC FUNCTIONS
4.1
EXPONENTIAL FUNCTIONS
1 , f 0 50 1, f 2 52 25, 1. The function f x 5x is an exponential function with base 5; f 2 52 25 and f 6 56 15,625.
2. (a) f x 2x is an exponential function with base 2. It has graph III. x (b) f x 2x 12 is an exponential function with base 12 . It has graph I.
(c) The graph of f x 2x is obtained from the graph of 2x by reflecting about the x-axis. Thus, it has graph II.
(d) The graph of f x 2x is obtained from the graph of 2x by reflecting about the x- and y-axes. Thus, it has graph IV. 3. (a) To obtain the graph of g x 2x 1 we start with the graph of f x 2x and shift it downward 1 unit. 4.
5.
6.
7. 8.
(b) To obtain the graph of h x 2x1 we start with the graph of f x 2x and shift it to the right 1 unit. nt for compound interest, the letters P, r, n, and t stand for principal, interest rate per In the formula A t P 1 nr year, number of times interest is compounded per year, and number of years, respectively, and A t stands for the amount after t years. So if $100 is invested at an interest rate of 6% compounded quarterly, then the amount after 2 years is 4 2 100 1 006 11265. 4 x The exponential function f x 12 has the horizontal asymptote y 0. This means that as x , we have x 1 0. 2 x The exponential function f x 12 3 has the horizontal asymptote y 3. This means that as x , we have x 1 3 3. 2 5 22195, f 2 0063, f 03 1516. f x 4x ; f 12 412 4 2, f f x 3x1 ; f 12 0577, f 25 5196, f 1 0111, f 14 0439.
x1 9. g x 13 ; g 12 0192, g 2 0070, g 35 15588, g 14 1552.
3x ; f 12 0650, f 6 8281, f 3 0075, f 43 3160. 10. f x 43
11.
f x 2x
y
12.
g x 8x
x
y
x
y
4
2
2
1 16 1 4
1
1 64 1 8
0
1
0
1
1
8
2
64
2
4
4
16
1 1
x
y
1 1
x
401
402
13.
CHAPTER 4 Exponential and Logarithmic Functions
x f x 13
y
x
y
x
y
2
9
5
0620921323
1
3
1
090
0
1
1
1 3 1 9
2 15.
14. h x 11x
y
1 x
1
g x 3 13x
0
1
5
161051
259374246 x 16. h x 2 14
y
y
x
y
2
1775
32
2308
2
1
1
8
30
0
2
1
1 2 1 8 1 32 1 128
1
39
2
507
3
6591
3
4
8568
4
1 1
x
2
x
1
10
x
0
1
y
2 1
x
x 18. f x 3x and g x 13
17. f x 2x and g x 2x y
y
g
f=g
f
1
1
x
1
x
1
x 20. f x 34 and g x 15x .
19. f x 4x and g x 7x . y
y
g f
f
2
5 1
x
21. From the graph, f 2 a 2 9, so a 3. Thus f x 3x .
g
0
1
x
SECTION 4.1 Exponential Functions
403
22. From the graph, f 1 a 1 15 , so a 5. Thus f x 5x . 1 , so a 1 . Thus f x 1 x . 23. From the graph, f 2 a 2 16 4 4 x 24. From the graph, f 3 a 3 8, so a 12 . Thus f x 12 . 25. The graph of f x 5x1 is obtained from that of y 5x by shifting 1 unit to the left, so it has graph II. 26. The graph of f x 5x 1 is obtained from that of y 5x by shifting 1 unit upward, so it has graph I.
27. g x 2x 3. The graph of g is obtained by shifting the graph of y 2x downward 3 units. Domain: .
Range: 3 . Asymptote: y 3.
x 28. h x 4 12 . The graph of h is obtained by shifting x the graph of y 12 upward 4 units. Domain: . Range: 4 . Asymptote: y 4.
y
y
1
(_1, 6)
1 (1, _1)
x 1 x
1
29. The graph of f x 3x is obtained by reflecting the graph of y 3x about the x-axis. Domain: . Range: 0. Asymptote: y 0. y
30. The graph of f x 10x is obtained by reflecting the graph of y 10x about the y-axis. Domain: . Range: 0 . Asymptote: y 0. y
1 x
1 (1, _3)
1 1
x
404
CHAPTER 4 Exponential and Logarithmic Functions
31. f x 10x3 . The graph of f is obtained by shifting the graph of y 10x to the left 3 units. Domain: .
Range: 0 . Asymptote: y 0.
y
Range: 0 . Asymptote: y 0.
(3, 1)
1 x
_1
33. y 5x 1. The graph of y is obtained by reflecting the
graph of y 5x about the x-axis and then shifting upward 1 unit. Domain: . Range: 1 . Asymptote: y 1.
graph of y 2x to the right 3 units. Domain: . y
100
(_3, 1)
32. g x 2x3 . The graph of g is obtained by shifting the
y
x
1
34. h x 6 3x . The graph of h is obtained by reflecting
the graph of y 3x about the x-axis and shifting upward 6 units. Domain: . Range: 6. Asymptote: y 6.
y
(1, 3) 1 x
1
(0, 2) 1 1
x
x 35. y 2 13 . The graph is obtained by reflecting the x graph of y 13 about the x-axis and shifting upward 2 units. Domain: . Range: 2. Asymptote: y 2.
36. y 5x 3. The graph is obtained by reflecting the
graph of y 5x about the y-axis, then shifting downward
3 units. Domain: . Range: 3 . Asymptote: y 3. y
y
1 0
1
x
1 0
1
x
SECTION 4.1 Exponential Functions
37. h x 2x4 1. The graph of h is obtained by shifting
the graph of y 2x to the right 4 units and upward 1 unit.
Domain: . Range: 1 . Asymptote: y 1. y
405
38. y 3 10x1 10x1 3. The graph of y is
obtained by reflecting the graph of y 10x about the
y-axis, then shifting to the right 1 unit and upward 3 units.
Domain: . Range: 3. Asymptote: y 3. y
2
(1, 2)
(4, 2)
1
1 1
1
x
x
x 39. g x 1 3x 3x 1. The graph of g is obtained 40. y 3 15 5x 3. The graph of y is obtained by reflecting the graph of y 3x about the x- and y-axes
and then shifting upward 1 unit. Domain: . Range: 1. Asymptote: y 1.
by reflecting the graph of y 5x about the x- and y-axes and then shifting upward 3 unit. Domain: . Range: 3. Asymptote: y 3.
y
y
1 (_1, _2)
41. (a)
1
(0, 2)
1
x
x
1
y
42. (a)
y
g f
1
1 1
x
(b) Since g x 3 2x 3 f x and f x 0, the
height of the graph of g x is always three times the height of the graph of f x 2x , so the graph of g
is steeper than the graph of f .
1
x
x2 (b) f x 9x2 32 32x2 3x g x. So f x g x, and the graphs are the same.
406
CHAPTER 4 Exponential and Logarithmic Functions
43.
f x x 3
x
y
g x 3x
0
0
1
1
1
3
2
8
9
3
27
27
4
64
81
6
216
729
8
512
6561
10
1000
59,049
x
g(x)=3
f(x)=x#
20 0
x
1
From the graph and the table, we see that the graph of g ultimately increases much more quickly than the graph of f . 44.
f x x 4
x
y
g x 4x
0
0
1
1
1
3
2
16
16
3
81
64
4
256
256
6
1296
4096
8
4096
65,536
10
10,000
1,048,576
g(x)=4
x
f(x)=x$
100 0
x
1
From the graph and the table, we see that the graph of g ultimately increases much more quickly than the graph of f . 45. (a) From the graphs below, we see that the graph of f ultimately increases much more quickly than the of g. graph (i) [0 5] by [0 20] (ii) [0 25] by 0 107 (iii) [0 50] by 0 108 20
10,000,000
f
8,000,000 6,000,000
10
g
f
1
2
4,000,000
3
4
80,000,000
f
g
40,000,000 20,000,000
0
5
100,000,000
60,000,000
2,000,000
0
g
10
20
0
20
40
(b) From the graphs in parts (a)(i) and (a)(ii), we see that the approximate solutions are x 12 and x 224. 46. (a) (i) [4 4] by [0 20]
(iii) [0 20] by 0 105
(ii) [0 10] by [0 5000]
20 g
f
4000
f
g
100,000
80,000 60,000
10
f
g
2000
40,000 20,000 -4
-3
-2
-1 0
1
2
3
4
0
2
4
6
8
10
0
10
(b) From the graphs in parts (i) and (ii), we see that the solutions of 3x x 4 are x 080, x 152 and x 717.
20
SECTION 4.1 Exponential Functions
47.
c=2 c=1
5
48.
10
c=0.5
4
c=2
c=1
8
3 c=4
c=4
c=0.5
6 c=0.25
2
4
1
c=0.25
2 -3
-2
-1
0 -1
1
2
3 0
-2
The larger the value of c, the more rapidly the graph of shifted horizontally 1 unit. This is because of our choice of c; each c in this exercise is of the form 2k . So f x 2k 2x 2xk . 2
2
4
The larger the value of c, the more rapidly the graph of x f x 2cx increases. In general, f x 2cx 2c ; x so, for example, f x 22x 22 4x .
f x c2x increases. Also notice that the graphs are just
49. y 10xx
407
50. y x2x 2 1 -4 -4
-2
0
2
(a) From the graph, we see that the function is increasing
on 144 and decreasing on 144.
(b) From the graph, we see that the range is
(b) From the graph, we see that the range is
approximately 0 178].
approximately [053 .
10xh 10x f x h f x 51. f x 10x , so 10x
52. f x 3x1 , so
2
(a) From the graph, we see that the function is increasing
on 050 and decreasing on 050 .
h
-2
4
h
10h 1 . h
3xh1 3x1 f x h f x 3x1 h h
3h 1 . h
53. (a) After 1 hour, there are 1500 2 3000 bacteria. After 2 hours, there are 1500 2 2 6000 bacteria. After 3 hours, there are 1500 2 2 2 12,000 bacteria. We see that after t hours, there are N t 1500 2t bacteria. (b) After 24 hours, there are N 24 1500 224 25,165,824,000 bacteria.
54. (a) Because the population doubles every year, there are N t 320 2t mice after t years. (b) After 8 years, there are approximately N 8 320 28 81,920 mice.
408
CHAPTER 4 Exponential and Logarithmic Functions
55. Using the formula A t P 1 ik with P 5000,
004 per month, and k 12 number 12 of years, we fill in the table:
i 4% per year
56. Using the formula A t P 1 ik with P 5000, rate per year i per month, and k 12 5 60 months, 12 we fill in the table:
Amount
Rate per year
Amount
1
$520371
1%
$525625
2
$541571
2%
$552539
3
$563636
3%
$580808
4
$586599
4%
$610498
5
$610498
5%
$641679
6
$635371
6%
$674425
Time (years)
2t 57. P 10,000, r 003, and n 2. So A t 10,000 1 003 10,000 10152t . 2
(a) A 5 10000 101510 11,60541, and so the value of the investment is $11,60541.
(b) A 10 10000 101520 13,46855, and so the value of the investment is $13,46855. (c) A 15 10000 101530 15,63080, and so the value of the investment is $15,63080.
365t . 58. P 2500, r 0025, and n 365. So A t 2500 1 0025 365 365 2 (a) A 2 2500 1 0025 262817, and so the value of the investment is $262817. 365 365 3 269470, and so the value of the investment is $269470. (b) A 3 2500 1 0025 365 365 6 290457, and so the value of the investment is $290457. (c) A 6 2500 1 0025 365
4t . 59. P 500, r 00375, and n 4. So A t 500 1 00375 4 4 51902, and so the value of the investment is $51902. (a) A 1 500 1 00375 4 8 53875, and so the value of the investment is $53875. (b) A 2 500 1 00375 4 40 72623, and so the value of the investment is $72623. (c) A 10 500 1 00375 4 4t . 60. P 4,000, r 00575, and n 4. So A t 4000 1 00575 4 16 502616, and so the amount due is $502616. (a) A 4 4000 1 00575 4 24 563410, and so the amount due is $563410. (b) A 6 4000 1 00575 4 32 631558, and so the amount due is $631558. (c) A 8 4000 1 00575 4
23 P 10456 10000 13023P P 767896. 61. We must solve for P in the equation 10000 P 1 009 2 Thus, the present value is $7,67896.
125 62. We must solve for P in the equation 100000 P 1 008 P 10066760 100000 14898P 12 P $67,12104.
SECTION 4.2 The Natural Exponential Function
409
r n 008 12 63. rAPY 1 1. Here r 008 and n 12, so rAPY 1 1 1006666712 1 0083000. n 12 Thus, the annual percentage yield is about 83%. r n 0055 4 1. Here r 0055 and n 4, so rAPY 1 1 0056145. Thus, the annual 64. rAPY 1 n 4 percentage yield is about 561%.
65. (a) In this case the payment is $1 million. (b) In this case the total pay is 2 22 23 230 230 cents $10,737,41824. Since this is much more than method (a), method (b) is more profitable. 66. Since f 40 240 1,099,511,627,776, it would take a sheet of paper 4 inches by 1,099,511,627,776 inches. Since there are 12 inches in a foot and 5,280 feet in a mile, 1,099,511,627,776 inches 174 million miles. So the dimensions of the sheet of paper required are 4 inches by about 174 million miles.
4.2
THE NATURAL EXPONENTIAL FUNCTION
1. The function f x e x is called the natural exponential function. The number e is approximately equal to 271828. 2. In the formula A t Pert for continuously compound interest, the letters P, r, and t stand for principal, interest rate per year, and number of years respectively, and A t stands for amount after t years. So, if $100 is invested at an interest rate of 6% compounded continuously, then the amount after 2 years is A 2 100 e0062 $11275. 3. h x e x ; h 1 2718, h 23141, h 3 0050, h
2 4113
4. h x e3x ; h 13 0368, h 15 0011, h 1 20086, h 12,391648 5.
f x 15e x x
y
6. g x 4e13x
y
x
y
2
020
3
1087
1
055
2
779
05
091
1
558
0
15
05
247
1
408
2
1108
2 0
1
x
0
4
1
287
2
205
3
147
y
2 0
1
x
410
CHAPTER 4 Exponential and Logarithmic Functions
7. g x 2 e x . The graph of g is obtained from the graph of y e x by shifting it upward 2 units. Domain: . Range: 2 . Asymptote: y 2. y
8. h x ex 3. The graph of h is obtained from the graph of y e x by reflecting it about the y-axis and
shifting it downward 3 units. Domain: . Range: 3 . Asymptote: y 3. y
5
5 0
1
0
x
9. y e x . The graph of y e x is obtained from the
graph of y e x by reflecting it about the x-axis. Domain: . Range: 0. Asymptote: y 0. y
1 1
1
10. f x 1 e x . The graph of f x 1 e x is obtained
by reflecting the graph of y e x about the x-axis and then shifting upward 1 unit. Domain: . Range: 1. Asymptote: y 1. y
x
1 1
11. y ex 1. The graph of y ex 1 is obtained from the graph of y e x by reflecting it about the y-axis then shifting downward 1 unit. Domain: . Range: 1 . Asymptote: y 1. y
x
x
12. f x ex . The graph of f x ex is obtained
by reflecting the graph of y e x about the y-axis and then about the x-axis. Domain: . Range: 0. Asymptote: y 0.
y 1 1
1 1
x
x
SECTION 4.2 The Natural Exponential Function
13. y e x2 . The graph of y e x2 is obtained from the
graph of y e x by shifting it to the right 2 units. Domain: . Range: 0 . Asymptote: y 0. y
411
14. f x e x3 4. The graph of f x e x3 4 is obtained by shifting the graph of y e x to the right
3 units, and then upward 4 units. Domain: . Range: 4 . Asymptote: y 4. y
(3, 5) 1
(2, 1) x
1
1 x
1
15. h x e x1 3. The graph of h is obtained from the graph of y e x by shifting it to the left 1 unit and
downward 3 units. Domain: . Range: 3 .
Asymptote: y 3.
y
16. g x e x1 2. The graph of g is obtained by shifting the graph of y e x to the right 1 unit, reflecting it about the x-axis, then shifting it downward 2 units.
Domain: . Range: 2. Asymptote: y 2.
y 1
1
1
(0, e-3)
(1, _3)
x
1
x
(_1, _2)
y
17. (a)
y
18. (a)
1
1 y=_ 2 e¨
y=sinh x
y=cosh x 1 y=_ 2 eШ
1
1
1 y=_ 2 e¨
1
x
ex e x ex ex 2 2 e x ex cosh x 2
(b) cosh x
x
1 y=_ _ 2 eШ
ex e x ex ex 2 2 ex e x e x ex sinh x 2 2
(b) sinh x
412
CHAPTER 4 Exponential and Logarithmic Functions a=1
19. (a)
a=1.5
10 a=0.5
20. 4
a=2
8
2
6 4
0
2 -4
-2
0
0 2
4
a xa e (b) As a increases the curve y exa 2 flattens out and the y intercept increases.
21. g x x x . Notice that g x is only defined for x 0. The graph of g x is shown in the viewing rectangle
[0 15] by [0 15]. From the graph, we see that there is a local minimum of about 069 when x 037.
20
40
x From the graph, we see that y 1 1x approaches e as x get large.
22. g x e x e2x . The graph of g x is shown in the
viewing rectangle [1 2] by [1 6]. From the graph, we see that there is a local minimum of about 189 when x 023. 5
1
0 0
1
-1
1
2
23. D t 50e02t . So when t 3 we have D 3 50e023 274 milligrams.
24. m t 13e0015t (a) m 0 13 kg.
(b) m 45 13e001545 13e0675 6619 kg. Thus the mass of the radioactive substance after 45 days is about 66 kg.
25. t 180 1 e02t
(a) 0 180 1 e0 180 1 1 0. (b) 5 180 1 e025 180 0632 11376 ft/s. So the
velocity after 5 s is about 1138 ft/s. 10 180 1 e0210 180 0865 1557 ft/s. So the
(c)
y 200
100
velocity after 10 s is about 1557 ft/s. (d) The terminal velocity is 180 ft/s.
5
x
SECTION 4.2 The Natural Exponential Function
26. (a) Q 5 15 1 e0045 15 01813 27345. Thus
(c) 20
approximately 27 lb of salt are in the barrel after 5 minutes. (b) Q 10 15 1 e00410 15 03297 4946Thus
10
approximately 49 lb of salt are in the barrel after 10 minutes.
0 0
(d) The amount of salt approaches 15 lb. This is to be expected, since
100
200
50 gal 03 lb/gal 15 lb. 1200 1 11e02t 1200 1200 100. (a) P 0 1 11 1 11e020 1200 1200 1200 (b) P 10 482. P 20 999. P 30 1168. 1 11e0210 1 11e0220 1 11e0230 1200 1200. The graph shown confirms this. (c) As t we have e02t 0, so P t 10
27. P t
28. n t
5600 05 275e0044t
(b) 10000
(a) n 0 5600 28 200 (c) From the graph, we see that n t approaches about 11,200 as t gets large.
0 0
29. P t
732 61 59e002t
(a) In the year 2200, t 2200 2000 200, and the population is
732 1179 billion. In 61 59e002200 732 1197 billion. 2300, t 300, and P 300 61 59e002300 predicted to be P 200
(c) As t increases, the denominator approaches 61, so according to this model, the world population approaches 732 61 12 billion people. 30. D t
(b)
200
P 14 12 10 8 6 4 2 0
100
200
300
400
500 t
54 54 . So D 20 1600 ft. 1 29e001t 1 29e00120
31. Using the formula A t Pert with P 7000 and r 3% 003, we fill in the table:
32. Using the formula A t Pert with P 7000 and t 10 years, we fill in the table:
Time (years)
Amount
Rate per year
1
$721318
1%
$773620
2
$743286
2%
$854982
3
$765922
3%
$944901
4
$789248
4%
$10,44277
5
$813284
5%
$11,54105
6
$838052
6%
$12,75483
Amount
413
414
CHAPTER 4 Exponential and Logarithmic Functions
33. We use the formula A t Pert with P 2000 and r 35% 0035. (a) A 2 2000e0035 2 $214502
(b) A 42 2000e0035 4 $230055 (c) A 12 2000e0035 12 $304392
34. We use the formula A t Pert with P 3500 and r 625% 00625. (a) A 3 3500e00625 3 $422181 (b) A 6 3500e00625 6 $509247 (c) A 9 3500e00625 9 $614269 35. (a) Using the formula A t P 1 ik with P 600, i 25% per year 0025, and k 10, we calculate A 10 600 102510 $76805.
0025 20 $76922. semiannually and k 10 2 20, so A 600 1 (b) Here i 0025 10 2 2 0025 40 $76982. quarterly and k 10 4 40, so A 10 600 1 (c) Here i 25% per year 0025 4 4 (d) Using the formula A t Pert with P 600, r 25% 0025, and t 10, we have A 10 600e0025 10 $77042.
36. We use the formula A t Pert with P 8000 and t 12.
(a) If r 2% 002, then A 12 8000e002 12 $10,16999.
(b) If r 3% 003, then A 12 8000e003 12 $11,46664.
(c) If r 45% 0045, then A 12 8000e0045 12 $13,72805. (d) If r 7% 007, then A 12 8000e007 12 $18,53094.
2 10252. 37. Investment 1: After 1 year, a $100 investment grows to A 1 100 1 0025 2 4 10227. Investment 2: After 1 year, a $100 investment grows to A 1 100 1 00225 4
Investment 3: After 1 year, a $100 investment grows to A 1 100e002 10202. We see that Investment 1 yields the highest return. 2 10519. 38. Investment 1: After 1 year, a $100 investment grows to A 1 100 1 005125 2 Investment 2: After 1 year, a $100 investment grows to A 1 100e005 10512. We see that Investment 1 yields a higher return.
39. (a) A t Pert 5000e009t (b) 20000
0 0
10
20
(c) A t 25,000 when t 1788 years.
4.3
LOGARITHMIC FUNCTIONS
1. log x is the exponent to which the base 10 must be raised in order to get x. x log x
103
102
101
100
101
102
103
1012
3
2
1
0
1
2
3
12
SECTION 4.3 Logarithmic Functions
415
2. The function f x log9 x is the logarithm function with base 9. So f 9 log9 91 1, f 1 log9 90 0, f 19 log9 91 1, f 81 log9 92 2, and f 3 log9 912 12 . 3. (a) 53 125, so log5 125 3.
(b) log5 25 2, so 52 25.
4. (a) f x log2 x is a logarithmic function with base 2. It has graph III.
(b) The graph of f x log2 x is obtained from that of y log2 x by reflecting about the y-axis. It has graph II.
(c) f x log2 x is obtained from that of y log2 x by reflecting about the x-axis. It has graph I.
(d) f x log2 x is obtained from that of y log2 x by reflecting about the x- and y-axes. It has graph IV.
5. The natural logarithmic function f x ln x has the vertical asymptote x 0. 6. The natural logarithmic function f x ln x 1 has the vertical asymptote x 1. 7.
8.
Logarithmic form
Logarithmic form
Exponential form
log8 8 1
81 8
log4 64 3
823 4
log4 8 32
log8 64 2
log8 4 23
log8 512 3
log8 81 1
1 2 log8 64
9. (a) 34 81 (b) 30 1 13. (a) 3x 5 (b) 72 3y
82 64 83 512
81 18 1 82 64
10. (a) 51 15 (b) 43 64 14. (a) 61 z (b) 102t 3
Exponential form 43 64
log4 2 12
412 2
1 2 log4 16 log4 21 12 1 5 log4 32 2
1 42 16 412 12
11. (a) 813 2 (b) 102 001 15. (a) e3y 5
432 8
1 452 32 1 12. (a) 53 125
(b) 823 4 16. (a) e2 x 1
(b) e1 t 1
(b) e4 x 1
17. (a) log10 10,000 4 1 2 (b) log5 25
18. (a) log6 36 2 1 1 (b) log10 10
19. (a) log8 81 1 (b) log2 18 3
20. (a) log4 0125 32
21. (a) log4 70 x
22. (a) log3 10 2x
23. (a) ln 2 x
24. (a) ln 05 x 1
(b) log3 5 25. (a) log2 2 1 (b) log5 1 log5 50 0
(b) log10 01 4x 26. (a) log3 37 7 (b) log4 64 log4 43 3
(c) log6 65 5 (c) log5 125 3 1 log 33 3 29. (a) log3 27 3 (b) log10 10 log10 1012 12 (c) log5 02 log5 15 log5 51 1
31. (a) 3log3 5 5
(b) ln y 3
(b) log7 343 3
(b) ln t 05x
27. (a) log6 36 log6 62 2 28. (a) log2 32 log2 25 5 (b) log9 81 log9 92 2
(b) log8 817 17
(c) log7 710 10
(c) log6 1 log6 60 0
30. (a) log5 125 log5 53 3 (b) log49 7 log49 4912 12 12 (c) log9 3 log9 312 log9 912 14 32. (a) eln 3 3
(b) 5log5 27 27
1 (b) eln1
(c) eln 10 10
(c) 10log 13 13
416
CHAPTER 4 Exponential and Logarithmic Functions
12 2 log4 212 log4 412 14 1 (b) log4 12 log4 21 log4 412 12 3 (c) log4 8 log4 23 log4 412 log4 432 32
33. (a) log8 025 log8 823 23
34. (a) log4
(b) ln e4 4 1 ln e1 1 (c) ln e
36. (a) log3 x 2 x 32 19
35. (a) log4 x 3 x 43 64 (b) log10 001 x 10x 001 x 2
(b) log5 125 x 5x 125 x 3
37. (a) ln x 3 x e3
38. (a) ln x 1 x e1 1e (b) ln 1e x x ln e1 ln e 1
(b) ln e2 x x 2 ln e 2 1 x 7x 1 x 2 39. (a) log7 49 49
40. (a) x log4 2 log4 412 12
41. (a) log2 12 x 2x 12 x 1
42. (a) logx 1000 3 x 3 1000 x 10
43. (a) logx 16 4 x 4 16 x 2
44. (a) logx 6 12 x 12 6 x 36
(b) log4 x 2 x 42 16
(b) log2 x 5 25 x x 32
(b) logx 25 2 x 2 25 x 5
1 (b) log10 x 3 103 x x 1000
(b) logx 8 32 x 32 8 x 823 4
(b) logx 3 13 x 13 3 x 27
45. (a) log 2 03010
46. (a) log 50 16990 (b) log 2 01505 (c) log 3 2 06276
(b) log 352 15465 (c) log 23 01761
48. (a) ln 27 32958
47. (a) ln 5 16094
(b) ln 739 20001
(b) ln 253 32308 (c) ln 1 3 10051
(c) ln 546 40000
y
49. x 1 33 1 32 1 3
f x 3
x
1
_3
1 43 1 42 1 4
0
2
_1
1
2
3
4
_2
1
1
0
3
1
32
2
y
50.
2
5
6
x
g x 3
2 1 0
2
_1
2
3
4
_2
1
_3
_4
1
0
_4
_5
4
1
_5
42
2
f x log3 x
1
g x log4 x
5
6
x
SECTION 4.3 Logarithmic Functions y
51. x 1 103 1 102 1 10
f x
y
52.
2
x
1
2
g x
2
_3
1 103 1 102 1 10
1
0
_4
1
1
_4
10
2
_5
10
2
_5
102
4
102
3
6
0
4
1
_1
2
3
4
5
6
x
_2
f x 2 log x
417
1
2
0
1
_1
1
2
3
4
5
6
x
_2
0
_3
g x log x 1
53. Since the point 5 1 is on the graph, we have 1 loga 5 a 1 5. Thus the function is y log5 x. 54. Since the point 12 1 is on the graph, we have 1 loga 12 a 1 12 a 2. Thus the function is y log2 x. 55. Since the point 3 12 is on the graph, we have 12 loga 3 a 12 3 a 9. Thus the function is y log9 x.
56. Since the point 9 2 is on the graph, we have 2 loga 9 a 2 9 a 3. Thus the function is y log3 x.
57. The graph of f x 2 ln x is obtained from that of y ln x by shifting it upward 2 units, as in graph I.
58. The graph of f x ln x 2 is obtained from that of y ln x by shifting it to the right 2 units, as in graph II. 59. The graph of y log4 x is obtained from that of y 4x by reflecting it in the line y x. y
60. The graph of y log3 x is obtained from that of y 3x by reflecting it in the line y x. y
y=4¨
y=3¨
y=x
y=log£ x
y=log¢ x
1
1 x
1
61. The graph of g x log5 x is obtained from that of y log5 x by reflecting it about the y-axis.
Domain: 0. Range: . Vertical asymptote: x 0.
y=x
1
x
62. The graph of f x log10 x is obtained from that of y log10 x by reflecting it about the x-axis.
Domain: 0 . Range: . Vertical asymptote: x 0.
y
y
1
1 1
x
1
x
418
CHAPTER 4 Exponential and Logarithmic Functions
63. The graph of f x log2 x 4 is obtained from that of 64. The graph of g x ln x 2 is obtained from that of
y ln x by shifting to the left 2 units. Domain: 2 .
y log2 x by shifting it to the right 4 units.
Range: . Vertical asymptote: x 2.
Domain: 4 . Range: .
y
Vertical asymptote: x 4. y
1 1
1
x
x
1
65. The graph of h x ln x 5 is obtained from that of
y ln x by shifting to the left 5 units. Domain: 5 .
Range: . Vertical asymptote: x 5. y
66. The graph of g x log6 x 3 is obtained from that of y log6 x by shifting to the right 3 units.
Domain: 3 . Range: . Vertical asymptote: x 3. y
1
1 1
x
67. The graph of y 2 log3 x is obtained from that of
y log3 x by shifting upward 2 units. Domain: 0 .
Range: . Vertical asymptote: x 0. y
x
1
68. The graph of y 1 log10 x is obtained from that of y log10 x by reflecting it about the x-axis, and then
shifting it upward 1 unit. Domain: 0 .
Range: . Vertical asymptote: x 0. y
(1, 2) 1
(1, 1)
1 1
x
1
x
SECTION 4.3 Logarithmic Functions
419
69. The graph of y log3 x 1 2 is obtained from that of 70. The graph of y 1 ln x is obtained from that of y log3 x by shifting to the right 1 unit and then
downward 2 units. Domain: 1 . Range: . Vertical asymptote: x 1. 1
y
y ln x by reflecting it about the y-axis and then shifting
it upward 1 unit. Domain: 0. Range: . Vertical asymptote: x 0.
y
x
1
(_1, 1) 1
(2, _2)
1
71. The graph of y ln x is obtained from that of y ln x
by reflecting the part of the graph for 0 x 1 about the
x
ln x if x 0 72. Note that y ln x ln x if x 0
The graph
of y ln x is obtained by combining the graph of
x-axis. Domain: 0 . Range: [0 .
y ln x and its reflection about the y-axis.
Vertical asymptote: x 0.
Domain: 0 0 . Range: . Vertical
y
asymptote: x 0.
y 1 x
1 1 1
x
73. f x log10 x 3. We require that x 3 0 x 3, so the domain is 3 . 74. f x log5 8 2x. Then we must have 8 2x 0 8 2x 4 x, and so the domain is 4. 75. g x log3 x 2 1 . We require that x 2 1 0 x 2 1 x 1 or x 1, so the domain is 1 1 . 76. g x ln x x 2 . Then we must have x x 2 0
x 1 x 0. Using the methods from Chapter 1 with the
endpoints 0 and 1, we get the table at right. Thus the domain is 0 1.
Interval
0
0 1
1
Sign of x
Sign of 1 x
Sign of x 1 x
77. h x ln x ln 2 x. We require that x 0 and 2 x 0 x 0 and x 2 0 x 2, so the domain is 0 2. 78. h x x 2 log5 10 x . Then we must have x 2 0 and 10 x 0 x 2 and 10 x 2 x 10. So the domain is [2 10.
420
CHAPTER 4 Exponential and Logarithmic Functions
79. y log10 1 x 2 has domain 1 1, vertical
asymptotes x 1 and x 1, and local maximum y 0 at x 0.
80. y ln x 2 x ln x x 1 has domain
0 1 , vertical asymptotes x 0 and x 1, and no local maximum or minimum. 10
-2
2 -10
10
-2
-10
81. y x ln x has domain 0 , vertical asymptote x 0, and no local maximum or minimum.
82. y x ln x2 has domain 0 , no vertical asymptote, local minimum y 0 at x 1, and local maximum y 054 at x 014. 10
2 5 -5 0
ln x has domain 0 , vertical asymptote x 0, x horizontal asymptote y 0, and local maximum y 037
83. y
at x 272.
2
4
84. y x log10 x 10 has domain 10 , vertical
asymptote x 10, and local minimum y 362 at x 587.
5 0 10
20 -10
10
-2 -5
85. f x 2x and g x x 1 both have domain , so f g x f g x 2gx 2x1 with domain and g f x g f x 2x 1 with domain . 2 86. f x 3x and g x x 2 1 both have domain , so f g x f g x 3x 1 with domain x 2 1 32x 1 with domain . and g f x g f x 3
87. f x log2 x has domain 0 and g x x 2 has domain , so f g x f g x log2 x 2 with domain 2 and g f x g f x log2 x 2 with domain 0 . 88. f x log x has domain 0 and g x x 2 has domain , so f g x f g x log x 2 with domain 0 0 and g f x g f x log x2 with domain 0 .
SECTION 4.3 Logarithmic Functions
89. The graph of g x
x grows faster than the graph of
90. (a)
f x ln x.
6 4 2
0
91. (a)
6
g
4
f
g
2
f
0
10
20
20
30
(b) From the graph, we see that the solution to the equation x 1 ln 1 x is x 1350.
30
c=4 c=3 c=2 c=1
2
10
92. (a)
c=4
6
c=3
4
c=2
1 2
0
421
10
20
30
0
(b) Notice that f x log cx log c log x, so
c=1 10
20
30
(b) As c increases, the graph of f x c log x
as c increases, the graph of f x log cx is
stretches vertically by a factor of c. shifted upward log c units. 93. (a) f x log2 log10 x . Since the domain of log2 x is the positive real numbers, we have: log10 x 0 x 100 1. Thus the domain of f x is 1 . y x (b) y log2 log10 x 2 y log10 x 102 x. Thus f 1 x 102 .
94. (a) f x ln ln ln x. We must have ln ln x 0 ln x 1 x e. So the domain of f is e . y
(b) y ln ln ln x e y ln ln x ee ln x ee 2x 2x .y y y2x 2x x 12 1 2x y y 2x y2x 2x 1 y 2x 1 y y x log2 . Thus 1y x f 1 x log2 . 1x
95. (a) f x
ey
x
e x. Thus the inverse function is f 1 x ee .
(b)
x 0. Solving this using the methods from 1x Chapter 1, we start with the endpoints, 0 and 1. Interval
0
0 1
1
Sign of x
Sign of 1 x x Sign of 1x
Thus the domain of f 1 x is 0 1.
I 2500 ln 07 89169 moles/liter. I0 D 8267 ln 073 2602 years. 97. Using D 073D0 we have A 8267 ln D0 log N50 log 20,000 98. Substituting N 1,000,000 we get t 3 3 4286 hours. log 2 log 2 ln 2 ln 2 116 years. When r 7% we have t 99 years. And when r 8% we have 99. When r 6% we have t 006 007 ln 2 87 years. t 008 96. Using I 07I0 we have C 2500 ln
422
CHAPTER 4 Exponential and Logarithmic Functions
100. Using k
025 and substituting C 09C0 we have C 025 ln 1 09 025 ln 01 058 hours. t 025 ln 1 C0 log 2 1005 log 40 log 2AW 532. Using A 100 and 101. Using A 100 and W 5 we find the ID to be log 2 log 2 log 2 log 2 10010 log 20 523 log 2AW 432. So the smaller icon is 123 times W 10 we find the ID to be log 2 log 2 log 2 432 harder.
102. (a) Since 2 feet 24 inches, the height of the graph is 224 1677216 inches. Now, since there are 12 inches per foot and ,216 5280 feet per mile, there are 12 5280 63,360 inches per mile. So the height of the graph is 1,677 63360 2648, or about 265 miles. (b) Since log2 224 24, we must be about 224 inches 265 miles to the right of the origin before the height of the
graph of y log2 x reaches 24 inches or 2 feet. 103. log log 10100 log 100 2 log log log 10googol log log googol log log 10100 log 100 2
104. Notice that loga x is increasing for a 1. So we have log4 17 log4 16 log4 42 2. Also, we have log5 24 log5 25 log5 52 2. Thus, log5 24 2 log4 17.
105. The numbers between 1000 and 9999 (inclusive) each have 4 digits, while log 1000 3 and log 10,000 4. Since log x 3 for all integers x where 1000 x 10,000, the number of digits is log x 1. Likewise, if x is an integer where 10n1 x 10n , then x has n digits and log x n 1. Since log x n 1 n log x 1, the number of digits in x is log x 1.
4.4
LAWS OF LOGARITHMS
1. The logarithm of a product of two numbers is the same as the sum of the logarithms of these numbers. So log5 25 125 log5 25 log5 125 2 3 5.
2. The logarithm of a quotient of two numbers is the same as the difference of the logarithms of these numbers. So 25 log 25 log 125 2 3 1. log5 125 5 5
3. The logarithm of a number raised to a power is the same as the power times the logarithm of the number. So log5 2510 10 log5 25 10 2 20. x2 y log x 2 y log z log x 2 log y log z 2 log x log y log z 4. log z x2 y 5. 2 log x log y log z log x 2 log y log z log x 2 y log z log . z
6. (a) Most calculators can find logarithms with base 10 and base e. To find logarithms with different bases we use the change 1079 log 12 of base formula. To find log7 12, we write log7 12 1277. log 7 0845 2485 ln 12 (b) Yes, the result is the same: log7 12 1277. ln 7 1946 7. (a) False. log A B log A log B. (b) True. log AB log A log B.
SECTION 4.4 Laws of Logarithms
A log A log B. B log A log A log B. (b) False. log B
8. (a) True. log
9. log 50 log 200 log 50 200 4 60 2 11. log2 60 log2 15 log2 15
10. log6 9 log6 24 log6 9 24 3 12. log3 135 log3 45 log3 135 45 1
13. 14 log3 81 14 4 1 15. log5 5 log5 512 12
17.
18. 19. 20. 21. 22.
14. 13 log3 27 13 3 1 16. log5 1 log5 532 32 125 2 6 log2 6 log2 15 log2 20 log2 15 log2 20 log2 5 20 log2 8 log2 23 3 100 log3 19 log3 32 2 log3 100 log3 18 log3 50 log3 18 50 100 log4 16100 log4 42 log4 4200 200 33 log2 833 log2 23 log2 299 99 log log 1010,000 log 10,000 log 10 log 10,000 1 log 10,000 log 104 4 log 10 4 200 ln e200 ln e ln e200 200 ln e 200 ln ln ee
23. log3 8x log3 8 log3 x
24. log6 7r log6 7 log6 r
25. log3 2x y log3 2 log3 x log3 y
26. log5 4st log5 4 log5 s log5 t 28. log t 5 log t 52 52 log t 30. ln ab 12 ln ab 12 ln a ln b 32. log3 x y log3 x log3 y log3 x 12 log3 y r ln r ln 3 ln s 34. ln 3s s5 5 log2 s log2 7 2 log2 t 36. log2 7t 2
27. ln a 3 3 ln a
29. log2 x y10 10 log2 x y 10 log2 x log2 y 31. log2 AB 2 log2 A log2 B 2 log2 A 2 log2 B
2x log3 2 log3 x log3 y y 3x 2 log5 3 2 log5 x 3 log5 y log5 y3 y3 3x 5 12 52 log3 x log3 y log3 38. log 3 log y 12 log 2 12 log x y 2x 3 4 x y log x 3 y 4 log z 6 3 log x 4 log y 6 log z log z6 x2 2 log log yz 3 2 loga x loga y 3 loga z 2 loga x loga y 3 loga z loga x a a yz 3 ln x 4 2 12 ln x 4 2 3 log x 2 4 13 log x 2 4 y y 1 ln x 2 ln ln x 12 ln y ln z ln x z z 3x 2 2 ln x 110 ln 3 2 ln x 10 ln x 1 ln ln 3x x 110
33. log3 35.
37. 39.
40. 41. 42. 43. 44.
423
424
CHAPTER 4 Exponential and Logarithmic Functions
x 2 y 2 14 log x 2 y 2 x log x log 3 1 x log x 13 log 1 x log 3 1x 2 x2 4 x2 4 1 1 2 2 3 log 2 2 log 2 2 log x 4 log x 1 x 7 x2 1 x3 7 x2 1 x3 7 12 log x 2 4 log x 2 1 2 log x 3 7 log x y z 12 log x y z 12 log x log y z 12 log x 12 log y z 12 log x 12 log y 12 log z 12 log x 14 log y 18 log z log4 6 2 log4 7 log4 6 log4 72 log4 6 72 log4 294
45. log 46.
47.
48.
49.
4
50. 12 log2 5 2 log2 7 log2 5 log2 49 log2 495
51. 2 log x 3 log x 1 log x 2 log x 13 log
x2 x 13
8x 2 52. 3 ln 2 2 ln x 12 ln x 4 ln 23 ln x 2 ln x 4 ln x 4 3 1 53. 4 log x 3 log x 2 1 2 log x 1 log x 4 log x 2 1 log x 12 x4 x 4 x 12 2 log x 1 log log 3 2 3 2 x 1 x 1
x2 1 x 1 x 1 log5 log5 x 1 54. log5 x 2 1 log5 x 1 log5 x 1 x 1 a 2 b2 55. ln a b ln a b 2 ln c ln a b a b ln c2 ln c2 2 x y2 x 2 y4 x y2 56. 2 log5 x 2 log5 y 3 log5 z 2 log5 3 log5 log5 6 3 z z z 2 x4 3 13 log x 2 12 log 57. 13 log x 23 12 log x 4 log x 2 x 6 2 x2 x 6 12 x2 x 2 x 2 x2 x4 log x 2 log log 2 log log log x x 3 x 3 x 2 x 3 x 2 [x 3 x 2]2 bd c 58. loga b c loga d r loga s loga bd c loga s r loga r s log 5 log 2 2321928 60. log5 2 0430677 59. log2 5 log 2 log 5
61. log3 16
log 16 2523719 log 3
62. log6 92
log 92 2523658 log 6
63. log7 261
log 261 0493008 log 7
64. log6 532
log 532 3503061 log 6
65. log4 125
log 125 3482892 log 4
66. log12 25
log 25 0368743 log 12
SECTION 4.4 Laws of Logarithms
67. log3 x
ln x 1 1 loge x ln x. The graph of y ln x is loge 3 ln 3 ln 3 ln 3
425
2
shown in the viewing rectangle [1 4] by [3 2]. 2
4
-2
68. Note that logc x
1 ln x (by the change of base formula). So ln c
a=2 a=e a=5 a=10
2
the graph of y logc x is obtained from the graph of y ln x by
1 depending ln c on whether ln c 1 or ln c 1. All of the graphs pass through 1 0 either shrinking or stretching vertically by a factor of
0
2
4
-2
because logc 1 0 for all c.
log 5 log 7 log 7 70. log2 5 log5 7 log2 7 log 2 log 5 log 2 1 x x2 1 x x2 1 1 ln ln x2 x2 1 x x2 1 x x2 1 x x2 1 ln x x 2 1
1 ln e ln 10 ln 10 71. ln x x 2 1 ln 69. log e
80 P0 . Substituting P0 80, t 24, and c 03 we have P 305. So the t 1c 24 103 student should get a score of 30. c c 73. (a) log P log c k log W log P log c log W k log P log P k. Wk W 8000 8000 (b) Using k 21 and c 8000, when W 2 we have P 21 1866 and when W 10 we have P 21 64. 2 10 74. (a) log S log c k log A log S log c log Ak log S log c Ak S c Ak . 72. From Example 5(a), P
(b) If A 2A0 when k 3 we get S c 2A0 3 c 23 A30 8 c A30 . Thus doubling the area increases the species eightfold.
75. (a) M 25 log BB0 25 log B 25logB0 .
(b) Suppose B1 and B2 are the brightness of two stars such that B1 B2 and let M1 and M2 be their respective magnitudes. Since log is an increasing function, we have log B1 log B2 . Then log B1 log B2 log B1 log B0 log B2 log B0 log B1 B0 log B2 B0 25 log B1 B0 25 log B2 B0 M1 M2 . Thus the brighter star has less magnitudes.
(c) Let B1 be the brightness of the star Albiero. Then 100B1 is the brightness of Betelgeuse, and its magnitude is M 25log 100B1 B0 25 log 100 log B1 B0 25 2 log B1 B0 5 25 log B1 B0 5 magnitude of Albiero
76. (a) False; log xy log x log y log x log y.
(b) False; log2 x log2 y log2 xy log2 x y. (c) True; the equation is an identity: log5 ab2 log5 a log5 b2 log5 a 2 log5 b. (d) True; the equation is an identity: log 2z z log 2.
(e) False; log P log Q log P Q log P log Q. (f) False; log a log b log ab log a log b.
426
CHAPTER 4 Exponential and Logarithmic Functions
x (g) False; x log2 7 log2 7x log2 7 .
(h) True; the equation is an identity. loga a a a loga a a 1 a.
(i) False; log x y log x log y. For example, 0 log 3 2 log 3 log 2. (j) True; the equation is an identity: ln 1A ln A1 1 ln A ln A.
77. The error is on the first line: log 01 0, so 2 log 01 log 01.
78. Let f x x 2 . Then f 2x 2x2 4x 2 4 f x. Now the graph of f 2x is the same as the graph of f shrunk
horizontally by a factor of 12 , whereas the graph of 4 f x is the same as the graph of f x stretched vertically by a factor of 4. Let g x e x . Then g x 2 e x2 e2 e x e2 g x. This shows that a horizontal shift of 2 units to the right is the same as a vertical stretch by a factor of e2 .
Let h x ln x. Then h 2x ln 2x ln 2 ln x ln 2 h x. This shows that a horizontal shrinking by a factor of 12 is the same as a vertical shift upward by ln 2.
4.5
EXPONENTIAL AND LOGARITHMIC EQUATIONS
1. (a) First we isolate e x to get the equivalent equation e x 25.
(b) Next, we take the natural logarithm of each side to get the equivalent equation x ln 25.
(c) Now we use a calculator to find x 3219.
2. (a) First we combine the logarithms to get the equivalent equation log 3 x 2 log x.
(b) Next, we write each side in exponential form to get the equivalent equation 3 x 2 x. (c) Now we find x 3.
3. Because the function 5x is one-to-one, 5x1 125 5x1 53 x 1 3 x 4. 2
4. Because the function e x is one-to-one, e x e9 x 2 9 x 3.
5. Because the function 5x is one-to-one, 52x3 1 52x3 50 2x 3 0 x 32 . 1 2x 3 1 x 1. 6. Because the function 10x is one-to-one, 102x3 10
7. Because the function 7x is one-to-one, 72x3 765x 2x 3 6 5x 3x 9 x 3. 8. Because the function e x is one-to-one, e12x e3x5 1 2x 3x 5 5x 6 x 65 .
2 2 9. Because the function 6x is one-to-one, 6x 1 61x x 2 1 1 x 2 2x 2 2 x 1.
2 2 10. Because the function 10x is one-to-one, 102x 3 109x 2x 2 3 9 x 2 3x 2 12 x 2.
11. (a) 10x 25 log 10x log 25 x log 10 log 25 x log 25 2 log 5 (b) x 1397940
12. (a) 10x 4 log 10x log 4 x log 4 x log 4 (b) x 0602060
13. (a) e5x 10 ln e5x ln 10 5x ln e ln 10 5x ln 10 x 15 ln 10 (b) x 0460517
14. (a) e04x 8 ln e04x ln 8 04x ln 8 x 52 ln 8 (b) x 5198604
3 log 3 15. (a) 21x 3 log 21x log 3 1 x log 2 log 3 1 x log log 2 x 1 log 2 (b) x 0584963
SECTION 4.5 Exponential and Logarithmic Equations
427
1 5 log 5 log 5 16. (a) 32x1 5 log 32x1 log 5 2x 1 log 3 log 5 2x 1 log log 3 2x 1 log 3 x 2 1 log 3 (b) x 1232487 10 17. (a) 3e x 10 e x 10 3 x ln 3 (b) x 1203973
17 1 17 18. (a) 2e12x 17 e12x 17 2 12x ln 2 x 12 ln 2 (b) x 0178339
10 10 12t 10 12t ln 41 ln 10 12t ln 3 t ln 3 19. (a) 300 102512t 1000 41 40 3 40 3 ln 41 12 ln 41 40 40 (b) x 4063202 10t ln 5 5 10t ln 11 20. (a) 10 137510t 50 11 8 8 ln 5 t 10 ln 11 8
(b) t 0505391
21. (a) e14x 2 1 4x ln 2 4x 1 ln 2 x 14 1 ln 2 (b) x 0076713 22. (a) e35x 16 3 5x ln 16 5x ln 16 3 x 15 ln 16 3 (b) x 0045482
ln 15 5 ln 15 ln 15 7x 5 x 23. (a) 257x 15 5 7x ln 2 ln 15 5 7x ln 2 ln 2 7 7 ln 2 (b) x 0156158 34 24. (a) 23x 34 log 23x log 34 3x log 2 log 34 x 3log log 2 (b) x 1695821 x log 01 log 3 log 01 x 14log 25. (a) 3x14 01 log 3x14 log 01 3 14 (b) x 29342646 x log 2 log 5 log 2 x 100 26. (a) 5x100 2 log 5x100 log 2 log 5 100 (b) x 43067656 27. (a) 4 1 105x 9 1 105x 94 105x 54 5x log 54 x 15 log 54
(b) x 0019382 ln 45 ln 45 28. (a) 2 5 3x1 100 5 3x1 50 3x1 45 x 1 ln 3 ln 45 x 1 x 1 ln 3 ln 3 (b) x 2464974 29. (a) 8 e14x 20 e14x 12 1 4x ln 12 x
1 ln 12 4
(b) x 0371227 30. (a) 1 e4x1 20 e4x1 19 4x 1 ln 19 x
ln 19 1 4
(b) x 0486110 ln 50 3 2x ln 2 ln 50 ln 3 x 31. (a) 4x 212x 50 22x 212x 50 22x 1 2 50 22x 50 3 ln 4 (b) x 2029447
428
CHAPTER 4 Exponential and Logarithmic Functions
32. (a) 125x 53x1 200 53x 53x1 200 53x 1 5 200 53x 100 3 3x ln 5 ln 100 ln 3 100 ln 3 x 3 ln 5 (b) x 0726249 33. (a) 5x 4x1 log 5x log 4x1 x log 5 x 1 log 4 x log 4 log 4 x log 5 x log 4 log 4 log 4 x log 5 log 4 log 4 x log 54 (b) x 6212567 34. (a) 101x 6x log 101x log 6x 1 x x log 6 1 x log 6 x 1 x log 6 1 x log 161 (b) x 0562382 35. (a) 23x1 3x2 log 23x1 log 3x2 3x 1 log 2 x 2 log 3 3x log 2 log 2 x log 3 2 log 3 3x log 2 x log 3 log 2 2 log 3 x 3 log 2 log 3 log 2 2 log 3 log 2 32 log 18 log 22 log 3 x 3 log 2log 3 log 23 3 log 8 3
(b) x 2946865
x x 36. (a) 7x2 51x log 7x2 log 51x log 7 1 x log 5 log 7 log 5 x log 5 2 2 x log 7 x log 5 log 5 x 12 log 7 log 5 log 5 x 1 log 5 2 2 log 7log 5 (b) x 0623235 50 4 50 4 4ex 46 4ex 115 ex ln 115 x x ln 115 1 ex (b) x 2442347
37. (a)
10 2 10 2 2ex 8 2ex 4 ex ln 4 x x ln 4 1 ex (b) x 1386294 39. e2x 3e x 2 0 e x 1 e x 2 0 e x 1 0 or e x 2 0. If e x 1 0, then e x 1 x ln 1 0. If 38. (a)
e x 2 0, then e x 2 x ln 2 06931. So the solutions are x 0 and x 06931. 40. e2x e x 6 0 e x 3 e x 2 0 e x 2 0 (impossible) or e x 3 0. If e x 3 0, then e x 3 x ln 3 10986. So the only solution is x 10986. 41. e4x 4e2x 21 0 e2x 7 e2x 3 0 e2x 7 or e2x 3. Now e2x 7 has no solution, since e2x 0
for all x. But we can solve e2x 3 2x ln 3 x 12 ln 3 05493. So the only solution is x 05493. 42. 34x 32x 6 0 32x 3 32x 2 0 32x 3 or 32x 2. The latter equation has no solution, so we solve
32x 3 2x 1 x 12 . 43. 2x 10 2x 3 0 2x 22x 10 3 2x 0 2x 2x 5 2x 2 0. The first two factors are positive
everywhere, so we solve 2x 2 0 x 1. 44. e x 15ex 8 0 ex e2x 15 8e x 0 ex e x 5 e x 3 0. The first factor is positive everywhere,
so the solutions occur where e x 5 or e x 3; that is, x ln 3 10986 or x ln 5 16094. 45. x 2 2x 2x 0 2x x 2 1 0 2x 0 (never) or x 2 1 0. If x 2 1 0, then x 2 1 x 1. So the only solutions are x 1.
SECTION 4.5 Exponential and Logarithmic Equations
429
46. x 2 10x x10x 2 10x x 2 10x x10x 2 10x 0 10x x 2 x 2 0 10x 0(never) or x 2 x 2 0. If x 2 x 2 0, then x 2 x 1 0 x 2, 1. So the only solutions are x 2, 1.
47. 4x 3 e3x 3x 4 e3x 0 x 3 e3x 4 3x 0 x 0 or e3x 0 (never) or 4 3x 0. If 4 3x 0, then
3x 4 x 43 . So the solutions are x 0 and x 43 . 48. x 2 e x xe x e x 0 e x x 2 x 1 0 e x 0 (impossible) or x 2 x 1 0. If x 2 x 1 0, then
5 . So the solutions are x 1 5 . x 1 2 2
49. log x log x 1 log 4x log [x x 1] log 4x x 2 x 4x x 2 5x 0 x x 5 0 x 0 or x 5. So the possible solutions are x 0 and x 5. However, when x 0, log x is undefined. Thus the only solution is x 5. 50. log5 x log5 x 1 log5 20 log5 x 2 x log5 20 x 2 x 20 x 2 x 20 0 x 5 x 4 0
x 5 or x 4. Since log5 5 is undefined, the only solution is x 4. 51. 2 log x log 2 log 3x 4 log x 2 log 6x 8 x 2 6x 8 x 2 6x 8 0 x 4 x 2 0 x 4 or x 2. Thus the solutions are x 4 and x 2. 52. ln x 12 ln 2 2 ln x ln 2 x 12 ln x 2 2x 1 x 2 x 2 2x 1 0 x 12 0 x 1. Thus the only solution is x 1.
53. log2 3 log2 x log2 5 log2 x 2 log2 3x log2 5x 10 3x 5x 10 2x 10 x 5
54. log4 x 2 log4 3 log4 5 log4 2x 3 log4 3 x 2 log4 5 2x 3 3x 6 10x 15 7x 21 x 3
55. ln x 10 x e10 22,026
56. ln 2 x 1 2 x e1 x e 2 07183 57. log x 2 x 102 001
58. log x 4 3 x 4 103 1000 x 1004
59. log 3x 5 2 3x 5 102 100 3x 95 x 95 3 316667
60. log3 2 x 3 2 x 33 27 x 25 x 25
61. 4 log 3 x 3 log 3 x 1 3 x 10 x 7 62. log2 x 2 x 2 2 x 2 x 2 22 4 x 2 x 6 0 x 3 x 2 0 x 3 or x 2. Thus the solutions are x 3 and x 2.
63. log2 x log2 x 3 2 log2 [x x 3] 2 x 2 3x 22 x 2 3x 4 0 x 4 x 1 x 1 or x 4. Since log 1 3 log 4 is undefined, the only solution is x 4. 64. log x log x 3 1 log [x x 3] 1 x 2 3x 10 x 2 3x 10 0 x 2 x 5 0 x 2 or x 5. Since log 2 is undefined, the only solution is x 5.
65. log9 x 5 log9 x 3 1 log9 [x 5 x 3] 1 x 5 x 3 91 x 2 2x 24 0 x 6 x 4 0 x 6 or4. However, x 4 is inadmissible, so x 6 is the only solution.
66. ln x 1 ln x 2 1 ln [x 1 x 2] 1 x 2 x 2 e x 2 x 2 e 0
x 1 142e 1 2 94e . Since x 1 0 when x 1 2 94e , the only solution is x 1 2 94e 17290 2 x 1 x 1 2 52 x 1 25x 25 24x 26 x 13 67. log5 x 1 log5 x 1 2 log5 12 x 1 x 1 x 15 x 15 x 15 2 32 9 x 15 9 x 1 68. log3 x 15 log3 x 1 2 log3 x 1 x 1 x 1 x 15 9x 9 8x 24 x 3
430
CHAPTER 4 Exponential and Logarithmic Functions
69. ln x 3 x ln x x 3 0. Let f x ln x x 3. We need to solve the equation f x 0. From the graph of f , we get x 221.
70. log x x 2 2 log x x 2 2 0. Let
f x log x x 2 2. We need to solve the equation
f x 0. From the graph of f , we get x 001 or
x 147.
2 0 2
4 0 1
2
-2
71. x 3 x log10 x 1 x 3 x log10 x 1 0.
Let f x x 3 x log10 x 1. We need to solve the
equation f x 0. From the graph of f , we get x 0 or x 114.
72. x ln 4 x 2 x ln 4 x 2 0. Let f x x ln 4 x 2 . We need to solve the equation f x 0. From the graph of f , we get x 196 or
x 106.
2
-1
1
2 -2
2
-2
73. e x x e x x 0. Let f x e x x. We need to
solve the equation f x 0. From the graph of f , we get x 057.
74. 2x x 1 2x x 1 0. Let
f x 2x x 1. We need to solve the equation
f x 0. From the graph of f , we get x 138.
2
2
0 -1.0
-0.5
-1 -2
75. 4x x 4x x 0. Let f x 4x x.
We need to solve the equation f x 0. From the graph of f , we get x 036.
1
2
-2 2
2
76. e x 2 x 3 x e x 2 x 3 x 0. Let 2
f x e x 2 x 3 x. We need to solve the equation
f x 0. From the graph of f , we get x 089 or
x 071.
2
2 0 1
2 -2
2
-2 -2
SECTION 4.5 Exponential and Logarithmic Equations
431
2 1 log 22 log5 x log 1 4 log5 x 12 x 512 1 04472 77. 22 log5 x 16 2 2 16 5 log5 x 78. log2 log3 x 4 log3 x 24 16 x 316 43,046,721 79. log x 2 log 9 x 1 log [x 2 9 x] 1 log x 2 11x 18 1 x 2 11x 18 101 0 x 2 11x 28 0 x 7 x 4. Also, since the domain of a logarithm is positive we must have
0 x 2 11x 18 0 x 2 9 x. Using the methods from Chapter 1 with the endpoints 2, 4, 7, 9 for the intervals, we make the following table: 2
2 4
4 7
7 9
9
Sign of x 7
Sign of x 4
Sign of x 2
Sign of 9 x
Interval
Sign of x 7 x 4 Sign of x 2 9 x
Thus the solution is 2 4 7 9.
80. 3 log2 x 4 23 x 24 8 x 16.
81. 2 10x 5 log 2 x log 5 03010 x 06990. Hence the solution to the inequality is approximately the interval 03010 06990. 82. x 2 e x 2e x 0 e x x 2 2 0 e x x 2 x 2 0. We use the methods of Chapter 1 with the endpoints 2 and 2, noting that e x 0 for all x. We make a table: Interval Sign of e x Sign of x 2 Sign of x 2 Sign of e x x 2 x 2
Thus 2 x 2.
2
2 2
2
ln y . 83. To find the inverse of f x 22x , we set y f x and solve for x. y 22x ln y ln 22x 2x ln 2 x 2 ln 2 ln x ln x Interchange x and y: y . Thus, f 1 x . 2 ln 2 2 ln 2 84. To find the inverse of f x 3x1 , we set y f x and solve for x. y 3x1 ln y ln 3x1 x 1 ln 3 x 1
ln y ln y ln x ln x x 1. Interchange x and y: y 1. Thus, f 1 x 1. ln 3 ln 3 ln 3 ln 3
85. To find the inverse of f x log2 x 1, we set y f x and solve for x. y log2 x 1 2 y 2log2 x1 x 1 x 2 y 1. Interchange x and y: y 2x 1. Thus, f 1 x 2x 1.
86. To find the inverse of f x log 3x, we set y f x and solve for x. y log 3x 10 y 3x x x and y: y
10x 10x . Thus, f 1 x . 3 3
10 y . Interchange 3
432
CHAPTER 4 Exponential and Logarithmic Functions
87. log x 3 log x log 3 log x 3 log 3x x 3 3x 2x 3 x 32 88. log x3 3 log x log x3 3 log x 0 log x log x2 3 log x 0 or log x2 3 0. Now log x 0 x 1. Also log x2 3 0 log x2 3 log x 3 x 10 3 , so x 10 3 539574 or
x 10 3 00185. Thus the solutions to the equation are x 1, x 10 3 539574 and x 10 3 00185. 0085 43 89. (a) A 3 5000 1 5000 10212512 643509. Thus the amount after 3 years is $6,43509. 4 0085 4t 5000 1021254t 2 1021254t log 2 4t log 102125 (b) 10000 5000 1 4 log 2 824 years. Thus the investment will double in about 824 years. t 4 log 102125
90. (a) A 2 6500e0062 $732873
91.
92.
93. 94.
95.
96.
006t ln 16 006t t 1 ln 16 346. So the investment doubles in (b) 8000 6500e006t 16 13 e 13 13 006 1 about 3 2 years. 0075 4t 8000 5000 1 5000 1018754t 16 1018754t log 16 4t log 101875 4 log 16 t 633 years. The investment will increase to $8000 in approximately 6 years and 4 months. 4 log 101875 00975 2t log 125 5000 4000 1 2344. So it 125 1048752t log 125 2t log 104875 t 2 2 log 104875 1 takes about 2 years to save $5000. 3 ln 2 2 e0085t ln 2 0085t t 815 years. Thus the investment will double in about 815 years. 0085 r 24 r 8 r r 143577 1000 1 8 143577 8 143577 1 143577 1 1 2 2 2 2 r 2 8 143577 1 00925. Thus the rate was about 925%. ln 3 15e0087t 5 e0087t 13 0087t ln 13 ln 3 t 126277. So only 5 grams remain after 0087 approximately 13 days. We want to solve for t in the equation 80 e02t 1 70 (when motion is downwards, the velocity is negative). Then ln 18 80 e02t 1 70 e02t 1 78 e02t 18 02t ln 18 t 104 seconds. Thus the 02 velocity is 70 ft/s after about 10 seconds. 10 7337, so there are approximately 7337 fish after 3 years. 1 4e083 10 08t 1 e08t 025 08t ln 025 (b) We solve for t. 5 1 4e08t 10 5 2 4e 1 4e08t ln 025 173. So the population will reach 5000 fish in about 1 year and 9 months. t 08
97. (a) P 3
98. (a) I 10e000830 10e024 787. So at 30 ft the intensity is 787 lumens. ln 12 (b) 5 10e0008x e0008x 12 0008x ln 12 x 866. So the intensity drops to 25 lumens 0008 at 866 ft.
SECTION 4.6 Modeling with Exponential Functions
99. (a) ln
P P0
433
P h ehk P P0 ehk . Substituting k 7 and P0 100 we get P 100eh7 . k P0
(b) When h 4 we have P 100e47 5647 kPa. T 20 T 20 011t e011t T 20 200e011t T 20 200e011t 100. (a) ln 200 200 (b) When t 20 we have T 20 200e01120 20 200e22 422 F. 13t5 13 I 1 e13t5 e13t5 1 13 I 13 t ln 1 13 I 1 e 101. (a) I 60 13 60 60 5 60 5 13 t 13 ln 1 60 I . 5 ln 1 13 2 0218 seconds. (b) Substituting I 2, we have t 13 60 102. (a) P M Cekt Cekt M P ekt 1 MP MP t ln kt ln C k C
MP C
(c)
(b) P t 20 14e0024t . Substituting M 20, C 14, k 0024, 1 MP , we have and P 12 into t ln k C 1 20 12 t ln 2332. So it takes about 23 months. 0024 14
20 10 0 0
100
200
103. Since 91 9, 92 81, and 93 729, the solution of 9x 20 must be between 1 and 2 (because 20 is between 9 and 81), whereas the solution to 9x 100 must be between 2 and 3 (because 100 is between 81 and 729). 1 log x 1, so 1 log x 101 for all x 0. So 104. Notice that log x 1 log x log x x 1 log x 5 has no solution, and x 1 log x k has a solution only when k 10.
10
This is verified by the graph of f x x 1 log x .
0 0
5
10
105. (a) x 1logx1 100 x 1 log x 1logx1 log 100 x 1 2 log x 1 log x 1 log 100 log x 1 log x 1 log x 1 2 0 log x 1 2 log x 1 1 0. Thus either log x 1 2 x 101 or log x 1 1 x 11 10 . (b) log2 x log4 x log8 x 11 log2 x log2 x log2 3 x 11 log2 x x 3 x 11 log2 x 116 11 6 11 6 log2 x 11 log2 x 6 x 2 64
2 ln 3 or 2x 1, which (c) 4x 2x1 3 2x 2 2x 3 0 2x 3 2x 1 0 either 2x 3 x ln 2 ln 3 has no real solution. So x is the only real solution. ln 2
4.6
MODELING WITH EXPONENTIAL FUNCTIONS
1. (a) Here n 0 10 and a 15 hours, so n t 10 2t15 10 22t3 . (b) After 35 hours, there will be n 35 10 22353 106 108 bacteria.
434
CHAPTER 4 Exponential and Logarithmic Functions
2t 3 ln 1000 ln 2 ln 1000 t 149, (c) n t 10 22t3 10,000 22t3 1000 ln 22t3 ln 1000 3 2 ln 2 so the bacteria count will reach 10,000 in about 149 hours. 2. (a) Here n 0 25 and a 5 hours, so n t 25 2t5 . (b) After 18 hours, there will be n 18 25 2185 303 bacteria. t (c) n t 25 2t5 1,000,000 2t5 40,000 ln 2t5 ln 40,000 ln 2 ln 40,000 5 5 ln 40,000 764, so the bacteria count will reach 1,000,000 in about 764 hours. t ln 2 3. (a) A model for the squirrel population is n t n 0 2t6 . We are given
n
(c)
1,000,000
that n 30 100,000, so n 0 2306 100,000
800,000
100,000 3125. Initially, there were approximately 25 3125 squirrels. n0
600,000 400,000 200,000
(b) In 10 years, we will have t 40, so the population will be n 40 3125 2406 317,480 squirrels.
4. (a) A model for the bird population is n t n 0 2t10 . We are given
0
20
30
40
50 t (years)
n
(c)
that n 25 13,000, so n 0 22510 13,000
60,000
n0
40,000
50,000
13,000 2298. Initially, there were approximately 2300 birds. 252
30,000
(b) In 5 years, we will have t 30, so the population will be
20,000 10,000
approximately n 30 2298 23010 18,384 18,400 birds.
5. (a) r 008 and n 0 18,000. Thus the population is given by the
10
10
0
(d)
20
30
40 t (years)
n 40,000
formula n t 18,000e008t .
30,000
(b) t 2021 2013 8. Then we have
20,000
n 8 18,000e0088 18,000e064 34,100. Thus there should
10,000
be about 34,100 foxes in the region by the year 2021.
1
2
3
4
5
6
7
8 t (years)
(c) Solving n t 25,000, we get 18,000e008t 25,000 1 ln 25 ln 18 41, so the fox 18e008t 25 ln 18e008t ln 25 ln 18 008t ln 25 t 008 population will reach 25,000 after about 41 years.
6. (a) Taking t 0 in the year 2010 and measuring n in millions, we have n 0 12 and r 0012. Therefore, an exponential model is
n t 12e0012t .
(b) In 2015, t 5, so the population was approximately
n 5 12e00125 12742, or 12,742,000 fish. (c) Solving n t 14, we get 12e0012t 14 ln 12e0012t ln 14
1 ln 14 ln 12 128, so the ln 12 0012t ln 14 t 0012 fish population will reach 14 million after about 128 years.
(d)
n (millions) 15 10 5 0
2020
2030 year
SECTION 4.6 Modeling with Exponential Functions
435
7. n t n 0 ert ; n 0 110 million, t 2036 2011 25.
(a) r 003; n 25 110,000,000e00325 110,000,000e075 232,870,000. Thus at a 3% growth rate, the projected population will be approximately 233 million people by the year 2036.
(b) r 002; n 25 110,000,000e00225 110,000,000e050 181,359,340. Thus at a 2% growth rate, the projected population will be approximately 181 million people by the year 2036. 8. (a) In this case, a model for the bacteria population is n t n 0 ert 22e012t , so after 24 hours the population is approximately n 24 22e01224 392 bacteria.
(b) In this case, a model is n t n 0 ert 22e005t , so after 24 hours the population is approximately n 24 22e00524 73 bacteria.
9. (a) The doubling time is 18 years and the initial population is 112,000, so
(c)
a model is n t 112,000 2t18 . (b) We need to find the relative growth rate r. Since the population is 2 112,000 224,000 when t 18, we have 224,000 112,000e18r 2 e18r ln 2 18r r ln182 00385. Thus, a model is
n 800,000 700,000 600,000 500,000 400,000 300,000 200,000 100,000 0
10 20 30 40 n t 112,000e00385t . t18 t18 (d) Using the model in part (a), we solve the equation n t 112,000 2 500,000 2 125 28 125 18 ln 28 t 125 3885. Therefore, it takes about 3885 years for the ln 2t18 ln 125 28 18 ln 2 ln 28 t ln 2 population to reach 500,000.
10. (a) The doubling time is 25 years and the initial population is 350,000, so a model is n t 350,000 2t25 . (b) r ln252 00277, so a model is n t 350,000e00277t .
(c)
t
n 2,000,000 1,000,000
(d) We solve the equation n t 350,000 2t25 2,000,000 t25 ln 40 t ln 2 ln 40 2t25 40 7 ln 2 7 7 25
0
20
40
60
t
25 ln 40 7 6286. Therefore, it takes about 629 years for the t
ln 2 population to reach 2,000,000.
11. (a) The deer population in 2010 was 20,000. (b) Using the model n t 20,000ert and the point 4 31000, we have 31,000 20,000e4r 155 e4r 4r ln 155 r 14 ln 155 01096. Thus n t 20,000e01096t
(c) n 8 20,000e010968 48,218, so the projected deer population in 2018 is about 48,000. ln 5 1468. Thus, it takes about 147 years (d) 100,000 20,000e01096t 5 e01096t 01096t ln 5 t 01096 for the deer population to reach 100,000. 12. (a) From the graph, we see that the initial bullfrog population was 100. (b) We use a model of the form n t n 0 ert with n 0 100. Because we know that the population was 225 at t 2, we
04055t . solve n 2 225 100e2r 225 r 12 ln 225 100 04055. Thus, a model is n t 100e (c) The estimated population after 15 years is n 15 100e0405515 43,800 frogs.
436
CHAPTER 4 Exponential and Logarithmic Functions
(d) The population will reach 75,000 when n t 100e04055t 75,000 e04055t 750 t
ln 750 1632. So it 04055
will take about 163 years for the population to reach 75,000. 13. (a) Using the formula n t n 0 ert with n 0 8600 and n 1 10000, we solve for r, giving 10000 n 1 8600er 50 r 01508t . 50 43 e r ln 43 01508. Thus n t 8600e (b) n 2 8600e015082 11627. Thus the number of bacteria after two hours is about 11,600. ln 2 (c) 17200 8600e01508t 2 e01508t 01508t ln 2 t 4596. Thus the number of bacteria will 01508 double in about 46 hours.
14. (a) Using n t n 0 ert with n 2 400 and n 6 25,600, we have n 0 e2r 400 and n 0 e6r 25,600. Dividing the
n e6r 25,600 second equation by the first gives 0 2r 64 e4r 64 4r ln 64 r 14 ln 64 104. Thus the 400 n0e relative rate of growth is about 104%.
(b) Since r 14 ln 64 12 ln 8, we have from part (a) n t n 0 e
1 2
ln 8 t
. Since n 2 400, we have 400 n 0 eln 8
400 n 0 ln 8 400 8 50. So the initial size of the culture was 50. e (c) Substituting n 0 50 and r 104, we have n t n 0 ert 50e104t .
(d) n 45 50e10445 50e468 53885, so the size after 45 hours is approximately 5400. ln 1000 664. Hence the population will (e) n t 50,000 50e104t e104t 1000 104t ln 1000 t 104 reach 50,000 after roughly 6 hours 40 minutes. 15. (a) Calculating dates relative to 1990 gives n 0 2976 and n 10 3387. Then n 10 2976e10r 3387
1 0012936t million e10r 3387 2976 11381 10r ln 11381 r 10 ln 11381 0012936. Thus n t 2976e people. ln 2 5358, so the population (b) 2 2976 2976e0012936t 2 e0012936t ln 2 0012936t t 0012936 doubles in about 54 years.
(c) t 2010 1990 20, so our model gives the 2010 population as n 20 2976e001293620 3855 million. The actual population was estimated at 3696 million in 2009. 16. (a) 2n 0 n 0 e002t 2 e002t 002t ln 2 t 50 ln 2 3466. So at the current growth rate, it will take approximately 3466 years for the population to double. (b) 3n 0 n 0 e002t 3 e002t 002t ln 3 t 50 ln 3 5493. So at the current growth rate, it will take approximately 5493 years for the population to double. 17. (a) Because the half-life is 1600 years and the sample weighs 22 mg initially, a suitable model is m t 22 2t1600 . ln 2 ln 2 (b) From the formula for radioactive decay, we have m t m 0 ert , where m 0 22 and r 0000433. h 1600 Thus, the amount after t years is given by m t 22e0000433t . (c) m 4000 22e00004334000 389, so the amount after 4000 years is about 4 mg.
9 e0000433t (d) We have to solve for t in the equation 18 22 e0000433t . This gives 18 22e0000433t 11 9 ln 11 9 t 0000433t ln 11 4634, so it takes about 463 years. 0000433
18. (a) Because the half-life is 30 years and the sample weighs 10 g initially, a suitable model is m t 10 2t30 . ln 2 ln 2 (b) Using m t m 0 ert with m 0 10 and h 30, we have r 00231. Thus m t 10e00231t . h 30
SECTION 4.6 Modeling with Exponential Functions
437
(c) m 80 10e0023180 16 grams.
19.
20.
21.
22.
ln 5 (d) 2 10e00231t 15 e00231t ln 15 00231t t 70 years. 00231 ln 2 , so m t 50e[ln 228]t . We need to solve for t in the By the formula in the text, m t m 0 ert where r h 28 ln 2 32 32 equation 32 50e[ln 228]t . This gives e[ln 228]t 32 50 28 t ln 50 t ln 2 ln 50 1803, so it takes about 18 years. ln 2 From the formula for radioactive decay, we have m t m 0 ert , where r . Since h 30, we have h ln 2 00231 and m t m 0 e00231t . In this exercise we have to solve for t in the equation 005m 0 m 0 e00231t r 30 ln 005 1297. So it will take about 130 s. e00231t 005 00231t ln 005 t 00231 ln 2 , in other words m t m 0 e[ln 2 h ]t . In By the formula for radioactive decay, we have m t m 0 ert , where r h ln 2 48 this exercise we have to solve for h in the equation 200 250e[ln 2 h ]48 08 e[ln 2 h ]48 ln 08 h ln 2 48 1491 hours. So the half-life is approximately 149 hours. h ln 08 ln 2 From the formula for radioactive decay, we have m t m 0 ert , where r . In other words, m t m 0 e[ln 2 h ]t . h (a) Using m 3 058m 0 , we have to solve for h in the equation 058m 0 m 3 m 0 e[ln 2 h ]3 . 3 ln 2 3 ln 2 Then 058m 0 m 0 e[3 ln 2 h ] e[3 ln 2 h ] 058 ln 058 h 38 days. Thus the h ln 058 half-life of radon-222 is about 38 days. (b) Here we have to solve for t in the equation 02m m e[ln 2382]t . So we have 02m m e[ln 2382]t 0
0
0
0
382 ln 02 ln 2 t ln 02 t 887. So it takes roughly 9 days for a sample of 02 e[ln 2382]t 382 ln 2 Radon-222 to decay to 20% of its original mass. ln 2 23. By the formula in the text, m t m 0 e[ln 2 h ]t , so we have 065 1 e[ln 25730]t ln 065 t 5730 5730 ln 065 3561. Thus the artifact is about 3560 years old. t ln 2 ln 2 ln 2 . Since h 5730, r 0000121 24. From the formula for radioactive decay, we have m t m 0 ert where r h 5730 and m t m 0 e0000121t . We need to solve for t in the equation 059m 0 m 0 e0000121t e0000121t 059 ln 059 43606. So the mummy was buried about 4360 years ago. 0000121t ln 059 t 0000121 25. (a) T 0 65 145e0050 65 145 210 F.
(b) T 10 65 145e00510 1529. Thus the temperature after 10 minutes is about 153 F.
(c) 100 65 145e005t 35 145e005t 02414 e005t ln 02414 005t t
ln 02414 284. 005
Thus the temperature will be 100 F in about 28 minutes. 26. (a) We use Newton’s Law of Cooling: T t Ts D0 ekt with k 01947, Ts 60, and D0 986 60 386 . So T t 60 386e01947t .
12 12 (b) Solve T t 72. So 72 60 386e01947t 386e01947t 12 e01947t 01947t ln 386 386 1 12 t ln 600, and the time of death was about 6 hours ago. 01947 386
438
CHAPTER 4 Exponential and Logarithmic Functions
kt 27. Using Newton’s Law of Cooling, T t Ts D0 ekt with Ts 75 and D0 185 75 110. So T t 75 110e . 15 (a) Since T 30 150, we have T 30 75 110e30k 150 110e30k 75 e30k 15 22 30k ln 22 1 ln 15 . Thus we have T 45 75 110e4530 ln1522 1369, and so the temperature of the turkey k 30 22
after 45 minutes is about 137 F.
25 5 (b) The temperature will be 100 F when 75 110et30 ln1522 100 et30 ln1522 22 110 5 ln 22 t 15 5 1161. So the temperature will be 100 F after about 2 hours. ln 22 ln 22 t 30 30 ln 15 22
28. We use Newton’s Law of Cooling: T t Ts D0 ekt , with Ts 20 and
D0 100 20 80. So T t 20 80ekt . Since T 15 75, we have 11 20 80e15k 75 80e15k 55 e15k 11 16 15k ln 16 1 ln 11 . Thus T 25 20 80e2515ln1116 628, and so the k 15 16 temperature after another 10 min is 63 C. The function
100 80 60 0
20
T t 20 80e115ln1116t is shown in the viewing rectangle [0 30] by
[50 100].
4.7
LOGARITHMIC SCALES
1. (a) pH log H log 50 103 23 (b) pH log H log 32 104 35 (c) pH log H log 50 109 83 2. pH log H log 31 108 75 and the substance is basic.
3. (a) pH log H 30 H 103 M (b) pH log H 65 H 1065 32 107 M 4. (a) pH log H 46 H 1046 M 25 105 M (b) pH log H 73 H 1073 M 50 108 M 5. 40 107 H 16 105 log 40 107 log H log 16 105 log 40 107 pH log 16 105 64 pH 48. Therefore the range of pH readings for cheese is approximately 48 to 64.
6. 28 pH 38 28 pH 38 1028 10pH 1038 158 103 H 158 104 . The range of H is 158 104 to 158 103 . 7. (a) For the California wine, we have pH log H log H 32 H 1032 63 104 M. For the Italian wine, pH log H log H 29 H 1029 13 103 M. (b) The California wine has lower hydrogen ion concentration.
SECTION 4.7 Logarithmic Scales
439
8. (a) pH log H log H 55 H 1055 32 106 M.
(b) As pH increases, hydrogen ion concentration decreases. So as Marco gets better, the pH of his saliva will increase and its hydrogen ion concentration will decrease. (c) The saliva was more acidic when he was sick.
3125 I with S 104 and I 3125, so M log 4 55. S 10 I I (b) M log 10 M I S 10 M . We have M 48 and S 104 , so I 104 1048 63. S S
9. (a) M log
I 721 with S 104 and I 721, so M log10 4 59. S 10 (b) I 104 1058 631.
10. (a) M log
11. Let I0 be the intensity of the smaller earthquake and I1 the intensity of the larger earthquake. Then I1 20I0 . 20I0 I I log 20 log I0 log S. Then Notice that M0 log 0 log I0 log S and M1 log 1 log S S S M1 M0 log 20 log I0 log S log I0 log S log 20 13. Therefore the magnitude is 13 times larger.
I 12. Let the subscript S represent the San Francisco earthquake and J the Japan earthquake. Then we have M S log S 83 S 83 I I 10 I S S 1083 and M J log J 49 I J S 1049 . So S 49 1034 25119, and so the San Francisco S IJ 10 earthquake was 2500 times more intense than the Japan earthquake.
I 13. Let the subscript J represent the Japan earthquake and S represent the San Francisco earthquake. Then M J log J 91 S 91 I I S 10 1008 63, and hence the Japan I J S 1091 and M S log S 83 I S S 1083 . So J S IS S 1083 earthquake was about six times more intense than the San Francisco earthquake. 14. Let the subscript N represent the Northridge, California earthquake and K the Kobe, Japan earthquake. Then I I I 1072 M N log N 68 I N S 1068 and M K log K 72 I K S 1072 . So K 68 1004 251, and S S IN 10 so the Kobe, Japan earthquake was 25 times more intense than the Northridge, California earthquake. 20 105 I 10 log 10 log 2 107 10 log 2 log 107 10 log 2 7 73. Therefore the 12 I0 10 10 intensity level was 73 dB.
15. 10 log
16. 10 log
32 102 I 10 log 10 log 32 1010 105 dB. I0 10 1012
I I 70 10 log log I 12 7 log I 5, so the intensity was 105 wattsm2 . I0 10 1012 I 18. 98 10 log 12 log I 1012 98 log I 98 log 1012 22 I 1022 63 103 . So the 10 intensity was 63 103 wattsm2 . 17. 10 log
19. (a) The intensity is 31 105 Wm2 , so 10 log
I 31 105 10 log 10 log 31 107 75 dB. 12 I0 10 10
I log I 12 9 I 103 Wm2 . 1012 103 Ie 323. (c) The ratio of the intensities is Is 31 105 (b) Here 90 dB, so 90 10 log
440
CHAPTER 4 Exponential and Logarithmic Functions
IM 20. Let the subscript M represent the power mower and C the rock concert. Then 106 10 log 1012 IC 12 120 log I 10 log I M 1012 106 I M 1012 10106 . Also 120 10 log C 1012
I 1012 IC 1012 10120 . So C 106 1014 2512, and so the ratio of intensity is roughly 25. IM 10 k I k k k 21. (a) 1 10 log 1 and I1 2 1 10 log 2 10 log 2 log d1 10 log 20 log d1 . Similarly, I0 I0 I0 d d I0 1
1
k 20 log d2 . Substituting the expression for 1 gives 2 10 log I0 k d 20 log d1 20 log d1 20 log d2 1 20 log d1 20 log d2 1 20 log 1 . 2 10 log I0 d2 d 2 120 20 log 02 106, and so the (b) 1 120, d1 2, and d2 10. Then 2 1 20 log 1 120 20 log 10 d2 intensity level at 10 m is approximately 106 dB.
CHAPTER 4 REVIEW 2 9739, f 25 55902 7 18775, f 55 135765 2. f x 3 2x ; f 22 0653, f
1. f x 5x ; f 15 0089, f
3. g x 4e x2 ; g 07 0269, g 1 1472, g 12527 3 26888, g 36 174098 4. g x 74 e x1 ; g 2 0644, g 5. f x 3x2 . domain , range 0 , asymptote y 0.
y
6. f x 2x1 . Domain , range 0 , asymptote y 0.
1
1 x
1
7. g x 3 2x . Domain , range 3 , asymptote y 3.
y
y
x
1
8. g x 5x 5. Domain , range 5 , asymptote y 5.
y
1 1
1 1
x
x
CHAPTER 4
9. F x e x1 1. Domain , range 1 , asymptote y 1.
y
Review
10. G x e x1 2. Domain , range 2, asymptote y 2.
y 1 x
1
1 x
1
11. f x log3 x 1. Domain 1 , range , asymptote x 1.
y
12. g x log x. Domain 0, range , asymptote x 0.
y
1
1 x
1
13. f x 2 log2 x. Domain 0 , range , asymptote x 0.
1
x
14. f x 3 log5 x 4. Domain 4 , range , asymptote x 4. y
y (1, 2) 1 1
x
(_3, 3) 1 1
x
441
442
CHAPTER 4 Exponential and Logarithmic Functions
15. g x 2 ln x. Domain 0 , range , asymptote x 0.
16. g x ln x 2 . Domain
x x 0 0 0 , range ,
y
asymptote x 0.
y
1 x
1
1 1
x
2
17. f x 10x log 1 2x. Since log u is defined only for u 0, we require 1 2x 0 2x 1 x 12 , and so the domain is 12 . 18. g x log 2 x x 2 . We must have 2 x x 2 0 (since log y is defined only for y 0) x 2 x 2 0 x 2 x 1 0. The endpoints of the intervals are 2 and 1.
1
1 2
2
Sign of x 2
Sign of x 1
2
2 2
2
Sign of x 2
Sign of x 2
Interval
Sign of x 2 x 1
Thus the domain is 1 2. 19. h x ln x 2 4 . We must have x 2 4 0 ( since ln y is defined only for y 0) x 2 4 0 x 2 x 2 0. The endpoints of the intervals are 2 and 2. Interval
Sign of x 2 x 2
Thus the domain is 2 2 .
20. k x ln x. We must have x 0. So x 0 x 0 or x 0. Since x 0 x 0, the domain is x 0 or x 0 which is equivalent to x 0. In interval notation, 0 0 . 21. log2 1024 10 210 1024
22. log6 37 x 6x 37
23. log x y 10 y x
24. ln c 17 e17 c
25. 26 64 log2 64 6
26. 4912 17 log49 17 12
27. 10x 74 log10 74 x log 74 x 29. log2 128 log2 27 7
28. ek m ln m k 30. log8 1 log8 80 0
31. 10log 45 45 33. ln e6 6
32. log 0000001 log 106 6 34. log4 8 log4 432 32
CHAPTER 4 1 log 33 3 35. log3 27 3 37. log5 5 log5 512 12
Review
443
36. 2log2 13 13 2 38. e2 ln 7 eln 7 72 49 40. log3 243 log3 352 52
39. log 25 log 4 log 25 4 log 102 2 23 3 41. log2 1623 log2 24 log2 292 92 42. log5 250 log5 2 log5 250 2 log5 125 log5 5 3 43. log8 6 log8 3 log8 2 log8 63 2 log8 4 log8 823 23 44. log10 log10 10100 log10 100 log10 102 2 45. log AB 2 C 3 log A 2 log B 3 log C 46. log2 x x 2 1 log2 x log2 x 2 1 log2 x 12 log2 x 2 1 1 x2 1 x2 1 1 ln x 2 1 ln x 2 1 1 ln x 1 x 1 ln x 2 1 ln 47. ln 2 2 2 x2 1 x2 1 12 ln x 1 ln x 1 ln x 2 1 4x 3 3 log y 2 x 15 log 4 3 log x 2 log y 5 log x 1 log 4x 48. log y 2 x 15 x 2 1 5x3/2 log5 x 2 1 5x3/2 log5 x x 2 1 2 log5 x 32 log5 1 5x 12 log5 x 3 x 49. log5 x3 x 2 log5 x 32 log5 1 5x 12 log5 x log5 x 2 1 2 log5 x 32 log5 1 5x 12 log5 x log5 x 1 log5 x 1 3 4 x 12 50. ln 13 ln x 4 12 ln x 16 12 ln x 3 x 16 x 3 51. log 6 4 log 2 log 6 log 24 log 6 24 log 96 52. log x log x 2 y 3 log y log x x 2 y y 3 log x 3 y 4 2 x y3/2 3 2 2 3 / 2 2 2 log2 53. 2 log2 x y 2 log2 x y log2 x y log2 x y 2 x 2 y2 2 x 1 54. log5 2 log5 x 1 13 log5 3x 7 log5 [2 x 1] log5 3x 713 log5 3 3x 7 x2 4 1 2 2 55. log x 2 log x 2 2 log x 4 log [x 2 x 2] log x 4 log x2 4 5 5 56. 12 ln x 4 5 ln x 2 4x 12 ln x 4 x 2 4x ln x 4 x 2 4x 57. 32x7 27 32x7 33 2x 7 3 2x 10 x 5
1 54x 53 4 x 3 x 7 58. 54x 125 59. 23x5 7 log2 23x5 log2 7 3x 5 log2 7 x 13 log2 7 5 . Using the Change of Base Formula, we 7 1 have log2 7 log log 2 2807, so x 3 2807 5 260. 60. 1063x 18 log10 1063x log10 18 6 3x log10 18 x 13 6 log10 18 158
444
CHAPTER 4 Exponential and Logarithmic Functions
61. 41x 32x5 log 41x log 32x5 1 x log 4 2x 5 log 3 log 4 5 log 3 2x log 3 x log 4 log 4 5 log 3 115 x log 3 log 4 log 4 5 log 3 x 2 log 3 log 4 62. e3x4 10 ln e3x4 ln 10 34 x ln 10 x 43 ln 10 307
63. x 2 e2x 2xe2x 8e2x e2x x 2 2x 8 0 x 2 2x 8 0 (since e2x 0) x 4 x 2 0 x 4 or x 2
64. 32x 3x 6 0 3x 3 3x 2 0 3x 3 or 3x 2 x 1 (the second equation has no solution)
65. log x log x 1 log 12 log x x 1 log 12 x x 1 12 x 2 x 12 0 x 3 x 4 0 x 4 or 3. Since log 4 is undefined, the only solution is x 3. 66. ln x 2 ln 3 ln 5x 7 ln 3 x 2 ln 5x 7 3 x 2 5x 7 2x 1 x 12 , but since ln x 2 is undefined for x 12 , there is no solution.
67. log2 1 x 4 1 x 24 x 1 16 15
68. ln 2x 3 1 0 ln 2x 3 1 2x 3 e1 x 12 1e 3 168 69. log3 x 8 log3 x 2 log3 x x 8 2 x x 8 9 x 2 8x 9 0 x 9 x 1 0 x 1 or 9. We reject 1 because it does not satisfy the original equation, so the only solution is x 9. 70. log8 x 5 log8 x 2 1 log8 x 3
x 5 x 5 1 8 x 5 8 x 2 x 5 8x 16 7x 21 x 2 x 2
3 log 063 2x log 5 log 063 x 0430618 71. 52x /3 063 3 2 log 5 log 7 2602452 72. 23x5 7 3x 5 log 2 log 7 x 13 5 log 2 73. 52x1 34x1 2x 1 log 5 4x 1 log 3 2x log 5 log 5 4x log 3 log 3 log 3 log 5 2303600 x 2 log 5 4 log 3 log 3 log 5 x 4 log 3 2 log 5 1 ln 10000 0614023 74. e15k 10000 15k ln 10000 k 15
75. y e xx2 . Vertical asymptote x 2, horizontal asymptote y 272, no maximum or minimum. 10
76. y 10x 5x . No vertical asymptote, horizontal
asymptote y 0, local minimum of about 013 at x 052.
2 5 1 -20
0
20 -2
2
CHAPTER 4
77. y log x 3 x . Vertical asymptotes x 1, x 0,
x 1, no horizontal asymptote, local maximum of about
041 when x 058.
Review
445
78. y 2x 2 ln x. Vertical asymptote x 0, no horizontal asymptote, local minimum of about 119 at x 050. 100 50
-1
1
2
0 0
5
10
-2
79. 3 log x 6 2x. We graph y 3 log x and y 6 2x in 80. 4 x 2 e2x . From the graphs, we see that the solutions the same viewing rectangle. The solution occurs where the two graphs intersect. From the graphs, we see that the
are x 064 and x 2.
5
solution is x 242. 10 -2 5
2
10
-10
81. ln x x 2We graph the function f x ln x x 2, and we see that the graph lies above the x-axis for
016 x 315. So the approximate solution of the given inequality is 016 x 315.
82. e x 4x 2 e x 4x 2 0. We graph the function
f x e x 4x 2 , and we see that the graph lies below the
x-axis for 041 071 431.
2
5 -10
0 5 -2
83. f x e x 3ex 4x. We graph the function f x,
and we see that the function is increasing on 0 and 110 and that it is decreasing on 0 110.
-2
84. The line has x-intercept at x e0 1. When x ea ,
y ln ea a. Therefore, using the point-slope equation,
a a0 we have y 0 a x 1 y a x 1. e 1 e 1
2 -5
85. log4 15
log 15 1953445 log 4
log 3 4 0147839 86. log7 34 log 7
446
CHAPTER 4 Exponential and Logarithmic Functions
87. log9 028
log 028 0579352 log 9
88. log100 250
log 250 1198970 log 100
89. Notice that log4 258 log4 256 log4 44 4 and so log4 258 4. Also log5 620 log5 625 log5 54 4 and so log5 620 4. Then log4 258 4 log5 620 and so log4 258 is larger. x x 90. f x 23 . Then y 23 log2 y 3x log3 log2 y x, and so the inverse function is f 1 x log3 log2 x . Since log3 y is defined only when y 0, we have log2 x 0 x 1. Therefore the domain is 1 , and the range is . r nt . 91. P 12,000, r 010, and t 3. Then A P 1 n 23
(a) For n 2, A 12,000 1 010 12,000 1056 $16,08115. 2 123 $16,17818. (b) For n 12, A 12,000 1 010 12 3653 $16,19764. (c) For n 365, A 12,000 1 010 365
(d) For n , A Pert 12,000e0103 $16,19831.
92. P 5000, r 0085, and n 2. 215 (a) For t 15, A 5000 1 0085 5000 104253 $566498. 2
7 (b) We want to find t such that A 7000. Then A 5000 104252t 7000 104252t 7000 5000 5 log 75 log 75 7 t 404, and so the investment will amount to $7000 2t log 10425 log 5 2t log 10425 2 log 10425 after approximately 4 years.
(c) In this case, we solve n t n 0 ert for t when n 0 5000, r 0085, and n t 7000: 7000 5000e0085t 7 7 e0085t ln 7 0085t t ln 5 396, so the investment will grow to $7000 in just under 4 years. 5 5 0085 r nt
with P 100,000, r 0052, n 365, and A 100,000 10,000 110,000, 365t 365t and solve for t: 110,000 100,000 1 0052 11 1 0052 log 11 365t log 1 0052 365 365 365
93. We use the formula A P 1
t
n
log 11 1833. The account will accumulate $10,000 in interest in approximately 18 years. 365 log 1 0052 365
94. We solve n t n 0 ert for n t 2n 0 and r 0045: 2n 0 n 0 e0045t ln 2 0045t t
ln 2 15403. The 0045
retirement savings plan will double in about 154 years. 00425 365 95. After one year, a principal P will grow to the amount A P 1 P 104341. The formula for simple 365 interest is A P 1 r. Comparing, we see that 1 r 104341, so r 004341. Thus the annual percentage yield is 4341%. 0032 12 96. A P 1 P 103247 P 1 r r 3247% 12 97. (a) Using the model n t n 0 ert , with n 0 30 and r 015, we have the formula n t 30e015t . (b) n 4 30e0154 55.
015t 015t ln 50 t 1 ln 50 1876. So the stray cat population will (c) 500 30e015t 50 3 e 3 3 015 reach 500 in about 19 years.
CHAPTER 4
Review
447
98. Using the model n t n 0 ert , with n 0 10000 and n 1 25000, we have 25000 n 1 10000er1 er 52 r ln 52 0916. So n t 10000e0916t . (a) Here we must solve the equation n t 20000 for t. So n t 10000e0916t 20000 e0916t 2 ln 2 0756. Thus the doubling period is about 45 minutes. 0916t ln 2 t 0916 (b) n 3 10000e09163 156250, so the population after 3 hours is about 156,250.
99. (a) From the formula for radioactive decay, we have m t 10ert , where r
5
the amount remaining is m 1000 10 e ln 2 2710 Therefore the amount remaining is about 997 mg.
ln 2 . So after 1000 years 27 105
1000 10eln 2 27102 10eln 2270 997.
5 5 (b) We solve for t in the equation 7 10e ln 2 2710 t . We have 7 10e ln 2 2710 t 5 ln 2 ln 07 27 105 138,93475. Thus it takes about t t 07 e ln 2 2710 t ln 07 ln 2 27 105 139,000 years.
ln 2 . So m t m 0 e[ln 2 h ]t . h (a) Using m 8 033m 0 , we solve for h. We have 033m 0 m 8 m 0 eln 2 h 033 e8 ln 2 h 8 ln 2 8 ln 2 ln 033 h 5002. So the half-life of this element is roughly 5 days. h ln 033 (b) m 12 m 0 e[ln 25]12 019m 0 , so about 19% of the original mass remains.
100. From the formula for radioactive decay, we have m t m 0 ert , where r
ln 2 00004359 and n t 150 e00004359t . 1590 (b) n 1000 150 e000043591000 9700, and so the amount remaining is about 9700 mg.
101. (a) From the formula for radioactive decay, r
(c) Find t so that 50 150 e00004359t . We have 50 150 e00004359t 13 e00004359t 1 ln 13 2520. Thus only 50 mg remain after about 2520 years. t 00004359
102. From the formula for radioactive decay, we have m t m 0 ert , where r
ln 2 ln 2 . Since h 4, we have r 0173 h 4
and m t m 0 e0173t . (a) Using m 20 0375, we solve for m 0 . We have 0375 m 20 m 0 e017320 003125m 0 0375 0375 12. So the initial mass of the sample was about 12 g. m0 003125 (b) m t 12e0173t . (c) m 3 12e01733 7135. So there are about 71 g remaining after 3 days.
(d) Here we solve m t 015 for t: 015 12e0173t 00125 e0173t 0173t ln 00125 ln 00125 t 253. So it will take about 25 days until only 15% of the substance remains. 0173 103. (a) Using n 0 1500 and n 5 3200 in the formula n t n 0 ert , we have 3200 n 5 1500e5r e5r 32 15 1 32 01515t . 5r ln 32 15 r 5 ln 15 01515. Thus n t 1500 e
(b) We have t 2020 2009 11 so n 11 1500e0151511 7940. Thus in 2009 the bird population should be about 7940.
448
CHAPTER 4 Exponential and Logarithmic Functions
104. We use Newton’s Law of Cooling: T t Ts D0 ekt with k 00341, Ts 60 and D0 190 60 130.
3 So 90 T t 60 130e00341t 90 60 130e00341t 130e00341t 30 e00341t 13 3 t ln 313 430, so the engine cools to 90 F in about 43 minutes. 00341t ln 13 00341 105. H 13 108 M. Then pH log H log 13 108 79, and so fresh egg whites are basic.
106. pH 19 log H . Then H 1019 126 102 M.
107. Let I0 be the intensity of the smaller earthquake and I1 be the intensity of the larger earthquake. Then I1 35I0 . Since I I M log , we have M0 log 0 65 and S S 35I0 I I1 log log 35 log 0 log 35 M0 log 35 65 804. So the magnitude on the M1 log S S S Richter scale of the larger earthquake is approximately 80. I I 108. Let the subscript J represent the jackhammer and W the whispering: J 132 10 log J log J 132 I0 I0 I I I 10132 J 10132 . Similarly W 1028 . So J 10104 251 1010 , and so the ratio of intensities is I0 I0 IW 1028
251 1010 .
CHAPTER 4 TEST y
1. (a)
y
(b)
10
4
8 6 (0, 5) 4
(_2, 0)
_4 _2 0 _2
2 _4 _2 0
2
2
4
6
8 x
f x 2x 4 has domain , range 4 ,
and horizontal asymptote y 4.
(0, 1) 2
4
6
4. (a) 10log 36 36
(b) ln e3 3 12 (c) log3 27 log3 33 log3 332 32 3 (d) log2 80 log2 10 log2 80 10 log2 8 log2 2 3
(e) log8 4 log8 823 23 (f) log6 4 log6 9 log6 4 9 log6 62 2
x
_4
g x log3 x 3 has domain 3 , range
, and vertical asymptote x 3.
2. (a) f t ln 2t 3 is defined where 2t 3 0 2t 3 t 32 , so its domain is 32 . (b) g x log x 2 1 is defined where x 2 1 0 x 1, so its domain is 1 1 .
3. (a) 62x 25 log6 62x log6 25 2x log6 25
8
(b) ln A 3 eln A e3 A e3
CHAPTER 4
5. (a) log
x y3 z2
Test
449
log x log y 3 log z 2 log x 3 log y 2 log z
x 12 x 1 x ln 12 ln x 12 ln y (b) ln ln y y 2 y 1 x 2 x 2 3 log 13 log x 2 4 log x log x 2 4 (c) log 3 x4 x2 4 x4 x2 4 1 3 log x 2 43 log x 13 log x 2 4
6. (a) log a 2 log b log a log b2 log ab2
x 2 25 x 5 x 5 ln ln x 5 (b) ln x 2 25 ln x 5 ln x 5 x 5 (c) log2 3 3 log2 x 12 log2 x 1 log2 3 log2 x 3 log2 x 112 log2
3 x 1 x3
7. (a) 34x 3100 4x 100 x 25
2 (b) e3x2 e x 3x 2 x 2 x 2 3x 2 0 x 1 x 2 0 x 1 or x 2
(c) 5x10 1 7 5x10 6 log5 5x10 log5 6 x10 log5 6 x 10 log5 6 1113
(d) 10x3 62x log 10x3 log 62x x 3 x
3 log6 10 539 2 log6 10
log6 62x log6 10 x 3 2x 2 log6 10 x 3 log6 10 log6 10
8. (a) log 2x 3 2x 103 x 12 1000 500
(b) log x 1 log 2 log 5x log 2 x 1 log 5x 2x 2 5x 3x 2 x 23
(c) 5 ln 3 x 4 ln 3 x 45 3 x e45 x 3 e45 0774
(d) log2 x 2 log2 x 1 2 log2 x 2 x 1 2 x 2 x 1 22 4 x 2 x 2 4
x 2 x 6 0 x 3 x 2 0 x 3 or 2. However, x 3 does not satisfy the original equation, so the only solution is x 2.
9. Using the Change of Base Formula, we have log12 27
log 27 1326. log 12
10. (a) From the formula for population growth, we have 8000 1000er1
y
(d)
8 er r ln 8 207944. Thus n t 1000e207944t .
(b) n 15 1000e20794415 22,600 (c) 15,000 1000e207944t 15 e207944t ln 15 207944t ln 15 13. Thus the population will reach 15,000 after t 207944 approximately 13 hours. 12t 11. (a) A t 12,000 1 0056 , where t is in years. 12
50,000
365t 0056 3653 . So A 12,000 1 $14,19506. (b) A t 12,000 1 0056 3 365 365
1
x
450
FOCUS ON MODELING
(c) A t 12,000e0056t . So 20,000 12,000e0056t 5 3e0056t ln 5 ln 3e0056t ln 5 ln 3 0056t 1 ln 5 ln 3 912. Thus, the amount will grow to $20,000 in approximately 912 years. t 0056
12. (a) The initial mass is m 0 3 and the half-life is h 10, so using the formula m t m 0 2t h , we have m t 3 2t10 .
(b) Using the radioactive decay model with m 0 3 and r lnh2 ln102 , we have m t 3e[ln 210]t 3e00693t . (c) After 1 minute 60 seconds, the amount remaining is m 60 3e0069360 0047 g. 106 106 106 00693t 6 00693t 00693t ln e 00693t ln 10 e ln (d) We solve 3e 3 3 3 1 106 ln 215, so there is 1 g of 91 Kr remaining after about 215 seconds, or 36 minutes. t 00693 3
IJ IJ 13. Let the subscripts J and P represent the two earthquakes. Then we have M J log 64 1064 S S I I I 1064 S 1064 S I J . Similarly, M P log P 31 1031 P 1031 S I P . So J 31 1033 19953, S S IP 10 S and so the Japan earthquake was about 1995 times more intense than the Pennsylvania earthquake.
FOCUS ON MODELING Fitting Exponential and Power Curves to Data
1. (a)
y
(b) Using a graphing calculator, we obtain the model y abt , where a 33349260 1015 and
300
b 10198444, and y is the population (in millions) in
250
the year t.
200
(c) Substituting t 2020 into the model of part (b), we get y ab2020 5776 million.
150
(d) According to the model, the population in 1965 was
100
y ab1965 1960 million.
50 0 1780
1820
1860
1900
Year
1940
1980
2020 x
Fitting Exponential and Power Curves to Data y
2. (a)
451
(b) We let t represent the time (in seconds) and y the
5
distance fallen (in meters). Using a calculator, we obtain the power model: y 49622t 20027 .
4
(c) When t 3 the model predicts that y 44792 m.
3 2 1
0
0.2
0.4
0.6
0.8
1
x
Time (s)
3. (a) Yes. (b) Year t
Health Expenditures E ($bn)
ln E
ln E
1970
743
430811
1980
2558
554440
1985
4446
609718
1987
5191
625210
5
1990
7243
658521
4
1992
8579
675449
3
1994
9727
688008
2
1996
10818
698638
1998
12089
709747
2000
13772
722781
2002
16380
740123
2005
20354
761845
2008
24117
778809
2010
25990
786288
2012
27934
793501
7 6
1 0
10
20
30
40
x
Year since 1970
Yes, the scatter plot appears to be roughly linear.
(c) Let t be the number of years elapsed since 1970 . Then ln E 47473926 008213193t, where E is expenditure in billions of dollars. (d) E e47473926008213193t 11528330e008213193t
(e) In 2020 we have t 2020 1970 50, so the estimated 2020 health-care expenditures are 11528330e00821319350 70023 billion dollars.
452
FOCUS ON MODELING
4. (a)
y
(b) Let t be the time (in hours) and y be amount of
5
iodine-131 (in grams). Using a calculator, we obtain the exponential model y abt , where a 479246 and
4.8
b 099642.
4.6
(c) To find the half-life of iodine-131, we must find the time when the sample has decayed to half its original
4.4
mass. Setting y 240 g, we get
4.2
240 479246 099642t
4 3.8 0
10
20
30
40
ln 240 ln 479246 t ln 099642 ln 240 ln 479246 1928 h. t ln 099642
50 x
Time (h)
5. (a) Using a graphing calculator, we find that
y
(b)
I0 227586444 and k 01062398.
14
(c) We solve 015 227586444e01062398x for x: 015 227586444e01062398x
12 10
0006590902 e01062398x 5022065 01062398x x 4727. So light
8 6
intensity drops below 015 lumens below around
4
4727 feet.
2 0
10
20
30
40
x
Depth (ft)
6. (a) Using a graphing calculator, we find the power function model y 4970030t 015437 and the exponential model y 4482418 099317t .
(c) The power function, y 4970030t 015437 , seems to provide a better model.
(b)
y 80 70 60 50 40 30
exponential power
20 10 0
20
40
60
80
Time (hours)
100
120 x
Fitting Exponential and Power Curves to Data
7. (a) Let A be the area of the cave and S the number of species of bat. Using a graphing calculator, we obtain
S
(b)
7
the power function model S 014A064 .
6 5
(c) According to the model, there are
4
S 014 205064 4 species of bat living in the El
3
Sapo cave.
2 1
0
100
200
300
500 A
400
Area (m@ ) The model fits the data reasonably well.
8. Let x be the reduction in emissions (in percent), and y be the cost (in
y
dollars). First we make a scatter plot of the data. A linear model does
600
not appear appropriate, so we try an exponential model. Using a calculator, we get the model y ab x , where a 2414 and
500 400
b 105452. This model is graphed on the scatter plot.
300 200 100 0 40
50
60
70
80
90
Reduction in emissions (%)
9. (a)
y
(b)
1.2 1 0.8 0.6 0.4 0.2 0
2
4
6
8
10
12
14
16
18 x
x
y
ln x
ln y
2
008
069315
252573
4
012
138629
212026
6
018
179176
171480
8
025
207944
138629
10
036
230259
102165
12
052
248491
065393
14
073
263906
031471
16
106
277259
005827
100 x
453
454
FOCUS ON MODELING ln y
ln y
(b) (cont’d)
2
0
4
6
8
10
12
14
16
18
_1
_1
_2
_2
y
2
2.5
3 ln x
(b)
800 700 600 500 400 300 200 100
(b) (cont’d)
1.5
(d) y a b x where a 0057697 and b 1200236.
(c) The exponential function.
0
1
_3
_3
10. (a)
0.5
0
x
20
40
60
100 x
80
ln y
ln y
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
0
20
40
(c) The power function is more appropriate.
60
80
100 x
y
ln x
ln y
10 20 30 40 50 60 70 80 90
29 82 151 235 330 430 546 669 797
230259 299573 340120 368888 391202 409434 424850 438203 449981
336730 440672 501728 545959 579909 606379 630262 650578 668085
1
2
3
4
5
ln x
(d) y ax n where a 0893421326 and n 150983.
11. (a) Using the Logistic command on a TI-83 we get y c 500855793.
0
x
c where a 4910976596, b 04981144989, and 1 aebx
c cN c c c aebt 1 we solve for t. So N 1aebt bt bt N N N 1 ae 1 ae cN 1 bt e bt ln c N ln a N t [ln a N ln c N ]. Substituting the values for a, b, and aN b 1 c, with N 400 we have t 04981144989 ln 1964390638 ln 100855793 1058 days.
(b) Using the model N
5
TRIGONOMETRIC FUNCTIONS: RIGHT TRIANGLE APPROACH
5.1
ANGLE MEASURE
1. (a) The radian measure of an angle is the length of the arc that subtends the angle in a circle of radius 1. . (b) To convert degrees to radians we multiply by 180
(c) To convert radians to degrees we multiply by 180 . 2. (a) If a central angle is drawn in a circle of radius r, the length of the arc subtended by is s r . (b) The area of the sector with central angle is A 12 r 2 .
3. (a) The angular speed of the point is
. t
s (b) The linear speed of the point is . t (c) The linear speed and the angular speed ome are related by the equation r.
4. No, if the common angular speed is , Object A has linear speed 2, while Object B has angular speed 5. Object B has greater linear speed. rad rad 0262 rad 5. 15 15 180 12
rad rad 0628 rad 6. 36 36 180 5
rad 3 rad 0942 rad 7. 54 54 180 10
rad 5 rad 1309 rad 8. 75 75 180 12
rad rad 0785 rad 9. 45 45 180 4
rad rad 0524 rad 10. 30 30 180 6
rad 5 rad 1745 rad 11. 100 100 180 9
rad 10 rad 3491 rad 12. 200 200 180 9
rad 50 rad 17453 rad 13. 1000 1000 180 9
rad 20 rad 62832 rad 14. 3600 3600 180
rad 7 rad 1222 rad 15. 70 70 180 18
rad 5 rad 2618 rad 16. 150 150 180 6
17. 53 53 180 300
18. 34 34 180 135
19. 56 56 180 150
20. 32 32 180 270
540 21. 3 3 180 1719
360 22. 2 2 180 1146
216 23. 12 12 180 688
612 24. 34 34 180 1948
180 18 25. 10 10
5 180 50 26. 518 18
2 180 24 13 180 27. 215 28. 13 15 12 12 195 29. 50 is coterminal with 50 360 410 , 50 720 770 , 50 360 310 , and 50 720 670 . (Other answers are possible.)
30. 135 is coterminal with 135 360 495 , 135 720 855 , 135 360 225 , and 135 720 585 . (Other answers are possible.) 31. 34 is coterminal with 34 2 114 , 34 4 194 , 34 2 54 , and 34 4 134 . (Other answers are possible.) 11 13 32. 116 is coterminal with 116 2 236 , 116 4 356 , 116 2 6 , and 6 4 6 . (Other answers are possible.)
455
456
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
7 15 9 17 33. 4 is coterminal with 4 2 4 , 4 4 4 , 4 2 4 , and 4 4 4 . (Other answers are possible.)
34. 45 is coterminal with 45 360 315 , 45 720 675 , 45 360 405 , and 45 720 765 . (Other answers are possible.) 35. Since 430 70 360 , the angles are coterminal.
36. Since 330 30 360 , the angles are coterminal.
37. Since 176 56 126 2; the angles are coterminal. 38. Since 323 113 213 7 is not a multiple of 2, the angles are not coterminal. 39. Since 875 155 720 2 360 , the angles are coterminal. 40. Since 340 50 290 is not a multiple of 360 , the angles are not coterminal.
41. Since 400 360 40 , the angles 400 and 40 are coterminal. 42. Since 375 360 15 , the angles 375 and 15 are coterminal.
43. Since 780 2 360 60 , the angles 780 and 60 are coterminal.
44. Since 100 260 360 is a multiple of 360 , the angles 100 and 260 are coterminal. 45. Since 800 3 360 280 , the angles 800 and 280 are coterminal.
46. Since 1270 190 1080 3 360 is a multiple of 360 , the angles 1270 and 190 are coterminal.
47. Since 196 2 76 , the angles 196 and 76 are coterminal. 5 48. Since 53 2 3 , the angles 3 and 3 are coterminal.
49. Since 25 12 2 , the angles 25 and are coterminal.
50. Since 10 2 3717, the angles 10 and 10 2 are coterminal.
17 51. Since 174 2 2 4 , the angles 4 and 4 are coterminal. 52. Since 512 32 24 12 2, the angles 512 and 32 are coterminal.
53. Using the formula s r, s 56 9 152 .
7 35 5 5 122. 54. Using the formula s r, the length of the arc is s 140 180 9 9 s 10 180 55. 2 rad 2 1146 r 5 8 s 56. Solving for r, we have r , so the radius of the circle is r 4. 2 57. Solving for s, we have s r , so the length of the arc is 5 3 15 cm. 8 58. Solving for s, we have s r 12 40 838 m. 180 3 14 14 180 s rad 891 . 59. Solving for , we have r 9 9 s 15 5 5 180 60. Solving for , we have rad 955 . r 9 3 3 15 18 s 573 m 61. r 56 s 20 72 62. r 2292 cm 50 180 32 4 128 4468 63. (a) A 12 r 2 12 82 80 180 9 9
(b) A 12 r 2 12 102 05 25 2A 2 12 , so the radius is 586. 64. (a) A 12 r 2 r 07
SECTION 5.1 Angle Measure
(b) r
2A
457
2 12 303 150 180
2 65. 23 rad and r 10 m, so A 12 r 2 12 102 23 100 3 1047 m . 1 2 1 2 29 29 456 ft2 . 29 66. 145 145 36 and r 6 ft, so A 2 r 2 6 36 2 180 7 1 7 2 rad. Thus, A 12 r 2 70 r 67. A 70 m2 and 140 140 180 9 2 9 6 5 9 76 m. r 2 70 7 4 6 12 5 1 5 2 2 2 68. A 20 m and 12 , so A 2 r 20 12 r r 2 20 55 m. 5
69. r 80 mi and A 1600 mi2 , so A 12 r 2 1600 12 802 12 rad. 600 . Thus, the area of the sector is 70. The area of the circle is r 2 600 m2 , so r 900 1 600 3 2865 m2 . A 12 r 2 2 71. Referring to the figure, we have AC 3 1 4, BC 1 2 3, and AB 2 3 5. Since
2
AB 2 AC 2 BC 2 , then by the Pythagorean Theorem, the
triangle is a right triangle. Therefore, 2 and 2 A 12 r 2 12 12 2 4 ft .
3 A
3
B 2
1 1 C
s 1 72. The triangle is equilateral, so 1 3 rad. To find 2 , we use the formula 2 r 1 1 rad. Thus, 1 2 3 1 0047 rad, or approximately 27 .
73. Between 1:00 P. M . and 1:00 P. M ., the minute hand traverses three-quarters of a complete revolution, or 34 2 32 rad, while the hour hand moves three-quarters of the way from 12 to 1, which is itself one-twelfth of a revolution. So the hour 1 2 rad. hand traverses 34 12 8
74. Between 1:00 P. M . and 6:45 P. M ., the minute hand traverses five complete revolutions plus three-quarters of a revolution;
that is, 5 2 34 2 232 rad, while the hour hand moves through five-twelfths of a revolution, plus three-quarters of 5 2 3 1 2 23 rad. the way from 6 to 7; that is, 12 4 12 24
75. The circumference of each wheel is d 28 in. If the wheels revolve 10,000 times, the distance traveled is 1 ft 1 mi 10,000 28 in. 1388 mi. 12 in. 5280 ft 76. Since the diameter is 30 in., we have r 15 in. In one revolution, the arc length (distance traveled) is s r 2 1 rev 15 30 in. The total distance traveled is 1 mi 5280 ft/mi 12 in/ft 63,360 in. 63,360 in. 67227 rev. 30 in Therefore the car wheel will make approximately 672 revolutions. rad rad. 77. We find the measure of the angle in degrees and then convert to radians. 405 255 15 and 15 180 12
Then using the formula s r , we have s roughly 1037 mi. rad 78. 35 30 5 5 180
12 3960 330 1036725 and so the distance between the two cities is
rad. Then using the formula s r , the length of the arc is 3
s 3 3960 110 345575. So the distance between the two cities is roughly 346 mi.
458
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
1 of its orbit which is 2 rad. Then s r 2 93,000,000 1,600,9113, so the 79. In one day, the earth travels 365 365 365 distance traveled is approximately 16 million miles.
80. Since the sun is so far away, we can assume that the rays of the sun are parallel when striking the earth. Thus, the angle 500 s 180 500 formed at the center of the earth is also 72 . So r 72 3980 mi, and the circumference 72 180 2 180 500 25,000 mi. 72 1 1 rad rad. Then s r 3960 1152, and so a 81. The central angle is 1 minute 60 60 180 10,800 10,800 nautical mile is approximately 1152 mi. 219,900 ft2 . 82. The area is A 12 r 2 12 3002 280 180 is c 2r
83. The area is equal to the area of the large sector (with radius 34 in.) minus the area of the small sector (with radius 14 in.) 1131 in.2 . Thus, A 12 r12 12 r22 12 342 142 135 180 84. The area available to the cow is shown in the diagram. Its area is the sum of four quarter-circles: A 14 1002 502 402 302 3750
11,781 ft2
50 ft 30 ft
20 ft 50 ft
100 ft
60 ft 50 ft
40 ft
45 2 rad 90 rad/min. 1 min 45 2 16 1440 in./min 45239 in./min. (b) The linear speed is 1 1000 2 rad 2000 rad/min. 86. (a) The angular speed is 1 min
85. (a) The angular speed is
(b) The linear speed is 87.
6 1000 2 12 50 ft/s 524 ft/s. 60 3
8 2 2 32 6702 ft/s. 15 15
600 2 11 12 ft 1 mi 60 min 125 mi/h 393 mi/h. 1 min 5280 ft 1 hr 1 day 1 2 3960 103957 mi/h. 89. 23 h 56 min 4 s 239344 hr. So the linear speed is 1 day 239344 hr
88.
50 mi/h 1h 5280 ft linear speed 2200 rad/min. radius 2 ft 60 min 1 mi angular speed 2200 rad/min (b) The rate of revolution is 350 rev/min. 2 2 100 2 020 m 2 91. 209 m/s. 60 s 3 40 2 4 linear speed of pedal 160 rad/min. 92. (a) The angular speed is radius of wheel sprocket 2 (b) The linear speed of the bicycle is angular speed radius 160 rad/min 13 in 2080 in/min 62 mi/h. 90. (a) The radius is 2 ft, so the angular speed is
SECTION 5.2 Trigonometry of Right Triangles
459
93. (a) The circumference of the opening is the length of the arc subtended by the angle on the flat piece of paper, that is, C s r 6 53 10 314 cm.
10 C 5 cm. 2 2 (c) By the Pythagorean Theorem, h 2 62 52 11, so h 11 33 cm. (d) The volume of a cone is V 13 r 2 h. In this case V 13 52 11 868 cm3 . (b) Solving for r, we find r
94. (a) With an arbitrary angle , the circumference of the opening is 3 92 C 2 2 , h 6 r 36 2 , and C 6, r 2 92 9 2 9 V 13 r 2 h 2 36 2 2 2 42 2 . 3
(b) 100 50 0 0
2
4
6
(c) The volume seems to be maximized for 513 rad or about 293 .
95. Answers will vary, although of course everybody prefers radians.
5.2 1. (a)
TRIGONOMETRY OF RIGHT TRIANGLES (b) sin
opposite adjacent
adjacent opposite opposite , cos , and tan . hypotenuse hypotenuse adjacent
(c) The trigonometric ratios do not depend on the size of the triangle because all right triangles with angle are similar.
¬
hypotenuse
2. The reciprocal identities state that csc
1 1 1 , sec , and cot . sin cos tan
3. sin 45 , cos 35 , tan 43 , csc 54 , sec 53 , cot 34
7 , cos 24 , tan 7 , csc 25 , sec 25 , cot 24 4. sin 25 7 7 25 24 24 9 2 5. The remaining side is obtained by the Pythagorean Theorem: 41 402 81 9. Then sin 40 41 , cos 41 , 41 41 9 tan 40 9 , csc 40 , sec 9 , cot 40
6. The hypotenuse is obtained by the Pythagorean Theorem: 17 17 8 tan 15 8 , csc 15 , sec 8 , cot 15
8 82 152 289 17. Then sin 15 17 , cos 17 ,
7. The remaining side is obtained by the Pythagorean Theorem:
32 22 13. Then sin 2
2 1313 , 13
cos 3 3 1313 , tan 23 , csc 213 , sec 313 , cot 32 13 8. The remaining side is obtained by the Pythagorean Theorem: 82 72 15. Then sin 78 , cos 815 ,
tan 7 7 1515 , csc 87 , sec 8 8 1515 , cot 715 15 15 2 2 9. c 5 3 34
(a) sin cos 3 3 3434 34 2 2 10. b 7 4 33
(b) tan cot 35
(c) sec csc 534
460
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
(a) sin cos 47
(b) tan cot 4
(c) sec csc 7
33
33
11. (a) sin 22 037461
(b) cot 23 235585
12. (a) cos 37 079864
(b) csc 48 134563
13. (a) sec 13 102630
(b) tan 51 123490
14. (a) csc 10 575877 x , we have x 25 sin 30 25 12 25 15. Since sin 30 2. 25 12 12 12 , we have x 1 12 2. 16. Since sin 45 x sin 45
(b) sin 46 071934
2
x 17. Since sin 60 , we have x 13 sin 60 13 23 132 3 . 13 4 4 4 1 4 3. 18. Since tan 30 , we have x x tan 30 3
19. 20. 21. 22. 23.
12 12 1651658. Since tan 36 , we have x x tan 36 25 25 Since sin 53 , we have x 3130339. x sin 53 y x cos x 28 cos , and sin y 28 sin . 28 28 4 4 x tan x 4 tan , and cos y 4 sec . 4 y cos 24. cos 12 tan 56 . Then the third side is x 52 62 61. 13 . The third side is y 132 122 25 5. The other five ratios are The other five ratios are sin 5 6161 , cos 6 6161 ,
csc 561 , sec 661 , and cot 65 . Ï61 ¬
5 , tan 5 , csc 13 , sec 13 , and sin 13 12 5 12
cot 12 5. 13
5
5
¬ 12
6
25. cot 1. Then the third side is r
12 12 2.
The other five ratios are sin 1 22 , 2 2 1 cos 2 , tan 1, csc 2, and 2 sec 2.
Ï2
¬ 1
1
26. tan
3. The third side is r 12 3 2. The other
five ratios are sin 23 , cos 12 , csc 2 , 3 sec 2, and cot 1 . 3
2
Ï3
¬ 1
SECTION 5.2 Trigonometry of Right Triangles
27. csc 11 6 . The third side is x
112 62 85. The 28. cot 53 . The third side is x 52 32 34. The
6 , cos 85 , other five ratios are sin 11 11
other five ratios are sin 3 3434 , cos 5 3434 ,
tan 6 8585 , sec 118585 , and cot 685 .
tan 35 , csc 334 and sec 534 .
11
Ï34
6
¬
¬ Ï85 1 3 1 3 29. sin cos 6 6 2 2 2
30. sin 30 csc 30 sin 30
461
1 1 sin 30
3
5
31. sin 30 cos 60 sin 60 cos 30 12 12 23 23 14 34 1 2 2 32. sin 60 2 cos 60 2 23 12 34 14 1
2 12 34 3 1 1 sin cos 2 34. sin cos 3 4 4 3 2 2 2 18 4 2 3 33. cos 30 2 sin 30 2
2 3 2
1 1 4 2
2 2 2 1 12 31 18 3 1 18 3 2 3 1 2 2 1 2 3 4
2 2 1 2 3 2 2 1 1 2 1 sin 35. cos 4 6 2 2 2 4 2 2 4 2 2 2 3 3 2 36. sin 3 tan 12 2 94 2 6 csc 4 2 3 2
37. This is an isosceles right triangle, so the other leg has length 16 tan 45 16, the hypotenuse has length 16 16 2 2263, and the other angle is 90 45 45 . sin 45 100 10352, and the other angle is 38. The other leg has length 100 tan 75 2679, the hypotenuse has length sin 75 90 75 15 . 35 5685, and the other angle is 39. The other leg has length 35 tan 52 4480, the hypotenuse has length cos 52 90 52 38 .
40. The adjacent leg has length 1000 cos 68 37461, the opposite leg has length 1000 sin 68 92718, and the other angle is 90 68 22 .
41. The adjacent leg has length 335 cos 8 3095, the opposite leg has length 335 sin 8 1282, and the other angle is 3 . 2 8 8
42. The opposite leg has length 723 tan 6 4174, the hypotenuse has length
723 8348, and the other angle is cos 6
. 2 6 3
43. The adjacent leg has length 3 . 2 5 10
106 106 14590, the hypotenuse has length 18034, and the other angle is tan sin 5 5
44. The adjacent leg has length 425 cos 38 16264, the opposite leg has length 425 sin 38 39265, and the other angle is 3 . 2 8 8
462
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
1 045 cos 2 089, tan 1 , csc 224, sec 224 112, cot 200. 45. sin 224 224 2 2 064 083, csc 40 156, sin 40 064, cos 40 077, tan 40 46. 077 sec 40 131, cot 40 120. 40¡
47. x
100 100 2309 tan 60 tan 30
48. Let d be the length of the base of the 60 triangle. Then tan 60 dx
85 85 x d 981. tan 30 tan 30
85 85 85 d 49075, and so tan 30 d tan 60 dx
50 h 50 h h 57735 sin 65 x 637 h sin 60 x sin 65 5 5 h h 10, and so tan 60 50. Let h be the hypotenuse of the top triangle. Then sin 30 h sin 30 x 10 h 58. x tan 60 tan 60 y x , so x y sin 10 sin tan . 10 From the diagram, sin and tan 51. y 10 49. Let h be the length of the shared side. Then sin 60
¬
¹ -¬ 2
y
x
¬
a b 1 d a sin , tan b tan , cos c sec , cos d cos 1 1 c 1 h h 5280 tan 11 1026 ft. 53. Let h be the height, in feet, of the Empire State Building. Then tan 11 5280 35,000 54. (a) Let r be the distance, in feet, between the plane and the Gateway Arch. Therefore, sin 22 r 35,000 r 93,431 ft. sin 22 (b) Let x be the distance, also in feet, between a point on the ground directly below the plane and the Gateway Arch. Then 35,000 35,000 x 86,628 ft. tan 22 x tan 22 h 55. (a) Let h be the distance, in miles, that the beam has diverged. Then tan 05 240,000 h 240,000 tan 05 2100 mi. 52. sin
56. 57. 58. 59.
(b) Since the deflection is about 2100 mi whereas the radius of the moon is about 1000 mi, the beam will not strike the moon. 200 200 x 471 ft. Let x be the distance, in feet, of the ship from the base of the lighthouse. Then tan 23 x tan 23 h h 20 sin 72 19 ft. Let h represent the height, in feet, that the ladder reaches on the building. Then sin 72 20 h h 600 sin 65 544 ft. Let h be the height, in feet, of the communication tower. Then sin 65 600 h Let h be the height, in feet, of the kite above the ground. Then sin 50 h 450 sin 50 345 ft. 450
SECTION 5.2 Trigonometry of Right Triangles
60.
hÁ
18¡ x 14¡
hª
463
Let h 1 be the height of the flagpole above elevation and let h 2 be the height below, h as shown in the figure. So tan 18 1 h 1 x tan 18 . Similarly, x h 2 x tan 14 . Since the flagpole is 60 feet tall, we have h 1 h 2 60, so x tan 18 tan 14 60 x
60 1045 ft. tan 18 tan 14
h 61. Let h 1 be the height of the window in feet and h 2 be the height from the window to the top of the tower. Then tan 25 1 325 h h 1 325 tan 25 152 ft. Also, tan 39 2 h 2 325 tan 39 263 ft. Therefore, the height of the window 325 is approximately 152 ft and the height of the tower is approximately 152 263 415 ft. 35¡
52¡
62.
52¡
5150
dÁ
Let d1 be the distance, in feet, between a point directly below the plane and one car, and d2 be the distance, in feet, between the same point and the other car. Then
dª
35¡
tan 52
5150 5150 5150 d1 402362 ft, and tan 35 d1 tan 52 d2
5150 735496 ft. So the distance between the two cars is about d1 d2 402362 735496 11,379 ft. tan 35 63. Let d1 be the distance, in feet, between a point directly below the plane and one car, and d2 be the distance, in feet, between d d the same point and the other car. Then tan 52 1 d1 5150 tan 52 65917 ft. Also, tan 38 2 5150 5150 d2 5150 tan 38 40236 ft. So in this case, the distance between the two cars is about 2570 ft. d2
64. Let x be the distance, in feet, between a point directly below the balloon and the first mile post. Let h be the height, h h in feet, of the balloon. Then tan 22 and tan 20 . So h x tan 22 x 5280 tan 20 x x 5280 5280 tan 20 47,977 ft. Therefore h 47,9769 tan 22 19,384 ft 37 mi. x tan 22 tan 20 65. Let x be the horizontal distance, in feet, between a point on the ground directly below the top of the mountain and h the point on the plain closest to the mountain. Let h be the height, in feet, of the mountain. Then tan 35 x h 1000 tan 32 and tan 32 . So h x tan 35 x 1000 tan 32 x 82942. Thus x 1000 tan 35 tan 32 h 82942 tan 35 5808 ft. Since the angle of elevation from the observer is 45 , the distance from the
66.
observer is h, as shown in the figure. Thus, the length of the leg in the smaller right h 75¡ h
ã600ã
h 600 h tan 75 h 600 h 600 tan 75 473 m. 600 tan 75 h 1 tan 75 h 1 tan 75 triangle is 600 h. Then tan 75
600-h
67. Let d be the distance, in miles, from the earth to the sun.
Then sec 8985
d 240,000
d 240,000 sec 8985 917 million miles. 68. (a) s r rs 6155 3960 15543 rad 8905
(b) Let d represent the distance, in miles, from the center of the earth to the moon. Since cos d
3960 3960 239,9615. So the distance AC is 239,9615 3960 236,000 mi. cos cos 8905
3960 , we have d
464
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
r 69. Let r represent the radius, in miles, of the earth. Then sin 60276 r600 r 600 sin 60276 r sin 60276 600 sin 60276 r 1 sin 60276 r 600 1sin 60276 3960099. So the earth’s radius is about 3960 mi.
,000 , we have 70. Let d represent the distance, in miles, from the earth to Alpha Centauri. Since sin 0000211 93,000 d 93,000,000 13 d sin 0000211 25,253,590,022,410. So the distance from the earth to Alpha Centauri is about 253 10 mi.
d d, so d sin 463 0723 AU. 1 72. If two triangles are similar, then their corresponding angles are equal and their corresponding sides are proportional. That is, if triangle ABC is similar to triangle A B C then AB r A B , AC r A C , and BC r B C . Thus when we express any trigonometric ratio of these lengths as a fraction, the factor r cancels out.
71. Let d be the distance, in AU, between Venus and the sun. Then sin 463
5.3
TRIGONOMETRIC FUNCTIONS OF ANGLES
1. If the angle is in standard position and P x y is a point on the terminal side of , and r is the distance from the origin to x y y P, then sin , cos , and tan . r r x 2. The sign of a trigonometric function of depends on the quadrant in which the terminal angle of lies. For example, if is in quadrant II, sin is positive. In quadrant III, cos is negative. In quadrant IV, sin is negative. 3. (a) If is in standard position, then the reference angle is the acute angle formed by the terminal side of and the x-axis. So the reference angle for 100 is 80 and that for 190 is 10 .
(b) If is any angle, the value of a trigonometric function of is the same, except possibly for sign, as the value of the trigonometric function of . So sin 100 sin 80 and sin 190 sin 10 .
4. The area A of a triangle with sides of lengths a and b and with included angle is given by the formula A 12 ab sin . So the area of the triangle with sides 4 and 7 and included angle 30 is 12 4 7 sin 30 7.
5. (a) The reference angle for 120 is 180 120 60 .
6. (a) The reference angle for 175 is 180 175 5 .
(b) The reference angle for 200 is 200 180 20 .
(b) The reference angle for 310 is 360 310 50 .
(c) The reference angle for 285 is 360 285 75 .
(c) The reference angle for 730 is 730 720 10 .
7. (a) The reference angle for 225 is 225 180 45 . (b) The reference angle for 810 is 810 720 90 . (c) The reference angle for 105 is 180 105 75 .
is 7 3 . 9. (a) The reference angle for 710 10 10
8. (a) The reference angle for 99 is 180 99 81 . (b) The reference angle for 199 is 199 180 19 .
(c) The reference angle for 359 is 360 359 1 . 10. (a) The reference angle for 56 is 56 6.
(b) The reference angle for 98 is 98 8.
(b) The reference angle for 109 is 109 9.
(c) The reference angle for 103 is 103 3 3.
(c) The reference angle for 237 is 237 3 27 .
11. (a) The reference angle for 57 is 57 27 .
12. (a) The reference angle for 23 is 23 2 03.
(b) The reference angle for 14 is 14 04.
(b) The reference angle for 23 is 23 084.
(c) The reference angle for 14 is 14 because 14 2.
(c) The reference angle for 10 is 10 10 0.
13. cos 150 cos 30 23
14. sin 240 sin 60 23
SECTION 5.3 Trigonometric Functions of Angles 15. tan 330 tan 30 33 17. cot 120 cot 60 33
16. sin 30 sin 30 12
18. csc 300 csc 60 2 3 3
3
19. csc 630 csc 90 sin190 1
20. cot 210 cot 30 tan130
21. cos 570 cos 30 23
22. sec 120 sec 60 cos160 2
23. tan 750 tan 30 1 33 3
25. sin 32 sin 2 1 27. tan 43 tan 3 3 29. csc 56 csc 6 2
465
24. cos 660 cos 60 12 1 26. cos 43 cos 3 2 3 28. cos 116 cos 6 2
2 3 30. sec 76 sec 6 3
1 31. sec 173 sec 3 cos 2 3 1 33. cot 4 cot 4 tan 1 4
32. csc 54 csc 4
1 2 sin 4
2 1 34. cos 74 cos 4 2 2
1 35. tan 52 tan 36. sin 116 sin 2 which is undefined. 6 2 37. Since sin 0 and cos 0, is in quadrant III. 38. Since both tan and sin are negative, is in quadrant IV. sin 0 sin 0 (since cos 0). Since sin 0 and cos 0, is 39. sec 0 cos 0. Also tan 0 cos in quadrant IV. 40. Since csc 0 sin 0 and cos 0, is in quadrant II. sin 1 cos2 2 . 41. Since sin is negative in quadrant III, sin 1 cos and we have tan cos cos 1 sin2 cos because cos 0 in quadrant II. 42. cot sin sin 43. cos2 sin2 1 cos 1 sin2 because cos 0 in quadrant IV. 1 1 44. sec because all trigonometric functions are positive in quadrant I. cos 1 sin2 45. sec2 1 tan2 sec 1 tan2 because sec 0 in quadrant II. 46. csc2 1 cot2 csc 1 cot2 because csc 0 in quadrant III. 47. sin 45 . Then x 52 42 9 3, since is in quadrant IV. Thus, cos 35 , tan 43 , csc 54 ,
sec 53 , and cot 34 . 48. tan 43 , so r 42 32 5. Since is in quadrant III, x 3 and y 4, and so sin 45 , cos 35 , csc 54 , sec 53 , and cot 34 . 2 7 95 7 7 49. cos 12 , so sin 1 12 1295 , tan 795 , csc 129595 , sec 12 , and cot 7 95 9 145 8 145 8 9 9 50. cot 9 , so r 82 92 145. Thus, sin 145 , cos 145 , tan 8 , csc 145 9 , and sec 145 8 .
51. csc 2. Then sin 12 and x and cot 3.
145
22 12 3. So sin 12 , cos 23 , tan 1 33 , sec 2 2 3 3 , 3 3
466
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
52. cot 14 . Then tan 4 and r 42 12 17. So sin 4 , cos 1 , tan 4, csc 417 , and 17 17 sec 17. 7 7 5 , sec 7 , 53. cos 27 . Then y 72 22 45 3 5, and so sin 3 7 5 , tan 3 2 5 , csc 15 2 3 5
2 2 5. and cot 15 3 5
54. tan 4. Then r cot 14 .
42 12 17, and so sin 4 , cos 1 , csc 417 , sec 17, and 17
17
2 3 and 2 sin 2 sin 3. 55. If , then sin 2 sin 3 3 2 3 2 sin 2 3 and sin 2 sin 2 0890. 56. If , then sin 3 3 4 9
57. The area is 12 7 9 sin 72 300.
58. The area is 12 10 22 sin 10 191. 59. The area is 12 102 sin 60 25 3 433.
60. The area is 12 132 sin 60 1694 3 732.
1 32 1154, or 61. Let the angle be . Then A 12 x y sin 16 12 5 7 sin sin 32 35 sin 35 approximately 661 . 48 1 1 5 2 2 62. Let the lengths of the equal sides be x. Then A 2 x sin 24 2 x sin 6 x 96 4 6 98 cm. 5 sin 6
4 . For the triangle defined by the two sides, 63. For the sector defined by the two sides, A1 12 r 2 12 22 120 180 3 A2 12 ab sin 12 2 2 sin 120 2 sin 60 3. Thus the area of the region is A1 A2 43 3 246.
64. The area of the entire circle is r 2 122 144, the area of the sector is 12 r 2 12 122 3 24, 1 1 2 and the area of the triangle is 2 ab sin 2 12 sin 3 36 3, so the area of the shaded region is A1 A2 A3 144 24 36 3 120 36 3 4393.
65. (a) tan
5280 ft h , so h tan 1 mile 5280 tan ft. 1 mile 1 mile
(b)
20
60
80
85
h
1922
9145
29,944
60,351
66. (a) Let the depth of the water be h. Then the cross-sectional area of the gutter is a trapezoid whose height is h 10 sin . The bases are 10 and 10 2 10 cos 10 20 cos . Thus, the area is b b2 10 10 20 cos A 1 h 10 sin 100 sin 100 sin cos . 2 2 (b)
(c) 200
130.0
100
129.5
0 0
1
129.0 1.00
1.05
1.10
From the graph, the largest area is achieved when 1047 rad 60 .
SECTION 5.3 Trigonometric Functions of Angles
67. (a) From the figure in the text, we express depth and width in terms of .
467
(b)
width depth and cos , we have depth 20 sin Since sin 20 20 and width 20 cos . Thus, the cross-section area of the beam is A depth width 20 cos 20 sin 400 cos sin .
200
0 0
(c) The beam with the largest cross-sectional area is the square beam, 10 2 by 10 2 (about 1414 by 1414).
1
68. Using depth 20 sin and width 20 cos , from Exercise 65, we have strength k width depth2 . Thus S k 20 cos 20 sin 2 8000k cos sin2 .
69. (a) On Earth, the range is R
122 sin 02 sin 2 3 9 3 3897 ft and the height is g 32 4
122 sin2 2 sin2 6 9 05625 ft. H 0 2g 2 32 16
2 2 122 sin 3 23982 ft and H 12 sin 6 3462 ft (b) On the moon, R 52 2 52 2000 70. Substituting, t 5 10 158 s. 16 sin 30
71. (a) W 302 038 cot 065 csc
10
(b) From the graph, it appears that W has its minimum value at about 0946 542 .
5 0 0
72. (a) We label the lengths L 1 and L 2 as shown in the
¬ 6 Lª
9 6 figure. Since sin and cos , we L1 L2 ¬
have L 1 9 csc and L 2 6 sec . Thus L L 1 L 2 9 csc 6 sec .
(c) The minimum value of L is 2107. (d) The minimum value of L is the shortest distance the pipe must pass through.
LÁ ¬
2
(b) 40 20 0 0.0
0.5
1.0
1.5
9
73. We have sin k sin , where 594 and k 133. Substituting, sin 594 133 sin sin 594 06472. Using a calculator, we find that sin1 06472 403 , so sin 133 4 2 4 403 2 594 424 . 74. sin 4 00697564737 but sin 4 07568024953. Your partner has found sin 4 instead of sin 4 radians. O P O P adj O P. Since QS is tangent to the circle at R, O R Q is a right triangle. Then 75. cos O R hyp 1 R Q O Q hyp opp R Q and sec O Q. Since SO Q is a right angle S O Q is a right tan O R O R adj adj O S S R adj hyp O S and cot S R. Summarizing, we have triangle and O S R . Then csc O R O R opp opp sin P R, cos O P, tan R Q, sec O Q, csc O S, and cot S R.
468
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
76. (a) sin2 cos2 1 sin2 cos2
1 1 1 tan2 1 sec2 2 cos cos2 1 1 (b) sin2 cos2 1 sin2 cos2 1 1 cot2 csc2 sin2 sin2
77. Let x represent the desired real number, in radians. We wish to find the smallest nonzero solution of the equation sin x sin 180x . (Since x is in radians, 180x is in degrees.) Now if a is in degrees with 0 a 180 and sin a b, then either a b or 180 a b. The graph shows that the first positive solution occurs when
y 1
180x y=sin ¹ y=sin x
0
x
2
_1 180x 180x 180, so we have 180 x. 180 180 3088. You can verify that the value of sin is the same whether Solving this equation, we get x 180 180 your calculator is set to radian mode or degree mode.
0
5.4
INVERSE TRIGONOMETRIC FUNCTIONS AND TRIANGLES
1. For a function to have an inverse, it must be one-to-one. To define the inverse sine function we restrict the domain of the sine function to the interval 2 2 . 2. (a) The function sin1 has domain [1 1] and range 2 2 . (b) The function cos1 has domain [1 1] and range [0 ]. (c) The function tan1 has domain R and range 2 2 .
8 sin1 4 6 cos1 3 (b) cos1 10 3. (a) sin1 10 5 5 5 , we let cos1 5 and 4. To find sin cos1 13 13
complete the right triangle as shown. We find that 5 12 . sin cos1 13 13
(c) tan1 68 tan1 43
13 ¬
12
5
3 3
5. (a) sin1 1 2
(b) cos1 0 2
(c) tan1
6. (a) sin1 0 0 7. (a) sin1 22 4
(b) cos1 1 (b) cos1 22 34
(c) tan1 0 0
8. (a) sin1 23 3 7. sin1 030 030469 13. tan1 3 124905
(b) cos1 12 23
(c) tan1 1 4
(c) tan1 3 3
10. cos1 02 177215 11. cos1 13 123096
14. tan1 4 132582 15. cos1 3 is undefined.
12. sin1 65 098511
16. sin1 2 is undefined.
6 3 , so sin1 3 369 . 17. sin 10 5 5
7 , so tan1 7 213 . 18. tan 18 18
9 , so tan1 9 347 . 19. tan 13 13
3 1 3 254 . 20. sin 30 7 70 7 , so sin
21. sin 47 , so sin1 47 348 .
22. cos 89 , so cos1 98 273 .
SECTION 5.4 Inverse Trigonometric Functions and Triangles
469
23. We use sin1 to find one solution in the interval 90 90 . sin 23 sin1 23 418 . Another solution with between 0 and 180 is obtained by taking the supplement of the angle: 180 418 1382 . So the solutions of the equation with between 0 and 180 are approximately 418 and 1382 . 24. One solution is given by cos1 43 414 . This is the only solution, because cos x is one-to-one on 0 180 .
25. One solution is given by cos1 25 1136 . This is the only solution, because cos x is one-to-one on 0 180 . 26. tan1 20 871 , so the only solution in 0 180 is approximately 180 871 929 . 27. tan1 5 787 . This is the only solution on 0 180 .
28. One solution is sin1 45 531 . Another solution on 0 180 is approximately 180 531 1269 . 29. To find cos sin1 45 , first let sin1 54 . Then is the number in the interval 2 2 whose sine is 45 . We draw a right triangle with as one of its acute
5
angles, with opposite side 4 and hypotenuse 5. The remaining leg of the triangle is
¬
found by the Pythagorean Theorem to be 3. From the figure we get cos sin1 54 sin 35 .
4
3
Another method: By the cancellation properties of inverse functions, sin sin1 54 is exactly 45 . To find cos sin1 45 , we
first write the cosine function in terms of the sine function. Let u sin1 45 . Since 0 u 2 , cos u is positive, and since 2 9 3 cos2 u sin2 u 1, we can write cos u 1 sin2 u 1 sin2 sin1 45 1 45 1 16 25 25 5 . Therefore, cos sin1 54 35 . 30. To find cos tan1 54 , we draw a right triangle with angle , opposite side 4, and adjacent side 3. From the figure we see that cos tan1 34 cos 35 .
12 , we draw a right triangle with 31. To find sec sin1 13
angle , opposite side 12, and hypotenuse 13. From the 12 sec 13 . figure we see that sec sin1 13 5 13 12
¬
5
5 ¬
4
3
7 , we draw a right triangle with 32. To find csc cos1 25
angle , opposite side 7, and hypotenuse 25. From the 7 csc 25 . figure we see that csc cos1 25 24 25
¬
24
7
470
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
1 2 , we draw a right triangle with angle , we draw a right triangle with angle 33. To find tan sin1 12 34. To find cot sin 13 3 , opposite side 12, and hypotenuse 13. From the figure 12 tan 12 . we see that tan sin1 13 5 13
¬
, opposite side 2, and hypotenuse 3. From the figure we see that cot sin1 32 cot 25 .
5
12
3
2
¬ Ï5
35. We want to find cos sin1 x . Let sin1 x, so sin x. We sketch a right triangle with an acute angle , opposite side x, and hypotenuse 1. By the Pythagorean Theorem, the remaining leg is 1 x 2 . From the figure we have cos sin1 x cos 1 x 2 .
1
x
¬ Ï1-x@
Another method: Let u sin1 x. We need to find cos u in terms of x. To do so, we write cosine in terms of sine. Note 1 x. Now cos u 1 sin2 u is positive because u lies in the interval . that 2 u 2 because u sin 2 2 Substituting u sin1 x and using the cancellation property sinsin1 x x gives cossin1 x 1 x 2 . 36. We want to find sin tan1 x . Let tan1 x, so tan x. We sketch a right Ïx@+1
triangle with an acute angle , opposite side x, and adjacent side 1. By the Pythagorean Theorem, the remaining leg is x 2 1. From the figure we have x . sin tan1 x sin x2 1 37. We want to find tan sin1 x . Let sin1 x, so sin x. We sketch a right
triangle with an acute angle , opposite side x, and hypotenuse 1. By the Pythagorean Theorem, the remaining leg is 1 x 2 . From the figure we have x . tan sin1 x tan 1 x2 38. We want to find cos tan1 x . Let tan1 x, so tan x. We sketch a right
¬
x
1
1
x
¬ Ï1-x@
Ïx@+1
x triangle with an acute angle , opposite side x, and adjacent side 1. By the ¬ Pythagorean Theorem, the remaining leg is x 2 1. From the figure we have 1 1 1 cos tan x cos . x2 1 39. Let represent the angle of elevation of the ladder. Let h represent the height, in feet, that the ladder reaches on the
6 03 cos1 03 1266 rad 725 . By the Pythagorean Theorem, h 2 62 202 building. Then cos 20 h 400 36 364 19 ft.
96 08 tan1 08 0675 387 . 40. Let be the angle of elevation of the sun. Then tan 120 41. (a) Solving tan h2 for h, we have h 2 tan .
(b) Solving tan h2 for we have tan1 h2.
42. (a) Solving tan 50s. Solving for , we have tan1 50s.
5 1 5 682 . (b) Set s 20 ft to get tan 50 20 2 . Solving for , we have tan 2
SECTION 5.5 The Law of Sines
471
43. (a) Solving sin h680 for we have sin1 h680. (b) Set h 500 to get sin1 500 680 0826 rad 473 .
3960 . Solving for , we get 44. (a) Since the radius of the earth is 3960 miles, we have the relationship cos h3960 3960 . cos1 h3960
(b) The arc length is s radius included angle 3960 2 7920. 3960 (c) s 7920 cos1 h3960 3960 (d) When h 100 we have s 7920 cos1 1003960 7920 cos1 3960 4060 17615 miles. 3960 3960 3960 2450 1 (e) When s 2450 we have 2450 7920 cos1 h3960 2450 7920 cos h3960 h3960 cos 7920 2450 h 3960 3960 sec 2450 7920 h 3960 sec 7920 3960 1973 mi. 1 1 1 1 sin sin1 08102 541 45. (a) sin 7 tan 10 2 3 1 tan 10 1 1 . For n 3, sin1 483 322 . For n 4, (b) For n 2, sin1 5 tan 15 7 tan 15 1 1 1 245 . n 0 and n 1 are outside of the domain for 15 , because 3732 sin 9 tan 15 tan 15 1 and 1244, neither of which is in the domain of sin1 . 3 tan 15 1 46. Let sec1 x. Then sec x, as shown in the figure. Then cos , so x x 1 1 . Thus, sec1 x cos1 , x 1. cos1 x x In particular, sec1 2 cos1 12 3.
1 Let csc1 x. Then csc x, as shown in the figure. Then sin , so x 1 1 sin1 . Thus, csc1 x sin1 , x 1. x x In particular, csc1 3 sin1 31 0340.
1 Let cot1 x. Then cot x, as shown in the figure. Then tan , so x 1 1 tan1 . Thus, cot1 x tan1 , x 1. x x In particular, cot1 4 tan1 41 0245.
5.5
THE LAW OF SINES
sin B sin C sin A . a b c 2. (a) The Law of Sines can be used to solve triangles in cases ASA or SSA.
1. In triangle ABC with sides a, b, and c the Law of Sines states that
(b) The Law of Sines can give ambiguous solutions in case SSA. 376 sin 57 31875. 3. C 180 984 246 57 . x sin 984
¬
1
x
1
º
1
x
472
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
17 sin 1144 4. C 180 375 281 1144 . x 254. sin 375 267 sin 52 5. C 180 52 70 58 . x 248. sin 58 6. sin
563 sin 67 0646. Then sin1 0646 403 . 802
7. sin C
36 sin 120 0693 C sin1 0693 44 . 45
185 sin 50 1449. 8. C 180 102 28 50 . x sin 102 65 sin 46 65 sin 20 9. C 180 46 20 114 . Then a 51 and b 24. sin 114 sin 114 2 sin 100 2 sin 30 257 and a 131. 10. B 180 30 100 50 . Then c sin 50 sin 50 12 sin 44 11. B 68 , so A 180 68 68 44 and a 899. sin 68 12. sin B
34 sin 80 65 sin 69 0515, so B sin1 0515 31 . Then C 180 80 31 69 and c 62. 65 sin 80
13. C 180 50 68 62 . Then 230 sin 50 230 sin 68 a 200 and b 242. sin 62 sin 62
14. C 180 110 23 47 . Then 50 sin 23 50 sin 110 a 267 and b 642. sin 47 sin 47 C
C
50¡
A
68¡ 230
23¡
A
110¡ 50
B
B
15. B 180 30 65 85 . Then 10 sin 65 10 sin 30 50 and c 9. a sin 85 sin 85
16. C 180 95 22 63 . Then 420 sin 63 420 sin 95 11169 and c 9990. b sin 22 sin 22
C 10
65¡
30¡
A
C 420 22¡
A
B
17. A 180 51 29 100 . Then 44 sin 51 44 sin 100 89 and c 71. a sin 29 sin 29
95¡
18. A 180 100 10 70 . Then 115 sin 10 115 sin 70 1097 and b 203. a sin 100 sin 100 C
C
A
44 A
10¡
100¡
51¡
29¡
B
B
115
B
SECTION 5.5 The Law of Sines
473
15 sin 110 0503 B sin1 0503 30 . Then 19. Since A 90 there is only one triangle. sin B 28 28 sin 40 C 180 110 30 40 , and so c 19. Thus B 30 , C 40 , and c 19. sin 110 40 sin 37 20. sin C 0802 C1 sin1 0822 534 or C2 180 534 1266 . 30 30 sin 896 If C1 534 , then B1 180 37 534 896 and b1 498. sin 37 30 sin 164 If C2 1266 , then B2 180 37 1266 164 and b2 141. sin 37 Thus, one triangle has B1 896 , C1 534 , and b1 498; the other has B2 164 , C2 1266 , and b2 141. 21. A 125 is the largest angle, but since side a is not the longest side, there can be no such triangle. 45 sin 38 22. sin B 0660 B1 sin1 0660 413 or B2 180 413 1387 . 42 42 sin 1007 67. If B1 413 , then A1 180 38 413 1007 and a1 sin 38 42 sin 33 If B2 1387 , then A2 180 38 1387 33 and a2 39. sin 38 Thus, one triangle has A1 1007 , B1 413 , and a1 67; the other has A2 33 , B2 1387 , and a2 39. 30 sin 25 23. sin C 0507 C1 sin1 0507 3047 or C2 180 3947 14953 . 25 25 sin 12453 4873. If C1 3047 , then A1 180 25 3047 12453 and a1 sin 25 25 sin 547 If C2 14953 , then A2 180 25 14953 547 and a2 564. sin 25 Thus, one triangle has A1 125 , C1 30 , and a1 49; the other has A2 5 , C2 150 , and a2 56. 100 sin 30 24. sin B 23 B1 sin1 32 418 or B2 180 418 1382 . 75 75 sin 1082 1425. If B1 418 , then C1 180 30 418 1082 and c1 sin 30 75 sin 118 If B2 1382 , then C2 180 30 1382 118 and c2 307. sin 30 Thus, one triangle has B1 418 , C1 1082 , and c1 1425; the other has B2 1382 , C2 118 , and c2 307. 100 sin 50 1532. Since sin 1 for all , there can be no such angle B, and thus no such triangle. 25. sin B 50 80 sin 135 26. sin B 0566 B1 sin1 0566 344 or B2 180 344 1456 . 100 100 sin 106 If B1 344 , then C 180 135 344 106 and c 259. sin 135 If B2 180 344 1456 , then A B2 135 1456 180 , so there is no such triangle. Thus, the only possible triangle is B 344 , C 106 , and c 259. 26 sin 29 27. sin A 0840 A1 sin1 0840 572 or A2 180 572 1228 . 15 15 sin 938 309. If A1 572 , then B1 180 29 572 938 and b1 sin 29 15 sin 281 146. If A2 1228 , then B2 180 29 1228 282 and b2 sin 29 Thus, one triangle has A1 572 , B1 938 , and b1 309; the other has A2 1228 , B2 282 , and b2 146.
474
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
82 sin 58 0953, so C1 sin1 0953 724 or C2 180 724 1076 . 73 73 sin 496 656. If C1 724 then A1 180 58 724 496 and a1 sin 58 73 sin 144 If C2 1076 , then A2 180 58 1076 144 and a2 214. sin 58 Thus, one triangle has A1 496 , C1 724 , and a1 656; the other has A2 144 , C2 1076 , and a2 214.
28. sin C
sin B 28 sin 30 sin 30 sin B 07, so 20 28 20 B sin1 07 44427 . Since BC D is isosceles, B B DC 44427 . Thus, BC D 180 2 B 91146 911 .
29. (a) From ABC and the Law of Sines we get
(b) From ABC we get BC A 180 A B 180 30 44427 105573 . Hence DC A BC A BC D 105573 91146 144 . 30. By symmetry, DC B 25 , so A 180 25 50 105 . Then by the Law of Sines, AD
12 sin 25 525. sin 105
31. (a) Let a be the distance from satellite to the tracking station A in miles. Then the subtended angle at the satellite is 50 sin 842 C 180 93 842 28 , and so a 1018 mi. sin 28 (b) Let d be the distance above the ground in miles. Then d 10183 sin 87 1017 mi. 32. (a) Let x be the distance from the plane to point A. Then x 5
sin 48 377 mi. sin 100
(b) Let h be the height of the plane. Then sin 32
sin 48 x sin 48 AB sin 180 32 48 sin 100
h h 377 sin 32 200 mi. x
AC AB AB sin 52 AC , so substituting we 33. C 180 82 52 46 , so by the Law of Sines, sin 52 sin 46 sin 46 200 sin 52 219 ft. have AC sin 46 312 sin 486 0444 ABC sin1 0444 264 , and so BC A 180 486 264 105 . 527 527 sin 105 6785 ft. Then the distance between A and B is AB sin 486
34. sin ABC
35.
We draw a diagram. A is the position of the tourist and C is the top of the tower.
C
B 90 56 844 and so C 180 292 844 664 . Thus, by 105 sin 292
5.6¡
the Law of Sines, the length of the tower is BC
29.2¡ B
C 165
The situation is illustrated in the diagram. 180
67¡ B
559 m.
A
105
36.
sin 664
A
D
AC 165 sin 67 1519 ft, so using the Pythagorean Theorem we can calculate B A 1652 15192 644 ft and AD 1802 15192 966 ft. Thus the anchor points are B A AD 644 966 161 ft apart.
SECTION 5.5 The Law of Sines
475
37. The angle subtended by the top of the tree and the sun’s rays is A 180 90 52 38 . Thus the height of the tree 215 sin 30 is h 175 ft. sin 38 C
38.
Let x be the length of the wire, as shown in the figure. Since 12 , other angles in ABC are 90 58 148 , and 180 12 148 20 .
x
Œ
Thus,
100 sin 148 x x 100 155 m. sin 148 sin 20 sin 20
º B 100
A 58¡
39. Call the balloon’s position R. Then in P Q R, we see that P 62 32 30 , and Q 180 71 32 141 . Q R P Q sin 30 Therefore, R 180 30 141 9 . So by the Law of Sines, Q R 60 192 m. sin 30 sin 9 sin 9 B
40. º
30 C
Label the diagram as shown, and let the hill’s angle of elevation be . Then applying the Law of Sines to ABC,
sin sin 8 120 30
sin 4 sin 8 055669 sin1 055669 338 . But from AB D,
B AD B 8 90 , so 90 8 338 482 . 8¡
A
Œ
120
D
41. Let d be the distance from the earth to Venus, and let be the angle formed by sun, Venus, and earth. By the Law sin 394 sin 0878, so either sin1 0878 614 or 180 sin1 0878 1186 . of Sines, 1 0723 0723 d d 1119 AU; in the second case, In the first case, sin 180 394 614 sin 394 d 0723 d 0427 AU. sin 180 394 1186 sin 394 sin 60 b sin 60 sin B or sin B . Similarly, applying the Law of b c c sin B sin 120 r sin 120 r b Sines to BC D gives or sin B . Since sin 120 sin 60 , we have r cd cd c cd c sin D sin 60 b sin 60 b (). Similarly, from ADC and the Law of Sines we have or sin D , and r cd b d d a sin 120 b sin 60 a sin 120 b d from B DC we have sin D . Thus, . Combining this with cd d cd a cd b c d cd b b ab r a b 1. Solving for r, we find 1 (), we get r a cd cd cd r a a b ab ab . r ab
42. (a) Applying the Law of Sines to ABC, we get
476
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
43 12 cm. 43 (c) If a b, then r is infinite, and so the face is a flat disk.
(b) r
43. By the area formula from Section 5.3, the area of ABC is A 12 ab sin C. Because we are given a and the three sin A a sin B sin B b . Thus, angles, we need to find b in terms of these quantities. By the Law of Sines, b a sin A 2 a sin B sin C a sin B A 12 ab sin C 12 a sin C . sin A 2 sin A 44. By the area formula from Section 5.3,
1 ab sin C sin C Area of ABC 12 , because a and b are the same for both Area of A B C sin C ab sin C 2
triangles. C
45. b
a
B
C b
B
A
a b: One solution
a
C b
a»
B
B
b a b sin A: Two solutions
A
C b
a B
a b sin A: One solution
A
a b sin A: No solution
A 30 , b 100, sin A 1 . If a b 100 then there is one triangle. If 100 a 100 sin 30 50, then there are 2
two possible triangles. If a 50, then there is one (right) triangle. And if a 50, then no triangle is possible.
5.6
THE LAW OF COSINES
1. For triangle ABC with sides a, b, and c the Law of Cosines states c2 a 2 b2 2ab cos C.
2. The Law of Cosines is required to solve triangles in cases SSS and SAS.
3. x 2 212 422 2 21 42 cos 39 441 1764 1764 cos 39 834115 and so x 834115 289. 4. x 2 152 182 2 15 18 cos 108 225 324 540 cos 108 715869 and so x 715869 268. 5. x 2 252 252 2 25 25 cos 140 625 625 1250 cos 140 2207556 and so x 2207556 47. 6. x 2 22 82 2 2 8 cos 88 4 64 32 cos 88 66883 and so x 66883 82. 7. 37832 68012 42152 2 6801 4215 cos . Then cos cos1 0867 2989 . 8. 15462 6012 12252 2 601 1225 cos . Then cos cos1 0359 111 .
37832 68012 42152 0867 2 6801 4215
15462 6012 12252 0359 2 601 1225
9. x 2 242 302 2 24 30 cos 30 576 900 1440 cos 30 228923 and so x 228923 15.
156 202 102 122 065 cos1 065 13054 . 2 10 12 240 11. c2 102 182 2 10 18 cos 120 100 324 360 cos 120 604 and so c 604 24576. Then 18 sin 120 sin A 0634295 A sin1 0634295 394 , and B 180 120 394 206 . 24576
10. 202 102 122 2 10 12 cos . Then cos
12. 122 402 442 2 40 44 cos B cos B sin A
122 402 442 0964 B cos1 0964 15 . Then 2 40 44
40 sin 155 0891 A sin1 0891 63 , and so C 180 15 63 102 . 12
SECTION 5.6 The Law of Cosines
477
4 sin 53 0983 13. c2 32 42 2 3 4 cos 53 9 16 24 cos 53 10556 c 10556 32. Then sin B 325 B sin1 0983 79 and A 180 53 79 48 . 14. a 2 602 302 2 60 30 cos 70 3600 900 3600 cos 70 326873 30 sin 70 a 326873 572. Then sin C 0493 572 C sin1 0493 295 , and B 180 70 295 805 .
2 252 222 15. 202 252 222 2 25 22 cos A cos A 20 22522 0644 A cos1 0644 50 . Then sin B 25 sin20499 0956 B sin1 0956 73 , and so C 180 50 73 57 . 2 122 162 078125 16. 102 122 162 2 12 16 cos A cos A 10 21216 A cos1 078125 386 . Then sin B 12 sin 386 0749 10
B sin1 0749 485 , and so C 180 386 485 929 .
sin 40 0833 C sin1 0833 564 or C 180 564 1236 . 17. sin C 162125 1 2 sin 836 193. If C1 564 , then A1 180 40 564 836 and a1 125sin 40 125 sin 164 549. If C2 1236 , then A2 180 40 1236 164 and a2 sin 40
Thus, one triangle has A 836 , C 564 , and a 193; the other has A 164 , C 1236 , and a 549.
52 1024. Since sin 1 for all , there is no such A, and hence there is no such triangle. 18. sin A 65 sin 50
55 1065. Since sin 1 for all , there is no such B, and hence there is no such triangle. 19. sin B 65 sin 50 sin 61 1094 and c 735 sin 83 1241. 20. A 180 61 83 36 . Then b 735 sin 36 sin 36 sin 35 2. 21. B 180 35 85 60 . Then x 3sin 60
22. x 2 10 18 2 10 18 cos 40 100 324 360 cos 40 148224 and so x 148224 122.
sin 30 23. x 50 sin 100 254
2 102 112 1 0932 213 . 205 24. 42 102 112 2 10 11 cos . Then cos 4 21011 220 0932 cos
25. b2 1102 1382 2 110 138 cos 38 12,100 19,044 30,360 cos 38 72200 and so b 850. Therefore, 2 852 1282 using the Law of Cosines again, we have cos 1102110138
8915 .
40 0803 sin1 0803 535 or 180 535 1265 , but 535 doesn’t fit the 26. sin 10 sin 8 picture, so 1265 .
27. x 2 382 482 2 38 48 cos 30 1444 2304 3648 cos 30 588739 and so x 243. sin 98 11808. 28. A 180 98 25 57 . Then x 1000 sin 57
18, so by Heron’s Formula the area is 29. The semiperimeter is s 91215 2 A 18 18 9 18 12 18 15 2916 54. 52 , so by Heron’s Formula the area is 30. The semiperimeter is s 122 2 15 A 52 52 1 52 2 52 2 15 16 4 0968.
12, so by Heron’s Formula the area is 31. The semiperimeter is s 789 2 A 12 12 7 12 8 12 9 720 12 5 268.
32. The semiperimeter is s 11100101 106, so by Heron’s Formula the area is 2 A 106 106 11 106 100 106 101 302,100 10 3021 550.
478
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
33. The semiperimeter is s 346 13 2 2 , so by Heron’s Formula the area is 13 3 13 4 13 6 455 455 533. A 13 2 2 2 2 16 4 6, so by Heron’s Formula the 34. Both of the smaller triangles have the same area. The semiperimeter of each is s 255 2 area of each is A 6 6 2 6 5 6 5 24 2 6, so the shaded area is 4 6 980. 35. We draw a diagonal connecting the vertices adjacent to the 100 angle. This forms two triangles. Consider the triangle with sides of length 5 and 6 containing the 100 angle. The area of this triangle is A1 12 5 6 sin 100 1477. To use Heron’s Formula to find the area of the second triangle, we need to find the length of the diagonal using the Law of Cosines: c2 a 2 b2 2ab cos C 52 62 2 5 6 cos 100 71419 c 845. Thus the second triangle has semiperimeter 8 7 845 117255 and area A2 117255 117255 8 117255 7 117255 845 2600. The area s 2 of the quadrilateral is the sum of the areas of the two triangles: A A1 A2 1477 2600 4077. 36. We draw a line segment with length x bisecting the 60 angle to create two triangles. By the Law of Cosines, 32 42 x 2 2 4x cos 30 x 2 4 3x 7 0. Using the Quadratic Formula, we find x 2 3 5. The minus sign provides the correct length of about 123 (the other solution is about 57, which corresponds to a convex quadrilateral 342 3 5 with the same side lengths), so the semiperimeter of each triangle is s and the total area of the figure 2 is A 2 s s 3 s 4 s 2 3 5 246. 37.
Label the centers of the circles A, B, and C, as in the figure. By the Law of C
6
5
6
B 5
4
4 A
Cosines, cos A
92 102 112 AB 2 AC 2 BC 2 13 A 7053 . 2 AB AC 2 9 10
Now, by the Law of Sines,
sin 7053 sin B sin C . So 11 AC AB
1 085710 5899 and sin B 10 11 sin 7053 085710 B sin 9 sin 7053 077139 C sin1 077139 5048 . The area of sin C 11
ABC is 12 AB AC sin A 12 9 10 sin 7053 42426. 2 7053 9848. Similarly, the areas of sectors B and C The area of sector A is given by S A R 2 4 360 360 are S B 12870 and SC 15859. Thus, the area enclosed between the circles is A ABC S A S B SC A 42426 9848 12870 15859 385 cm2 .
38. By the Law of Cosines we have a 2 62 42 2 6 4 cos 45 52 24 2, b2 62 x 2 2 6 x cos 30 x 2 6 3x36, and c2 x 2 42 2 x 4 cos 30 45 x 2 2 2 6 x16. By the Pythagorean Theorem, a 2 b2 c2 , so we have 52 24 2 x 2 6 3x 36 x 2 2 2 6 x 16 12 3 2 26 12 2 3 3x 18 2 6 x 8 36 12 2 3 3 2 6 x, so x . 3 3 2 6 39. Let c be the distance across the lake, in miles. Then c2 2822 3562 2 282 356 cos 403 5313 c 230 mi.
SECTION 5.6 The Law of Cosines D
40. Suppose ABC D is a parallelogram with AB DC 5, AD BC 3, and A 50 (see the figure). Since opposite angles are equal in a
parallelogram, it follows that C 50 , and
260 B D 360 100 260 . Thus, B D 130 .
479
C 3
50¡l
A
B
5
2 By the Law of Cosines, AC 2 32 52 2 3 5 cos 130 AC 9 25 30 cos 130 73. Similarly, B D 32 52 2 3 5 cos 50 38. 41. In half an hour, the faster car travels 25 miles while the slower car travels 15 miles. The distance between them is given by the Law of Cosines: d 2 252 152 2 25 15 cos 65 d 252 152 2 25 15 cos 65 5 25 9 30 cos 65 231 mi. 42. Let x be the car’s distance from its original position. Since the car travels
x
at a constant speed of 40 miles per hour, it must have traveled 40 miles
º 45¡
east, and then 20 miles northeast (which is 45 east of “due north”). From 40
the diagram, we see that 135 , so x 202 402 2 20 40 cos 135 10 4 16 16 cos 135 560 mi. 43. The pilot travels a distance of 625 15 9375 miles in her original
direction and 625 2 1250 miles in the new direction. Since she makes a course correction of 10 to the right, the included angle is
20
10¡
937.5
1250
d
180 10 170 . From the figure, we use the Law of Cosines to get
the expression d 2 93752 12502 2 9375 1250 cos 170 4,749,54942, so d 2179 miles. Thus, the pilot’s distance from her original position is approximately 2179 miles.
44. Let d be the distance between the two boats in miles. After one hour, the boats have traveled distances of 30 miles and 26 miles. Also, the angle subtended by their directions is 180 50 70 60 . Then d 2 302 262 2 30 26 cos 60 796 d 796 282. Thus the distance between the two boats is about 28 miles. 45. (a) The angle subtended at Egg Island is 100 . Thus using the Law of
Forrest Island
Cosines, the distance from Forrest Island to the fisherman’s home port is x 2 302 502 2 30 50 cos 100 900 2500 3000 cos 100 3920945
10¡
and so x 3920945 6262 miles.
S 182 E.
x
80¡
(b) Let be the angle shown in the figure. Using the Law of Sines, 50 sin 100 07863 sin1 07863 518 . Then sin 6262 90 20 518 182 . Thus the bearing to his home port is
50
Egg Island
20¡ 30
¬
70¡ Home Port
480
CHAPTER 5 Trigonometric Functions: Right Triangle Approach B
46. (a) In 30 minutes the pilot flies 100 miles due east, so using the Law of Cosines we have x 2 1002 3002 2 100 300 cos 40
300
1002 1 9 6 cos 40 1002 5404. Thus, x
1002 5404 2325, and so the pilot is 2325 miles from
A
his destination.
(b) Using the Law of Sines, sin
50¡ 40¡
x ¬
100
300 sin 40 0829 2325
sin1 0829 56 . However, since 90 , the angle we seek is
180 56 124 . Hence the bearing is 124 90 34 , that is, N 34 E.
47. The largest angle is the one opposite the longest side; call this angle . Then by the Law of Cosines, 442 362 222 2 36 22 cos cos
362 222 442 009848 cos1 009848 96 . 2 36 22
48. Let be the angle formed by the cables. The two tugboats and the barge form a triangle: the side opposite has a length of 120 ft and the other two sides have lengths of 212 and 230 ft. Therefore, 1202 2122 2302 2 212 230 cos cos
2122 2302 1202 cos 08557 cos1 08557 31 . 2 212 230
49. Let d be the distance between the kites. Then d 2 3802 4202 2 380 420 cos 30 d 3802 4202 2 380 420 cos 30 211 ft. 50. Let x be the length of the wire and the angle opposite x, as shown in the figure. Since the mountain is inclined 32 , we must have
125
x
180 90 32 122 . Thus, x 552 1252 2 55 125 cos 122 161 ft.
¬ 55 32¡
51. Solution 1: From the figure, we see that 106 and sin 74
3400 b
3400 3537. Thus, x 2 8002 35372 2 800 3537 cos 106 sin 74 x 8002 35372 2 800 3537 cos 106 x 3835 ft.
b
Solution 2: Notice that tan 74
3400 a
a
Pythagorean Theorem, x 2 a 8002 34002 . So x 9749 8002 34002 3835 ft.
x
3400 9749. By the tan 74
b
3400
74¡
800
52. Let the woman be at point A, the first landmark (at 62 ) be at point B, and the other landmark be at point C. We 1150 1150 1150 AB 2450. Similarly, cos 54 want to find the length BC. Now, cos 62 AB cos 62 AC 1150 1956. Therefore, by the Law of Cosines, BC 2 AB 2 AC 2 2 AB AC cos 43 AC cos 54 BC 24502 19562 2 2450 1956 cos 43 BC 1679. Thus, the two landmarks are roughly 1679 feet apart.
CHAPTER 5
Review
481
112 148 190 abc 225. Thus, 53. By Heron’s formula, A s s a s b s c, where s 2 2 A 225 225 112 225 148 225 190 82777 ft2 . Since the land value is $20 per square foot, the value of the lot is approximately 82777 20 $165,554. sin 465 sin B 54. Having found a 132 using the Law of Cosines, we use the Law of Sines to find B: 105 132 105 sin 465 sin B 0577. Now there are two angles B between 0 and 180 which have sin B 0577, namely 132 B 352 and B 1448 . But we must choose B , since otherwise A B 180 . 1 2 1 sin 465 180 sin 465 sin C sin C 0989, so either C 815 Using the Law of Sines again, 180 132 132 or C 985 . In this case we must choose C 985 so that the sum of the angles in the triangle is A B C 465 352 985 180 . (The fact that the angles do not sum to exactly 180 , and the discrepancies between these results and those of Example 3, are due to roundoff error.) The method in this exercise is slightly easier computationally, but the method in Example 3 is more reliable. 55. In any ABC, the Law of Cosines gives a 2 b2 c2 2bccos A, b2 a 2 c2 2accos B, and c2 a 2 b2 2abcos C. Adding the second and third equations gives b2 a 2 c2 2ac cos B
c2 a 2 b2 2ab cos C
b2 c2 2a 2 b2 c2 2a c cos B b cos C Thus 2a 2 2a c cos B b cos C 0, and so 2a a c cos B b cos C 0. Since a 0 we must have a c cos B b cos C 0 a b cos C c cos B . The other laws follow from the symmetry of a, b, and c.
CHAPTER 5 REVIEW 052 rad 1. (a) 30 30 180 6
7 183 rad 2. (a) 105 105 180 12
5 262 rad (b) 150 150 180 6
2 126 rad (b) 72 72 180 5
035 rad (c) 20 20 180 9
9 707 rad (c) 405 405 180 4
5 393 rad (d) 225 225 180 4
7 550 rad (d) 315 315 180 4
3. (a) 56 rad 56 180 150
4. (a) 53 rad 53 180 300
180 (b) 9 rad 9 20
(b) 109 rad 109 180 200
(c) 43 rad 43 180 240
900 (c) 5 rad 5 180 2865
720 (d) 4 rad 4 180 2292
(d) 113 rad 113 180 660
5. r 10 m, 25 rad. Then s r 10 25 4 126 m. 6. s 7 cm, r 25 cm. Then
s 7 28 rad 1604 25 r
5 rad. Then r s 25 18 90 286 ft. 7. s 25 ft, 50 50 180 18 5 13 rad. Then r s 13 18 18 m. 8. s 13 m, 130 130 180 18 13
482
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
9. Since the diameter is 28 in, r 14 in. In one revolution, the arc length (distance traveled) is s r 2 14 28 in. The total distance traveled is 60 mi/h 05 h 30 mi 30 mi 5280 ft/mi 12 in./ft 1,900,800 in. The number of 1 rev revolution is 1,900,800 in 216087 rev. Therefore the car wheel will make approximately 21,609 revolutions. 28 in. 10. r 3960 miles, s 2450 miles. Then
2450 s 35448 and so the angle is 0619 rad 0619 180 r 3960
approximately 354 . 11. r 5 m, 2 rad. Then A 12 r 2 12 52 2 25 m2 . 18,151 ft2 12. A 12 r 2 12 2002 52 180
2A 2 125 250 625 04 rad 229 r2 252 2A 100 11 42 m. 14. A 50 m2 and 116 rad. Thus, r 13. A 125 ft2 , r 25 ft. Then
6
150 2 rad 300 rad/min 9425 rad/min. The linear speed is 1 min
15. The angular speed is
150 2 8 2400 in./min 75398 in./min 6283 ftmin. 1
rad 7000 rad/min 21,9911 rad/min. 16. (a) The angular speed of the engine is e 35002 1 min
e 7000rad/min 09 (b) To find the angular speed of the wheels, we calculate g 77778 rad/min 24,4346 rad/min.
(c) The speed of the car is the angular speed of the wheels times their radius: 77778 rad 11 in 60 min 1 mile 2545 mi/h. min 1 hr 63,360 in.
52 72 74. Then sin 5 , cos 7 , tan 57 , csc 574 , sec 774 , and cot 75 . 74 74 3 , cos 91 , tan 3 , csc 10 , sec 10 , and cot 91 . 18. x 102 32 91. Then sin 10 10 3 3
17. r
91
x 19. cos 40 5
20. cos 35
x 5 cos 40 383, and 5y sin 40
91
y 5 sin 40 321.
2 x cos235 244, and tan 35 2y y 2 tan 35 140. x
1 2924 311. sin 20 x sin120 292, and xy cos 20 y cosx20 09397 x x 22. cos 30 x 4 cos 30 346, and sin 30 xy y x sin 30 346 05 173. 4
21.
23. A 90 20 70 , a 3 cos 20 2819, and b 3 sin 20 1026. A
C
b
24. C 90 60 30 , cos 60 20a
a 20 cos 60 40, and tan 60 b20 b 20 tan 60 3464. C
70¡ 3
20¡
a b B A
30¡
a 60¡ 20
B
CHAPTER 5
25. c
7 02960 170 , 252 72 24, A sin1 24
and C sin1 24 25 12870 737 . 25 A
c
26. b
Review
483
5 03948 226 , 122 52 13, A sin1 13
and C sin1 12 13 11760 674 .
C
b
7 B
A
12
C 5 B
1 1 1 1 a cot , sin b csc a tan b sin h 28. Let h be the height of the tower in meters. Then tan 2881 h 1000 tan 2881 550 m. 1000
27. tan
One side of the hexagon together with radial line segments through its endpoints
29. x
8
forms a triangle with two sides of length 8 m and subtended angle 60 . Let x be the
60¡
length of one such side (in meters). By the Law of Cosines,
8
x 2 82 82 2 8 8 cos 60 64
x 8. Thus the perimeter of the
hexagon is 6x 6 8 48 m.
y
30. As the crankshaft moves in its circular pattern, point Q is
y
determined by the angle , namely it has coordinates Q 2 cos , 2 sin . We split the triangle into two right triangles O Q R and P Q R, as shown in the figure. Let h be the height of
the piston. We consider two cases, 0 180 and
180 360 . If 0 180 , then h is the sum of O R and R P. Using the Pythagorean Theorem, we find R P
82 2 cos 2 , while
8
h
h
8 R ¬ O 2
Q(2 cos ¬, 2 sin ¬) x
Q(2 cos ¬, 2 sin ¬)
¬
O 2 R
x
O R is the y-coordinate of the point Q, 2 sin . Thus h 64 4 cos2 2 sin .
If 180 360 , then h is the difference between R P and R O. Again, R P 64 4 cos2 and O R is the y-coordinate of the point Q, 2 sin . Thus h 64 4 cos2 2 sin . Since sin 0 for 180 360 , this also reduces to h 64 4 cos2 2 sin . Since we get the same result in both cases, the height of the piston in terms of is h 64 4 cos2 2 sin . r 31. Let r represent the radius, in miles, of the moon. Then tan , 0518 r r 236,900tan 0259 2 r AB 236,900 tan 0259 r 1 tan 0259 236,900 tan 0259 r 1076 and so the radius of the moon is 1 tan 0259 roughly 1076 miles. 32. Let d1 represent the horizontal distance from a point directly below the plane to the closer ship in feet, and d2 represent the 35,000 35,000 35,000 d1 , and similarly tan 40 horizontal distance to the other ship in feet. Then tan 52 d1 tan 52 d2 35,000 35,000 35,000 . So the distance between the two ships is d2 d1 14,400 ft. d2 tan 40 tan 40 tan 52 34. csc 94 csc 33. sin 315 sin 45 1 22 4 2 2
35. tan 135 tan 45 1
3 36. cos 56 cos 6 2
484
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
1 3 37. cot 223 cot 23 cot 3 3 3
38. sin 405 sin 45 1 22 2
39. cos 585 cos 225 cos 45 1 22
40. sec 223 sec 43 sec 3 2
2 3 2 41. csc 83 csc 23 csc 3 3 3
2 3 42. sec 136 sec 6 3
2
43. cot 390 cot 30 cot 30 3 44. tan 234 tan 34 tan 4 1 5 12 13 13 45. r 52 122 169 13. Then sin 12 13 , cos 13 , tan 5 , csc 12 , sec 5 , and 5 . cot 12
46. If is in standard position, then the terminal point of on the unit circle is simply cos sin . Since the terminal point is given as 23 12 , sin 12 . 47. y 3x 1 0 y 3x 1, so the slope of the line is m 3. Then tan m 3 60 . 48. 4y 2x 1 0 y 12 x 14 . The slope of the line is m 12 . Then tan m 12 and r 12 22 5. So sin 1 , cos 2 , tan 12 , csc 5, sec 25 , and cot 2. 5 5 sin 1 cos2 49. Since sin is positive in quadrant II, sin 1 cos2 and we have tan . cos cos 1 1 50. sec (because cos 0 in quadrant III). cos 1 sin2 51. tan2
sin2 sin2 cos2 1 sin2
52. csc2 cos2
1
sin2
cos2
1 sin2 sin2
1
1
sin2 53. tan 37 , sec 43 . Then cos 34 and sin tan cos 47 , csc 4 4 7 7 , and cot 3 3 7 7 . 7 7 9 sin 41 41 9 40 9 40 54. sec 41 40 , csc 9 . Then sin 41 , cos 41 , tan cos 40 40 , and cot 9 . 41 55. sin 35 . Since cos 0, is in quadrant II. Thus, x 52 32 16 4 and so cos 45 , tan 34 , csc 53 , sec 54 , cot 43 . 5 56. sec 13 5 and tan 0. Then cos 13 , and must be in quadrant III sin 0. Therefore,
25 12 , csc 13 , tan sin 12 , and cot 5 . sin 1 cos2 1 169 13 12 5 12 cos 4 cos 45 2 since 57. tan 12 . sec2 1 tan2 1 14 54 cos2 5 5 sin 1 1 1 2 1 . Therefore, 2 sin 2 cos 2 cos 0 in quadrant II. But tan 5 5 cos 5 1 2 1 sin cos 5 . 5 5 5 1 1 1 sin 1 sin 1 sin 2 3 3. 58. sin 12 for in quadrant I. Then tan sec 2 cos cos cos 2 1 sin2 1 1 2
59. By the Pythagorean Theorem, sin2 cos2 1 for any angle .
3 5 10 5 5 60. cos 23 and 2 . Then 6 2 6 3 . So sin 2 sin 3 sin 3 2 .
61. sin1 23 3
62. tan1 33 6
CHAPTER 5
Review
485
63. Let u sin1 25 and so sin u 25 . Then from the triangle, 64. Let u cos1 83 then cos u 38 . From the triangle, we tan sin1 25 tan u 2 . have sin cos1 38 sin u 855 . 21
u
5
8
Ï21
u
2
65. Let tan1 x
tan x. Then from the x . triangle, we have sin tan1 x sin 1 x2 Ï1+x@ ¬
3
66. Let sin1 x. Then sin x. From the triangle, we 1 have sec sin1 x sec . 1 x2
x
1
x
¬
1
Ï1-x@
x x tan1 2 2 10 sin 30 532. 69. B 180 30 80 70 , and so by the Law of Sines, x sin 70 2 sin 45 146 70. x sin 105 67. cos
x x cos1 3 3
Ï55
68. tan
71. x 2 1002 2102 2 100 210 cos 40 21,926133 x 14807 20 sin 60 72. sin B 0247 B sin1 0247 1433 . Then C 180 60 1433 10567 , and so 70 70 sin 10567 x 7782. sin 60 73. x 2 22 82 2 2 8 cos 120 84 x 84 917 6 sin 3121 4 sin 110 0626 B 3879 . Then C 180 110 3879 3121 , and so x 33. 74. sin B 6 sin 110 sin 25 23 sin 25 sin 23 sin 25 sin sin1 541 or 75. By the Law of Sines, 23 12 12 12 180 541 1259 . sin sin 80 4 sin 80 4 sin 80 76. By the Law of Sines, sin sin1 520 . 4 5 5 5 77. By the Law of Cosines, 1202 1002 852 2 100 85 cos , so cos cos1 016618 804 .
1202 1002 852 016618. Thus, 2 100 85
sin 10 5 sin 10 sin A sin A 02894, so A sin1 02894 168 5 3 3 or A 180 168 1632 . Therefore, 180 10 168 1532 or 180 10 1632 68 .
78. We first use the Law of Sines to find A:
79. After 2 hours the ships have traveled distances d1 40 mi and d2 56 mi. The subtended angle is 180 32 42 106 . Let d be the distance between the two ships in miles. Then by the Law of Cosines, d 2 402 562 2 40 56 cos 106 5970855 d 773 miles.
486
CHAPTER 5 Trigonometric Functions: Right Triangle Approach
80. Let h represent the height of the building in feet, and x the horizontal distance from the building to point B. Then h h h tan 241 and tan 302 x h cot 302 . Substituting for x gives tan 241 x 600 x h cot 302 600 600 tan 241 h tan 241 h cot 302 600 h 1160 ft. 1 tan 241 cot 302 81. Let d be the distance, in miles, between the points A and B . Then by the Law of Cosines, d 2 322 562 2 32 56 cos 42 14966 d 39 mi.
120 sin 689 82. C 180 423 689 688 . Then b 12008 miles. Let d be the shortest distance, in miles, to sin 688 the shore. Then d b sin A 12008 sin 423 808 miles.
83. A 12 ab sin 12 8 14 sin 35 3212 568 abc 95. Thus, 84. By Heron’s Formula, A s s a s b s c, where s 2 2 A 95 95 5 95 6 95 8 1498.
CHAPTER 5 TEST 11 rad. 135 135 3 rad. 1. 330 330 180 6 180 4 180 234 2. 43 rad 43 180 240 . 13 rad 13 745
3. (a) The angular speed is
120 2 rad 240 rad/min 75398 rad/min. 1 min
(b) The linear speed is 120 2 16 3840 ft/min 1 12,0637 ft/min
4. (a) sin 405 sin 45 1 22 2
(b) tan 150 tan 30 1 33 3 (c) sec 53 sec 3 2 (d) csc 52 csc 2 1
137 mi/h
2 13 3 2 2 26 6 13 . 5. r 32 22 13. Then tan sin 3 39 13 3 13 b a a 24 sin . Also, cos b 24 cos . 6. sin 24 24 7. cos 13 and is in quadrant III, so r 3, x 1, and y 32 12 2 2. Then
1 23 43 2. 3 2 csc 1 4 2 2 2 2 tan 1 sin 1 5 13 13 1 5 , tan 5 . Then sec 8. sin 13 tan . 12 cos cos sin sin 12 5 12 9. sec2 1 tan2 tan sec2 1. Thus, tan sec2 1 since tan 0 in quadrant II.
tan cot csc tan
h h 6 tan 73 196 ft. 6 x x 11. (a) tan tan1 4 4 3 3 (b) cos cos1 x x 10. tan 73
Surveying 9 so tan u 9 . From the triangle, r 12. Let u tan1 40 40 40 9 1 cos tan 40 cos u 41 .
92 402 41. So
41 u
487
9
40
13. By the Law of Cosines, x 2 102 122 2 10 12 cos 48 8409 x 91.
230 sin 69 14. C 180 52 69 59 . Then by the Law of Sines, x 2505. sin 59 h x h h 50 tan 20 and tan 28 x h 50 tan 28 15. Let h be the height of the shorter altitude. Then tan 20 50 50 x 50 tan 28 h 50 tan 28 50 tan 20 84. 15 sin 108 16. Let A and X be the other angles in the triangle. Then sin A 0509 A 3063 . Then 28 28 sin 4137 X 180 108 3063 4137 , and so x 195. sin 108 17. By the Law of Cosines, 92 82 62 2 8 6 cos cos
82 62 92 01979, so cos1 01979 786 . 2 8 6
18. We find the length of the third side x using the Law of Cosines: x 2 52 72 2 5 7 cos 75 5588 x 7475. sin 75 5 sin 75 sin sin 06461, so sin1 06461 402 . Therefore, by the Law of Sines, 5 7475 7475 50 72 . A triangle 1 r r sin 1 102 sin 72 . Thus, the area of the 19. (a) A sector 12 r 2 12 102 72 180 180 2 2 153 m2 . shaded region is A shaded A sector A triangle 50 72 180 sin 72
(b) The shaded region is bounded by two pieces: one piece is part of the triangle, the other is part of the circle. The first part has length l 102 102 2 10 10 cos 72 10 2 2 cos 72 . The second has length 4. Thus, the perimeter of the shaded region is p l s 102 2 cos 72 4 243 m. s 10 72 180
20. (a) If is the angle opposite the longest side, then by the Law of Cosines cos
92 132 202 06410. Therefore, 2 9 20
cos1 06410 1299 .
(b) From part (a), 1299 , so the area of the triangle is A 12 9 13 sin 1299 449 units2 . Another way to find 9 13 20 abc 21. Thus, the area is to use Heron’s Formula: A s s a s b s c, where s 2 2 A 21 21 20 21 13 21 9 2016 449 units2 . 21. Label the figure as shown. Now 85 75 10 , so by the Law of Sines, 100 x sin 75 sin 10
x 100
sin 75 h . Now sin 85 sin 10 x
sin 75 h x sin 85 100 sin 85 554. sin 10
º
x 75¡ 100
h
85¡
FOCUS ON MODELING Surveying 1. Let x be the distance between the church and City Hall. To apply the Law of Sines to the triangle with vertices at City Hall, the church, and the first bridge, we first need the measure of the angle at the first bridge, which is 180 25 30 125 . x 086 086 sin 125 Then x 14089. So the distance between the church and City Hall is about sin 125 sin 30 sin 30 141 miles.
488
FOCUS ON MODELING
2. To find the distance z between the fire hall and the school, we use the distance found in the text between the bank and the cliff. To find z we first need to find the length of the edges labeled x and y.
bank 50¡
In the bank-second bridge-cliff triangle, the third angle is
1.55
155 sin 50 180 50 60 70 , so x 126.
sin 70 In the second bridge-school-cliff triangle, the third angle is
60¡
cliff
126 sin 45 109. sin 55 Finally, in the school-fire hall-cliff triangle, the third angle is
80¡
180 80 55 45 , so y
180 45 80 55 , so z
109 sin 80
131. sin 55 Thus, the fire hall and the school are about 131 miles apart.
fire hall
55¡
z
70¡
x
second bridge
45¡
80¡ y 45¡
55¡
school
3. First notice that D BC 180 20 95 65 and D AC 180 60 45 75 . AC 20 20 sin 45 AC 146 . From BC D we get From AC D we get sin 45 sin 75 sin 75 BC 20 20 sin 95 BC 220. By applying the Law of Cosines to ABC we get sin 95 sin 65 sin 65 AB2 AC2 BC2 2 AC BC cos 40 1462 2202 2146220cos 40 205, so AB 205 143 m. Therefore, the distance between A and B is approximately 143 m.
4. Let h represent the height in meters of the cliff, and d the horizontal distance to the cliff. The third horizontal 200 sin 516 h angle is 180 694 516 59 and so d 182857. Then tan 331 sin 59 d h d tan 331 182857 tan 331 1192 m.
5. (a) In ABC, B 180 , so C 180 180 . By the Law of Sines,
AB BC sin sin
d sin sin . sin sin d sin h h (b) From part (a) we know that BC . But sin . Therefore, BC BC sin sin d sin h d sin sin BC h . sin sin sin 800 sin 25 sin 29 d sin sin (c) h 2350 ft sin sin 4 BC AB
6. Let the surveyor be at point A, the first landmark (with angle of depression 42 ) be at point B, and the other landmark 2430 2430 2430 36316. Similarly, sin 39 be at point C. We want to find BC. Now sin 42 AB AB AC sin 42 2430 38613. Therefore, by the Law of Cosines, BC2 AB2 AC2 2 AB AC cos 68 AC sin 39 BC 363162 386132 2 36316 38613 cos 68 4194. Thus, the two landmarks are approximately 4194 ft apart.
Surveying
489
7. We start by labeling the edges and calculating the remaining angles, as shown in the first figure. Using the Law of Sines, a 150 150 sin 29 b 150 150 sin 91 we find the following: a 8397, b 17318, sin 29 sin 60 sin 60 sin 91 sin 60 sin 60 c 17318 17318 sin 32 d 17318 17318 sin 61 c 9190, e 15167, sin 32 sin 87 sin 87 sin 61 sin 87 sin 87 15167 15167 sin 41 f 15167 15167 sin 88 e e 12804, f 19504, sin 41 sin 51 sin 51 sin 88 sin 51 sin 51 19504 19504 sin 50 h 19504 19504 sin 38 g g 14950, and h 12015. Note that sin 50 sin 92 sin 92 sin 38 sin 92 sin 92 we used two decimal places throughout our calculations. Our results are shown (to one decimal place) in the second figure.
29¡ 150
c
61¡
87¡ 88¡
91¡
60¡ a
32¡ 41¡
128.0
51¡ 38¡
d
b
91.9
e
g
f 50¡ h
92¡
150
173.2
84
151.7
195.0
149.5
120.2
8. Answers will vary. Measurements from the Great Trigonometric Survey were used to calculate the height of Mount Everest to be exactly 29,000 ft, but in order to make it clear that the figure was considered accurate to within a foot, the height was published as 29,002 ft. The accepted figure today is 29,029 ft.
6
TRIGONOMETRIC FUNCTIONS: UNIT CIRCLE APPROACH
6.1
THE UNIT CIRCLE
1. (a) The unit circle is the circle centered at 0 0 with radius 1. (b) The equation of the unit circle is x 2 y 2 1.
(c) (i) Since 12 02 1, the point is P 1 0.
(ii) P 0 1
(iii) P 1 0
(iv) P 0 1
2. (a) If we mark off a distance t along the unit circle, starting at 1 0 and moving in a counterclockwise direction, we arrive at the terminal point determined by t. (b) The terminal points determined by 2 , , 2 , 2 are 0 1, 1 0, 0 1, and 1 0, respectively. 2 2 9 16 1, P 3 4 lies on the unit circle. 3. Since 35 45 25 25 5 5 2 7 2 576 49 1, P 24 7 lies on the unit circle. 25 4. Since 24 25 625 25 25 25
2 2 9 7 1, P 3 7 lies on the unit circle. 5. Since 34 47 16 16 4 4 2 2 24 1, P 5 2 6 lies on the unit circle. 6. Since 57 2 7 6 25 7 7 49 49
2 2 7. Since 35 23 59 49 1, P 35 23 lies on the unit circle. 2 2 11 5 lies on the unit circle. 25 1, P 8. Since 611 56 11 36 36 6 6 2 9 y 2 16 y 4 . Since P x y is in quadrant III, y is negative, so the point is 9. 35 y 2 1 y 2 1 25 25 5 P 35 45 . 7 2 1 x 2 1 49 x 2 576 x 24 . Since P is in quadrant IV, x is positive, so the point is 10. x 2 25 625 625 25 24 7 P 25 25 . 2 1 x 2 1 19 x 2 89 x 2 3 2 . Since P is in quadrant II, x is negative, so the point is 11. x 2 13 P 2 3 2 13 . 2 4 y 2 21 y 21 . Since P is in quadrant I, y is positive, so the point is 12. 25 y 2 1 y 2 1 25 25 5 21 2 P 5 5 . 2 4 x 2 45 x 3 5 . Since P x y is in quadrant IV, x is positive, so the point is 13. x 2 27 1 x 2 1 49 7 49 3 5 2 P 7 7 . 2 14. 23 y 2 1 y 2 1 49 y 2 59 y 35 . Since P is in quadrant II, y is positive, so the point is P 23 35 . 2 5 25 y 2 144 y 12 . Since its y-coordinate is negative, the point is P 5 12 . y 2 1 y 2 1 169 15. 13 169 13 13 13 491
492
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
2 9 x 2 16 x 4 . Since its x-coordinate is positive, the point is P 4 3 . 16. x 2 35 1 x 2 1 25 25 5 5 5
2 17. x 2 23 1 x 2 1 49 x 2 59 x 35 . Since its x-coordinate is negative, the point is P 35 23 . 2 5 x 2 20 x 2 5 . Since its x-coordinate is positive, the point is 1 x 2 1 25 18. x 2 55 25 5 2 5 5 P 5 5 .
2 19. 32 y 2 1 y 2 1 29 y 2 79 y 37 . Since P lies below the x-axis, its y-coordinate is negative, so the point is P 32 37 .
2 4 y 2 21 y 21 . Since P lies above the x-axis, its y-coordinate is positive, so 20. 25 y 2 1 y 2 1 25 25 5 the point is P 25 521 .
21.
22. t
Terminal Point
0
1 0 2 2 2 2
4
2 3 4
t
Terminal Point
t
Terminal Point
1 0 22 22
0
1 0 3 1 2 2 1 3 2 2
5 4 3 2 7 4
0 1
22 22 1 0
2
0 1
2 2 2 2
1 0
6
3
2 2 3 5 6
23. t 0, so t 4 corresponds to P x y 1 0. y
t=0 P(1, 0) t=4¹ x
3 25. t 2 , so t 2 corresponds to P x y 0 1. y ¹ Q(0, 1) t= 2
t
Terminal Point
1 0 23 12 12 23
7 6 4 3 3 2 5 3 11 6
0 1 12 23 23 12 1 0
1 0
2
24. t 0, so t 3 corresponds to P x y 1 0. y
P(_1, 0)
t=_3¹
Q(1, 0)
t=0
x
5 26. t 2 , so t 2 corresponds to P x y 0 1. y P(0, 1)
5¹
t= 2
¹
t= 2 x
3¹
t= 2
P(0, _1)
0 1
1 3 2 2 31 2 2
x
SECTION 6.1 The Unit Circle
27. t 6 corresponds to P x y y
31 . 2 2
493
7 28. t 6 and t 6 corresponds to P x y 23 12 . y
¹
t=_ 6
7¹ t= 6
x
(
)
P Ï3 , _21 2
x 1 P(_ Ï3 2 , _ 2)
5 29. t 4 and t 4 corresponds to P x y 22 22 .
4 30. t 3 and t 3 corresponds to P x y 12 23 .
y
y
4¹
t= 3 x
5¹
t= 4
P(_ Ï2 , _ Ï2 ) 2 2
x P (_ 21 , _ Ï3 2)
5 corresponds to P x y 1 3 . 32. t and t 3 3 2 2
7 31. t 6 and t 6 corresponds to P x y 23 12 .
y
y
P(_ 2 , 2 ) Ï3 1
x
x
5¹
t= 3
P ( 21 , _ Ï3 2 )
7¹
t=_ 6
7 33. t 4 and t 4 corresponds to P x y 22 22 .
4 34. t 3 and t 3 corresponds to P x y 12 23 .
y
7¹
t=_ 4
P( Ï2 , Ï2) 2 2
P (_ 21 , Ï3 2)
y
4¹
t=_ 3 x
x
494
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
3 35. t 4 and t 4 corresponds to P x y 22 22 . y
11 36. t 6 and t 6 corresponds to P x y 23 12 .
¹ 4
(
P _ Ï2 , _ Ï2 2 2
y
Q(Ï2 , Ï2 2 2 )
¹ 6
x
)
11¹
3¹ t=_ 4
37. (a) t 43 3 (b) t 2 53 3 7 (c) t 6 6 (d) t 35 036 39. (a) t 57 27
t= 6
38. (a) t 9 9 0 (b) t 54 4 (c) t 56 6
(d) t 4 086 40. (a) t 115 2 5
(b) t 79 29
(b) t 97 27
(c) t 3 0142
(c) t 2 6 0283
(d) t 2 5 1283
(d) t 7 2 0717
41. (a) t 2 116 6 3 1 (b) P 2 2 43. (a) t 43 3 3 1 (b) P 2 2 45. (a) t 23 3 3 1 (b) P 2 2
42. (a) t 23 3 3 1 (b) P 2 2
44. (a) t 2 53 3 3 1 (b) P 2 2
46. (a) t 76 6 3 1 (b) P 2 2
47. (a) t 134 3 4 2 2 (b) P 2 2
48. (a) t 136 2 6 3 1 (b) P 2 2
51. (a) t 4 113 3 3 1 (b) P 2 2 53. (a) t 163 5 3 3 1 (b) P 2 2
52. (a) t 316 5 6 3 1 (b) P 2 2 54. (a) t 10 414 4 2 2 (b) P 2 2 56. t 25
P 08 06
57. t 11 05 09
58. t 42
P 05 09
49. (a) t 7 416 6 3 1 (b) P 2 2
55. t 1 05 08
50. (a) t 174 4 4 2 2 (b) P 2 2
Q(Ï3 ,1 2 2) x
(
P Ï3 , _ 21 2
)
SECTION 6.2 Trigonometric Functions of Real Numbers
59. Let Q x y 35 45 be the terminal point determined by t. y
¹-t
¹+t
2¹+t
(b) Q( 35 , 45)
(c)
t
(d) x
_t
60. Let Q x y 34 47 be the terminal point determined by t. y ¹-t 4¹+t t-¹
(a)
495
t determines the point P x y 35 45 . t determines the point P x y 35 45 . t determines the point P x y 35 45 . 2 t determines the point P x y 35 45 .
Q( 34 , Ï7 4) t _t
x
(a) t determines the point P x y 34 47 . (b) 4 t determines the point P x y 34 47 . (c) t determines the point P x y 34 47 . (d) t determines the point P x y 34 47 .
61. The distances P Q and P R are equal because they both subtend arcs of length 3 . Since P x y is a point on the unit circle, x 2 y 2 1. Now d P Q x x2 y y2 2y and d R S x 02 y 12 x 2 y 2 2y 1 2 2y (using the fact that x 2 y 2 1). Setting these equal gives 2y 2 2y 4y 2 2 2y 4y 2 2y 2 0 2 2y 1 y 1 0. So y 1 or y 12 . Since 2 P is in quadrant I, y 12 is the only viable solution. Again using x 2 y 2 1 we have x 2 12 1 x 2 34 x 23 . Again, since P is in quadrant I the coordinates must be 23 12 . 62. P is the reflection of Q about the line y x. Since Q is the point Q
6.2
3 1 2 2
it follows that P is the point P 12 23 .
TRIGONOMETRIC FUNCTIONS OF REAL NUMBERS
1. If Px y is the terminal point on the unit circle determined by t, then sin t y, cos t x, and tan t yx. 2. If Px y is on the unit circle, then x 2 y 2 1. So for all t we have sin2 t cos2 t 1.
496
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
3.
4. t
sin t
cos t
t
sin t
cos t
t
sin t
cos t
0
0
1
0
0
1
0
1 2 3 2
3 2 1 2
12 23
1
23
1
0
7 6 4 3 3 2 5 3 11 6
1
0
23 12
1 2 3 2
2
0
1
4
(b) cos 54
0
22
0
(c) cos 74 1 13. (a) cos 3 2 (b) sec 3 2 3 (c) sin 3 2
17. (a) csc 76 2 23 (b) sec 6 3 (c) cot 56 3 (b) cos 14 1 (c) tan 15 0
22
1
3
1
22
2 2 3 5 6
3 2 1 2
0
0
22
2 2
0
1
3 3 22 22 2 2
21. (a) sin 13 0
6
1
(b) cos 176 23
9. (a) cos 34
2 2
2
(c) tan 76
2 2
2 3 4 5 4 3 2 7 4
5. (a) sin 76 12
2 2
6. (a) sin 53 23 (b) cos 113 12 (c) tan 53 3
10. (a) sin 34 22
12
23
1
7. (a) sin 114 22 2 (b) sin 4 2
(c) sin 54 22
11. (a) sin 73 23
(b) sin 54 22
(b) csc 73 2 3 3
(c) sin 74 22 14. (a) tan 4 1 (b) csc 4 2 (c) cot 4 1
(c) cot 73 33 3 15. (a) cos 6 2 2 3 (b) csc 3 3 (c) tan 6 33
18. (a) sec 34 2 (b) cos 23 12 (c) tan 76 33
19. (a) sin 43 23
(b) sec 116 2 3 3 3 (c) cot 3 3
24. t 2 sin t 1, cos t 0, csc t 1, cot t 0, tan t and sec t are undefined. 25. t sin t 0, cos t 1, tan t 0, sec t 1, csc t and cot t are undefined.
26. t 32 sin t 1, cos t 0, csc t 1, cot t 0, tan t and sec t are undefined. 4 2 2 9 16 1. So sin t 4 , cos t 3 , and tan t 5 4 . 27. 35 45 25 25 5 5 3 35
3 2 2 28. 12 23 14 34 1. So sin t 23 , cos t 12 , and tan t 21 3. 2
12
8. (a) cos 196 23 (b) cos 76 23 3 (c) cos 6 2 12. (a) csc 54 2 (b) sec 54 2 (c) tan 54 1 2 16. (a) sin 4 2 (b) sec 4 2 (c) cot 6 3
20. (a) csc 23 2 3 3 (b) sec 53 2 (c) cos 103 12
22. (a) sin 252 sin 2 24 1 (b) cos 252 cos 2 24 0 (c) cot 252 cot 2 24 0
23. t 0 sin t 0, cos t 1, tan t 0, sec t 1, csc t and cot t are undefined.
SECTION 6.2 Trigonometric Functions of Real Numbers
29.
30.
31.
32.
33.
34.
35.
36.
497
2 2 2 2 13 2 3 2 19 89 1. So sin t 2 3 2 , cos t 13 , and tan t 31 2 2. 3 2 2 256 2 6 2 6 1 1 24 1 5 25 25 1. So sin t 5 , cos t 5 , and tan t 1 2 6. 5 5 13 2 2 7 13 . 13 13 6 67 713 36 49 49 1. So sin t 7 , cos t 7 , and tan t 6 6 7 9 2 2 41 40 9 81 9 40 9 41 1600 41 1681 1681 1. So sin t 41 , cos t 41 , and tan t 40 40 . 41 12 2 2 5 25 144 1. So sin t 12 , cos t 5 , and tan t 13 12 . 12 13 13 169 169 13 13 5 5 13 2 5 2 2 5 5 20 1. So sin t 2 5 , cos t 5 , and tan t 5 2. 2 5 5 25 5 25 5 5 5 5 21 2 2 29 441 21 20 20 21 400 21 29 29 841 841 1. So sin t 29 , cos t 29 , and tan t 20 . 20 29 7 2 24 7 2 576 49 1. So sin t 7 , cos t 24 , and tan t 25 7 . 25 25 625 625 25 25 24 24 25
37. (a) 08
38. (a) 07
(b) 084147 41. (a) 10
(b) 069671 42. (a) 36
(b) 102964
(b) 360210
39. (a) 09
40. (a) 03
(b) 093204
(b) 028366
43. (a) 06
44. (a) 09
(b) 057482
(b) 088345
45. sin t cos t. Since sin t is positive in quadrant II and cos t is negative in quadrant II, their product is negative. 46. tan t sec t is negative in quadrant IV because tan t is negative and cos t is positive in quadrant IV. 47.
1 tan t sin t tan t sin t tan t tan t sin t tan2 t sin t. Since tan2 t is always positive and sin t is negative in cot t cot t quadrant III, the expression is negative in quadrant III.
48. cos t sec t is positive in any quadrant, since cos t sec t cos t 49. Quadrant II 53. sin t 1 cos2 t
sin t sin t cos t 1 sin2 t 57. sec t 1 tan2 t 55. tan t
59. tan t sec2 t 1 61. tan2 t
sin2 t sin2 t cos2 t 1 sin2 t
50. Quadrant III
1 1, provided cos t 0. cos t
51. Quadrant II 54. cos t 1 sin2 t 1 cos2 t 56. tan t cos t 58. csc t 1 cot2 t
52. Quadrant II
1 sec2 t 1 sec t sec2 t (sin t 0 and sec t 0 in quadrant IV)
2 60. sin t 1 cos t 1
62. sec2 t sin2 t
1 1 2t 1 cos 1 2 cos t cos2 t
498
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
63. sin t 45 and the terminal point of t is in quadrant IV, so the terminal point determined by t is P x 45 . Since P is 2 9 3 on the unit circle, x 2 45 1. Solving for x gives x 1 16 25 25 5 . Since the terminal point is in quadrant IV, x 35 . Thus the terminal point is P 35 45 . Thus, cos t 35 , tan t 43 , csc t 54 , sec t 53 , cot t 34 .
7 and the terminal point of t lies in quadrant III, so the terminal point determined by t is P 7 y . Since 64. cos t 25 25 7 2 y 2 1. Solving for y gives x 1 49 576 24 . Since the terminal P is on the unit circle, 25 625 625 25 24 7 24 24 25 point is in quadrant III, y 25 . Thus the terminal point is P 25 25 . Thus, sin t 25 , tan t 24 7 , csc t 24 , 7 sec t 25 7 , cot t 24 .
65. sec t 3 and the terminal point of t lies in quadrant IV. Thus, cos t 13 and the terminal point determined by t is P 13 y . 2 Since P is on the unit circle, 13 y 2 1. Solving for y gives y 1 19 89 2 3 2 . Since the terminal point is in quadrant IV, y 2 3 2 . Thus the terminal point is P 13 2 3 2 . Therefore, sin t 2 3 2 , cos t 13 , 3 3 2 , cot t 1 2. tan t 2 2, csc t 4 4 2 2
2 2
2 1 1 17 . 66. tan t 14 and the terminal point of t lies in quadrant III. Since sec2 t tan2 t 1 we have sec2 t 14 1 16 16 17 17 Thus sec t 17 16 4 . Since sec t 0 in quadrant III we have sec t 4 , so 4 17 1 4 1 1 1717 . . Since tan t cos t sin t we have sin t 14 4 cos t 17 17 sec t 17 17 417 Thus, the terminal point determined by t is P 4 1717 1717 . Therefore, sin t 1717 , cos t 4 1717 , csc t 17,
sec t 417 , cot t 4.
12 2 1 169 , so because 2 2 2 67. tan t 12 5 and sin t 0, so t is in quadrant II. Since sec t tan t 1 we have sec t 5 25 1 13 5 5 secant is negative in quadrant II, sec t 169 25 5 . Thus, cos t sec t 13 and we have P 13 y . Since 5 12 . Thus, the terminal point determined by t is P 5 12 , and so tan t cos t sin t we have sin t 12 5 13 13 13 13 5 13 13 5 sin t 12 13 , cos t 13 , csc t 12 , sec t 5 , cot t 12 .
68. csc t 5 and cos t 0, so t is in quadrant II. Thus, sin t 15 and the terminal point determined by t is P x 15 . Since 2 1 2 6 . Since the terminal point is in P is on the unit circle, x 2 15 1. Solving for x gives x 1 25 5 2 6 2 6 1 1 1 6, quadrant II, x 5 and the terminal point is P 5 5 . Thus, sin t 5 , cos t 2 5 6 , tan t 12 2 6 5 6 5 sec t 12 , cot t 2 6. 2 6
SECTION 6.2 Trigonometric Functions of Real Numbers
499
69. sin t 14 , sec t 0, so t is in quadrant III. So the terminal point determined by t is P x 14 . Since P is on the unit 2 1 15 15 . Since the terminal point is in quadrant III, circle, x 2 14 1. Solving for x gives x 1 16 16 4 15 . Thus, the terminal point determined by t is P 415 14 , and so cos t 415 , tan t 1 1515 , x 15 4 4 15 4 csc t 4, sec t 15 , cot t 15.
15
70. tan t 4 and the terminal point of t lies in quadrant II. Since sec2 t tan2 t 1 we have sec2 t 42 1 161 17. 1 17 Thus sec t 17. Since sec t 0, we have sec t 17and cos t 1 . Since tan t cos t sin t 17 sec t 17 we have sin t 4 1717 4 1717 . Thus, the terminal point determined by t is P 1717 4 1717 . Thus, sin t 4 1717 , cos t 1717 , csc t 417 , sec t 17, cot t 14 .
71. f x x2 sin x x 2 sin x f x, so f is odd.
72. f x x2 cos 2 x x 2 cos 2x f x, so f is even.
73. f x sin x cos x sin x cos x f x, so f is odd.
74. f x sin x cos x sin x cos x which is neither f x nor f x, so f is neither even nor odd. 75. f x x cos x x cos x f x, so f is even.
76. f x x sin3 x x [sin x]3 x sin x3 x sin3 x f x, so f is even.
77. f x x3 cos x x 3 cos x which is neither f x nor f x, so f is neither even nor odd. 78. f x cos sin x cos sin x cos sin x f x, so f is even. 79. t
0
025
y t 4 283
050 075 100 125 0
80. (a) B 6 80 7 sin 2 87 mmHG (b) B 105 80 7 sin 105 12 827 mmHG
283 4 283
(c) B 12 80 7 sin 80 mmHG
7 3 (d) B 20 80 7 sin 20 12 80 2 739 mmHG
81. (a) I 01 08e03 sin 1 0499 A
82.
(b) I 05 08e15 sin 5 0171 A
t
0
1
2
4
6
8
12
H t 175 1504 100 386 100 1503 588
83. Notice that if P t x y, then P t x y. Thus, (a) sin t y and sin t y. Therefore, sin t sin t.
(b) cos t x and cos t x. Therefore, cos t cos t. y y sin t sin t tan t. (c) tan t cos t x x cos t
84. To prove that AO B C DO, first note that O B O D 1 and O AB OC D 2 . Now C O D AO B C O D AO B AB O. Since we know two angles and one side to be 2 2 equal, the triangles are (SAA) congruent. Thus AB OC and O A C D, so if B has coordinates x y, then D has coordinates y x. Therefore, (a) sin t 2 x cos t (b) cos t 2 y, and sin t y. Therefore, cos t 2 sin t. x cos t x (c) tan t 2 y y sin t cot t
500
6.3
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
TRIGONOMETRIC GRAPHS
1. If a function f is periodic with period p, then f t p f t for every t. The trigonometric functions y sin x and cos x are periodic, with period 2 and amplitude 1. y
y
1
1
2¹
2¹ x
¹
_1
x
¹
_1
2. To obtain the graph of y 5 sin x, we start with the graph of y sin x, then shift it 5 units upward. To obtain the graph of y cos x, we start with the graph of y cos x, then reflect it in the x-axis. 3. The sine and cosine curves y a sin kx and y a cos kx, k 0, have amplitude a and period 2k. The sine curve y 3 sin 2x has amplitude 3 3 and period 22 . 4. The sine curve y a sin k x b has amplitude a, period 2k, and horizontal shift b. The sine curve 2 y 4 sin 3 x 6 has amplitude 4 4, period 3 , and horizontal shift 6 . 6. f x 2 cos x
5. f x 2 sin x
y 1
y
2 ¹
2¹
¹
2¹
x
¹
2¹
x
x _2
7. f x sin x
8. f x 2 cos x
y
y
1 ¹
2¹
x 1
SECTION 6.3 Trigonometric Graphs
9. f x 2 sin x
10. f x 1 cos x
y
y
¹
2¹
1
x
_1
11. g x 3 cos x
12. g x 2 sin x
y
2
¹
2¹
x
¹
2¹
x
¹
2¹
¹
2¹
y
2 ¹
2¹
x
13. g x 12 sin x
14. g x 23 cos x y
y
1
1 ¹
2¹
x
x
15. g x 3 3 cos x
16. g x 4 2 sin x
y
y
2
2 ¹
2¹
x
x
501
502
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
17. h x cos x
18. h x sin x
y
1
y
1 ¹
2¹
x
19. y cos 2x has amplitude 1 and period .
¹
2¹
x
20. y sin 2x has amplitude 1 and period .
y
y
1
1 ¹
2¹
x
21. y sin 3x has amplitude 1 and period 23 .
¹
2¹
x
22. y cos 4x has amplitude 1 and period 12 . y
y
1
1
¹
2¹
x
1
1 2
x
_1
23. y 2 cos 3x has amplitude 2 and period 23 . y 2
_1
24. y 3 sin 6x has amplitude 3 and period 3. y
1
x
3
¹ 6
_2 _3
x
SECTION 6.3 Trigonometric Graphs
25. y 10 sin 12 x has amplitude 10 and period 4.
26. y 5 cos 14 x has amplitude 5 and period 8.
y
y
10
5 2¹
4¹
x
27. y 13 cos 13 x has amplitude 13 and period 6. y
1 2
4¹
8¹
x
28. y 4 sin 2x has amplitude 4 and period . y
4 x
3¹
_
¹
2¹
x
1 2
29. y 2 sin 2x has amplitude 2 and period 1. y
2
30. y 3 sin x has amplitude 3 and period 2. y
2 1
2
x
31. y 1 12 cos x has amplitude 12 and period 2. y
1
2
x
32. y 2 cos 4x has amplitude 1 and period 12 . y 1
1
_1 1
2
x
2
x
503
504
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
33. y cos x 2 has amplitude 1, period 2, and horizontal shift 2.
34. y 2 sin x 3 has amplitude 2, period 2, and horizontal shift 3.
y
y
1
2 4¹ 3
¹
2¹
¹ 3
x
35. y 2 sin x 6 has amplitude 2, period 2, and horizontal shift 6.
7¹ 3
¹
2¹
x
36. y 3 cos x 4 has amplitude 3, period 2, and horizontal shift 4. y
y
3
2 ¹ 6
7¹ 6
¹
13¹ 6
2¹
_
¹ 4
3¹ 4
x
37. y 4 sin 2 x 2 has amplitude 4, period , and horizontal shift 2.
¹
y
y 1
¹
2¹
x
39. y 5 cos 3x 4 5 cos 3 x 12 has amplitude 5, . period 23 , and horizontal shift 12
4¹
x
period 3, and horizontal shift 4. y 2
x
15¹ 4
2 40. y 2 sin 23 x 6 2 sin 3 x 4 has amplitude 2,
y
3¹ 4
7¹ 4
¹
_4
5
5¹ 12
x
horizontal shift 4.
4
¹ 2
2¹
38. y sin 12 x 4 has amplitude 1, period 4, and
_2
¹ 12
7¹ 4
¹ 4
7¹ 4
13¹ 4
x
SECTION 6.3 Trigonometric Graphs
1 1 41. y 12 12 cos 2x 3 2 2 cos 2 x 6 has amplitude 12 , period , and horizontal shift 6. y
505
42. y 1 cos 3x 2 1 cos 3 x 6 has
amplitude 1, period 23 , and horizontal shift 6. y
1
2
1 2
1
¹ 6
2¹ 3
7¹ 6
x
43. y 3 cos x 12 has amplitude 3, period 2, and horizontal shift 12 .
¹
¹ 6
_6
x
¹ 2
44. y 3 2 sin 3 x 1 has amplitude 2, period 23 , and horizontal shift 1.
y
y 3
3 1
1 2
_2
x
3 2
1
2¹ -1 3
_1
x
45. y sin 3x sin 3 x 3 has amplitude 1, period 46. y cos 2 x cos x 2 has amplitude 1, period 2 , and horizontal shift 2, and horizontal shift 2. 3
3
y
y
1
1
¹
_3
¹ 3
x
¹ 2
3¹ 2
5¹ 2
47. (a) This function has amplitude a 4, period 2k 2, and horizontal shift b 0 as a sine curve. (b) y a sin k x b 4 sin x
48. (a) This curve has amplitude a 2, period 2k , and horizontal shift b 0 as a cosine curve. (b) y a cos k x b 2 cos 2x
49. (a) This curve has amplitude a 32 , period 2k 23 , and horizontal shift b 0 as a cosine curve. (b) y a cos k x b 32 cos 3x
50. (a) This curve has amplitude a 3, period 2k 4, and horizontal shift b 0 as a sine curve. (b) y 3 sin 12 x
51. (a) This curve has amplitude a 12 , period 2k , and horizontal shift b 3 as a cosine curve. 1 (b) y 2 cos 2 x 3
1 , period 2 , and horizontal shift b 3 as a cosine curve. 52. (a) This curve has amplitude a 10 k 4 1 cos 2 x 3 (b) y 10 4
x
506
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
53. (a) This curve has amplitude a 4, period 2k 32 , and horizontal shift b 12 as a sine curve. (b) y 4 sin 43 x 12 54. (a) This curve has amplitude a 5, period 2k 1, and horizontal shift b 14 as a sine curve. (b) y 5 sin 2 x 14 55. f x cos 100x, [01 01] by [15 15]
56. f x 3 sin 120x, [01 01] by [4 4]
1
-0.1
0.1
-0.1
0.1
-1
57. f x sin
x , [250 250] by [15 15] 40
58. f x cos
x , [0 500] by [1 1] 80 1
1
0 -200
200
200
400
-1 -1
59. y tan 25x, [02 02] by [3 3]
60. y csc 40x, [01 01] by [10 10] 10
2
-0.2
0.2
-0.1
0.1
-2 -10
61. y sin2 20x, [05 05] by [02 12]
62. y
tan 10x, [02 02] by [1 4] 4
1.0
2
0.5
-0.5
0.5
-0.2
0.2
SECTION 6.3 Trigonometric Graphs
63. f x x, g x sin x
64. f x sin x, g x sin 2x 6
g
f+g
2 -4
-2
2
f
4
-6
507
0 -2
2
4
f
g 1
6
-6
-4
-4
f+g
-2
0
2
4
6
-1
-6
-2
65. f x sin 3x, g x cos 12 x
_5
66. f x 05 sin 5x, g x cos 2x.
2
f+g
2
1
g
1
0 _1 _2
5
f+g f
_2
2 _1
f
g
_2
67. y x 2 sin x is a sine curve that lies between the graphs of y x 2 and y x 2 .
68. y x cos x is a cosine curve that lies between the graphs of y x and y x.
200
5
-10
-5
10
5 -5
-200
69. y
x sin 5x is a sine curve that lies between the graphs of y x and y x. 2
cos 2x is a cosine curve that lies between the 1 x2 1 1 and y . graphs of y 1 x2 1 x2
70. y
1 2 -2
4 -5
5 -1
508
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
71. y cos 3x cos 21x is a cosine curve that lies between the graphs of y cos 3x and y cos 3x.
72. y sin 2x sin 10x is a sine curve that lies between the graphs of y sin 2x and y sin 2x.
1
1
-0.5
0.5
-0.5
-1
0.5 -1
73. y sin x sin 2x. The period is 2, so we graph the
function over one period, . Maximum value 176 when x 094 2n, minimum value 176 when x 094 2n, n any integer.
74. y x 2 sin x, 0 x 2. Maximum value 697 when x 524, minimum value 068 when x 105.
5
2 0 2 -2
4
6
2 -2
75. y 2 sin x sin2 x. The period is 2, so we graph the
function over one period, . Maximum value 300 when x 157 2n, minimum value 100 when x 157 2n, n any integer.
cos x . The period is 2, so we graph the function 2 sin x over one period. Maximum value 058 when
76. y
x 576 2n (exact value x 116 2n); Minimum value 058 when x 367 2n (exact value x 76 2n) for any integer n.
2
1 -2
2 0
-2
2
4
6
-1
77. cos x 04, x [0 ]. The solution is x 116.
78. tan x 2, x [0 ]. The solution is x 111.
1 2 0 1
2
3 0
-1
0
1
2
3
SECTION 6.3 Trigonometric Graphs
79. csc x 3, x [0 ]. The solutions are x 034, 280.
509
80. cos x x, x [0 ]. The solution is x 074.
4
2
2 0 0
1 0
81. f x
1
2
2
3
3
1 cos x x
(a) Since f x is odd.
1 cos x 1 cos x f x, the function x x
(c) 1
(b) The function is undefined at x 0, so the x-intercepts occur when
-20
1 cos x 0, x 0 cos x 1, x 0 x 2, 4, 6,
20 -1
(d) As x , f x 0. (e) As x 0, f x 0.
82. f x
sin 4x 2x
(a) f x
sin 4x sin 4x f x, so the function is even. 2 x 2x
(b) The x-intercepts occur when sin 4x 0, x 0 4x n x 14 n, n any nonzero integer.
(c) 2
1
(d) As x , f x 0.
-2
(e) As x 0, f x 2.
2
2 20 seconds. 10 (b) Since h 0 3 and h 10 3, the wave height is 3 3 6 feet.
83. (a) The period of the wave is
84. (a) The period of the vibration is (b) Since each vibration takes
1 2 second. 880 440
1 of a second, there are 440 vibrations 440
(c)
v 1
per second. 0
_1
0.002 0.004 0.006 0.008
t
510
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
85. (a) The period of p is
1 2 minute. 160 80
p
(c)
160
(b) Since each period represents a heart beat, there are 80 heart beats per
140
minute.
120 115 100 90 80
(d) The maximum (or systolic) is 115 25 140 and the minimum (or
diastolic) is 115 25 90. The read would be 14090 which higher than normal.
0
86. (a) The period of R Leonis is
2 312 days. 156
2 t (s)
1
b
(c)
10 8
(b) The maximum brightness is 79 21 10; the minimum brightness
6
is 79 21 58.
4 2 0
87. (a) y sin
x . This graph looks like a sine function
which has been stretched horizontally (stretched
more for larger values of x). It is defined only for x 0, so it is neither even nor odd.
200
400
600
t
(b) y sin x 2 . This graph looks like a graph of sin x which
has been shrunk for x 1 (shrunk more for larger values of x) and stretched for x 1. It is an even function, whereas
sin x is odd.
1
1
0 200
400
-5
-1
5 -1
88. (a) This function is periodic with period 2. (b) This function is not periodic. f 25 f 45.
(c) This function is not periodic; its graph on the interval [2 0] is different from its graph on the interval [0 2].
(d) This function is periodic with period 3. 89. (a) The graph of y sin x is shown in the viewing rectangle [628 628] by [05 15]. This
(b) The graph of y sin x is shown in the viewing rectangle
[10 10] by [15 15]. The function is not periodic. Note that while sin x 2 sin x for many values of x, it is false for x 2 0. For example sin 2 sin 2 1 3 while sin 2 2 sin 2 1.
function is periodic with period.
1
1
-5
5 -10
10 -1
SECTION 6.4 More Trigonometric Graphs
(c) The graph of y 2cos x is shown in the viewing rectangle [10 10] by [1 3]. This function is periodic with period 2.
511
(d) The graph of y x [[x]] is shown in the viewing rectangle [75 75] by [05 15]. This function is periodic with period1. Be sure to turn off “connected” mode when graphing functions with gaps in their graph.
2 1 -10
10 -5
5
90. From the graph we see that the amplitude is 5 and the period is 2. In Exercises 25 and 26 in Section 7.2, we will prove the reduction formulas sin x sin x and cos x cos x. So starting with y 5 sin x we may also represent the curve by y 5 sin x b where b 2n, where n is an integer; y 5 sin x b where b 2n 1 , where n is an integer. Then starting with y 5 cos x 2 we may also represent the curve by y 5 cos x b where b 2 2n, where n is an integer; y 5 cos x b where b 2 2n 1 , where n is an integer.
6.4
MORE TRIGONOMETRIC GRAPHS
1. The trigonometric function y tan x has period and asymptotes x 2 n, n an integer.
2. The trigonometric function y csc x has period 2 and asymptotes x n, n an integer. y
y 10
5
¹/2 x
_¹/2
_10
_¹
0
¹ x
_5
3 3. f x tan x 4 corresponds to Graph II. f is undefined at x 4 and x 4 , and Graph II has the shape of a graph of a tangent function. 3 4. f x sec 2x corresponds to Graph III. f is undefined at x 4 and x 4 , and Graph III has the shape of a graph of a secant function.
5. f x cot 4x corresponds to Graph VI.
6. f x tan x corresponds to Graph I.
7. f x 2 sec x corresponds to Graph IV.
8. f x 1 csc x corresponds to Graph V.
512
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
9. y 4 tan x has period .
_2¹
10. y 3 tan x has period .
y 10
_2¹
0
_¹
_¹
0
x
¹
_10
¹
x
¹
x
_10
11. y 32 tan x has period .
12. y 34 tan x has period . y 5
_2¹
y 10
y 4 2
_¹
0
x
¹
_2¹
0
_¹
_2 _4
_5
13. y 2 cot x has period .
14. y 2 cot x has period .
y
y 10
4
5 _2¹
0
_¹
x
¹
_¹
x
¹
_5 _10
15. y 2 csc x has period 2.
16. y 12 csc x has period 2.
y 5
_¹
y 1
¹
x
_¹
¹
x
SECTION 6.4 More Trigonometric Graphs
17. y 3 sec x has period 2.
18. y 3 sec x has period 2.
y
y
5
5
_¹
¹
19. y tan 3x has period 3. y
y
4
4 2 ¹
0
_¹/2
x
¹/2
_¹
_2
_4
_4
0
1
2
3 x
_1
0
_1/2
_5
_20
_10
23. y 2 cot 3x has period 13 .
24. y 3 cot 2x has period 12 .
_1/3
y 10
y 10
5
5
0
1/2
1 x
1/2
1 x
5
_10
_2/3
x
y 10
10 _1
¹ ¹/2
22. y 3 tan 4x has period 14 .
y 20
_2
0
_¹/2
_2
21. y 5 tan x has period 1.
_3
¹
20. y tan 4x has period 4.
2 _¹
x
_¹
x
1/3
2/3
1 x
_1
_1/2
0
_5
_5
_10
_10
513
514
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
25. y tan 4 x has period 4.
26. y cot 2 x has period 2.
4
2
y
y
2
2
_2
2
x
_1
1
28. y 2 tan 2 x has period 2.
27. y 2 tan 3x has period 13 .
2
y
y
4 _0.5
4 0.5
x
29. y csc 4x has period 24 2.
_1
x
¹
x
y
2
10
_¹ 2
¹ 2
_¹
x
2 . 2
32. y 12 sec 4x has period 12 . y 3
y
2
2 _¹
1
30. y 5 csc 3x has period 23 .
y
31. y sec 2x has period
x
1
¹
x
_1
_1/2
0 1/8 _1 _2 _3
1/2
1 x
SECTION 6.4 More Trigonometric Graphs
2 4 33. y 5 csc 32 x has period 3 . 3
34. y 5 sec 2x has period 22 1. y
2
y
10
10 _1
1
_0.5
x
35. y tan x 4 has period .
0.5
36. y tan x 4 has period .
y 4
y 4
2 _2¹
_¹
2
0 ¹/4
2¹ x
¹
_2
_2¹
_¹
38. y 2 cot x 3 has period . 4
0
y
2
2
_2¹
¹ 3¹/4
_¹
2¹ x
0 ¹/3
_2
_2
_4
_4
39. y csc x 4 has period 2.
¹
2¹
40. y sec x 4 has period 2.
y 4
y
2 _¹
2
0 ¹/4
_2
2¹ x
¹
_¹
¹
x
¹
x
_4
41. y 12 sec x 6 has period 2.
42. y 3 csc x 2 has period 2. y
y
6
1 _¹
x
_4
y 4
_2¹
2¹
_2
37. y cot x 4 has period .
_¹
¹ 3¹/4
0
_4
_2¹
x
¹
x
_¹
x
515
516
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
43. y tan 2 x 3 has period 2 .
44. y cot 2x 4 has period 2 . y 4
y 4
2
2 _¹/2
0¹/12
_2
¹/2 x
_¹
_¹/2
46. y 4 tan 4x 2 has period 4. y 10
y 10
5
5 0 ¹/6 ¹/2
¹ x
0 ¹/8
_¹/2
_5
_5
_10
_10
47. y cot 2x 2 cot 2 x 4 has period 2 . y
_¹
y
1 ¹ x
_0.5
1 49. y 2 csc x 3 2 csc x 3 has period y
0.5
y 10
_ 16¹ 3
1
x
x
1 2 50. y 3 sec 14 x 6 3 sec 4 x 3 has period 8. 5
4 1
¹/2 x
48. y 12 tan x 12 tan x 1 has period 1.
2
2 2.
x
_4
45. y 5 cot 3x 2 has period 3 .
_¹/2
¹
_2
_4
_¹
¹/2
0 ¹/8
_ 16¹ 3
_ 4¹ 3
_5
_10
0
8¹ 3
20¹ 3
x
SECTION 6.4 More Trigonometric Graphs
51. y sec 2 x 4 has period .
52. y csc 2 x 2 has period .
y 5
_¹
_¹/2
0
y 5
¹ x
¹/2
0
_¹
_5
x
¹
_5
2 53. y 5 sec 3x 2 5 sec 3 x 6 has period 3 . y
54. y 12 sec 2x 12 sec 2 x 12 has period 2 1. 2
y
10 _¹/2
¹/2
1
x
_0.5
2 55. y tan 23 x 6 tan 3 x 4 has period 23 32 .
0.5
x
56. y tan 12 x 4 has period 1 2. 2
y
y
2 2
5¹
_¹
¹
x
57. y 3 sec x 12 has period 2 2. y
x
2 58. y sec 3x 2 sec 3 x 6 has period 3 . y
2
6 1
3¹ 4
_4
1
x
_¹/2
¹/2
x
517
518
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
59. y 2 tan 2x 3 2 tan 2 x 6 has period 2 . y
60. y 2 cot 3x 3 2 cot 3 x 1 has period 13 . y 4
4 _¹/3
¹/6
2 2¹/3 x
0
_1
1 x
_2 _4
61. (a) d t 3 tan t, so d 015 153,
d
(b)
d 025 300, and d 045 1894.
10
(c) d as t 12 .
0
t , so S 2 1039, S 6 0, 62. (a) S t 6 cot 12 S 8 346, and S 1175 9154.
(b)
0.1
0.2
0.3
0.4
0.5 t
S 10
(c) From the graph, it appears that S t 6 at
approximately t 3 and t 9, corresponding to 9 A . M . and 3 P. M .
(d) As t 12 , the sun approaches the horizon and the
0
10
t
man’s shadow grows longer and longer.
63. (a) If f is periodic with period p, then by the definition of a period, f x p f x for all x in the domain of f . 1 1 1 for all f x 0. Thus, is also periodic with period p. Therefore, f x p f x f 1 also has period 2. Similarly, since cos x has (b) Since sin x has period 2, it follows from part (a) that csc x sin x 1 also has period 2. period 2, we conclude sec x cos x 64. If f and g are periodic with period p, then f x p f x for all x and g x p g x for all x. Thus f x f x p for all x [unless g x is undefined, in which case both are undefined.] But consider f x sin x g x p g x f x sin x (period 2) and g x cos x (period 2). Their quotient is tan x, whose period is . g x cos x 65. The graph of y cot x is the same as the graph of y tan x shifted 2 units to the right, and the graph of y csc x is the same as the graph of y sec x shifted 2 units to the right.
SECTION 6.5 Inverse Trigonometric Functions and Their Graphs
6.5
INVERSE TRIGONOMETRIC FUNCTIONS AND THEIR GRAPHS
1. (a) To define the inverse sine function we restrict the domain of sine to the interval 2 2 . On this interval the sine function is one-to-one and its inverse function sin1 is defined by sin1 x y sin y x. For example, 1 sin1 21 6 because sin 6 2 . (b) To define the inverse cosine function we restrict the domain of cosine to the interval [0 ]. On this interval the
cosine function is one-to-one and its inverse function cos1 is defined by cos1 x y cos y x. For example,
1 cos1 12 3 because cos 3 2 . 1 sin ; 2. The cancellation property sin1 sin x x is valid for x in the interval 2 2 . Therefore, (i) sin 3 3 1 sin . (ii) sin1 sin 103 103 because 103 ; and (iii) sin 2 4 4 3. (a) sin1 1 2 because sin 2 1 and 2 lies in 2 2 . 3 (b) sin1 23 3 because sin 3 2 and 3 lies in 2 2 .
(c) sin1 2 is undefined because there is no real number x such that sin x 2.
4. (a) sin1 1 2
(b) sin1 22 4
5. (a) cos1 1 6. (a) cos1 22 4
(b) cos1 21 3
7. (a) tan1 1 4 8. (a) tan1 0 0 9. (a) cos1 12 23
10. (a) cos1 0 2
(b) cos1 1 0 3 3 1 3 (b) tan 3 (b) sin1 22 4 (b) tan1
(b) sin1 0 0
11. sin1 23 072973 13. cos1 37 201371
(c) sin1 2 is undefined. (c) cos1 23 56 (c) cos1 22 34
(c) tan1 33 6 3 1 3 (c) tan 6
(c) tan1 1 4 (c) sin1 12 6
12. sin1 89 109491 14. cos1 49 111024
15. cos1 092761 275876
16. sin1 013844 013889
17. tan1 10 147113
18. tan1 26 153235
19. tan1 123456 088998
20. cos1 123456 is undefined because 123456 1.
21. sin1 025713 026005 23. sin sin1 14 14 25. tan tan1 5 5 27. sin sin1 32 is undefined because 32 1. 29. cos cos1 15 15 4 31. sin1 sin 4 33. sin1 sin 34 4
22. tan1 025713 025168 24. cos cos1 32 23 26. sin sin1 5 is undefined because 5 1. 28. tan tan1 32 32 30. sin sin1 34 34 4 32. cos1 cos 4 34 34. cos1 cos 34
519
520
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
56 35. cos1 cos 56
36. sin1 sin 56 6
56 37. cos1 cos 76
38. sin1 sin 76 6
4 39. tan1 tan 4
40. tan1 tan 3 3
2 41. tan1 tan 23 3 because tan 3 tan 3 and 3 lies in 2 2 . 42. sin1 sin 114 4 3 43. tan sin1 12 tan 6 3
44. cos sin1 0 cos 0 1
1 45. cos sin1 23 cos 3 2
46. tan sin1 22 tan 4 1
2 47. sin tan1 1 sin 4 2
3 48. sin tan1 3 sin 3 2
From the graph of y sin1 x cos1 x, it appears that y 157. We suspect
49. (a)
that the actual value is 2. 2
-1
(b) To show that sin1 x cos1 x 2 , start with the identity sin a 2 cos a and take arcsin of both sides to obtain
1 cos a. Now let a cos1 x. Then a 2 sin 1 cos cos1 x cos1 x sin1 x sin1 x, so 2 sin
1
sin1 x cos1 x 2.
From the graph of y tan1 x tan1 1x , it appears that y 157 for x 0
50. (a) 2
-2
and y 157 for x 0. We suspect that the actual values are 2.
2
-2
(b) To show that tan1 x tan1 x1 2 for x 0, we start with the identity a and take arccot of both sides to obtain tan a cot 2 cot1 tan a a. Now use the identity cot1 x tan1 x1 to write 2 1 1 , we have tan1 tan1 a x 2 a. Substituting a tan 1 tan1 1 tan1 1 tan1 1 tan1 1 x x 2 1x 2 tan tan
1x
. For the case x 0, simply note that tan1 x tan1 x, so for positive x, tan1 x tan1 x1 2 1 tan1 x tan1 x1 tan1 x tan1 x 2.
SECTION 6.6 Modeling Harmonic Motion
521
51. The domain of f x sin sin1 x is the same as that of sin1 x, [1 1], and the graph of f is the same as that of y x on [1 1].
The domain of g x sin1 sin x is the same as that of sin x, , because for all x, the value of sin x lies within
the domain of sin1 x. g x sin1 sin x x for 2 x 2 . Because the graph of y sin x is symmetric about the 3 line x 2 , we can obtain the part of the graph of g for 2 x 2 by reflecting the graph of y x about this vertical line. The graph of g is periodic with period 2. y
y
¹/2
1
y=g(x)
y=f(x) _1
1
x
_3¹/2
_¹
_¹/2
_1
6.6
¹/2
¹
3¹/2 x
_¹/2
MODELING HARMONIC MOTION
1. (a) Because y 0 at time t 0, y a sin t is an appropriate model.
(b) Because y a at time t 0, y a cos t is an appropriate model.
2. (a) Because y 0 at time t 0, y kect sin t is an appropriate model.
(b) Because y a at time t 0, y kect cos t is an appropriate model.
2 , and the phase 3. (a) For an object in harmonic motion modeled by y A sin kt b the amplitude is A, the period is k b is b. To find the horizontal shift, we factor k to get y A sin k t . From this form of the equation we see that the k b horizontal shift is . k (b) For an object in harmonic motion modeled by y 5 sin 4t the amplitude is 5, the period is 2 , the phase is , and the horizontal shift is 4.
4. Objects A and B are in harmonic motion modeled by y 3 sin 2t and y 3 sin 2t 2 . The phase of A is and the phase of B is 2 . The phase difference is 2 , so the objects are moving out of phase. 6. y 3 cos 12 t
5. y 2 sin 3t 1 (a) Amplitude 2, period 23 , frequency period 23 .
(b)
2 4, frequency (a) Amplitude 3, period 12
y
1 1 period 4 .
2
(b) ¹ 3
2¹ 3
y 3
x 2¹
4¹ x
522
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
7. y cos 03t
8. y 24 sin 36t
20 , frequency 3 . (a) Amplitude 1, period 203 3 20 y
(b)
5 , frequency 9 . (a) Amplitude 24, period 236 9 5 y
(b)
2.4
1
5¹ 18
5¹ 9
x
20¹ x 3
10¹ 3
_1
3t 025 cos 9. y 025 cos 15t 3 2 3 025 cos 32 t 29
2 4 , frequency 3 . (a) Amplitude 025, period 32 3 4
(b)
10. y 32 sin 02t 14 32 sin 02 t 7 10, frequency 1 . (a) Amplitude 32 , period 202 10 y
(b)
1.5
y 1 -7 2¹ 9
8¹ 9
x
14¹ 9
5¹-7
x
10¹-7
_1.5
_1
11. y 5 cos 23 t 34 5 cos 23 t 98
12. y 16 sin t 18
2 3, frequency 1 . (a) Amplitude 5, period 23 3 y
(b)
(a) Amplitude 16, period 2, frequency 21 . (b)
y 1.6
5
¹
3¹
1.8+2 9
_8
3¹ 9 2 _8
3¹ _ 9 8
x
1.8
1.8+2 1.8+¹
13. The amplitude is a 10 cm, the period is 2k 3 s, and f 0 0, so f t 10 sin 23 t. 14. The amplitude is 24 ft, the period is 2k 2 min, and f 0 0, so f t 24 sin t. 5 Hz, and f 0 0, so f t 6 sin 10t. 15. The amplitude is 6 in., the frequency is 2k
16. The amplitude is 12 m, the frequency is 2k 05 Hz, and f 0 0, so f t 12 sin t. 17. The amplitude is 60 ft, the period is 2k 05 min, and f 0 60, so f t 60 cos 4t. 18. The amplitude is 35 cm, the period is 2k 8 s, and f 0 35, so f t 35 cos 4 t. 19. The amplitude is 24 m, the frequency is 2k 750 Hz, and f 0 24, so f t 24 cos 1500t. 20. The amplitude is 625 in., the frequency is 2k 60 Hz, and f 0 625, so f t 625 cos 120t.
x
1.8+2¹
SECTION 6.6 Modeling Harmonic Motion
21. (a) k 2, c 15, and f 3 6, so we have
22. (a) k 15, c 025, and f 06 12, so we
y 2e15t cos 6t. (b)
have y 15e025t cos 12t.
y 2
(b)
23. (a) k 100, c 005, and p 4 2 , so we have y 100e005t cos 2 t. y 100
24. (a) k 075, c 3, and p 3 23 , so we have (b)
y 0.75
x
25. (a) k 7, c 10, and p 6 12, so we have
26. (a) k 1, c 1, and p 1 2, so we have y et sin 2t. y
(b)
y
x
¹/5
y 7e10t sin 12t.
0.5
1
27. (a) k 03, c 02, and f 20 40, so we
x
1
x
¹/30
28. (a) k 12, c 001, and f 8 16, so we
have y 03e02t sin 40t. (b)
x
5
y 075e3t cos 23 t.
4
(b)
y 15
x
1
(b)
have y 12e001t sin 16t.
y
(b)
y 10
0.2 0.2
523
0.4
0.6
0.8
x
29. y 5 sin 2t 2 has amplitude 5, period , phase 2 , and horizontal shift 4 . 30. y 10 sin t 3 has amplitude 10, period 2, phase 3 , and horizontal shift 3 .
31. y 100 sin 5t has amplitude 100, period 25 , phase , and horizontal shift 5. 2 32. y 50 sin 12 t 5 has amplitude 50, period 4, phase 5 , and horizontal shift 5 .
1
2
x
524
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
33. y 20 sin 2 t 4 has amplitude 20, period , phase 2 , and horizontal shift 4 . has amplitude 8, period , phase , and horizontal shift . 34. y 8 sin 4 t 12 2 3 12 5 35. y1 10 sin 3t ; y 10 sin 3t 2 2 2 5 (a) y1 has phase 2 and y2 has phase 2 .
y 10
(d)
5 (b) The phase difference is 2 2 2.
yÁ, yª
5
(c) Because the phase difference is a multiple of 2, the curves are in
_¹
phase.
0
¹ x
_5 _10
36. y1 15 sin 2t 3 ; y2 15 sin 2t 6 (a) y1 has phase 3 and y2 has phase 6 .
y
(d)
(b) The phase difference is 3 6 6.
10 _¹
(c) Because the phase difference is not a multiple of 2, the curves are out of phase.
yª
yÁ
0
¹ x
_10
80 sin 5t ; y 80 sin 5t 37. y1 80 sin 5 t 10 2 2 3 (a) y1 has phase 2 and y2 has phase 3 .
y 80
(d)
(b) The phase difference is 2 3 6.
(c) Because the phase difference is not a multiple of 2, the curves are
_¹
out of phase.
yª
0
yÁ
¹ x
_80
3 20 sin 2t 3 38. y1 20 sin 2 t 2 20 sin 2t ; y2 20 sin 2 t 2 (a) y1 has phase and y2 has phase 3.
y 20
(d)
(b) The phase difference is 3 2. (c) Because the phase difference is a multiple of 2, the curves are in phase.
yÁ, yª _¹
0
_20
¹ x
SECTION 6.6 Modeling Harmonic Motion
40. y a sin 2 915 107 t . The period is
39. y 02 cos 20t 8 (a) The frequency is 20 2 10 cycles/min.
1 2 107 109 108 and the 915 2 915 107 2 915 107 915 107 . frequency is 2
y
(b)
525
8.4 8.2 8 7.8
0
0.2
0.4
t
(c) Since y 02 cos 20t 8 02 1 8 82 and when t 0, y 82, the maximum displacement is 82 m.
41. p t 115 25 sin 160t
1 2 1 00125, frequency 80. (a) Amplitude 25, period 160 80 period
(b)
(c) The period decreases and the frequency increases.
P 140 120 100 80 0
0.01
0.02
t
42. (a) When t 0, , 2, we have cos 2t 1 so y is maximized at y 8900.
(b) The length of time between successive periods of maximum population is the length of a period which is 2 314 years. 2
2 2 5 43. The graph resembles a sine wave with an amplitude of 5, a period of 25 , and no phase shift. Therefore, a 5, 5, and a formula is d t 5 sin 5t. 44. From the graph we see that the amplitude is 6 feet and the period is 12 hours. Also, the sine curve is shifted 6 hours (exactly half its period) to the right, which is equivalent to reflecting the curve about the t-axis. Thus, the equation is t 6 sin t. y 6 sin 212 6
1 cycle/hour 1 12 45. a 21, f 12 6 . So, 2 y 21 sin 6 t (assuming the tide is at mean level and rising when t 0).
y
20
0
2
4
6
8
10
12 t
_20
2 1 2. So y 2 cos 2t. 47. Since the mass travels from its highest point (compressed spring) to its lowest point in 12 s, it completes half a period in 12 s.
46. a 2,
So, 12 one period 12 s
1 2 1 2 2
2. Also, a 5. So y 5 cos 2t.
526
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
48. (a) m 10 g, k 3, a 5 cm. Then f t 5 cos
3 10 t .
(b) The function f t a cos kmt describes an object oscillating in simple harmonic motion, so by comparing it with the general equation y a cos t we see that km. This means the frequency is 1 km k . f 2 2 2 m k (c) If the mass is increased, then the denominator of increases, so overall the frequency decreases. If the frequency m has decreased, then by definition the oscillations are slower. k increases. Thus, overall the frequency increases. If (d) If a stiffer spring is used, k is larger and so the numerator of m the frequency has increased, then by definition the oscillations are faster. 49. Since the Ferris wheel has a radius of 10 m and the bottom of the wheel is 1 m above the ground, the minimum height is 2 , and so y 11 10 sin t , where t 1 m and the maximum height is 21 m. Then a 10 and 20 s 10 10 is in seconds. 50. Let f t represent the measure of the angle at time t. The amplitude of this motion is 10. Since the period is 2s, we have 2 2 . Thus f t 10 sin t. 2 51. a 02, 10 . Then y 38 02 sin 5t . 5 1 2 2 . So R t 20 15 sin t , where R is in millions of miles and t is in days. 52. a 15, 2 54 54 54 53. The amplitude is 12 100 80 10 mmHG, the period is 24 hours, and the phase shift is 8 hours, so f t 10 sin 12 t 8 90. 54. E 0 310, frequency is 100
100 2
200. Then, E t 310 cos 200t. The maximum voltage
310 produced occurs when cos 2t 1, and hence is E max 310 V. The rms voltage is 219 V. 2 55. (a) The maximum voltage is the amplitude, that is, Vmax a 45 V.
(b) From the graph we see that 4 cycles are completed every 01 seconds, or equivalently, 40 cycles are completed every second, so f 40. f 40. (c) The number of revolutions per second of the armature is the frequency, that is, 2 (d) a 45, f 40 80. Then V t 45 cos 80t. 2 1130 56. (a) As the car approaches, the perceived frequency is f 500 1130110 5539 Hz. As it moves away, the perceived 1130 frequency is f 500 1130110 4556 Hz. 1130 1130 (b) The frequency is 500 1130110 , so 1000 1130110 11078 (approaching) or 9113 (receding). 2 Thus, models are y A sin 11078t and A sin 9113t. 1 . Since f 0 0, f t e09t sin t. 57. k 1, c 09, and 2 2 58. k 6, c 28, and 2 4. 2 (a) Since f 0 6, f t 6e28t cos 4t.
CHAPTER 6 1 (b) The amplitude of the vibration is 05 when 6e28t 05 e28t 12 1 ln 12 089 s. t 12 ln 12 28
59.
kect 4 ectct3 4 kect3
60. (a)
3 kec0 5 06 kec2
e2c 5
e3c 4
2c ln 5
3c ln 4
1 28t ln 12
Review
527
c 13 ln 4 046.
c 12 ln 5 080.
(b) f t 3e08t cos t. If the frequency is 165 cycles per second, then
165 2
330. Thus,
f t 3e08t cos 330t. 61. (a) For fan A, the amplitude is 1, the frequency is 100, and the phase is 0, so
100 200 and an equation is 2
y sin 200t.
For fan B, the amplitude is 1, the frequency is 100, and the phase is 34 , so again 200 and an equation is y sin 200t 34 .
(b) The phase difference is 34 , so the fans are out of phase. If fan A were rotated 34 counterclockwise, the phase difference would become 0 and the fans would be in phase. 62. (a) E I 50 sin 120t, so the voltage phase is 0. E I I 50 sin 120t 54 , so the voltage phase is 54 .
(b) The phase difference is 54 , so the generators are out of phase. If the armature in the second generator were rotated 34 , then the phase difference would become 54 34 2, and the generators would produce voltage in phase.
1 63. From left to right: at t 6 , sin t 2 and the values are increasing; at t 2 , sin t 1 and the values reach a maximum;
at t 56 , sin t 12 and the values are decreasing; at t , sin t 0 and the values are decreasing; at t 76 , sin t 12 and the values are decreasing; at t 32 , sin t 1 and the values reach a minimum, and at t 116 , sin t 12 and the values are increasing.
64. From left to right: new moon, waxing crescent moon, waxing gibbous moon, full moon, waning gibbous moon, third quarter moon, waning crescent moon, new moon. Tides and werewolf sightings are in phase with the lunar cycle; paying rent and cellphone bills are out of phase.
CHAPTER 6 REVIEW 2 2 1. (a) Since 23 12 34 14 1, the point P 23 12 lies on the unit circle.
(b) sin t 12 , cos t 23 , tan t
1 2 3 . 3 23
2 2 9 1 1, the point P 3 4 lies on the unit circle. 2. (a) Since 35 45 25 2 5 5 4 (b) sin t 45 , cos t 35 , tan t 35 43 . 5
528
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
3. t 23
4. t 53
(a) t 23 3 3 1 (b) P 2 2
(c) sin t 23 , cos t 12 , tan t 3, csc t 2 3 3 ,
(c) sin t 23 , cos t 12 , tan t 3,
csc t 2 3 3 , sec t 2, and cot t 33 .
sec t 2, and cot t 33 . 5. t 114
6. t 76
(a) t 3 114 4 (b) P 22 22
(c) sin t 22 , cos t 22 , tan t 1, csc t 2, sec t 2, and cot t 1.
2 7. (a) sin 34 sin 4 2
(a) t 2 53 3 3 1 (b) P 2 2
(a) t 76 6 3 1 (b) P 2 2
(c) sin t 12 , cos t 23 , tan t 33 , csc t 2, sec t 2 3 3 , and cot t 3.
8. (a) tan 3 3 (b) tan 3 3
2 (b) cos 34 cos 4 2
10. (a) cos 5 080902 (b) cos 5 080902
9. (a) sin 11 089121 (b) cos 11 045360 11. (a) cos 92 cos 2 0
12. (a) sin 7 043388
(b) sec 92 is undefined
(b) csc 7 230476
13. (a) tan 52 is undefined
14. (a) sin 2 0
(b) cot 52 cot 2 0
(b) csc 2is undefined.
15. (a) tan 56 33 (b) cot 56 3
1 16. (a) cos 3 2 1 (b) sin 6 2
sin t tan t cos t sin t sin t 17. cos t cos t cos2 t 1 sin2 t
sin2 t 1 cos2 t 1 1 cos t cos2 t cos t cos3 t sin t sin t sin t (because t is in quadrant IV, cos t is positive). 19. tan t 2 cos t 1 sin t 1 sin2 t 18. tan2 t sec t tan2 t
20. sec t
1 1 1 (because t is in quadrant II, cos t is negative). 2 cos t 1 sin t 1 sin2 t 5
5 , cos t 12 . Then tan t 13 5 , csc t 13 , sec t 13 , and cot t 12 . 21. sin t 13 13 12 5 12 5 12 13
22. sin t 12 , cos t 0. Since sin t is negative and cos t is positive, t is in quadrant IV. Thus, t determines the terminal point 2 3 3 , csc t 2, sec t , cot t 3. P 23 12 , and cos t 23 , tan t 3 3
CHAPTER 6
Review
529
1 cos t 5 2 2 5 . Now cot t , so 2 . Since csc t sin t , we know sin t 5 5 sin t 1 1 1 5 5. cos t sin t cot t 2 5 5 12 55 , and tan t 2 while sec t 5 1 5 cos t 2 5
23. cot t 12 , csc t
24. cos t 35 , tan t 0. Since cos t and tan t are both negative, t is in quadrant II. Thus, t determines the terminal point 5 P 35 , 45 , and sin t 45 , tan t 43 , csc t 54 , sec t , cot t 34 . 3 2 1 sin t sin t 14 cos t. Thus, cos2 t sin2 t 1 cos2 t 14 cos t 1 25. tan t 14 , so cot t 4 and cos t 4
17 17 2 cos2 t 16 17 sec t 16 . Because cosine and secant are negative in Quadrant III, we have sec t 4 , and thus,
sec t cot t 4 417 . 8 , so csc t 17 and cos2 t 1 sin2 t 1 8 2 225 sec2 t 289 . Because cosine and secant 26. sin t 17 8 17 289 225 17 17 119 are positive in Quadrant IV, sec t 17 15 , and so csc t sec t 8 15 120 .
27. cos t 35 , so because the terminal point is in Quadrant I, sin t 45 . Thus, tan t 43 and sec t 53 , so tan t sec t 43 53 3. 28. sin2 t cos2 t 1 for any value of t. 29. y 10 cos 12 x
30. y 4 sin 2x
2 (a) This function has amplitude 10, period 1 4,
(a) This function has amplitude 4, period 22 1, and phase shift 0.
2
and phase shift 0.
y
(b)
y
(b)
4 10 2¹
4¹
0.5
x
31. y sin 12 x 2 4, and (a) This function has amplitude 1, period 12
(a) This function has amplitude 2, period 2, and phase shift 4.
(b)
y
x
32. y 2 sin x 4
phase shift 0. (b)
1
y
2
1
2¹ 2¹
4¹
x
¹/4
¹
x
530
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
33. y 3 sin 2x 2 3 sin 2 x 1 (a) This function has amplitude 3, period 22 , and
(a) This function has amplitude 1, period 22 , and phase shift 2.
phase shift 1. (b)
34. y cos 2 x 2
y
y
(b)
3 x 1-¹/2
1
1+¹/2
1
¹/2
cos x 1 35. y cos x 2 6 2 3
(a) This function has amplitude 1, period 2 2 4, and phase shift 13 . y
(b)
x
¹
36. y 10 sin 2x 2 10 sin 2 x 4
(a) This function has amplitude 10, period 22 , and phase shift 4.
(b)
y
1
_13/3
11/3 x
_1/3
10 x ¹/2
¹
37. From the graph we see that the amplitude is 5, the period is 2 , and there is no phase shift. Therefore, the function is y 5 sin 4x. 38. From the graph we see that the amplitude is 2, the period is 4 since 14 of the period has been completed at 1, 2, and there is 2 4k no phase shift. Thus, 2 . Therefore, the function is y 2 sin 2 x. k 39. From the graph we see that the amplitude is 12 , the period is 1, and there is a phase shift of 13 . Therefore, the function is y 12 sin 2 x 13 .
40. From the graph we see that the amplitude is 4, the period is 43 . Thus, 2k 43 k 32 . The phase shift is 3. 3 Therefore, the function is y 4 sin 2 x 3 . 41. y 3 tan x has period .
42. y tan x has period 1.
y
y
10
_¹
¹
2
x _1
1
x
CHAPTER 6
43. y 2 cot x 2 has period .
Review
1 44. y sec 12 x 2 sec 2 x has period
y
2 4. 1 2
4
y
x
¹
2 _2¹
45. y 4 csc 2x 4 csc 2 x 2 has period 2 . 2
2¹
x
5¹/6
x
46. y tan x 6 has period . y
y
2 5
_¹
_¹/6 ¹
x
1 47. y tan 12 x 8 tan 2 x 4 has period 1 2. 2
1 48. y 4 sec 4x has period 24 2. y
y
10 2 ¹/4
5¹/4
x
_0.5 _0.25
0.25
x
0.5
1 1 3 50. cos1 12 23 51. sin1 sin 136 52. cos 6 2 . 53. y 100 sin 8 t 16 100 sin 8t 2 has amplitude 100, period 4 , phase 2 , and horizontal shift 16 3 has amplitude 80, period 2 , phase 3 , and horizontal shift . 80 sin 3t 54. y 80 sin 3 t 2 2 3 2 2
49. sin1 1 2
3 5 55. y1 25 sin 3 t 2 25 sin 3t 2 ; y2 10 sin 3t 2 (a) y1 has phase 32 and y2 has phase 52 .
y
(d)
20
(b) The phase difference is 32 52 . (c) Because the phase difference is not a multiple of 2, the curves are out of phase.
_¹
0 _20
yÁ
yª ¹ x
531
532
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
56. y1 50 sin 10t 2 ; y2 50 sin 10 t 20 50 sin 10t 2 (a) y1 has phase 2 and y2 has phase 2 .
y
(d)
yÁ, yª
40
(b) The phase difference is 2 2 0.
(c) Because the phase difference is a multiple of 2, the curves are in
0
_¹/2
phase.
¹/2 x
_40
58. (a) y sin cos x
57. (a) y cos x
1
1 -5 -5
5 -1
5
(b) This function has period .
(b) This function has period 2.
(c) This function is even.
(c) This function is even.
59. (a) y cos 201x
60. (a) y 1 2cos x 4 1 2
-50
50 -1
-10
10
(b) This function is not periodic.
(b) This function has period 2.
(c) This function is neither even nor odd.
(c) This function is even.
61. (a) y x cos 3x
62. (a) y
x sin 3x (x 0)
5
0 -5
5
5
-5
(b) This function is not periodic.
(b) This function is not periodic.
(c) This function is even.
(c) This function is neither even nor odd.
10
CHAPTER 6
63. y x sin x is a sine function whose graph lies between those of y x and y x.
Review
533
64. y 2x cos 4x is a cosine function whose graph lies between the graphs of y 2x and y 2x .
10
5
-10
10
-2
-10
2 -5
65. y x sin 4x is the sum of the two functions y x and y sin 4x.
66. y sin2 x cos2 x is the sum of the two functions
y sin2 x and y cos2 x. Note that sin2 x cos2 x 1
for all x. 2
1 -2
2 -2 -5
67. We graph y f x cos x sin 2x in the viewing
rectangle [0 2] [2 2] and see that the function has local maxima of approximately f 063 176 and
5
68. We graph y f x cos x sin2 x in the viewing
rectangle [0 2] [2 2] and see that the function has local maxima of approximately
f 414 037 and local minima of approximately
f 105 f 524 125 and local minima of
periodic with period 2.
periodic with period 2.
f 251 176 and f 528 037. The function is
f 0 f 2 1 and f 1. The function is 2
2
0
0 2
4
2
6
4
6
-2
-2
69. We want to find solutions to sin x 03 in the interval [0 2], so we plot the functions y sin x and
y 03 and look for their intersection. We see that the
graphs intersect at x 0305 and at x 2837.
70. We want to find solutions to cos 3x x in the interval [0 ], so we plot the functions y cos 3x and y x and
look for their intersection. We see that the graphs intersect at x 0390. 1
1 0
0 2 -1
4
1
6 -1
2
3
534
CHAPTER 6 Trigonometric Functions: Unit Circle Approach
71. y1 cos sin x, y2 sin cos x (a)
1
yÁ
(b) y1 has period , while y2 has period 2. (c) sin cos x cos sin x for all x.
-10
0
10
-1
yª
72. The amplitude is a 50 cm. The frequency is 8 Hz, so 8 2 16. Since the mass is at its maximum displacement when t 0, the motion follows a cosine curve. So a function describing the motion of P is f t 50 cos 16t. 73. The amplitude is 12 100 50 cm, the frequency is 4 Hz, so 4 2 8. Since the mass is at its lowest point when t 0, a function describing the distance of the mass from its rest position is f t 50 cos 8t. (b)
74. (a) The initial amplitude is 16 cm and the frequency is 14 Hz, so
20
a function describing the motion is y 16e072t cos 28t. (c) When t 10 s, y 16e72 cos 28 00119 cm.
0 2 -20
CHAPTER 6 TEST
1. Since P x y lies on the unit circle, x 2 y 2 1 y 1 quadrant. Therefore y is negative
y 56 .
2 11 6
5 25 36 6 . But P x y lies in the fourth
2 9 , and so x 3 . From the diagram, 2. Since P is on the unit circle, x 2 y 2 1 x 2 1 y 2 . Thus, x 2 1 45 25 5 x is clearly negative, so x 35 . Therefore, P is the point 35 45 . (a) sin t 45
(b) cos t 35
4
(c) tan t 53 43 5
(d) sec t 53 13 22 4 3 (d) csc 2 1
3. (a) sin 76 05 (c) tan 53 3
(b) cos
sin t sin t . But t is in quadrant II cos t 1 sin2 t sin t . Thus, tan t 1 sin2 t
4. tan t
8 , t in quadrant III 5. cos t 17
quadrant III) 1
cos t is negative, so we choose the negative square root.
tan t cot t csc t 1
1 1 2 . 1 15 15 64 1 289 17
1 (since t is in 1 cos2 t
CHAPTER 6
6. y 5 cos 4x (a) This function has amplitude5, period 24 2, phase 0, and horizontal shift 0.
535
1 7. y 2 sin 12 x 6 sin 2 x 3
2 (a) This function has amplitude2, period 1 4, 2
phase 6 , and horizontal shift 3 .
y
(b)
Test
5
y 2
(b)
0
¹/2 x
¹/4
¹/3
9. y tan 2 x 4 has period 2 .
8. y csc 2x has period 22 . y
y
2 _¹/2
2 ¹/2
_¹/2
¹/2
x
(b) cos1 23 56
10. (a) tan1 1 4
13¹/3 x
7¹/3
x
(c) tan1 tan 3 0
(d) sin sin1 21 12
2 11. From the graph, we see that the amplitude is 2 and the phase shift is 3 . Also, the period is , so k k 2 2. Thus, the function is y 2 sin 2 x 3 .
12. y1 30 sin 6t 2 ; y2 30 sin 6t 3 (a) y1 has phase 2 and y2 has phase 3 .
y 30
(d)
(b) The phase difference is 2 3 6.
(c) Because the phase difference is not a multiple of 2, the curves are out of phase.
_¹/2
0
yª
yÁ
¹/2 x
_30
13. y (a)
cos x 1 x2
(b) The function is even. 1.0
(c) The function has a minimum value of approximately 011 when x 254 and a maximum value of 1 when x 0.
0.5
-10
10
14. The amplitude is 12 10 5 cm and the frequency is 2 Hz. Assuming that the mass is at its rest position and moving upward when t 0, a function describing the distance of the mass from its rest position is f t 5 sin 4t.
536
FOCUS ON MODELING
(b)
15. (a) The initial amplitude is 16 in. and the frequency
20
is 12 Hz, so a function describing the motion is y 16e01t cos 24t.
0 0.5
1.0
-20
FOCUS ON MODELING Fitting Sinusoidal Curves to Data
1. (a) See the graph in part (c). (b) Using the method of Example 1, we find the vertical shift b 12 maximum value minimum value 12 21 21 0, the amplitude 2 2 6 0 12 (so a 12 maximum value minimum value 12 21 21 21, the period 05236), and the phase shift c 0. Thus, our model is y 21 cos 6 t. (c)
y
(d) Using the SinReg command on the TI-83, we find
2
y 2048714222 sin 05030795477t 1551856108
1
00089616507.
y 0
2
4
6
8
10
12
2
14 x
_1
1
_2
0
The curve fits the data quite well.
2
4
6
8
10
12
14 x
_1 _2
(e) Our model from part (d) is equivalent to y 205 cos 050t 155 2 001 205 cos 050t 002 001. This is the same as the function in part (b), correct to one decimal place.
2. (a) See the graph in part (c). (b) Using the method of Example 1, we find the vertical shift b 12 maximum value minimum value 12 200 95 1475, the amplitude 2 2 275 125 300 (so a 12 maximum value minimum value 12 200 95 525, the period 002094), and the phase shift c 25. Thus, our model is y 525 cos 002094 t 25 1475.
Fitting Sinusoidal Curves to Data y
(c)
(d) Using the SinReg command on the TI-83, we find
240
y 4970329025 sin 00200800412x 2093611012
200
1491294508.
y
160
240
120
200
80
160
40 0
537
120
400 x
200
80
The curve fits the data quite well.
40 0
200
400
x
(e) Our model from part (d) is equivalent to y 4970 cos 002t 209 2 14913 4970 cos 002t 052 14913. This is close but not identical to the function in part (b).
3. (a) See the graph in part (c). (b) Using the method of Example 1, we find the vertical shift b 12 maximum value minimum value 12 251 10 1305, the amplitude 2 2 15 09 12 (so a 12 maximum value minimum value 12 251 10 1205, the period 5236), and the phase shift c 03. Thus, our model is y 1205 cos 5236 t 03 1305. (c)
y 30
(d) Using the SinReg command on the TI-83, we find y 1171905062 sin 5048853286t 02388957877
10
0
1296070536.
y 30
20
20
1
x
10
The curve fits the data fairly well. 0
1
x
(e) Our model from part (d) is equivalent to y 1172 cos 505t 024 2 1296 1172 cos 505t 1331296. This is close but not identical to the function in part (b). 4. (a) See the graph in part (c). (b) Using the method of Example 1, we find the vertical shift b 12 maximum value minimum value 12 063 010 0265, the amplitude 2 2 55 25 6 (so a 12 maximum value minimum value 12 063 010 0365, the period 1047), and the phase shift c 05. Thus, our model is y 0365 cos 1047 t 05 0265.
538
FOCUS ON MODELING
(c)
y
(d) Using the SinReg command on the TI-83, we find
0.6
y 0327038879 sin 1021164911t 2118186963
0.4
0.6
0.2 0
02896017397.
y
2
4
6
0.4
x
0.2
_0.2
The curve fits the data reasonably well. 0
2
4
6
x
_0.2
(e) Our model from part (d) is equivalent to 033 cos 102t 212 2 029 033 cos 102t 055 029. This is the same as the function in part (b), correct to one decimal place.
5. (a) See the graph in part (c). (b) Let t be the time since midnight. We find a function of the form y a sin t c b. a 12 374 366 04. 026. b 1 374 366 37. Because the maximum value occurs at t 16, we The period is 24 and so 224 2 get c 16. Thus the function is y 04 cos 026 t 16 37. y
(c)
37
36 0
10
20
x
Time (d) Using the SinReg command on the TI-83 we obtain the function y a sin bt c d, where a 04, b 026, c 264, and d 370. Thus we get the model y 04 sin 026t 264 370.
6. (a) See the graph in part (c). (b) Let t be the time years. We find a function of the form y a sin t c b, where y is the owl population.
052. b 1 80 20 50. Because the a 12 80 20 30. The period is 2 9 3 12 and so 212 2 values start at the middle we have c 0. Thus the function is y 30 sin 052t 50.
Fitting Sinusoidal Curves to Data
539
y
(c)
80 60 40 20
0
2
4
6
8
10
12 x
Year (d) Using the SinReg command on the TI–83 we find that for the function y a sin bt c d, where a 258, b 052, c 002, and d 506. Thus we get the model y 258 sin 052t 002 506.
7. (a) See the graph in part (c). (b) Let t be the time since 1985. We find a function of the form y a sin t c b. a 12 63 22 205. The 052. b 1 63 22 425. The average value occurs in the ninth year, periodis 2 15 9 12 and so 220 2 so c 6. Thus, our model is y 205 sin 052 t 6 425. y
(c)
60
40
20
0
2
4
6
8
10
12
14
x
Year since 1985 (d) Using the SinReg command on the TI-83, we find that for the function y a sin bt c d, where a 178, b 052, c 311, and d 424. Thus we get the model y 178 sin 052t 311 424.
8. (a) See the graph in part (c). (b) Let t be the number of years since 1968. We find a function of the form y a cos t c b. We have
a 12 157 3 77. The number of sunspots varies over a cycle of approximately 11 years (as the problem 057. Also, we have b 1 157 3 80. The first maximum value seems to occur at indicates), so 211 2
approximately c 1. Thus, our model is y 77 cos 057 t 1 80, where y is the average daily sunspot count and t is the number of years since 1968.
540
FOCUS ON MODELING
(c)
y 160 140 120
(c)
100 80 60
(d)
40 20 0 1970
1980
1990
2000
2010
Year
(d) Using the SinReg command on the TI-83, we find the function y a sin bt c d, where a a 5527740309, b 05772889486, c 1032016596, and d 6397239705. Thus we get the model y 5527sin 058t 103 640. Converting this to a cosine function for comparison, we have y 5527cos 058t 103 2 640 5527 cos 058t 054 640. This model is less accurate for the peaks around 1980 and 1990, but more accurate for the early 2000s.
7
ANALYTIC TRIGONOMETRY
7.1
TRIGONOMETRIC IDENTITIES
1. An equation is called an identity if it is valid for all values of the variable. The equation 2x x x is an algebraic identity and the equation sin2 x cos2 x 1 is a trigonometric identity.
2. The fact that cos x has the same value as cos x for any x can be expressed as the identity cos x cos x. sin t 3. cos t tan t cos t sin t cos t 1 4. cos t csc t cos t cot t sin t 1 5. sin sec sin tan cos 1 1 sin sec 6. tan csc cos sin cos sin2 x 1 sin2 x 1 cos2 x 7. tan2 x sec2 x 1 2 2 2 cos x cos x cos x cos2 x 1 sec x cos x sin x tan x 8. 1 csc x cos x sin x cos u sin2 u cos2 u 1 9. sin u cot u cos u sin u cos u csc u sin u sin u sin u sin2 cos2 sin2 1 10. cos2 1 tan2 cos2 1 cos2
1 cos 1 cos2 sin2 sin sec cos cos tan 11. sin sin sin cos sin cos cos cos cos cot cos sin sec 12. 2 1 csc sin cos2 1 sin sin sin 1 sin x sin x sec x cos x 1 13. cos x tan x sin x 1 cos x cos x sec x cos x sin x 14. cos x cot x cos x sin x sin t sin t tan t tan t cos t 1 15. sin t tan t tan t cos t cos A 1 cot A 1 cot A sin A sin A sin A cot A sin A sin A sin A cos A 16. csc A sin A 17. cos3 x sin2 x cos x cos x cos2 x sin2 x cos x
541
542
CHAPTER 7 Analytic Trigonometry
18. sin4 cos4 cos2 sin2 cos2 sin2 cos2 cos2 sin2 19.
sec2 x 1 tan2 x sin2 x cos2 x sin2 x. sec2 x sec2 x cos2 x 1 sec2 x 1 1 1 cos2 x sin2 x Another method: sec2 x sec2 x
cos2 x 1 2 2 sec x cos x cos x 1 cos x sin x sin x cos x 20. sin x tan x sin x sin x cos x
21.
22.
23.
1 cos y cos y 1 cos y 1 cos y 1 cos y cos y 1 cos y 1 1 sec y 1 cos y 1 1 cos y cos y 1 sin y 1 sin y 1 sin y sin y 1 1 1 csc y 1 1 sin y sin y sin y 1 sin u cos u 1 2 sin u sin2 u cos2 u 1 2 sin u 1 1 sin u2 cos2 u cos u 1 sin u cos u 1 sin u cos u 1 sin u cos u 1 sin u 2 2 sin u 2 1 sin u 2 2 sec u cos u 1 sin u cos u 1 sin u cos u
24. Note that because sin2 t 1 cos2 t 1 cos t 1 cos t, we can write sin t 1 cos t 1 cos t csc t cot t. 1 cos t sin t sin t sin t
25.
cos x sec x tan x
cos x 1 sin2 x cos2 x 1 sin x 1 sin x 1 sin x 1 sin x 1 sin x 1 sin x 1 sin x cos x cos x
26. Because tan A tan A,
27.
28.
1 cos t sin t . Thus, 1 cos t sin t
cot A 1 tan A 1 tan A cot A 1 cot A 1 tan A 1 tan A tan A tan A 1 tan A
1 1 1 sin 1 sin 2 2 sec2 2 1 sin 1 sin cos2 1 sin 2 tan2 x 1 1 tan2 x 1 1 tan2 x 1 sec2 x 1 1 1 1 2 2 2 2 2 sec x sec x sec x sec x sec x sec2 x 1 1 11 cos2 x sec2 x sec2 x
SECTION 7.1 Trigonometric Identities
cos x 29. (a) sec x sin x
1 sin2 x cos2 x sin x sin x
cos x
1 sin x cos x sin2 x 1 csc x sin x sin x sin x
sin y tan y 1 cos2 y sin2 y cos y 30. (a) 1 csc y cos y cos y sin y
(b) We graph each side of the equation and see that the cos x and y csc x sin x are graphs of y sec x sin x identical, confirming that the equation is an identity.
cos2 y 1 sec y cos y cos y cos y
(b) We graph each side of the equation and see that the tan x graphs of y and y sec x cos x are csc x identical, confirming that the equation is an identity.
1
1
-5
5
-5
-1
sin sin sin sin tan cos 1 cos u sec u cos u 33. tan u cos u sin y tan y sin2 y cos y 35. 1 csc y cos y sin y 31.
5 -1
cos cos sin
32.
tan x sin x cos x sin x sec x cos x
cot u cot u
34.
cot x sec x cos x 1 sin x 1 csc x sin x cos x
543
1 cos2 y 1 1 1 cos y cos y cos y cos y sec y
1 sin2 1 cos2 sin csc sin sin sin sin 37. cos x sin x cos x sin x cos x sin x 36.
38. cot cos sin 39. tan cot
cos2 sin2 1 cos cos sin csc sin sin sin
cos sin2 cos2 1 sin sec csc cos sin cos sin cos sin
40. sin x cos x2 sin2 x 2 sin x cos x cos2 x 1 2 sin x cos x 1 41. 1 cos 1 cos 1 cos2 sin2 csc2 cos x sin x 42. cos2 x sin2 x 1 sec x csc x 1 1 43. sec2 y 1 tan2 y 2 cos2 y 1 sin y 44. csc x sin x
1 1 sin2 x cos2 x sin x cos x cot x sin x sin x sin x
45. tan x cot x2 tan2 x 2 tan x cot x cot2 x tan2 x 2 cot2 x tan2 x 1 cot2 x 1 sec2 x csc2 x 46. tan2 x cot2 x sec2 x 1 csc2 x 1 sec2 x csc2 x 2 2 47. 1 sin2 t cos2 t 4 sin2 t cos2 t 2 cos2 t 4 sin2 t cos2 t 4 cos2 t cos2 t sin2 t 4 cos2 t
544
48.
CHAPTER 7 Analytic Trigonometry
2 sin x cos x sin x cos x2 1
2 sin x cos x sin2 x cos2 x 2 sin x cos x 1 2 sin x cos x 1 2 sin x cos x 1 1
2 sin x cos x 2 sin x cos x sin2 x cos2 x 1
sin2 x 1 cos2 x csc x sin x sin x sin x cos2 t 1 2 t cos2 t 50. cot2 t cos2 t cos 1 cos2 t csc2 t 1 cot2 t cos2 t 2 2 sin t sin t
49. csc x cos2 x sin x
sin x cos x sin x cos x sin x cos x sin2 x cos2 x sin x cos x2 sin x cos x sin x cos x sin x cos x sin x cos x sin x cos x sin x cos x2 2 2 52. sin x cos x4 sin x cos x2 sin2 x 2 sin x cos x cos2 x 1 2 sin x cos x2
51.
sin x cos x2 sin2 x cos2 x
1 1 cos t cos t cos t 1 cos2 t sec t cos t cos t cos t sin2 t 53. 1 1 sec t cos t 1 cos t cos t cos2 x cos x cos x 1 54. cot x csc x cos x 1 cot x cos x cot x csc x cos x csc x sin x sin x sin x sin x cos2 x 1 sin2 x sin x sin x sin x 55. cos2 x sin2 x cos2 x 1 cos2 x 2 cos2 x 1 56. 2 cos2 x 1 2 1 sin2 x 1 2 2 sin2 x 1 1 2 sin2 x 2 2 57. sin4 cos4 sin2 cos2 sin2 cos2 sin2 cos2 sin2 cos2 cos2 x 2 2 2 sin2 x cos2 x 1 58. 1 cos x 1 cot x sin x 1 sin2 x
sin2 t 2 sin t cos t cos2 t sin2 t cos2 t 2 sin t cos t 1 sin t cos t2 2 2 sec t csc t sin t cos t sin t cos t sin t cos t sin t cos t sin t cos t 1 cos t 1 1 1 sin t 2 sec2 t csc2 t 60. sec t csc t tan t cot t cos t sin t cos t sin t cos2 t sin t
59.
61.
62.
63.
64.
65.
sin2 u sin2 u 1 2 2 2 1 cos2 u cos2 u cos u cos u sin u 2 2 2 2 2 1 tan u cos u sin2 u cos2 u sin2 u sin u sin u cos u 1 1 cos2 u cos2 u 2 2 1 sec x 1 sec x 1 1 cos2 x 1 1 tan2 x sec2 x sec2 x 1 1 1 1 sec x csc x cos x sin x cos x sin x sin x cos x sin x cos x sin x cos x cos x cos x sin x cos x sin x sin x tan x cot x sin2 x cos2 x cos x sin x cos x sin x sin x cos x sin x cos x sin x cos x cos x sin x cos x sin x sin x cos x 1 1 sin x cos x sec x csc x sin x cos x cos x sin x cos x sin x 1 cos x sin x 1 cos x 1 cos x sin x sin x 1 2 cos x cos2 x sin2 x sin x 1 cos x sin x 1 cos x 1 cos x sin x sin x 1 cos x 2 2 cos x 2 1 cos x 2 csc x sin x 1 cos x sin x 1 cos x 1 tan2 u
1
SECTION 7.1 Trigonometric Identities
cos x cos x 1 1 csc x cot x sin x sin x sin x sin x cos x cos x 1 cos x cos x cot x sin x 66. 1 1 sec x 1 sin x cos x sin x 1 cos x sin x 1 1 cos x cos x sin2 u cos2 u sin2 u sin2 u 67. tan2 u sin2 u 1 cos2 u tan2 u sin2 u 2 2 2 cos u cos u cos u 4 4 2 2 2 68. sec x tan x sec x tan x sec x tan2 x 1 sec2 x tan2 x sec2 x tan2 x
sin x 1 1 tan x cos x cos x cos x sin x 69. sin x cos x 1 tan x cos x sin x 1 cos x
sin2 1 1 sin csc cos2 cos sin sin 70. cos cos sin cos cos cot cos sin 1 1 sin cos sin sin 1 sec x tan x sec x tan x 2 sec x 2 sec x 1 2 sec x 71. sec x tan x sec x tan x 1 sec x tan x sec x tan x sec2 x tan2 x sin
72.
cos2 t tan2 t 1 sin2 t
sin2 t tan2 t sin2 t
1
1 sin2 t 1 sec2 t tan2 t cos2 t sin2 t
1 sin x 1 2 sin x sin2 x 1 2 sin x sin2 x 1 sin x 1 sin x2 1 sin x2 1 sin x 1 sin x 1 sin x 1 sin x 1 sin2 x 4 sin x 1 sin x 4 tan x sec x 4 cos x cos x cos2 x sin x sin x cos y cos x sin y sin y tan x tan y cos x cos y cos x cos y cos x 74. cos y cos x sin y sin x cos y cot x cot y sin x sin y sin x sin y sin x cos y cos x sin y sin x sin y sin x sin y tan x tan y cos x cos y cos x sin y sin x cos y cos x cos y sin x cos x sin2 x sin x cos x cos2 x sin3 x cos3 x sin2 sin x cos x cos2 x 1 sin x cos x 75. sin x cos x sin x cos x tan cot 1 tan cot 1 sin cos 76. cos sin cos sin tan cot tan cot tan cot tan2 cot2 cos sin sin cos sin cos sin2 cos2
73.
77.
1 cos 1 cos 1 cos2 sin2 sin 1 cos sin sin 1 cos sin 1 cos sin 1 cos 1 cos
sin x 1 sin x 1 sin2 x 1 cos2 x sin x 1 2 sin x 1 sin x 1 sin x 1 sin x 1 sin x 12 1 sin sin cos tan 79. 1 sin cos sin cos 1 tan cos sin A sin A 1 cos A cos A sin A 1 cos A sin A 1 cos A 80. cot A cot A cot A 1 cos A 1 cos A 1 cos A sin A 1 cos2 A sin2 A 1 cos A cos A 1 csc A sin A sin A sin A sin A
78.
545
546
81.
CHAPTER 7 Analytic Trigonometry
sec x sec x tan x sec x sec x tan x sec x sec x sec x tan x sec x sec x tan x 2 2 sec x tan x sec x tan x sec x tan x 1 sec x tan x
82. sec tan sec tan 83. 84.
85.
sec tan sec2 tan2 1 sec tan sec tan sec tan
cos 1 sin cos 1 sin sin cos 1 sin 1 cos sec tan 1 sin 1 sin 1 sin cos cos cos2 1 sin2 tan sin tan sin tan sin tan sin tan sin tan sin tan sin 2 2 tan sin tan sin tan sin tan sin sin2 sec2 1 tan sin tan sin tan sin 2 2 tan sin tan sin 1 sin x 1 sin x 1 2 sin x sin2 x 1 2 sin x sin2 x 1 2 sin x sin2 x 1 sin x 1 sin x 1 sin x 1 sin x cos2 x cos2 x cos2 x cos2 x 1 sin2 x sec2 x 2 sec x tan x tan2 x sec x tan x2
86.
1 sin x 1 sin x 1 sin x 1 sin x2 1 sin x2 1 sin x 1 sin x 1 sin x cos2 x 1 sin2 x
1 sin x 2 tan x sec x2 cos x
cos x 1 cos x 1 cos x 1 cos x 1 cos2 x sin2 x 1 sin x sin x sin x sin x 1 cos x sin x 1 cos x sin x 1 cos x 1 1 1 1 1 cos x csc x cot x 1 cos x sin x sin x sin x
87. csc x cot x
1 sin u 1 sin u sin u tan u sin u sec u 1 cos u cos u 88. 1 sin u sec u 1 sin u tan u sin u 1 sin u cos u cos u sin sin x sin tan (since cos 0 for 0 89. x sin ; then 2 ). 2 2 2 cos cos 1x 1 sin 90. x tan ; then 1 x 2 1 tan2 1 sec2 1 sec2 sec (since sec 0 for 0 2 ). 91. x sec ; then
x2 1
sec2 1 tan2 1 1 tan2 tan
(since tan 0 for 0 2)
92. x 2 tan ; then 1 1 1 1 2 2 2 2 2 4 tan 2 sec2 x 4x 4 tan 4 1 tan 2 tan 2 4 2 tan 2 1 1 cos2 cos 18 cot2 cos 2 8 8 tan sec sin2 93. x 3 sin ; then 9 x 2 9 3 sin 2 9 9 sin2 9 1 sin2 3 cos2 3 cos
(since
cos 0 for 0 2 ).
94. x 5 sec ; then
x 2 25 x
tan 0 for 0 2 ).
5 sec 2 25 5 sec
25 sec2 1 5 sec
sin 5 tan2 tan cos sin (since 1 5 sec sec cos
SECTION 7.1 Trigonometric Identities
95. f x cos2 x sin2 x, g x 1 2 sin2 x. From the graph, f x g x this appears to be an identity. Proof:
1
f x cos2 x sin2 x cos2 x sin2 x 2 sin2 x 1 2 sin2 x g x. Since
f x g x for all x, this is an identity.
-5
5 -1
96. f x tan x 1 sin x, g x
sin x cos x . From the graph, f x g x does 1 sin x
not appear to be an identity. In order to show this, let x 4 . Then, 21 . However, f 4 1 1 1 2
2
1
-5
1 1 1 1 g 4 2 2 g 22 . Since f 4 4 , this is not an 1 21 2 1 2 2
5 -1
identity.
97. f x sin x cos x2 , g x 1. From the graph, f x g x does not
appear to be an identity. In order to show this, we can set x 4 . Then we have 2 2 2 1 1 2 2 2 1 g f 4 4 . Since 2 2 2 f 4 g 4 , this is not an identity.
1
-5
5 -1
98. f x cos4 x sin4 x, g x 2 cos2 x 1. From the graph, f x g x appears to be an identity. In order to prove this, simplify the expression f x: f x cos4 x sin4 x cos2 x sin2 x cos2 x sin2 x cos2 x sin2 x 1 2 cos2 x cos2 x sin2 x 2 cos2 x cos2 x sin2 x 2 cos2 x 1 g x
1
-5
5 -1
Since f x g x for all x, this is an identity.
99. sin x sin y cos x cos y sin x sin y cos x cos y sin x sin y2 cos x cos y2 1 cos2 x sin2 y cos2 x 1 sin2 y sin2 y cos2 x
1 cos2 x sin2 x 2 cos x sin x cos x sin x 1 cos x sin x 1 cos x sin x 1 cos x sin x 1 cos x sin x 1 cos x sin x 1 cos x sin x 1 cos2 x sin2 x 2 cos x 2 2 cos x 2 sin x 2 sin x cos x 1 sin x 1 cos x sin x 1 cos x cos x 1 cos x cos x 2 cos2 x 2 cos x 4 4 2 4 cos x sin x sin x cos2 x 1 101. tan x cot x4 sec4 x csc4 x cos x sin x sin x cos x sin x cos x cos 1 1 sin cos sin 1 cos 1 102. sin tan cos cot sin cos sin cos sin 1 1 cos 1 sin 1 cos 1 sin 1 cos sin
100.
547
548
CHAPTER 7 Analytic Trigonometry
sin y csc y sin2 y sin y csc y csc2 y sin3 y csc3 y sin2 y csc2 y 1 103. sin y csc y sin y csc y
104. sin6 cos6 sin2 cos2 sin4 sin2 cos2 cos4 sin2 1 cos2 sin2 cos2 cos2 1 sin2 sin2 cos2 3 sin2 cos2 1 3 sin2 cos2
sin x ln sin x ln sin x ln cos x ln sin x 105. ln tan x sin x ln tan x ln sin x ln cos x 1 2 ln sin x ln sec x 2 ln sin x ln cos x 106. ln tan x ln cot x ln tan x cot x ln 1 0
2 2 2 2 2 2 2 2 107. esin x etan x e1cos x esec x1 e1cos xsec x1 esec x e cos x
108. e x2 lnsin x e x e2 lnsin x e x elnsin x elnsin x e x sin2 x 2 2 109. This is an identity: LHS esin xcos x e1 RHS.
110.
x 1 x x x 12 x 2 x 1 0, which has no real solution. x 1
111. Squaring both sides, we have sin2 x 1 sin2 x 2 sin2 x 1 sin x 0, which has solutions x k, k an integer. 2
112. xeln x x x 2 x 3 is true for all x 0.
113. LHS R cos sin 2 R sin sin 2 R cos 2 R sin 2 cos2 sin2 R cos 2 R sin 2 R cos 2 RHS
114. (a) x y2 x 2 2x y y 2 is an identity.
(b) x 2 y 2 1 is not an identity. For example, 12 12 1. (c) x y z x y xz is the Distributive Law, an identity.
(d) t 2 cos2 t t cos t t cos t (difference of squares).
(e) sin t cos t 1 is not an identity. For example, sin 4 cos 4
2.
(f) x 2 tan2 x 0 is not an identity.
115. (a) Choose x 2 . Then sin 2x sin 0, whereas 2 sin x 2 sin 2 2.
1 1 2 (b) Choose x 4 and y 4 . Then sin x y sin 2 1, whereas sin x sin y sin 4 sin 4 2 2 2 . Since these are not equal, the equation is not an identity. 2 2 2 2 (c) Choose 2 2 4 1. 4 . Then sec csc
1 1 1 1 whereas (d) Choose x 4 . Then sin x cos x sin cos 1 1 2 4 4 2 2 csc x sec x csc 4 sec 4 2 2. Since these are not equal, the equation is not an identity.
SECTION 7.2 Addition and Subtraction Formulas
549
116. No. All this proves is that f x g x for x in the range of the viewing rectangle. It does not prove that these functions
x4 x6 x2 and g x cos x. In the first viewing rectangle 2 24 720 the graphs of these two functions appear identical. However, when the domain is expanded in the second viewing rectangle, you can see that these two functions are not identical. are equal for all values of x. For example, let f x 1
1 -10 -2
10 -5
2 -1
-10
117. Answers will vary. 118. Label a the side opposite , b the side opposite u, and c the hypotenuse. Since u 2 , we must have u 2 a 2 u. Next we express all six trigonometric function for each angle: cos u c sin cos u sin 2 u , b a sin u bc cos sin u cos 2 u , tan u a cot tan u cot 2 u , cot u b tan c cot u tan 2 u , sec u ac csc sec u csc 2 u , and csc u b sec csc u sec 2 u .
7.2
ADDITION AND SUBTRACTION FORMULAS
1. If we know the values of the sine and cosine of x and y we can find the value of sin x y using the addition formula for sine, sin x y sin x cos y cos x sin y. 2. If we know the values of the sine and cosine of x and y we can find the value of cos x y using the subtraction formula for cosine, cos x y cos x cos y sin x sin y.
6 2 4 2 3 2 6 2 1 sin 15 sin 45 30 sin 45 cos 30 cos 45 sin 30 2 2 2 2 4 6 cos 105 cos 60 45 cos 60 cos 45 sin 60 sin 45 12 22 23 22 2 4 2 cos 195 cos 15 cos 45 30 cos 45 cos 30 sin 45 sin 30 22 23 22 12 6 4 3 tan 30 1 3 3 tan 45 3 tan 15 tan 45 30 2 3 3 1 tan 45 tan 30 3 3 11 3 3 1 tan 45 tan 30 33 3 tan 165 tan 15 tan 45 30 32 1 tan 45 tan 30 3 3 1 1 33 3 6 2 2 1 2 7 sin 19 12 sin 12 sin 4 3 sin 4 cos 3 cos 4 sin 3 2 2 2 2 4 cos 5 cos cos cos sin sin 2 3 2 1 2 6 cos 17 12 12 4 6 4 6 4 6 2 2 2 2 4
3. sin 75 sin 45 30 sin 45 cos 30 cos 45 sin 30 22 23 22 12 4. 5. 6. 7.
8. 9. 10.
tan tan 1 3 3 4 11. tan 12 tan 12 tan 3 4 32 1 tan 1 3 3 tan 4 sin 5 sin sin cos cos sin 2 3 2 1 6 2 12. sin 512 12 4 6 4 6 4 6 2 2 2 2 4 3 6 2 2 2 1 13. cos 11 12 cos 12 cos 3 4 cos 3 cos 4 sin 3 sin 4 2 2 2 2 4
550
CHAPTER 7 Analytic Trigonometry
3 1 3 3 3 7 5 14. tan 12 tan 12 tan 4 6 2 3 3 1 tan tan 3 3 4 6 11 3
tan 4 tan 6
15. sin 18 cos 27 cos 18 sin 27 sin 18 27 sin 45 1 22 2 16. cos 10 cos 80 sin 10 sin 80 cos 10 80 cos 90 0 sin 3 sin 2 cos 3 2 cos 7 cos 1 17. cos 37 cos 221 7 7 21 21 21 3 2 18. 19. 20. 21. 22. 23. 24. 25. 26.
tan tan 18 9
tan 3 tan 318 tan 18 9 6 3 1 tan 18 tan 9
tan 73 tan 13 tan 73 13 tan 60 3 1 tan 73 tan 13 13 13 2 1 cos 10 cos 13 15 cos 5 sin 15 sin 5 cos 15 5 15 cos 3 2 sin 2 u sin 2 cos u cos 2 sin u 1 cos u 0 sin u cos u cot u tan u 2 cos 2 cos u sin 0 cos u 1 sin u sin u cos 2 u 2 sin u cos 2 u cos 2 cos u sin 2 sin u 0 cos u 1 sin u sin u cot 2 u sin u sin cos u cos sin u 1 cos u 0 sin u cos u tan u 2 2 2 1 1 1 1 sec 2 u cos u sin sin u 0 cos u 1 sin u sin u csc u cos cos u 2 2 2 1 1 1 1 csc 2 u cos u cos sin u 1 cos u 0 sin u cos u sec u sin sin u 2 2 2 cos x sin 0 sin x 1 cos x cos x sin x cos sin x 2 2 2 cos x 2 cos x cos 2 sin x sin 2 0 cos x 1 sin x sin x
27. sin x sin x cos cos x sin 1 sin x 0 cos x sin x 28. cos x cos x cos sin x sin 1 cos x 0 sin x cos x tan x 0 tan x tan tan x 29. tan x 1 tan x tan 1 tan x 0 sin x sin x cos cos x 2 2 cos x sin 2 30. tan x 2 cos x cos x cos sin x sin sin x cot x 2 2 2 cos x cos sin x 1 cos x 0 sin x cos x and 31. LHS sin x sin 2 2 2 cos x cos sin x 1 cos x 0 sin x cos x. Therefore, LHS RHS. x sin RHS sin 2 2 2 1 cos x 3 sin x 3 sin x 1 cos x 0 32. cos x sin x 3 6 2 2 2 2 tan x tan 3 tan x 3 33. tan x 3 1 tan x tan 1 3 tan x 3 tan x tan 4 tan x 1 tan x 1 34. tan x 4 1 tan x tan 1 tan x 1 tan x 1 4
35. sin x y sin x y sin x cos y cos x sin y sin x cos y cos x sin y 2 cos x sin y 36. cos x y cos x y cos x cos y sin x sin y cos x cos y sin x sin y 2 cos x cos y 1 1 1 1 1 tan x tan y cot x cot y 1 cot x cot y cot x cot y 37. cot x y 1 1 tan x y tan x tan y cot x cot y cot y cot x cot x cot y 1 1 1 1 tan x tan y cot x cot y 1 1 cot x cot y cot x cot y 38. cot x y 1 1 tan x y tan x tan y cot x cot y cot x cot y cot x cot y
SECTION 7.2 Addition and Subtraction Formulas
39. tan x tan y
551
sin y sin x cos y cos x sin y sin x y sin x cos x cos y cos x cos y cos x cos y
40. 1 tan x tan y 1
cos x cos y sin x sin y cos x y sin x sin y cos x cos y cos x cos y cos x cos y
41.
sin x y sin x cos y cos x sin y tan x tan y tan x tan y cos x cos y 1 tan x tan y cos x cos y sin x sin y cos x y 1 tan x tan y cos x cos y
42.
sin x cos y cos x sin y sin x cos y cos x sin y 2 cos x sin y sin x y sin x y tan y cos x y cos x y cos x cos y sin x sin y cos x cos y sin x sin y 2 cos x cos y
43. cos x y cos x y cos x cos y sin x sin y cos x cos y sin x sin y cos2 x cos2 y sin2 x sin2 y cos2 x 1 sin2 y 1 cos2 x sin2 y cos2 x sin2 y cos2 x sin2 y cos2 x sin2 y cos2 x sin2 y
44. cos x y cos y sin x y sin y cos x cos y sin x sin y cos y sin x cos y cos x sin y sin y cos2 y cos x sin2 y cos x cos2 y sin2 y cos x cos x 45. sin x y z sin x y z sin x y cos z cos x y sin z
cos z sin x cos y cos x sin y sin z cos x cos y sin x sin y
sin x cos y cos z cos x sin y cos z cos x cos y sin z sin x sin y sin z 46. The addition formula for the tangent function can be written as tan A tan B tan A B 1 tan A tan B. Also note that tan A tan A. Using these facts, we get tan x y tan y z tan z x tan x y y z 1 tan x y tan y z tan z x tan x z 1 tan x y tan y z tan z x tan x z tan z x tan x y tan y z tan x z
tan x z tan x z tan x y tan y z tan x z 0 tan x y tan y z tan x z tan x y tan y z tan z x 47. We want to write cossin1 x tan1 y in terms of x and y only. We let sin1 x and tan1 y and sketch triangles with angles and such that sin x and tan y. From the triangles, we have cos 1 x 2 ,
1 ¬
x
Ï1+y@ ú
1
y
Ï1-x@ 1 y cos , and sin . tan y sin x 1 y2 1 y2 From the subtraction formula for cosine we have 1 x2 x y 1 y cos sin1 x tan1 y cos cos cos sin sin 1 x 2 x 1 y2 1 y2 1 y2
552
CHAPTER 7 Analytic Trigonometry
x 48. Let sin1 x and cos1 y. From the triangles, tan and 1 x2 1 y2 , so using the addition formula for tangent, we have tan y x 1 y2 y 1 x2 tan sin1 x cos1 y tan 1 y2 x 1 y 1 x2 x y 1 x 2 1 y2 y 1 x 2 x 1 y2 1 , 49. Let tan1 x and tan1 y. From the triangles, cos 1 x2 x 1 y sin , cos , and sin , so using the 2 2 1y 1 y2 1x subtraction formula for sine, we have sin tan1 x tan1 y sin sin cos cos sin
1
x
1
¬
ú
Ï1-x@
sin x
Ï1-y@ y
cos y
Ï1+x@
Ï1+y@
x
ú
¬
y
1
1
tan x
tan y
x 1 1 y xy 1 y2 1 y2 1 x2 1 x2 1 x 2 1 y2
50. Let sin1 x and cos1 y. From the triangles, cos 1 x 2 and sin 1 y 2 , so using the addition formula for sine, we have sin sin1 x cos1 y sin sin cos cos sin x y 1 x 2 1 y2 x y 1 x 2 1 y2
1
1
x
ú
¬
Ï1-x@
Ï1-y@
y
cos y
sin x
1 1 , so the addition formula for sine gives 51. We know that cos1 21 3 and tan 4 3 2 2 6 2. 1 1 1 1 sin cos 2 tan 1 sin 3 4 sin 3 cos 4 cos 3 sin 4 2 2 2 2 4 1 3 , so the addition formula for cosine gives 52. We know that sin1 23 3 and cot 6 3 sin sin 1 3 3 1 0. 1 1 cos cot 3 cos cos cos sin 2 3 6 3 6 3 6 2 2 2 2
53. We sketch triangles such that sin1 43 and cos1 31 . From the triangles, we have tan 3 and tan 2 2, so the subtraction formula for 7
4
tangent gives 3 2 2 tan tan 7 tan sin1 34 cos1 13 1 tan tan 1 3 2 2 7 3 2 14 76 2
¬ Ï7
sin 34
3
3
ú
2Ï2
1
cos 13
SECTION 7.2 Addition and Subtraction Formulas
553
54. We sketch triangles such that cos1 23 and tan1 12 . From the
triangles, we have sin 35 , cos 2 , and sin 1 , so the addition 5 5
3
formula for sine gives sin cos1 32 tan1 12 sin cos cos sin 35 2 23 1 5 5
¬
remaining sides using the Pythagorean Theorem. To find cos ,
x
ú 1
(3, _4)
cos cos cos sin sin 4 3 35 12 45 23 3 10
y 3
have sin sin cos cos sin 4 3 10 10 3 15 10 5 10 5 10 50 3 10 10
4
y
¬ x
ú
3Ï10
5
Ï10
10
sin 1010
tan 43
(12, 5)
13 ¬
(_2Ï5, Ï5)
x
12
y
y
2Ï2
x 3
(_1, _2Ï2)
(_Ï15, 1) 1
4 Ï15
sin 14
cos 13
2 3 12 4 2. Thus, sin 12 and cos 2 3 56 , so 3 sin x cos x k sin x 2 sin x 56 .
59. k
A2 B 2
12 12 2 and satisfies sin 1 , cos 1 4 . Thus, 2 2 sin x cos x k sin x 2 sin x 4 .
60. k
A2 B 2
x
cos 2 5 5
¬ 1
ú
2Ï5
5 sin 13
have
y 5
Ï5
5
58. Using the addition formula for tangent and the triangles shown, we
x
(3Ï10, _Ï10)
(_3, _4)
y
x
tan 3
cos 35
56. Using the subtraction formula for sine and the triangles shown, we
1 2 2 tan tan 15 tan 1 1 tan tan 12 2 15 2 30 1 15 2 2
2
Ï3
4
sketched:
sin sin cos cos sin 5 2 5 12 2 5 5 13 5 13 5 65
y
(_1, Ï3)
5
we use the addition formula for sine and the triangles we have
57. Using the addition formula for sine and the triangles shown, we have
tan 12
y 3
1
2
2
cos 23
55. As in Example 7, we sketch the angles and in standard position with terminal sides in the appropriate quadrants and find the
Ï5 ú
¬
2 10 2 5 23 15 3 5
Ï5
ú x
554
CHAPTER 7 Analytic Trigonometry
5 1 and cos 5 1 7 , so 52 52 50 5 2. Thus, sin 4 5 2 2 5 2 2 7 5 sin 2x cos 2x k sin 2x 5 2 sin 2x 4 .
61. k
A2 B 2
2 32 3 3 36 6 and satisfies sin 3 6 3 23 , cos 36 12 3 . Thus, 1 3 sin x 3 3 cos x k sin x 6 sin x 3 6 sin x 3 .
62. k
A2 B 2
63. (a) g x cos 2x 3 sin 2x 2 k 12 3 4 2, and satisfies
(b) This is a sine curve with amplitude 2, period , and . phase shift 12 y
sin 12 , cos 23 6 . Thus, we can
2 1
write
g x k sin 2x 2 sin 2x 6 2 sin 2 x 12 .
_2¹
0
_¹
¹
2¹ x
_1 _2
64. (a) f x sin x cos x k
12 12 2, and
satisfies sin cos 1 4 . Thus, 2
(b) This is a sine curve with amplitude
2, period 2,
and phase shift 4. y
we can write
2
f x k sin x 2 sin x 4 .
1 _2¹
_¹
0
¹
_1 _2
65. f x cos x. Now cos x h cos x cos x cos h sin x sin h cos x f x h f x h h h cos x 1 cos h sin h sin x 1 cos h sin h cos x sin x h h h 66. g x sin x. Now sin x h sin x sin x cos h cos x sin h sin x g x h g x h h h sin h cos x sin x 1 cos h sin h 1 cos h cos x sin x h h h
2¹ x
SECTION 7.2 Addition and Subtraction Formulas
2 67. (a) y sin2 x 4 sin x 4 . From the graph we see that the value of y seems to always be equal to 1.
555
1
-5
5
2 sin x cos cos x sin 2 2 (b) y sin2 x 4 sin x 4 sin x cos 4 cos x sin 4 4 4 2 2 1 1 1 2 2 sin x cos x sin x cos x 2 sin x cos x sin x cos x 2 2 12 sin2 x 2 sin x cos x cos2 x sin2 x 2 sin x cos x cos2 x 12 [1 2 sin x cos x 1 2 sin x cos x] 12 2 1
68. (a) y 12 [cos x cos x ]. The graph of y appears to be the same as 1
that of cos x.
-5
5 -1
(b) y 12 [cos x cos x ]
12 [cos x cos sin x sin cos x cos sin x sin ]
12 [ cos x 0 cos x 0] 12 2 cos x cos x
69. If 2 , then 2 . Now if we let y x , then sin x cos x 2 sin y cos y 2 sin y sin y 0. Therefore, sin x cos x 0. 70. Let A and B be the two angles shown in the diagram. Then
180 A B, 90 A, and 90 B. Subtracting the second and third equation from the first, we get
180 90 90 A B A B
. Then
6 Œ
4
3
A
B
º
4
4 3 8 9 tan tan 17 6 4 4 3 12 112 2 17 12 6 . 1 tan tan 1 6 4 1 2 xy tan u tan tan1 tan1 tan u u tan1 x tan1 y 71. tan1 1 xy 1 tan u tan cot cot u 1 x 1x 1 1 72. cot u 0, so tan1 x tan1 u cot1 cot u cot1 0 . cot cot u x 1x x 2
tan tan
y y and tan . Thus, m tan . x x tan 2 tan 1 (b) tan tan 2 1 . From part (a), we have m 1 tan 1 and m 2 tan 2 . Then by 1 tan 2 tan 1 m m1 substitution, tan 2 . 1 m1m2
73. (a) By definition, m
556
CHAPTER 7 Analytic Trigonometry
(c) Let be the unknown angle as in part (b). Since m 1 13 and m 2 12 , 1 1 1 m m1 tan 2 2 3 67 17 tan1 17 0142 rad 81 . 1 m1m2 1 13 12 6
(d) From part (b), we have cot have 0
1 m1m2 . If the two lines are perpendicular then 90 and so cot 0. Thus we m2 m1
1 m1m2 0 1 m 1 m 2 m 1 m 2 1 m 2 1m 1 . Thus m 2 is the negative reciprocal of m 1 . m2 m1 1
1
tan A tan B 3 2 1. Thus A B , so A B C . 74. Clearly C 4 . Now tan A B 1 tan A tan B 4 4 4 2 1 13 12 A2 B 2 52 52 5 2. Therefore, B 5 1 sin and 2 2 5 2 2 A B 1 A . Thus, cos 4. 2 A2 B 2
75. (a) y f 1 t f 2 t 5 sin t 5 cos t 5 sin t cos t
(b) k
10
-5
5
-10
76. (a) f t C sin t C sin t C sin t C sin t cos cos t sin C 1 cos sin t C sin cos t A sin t B cos t
where A C 1 cos and B C sin . (b) In this case, f t 10 1 cos 3 sin t 10 sin 3 cos t 15 sin t 5 3 cos t. Thus 2 3 and sin 5 3 1 , so . Therefore, k 152 5 3 10 3 and has cos 15 2 2 6 10 3 10 3 . f t 10 3 sin t 6 77. sin s t cos 2 s t cos 2 s t cos 2 s cos t sin 2 s sin t sin s cos t cos s sin t. The last equality comes from again applying the cofunction identities. sin s t sin s cos t cos s sin t cos s t cos s cos t sin s sin t 1 sin s sin s cos t cos s sin t cos s cos t cos s sin s 1 cos s cos t sin s sin t 1 cos s cos t cos s
78. tan s t
7.3
sin t cos t tan s tan t sin t 1 tan s tan t cos t
DOUBLE-ANGLE, HALF-ANGLE, AND PRODUCT-SUM FORMULAS
1. If we know the values of sin x and cos x, we can find the value of sin 2x using the Double-Angle Formula for Sine: sin 2x 2 sin x cos x.
x x 2. If we know the value of cos x and the quadrant in which lies, we can find the value of sin using the Half-Angle Formula 2 2 x 1 cos x . for Sine: sin 2 2
SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas
557
5 , x in quadrant I cos x 12 and tan x 5 . Thus, sin 2x 2 sin x cos x 2 5 12 120 , 3. sin x 13 13 12 13 13 169 120 2 2 sin 2x 169 5 119 120 169 120 cos 2x cos2 x sin2 x 12 13 14425 13 169 169 , and tan 2x cos 2x 119 169 119 119 . 169
4. tan x 43 . Then sin x 45 and cos x 35 (x is in quadrant II). Thus, sin 2x 2 sin x cos x 2 45 35 24 25 , 24 2 2 7 , and tan 2x sin 2x 25 24 25 24 . cos 2x cos2 x sin2 x 35 45 916 7 25 25 25 7 cos 2x 7 25
5. cos x 45 . Then sin x 35 (csc x 0) and tan x 34 . Thus, sin 2x 2 sin x cos x 2 35 45 24 25 , 24 2 2 7 , and tan 2x sin 2x 25 24 25 24 . cos 2x cos2 x sin2 x 45 35 169 7 25 25 25 7 7 cos 2x
25 6. csc x 4. Then sin x 14 , cos x 415 , and tan x 1 (tan x 0). Thus, 15 2 2 7 sin 2x 2 sin x cos x 2 14 415 815 , cos 2x cos2 x sin2 x 415 14 151 16 8 , and 15 sin 2x 78 815 87 715 . tan 2x cos 2x 8
7. sin x 35 . Then, cos x 45 and tan x 34 (x is in quadrant III). Thus, sin 2x 2 sin x cos x 2 35 45 24 25 , 24 2 2 7 , and tan 2x sin 2x 25 24 25 24 . cos 2x cos2 x sin2 x 45 35 169 7 25 25 25 7 7 cos 2x 25 8. sec x 2. Then cos x 12 , sin x 23 , and tan x 3 (x is in quadrant IV). Thus, 2 1 3 , cos 2x cos2 x sin2 x 1 2 3 sin 2x 2 sin x cos x 2 23 12 , and 2 2 2 2 23 sin 2x tan 2x 3. 1 cos 2x 2
9. tan x 13 and cos x 0, so sin x 0. Thus, sin x 1 and cos x 3 . Thus, 10 10 6 3 , cos 2x cos2 x sin2 x 3 2 1 2 8 4 , 3 10 sin 2x 2 sin x cos x 2 1 5 10 5 10 10 10 10 and tan 2x
3 sin 2x 45 35 54 34 . cos 2x 5
10. cot x 23 .
Then tan x 32 , sin x 3 (sin x 0), and cos x 2 . Thus, 13 13 2 2 5 2 12 , cos 2x cos2 x sin2 x 2 3 49 sin 2x 2 sin x cos x 2 3 13 13 13 , and 13 13 13 13 12 sin 2x 135 12 12 13 tan 2x 13 5 5. cos 2x 13
2 1 cos 2x 2 11. sin4 x sin2 x 14 12 cos 2x 14 cos2 2x 2 1 cos 4x 14 12 cos 2x 14 14 12 cos 2x 18 18 cos 4x 38 12 cos 2x 18 cos 4x 2 12 34 cos 2x 14 cos 4x
558
CHAPTER 7 Analytic Trigonometry
2 1 cos 2x 2 12. cos4 x cos2 x 14 12 cos 2x 14 cos2 2x 2 1 cos 4x 14 12 cos 2x 14 14 12 cos 2x 18 18 cos 4x 38 12 cos 2x 18 cos 4x 2 12 34 cos 2x 14 cos 4x
13. We use the result of Example 4 to get 1 1 cos 2x cos 4x cos 2x cos 4x. cos2 x sin4 x sin2 x cos2 x sin2 x 18 18 cos 4x 12 12 cos 2x 16 14. Using Example 4, we have 1 1 cos 4x cos 2x cos 2x cos 4x. cos4 x sin2 x cos2 x cos2 x sin2 x 12 1 cos 2x 18 1 cos 4x 16 2 15. Since sin4 x cos4 x sin2 x cos2 x we can use the result of Example 4 to get
sin4 x cos4 x
1 1 cos 4x 2 1 1 cos 4x 1 cos2 4x 8 8 64 32 64
1 1 cos 4x 1 1 1 cos 8x 1 1 cos 4x 1 1 cos 8x 64 32 64 2 64 32 128 128 3 1 1 1 3 1 128 32 cos 4x 128 cos 8x 32 4 cos 4x 4 cos 8x
16. Using the result of Exercise 12, we have 2 cos6 x cos2 x cos2 x 38 12 cos 2x 18 cos 4x 12 12 cos 2x
3 1 cos 2x 1 cos 4x 3 cos 2x 1 cos2 2x 1 cos 2x cos 4x 16 4 16 16 4 16
1 cos 4x 1 1 cos 4x , the last expression is equal to Because 14 cos2 2x 4 2 8 8 3 1 1 3 1 1 1 16 4 cos 2x 16 cos 4x 16 cos 2x 8 8 cos 4x 16 cos 2x cos 4x 5 7 cos 2x 3 cos 4x 1 cos 2x cos 4x 1 5 7 cos 2x 3 cos 4x cos 2x cos 4x 16 16 16 16 16 17. sin 15 12 1 cos 30 12 1 23 14 2 3 12 2 3
1 23 1 cos 30 2 3 18. tan 15 1 sin 30 2 2 1 1 cos 45 2 21 19. tan 225 2 sin 45 2 20. sin 75 12 1 cos 150 12 1 23 14 2 3 12 2 3
21. cos 165 12 1 cos 330 12 1 cos 30 12 1 23 12 2 3 1 1 22. cos 1125 2 1 cos 225 2 1 cos 45 12 1 22 12 2 2
2 1 cos 4 1 2 21 23. tan 8
sin 4
24. cos 38 cos 12 34
2 2
1 cos 34 2
1 22 2
root because 38 is in quadrant I, so cos 38 0.
2 2 12 2 2. Note that we have chosen the positive 4
SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas
559
1 25. cos 12 2 1 cos 6 12 1 23 12 2 3 1 cos 56 1 23 5 2 3 26. tan 12 5 1 sin 6 2
27. sin 98 12 1 cos 94 12 1 22 12 2 2. We have chosen the negative root because 98 is in quadrant III, so sin 98 0.
28. sin 11 12
1 1 cos 11 1 1 3 1 2 3. We have chosen the positive root because 11 is in 2 6 2 2 2 12
quadrant II, so sin 11 12 0.
2 tan 7 tan 14 1 tan2 7 2 tan 7 (b) tan 14 1 tan2 7
29. (a) 2 sin 18 cos 18 sin 36
30. (a)
(b) 2 sin 3 cos 3 sin 6
sin2 cos 2 2 (b) 2 sin cos sin 2 2 1 cos 30 34. (a) sin 15 2 1 cos 8 (b) sin 4 2
31. (a) cos2 34 sin2 34 cos 68
32. (a) cos2
(b) cos2 5 sin2 5 cos 10 sin 8 8 tan tan 4 1 cos 8 2 4 1 cos 4 tan tan 2 (b) sin 4 2
33. (a)
35. sin x x sin x cos x cos x sin x 2 sin x cos x
36. tan x x
2 tan x tan x tan x 1 tan x tan x 1 tan2 x
37. sin x 35 . Since x is in quadrant I, cos x 45 and x2 is also in quadrant I. Thus, 10 x 1 1 4 1 x 1 sin 2 2 1 cos x 2 1 5 10 , cos 2 2 1 cos x 12 1 45 3
3 1010 , and 10
10
sin x2 1 310 13 . tan x2 x 10 cos 2
38. cos x 45 . Since x is in quadrant III, sin x 35 and tan x 34 . Also, since 180 x 270 , 90 x2 135 and so x2 is in quadrant II. Thus, sin x2 12 1 cos x 12 1 45 3 3 1010 , 10 sin x2 10 3. x 1 3 1 cos 2 2 1 cos x 12 1 45 1 1010 , and tan x2 x 10 10 cos 2
2 2 . Since 90 x 180 , we have 39. csc x 3. Then, sin x 13 and since x is in quadrant II, cos x 3 45 x2 90 and so x2 is in quadrant I. Thus, sin x2 12 1 cos x 12 1 2 3 2 16 3 2 2 , cos x2
1 1 cos x 2
x 1 1 2 2 1 3 2 2 , and tan x sin 2 322 3 2 2. x 2 3 6 2 32 2 cos 2
560
CHAPTER 7 Analytic Trigonometry 2 and cos x 2 2 2 , since x is in quadrant I. Also, since 0 x 90 , 0 x2 45 and so x2 is also in quadrant I. Thus, sin x2 12 1 cos x 12 1 22 12 2 2, 1 22 1 cos x 2 x 1 1 1 x 2 1. cos 2 2 1 cos x 2 1 2 2 2 2, and tan 2 2 sin x 2
40. tan x 1. Then sin x
41. sec x 32 . Then cos x 23 and since x is in quadrant IV, sin x 35 . Since 270 x 360 , we have 135 x2 180 and so x2 is in quadrant II. Thus, sin x2 12 1 cos x 12 1 23 1 66 , 6 x sin 30 2 1 6 1 5 . cos x2 12 1 cos x 12 1 23 5 , and tan x2 x 5 6 6 5 5 6 cos 2 42. cot x 5.
Then, cos x 5
26
and sin x 1 (csc x 0). 2
Since cot x 0 and
csc x 0, it follows that x is in quadrant III. Thus 180 x 270 and so 90 x2 135 . 26 , 1 1 cos x 1 1 5 Thus x2 is in quadrant II. sin x2 12 265 2 2 13 26
1 5 26 26 , and tan x 1 cos x cos x2 12 1 cos x 12 1 5 12 265 5 26. 13 2 1 26 sin x 2
43. To write sin 2 tan1 x as an algebraic expression in x, we let tan1 x
x and sketch a suitable triangle. We see that sin and 1 x2 1 cos , so using the double-angle formula for sine, we have 1 x2 x 1 2x sin 2 tan1 x sin 2 2 sin cos 2 . 2 2 1 x2 1x 1x
Ï1+x@ ¬
x
1
44. To write tan 2 cos1 x as an algebraic expression in x, we let cos1 x 1 1 x2 , so using the and sketch a suitable triangle. We see that tan ¬ x x double-angle formula for tangent, we have 1 x2 2 2 tan 2 1 x2 2x 1 x 2 x 1 tan 2 cos x tan 2 . 2 1 tan2 2x 2 1 1 x2 1 x2 x 1 1 x2 x
Ï1-x@
1 cos cos1 x 1x . Because cos1 45. Using the half-angle formula for sine, we have sin 12 cos1 x 2 2 1 x 1 1 1 1 1 1 . has range [0 ], 2 cos x lies in 0 2 and so sin 2 cos x is positive. Thus, sin 2 cos x 2
46. Using a double-angle formula for cosine, we have cos 2 sin1 x 1 2 sin2 sin1 x 1 2x 2 .
SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas 7 and find that sin 24 . Thus, using 47. We sketch a triangle with cos1 25 25 the double-angle formula for sine, 7 sin 2 2 sin cos 2 24 7 336 . sin 2 cos1 25 25 25 625
25
¬ 7
24
7 cos 25
12 48. We sketch a triangle with tan1 12 5 and find that sin 13 . Thus, using a
¬
13
double-angle formula for cosine, 2 2 1 2 12 cos 2 1 2 sin 119 cos 2 tan1 12 5 13 169 .
5
12
tan 12 5
49. Rewriting the given expression and using a double-angle formula for cosine, we have 1 8 1 1 sec 2 sin1 14 2 . 1 1 1 2 1 7 cos 2 sin 4 1 2 sin sin 4 12 1
4 50. We sketch a triangle with cos1 23 and find that sin 35 . Thus, using a 5 3
sin 5 1 1 2 1 . tan 2 cos 3 tan 2 2 1 cos 5 1 3
2
Ï5
2 7. 51. Using a double-angle formula for cosine, we have cos 2 1 2 sin2 1 2 35 25
cos 23
y ¬
52. To evaluate sin 2, we first sketch the angle in standard position with
terminal side in quadrant IV and find the remaining side using the Pythagorean Theorem. Using the half-angle formula for sine, we have 1 12 1 cos 13 26 . Because 2 lies in sin 26 2 2 2
¬
3
half-angle formula for tangent,
12 13
x
5
(12, _5)
quadrant II, where sine is positive, we take the positive value. Thus, sin 2 2626 . 53. To evaluate sin 2, we first sketch the angle in standard position with terminal side in quadrant II and find the remaining side using the Pythagorean Theorem. Using the double-angle formula for sine, we have sin 2 2 sin cos 2 17 4 7 3 8493 .
54. To evaluate tan 2, we first sketch the angle in standard position with terminal
1
y
(_4Ï3, 1)
y
(3, 4) 5
Using the double-angle formula for tangent, we have 2 tan tan 2 1 tan2
24 2 . 7 1 4 3
55. sin 2x cos 3x 12 [sin 2x 3x sin 2x 3x] 12 sin 5x sin x 56. sin x sin 5x 12 [cos x 5x cos x 5x] 12 cos 4x cos 6x 57. cos x sin 4x 12 [sin 4x x sin 4x x] 12 sin 5x sin 3x 58. cos 5x cos 3x 12 [cos 5x 3x cos 5x 3x] 12 cos 8x cos 2x
x
4Ï3
side in quadrant I and find the remaining side using the Pythagorean Theorem. 2 43
¬
7
¬
3
4 x
561
562
CHAPTER 7 Analytic Trigonometry
59. 3 cos 4x cos 7x 3 12 [cos 4x 7x cos 4x 7x] 32 cos 11x cos 3x 3x x 60. 11 sin x2 cos x4 11 12 sin x2 x4 sin x2 x4 11 2 sin 4 sin 4 5x 3x 5x 3x 61. sin 5x sin 3x 2 sin cos 2 sin 4x cos x 2 2 x 4x x 4x 5x 3x 5x 3x 62. sin x sin 4x 2 cos sin 2 cos sin 2 cos sin 2 2 2 2 2 2 4x 6x 4x 6x sin 2 sin 5x sin x 2 sin 5x sin x 63. cos 4x cos 6x 2 sin 2 2 11x 7x 9x 2x 9x 2x 64. cos 9x cos 2x 2 cos cos 2 cos cos 2 2 2 2 9x 5x 9x 5x 2x 7x 2x 7x sin 2 cos sin 2 cos sin 65. sin 2x sin 7x 2 cos 2 2 2 2 2 2 7x x 7x x 3x 4x 3x 4x cos 2 sin cos 2 sin cos 66. sin 3x sin 4x 2 sin 2 2 2 2 2 2 67. 2 sin 525 sin 975 2 12 cos 525 975 cos 525 975 cos 45 cos 150 cos 45 cos 150 22 23 12 2 3 68. 3 cos 375 cos 75 32 cos 45 cos 30 32 22 23 34 2 3 21 69. cos 375 sin 75 12 sin 45 sin 30 12 22 12 14 75 15 75 15 cos 2 sin 45 cos 30 2 22 23 26 70. sin 75 sin 15 2 sin 2 2 255 195 255 195 sin 2 sin 225 sin 30 2 22 12 22 71. cos 255 cos 195 2 sin 2 2 3 6 2 cos 5 2 cos 1 5 cos 1 5 2 cos 72. cos 12 12 2 12 12 2 12 12 4 cos 6 2 2 2 2 73. cos2 5x sin2 5x cos 2 5x cos 10x 74. sin 8x sin 2 4x 2 sin 4x cos 4x
75. sin x cos x2 sin2 x 2 sin x cos x cos2 x 1 2 sin x cos x 1 sin 2x 76. cos4 x sin4 x cos2 x sin2 x cos2 x sin2 x cos2 x sin2 x cos 2x 77.
78. 79. 80. 81. 82.
2 tan x sin x 2 tan x 2 cos2 x 2 sin x cos x sin 2x cos x 1 tan2 x sec2 x 1 1 2 sin2 x 2 sin2 x 1 cos 2x tan x sin 2x 2 sin x cos x 2 sin x cos x x x 1 cos x 1 cos x cos x cos2 x sin2 x 1 cos x tan cos x tan cos x sin x 2 2 sin x sin x sin x sin x x 1 cos x 1 2 cos x tan csc x 2 sin x sin x sin x 2 sin 2x cos 2x 2 2 sin x cos x cos 2x sin 4x 4 cos x cos 2x sin x sin x sin x 1 sin 2x 1 2 sin x cos x 1 1 1 12 csc x sec x sin 2x 2 sin x cos x 2 sin x cos x
SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas
563
2 2 tan x cot x 2 2 tan x cot x 2 2 cos x sin x tan x cot x tan x cot x tan x cot x tan x cot x cos x sin x 2 sin x cos x 2 sin x cos x 2 sin x cos x sin 2x cos x sin x cos x sin x sin2 x cos2 x cos x sin x 2 sin x cos x sin 2x tan x 84. 1 cos 2x 1 2 cos2 x 1
83.
1 tan2 x 1 2 tan x 2 tan x 1 tan2 x 3 86. 4 sin6 x cos6 x 4 sin2 x cos2 x 3 sin4 x cos2 x sin2 x cos4 x 4 1 3 sin2 x cos2 x sin2 x cos2 x 4 12 sin2 x cos2 x 85. cot 2x
1 tan 2x
4 3 2 sin x cos x2 4 3 sin2 2x
2 tan x tan x 2 tan x tan x 1 tan2 x 2x tan 2x tan x 1 tan 87. tan 3x tan 2x x 2 tan x 1 tan 2x tan x 1 tan2 x 2 tan x tan x 1 tan x 1 tan2 x 3 tan x tan3 x 1 3 tan2 x 88.
sin 3x cos 3x sin 2x cos x cos 2x sin x cos 2x cos x sin 2x sin x cos x sin x cos x sin x cos2 x sin2 x cos x sin x sin 2x cos x sin x cos 2x cos x sin x sin 2x cos x sin x cos x sin x sin 2x cos x sin x2 1 4 sin x cos x
89. 90. 91.
2 sin 3x cos 2x sin 3x sin x sin 5x tan 3x cos x cos 5x 2 cos 3x cos 2x cos 3x
2 sin 5x cos 2x cos 2x sin 3x sin 7x cot 2x cos 3x cos 7x 2 sin 5x sin 2x sin 2x sin 10x 2 sin 5x cos 5x cos 5x sin 9x sin x 2 sin 5x cos 4x cos 4x
sin x sin 5x sin 3x 2 sin 3x cos 2x sin 3x sin 3x 2 cos 2x 1 sin x sin 3x sin 5x tan 3x cos x cos 3x cos 5x cos x cos 5x cos 3x 2 cos 3x cos 2x cos 3x cos 3x 2 cos 2x 1 xy xy xy cos 2 sin sin sin x sin y xy 2 2 2 tan 93. xy xy xy cos x cos y 2 cos 2 cos cos 2 2 2 2x xyxy xyxy 2y 2 cos 2 cos sin sin sin y sin x y sin x y 2 2 2 2 tan y 94. x yx y 2y xyxy 2x cos x y cos x y cos y cos cos 2 cos 2 cos 2 2 2 2 x 2y x . Then 95. Let y 2 4 2 x 1 cos x 1 cos 2y 2 1 sin x 1 sin x tan2 tan2 y 2 4 1 cos 2y 1 sin x 1 sin x 1 cos x 2 92.
564
CHAPTER 7 Analytic Trigonometry
96. 1 cos 4x 2 tan2 x cot2 x 1 1 2 sin2 2x tan2 x 1 cot2 x 1 2 sin2 2x sec2 x csc2 x 2 4 cos2 x sin2 x sec2 x csc2 x 8 sin2 x cos2 x 8 130 110 130 110 sin 2 cos 120 sin 10 2 12 sin 10 sin 10 97. sin 130 sin 110 2 cos 2 2 200 200 100 100 sin 2 sin 150 sin 50 98. cos 100 cos 200 2 sin 2 2 2 12 sin 50 sin 50 45 15 45 15 cos 2 sin 30 cos 15 2 12 cos 15 99. sin 45 sin 15 2 sin 2 2 cos 15 sin 90 15 sin 75 (applying the cofunction identity) 87 33 87 33 cos 2 cos 60 cos 27 2 2 2 12 cos 27 cos 27 sin 90 27 sin 63
100. cos 87 cos 33 2 cos
sin x sin 2x sin 3x sin 4x sin 5x sin x sin 5x sin 2x sin 4x sin 3x cos x cos 2x cos 3x cos 4x cos 5x cos x cos 5x cos 2x cos 4x cos 3x sin 3x 2 cos 2x 2 cos x 1 2 sin 3x cos 2x 2 sin 3x cos x sin 3x tan 3x 2 cos 3x cos 2x 2 cos 3x cos x cos 3x cos 3x 2 cos 2x 2 cos x 1 102. n 1: sin 21 x 2 sin x cos x 21 sin x cos 20 x n 2: sin 22 x sin 4x 2 sin 2x cos 2x 2 2 sin x cos x cos 2x 4 sin x cos x cos 2x 22 sin x cos x cos 21 x n 3: sin 23 x sin 8x 2 sin 4x cos 4x 2 4 sin x cos x cos 2x cos 4x 8 sin x cos x cos 2x cos 4x 23 sin x cos x cos 2x cos 22 x
101.
In general, for n 0 we have sin 2n x sin 2 2n1 x 2 sin 2n1 x cos 2n1 x 2 2n1 sin x cos x cos 2x cos 4x cos 8x cos 2n2 x cos 2n1 x 2n sin x cos x cos 2x cos 4x cos 8x cos 2n2 x cos 2n1 x
103. With u sin1 x for 0 x 1, we can write cos1 1 2x 2 cos1 1 2 sin2 u cos1 cos 2u 2u 2 sin1 x. x2 1 1 1 cos 2u 2u 2 tan1 1 , we can write cos1 cos 104. With u tan1 x x x2 1 105. (a) f x
sin 3x cos 3x sin x cos x
cos 3x sin 3x sin x cos x sin 3x cos x cos 3x sin x sin 3x x sin x cos x sin x cos x 2 sin x cos x sin 2x 2 sin x cos x sin x cos x for all x for which the function is defined.
(b) f x
2 1
-5
5
The function appears to have a constant value of 2 wherever it is defined.
SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas
106. (a) f x cos 2x 2 sin2 x
565
(b) f x cos 2x 2 sin2 x cos2 x sin2 x 2 sin2 x cos2 x sin2 x 1
1
-5
5
The function appears to have a constant value of 1.
(b) By a sum-to-product formula,
107. (a) y sin 6x sin 7x 2
-5
5 -2
y sin 6x sin 7x 6x 7x 6x 7x 2 sin cos 2 2 1 2 sin 13 2 x cos 2 x
(c) We graph y sin 6x sin 7x, y 2 cos 12 x , and y 2 cos 12 x . 2
1 2 sin 13 2 x cos 2 x
-5
5 -2
The graph of y f x lies
between the other two graphs.
1 3 108. From Example 2, we have cos 3x 4 cos3 x 3 cos x. If 3x 3 , then cos 3 2 4 cos x 3 cos x 1 8 cos3 x 6 cos x 8 cos3 x 6 cos x 1 0. Substituting y cos x gives 8y 3 6y 1 0.
2 109. (a) cos 4x cos 2x 2x 2 cos2 2x 1 2 2 cos2 x 1 1 8 cos4 x 8 cos2 x 1. Thus the desired polynomial is P t 8t 4 8t 2 1.
(b) cos 5x cos 4x x cos 4x cos x sin 4x sin x cos x 8 cos4 x 8 cos2 x 1 2 sin 2x cos 2x sin x 8 cos5 x 8 cos3 x cos x 4 sin x cos x 2 cos2 x 1 sin x [from part (a)] 8 cos5 x 8 cos3 x cos x 4 cos x 2 cos2 x 1 sin2 x 8 cos5 x 8 cos3 x cos x 4 cos x 2 cos2 x 1 1 cos2 x 8 cos5 x 8 cos3 x cos x 8 cos5 x 12 cos3 x 4 cos x 16 cos5 x 20 cos3 x 5 cos x
Thus, the desired polynomial is P t 16t 5 20t 3 5t.
566
CHAPTER 7 Analytic Trigonometry
110. Let c1 and c2 be the lengths of the segments shown in the figure. By the Law of
C b b sin 2x c or c . Also by the x sin 2x sin B sin B b a x s c1 s and Law of Sines applied to BC D and AC D we have sin B sin x B A cÁ cª D s c2 s sin x s sin x . So c1 and c2 . Since c c1 c2 , we have sin A sin x sin B sin A s sin x s sin x 1 b a b sin 2x 1 s sin x . By applying the Law of Sines to ABC, sin B sin B sin A sin B sin A sin B sin A b b sin 2x 1 b sin x b 1 . Substituting we have s sin x s 1 sin A a sin B sin B sin B a sin B sin B a b b ab 2ab cos x b 2b sin x cos x sin x 1 2b cos x s 1 s s . b sin 2x s sin x 1 a a a a ab
Sines applied to ABC, we have
111. Using
a
product-to-sum
formula, RHS 4 sin A sin B sin C 4 sin A 12 [cos B C cos B C] 2 sin A cos B C 2 sin A cos B C.
Using another product-to-sum formula, this is equal to 2 12 [sin A B C sin A B C] 2 12 [sin A B C sin A B C]
sin A B C sin A B C sin A B C sin A B C Now A B C , so A B C 2C, A B C 2B, and A B C 2A . Thus our expression simplifies to sin A B C sin A B C sin A B C sin A B C sin 2C sin 2B 0 sin 2A sin 2C sin 2B sin 2A LHS
112. (a) The length of the base of the inscribed rectangle twice the length of the adjacent side which is 2 5 cos and the length of the opposite side is 5 sin . Thus the area of the rectangle is modeled by A 2 5 cos 5 sin 25 2 sin cos 25 sin 2.
(b) The function y sin u is maximized when u 2 . So 2 2 4 . Thus the maximum cross-sectional area is A 4 25 sin 2 4 25 cm2 . 5 2 354 cm. (c) The length of the base is 2 5 cos 5 2 707 cm and the width of the rectangle is 5 sin 4 4 2
113. (a) In both logs the length of the adjacent side is 20 cos and the length of the opposite side is 20 sin . Thus the cross-sectional area of the beam is modeled by A 20 cos 20 sin 400 sin cos 200 2 sin cos 200 sin 2.
(b) The function y sin u is maximized when u 2 . So 2 2 4 . Thus the maximum cross-sectional area is A 4 200 sin 2 4 200.
114. We first label the figure as shown. Because the sheet of paper is folded over, E AC C AB . Thus BC A AC E 90 . It follows that EC D 180 BC A AC E 180 90 90 2. Also,
from the figure we see that BC L sin and C E L sin , so
A ¬ L
E
DC EC cos 2 L sin cos 2. Thus 6 D B DC
C B L sin cos 2 L sin L sin 1 cos 2 L sin 2 cos2 . So L
3 6 . 2 sin cos2 sin cos2
D
C
B
SECTION 7.4 Basic Trigonometric Equations
115. (a) y f 1 t f 2 t cos 11t cos 13t
567
(c) We graph y cos 11t cos 13t, y 2 cos t, and y 2 cos t.
2
2 -5
5 -5
-2
5 -2
(b) Using the identity
y cos cos y 2 cos 2
The graph of f lies between the graphs
y cos , we have 2 11t 13t 11t 13t f t cos 11t cos 13t 2 cos cos 2 2 2 cos 12t cos t 2 cos 12t cos t
116. (a) f 1 770 Hz and f 2 1209 Hz, so
of y 2 cos t and y 2 cos t. Thus,
the loudness of the sound varies between y 2 cos t.
(c) 2
y sin 2 770t sin 2 1209t sin 1540t sin 2418t
0 0.005
(b) Using a sum-to-product formula, we have 1540t 2418t 1540t 2418t cos 2 2 2 sin 1979t cos 439t
-2
y 2 sin
117. We find the area of ABC in two different ways. First, let AB be the base and C D
C
be the height. Since B OC 2 we see that C D sin 2. So the area is
1 base height 1 2 sin 2 sin 2. On the other hand, in ABC we see 2 2 that C is a right angle. So BC 2 sin and AC 2 cos , and the area is 1 base height 1 2 sin 2 cos 2 sin cos . Equating the two 2 2
1
A
¬
1
O
D
B
expressions for the area of ABC, we get sin 2 2 sin cos .
7.4
BASIC TRIGONOMETRIC EQUATIONS
1. Because the trigonometric functions are periodic, if a basic trigonometric equation has one solution, it has infinitely many solutions. 2. The basic equation sin x 2 has no solution (because the sine function has range [1 1]), whereas the basic equation sin x 03 has infinitely many solutions. 3. We can find some of the solutions of sin x 03 graphically by graphing y sin x and y 03. The solutions shown are x 97, x 60, x 34, x 03, x 28, x 66, and x 91. 4. (a) To find one solution of sin x 03 in the interval [0 2, we take sin1 to get x sin1 03 030 The other solution in this interval is x sin1 03 284.
(b) To find all solutions, we add multiples of 2 to the solutions in [0 2. The solutions are x 030 2k and x 284 2k.
568
CHAPTER 7 Analytic Trigonometry y 1
5. Because sine has period 2, we first find the solutions in the interval [0 2. From the unit circle shown, we see that sin 23 in quadrants I 2 and II, so the solutions are 3 and 3 . We get all solutions of the
5¹
¬= 3
equation by adding integer multiples of 2 to these solutions: 2 3 2k and 3 2k for any integer k.
y=Ï3/2
_1
0
¹
¬= 3
1
x
6. The sine function is negative in quadrants III and IV, so solutions of sin 22 on the interval [0 2 are 54 and
74 . Adding integer multiples of 2 to these solutions gives all solutions: 54 2k, 74 2k for any integer k. 7. The cosine function is negative in quadrants II and III, so the solution of cos 1 on the interval [0 2 is . Adding integer multiples of 2 to this solution gives all solutions: 2k 2k 1 for any integer k.
8. The cosine function is positive in quadrants I and IV, so the solutions of cos 23 on the interval [0 2 are 6 and 11 116 . Adding integer multiples of 2 to these solutions gives all solutions: 6 2k, 6 2k for any integer k.
9. The cosine function is positive in quadrants I and IV, so the solutions of cos 14 on the interval [0 2 are cos1 41 132 and 2 cos1 41 497. Adding integer multiples of 2 to these solutions gives all solutions: 132 2k, 497 2k for any integer k.
10. The sine function is negative in quadrants III and IV, so the solutions of sin 03 on the interval [0 2 are
sin1 03 345 and 2 sin1 03 598. Adding integer multiples of 2 to these solutions gives all solutions, 345 2k and 598 2k for any integer k.
11. The sine function is negative in quadrants III and IV, so the solutions of sin 045 on the interval [0 2 are
sin1 045 361 and 2 sin1 045 582. Adding integer multiples of 2 to these solutions gives all solutions, 361 2k, 582 2k for any integer k.
12. The cosine function is positive in quadrants I and IV, so the solutions of cos 032 on the interval [0 2 are
cos1 032 125 and 2 cos1 032 504. Adding integer multiples of 2 to these solutions gives all solutions, 125 2k, 504 2k for any integer k. 13. We first find one solution by taking tan1 of each side of the equation: tan1 3 3 . By definition, this is the only solution in the interval 2 2 . Since tangent has period , we get all solutions of the equation by adding integer multiples of : 3 k for any integer k.
14. One solution of tan 1 is tan1 1 4 . Adding integer multiples of to this solution gives all solutions: 4 k for any integer k. 15. One solution of tan 5 is tan1 5 137. Adding integer multiples of to this solution gives all solutions: 137 k for any integer k. 16. One solution of tan 13 is tan1 13 032. Adding integer multiples of to this solution gives all solutions: 032 k for any integer k.
17. One solution of cos 23 is cos1 23 56 and another is 2 56 76 . All solutions are
56 2k, 76 2k for any integer k. Specific solutions include 56 2 76 , 76 2 56 , 56 , 76 , 56 2 176 , and 76 2 196 .
18. One solution of cos 12 is cos1 12 3 and another is 2 3 53 2k for any integer k. Specific solutions include 53 , 3, 3,
5 . All solutions are 2k and 3 3 5 , 7 , and 11 . 3 3 3
SECTION 7.4 Basic Trigonometric Equations
569
3 19. One solution of sin 22 is sin1 22 4 and another is 4 4 . All solutions are 4 2k and 3 9 11 34 2k for any integer k. Specific solutions include 74 , 54 , 4 , 4 , 4 , and 4 . 20. One solution of sin 23 is sin1 23 43 and another is 2 sin1 23 53 . All solutions are 4 5 10 11 43 2k and 53 2k for any integer k. Specific solutions include 23 , 3 , 3 , 3 , 3 , and 3 . 21. One solution of cos 028 is cos1 028 129 and another is 2 cos1 028 500. All solutions are
129 2k and 500 2k for any integer k. Specific solutions include 500, 129, 129, 500, 757, and 1128.
22. One solution of tan 25 is tan1 25 119. All solutions are 119 k for any integer k. Specific solutions include 509, 195, 119, 433, 747, and 1061.
23. One solution of tan 10 is tan1 10 147. All solutions are 147 k for any integer k. Specific solutions include 775, 461, 147, 167, 481, and 795.
24. One solution of sin 09 is sin1 09 426 and another is 2 sin1 09 516. All solutions are 426 2k and 516 2k for any integer k. Specific solutions include 202, 112, 426, 516, 1054, and 1144. 25. cos 1 0 cos 1. In the interval [0 2 the only solution is Thus the solutions are 2k 1 for any integer k. 26. sin 1 0 sin 1. In the interval [0 2 the only solution is 32 . Therefore, the solutions are 32 2k for any integer k. 27. 2 sin 1 0 2 sin 1 sin 1 The solutions in the interval [0 2 are 54 , 74 . Thus the
2 5 7 solutions are 4 2k, 4 2k for any integer k. 7 28. 2 cos 1 0 2 cos 1 cos 1 . The solutions in the interval [0 2 are 4 , 4 . Thus the solutions 2 7 are 4 2k, 4 2k for any integer k. 29. 5 sin 1 0 sin 15 . The solutions in the interval [0 2 are sin1 15 020 and sin1 51 294.
Thus the solutions are 020 2k, 294 2k for any integer k.
30. 4 cos 1 0 cos 14 . The solutions in the interval [0 2 are cos1 14 182 and 2 cos1 14 446. Thus the solutions are 182 2k, 446 2k for any integer k.
31. 3 tan2 1 0 tan2 13 tan 33 . The solutions in the interval 2 2 are 6 , so all solutions are 6 k, 6 k for any integer k.
32. cot 1 0 cot 1. The solution in the interval 0 is 34 . Thus, the solutions are 34 k for any integer k.
3 5 7 33. 2 cos2 1 0 cos2 12 cos 1 4 , 4 , 4 , 4 in [0 2. Thus, the solutions are 4 k, 2
3 k for any integer k. 4
2 4 5 34. 4 sin2 3 0 sin2 34 sin 23 3 , 3 , 3 , 3 in [0 2. Thus, the solutions are 3 k, 2 k for any integer k. 3 35. tan2 4 0 tan2 4 tan 2 tan1 2 111 or tan1 2 111 in 2 2 . Thus, the
solutions are 111 k, 111 k for any integer k.
36. 9 sin2 1 0 sin2 19 sin 13 sin1 31 034, sin1 31 280, sin1 13 348, or 2 sin1 13 594 on [0 2. Thus, the solutions are 034 k, 280 k for any integer k.
570
CHAPTER 7 Analytic Trigonometry
3 5 7 37. sec2 2 0 sec2 2 sec 2. In the interval [0 2 the solutions are 4 , 4 , 4 , 4 . Thus, the solutions are 2k 1 4 for any integer k. 5 7 11 38. csc2 4 0 csc2 4 csc 2. In the interval [0 2 the solutions are 6 , 6 , 6 , 6 . So the 5 solutions are 6 k, 6 k for any integer k. 39. tan2 4 2 cos 1 0 tan2 4 or 2 cos 1. From Exercise 35, we know that the first equation has solutions 111 k, 111 k for any integer k. 2 cos 1 cos 12 has solutions cos1 12 23
and 43 on [0 2, so all solutions are 23 2k and 43 2k for any integer k. Thus, the original equation has solutions 111 k, 111 k, 23 2k, and 43 2k for any integer k. 1 . The first equation has solution tan1 2 111 40. tan 2 16 sin2 1 0 tan 2 or sin2 16 1 1 1 025, on 2 2 . The second equation is equivalent to sin 4 , which has solutions sin 4 sin1 41 289, sin1 41 339, and 2 sin1 41 603 on [0 2.
Thus, the original equation has solutions 111 k, 025 2k, 289 2k, 339 2k, and 603 2k for any integer k. 5 41. 4 cos2 4 cos 1 0 2 cos 12 0 2 cos 1 0 cos 12 3 2k, 3 2k for any
integer k.
42. 2 sin2 sin 1 0 2 sin 1 sin 1 0 2 sin 1 0 or sin 1 0. Since 2 sin 1 0 7 2 sin 12 76 , 116 in [0 2 and sin 1 2 in [0 2. Thus the solutions are 6 2k, 11 2k, 2k for any integer k. 6 2
43. 3 sin2 7 sin 2 0 3 sin 1 sin 2 0 3 sin 1 0 or sin 2 0. Since sin 1,
sin 2 0 has no solution. Thus 3 sin 1 0 sin 13 033984 and 033984 280176 are the solutions in [0 2, and all solutions are 033984 2k, 280176 2k for any integer k. 44. tan4 13 tan2 36 0 tan2 4 tan2 9 0 tan 2 or tan 3 110715, 110715, 124905, 124905 in 2 2 . Since the period for tangent is , the solutions are 124905 k, 110715 k for any integer k.
45. 2 cos2 7 cos 3 0 2 cos 1 cos 3 0 cos 12 or cos 3 (which is inadmissible) 3, 5 . Therefore, the solutions are 2k, 5 2k for any integer k. 3 3 3
46. sin2 sin 2 0 sin 2 sin 1 0 sin 2 (inadmissible) or sin 1. Thus, the solutions are 32 2k for any integer k.
47. cos2 cos 6 0 cos 2 cos x 3 0 cos x 2 or cos x 3, neither of which has a solution. Thus, the original equation has no solution. 48. 2 sin2 5 sin 12 0 sin x 4 2 sin x 3 0 sin x 4 or sin x 32 , neither of which has a solution. Thus, the original equation has no solution. 49. sin2 2 sin 3 sin2 2 sin 3 0 sin 3 sin 1 0 sin 3 0 or sin 1 0. Since
sin 1 for all , there is no solution for sin 3 0. Hence sin 1 0 sin 1 32 2k for any integer k. 50. 3 tan3 tan 3 tan3 tan 0 tan 3 tan2 1 0 tan 0 or 3 tan2 1 0. Now tan 0
5 k and 3 tan2 1 0 tan2 13 tan 1 6 k, 6 k. Thus the solutions are k, 3
k, 5 k for any integer k. 6 6
SECTION 7.5 More Trigonometric Equations
571
7 11 51. cos 2 sin 1 0 cos 0 or sin 12 2 k, 6 2k, 6 2k for any integer k. 52. sec 2 cos 2 0 sec 0 or 2 cos 2 0. Since sec 1, sec 0 has no solution. Thus 7 7 2 cos 2 0 2 cos 2 cos 22 4 4 in [0 2. Thus 4 2k, 4 2k for any
integer k.
53. cos sin 2 cos 0 cos sin 2 0 cos 0 or sin 2 0. Since sin 1 for all , there is no 3 solution for sin 2 0. Hence, cos 0 2 2k, 2 2k 2 k for any integer k.
54. tan sin sin 0 sin tan 1 0 sin 0 or tan 1 0. Now sin 0 when k and tan 1 0 tan 1 34 k. Thus, the solutions are k, 34 k for any integer k.
55. 3 tan sin 2 tan 0 tan 3 sin 2 0 tan 0 or sin 23 . tan 0 has solution 0 on 2 2 and sin 23 has solutions sin1 23 073 and sin1 32 241 on [0 2, so the original equation has solutions k, 073 2k, 241 2k for any integer k.
3 56. 4 cos sin 3 cos 0 cos 4 sin 3 0 cos 0 or sin 34 . cos 0 has solutions 2 , 2 on [0 2, while sin 34 has solutions sin1 43 399 and 2 sin1 43 544 on [0 2. Thus, the
original equation has solutions 2 k, 399 2k, 544 2k for any integer k. sin 70 sin 70 07065 133 sin 2 57. We substitute 1 70 and 1 133 into Snell’s Law to get 2 sin 2 133 2 4495 . sin 1 1 0658, so we substitute 2 90 into Snell’s Law to get 0658 58. The index of refraction from glass to air is 152 sin 90 sin 1 0658 1 411 .
59. (a) F 12 1 cos 0 cos 1 0
(b) F 12 1 cos 025 1 cos 05 cos 05 60 or 120 (c) F 12 1 cos 05 1 cos 1 cos 0 90 or 270
(d) F 12 1 cos 1 1 cos 2 cos 1 180 60. Statement A is true: every identity is an equation. However, Statement B is false: not every equation is an identity. The difference between an identity and an equation is that an identity is true for all values in the domain, whereas an equation may be true only for certain values in the domain and false for others. For example, x 0 is an equation but not an identity, because it is true for only one value of x.
7.5
MORE TRIGONOMETRIC EQUATIONS
1. Using a Pythagorean identity, we calculate sin x sin2 x cos2 x 1 sin x 1 1 sin x 0, whose solutions are x k for any integer k.
2. Using a double-angle formula, we we see that the equation sin x sin 2x 0 is equivalent to the equation sin x 2 sin x cos x 0. Factoring the left-hand side as sin x 1 2 cos x, we see that solving this equation is equivalent to solving the two basic equations sin x 0 and 1 2 cos x 0. 3. 2 cos2 sin 1 2 1 sin2 sin 1 0 2 sin2 sin 1 0 2 sin2 sin 1 0. From Exercise 7.4.42, the solutions are 76 2k, 116 2k, 2 2k for any integer k.
4. sin2 4 2 cos2 sin2 cos2 cos2 4 1 cos2 4 cos2 3 Since cos 1 for all , it follows that the original equation has no solution.
572
CHAPTER 7 Analytic Trigonometry
5. tan2 2 sec 2 sec2 1 2 sec 2 sec2 2 sec 3 0 sec 3 sec 1 0 sec 3 or sec 1. If sec 3, then cos 13 , which has solutions cos1 31 123 and 2 cos1 31 505
on [0 2. If sec 1, then cos 1, which has solution on [0 2. Thus, solutions are 2k 1 , 123 2k, 505 2k for any integer k.
6. csc2 cot 3 1 cot2 cot 3 cot2 cot 2 0 cot 2 cot 1 0 cot 2 or cot 1. If cot 2, then tan 12 , which has solution tan1 12 046 on 2 2 , and if cot 1, then tan 1, which has solution 4 on 2 2 . Thus, solutions are 4 k, 046 k for any integer k. 7. 2 sin 2 3 sin 0 2 2 sin cos 3 sin 0 sin 4 cos 3 0 sin 0 or cos 34 . The first equation has solutions 0, on [0 2, and the second has solutions cos1 34 072 and 2 cos1 34 556 on [0 2. Thus, solutions are k, 072 2k, 556 2k for any integer k.
8. 3 sin 2 2 sin 0 3 2 sin cos 2 sin 0 2 sin 3 cos 1 0 sin 0 or cos 13 . The first equation has solutions 0, on [0 2, and the second has solutions cos1 13 123 and 2 cos1 13 505 on [0 2. Thus, solutions are k, 123 2k, 505 2k for any integer k.
9. cos 2 3 sin 1 1 2 sin2 3 sin 1 2 sin2 3 sin 2 0 sin 2 2 sin 1 0
sin 2 or sin 12 . The first equation has no solution and the second has solutions sin1 21 6 and 5 5 1 1 sin 2 6 on [0 2, so the original equation has solutions 6 2k, 6 2k for any integer k.
10. cos 2 cos2 12 2 cos2 1 cos2 12 cos2 12 cos 22 . Thus, the solutions are 4 k, 3 k for any integer k. 4 11. 2 sin2 cos 1 2 1 cos2 cos 1 0 2 cos2 cos 1 0 2 cos 1 cos 1 0 5 2 cos 1 0 or cos 1 0 cos 12 or cos 1 3 2k, 3 2k, 2k 1 for any integer k.
12. tan 3 cot 0
3 cos sin2 3 cos2 sin2 cos2 4 cos2 sin 0 0 0 cos sin cos sin cos sin
1 4 cos2 2 0 1 4 cos2 0 4 cos2 1 cos 12 3 k, 3 k for any integer k. cos sin 13. sin 1 cos sin cos 1. Squaring both sides, we have sin2 cos2 2 sin cos 1 sin 2 0,
3 which has solutions 0, 2 , , 2 in [0 2. Checking in the original equation, we see that only 2 and are valid. (The extraneous solutions were introduced by squaring both sides.) Thus, the solutions are 2k 1 , 2 2k for any integer k.
14. Square both sides of cos sin 1 to get cos2 sin2 2 sin cos 1 sin 2 0, which has solutions 0, , , 3 on [0 2. Checking in the original equation, we see that only 0 and 3 are valid. Thus, the solutions 2 2 2 are 2k, 3 2k for any integer k. 2
1 sin 1 sin cos 1. Squaring both sides, we have sin2 cos2 2 sin cos 1 cos cos 3 sin 2 0, which has solutions 0, 2 , , 2 on [0 2. Checking in the original equation, we see that only 0
15. tan 1 sec
is valid. Thus, the solutions are 2k for any integer k. 16. 2 tan sec2 4 2 tan 1 tan2 4 tan2 2 tan 3 0 tan 3 tan 1 0 tan 3 or tan 1. The first equation has the solution tan1 3 125 on 2 2 , and the second has solution 4 on 2 2 . Thus, the solutions are 125 k, 4 k for any integer k.
SECTION 7.5 More Trigonometric Equations
573
5 2 5 2 17. (a) 2 cos 3 1 cos 3 12 3 3 , 3 for 3 in [0 2. Thus, solutions are 9 3 k, 9 3 k for any integer k. 5 7 11 13 17 (b) We take k 0, 1, 2 in the expressions in part (a) to obtain the solutions 9 , 9 , 9 , 9 , 9 , 9 in [0 2.
5 5 18. (a) 2 sin 2 1 sin 2 12 2 6 , 6 for 2 in [0 . Thus, solutions are 12 k, 12 k for any integer k.
, 5 , 13 , 17 in [0 2. (b) Take k 0, 1 in the expressions in part (a) to obtain the solutions 12 12 12 12
2 19. (a) 2 cos 2 1 0 cos 2 12 2 23 2k, 43 2k 3 k, 3 k for any integer k. 2 4 5 (b) The solutions in [0 2 are 3, 3 , 3 , 3 .
2 k, 11 2 k for any integer k. 20. (a) 2 sin 3 1 0 2 sin 3 1 sin 3 12 , which has solutions 718 3 18 3
, 11 , 19 , 23 , 31 , 35 . (b) The solutions in [0 2 are 718 18 18 18 18 18 1 k for any integer k. 21. (a) 3 tan 3 1 0 tan 3 1 3 56 k 518 3 3 , 11 , 17 , 23 , 29 , 35 . (b) The solutions in [0 2 are 518 18 18 18 18 18
5 1 5 1 22. (a) sec 4 2 0 sec 4 2 4 3 2k, 3 2k 12 2 k, 12 2 k for any integer k.
, 5 , 7 , 11 , 13 , 17 , 19 , 23 . (b) The solutions in [0 2 are 12 12 12 12 12 12 12 12
23. (a) cos 2 1 0 cos 2 1 2 2k 4k for any integer k.
(b) The only solution in [0 2 is 0. 24. (a) tan 4 3 0 tan 4 3 4 23 k 83 4k for any integer k. (b) There is no solution in [0 2. 25. (a) 2 sin 3 3 0 2 sin 3 3 sin 3 23 3 43 2k, 53 2k 4 6k, 5 6k for any integer k. (b) There is no solution in [0 2. 26. (a) sec 2 cos 2 cos2 2 1 cos 2 1 2 k 2k for any integer k. (b) The only solution in [0 2 is 0. 1 27. (a) sin 2 3 cos 2 tan 2 3 12 tan1 3 062 on 4 4 . Thus, solutions are 062 2 k for any integer k.
(b) The solutions in [0 2 are 062, 219, 376, 533. 1 5 sin 3 1 5 sin2 3 sin 3 55 , which has solutions 28. (a) csc 3 5 sin 3 sin 3 5 5 089, 1 sin1 5 120, and 1 1 1 1 015, 3 sin 5 3 3 sin 5 3 3 5 23 13 sin1 55 194 on 0 23 . Thus, solutions are 015 13 k, 089 13 k for any integer k. (b) The solutions in [0 2 are 015, 089, 120, 194, 224, 298, 329, 403, 434, 508, 539, 613.
29. (a) 1 2 sin cos 2 1 2 sin 1 2 sin2 2 sin2 2 sin 0 2 sin sin 1 0 sin 0 or sin 1 0, , or 2 in [0 2. Thus, the solutions are k, 2 2k for any integer k. (b) The solutions in [0 2 are 0, 2 , .
30. (a) tan 3 1 sec 3 tan 3 12 sec2 3 tan2 3 2 tan 3 1 sec2 3 sec2 3 2 tan 3 sec2 3 2 tan 3 0 3 k for any integer k. Because squaring both sides is an operation that can introduce extraneous solutions, we must check each of the possible solutions in the original equation, and we see that only 23 k are valid solutions.
(b) The solutions in [0 2 are 0, 23 , 43 .
574
CHAPTER 7 Analytic Trigonometry
31. (a) 3 tan3 3 tan2 tan 1 0 tan 1 3 tan2 1 0 tan 1 or 3 tan2 1 tan 1 or 5 tan 1 6 k, 4 k, 6 k for any integer k. 3
5 7 5 11 (b) The solutions in [0 2 are 6, 4, 6 , 6 , 4 , 6 .
32. (a) 4 sin cos 2 sin 2 cos 1 0 2 sin 1 2 cos 1 0 2 sin 1 0 or 2 cos 1 0 5 2 4 sin 12 or cos 12 6 2k, 6 2k, 3 2k, 3 2k for any integer k.
5 2 4 (b) The solutions in [0 2 are 6, 6 , 3 , 3 .
33. (a) 2 sin tan tan 1 2 sin 2 sin tan tan 2 sin 1 0 2 sin 1 tan 1 0
5 3 2 sin 1 0 or tan 1 0 sin 12 or tan 1 6 2k, 6 2k, 4 k for any integer k.
3 5 7 (b) The solutions in [0 2 are 6, 4 , 6 , 4 .
cos sin 1 sin cos2 cos sin sin . cos cos sin sin cos2 Multiplying both sides by the common denominator cos2 sin gives sin2 cos4 sin2 cos2 sin2 cos4 1 cos2 cos2 sin2 cos4 cos2 cos4 sin2 cos2 sin2 1 sin2
34. (a) sec tan cos cot sin
3 2 sin2 1 sin 1 4 k, 4 k for any integer k. Since we multiplied the original 2
equation by cos2 sin (which could be zero) we must check to see if we have introduced extraneous solutions. However, each of the values of does indeed satisfy the original equation. 3 5 7 (b) The solutions in [0 2 are 4, 4 , 4 , 4 .
(b) f x 3 cos x 1; g x cos x 1. f x g x when
35. (a) 4
3 cos x 1 cos x 1 2 cos x 2 cos x 1 x 2k 2k 1 . The points of intersection are
2
2k 1 2 for any integer k.
-6
-4
-2
2
4
6
-2
The points of intersection are approximately 314 2. (b) f x sin 2x 1; gx 2 sin 2x 1. f x g x when
36. (a)
sin 2x 1 2 sin 2x 1 sin 2x 0 x 12 k. The points of intersection are 12 k 1 for any integer k.
2
-6
-4
-2
2
4
6
The points of intersection are approximately 628 1, 471 1, 314 1, 157 1, and 0 1.
SECTION 7.5 More Trigonometric Equations
575
3. f x g x when tan x 3 x 3 k. The intersection points are 3 k 3 for any integer k.
(b) f x tan x; g x
37. (a) 10
-1
1 -10
The point of intersection is approximately 104 173. (b) f x sin x 1; g x cos x. f x g x when sin x 1 cos x
38. (a)
sin x 12 cos2 x sin2 x 2 sin x 1 cos2 x
-6
-4
-2
2
4
6
sin2 x 2 sin x 1 cos2 x 0 2 sin2 x 2 sin x 0
2 sin x sin x 1 0 sin x 0 or sin x 1 x k, 2 2k. However, x k is not a solution when k is even. (The extraneous
-2
solutions were introduced by squaring both sides.) So the solutions are
The points of intersection are approximately 471 0, 314 1, 157 0, and 314 1.
x 2k 1 , 2 2k, and the intersection points are 2k 1, 2 2k 0 for any integer k.
3 5 7 9 11 13 15 39. cos cos 3 sin sin 3 0 cos 3 0 cos 4 0 4 2 , 2 , 2 , 2 , 2 , 2 , 2 , 2 in 3 5 7 9 11 13 15 [0 8 8 , 8 , 8 , 8 , 8 , 8 , 8 , 8 in [0 2. 5 40. cos cos 2 sin sin 2 12 cos 2 12 cos 12 cos 12 3 , 3 in [0 2.
2 41. sin 2 cos cos 2 sin 23 sin 2 23 sin 23 3 , 3 in [0 2.
42. sin 3 cos cos 3 sin 0 sin 3 0 sin 2 0 2 0, , 2, 3, 4 in [0 4 0, 2 , , 3 in [0 2. 2
7 43. sin 2 cos 0 2 sin cos cos 0 cos 2 sin 1 0 cos 0 or sin 12 2, 6 , 3 , 11 in [0 2. 2 6
sin sin 0 sin 0 sin sin 1 cos 0 (and cos 1 ) 2 1 cos 3 sin cos 0 sin 0 or cos 0 0, 2 , 2 in [0 2 ( is inadmissible).
44. tan
45. cos 2 cos 2 2 cos2 1 cos 2 0 2 cos2 cos 3 0 2 cos 3 cos 1 0
2 cos 3 0 or cos 1 0 cos 32 (which is impossible) or cos 1 0 in [0 2. cos sin cos sin 8 sin cos sin cos 8 sin cos sin cos 46. tan cot 4 sin 2 cos sin cos sin sin2 cos2 8 sin2 cos2 1 2 2 sin cos 2 sin 22 12 sin 2 1 . Therefore, 2 4 k 2
k 3 k 3 5 7 or 2 34 k 8 2 or 8 2 . Thus on the interval [0 2 the solutions are 8 , 8 , 8 , 8 , 9 , 11 , 13 , 15 . 8 8 8 8
47. cos 2 cos2 0 2 cos2 1 cos2 0 cos2 1 k for any integer k. On [0 2, the solutions are 0, .
576
CHAPTER 7 Analytic Trigonometry
48. 2 sin2 2 cos 2 2 sin2 2 1 2 sin2 4 sin2 3 sin 23 23 k, 43 k for any 2 4 5 integer k. On [0 2, the solutions are 3, 3 , 3 , 3 .
49. cos 2 cos 4 0 cos 2 2 cos2 2 1 0 cos 2 1 2 cos 2 1 0. The first factor has zeros at
2 4 5 2 0, and the second has zeros at 3 , 3 , 3 , 3 . Thus, solutions of the original equation are are 0, 3 , 3 , , 43 , 53 in [0 2.
50. sin 3 sin 6 0 sin 3 2 sin 3 cos 3 0 sin 3 1 2 cos 3 0. The first factor has zeros at 0, 3,
2 , , 4 , 5 and the second has zeros at , 5 , 7 , 11 , 13 , 17 , so these are all solutions of the original equation 3 3 3 9 9 9 9 9 9
in [0 2. 1 cos . Squaring both sides, we have 51. cos sin 2 sin 2 cos sin 2 2 cos2 sin2 2 sin cos 1 cos 1 2 sin cos 1 cos either 2 sin 1 or cos 0 6, , 5 , 3 in [0 2. Of these, only and 3 satisfy the original equation. 2 6 2 6 2
52. Square both sides of sin cos 12 to get sin2 cos2 2 sin cos 14 1 sin 2 14 sin 2 34 . 1 1 3 115, 1 sin1 3 357, and Thus, the possible solutions in [0 2 are 12 sin1 34 042, 2 2 sin 4 2 4 3 1 sin1 3 429. We find that only valid solutions to the original equation on [0 2 are 115, 357. 2 2 4
53. sin sin 3 0 2 sin 2 cos 0 2 sin 2 cos 0 sin 2 0 or cos 0 2 k or k 2 12 k for any integer k.
54. cos 5 cos 7 0 2 sin 6 sin 0 sin 6 sin 0 sin 6 0 or sin 0 6 k or k 16 k for any integer k.
55. cos 4 cos 2 cos 2 cos 3 cos cos cos 2 cos 3 1 0 cos 0 or cos 3 12 2 or
5 7 11 13 17 2 5 2 3 3 2k, 3 2k, 3 2k, 3 2k, 3 2k, 3 2k 2 k, 9 3 k, 9 3 k for any integer k.
56. sin 5 sin 3 cos 4 2 cos 4 sin cos 4 cos 4 2 sin 1 0 cos 4 0 or sin 12 5 1 5 4 2 k or 6 2k, 6 2k 8 4 k, 6 2k, 6 2k for any integer k. 58. cos x
57. sin 2x x
x 3
1
-1
1 1
-1
The three solutions are x 0 and x 095.
-4
-2
2 -1
The three solutions are x 117, 266, and 294.
SECTION 7.5 More Trigonometric Equations
59. 2sin x x
60. sin x x 3 1
2 -1 2
1 -1
4
The only solution is x 192. 61.
577
The three solutions are x 0 and x 093. 62. cos x 12 e x ex
cos x x2 1 x2
1
1
-1
-1
1
The two solutions are x 071.
1
The only solution is x 0.
x 2x tan u tan 63. With u tan1 x and tan1 2x, we have u 4 tan u 1 1 tan u tan 1 1 x 2x 1 3 17 3 32 4 2 1 2 2 x 2x 1 2x 2x 3x 1 0 x . Because tan x is not one-to-one, 2 2 4 we must check both roots, and find that only
173 is a solution to the original equation. 4
64. With u sin1 x and cos1 x, we have 2u cos 2u 1 cos 2u cos sin 2u sin 1 1 2 sin2 u cos 2 sin u cos u sin 1. Referring to the diagram, this becomes 1 2x 2 x 2x 1 x 2 1 x 2 1 x 2x 3 2x 2x 3 1 x 1.
22002 sin 2 5000 151250 sin 2 32 sin 2 003308 2 189442 or 2 180 189442 17810558 . If 2 189442 , then 094721 , and if 2 17810558 , then 8905279 .
65. We substitute 0 2200 and R 5000 and solve for . So 5000
66. Since 4e3t 0, we have 0 4e3t sin 2t 0 sin 2t 2t 0, , 2, t 0, 12 , 1, 32 , . 67. (a) 10 12 283 sin 23 t 80 283 sin 23 t 80 2 sin 23 t 80 070671. Now
sin 070671 and 078484. If 23 t 80 078484 t 80 456 t 344. Now in the interval [0 2, we have 078484 392644 and 2 078484 549834. If 23 t 80 392644
t 80 2281 t 3081. And if 23 t 80 549834 t 80 3194 t 3994 3994 365 344. So according to this model, there should be 10 hours of sunshine on the 34th day (February 3) and on the 308th day (November 4). (b) Since L t 12 283 sin 23 t 80 10 for t [34 308], the number of days with more than 10 hours of daylight is 308 34 1 275 days.
578
CHAPTER 7 Analytic Trigonometry
. The part of the belt touching the larger 2 2 pulley has length 2 R R and similarly the part
68. (a) First note that
touching the smaller belt has length r. To calculate a and b, we write cot
a b a R cot and b r cot , so 2 R r 2 2
R
a Œ
¬l
Œ
b
R
r r
the length of the straight parts of the belt is 2a 2b 2 R r cot . Thus, the total length of the belt is 2 L and so 2 cot L R r 2 R r cot R r 2 cot 2 2 2 R r L . 2 cot 2 R r L 2778 (b) We plot 2 cot and 45113 in the same 2 R r 242 121 10 viewing rectangle. The solution is 1047 rad 60 . 5 0 0
2 y
69. sin cos x is a function of a function, that is, a composition of trigonometric functions (see Section 2.6). Most of the other equations involve sums, products,
1
differences, or quotients of trigonometric functions. sin cos x 0 cos x 0 or cos x . However, since cos x 1, the only
solution is cos x 0 x 2 k. The graph of f x sin cos x is shown.
_2¹
0
_¹
_1
CHAPTER 7 REVIEW
sin cos 1. sin cot tan sin sin cos
cos
sin2 cos2 sin2 1 sec cos cos cos
2. sec 1 sec 1 sec2 1 tan2 3. cos2 x csc x csc x 1 sin2 x csc x csc x csc x sin2 x csc x csc x sin2 x 4.
1
1 sin2 x
1 sec2 x 1 tan2 x cos2 x
5.
cos2 x tan2 x
6.
1 1 cos x 1 cos2 x sin2 x 1 sec x 1 1 cos x 1 cos x sec x sec x 1 cos x 1 cos x 1 cos x
7.
sin2 x
cos2 x 1 sin x
1 sin x sin x
cos2 x sin2 x
tan2 x sin2 x
cot2 x
1 cot2 x sec2 x cos2 x
cos x cos x cos x sin x 1 1 sec x tan x 1 sin x cos x cos x cos x
8. 1 tan x 1 cot x 1 cot x tan x tan x cot x 2 cot x tan x 2 2
1 cos2 x sin2 x 2 2 sec x csc x cos x sin x cos x sin x
sin x cos x sin x cos x
¹
x
CHAPTER 7
579
sin2 x cos2 x sin2 x 1 cos2 x sin2 x 2 2 cos x 2 sin2 x cos2 x sin x 1 2 10. tan x cot x sec x csc x2 csc2 x sec2 x cos x sin x cos x sin x cos x sin x 9. sin2 x cot2 x cos2 x tan2 x sin2 x
11. 12. 13. 14. 15. 16.
cos2 x
Review
cos2 x
2 sin x cos x 2 sin x cos x 2 sin x sin 2x tan x 1 cos 2x 2 cos x 1 2 cos2 x 1 2 cos2 x cos x cos y sin x sin y cos x cos y sin x sin y cos x y cos x sin y cos x sin y cos x sin y cos x sin y x 1 cos x csc x tan csc x csc x csc x cot x cot x 2 sin x x sin x 1 cos x 1 cos x 1 1 tan x tan 1 1 1 2 cos x sin x cos x cos x cos 2x 2 sin x cos x 2 cos2 x 1 sin 2x 2 cos x 2 cos x sin x cos x sin x cos x tan x tan 1 tan x 4 tan x 4 1 tan x tan 1 tan x 4
cos y sin x cot y tan x sin y cos x
1
1 sec x cos x
1 sec x cos x
1 1 1 1 cos x sec x 1 1 cos x x cos x cos x 17. tan 1 1 sin x sec x cos x sin x 2 sin x sin x cos x cos x
18. cos x cos y2 sin x sin y2 cos2 x 2 cos x cos y cos2 y sin2 x 2 sin x sin y sin2 y cos2 x sin2 x sin2 y cos2 y 2 cos x cos y sin x sin y 2 2 cos x y x x x 2 x x x x x x x x 19. cos sin 1 sin x cos2 2 sin cos sin2 sin2 cos2 2 sin cos 1 sin 2 2 2 2 2 2 2 2 2 2 2 2 3x 7x 3x 7x 2 sin sin 2 sin 5x sin 2x sin 2x cos 3x cos 7x 2 sin 5x sin 2x 2 2 tan 2x 20. 3x 7x 3x 7x sin 3x sin 7x 2 sin 5x cos 2x 2 sin 5x cos 2x cos 2x cos 2 sin 2 2 x y x y x y x y 2 sin cos 2 sin x cos y sin x sin x y sin x y 2 2 tan x 21. x y x y x y x y cos x y cos x y 2 cos x cos y cos x cos 2 cos 2 2 1 22. sin x y sin x y 2 cos x y x y cos x y x y 12 cos 2y cos 2x 12 1 2 sin2 y 1 2 sin2 x 12 2 sin2 x 2 sin2 y sin2 x sin2 y
2 23. (a) f x 1 cos x2 sin x2 , g x sin x
5 -1
prove this, expand f x and simplify, using the double-angle formula for sine:
1
-5
(b) The graphs suggest that f x g x is an identity. To
2 f x 1 cos x2 sin x2 1 cos2 x2 2 cos x2 sin x2 sin2 2x 1 2 cos x2 sin x2 cos2 x2 sin2 2x 1 sin x 1 sin x g x
580
CHAPTER 7 Analytic Trigonometry
24. (a) f x sin x cos x, g x 2
sin2 x cos2 x
-5
5
(b) The graphs suggest that f x g x in general. For example, choose x 6 and evaluate the functions: 1 3 f 6 2 2 12 3 , whereas 1 3 g 6 4 4 1 1, so f x g x.
-2
25. (a) f x tan x tan x2 , g x
1 cos x
(b) The graphs suggest that f x g x in general. For example, choose x 3 and evaluate: f 3 tan 3
5
-5
tan 6
5
1 3 1 1, whereas g 3 1 2, so 3
2
f x g x.
-5
26. (a) f x 1 8 sin2 x 8 sin4 x, g x cos 4x 1
-5
5
(b) The graphs suggest that f x g x is an identity. To
show this, expand g x by using double-angle identities: g x cos 4x cos 2 2x 1 2 sin2 2x 1 2 2 sin x cos x2 1 2 4 sin2 x cos2 x 1 8 sin2 x 1 sin2 x
-1
1 8 sin2 x 8 sin4 x f x
27. (a) f x 2 sin2 3x cos 6x
(b) The graph suggests that f x 1 for all x. To prove this, we use the double angle formula to note that
1
-5
5
cos 6x cos 2 3x 1 2 sin2 3x, so f x 2 sin2 3x 1 2 sin2 3x 1.
-1
x 28. (a) f x sin x cot , g x cos x 2
(b) Proof: f x sin x cot x2 sin x
2
-5
Conjecture: f x g x 1
5
2 sin x2 cos x2
cos x2 sin x2
cos x2 sin x2
Now subtract and add 1, so f x 2 x cos2 1 1 cos x 1 g x 1. 2
2 cos2 x2
CHAPTER 7
Review
581
29. 4 sin 3 0 4 sin 3 sin 34 sin1 34 08481 or sin1 34 22935. 30. 5 cos 3 0 5 cos 3 cos 35 cos1 35 22143 or 2 cos1 35 40689.
31. cos x sin x sin x 0 sin x cos x 1 0 sin x 0 or cos x 1 x 0, or x 0. Therefore, the solutions are x 0 and . 5 32. sin x 2 sin2 x 0 sin x 1 2 sin x 0 sin x 0 or sin x 12 x 0, or x 6 , 6 . Therefore, the 5 solutions in [0 2 are x 0, 6 , 6 , .
33. 2 sin2 x 5 sin x 2 0 2 sin x 1 sin x 2 0 sin x 12 or sin x 2 (which is inadmissible) x 6, 5 . Thus, the solutions in [0 2 are x and 5 . 6 6 6 34. sin x cos x tan x 1 sin x cos x cos2 x sin x cos x sin x cos x sin x cos2 x cos x 0 sin x cos x 1 cos x cos x 1 0 sin x cos x cos x 1 0 sin x cos x or cos x 1 tan x 1 or 5 5 cos x 1 x 4 , 4 or x 0. Therefore, the solutions in [0 2 are x 0, 4 , 4 .
35. 2 cos2 x 7 cos x 3 0 2 cos x 1 cos x 3 0 cos x 12 or cos x 3 (which is inadmissible) x 3, 5 . Therefore, the solutions in [0 2 are x , 5 . 3 3 3 36. 4 sin2 x 2 cos2 x 3 2 sin2 x 2 sin2 x cos2 x 3 0 2 sin2 x 2 3 0 2 sin2 x 1 sin x 1 . 2
3 5 7 So the solutions in [0 2 are x 4, 4 , 4 , 4 .
37. Note that x is not a solution because the denominator is zero. 4 cos x 2 cos x 12 x 23 , 43 in [0 2.
1 cos x 3 1 cos x 3 3 cos x 1 cos x
38. sin x cos 2x sin x 1 2 sin2 x 2 sin2 x sin x 1 0 2 sin x 1 sin x 1 0 sin x 12 or 5 3 3 5 sin x 1 x 6 , 6 or x 2 . Thus, the solutions in [0 2 are x 6 , 2 , 6 . 39. Factor by grouping: tan3 x tan2 x 3 tan x 3 0 tan x 1 tan2 x 3 0 tan x 1 or tan x 3 2 4 5 2 3 4 5 7 x 34 , 74 or x 3 , 3 , 3 , 3 . Therefore, the solutions in [0 2 are x 3 , 3 , 4 , 3 , 3 , 4 . 40. cos 2x csc2 x 2 cos 2x cos 2x csc2 x 2 cos 2x 0 cos 2x csc2 x 2 0 cos 2x 0 or csc2 x 2
cos 2x 0or sin2 x 12 cos 2x 0 or sin x 1 . 2
3 5 7 3 5 7 For cos 2x 0, the solutions in [0 4 are 2x 2 , 2 , 2 , 2 the solutions in [0 2 are x 4 , 4 , 4 , 4 . 3 5 7 For sin x 1 , the solutions in [0 2 are x 4, 4 , 4 , 4 . 2
3 5 7 Thus, the solutions of the equation in [0 2 are x 4, 4 , 4 , 4 . 1 cos x 1 4 sin x cos x 1cos x 4 sin2 x cos x 1 4 sin2 x cos x cos x 0 41. tan 12 x 2 sin 2x csc x sin x sin x 3 5 7 11 cos x 4 sin2 x 1 0 cos x 0 or sin x 12 x 2 , 2 or x 6 , 6 , 6 , 6 . Thus, the solutions in
5 7 3 11 [0 2 are x 6, 2, 6 , 6 , 2 , 6 . 42. cos 3xcos 2xcos x 0 cos 2x cos xsin 2x sin xcos 2xcos x 0 cos 2x cos xcos 2xsin 2x sin xcos x 0 cos 2x cos x 1 2 sin2 x cos x cos x 0 cos 2x cos x 1 cos x 1 2 sin2 x 0
cos 2x cos x 1 cos x cos 2x 0 cos 2x cos x 1 cos x 0 cos 2x 2 cos x 1 0 cos 2x 0 or
3 5 7 2 4 cos x 12 2x 2 , 2 , 2 , 2 (in [0 4) or x 3 , 3 (in [0 2). Thus, the solutions in [0 2 are x 4 , 2 , 3 , 5 , 4 , 7 . 3 4 4 3 4
582
CHAPTER 7 Analytic Trigonometry
1 sin x 3 sin x 1 3 cos x 3 cos x sin x 1 23 cos x 12 sin x 12 3 cos x cos x sin x 1 cos x 1 x , 5 x , 3 . However, x 3 is cos cos x sin 6 6 2 6 2 6 3 3 6 2 2 inadmissible because sec 32 is undefined. Thus, the only solution in [0 2 is x . 6 sin x 2 0 2 cos x 3 sin x 0 (cos x 0) 2 1 sin2 x 3 sin x 0 44. 2 cos x 3 tan x 0 2 cos x 3 cos x 2 sin2 x 3 sin x 2 0 2 sin x 1 sin x 2 0 sin x 12 or sin x 2 (which has no solution) x 6,
43. tan x sec x
5 . 6
45. We graph f x cos x and g x x 2 1 in the viewing 46. We graph f x esin x and g x x in the viewing rectangle [0 65] by [2 2]. The two functions intersect
rectangle [0 65] by [1 3]. The two functions intersect
at only one point, x 118.
at only one point, x 222.
2 2 0 2
4
6 0 2
-2
4
6
4002 sin2 sin2 08 sin 08944 634 64 4002 sin2 2500 sin2 2500. Therefore it is impossible for the projectile to reach a height of 3000 ft. (b) 64 (c) The function M 2500 sin2 is maximized when sin2 1, so 90 . The projectile will travel the highest when it is shot straight up. k 48. Since e02t 0 we have f t e02t sin 4t 0 sin 4t 0 4t k t where k is any integer. Thus 4 the shock absorber is at equilibrium position every quarter second. 1 cos 30 2 3 2 3 49. Since 15 is in quadrant I, cos 15 . 2 4 2
47. (a) 2000
2 , which is Another method: cos 15 cos 45 30 cos 45 cos 30 sin 45 sin 30 22 23 22 12 6 4 equal to 12 2 3. 3 5 1 1 cos 2 3 2 3 2 6 is in quadrant I, sin 5 . 50. Since 512 12 2 2 4 2 sin sin cos cos sin 1 3 1 1 31 6 2 , which is Another method: sin 512 4 6 4 6 4 6 2 2 4 2 2 2 2 1 equal to 2 2 3.
51. tan 8
1 1 1 cos 2 4 1 1 2 21 1 2 sin 4 2
cos sin 2 sin 1 52. 2 sin 12 12 12 6 2
53. sin 5 cos 40 cos 5 sin 40 sin 5 40 sin 45 1 22 2 tan 66 tan 6 54. tan 66 6 tan 60 3 1 tan 66 tan 6
CHAPTER 7
Review
583
2 cos 2 cos 1 2 55. cos2 sin 8 8 8 4 2 2 3 sin sin cos cos sin sin sin 1 2 56. 12 cos 12 2 12 6 12 6 12 6 12 4 2 2 57. We use a product-to-sum formula: cos 375 cos 75 12 cos 45 cos 30 12 22 23 14 2 3 .
1 1 cos 45 675 225 675 225 cos 2 cos 45 cos 225 2 2 2 2 2 1 cos 45 1 1 22 2
58. cos 675 cos 225 2 cos
2
In Exercises 59–64, x and y are in quadrant I, so we know that sec x 32 cos x 23 , so sin x 35 and tan x 25 .
1 2. Also, csc y 3 sin y 13 , and so cos y 2 3 2 , and tan y 4 2 2 59. sin x y sin x cos y cos x sin y 35 2 3 2 23 13 29 1 10 .
60. cos x y cos x cos y sin x sin y 23 2 3 2 35 13 19 4 2 5 .
5 2 5 2 2 2 5 2 8 10 tan x tan y 8 2 5 61. tan x y 23 2 4 2 4 2 2 1 tan x tan y 8 8 10 8 10 1 25 1 25 4 4
62. sin 2x 2 sin x cos x 2 35 23 4 9 5 . 1 2 2 y 1 cos y 32 2 3 63. cos (since cosine is positive in quadrant I) 2 2 2 6 1 232 1 cos y 32 2 y 32 2 64. tan 1 2 sin y 1 3
65. We sketch a triangle such that cos1 73 . We see that tan 2 310 , and the double-angle formula for tangent gives 4 10 2 2 310 2 tan 12 10 3 tan 2 . 2 31 1 tan2 1 40 9 1 2 310
7 ¬
sin sin cos cos sin 5 4 12 63 35 13 5 13 65
67. The double-angle formula for tangent gives tan 2 tan1 x
3
cos 37
5 . From the 66. We sketch triangles such that tan1 43 and cos1 13
triangles, we see that sin 35 , cos 45 , and sin 12 13 , so the addition formula for sine gives
2Ï10
5
¬
3
tan 34
2 tan tan1 x 2x . 1 tan2 tan1 x 1 x2
4
13 12
5 cos 13
ú
5
584
CHAPTER 7 Analytic Trigonometry
68. Let sin1 x and cos1 y. From the triangles, cos 1 x 2 and sin 1 y 2 , so using the addition formula for cosine, we have cos cos cos sin sin 1 x 2 y x 1 y 2 y 1 x 2 x 1 y2
10 10 tan1 x x 10 , for x 0. Since the road sign can first be seen when 2 , (b) tan1 x 10 10 x we have 2 tan1 2864 ft. Thus, the sign can first be x tan 2
69. (a) tan
1
1
x
ú
¬
Ï1-x@
y
cos y
sin x 2 1 0 20
seen at a height of 2864 ft.
40
70. (a) Let be the angle formed by the top of the tower, the car, and the base of the
40
building, and let be the angle formed by base of the tower, the car, and the base 420 420 of the building, as shown in the diagram. Then tan tan1 x x 380 380 and tan tan1 . Thus, x x 420 380 tan1 tan1 . x x (b) We graph x and find that is maximized when x 400.
Ï1-y@
380
¬
º x
0.055 0.050 0.045 300
400
500
CHAPTER 7 TEST 1. tan sin cos
sin sin2 cos2 1 sin cos sec cos cos cos cos
sin x 1 cos x tan x 1 cos x tan x 1 cos x 1 cos x 1 tan x cos x csc x 1 sec x 2. 2 1 cos x 1 cos x 1 cos x sin x cos x 1 cos2 x sin x 2 tan x 2 tan x 2 sin x cos2 x 2 sin x cos x sin 2x 2 2 cos x 1 tan x sec x x 1 cos x sin x 1 cos x 4. sin x tan 2 sin x 3.
5. 2 sin2 3x 1 cos 2 3x 1 cos 6x
6. cos 4x 1 2 sin2 2x 1 2 2 sin x cos x2 1 8 sin2 x 1 sin2 x 1 8 sin2 x 8 sin4 x
CHAPTER 7
Test
585
2 x 2 x 1 cos x 1 cos x 1 cos x 1 cos x 1 cos2 x cos 2 1 sin x 7. sin 2 2 2 2 2 2 4 x x x x 2 x x x Another method: sin sin2 cos 2 sin cos cos2 1 2 sin 1 sin x 2 2 2 2 2 2 2 sin 2 sin 2 sin x 2 sin sin sin tan (because 8. 2 2 2 cos cos cos 4 x2 4 4 sin 2 1 sin 4 2 sin 2 cos 0 for 2 2)
9. (a) sin 8 cos 22 cos 8 sin 22 sin 8 22 sin 30 12
(b) sin 75 sin 45 30 sin 45 cos 30 cos 45 sin 30 22 23 22 12 14 6 2 1 3 1 cos 150 2 3 2 3 2 Another method: Since 75 is in quadrant I, sin 75 , 2 2 4 2 which is equal to 14 6 2 . 3 1 cos 1 2 3 6 2 12 2 3 (c) sin 12 2 2 4
is in quadrant I, Another method: Since 12 sin sin cos cos sin 3 2 1 2 1 6 2 , which is equal to sin 12 3 4 3 4 3 4 2 2 2 2 4 1 2 3. 2
52 102 5 . 10. From the figures, we have cos cos cos sin sin 2 35 1 23 2 15 5 5 3 5
11. sin 3x cos 5x 12 [sin 3x 5x sin 3x 5x] 12 sin 8x sin 2x 2x 5x 7x 3x 2x 5x sin 2 cos sin 12. sin 2x sin 5x 2 cos 2 2 2 2
1 35 1 35 1 cos 53 2. 13. sin 45 . Since is in quadrant III, cos 35 . Then tan 4 4 2 sin 4 5 5 14. 3 sin 1 0 3 sin 1 sin 13 sin1 13 034 or sin1 31 280 on [0 2. 15. 2 cos 1 sin 1 0 cos 12 or sin 1. The first equation has solutions 3 105 and 5 3 524 on [0 2, while the second has the solution 2 157. 16. 2 cos2 5 cos 2 0 2 cos 1 cos 2 0 cos 12 or cos 2 (which is impossible). So in the interval [0 2, the solutions are 23 209, 43 419.
3 17. sin 2 cos 0 2 sin cos cos 0 cos 2 sin 1 0 cos 0 or sin 12 2 , 2 or 5 5 3 6 , 6 . Therefore, the solutions in [0 2 are 6 052, 2 157, 6 262, 2 471.
18. 5 cos 2 2 cos 2 25 2 cos1 04 1159279. The solutions in [0 4 are 2 1159279, 2 1159279, 2 1159279, 4 1159279 2 1159279, 5123906, 7442465, 11407091 057964, 256195, 372123, 570355 in [0 2. 19. 2 cos2 x cos 2x 0 2 cos2 x 2 cos2 x 1 0 cos x 12 . The solutions in [0 4 are x 3 105, x 23 209, x 43 419, and x 53 524.
586
FOCUS ON MODELING
20. 2 tan
x 2
csc x 0 2
x 53 524.
1 cos x sin x
1 0 cos x 12 . The solutions in [0 4 are x 3 105 and sin x
9 so tan u 9 . From the triangle, cos u 40 , so using a 21. Let u tan1 40 40 41 2 1 1519 double-angle formula for cosine, cos 2u 2 cos2 u 1 2 40 41 1681 .
41 u
22. We sketch triangles such that cos1 x and tan1 y. From the y 1 triangles, we have sin 1 x 2 , sin , and cos , 2 1y 1 y2
1 ¬
sin sin cos cos sin
1 y 1 x2 x 2 1y 1 y2
40
Ï1-x@
Ï1+y@ ú
x
so the addition formula for sine gives
9
cos x
y
1
tan y
1 x2 x y 1 y2
FOCUS ON MODELING Traveling and Standing Waves 1. (a) Substituting x 0, we get y 0 t 5 sin 2 0 2 t 5 sin 2 t 5 sin 2 t. (b)
y
6
t=0
0.8
1.6
4 2
x
¹ ¹/2
_2
3¹/2
_4 t=0.4
_6
1.2
(c) We express the function in the standard form y x t A sin k x t: y x t 5 sin 2x 2 t 5 sin 2 x 4 t . Comparing this to the standard form, we see that the velocity of the wave is 4. 2. (a) y 02 sin 1047x 0524t 02 sin 1047x 0524t 2 02 sin 1047x cos 0524t 04 sin 1047x cos 0524t. m 3m. So the nodes are at 3, 6, 9, 12, . The nodes occur when 1047x m x 1047 (b)
y
t=0
0.4
t=1
0.2
_0.2 _0.4
t=2 1
2
3
4
t=3l t=4
5
6
x
t=5 t=6
Note that when t 3, cos 0524 3 0001, so y x 3 00004. Thus, the graph of y x 3 cannot be distinguished from the x-axis in the diagram. Yes, this is a standing wave. 068. Since 6, we have 3. From the graph, we see that the amplitude is A 27 and the period is 92, so k 292 6 410, so the equation we seek is y x t 27 sin 068x 410t. k 292
Traveling and Standing Waves
587
4. (a) We are given A 5, period 23 , and 05. Since the period is 23 , we have k 223 3. Thus, expressing the function in standard form, we have y x t 5 sin 3 x 05t 5 sin 3x 15t. (b)
y
t=0
1
2
4 2
_2
x
¹ ¹/2
3¹/2
_4 t=0.5
1.5
5. From the graphs, we see that the amplitude is A 06. The nodes occur at x 0, 1, 2, 3. Since sin x 0 when x k (k any integer), we have . Then since the frequency is 2, we get 20 2 40. Thus, an equation for this model is f x t 06 sin x cos 40t. 6. From the graph, we see that the amplitude is A 7. Now sin x 0 when x k (k an integer). So for k 1, we must 1 have 2 2. Then since the period is 4, we have 2 4 2 . Thus, an equation for this model is f x t 7 sin 2x cos 12 t.
7. (a) The first standing wave has 1, the second has 2, the third has 3, and the fourth has 4.
(b) is equal to the number of nodes minus 1. The first string has two nodes and 1; the second string has three nodes and 2, and so forth. Thus, the next two values of would be 5 and 6, as sketched below.
(c) Since the frequency is 2, we have 440 2 880.
(d) The first standing wave has equation y sin x cos 880t, the second has equation y sin 2x cos 880t, the third has equation y sin 3x cos 880t, and the fourth has equation y sin 4x cos 880t.
8. (a) The nodes of the tube occur when cos 12 x 0 and 0 x 377. So 12 x 2k 1 2 x 2k 1 . Thus, the nodes are at x , 3, 5, 7, 9, and 11. We stop there since 13 377. Note that the endpoints of the tube (x 0 and x 377) are not nodes. (b) In the function y A cos x cos t, the frequency is 2. In this case, the frequency is 50 2 25 Hz.
8
POLAR COORDINATES AND PARAMETRIC EQUATIONS
8.1
POLAR COORDINATES
1. We can describe the location of a point in the plane using different coordinate systems. The point P shown in the figure has rectangular coordinates 1 1 and polar coordinates 2 4 .
2. (a) If a point P in the plane has polar coordinates r then it has rectangular coordinates x y where x r cos and y r sin . y (b) If P has rectangular coordinates x y then it has polar coordinates r where r 2 x 2 y 2 and tan . x 7 3. Yes; both 2 3 1 in Cartesian coordinates. 6 and 2 6 correspond to the point 4. No; adding a multiple of 2 to gives the same point, as does adding an odd multiple of and reversing the sign of r .
5.
O
8.
6.
(4, ¹4 )
O
(1, 0)
7.
(6, _ 7¹ 6 ) O
¹ 4
7¹
_ 6 O
9. 2¹ _ 3
(_2, 4¹ 3) 4¹ 3
(3, _ 2¹ 3 )
(3, ¹2 )
5 13. 1 76 has polar coordinates 1 6 or 1 6 .
O
O
1
_ 17¹ 6
12. 2 34 has polar coordinates 2 114 or 2 74 .
(2, 3¹ 4 ) O
1
(_1, 7¹ 6 )
(_5, _ 17¹ 6 )
O
Answers to Exercises 11–16 will vary. 5 or 3 3 11. 3 has polar coordinates 3 2 2 2
O
10.
1
2 5 14. 2 3 has polar coordinates 2 3 or 2 3 .
(_2, _ ¹3 ) O
1
589
590
CHAPTER 8 Polar Coordinates and Parametric Equations
15. 5 0 has polar coordinates 5 or 5 2.
16. 3 1 has polar coordinates 3 1 2 or 3 1 . (3, 1)
(_5, 0)
O
1
O
1
17. Q has coordinates 4 34 . 3 19. Q has coordinates 4 4 4 4 . 21. P has coordinates 4 234 4 4 . 4 5 4 . 23. P has coordinates 4 101 4 4 4 25.
26. 27.
28.
18. R has coordinates 4 34 4 54 . 20. P has coordinates 4 134 4 54 4 4 . 22. Q has coordinates 4 234 4 4 4 34 . 4 7 . 24. S has coordinates 4 103 4 4 P 3 3 in rectangular coordinates, so r 2 x 2 y 2 32 32 18 and we can take r 3 2. y 3 tan 1, so since P is in quadrant 2 we take 34 . Thus, polar coordinates for P are 3 2 34 . x 3 Q 0 3 in rectangular coordinates, so r 3 and 32 . Polar coordinates for Q are 3 32 . Here r 5 and 23 , so x r cos 5 cos 23 52 and y r sin 5 sin 23 5 2 3 . R has rectangular coordinates 52 5 2 3 . r 2 and 56 , so S has rectangular coordinates r cos r sin 2 cos 56 2 sin 56 3 1 .
4 3 23 and y r sin 4 sin 4 1 2. Thus, the rectangular . So x r cos 4 cos 29. r 4 6 6 2 6 2 coordinates are 2 3 2 . 30. r 6 23 . So x r cos 6 cos 23 6 12 3 and y r sin 6 sin 23 6 23 3 3. Thus, the rectangular coordinates are 3 3 3 . 2 2 cos 2 1 1, and 31. r 4 . So x r cos 4 2 1 y r sin 2 sin 1. Thus, the rectangular coordinates are 1 1. 2 4 2 32. r 1 52 . So x r cos 1 cos 52 1 0 0 and y r sin 1 sin 52 1 1 1. Thus, the
rectangular coordinates are 0 1.
33. r 5 5. So x r cos 5 cos 5 5, and y r sin 5 sin 5 0. Thus, the rectangular coordinates are 5 0. 34. r 0 13. So x y 0 0 because r 0. 35. r 6 2 116 . So x r cos 6 2 cos 116 3 6 and y r sin 6 2 sin 116 3 2. Thus, the rectangular coordinates are 3 6 3 2 . 3 53 . So x r cos 3 cos 53 23 and y r sin 3 sin 53 32 . Thus, the 36. r rectangular coordinates are 23 32 .
SECTION 8.1 Polar Coordinates
591
1 y 1, so, since 37. x y 1 1. Since r 2 x 2 y 2 , we have r 2 12 12 2, so r 2. Now tan x 1 the point is in the second quadrant, 34 . Thus, polar coordinates are 2 34 . 2 38. x y 3 3 3 . Since r 2 x 2 y 2 , we have r 2 3 3 32 36, so r 6. Now y 1 3 , so, since the point is in the fourth quadrant, 116 . Thus, polar coordinates are 6 116 . tan 3 3 x 3 2 2 y 8 8 . Since r 2 x 2 y 2 , we have r 2 8 8 16, so r 4. Now tan 8 1, so, 39. x y 8 x . . Thus, polar coordinates are 4 since the point is in the first quadrant, 4 4 2 2 2 2 2 2 40. x y 6 2 . Since r x y , we have r 6 2 8, so r 2 2. Now 2 y tan 1 , so, since the point is in the third quadrant, 76 . Thus, polar coordinates are 2 2 76 . 3 x 6 y 41. x y 3 4. Since r 2 x 2 y 2 , we have r 2 32 42 25, so r 5. Now tan 43 , so, since the point is in x the first quadrant, tan1 34 . Thus, polar coordinates are 5 tan1 43 .
2 y 2, and 5. Now, tan x 1 since the point is in the fourth quadrant, 2 tan1 2 (since we need 0 2). Thus, polar coordinates are 5 2 tan1 2 . y 43. x y 6 0. r 2 62 36, so r 6. Now tan 0, so since the point is on the negative x-axis, . x Thus, polar coordinates are 6 . 44. x y 0 3 . r 3 and since the point is on the negative y-axis, 32 . Thus, polar coordinates are 3 32 . 42. x y 1 2. Since r 2 x 2 y 2 , we have r 2 12 22 2, so r
45. x y r cos r sin tan 1, and so 4.
46. x 2 y 2 9. By substitution, r cos 2 r sin 2 9 r 2 cos2 sin2 9 r 2 9 r 3.
47. y x 2 . We substitute and then solve for r: r sin r cos 2 r 2 cos2 sin r cos2 sin tan sec . r cos2 5 5 csc . 48. y 5. By substitution, r sin 5 r sin 4 4 sec . 49. x 4. We substitute and then solve for r: r cos 4 r cos 50. x 2 y 2 1. By substitution, r cos 2 r sin 2 1 r 2 cos2 sin2 1 r 2 cos 2 1 r2
1 sec 2. cos 2
51. r 7. But r 2 x 2 y 2 , so x 2 y 2 r 2 49. Hence, the equivalent equation in rectangular coordinates is x 2 y 2 49. 52. r 3 x 2 y 2 r 2 9, so an equivalent equation in rectangular coordinates is x 2 y 2 9.
53. 2 cos 0, so an equivalent equation in rectangular coordinates is x 0. y 54. tan 0 0 y 0. x 55. r cos 6. But x r cos , and so x 6 is an equivalent rectangular equation. 2 56. r 2 csc r r sin 2. But r sin y, so y 2 is an equivalent rectangular equation. sin
592
CHAPTER 8 Polar Coordinates and Parametric Equations
57. r 4 sin r 2 4r sin . Thus, x 2 y 2 4y is an equivalent rectangular equation. Completing the square, it can be written as x 2 y 22 4.
58. r 6 cos r 2 6r cos . By substitution, x 2 y 2 6x x 2 6x 9 y 2 9 x 32 y 2 9.
59. r 1 cos . If we multiply both sides of this equation by r we get r 2 r r cos . Thus r 2 r cos r, and squaring 2 2 both sides gives r 2 r cos r 2 , or x 2 y 2 x x 2 y 2 in rectangular coordinates.
60. r 3 1 sin 3 3 sin r 2 3r 3r sin r 2 3r sin 3r. Squaring both sides gives 2 2 r 2 3r sin 9r 2 , or x 2 y 2 3y 9 x 2 y 2 in rectangular coordinates.
61. r 1 2 sin . If we multiply both sides of this equation by r we get r 2 r 2r sin . Thus r 2 2r sin r, and 2 2 squaring both sides gives r 2 2r sin r 2 , or x 2 y 2 2y x 2 y 2 in rectangular coordinates. 62. r 2 cos r 2 2r r cos r 2 r cos 2r 2 x 2 y2 x 4 x 2 y2 1 sin cos y x 1.
63. r
2 2 r 2 r cos 2r2 r 2 r cos 4r 2
r sin cos 1 r sin r cos 1, and since r cos x and r sin y, we get
1 r 1 sin 1 r r sin 1. Thus r 1 r sin , and squaring both sides gives 1 sin r 2 1 r sin 2 x 2 y 2 1 y2 1 2y y 2 x 2 2y 1 0.
64. r
4 r 1 2 sin 4 r 2r sin 4. Thus r 4 2r sin . Squaring both sides, we get 1 2 sin r 2 4 2r sin 2 . Substituting, x 2 y 2 4 2y2 x 2 y 2 16 16y 4y 2 x 2 3y 2 16y 16 0.
65. r
2 r 1 cos 2 r r cos 2 r 2 r cos . Squaring both sides, we get r 2 2 r cos 2 . 1 cos Substituting, x 2 y 2 2 x2 4 4x x 2 y 2 4x 4. y y 67. r 2 tan . Substituting r 2 x 2 y 2 and tan , we get x 2 y 2 . x x 2 68. r 2 sin 2 2 sin cos r 4 2r 2 sin cos 2 r cos r sin . By substitution, x 2 y 2 2x y 66. r
x 4 2x 2 y 2 y 4 2x y 0.
y 2 2 2 2 69. sec 2 cos 12 3 tan 3 x 3 y 3x y 3x y 3x 0. y 70. cos 2 1 means that 2 0 0 tan 0 0 y 0 x 71. (a) In rectangular coordinates, the points r1 1 and r2 2 are x1 y1 r1 cos 1 r1 sin 1 and x2 y2 r2 cos 2 r2 sin 2 . Then, the distance between the points is D x1 x2 2 y1 y2 2 r1 cos 1 r2 cos 2 2 r1 sin 1 r2 sin 2 2 r12 cos2 1 sin2 1 r22 cos2 2 sin2 2 2r1r2 cos 1 cos 2 sin 1 sin 2
r12 r22 2r1r2 cos 2 1
SECTION 8.2 Graphs of Polar Equations
593
(b) The distance between the points 3 34 and 1 76 is D
340 32 12 2 3 1 cos 76 34 9 1 6 cos 512
72. (a) Because streets are laid out in a grid, rectangular coordinates are more appropriate when giving directions to a taxi driver. (b) Descartes’ famous declaration cogito ergo sum does not apply to pigeons, so polar coordinates are probably more useful in this case.
8.2
GRAPHS OF POLAR EQUATIONS
1. To plot points in polar coordinates we use a grid consisting of circles centered at the pole and rays emanating from the pole. 2. (a) To graph a polar equation r f we plot all the points r that satisfy the equation.
(b) The graph of the polar equation r 3 is a circle with radius 3 centered at the pole. The graph of the polar equation 4 is a line passing through the pole with slope 1.
3
O
3. VI
4. III
5. II
O
6. IV
7. I
8. V
9. Polar axis: 2 sin 2 sin r, so the graph is not symmetric about the polar axis. Pole: 2 sin 2 sin cos cos sin 2 sin 2 sin r, so the graph is not symmetric about the pole. Line 2 : 2 sin 2 sin cos cos sin 2 sin r, so the graph is symmetric about 2 . 10. Polar axis: 4 8 cos 4 8 cos r, so the graph is symmetric about the polar axis. Pole: 4 8 cos 4 8 cos cos sin sin 4 8 cos r, so the graph is not symmetric about the pole. Line 2 : 4 8 cos 4 8 cos cos sin sin 4 8 cos r, so the graph is not symmetric about 2.
11. Polar axis: 3 sec 3 sec r, so the graph is symmetric about the polar axis. 3 1 3 Pole: 3 sec 3 sec r , so the graph is not symmetric cos cos cos sin sin cos about the pole. 1 3 3 Line 2 : 3 sec cos cos cos sin sin cos 3 sec r, so the graph is not symmetric about 2.
594
CHAPTER 8 Polar Coordinates and Parametric Equations
12. Polar axis: 5 cos csc 5 cos csc r, so the graph is not symmetric about the polar axis. 1 Pole: 5 cos csc 5 cos sin 1 1 5 cos cos sin sin 5 cos 5 cos csc r , sin cos cos sin sin so the graph is symmetric about the pole. 1 Line 2 : 5 cos csc 5 cos sin 1 1 5 cos cos sin sin 5 cos 5 cos csc r , sin cos cos sin sin so the graph is not symmetric about 2 . 4 4 r, so the graph is not symmetric about the polar axis. 3 2 sin 3 2 sin 4 4 4 4 Pole: r, so the graph is not 3 2 sin 3 2 sin cos cos sin 3 2 sin 3 2 sin symmetric about the pole. 4 4 4 Line 2 : 3 2 sin 3 2 sin cos cos sin 3 2 sin r, so the graph is symmetric about 2.
13. Polar axis:
5 5 r , so the graph is symmetric about the polar axis. 1 3 cos 1 3 cos 5 5 5 5 Pole: r, so the graph is not 1 3 cos 1 3 cos cos sin sin 1 3 cos 1 3 cos symmetric about the pole. 5 5 5 Line 2 : 1 3 cos 1 3 cos cos sin sin 1 3 cos r, so the graph is not symmetric about 2.
14. Polar axis:
15. Polar axis: 4 cos 2 4 cos 2 r 2 , so the graph is symmetric about the polar axis. Pole: r2 r 2 , so the graph is symmetric about the pole.
2 Line 2 : 4 cos 2 4 cos 2 2 4 cos 2 4 cos 2 r , so the graph is symmetric about 2 .
16. Polar axis: 9 sin 9 sin r 2 , so the graph is not symmetric about the polar axis. Pole: r2 r 2 , so the graph is symmetric about the pole.
2 Line 2 : 9 sin 9 sin cos cos sin 9 sin r , so the graph is symmetric about 2 .
17. r 2 r 2 4 x 2 y 2 4 is an equation of a circle with radius 2 centered at the origin.
O
1
18. r 1 r 2 1 x 2 y 2 1 is an equation of a circle with radius 1 centered at the origin.
O
1
SECTION 8.2 Graphs of Polar Equations
19. 2 cos 0 x 0 is an equation of a vertical line.
O
1
O
21. r 6 sin r 2 6r sin x 2 y 2 6y
x 2 y 32 9, a circle of radius 3 centered at 0 3.
O
1
22. r cos r 2 r cos x 2 y 2 x 2 x 12 y 2 14 , a circle of radius 12 centered at 10 . 2
O
1
23. r 2 cos . Circle.
1
24. r 3 sin . Circle. (3, ¹2 ) O
1
O
25. r 2 2 cos . Cardioid.
595
y 20. 56 tan 33 33 y 33 x, a x 3 line with slope 3 passing through the origin.
1
26. r 1 sin . Cardioid.
1
1
596
CHAPTER 8 Polar Coordinates and Parametric Equations
27. r 3 1 sin . Cardioid.
28. r cos 1. Cardioid. 1 1
29. r sin 2
30. r 2 cos 3
O
O
1
31. r cos 5
1
32. r sin 4 1
1
_1
_1
1
O
1
_1
_1
33. r 2 sin 5
O
34. r 3 cos 4
(5, ¹2 )
(_3, 0)
O
(_3, 3¹ 2 )
(_3, ¹)
O
(_3, ¹2 )
35. r
3 2 sin
_2
_1 O
1
2
36. r 2 sin
_1 _2 _3 _4
O
1
SECTION 8.2 Graphs of Polar Equations
37. r
3 cos
38. r 1 2 cos
2 1 _1
O
1
2
1
O
3
_1 _2
39. r 2 2 2 cos
40. r 3 6 sin
(9, ¹2 )
(2, ¹2 ) (2-2Ï2, 0) (2+2Ï2, ¹)
(_3, 3¹ 2 )
O
(2, 3¹ 2 )
41. r 2 cos 2
(3, ¹)
(3, 0)
O
42. r 2 4 sin 2
O
O
1
1
44. r 1, 0
43. r , 0
O
10 0.2
45. r 2 sec
46. r sin tan 2 O
1
O
1
597
598
CHAPTER 8 Polar Coordinates and Parametric Equations
47. r cos
, [0 4] 2
48. r sin
8 , [0 10] 5
1
1
-1
1
-1
-1
49. r 1 2 sin
1 -1
, [0 4] 2
50. r
1 08 sin2 , [0 2] 1
2
-1 -2
1 -1
51. r 1 sin n. There are n loops. 2
-2
2
2
-2
-2
n1 52. r 1 c sin 2 2
-2
2 -2
2
2
-2
2
2
-2
2
2
-2
2
-2
-2
-2
-2
n2
n3
n4
n5
2
2
2
2
-2
2 -2
-2
2 -2
-2
2 -2
-2
2 -2
c 03 c 06 c1 c 15 c2 3 As c increases, the graph becomes “pinched” along the line 4 , eventually growing more “leaves” with that line as their axis. 1 1 , and so on. is IV, since the graph must contain the points 0 0 53. The graph of r sin 2 2 2 1 54. The graph of r is I, since as increase, r decreases ( 0). So this is a spiral into the origin. 5 7 5 7 55. The graph of r sin is III, since for 2 2 2 the values of r are also 2 2 2 . Thus the graph must cross the vertical axis at an infinite number of points. 56. The graph of r 1 3 cos 3 is II, since when 0 r 1 3 4 and when r 1 3 2, so there should be two intercepts on the positive-axis.
SECTION 8.2 Graphs of Polar Equations
57.
x 2 y2
3
3 4x 2 y 2 r 2 4 r cos 2 r sin 2
r 6 4r 4 cos2 sin2 r 2 4 cos2 sin2 r 2 cos sin sin 2. The equation is r sin 2, a rose.
58.
x 2 y2
3
O
1
2 2 3 x 2 y 2 r 2 r cos 2 r sin 2
2 2 r 6 r 2 cos2 r 2 sin2 r 6 r 2 cos2 sin2
2 r 6 r 4 cos2 sin2 r 2 cos 22 r cos 2. The equations
1
r cos 2 and r cos 2 have the same graph, a rose.
59.
2 x 2 y 2 r 2 r cos 2 r sin 2 r 4 r 2 cos2 r 2 sin2 r 4 r 2 cos2 sin2
x 2 y2
2
r 2 cos2 sin2 cos 2. The graph is r 2 cos 2, a leminiscate.
O
1
2 2 60. x 2 y 2 x 2 y 2 x r 2 r 2 r cos r 2 [r r cos ]2 r 2 r 2 r cos 2 1 r cos 2 1 r cos r cos 1.
Both r cos 1 and r cos 1 give the same graph. To see this, replace
1
with : cos 1 cos 1 cos 1. This is a cardioid.
61. (a) r a cos b sin r 2 ar cos br sin
(b) r 2 sin 2 cos has center 1 1 and radius 12 22 22 2.
x 2 y 2 ax by x 2 ax y 2 by 0
x 2 ax 14 a 2 y 2 by 14 b2 14 a 2 14 b2 2 2 x 12 a y 12 b 14 a 2 b2 . Thus, in rectangular coordinates the center is 12 a 12 b and the radius is 12 a 2 b2 .
(b) r tan sec
62. (a)
y x 2.
5
-2
2
O
sin cos
1 cos
yr xr2
1
ry y 2 2 1 x x
599
600
CHAPTER 8 Polar Coordinates and Parametric Equations
64. (a)
63. (a) 5000
-5000
5000
5000
-5000
-5000
5000
-5000
At 0, the satellite is at the “rightmost” point in
The satellite starts out at the same point, 5625 0,
its orbit, 5625 0. As increases, it travels
but its orbit decays. It narrowly misses the earth at its
counterclockwise. Note that it is moving fastest
first perigee (closest approach to the earth), and
when .
crashes during its second orbit.
(b) The satellite is closest to earth when . Its height above the earth’s surface at this point is
(b) The satellite crashes when 837 rad 480 .
22,500 4 cos 3960 45003960 540 mi. 65. The graphs of r 1 sin 6 and r 1 sin 3 have
2
, respectively. Similarly, the graph of r f is the 3
1
the same shape as r 1 sin , rotated through angles of 6 and graph of r f rotated by the angle .
-2
2 -1
66. The circle r 2 (in polar coordinates) has rectangular coordinate equation x 2 y 2 4. The polar coordinate
form is simpler. The graph of the equation r sin 2 is a four-leafed rose. Multiplying both sides by r 2 , we get 32 r 3 r 2 2 cos sin 2 r cos r sin which is x 2 y 2 2x y in rectangular form. The polar form is definitely simpler.
67. y 2 r sin 2 r 2 csc . The rectangular coordinate system gives the simpler equation here. It is easier to study lines in rectangular coordinates.
8.3
POLAR FORM OF COMPLEX NUMBERS; DE MOIVRE’S THEOREM
1. A complex number z a bi has two parts: a is the real part and b is the imaginary part. To graph a bi we graph the ordered pair a b in the complex plane. 2. (a) The modulus of z is r a 2 b2 and an argument of z is an angle satisfying tan ba.
(b) We can express z in polar form as z r cos i sin where r is the modulus of z and is the argument of z. 3. (a) The complex number z 1 i in polar form is z 2 cos 34 i sin 34 . (b) The complex number z 2 cos 6 i sin 6 in rectangular form is z 3 i. (c) The complex number z can be expressed in rectangular form as 1 i or in polar form as 2 cos 4 i sin 4 .
SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem
4. A nonzero complex number has n different nth roots. The number 16 has four
Im
fourth roots. These roots are 2, 2, 2i, and 2i. In the complex plane, these
2i
roots all lie on a circle of radius 2.
_2
2
0 _2i
5. 4i
02 42 4
6. 3i
09 3
Im
Im 4i
1 0
1 1
0
Re
1
Re
_3i
7. 2
402
8. 6 6 Im
Im
1 _2
9. 5 2i
0
1 1
0
Re
52 22 29
10. 7 3i
Im
6
1
Re
49 9 58 Im
5+2i 1 0
1 1
Re
0
1
Re 7-3i
11. 3 i 3 1 2
12. 1 33 i 1 13 43 2 3 3
Im
Im
0
1
Ï3+i
1 1
0
Re _1- Ï3 i 3
1
Re
Re
601
602
CHAPTER 8 Polar Coordinates and Parametric Equations
3 4i 9 16 13. 25 25 1 5
2 i 2 14. 12 12 1 2
Im
Im
1
3+4i 5
0
1
Im 2z
0
0
Re
15. z 1 i, 2z 2 2i, z 1 i, 12 z 12 12 i
1
1
_Ï2+iÏ2 2
16. z 1
1
Re
3i, 2z 2 2 3i, z 1 3i,
1z 1 3i 2 2 2
Im
z
2z
1z 2
1
z
Re
_z
1z 1 2
0
1
Re
_z
17. z 8 2i, z 8 2i
18. z 5 6i, z 5 6i
Im
z
z
1 0
Im
1
1 0
Re
1
Re
z
z
19. z 1 2 i, z 2 2 i, z 1 z 2 2 i 2 i 4, z 1 z 2 2 i 2 i 4 i 2 5
20. z 1 1 i, z 2 2 3i,
z 1 z 2 1 i 2 3i 1 2i,
z 1 z 2 1 i 2 3i 2 3i 2i 3i 2 1 5i
Im
Im 1 0
zÁzª
zª 1
zÁzª zÁ+zª
zÁ
Re
zÁ 1 0
1
zÁ+zª zª
Re
SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem
22. z a bi a 1 b 1
21. z a bi a 0 b 0
Im
Im
1
1 0
23. z z 3
1
0
Re
24. z z 1
Im
25. z z 2
1
0
Re
26. z 2 z 5
Im
1
Re
Im
1
0
1
0
Re
28. z a bi a b
Im
1
Re
1
Re
Im
1
1 0
Re
1
1
27. z a bi a b 2
1
Im
1 0
603
1
0
Re
29. 1 i. Then tan 11 1 with in quadrant I 4 , and r
12 12 2. Hence, 1 i 2 cos 4 i sin 4 .
7 30. 1 i. Then tan 1 1 1 with in quadrant IV 4 , and r 1 i 2 cos 74 i sin 74 .
2 1 with in quadrant II 3 , and r 31. 2 2i. Then tan 2 4 2 2i 2 2 cos 34 i sin 34 .
12 12 2. Hence,
22 22 2 2. Hence,
604
CHAPTER 8 Polar Coordinates and Parametric Equations
32. 2 2i. Then tan 2 1 with in quadrant III 54 , and r 2 5 2 2i 2 cos 4 i sin 54 .
3 with in quadrant III 7 , and r 33. 3 i. Then tan 1 3 6 3 7 7 3 i 2 cos 6 i sin 6 .
2 2 2 2 2. Hence, 2 3 12 2. Hence,
34. 5 5 3i. Then tan 553 3 with in quadrant II 23 , and r 5 5 3i 10 cos 23 i sin 23 .
2 52 5 3 10. Hence,
2 3 with in quadrant IV 11 , and r 35. 2 3 2i. Then tan 2 2 3 22 4. Hence, 3 6 2 3 11 2 3 2i 4 cos 6 i sin 116 . 2 36. 3 3 3i. Then tan 3 3 3 3 with in quadrant I , and r 32 3 3 6. Hence, 3 3 3 3i 6 cos 3 i sin 3 . 37. 2i. Then tan is undefined, 2 , and r 2. Hence, 2i 2 cos 2 i sin 2 . 38. 5i. Then tan is undefined, 32 , and r 5. Hence, 5i 5 cos 32 i sin 32 .
39. 3. Then tan 0, , and r 3. Hence, 3 3 cos i sin . 40. 2. Then tan 0, 0, and r 2. Hence, 2 2 cos 0 i sin 0.
41. 6 2i. Then tan 2 33 with in quadrant II 56 , and r 6 6 2i 2 2 cos 56 i sin 56 .
2 2 6 2 2 2. Hence,
42. 5 15i. Then tan 15 3 with in quadrant III 43 , and r 5 Hence, 5 15i 2 5 cos 43 i sin 43 .
43. 4 3i. Then tan 4 3i 5 cos tan1
2 2 5 15 2 5.
3 with in quadrant I tan1 3 06435, and r 4 4 3 i sin tan1 3 . 4 4
44. 3 2i. Then tan 23 with in quadrant I tan1 23 05880, and r 3 2i 13 cos tan1 23 i sin tan1 23 .
42 32 5. Hence,
32 22 13. Hence,
2 3 with in quadrant IV 11 , and r 4 3 42 8. 3 i 4 3 4i. Then tan 4 3 6 4 3 11 3 i 8 cos 6 i sin 116 . Hence, 4 46. i 2 6i 6 2i. Then tan 2 33 with in quadrant I tan1 33 6 , and 6 2 2 6 2 2 2. Hence, i 2 6i 2 2 cos r 6 i sin 6 . 3 1 with in quadrant II 3 , and r 32 32 3 2. Hence, 47. 3 1 i 3 3i. Then tan 3 4 3 1 i 3 2 cos 34 i sin 34 .
45. 4
SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem
605
2 1 with in quadrant II 3 , and r 22 22 2 2. Hence, 48. 2i 1 i 2 2i. Then tan 2 4 2i 1 i 2 2 cos 34 i sin 34 .
6 cos i sin 49. z 1 3 cos 3 i sin 3 , z 2 2 cos 6 i sin 6 , z 1 z 2 3 2 cos 3 6 i sin 3 6 2 2 i sin 3 cos i sin z 1 z 2 32 cos 3 6 3 6 2 6 6 3 cos 54 i sin 54 , z 2 2 cos i sin , z1 z2 3 2 cos 54 i sin 54 2 3 cos 4 i sin 4 , z 1 z 2 23 cos 54 i sin 54 23 cos 4 i sin 4
50. z 1
2 cos 53 i sin 53 , z 2 2 2 cos 32 i sin 32 , 4 cos 76 i sin 76 , 2 2 2 cos 53 32 i sin 53 32 z1 z2 z 1 z 2 2 cos 53 32 i sin 53 32 12 cos 6 i sin 6
51. z 1
2 2
, z z cos 3 i sin 3 cos 13 i sin 13 , i sin 52. z 1 cos 34 i sin 34 , z 2 cos 1 2 3 3 4 3 4 3 12 12 3 3 5 5 z 1 z 2 cos 4 3 i sin 4 3 cos 12 i sin 12 53. z 1 4 cos 120 i sin 120 , z 2 2 cos 30 i sin 30 , z 1 z 2 4 2 cos 120 30 i sin 120 30 8 cos 150 i sin 150 , z 1 z 2 42 cos 120 30 i sin 120 30 2 cos 90 i sin 90
54. z 1 2 cos 75 i sin 75 , z 2 3 2 cos 60 i sin 60 , z 1 z 2 2 3 2 cos 75 60 i sin 75 60 6 cos 135 i sin 135 , z 1 z 2 2 cos 75 60 i sin 75 60 13 cos 15 i sin 15 3 2
55. z 1 4 cos 200 i sin 200 , z 2 25 cos 150 i sin 150 , z 1 z 2 4 25 cos 200 150 i sin 200 150 100 cos 350 i sin 350 , 4 cos 200 150 i sin 200 150 4 cos 50 i sin 50 z 1 z 2 25 25
56. z 1 45 cos 25 i sin 25 , z 2 15 cos 155 i sin 155 , 4 cos 180 i sin 180 , z 1 z 2 45 15 cos 25 155 i sin 25 155 25 z 1 z 2 45 15 cos 25 155 i sin 25 155 4 cos 130 i sin 130 4 cos 130 i sin 130 3 i, so tan 1 1 with 1 in quadrant I 1 6 , and r1 3 1 2. 3 z 2 1 3i, so tan 2 3 with 2 in quadrant I 2 3 , and r1 1 3 2. Hence, z 1 2 cos 6 i sin 6 and z 2 2 cos 3 i sin 3 . 4 cos i sin , Thus, z 1 z 2 2 2 cos 6 3 i sin 6 3 2 2 i sin cos i sin , and 1z 1 cos i sin . z 1 z 2 22 cos 1 6 3 6 3 6 6 2 6 6
57. z 1
606
CHAPTER 8 Polar Coordinates and Parametric Equations
2 2i, so tan 1 1 with 1 in quadrant IV 1 74 , and r1 2 2 2. z 2 1 i, so tan 2 1 with 2 in quadrant IV 2 74 , and r2 1 1 2. Hence, z 1 2 cos 74 i sin 74 and z 2 2 cos 74 i sin 74 . Thus, z 1 z 2 2 2 cos 74 74 i sin 74 74 2 2 cos 72 i sin 72 2 2 cos 32 i sin 32 , z 1 z 2 2 cos 74 74 i sin 74 74 2 cos 0 i sin 0, and 2 1z 1 12 cos 74 i sin 74 12 cos 4 i sin 4 .
58. z 1
1 with 1 in quadrant IV 1 11 , and r1 12 4 4. 59. z 1 2 3 2i, so tan 1 2 6 2 3 3 z 2 1 i, so tan 2 1 with 2 in quadrant II 2 34 , and r2 1 1 2. Hence, z 1 4 cos 116 i sin 116 and z 2 2 cos 34 i sin 34 . i sin 7 , Thus, z 1 z 2 4 2 cos 116 34 i sin 116 34 4 2 cos 712 12 4 13 z 1 z 2 cos 116 34 i sin 116 34 2 2 cos 13 12 i sin 12 , and 2 1 14 cos 1z 1 4 cos 116 i sin 116 6 i sin 6 . 60. z 1 2i, so 1 32 , and r1 2. z 2 3 3 3i, so tan 2 3 with 2 in quadrant IV 2 43 , and r2 9 27 6. Hence, z 1 2 cos 32 i sin 32 and z 2 6 cos 43 i sin 43 . Thus, z 1 z 2 6 2 cos 56 i sin 56 , 2 6 cos 32 43 i sin 32 43 z 1 z 2 62 cos 32 43 i sin 32 43 62 cos 6 i sin 6 , and 1z 1 1 cos 32 i sin 32 22 cos 2 i sin 2 . 2
61. z 1 5 5i, so tan 1 55 1with 1 in quadrant I 1 4 , and r1 25 25 5 2. z 2 4, so 2 0, and r2 4. Hence, z 1 5 2 cos 4 i sin 4 and z 2 4 cos 0 i sin 0. i sin , z z 5 2 cos i sin , and 0 i sin 0 20 2 cos Thus, z 1 z 2 5 2 4 cos 1 2 4 4 4 4 4 4 4 1 cos i sin 2 cos i sin . 1z 1 4 4 10 4 4 5 2
3 with 1 in quadrant III 1 11 , and r1 48 16 8. 62. z 1 4 3 4i, so tan 1 4 3 6 4 3
, and r2 8. z 2 8i, so 2 2 Hence, z 1 8 cos 116 i sin 116 and z 2 8 cos 2 i sin 2 . 11 64 cos Thus, z 1 z 2 8 8 cos 116 2 i sin 6 2 3 i sin 3 , 11 z 1 z 2 88 cos 116 cos 43 i sin 43 , and 1z 1 18 cos 2 i sin 6 2 6 i sin 6 .
SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem
607
63. z 1 20, so 1 , and r1 20. z 2 3 i, so tan 2 1 with 2 in quadrant I 2 6 , and r2 3 1 2. 3 Hence, z 1 20 cos i sin and z 2 2 cos 6 i sin 6 . 40 cos 7 i sin 7 , Thus, z 1 z 2 20 2 cos 6 i sin 6 6 6 10 cos 5 i sin 5 , and z 1 z 2 20 2 cos 6 i sin 6 6 6 1 [cos i sin ] 1 cos i sin . 1z 1 20 20
64. z 1 3 4i, so tan 1 43 with 1 in quadrant I 1 tan1 43 , and r1 9 16 5. z 2 2 2i, so tan 2 1 with 2 in quadrant IV 2 74 , and r2 4 4 2 2. Hence, z 1 5 cos tan1 43 i sin tan1 43 5 cos 0927 i sin 0927 and z 2 2 2 cos 74 i sin 74 . Thus, z 1 z 2 5 2 2 cos tan1 43 74 i sin tan1 34 74 10 2 cos 0142 i sin 0142, 5 cos tan1 34 74 i sin tan1 34 74 5 4 2 cos tan1 43 74 i sin tan1 43 74 z 1 z 2 22
5 4 2 [cos 457 i sin 457], and 1z 1 15 cos tan1 43 i sin tan1 43 15 [cos 0927 i sin 0927].
6 65. 3 i 2 cos 56 i sin 56 , so 3 i 26 cos 306 i sin 306 64 cos 5 i sin 5 64. 66. 1 i
10 cos 704 i sin 704 32 cos 32 i sin 32 32i. 2 cos 74 i sin 74 , so 1 i10 2
5 67. 2 2i 2 cos 54 i sin 54 , so 2 2i 25 cos 254 i sin 254 16 2 16 2i.
7 7 68. 1 i 2 cos cos 74 i sin 74 8 8i. 2 4 i sin 4 , so 1 i 1 1 1 and tan 1 . Thus 2 2 i cos i sin . Therefore, 69. r 2 2 4 2 2 4 4 12 2 2 cos 12 2 2 i 4 i sin 12 4 cos 3 i sin 3 1. 1 with in quadrant IV 11 . Thus 3 i 2 cos 11 i sin 11 , so 70. r 3 1 2 and tan 6 6 6 3 10 10 3 i sin 110 1 cos i sin 1 1 1 cos 110 3i 12 6 6 1024 3 3 1024 2 2 i 2048 1 3i . 71. r 4 4 4 2 and tan 1 with in quadrant IV 74 . Thus 2 2i 2 2 cos 74 i sin 74 , so 8 2 2i8 2 2 cos 14 i sin 14 4096 1 0i 4096. 72. r 14 34 1 and tan 3 with in quadrant III 43 . Thus 12 23 i cos 43 i sin 43 , so 15 12 23 i cos 603 i sin 603 cos 20 i sin 20 1. 73. r 1 1 2 and tan 1 with in quadrant III 54 . Thus 1 i 2 cos 54 i sin 54 , so 7 cos 354 i sin 354 8 2 cos 34 i sin 34 8 2 1 i 1 8 1 i. 2 1 i7 2
2
74. r 9 3 2 3 and tan 33 with in quadrant I 6 . Thus 3 3i 2 3 cos 6 i sin 6 , so 4 3 3i 144 cos 23 i sin 23 144 12 23 i 72 1 3i .
608
CHAPTER 8 Polar Coordinates and Parametric Equations
2 1 . Thus 2 3 2i 4 cos i sin , so 75. r 12 4 4 and tan 6 6 6 2 3 3 5 5 1 1 2 3 2i cos 56 i sin 56 1024 23 12 i 2048 3i 14
1 1 2 and tan 1 with in quadrant IV 74 . Thus 1 i 2 cos 74 i sin 74 , so 1 cos 56 i sin 56 1 cos 14 i sin 14 1 . 1 i8 16 4 4 16 16
76. r
4 1 . Thus 77. r 48 16 8 and tan 6 4 3 3 4 3 4i 8 cos 6 i sin 6 . So, 12 6 2k 6 2k i sin for k 0, 1. 8 cos 4 3 4i 2 2 i sin and Thus the two roots are 0 2 2 cos 12 12 13 13 1 2 2 cos 12 i sin 12 .
Im
1 0 wÁ
4 1 . Thus 78. r 48 16 8 and tan 6 4 3 3 4 3 4i 8 cos 6 i sin 6 . So 13 6 2k 6 2k i sin for k 0, 1, 2. 2 cos 4 3 4i 3 3 i sin , Thus the three roots are 0 2 cos 18 18 13 13 i sin 25 . 1 2 cos 18 i sin 8 , and 2 2 cos 25 18 18 79. 81i 81 cos 32 i sin 32 . Thus, 32 2k 32 2k i sin for k 0, 1, 81i14 8114 cos 4 4 2, 3. The four roots are 0 3 cos 38 i sin 38 , 1 3 cos 78 i sin 78 , 2 3 cos 118 i sin 118 , and 3 3 cos 158 i sin 158 . 2k 2k cos i sin for k 0, 1, 2, 3, 5 5 4. Thus the five roots are 0 2 cos 0 i sin 0, 1 2 cos 25 i sin 25 , 2 2 cos 45 i sin 45 , 3 2 cos 65 i sin 65 , and 4 2 cos 85 i sin 85 .
wü 1
Re
Im
wÁ 1 0
wü 1
Re
wª
wü
Im
wÁ
1 0
1
Re
w£ wª
80. 32 32 cos 0 i sin 0. Thus, 3215
Im
wÁ wª
1 0
w£
wü 1
w¢
Re
SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem
2k 2k i sin , for k 0, 1, 2, 3, 4, 5, 6, 8 8 7. So the eight roots are 0 cos 0 i sin 0 1,
81. 1 cos 0 i sin 0. Thus, 118 cos 1 cos 3 cos 5 cos 7 cos
i sin 2 i 2 , cos 2 4 4 2 2 3 i sin 3 2 i 2 , 4 4 4 2 2 5 i sin 5 2 i 2 , 6 4 4 2 2 7 i sin 7 2 i 2 . 4 4 2 2
Im
wª
w£
i sin i, 2 2
cos 32 i sin 32 i, and
wÁ wü 1
Re
w¦
w°
2 cos 4 i sin 4 . So 13 4 2k 4 2k cos i sin for k 0, 1, 2 1 i13 3 3 i sin , 2. Thus the three roots are 0 216 cos 12 12 i sin 9 , and 216 cos 17 i sin 17 . 1 216 cos 912 2 12 12 12
1
0
w¢
cos i sin 1,
609
w§
82. 1 i
Im
wÁ
1 wü 0
1
Re
wª
13 cos 83. i cos 2 i sin 2 , so i
2 2k 3
i sin
2 2k 3
for
Im
3 1 k 0, 1, 2. Thus the three roots are 0 cos 6 i sin 6 2 2 i,
1
wÁ
wü
1 cos 56 i sin 56 23 12 i, and 2 cos 32 i sin 32 i.
0
1
Re
wª
15 cos 84. i cos 2 i sin 2 , so i
2 2k 5
i sin
2 2k 5
i sin , k 0, 1, 2, 3, 4. Thus the five roots are 0 cos 10 10
for
Im
1
9 9 13 13 1 cos 2 i sin 2 , 2 cos 10 i sin 10 , 3 cos 10 i sin 10 , and
wü
wª 0
17 4 cos 17 10 i sin 10 .
2k 4
2k i sin 4
2 2 k 0, 1, 2, 3. So the four roots are 0 cos 4 i sin 4 2 i 2 ,
for
1 cos 34 i sin 34 22 i 22 , 2 cos 54 i sin 54 22 i 22 , and 3 cos 74 i sin 74 22 i 22 .
1
Re
1
Re
w¢
w£
85. 1 cos i sin . Then 114 cos
wÁ
Im
wÁ
1
0 wª
wü
w£
610
CHAPTER 8 Polar Coordinates and Parametric Equations 162 1 3 32 and tan 1616 3 3 with in quadrant III 43 . Thus 16 16 3i 32 cos 43 i sin 43 . So 15 43 2k 43 2k 16 16 3i i sin for 3215 cos 5 5 i sin 4 , k 0, 1, 2, 3, 4. The five roots are 0 2 cos 415 15 i sin 16 , 1 2 cos 23 i sin 23 , 2 2 cos 16 15 15 22 22 28 3 2 cos 15 i sin 15 , and 4 2 cos 15 i sin 28 15 .
86. r
Im
wÁ
wü 1
wª
0
1
Re
w¢ w£
87. z 4 1 0 z 114 z 22 22 i, z 22 22 i (from Exercise 85) , then, z i 18 cos 2 2k i sin 2 2k for i sin 88. z 8 i 0 x i 18 . Since i cos 2 2 8 8
i sin , cos 5 i sin 5 , cos 9 i sin 9 , k 0, 1, 2, 3, 4, 5, 6, 7. Thus there are eight solutions: z cos 16 16 16 16 16 16 i sin 13 , cos 17 i sin 17 , cos 21 i sin 21 , cos 25 i sin 25 , and cos 29 i sin 29 . cos 13 16 16 16 16 16 16 16 16 16 16 13 89. z 3 4 3 4i 0 z 4 3 4i . Since 4 3 4i 8 cos 6 i sin 6 , 13 6 2k 6 2k i sin , for k 0, 1, 2. Thus the three roots are 813 cos 4 3 4i 3 3 i sin , z 2 cos 13 i sin 13 , and z 2 cos 25 i sin 25 . z 2 cos 18 18 18 8 18 18
90. z 6 1 0 z 116 . Since 1 cos 0 i sin 0 z 116 cos 2k6 i sin 2k6 for k 0, 1, 2, 3, 4, 5. Thus there are
six solutions: z 1, 12 23 i, 12 23 i.
2 cos 54 i sin 54 , 91. z 3 1 i z 1 i13 . Since 1 i 54 2k 54 2k i sin for k 0, 1, 2. Thus the three solutions to this z 1 i13 216 cos 3 3 i sin 5 , 216 cos 13 i sin 13 , and 216 cos 21 i sin 21 . equation are z 216 cos 512 12 12 12 12 12
2k 2k 92. z 3 1 0 z 113 . Since 1 cos 0 i sin 0 z 113 cos i sin for k 0, 1, 2. Thus the three 3 3
solutions to this equation are z cos 0 i sin 0, cos 23 i sin 23 , and cos 43 i sin 43 or z 1, 12 23 i, 12 23 i. i i2 4 1 1 i 5 1 5 2 i 93. z i z 1 0 z 2 1 2 2 i 9 i i 2 4 1 2 2i or i 94. z 2 i z 2 0 z 2 2 2i 2i2 4 1 2 2i 4 95. z 2 2i z 2 0 z i 1 2 2 96. z 2 1 i z i z i z 1, so z 1 or i. 97. 1 zk z1
1 cos 0 i sin 0, so by De Moivre’s Theorem, its n roots are 2k 2k 0 2k 0 2k i sin cos i sin for k 0 1 2 n 1. So z 0 1, 11n cos n n n n 2 4 2 4 i sin , z 2 cos i sin 2 , and so on. cos n n n n
SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem
611
n 98. From Exercise 97, k 1 for 0 k n 1. Multiplying both sides by s n and noting that s n z, we have n n s n k s n sk z for 0 k n 1, showing that 1 s s2 sn1 are nth roots of z.
99. The cube roots of 1 are 0 1, 1 cos 23 i sin 23 12 23 i, and 2 cos 43 i sin 43 12 23 i, so their sum is 0 1 2 1 12 23 i 12 23 i 0.
The fourth roots of 1 are 0 1, 1 i, 2 1, and 3 i, so their sum is 0 1 2 3 1 i 1 i 0.
The fifth roots of 1 are 0 1, 1 cos 25 i sin 25 , 2 cos 45 i sin 45 , 3 cos 65 i sin 65 , and 4 cos 85 i sin 85 , so their sum is 1 cos 25 i sin 25 cos 45 i sin 45 cos 65 i sin 65 cos 85 i sin 85 5 5 1 1 1 2 cos 25 2 cos 65 (most terms cancel) 1 2 0 4 4 4 4
3 3 1 2 2 1 2 The sixth roots of 1 are 0 1, 1 cos 3 i sin 3 2 2 i, cos 3 i sin 3 2 2 i,
3 3 5 5 1 5 2 i, and cos 3 i sin 3 2 2 i, so their sum is 3 3 1 2 i 2 2 i 0. 2 2 2 2 3 3 2 3 The eight roots of 1 are 0 1, 1 cos 4 i sin 4 2 2 i, i, cos 4 i sin 4 2 2 i, 4 1, 5 cos 54 i sin 54 22 22 i, 6 i, and 7 cos 74 i sin 74 22 22 i, so their sum is 1 22 22 i i 22 22 1 22 22 i i 22 22 i 0.
3 1, 4 cos 43 i sin 43 12 1 1 12 23 i 23 i 1 12 2
It seems that the sum of any set of nth roots is 0. To prove this, factor n 1 1 1 1 1 2 3 n1 . Since this is 0 and 1 0, we must have
1 1 2 3 n1 0.
100. The cube roots of 1 are 0 1, 1 cos 23 i sin 23 , and 2 cos 43 i sin 43 , so their product is 0 1 2 1 cos 23 i sin 23 cos 43 i sin 43 cos 2 i sin 2 1. The fourth roots of 1 are 0 1, 1 i, 2 1, and 3 i, so their product is 0 1 2 3 1 i 1 i i 2 1.
The fifth roots of 1 are 0 1, 1 cos 25 i sin 25 , 2 cos 45 i sin 45 , 3 cos 65 i sin 65 , and 4 cos 85 i sin 85 , so their product is 1 cos 25 i sin 25 cos 45 i sin 45 cos 65 i sin 65 cos 85 i sin 85 cos 4 i sin 4 1.
3 2 2 1 2 3 The sixth roots of 1 are 0 1, 1 cos 3 i sin 3 , cos 3 i sin 3 2 2 i, 1,
4 cos 43 i sin 43 12 23 i, and 5 cos 53 i sin 53 12 23 i, so their product is cos 2 i sin 2 1 cos 4 i sin 4 5 i sin 5 cos 5 i sin 5 1. cos 1 cos i sin 3 3 3 3 3 3 3 3
3 3 2 3 The eight roots of 1 are 0 1, 1 cos 4 i sin 4 , i, cos 4 i sin 4 , 4 1, 5 cos 54 i sin 54 , 6 i, 7 cos 74 i sin 74 , so their product is i cos 3 i sin 3 1 cos 5 i sin 5 i cos 7 i sin 7 i2 i sin 1 cos 4 4 4 4 4 4 4 4
cos 2 i sin 2 1.
612
CHAPTER 8 Polar Coordinates and Parametric Equations
The product of the nth roots of 1 is 1 if n is even and 1 if n is odd. m m 1 . The proof requires the fact that the sum of the first m integers is 2 2 2 2k 2k Let cos i sin . Then k cos i sin for k 0 1 2 n 1. The argument of the n n n n product of the n roots of unity can be found by adding the arguments of each k . So the argument of the product is 2 n 2 2 n 1 2 2 1 2 2 2 3 [0 1 2 3 n 2 n 1]. 0 n n n n n n 2 n 1 n n 1 . Thus the product of the n roots of Since this is the sum of the first n 1 integers, this sum is n 2 unity is cos n 1 i sin n 1 1 if n is even and 1 if n is odd. 101.
z1 r cos 1 i sin 1 cos 2 i sin 2 r cos 1 cos 2 i 2 sin 1 cos 2 i sin 1 cos 2 i sin 2 cos 1 1 1 z2 r2 cos 2 i sin 2 cos 2 i sin 2 r2 cos2 2 i 2 sin2 2 r1 cos 1 2 i sin 1 2 r2
8.4
PLANE CURVES AND PARAMETRIC EQUATIONS
1. (a) The parametric equations x f t and y g t give the coordinates of a point x y f t g t for appropriate values of t. The variable t is called a parameter. (b) When t 0 the object is at 0 02 0 0 and when t 1 the object is at 1 12 1 1. (c) If we eliminate the parameter in part (b) we get the equation y x 2 . We see from this equation that the path of the moving object is a parabola.
2. (a) It is true that the same curve can be described by parametric equations in many different ways. For instance, the parabola y x 2 can be represented by x t, y t 2 or by x t, y t 2 or by x t, y t. (b) The parametric equations x 2t, y 2t2 model the position of a
y
moving object at time t. When t 0 the object is at 0 0, and when t 1 the object is at 2 4.
t=1 [Ex. 2(b)]
(c) If we eliminate the parameter we get the equation y x 2 , which is the same equation as in Exercise 1(b). So the objects in Exercises 1(b) and 2(b) move along the same path, but traverse the path differently.
1
t=1 [Ex. 1(b)]
0
1
2
t=0 [1(b) and 2(b)]
3. (a) x 2t, y t 6
y
y
1 1 2
x 2y 12 0.
x
4. (a) x 6t 4, y 3t, t 0
1
(b) Since x 2t, t
3
x
x
x x and so y 6 2 2
y (b) Since y 3t, t and so 3 y 4 2y 4 x 2y 4 0, y 0. x 6 3
SECTION 8.4 Plane Curves and Parametric Equations
613
2 6. (a) x 2t 1, y t 12
5. (a) x t 2 , y t 2, 2 t 4 y
y
1
1 x
4
t y 2, we have x t 2
(b) Since y t 2
x y 22 , and since 2 t 4, we have
4 x 16.
7. (a) x
1
2 2 (b) Since y t 12 , 4y 4 t 12 2t 12 ,
and since x 2t 1, we have x 2 2t 12 4y
t, y 1 t t 0 y
1
x
y 14 x 2 .
8. (a) x t 2 , y t 4 1 y
x
20 10
_10
x
1
(b) Since x
t, we have x 2 t, and so y 1 x 2
with x 0.
9. (a) x
1 , y t 1 t
x 0.
10. (a) x t 1, y y
1 1
(b) Since x
(b) Since x t 2 , we have x 2 t 4 and so y x 2 1,
x
1 1 1 we have t and so y 1. t x x
t t 1
y
1 1
x
(b) Since x t 1, we have t x 1, so y
x 1 . x
614
CHAPTER 8 Polar Coordinates and Parametric Equations
11. (a) x 4t 2 , y 8t 3
12. (a) x t, y 1 t
y
y
1 x
1
1 x
1
(b) Since y 8t 3
have y 2 x 3 .
3 y 2 64t 6 4t 2 x 3 , we
13. (a) x 2 sin t, y 2 cos t, 0 t
(b) Since x t, we have y 1 x, where x 0.
14. (a) x 2 cos t, y 3 sin t, 0 t 2
y
y
1
1 x
1
(b) x 2 2 sin t2 4 sin2 t and y 2 4 cos2 t. Hence, x 2 y 2 4 sin2 t 4 cos2 t 4 x 2 y 2 4,
where x 0.
15. (a) x sin2 t, y sin4 t
(b) We have cos t x2 and sin t y3, so
x22 y32 cos2 t sin2 t 1 1 x 2 1 y 2 1. 4 9
16. (a) x sin2 t, y cos t
y
x
1
y 1
1
1
1
x
(b) Since x sin2 t we have x 2 sin4 t and so y x 2 . But since 0 sin2 t 1 we only get the part of this parabola for which 0 x 1.
x
(b) Since y cos t, we have y 2 cos2 t and so
x y 2 sin2 t cos2 t 1 or x 1 y 2 (actually y 1 x), where 0 x 1.
SECTION 8.4 Plane Curves and Parametric Equations
17. (a) x cos t, y cos 2t
18. (a) x cos 2t, y sin 2t
y 1
y 1
1
x
(b) Since x cos t we have x 2 cos2 t, so
2x 2 1 2 cos2 t 1 cos 2t y. Hence, the
rectangular equation is y 2x 2 1, 1 x 1.
19. (a) x sec t, y tan t, 0 t 2 x 1 and y 0.
1
x
(b) x 2 cos2 2t and y 2 sin2 2t. Then
x 2 y 2 cos2 2t sin2 2t 1 x 2 y 2 1.
20. (a) x cot t, y csc t, 0 t so y 1. y
y
1 1
1
x
x
1
(b) x 2 sec2 t, y 2 tan2 t, and
y 2 1 tan2 t 1 sec2 t x 2 . Therefore, y 2 1 x 2 x 2 y 2 1, x 1, y 0.
21. (a) x tan t, y cot t, 0 t 2 x 0.
(b) x 2 cot2 t, y 2 csc2 t, and so
x 2 1 cot2 t 1 csc2 t y 2 . Therefore,
y 2 x 2 1, with y 1. This is the top half of the
hyperbola y 2 x 2 1.
22. (a) x et , y et , so y 0.
y
y
1
1 1
x
(b) x y tan t cot t 1, so y 1x for x 0.
1
(b) x y et et 1, so y 1x, x 0.
x
615
616
CHAPTER 8 Polar Coordinates and Parametric Equations
24. (a) x sec t, y tan2 t, 0 t 2 , so y 0 and x 1.
23. (a) x e2t , y et , so y 0. y
y
2 1 _1 0
1
2
3
4
5
7 x
6
2
2 (b) x e2t et y 2 , y 0.
x
1
(b) x 2 sec2 t 1 tan2 t, so x 2 1 y, x 1, y 0.
25. (a) x cos2 t, y sin2 t, so 0 x 1 and 0 y 1. y
26. (a) x cos3 t, y sin3 t, 0 t 2 y
1
1
1
x
(b) x y cos2 t sin2 t 1. Hence, the equation is x y 1 for 0 x 1 and 0 y 1.
1
x
(b) x 23 cos2 t and y 23 sin2 t, and so x 23 y 23 1.
27. x 3 cos t, y 3 sin t. The radius of the circle is 3, the position at time 0 is x 0 y 0 3 cos 0 3 sin 0 3 0 and the orientation is counterclockwise (because x is decreasing and y is increasing initially). x y 3 0 again when t 2, so it takes 2 units of time to complete one revolution. 28. x 2 sin t, y 2 cos t. The radius is 2, the position at time 0 is 0 2, the orientation is clockwise (because x is increasing and y is decreasing initially), and it takes 2 units of time to complete one revolution. 29. x sin 2t, y cos 2t. The radius of the circle is 1, the position at time 0 is x 0 y 0 sin 0 cos 0 0 1 and the orientation is clockwise (because x is increasing and y is decreasing initially). x y 0 1 again when t , so it takes units of time to complete one revolution. 30. x 4 cos 3t, y 4 sin 3t. The radius is 4, the position at time 0 is 4 0, the orientation is counterclockwise, and it takes 2 units of time to complete one revolution. 3
31. Since the line passes through the point 4 1 and has slope 12 , parametric equations for the line are x 4 t, y 1 12 t.
32. Since the line passes through the point 10 20 and has slope 2, parametric equations for the line are x 10 t, y 20 2t. 87 1. Thus, parametric equations for the line are 33. Since the line passes through the points 6 7 and 7 8, its slope is 76 x 6 t, y 7 t. 70 7 34. Since the line passes through the points 0 0 and 12 7, its slope is . Thus, parametric equations for the line 12 0 12 7 t. are x t, y 12 35. Since cos2 t sin2 t 1, we have a 2 cos2 t a 2 sin2 t a 2 . If we let x a cos t and y a sin t, then x 2 y 2 a 2 . Hence, parametric equations for the circle are x a cos t, y a sin t.
SECTION 8.4 Plane Curves and Parametric Equations
617
a 2 cos2 t b2 sin2 t x2 y2 1. If we let x a cos t and y b sin t, then 1. a2 b2 a2 b2 Hence, parametric equations for the ellipse are x a cos t, y b sin t.
36. Since cos2 t sin2 t 1, we have
37. x 0 cos t, y 0 sin t 16t 2 . From the equation for x, t
x . Substituting into the equation for y gives 0 cos
2 x 16x 2 x . Thus the equation is of the form y c1 x c2 x 2 , 16 x tan 2 0 cos 0 cos 0 cos2 where c1 and c2 are constants, so its graph is a parabola. y 0 sin
38. 0 2048 fts and 30 . (a) The bullet will hit the ground when y 0. Since y 0 sin t 16t 2 , we have 0 2048 sin 30 t 16t 2 0 1024t 16t 2
16t t 64 0 t 0 or t 64. Hence, the bullet will hit the ground after 64 seconds. (b) After 64 seconds, the bullet will hit the ground at x 2048 cos 30 64 65,536 3 113,5117 ft 215 miles. (c) The maximum height attained by the bullet is the maximum value of y. Since y 1024t 16t 2 16 t 2 64t 16 t 2 64t 1024 16,384 16 t 322 16,384, y is maximized when t 32, and therefore the maximum height is 16,384 ft 31 miles. 40. x 2 sin t, y cos 4t
39. x sin t, y 2 cos 3t 2
1 -1
-2
1
2 -1
-2
41. x 3 sin 5t, y 5 cos 3t
42. x sin 4t, y cos 3t 5
1
-1
1 -1
-5
43. x sin cos t, y cos t 32 , 0 t 2
44. x 2 cos t cos 2t, y 2 sin t sin 2t 2
1
-1
1 -1
-2
2 -2
618
CHAPTER 8 Polar Coordinates and Parametric Equations
45. (a) r 212 , 0 4 x 2t12 cos t, y 2t12 sin t
46. (a) r sin 2 cos x sin t 2 cos t cos t, y sin t 2 cos t sin t
(b)
(b)
2 2
-2
2 -2
47. (a) r
1
4 cos t 4 sin t 4 x ,y 2 cos 2 cos t 2 cos t
(b)
2
48. (a) r 2sin x 2sin t cos t, y 2sin t sin t (b)
2
2
-2
2
4
-2
-2
2
2 49. x t 3 2t, y t 2 t is Graph III, since y t 2 t t 2 t 14 14 t 12 14 , and so y 14 on this curve, while x is unbounded.
50. x sin 3t, y sin 4t is Graph IV, since when t 0 and t the curve passes through 0 0. Thus this curve must pass through 0 0 twice. 51. x t sin 2t, y t sin 3t is Graph II, since the values of x and y oscillate about their values on the line x t, y t y x. 52. x sin t sin t, y cos t cos t is Graph I, since this curve does not pass through the point 0 0. 53. (a) It is apparent that x O Q and
(b) The curve is graphed with a 3 and b 2.
y Q P ST . From the diagram,
x O Q a cos and y ST b sin .
2
Thus, parametric equations are x a cos and y b sin .
-4 y
a
b O
-2
2
4
-2 S ¬ T
(c) To eliminate we rearrange: sin yb
P Q
x
sin2 yb2 and cos xa
cos2 xa2 . Adding the two equations:
sin2 cos2 1 x 2 a 2 y 2 b2 . As
indicated in part (b), the curve is an ellipse.
SECTION 8.4 Plane Curves and Parametric Equations
619
54. (a) A has coordinates a cos a sin . Since O A is perpendicular to AB, 2
O AB is a right triangle and B has coordinates a sec 0. It follows that P has coordinates a sec b sin . Thus, the parametric equations
0
are x a sec , y b sin .
5
(b) The right half of the curve is graphed with a 3 and b 2.
55. (a) If we modify Figure 8 so that PC b, then by the
10
-2
(b) 5
same reasoning as in Example 6, we see that x OT P Q a b sin and y T C C Q a b cos .
We graph the case where a 3 and b 2.
-20
20
56. From the figure, we see that x OT P Q a b sin and y T C C Q a b cos . When a 1 and b 2, the parametric equations become x 2 sin , y 1 2 cos . The curve is graphed for 0 4. y
(b)
C(a¬,a)
b
P(x, y) O
¬
a¬
-10
Q
T
10
x
x2 y y2 sec2 2 . Since . Also, y b sec sec 2 b a b 2 2 2 2 x y y x tan2 sec2 1, we have 2 2 1 2 2 1, which is the equation of a hyperbola. a b b a
57. x a tan tan
x a
tan2
58. Substituting the given values for x and y into the equation we derived in Exercise 57, we get 2 2 a t b t 1 t 1 t 1. Thus the points on this curve satisfy the equation, which is that of a hyperbola. 2 b a2 y2 x2 However, this hyperbola is only the part of 2 2 1 for which x 0 and y 0. b a 59. x t cos t, y t sin t, t 0 t
x
y
t
0
0
0
5 4 3 2 7 4
4
2 3 4
2 8
2 8
0
38 2
2 3 2 8
0
2
x
y
58 2 58 2 0 32 7 2 7 2 8 8
2
0
60. x sin t, y sin 2t
y
y
1
1
1 1
x
x
620
CHAPTER 8 Polar Coordinates and Parametric Equations
61. x
3t 3t 2 , y , t 1 1 t3 1 t3 t
09
x
y
996 897
075 389 292 05
171 086 0
t 2
y
x
y
067 133
t 11
x
y
t
997 1097
4
x
y
25 045 113
125 393
492
45 015 067
3
032 096
15
189
284
5
012 060
4
018 074
2
086
171
6
008 050
0
0
05
133 067
5
012 060
25
051
128
7
006 043
1
15
6
008 050
3
035
104
8
005 038
15
103 154
35
025
087
15
1
019 076
1
x
As t 1 we have x and y . As t 1 we have x and y . As t we have x 0 and y 0 . As t we have x 0 and y 0 . 62. x cot t, y 2 sin2 t, 0 t
y
2
1
1
x
63. (a) We first note that the center of circle C (the small circle) has coordinates
y
[a b] cos [a b] sin . Now the arc P Q has the same length as the arc
a ab a , and so . Thus the b b b x-coordinate of P is the x-coordinate of the center of circle C plus ab b cos b cos , and the y-coordinate of P is the b ab y-coordinate of the center of circle C minus b sin b sin . b ab So x a b cos b cos and b ab y a b sin b sin . b P Q, so b a
(b) If a 4b, b
a , and x 34 a cos 14 a cos 3, y 34 a sin 14 a sin 3. 4
¬ ú P»
¬
Q
(a, 0) P x
y
a
From Example 2 in Section 7.3, cos 3 4 cos3 3 cos . Similarly, one can prove that sin 3 3 sin 4 sin3 . Substituting, we get x 34 a cos 14 a 4 cos3 3 cos a cos3 y 34 a sin 14 a 3 sin 4 sin3 a sin3 . Thus,
x 23 y 23 a 23 cos2 a 23 sin2 a 23 , so x 23 y 23 a 23 .
a x
SECTION 8.4 Plane Curves and Parametric Equations
621
64. The coordinates of C are [a b] cos [a b] sin . Let be OC P as shown in the figure. Then the arcs traced a out by rolling the circle along the outside are equal, that is, b a . P is displaced from C by amounts b equal to the legs of the right triangle C PT , where C P is the hypotenuse. Since OC Q is a right triangle, it follows that a ab OC Q . Thus, . So 2 2 b 2 b 2 ab ab a b cos b sin x a b cos b sin b 2 2 b ab a b cos b cos b and ab ab a b sin b cos y a b sin b cos b 2 2 b ab a b sin b sin b Therefore, parametric equations for the epicycloid are ab ab x a b cos b cos and y a b sin b sin . b b
y
C a ¬
ú
Œ
T
P Q
x
65. A polar equation for the circle is r 2a sin . Thus the coordinates of Q are x r cos 2a sin cos and
y r sin 2a sin2 . The coordinates of R are x 2a cot and y 2a. Since P is the midpoint of Q R, we use the midpoint formula to get x a sin cos cot and y a 1 sin2 .
66. (a) C 2a cot 2a, so the x-coordinate of P is x 2a cot . Let B 0 2a. Then O AB is a right angle and O B A , so O A 2a sin and A 2a sin cos 2a sin2 . Thus, the
5
y-coordinate of P is y 2a sin2 .
(b) The curve is graphed with a 1.
-10
10
ay cos 67. We use the equation for y from Example 6 and solve for . Thus for 0 , y a 1 cos a ay ay ay cos1 . Substituting into the equation for x, we get x a cos1 sin cos1 . a a a 2ay y 2 2ay y 2 ay ay ay 2 1 1 However, sin cos 1 . Thus, x a cos , a a a a a y 2ay y 2 x 2ay y 2 x ay and we have cos1 1 cos a a a a 2ay y 2 x . y a 1 cos a
622
CHAPTER 8 Polar Coordinates and Parametric Equations
68. (a) 2
5
2
-2
-2
2
2
-5
-5
5
-2
-2
5 5 -5
-5
R 05 R1 R3 (b) The graph with R 5 seems to most closely resemble the profile of the engine housing.
R5
69. (a) In the figure, since O Q and QT are perpendicular and OT and T D are perpendicular, the angles formed by their intersections are equal, that is, DT Q. Now the coordinates of T are cos sin . Since T D is the length of the string that has been unwound from the circle, it must also have arc length , so T D . Thus the x-displacement from T to D is sin while the y-displacement from T to D is cos . So the coordinates of D are x cos sin and y sin cos . (b) y 10
T 1 ¬
D Q
1
-20
20
x
-10
70. (a) It takes 2 units of time. The parametric equations of the particle that moves twice as fast around the circle are x sin 2t, y cos 2t.
(b) From the first table, we see that the particle travels counterclockwise. If we want the particle to travel in a clockwise direction, then we want the second table to apply. Possible parametric equations for clockwise traversal of the unit circle are x sin t, y cos t. y
t
x sin t
0
0
1
2
0
0
0
1
1
0
2
0
1
71. C: x t, y t 2 ; D: x t, y t, t 0 (a) For C, x t, y t 2 y x 2 . For D, x t, y t y x 2 . For F, x et equation.
t
1
3 2
For E, x sin t
¹
1 t= 2
y cos t
t=0
t=¹
1
3¹ t= 2
E: x sin t, y 1 cos2 t
x 2 sin2 t 1 cos2 t y and so y x 2 .
x
2
x
y
0
1
1
0
0
3 2
1
1
0
2
0
1
F: x et , y e2t
x 2 e2t y and so y x 2 . Therefore, the points on all four curves satisfy the same rectangular
CHAPTER 8
Review
623
(b) Curve C is the entire parabola y x 2 . Curve D is the right half of the parabola because t 0 and so x 0. Curve E is the portion of the parabola for 1 x 1. Curve F is the portion of the parabola where x 0, since et 0 for all t. y
y
y
y
1
1
1
1
0
1
x
C
0
1
0
x
D
1
0
x
E
1
x
F
CHAPTER 8 REVIEW (12, ¹6)
1. (a)
O
(8, _ 3¹ 4)
¹/6
3 (b) x 12 cos 6 12 2 6 3, y 12 sin 12 1 6. Thus, the rectangular 6
2
coordinates of P are 6 3 6 .
3. (a)
(
7¹ _3, 4
7¹ 4
)
O
2. (a)
(b) x 8 cos 34 8 22 4 2, y 8 sin 34 8 22 4 2. Thus, the rectangular coordinates of P are 4 2 4 2 . 2¹ 3
4. (a) O
O
(b) x 3 cos 74 3 22 3 2 2 , y 3 sin 74 3 22 3 2 2 . Thus, the rectangular coordinates of P are 3 2 2 3 2 2 .
5. (a)
(_Ï3, 2¹ 3 ) (b) x 3 cos 23 3 12 23 , y 3 sin 23 3 23 32 . Thus, the rectangular coordinates of P are 23 32 . 6. (a)
(4Ï3, _ 5¹ 3) _ 5¹ 3
O
(b) x 4 3 cos 53 4 3 12 2 3, y 4 3 sin 53 4 3 23 6. Thus, the rectangular coordinates of P are 2 3 6 .
_ 3¹ 4
(_6Ï2, _ ¹4 ) O
_¹ 4
2 6, (b) x 6 2 cos 4 6 2 2 2 6. Thus, y 6 2 sin 4 6 2 2 the rectangular coordinates of P are 6 6.
624
CHAPTER 8 Polar Coordinates and Parametric Equations y
7. (a)
y
8. (a) P
P
1
2 x
2
(b) r
82 82 128 8 2 and tan1 88 .
Since P is in quadrant I, 4 . Polar coordinates for P are 8 2 4 . (c) 8 2 54
y
9. (a)
1 x
2 2 2 (b) r 6 8 2 2 and tan1
6 tan1 3 . Since x is 3 2
negative and y is positive, 23 . Polar coordinates for P are 2 2 23 . (c) 2 2 53 y
10. (a)
2 2
x
P
P
1 x
1
(b) r
2 2 6 2 6 2 144 12 and
(b) r
tan1 62 4 . Since P is in quadrant III, 6 2 54 . Polar coordinates for P are 12 54 . (c) 12 4
y
11. (a)
3 tan1 3 . Since P is in tan1 3 6 3 3
quadrant I, 6 . Polar coordinates for P are 6 6 . (c) 6 76 12. (a)
P
1 x
32
x
1 1
y 1
P
(b) r
2 3 3 32 36 6 and
2 3 12 2 3 and
tan1 33 . Since P is in quadrant II, 56 . Polar coordinates for P are 2 3 56 . (c) 2 3 6
(b) r
42 42 32 4 2 and
7 tan1 4 4 . Since P is in quadrant IV, 4 . Polar coordinates for P are 4 2 74 . (c) 4 2 34
CHAPTER 8
13. (a) x y 4 r cos r sin 4
4 r cos sin 4 r cos sin
14. (a) x y 1 r cos r sin 1 r 2 cos sin 1 r 2
(b) The rectangular equation is easier to graph. y
1
1
15. (a) x 2 y 2 4x 4y r 2 4r cos 4r sin r 2 r 4 cos 4 sin r 4 cos 4 sin
x
1
x
1
2 2 16. (a) x 2 y 2 2x y r 2 2 r cos r sin r 4 2r 2 cos sin r 2 2 cos sin
r 2 sin 2
(b) The polar equation is easier to graph.
(b) The polar equation is easier to graph.
O
1
O
17. (a)
1
18. (a)
2 O
(b) r 3 3 cos r 2 3r 3r cos , which gives x 2 y 2 3 x 2 y 2 3x x 2 3x y 2 3 x 2 y 2 . Squaring both sides 2 gives x 2 3x y 2 9 x 2 y 2 .
625
2 1 or cos sin sin 2
r 2 2 csc 2.
(b) The rectangular equation is easier to graph. y
Review
1
(b) r 3 sin r 2 3r sin , so x 2 y 2 3y 2 x 2 y 2 3y 94 94 x 2 y 32 94 .
626
CHAPTER 8 Polar Coordinates and Parametric Equations
19. (a)
20. (a)
O
O
2
2
(b) r 4 cos 3 4 cos 2 cos sin 2 sin 4 cos cos2 sin2 2 sin2 cos r 4 cos3 3 sin2 cos r 4 4r 3 cos3 3 sin2 cos , which gives
(b) r 2 sin 2 r 2 2 sin cos
r 3 4r 2 sin cos 32 r2 4 r sin r cos and so, since x r cos and y r sin , we get 3 x 2 y 2 16x 2 y 2 .
x 2 y2
2
4x 3 12x y 2 .
22. (a)
21. (a)
O
1
O
1 1 cos 2 cos2 sin2 r 2 cos2 sin2 1
(b) r 2 sec 2
1
(b) r 2 4 sin 2 8 sin cos 2 r 4 8r 2 sin cos , so x 2 y 2 8x y.
r 2 cos2 r 2 sin2 1
r cos 2 r sin 2 1 x 2 y 2 1. (b) r sin cos r 2 r sin r cos , so x 2 y 2 y x 2 2 x 2 x 14 y 2 y 14 12 x 12 y 12
23. (a)
O
1. 2
1
(b) r
24. (a)
O
1
4 2r r cos 4, so 2 x 2 y 2 x 4 2 cos
2 x 2 y 2 4 x. Squaring both sides gives 4 x 2 y 2 4 x2 16 8x x 2
3x 2 8x 4y 2 16.
CHAPTER 8
25. r cos 3, [0 3].
26. r sin 94, [0 8]
1
1
-1
1
-1
-1
27. r 1 4 cos 3, [0 6].
28. r sin , [3 3]. This graph is not bounded.
5
20
5
-20
20
-20
-5
30. (a)
Im
Im 2 2
4+4i 1 1
(b) 4 4i has r
16 16 4 2, and
(b) 10i has r 10 and 32 . (c) 10i 10 cos 32 i sin 32
32. (a)
Im
Re
_10i
Re
tan1 44 4 (in quadrant I). (c) 4 4i 4 2 cos 4 i sin 4
31. (a)
Im 1+Ï3 i 1
5+3i 1 1
(b) 5 3i. Then r
627
1
-1
29. (a)
Review
1
Re
25 9 34, and
tan1 53 . (c) 5 3i 34 cos tan1 53 i sin tan1 53
Re
3i has r 1 3 2, and tan1 3 3 (since is in quadrant I). (c) 1 3i 2 cos 3 i sin 3
(b) 1
628
CHAPTER 8 Polar Coordinates and Parametric Equations
33. (a)
34. (a)
Im
_1+i
5
1 _20
_1
(b) 1 i has r
Im
5
Re
Re
1 with 1 1 2 and tan 1
in quadrant II 34 . (c) 1 i 2 cos 34 i sin 34
(b) 20 has r 20 and . (c) 20 20 cos i sin
1 3 2 and tan 1 3 3 with in quadrant III 53 . Therefore, 1 3i 2 cos 53 i sin 53 , and so 4 1 3i 24 cos 203 i sin 203 16 cos 23 i sin 23 16 12 i 23 8 1 i 3 .
35. 1
3i has r
11 2 and 2 cos 36. 1 i has r 4 . Thus, 1 i 4 i sin 4 , and so 8 2 cos 2 i sin 2 16 1 i 0 16. 1 i8
37.
3 i has r 3 1 2 and tan 1 with in quadrant I 6 . Therefore, 3 i 2 cos 6 i sin 6 , 3 and so 4 1 cos 2 i sin 2 1 1 i 3 3i 24 cos 46 i sin 46 16 3 3 16 2 2 1 1 i 3 1 1 i 3 32 32
32 3 1 38. 12 23 i has r 14 34 1 and tan1 12 3 (since is in quadrant I). So 2 2 i cos 3 i sin 3 . Thus 20 1 3i 20 cos 20 i sin 20 cos 2 i sin 2 1 3 i 1 1 i 3 1 2 2 3 3 3 3 2 2 2
39. 16i has r 16 and 32 . Thus, 16i 16 cos 32 i sin 32 and so 3 4k 3 4k 12 12 cos i sin for k 0, 1. Thus 16 16i 4 4 4 1 i 1 2 2 1 i and the roots are 0 4 cos 34 i sin 34 2 2 1 4 cos 74 i sin 74 4 1 i 1 2 2 1 i. 2
2
40. 4 4 3i has r 16 48 8 and tan1 4 4 3 3 . Therefore, 4 4 3i 8 cos 3 i sin 3 . 13 6k 6k Thus 4 4 3i i sin for k 0, 1, 2. Thus the three roots are 3 8 cos 9 9 , 2 cos 7 i sin 7 , and 2 cos 13 i sin 13 . i sin 0 2 cos 1 2 9 9 9 9 9 9
2k 2k 41. 1 cos 0 i sin 0. Then 116 1 cos i sin for k 0, 1, 2, 3, 4, 5. Thus the six roots are 6 6 3 1 2 2 1 i 3 , 0 1 cos 0 i sin 0 1, 1 1 cos 3 i sin 3 2 i 2 , 2 1 cos 3 i sin 3 2 2 3 1 cos i sin 1, 4 1 cos 43 i sin 43 12 i 23 , and 5 1 cos 53 i sin 53 12 i 23 .
CHAPTER 8 18 cos 42. i cos 2 i sin 2 . Then i
2 2k 8
i sin
2 2k 8
cos
4k 16
i sin
Review
4k 16
629
for
i sin , cos 5 i sin 5 , cos 9 i sin 9 , k 0, 1, 2, 3, 4, 5, 6, 7. Thus the eight roots are 0 cos 16 1 2 16 16 16 16 16
13 17 17 21 21 25 25 3 cos 13 16 i sin 16 , 4 cos 16 i sin 16 , 5 cos 16 i sin 6 , 6 cos 16 i sin 16 , and 29 7 cos 29 16 i sin 16 . y
43. (a)
y
44. (a)
1 1 x
1 x
1
(b) x 1 t 2 , y 1 t t y 1. Substituting for t gives x 1 y 12 x 2y y 2 in
(b) x t 2 1, y t 2 1
y
t 2 y 1.
Substituting for t 2 gives x y 1 1 y x 2 where x 1 and y 1.
rectangular coordinates. 45. (a)
y
46. (a)
1 1
x
1
1
(b) x 1 cos t cos t x 1, and y 1 sin t sin t 1 y. Since cos2 t sin2 t 1, it follows that x 12 1 y2 1
x 12 y 12 1. Since t is restricted by
0t 2 , 1 cos 0 x 1 cos 2 1 x 2, and similarly, 0 y 1. (This is the
lower right quarter of the circle.)
(b) x
x
1 2 1 2 x 2, y 2 . Substituting for t t t
1 gives y 2 x 22 . Since t is restricted by t 1 1 0 t 2, we have , so x 52 and y 12 . t 2 The rectangular coordinate equation is y 2 x 22 , x 52 . 48. x sin t cos 2t, y cos t sin 3t
47. x cos 2t, y sin 3t 1
-1
1
1 -1
-1
1 -1
49. The coordinates of Q are x cos and y sin . The coordinates of R are x 1 and y tan . Hence, the midpoint P 1 cos sin tan 1 cos sin tan , so parametric equations for the curve are x and y . is 2 2 2 2
630
CHAPTER 8 Polar Coordinates and Parametric Equations
CHAPTER 8 TEST 1. (a) x 8 cos 54 8 22 4 2, y 8 sin 54 8 22 4 2. So the point has rectangular coordinates 4 2 4 2 . (b) P 6 2 3 in rectangular coordinates. So tan 263 and the reference angle is 6 . Since P is in 2 quadrant II, we have 56 . Next, r 2 62 2 3 36 12 48, so r 4 3. Thus, polar coordinates for the point are 4 3 56 or 4 3 116 . (b) r 8 cos
2. (a)
x 2 8x y 2 0
O
x 42 y 2 16
1
r 2 8r cos
x 2 y 2 8x
x 2 8x 16 y 2 16
The curve is a circle.
3.
The curve is a limaçon.
O
4. (a)
2
1+Ï3 i 1
1
3i has r 1 3 2 and tan1 3 3 . So, in trigonometric form, 1 3i 2 cos 3 i sin 3 . (c) z 1 3i 2 cos 3 i sin 3 z 9 29 cos 93 i sin 93 512 cos 3 i sin 3
(b) 1
Im
Re
512 1 i 0 512
i sin 7 and z 2 cos 5 i sin 5 . 5. z 1 4 cos 712 2 12 12 12 7 5 7 5 i sin 8 cos i sin Then z 1 z 2 4 2 cos 12 12 7 5 7 5 3 2 i sin z 1 z 2 42 cos i sin 2 cos 6 6 2 12 12
8 and 1i 2
3 i.
The Path of a Projectile
6. 27i has r 27 and 2 , so 27i 27 cos 2 i sin 2 . Thus, 2k 2k 3 2 2 13 i sin 27 cos 27i 3 3 4k 4k i sin 3 cos 6 6
Im wÁ
wü 1 0
for k 0, 1, 2. Thus, the three roots are 3 1 i 3 3 i , 3 i sin 0 3 cos 6 6 2 2 2 1 3 cos 56 i sin 56 3 23 12 i 32 3 i , and 2 3 cos 96 i sin 96 3i. 7. (a) x 3 sin t 3, y 2 cos t, 0 t . From the
work of part (b), we see that this is the half-ellipse
Squaring both sides gives y
y 2 cos t
1 1
x
1
Re
wª
(b) x 3 sin t 3 x 3 3 sin t
shown.
631
x 3 sin t. 3
x 32 sin2 t. Similarly, 9
y cos t, and squaring both sides gives 2
y2 cos2 t. Since sin2 t cos2 t 1, it follows that 4 y2 x 32 1. Since 0 t , sin t 0, so 9 4 3 sin t 0 3 sin t 3 3, and so x 3. Thus the curve consists of only the right half of the ellipse.
8. We start at the point 3 5. Because the line has slope 2, for every 1 unit we move to the right, we must move up 2 units. Therefore, parametric equations are x 3 t, y 5 2t. 9. (a) x 3 sin 2t, y 3 cos 2t. The radius is 3 and the position at time t 0 is 3 sin 2 0 3 cos 2 0 0 3. Initially x is increasing and y is decreasing, so the motion is clockwise. At time t the object is back at 0 3, so it takes units of time to complete one revolution. (b) If the speed is doubled, the time to complete one revolution is halved, to 2 . Parametric equations modeling this motion are x 3 sin 4t, y 3 cos 4t. (c) x 2 y 2 3 sin 2t2 3 cos 2t2 9 sin2 2t cos2 2t 9, so an equation in rectangular coordinates is x 2 y 2 9.
(d) In polar coordinates, an equation is r 3.
FOCUS ON MODELING The Path of a Projectile x . Substituting this value for t into the equation for y, we get 0 cos 2 x x g x 2 . This shows y 0 sin t 12 gt 2 y 0 sin y tan x 2 12 g 0 cos 0 cos 2 cos2
1. From x 0 cos t, we get t
0
that y is a quadratic function of x, so its graph is a parabola as long as 90 . When 90 , the path of the projectile is a straight line up (then down).
632
FOCUS ON MODELING
2. (a) Applying the given values, we get x 0 cos t 15t and y 4 0 sin t 12 gt 2 4 15 3t 16t 2 as the parametric equations for the path of the baseball.
y 10
0
10
x
20
(b) The baseball will hit the ground when y 0 4 2598t 16t 2 . Using the Quadratic Formula, 2598 25982 4164 t 177 seconds (since t must be positive). So the baseball travels 32 x 15 177 265 ft before hitting the ground after 177 s.
3. (a) We find the time at which the projectile hits the ground using the equation t 2 0 gsin , where 0 1000 and g 16. Since the rocket is fired 5 from the vertical axis and is measured from the horizontal, we have sin 85 6226, so the rocket is in the air for 6226 seconds. 90 5 85 . Then t 200032
(b) Substituting the given values into y 0 sin t 12 gt 2 , we get
(d)
y 1000 sin 85 t 16t 2 996t 16t 2 . Then
y f t at 2 bt c, where a 16, b 996, and c 0. So y is a
quadratic function whose maximum value is attained at
b 996 31125, and the maximum value is t 2a 216
f 31125 996 31125 16 311252 15,500 (see Section 3.1 for a
guide to finding the maximum value of a quadratic function). Thus, the
y 15,000 10,000 5000
0
2000
4000
6000 x
rocket reaches a maximum height of 15,500 feet. (c) We use the equation x 0 cos t to find the horizontal distance traveled after t seconds. Since the rocket hits the ground after
6226 seconds, substituting into the equation for horizontal distance gives x 1000 cos 85 6226 5426. Thus, the rocket travels a horizontal
distance of 5426 feet.
4. (a) The missile hits the ground when t x 0 cos
2 0 sin , so the missile travels g
2 2 0 sin 0 sin 2 meters. Substituting 0 330 ms, g g
3302 sin 2 sin 2 08999. 98 So 2 6415 3208 or 2 11585 5793 . x 10 km, and g 98 m/s2 , we get 10000
y 4000 3000 2000 1000 0
4000
8000
x
(b) The missile fired at 3208 will hit the target in 357 seconds, while the missile fired at 5793 reaches the target in 571 seconds.
The Path of a Projectile
633
5. We use the equation of the parabola from Exercise 1 and find its vertex: 2 2 sin cos x g g x2 y 2 x2 0 y tan x 2 g 2 0 cos2 2 0 cos2 2 2 sin cos x 2 sin cos 2 02 sin cos 2 g g 0 0 2 y 2 2 x g g g 2 0 cos2 2 0 cos2 2 02 sin cos 02 sin cos 02 sin2 02 sin2 g x y 2 . Thus the vertex is at , so the maximum g 2g g 2g 2 0 cos2 2 sin2 . height is 0 2g 6. Since the horizontal component of the projectile’s velocity has been reduced by , the parametric equations become x 0 cos t, y 0 sin t 12 gt 2 . y
7. In Exercise 6 we derived the equations x 0 cos t,
y 0 sin t 12 gt 2 . We plot the graphs for the given values of
10 ¬=60¡
0 , , and in the figure to the right. The projectile will be blown
8
backwards if the horizontal component of its velocity is less than the
6
speed of the wind, that is, 32 cos 24 cos 34 414 .
4 ¬=45¡
¬=40¡ ¬=30¡
2
_14 _12 _10 _8 _6 _4 _2 0
The optimal firing angle appears to be between 15 and 30 . We
y
graph the trajectory for 20 , 23 , and 25 . The solution appears to be close to 23 .
3
2
4 x
¬=15¡ ¬=5¡
¬=25¡ ¬=23¡
2
¬=20¡ 1
0
1
2
3
4
x
8. (a) Answers will vary. (b) Both projectiles land at the same spot. The projectile fired with 30 flies lower and lands first. 2 2 o o o 0 . Thus, when 0 t 49 t 2 t (c) When 90 , we have y 0 t 49t 2 49 t 2 49 49 98 196 is doubled the maximum height of the ball increases by a factor of 4.
9
VECTORS IN TWO AND THREE DIMENSIONS
9.1
VECTORS IN TWO DIMENSIONS
1. (a) The vector u has initial point A and terminal point B.
(b) The vector u has initial point 2 1 and terminal point 4 3. In component form we write u 2 2 and v 3 6. Then 2u 4 4 and u v 1 8.
v
u
u+ v
2u
u
2. (a) The length of a vector w a1 a 2 is w u 22 22 8 2 2.
a12 a22 , so the length of the vector u in Figure II is
(b) If we know the length w and direction of a vector w then we can express the vector in component form as w w cos w sin .
3. 2u 2 2 3 4 6
4. v 3 4 3 4
y
y
2u v
u 1
1
v
u 1
x
5. u v 2 3 3 4 2 3 3 4 1 7 y
1
_v
x
6. u v 2 3 3 4 2 3 3 4 5 1 y
v u+v
_v
u 1
1
x
u 1 u-v
1
x
635
636
CHAPTER 9 Vectors in Two and Three Dimensions
7. v 2u 3 4 2 2 3 3 2 2 4 2 3
8. 2u v 2 2 3 3 4 2 2 3 2 3 4 1 10
7 2 y
y
v
2u+v _2u
v
1
2u
x
1
1
v-2u
x
1
In Solutions 9–18, v represents the vector with initial point P and terminal point Q.
9. P 2 1, Q 5 4. v 5 2 4 1 3 3
10. P 2 1, Q 3 4. v 3 2 4 1 5 3
11. P 1 2, Q 4 1. v 4 1 1 2 3 1
12. P 3 1, Q 1 2. v 1 3 2 1 2 3
13. P 3 2, Q 8 9. v 8 3 9 2 5 7
14. P 1 1, Q 9 9. v 9 1 9 1 8 8
15. P 5 3, Q 1 0. v 1 5 0 3 4 3
16. P 1 3, Q 6 1. v 6 1 1 3 5 4
17. P 1 1, Q 1 1.
18. P 8 6, Q 1 1.
v 1 1 1 1 0 2 y
19.
v 1 8 1 6 7 5
(6, 7)
y
20.
(3, 5)
u
u (4, 3)
(4, 3)
1
1 x
1
The terminal point is 4 2 3 4 6 7. 21.
y
The terminal point is 4 1 3 2 3 5. y
22.
(4, 3) 1 1
(4, 3)
u u
(_4, 2) (8, 0) x
The terminal point is 4 4 3 3 8 0.
x
1
1 1
x
The terminal point is 4 8 3 1 4 2.
SECTION 9.1 Vectors in Two Dimensions y
23.
637
y
24. (_3, 5)
(2, 3)
u u 1
u
(_3, 5)
u
1
x
1
y
u
x
u
1
25.
u
(2, 3)
26.
(_3, 5)
y (2, 3)
(2, 3)
u u
(_3, 5)
u
1 1
1 1
x
u
x
u
27. u 1 4 i 4j
28. u 2 10 2i 10j
29. u 3 0 3i
30. u 0 5 5j
31. u 2 7, v 3 1. 2u 2 2 7 4 14; 3v 3 3 1 9 3; u v 2 7 3 1 5 8; 3u 4v 6 21 12 4 6 17
32. u 2 5 v 2 8. 2u 2 2 5 4 10; 3v 3 2 8 6 24; u v 2 5 2 8 0 3; 3u 4v 6 15 8 32 14 47
33. u 0 1, v 2 0. 2u 20 1 0 2; 3v 32 0 6 0; uv 0 12 0 2 1; 3u 4v 0 3 8 0 8 3
34. u i, v 2j. 2u 2i; 3v 3 2j 6j; u v i 2j; 3u 4v 3i 8j
35. u 2i, v 3i 2j. 2u 2 2i 4i; 3v 3 3i 2j 9i 6j; u v 2i 3i 2j 5i 2j; 3u 4v 3 2i 4 3i 2j 6i 8j
36. u i j, v i j. 2u 2i 2j 3v 3i 3j u v i j i j 2i; 3u 4v 3 i j 4 i j i 7j 37. u 2i j v 3i 2j. Then u 22 12 5; v 32 22 13; 2u 4i 2j; 2u 42 22 2 5; 2 1 v 3 ij; 1 v 3 12 12 13; uv 5ij; u v 52 12 26; uv 2ij3i2j i3j; 2 2 2 2 u v 12 32 10; u v 5 13 38. u 2i 3j v i 2j. Then u 4 9 13; v 1 4 5; 2u 4i 6j; 2u 16 36 2 13; 1 v 1 i j; 1 v 1 1 1 5; u v i j; u v 1 1 2; u v 3i 5j; 2 2 2 4 2 u v 9 25 34; u v 13 5 39. u 10 1, v 2 2. Then u 102 12 101; v 22 22 2 2; 2u 20 2; 2u 202 22 404 2 101; 12 v 1 1; 12 v 12 12 2; u v 8 3; u v 82 32 73; u v 12 1; u v 122 12 145; u v 101 2 2
638
CHAPTER 9 Vectors in Two and Three Dimensions
40. u 6 6, v 2 1. Then u 36 36 6 2; v 4 1 5; 2u 12 12; 2u 144 144 12 2; 12 v 1 12 ; 12 v 1 14 12 5; u v 8 5; u v 64 25 89; u v 4 7; u v 16 49 65; u v 6 2 5 In Solutions 41–46, x represents the horizontal component and y the vertical component. 41. v 40, direction 30 . x 40 cos 30 20 3 and y 40 sin 30 20. Thus, v xi yj 20 3i 20j. 42. v 50, direction 120 . x 50 cos 120 25 and y 50 sin 120 25 3. Thus, v xi yj 25i 25 3j. 43. v 1, direction 225 . x cos 225 1 and y sin 225 1 . Thus, v xi yj 1 i 1 j 22 i 22 j. 2 2
2
2
44. v 800, direction 125 . x 800 cos 125 45886 and y 800 sin 125 65532. Thus, v xi yj 800 cos 125 i 800 sin 125 j 45886i 65532j. 45. v 4, direction 10 . x 4 cos 10 394 and y 4 sin 10 069. Thus, v xi yj 4 cos 10 i 4 sin 10 j 394i 069j. 46. v 3, direction 300 . x 3 cos 300 23 and y 3 sin 300 32 . Thus, v xi yj 23 i 32 j. 47. v 3 4. The magnitude is v 32 42 5. The direction is , where tan 43 tan1 43 5313 .
48. v 22 22 . The magnitude is v 12 12 1. The direction is , where tan 1 with in quadrant III 180 tan1 1 225 .
5 with in 49. v 12 5. The magnitude is v 122 52 169 13. The direction is , where tan 12 5 15738 . quadrant II tan1 12 50. v 40 9. The magnitude is v 9 1268 . tan1 40
9 with in quadrant I 1600 81 41. The direction is , where tan 40
2 3j. The magnitude is v 12 3 2. The direction is , where tan 3 with in quadrant I tan1 3 60 . 52. v i j. The magnitude is v 1 1 2. The direction is , where tan 1 with in quadrant I
51. v i
tan1 1 45 .
3 2598, y 30 sin 30 15. So the horizontal component of 2 force is 15 3 lb and the vertical component is 15 lb.
53. v 30, direction 30 . x 30 cos 30 30
54. v 500, direction 70 . x 500 cos 70 17101, y 500 sin 70 46985. So the east component of the velocity is 17101 mi/h and the north component is 46985 mi/h. 55. The flow of the river can be represented by the vector v 3j and the swimmer can be represented by the vector u 2i. Therefore the true velocity is u v 2i 3j. 56. If the current is 12 mi/h, it can be represented by the vector v 12j. If the swimmer heads in a direction north of east, his velocity relative to the water is u 2 cos i 2 sin j and so his velocity relative to land is u v 2 cos i 2 sin 12 j. We want the y-component of his velocity to be 0, so we calculate 2 sin 12 0 sin 35 sin1 53 369 . Therefore, he should swim 369 north of east, or N 531 E.
SECTION 9.1 Vectors in Two Dimensions
u
57. The speed of the airplane is 300 mi/h, so its velocity relative to the air is v 300 cos i 300 sin j. The wind has velocity w 30j, so
639
w
¬
v
the true course of the airplane is given by
u v+w 300 cos i 300 sin 30 j. We want the y-component of the airplane’s velocity to be 0, so we
1 574 . Therefore, the airplane should head in the direction 18574 (or solve 300 sin 30 0 sin 10 S 8426 W). 58. The ocean currents can be represented by the vector v 3i and the salmon can be represented by the vector u 5 2 2 i 5 2 2 j. Therefore u v 5 2 2 3 i 5 2 2 j represents the true velocity of the fish.
59. (a) The velocity of the wind is 40j.
(b) The velocity of the jet relative to the air is 425i. (c) The true velocity of the jet is v 425i 40j 425 40. 40 54 is (d) The true speed of the jet is v 4252 402 427 mi/h, and the true direction is tan1 425 N 846 E.
55 3 60. (a) The velocity of the wind is w 55 cos 60 55 sin 60 55 2 2 . (b) The velocity of the jet is w 765 cos 45 765 sin 45 7652 2 7652 2 . 765 2 55 3 765 2 56844 58857. (c) The true velocity of the jet is v w 55 2 2 2 2 (d) The true speed of the jet is w v 568442 588572 818 mih. The direction of the vector w v is tan1 58857 56844 46 . Thus the true direction of the jet is approximately N 44 E. 61. If the direction of the plane is N 30 W, the airplane’s velocity is u ux u y where ux 765 cos 60 3825,
and u y 765 sin 60 66251. If the direction of the wind is N 30 E, the wind velocity is w x y where x 55 cos 60 275, and y 55 sin 60 4763. Thus, the actual flight path is v u w 3825 275 66251 4763 355 71014, and so the true speed is v 3552 710142 794 mih, and the true direction is tan1 71014 1166 so is N 266 W. 355
62. Let v be the velocity vector of the jet and let be the direction of this vector. Thus v 765 cos i 765 sin j. If w 55 3 j. is the velocity vector of the wind then the true course of the jet is v w 765 cos 55 2 i 765 sin 2 To achieve a course of true north, the east-west component (the i component) of the jet’s velocity vector must be 0. That 275 1 275 921 . Thus the pilot should head his plane in the is 765 cos 55 2 0 cos 765 cos 765
direction N 21 W.
63. (a) The velocity of the river is represented by the vector r 10 0.
(b) Since the boater direction is 60 from the shore at 20 mih, the velocity of the boat is represented by the vector b 20 cos 60 20 sin 60 10 10 3 10 1732. (c) w r b 10 10 0 1732 20 1732 (d) The true speed of the boat is w 202 17322 265 mih, and the true direction is 409 N 491 E. tan1 1732 20
64. Let w be the velocity vector of the water, let v be the velocity vector of the boat, and let be the direction of v. Then v 20 cos i 20 sin j and w 10i. The true course of the boat is v w 20 cos 10 i 20 sin j. To achieve a course of true north, the east-west component of the boat’s velocity vector must be 0. Thus, 20 cos 10 0 10 12 cos1 12 120 . Thus the boater should head her boat in the direction N 30 W. cos 20
640
CHAPTER 9 Vectors in Two and Three Dimensions
65. (a) Let b bx b y represent the velocity of the boat relative to the water. Then b 24 cos 18 24 sin 18 228 74. (b) Let w x y represent the velocity of the water. Then w 0 where is the speed of the water. So the true velocity of the boat is b w 24 cos 18 24 sin 18 . For the direction to be due east, we must have 24 sin 18 0 742 mih. Therefore, the true speed of the water is 74 mi/h. Since b w 24 cos 18 0, the true speed of the boat is b w 24 cos 18 228 mih. 66. Let w represent the velocity of the woman and l the velocity of the ocean liner. Then w 2 0, and l 0 25, and so r 2 0 0 25. Hence, relative to the water, the woman’s speed is r 4 625 2508 mih, and her direction 25 9457 or approximately N 457 W. is tan1 2
67. F1 2 5 and F2 3 8. (a) F1 F2 2 3 5 8 5 3
(b) The additional force required is F3 0 0 5 3 5 3.
68. F1 3 7, F2 4 2, and F3 7 9. (a) F1 F2 F3 3 4 7 7 2 9 0 0 (b) No additional force is required.
69. F1 4i j, F2 3i 7j, F3 8i 3j, and F4 i j. (a) F1 F2 F3 F4 4 3 8 1 i 1 7 3 1 j 0i 4j 4j (b) The additional force required is F5 0i 0j 0i 4j 4j.
70. F1 i j, F2 i j, and F3 2i j. (a) F1 F2 F3 1 1 2 i 1 1 1 j 0i j j
(b) The additional force required is F4 0i 0j 0i j j.
71. F1 10 cos 60 10 sin 60
5 5 3 , F2 8 cos 30 8 sin 30 4 3 4 , and
F3 6 cos 20 6 sin 20 5638 2052. (a) F1 F2 F3 5 4 3 5638 5 3 4 2052 757 1061.
(b) The additional force required is F4 0 0 757 1061 757 1061. 72. F1 3 1, F2 1 2, F3 2 1, and F4 0 4. (a) F1 F2 F3 F4 3 1 2 0 1 2 1 4 2 4 (b) The additional force required is F5 0 0 2 4 2 4.
73. From the figure we see that T1 T1 cos 50 i T1 sin 50 j and T2 T2 cos 30 i T2 sin 30 j. Since T1 T2 100j we get T1 cos 50 T2 cos 30 0 and T1 sin 50 T2 sin 30 100. From the first cos 50 cos 50 sin 30 , and substituting into the second equation gives T1 sin 50 T1 100 equation, T2 T1 cos 30 cos 30 T1 sin 50 cos 30 cos 50 sin 30 100 cos 30 T1 sin 50 30 100 cos 30 cos 30 T1 100 879385. sin 80 cos 30 Similarly, solving for T1 in the first equation gives T1 T2 and substituting gives cos 50 cos 30 sin 50 T2 T2 sin 30 100 T2 cos 30 sin 50 cos 50 sin 30 100 cos 50 cos 50 100 cos 50 T2 652704. Thus, T1 879416 cos 50 i 879416 sin 50 j 565i 674j and sin 80 T2 652704 cos 30 i 652704 sin 30 j 565i 326j.
SECTION 9.2 The Dot Product
641
74. From the figure we see that T1 T1 cos 223 i T1 sin 223 j and T2 T2 cos 415 i T2 sin 415 j. Since T1 T2 18 278j, we get T1 cos 223 T2 cos 415 0 and T1 sin 223 T2 sin 415 18 278. From the cos 223 first equation, T2 T1 , and substituting into the second equation gives cos 415 cos 223 sin 415 T1 sin 223 T1 18 278 cos 415 T1 sin 223 cos 415 cos 223 sin 415 18 278 cos 415 cos 415 T1 sin 223 415 18,278 cos 415 T1 18,278 15 257. sin 638 cos 415 Similarly, solving for T1 in the first equation gives T1 T2 and substituting gives cos 223 cos 415 sin 223 T2 T2 sin 415 18,278 T2 sin 223 cos 415 sin 415 cos 223 18,278 cos 223 cos 223 18,278 cos 223 T2 18,847. sin 638 Thus, T1 15,257 cos 223 i 15,257 sin 223 j 14,116i 5,789j and T2 18,847 cos 415 i 18,847 sin 415 j 14,116i 12,488j. 75. When we add two (or more vectors), the resultant vector can be found by first placing the initial point of the second vector at the terminal point of the first vector. The resultant vector can then found by using the new terminal point of the second vector and the initial point of the first vector. When the n vectors are placed head to tail in the plane so that they form a polygon, the initial point and the terminal point are the same. Thus the sum of these n vectors is the zero vector.
9.2
THE DOT PRODUCT
1. The dot product of u a1 a2 and v b1 b2 is defined by u v a1 a2 b1 b2 . The dot product of two vectors is a real number, or scalar, not a vector. uv . So if u v 0, the vectors are perpendicular. 2. The angle satisfies cos u v u 3. (a) The component of u along v is the scalar u cos and can be uv expressed in terms of the dot product as . v projv u ¬ v uv compv u v. (b) The projection of u onto v is the vector projv u v2 4. The work done by a force F in moving an object along a vector D is W F D. 5. (a) u v 2 0 1 1 2 0 2 6. (a) u v 1 3 3 1 3 3 0 uv uv 2 1 45 (b) cos (b) cos 0 90 2 2 2 u v u v
7. (a) u v 2 7 3 1 6 7 13 uv 13 56 (b) cos 53 10 u v 9. (a) u v 3 2 1 2 3 4 1 uv (b) cos 1 97 13 5 u v 11. (a) u v 0 5 1 3 0 5 3 5 3 uv (b) cos 5523 23 30 u v
8. (a) u v 6 6 1 1 6 6 12 uv 12 1 180 (b) cos 6 2 2 u v 10. (a) u v 2 1 3 2 6 2 4 uv (b) cos 4 60 5 13 u v 12. (a) u v 1 1 1 1 1 1 0 uv 0 0 90 (b) cos 2 2 u v
642
CHAPTER 9 Vectors in Two and Three Dimensions
13. (a) u v i 3j 4i j 4 3 1 uv 1 856 (b) cos 10 17 u v
14. (a) u v 3i 4j 2i j 6 4 10 uv 10 (b) cos 5 5 u v 1534 cos1 10 5 5
15. u v 12 12 0 vectors are orthogonal
16. u v 0 0 0 vectors are orthogonal
17. u v 8 12 4 0 vectors are not orthogonal
18. u v 0 0 0 vectors are orthogonal
19. u v 24 24 0 vectors are orthogonal
20. u v 4 0 4 vectors are not orthogonal
21. u v u w 2 1 1 3 2 1 3 4
22. u v w 2 1 [1 3 3 4]
23649 23. u v u v [2 1 1 3] [2 1 1 3] 3 2 1 4 3 8 5 25. 27. 29.
30.
31.
32.
33.
34.
2 1 4 1 8 1 9 24. u v u w 2 1 1 3 2 1 3 4 1 10 10
3 5 12 24 uv 12 uv 2 2 2 x 26. x 2 v v 5 5 1 2 uv uv 0 24 56 0 28 x 24 28. x 5 v v 1 10 2 4 1 1 uv 1 1 1 1. v (a) u1 projv u 12 12 v2 (b) u2 u u1 2 4 1 1 3 3. We resolve the vector u as u1 u2 , where u1 1 1 and u2 3 3. 7 4 2 1 uv 2 1 2 2 1 4 2 v (a) u1 projv u 22 12 v2 (b) u2 u u1 7 4 4 2 3 6. We resolve the vector u as u1 u2 , where u1 4 2 and u2 3 6. 1 2 1 3 uv 1 1 3 1 3 1 v 3 (a) u1 projv u 2 2 2 v2 12 32 (b) u2 u u1 1 2 12 32 32 12 . We resolve the vector u as u1 u2 , where u1 12 32 and u2 32 12 . 11 3 3 2 uv 3 2 3 3 2 9 6 (a) u1 projv u v v2 32 22 (b) u2 u u1 11 3 9 6 2 3. We resolve the vector u as u1 u2 , where u1 9 6 and u2 2 3. 2 9 3 4 uv 24 3 4 65 3 4 18 v (a) u1 projv u 5 5 2 3 2 v 3 4 18 24 21 18 24 (b) u2 u u1 2 9 5 5 28 5 5 . We resolve the vector u as u1 u2 , where u1 5 5 and 21 u2 28 5 5 . 1 1 2 1 uv 2 1 15 2 1 25 15 v (a) u1 projv u 2 2 v 22 1 (b) u2 u u1 1 1 25 15 35 65 . We resolve the vector u as u1 u2 , where u1 25 15 and u2 35 65 .
35. W F d 4 5 3 8 28
36. W F d 400 50 201 0 80 400
37. W F d 10 3 4 5 25
38. W F d 4 20 5 15 280
39. Let u u1 u2 and v v1 v2 . Then u v u 1 u 2 1 2 u 1 1 u 2 2 1 u 1 2 u 2 1 2 u 1 u 2 v u
SECTION 9.2 The Dot Product
643
40. Let u u1 u2 and v v1 v2 . Then cu v c u 1 u 2 1 2 cu 1 cu 2 1 2 cu 1 1 cu 2 2 c u 1 1 u 2 2 c u v u 1 c 1 u 2 c 2 u 1 u 2 c 1 c 2 u cv
41. Let u u 1 u 2 , v 1 2 , and w 1 2 . Then u v w u 1 u 2 1 2 1 2 u 1 1 u 2 2 1 2
u 1 1 1 1 u 2 2 2 2 u 1 1 u 2 2 1 1 2 2 u 1 u 2 1 2 1 2 1 2 u w v w
42. Let u u 1 u 2 and v 1 2 . Then u v u v u 1 u 2 1 2 u 1 u 2 1 2 u 1 1 u 2 2 u 1 1 u 2 2 2 2 u 21 u 22 12 22 u2 v2 u 21 12 u 22 22 u 21 u 22 12 22 43. We use the definition that projv u projv u u projv u
uv
v. Then v2 uv uv uv uv uv v u v v v v u v2 v2 v2 v2 v2
u v2 v2
u v2 v4
v2
Thus u and u projv u are orthogonal. uv u v v v v= u v. 44. v projv u v v2 v2
u v2 v2
u v2 v2
0
45. W F d 4 7 4 0 16 ft-lb
46. The displacement of the object is D 11 13 2 5 9 8. Hence, the work done is W F D 2 8 9 8 18 64 82 ft-lb.
47. The distance vector is D 200 0 and the force vector is F 50 cos 30 50 sin 30 . Hence, the work done is W F D 200 0 50 cos 30 50 sin 30 200 50 cos 30 8660 ft-lb.
48. W F d, and in this problem F 0 2500, d 500 cos 12 500 sin 12 Thus,
W 0 2500 500 cos 12 500 sin 12 0 2500 1040 260,000 ft-lb. This is the work done by gravity; the
car does (positive) work in overcoming the force of gravity. So the work done by the car is 260,000 ft-lb. 490 28218, and thus the weight of the car 49. (a) Since the force parallel to the driveway is 490 w sin 10 w sin 10 is about 2822 lb. (b) The force exerted against the driveway is 28218 cos 10 2779 lb.
50. Since the weight of the car is 2755 lb, the force exerted perpendicular to the earth is 2755 lb. Resolving this into a force u perpendicular to the driveway gives u 2766 cos 65 1164 lb. Thus, a force of about 1164 lb is required.
51. Since the force required parallel to the plane is 80 lb and the weight of the package is 200 lb, it follows that 80 200 sin , 80 2358 , and so the angle of inclination is where is the angle of inclination of the plane. Then sin1 200 approximately 236 .
52. Let R represent the force exerted by the rope and d the force causing the cart to roll down the ramp. Gravity acting on the cart exerts a force w of 40 lb directly downward. So the magnitude of the part of that force causing the cart to roll down the ramp is d 40 sin 15 . The angle between R and d is 45 , so the magnitude of the force holding the cart up is R cos 45 . Equating these two, we have R cos 45 40 sin 15 , so R
40 sin 15 1464 lb. cos 45
644
CHAPTER 9 Vectors in Two and Three Dimensions
53. (a) 2 0 4 2 8, so Q 0 2 lies on L. 2 2 4 1 4 4 8,
(c) From the graph, we can see that u w is
so R 2 1 lies on L. (b) u Q P 0 2 3 4 3 2. v Q R 0 2 2 1 2 1. 3 2 2 1 uv 2 1 w projv u v 2 2 v 2 2 1 8 85 2 1 16 5 5
9.3
orthogonal to v (and thus to L). Thus, the
distance from P to L is u w. y
P
L
u Q
u-w
w v
R x
0
THREE-DIMENSIONAL COORDINATE GEOMETRY
z
1. In a three-dimensional coordinate system the three mutually perpendicular axes are called the x-axis, the y-axis, and the z-axis. The point P has coordinates 5 2 3. The equation of the plane passing through P and parallel to the x z-plane is y 2.
0
P
y
x
2. The distance between the point P x1 y1 z 1 and Q x2 y2 z 2 is given by the formula d P Q x2 x1 2 y2 y1 2 z 2 z 1 2 . The distance between the point P in the figure and the origin is 5 02 2 02 3 02 38. The equation of the sphere centered at P with radius 3 is x 52 y 22 z 32 9.
z
3. (a)
z
4. (a) 0
P(3, 1, 0)
x
y
P(5, 0, 10) Q(3, _6, 7)
Q(_1, 2, _5)
1 32 2 12 5 02 42
0
(b) d P Q
x
(b) d P Q
x
3 52 6 02 7 102 7
SECTION 9.3 Three-Dimensional Coordinate Geometry
5. (a)
645
6. (a) Q(8, _7, 4)
Q(_12, 3, 0) P(_2, _1, 0) x
0
z 0
y
(b) d P Q 12 22 3 12 0 02 2 29 7. x 4 is a plane parallel to the yz-plane.
P(5, _4, _6)
(b) d P Q 8 52 7 42 4 62 118 8. y 2 is a plane parallel to the xz-plane.
z
z
0
y
4
_2
x
0
y
x
9. z 8 is a plane parallel to the x y-plane.
10. y 1 is a plane parallel to the xz-plane.
z 8
z
0
x
y
_1
0
y
x
2 11. A sphere with radius r 5 and center C 2 5 3 has equation x 22 y 5 z 32 52 , or x 22 y 52 z 32 25.
12. A sphere with radius r 3 and center C 1 4 7 has equation x 12 y 42 z 72 9. 13. A sphere with radius r 6 and center C 3 1 0 has equation x 32 y 12 z 2 6. 14. A sphere with radius r 11 and center C 10 0 1 has equation x 102 y 2 z 12 11. 15. We complete the squares in x, y, and z: x 2 y 2 z 2 10x 2y 8z 9 x 2 10x 25 y 2 2y 1 z 2 8z 16 9 25 1 16 x 52 y 12 z 42 51. This is an equation of a sphere with center 5 1 4 and radius 51.
646
CHAPTER 9 Vectors in Two and Three Dimensions
16. We complete the squares in x, y, and z: x 2 y 2 z 2 4x 6y 2z 10 x 2 4x 4 y 2 6y 9 z 2 2z 1 10 4 9 1 x 22 y 32 z 12 24. This is an equation of a sphere with center 2 3 1 and radius 2 6. 17. We complete the squares in x, y, and z: x 2 y 2 z 2 12x 2y x 2 12x 36 y 2 2y 1 z 2 36 1 x 62 y 12 z 2 37. This is an equation of a sphere with center 6 1 0 and radius 37. 18. We complete the squares in x, y, and z: x 2 y 2 z 2 14y 6z x 2 y 2 14y 49 z 2 6z 9 49 9 x 2 y 72 z 32 58. This is an equation of a sphere with center 0 7 3 and radius 58.
19. (a) To find the trace in the yz-plane, we set x 0: 0 12 y 22 z 102 100 y 22 z 102 99. This represents a circle with center 0 2 10 and radius 3 11. (b) We set x 4 and find 4 12 y 22 z 102 100 y 22 z 102 75. This represents a circle with center 4 2 10 and radius 5 3.
20. (a) To find the trace in the xz-plane, we set y 0: x 2 0 42 z 32 144 x 2 z 32 128. This represents a circle with center 0 0 3 and radius 8 2.
(b) We set z 2 and find x 2 y 42 2 32 144 x 2 y 42 119. This is a circle with center 0 4 2 and radius 119.
21. With the origin at its center, an equation of the tank is x 2 y 2 z 2 25. The metal circle is the trace in the plane z 4, so its equation is x 2 y 2 42 25 or x 2 y 2 9. Therefore, its radius is 3.
22. (a) With the origin at the center of the sphere, an equation is x 2 y 2 z 2 4.
(b) Six inches of the buoy are submerged, so the waterline is 2 12 or 32 ft below the center of the sphere. Therefore, an 2 equation of the circle is x 2 y 2 32 4 x 2 y 2 74 .
23. We use the Distance Formula to write the condition that X x y z is equidistant from P 0 0 0 and Q 0 3 0. (It is easier to equate the squares of the distances.) X P2 X Q2 x 02 y 02 z 02 x 02 y 32 z 02
y 2 y 32 y 2 y 2 6y 9 6y 9 y 32 . This is an equation of a plane parallel to the xz-plane. 24. Using the Distance Formula (squaring both sides first), we have x 2 y 32 z 2 2 x 2 y 2 z 2
x 2 y 32 z 2 2x 2 2y 2 2z 2 x 2 2y 2 y 32 z 2 0 x 2 y 2 6y 9 z 2 0. Completing
the square in y, we have x 2 y 32 z 2 18. This is an equation of the sphere with center 0 3 0 and radius 18 3 2.
9.4
VECTORS IN THREE DIMENSIONS
1. A vector v a1 a2 a3 in three dimensions can be written in terms of the unit vectors i, j, and k as v a1 i a2 j a3 k. The magnitude of the vector v is v a12 a22 a32 . So 4 2 4 4i 2 j 4k and 7j 24k 0 7 24. uv 2. The angle between the vectors u and v satisfies cos . So if u and v are perpendicular then u v 0. If u v u 4 5 6 and v 3 0 2 then u v 4 3 5 0 6 2 0, so u and v are perpendicular. 3. The vector with initial point P 1 1 0 and terminal point Q 0 2 5 is v 0 1 2 1 5 0 1 1 5. 4. The vector with initial point P 1 2 1 and terminal point Q 3 1 2 is v 3 1 1 2 2 1 2 3 3. 5. The vector with initial point P 6 1 0 and terminal point Q 0 3 0 is v 0 6 3 1 0 0 6 2 0. 6. The vector with initial point P 1 1 1 and terminal point Q 0 0 1 is v 0 1 0 1 1 1 1 1 0.
SECTION 9.4 Vectors in Three Dimensions
647
7. If the vector v 3 4 2 has initial point P 2 0 1, its terminal point is 2 3 0 4 1 2 5 4 1.
8. If the vector v 0 0 1 has initial point P 0 1 1, its terminal point is 0 0 1 0 1 1 0 1 0.
9. If the vector v 2 0 2 has initial point P 3 0 3, its terminal point is 3 2 0 0 3 2 1 0 1.
10. If the vector v 23 5 12 has initial point P 6 4 2, its terminal point is 6 23 4 5 2 12 17 1 14. 11. 2 1 2 22 12 22 3 12. 5 0 12 52 02 122 13 2 2 2 2 14. 1 6 2 2 12 62 2 2 3 5 13. 3 5 4 3 5 4 5 2
15. If u 2 7 3 and v 0 4 1, then u v 2 0 7 4 3 1 2 3 2, u v 2 0 7 4 3 1 2 11 4, and 3u 12 v 3 2 12 0 3 7 12 4 3 3 12 1 6 23 19 2 .
16. If u 0 1 3 and v 4 2 0, then u v 0 4 1 2 3 0 4 3 3, u v 0 4 1 2 3 0 4 1 3, and 3u 12 v 3 0 12 4 3 1 12 2 3 3 12 0 2 2 9.
17. If u i j and v j 2k, then u v i j j 2k i 2k, u v i j j 2k i 2j 2k, and 3u 12 v 3 i j 12 j 2k 3i 72 j k.
18. If u a 2b 3c and v 4a b 2c, then u v a 4a 2b b 3c 2c 3a 3b c, u b 10c . v a 4a 2b b 3c 2c 5a b 5c, and 3u 12 v 3a 12 4a 6b 12 b 9c 12 2c 5a 11 2 19. 12 0 2 12i 2k 21. 3 3 0 3i 3j 23. (a) 2u 3v 2 0 2 1 3 1 1 0 3 1 2
20. 0 3 5 3j 5k 22. a 13 a 4 ai 13 aj 4k
24. (a) 2u 3v 2 3 1 0 3 3 0 5 3 2 15 25. u v 2 5 0 12 1 10 2 12 5 1 0 10 4 26. u v 3 0 4 2 4 12 3 2 0 4 4 12 4 27. u v= 6i 4j 2k 56 i 32 j k 6 56 4 32 2 1 1 28. u v 3j 2k 56 i 53 j 0 56 3 53 2 0 5
(b) 2u 3v 3i j 2k (b) 2u 3v 3i 2j 15k
29. 4 2 4 1 2 2 4 1 2 2 4 2 0, so the vectors are perpendicular.
30. 4j k i 2j 9k 0 1 4 2 1 9 1, so the vectors are not perpendicular.
31. 03 12 09 10 5 10 03 10 12 5 09 10 12, so the vectors are not perpendicular.
32. x 2x 3x 5 7 3 x 5 2x 7 3x 3 0, so the vectors are perpendicular (regardless of the value of x). 2 2 1 1 2 2 2 1 2 2 1 2 4 uv , so 33. cos u v 2 2 1 1 2 2 9 22 22 12 12 22 22 cos1 49 1164 .
4 0 2 2 1 0 4 2 4 uv , so cos1 45 369 . u v 4 0 2 2 1 0 5 2 2 2 2 4 2 2 1 uv 1 2 1 3 28 j k i 2j 3k 35. cos , so cos1 2828 1009 . u v j k i 2j 3k 28 12 12 12 22 32
34. cos
648
CHAPTER 9 Vectors in Two and Three Dimensions
1 4 2 3 uv 2 i 2j 2k 4i 3k , so cos1 23 482 . u v i 2j 2k 4i 3k 3 2 2 2 2 2 1 2 2 4 3 3 , 37. The length of the vector 3i 4j 5k is 32 42 52 5 2, so by definition, its direction angles satisfy cos 36. cos
4 , and cos 1 . Thus, cos1 3 65 , cos1 2 2 56 , and cos1 1 45 . cos 5 5 2 2 5 2 2
5 2
2 , and cos 1 cos1 1 66 , 38. i 2j k 12 22 12 6, so cos 1 , cos 6 6 6 6 2 1 1 1 cos 145 , and cos 114 . 6 6 1 2 73 , 39. 2 3 6 22 32 62 7, so cos 27 , cos 37 , and cos 6 7 cos 7 cos1 73 65 , and cos1 67 149 . 2 1 2 48 , 40. 2 1 2 22 12 22 3, so cos 23 , cos 1 3 , and cos 3 cos 3 1 2 1 1 3 109 , and cos cos 3 48 .
2 2 2 2 41. We are given that 3 , 3 , and is acute. Using the property of direction cosines cos cos cos 1, 2 2 2 2 we have cos cos2 cos 23 1 12 cos2 12 1 cos2 12 cos 1 . Because 3 2
is acute, we have cos1 1 45 . 2
2 2 1 2 2 2 2 2 22 14 cos 12 . 42. 23 and 4 cos cos 3 cos 4 1 cos 1 2 Because is acute, we have cos1 21 60 .
43. We are given that 60 , 50 , and is obtuse, so cos2 60 cos2 50 cos2 1 cos2 1 cos2 60 cos2 50 0337. Because is obtuse, cos1 0337 125 .
44. 75 and 15 cos2 75 cos2 cos2 15 1 cos2 1 cos2 75 cos2 15 . Using double-angle 1 cos 30 1 cos 150 0. Therefore, cos 0, and so identities, we can write the right-hand side as 1 2 2 90 .
45. Here cos2 cos2 cos2 20 cos2 45 138 1, so there is no angle satisfying the property of direction cosines cos2 cos2 cos2 1.
46. Here cos2 cos2 cos2 150 cos2 25 157 1, so there is no angle satisfying cos2 cos2 cos2 1. 47. (a) We solve v au 6 4 8 a 3 2 4 a 2. Therefore, the vectors are parallel and v 2u. (b) v au 12 8 16 a 9 6 12 a 43 , so the vectors are parallel and v 43 u.
(c) v au 2i 2j 2k a i j k has no solution, so the vectors are not parallel. 1 1 1 1 48. (a) If v has magnitude m, then v v m 1. Therefore, v is a unit vector. m m m m (b) 1 2 2 12 22 22 3, so 13 1 2 2 13 23 23 is a unit vector in the same direction as 1 2 2.
6 8 10
62 82 102 10 2, so
the same direction as 6 8 10. 6 5 9 62 52 92 142, so 1
direction as 6 5 9.
1 6 8 10 3 2 2 2 2 is a unit vector in 10 5 2 10 2
142 9 142 is a unit vector in the same 6 5 9 3 71142 5 142 142 142
SECTION 9.5 The Cross Product
649
49. (a) The second and third forces are F2 24j and F3 25k. Therefore, F1 F2 F3 F4 0 7i 24j 25k F4 0 F4 7i 24j 25k. (b) F4 72 242 252 25 2
50. (a) The side lengths are AB 0 12 1 02 2, BC 0 12 1 02 2, C D 1 02 1 02 2, and D A 1 12 0 12 0 12 2. Because the side lengths are equal, the tetrahedron is regular.
(b) We calculate
1 1 1 1 1 1 1 E A E B 1 4 2 2 2 2 2 2 , so the central angle is cos AE B 4 3 3 1 1 1 1 1 1 E A E B 4 4 4 4 4 4 cos1 13 1095 .
51. (a) r u r v 0 x 2 y 2 z 2 x 2 y 2 z 0 0 x 2 x 2
y 2 y 2 z 2 z 0 x 2 4 y 2 4 z 2 2z 0 x 2 y 2 z 12 4 4 1 9.
(b) The sphere with equation x 2 y 2 z 12 9 has center 0 0 1 and radius 3.
(x, y, z)
(c) The diagram shows the plane determined by u, v, and r, along with the trace of the sphere in that plane. We see that the equation
r-v
r u r v 0 states the fact that lines from the ends of a diameter of a circle to any point on its surface meet at right angles.
r-u
r v
(0, 0, 1)
u
(d) Let u 0 1 3 and v 2 1 4. Then r u r v 0 x y 1 z 3 x 2 y 1 z 4 0 x x 2 y 1 y 1 z 3 z 4 2 x 12 y 2 z 72 1 12 1 49 4
9.5
0 x 2 2x y 2 1 z 2 7x 12 0 1 , an equation of a circle with center 1 0 7 and radius 1 . 4 2 2
THE CROSS PRODUCT
1. The cross product of the vectors u a1 a2 a3 and v b1 b2 b3 is the vector i j k u v a1 a2 a3 a2 b3 a3 b2 i a3 b1 a1 b3 j a1 b2 a2 b1 k. So the cross product of u 1 0 1 and b1 b2 b3 i j k v 2 3 0 is u v 1 0 1 3i 2j 3k. 2 3 0 2. The cross product of two vectors u and v is perpendicular to u and to v. Thus if both vectors u and v lie inside a plane, the vector u v is perpendicular to the plane.
i j k 3. u v 1 0 3 9i 6j 3k 2 3 0
i j k 4. u v 0 4 1 1 1 2
7i j 4k
650
CHAPTER 9 Vectors in Two and Three Dimensions
i j k k 6. u v 2 3 4 0 6 2 8 0 1 1 1 9 3 12 6 4 3 i j k i j k 8. u v 3 1 0 i 3j 9k 1 1 1 4i 7j 3k 0 3 1 3 0 4 i j k 9. (a) u v 1 1 1 0 2 2 is perpendicular to both u and v. 1 1 1 0 2 2 uv 0 22 22 is a unit vector perpendicular to both u and v. (b) u v 22 22 i j k 10. (a) u v 2 5 3 1 11 19 is perpendicular to both u and v. 3 2 1 5. u v 7. u v
i
j
1 11 19 uv 483 11 483 19 483 is a unit vector perpendicular to both u and v. 483 483 483 u v 12 112 192 i j k 11. (a) u v 12 1 23 14 7 0 is perpendicular to both u and v. 6 12 6 14 7 0 uv (b) 2 5 5 55 0 is a unit vector perpendicular to both u and v. u v 142 72 i j k 12. (a) u v 0 3 5 6 5 3 is perpendicular to both u and v. 1 0 2
(b)
6 5 3 uv 3 3570 1470 3 7070 is a unit vector perpendicular to both u and v. u v 62 52 32 14. u v u v sin 4 5 sin 30 10 13. u v u v sin 6 12 sin 60 3 2 3
(b)
15. u v u v sin 10 10 sin 90 100
16. u v u v sin 012 125 sin 75 0145
17. P Q P R 1 1 1 2 0 0 0 2 2 is perpendicular to the plane passing through P, Q, and R. 18. P Q P R 2 2 2 1 3 1 4 0 4 is perpendicular to the plane passing through P, Q, and R. 19. P Q P R 1 1 5 1 1 5 10 10 0 is perpendicular to the plane passing through P, Q, and R. 20. P Q P R 3 2 5 5 0 6 12 43 10 is perpendicular to the plane passing through P, Q, and R. 21. The area of the parallelogram determined by u 3 2 1 and v 1 2 3 is u v 4 8 4 4 6. 22. The area of the parallelogram determined by u 0 3 2 and v 5 6 0 is u v 12 10 15 469. 9 5 14 . 23. The area of the parallelogram determined by u 2i j 4k and v 12 i 2j 32 k is u v 13 5 2 2 2 24. The area of the parallelogram determined by u i j k and v i j k is u v 0 2 2 2 2.
SECTION 9.5 The Cross Product
651
25. The area of triangle P Q R is one-half the area of the parallelogram determined by P Q and P R, that is, 1 1 1 2 2 2 2 P Q P R 2 1 1 1 1 3 3 2 6 2 4 14.
26. The area of triangle P Q R is one-half the area of the parallelogram determined by P Q and P R, that is, 1 P Q P R 1 2 1 1 6 1 0 1 12 62 82 101 . 2 2 2 2
27. The area of triangle P Q R is one-half the area of the parallelogram determined by P Q and P R, that is, 1 P Q P R 1 6 6 0 6 0 6 1 362 362 362 18 3. 2 2 2
28. The area of triangle P Q R is one-half the area of the parallelogram determined by P Q and P R, that is, 1 P Q P R 1 4 2 12 0 6 0 1 722 02 242 12 10. 2 2 2 29. (a) u v w 1 2 3 3 2 1 0 8 10 1 2 3 12 30 24 0 (b) Because their scalar triple product is 0, the vectors are coplanar.
30. (a) u v w 3 0 4 1 1 1 7 4 0 3 0 4 4 7 3 0 (b) Because their scalar triple product is 0, the vectors are coplanar.
31. (a) u v w 2 3 2 1 4 0 3 1 3 2 3 2 12 3 11 55
(b) Because their scalar triple product is nonzero, the vectors are not coplanar. The volume of the parallelepiped that they determine is u v w 55.
32. (a) u v w 1 1 0 1 0 1 0 1 1 1 1 0 1 1 1 0 (b) Because their scalar triple product is 0, the vectors are coplanar.
33. (a) u v w 1 1 1 0 1 1 1 1 1 1 1 1 2 1 1 2
(b) Because their scalar triple product is nonzero, the vectors are not coplanar. The volume of the parallelepiped that they determine is u v w 2.
34. (a) u v w 2 2 3 3 1 1 6 0 0 2 2 3 0 6 6 6
(b) Because their scalar triple product is nonzero, the vectors are not coplanar. The volume of the parallelepiped that they determine is u v w 6.
35. (a) We have u 120 cm, v 150 cm, w 300 cm, the angle between v and w is 90 30 60 , and the angle between u and v w is 0 (because u is perpendicular to both v and w). Therefore, u v w u v w cos 0 120 150 300 sin 60 2,700,000 3 4,676,537.
4,676,537 cm3 4677 liters. 1000 cm3 L 36. (a) From Exercise 9.4.48, the object with vertices A 1 0 0, B 0 1 0, C 0 0 1, and D 1 1 1 is a Rubik’s Tetrahedron. The volume of the parallelepiped determined by the vectors AB, AC, and AD is calculated as AB AC AD 1 1 0 1 0 1 0 1 1 1 1 0 1 1 1 2. Therefore, the volume of (b) The capacity in liters is approximately
the Rubik’s Tetrahedron is 16 2 13 .
37. (a) u v w 0 1 1 1 0 1 1 1 0 0 1 1 1 1 1 2, u w v 0 1 1 1 1 0 1 0 1 0 1 1 1 1 1 2, v u w 1 0 1 0 1 1 1 1 0 1 0 1 1 1 1 2, v w u 1 0 1 1 1 0 0 1 1 1 0 1 1 1 1 2, w u v 1 1 0 0 1 1 1 0 1 1 1 0 1 1 1 2, and w v u 1 1 0 1 0 1 0 1 1 1 1 0 1 1 1 2. (b) It appears that u v w v w u w u v u w v v u w w v u.
(c) We know that the absolute values of the six scalar triple products must be equal because they all represent the volume of the parallelepiped determined by u, v, and w. The fact that a b b a completes the proof.
652
CHAPTER 9 Vectors in Two and Three Dimensions
9.6
EQUATIONS OF LINES AND PLANES
1. A line in space is described algebraically by using parametric equations. The line that passes through the point P x0 y0 z 0 and is parallel to the vector v a b c is described by the equations x x0 at, y y0 bt, z z 0 ct. 2. The plane containing the point P x0 y0 z 0 and having the normal vector n a b c is described algebraically by the equation a x x0 b y y0 c z z 0 0. 3. The line passing through P 1 0 2 parallel to v 3 2 3 has parametric equations x 1 3t, y 2t, z 2 3t. 4. The line passing through P 0 5 3 parallel to v 2 0 4 has parametric equations x 2t, y 5, z 3 4t. 5. x 3, y 2 4t, z 1 2t
6. x 4t, y 3t, z 5t
7. x 1 2t, y 0, z 2 5t
8. x 1 t, y 1 t, z 1 t
9. We first find a vector determined by P 1 3 2 and Q 2 1 1: v 2 1 1 3 1 2 1 4 3. Now we use v and the point 1 3 2 to find parametric equations: x 1 t, y 3 4t, z 2 3t where t is any real number. 10. We first find a vector determined by P 2 1 2 and Q 0 1 3: v 0 2 1 1 3 2 2 2 1. Now we use v and the point 2 1 2 to find parametric equations: x 2 2t, y 1 2t, z 2 t. 11. A vector determined by P 1 1 0 and Q 0 2 2 is 1 1 2, so parametric equations are x 1 t, y 1 t, z 2t. 12. A vector determined by P 3 3 3 and Q 7 0 0 is 4 3 3, so parametric equations are x 3 4t, y 3 3t, z 3 3t. 13. A vector determined by P 3 7 5 and Q 7 3 5 is 4 4 0, so parametric equations are x 3 4t, y 7 4t, z 5. 14. A vector determined by P 12 16 18 and Q 12 6 0 is 0 22 18, so parametric equations are x 12, y 16 22t, z 18 18t. 15. (a) An equation of the plane with normal vector
16. (a) An equation of the plane with normal vector
n 1 1 1 that passes through P 0 2 3 is
n 3 2 0 that passes through P 1 2 7 is
x y z 5.
3x 2y 7.
1 x 0 1 y 2 1 [z 3] 0 or
(b) Setting y z 0, we find x 5, so the x-intercept
is 5. Similarly, the y-intercept is 5 and the z-intercept
is 5.
x
(b) Setting y z 0, we find 3x 7, so the x-intercept is 73 . Similarly, the y-intercept is 72 . If we set
x y 0, we have 3 0 2 0 7, which has no
z
solution, and so there is no z-intercept. 0
5
3 x 1 2 y 2 0 z 7 0 or
z
5 y
_5
7 3
x
0
7 2
y
SECTION 9.6 Equations of Lines and Planes
17. (a) 3 x 2 12 z 8 0 6x z 4 (b) x-intercept 23 , no y-intercept, z-intercept 4 z
18. (a) 23 x 6 13 y 0 z 3 0 2 x 6 y 3 z 3 0 2x y 3z 3 (b) x-intercept 32 , y-intercept 3, z-intercept 1 z
0
2 3
x
19. (a) 3x y 2 2 z 3 0 3x y 2z 8 (b) x-intercept 83 , y-intercept 8, z-intercept 4 z
_3
y
_4
y
0
3 2
_1
_4
x
20. (a) x 1 4y 0 x 4y 1 (b) x-intercept 1, y-intercept 14 , no z-intercept z
_83 0
x
653
8
0
y
1
1 4
y
x
21. The vector P Q P R 1 1 2 1 2 1 5 3 1 is perpendicular to both P Q and P R and is therefore perpendicular to the plane through P, Q, and R. Using the formula for an equation of a plane with the point P, we have 5 x 6 3 y 2 z 1 0 5x 3y z 35. 22. The vector P Q P R 2 2 2 1 3 1 4 0 4 is perpendicular to both P Q and P R and is therefore perpendicular to the plane through P, Q, and R. Using the formula for an equation of a plane, we have 4 x 3 4 z 5 0 x z 2. 1 23. P Q P R 1 3 2 1 13 6 83 8 0 , so an equation is 83 x 3 8 y 13 0 x 3y 2. 24. P Q P R 2 2 2 2 4 4 0 4 4, so an equation is 4 y 4 4 z 2 0 y z 2. 25. P Q P R 3 1 1 6 1 1 2 3 9, so an equation is 2 x 6 3 y 1 9 z 1 0 2x 3y 9z 0 26. P Q P R 2 2 2 2 0 4 8 12 4, so an equation is 8 x 2 12y 4z 0 2x 3y z 4.
27. The line passes through 0 0 4 and 2 5 0. A vector determined by these two points is v 2 0 5 0 0 4 2 5 4. Now we use v and the point 0 0 4 to find parametric equations: x 2t, y 5t, z 4 4t, where t is any real number. 28. The line passes through 2 0 0 and 0 0 10, so a direction vector is v 2 0 10 and equations are x 2 2t, y 0, z 10t, where t is any real number.
29. The line passes through 2 1 5 and is parallel to j, so equations are x 2, y 1 t, z 5, where t is any real number.
30. The line passes through 3 0 2 and is parallel to j, so equations are x 3, y t, z 2, where t is any real number. 31. The plane passes through P 1 0 0, Q 0 3 0, and R 0 0 4. The vector P Q P R 1 3 01 0 4 12 4 3 is perpendicular to both P Q and P R and is therefore perpendicular to the plane through P, Q, and R. Using the formula for an equation of a plane, we have 12 x 1 4y 3z 0 12x 4y 3z 12.
32. This plane has the same normal vector as x 2y 4z 6, and because it contains the origin, its equation is x 2y 4z 0.
654
CHAPTER 9 Vectors in Two and Three Dimensions
33. If the point x y z is equidistant from P 3 2 5 and Q 1 1 4, then it satisfies
x 32 y 22 z 52 x 12 y 12 z 42 6x 94y 410z 25 2x 12y 18z 16 8x 6y 2z 20 4x 3y z 10.
34. Taking t 0 and t 1 in the equation of the given line, we see that the points Q 1 2 0 and R 0 3 3 lie on the plane. P Q P R 1 2 6 2 3 3 12 9 1, so an equation is 12 x 2 9y z 6 0 12x 9y z 30. 35. (a) To find the point of intersection, we substitute the parametric equations of the line into the equation of the plane: 5 2 t 2 3t 2 5 t 1 10 5t 6t 10 2t 1 t 1. (b) The parameter value t 1 corresponds to the point 3 3 4.
36. (a) If the line is perpendicular to the plane, then it is parallel to the plane’s normal vector, so v is parallel to n. (b) If the line is parallel to the plane, then it is perpendicular to the plane’s normal vector, so v is perpendicular to n. (c) The plane’s normal vector is n 1 1 4. Line 1 is parallel to 2 2 8 2 1 1 4, so, as in part (a), it is perpendicular to the given plane. Line 2 is parallel to v 2 2 1, and v n 2 2 4 0, so as in part (b), it is parallel to the given plane. 37. (a) Setting t 0 in the equation for Line 1 gives the point P 1 0 6. Setting t 1 gives Q 0 3 1. If we set t 1 in the equation for Line 2, we have the point P 1 0 6 P, and if we set t 12 , we get Q 0 3 1 Q. (b) Setting t 0 in the equation for Line 1 gives the point 0 3 5. But if a point on Line 4 has x-coordinate 0, it must have x 8 2t 0 t 4. But this parameter value gives the point 0 3 2 on Line 2, so the two lines are not the same.
CHAPTER 9 REVIEW 1. u 2 3, v 8 1. u
22 32 13, u v 2 8 3 1 6 4,
u v 2 8 3 1 10 2, 2u 2 2 2 3 4 6, and 3u 2v 3 2 2 8 3 3 2 1 22 7. 2. u 5 2, v 3 0. u 52 22 29, u v 5 3 2 0 2 2,
u v 5 3 2 0 8 2, 2u 2 5 2 2 10 4, and 3u 2v 3 5 2 3 3 2 2 0 21 6. 3. u 2i j, v i 2j. u 22 12 5, u v 2 1 i 1 2 j 3i j, u v 2 1 i 1 2 j i 3j, 2u 4i 2j, and 3u 2v 3 2i j 2 i 2j 4i 7j. 4. u 3j, v i 2j. u 32 3, u v 3j i 2j i 5j, u v 3j i 2j i j, 2u 6j, and 3u 2v 3 3j 2 i 2j 2i 5j. 5. The vector with initial point P 0 3 and terminal point Q 3 1 is 3 0 1 3 3 4.
6. If the vector 5i 8j is placed in the plane with its initial point at P 5 6, its terminal point is 5 5 6 8 10 2. 2 2 3 2 7. u 2 2 3 has length 2 2 3 4. Its direction is given by tan 3 with in quadrant II, 2 so tan1 3 120 . 8. v 2i 5j has length 22 52 29. Its direction is given by tan 52 with in quadrant IV, so 2 tan1 52 2918 . 9. u u cos u sin 20 cos 60 20 sin 60 10 10 3 .
10. u u cos u sin 135 cos 125 135 sin 125 135 cos 125 135 sin 125 77 111
CHAPTER 9
Review
655
11. (a) The force exerted by the first tugboat can be expressed in component form as u 20 104 cos 40 20 104 sin 40 15321 12856, and that of the second tugboat is 34 104 cos 15 34 104 sin 15 32841 8800. Therefore, the resultant force is
12.
13. 14. 15. 16. 17. 18.
w u v 15321 32841 12856 8800 48162 4056. (b) The magnitude of the resultant force is 481622 40562 48,332 lb. Its direction is given by 4056 0084, so tan1 0084 48 or N 852 E. tan 48,162 (a) The velocity of the airplane is the sum of its velocity relative to the air, which is u 600 cos 30 600 sin 30 300 3 300 , and the wind velocity v 50 cos 120 50 sin 120 25 25 3 . Thus, its velocity is w 300 3 25 300 25 3 . 2 2 300 3 25 300 25 3 602 mi/h, and its direction is given by (b) The true speed of the airplane is w 300 25 3 tan 0694, so tan1 0694 348 , or N 552 E. 300 3 25 u 4 3, v 9 8. u 42 32 5, u u 42 32 25, and u v 4 9 3 8 60. u 5 12, v 10 4. u 52 122 13, u u 52 122 169, and u v 5 10 12 4 2. u 2i 2j, v i j. u 22 22 2 2, u u 22 22 8, and u v 2 1 2 1 0. u 10j, v 5i 3j. u 102 10, u u 102 100, and u v 10 3 30. u v 4 2 3 6 4 3 2 6 0, so the vectors are perpendicular. u v 5 3 2 6 5 2 3 6 8 0, so the vectors are not perpendicular. The angle between them is given by uv 340 8 340 775 . cos , so cos1 85 u v 85 52 32 22 62
19. u v 2i j i 3j 2 1 1 3 5, so the vectors are not perpendicular. The angle between them is given by uv 2 5 cos , so cos1 22 45 . u v 2 22 12 12 32
20. u v i j i j 1 1 1 1 0, so the vectors are perpendicular.
3 6 1 1 17 37 uv . 21. (a) u 3 1, v 6 1. The component of u along v is v 37 62 12 uv 17 6 1 102 17 v (b) projv u 37 37 37 v2 17 and u u proj u 3 1 102 17 9 54 . (c) u1 projv u 102 2 v 37 37 37 37 37 37
22. (a) u 8 6, v 20 20. The component of u along v is (b) projv u
uv v2
v
40 20 20 1 1 800
uv 8 20 6 20 2. v 2 2 20 20
(c) u1 projv u 1 1 and u2 u projv u 8 6 1 1 7 7. 23. (a) u i 2j, v 4i 9j. The component of u along v is (b) projv u
uv v2
56 126 v 14 97 4i 9j 97 i 97 j
1 4 2 9 14 97 uv . v 97 42 92
656
CHAPTER 9 Vectors in Two and Three Dimensions
126 j and u u proj u i 2j 56 i 126 j 153 i 68 j. (c) u1 projv u 56 i 2 v 97 97 97 97 97 97 24. (a) u 2i 4j, v 10j. The component of u along v is
4 10 uv 4. v 102
40 10j 4j v 100 v2 (c) u1 projv u 4j and u2 u projv u 2i 4j 4j 2i.
(b) projv u
uv
z
25.
z
26.
P(0, 2, 4)
Q(3, _2, 3) P(1, 0, 2)
0
y
0
x
The distance between P and Q is 3 12 2 02 3 22 3.
x
Q(1, 3, 0)
y
The distance between P and Q is 1 02 3 22 0 42 3 2.
27. The sphere with radius r 6 and center C 0 0 0 has equation x 02 y 02 z 02 62 x 2 y 2 z 2 36. 2 28. The sphere with radius r 2 and center C 1 2 4 has equation x 12 y 2 z 42 22 x 12 y 22 z 42 4.
29. We complete the squares to find x 2 y 2 z 2 2x 6y 4z 2 x 2 2x 1 y 2 6y 9 z 2 4z 4 2
1 9 4 x 12 y 32 z 22 16, an equation of the sphere with center 1 3 2 and radius 4. 30. We complete the squares to find x 2 y 2 z 2 4y 4z x 2 y 2 4y 4 z 2 4z 4 4 4 x 2 y 22 z 22 8, an equation of the sphere with center 0 2 2 and radius 2 2. 31. u 4 2 4 and v 2 3 1, so u 42 22 42 6, u v 4 2 2 3 4 1 6 1 3, u v 4 2 2 3 4 1 2 5 5, and 34 u 2v 34 4 2 4 2 2 3 1 1 15 2 5 . 32. u 6i 8k and v i j k, so u 62 82 10, u v 7i j 7k, u v 5i j 9k, and 3 u 2v 3 6i 8k 2 i j k 5 i 2j 8k. 4 4 2
33. (a) u v 3 2 4 3 1 2 3 3 2 1 4 2 1
(b) u v 0, so the vectors are not perpendicular. The angle between them is given by 406 1 uv 406 928 . , so cos1 406 cos u v 406 2 2 2 2 2 2 3 2 4 3 1 2 34. (a) u v 2 6 5 1 12 1 2 1 6 12 5 1 0 (b) u v 0, so the vectors are perpendicular.
35. (a) u v 2i j 4k 3i 2j k 2 3 1 2 4 1 0 (b) u v 0, so the vectors are perpendicular.
36. (a) u v j k i j 0 1 1 1 1 0 1 (b) u v 0, so the vectors are not perpendicular. The angle between them is given by cos cos1 21 60 .
1 1 uv , so u v 2 2 2
CHAPTER 9
Review
657
37. (a) u v 1 1 3 5 0 2 2 0 i 2 15 j 0 5 k 2 17 5 2 17 5 uv 318 17 318 5 318 . 159 (b) A unit vector perpendicular to u and v is 318 318 u v 22 172 52 38. (a) u v 2 3 0 0 4 1 3 2 8 3 2 8 uv (b) A unit vector perpendicular to u and v is 3 7777 2 7777 8 7777 . u v 32 22 82 39. (a) u v i j 2j k 1 0 i 1 0 j 2 0 k i j 2k uv i j 2k (b) A unit vector perpendicular to u and v is 66 i 66 j 36 k. u v 12 12 22
40. (a) u v i j k i j k 1 1 i 1 1 j 1 1 k 2j 2k uv 2j 2k (b) A unit vector perpendicular to u and v is 22 j 22 k. u v 22 22 41. The area of triangle P Q R is one-half the area of the parallelogram determined by P Q and P R, that is, 1 P Q P R 1 1 6 i 2 8 j 6 4 k 1 52 102 102 1 225 15 . 2 2 2 2 2
42. The area of the parallelogram determined by u 4 1 1 and v 1 2 2 is the absolute value of their cross product: A u v 0 9 9 92 92 9 2.
43. The volume of the parallelepiped determined by u 2i j, v 2j k, and w 3i j k is the absolute value of their scalar triple product: V u v w 2i j 3i 3j 6k 6 3 9. 44. This parallelepiped is determined by the vectors u 0 2 2, v 3 1 1, and w 1 4 1, so its volume is V u v w 0 2 2 5 4 11 8 22 14.
45. The line that passes through P 2 0 6 and is parallel to v 3 1 0 has parametric equations x 2 3t, y t, z 6.
46. The line that passes through P 5 2 8 and is parallel to v 2i j 5k has parametric equations x 5 2t, y 2 t, z 8 5t.
47. A vector determined by P 6 2 3 and Q 4 1 2 is 2 3 1, so parametric equations are x 6 2t, y 2 3t, z 3 t. 48. A vector determined by P 1 0 0 and Q 3 4 2 is 2 4 2 or 1 2 1, so parametric equations are x 1 t, y 2t, z t.
49. Using the formula for an equation of a plane, the plane with normal vector n 2 3 5 passing through P 2 1 1 has equation 2 x 2 3 y 1 5 z 1 0 2x 3y 5z 2.
50. Using the formula for an equation of a plane, the plane with normal vector n i 2j 7k passing through P 2 5 2 has equation [x 2] 2 y 5 7 z 2 0 x 2y 7z 6.
51. The plane passes through P 1 1 1, Q 3 4 2, and R 6 1 0. i j k P Q P R 2 5 1 5 2 1 2 5 1 7i 7 j 21k i j 3k, so an equation is 5 2 1 1 x 1 1 y 1 3 z 1 0 x y 3z 5.
52. The plane passes through P 4 0 0, Q 0 3 0, and R 0 0 5. P Q P R 4 3 0 4 0 5 15 20 12, so an equation is 15 x 4 20y 12z 0 15x 20y 12z 60.
53. The line passes through the points 2 0 0 and 0 0 4. A vector determined by these points is 2 0 4 or 1 0 2, so parametric equations are x 2 t, y 0, z 2t.
658
CHAPTER 9 Vectors in Two and Three Dimensions
54. We are given P 5 3 0. We find two points on the given line by substituting t 0 and t 1 in its parametric equations, finding Q 2 0 6 and R 4 4 6. P Q P R 3 3 6 1 1 6 24 12 6, so an equation is 24 x 5 12 y 3 6z 0 4x 2y z 14.
CHAPTER 9 TEST 1. (a)
y
(b) u 3 3 i [9 1] j 6i 10j (c) u 62 102 2 34
(_3, 9)
u 1 1
(3, _1)
x
2. (a) u 3v 1 3 3 6 2 1 3 6 3 3 2 19 3 (b) u v 1 3 6 2 5 5 52 52 5 2 (c) u v 1 3 6 2 1 6 3 2 0
(d) Because u v 0, u and v are perpendicular.
2 4 3 42 8. Its direction is given by (b) The length of u is u
y
3. (a) (_4Ï3, 4)
4 3 with in quadrant II, so 3 4 3 180 tan1 33 150 .
tan
u 1 1
x
4. (a) The river’s current can be represented by the vector u 8 0 and the motorboat’s velocity relative to the water by the vector v 12 cos 60 12 sin 60 6 6 3 . Thus, the true velocity is w u v 14 6 3 . 2 6 13 2 0742, so (b) The true speed is w 14 6 3 174 mi/h. The direction is given by tan 14 tan1 0742 366 or N 534 E. uv 338 3 5 2 1 338 cos1 2 45 . , so cos1 26 5. (a) cos 2 u v 26 32 22 52 12 uv 13 26 (b) The component of u along v is . v 2 26 uv 5 1 v 13 (c) projv u 26 5i j 2 i 2 j v2 6. The work is W F d 3i 5j 7 2 i 13 2 j 3i 5j 5i 15j 3 5 5 15 90. 7. (a) The distance between P 4 3 1 and Q 6 1 3 is d 6 42 1 32 [3 1]2 6.
Vector Fields
659
(b) An equation is x 42 y 32 [z 1]2 62 x 42 y 32 z 12 36. (c) u 6 4 1 3 3 1 2 4 4 2i 4j 4k
8. u i j 2k, v 3i 2j k, and w j 5k. (a) 2u 3v [2 1 3 3] i [2 1 3 2] j [2 2 3 1] k 11i 4j k (b) u 12 12 22 6 (c) u v 1 3 1 2 2 1 1 i j k (d) u v 1 1 2 3i 7j 5k 3 2 1 i j k (e) v w 3 2 1 9i 15j 3k 92 152 32 3 35 0 1 5
(f) u v w i j 2k 9i 15j 3k 1 9 1 15 2 3 18 84 1 uv , so cos1 8484 96 . (g) cos u v 84 6 32 22 12
9. A vector perpendicular to both u j 2k and v i 2j 3k is u v 0 1 2 1 2 3 7 2 1, so two unit 7 2 1 uv vectors perpendicular to u and v are 7186 96 186 and 7186 96 186 . u v 72 22 12
10. (a) A vector perpendicular to the plane that contains the points P 1 0 0, Q 2 0 1, and R 1 4 3 is P Q P R 1 0 1 0 4 3 4 3 4. (b) An equation of the plane is 4 x 1 3y 4z 0 4x 3y 4z 4.
(c) The area of triangle P Q R is half the area of the parallelogram determined by P Q and P R, that is, 1 P Q P R 1 42 32 42 41 . 2 2 2
11. A vector determined by the two points is P Q 2 1 2, so parametric equations are x 2 2t, y 4 t, z 7 2t.
FOCUS ON MODELING Vector Fields 1. F x y 12 i 12 j
All vectors point in the same direction and have length 22 . y
0
x
2. F x y i xj
The vectors lie on lines with slope x. y
0
x
660
FOCUS ON MODELING
3. F x y yi 12 j
4. F x y x y i xj
The vectors point to the left for y 0 and to the right for
The vectors form a spiral pattern. y
y 0. y
0 0
x
x
yi xj 5. F x y x 2 y2
yi xj 6. F x y x 2 y2
All the vectors F x y are unit vectors tangent to circles centered at the origin with radius x 2 y 2 .
yi xj The length of the vector is 1. x 2 y2 y
0
y
x
7. F x y z j
All vectors in this field are parallel to the y-axis and have length 1.
0
x
8. F x y z j k All vectors are parallel to 0 1 1 and have length z
z
x
0
x y
0
y
2.
Vector Fields
9. F x y z zj
At each point x y z, F x y z is a vector of length z. For z 0, all point in the direction of the positive y-axis
661
10. F x y z yk
At each point x y z, F x y z is a vector of length y. For y 0, all point in the direction of the positive z-axis
while for z 0, all are in the direction of the negative
while for y 0, all are in the direction of the negative
y-axis.
z-axis. z z
x
0
y
0
y
x
11. F x y y x corresponds to graph II, because in the first quadrant all the vectors have positive x- and y-components, in the second quadrant all vectors have positive x-components and negative y-components, in the third quadrant all vectors have negative x- and y-components, and in the fourth quadrant all vectors have negative x-components and positive y-components.
12. F x y 1 sin y corresponds to graph IV because the x-component of each vector is constant, the vectors are independent of x (vectors along horizontal lines are identical), and the vector field appears to repeat the same pattern vertically.
13. F x y x 2 x 1 corresponds to graph I because the vectors are independent of y (vectors along vertical lines are identical) and, as we move to the right, both the x- and the y-components get larger.
14. F x y y 1x corresponds to graph III. As in Problem 11, all the vectors in the first quadrant have positive x- and y-components, in the second quadrant all vectors have positive x-components and negative y-components, in the third quadrant all vectors have negative x- and y-components, and in the fourth quadrant all vectors have negative x-components and positive y-components. Also, the vectors become longer as we approach the y-axis.
15. F x y z i 2j 3k corresponds to graph IV, since all vectors have identical length and direction. 16. F x y z i 2j zk corresponds to graph I, since the horizontal vector components remain constant, but the vectors above the x y-plane point generally upward while the vectors below the x y-plane point generally downward.
17. F x y z xi yj 3k corresponds to graph III; the projection of each vector onto the x y-plane is xi yj, which points away from the origin, and the vectors point generally upward because their z-components are all 3.
18. F x y z xi yj zk corresponds to graph II; each vector F x y z has the same length and direction as the position vector of the point x y z, and therefore the vectors all point directly away from the origin.
662
FOCUS ON MODELING y 5
19.
4
(a)
3 2 (b)
1
_5 _4 _3 _2 _1 0 _1 (c) _2 _3 _4 _5
1
2
3
4
5 x
10
SYSTEMS OF EQUATIONS AND INEQUALITIES
10.1 SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES 1. The given system is a system of two equations in the two variables x and y. To check if 5 1 is a solution of this system, we check if x 5 and y 1 satisfy each equation in the system. The only solution of the given system is 2 1.
2. A system of equations in two variables can be solved by the substitution method, the elimination method, or the graphical method. 3. A system of two linear equations in two variables can have one solution, no solution, or infinitely many solutions. 4. For the given system, the graph of the first equation is the same as the graph of the second equation, so the system has x t where t is any real number. Some of the infinitely many solutions. We express these solutions by writing y 1t
solutions of this system are 1 0, 3 4, and 5 4. x y 1 Solving the first equation for x, we get x y 1, and substituting this into the second equation gives 5. 4x 3y 18
4 y 1 3y 18 7y 4 18 7y 14 y 2. Substituting for y we get x y 1 2 1 3. Thus, the solution is 3 2. 3x y 1 6. Solving the first equation for y, we get y 1 3x, and substituting this into the second equation gives 5x 2y 1
5x 2 1 3x 1 x 2 1 x 1. Substituting for x we get y 1 3 1 2. Thus, the solution is 1 2. x y2 7. Solving the first equation for x, we get x y 2, and substituting this into the second equation gives 2x 3y 9
2 y 2 3y 9 5y 4 9 5y 5 y 1. Substituting for y we get x y 2 1 2 3. Thus, the solution is 3 1. 2x y 7 Solving the first equation for y, we get y 7 2x, and substituting this into the second equation gives 8. x 2y 2
x 2 7 2x 2 x 14 4x 2 3x 12 x 4. Substituting for x we get y 7 2x 7 2 4 1. Thus, the solution is 4 1. 3x 4y 10 9. Adding the two equations, we get 4x 8 x 2, and substituting into the first equation in the x 4y 2
original system gives 3 2 4y 10 4y 4 y 1. Thus, the solution is 2 1. 2x 5y 15 4x 10y 30 10. Multiplying the first equation by 2 gives the system 4x y 21 4x y 21
Subtracting the second
equation from the first gives 9y 9 y 1, and substituting into the original second equation gives 4x 1 21 4x 20 x 5. Thus, the solution is 5 1. 3x 2y 13 6x 4y 26 11. Multiplying the first equation by 2 gives the system Adding, we get y 2, 6x 5y 28 6x 5y 28
and substituting into the first equation in the original system gives 3x 2 2 13 3x 9 x 3. The solution is 3 2. 663
664
12.
CHAPTER 10 Systems of Equations and Inequalities
2x 5y 18
Multiplying the first equation by 3 and the second by 2 gives the 3x 4y 19 6x 15y 54 Subtracting the equations gives 23y 92 y 4. Substituting this value into the first system 6x 8y 38 equation in the original system gives 2x 5 4 18 2x 2 x 1. Thus, the solution is 1 4.
13.
2x y 1
By inspection of the graph, it appears that 2 3 is the solution to the system. We check this in both
x 2y 8
equations to verify that it is a solution. 2 2 3 4 3 1 and 2 2 3 2 6 8. Since both equations are satisfied, the solution is 2 3.
14.
xy2
2x y 5
By inspection of the graph, it appears that 3 1 is the solution to the system. We check this in both
equations to verify that it is a solution. 3 1 2 and 2 3 1 6 1 5. Since both equations are satisfied, the solution is 3 1.
15.
xy4
16.
2x y 2
2x y 4
3x y 6
The solution is x 2, y 0.
The solution is x 2, y 2. y
y
1 1
17.
1
x
2x 3y 12
x 32 y 4 The lines are parallel, so there is no intersection and hence no solution.
1
18.
x
2x 6y 0
3x 9y 18
The lines are parallel, so there is no intersection and hence no solution.
y
y
1 1
x
1 1
x
SECTION 10.1 Systems of Linear Equations in Two Variables
x 1 y 5 2 19. 2x y 10
20.
There are infinitely many solutions. y
12x 15y 18
2x 52 y 3 There are infinitely many solutions. The lines are the same. y
1 1
21.
xy4
x y 0
665
x
1 1
x
Adding the two equations gives 2y 4 y 2. Substituting for y in the first equation gives
x 2 4 x 2. Hence, the solution is 2 2. x y3 Subtracting the first equation from the second equation gives 4y 4 y 1. Substituting, we have 22. x 3y 7 x 1 3 x 4. Hence, the solution is 4 1. 2x 3y 9 23. Adding the two equations gives 6x 18 x 3. Substituting for x in the second equation gives 4x 3y 9 4 3 3y 9 12 3y 9 3y 3 x 1. Hence, the solution is 3 1. 3x 2y 0 24. Adding the two equations gives 2x 8 x 4. Substituting for x in the second equation gives x 2y 8
3 4 2y 0 12 2y 0 y 6. Hence, the solution is 4 6. x 3y 5 Solving the first equation for x gives x 3y 5. Substituting for x in the second equation gives 25. 2x y 3
2 3y 5 y 3 6y 10 y 3 7y 7 y 1. Then x 3 1 5 2. Hence, the solution is 2 1. x y 7 26. Adding 3 times the first equation to the second equation gives 5x 20 x 4. So 2x 3y 1 27. 28. 29. 30.
4 y 7 y 3, and the solution is 4 3. x y 2 y x 2. Substituting for y into 4x 3y 3 gives 4x 3 x 2 3 4x 3x 6 3 x 3, and so y 3 2 5. Hence, the solution is 3 5. 9x y 6 y 9x 6. Substituting for y into 4x 3y 28 gives 4x 3 9x 6 28 23x 46 x 2, and so y 9 2 6 12. Thus, the solution is 2 12. x 2y 7 x 72y. Substituting for x into 5x y 2 gives 5 7 2y y 2 3510y y 2 11y 33 y 3, and so x 7 2 3 1. Hence, the solution is 1 3. 4x 12y 0 x 3y. Substituting for x into 12x 4y 160 gives 12 3y 4y 160 40y 160 y 4, and so x 3 4 12. Therefore, the solution is 12 4.
31. 13 x 16 y 1 2x y 6 y 2x 6. Substituting for y into 23 x 16 y 3 gives 23 x 16 2x 6 3 4x 2x 6 18 2x 12 x 6, and so y 2 6 6 6. Hence, the solution is 6 6. 32. 34 x 12 y 5 y 10 32 x. Substituting for y into 14 x 32 y 1 gives 14 x 32 10 32 x 1 x 60 9x 4 8x 64 x 8, and so y 10 32 8 2. Hence, the solution is 8 2.
666
CHAPTER 10 Systems of Equations and Inequalities
33. 12 x 13 y 2 x 23 y 4 x 4 23 y. Substituting for x into 15 x 23 y 8 gives 15 4 23 y 23 y 8
4 2 y 10 y 8 12 2y 10y 120 y 9, and so x 4 2 9 10. Hence, the solution is 10 9. 5 15 15 3
34. 02x 02y 18 x y9. Substituting for x into 03x 05y 33 gives 03 y 905y 33 02y 06 y 3, and so x 3 9 6. Hence, the solution is 6 3. 3x 2y 8 3x 2y 8 Subtracting the second Multiplying the second equation by 3 gives the system 35. 3x 6y 0 x 2y 0 equation from the first gives 8y 8 y 1. Substituting into the first equation we get 3x 2 1 8 3x 6 x 2. Thus, the solution is 2 1. 4x 2y 16 36. Adding the first equation to 4 times the second equation gives 22y 264 y 12, so x 5y 70
4x 2 12 16 x 10, and the solution is 10 12. x 4y 8 Adding 3 times the first equation to the second equation gives 0 22, which is never true. Thus, 37. 3x 12y 2 the system has no solution. 3x 5y 2 38. Adding 3 times the first equation to the second equation gives 0 12, which is false. Therefore, 9x 15y 6
there is no solution to this system. 2x 6y 10 39. Adding 3 times the first equation to 2 times the second equation gives 0 0. Writing the equation 3x 9y 15
in slope-intercept form, we have 2x 6y 10 6y 2x 10 y 13 x 53 , so the solutions are all pairs of the form x 13 x 53 where x is a real number. 2x 3y 8 40. Adding 7 times the first equation to 1 times the second equation gives 0 59, which is false. 14x 21y 3 Therefore, there is no solution to this system. 6x 4y 12 41. Adding 3 times the first equation to 2 times the second equation gives 0 0. Writing the equation in 9x 6y 18
slope-intercept form, we have 6x 4y 12 4y 6x 12 y 32 x 3, so the solutions are all pairs of the form x 32 x 3 where x is a real number. 25x 75y 100 1 times the first equation to 1 times the second equation gives 0 0, which is always 42. Adding 25 10 10x 30y 40 true. We now put the equation in slope-intercept form. We have x 3y 4 3y x 4 y 13 x 43 , so the solutions are all pairs of the form x 13 x 43 where x is a real number. 8s 3t 3 43. Adding 2 times the first equation to 3 times the second equation gives s 3, so 5s 2t 1 8 3 3t 3 24 3t 3 t 7. Thus, the solution is 3 7. u 30 5 Adding 3 times the first equation to the second equation gives 10 10 1, so 44. 3u 80 5 u 30 1 5 u 25. Thus, the solution is u 25 1.
SECTION 10.1 Systems of Linear Equations in Two Variables
667
1x 3y 3 2 5 45. 5 x 2y 10 3
Adding 10 times the first equation to 3 times the second equation gives 0 0. Writing the equation
3x 1y 1 2 3 2 46. 2x 1 y 1 2 2
Adding 6 times the first equation to 4 times the second equation gives x 5 x 5. So
in slope-intercept form, we have 12 x 35 y 3 35 y 12 x 3 y 56 x 5, so the solutions are all pairs of the form x 56 x 5 where x is a real number.
9 5 2y 3 y 21. Thus, the solution is 5 21.
47.
04x 12y 14 12x
Adding 30 times the first equation to 1 times the second equation gives 41y 410 y 10, so
5y 10
12x 5 10 10 12x 60 x 5. Thus, the solution is 5 10.
48.
26x 10y 4
06x 12y
Adding 3 times the first equation to 25 times the second equation gives 63x 63 x 1, so
3
26 1 10y 4 10y 30 y 3. Thus, the solution is 1 3. 1x 1y 2 3 4 49. 8x 6y 10
Adding 24 times the first equation to the second equation gives 0 58, which is never true. Thus,
the system has no solution.
1 x 1y 4 10 2 50. 2x 10y 80
Adding 20 times the first equation to the second equation gives 0 0, which is always true. We
now put the equation in slope-intercept form. We have 2x 10y 80 10y 2x 80 y 15 x 8, so the solutions are all pairs of the form x 15 x 8 where x is a real number.
51.
021x 317y 951
52.
235x 117y 589
The solution is approximately 387 274.
1872x 1491y 1233
621x 1292y 1782
The solution is approximately 071 172.
5
-5
5
5 -5
-5
5 -5
668
53.
CHAPTER 10 Systems of Equations and Inequalities
2371x 6552y 13,591
54.
9815x 992y 618,555
The solution is approximately 6100 2000.
435x 912y
0
132x 455y 994
The solution is approximately 285 136.
5
20 -5
5
0
-5
50
1 , a 1. So 55. Subtracting the first equation from the second, we get ay y 1 y a 1 1 y a1 1 1 1 1 1 0x . Thus, the solution is . x a1 1a a1 a1 a1 a , a b. So x 56. Subtracting the first equation from a times the second, we get a b y a y ab b a b . Hence, the solution is . x ba ba ab
a ab
1
ab 1 57. Subtracting b times the first equation from a times the second, we get a 2 b2 y a b y 2 , ab a b2 a 1 1 b 1 1 ax x . Thus, the solution is . a 2 b2 0. So ax ab ab ab ab ab 1 b 58. Subtracting a times the first equation from the second, we get b b a y 1 y . So ax 0 b b a b b a 1 1 1 1 1 . Hence, the solution is 2 . x 2 a b a a b a b b a a ab b ab 59. Let the two numbers be x and y. Then
x y 34 x y 10
Adding these two equations gives 2x 44 x 22. So
22 y 34 y 12. Therefore, the two numbers are 22 and 12. 60. Let x be the larger number and y be the other number. This gives
x y 2 x y x 6 2y
two equations gives y 6, so x 6 2 6 18. Therefore, the two numbers are 18 and 6. 61. Let d be the number of dimes and q be the number of quarters. This gives
d
x 3y 0
x 2y 6
q 14
010d 025q 275
Adding these
Subtracting the
first equation from 10 times the second gives 15q 135 q 9. So d 9 14 d 5. Thus, the number of dimes is 5 and the number of quarters is 9. 62. Let c be the number of children and a be the number of adults. This gives
c
a 2200
150c 400a 5050
Subtracting 3 times
the first equation from 2 times the second gives 5a 3500 a 700, so c 700 2200 c 1500. Therefore, the number of children admitted was 1500 and the number of adults was 700.
SECTION 10.1 Systems of Linear Equations in Two Variables
63. Let r be the amount of regular gas sold and p the amount of premium gas sold. Then
r
p 280
220r 300 p 680
669
Subtracting
the second equation from three times the first equation gives 3r 22 3 280 680 08r 160 r 200. Substituting this value of r into the original first equation gives 200 p 280 p 80. Thus, 200 gallons of regular gas and 80 gallons of premium were sold. 64. Let s be the number of boxes of regular strawberries sold and d the number of deluxe strawberries sold. Then s d 135 Subtracting 7 times the first equation from the second equation gives 10d 7d 1110 7 135 7s 10d 1110
3d 165 d 55. Substituting this value of d into the original first equation gives s 55 135 s 80. Thus, 80 boxes of standard strawberries and 55 boxes of deluxe strawberries were sold. 2x 2y 180 65. Let x be the speed of the plane in still air and y be the speed of the wind. This gives Subtracting 12x 12y 180
6 times the first equation from 10 times the second gives 24x 2880 x 120, so 2 120 2y 180 2y 60 y 30. Therefore, the speed of the plane is 120 mi/h and the wind speed is 30 mi/h. Downriver: x y 20 66. Let x be speed of the boat in still water and y be speed of the river flow. Adding Upriver: 25x 25y 20
5 times the first equation to 2 times the second gives 10x 140 x 14, so 14 y 20 y 6. Therefore, the boat’s speed is 14 mi/h and the current in the river flows at 6 mi/h. 012a 020b 32 Subtracting 250 times the 67. Let a and b be the number of grams of food A and food B. Then 100a 50b 22,000 first equation from the second, we get 70a 14,000 a 200, so 012 200 020b 32 020b 8 b 40. Thus, she should use 200 grams of food A and 40 grams of food B.
68. Let x be the number of pounds of Kenyan coffee and y be the number of pounds of Sri Lankan coffee. This gives 350x 560y 1155 Adding 10 times the first equation and 35 times the second gives 21y 105 y 05, x y 3 so x 25 3 x 25. Thus, 25 pounds of Kenyan coffee and 05 pounds of Sri Lankan coffee should be mixed.
69. Let x and y be the sulfuric acid concentrations in the first and second containers. 300x 600y 900 015 Subtracting the first equation from 3 times the second gives 900y 90 y 010, so 100x 500y 600 0125 100x 500 010 75 x 025. Thus, the concentrations of sulfuric acid are 25% in the first container and 10% in the second.
70. Let x and y be the number of milliliters of the two brine solutions. Quantity: x y 1000 Subtracting the first equation from 20 times the second gives 3y 1800 Concentrations: 005x 020y 014 y 600, so x 600 1000 x 400. Therefore, 400 milliliters of the 5% solution and 600 milliliters of the 20% solution should be mixed. Total invested: x y 20,000 71. Let x be the amount invested at 5% and y the amount invested at 8%. Interest earned: 005x 008y 1180
Subtracting 5 times the first equation from 100 times the second gives 3y 18,000 y 6,000, so x 6,000 20,000 x 14,000. She invests $14,000 at 5% and $6,000 at 8%.
670
CHAPTER 10 Systems of Equations and Inequalities
72. Let x be the amount invested at 6% and y the amount invested at 10%. The ratio of the amounts invested gives x 2y. Then the interest earned is 006x 010y 3520 6x 10y 352,000. Substituting gives 6 2y 10y 352,000 22y 352,000 y 16,000. Then x 2 16,000 32,000. Thus, he invests $32,000 at 6% and $16,000 at 10%. 73. Let x be the length of time John drives and y be the length of time Mary drives. Then y x 025, so x y 025, and multiplying by 40, we get 40x 40y 10. Comparing the distances, we get 60x 40y 35, or 60x 40y 35. This 40x 40y 10 gives the system Adding, we get 20x 45 x 225, so y 225 025 25. Thus, John 60x 40y 35 drives for 2 14 hours and Mary drives for 2 12 hours.
74. Let x be the cycling speed and y be the running speed. (Remember to divide by 60 to convert minutes to decimal hours.) 05x 05y 125 We have Subtracting 2 times the first equation from 5 times the second, we get 275x 55 075x 02y 16 x 20, so 20 y 25 y 5. Thus, the cycling speed is 20 mi/h and the running speed is 5 mi/h. xy7 75. Let x be the tens digit and y be the ones digit of the number. Adding 9 times the first 10y x 27 10x y
equation to the second gives 18x 36 x 2, so 2 y 7 y 5. Thus, the number is 25. 76. First let us find the intersection point of the two lines. The y-coordinate of the intersection point is the height of the triangle. y 2x 4 We have Adding 2 times the first equation to the second gives 3y 12, so the triangle has height 4. y 4x 20 Furthermore, y 2x 4 intersects the x-axis at x 2, and y 4x 20 intersects the x-axis at x 5. Thus the base has
length 5 2 3. Therefore, the area of the triangle is A 12 bh 12 3 4 6. 77. n 5, so nk1 xk 1 2 3 5 7 18, nk1 yk 3 5 6 6 9 29, n k1 xk yk 1 3 2 5 3 6 5 6 7 9 124, and n 2 2 2 2 2 2 k1 xk 1 2 3 5 7 88. Thus we get the system 18a 5b 29 Subtracting 18 times the first equation from 5 times the 88a 18b 124 second, we get 116a 98 a 0845. Then
b 15 [18 0845 29] 2758. So the regression line is y 0845x 2758.
y 10 8 6 4 2 0
1
2 3 4 5 6 7 x
10.2 SYSTEMS OF LINEAR EQUATIONS IN SEVERAL VARIABLES 1. If we add 2 times the first equation to the second equation, the second equation becomes x 3z 1. 2. To eliminate x from the third equation, we add 3 times the first equation to the third equation. The third equation becomes 4y 5z 4. 3. The equation 6x 3y 12 z 0 is linear. 4. The equation x 2 y 2 z 2 4 is not linear, since it contains squares of variables. x y 3y z 5 is not a linear system, since the first equation contains a product of variables. In fact 5. The system x y 2 5z 0 2x yz 3 both the second and the third equation are not linear.
SECTION 10.2 Systems of Linear Equations in Several Variables
x 2y 3z 10 6. The system 2x 5y 2 y 2z 4
x 3y z 0 7. yz 3 z 2
671
is linear.
Substituting z 2 into the second equation gives y 2 3 y 1. Substituting z 2
and y 1 into the first equation gives x 3 1 2 0 x 5. Thus, the solution is 5 1 2. 3x 3y z 0 8. Substituting z 3 into the second equation gives y 4 3 10 y 2. Substituting z 3 y 4z 10 z 3 and y 2 into the first equation gives 3x 3 2 3 0 x 3. Thus, the solution is 3 2 3. x 2y z 7 Solving we get 2z 6 z 3. Substituting z 3 into the second equation gives 9. y 3z 9 2z 6
y 3 3 9 y 0. Substituting z 3 and y 0 into the first equation gives x 2 0 3 7 x 4. Thus, the solution is 4 0 3. x 2y 3z 10 Solving we get 3z 12 z 4. Substituting z 4 into the second equation gives 10. 2y z 2 3z 12
2y 4 2 y 3. Substituting z 4 and y 3 into the first equation gives x 2 3 3 4 10 x 4. Thus, the solution is 4 3 4. 2x y 6z 5 11. Solving we get 2z 1 z 12 . Substituting z 12 into the second equation gives y 4z 0 2z 1 y 4 12 0 y 2. Substituting z 12 and y 2 into the first equation gives 2x 2 6 12 5 x 5. Thus, the solution is 5 2 12 .
12.
4x
2y
3z 10
z 6
1z 2
2
Solving we get 12 z 4 z 8 . Substituting z 8 into the second equation gives
2y 8 6 2y 2 y 1. Substituting z 8 into the first equation gives 4x 3 8 10 4x 14 x 72 . Thus, the solution is 72 1 8 .
3x y z 4 Add the third equation to the second equation: y z 1 x 2y z 1 3x y z 4 Or, add the first equation to three times the second equation: 4y 7z 4 x 2y z 1
3x y z 4 13. x y 2z 0 x 2y z 1
672
CHAPTER 10 Systems of Equations and Inequalities
3 5x 2y 3z 14. 10x 3y z 20 x 3y z 8
3 5x 2y 3z Add twice the first equation to the second equation: y 5z 14 x 3y z 8 5x 2y 3z 3 Or, add 10 times the third equation to the second equation: 27y 11z 60 x 3y z 8 2x y 3z 5 Add 3 times the first equation to the third equation: 2x 3y z 13 8y 8z 8 5 2x y 3z Or, add 3 times the second equation to the third equation: 2x 3y z 13 14y 4z 32
2x y 3z 5 15. 2x 3y z 13 6x 5y z 7
x 3y 2z 1 Add 2 times the first equation to 3 times the third equation: y z 1 2x z 1 x 3y 2z 1 Or, add 2 times the second equation to the third equation: y z 1 3z 3
x 3y 2z 1 16. y z 1 2y z 1
x y z 4 x y z 4 17. 2y z 1 2y z 1 x y 2z 5 3z 9
Eq. 1 Eq. 3
So z 3 and 2y 3 1 2y 2 y 1. Thus, x 1 3 4 x 2. So the solution is 2 1 3.
x y z 0 x y z 0 18. y 2z 2 y 2z 2 x y z 2 2y 2z 2
1 Eq. 1 Eq. 3
x y z 0 y 2z 2 6z 6
2 Eq. 2 Eq. 3
So z 1 and y 2 1 2 y 0. Then x 0 1 0 x 1. So the solution is 1 0 1.
x 2y z 6 x 2y z 6 19. y 3z 16 y 3z 16 x 3y 2z 14 5y 3z 20 x 2y z 6 y 3z 16 12z 60 5 Eq. 2 Eq. 3
Eq. 1 1 Eq. 3
So z 5 and y 3 5 16 y 1. Then x 2 1 5 6 x 1. So the solution is 1 1 5.
x 2y 3z 10 x 2y 3z 10 20. 3y z 7 3y z 7 x y z 3y 4z 17 7
1 Eq. 1Eq. 3
x 2y 3z 10 3y z 7 5z 10
Eq. 2 1 Eq. 3
So z 2 and 3y 2 7 y 3. Then x 2 3 3 2 10 x 2. So the solution is 2 3 2.
SECTION 10.2 Systems of Linear Equations in Several Variables
x y z 4 x y z 4 21. x 3y 3z 10 2y 2z 6 2x y z 3 y 3z 5 x y z 4 y 3z 5 4z 4 2 Eq. 2 Eq. 3
1 Eq. 1 Eq. 2 2 Eq. 1 1 Eq. 3
x y z 4 y 3z 5 2y 2z 6
Eq. 3
673
Eq. 2
So z 1 and y 3 1 8 y 2. Then x 2 1 4 x 1. So the solution is 1 2 1. x y z 0 x y z 0 x y z 0 22. x 2y 5z 3 y 2z 1 y 2z 1 13 Eq. 1 13 Eq. 2 3x y 5z 10 4 Eq. 2 Eq. 3 6 4y 3z 6 3 Eq. 1 Eq. 3 So z 2 and y 2 2 1 y 3. Then x 3 2 0 x 1. So the solution is 1 3 2. 4z 1 4z 1 4z 1 x x x 23. 2x y 6z 4 y 2z 2 y 2z 2 2 Eq. 1 Eq. 2 2x 3y 2z 8 3y 6z 6 2 Eq. 1 Eq. 3 12z 12 3 Eq. 2 Eq. 3
So z 1 and y 2 1 2 y 0. Then x 4 1 1 x 5. So the solution is 5 0 1. 2 2 x y 2z x y 2z 2 x y 2z 24. y 6z 11 3x y 5z 8 y 6z 11 3 Eq. 1 Eq. 3 2x y 2z 7 4y z 23z 46 4 Eq. 2 Eq. 2 2 2 Eq. 1 Eq. 2
So z 2 and y 6 2 11 y 1. Then x 1 2 2 x 1. So the solution is 1 1 2. 2x 4y z 2 2x 4y z 2 x 2y 3z 4 25. 5z 10 x 2y 3z 4 Eq. 1 2 Eq. 2 14y 5z 4 Eq. 2 Eq. 3 14y 5z 4 3 Eq. 1 2 Eq. 3 3x y z 1 5z 10 Eq. 2 Eq. 3 So z 2 and 14y 5 2 4 y 1. 2x y 2x y z 8 26. 3y x y z 3 2x 4z 18 y
Then x 2 1 3 2 4 x 0. So the solution is 0 1 2. z 8 2x y z 8 z 2 Eq. 1 2 Eq. 2 3y z 2 8z 32 Eq. 2 3 Eq. 3 3z 10 Eq. 1 Eq. 3
So z 4 and 3y 4 2 y 2. Then 2x 2 4 8 x 1. So the solution is 1 2 4. Eq. 2 2y 4z 1 2x y 2z 1 27. 2x y 2z 1 2y 4z 1 Eq. 1 4x 2y 0 4z 2 2Eq. 2 Eq. 3 So z 12 and 2y 4 12 1 y 12 . Then 2x 12 2 12 1 x 14 . So the solution is 14 12 12 . Eq. 2 y z 1 2 z 2 6x 2y z 6x 2y 28. 6x 2y z 2 y z 1 Eq. 1 y z 1 x y 3z 2 21z 14 4 Eq. 2 Eq. 3 4y 17z 10 Eq. 2 6 Eq. 3 So z 23 and y 23 1 y 13 . Then 6x 2 13 23 2 x 13 . So the solution is 13 13 23 . x 2y z 1 x 2y z 1 29. 2x 3y 4z 3 Since 0 1 is false, this system is inconsistent. y 2z 5 2 Eq. 1 Eq. 2 3x 6y 3z 4 0 1 3 Eq. 1 Eq. 3
674
CHAPTER 10 Systems of Equations and Inequalities
2z 0 2z 0 Eq. 2 x x 2y 5z 4 x 30. 2y 3z 4 x 2z 0 x 2y 5z 4 Eq. 1 2y 3z 2 4x 2y 11z 2 4x 2y 11z 2 2z 0 x Since 0 6 is false, this system is inconsistent. 2y 3z 4 0 6 Eq. 2 Eq. 3
Eq. 1 Eq. 2
4 Eq. 1 Eq. 3
3 Eq. 2 3 x 2y x 2y 2x 3y z 1 31. y z 5 Eq. 2 2 Eq. 1 x 2y 3 2x 3y z 1 Eq. 1 x 3y z 4 x 3y z 4 yz 1 Eq. 3 Eq. 1 3 x 2y Since 0 4 is false, this system is inconsistent. y z 5 0 4 Eq. 2 Eq. 3 5 x 2y 3z 5 x 2y 3z 32. 2x y z 5 5y 5z 5 4x 3y 7z 5 5y 5z 5
Eq. 2 2 Eq. 1 2 Eq. 2 Eq. 3
5 x 2y 3z 5y 5z 5 0 10
Since 0 10 is false, this system is inconsistent. x y z 0 x y z 0 x y z 0 33. y 2z 3 x 2y 3z 3 y 2z 3 Eq. 2 Eq. 1 2x 3y 4z 3 00 y 2z 3 2 Eq. 1 Eq. 3
Eq. 3 Eq. 2
Eq. 2 Eq. 3
So z t and y 2t 3 y 2t 3. Then x 2t 3 t 0 x t 3. So the solutions are t 3 2t 3 t, where t is any real number. x 2y z 3 x 2y z 3 x 2y z 3 34. 2x 5y 6z 7 y 4z 1 Eq. 2 2 Eq. 1 y 4z 1 2x 3y 2z 5 0 0 Eq. 2 Eq. 3 y 4z 1 Eq. 3 2 Eq. 1
So z t and we have y 4t 1 y 4t 1. Substituting into the first equation, we have x 2 4t 1 t 3 x 7t 1. So the solutions are 7t 1 4t 1 t, where t is any real number. x 3y 2z 0 x 3y 2z 0 x 3y 2z 0 35. 2x 6y 8z 4 6y 8z 4 Eq. 2 2 Eq. 1 4z 4 4x 6y 0 0 Eq. 2 Eq. 3 4 6y 8z 4 Eq. 3 4 Eq. 1 So z t and 6y 8t 4 6y 8t 4 y 43 t 23 . Then x 3 43 t 23 2t 0 x 2t 2. So the solutions are 2t 2 43 t 23 t , where t is any real number. 2x 4y z 3 x 2y 2z 0 Eq. 3 x 2y 2z 0 36. x 2y 4z 6 2x 4y z 3 Eq. 1 3z 3 2 Eq. 1 Eq. 2 x 2y 2z 0 x 2y 4z 6 Eq. 2 6z 6 Eq. 3 Eq. 1 x 2y 2z 0 3z 3 0 0 2 Eq. 2 Eq. 3
So z 1 and y t, and substituting into the first equation we have x 2t 2 1 0 x 2t 2. Thus, the solutions are 2t 2 t 1, where t is any real number.
SECTION 10.2 Systems of Linear Equations in Several Variables
37.
x
z 2
y 2z
6
3
x
z 2
y 2z
675
6
3
2y 2z 2 8 x 2y z 2 Eq. 3 Eq. 1 2x y 3z 2 0 y z 6 12 Eq. 4 2 Eq. 1 x z 2 6 x z 2 6 y 2z y 2z 3 3 1 Eq. 3 z 1 2z 2 2 Eq. 3 2 Eq. 2 2 3z 6 9 Eq. 4 Eq. 2 6 12 3 Eq. 3 2 Eq. 4
So 2 and z 2 1 z 1. Then y 2 1 3 y 1 and x 1 2 2 6 x 1. Thus, the solution is 1 1 1 2.
x y z 0 x y z 0 x y 2z 2 0 z 0 38. 2x 2y 3z 4 1 z 2 1 2x 3y 4z 5 2 y z 1 x y z 0 yz 1 z 0 1 Eq. 4 Eq. 3
Eq. 2 Eq. 1 Eq. 3 2 Eq. 1 Eq. 4 Eq. 3
x y z 0 yz 1 z 0 z 2 1
Eq. 4 Eq. 2
Eq. 3
So 1 and z 1 0 z 1. Then y 1 1 1 y 1 and x 1 1 1 0 x 1. So the solution is 1 1 1 1.
39. Let x be the amount invested at 4%, y the amount invested at 5%, and z the amount invested at 6%. We set x y z 100,000 Total money: up a model and get the following equations: Annual income: 004x 005y 006z 0051 100,000 Equal amounts: xy x y z 100,000 x y z 100,000 4x 5y 6z 510,000 y 2z 110,000 Eq. 2 4 Eq. 1 x y 0 2y z 100,000 Eq. 3 Eq. 1 x y z 100,000 y 2z 110,000 3z 120,000 2 Eq. 2 Eq. 3
So z 40,000 and y 2 40,000 110,000 y 30,000. Since x y, x 30,000. Mark should invest $30,000 in short-term bonds, $30,000 in intermediate-term bonds, and $40,000 in long-term bonds.
676
CHAPTER 10 Systems of Equations and Inequalities
40. Let x be the amount invested at 3%, y the amount invested at 5 12 %, and z the amount invested at 9%. We x y z 50,000 Total investment: set up a model and get the following equations: Annual income: 003x 0055y 009z 2540 Twice as much: x 2z x y z 50,000 x y z 50,000 6x 11y 18z 508,000 200 Eq. 2 5x 7z 42,000 Eq. 2 11 Eq. 1 x x 2z 0 2z 0 x y z 50,000 5x 7z 42,000 3z 42,000 Eq. 2 5 Eq. 3
So z 14,000 and x 2z 28,000. Since x y z 50,000, we have y 50,000 14,000 28,000 8000. Cyndee should invest $28,000 in the least risky account, $8,000 in the intermediate account, and 14,000 in the highest-yielding account.
41. Let x, y, and z be the number of acres of land planted with corn, wheat, and soybeans. We set up a model and x y z 1200 Total acres: Substituting 2x for y, we get get the following equations: Market demand: 2x y Total cost: 45x 60y 50z 63,750 x 2x z 1200 z 1200 z 1200 3x 3x 2x y 2x y 0 2x y 0 45x 60 2x 50z 63,750 165x 50z 63,750 15x 3750 Eq. 3 50 Eq. 1
So 15x 3,750 x 250 and y 2 250 500. Substituting into the original equation, we have 250 500 z 1200 z 450. Thus the farmer should plant 250 acres of corn, 500 acres of wheat, and 450 acres of soybeans.
42. Let a, b, and c be the number of gallons of Regular, Performance Plus, and Premium gas sold. The information provided b c 6500 a b c 6500 a gives the following system: 300a 320b 330c 20,050 02b 03c 550 Eq. 2 3 Eq. 1 a 3c 0 b 4c 6500 Eq. 1 Eq. 3 b c 6500 a 02b 03c 550 5c 7500 10 Eq. 2 2 Eq. 3
Thus, c 1500, so 02b 03 1500 550 b 500 and a 500 1500 6500 a 4500. The gas station sold 4500 gallons of Regular, 500 gallons of Performance Plus, and 1500 gallons of Premium gas.
43. Let a, b, and c be the number of ounces of Type A, Type B, and Type C pellets used. The 2a 3b c 9 requirements for the different vitamins gives the following system: 3a b 3c 14 8a 5b 7c 32 2a 3b c 9 Equations 2 and 3 are inconsistent, so there is no solution. 7b 3c 1 2 Eq. 2 3 Eq. 1 7b 3c 4 Eq. 3 4 Eq. 1
SECTION 10.2 Systems of Linear Equations in Several Variables
677
44. Let a, b, and c represent the number of servings of toast, cottage cheese, and fruit, respectively. Then 2c 6 2a the given dietary requirements give rise to the following system: a 5b 11 100a 120b 60c 460 2a 2c 6 c 6 2a 10b 2c 16 10b 2c 16 2 Eq. 2 Eq. 1 16c 32 Eq. 3 12 Eq. 2 120b 40c 160 Eq. 3 50 Eq. 1
Thus, c 2, so 10b 2 2 16 10b 20 b 2 and 2a 2 2 6 a 1. Nicole should eat one serving of toast and two each of cottage cheese and fruit.
45. Let a, b, and c represent the number of Midnight Mango, Tropical Torrent, and Pineapple Power smoothies sold. The 8a 6b 2c 820 8a 6b 2c 820 given information leads to the system 3a 5b 8c 690 22b 58c 3060 8 Eq. 2 3 Eq. 1 3a 3b 4c 450 2b 4c 240 Eq. 2 Eq. 3 8a 6b 2c 820 22b 58c 3060 14c 420 Eq. 2 11 Eq. 3
Thus, c 30, so 22b 58 30 3060 22b 1320 b 60 and 8a 6 60 2 30 820 a 50. Thus, The Juice Company sold 50 Midnight Mango, 60 Tropical Torrent, and 30 Pineapple Power smoothies on that particular day.
46. Let a, b, and c represent the number of days required at each plant. The given information leads 8a 10b 14c 110 8a 10b 14c 110 to the system 16a 12b 10c 150 8b 18c 70 2 Eq. 1 1 Eq. 2 10a 18b 6c 114 84b 2c 162 5 Eq. 2 8 Eq. 3 8a 10b 14c 110 8b 18c 70 382c 1146 21 Eq. 2 2 Eq. 3
Thus, c 3, so 8b 18 3 70 8b 16 b 2 and 8a 10 2 14 3 110 a 6. Thus, Factory A should be scheduled for 6 days, Factory B for 2 days, and Factory C for 3 days.
47. Let a, b, and c be the number of shares of Stock A, Stock B, and Stock C in the investor’s portfolio. Since the total value remains unchanged, we get the following system: 10a 25b 29c 74,000 10a 25b 29c 74,000 12a 20b 32c 74,000 50b 14c 74,000 6 Eq. 1 5 Eq. 2 16a 15b 32c 74,000 125b 72c 222,000 8 Eq. 1 5 Eq. 3 10a 25b 29c 74,000
50b 14c
74,000
74c 74,000
5 Eq. 2 2 Eq. 3
So c 1,000. Back-substituting we have 50b 14 1000 74,000 50b 60,000 b 1,200 . And finally 10a 25 1200 29 1000 74,000 10a 30,000 29,000 74,000 10a 15,000 a 1,500. Thus the portfolio consists of 1,500 shares of Stock A, 1,200 shares of Stock B, and 1,000 shares of Stock C.
678
CHAPTER 10 Systems of Equations and Inequalities
I1 I2 I3 0 48. 16I1 8I2 4 24I2 16I3 4 16 Eq. 1 Eq. 2 8I2 4I3 5 8I2 4I3 5 I2 I3 0 I1 24I2 16I3 4 28I3 19 Eq. 2 3 Eq. 3 19 2 2 19 11 So I3 19 28 068 and 24I2 16 28 4 I2 7 029. Then I1 7 28 0 I1 28 039.
I 1 I2 I3 0
x x1 y0 y1 z z1 49. (a) We begin by substituting 0 , , and 0 into the left-hand side of the first equation: 2 2 2 x0 x1 y0 y1 z z1 b1 c1 0 12 a1 x0 b1 y0 c1 z 0 a1 x1 b1 y1 c1 z 1 a1 2 2 2 12 d1 d1 d1
Thus the given ordered triple satisfies the first equation. We can show that it satisfies the second and the third in exactly the same way. Thus it is a solution of the system.
(b) We have shown in part (a) that if the system has two different solutions, we can find a third one by averaging the two solutions. But then we can find a fourth and a fifth solution by averaging the new one with each of the previous two. Then we can find four more by repeating this process with these new solutions, and so on. Clearly this process can continue indefinitely, so there are infinitely many solutions.
10.3 PARTIAL FRACTIONS 1. (iii): r x 2. (ii): r x 3. 5.
7. 8.
4 x x 22
B C A x x 2 x 22
Bx C A 2x 8 2 x 1 x 4 x 1 x 2 4
B A 1 x 1 x 2 x 1 x 2 x 2 3x 5
x 22 x 4
B C A x 2 x 22 x 4
x A B x 4. 2 x 1 x 4 x 1 x 4 x 3x 4 1 B 1 C D A 6. 4 3 2 3 x x 1 x x3 x x 1 x x
Bx C x2 A 2 x 3 x 4 x 3 x 2 4
B Cx D 1 1 A 1 2 2 x 1 x 1 x 1 x2 1 x4 1 x 1 x 1 x 1 x 2 1
Ax B Cx D x 3 4x 2 2 2 9. 2 x 1 x2 2 x2 1 x 2
10.
B A Cx D x4 x2 1 Ex F 2 x 2 2 2 2 2 x x 4 x2 4 x x 4
B C x3 x 1 A D Ex F Gx H 2 2 x 2x 5 2 x 2x 5 2x 52 2x 53 x 2 2x 5 x 2x 53 x 2 2x 5 12. Since x 3 1 x 2 1 x 1 x 2 x 1 x 1 x 1 x 12 x 1 x 2 x 1 , we have 11.
B Dx E C 1 A 1 2 . x 1 x 12 x 1 x3 1 x2 1 x x 1 x 12 x 1 x 2 x 1
SECTION 10.3 Partial Fractions
13.
B 2 A . Multiplying by x 1 x 1, we get 2 A x 1 B x 1 x 1 x 1 x 1 x 1 AB 0 Adding we get 2A 2 A 1. Now A B 0 B A, so 2 Ax A Bx B. Thus AB 2 B 1. Thus, the required partial fraction decomposition is
14.
2 1 1 . x 1 x 1 x 1 x 1
B A 2x . Multiplying by x 1 x 1, we get 2x A x 1 B x 1 x 1 x 1 x 1 x 1 AB 2 2x Ax A Bx B. Thus Adding we 2A 2 A 1. Since A B 0 B A, B 1. The AB 0 required partial fraction decomposition is
15.
1 2x 1 . x 1 x 1 x 1 x 1
5 B A . Multiplying by x 1 x 4, we get 5 A x 4 B x 1 x 1 x 4 x 1 x 4 AB 0 5 Ax 4A Bx B. Thus Now A B 0 B A, so substituting,we get 4A A 5 4A B 5 5A 5 A 1and B 1. The required partial fraction decomposition is
16.
1 1 5 . x 1 x 4 x 1 x 4
A B x 6 . x x 3 x x 3
Multiplying by x x 3we get x 6 A x 3 x B AB1 x 6 Ax 3A Bx A B x 3A. Thus Now 3A 6 A 2, and 2 B 1 B 1. 3A 6 The required partial fraction decomposition is
17.
679
x 6 2 1 . x x 3 x x 3
B 12 A 12 . Multiplying by x 3 x 3, we get 12 A x 3 B x 3 x 3 x 3 x 3 x 3 x2 9 A B 0 AB 0 12 Ax 3A Bx 3B. Thus Adding, we get 2A 4 A 2. So 2 B 0 3A 3B 12 AB 4 2 2 12 . B 2. The required partial fraction decomposition is 2 x 3 x 3 x 9
18.
x 12 A B x 12 . Multiplying by x x 4, we get x x 4 x x 4 x 2 4x x 12 A x 4 Bx Ax 4A Bx A B x 4A. Thus we must solve the system 2 3 x 12 . gives 4A 12 A 3, and 3 B 1 B 2. Thus 2 x x 4 x 4x
19.
AB 4A
4 B 4 A . Multiplying by x 2 4, we get x 2 x 2 x 2 x 2 x2 4 A B 0 AB 0 4 A x 2 B x 2 A B x 2A 2B, and so 2A 2B 4 AB 2 1 1 4 . A 1, and B 1. Therefore, 2 x 2 x 2 x 4
1
This
12
Adding we get 2A 2
680
20.
CHAPTER 10 Systems of Equations and Inequalities
2x 1
x2 x 2 and so
B 2x 1 A . Thus, 2x 1 A x 1 B x 2 A B x A 2B, x 2 x 1 x 2 x 1
A B 2
A 2B 1
Adding the two equations, we get 3B 3 B 1. Thus A 1 2 A 1. Therefore,
2x 1 1 1 . x 2 x 1 x2 x 2 21.
x 14 B x 14 A . Hence, x 14 A x 2 B x 4 A B x 2A 4B, x 4 x 2 x 4 x 2 x 2 2x 8 A B 1 2A 2B 2 and so Adding, we get 3A 9 A 3. So 3 B 1 B 2. 2A 4B 14 A 2B 7 2 3 x 14 . Therefore, 2 x 4 x 2 x 2x 8
22.
23.
A B 8x 3 8x 3 . Hence, 8x 3 A 2x 1 Bx 2A B x A, giving x 2x 1 x 2x 1 2x 2 x 2A B 8 2 3 8x 3 . So A 3 A 3, and 2 3 B 8 B 2. Therefore, 2 x 2x 1 2x x A 3 B x A x . Hence, 4x 3 2x 1 4x 3 2x 1 8x 2 10x 3 x A 2x 1 B 4x 3 2A 4B x A 3B, and so Adding, we get 2B 1 B 12 , and A 32 . Therefore,
24.
2A 4B 1
A 3B 0
3 2
2A 4B 1
2A 6B 0
1 x 2 . 4x 3 2x 1 8x 2 10x 3
A B C 7x 3 7x 3 . Hence, x x 3 x 1 x x 3 x 1 x 3 2x 2 3x 7x 3 A x 3 x 1 Bx x 1 C x x 3 A x 2 2x 3 B x 2 x C x 2 3x
Thus
A B C x 2 2A B 3C x 3A AB C
0
2A B 3C 7 3A 3
1 B C 0
2 B 3C 7
Coefficients of x 2 Coefficients of x Constant terms
So 3A 3 A 1. Substituting, the system reduces to
Adding these two equations, we get 3 4C 7 C 1. Thus 1 B 1 0 B 2.
7x 3 2 1 1 Therefore, 3 . x x 3 x 1 x 2x 2 3x 25.
B C A 9x 2 9x 6 9x 2 9x 6 . Thus, x 2 x 2 2x 1 x 2 x 2 2x 1 2x 3 x 2 8x 4 9x 2 9x 6 A x 2 2x 1 B x 2 2x 1 C x 2 x 2 A 2x 2 3x 2 B 2x 2 5x 2 C x 2 4 2A 2B C x 2 3A 5B x 2A 2B 4C
SECTION 10.3 Partial Fractions
681
Coefficients of x 2 2A 2B C 9 2A 2B C 9 This leads to the system 16B 3C 45 3A 5B 9 Coefficients of x 2A 2B 4C 6 4B 3C 15 Constant terms 9 2A 2B C Hence, 15C 15 C 1; 16B 3 45 B 3; and 2A 6 1 9 A 2. 16B 3C 45 15C 15 Therefore,
26.
9x 2 9x 6
2x 3 x 2 8x 4
3 1 2 . x 2 x 2 2x 1
B C 3x 2 3x 27 A 3x 2 3x 27 . Thus, 2 x 2 2x 3 x 3 x 2 2x 3 x 3 x 2 2x 3x 9
3x 2 3x 27 A 2x 3 x 3 B x 2 x 3 C x 2 2x 3 A 2x 2 3x 9 B x 2 5x 6 C 2x 2 x 6
2A B 2C 3 So 3A 5B C 3 9A 6B 6C 27
2A B 2C x 2 3A 5B C x 9A 6B 6C Coefficients of x 2 2A B 2C 3 2A B 2C 3 7B 4C 3 7B 6C 3 Coefficients of x 21B 6C 27 24C 24 Constant terms
Hence, 24C 24 C 1; then 7B 4 3 B 1; and 2A 1 2 3 A 3. Therefore,
27.
1 1 3 3x 2 3x 27 . 2 x 2 2x 3 x 3 x 2 2x 3x 9
x2 1 B x2 1 C A . Hence, 2 x x 1 x3 x2 x 2 x 1 x
x 2 1 Ax x 1 B x 1 C x 2 A C x 2 A B x B, and so B 1; A 1 0 A 1; and
1 x2 1 1 2 2 . 1 C 1 C 2. Therefore, 3 2 x x 1 x x x 28.
3x 2 5x 13 B C A 3x 2 5x 13 . Thus, 2 2 3x 2 x 2 3x 2 x 4x 4 3x 2 x 2 x 22
3x 2 5x 13 A x 22 B 3x 2 x 2 C 3x 2 A x 2 4x 4 B 3x 2 4x 4 C 3x 2
A 3B x 2 4A 4B 3C x 4A 4B 2C A 3B 3 Coefficients of x 2 A 3B This leads to the following system: 4A 4B 3C 5 Coefficients of x 8B 4A 4B 2C 13 Constant terms 8B 3 A 3B Hence, 8C 9 C 98 ; 8B 3 98 8B 27 8B 3C 17 8 17 B 8C 9
3
3C 17
5C 8
109 ; and 64
135 109 9 2 64 64 8 327 3 A 135 . Therefore, 3x 5x 13 A 3 109 A . 64 64 64 3x 2 x 2 x 22 3x 2 x 22
682
29.
CHAPTER 10 Systems of Equations and Inequalities
B A . Hence, 2x A 2x 3 B 2Ax 3A B. So 2A 2 2x 3 2x 32 2x 3 1 A 1; and 3 1 B 0 B 3. Therefore, 2 . 2x 3 2x 32 4x 12x 9 2x
4x 2 12x 9
2x
2x 32
B A . Hence, x 4 A 2x 5 B 2Ax 5A B, and so 30. 2 2x 5 2x 5 2x 52 x 4
A 12 and 5 12 B 4 B 32 . Therefore,
31.
x 4
1
3
2A
1
5A B 4
2 2 . 2x 5 2x 52 2x 52
B 4x 2 x 2 C D A 4x 2 x 2 . Hence, 3 2 3 4 3 x x 2 x 2x x x 2 x x 4x 2 x 2 Ax 2 x 2 Bx x 2 C x 2 Dx 3 A D x 3 2A B x 2 2B C x 2C So 2C 2 C 1; 2B 1 1 B 0; 2A 0 4 A 2; and 2 D 0 D 2. Therefore, 1 2 2 4x 2 x 2 . 3 4 3 x x 2 x 2x x
32.
x 3 2x 2 4x 3 B A C D 2 3 4 . Hence, x 3 2x 2 4x 3 Ax 3 Bx 2 C x D. Thus A 1; B 2; 4 x x x x x C 4; and D 3. Therefore,
33.
10x 2 27x 14 x 13 x 2
2 1 4 3 x 3 2x 2 4x 3 2 3 4. x x4 x x x
B C D A . Thus, x 2 x 1 x 12 x 13
10x 2 27x 14 A x 13 B x 2 x 12 C x 2 x 1 D x 2
A x 3 3x 2 3x 1 B x 2 x 2 2x 1 C x 2 x 2 D x 2 A x 3 3x 2 3x 1 B x 3 3x 2 C x 2 x 2 D x 2
A B x 3 3A C x 2 3A 3B C D x A 2B 2C 2D
SECTION 10.3 Partial Fractions
683
which leads to the system A B 3A C 3A 3B C D A 2B 2C 2D
A B 3B C 3C D 3C 8D
0
10 27
14
A B 0 2 3B C 10 Coefficients of x 3B 2C D 17 Coefficients of x 3B 5C 7D 15 Constant terms Coefficients of x 3
A B 10 3B C 7 3C D 2 9D
0
0
10
7 9
Hence, 9D 9 D 1, 3C 1 7 C 2, 3B 2 10 B 4, and A 4 0 A 4. Therefore, 10x 2 27x 14 x 13 x 2
34.
4 1 4 2 . x 2 x 1 x 12 x 13
2x 2 5x 1 B C 2x 2 5x 1 2x 2 5x 1 D A . Thus, x 1 x 1 x 12 x 4 2x 3 2x 1 x 1 x 3 x 2 x 1 x 13 x 1 x 13 2x 2 5x 1 A x 13 B x 1 x 12 C x 1 x 1 D x 1 A x 3 3x 2 3x 1 B x 1 x 2 2x 1 C x 2 1 D x 1 A x 3 3x 2 3x 1 B x 3 x 2 x 1 C x 2 1 D x 1
A B x 3 3A B C x 2 3A B D x A B C D
AB 0 3ABC 2 which leads to the system 3AB D 5 ABCD 1 A B A B 0 0 2B C 2B C 2 2 2C D 1 2C D 1 2C5D 5 6D 6
A B 2B C Coefficients of x 2 2B C D Coefficients of x 2B3C4D Constant terms Coefficients of x 3
0 2 3
2
Hence, 6D 6 D 1, 2C 1 1 C 0, 2B 0 2
2x 2 5x 1 1 1 1 B 1, and A 1 0 A 1. Therefore, 4 . x 1 x 1 x 13 x 2x 3 2x 1
684
35.
CHAPTER 10 Systems of Equations and Inequalities
3x 3 22x 2 53x 41 x 22 x 32
B D A C . Thus, x 2 x 22 x 3 x 32
3x 3 22x 2 53x 41 A x 2 x 32 B x 32 C x 22 x 3 D x 22 A x 3 8x 2 21x 18 B x 2 6x 9
C x 3 7x 2 16x 12 D x 2 4x 4
A C x 3 8A B 7C D x 2
21A 6B 16C 4D x 18A 9B 12C 4D A C 3 A C 3 Coefficients of x 3 8A B 7C D 22 2 B C D 2 Coefficients of x so we must solve the system 6B 5C 4D 10 21A 6B 16C 4D 53 Coefficients of x 9B 6C 4D 13 18A 9B 12C 4D 41 Constant terms A C 3 A C 3 B C D 2 B C D 2 Hence, D 1, C 2 2 C 0, B 0 1 2 C 2D 2 C 2D 2 3C 5D 5 D 1 B 1, and A 0 3 A 3. Therefore,
36.
3x 3 22x 2 53x 41 x 22 x 32
1 3 1 . x 2 x 22 x 32
B D 3x 2 12x 20 A C 3x 2 12x 20 3x 2 12x 20 . Thus, 2 2 x 22 2 x 2 x 2 2 x 4 8x 2 16 2 2 22 x x x x 4
3x 2 12x 20 A x 2 x 22 B x 22 C x 22 x 2 D x 22 A x 3 2x 2 4x 8 B x 2 4x 4 C x 3 2x 2 4x 8 D x 2 4x 4
A C x 3 2A B 2C D x 2 4A 4B 4C 4D x 8A 4B 8C 4D A C 0 Coefficients of x 3 A C 0 2A B 2C D 2 B 4C D 3 3 Coefficients of x which leads to the system 4A 4B 4C 4D 12 6B 8C 2D 6 Coefficients of x 4B 16C 12D 4 8A 4B 8C 4D 20 Constant terms A C 0 A C 0 B 4C D 3 B 4C D 3 Hence, 48D 48 D 1, 16C 8D 24 16C 8D 24 32C 32D 0 48D 48
16C 8 24 C 1; B 4 1 3 B 2, and A 1 0 A 1. Therefore, 3x 2 12x 20 2 1 1 1 . x 2 x 22 x 2 x 22 x 4 8x 2 16
37.
Bx C x 3 A x 3 2 2 . Hence, x 3 A x 2 3 Bx 2 C x A B x 2 C x 3A. So 3 x x 3x x x 3 x 3 x 3 x 1 1 3A 3 A 1; C 1; and 1 B 0 B 1. Therefore, 3 2 . x x 3x x 3
SECTION 10.3 Partial Fractions
38.
685
3x 2 2x 8 3x 2 2x 8 Ax B C . Thus, 2 x 1 x 2 2 x 1 x 3 x 2 2x 2 x 2 3x 2 2x 8 Ax B x 1 C x 2 2 A C x 2 A B x B 2C, which leads to the system 2 C3 Coefficients of x 2 A C 3 Coefficients of x A A B 2 Coefficients of x B C1 Coefficients of x B 2C 8 Constant terms 2C 6 Constant terms Hence, 2C 6 C 3, B 3 1 B 2; and A 3 3 A 0. Therefore, 3x 2 2x 8 2 3 . 2 x 3 x 2 2x 2 x 2 x 1
Ax B Cx D 2x 3 7x 5 2 . Thus, 39. 2 2 2 x x 2 x 1 x x 2 x 1 2x 3 7x 5 Ax B x 2 1 C x D x 2 x 2
Ax 3 Ax Bx 2 B C x 3 C x 2 2C x Dx 2 Dx 2D
A C x 3 B C D x 2 A 2C D x B 2D
We must solve the system A C B C D A 2C D B 2D
2 0 7 5
A C A C 2 2 BC D BC D 0 Coefficients of x C D C D 5 Coefficients of x 2D C D 5 Constant terms Coefficients of x 3
2 0 5 10
Hence, 2D 10 D 5, C 5 5 C 0, B 0 5 0 B 5, and A 0 2 A 2. Therefore,
40.
2x 5 5 2x 3 7x 5 . x2 x 2 x2 1 x2 x 2 x2 1
x2 x 1 Ax B Cx D x2 x 1 2 . Thus, 2x 4 3x 2 1 2x 2 1 x 1 2x 2 1 x 2 1 x 2 x 1 Ax B x 2 1 C x D 2x 2 1 Ax 3 Ax Bx 2 B 2C x 3 2Dx 2 C x D A 2C x 3 B 2D x 2 A C x B D
which leads to the system A 2C 0 B 2D 1 C 1 A B D1
A 2C 2 B 2D Coefficients of x C Coefficients of x D Constant terms Coefficients of x 3
0 1
1
0
Hence, D 0, C 1, B 0 1 B 1, and A 2 0 A 2. Therefore,
x2 x 1 2x 1 x 2 2 . 4 2 2x 3x 1 2x 1 x 1
686
41.
CHAPTER 10 Systems of Equations and Inequalities
Bx C x4 x3 x2 x 1 A Dx E 2 2 2 . Hence, x 2 x 1 x2 1 x x 1
2 x 4 x 3 x 2 x 1 A x 2 1 Bx C x x 2 1 x Dx E A x 4 2x 2 1 Bx 2 C x x 2 1 Dx 2 E x A x 4 2x 2 1 Bx 4 Bx 2 C x 3 C x Dx 2 E x A B x 4 C x 3 2A B D x 2 C E x A
So A 1, 1 B 1 B 0; C 1; 2 0 D 1 D 1; and 1 E 1 E 2. Therefore, x4 x3 x2 x 1 1 1 x 2 2 2 2 . x x 1 x2 1 x x2 1
42.
Ax B Cx D 2x 2 x 8 2 2 2 . Thus, x 4 x2 4 x2 4 2x 2 x 8 Ax B x 2 4 C x D Ax 3 4Ax Bx 2 4B C x D Ax 3 Bx 2 4A C x 4B D
2x 2 x 8 1 2 and so A 0, B 2, 0 C 1 C 1, and 8 D 8 D 0. Therefore, 2 2 2 . x 4 x2 4 x2 4
43. We must first get a proper rational function. Using long division, we find that
x 5 2x 4 x 3 x 5 x2 x 3 2x 2 x 2
2 2x 2 x 5 2 2x x 5 x 2 A Bx C . Hence, x x 2 x 3 2x 2 x 2 x2 1 x 2 x 2 1 2x 2 x 5 A x 2 1 Bx C x 2 Ax 2 A Bx 2 C x 2Bx 2C
A B x 2 C 2B x A 2C
Equating coefficients, we get the system 2 Coefficients of x 2 2 2 A B A B A B 2B C 1 Coefficients of x 2B C 1 2B C 1 A 2C 5 Constant terms B 2C 3 5C 5
Therefore, 5C 5 C 1, 2B 1 1 B 1, and A 1 2 A 3, so x 1 3 x 5 2x 4 x 3 x 5 2 x2 . 3 2 x 2 x 1 x 2x x 2
SECTION 10.3 Partial Fractions
44.
687
x 5 3x 4 3x 3 4x 2 4x 12 x 5 3x 4 3x 3 4x 2 4x 12 . We use long division to get a proper rational x 4 4x 3 6x 2 8x 8 x 22 x 2 2
function:
x 4 4x 3 6x 2 8x 8
x
1
x 5 3x 4 3x 3 4x 2 4x 12 x 5 4x 4 6x 3 8x 2 8x x 4 3x 3 4x 2 4x 12 x 4 4x 3 6x 2 8x 8 x 3 2x 2 4x 4
B x 5 3x 4 3x 3 4x 2 4x 12 Cx D x 3 2x 2 4x 4 A , so 2 x 1 x 1 2 2 2 2 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 2x 2 4x 4 A x 2 x 2 2 B x 2 2 C x D x 22
Thus,
A x 3 2x 2 2x 4 B x 2 2 C x D x 2 4x 4
Ax 3 2Ax 2 2Ax 4A Bx 2 2B C x 3 4C x 2 4C x Dx 2 4Dx 4D
A C x 3 2A B 4C D x 2 2A 4C 4D x 4A 2B 4D A C 1 Coefficients of x 3 2A B 4C D 2 Coefficients of x 2 which leads to the system 2A 4C 4D 4 Coefficients of x 4A 2B 4D 4 Constant terms A C 1 A C 1 B 2C D 0 2 Eq. 1 Eq. 2 B 2C D 0 2C 4D 2 B 3D 2 Eq. 2 Eq. 3 Eq. 2 Eq. 3 2B 8C 4D 12 2 Eq. 3 Eq. 4 8C 2D 8 2 Eq. 3 Eq. 4 A C 1 B 2C D 0 Then D 0; 2C 2 C 1; B 2 0 B 2; and A 1 1 2C 4D 2 18D 0 4 Eq. 3 Eq. 4 A 0. Therefore,
45.
x x 5 3x 4 3x 3 4x 2 4x 12 2 2 . x 1 2 2 2 x 2 x 2 x 2 x 2
A B ax b . Hence, ax b A x 1 B x 1 A B x A B. x 1 x 1 x2 1 AB a ab . So Adding, we get 2A a b A 2 AB b Substituting, we get B a A
ab ab ab ab 2a . Therefore, A and B . 2 2 2 2 2
688
CHAPTER 10 Systems of Equations and Inequalities
ax 3 bx 2 Ax B Cx D 46. 2 2 2 . Hence, 2 x 1 x 1 x2 1 ax 3 bx 2 Ax B x 2 1 C x D Ax 3 Ax Bx 2 B C x D Ax 3 Bx 2 A C x B D
and so A a, B b, a C 0 C a, and b D 0 D b. Therefore, A a, B b, C a, and D b. 1 x is already a partial fraction decomposition. The denominator in the first term is a 47. (a) The expression 2 x 1 x 1 quadratic which cannot be factored and the degree of the numerator is less than 2. The denominator of the second term is linear and the numerator is a constant. x can be decomposed further, since the numerator and denominator both have linear factors. (b) The term x 12 B A x . Hence, x A x 1 B Ax A B. So A 1, B 1, and x 1 x 12 x 12 x 1 1 . x 1 x 12 x 12
2 1 is already a partial fraction decomposition, since each numerator is constant. x 1 x 12 x 2 (d) The expression 2 is already a partial fraction decomposition, since the denominator is the square of a quadratic x2 1 which cannot be factored,and the degree of the numerator is less than 2. (c) The expression
48. Combining the terms, we have 2 x2 1 1 x 2 2x 1 1 1 1 x 1 2 x 1 x 12 x 1 x 12 x 1 x 12 x 1 x 12 x 1
2x 2 2 x 1 x 2 2x 1 x 12 x 1 3x 2 x
3x 2 x
x 12 x 1
B C A , and so x 1 x 12 x 1 3x 2 x A x 1 x 1 B x 1 C x 12 A x 2 1 B x 1 C x 2 2x 1
Now to find the partial fraction decomposition, we have
Ax 2 A Bx B C x 2 2C x C 3 A which result in the system B 2C 1 A B C 0
A
C
3
4C
4
x 12 x 1
C A C x 2 B 2C x A B C Coefficients of x 2 C 3 A Coefficients of x B 2C 1 Constant terms B 2C 3 Eq. 1 Eq. 3
so 4C 4 C 1, B 2 1 1 B 1, and A 1 3 A 2.
B 2C 1
1 Eq. 2 Eq. 3
Therefore, we get back the same expression:
3x 2 x
x 12 x 1
1 1 2 . 2 x 1 x 1 x 1
SECTION 10.4 Systems of Nonlinear Equations
689
10.4 SYSTEMS OF NONLINEAR EQUATIONS 1. The solutions of the system are the points of intersection of the two graphs, namely 2 2 and 4 8. 2. For 2 2: 2y x 2 2 2 22 0 and y x 2 2 4, so 2 2 is a solution.
For 4 8: 2y x 2 2 8 42 0 and y x 8 4 4, so 4 8 is a solution. y x2 Substituting y x 2 into the second equation gives x 2 x 12 3. y x 12
0 x 2 x 12 x 4 x 3 x 4 or x 3. So since y x 2 , the solutions are 3 9 and 4 16. x 2 y 2 25 Substituting for y in the first equation gives x 2 2x2 25 5x 2 25 x 2 5 x 5 4. y 2x 5 2 5 and When x 5 then y 2 5, and when x 5 then y 2 5 2 5. Thus the solutions are 5 2 5 .
5.
x 2 y2 8 x y0
Solving the second equation for y gives y x, and substituting this into the first equation gives
x 2 x2 8 2x 2 8 x 2. So since y x, the solutions are 2 2 and 2 2. x2 y 9 Solving the first equation for y, we get y 9 x 2 . Substituting this into the second equation gives 6. x y 3 0 x 9 x 2 3 0 x 2 x 6 0 x 3 x 2 0 x 3 or x 2. If x 3, then y 9 32 0,
and if x 2, then y 9 22 5. Thus the solutions are 3 0 and 2 5. x y2 0 7. Solving the first equation for x gives x y 2 , and substituting this into the second equation gives 2x 5y 2 75 2 y 2 5y 2 75 3y 2 75 y 2 25 y 5. So since x y 2 , the solutions are 25 5 and 25 5. 8.
x2 y 1
Solving the first equation for y, we get y x 2 1. Substituting this into the second equation gives 2x 2 3y 17 2x 2 3 x 2 1 17 2x 2 3x 2 3 17 5x 2 20 x 2 4 x 2. If x 2, then y 22 1 3,
and if x 2, then y 22 1 3. Thus the solutions are 2 3 and 2 3. x 2 2y 1 9. Subtracting the first equation from the second equation gives 7y 28 y 4. Substituting y 4 into x 2 5y 29
the first equation of the original system gives x 2 2 4 1 x 2 9 x 3. The solutions are 3 4 and 3 4. 3x 2 4y 17 10. Multiplying the first equation by 2 and the second by 3 gives the system 2x 2 5y 2 6x 2 8y 34 Adding we get 7y 28 y 4. Substituting this value into the second equation gives 6x 2 15y 6 2x 2 5 4 2 2x 2 22 x 2 11 x 11. Thus the solutions are 11 4 and 11 4 .
690
11.
CHAPTER 10 Systems of Equations and Inequalities
3x 2 y 2 11
Multiplying the first equation by 4 gives the system
x 2 4y 2 8
12x 2 4y 2 44 x 2 4y 2 8
Adding the equations
gives 13x 2 52 x 2. Substituting into the first equation we get 3 4 y 2 11 y 1. Thus, the solutions are 2 1, 2 1, 2 1, and 2 1.
12.
2x 2 4y 13
Multiplying the second equation by 2 gives the system
x 2 y 2 72
2x 2 4y 13
2x 2 2y 2 7
Subtracting the
equations gives 4y 2y 2 6 y 2 2y 3 0 y 3 y 1 0 y 3, y 1. If y 3, then 5 2 2x 2 4 3 13 x 2 25 . If y 1, then 2x 2 4 1 13 x 2 92 x 3 2 2 . Hence, the x 2 2 5 2 3 2 solutions are 2 3 and 2 1 .
13.
14.
x y2 3 0
Adding the two equations gives 2x 2 x 1 0. Using the Quadratic Formula we have 2x 2 y 2 4 0 1 1 4 2 1 1 9 1 3 1 3 1 3 x . So x 1 or x 12 . Substituting x 1 2 2 4 4 4 4 into the first equation gives 1 y 2 3 0 y 2 2 y 2. Substituting x 12 into the first equation gives 1 y 2 3 0 y 2 7 y 7 . Thus the solutions are 1 2 and 1 7 . 2 2 2 2 2 x 2 y2 1
2x 2 y 2 x 3
Subtracting the first equation from the second equation gives x 2 x 2
x 2 x 2 0 x 2 x 1 0 x 2, x 1. Solving the first equation for y 2 we have y 2 x 2 1. When x 1, y 2 12 1 0 so y 0 and when x 2, y 2 22 1 3 so y 3. Thus, the solutions are 1 0, 2 3 , and 2 3 .
15.
x2 y
8
x 2y 6
By inspection of the graph, it appears that 2 4 is a solution, but is difficult to get accurate values
for the other point. Multiplying the first equation by 2 gives the system
2x 2 2y 16 x 2y 6
Adding the equations gives
2x 2 x 10 2x 2 x 10 0 2x 5 x 2 0. So x 52 or x 2. If x 52 , then 52 2y 6 2y 72 y 74 , and if x 2, then 2 2y 6 2y 8 y 4. Hence, the solutions are 52 74 and
2 4.
16.
x y 2 4
x y
2
By inspection of the graph, it appears that 0 2 and 5 3 are solutions to the system. We check
each point in both equations to verify that it is a solution. For 0 2: 0 22 4 and 0 2 2. For 5 3: 5 32 5 9 4 and 5 3 2. Thus, the solutions are 0 2 and 5 3.
SECTION 10.4 Systems of Nonlinear Equations
17.
x2
y0
x 3 2x y 0
691
By inspection of the graph, it appears that 2 4, 0 0, and 1 1 are solutions to the system.
We check each point in both equations to verify that it is a solution. For 2 4: 22 4 4 4 0 and 23 2 2 4 8 4 4 0.
For 0 0: 02 0 0 and 03 2 0 0 0.
For 1 1: 12 1 1 1 0 and 13 2 1 1 1 2 1 0. Thus, the solutions are 2 4, 0 0, and 1 1. x 2 y 2 4x 18. By inspection of the graph, it appears that 0 0 is a solution, but is difficult to get accurate values for x y2
the other points. Substituting for y 2 we have x 2 x 4x x 2 3x 0 x x 3 0. So x 0 or x 3. If x 0, then y 2 0 so y 0. And is x 3 then y 2 3 so y 3. Hence, the solutions are 0 0, 3 3 , and 3 3 . y x 2 4x Subtracting the second equation from the first equation gives x 2 4x 4x 16 19. y 4x 16
x 2 8x 16 0 x 42 0 x 4. Substituting this value for x into either of the original equations gives y 0. Therefore, the solution is 4 0. x y2 0 x y2 20. Substituting for y in the Solving the first equation for x and the second equation for y gives 2 yx 0 y x2 first equation gives x x 4 x x 3 1 0 x 0, x 1. Thus, the solutions are 0 0 and 1 1. x 2y 2 21. Now x 2y 2 x 2y 2. Substituting for x gives y 2 x 2 2x 4 y 2 x 2 2x 4
y 2 2y 22 2 2y 2 4 y 2 4y 2 8y 4 4y 4 4 y 2 4y 4 0 y 22 0 y 2. Since x 2y 2, we have x 2 2 2 2. Thus, the solution is 2 2. y 4 x2 22. Setting the two equations equal, we get 4 x 2 x 2 4 2x 2 8 x 2. Therefore, the solutions y x2 4 are 2 0 and 2 0. xy 4 23. Now x y 4 x 4 y. Substituting for x gives x y 12 4 y y 12 y 2 4y 12 0 x y 12 y 6 y 2 0 y 6, y 2. Since x 4 y, the solutions are 2 6 and 6 2. x y 24 24 Since x 0 is not a solution, from the first equation we get y 24. . Substituting into the 2 2 x 2x y 4 0 2 24 2x 4 4x 2 576 x 4 2x 2 288 0 x 2 18 x 2 16 0. second equation, we get 2x 2 4 x
Since x 2 18 cannot be 0 if x is real, we have x 2 16 0 x 4. When x 4, we have y 24 4 6 and when 24 6. Thus the solutions are 4 6 and 4 6. x 4, we have y 4 x 2 y 16 16 16 . Substituting for x 2 gives 4y 16 0 4y 2 16y 16 0 25. Now x 2 y 16 x 2 y y x 2 4y 16 0 y 2 4y 4 0 y 22 0 y 2. Therefore, x 2 has no solution.
16 8, which has no real solution, and so the system 2
692
26.
27.
CHAPTER 10 Systems of Equations and Inequalities
x
y 0 Solving the first equation for x, we get x y. Substituting for x gives y 2 4x 2 12 2 y 2 4 y 12 y 2 4y 12 0 y 6 y 2 0 y 6, y 2. Since x 2 is not a real solution, the only solution is 6 6 .
x 2 y2 9
x 2 y2 1
Adding the equations gives 2x 2 10 x 2 5 x 5. Now x 5 y 2 9 5 4
y 2, and so the solutions are
28.
x 2 2y 2 2
5 2 , 5 2 , 5 2 , and 5 2 .
Multiplying the first equation by 2 gives the system
2x 2 3y 15
equations gives 4y 2 3y 11 4y 2 3y 11 0 y
3
2x 2 4y 2 4
2x 2 3y 15
Subtracting the two
9 4 4 11 which is not a real number. 2 4
Therefore, there are no real solutions.
29.
2x 2 8y 3 19
4x 2 16y 3 34
Multiplying the first equation by 2 gives the system
4x 2 16y 3 38 4x 2 16y 3 34
Adding the two
equations gives 8x 2 72 x 3, and then substituting into the first equation we have 2 9 8y 3 19 y 3 18 y 12 . Therefore, the solutions are 3 12 and 3 12 . 30.
x 4 y 3 15
3x 4 5y 3 53
Multiplying the first equation by 3 gives the system
3x 4 3y 3 45
3x 4 5y 3 53
Subtracting the equations
gives 8y 3 8 y 3 1 y 1, and then x 4 1 15 x 2. Therefore, the solutions are 2 1 and 2 1.
2 3 1 x y 31. 4 7 1 x y
1 1 If we let u and ,the system is equivalent to x y
equation by 4 gives the system
4u 6 2
4u 7 1
2u 3 1
4u 7 1
Multiplying the first
Adding the equations gives 3, and then substituting into the first
equation gives 2u 9 1 u 5. Thus, the solution is 15 13 .
4 6 7 2 4 2 x y 32. 1 2 0 2 x y4
4u 6 7 1 1 2 , and multiplying the If we let u 2 and 4 , the system is equivalent to u 2 0 x y
4u 6 7 2 second equation by 3, gives 3u 6 0
Adding the equations gives 7u 72 u 12 , and 14 . Therefore,
1 1 2 2 , 2 2 , 2 2 , x 2 2 x 2, and y 4 4 y 2. Thus, the solutions are u and 2 2 .
SECTION 10.4 Systems of Nonlinear Equations
33.
y x 2 8x
34.
y 2x 16
The solutions are 8 0 and 2 20.
y x 2 4x
2x y 2
y x 2 4x
y 2x 2
The solutions are approximately 035 130 and 565 930.
20 10 -10
10 -20
35.
x 2 y 2 25
5
y 25 x 2
y 13 x 23 The solutions are 451 217 and 491 097. x 3y 2
36.
x 2 y 2 17
x 2 2x y 2 13
y 17 x 2 y 13 2x x 2
The solutions are approximately 2 361.
5
5
-5
5
-5
-5
5 -5
2 y2 x y 18 2x 2 1 37. 9 18 y x 2 6x 2 y x 2 6x 2
38.
x 2 y2 3
y x 2 2x 8
y x2 3
y x 2 2x 8
The solutions are approximately 222 140,
The solutions are 123 387 and 035 421.
188 072, 345 299, and 465 431.
5
5
-5
5 -5
5
-5 -5
39.
4 32 x 4 y 2 x 2 2x y 0 y x 2 2x x 4 16y 4 32
The solutions are 230 070 and 048 119.
40.
y e x ex y 5 x2
119 359 and 119 359. 5
2
-2
The solution are approximately
2 -2
-2
2
693
694
CHAPTER 10 Systems of Equations and Inequalities
log x log y 3 2 41. 2 log x log y 0
Adding the two equations gives 3 log x 32 log x 12 x 10. Substituting into the
second equation we get 2 log 1012 log y 0 log 10 log y 0 log y 1 y 10. Thus, the solution is 10 10 .
42.
2x 2 y 10
4x 4 y 68
2x 2 y 10
22x 22y 68
If we let u 2x and 2 y , the system becomes
u 10
u 2 2 68
Solving the first equation for u, and substituting this into the second equation gives u 10 u 10 , so 10 2 2 68 100 20 2 2 68 2 10 16 0 8 2 0 2 or 8. If 2, then u 8, and so y 1 and x 3. If 8, then u 2, and so y 3 and x 1. Thus, the solutions are 1 3 and 3 1. xy 3 43. Solving the first equation for x gives x 3 y and using the hint, x 3 y 3 387 3 x y 3 387 x y x 2 x y y 2 387. Next, substituting for x, we get 3 3 y2 y 3 y y 2 387 9 6y y 2 3y y 2 y 2 129 3y 2 9y 9 129 y 8 y 5 0 y 8 or y 5. If y 8, then
x 3 8 5, and if y 5, then x 3 5 8. Thus the solutions are 5 8 and 8 5. x2 xy 1 44. Adding the equations gives x 2 x yx y y 2 4 x 2 2x y y 2 4 x y2 4 x y 2. x y y2 3 If x y 2, then from the first equation we get x x y 1 x 2 1 x 12 , and so y 2 12 32 . If x y 2, then from the first equation we get x x y 1 x 2 1 x 12 , and so y 2 12 32 . Thus the solutions are 12 32 and 12 32 . 45. Let and l be the lengths of the sides, in cm. Then we have the system
l 180
2l 2 54
We solve the second equation
for giving, 27 l, and substitute into the first equation to get l 27 l 180 l 2 27l 180 0
l 15 l 12 0 l 15 or l 12. If l 15, then 27 15 12, and if l 12, then 27 12 15. Therefore, the dimensions of the rectangle are 12 cm by 15 cm.
46. Let b be the length of the base of the triangle, in feet, and h be the height of the triangle, 1 bh 84 168 2 . By substitution, The first equation gives b in feet. Then b2 h 2 252 625 h
168 2 h 2 625 h 4 625h 2 1682 0 h 2 49 h 2 576 0 h 7 or h 24. Thus, the lengths of h
the other two sides are 7 ft and 24 ft.
47. Let l and be the length and width, respectively, of the rectangle. Then, the system of equations is 2l 2 70 Solving the first equation for l, we have l 35 , and substituting into the second gives l 2 2 25 l 2 2 25 l 2 2 625 35 2 2 625 1225 70 2 2 625 22 70 600 0 15 20 0 15 or 20. So the dimensions of the rectangle are 15 and 20.
SECTION 10.4 Systems of Nonlinear Equations
695
48. Let be the width and l be the length of the rectangle, in inches. From the figure, the diagonals of the rectangle are l 160 1602 160 simply diameters of the circle. Then, . By substitution, 2 l 2 400 2 2 2 l l l 20 400 l 4 400l 2 1602 0 l 2 80 l 2 320 0 l 80 4 5 or l 320 8 5. Therefore, the dimensions of the rectangle are 4 5 in. and 8 5 in..
y 1x 2 49. At the points where the rocket path and the hillside meet, we have Substituting for y in the second y x 2 401x 801 0 x 0, x 801 . When x 0, the rocket has x 0 x x equation gives 12 x x 2 401x x 2 801 2 2 2 801 1 801 801 801 not left the pad. When x 2 , then y 2 2 4 . So the rocket lands at the point 801 2 4 . The distance from the base of the hill is
801 2 2
801 2 44777 meters. 4
50. Let x be the circumference and y be length of the stove pipe. Using the circumference we can determine x 2 1 2 x . Thus the volume is x y. So the system is given by y the radius, 2r x r 2 2 4 x y 1200 1 1 1 2 x y x x y x 1200 600 Substituting for x y in the second equation gives 1 4 4 4 x 2 y 600 4 1200 600 600 1200 . Thus the dimensions of the sheet metal are 2 63 in and 1910 in. x 2. So y x 2
696
CHAPTER 10 Systems of Equations and Inequalities
51. The point P is at an intersection of the circle of radius 26 centered at A 22 32
y
40 and the circle of radius 20 centered at B 28 20. We have the system A B x 222 y 322 262 20 2 2 2 x 28 y 20 20 0 _20 20 40 x x 2 44x 484 y 2 64y 1024 676 2 2 x 56x 784 y 40y 400 400 _20 x 2 44x y 2 64y 832 Subtracting the two equations, we get 12x 24y 48 x 2y 4, x 2 56x y 2 40y 784
which is the equation of a line. Solving for x, we have x 2y 4. Substituting into the first equation gives
2y 42 44 2y 4 y 2 64y 832 4y 2 16y1688y176 y 2 64y 832 5y 2 168y192 832 2 451024 5y 2 168y 1024 0. Using the Quadratic Formula, we have y 168 16825 16810 7744 16888 10
y 8 or y 2560. Since the y-coordinate of the point P must be less than that of point A, we have y 8. Then x 2 8 4 12. So the coordinates of P are 12 8.
To solve graphically, we must solve each equation for y. This gives x 222 y 322 262 y 322 262 x 222 y 32 676 x 222 y 32 676 x 222 . We use the function y 32 676 x 222 because the intersection we at interested in is below the point A. Likewise, solving the second equation for y, we would get the function y 20 400 x 282 . In a three-dimensional situation, you would need a minimum of three satellites, since a point on the earth can be uniquely specified as the intersection of three spheres centered at the satellites.
52. The graphs of y x 2 and y x k for various values of k are shown. If we solve y x2 the system we get x 2 x k 0. Using the Quadratic Formula, y x k 1 1 4k . So there is no solution if 1 4k is undefined, that we have x 2
y k=2 1
k=_ 4 1
1
x
k=_3
is, if 1 4k 0 k 14 . There is exactly one solution if 1 4k 0 k 14 , and there are two solutions if 1 4k 0 k 14 .
10.5 SYSTEMS OF INEQUALITIES 1. If the point 2 3 is a solution of an inequality in x and y, then the inequality is satisfied when we replace x by 2 and y by 3. Because 4 2 2 3 8 6 2 1, the point 2 3 is a solution of the inequality 4x 2y 1.
SECTION 10.5 Systems of Inequalities
697
y
2. To graph an inequality we first graph the corresponding equation. So to graph y x 1, we first graph the equation y x 1. To decide which
side is the graph of the inequality we use test points. Test Point
Inequality y x 1 ?
0 0
001X
Part of graph
201
Not part of graph
?
0 2
1
x
1
y=x+1
Conclusion
3. If the point 2 3 is a solution of a system of inequalities in x and y, then each inequality is satisfied when we replace x by 2 and y by 3. Because 2 2 4 3 16 17 and 6 2 5 3 27 29, the point 2 3 is a solution of the given system. 4. (a)
y
(b) x-y=0
xy0 xy2
(c)
y
xy0 xy2
xy0 xy2
y
y
x-y=0
1
(d)
x-y=0
x-y=0
x+y=2 x
1
1
1
x+y=2 1
5.
x
1
x+y=2 x
1
x+y=2 1
6. Test Point Inequality x 5y 3 ?
1 2 1 5 2 3 X ?
Conclusion Solution
1 2
1 5 2 3 X
Solution
1 2
1 5 2 3
Not a solution
8 5 1 3
Not a solution
8 1
? ?
7. Test Point System
0 0
1 2
1 1
3 1
Test Point Inequality 3x 2y 2 2 1 1 3 1 3 0 1
?
3 2 2 1 2 X ?
3 1 2 3 2 ?
2x y 3
? 3 0 2 0 5X ? 2 0 0 3 ? 3 1 2 2 5X ? 2 1 2 3X ? 3 1 2 1 5 X ? 2 1 1 3X ? 3 3 2 1 5 ? 2 3 1 3X
Conclusion
0 0
Solution
1 3
Solution
3 0
Not a solution
1 2
Not a solution Solution
3 0 2 1 2 X
Solution
?
Test Point System
Not a solution
Solution
3 1 2 3 2 X
8. 3x 2y 5
Conclusion
x 2y 4
4x 3y 11
Conclusion
?
0 2 0 4
? 4 0 3 0 11 ? 1 2 3 4 X ? 4 1 3 3 11 X ? 3 2 0 4 ? 4 3 3 0 11 ? 1 2 2 4 X ? 4 1 3 2 11
Not a solution
Solution
Not a solution
Not a solution
x
698
CHAPTER 10 Systems of Equations and Inequalities
9. y 2x. The test point
10. y 3x. The test point
1 0 satisfies the
11. y 2. The test point
1 0 satisfies the
inequality.
inequality.
2 0 satisfies the
inequality.
y
y
12. x 1. The test point
0 3 satisfies the
inequality. y
y y=3x
1
1
x
1
x
1
1
x=_1
y=2
1
x
1
x
1
y=_2x
13. x 2. The test point
14. y 1. The test point
0 0 satisfies the
0 2 satisfies the
inequality.
inequality.
x=2
1 1
1
inequality.
1
1
point 0 6 satisfies the
inequality. y
10
x
_x@+y=5 1
21. x 2 y 2 9. The test point 0 4 satisfies the inequality.
1
22. x 2 y 22 4. The test point 0 1 satisfies the inequality.
x@+(y-2)@=4
1
y=x@+1
1
x
y
x@+y@=9 1
point 0 2 satisfies the
y
3x-y-9=0
y
20. y x 2 1. The test
inequality.
2 2
x
x y=1-x
1
19. x 2 y 5. The test
y
y
1
x
1
x
point 0 0 satisfies the
inequality.
y y=x-3
y=1
18. 3x y 9 0. The test
point 0 0 satisfies the
inequality.
y
1
17. 2x y 4. The test
point 0 0 satisfies the
inequality.
1
x
16. y 1 x. The test
point 0 0 satisfies the
y
y
2x-y=_4
15. y x 3. The test
x 1 1
x
x
SECTION 10.5 Systems of Inequalities
23. 3x 2y 18
699
24. 4x 3y 9 8
4 -2 0
2
4
6
8
4
10 12
-4
-4
-8
-2 0
2
4
6
8
-4
-12
25. 5x 2y 8
26. 5x 3y 15 8
4
4 0 -2
0 -4
2
4
6
5
-4
-8
-8
27. The boundary is a solid curve, so we have the inequality y 12 x 1. We take the test point 0 2 and verify that it satisfies the inequality: 2 12 0 1. 28. The boundary is a solid curve, so we have the inequality y x 2 2. We take the test point 0 0 and verify that it satisfies the inequality: 0 02 2.
29. The boundary is a broken curve, so we have the inequality x 2 y 2 4. We take the test point 0 4 and verify that it satisfies the inequality: 02 42 4.
30. The boundary is a solid curve, so we have the inequality y x 3 4x. We take the test point 1 1 and verify that it satisfies the inequality: 1 13 4 1.
31.
xy4 yx
The vertices occur where
xy4 yx
y
Substituting, we have
y=x 1
2x 4 x 2. Since y x, the vertex is 2 2, and the solution set is not
32.
2x 3y 12
2x 3y 12
3x y 21
The vertices occur where
2x 3y 12 3x y 21
x+y=4
and adding the two equations gives 11x 75 x 75 11 .
9x 3y 63 132150 18 y 6 . Therefore, the Then 2 75 11 3y 12 3y 11 11 11 75 6 vertex is 11 11 , and the solution set is not bounded. The test point 0 5 satisfies each inequality.
x
1
bounded. The test point 0 1 satisfies each inequality.
y 2x+3y=12 1 1 3x-y=21
x
700
CHAPTER 10 Systems of Equations and Inequalities
y 1x 2 4 33. y 2x 5
y 1x 2 4 The vertex occurs where y 2x 5
y
Substituting for y y=41 x+2
gives 14 x 2 2x 5 74 x 7 x 4, so y 3. Hence, the vertex is 4 3,
1
34.
xy0
4 y 2x
The vertices occur where
y x
4 y 2x
y
set is not bounded. The test point 3 0 satisfies each inequality.
One vertex occurs where
y=2x-5
Substituting for y
gives 4 x 2x x 4, so y 4. Hence, the vertex is 4 4, and the solution
y 2x 8 35. y 12 x 5 x 0, y 0
x
1
and the solution is not bounded. The test point 0 1 satisfies each inequality.
y 2x 8
y 12 x 5
4+y=2x 2 x-y=0
x
2
y
Substituting for
y=_2x+8 1
y=_ 2 x+5
y gives 2x 8 12 x 5 32 x 3 x 2, so y 2 2 8 4. y 2x 8 Hence, this vertex is 2 4. Another vertex occurs where y0
x
1 1
2x 8 0 x 4; this vertex is 4 0. Another occurs where y 12 x 5 y 5; this gives the vertex 0 5. The origin is another x 0 vertex, and the solution set is bounded. The test point 1 1 satisfies each inequality. 4x 3y 18 36. 2x y 8 x 0, y 0
One vertex occurs where
4x 3y 18 2x y 8
y
Subtracting twice 2x+y=8
the second equation from the first gives y 2, so 2x 2 8 x 3 and the
4x+3y=18
1
vertex is 3 2. Other vertices occur at the x-intercept of 2x y 8 and the
x
1
y-intercept of 4x 3y 18; these are 4 0 and 0 6, respectively. The origin is
another vertex, and the solution set is bounded. The test point 1 1 satisfies each inequality. x y 37. 3x 5y 3x 2y
0 0 15
From the graph, the points 3 0, 0 3 and 0 0 are vertices,
y 3x+2y=9
9
and the fourth vertex occurs where the lines 3x 5y 15 and 3x 2y 9
intersect. Subtracting these two equations gives 3y 6 y 2, and so x 53 . Thus, the fourth vertex is 53 2 , and the solution set is bounded. The test point 1 1 satisfies each inequality.
3x+5y=15
1 1
x
SECTION 10.5 Systems of Inequalities
38.
y
x 2 y 12
2x 4y 8
From the graph, the vertices occur at 2 1 and 28 12.
39.
y 9 x2
x 0, y 0
y=12
6
2x-4y=8 x
6
The solution set is not bounded. The test point 6 0 satisfies each inequality.
x=2
y
From the graph, the vertices occur at 0 0, 3 0, and 0 9.
y=9-x@
The solution set is bounded. The test point 1 1 satisfies each inequality. 1
y=0
x
1 x=0
2 y x 40. y 4 x 0
y x=0
From the graph, the vertices occur at 0 0, 0 2, and 2 4. The
y=4
solution set is bounded. The test point 1 3 satisfies each inequality. y=x@
1 1
41.
y 9 x2 y x 3
The vertices occur where
y 9 x2
y x 3
Substituting for y
y=9-x@
y y=x+3
gives 9 x 2 x 3 x 2 x 6 0 x 2 x 3 0 x 3, x 2. Therefore, the vertices are 3 0 and 2 5, and the solution set is
1
42.
y x2
xy6
The vertices occur where
y x2
xy6
x
1
bounded. The test point 0 4 satisfies each inequality.
y
Substituting for y y=x@
gives x 2 x 6 x 2 x 6 0 x 3 x 2 0 x 3, x 2. Since y x 2 , the vertices are 3 9 and 2 4, and the solution set is not bounded. The test point 0 7 satisfies each inequality.
43.
x 2 y2 4 xy0
The vertices occur where
x 2 y2 4 xy0
x
x+y=6
2 1 y
Since x y 0
x y, substituting for x gives y 2 y 2 4 y 2 2 y 2, and x 2. Therefore, the vertices are 2 2 and 2 2 , and the
solution set is bounded. The test point 1 0 satisfies each inequality.
x
x-y=0
1 1 x@+y@=4
x
701
702
44.
CHAPTER 10 Systems of Equations and Inequalities
y
x 0
y 0 x y 10 2 x y2 9
From the graph, the vertices are 0 3, 0 10, 3 0, and
1
10 0. The solution set is bounded. The test point 4 1 satisfies each inequality.
45.
x2 y 0
2x 2 y 12 2x 2 2y 0
2x 2 y 12
The vertices occur where
x2 y 0
2x 2 y 12
46.
x2 y 8
x@+y@=9
y
x@-y=0
Subtracting the equations gives 3y 12 y 4, and
The vertices occur where
2x 2 y 4 x2 y 8
y
Adding the
2 4 and 2 4. The solution set is not bounded. The test point 0 6 satisfies each inequality.
47.
x 2 y2 9
2x y 2 1
The vertices occur where
x 2 y2 9
2x y 2 1
Subtracting the
equations gives x 2 2x 8 x 2 2x 8 x 2 x 4 0. Therefore,
2 x@-y=8
point 0 0 satisfies each inequality.
48.
x 2 y2 4
x 2 2y 1
The vertices occur where
2x@+y=4
(_2, _4)
1
y
2x+y@=1
x (2, _4)
x@+y@=9
(_2, Ï5)
the vertices are 2 5 and 2 5 . The solution set is bounded. The test
x
1
equations gives 2x 2 x 2 12 x 2 4 x 2. Therefore, the vertices are
2x@+y=12
1
The test point 0 1 satisfies each inequality.
2x 2 y 4
x
y=0
1
x 2. Thus, the vertices are 2 4 and 2 4, and the solution set is bounded.
x+y=10
x=0
1 x
1
(_2, _Ï5)
x 2 y2 4
x 2 2y 1
equations gives y 2 2y 3 y 2 2y 3 y 3 y 1 0. However,
y 3 is extraneous, and the vertices are 3 1 and 3 1 . The solution
set is bounded. The test point 0 2 satisfies each inequality.
y
Subtracting the
x@-2y=1
(_Ï3, 1)
(Ï3, 1)
1 1
x x@+y@=4
SECTION 10.5 Systems of Inequalities
x 2y 14 49. 3x y 0 x y 2
We find the vertices of the region by solving pairs of the
corresponding equations:
3x y 0 xy2
x 2y 14 x y 2
3x y
0
2x y 2
x 2y 14
3y 12
x+2y=14
3x-y=0 1
y 4 and x 6.
x 1 and y 3. Therefore, the vertices
are 6 4 and 1 3, and the solution set is not bounded. The test point 0 4 satisfies each inequality. y x 6 50. 3x 2y 12 To find where the line y x 6 intersects the lines x 2y 2
point 4 2 satisfies each inequality. x 0, y 0 51. The points of intersection are 0 7, 0 0, 7 0, 5 2, x 5, x y 7 and 5 0. However, the point 7 0 is not in the solution set. Therefore, the
x-y=2
y 3x+2y=12 y=x+6
52.
x 0, y 0
y 4, 2x y 8
y x=5 x=0 x+y=7
1
y
2x+y=8 y=4
x=0 1 1
point 1 1 satisfies each inequality.
53.
x 2y 12 x 1 0
x
1
y=0
vertices are 0 4, 2 4, 4 0, and 0 0. The solution set is bounded. The test
y x 1
y
We find the vertices of the region by solving pairs of the
corresponding equations. Using x 1 and substituting for x in the line y x 1 gives the point 1 0. Substituting for x in the line x 2y 12 gives y x 1 13 the point 1 2 . x y 1 and y 1 2y 12 x 2y 12 10 13 3y 13 y 13 3 and x 3 . So the vertices are 1 0, 1 2 , and 10 13 , and none of these vertices is in the solution set. The solution set is 3 3 bounded. The test point 0 2 satisfies each inequality.
x
2
The points of intersection are 0 8, 0 4, 4 0, 2 4,
and 3 2. However, the point 0 8 is not in the solution set. Therefore, the
2
x-2y=2
vertices are 0 7, 0 0, 5 0, and 5 2, and the solution set is bounded. The test point 1 1 satisfies each inequality.
x
1
3x 2y 12 and x 2y 2, we substitute for y: 3x 2 x 6 12 x 0 and y 6; x 2 x 6 2 x 14 and y 8. Next, adding the equations 3x 2y 12 and x 2y 2 gives 4x 14 x 72 , so these lines intersect at the point 72 34 . Since the vertex 14 8 is not part of the solution set, the vertices are 0 6 and 72 34 , and the solution set is not bounded. The test
y
y=0
x
x+2y=12
x+1=0 y=x+1
1 1
x
703
704
CHAPTER 10 Systems of Equations and Inequalities
x y 12 54. y 12 x 6 3x y 6
y
Graphing these inequalities, we see that there are no points 3x+y=6
that satisfy all three, and hence the solution set is empty.
x+y=12 2
y=21 x-6
55.
x 2 y2 8
x 2, y 0
The intersection points are 2 2, 2 0, and 2 2 0 .
x@+y@=8
x
2 1 2
y
x=2
1
However, since 2 2 is not part of the solution set, the vertices are 2 2, 2 0, and 2 2 0 . The solution set is bounded. The test point 21 1 satisfies each
y=0
x
1
inequality.
2 x y 0 56. xy6 xy6
x+y=6
y
x@-y=0
Adding the equations x y 6 and x y 6 yields 2x 12 2
x 6. So these curves intersect at the point 6 0. To find where x 2 y 0
1
x-y=6
and x y 6 intersect, we solve the first for y, giving y x 2 , and then substitute
x
into the second equation to get x x 2 6 x 2 x 6 0 x 3 x 2 0 x 3 or x 2. When x 3, we have
y 9, and when x 2, we have y 4, so the points of intersection are 3 9 and 2 4. Substituting y x 2 into the
equation x y 6 gives x x 2 6, which has no solution. Thus the vertices (which are not in the solution set) are 6 0, 3 9, and 2 4. The solution set is not bounded. The test point 1 0 satisfies each inequality.
57.
x 2 y2 9
x y 0, x 0
x+y=0 y
Substituting x 0 into the equations x 2 y 2 9 and
x y 0 gives the vertices 0 3 and 0 0. To find the points of intersection
for the equations x 2 y 2 9 and x y 0, we solve for x y and substitute
1
x=0 1
x
into the first equation. This gives y2 y 2 9 y 3 2 2 . The points x@+y@=9 0 3and 3 2 2 3 2 2 lie away from the solution set, so the vertices are 0 0, 0 3, and 3 2 2 3 2 2 . Note that the vertices are not solutions in this case. The solution set is bounded. The test point
0 1 satisfies each inequality.
SECTION 10.5 Systems of Inequalities
58.
y x3
y 2x 4 x y 0
y
The curves y x 3 and x y 0 intersect when
y=2x+4
x 3 x 0 x 0 y 0. The lines x y 0 and y 2x 4 intersect
y=x#
1
when x 2x 4 3x 4 x 43 y 43 . To find where y x 3 and y 2x 4 intersect, we substitute for y and get x 3 2x 4 x 3 2x 4 0 x 2 x 2 2x 2 0 x 2 (the other factor has no
x
1 x+y=0
real solution). When x 2 we have y 8Thus, the vertices of the region are 0 0, 43 43 , and 2 8. The solution set is bounded. The test point 0 2 satisfies each inequality.
x 2y 14 59. 3x y 0 x y2
14 x 6x x 2 and y 6. The lines 3x y 0 and x y 2 intersect where 3x 2 x x 1 and y 3. The lines x 2y 14 and x y 2 intersect where 14 2y 2 y y 4 and x 6. Thus, the vertices of the region are 1 3, 2 6, and 6 4. The solution set is bounded. The test point 1 0 satisfies each inequality.
x 2y 14 60. 3x y 0 x y2
y
The lines x 2y 14 and 3x y 0 intersect when
(2, 6) 2 (_1, _3)
(2, 6) 2 (_1, _3)
The lines x y 12 and y 12 x 6 intersect where
(6, 4)
2
x
y (2, 12)
12 x 12 x 6 x 12 and y 0. The lines y 12 x 6 and y 2x 6 intersect where 12 x 6 2x 6 x 8 and y 10. The lines x y 12
and y 2x 6 intersect where 12 x 2x 6 x 2 and y 12. The solution set is not bounded. The test point 0 8 satisfies each inequality.
x
y
The vertices are 1 3, 2 6, and 6 4. In this case the
solution set is not bounded. The test point 3 0 satisfies each inequality.
x y 12 61. y 12 x 6 y 2x 6
(6, 4)
2
2 2 (_8, _10)
(12, 0)
x
705
706
62.
CHAPTER 10 Systems of Equations and Inequalities
y x 1
x 2y 12
x 1 0
The lines y x 1 and x 2y 12 intersect where
(_1, 132 )
y
( 103 , 133 )
10 y 1 12 2y y 13 3 and x 3 . The lines y x 1 and x 1 0 intersect at 1 0, and the lines x 2y 12 and x 1 0 intersect where
12 2y 1 y 13 2 and x 1. Thus, the vertices of the region are 10 13 1 0, 1 13 2 , and 3 3 . The solution set is bounded. The test point
1
(_1, 0)
3x y 5 30x 10y 50 x 2y 5 10x 20y 50 63. x 6y 9 10x 60y 90 x 0, y 0 x 0 y 0
y
y x 3 65. y 2x 6 y 8
Using a graphing calculator, we find the region shown. The
vertices are 3 0, 1 8, and 11 8.
x y 12 66. 2x y 24 x y 6
Using the graphing calculator we find the region shown. The
(3, 1)
1
(9, 0)
1 y
7 The vertices are 1 5, 17 4 4 , and 8 1. The solution set is bounded. The test point 4 3 satisfies each inequality.
(0, 5)
(1, 2)
The vertices are 0 5, 1 2, 3 1, and 9 0. The solution set is not bounded. The test point 1 3 satisfies each inequality.
x y6 4x 7y 39 64. x 5y 13 x 0 y 0
x
1
0 2 satisfies each inequality.
1
x
(1, 5)
( 174 , 74 )
(8, 1) x
1
10 8 6 4 2 0 -2 -4
10
10
vertices are 3 9, 6 12, and 12 0.
67.
y 6x x 2
xy4
0
Using a graphing calculator, we find the region shown. The
vertices are 06 34 and 64 24.
10
10 8 6 4 2 -4 -2 -2 0 2 -4
4
6
8 10
SECTION 10.5 Systems of Inequalities
68.
y x3
2x y 0
707
12 10 8 6 4 2
Using a graphing calculator, we find the region shown. The
y 2x 6
vertices are 0 0, 22 103 and 15 3. -4
-2
0 -2
2
4
y
69. (a) Let x and y be the numbers of acres of potatoes and corn, respectively. x y 500 90x 50y 40,000 A system describing the possibilities is 30x 80y 30,000 x 0 y 0
(0, 375) (200, 300) (375, 125)
100
x ( 4000 9 , 0)
100
(b) Because the point 300 180 lies in the feasible region, the farmer can plant 300 acres of potatoes and 180 acres of corn. (c) Because the point 150 325 lies outside the feasible region, the farmer cannot plant this combination of crops. y
70. (a) Let x and y be the numbers of acres of cauliflower and cabbage, respectively. x y 300 70x 35y 17,500 A system describing the possibilities is 25x 55y 12,000 x 0 y 0
(0, 2400 11 ) (150, 150)
(200, 100)
50 50
(250, 0)
(b) Because the point 155 115 lies in the feasible region, the farmer can plant 155 acres of cauliflower and 115 acres of cabbage. (c) Because the point 115 175 lies outside the feasible region, the farmer cannot plant this combination of crops. 71. Let x be the number of fiction books published in a year and y the number of nonfiction books. Then the following system of inequalities holds: x 0, y 0 From the graph, we see that the vertices are 50 50, 80 20 x y 100 y 20, x y and 20 20.
72. Let x be the number of chairs made and y the number of tables. Then the following 2x 3y 12 system of inequalities holds: 2x y 8 The intersection points are 0 4, x 0, y 0 0 8, 6 0, 4 0, 0 0, and 3 2. Since the points 0 8 and 6 0 are not in the solution set, the vertices are 0 4, 0 0, 4 0, and 3 2.
y
120 100 80
x+y=100 x=y
60 40
y=20
20
x
20 40 60 80 100
y 2x+y=8
2x+3y=12
1 1
x
x
708
CHAPTER 10 Systems of Equations and Inequalities y 200
73. Let x be the number of Standard Blend packages and y be the number of Deluxe Blend packages. Since there are 16 ounces per pound, we get the following system
3 x+38 y=90 4
of inequalities:
1 x+58 y=80 4
100
x 0
y 0 1 x 5 y 80 4 8 3 3 y 90 x 4 8
200 x
100
From the graph, we see that the vertices are 0 0, 120 0, 70 100 and 0 128. y
74. Let x be the amount of fish and y the amount of beef (in ounces) in each can. Then the following system of inequalities holds: 12x 6y 60 3x 9y 45 x 0, y 0
12x+6y=60 3x+9y=45
2
x
2
From the graph, the vertices are 15 0, 3 4, and 0 10.
75. x 2y 4, x y 1, x 3y 9, x 3. Method 1: We shade the solution to each inequality with lines perpendicular to the boundary. As you can see, as the number of inequalities in the system increases, it gets harder to locate the region where all of the shaded parts overlap. Method 2: Here, if a region is shaded then it fails to satisfy at least one inequality. As a result, the region that is left unshaded satisfies each inequality, and is the solution to the system of inequalities. In this case, this method makes it easier to identify the solution set. To finish, we find the vertices of the solution set. The line x 3 intersects the line x 2y 4 at 3 12 and the line
x 3y 9 at 3 2. To find where the lines x y 1 and x 2y 4 intersect, we add the two equations, which gives
3y 5 y 53 , and x 23 . To find where the lines x y 1 and x 3y 9 intersect, we add the two equations, 5 3 1 2 5 3 5 which gives 4y 10 y 10 4 2 , and x 2 . The vertices are 3 2 , 3 2, 3 3 , and 2 2 , and the solution set is bounded.
_x+y=1
y
1
x+3y=9 x
1 x=3
Method 1
Method 2
Solution Set
x+2y=4
CHAPTER 10
Review
CHAPTER 10 REVIEW 3x y 5 1. 2x y 5
y
Adding, we get 5x 10 x 2. So 2 2 y 5 y 1. 1
Thus, the solution is 2 1.
y 2x 6 2. y x 3
x
1
x
y
Subtracting the second equation from the first, we get 1
0 3x 3 x 1. So y 1 3 4. Thus, the solution is 1 4.
2x 7y 28 2x 7y 28 3. 2x 7y 28 y 27 x 4
y
Since these equations represent the
2
same line, any point on this line will satisfy the system. Thus the solution are x 27 x 4 , where x is any real number.
6x 8y 15 6x 8y 15 4. 3 x 2y 4 6x 8y 16 2
2
x
y
Adding gives 0 1 which is
false. Hence, there is no solution. The lines are parallel.
2x y 1 5. x 3y 10 3x 4y 15
1
1 1
x
1
x
y
Solving the first equation for y, we get y 2x 1.
Substituting into the second equation gives x 3 2x 1 10 5x 7 7 12 x 75 . So y 75 1 12 5 Checking the point 5 5 in the third ? 21 48 equation we have 3 75 4 12 5 15 but 5 5 15. Thus, there is no
solution, and the lines do not intersect at one point.
1
709
710
CHAPTER 10 Systems of Equations and Inequalities
2x 5y 9 6. x 3y 1 7x 2y 14
y
Adding the first equation to twice the second equation gives
11y 11 y 1. Substituting back into the second equation, we get
1
x 3 1 1 x 4. Checking point 4 1 in the third equation gives
7 4 2 1 26 14. Thus there is no solution, and the lines do not intersect at
x
1
one point.
y x 2 2x 7. y 6x
Substituting for y gives 6 x x 2 2x x 2 x 6 0. Factoring, we have x 2 x 3 0.
Thus x 2 or 3. If x 2, then y 8, and if x 3, then y 3. Thus the solutions are 3 3 and 2 8.
x 2 y2 8 8. y x 2
Substituting for y in the first equation gives x 2 x 22 8
2x 2 4x 4 0 2 x 2 2x 2 0. Using the Quadratic Formula, we have 2 2 3 2 4 8 1 3. If x 1 3. then y 1 3 2 1 3, and if x 1 3, x 2 2 then y 1 3 2 1 3. Thus, the solutions are 1 3 1 3 and 1 3 1 3 .
4 3x 6 y 9. 8 x 4 y
16 8 Adding twice the first equation to the second gives 7x 16 x 16 7 . So 7 y 4
16 14 16y 56 28y 12y 56 y 14 3 . Thus, the solution is 7 3 .
10.
x 2 y 2 10
x 2 2y 2 7y 0
Subtracting the first equation from the second gives y 2 7y 10
y 2 7y 10 0 y 2 y 5 0 y 2, y 5. If y 2, then x 2 4 10 x 2 6 x 6, and if y 5, then x 2 25 10 x 2 15, which leads to no real solution. Thus the solutions are 6 2 and 6 2 .
32x 032x 043y 0 y 43 11. 7x 12y 341 y 7x 341 12 The solution is approximately 2141 1593. 20
22
24
12x 32y 660 y 6 x 1102 3 12. 7137x 3931y 20,000 y 7137 x 20,000 3931 3931 The solution is approximately 6104 10573. 55
-10
-100
-15
-105
-20
-110
60
65
CHAPTER 10
x y 2 10 y x 10 13. 1 y 12 y 22 x 12 x 22
y 5x x 14. y x5 5
The solutions are 1194 139 and 1207 144.
Review
711
The solutions are approximately
145 135, 1 6, and 151 1293.
5 10 0 10
15 -2
-5
15.
x 2y z
8
x
4x z 9 2x y z 8
8 x 2y z 8y 3z 23 8y 3z 23 3y z 8 z 5 2y z
2
8
Therefore, z 5, 8y 3 5 23
y 1, and x 2 1 5 8 x 1. Hence, the solution is 1 1 5. x y 3z 4 x y 3z 4 x y 3z 4 16. 4x 2y z 11 6y 13z 27 6y 13z 27 5x y z 16 3z 9 6y 16z 36
Therefore, z 3,
6y 13 3 27 y 2, and x 2 3 3 4 x 3. Hence, the solution is 3 2 3. x y 2z 6 x y 2z 6 x y 2z 6 17. 2x Therefore, 3z 6 z 2, 2y 2 0 5z 12 2y z 0 2y z 0 x 2y 3z 9 4y z 6 3z 6 y 1, and x 1 2 2 6 x 1. Hence, the solution is 1 1 2. x 2y 3z 1 x 2y 3z 1 x 2y 3z 1 18. y 4z 1 y 4z 1 x 3y z 0 2x 28z 0 6y 4z 6 6z 6
x 3, and so the solution is 3 1 0. x 2y 3z 1 x 2y 3z 1 x 2y 3z 1 19. 2x y z 3 3y 5z 1 3y 5z 1 2x 7y 11z 2 6y 10z 1 0 1
system has no solution. x y z 2 2x 3z 5 20. 4 9 x 2y x y 2z 3 5 x y z 2 2y 5z 2 1 13z 12 11 14 28
Thus, z 0, y 1, and x 2 1 3 0 1
which is impossible. Therefore, the
x y z 2 x y z 2 2y 5z 2 1 2y 5z 2 1 3y z 3 7 13z 12 11 z 2 3 z 2 3 Therefore, 14 28 2, 13z12 2 11 z 1; 2y5 12 2 1
2y 1 1 y 0; and x 1 2 2 x 1. So the solution is 1 0 1 2.
712
CHAPTER 10 Systems of Equations and Inequalities
x 3y z 4 x 3y z 4 21. 4x y 15z 5 y z 1
Thus, the system has infinitely many solutions given by z t,
y t 1 y 1 t, and x 3 1 t t 4 x 1 4t. Therefore, the solutions are 1 4t 1 t t, where t is any real number.
2x 3y 4z 3 2x 3y 4z 3 3 2x 3y 4z 22. 4x 5y 9z 13 y z 7 y z 7 2x 7z 0 3y 3z 3 0 24
Since this last equation is impossible, the system is inconsistent and has no solution.
x z 2 x z 2 x z 2 2x y y 2z 4 8 y 2z 4 8 2 12 23. 5z 13 20 3y z 4 3y z 4 x yz z 4 20 y 8 10 z 2 x y 2z 4 8 20 z 60 , 5z 13 40 Therefore, 33 120 40 11 11 11 , 5z 13 20 33 120 40 8 y 48 ; and x 60 40 2 x 2 . Hence, the solution is y 2 60 4 11 11 11 11 11 11 2 48 60 40 . 11 11 11 11 x 4y z 8 x 4y z 8 x 4y z 8 24. 2x 6y z 9 2y 3z 7 2y 3z 7 x 6y 4z 15 6y 9z 21 00
Thus, the system has infinitely many solutions. Letting z t, we find 2y 3t 7 y 72 32 t, and x 4 72 32 t t 8 x 6 5t. Therefore, the solutions are 6 5t 72 32 t t , where t is any real number.
25. Let k and s be be Kieran’s and Siobhan’s ages. From the given information, we have the system k s 4 Subtracting the first equation from the second gives s 22 s 4 2s 18 s 9, so k s 22 k 22 9 13. Thus, Kieran is 13 and Siobhan is 9.
26. Let x be the amount in the 6% account and y the amount in the 7% account. The system is
y 2x 006x 007y 600
Substituting gives 006x 007 2x 600 02x 600 x 3000, so y 2 3000 6000. Hence, the man has $3000 invested at 6% and $6000 invested at 7%.
CHAPTER 10
Review
713
27. Let n be the number of nickels, d the number of dimes, and q the number of quarter in the piggy bank. We get the following 50 n d q Since 10d 25n, we have d 52 n, so substituting into the first equation system: 5n 10d 25z 560 10d 5 5n we get n 52 n q 50 72 n q 50 q 50 72 n. Now substituting this into the second equation we have 115 115 5n 10 52 n 25 50 72 n 560 5n 25n 1250 175 2 n 560 1250 2 n 560 2 n 650
n 12. Then d 52 12 30 and q 50 n d 50 12 30 8. Thus the piggy bank contains 12 nickels, 30 dimes, and 8 quarters. 28. Let c, s, and p be the number of each kind of salmon caught. c s c s p 25 c s p 25 c s p3 c s p 3 2s c 2s 0 3s c 2s
The given information leads to the system p 25 c s p 25 2 p 22 2s 2 p 22 p 4, so 4 p 16 p 25
2s 2 4 22 s 7 and c 7 4 25 c 14. Tornie caught 14 coho, 7 sockeye, and 4 pink salmon.
29.
3x 1
B 3x 1 A .Thus, 3x 1 A x 3 B x 5 x A B 3A 5B, x 5 x 3 x 5 x 3 3A 3B 9 AB 3 Adding, we have 8B 8 B 1, and A 2. Hence, and so 3A 5B 1 3A 5B 1 x 2 2x 15
3x 1 1 2 . x 5 x 3 x 2 2x 15
30.
A B C 8 8 8 2 4 Bx x 2 . Then 8 A x x x 2 x 2 x x 2 x 2 x 3 4x x x2 4
C x x 2 x 2 A B C x 2B 2C 4A. Thus, 4A 8 A 2, 2B 2C 0 B C, and 1 1 2 8 . 2 2B 0 B 1, so C 1. Therefore, 3 x x 2 x 2 x 4x 31.
2x 4
x x 12
B C A . Then 2x 4 A x 12 Bx x 1 C x Ax 2 2Ax A Bx 2 x x 1 x 12
Bx C x x 2 A B x 2A B C A. So A 4, 4 B 0 B 4, and 8 4 C 2 C 2. 2x 4 4 2 4 Therefore, . 2 x x 1 x 12 x x 1 32.
x 6
x 6 Ax B C . Thus, 2 2 x 2 x 4 x 4 x 2 x 6 Ax B x 2 C x 2 4 Ax 2 2Ax Bx 2B C x 2 4C
x 3 2x 2 4x 8
x 2 A C x 2A B 2B 4C A C0 C0 C0 A A and we get the system 2A B 1 B 2C 1 B 2C 1 2B 4C 6 2B 4C 6 8C 8
x 6 x 1 1 Thus, 8C 8 C 1, B 2 1 B 1, and A 1 0 A 1. So 3 . 2 2 x 2x 4x 8 x 4 x 2
714
33.
CHAPTER 10 Systems of Equations and Inequalities
2x 1 A Bx C 2x 1 2 . Then 2x 1 A x 2 1 Bx C x Ax 2 A Bx 2 C x A B x 2 3 2 x x x x x 1 x 1 x 2 1 2x 1 2 . C x A. So A 1, C 2, and A B 0 gives us B 1. Thus 3 x x x x 1
B Cx D 5x 2 3x 10 A 2 . 34. Since x 4 x 2 2 x 2 1 x 2 2 x 1 x 1 x 2 2 , we have 4 2 x 1 x 1 x x 2 x 2 Thus 5x 2 3x 10 x 1 x 2 2 A x 1 x 2 2 B x 2 1 C x D Ax 3 Ax 2 2Ax 2A Bx 3 Bx 2 2Bx 2B C x 3 Dx 2 C x D
A B C x 3 A B D x 2 2A 2B C x 2A 2B D A BC 0 Coefficients of x 3 A B C 0 A B 2 2B C D 5 D 5 Coefficients of x This leads to the system 3C 3 2A 2B C 3 Coefficients of x 2A 2B 4B C D 13 D 10 Constant terms A B C 0 2B C D 5 Thus C 1, 3 1 3D 3 D 0, 2B 1 0 5 B 3, and C 1 3C 3D 3 A 3 1 0 A 2. Thus,
35.
3 x 2 5x 2 3x 10 2 . x 1 x 1 x4 x2 2 x 2
Cx D Ax B 3x 2 x 6 2 2 . Thus x 2 22 x 2 x2 2 3x 2 x 6 x 2 2 Ax B C x D Ax 3 Bx 2 2Ax 2B C x D
x 3 A x 2 B x 2A C 2B D A A 0 B B 3 This leads to the system C C 1 2A D 2B D6
3x 2 x 6 x 3 Thus 2 2 . 2 2 2 x 2 x 2 x 2
0 3
1
0
CHAPTER 10
Review
715
Bx C x2 x 1 A Dx E 2 2 . Thus x xx 2 12 x 1 x2 1 2 x 2 x 1 x 2 1 A x x 2 1 Bx C x Dx E Ax 4 2Ax 2 A Bx 4 C x 3 Bx 2 C x Dx 2 E x 36.
x 4 A B x 3 C x 2 2A B D x C E A AB 0 C 0 Thus A 1, B 1, C 0, D 0, and E 1, so This leads to the system 2A B D 1 C E 1 A 1 x2 x 1 x 1 1 2 2 . x 2 xx 2 12 x 1 x 1
2x 3y 7 37. x 2y 0
By inspection of the graph, it appears that 2 1 is the solution to the system. We check this in both
equations to verify that it is the solution. 2 2 3 1 4 3 7 and 2 2 1 2 2 0. Since both equations are satisfied, the solution is indeed 2 1.
3x y 8 38. y x 2 5x
By inspection of the graph, it appears that 2 14 and 4 4 are solutions to the system.
We check each possible solution in both equations to verify that it is a solution: 3 2 14 6 14 8 and
22 5 2 4 10 14; also 3 4 4 12 4 8 and 42 5 4 16 20 4. Since both points satisfy both equations, the solutions are 2 14 and 4 4.
39.
x2 y 2
x 2 3x y 0
By inspection of the graph, it appears that 2 2 is a solution to the system, but is difficult
to get accurate values for the other point. Adding the equations, we get 2x 2 3x 2 2x 2 3x 2 0 2 2x 1 x 2 0. So 2x 1 0 x 12 or x 2. If x 12 , then 12 y 2 y 74 . If x 2, then 22 y 2 y 2. Thus, the solutions are 12 74 and 2 2. 40.
x y 2 x 2 y 2 4y 4
By inspection of the graph, it appears that 2 0 and 2 4 are solutions to the system. We
check each possible solution in both equations to verify that it is a solution: 2 0 2 and 22 02 4 0 4;
also 2 4 2 and 22 42 4 4 4 16 16 4. Since both points satisfy both equations, the solutions are 2 0 and 2 4.
41. The boundary is a solid curve, so we have the inequality x y 2 4. We take the test point 0 0 and verify that it satisfies the inequality: 0 02 4.
42. The boundary is a solid curve, so we have the inequality x 2 y 2 8. We take the test point 0 3 and verify that it satisfies the inequality: 02 32 8.
716
CHAPTER 10 Systems of Equations and Inequalities
44. y x 2 3
43. 3x y 6
y
45. x 2 y 2 9
y
46. x y 2 4
y
y
1 1 1
1
x
y
y
1
x
0
1
x
y 4
1 1
x 2 y2 9 51. xy0
y 2x 50. y 2x y 1x 2 2
y
1
1
4
x
y
The vertices occur where y x. By substitution, 1
x 2 x 2 9 x 3 , and so y 3 . Therefore, the vertices are 2 2 3 3 3 3 and and the solution set is bounded. The test point 2
2
x
1
x y 2 49. y x 2 x 3
y x 1 48. x 2 y2 1
3
1
x
x
1
y x 2 3x 47. y 1x 1 0
1
2
x
1
2
1 1 satisfies each inequality.
y x2 4 52. y 20
y
The vertices occur where y x 2 4 and y 2. By
substitution, x 2 4 20 x 2 16 x 4 and y 20. Thus, the vertices
(_4, 20)
(4, 20)
are 4 20 and the solution set is bounded. The test point 0 10 satisfies each inequality.
2 2
x 0, y 0 53. x 2y 12 y x 4
The intersection points are 4 0, 0 4,
y
4 16 , 0 6, 3 3
0 0, and 12 0. Since the points 4 0 and 0 6 are not in the solution set, the vertices are 0 4, 43 16 3 , 12 0, and 0 0. The solution set is bounded. The test point 1 1 satisfies each inequality.
x
y=x+4
x+2y=12 1 1
x
x
CHAPTER 10
54.
x4 x y 24
y
Test
717
x=4
The lines x y 24 and x 2y 12 intersect at the
x 2y 12
point 20 4. The lines x 4 and x 2y 12 intersect at the point 4 4, but this vertex does not satisfy the other inequality. The lines x y 24 and x 4 intersect at the point 4 20. Hence, the vertices are 4 20 and 20 4. The solution set is not bounded. The test point 12 12 satisfies each inequality.
x+y=24 1 1
x=2y+12 x
x y z a x y z a ab ac ab ac ,z , and x a 55. x yz b 2z a b Thus, y 2 2 2 2 x yz c 2y ac bc bc ac ab . The solution is . x 2 2 2 2 ax by cz a b c ax by cz a b c 56. bx by cz a b c 0 bx by cz Subtracting the second equation from c c cx cy cz c x yz 1 the first gives a b x a b x 1Subtracting the third equation from the second, b c x b c y 0 y x 1. So 1 1 z 1 z 1, and the solution is 1 1 1.
57. Solving the second equation for y, we have y kx. Substituting for yin the first equation gives us 12 x kx 12 1 k x 12 x . Substituting for y in the third equation gives us kx x 2k k1 2k 12 2k . These points of intersection are the same when the x-values are equal. Thus, k 1 x 2k x k 1 k1 k 1 12 k 1 2k k 1 12k 12 2k 2 2k 0 2k 2 10k 12 2 k 2 5k 6 2 k 3 k 2. Hence, k 2 or k 3.
58. The system will have infinitely many solutions when the system has solutions other than 0 0 0. So we solve the system with x 0, y 0, and z 0: kx y z0 kx y z 0 kx y z 0 2y k 3 z 0 x 2y kz 0 2y k 3 z 0 x y 3k 1 z 0 3z 0 [k 3 2 3k 1] z 0 Since z 0, we must have k 3 2 3k 1 5k 1 0 k 15 .
CHAPTER 10 TEST 1. (a) The system is linear. x 3y 7 (b) Multiplying the first equation by 5 and then adding gives 13y 39 y 3. So 5x 2y 4 x 3 3 7 x 2. Thus, the solution is 2 3.
2. (a) The system is nonlinear.
718
CHAPTER 10 Systems of Equations and Inequalities
6x y 2 10 6x y 2 10 (b) 3x y 5 y 2 2y 0
Thus y 2 2y y y 2 0, so either y 0 or y 2. If y 0, then
3x 5 x 53 and if y 2, then 3x 2 5 3x 3 x 1. Thus the solutions are 53 0 and 1 2.
3. (a) The system is nonlinear. x 2 y 2 100 Substituting y 3x into the first equation gives x 2 3x2 100 x 2 10 x 10. (b) y 3x If x 10, then y 3 10, and if x 10, then y 3 10. We can verify that 10 3 10 and 10 3 10 are valid solutions to the first equation in the given system. x 2y 1 4. y x 3 2x 2
2
The solutions are approximately 055 078, 043 029, and 212 056. -2
2 -2
x 2y z 3 x 2y z 3 5. (a) x 3y 2z 3 y z 0 2x 3y z 8 y 3z 2
Eq. 1 Eq. 2 2 Eq. 1 Eq. 3
x 2y z 3 y z 0 2z 2
Eq. 2 Eq. 3
z 1, so y 1 0 y 1 and z 2 1 1 3 x 2. Thus, the solution is 2 1 1.
(b) The system is neither inconsistent nor dependent.
x y 9z 3 x y 9z 3 6. (a) x 4z 7 y 13z 10 3x y z 5 2y 26z 14
Eq. 1 Eq. 2 3 Eq. 1 Eq. 3
The last equation cannot be satisfied, so the system has no solution.
(b) The system is inconsistent.
2x y z 0 2x y z 0 7. (a) 3x 2y 3z 1 7y 9z 2 x 4y 5z 1 7y 9z 2
x y 9z 3 y 13z 10 0 6
3 Eq. 1 2 Eq. 2
Eq. 1 2 Eq. 3
2x y z 0 7y 9z 2 00
2 Eq. 2 Eq. 3
Eq. 2 Eq. 3
Letting z t, we have 7y 9t 2 y 27 97 t, so 2x 27 97 t t 0 x 17 17 t. The solutions are 1 1 t 2 9 t t where t is any real number. 7 7 7 7
(b) The system is dependent.
x y 2z 8 x y 2z 8 8. (a) 2x y 20 3y 4z 4 2x 2y 5z 15 z 1
2 Eq. 1 Eq. 2 2 Eq. 1 Eq. 3
x 0 2 1 8 x 10. Thus, the solution is 10 0 1.
(b) The system is neither inconsistent nor dependent.
z 1, so 3y 4 1 4 y 0 and
CHAPTER 10
Test
719
9. Let be the speed of the wind and a the speed of the airplane in still air, in kilometers per hour. Then the speed of the of the plane flying against the wind is a and the speed of the plane flying with the wind is a . Using distance rate time, 600 25 a 240 a Adding the two equations, we get 600 2a a 300. we get the system 50 300 360 a a 60
So 360 300 60Thus the speed of the airplane in still air is 300 km/h and the speed of the wind is 60 km/h.
10. Let x, y, and z represent the price in dollars for coffee, juice, and donuts respectively. Then the system of equations is 2x y 2z 625 Anne 2x y 2z 625 2x y 2z 625 x 3z 375 Barry y 4z 125 y 4z 125 3x y 4z 925 Cathy y 5z 200 z 075
Thus, z 075, y 4 075 125 y 175, and 2x 175 2 075 625 x 15. Thus coffee costs $150, juice costs $175, and donuts cost $075.
11. 3x 4y 6
12. x 2 y 3
y
y
1 1
x 1 x
1
2x y 8 13. x y 2 x 2y 4
y
From the graph, the points 4 0 and 0 2 are vertices. The
third vertex occurs where the lines 2x y 8 and x y 2 intersect. Adding these two equations gives 3x 6 x 2, and so y 8 2 2 4 . Thus, the
third vertex is 2 4.
1 x
1
x2 5 y 0 14. y 5 2x
Substituting y 5 2x into the first equation gives
y x@-5+y=0
y=5+2x
x 2 5 5 2x 0 x 2 2x 0 x x 2 0 x 0 or x 2. If
x 0, then y 5 2 0 5, and if x 2, then y 5 2 2 1. Thus, the
vertices are 0 5 and 2 1.
1 1
15.
4x 1
x 12 x 2
B C A . Thus, x 1 x 12 x 2
x
4x 1 A x 1 x 2 B x 2 C x 12 A x 2 x 2 B x 2 C x 2 2x 1 A C x 2 A B 2C x 2A 2B C
720
FOCUS ON MODELING
which leads to the system of equations A C 0 A C 0 C 0 A A B 2C 4 B 3C 4 B 3C 4 2A 2B C 1 2B 3C 1 9C 9
Therefore, 9C 9 C 1, B 3 1 4 B 1, and A 1 0 A 1. Therefore, 4x 1
x 12 x 2
16.
1 1 1 . x 1 x 12 x 2
2x 3 Bx C 2x 3 A 2 . Then x x 3 3x x x 3 x2 3 2x 3 A x 2 3 Bx C x Ax 2 3A Bx 2 C x A B x 2 C x 3A.
x 2 1 2x 3 2 . So 3A 3 A 1, C 2 and A B 0 gives us B 1. Thus 3 x x x x 3
FOCUS ON MODELING Linear Programming
2.
1. Vertex 0 2 0 5 4 0
M 200 x y
200 0 2 198 200 0 5 195 200 4 0 196
Thus, the maximum value is 198 and the minimum value is 195.
x 0, y 0 3. 2x y 10 2x 4y 28
Vertex 1 0 1 1 2 2 2 2 4 0
N 12 x 14 y 40 1 1 1 0 40 405 4 2 1 1 1 1 40 40375 2 2 4 2 1 2 1 2 40 415 2 4 1 4 1 0 40 42 2 4
Thus, the maximum value is 42 and the minimum value is 40375.
The objective function is P 140 x 3y. From the graph,
y
2x+y=10
the vertices are 0 0, 5 0, 2 6, and 0 7. Vertex 0 0 5 0 2 6 0 7
P 140 x 3y
140 0 3 0 140 140 5 3 0 135 140 2 3 6 156 140 0 3 7 161
Thus the maximum value is 161, and the minimum value is 135.
2x+4y=28 1 1
x
Linear Programming
x 0, y 0 x 10, y 20 4. xy5 x 2y 18
The objective function is Q 70x 82y. From the graph,
y
x+2y=18
the vertices are at 0 9, 0 5, 5 0, 10 0, and 10 4. Note that the restriction y 20 is irrelevant, superseded by x 2y 18 and x 0. Vertex 0 9 0 5 5 0 10 0 10 4
721
Q 70x 82y
x+y=5
1
70 0 82 9 738
x=10
x
1
70 0 82 5 410 70 5 82 0 350
70 10 82 0 700
70 10 82 4 1028
Thus, the maximum value of Q is 1028 and the minimum value is 350.
5. Let t be the number of tables made daily and c be the number of chairs made daily. Then the data given can be summarized by the following table: Tables t
Chairs c
Available time
2h
3h
108 h
Finishing
1h
1 h 2
20 h
Profit
$35
$20
Carpentry
2t 3c 108 Thus we wish to maximize the total profit P 35t 20c subject to the constraints t 12 c 20 t 0, c 0 From the graph, the vertices occur at 0 0, 20 0, 0 36, and 3 34.
y
Vertex 0 0 20 0 0 36 3 34
P 35t 20c
35 0 20 0
0
1
t+2 c=20
35 20 20 0 700
35 0 20 36 720 35 3 20 34 785
2t+3c=108 10 10
Hence, 3 tables and 34 chairs should be produced daily for a maximum profit of $785.
x
722
FOCUS ON MODELING
6. Let c be the number of colonial homes built and r the number of ranch homes built. Since there are 100 lots available, c r 100. From the capital restriction, we get 30,000c 40,000r 3,600,000, or 3c 4r 360. Thus, we wish to c 0, r 0 maximize the profit P 4000c 8000r subject to the constraints c r 100 3c 4r 360 From the graph, the vertices occur at 0 0, 100 0, 40 60, and 0 90.
y
Vertex 0 0 100 0 40 60 0 90
P 4000c 8000r
4000 0 8000 0
c+r=100
0
4000 100 8000 0 400,000
4000 40 8000 60 640,000
3c+4r=360
20
4000 0 8000 90 720,000
x
20
Therefore, he should build no colonial style and 90 ranch style houses for a maximum profit of $720,000. Note that ten of the lots will be left vacant.
7. Let x be the number of crates of oranges and y the number of crates of grapefruit. Then the data given can be summarized by the following table: Oranges
Grapefruit
Available
Volume
4 ft3
6 ft3
300 ft3
Weight
80 lb
100 lb
5600 lb
Profit
$250
$400
In addition, x y. Thus we wish to maximize the total profit P 25x 4y subject to the constraints x 0, y 0, x y 4x 6y 300 80x 100y 5600 From the graph, the vertices occur at 0 0, 30 30, 45 20, and 70 0.
y
Vertex 0 0 30 30 45 20 70 0
80x+100y=5600
P 25x 4y
25 0 4 0
0
25 30 4 30 195
25 45 4 20 1925 25 70 4 0 175
x=y
4x+6y=300 10 10
Thus, she should carry 30 crates of oranges and 30 crates of grapefruit for a maximum profit of $195.
x
Linear Programming y
8. Let x be the daily production of standard calculators and y the daily production of x 100, y 80 scientific calculators. Then the inequalities x 200, y 170 describe the x y 200
180 120 90
100 100 100 170 200 170 200 80
C 5x 7y
30 60 90 120 150 180 x
Vertex 100 100
5 200 7 170 2190
200 170
5 100 7 170 1690
100 170
5 200 7 80 1560
200 80
5 120 7 80 1160
x+y=200
30
(b) Maximize the objective function P 2x 5y:
5 100 7 100 1200
120 80
y=80
60
200 170, 200 80, and 120 80.
Vertex
x=200
150
constraints. From the graph, the vertices occur at 100 100, 100 170,
(a) Minimize the objective function C 5x 7y:
x=100 y=170
723
P 2x 5y
2 100 5 100 300 2 100 5 170 650 2 200 5 170 450 2 200 5 80
0
2 120 5 80 160
120 80
So to minimize cost, they should produce
So to maximize profit, they should produce
120 standard and 80 scientific calculators.
100 standard and 170 scientific calculators.
9. Let x be the number of stereo sets shipped from Long Beach to Santa Monica and y the number of stereo sets shipped from Long Beach to El Toro. Thus, 15 x sets must be shipped to Santa Monica from Pasadena and 19 y sets to El Toro from Pasadena. Thus, x 0, y 0, 15 x 0, 19 y 0, x y 24, and 15 x 19 y 18. Simplifying, we get x 0, y 0 x 15, y 19 the constraints x y 24 x y 16 The objective function is the cost C 5x 6y 4 15 x 55 19 y x 05y 1645, which we wish to
minimize. From the graph, the vertices occur at 0 16, 0 19, 5 19, 15 9, and 15 1. Vertex 0 16 0 19 5 19 15 9 15 1
C x 05y 1645
y
x=15
0 05 16 1645 1725 0 05 19 1645 174 5 05 19 1645 179 15 05 9 1645 184 15 05 1 1645 180
10 x+y=16 10
x+y=24 x
The minimum cost is $17250 and occurs when x 0 and y 16. Hence, no stereo should be shipped from Long Beach to Santa Monica, 16 from Long Beach to El Toro, 15 from Pasadena to Santa Monica, and 3 from Pasadena to El Toro.
724
FOCUS ON MODELING
10. Let x be the number of sheets shipped from the east side store to customer A and y be the number of sheets shipped from the east side store to customer B. Then 50 x sheets must be shipped to customer A from the west side store and 70 y sheets must be shipped to customer B from the west side store. Thus, we obtain the constraints x 0, y 0 x 0, y 0 x 50, y 70 x 50, y 70 x y 80 x y 80 x y 75 50 x 70 y 45
The objective function is the cost C 05x 06y 04 50 x 055 70 y 01x 005y 585, which we wish to minimize. From the graph, the vertices occur at 5 70, 10 70, 50 30, and 50 25. y
Vertex 5 70 10 70 50 30 50 25
x=70
C 01x 005y 585
01 5 005 70 585 625
y=70
01 10 005 70 585 63
x+y=75
01 50 005 30 585 65
x+y=80
10
01 50 005 25 585 6475
x
10
Therefore, the minimum cost is $6250 and occurs when x 5 and y 70. So 5 sheets should be shipped from the east side store to customer A, 70 sheets from the east side store to customer B, 45 sheets from the west side store to customer A, and no sheet from the west side store to customer B.
11. Let x be the number of bags of standard mixtures and y be the number of bags of deluxe mixtures. Then the data can be summarized by the following table: Standard
Deluxe
Available
Cashews
100 g
150 g
15 kg
Peanuts
200 g
50 g
20 kg
Selling price
$195
$220
Thus the total revenue, which we want to maximize, is given by R 195x 225yWe have the constraints x 0, y 0, x y x 0, y 0, x y 01x 015y 15 10x 15y 1500 02x 005y 20 20x 5y 2000
From the graph, the vertices occur at 0 0, 60 60, 90 40, and 100 0.
y
Vertex 0 0 60 60 90 40 100 0
R 195x 225y
196 0 225 0 0
10x+15y=1500
195 60 225 60 252
195 90 225 40 2655 195 100 225 0 195
x=y 20x+5y=2000
10 10
x
Hence, he should pack 90 bags of standard and 40 bags of deluxe mixture for a maximum revenue of $26550.
Linear Programming
725
12. Let x be the quantity of type I food and y the quantity of type II food, in ounces. Then the data can be summarized by the following table: Type I
Type II
Required
Fat
8g
12 g
24 g
Carbohydrate
12 g
12 g
36 g
Protein
2g
1g
4g
Cost
$020
$030
x 0, y 0, Also, the total amount of food must be no more than 5 oz. Thus, the constraints are x y 5, 8x 12y 24 12x 12y 36, 2x y 4
The objective function is the cost C 02x 03y, which we wish to minimize. From the graph, the vertices occur at 1 2, 0 4, 0 5, 5 0, and 3 0. Vertex 1 2 0 4 0 5 5 0 3 0
y
C 02x 03y
x+y=5
02 1 03 2 08 02 0 03 4 12 02 0 03 5 15
2x+y=4 12x+12y=36
02 5 03 0 10 02 3 03 0 06
8x+12y=24 1 x
1
Hence, the rabbits should be fed 3 oz of type I food and no type II food, for a minimum cost of $060.
13. Let x be the amount in municipal bonds and y the amount in bank certificates, both in dollars. Then 12000 x y is the amount in high-risk bonds. So our constraints can be stated as x 0, y 0, x 3y x 0, y 0, x 3y x y 12,000 12,000 x y 0 12,000 x y 2000 x y 10,000
From the graph, the vertices occur at 7500 2500, 10000 0, 12000 0, and 9000 3000. The objective function is P 007x 008y 012 12000 x y 1440 005x 004y, which we wish to maximize. y
Vertex 7500 2500 10000 0
1440 005 10,000 004 0 940
12000 0 9000 3000
x+y=12,000
P 1440 005x 004y
1440 005 7500 004 2500 965
1440 005 12,000 004 0 840
1440 005 9000 004 3000 870
x+y=10,000
x=3y
2000 2000
x
Hence, she should invest $7500 in municipal bonds, $2500 in bank certificates, and the remaining $2000 in high-risk bonds for a maximum yield of $965.
726
FOCUS ON MODELING
14. The only change that needs to be made to the constraints in Exercise 13 is that the 2000 in the last inequality becomes 3000. Then we have x 0, y 0, x 3y x 0, y 0, x 3y x y 12,000 12,000 x y 0 12,000 x y 3000 x y 9000
From the graph, the vertices occur at 6750 2250, 9000 0, 12000 0, and 9000 3000. The objective function is Y 007x 008y 012 12000 x y 1440 005x 004y, which we wish to maximize. y
Vertex 6750 2250
1440 005 9000 004 0 990
9000 0 12000 0 9000 3000
x+y=12,000
Y 1440 005x 004y
1440 005 6750 004 2250 10125 1440 005 12,000 004 0 840
1440 005 9000 004 3000 870
x=3y
x+y=9000 2000
x
2000
Hence, she should invest $6750 in municipal bonds, $2250 in bank certificates, and $3000 in high-risk bonds for a maximum yield of $101250, which is an increase of $4750, over her yield in Exercise 13.
15. Let g be the number of games published and e be the number of educational programs published. Then the number of utility programs published is 36 g e. Hence we wish to maximize profit, P 5000g 8000e 6000 36 g e 216,000 1000g 2000e, subject to the constraints g 4, e 0 g 4, e 0 36 g e 0 g e 36 36 g e 2e g 3e 36.
From the graph, the vertices are at 4 32 3 , 4 32, and 36 0. The objective function is P 216,000 1000g 2000e. e
Vertex 4 32 3 4 32 36 0
g=4
P 216,000 1000g 2000e 216,000 1000 4 2000 32 3 233,33333 216,000 1000 4 2000 32 276,000 216,000 1000 36 2000 0 180,000
g+e=36 g+3e=36
10 10
g
So, they should publish 4 games, 32 educational programs, and no utility program for a maximum profit of $276,000 annually.
Linear Programming
x 0, x y 16. x 2y 12 x y 10 (a)
y
(b) x+y=10
y
x=y P=40 P=36 P=32 P=28
(5, 5) x+2y=12
1 1
1
x
1
(c) These lines will first touch the feasible region at the top vertex, 5 5. (d) Vertex 0 0 10 0 5 5 8 2
P x 4y
0 4 0 0
10 4 0 10 5 4 5 25 8 4 2 16
The maximum value of the objective function occurs at the vertex 5 5.
x
727
11
MATRICES AND DETERMINANTS
11.1 MATRICES AND SYSTEMS OF LINEAR EQUATIONS 1. A system of linear equations with infinitely many solutions is called dependent. A system of linear equations with no solution is called inconsistent. 1 1 1 1 x y z 1 2. x 2z 3 1 0 2 3 2y z 3 0 2 1 3
3. (a) The leading variables are x and y. (b) The system is dependent.
(c) The solution of the system is x 3 t y 5 2t z t.
x 2 The solution is 2 1 3. 0 1 y1 z3 1 3 1 2 x z2 The solution is z t, y 1 t, x 2 t. 1 1 yz1 0 0 0 2 2 x This system is inconsistent and has no solution. 0 1 y 1 03 0 3
1 0 0 2
4. (a) 0 0 1 (b) 0 0 1 (c) 0 0
1 0 0 1 0 0 1 0
5. 3 2
3
6. 2 4 1 1 2
7. 2 1
8. 3 1
1
9. 1 3 0
1 1
10. 2 2
11. 2 1 0 1 1 0 1 3
12.
13. (a) Yes, this matrix is in row-echelon form.
14. (a) Yes, this matrix is in row-echelon form.
(b) Yes, this matrix is in reduced row-echelon form. x 3 (c) y 5
0
3 2
1 1
3
7 3
(b) No, this matrix not in reduced row-echelon form. The entry above the leading 1 in the second row is not 0. x 3y 3 (c) y 5
729
730
CHAPTER 11 Matrices and Determinants
16. (a) Yes, the matrix is in row-echelon form.
15. (a) Yes, this matrix is in row-echelon form.
(b) Yes, the matrix is in reduced row-echelon form. 7z 0 x (c) y 3z 0 01
(b) No, this matrix is not in reduced row-echelon form, since the leading 1 in the second row does not have a zero above it. x 2y 8z 0 (c) y 3z 2 00 17. (a) No, this matrix is not in row-echelon form, since the row of zeros is not at the bottom.
18. (a) Yes, the matrix is in row-echelon form. (b) Yes, the matrix is in reduced row-echelon form. 1 x (c) y 2 z3
(b) No, this matrix is not in reduced row-echelon form. 0 x (c) 00 y 5z 1 19. (a) Yes, this matrix is in row-echelon form.
20. (a) No, this matrix is not in row-echelon form, since the fourth column has the leading 1 of two rows.
(b) Yes, this matrix is in reduced row-echelon form. x 3y 0 z 2 0 (c) 01 00
(b) No, this matrix is not in reduced row-echelon form. x 3y 0 y 4 0 (c) u2 0
Notice that this system has no solution.
1
1
2
3
1
1
5
2
1 3
21.
0
4 1 2 1 1 2 3
22. 10 3 1 3
3
1 3
5
2 1
1 1 2 1 1 1
1 20 1 8
23. 2 3 1 13 6 5 1 7 24. 0 0
3R1 R2 R2
2R2 R3 R3
1
2
0
4
7
0
3
4 1 2 1 1
2R1 R2 R2
3R1 R3 R3
1
5 2 3
0 1 5 14 1 3 1 8
2
2 3 0 8
5
1 13 8 8
1 3
2 1
0 0
1 3
1 1 0 3 3 1
SECTION 11.1 Matrices and Systems of Linear Equations
x 2y 4z 3 25. (a) y 2z 7 z2
(b) y 2 2 7 y 3, so x 2 3 4 2 3
(b) y 3 1 5 y 2, so x 2 3 1 8
x 1. The solution is 1 3 2.
x 2y 3z y 2z 27. (a) z 2
x 3. The solution is 3 2 1.
x 2z 2 5 y 3z 1 28. (a) z 0 1
7 5 5 3
(b) z 1 0 z 1, so y 3 1 1
(b) z 2 3 5 z 1, so y 2 1 5
y 2 and x 2 1 2 1 5 x 5. The
y 3 and x 2 3 3 1 3 7 x 7. The
solution is 5 2 1 1.
solution is 7 3 1 3.
1 2 1 1
29. 0 1
1 2 5 1 3 8
1 2 1 1
0 0
R3 R1 R3
731
x y 3z 8 26. (a) y 3z 5 z 1
1 2 5 3 2 7
1 2
1
1
2
0 0
R3 3R2 R3
1
5 . Thus, 4z 8 0 4 8
z 2; y 2 2 5 y 1; and x 2 1 2 1 x 1. Therefore, the solution is 1 1 2.
1 1 6 3
30. 1 1 3 3 1 2 4 7
R2 R1 R2
1 1
6 3
0 0 3 0 1 2 4 7
13 R2
R3 R1 R3
1 1
6 3
0 0 1 0 0 1 2 4
1 1
6 3
0 1 2 4 . 0 0 1 0
R3 R2
Thus, z 0; y 2z y 0 y 4; and x y 6z 3 x 4 0 3 x 1. Therefore, the solution is 1 4 0.
1
1
1 2
31. 2 3 2 4 4 1 3 1
R2 2R1 R2 R3 4R1 R3
1
1
1
2
0 5 0 0 0 3 7 7
R3 35 R2 R3
1
1
1
2
0 5 0 0 . 0 0 7 7
Thus, 7z 7 z 1; 5y 0 y 0; and x 0 1 2 x 1. Therefore, the solution is 1 0 1.
1
1 1
4
32. 1 2 3 17 2 1 0 7
R2 R1 R2 R3 2R1 R3
1
1
1
4
0 3 4 21 0 3 2 15
1 1 1
4
0 3 4 21 . 0 0 2 6
R3 R2 R3 R3 2R1 R3
Thus, 2z 6 z 3; 3y 4 3 21 y 3; and x 3 3 4 x 2. The solution is 2 3 3.
1
33. 1 2 1 0 0
2 1 2
1
2 1 2
R2 R1 R2 0 2 2 2 0 R3 2R1 R3 0 5 1 1 1 1 3 2 1 2 1 1 1 . Thus, 6z 6 z 1; y 1 1 0 6 6 0
1
solution is 1 0 1.
12 R2
1
2 1 2
0 1 0 5
1 1
1 1
R3 5R2 R3
y 0; and x 2 0 1 2 x 1. Therefore, the
732
CHAPTER 11 Matrices and Determinants
0 2
1
4
34. 1 1 0 4 3 3 1 10
R1 R2
1 1
0
4
0 2 1 4 3 3 1 10
1 1
0
4
0 2 1 4 . 0 0 1 2
R3 3R1 R3
Thus, z 2 z 2; 2y 2 4 y 1; and x 1 4 x 3. The solution is 3 1 2
1 2 1
9
35. 2 0 1 2 3 5 2 22
1
2 1
0 4 0 1
R2 2R1 R2 R3 3R1 R3
9
1 20 5 5
1
2 1
9
0 4 1 20 . 0 0 19 0
4R3 R2 R3
Thus, 19x3 0 x3 0; 4x2 20 x2 5; and x1 2 5 9 x1 1. Therefore, the solution is 1 5 0.
2
1 0
7
36. 2 1 1 6 3 2 4 11
2
1 0
7
0 2 1 1 0 7 8 1
R2 R1 R2 2R3 3R1 R3
2
1 0
7
0 2 1 1 . Then, 9x3 9 0 0 9 9
2R3 7R2 R3
x3 1; 2x2 1 1 x2 1; and 2x1 1 7 x1 3. Therefore, the solution is x1 x2 x3 3 1 1.
2 3 1 13
37. 1 2 5 6 5 1 1 49
2 3
1 13
0 1 11 25 0 13 3 33
2R2 R1 R2 2R3 5R1 R3
2 3
0 0
R3 13R2 R3
1
13
25 . 0 146 292 1 11
Thus, 146z 292 z 2; y 11 2 25 y 3; and 2x 3 3 2 13 x 10. Therefore, the solution is 10 3 2.
10 10 20
60
38. 15 20 30 25 5 30 10 45
2R2 3R1 R2 2R3 R1 R3
10 10 20
60
0 10 120 230 0 70 40 150
10 10
R3 7R2 R3
20
60
120 230 . 0 880 1760
0 10
0
Thus, 880z 1760 z 2; 10y 120 2 230 y 1; and 10x 10 40 60 x 1. Therefore, the solution is 1 1 2.
1 1
1 2
39. 0 1 3 1 2 1 5 0
R3 2R1 R3
1
1
1
2
0 1 3 1 0 1 3 4
R3 R2 R3
1 1
1
3
0 1 3 1 . The third row of the 0 0 0 3
matrix states 0 3, which is impossible. Hence, the system is inconsistent, and there is no solution.
1
0
3 3
40. 2 1 2 5 0 1 8 8
1
0
3
3
0 1 8 1 0 1 8 8
R2 2R1 R2
41.
2 3 9 5
R R 1 2 2 3 1 4 3 1 0 3 2 1 R R R 3 2 3 0 1 5 3 0 0 1 5 3 0 1
0
3
1
0
3
2
1 0
1
0
3
3
0 1 8 1 . 0 0 0 7
R3 R2 R3
The system is inconsistent, and there is no solution.
3
2
R2 2R1 R2 1 R2 3 0 3 15 9 2 3 9 5 R3 3R1 R3 3 1 4 3 0 1 5 3 0 3 2 1 5 3 . Therefore, this system has infinitely many solutions, given by x 3t 2 0 0 0
x 2 3t, and y 5t 3 y 3 5t. Hence, the solutions are 2 3t 3 5t t, where t is any real number.
SECTION 11.1 Matrices and Systems of Linear Equations
1 2 5 3 42. 2 6 11 1 3 16 20 26
1 2 5 3 0 2 1 7 0 10 5 35
R2 2R1 R2 R3 3R1 R3
R3 5R2 R3
733
1 2
5 3 2 1 7 . 0 0 0
0 0
The system is dependent; there are infinitely many solutions, given by 2y t 7 2y t 7 y 12 t 72 ; and x 2 12 t 72 5t 3 x t 7 5t 3 x 4t 10. The solutions are 4t 10 12 t 72 t , where t is any real number.
1 1
3
3
43. 4 8 32 24 2 3 11 4
1 1
3
3
0 4 20 12 0 1 5 2
R2 4R1 R2 R3 2R1 R3
14 R2
1 1
3
3
1 5 3 . The third row of the 0 0 5
0 R3 R2 R3
0
matrix states 0 5, which is impossible. Hence, the system is inconsistent, and there is no solution.
44.
2
6 2 12
1 3
2
3
2
1
1 3 1
6
1 3 1
1 R1 R2 R1 R2 0 2 10 1 3 2 10 R3 R1 R3 6 1 3 2 6 0
6
1 3 1 6
R 3R R 3 2 3 4 0 0 3 12 0 0 1
0 1 4 . 0 0 0
The system is dependent; there are infinitely many solutions, given by z 4, and x 3y z 6, so x 3t 4 6 x 3t 2. The solutions are 3t 2 t 4, where t is any real number.
1
4 2 3
45. 2 1 5 12 8 5 11 30
1
4 2 3
0 9 9 18 0 27 27 54
R2 2R1 R2 R3 8R1 R3
R3 3R2 R3
1
4 2 3
0 9 0 0
9 18 . 0 0
Therefore, this system has infinitely many solutions, given by 9y 9t 18 y 2 t, and x 4 2 t 2t 3 x 5 2t. Hence, the solutions are 5 2t 2 t t, where t is any real number.
3
2 3 10
1 1 1 5
1 1 1 5
R R R2 3R1 R2 1 2 0 5 0 25 46. 1 1 1 5 3 2 3 10 R3 R1 R3 1 4 1 20 1 4 1 20 0 5 0 25 1 1 1 5 0 5 0 25 . The system is dependent; there are infinitely many solutions, given by 5s 0 0 0 0 r 5 t 5 r t. Hence, the solutions are t 5 t, where t is any real number.
2
1 2 12
1 47. 1 6 1 2 3 3 18 3 2
R1 R2 R1
1 12 1
6
2 1 2 12 3 32 3 18
R2 2R1 R2 R3 3R1 R3
1 12 1 6
0 0 0 0
R3 R2 R3
25 s 5, and
0 0 . 0 0
Therefore, this system has infinitely many solutions, given by x 12 s t 6 x 6 12 s t. Hence, the solutions are 6 12 s t s t , where s and t are any real numbers.
0 1 5 7 R R 2 1 48. 3 2 0 12 3 0 10 80
3 2 0 12 3 R1 R3 0 1 5 7 R 3 0 10 80
Therefore, the system is inconsistent and has no solution.
3 2 0 12 3 2R2 R3 0 1 5 7 R 0 2 10 68
3 2 0 12 0 1 5 7 . 0 0 0 82
734
CHAPTER 11 Matrices and Determinants
4 3
1 8
49. 2 1 3 4 1 1 2 3 1 1 2 3 0 1 1 2 0 1 7 20
1 1
2
3
2 1 3 4 4 3 1 8 1 1 2 3 R3 R2 R3 0 1 1 2 0 0 6 18 R1 R3
1 1
2
3
0 1 1 2 0 1 7 20
R2 2R1 R2 R3 4R1 R3
R2
. Therefore, 6z 18 z 3; y 3 2
y 1; and x 1 2 3 3 x 2. Hence, the solution is 2 1 3.
2 3
5
14
1
1 1
3
1
1 1 3
2 4R1 R2 R1 R3 4 1 2 17 R 0 5 50. 4 1 2 17 R3 2R1 R3 R1 2 3 5 14 0 5 1 1 1 3
1
1 1 3
0
0
R R R 3 2 3 2 5 0 5
7 20
2 5 . 5 25
Thus 5z 25 z 5; 5y 2 5 5 5y 15 y 3; and x 3 5 3 x 1. Hence, the solution is 1 3 5.
2 1
3
9
51. 1 0 7 10 3 2 1 4
2 1
3
9
0 1 11 29 0 2 22 34
2R2 R1 R2 R3 3R2 R3
52.
4 1 36 24 1 2
9
1
6
2
3 6
4R2 R1 R2 R3 2R2 R3
4 1 36 24
0 9 72 36 0 3 24 12
3
9
0 1 11 29 . 0 0 0 24
2R2 R3 R3
Therefore, the system is inconsistent and there is no solution.
2 1
19 R2
3R3 R2 R3
4 1 36 24
1 8 4 . Therefore, 0 0 0
0 0
the system is dependent. Let z t. Then y 8t 4 y 4 8t and 4x y 36t 24
x 14 [24 4 8t 36t] 5 7t. The solutions are 5 7t 4 8t t where t is any real number.
1
2 3
5
53. 2 4 6 10 3 7 2 13
1 2
3 5
0 0 12 0 1 7
R2 2R1 R2 R3 3R1 R3
0 2
1 2
3 5
0 1 7 0 0 12
R2 R3
2 . Therefore, 0
12z 0 z 0; y 7 0 2 y 2; and x 2 2 3 0 5 x 9. Hence, the solution is 9 2 0.
3 1 0 2
54. 4 3 1 4 2 5 1 0
3R2 4R1 R2 2R3 R2 R3
3
1 0
2
0 13 3 20 0 13 3 4
R2 R3 R3
Therefore, the system is inconsistent and there is no solution.
1 1
6
0
1
55. 1 1
8
5 3 14 4
R2 R1 R2 R3 R1 R3
1 1
6
1
5
0 0
8
3 4 20 12
3
1 0
2
0 13 3 20 . 0 0 0 16
R3 4R2 R3
1 1
0 0
6
8
1 5 3 . Therefore, 0 0 0
the system is dependent. Let z t. Then y 5t 3 y 35t and x y 6t 8 x 83 5t6t 5t. The solutions are 5 t 3 5t t, where t is any real number.
SECTION 11.1 Matrices and Systems of Linear Equations
56.
3 1
2 1 1 7 1 3 2 1 1 3 2 1 0 2 1 1 0 10 7 11 4 2
735
1 3 2 1 1 3 2 1 R2 1 R3 2 4R1 R2 4 2 1 7 R 4 0 10 7 11 R3 3R1 R3 3 1 2 1 0 8 4 4 1 3 2 1 R3 5R2 R3 0 2 1 1 . Thus 2z 6 z 3; 2y 3 1 2y 2 0 0 2 6 R1 R3
R1
y 1; and x 3 1 2 3 1 x 2. Hence, the solution is 2 1 3. 1 2 1 3 3 1 2 1 3 3 R2 3R1 R2 3 4 1 1 9 3 4 1 1 9 R1 R R1 R3 57. 3 1 1 1 1 0 1 1 1 1 0 R4 2R1 R4 2 1 4 2 3 2 1 4 2 3 1 2 1 3 3 1 2 1 3 3 0 1 2 4 9 R 3R R 0 1 2 4 1 9 323 2 R2 0 3 0 4 3 R4 5R2 R4 0 0 6 8 24 0 5 6 8 9 0 0 4 12 36 1 2 1 3 3 0 1 2 4 9 . Therefore, 20 60 3; 6z 24 24 z 0 0 6 8 24 0 0 0 20 60
1 2 1
0 2 0 3 0 5
3 3
4 8 18 0 4 3 6 8
9
3R4 2R3 R4
0. Then y 12 9 y 3 and
x 6 9 3 x 0. Hence, the solution is 0 3 0 3. 1 1 1 1 6 1 1 1 1 6 R2 R4 R2 R2 2R1 R2 2 0 1 3 0 2 3 1 4 8 R3 R4 R3 R3 R1 R3 58. 1 1 1 0 4 10 0 2 1 5 16 R4 3R1 R4 2 R4 3 5 1 1 20 0 2 2 2 2 1 1 1 1 6 1 1 1 1 6 0 1 1 1 0 1 1 1 1 1 R4 R2 5 R 3 R R 4 3 4 . R3 R4 0 0 5 1 2 0 0 5 1 2 0 0 3 7 14 0 0 0 32 64
1 1 1 1
0 0 0 0 0 1
5 3 1
2 7 14 1 1 1
Thus 32 64 2; 5z 2 2 5z 0 z 0; y 0 2 1 y 3; and x 3 0 2 6 x 1. Hence the solution is 1 3 0 2. 1 1 2 1 2 1 1 2 1 2 1 1 2 1 2 0 3 1 2 2 0 3 1 2 2 0 3 1 2 2 R3 R1 R3 R R2 R4 4 59. 1 1 0 3 2 R4 3R1 R4 0 0 2 4 4 0 0 2 4 4 0 3 7 1 1 0 0 6 3 3 3 0 1 2 5 1 1 2 1 2 0 3 1 2 2 R4 3R3 R4 . Therefore, 9 9 1; 2z 4 1 4 z 0. Then 0 0 2 4 4 0 0 0 9 9 3y 0 2 1 2 y 0and x 0 2 0 1 2 x 1. Hence, the solution is 1 0 0 1.
6
736
CHAPTER 11 Matrices and Determinants
1 3 2
1
2
1 2 0 2 10 60. 0 0 1 5 15 3 0 2 1 3 1 3 0 1 R4 14R3 R4 0 0 0 0
R2 R1 R2 R4 3R1 R4
2
1
2
3
2
1 3
0 0 0
2
1 2
1 2 3 8 0 1 5 15 9 4 2
R4 9R2 R4
3
1 3
0 0 0
2
1 2
1 2 3 8 0 1 5 15
0 14 25 75
8 . Thus, 45 135 3; z 5 3 15 z 0; 1 5 15 0 45 135
y 2 0 3 3 8 y 1; and x 3 1 2 0 3 2 x 2. Hence, the solution is 2 1 0 3.
1 1
0 1 0
61. 3 0 1 2 0 1 4 1 2 0
1 1
0
1 0
0 3 1 1 0 0 3 1 1 0
R2 3R1 R2 R3 R1 R3
R3 R2 R3
1 1
0 0
0
1 0
3 1 1 0 . 0 0 0 0
Therefore, the system has infinitely many solutions, given by 3y s t 0 y 13 s t and x 13 s t t 0 x 13 s 2t. So the solutions are 13 s 2t 13 s t s t , where s and t are any real numbers.
2 1 2 1 5 R1 R2 62. 1 1 4 1 3 R1 3 2 1 0 0 1 1 4 1 3 R3 R2 R3 0 1 10 1 11 0 0 1 2 2
1 1 4 1 3
2 1 2 1 3 2 1 0
5 0
R2 2R1 R2 R3 3R1 R3
1 1 4
0 0
1 3
1 10 1 11 1 11 3 9
. Thus, the system has infinitely many solutions, given by z 2t 2
z 2 2t; y 10 2 2t t 11 y 31 19t; and x 31 19t 4 2 2t t 3 x 20 12t. Hence, the solutions are 20 12t 31 19t 2 2t t, where t is any real number.
1
0
1 1
4
1
0
1 1
4
1 0
1 1
4
0 1 1 0 4 0 1 1 0 4 0 1 1 0 4 R3 R1 R3 R 2 R R 3 2 3 63. 1 2 3 1 12 R4 2R1 R4 0 2 2 0 8 0 0 0 0 0 2 0 2 5 1 0 0 4 3 9 0 0 4 3 9 1 0 1 1 4 0 1 1 0 4 R3 R4 . Therefore, 4z 3t 9 4z 9 3t z 94 34 t. Then we have 0 0 4 3 9 0 0 0 0 0 3 9 3 7 7 y 94 34 t 4 y 7 4 4 t and x 4 4 t t 4 x 4 4 t. Hence, the solutions are 7 7 t 7 3 t 9 3 t t , where t is any real number. 4 4 4 4 4 4
SECTION 11.1 Matrices and Systems of Linear Equations
0 R2 R3 R2 0 R4 R3 R4 1 2 0 0 4 12 2 R3 2 0 2 5 6 1 0 0 2 6 0 R2 2R3 R2 0 1 1 2 R2 R3 0 0 2 9 18 0 0 2 9 18
64.
0 1 1 2
3 2
0 1
0 1 1 2 0 1 2 0 3 12 R1 R3 R 1 0 0 2 2 R1 R2 6 0 0 2 9 18 1 0 0 2 6 0 1 1 2 0 R4 R3 R4 0 0 2 9 18 0 0 0 0 0
737
1 0 0 2 6 0 2 0 5 18 0 1 1 2 0 0 0 2 9 18
.
Thus, the system has infinitely many solutions given by 2z 9t 18 z 92 t 9; y 92 t 9 2t 0 y 52 t 9; and x 2t 6 x 2t 6. So the solutions are 2t 6 52 t 9 92 t 9 t , where t is any real number.
075 375 295 40875
65. Using rref on the matrix 095 875 0
3375 , we find that the solution is x 125, y 025, z 075. 125 015 275 36625 131 272 371 139534
66. Using rref on the matrix 021 0 0
z 381.
42 31
49 27 52
6 67. Using rref on the matrix 35 0
373
0 42
04
45 , we find that the solution is x 12, y 34, z 52, 0 67 32 3488 31 48 52 766 0
0
9
0 145
0 27 0 43 1187 68. Using rref on the matrix 0 31 42 0 721 73 54 0 0 1327 32.
134322 , we find that the solution is x 371, y 172, 234 456 213984
13.
, we find that the solution is x 13, y 07, z 12,
69. Let x, y, z represent the number of VitaMax, Vitron, and VitaPlus pills taken daily. The matrix representation for the system of equations is 1 5 10 15 50 1 2 3 10 1 2 3 10 1 2 3 10 5 R1 1 2 2 3R1 R2 3 R2 R3 15 20 0 50 5R 3 4 0 10 R 0 2 9 20 R 1 0 2 9 20 . R3 2R1 R3 5 R3 10 10 10 50 2 2 2 10 0 2 4 10 0 0 5 10 Thus, 5z 10 z 2; 2y 18 20 y 1; and x 2 6 10 x 2. Hence, he should take 2 VitaMax, 1 Vitron, and 2 VitaPlus pills daily.
738
CHAPTER 11 Matrices and Determinants
70. Let x be the quantity, in mL, of 10% acid, y the quantity of 20% acid, and z the quantity of 40% acid. Then 1 1 1 100 10R2 01x 02y 04z 18 R2 R1 R2 Writing this equation in matrix form, we get x y z 100 01 02 04 18 R 3 R1 R3 x 4z 0 1 0 4 0 1 1 1 100 1 1 1 100 R3 R2 R3 0 1 3 80 0 1 3 80 . Thus 2z 20 z 10; y 30 80 y 50; and R3 0 1 5 100 0 0 2 20
x 50 10 100 x 40. So he should mix together 40 mL of 10% acid, 50 mL of 20% acid, and 10 mL of 40% acid.
71. Let x, y, and z represent the distance, in miles, of the run, swim, and cycle parts of the race respectively. Then, since distance , we get the following equations from the three contestants’ race times: time speed y x z 10 4 20 25 2x 5y z 50 x y z 3 which has the following matrix representation: 4x 5y 2z 90 75 6 15 x y z 175 8x 40y 3z 210
15
2
3
5 1
50
4 5 2 90 8 40 3 210
40
R2 2R1 R2 R3 4R1 R3
2
5
1
50
0 5 0 10 0 20 1 10
2
5
1
50
0 5 0 10 . 0 0 1 30
R3 4R2 R3
Thus, z 30 z 30; 5y 10 y 2; and 2x 10 30 50 x 5. So the race has a 5 mile run, 2 mile swim, and 30 mile cycle.
72. Let a, b, and c be the number of students in classrooms A, B, and C, respectively, where a, b, c 0. Then, the a b c 100 a b c 100 1a 1b system of equations is By substitution, a 52 a 32 a 100 a 20; b 52 a 2 5 c 3b 3a 1b 1c 5 3 5 2 b 52 20 50; and c 32 20 30. So there are 20 students in classroom A, 50 students in classroom B, and 30 students in classroom C.
73. Let t be the number of tables produced, c the number of chairs, and a the number of armoires. Then, the system of equations 1 t 2c 2a 600 2 t c a 300 1 3 and a matrix representation is is 2 t 2 c a 400 t 3c 2a 800 2t 3c 4a 1180 t 3 c 2a 590
2
1 2 2
600
1 3 2 800 2 3 4 1180
R2 R1 R2 R3 2R1 R3
1
2 2 600
0 1 0 200 0 1 0 20
R3 R2 R3
1 2 2 600
0 1 0 200 . 0 0 0 180
The third row states 0 180, which is impossible, and so the system is inconsistent. Therefore, it is impossible to use all of the available labor-hours.
SECTION 11.1 Matrices and Systems of Linear Equations
739
74. The number of cars entering each intersection must equal the number of cars leaving that intersection. This leads 200 180 x z x 70 20 to the following equations: Simplifying and writing this in matrix form, we get 200 y 30 y z 400 200 1 0 1 0 380 1 0 1 0 380 1 0 1 0 380 1 0 0 1 50 R R R 0 0 1 1 430 0 1 0 1 170 2 1 2 R R 2 3 . 0 1 0 1 170 R4 R3 R4 0 1 0 1 170 0 0 1 1 430 0 1 1 0 600 0 0 1 1 430 0 0 1 1 430
Therefore, z t 430 z 430 t; y t 170 y 170 t; and x 430 t 380 x t 50. Since x, y, z, 0, it follows that 50 t 430, and so the solutions are t 50 170 t 430 t t, where 50 t 430.
75. Line containing the points 0 0 and 1 12: Using the general form of a line, y ax b, we substitute for x and y and solve for a and b. The point 0 0 gives 0 a 0 b b 0; the point 1 12 gives 12 a 1 b a 12. Since a 12 and b 0, the equation of the line is y 12x. Quadratic containing the points 0 0, 1 12, and 3 6: Using the general form of a quadratic, y ax 2 bx c, we substitute for x and y and solve for a, b, and c. The point 0 0 gives 0 a 02 b 0 c c 0; the point 1 12
gives 12 a 12 b 1 c a b 12; the point 3 6 gives 6 a 32 b 3 c 9a 3b 6. Subtracting the third equation from 3 times the third gives 6a 30 a 5. So a b 12 b 12 a b 17. Since a 5, b 17, and c 0, the equation of the quadratic is y 5x 2 17x.
Cubic containing the points 0 0, 1 12, 2 40, and 3 6: Using the general form of a cubic, y ax 3 bx 2 cx d,
we substitute for x and y and solve for a, b, c, and d. The point 0 0 gives 0 a 03 b 02 c 0 d d 0; the point the point 1 12 gives 12 a 13 b 12 c 1 d a b c d 12; the point 2 40 gives
40 a 23 b 22 c 2 d 8a 4b 2c d 40; the point 3 6 gives 6 a 33 b 32 c 3 d a b c 12 27a 9b 3c d 6. Since d 0, the system reduces to 8a 4b 2c 40 which has representation 27a 9b 3c 6
1 1 1 12
1
1
1
12
12 R2
1 1 1 12
1 1
1 12
R 3R R 2 8R1 R2 3 2 3 8 4 2 40 R 0 4 6 56 0 2 3 28 2 3 28 0 2 . R3 1 27R1 R3 6 R3 0 18 24 318 0 3 4 53 0 0 1 22 27 9 3 6
So c 22 and back-substituting we have 2b 3 22 28 b 47 and a 47 22 0 a 13. So the
cubic is y 13x 3 47x 2 22x. Fourth-degree polynomial containing the points 0 0, 1 12, 2 40, 3 6, and 1 14: Using the general form of a
fourth-degree polynomial, y ax 4 bx 3 cx 2 dx e, we substitute for x and y and solve for a, b, c, d, and e. The point
0 0 gives 0 a 04 b 03 c 02 d 0e e 0; the point 1 12 gives 12 a 14 b 13 c 12 d 1e;
the point 2 40 gives 40 a 24 b 23 c 22 d 2e; the point 3 6 gives 6 a 34 b 33 c 32 d 3e; the point 1 14 gives 14 a 14 b 13 c 12 d 1 e.
740
CHAPTER 11 Matrices and Determinants
Because the first equation is e 0, we eliminate e from the other equations to get a b c d 12 1 1 1 1 12 1 1 1 1 16a 8b 4c 2d 40 16 8 4 2 40 R2 16R1 R2 0 8 12 14 R3 81R1 R3 6 6 81a 27b 9c 3d 81 27 9 3 R4 R1 R4 0 54 72 78 a b c d 14 1 1 1 1 14 0 2 0 2 1 1 1 1 12 1 1 1 1 12 1 1 0 1 0 1 13 R3 8R2 R3 0 1 0 1 13 R4 6R3 R4 0 1 0 8 12 14 152 R4 54R2 R4 0 0 12 6 48 0 0 0 54 72 78 966 0 0 72 24 264 0 0
So d 2. Then 12c 6 2 48 c 3and b 2 13 b 11.
Finally, a 11 3 2 12 a 4. So the fourth-degree polynomial
12
1 R4 R2 2 152 R2 R3 966 R3 R4 26 1 1 12 0 1 13 . 12 6 48 0 12 24
40
containing these points is y 4x 4 11x 3 3x 2 2x.
20
-2
2
4
-20
11.2 THE ALGEBRA OF MATRICES 1. We can add (or subtract) two matrices only if they have the same dimension. 2. (a) We can multiply two matrices only if the number of columns in the first matrix is the same as the number of rows in the second matrix. (b) If A is a 3 3 matrix and B is a 4 3 matrix, then (ii) B A and (iii) A A are defined, but (i) AB and (iv) B B are not.
3. (i) A A and (ii) 2A exist for all matrices A, but (iii) A A is not defined when A is not square.
4. The entry in the first row, second column is a12 3 3 1 2 2 1 9; the entry in the second row, third column is a23 1 2 2 1 0 0 0; the entry in the third row, first column is a31 1 1 3 3 2 2 4, and 3 1 2 1 3 2 4 9 7 so on: 1 2 0 3 2 1 7 7 0 1 3 2 2 1 0 4 5 5 5. The matrices have different dimensions, so they cannot be equal. 6. Because 14 025, ln 1 0, 2 4, and 3 62 , the corresponding entries are equal, so the matrices are equal.
7. All corresponding entries must be equal, so a 5 and b 3.
8. All corresponding entries must be equal, so a 5 and b 7. 2 6 1 3 1 3 0 1 1 2 1 1 2 0 2 9. 10. 5 3 6 2 1 5 1 1 0 1 3 2 0 2 2 1 1 0 1 1 1 2 3 6 11. 3 12. 2 1 0 1 2 1 is undefined because these 4 1 12 3 0 1 1 3 1 1 0 3 0 matrices have incompatible dimensions.
SECTION 11.2 The Algebra of Matrices
2 6 1 2 3 6 is undefined because these 13. 1 3 2 4 2 0
14. 6 3 4
matrices have incompatible dimensions.
2 1 2
2 3
1 2 6 7 6 0
1 2 3 2
741
7
5 1 1 1 4 2 2 1 7 10 7 1 7 1 2 1 2 5 4 6 1 2 1 1 12 1 17. 2X A B X 2 B A . 2 2 1 2 3 7 2 4 1 3
15.
1 2
1 2
3
5
2
1
16. 0
18. 3X B C 3X C B. but C is a 3 2 matrix and B is a 2 2 matrix, so C B is impossible. Thus, there is no solution. 19. 2 B X D. Since B is a 2 2 matrix, B X is defined only when X is a 2 2 matrix, so 2 B X is a 2 2 matrix. But D is a 3 2 matrix. Thus, there is no solution. 20. 5 X C D X C 15 D 10 20 2 3 2 4 2 3 4 7 1 X 15 D C 30 20 1 0 6 4 1 0 7 4 . 5 10 0 0 2 2 0 0 2 2 2 21. 15 X D C X D 5C 2 3 10 20 10 15 X 5C D 5 1 0 30 20 5 0 0 2 10 0 0 10 22. 2A B 3X 2A B 3X
10 20
0
5
30 20 25 20 . 10 0 10 10
X 13 2A B 1 8 12 2 5 1 6 7 2 73 1 4 6 2 5 2 1 1 3 3 3 1 1 1 3 3 7 2 6 3 7 3 3
In Solutions 23–36, the matrices A, B, C, D, E, F, G, and H are defined as follows: 0 2 5 2 52 3 12 5 A C B 1 1 3 0 2 3 0 7 1 1 0 0 5 3 10 E F G 2 0 1 0 6 1 0 0 0 0 1 5 2 2 0 3 12 5 2 52 5 2 5 23. (a) B C 1 1 3 0 2 3 1 1 0
(b) B F is undefined because B 2 3 and F 3 3 don’t have the same dimensions. 0 3 12 5 1 3 5 2 52 24. (a) C B 0 2 3 1 1 3 1 3 6
D
7 3
3
1
H
2 1
742
CHAPTER 11 Matrices and Determinants
0 2 52 3 12 5 14 8 30 6 (b) 2C 6B 2 0 2 3 1 1 3 6 10 24
2 5
25. (a) 5A 5
0
7
10 25 0
35
(b) C 5A is undefined because C 2 3 and A 2 2 don’t have the same dimensions. 2 52 0 13 72 15 3 12 5 2 26. (a) 3B 2C 3 1 1 3 0 2 3 3 1 3
(b) 2H D is undefined because 2H 2 2 and D 1 2 don’t have the same dimensions.
27. (a) AD is undefined because A 2 2 and D 1 2 have incompatible dimensions. 2 5 14 14 (b) D A 7 3 0 7 28. (a) D H
7 3
3
1
2 1
27 4
(b) H D is undefined because H and D have incompatible dimensions. 3 1 4 7 2 5 29. (a) AH 2 1 14 7 0 7 3 1 2 5 6 8 (b) 2 1 0 7 4 17
30. (a) BC is undefined because B 2 3 and C 2 3 have incompatible dimensions. 1 0 0 1 3 12 5 3 2 5 0 1 0 (b) B F 1 1 3 1 1 3 0 0 1
31. (a) G F
(b) G E
5 3 10 6
1
5
2
5 3 6
1
5
2
0 2 10 0 2
1 0 0
5 3 10
0 1 0 6 0 0 1 5 1 1 2 8 0 1
32. (a) B 2 is undefined because B 2 3 is not square. 1 0 0 1 0 0 1 0 0 (b) F 2 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1
33. (a) A2
2 5 0
7
2 5 0
7
4 45 0
49
1 2
0 2
SECTION 11.2 The Algebra of Matrices
(b) A3
2 5 0
34. (a) D A B (b) D AB
35. (a) AB E
7
0
7
0
2 7 3 0 7
2 5
7 3
2 5 0
2 5
3 1
2 5
0
7 3
4 45 0
49
2 5 0
7
1 5 3 2 14 14
8 335 0
343
743
1 5 2 28 21 28
1 1 3 1 6 5 5 3 12 5 7 3 28 21 28 1 1 3 7 7 21 7 1 1 1 5 1 6 5 13 2 2 2 1 3 7 7 21 7 0 0 7
1 1 3
(b) AH E is undefined because the dimensions of AH 2 2 and E 3 1 are incompatible. 5 2 5 0 2 52 3 12 5 7 3 38 11 35 36. (a) D BDC D B C 7 3 1 1 3 0 2 3 1 1 0 (b) B F F E is undefined because the dimensions 2 3 3 3 2 3 and 3 3 3 1 3 1 are incompatible.
In Solutions 37–42, the matrices A, B, and C are defined as follows: 12 01 03 11 24 02 02 01 C B A 0 05 09 01 04 11 21 21 05 21 07 03 05 156 562 37. AB 128 088 38. B has 2 columns and A has 3 rows, so B A is undefined. 109 097 035 003 033 019 029 39. BC 40. C B 055 105 105 027 325 241 431 446 06 094 004 41. B and C have different dimensions, so B C is undefined. 42. A2 112 192 01 041 095 131 x 2 2y 2 x 2y 2 2 . Thus we must solve the system So x 2 and 2y 2 y 1. Since 43. 4 2x 4 6 2x 6y 6 6y these values for x and y also satisfy the last two equations, the solution is x 2, y 1.
3x x y 6 9 x y 3x 3y 3y . Since 3 , we must solve the system 44. 3 y x 9 6 y x 3y 3x 3x 3y
x 2 and 3y 9 y 3. Thus, the solution is x 2, y 3.
6
9
6
9
So 3x 6
744
CHAPTER 11 Matrices and Determinants
45. 2
x
y
2 4
x
y
2x
2y
. Since 2 , Thus we must solve the xy xy 2 6 xy xy 2 x y 2 x y 2x 2 2y 4 So x 1 and y 2. Since these values for x and y also satisfy the last two equations, the system 2 x y 2 2 x y 6
solution is x 1, y 2. x y y x 4 4 x y yx 4 4 . Thus we must solve the system 46. y x x y 6 6 x y x y 6 6 xy 4 y x 4 Adding the first equation to the last equation gives 2x 10 x 5 and 5 y 6 y 1. So the x y 6 xy 6
solution is x 5, y 1. 2x 5y 7 2 5 x 7 . 47. written as a matrix equation is 3x 2y 4 3 2 y 4 6x y z 12 48. 2x z 7 y 2z 4
2 2 2 2
1
x
12
0
0
written as a matrix equation is 1 0 1 0 0 3 1 1
written as a matrix equation is
1 y 7 . 4 1 2 z
written as a matrix equation is 2
3x1 2x2 x3 x4 0 49. x1 x3 5 3x2 x3 x4 4 x y z 4x 2y z 50. x y 5z x y z
6 1
1 1
3 2 1
1
1
x1 x2 x3 x4 2
x 4 2 1 2 y 2 1 1 5 z 1 1 1 2 1
0 5 . 4
.
0 . ABC is undefined because the dimensions of A (2 4) 1 7 9 2 , and C 1 2 12 4 0 2 3 3 21 27 6 1 7 9 2 . B AC is undefined because and B (1 4) are not compatible. AC B 2 2 14 18 4
51. A
1 0 6 1
, B
the dimensions of B (1 4) and A (2 4) are not compatible. BC A is undefined because the dimensions of C (4 1) and A (2 4) are not compatible. C AB is undefined because the dimensions of C (4 1) and A (2 4) are not compatible. C B A is undefined because the dimensions of B (1 4) and A (2 4) are not compatible.
SECTION 11.2 The Algebra of Matrices
52. (a) Let A
a b c d
and B
e f g h
. Then A B
A2
a b
AB
a b
c d
c d
cg d h
cg d h
a e2 b f c g
c g a e d h c g
a b
e f g h
A2 AB B A B 2
c d
cg d h
ae b f ae b f A B2
ae b f
ac cd bc d 2
ae bg a f bh ce dg c f dh
, and
a e b f b f d h
a 2 bc ab bd
c g b f d h2
; B 2
; B A
e f g h e f g h
e f g h a b c d
;
e2 f g e f f h
ae c f be d f
eg gh f g h 2
ag ch bg dh
a 2 bc ae bg ae c f e2 f g ab bd e f f h a f bh be d f
ac cd eg gh ce dg ag ch bc d 2 f g h 2 c f dh bg dh a 2 2ae e2 b c g f c g
745
;
. Then
a b f e b f b d h f d h
c b f g b f d 2 2dh h 2 a e b f b f d h a e2 b f c g A B2 c g a e d h c g c g b f d h2 c a e g a e d c g h c g
(b) No. From part (a), A B2 A B A B A2 AB B A B 2 A2 2AB B 2 unless AB B A which is not true in general, as we saw in Example 5. 5 075 010 0 4 53. (a) AB 025 070 070 20 22 7 0 020 030 10
(b) Five members of the group have no postsecondary education, 22 have 1 to 4 years, and seven have more than 4 years. 075 020 005 80 96 54. (a) AB 060 030 010 170 103 040 030 030 40 95
(b) 96 students slept less than 4 hours, 103 slept 4 to 7 hours, and 95 slept more than 7 hours. 50 20 15 350 35375 55. (a) AB 40 75 20 575 65625 35 60 100 425 89250 (b) The total revenue for Monday is the 1 1th entry of the product matrix, $35375.
(c) The total revenue is the sum of the three entries in the product matrix, $190250. 4000 1000 3500 $4690 $1690 $13,210 56. (a) B A $090 $080 $110 400 300 200 700 500 9000
746
CHAPTER 11 Matrices and Determinants
(b) The entries in the product matrix represent the total food sales in Santa Monica, Long Beach, and Anaheim, respectively.
12 10
57. (a) AB 4 8
0
$1000
$500
$32,000 $18,000
4 20 $2000 $1200 $42,000 $26,800 9 12 $1500 $1000 $44,000 $26,800
(b) The daily profit in January from the Biloxi plant is the 2 1 matrix entry, namely $42,000. (c) The total daily profit from all three plants in February was $18,000 $26,800 $26,800 $71,600.
58. (a) AB
2000 2500
3000 1500 105,000 58,000 6 10 14 28 2500 1000 1000 500
(b) That day they canned 105,000 ounces of tomato sauce and 58,000 ounces of tomato paste.
120 50 60
010
9700
59. (a) AC 40 25 30 050 4650 60 30 20 100 4100
Amy’s stand sold $97 worth of produce on Saturday, Beth’s stand
sold $4650 worth, and Chad’s stand sold $41 worth. 100 60 30 010 7000 (b) BC 35 20 20 050 3350 Amy’s stand sold $70 worth of produce on Sunday, Beth’s stand 60 25 30 100 4850
sold $3350 worth, and Chad’s stand sold $4850 worth. 220 110 90 100 60 30 120 50 60 (c) A B 40 25 30 35 20 20 75 45 50 120 55 50 60 25 30 60 30 20
This represents the melons, squash, and
tomatoes they sold during the weekend. 120 50 60 100 60 30 010 220 110 90 010 16700 (d) A B C 40 25 30 35 20 20 050 75 45 50 050 8000 60 30 20 60 25 30 100 120 55 50 100 8950
During the weekend, Amy’s stand sold $167 worth, Beth’s stand sold $80 worth, and Chad’s stand sold $8950 worth of 9700 7000 16700 produce. Notice that A B C AC BC 4650 3350 8000 . 4100 4850 8950
60. (a)
1 0 1 0 1 1
0 3 0 1 2 1 1 2 0 0 3 0 1 3 2 3 2 0 0 3 0 0 2 1 1 2 0 1 3 1
(b)
2 1 2 1 2 2
1 3 1 2 3 2 2 3 1 1 3 1 2 3 3 3 3 1 1 3 1 1 3 2 2 3 1 2 3 2
(c)
2 3 2 3 2 2
3 0 3 2 1 2 2 1 3 3 0 3 2 0 1 0 1 3 3 0 3 3 1 2 2 1 3 2 0 2
SECTION 11.2 The Algebra of Matrices
(d)
0 0 0 0 0 0
747
0 3 0 0 3 0 0 3 0 0 3 0 0 3 3 3 3 0 0 3 0 0 3 0 0 3 0 0 3 0
(e)
61. Suppose A is n m and B is i j. If the product AB is defined, then m i. If the product B A is defined, then j n. Thus if both products are defined and if A is n m, then B must be m n.
62. A
1 1 0 1
0 1 0 1 0 1 0 1 0 1 1 1 1 3 1 n 1 4 . Therefore, it seems that An . A4 A A3 0 1 0 1 0 1 0 1
1 1 1 1 1 1 1 2 1 2 1 3 ; A2 ; A3 A A2 ; 0 1
1 1
1 1
2 2
1 1
2 2
4 4
; A3 A A2 ; 1 1 1 1 1 1 2 2 2 2 4 4 n1 2n1 1 1 4 4 8 8 2 . From this pattern, we see that An . A4 A A3 1 1 4 4 8 8 2n1 2n1
63. A
1 1
; A2
1 1
748
CHAPTER 11 Matrices and Determinants
64. Let A
a b c d
a b c d
.
For the first matrix, we have A2
a 2 bc ab bd
ac cd bc d 2
a 2 bc b a d
c a d bc d 2
a b
c d
4 0 . So A2 0 9
a 2 bc 4 b a d 0
c a d 0 bc d 2 9
If a d 0, then a d, so 4 a 2 bc d2 bc d 2 bc 9, which is a contradiction. Thus a d 0. Since b a d 0 and c a d 0, we must have b 0 and c 0. So the first equation becomes a 2 4 a 2, and the
fourth equation becomes d 2 9 d 3. 4 0 2 0 2 0 2 0 2 0 are A1 , A2 , A3 , and A4 . Thus the square roots of 0 9 0 3 0 3 0 3 0 3 a 2 bc 1 b a d 5 1 5 For the second matrix, we have A2 Since a d 0 and c a d 0, we must have c a d 0 0 9 bc d 2 9 a2 1 a 1 c 0. The equations then simplify into the system b a d 5 b a d 5 d2 9 d 3 We consider the four possible values of a and d. If a 1 and d 3, then b a d 5 b 4 5 b 54 . If a 1 and d 3, then b a d 5 b 2 5 b 52 . If a 1 and d 3, then b a d 5 b 2 5 1 5 are b 52 . If a 1 and d 3, then b a d 5 b 4 5 b 54 . Thus, the square roots of 0 9 1 54 1 52 1 52 1 54 , A2 , A3 , and A4 . A1 0 3 0 3 0 3 0 3
11.3 INVERSES OF MATRICES AND MATRIX EQUATIONS
1. (a) The matrix I
1 0 0 1
is called an identity matrix.
(b) If A is a 2 2 matrix then A I A and I A A.
(c) If A and B are 2 2 matrices with AB I then B is the inverse of A.
2. (a)
5 3 3 2
x y
4 3
d b 2 3 2 3 1 1 . (b) The inverse of A is A1 ad bc c a 5 2 3 3 3 5 3 5
SECTION 11.3 Inverses of Matrices and Matrix Equations
(c) The solution of the matrix equation is X A1 B
2 3 5
3
(d) The solution of the system is x 1, y 3.
4 1
3. A
2 1
4
3
1 3
749
.
; B . 7 2 7 4 1 0 4 1 2 1 1 0 2 1 4 1 . and B A AB 0 1 7 2 7 4 0 1 7 4 7 2
4. A
7 3 2 . ; B 2
2 3
2 1 7 3 2 3 1 0 2 2 . BA 2 1 4 7 0 1
5. A
4 7
1
3 1
1
4
8 3
AB
4
0 ; B 2 1 1 . 1 3 2 1 0 1 8 3 4 1 3 1 1 BA 2 1 1 1 4 0 0 1 0 1 1 3 2 0
3 2
4
9 10 8
6. A 1 1 6 ; B 12 14 11 . 1 1 2 1 12 12 2 2 9 10 8 3 2 4 1 1 1 6 0 BA 12 14 11 1 1 12 2 1 12 0 2 2
2 3 4 7
AB 0 0
8 3
4
9 10 8
7 3 2 1 0 and 2
2 1
1
3 1
1
4
1 3
0 2 2 1
0 1
1 0 0
1 1 0 1 0 and 0 0 1 0 1
1 0 0 1
3 2
4
AB 1 1 6 12 2 1 12 12 0 0 1 0 . 0 1
1 0 0
14 11 0 1 0 and 1 1 0 0 1 2 2
2 4 7 4 1 2 1 0 1 2 1 . Then, A A1 A1 7. A 14 12 3 7 3 2 32 72 0 1 3 2 32 27 1 2 7 4 1 0 . and A1 A 3 7 2 2 3 2 0 1
7 4
750
CHAPTER 11 Matrices and Determinants
1
8. B
3 2
2 2 We begin with a 3 6 matrix whose left half is B and whose right half is I3 . 2 1 0 1 3 2 1 0 0 1 3 2 1 0 0 1 3 2 1 0 0 R 2R R 2R 5R R 2 3 0 2 2 0 1 0 313 0 2 2 0 1 0 3 0 2 2 0 1 0 2 1 0 0 0 1 0 5 4 2 0 1 0 0 2 4 5 2 0 1 3 2 1 0 0 1 0 1 1 32 1 R1 3R2 R1 R R R 2 R2 1 3 1 0 1 1 0 1 0 1 0 2 2 0 1 R2 2 R3 R2 12 R3 0 0 1 2 52 1 0 0 1 2 52 1 1 1 1 1 1 1 1 0 0 1 1 1 1 3 2 1 0 0 0 1 0 2 2 1 . Then B 1 2 2 1 ; B 1 B 2 2 1 0 2 2 0 1 0 ; 2 52 1 2 52 1 0 0 1 2 52 1 2 1 0 0 0 1 1 3 2 1 1 1 1 0 0 and B B 1 0 2 2 2 2 1 0 1 0 . 2 1 0 2 52 1 0 0 1 1 1 2 and verify that A1 A A A1 I . 9. Using a calculator, we find A1 3 2 2 2 1 1 0 1 1 10. Using a calculator, we find B 1 33 31 3 and verify that B B B B I3 . 13 12 1 1 3 5 3 5 3 5 1 11. 9 10 2 3 2 3 2 3
12.
3 4
13.
2
7 9
0
1
5
5 13
9 4 9 4 9 4 1 27 28 7 3 7 3 7 3
1
13 5 13 5 1 26 25 5 2 5 2
5 4 53 43 1 14. 35 32 8 7 83 73 8 5 1 4 3 6 3 1 , which is not defined, and so there is no inverse. 15. 24 24 8 6 8 4
1 16. 2
17.
4
7
1
1 1 3
5 4
04 12 03
06
4 13 12 1 1 2 53 5 15 32 2
1
1
06 12 1 2 1 024 036 03 04 12 23
SECTION 11.3 Inverses of Matrices and Matrix Equations
4 18. 3 1 1 0 0
2 3 1 0 0
4 2 3 1 0 0 R1 R3 R1 0 6 1 3 4 0 3R3 R2 R3 0 2 1 1 0 4 0 1 0 0 1 1 0 0 3 2 5 1 0 1 R1 R3 R1 0 1 6 R2 0 6 0 6 6 1 3 4 0 6 6 R2 R3 R2 0 1 3 2 6 0 0 1 3 2 6 0 0 3 2 5 Therefore, the inverse matrix is 1 1 1 . 3 2 6
3 2 0 1 0 0 1 0 0 1
2 4
4R2 3R1 R2 4R3 R1 R3
1 1 0 0
19. 1 1 1 0 1 0 1 4 0 0 0 1 6 R1 5R3 R1 0 R2 R3 R2 0
2 4
1
1 0 0
4 0 4 0 0 4 0 6 1 3 4 0 0 0 2 6 4 12 0 3 2 5 0 1 1 1 . 1 3 2 6
6 0
5
751 1 4 R1
1 2 R3
1 4 0
3R3 2R2 R3 0 6 1 1 2 0 0 6 1 1 2 0 3R1 2R2 R1 0 4 1 1 0 2 0 0 1 5 4 6 0 0 24 24 30 1 0 0 4 4 5 1 6 R1 0 1 0 1 1 1 . 6 0 6 6 6 1 6 R2 , R3 0 1 5 4 6 0 0 1 5 4 6 4 4 5 Therefore, the inverse matrix is 1 1 1 . 5 4 6
5
7 4 1 0 0
2R2 R1 R2 2R3 R1 R3
1
0 1 1 0 1
R3 R1 R3 3 1 3 0 1 0 20. 3 1 3 0 1 0 R1 R3 6 7 5 0 0 1 5 7 4 1 0 0 1 0 1 1 0 1 1 R3 R1 R3 7R2 R3 0 1 0 3 1 3 R R 2 , R3 0 0 1 27 7 26 26 7 25 Therefore, the inverse matrix is 3 3 1 . 27 7 26
1 2 3 1 0 0 21. 4 5 1 0 1 0 1 1 10 0 0 1
R2 4R1 R2 R3 R1 R3
1
0
1 1 0
0 1 0 0 7 1 1 0 0 26 7 25 0 1 0 3 1 3 . 0 0 1 27 7 26 R2 3R1 R2 R3 5R1 R3
1 2 3 1 0 0 0 3 13 4 1 0 0 3 13 1 0 1
R3 R2 R3
Since the left half of the last row consists entirely of zeros, there is no inverse matrix.
1
3 1 3 6 0 5
1 2 3 1 0 0 0 3 13 4 1 0 . 0 0 0 3 1 1
752
CHAPTER 11 Matrices and Determinants
2 1 0 1 0 0
22. 1 1 4 0 1 0 2 1 2 0 0 1 1 0 4 1 0 1 8 1 0 0 2 1
1 1 4 0 1 0
2 1 0 1 0 0 2 1 2 0 0 1 1 0 1 0 R1 2R3 R1 0 1 2 0 R2 4R3 R2 0 1 0 0 1 1 2 Therefore, the inverse matrix is 3 2 4 . 12 0 12
0 2 2 1 0 0
R1 R2
1 2 3 0 0 1
R R 1 3 23. 3 1 3 0 1 0 3 0 1 2 3 0 0 1 1 0 1 1 0 1 R1 R3 R1 0 1 0 3 1 3 R2 3R3 R2 0 2 2 1 0 0 1 0 1 1 0 1 1 R3 R1 0 1 0 3 1 3 R 0 0 1 72 1 3 92 1 4 3 1 3 . 7 1 3 2
3 2 0 1 0 0
24. 5 2 1 0 0
1
R2 2R1 R2 R3 2R1 R3
0 1 1 2
0 3 2 4 2 1 0 1
1 3 0 1 0 2 2 1 0 0
1
1
4 0
0 1 8 0 1 6 1 R2 0 1 2 R3 0
1 0
1 2 0 0 2 1
R1 R2 R1 R3 R2 R3
0 0 1 1
1 0
0 1 12
3 0 0
0 0
1 0 1
2 4 . 1 0 2
3
1 2
2
1
0 7 6 0 1 3 0 2 2 1 0 0 1 0 1 1 0 1 1 R3 2R2 R3 0 1 0 3 1 3 2 R3 0 0 2 7 2 6 1 0 0 92 1 4 0 1 0 3 1 3 . Therefore, the inverse matrix is 7 0 0 1 1 3 2
0 0 1 0 1
R2 3R1 R2
1
R2 5R1 R2 R R R 1 3 1 0 1 1 5 1 5 1 1 0 1 0 5 1 1 0 1 0 R3 2R1 R3 2 2 0 0 0 1 0 2 0 2 0 3 2 0 0 0 1 0 0 1 0 1 1 0 0 1 0 1 1 1 2 R3 3 1 1 5 1 5 0 1 0 1 0 2 . Therefore, the inverse matrix is 1 R2 R3 R2 0 0 1 6 1 13 6 0 2 12 2 13 2
R3 2R2 R3
0 1
0 32 . 13 1 2
SECTION 11.3 Inverses of Matrices and Matrix Equations
1 2 0 3 1 0 0 0 1 0 1 1 1 0 1 0 0 R R R 0 3 2 3 25. 0 1 0 1 0 0 1 0 R4 R1 R4 0 1 2 0 2 0 0 0 1 0 1 2 0 3 1 0 0 0 0 1 1 1 0 1 0 0 R 2R R 121 R2 R3 R2 0 0 1 0 0 1 1 0 0 0 0 1 1 0 0 1 1 0 0 1 1 0 2 0 0 1 0 0 1 0 1 1 R1 R1 R4 0 0 1 0 0 1 1 0 0 0 0 1 1 0 0 1 0 0 2 1 1 0 1 1 . 0 1 1 0 1 0 0 1
1 0 0 1 26. 1 1 1 1
1 0 1 0 0 0
0 1 0 1 0 0 4 R1 R4 R R3 R1 R3 1 0 0 0 1 0 1 1 0 0 0 1
Therefore, there is no inverse matrix.
27.
29.
3
2
3
3 0 2
0 1 1
1 4 1 0 2
0
1
2 3
1
1 0 0 0
1
0 2 0 0 33. 0 0 4 0 0 0 0 7
3 1 2
2
1 0 0 0
0 1 0 0 2 0 0 1 0 4 0 0 0 17
1
1
0
0 1
0
1 0
0 0 0
1 0 0 0 1 1 0
0 1 1
0 0 1
1 0 2 1 1 2 0 1 0 1 0 0 0 1 1 0 0
R3 R4
0
0 0 1 1 0
0
0 1
1
1 0 0 0
0
0
5
2
5
1
3
28. 30.
1
1
1 0 0 1 0 0 0 0
1
0 0 1 2 0 1
1
3
1
2 3 1 0 0 0
1 0
1
2 5 0 0 32. 4 2 3 0 5 1 2 1 0
0
1
0
1
0
1
0 0 0
2 5 15
1 3 25 25 2 1 5 5 1 5 35
1 16 16 1 1 0 0 4 2 1 1 3 3 3 13 1 2 56 6 6 5 6
1
0
1 0
0
0 0
1 2 0 0 5 5 16 2 1 0 15 3 15 1 2 1 37 15 15 3
1 0 0 . 0 1 1 1 1 0 0 0 1 1 0 1
2 25
0
1
0 0 0 3 0 2
1
1
1 1 0 1
3 1
34.
0 1 2R3 R1 R R 2 R4 R2 1 1 0 0 0 1 0 2 1 0 1 1 . Therefore, the inverse matrix is 1 1 0 0 0 1 0
0 1 0 1 0 0 4 R2 R4 R R3 R2 R3 0 0 1 0 1 0 0 1 1 0 0 1
0 1 0 1 2 2 2 1 2 1 1 1 0
3
1
0 1 0 0 0 0 1 0 0 1 0 0 0 0
0
1 0
4 3 3
1 1 72 6 1 1 0 2 6 1 0 0 3
1
1 0 0 1 0 1 0 1
1 1 3 1 2 1 3 3
0 4 1 2 2 2 0 0 1
1 7 3 31. 0 2 1 0 0 3
1
2
0 0
0 0 13 1 2
753
754
CHAPTER 11 Matrices and Determinants
In Solutions 35–38, the matrices A and B are defined as follows: 1 0 2 A 0 2 1 4 2 1
14
3 3 4 4 7 23 3 35. A1 B 16 16 16 7 1 5 8 8 8
7 3 4
22 2 37. B AB 1 7 7 50 7
26 7
16 7 37 7
B
2 1 2 0
3
1
0
1 2
0
0
1
25 7
13 7
22 7
1 5 2 36. AB 1 7 7 7
4 8 13 7 7 7 1 8 17 6 38. B AB 7 7 7 31 9 10 7 7 7
3x 5y 4 3 5 x 4 . Using the inverse from Exercise 11, 39. is equivalent to the matrix equation 2x 3y 0 2 3 y 0 x 3 5 4 12 . Therefore, x 12 and y 8. y 2 3 0 8 3x 4y 10 3 4 x 10 is equivalent to the matrix equation 40. 7x 9y 20 7 9 y 20 x 9 4 10 10 . Therefore, x 10 and y 10. y 7 3 20 10
2x 5y 2 2 5 x 2 . Using the inverse from 41. is equivalent to the matrix equation 5x 13y 20 5 13 y 20 x 13 5 2 126 . Therefore, x 126 and y 50. Exercise 13, y 5 2 20 50 7x 4y 0 7 4 x 0 42. is equivalent to the matrix equation 8x 5y 100 8 5 y 100 0 400 x 53 43 3 . Therefore, x 400 and y 700 . 3 3 83 73 700 100 y 3
2 4 1 x 7 2x 4y z 7 43. x y z 0 is equivalent to the matrix equation 1 1 1 y 0 . Using the inverse from x 4y 2 1 4 0 z 2 x 4 4 5 7 38 Exercise 19, y 1 1 1 0 9 . Therefore, x 38, y 9, and z 47. z 5 4 6 2 47
SECTION 11.3 Inverses of Matrices and Matrix Equations
5x 7y 4z 1 5 44. 3x y 3z 1 is equivalent to the matrix equation 3 6x 7y 5z 1 6 x 26 7 25 1 8 y 3 1 1 1 . Therefore, x 8, 3 z 27 7 26 1 8 2y 2z 12 0 45. 3x y 3z 2 is equivalent to the matrix equation 3 x 2y 3z 8 1
x
92 1
4
12
20
7 4 x 1 y 1 1 3 7 5 z 1 y 1, and z 8. 2 2
x
12
1 3 y 2 . Using the inverse from 8 2 3 z
Exercise 23, y 3 1 3 2 10 . Therefore, x 20, y 10, and z 16. 7 z 1 3 8 16 2 0 x 2y 3 0 1 2 0 3 x yz 1 1 0 1 1 1 y 46. is equivalent to the matrix equation 2 0 1 0 1 z y 2 x 2y 3 2 3 1 2 0 2 1 0 0 2 1 0 x y 1 0 1 1 1 5 Therefore, x 1, y 5, z 1, and 3. z 0 1 1 0 2 1 3 1 0 0 1 3 47. Using a calculator, we get the result 3 2 1.
48. Using a calculator, we get the result 1 2 3.
49. Using a calculator, we get the result 3 2 2.
50. Using a calculator, we get the result 6 12 24.
51. Using a calculator, we get the result 8 1 0 3.
755
52. Using a calculator, we get the result 8 4 2 1. 53. This has the form M X C, so M 1 M X M 1 C and M 1 M X M 1 M X X. 1 3 2 3 2 3 2 1 1 . Since X M 1 C, we get Now M 98 4 3 4 3 4 3 x y z 3 2 1 0 1 7 2 3 . u 4 3 2 1 3 10 3 5 39 3 6 39 92 1 4 2 54. Using the inverse matrix from Exercise 23, we see that 3 1 3 6 12 15 30 . 7 33 1 3 0 0 33 2 2 39 39 x u 2 Hence, y 15 30 . 33 z 33 2 1 a a a a 1 1 a a 1 1 1 55. 2a 1 1 a 2 a 2 2a 2 a a a a a a
756
CHAPTER 11 Matrices and Determinants
a 0 0 0 1 0 0 0
2 x
0 b 0 0 0 1 0 0 56. 0 0 c 0 0 0 1 0 0 0 0 d 0 0 0 1 a 0 0 0 0 b 0 0 Thus the matrix 0 0 c 0 0 0 0 d 57.
x x2
1
1 a R1 1 b R2
1 0 0 0 1a
0
0
0
0 1 0 0 0 1b 0 0 1 0 1c 0 c R3 0 0 1 0 0 1 d R4 0 0 0 1 0 0 0 1d 1a 0 0 0 0 1b 0 0 . has inverse 0 0 1c 0 0 0 0 1d
.
1 1x x 2 x x 2 x 1 1 . 2 2x x 2 x 2 x 2 x 2 1x 2x 2
The inverse does not exist when x 0.
58.
e x e2x
e2x 1
e3x
ex
1
x e2x e3x e2x e 1 1 . The inverse exists for all x. 4x 2 e e4x e2x e x e2x e3x
0 1 0 0
1
ex
0
1
0 0
12 e2x R2 0 2e2x 0 e x 1 0 1 2 R3 0 0 2 0 0 1 1 1 ex 1 0 0 0 1 0 0 2 2 R ex R R 1 ex 1 e2x 0 1 x 1 e2x 0 . 1 2 1 0 1 0 2 e 2 2 2 1 1 0 0 0 0 1 0 0 2 2 1 1 ex 0 2 2 1 ex 1 e2x 0 . The inverse exists for all x. Therefore, the inverse matrix is 2 2 1 0 0 2
x 2x 59. e e 0 0 x 1 e 0 0 1 0 0 0 1
0 0 1 0 2 0 0 1
1
R2 e x R1 R2
1 x 1 1 1 1 1 1 x 1 x 1 x2 x 1 x2 . 60. x 1 x 1 x 2 1 x x x x x x x x 1 x 1 x x The inverse exists for x 0, 1.
x
1
3 1 3 1 0 0
4 2 4
0 1 0
1 1 1 1 1 0
R3 R1 R3 1 R2 R1 3 1 3 1 0 0 R 61. (a) 4 2 4 0 1 0 3 1 3 1 0 0 R1 R2 3 2 4 0 0 1 0 1 1 1 0 1 0 1 1 1 0 1 R1 12 R2 R1 1 1 1 1 1 0 1 0 1 1 12 0 R R R 12 R2 R2 3R1 R2 3 0 1 3 1 0 2 0 4 3 0 0 1 0 2 2 1 R3 2 R2 R3 0 0 1 1 32 1 0 1 1 1 0 1 0 1 1 1 0 0 0 1 1 3 3 0 1 0 2 0 0 . Therefore, the inverse of the matrix is 2 . 2 2 3 3 0 0 1 1 2 1 2 1 1
SECTION 11.3 Inverses of Matrices and Matrix Equations
757
A 0 1 1 10 1 3 2 14 1 . (b) B 0 2 C 1 32 1 13 2 Therefore, he should feed the rats 1 oz of food A, 1 oz of food B, and 2 oz of food C. A 0 1 1 9 2 3 (c) 0 12 0 B 2 . 2 3 C 1 2 1 10 1 Therefore, he should feed the rats 2 oz of food A, no food B, and 1 oz of food C. 2 7 A 0 1 1 3 (d) 0 4 2 . B 2 2 1 32 1 11 7 C Since A 0, there is no combination of foods giving the required supply.
3 1 4 1 0 0
62. 4 2 6 0 1 0 3 2 5 0 0 1 1 1 2 1 0 2 2 4 0 1 1 1
3 0 0 1
3 1 4
1 0 0
1 R2 1 1 2 1 1 0 R 0 1 1 1 0 1 1 1 2 1 1 0 R3 12 R2 R3 0 2 2 4 3 0 0 0 0 1 32 1
R2 R1 R2 R3 R1 R3
1 0
1 1 2 1 1 0
3 1 4 1 0 0 0 1 1 1 0 1
R2 3R1 R2
.
Since the inverse matrix does not exist, it would not be possible to use matrix inversion in the solutions of parts (b), (c), and (d).
9 11 8 x 740 9x 11y 8z 740 63. (a) 13x 15y 16z 1204 (b) 13 15 16 y 1204 8x 7y 14z 828 8 7 14 z 828 9 11 8 1 0 0 9 11 8 1 0 0 9R2 13R1 R2 R3 31R2 R3 0 8 40 13 9 0 20 (c) 13 15 16 0 1 0 9R3 8R1 R3 8 7 14 0 0 1 0 25 62 8 0 9 9 11 8 1 0 0 9 11 8 1 0 0 1 R R 252 R2 2 3 0 8 40 13 9 0 0 252 0 243 279 180 63 R3 2R2 R3 0 252 0 243 279 180 0 8 40 13 9 0 9 11 8 1 0 0 9 11 8 1 0 0 1 R 11R 8R R R3 27 5 27 5 31 31 1 2 3 1 0 1 2520 7 0 28 0 1 0 28 28 7 28 1 25 0 0 2520 1305 1125 360 0 0 1 29 56 56 7 63 63 7 7 9 1 9 0 0 1 0 0 4 4 4 4 1 31 5 0 1 0 27 0 1 0 27 31 5 . 9 R1 7 7 28 28 28 28 25 1 25 1 0 0 1 29 0 0 1 29 7 7 56 56 56 56
758
CHAPTER 11 Matrices and Determinants
7 7 1 4 4 27 31 5 . (You could also use a calculator to find the inverse.) Therefore, Thus, the inverse of the matrix is 28 28 7 25 1 29 7 56 56
7 7 1 x 4 4 y 27 31 5 28 28 7 25 1 29 z 7 56 56
740 16 1204 28 . She earns $16 on a standard phone, $28 on a deluxe phone, and 828 36
$36 on a super deluxe phone.
64. No. Consider the following counterexample: A
0 1 0 0
and B
0 2 0 0
. Then, AB O, but neither A O nor
B O. There are infinitely many matrices for which A2 O. One example is A
0 1 0 0
. Then, A2 O, but A O.
11.4 DETERMINANTS AND CRAMER’S RULE 1. True. det A is defined only for a square matrix A. 2. True. det A is a number not a matrix. 3. True. If det A 0 then A is not invertible. 2 1 2 4 1 3 11 4. (a) 3 4 1 0 2 (b) 3 2 1 1 [2 4 1 3] 0 [3 4 1 0] 2 [3 3 2 0] 8 3 2 9 7 0 3 4 2 0 has determinant D 2 3 0 0 6. 5. The matrix 0 3 0 1 has determinant D 0 0 1 2 2. 6. The matrix 2 0 3 1 has determinant D 3 2 1 1 0. 7. The matrix 2 2 3 1 23 02 04 has determinant D 02 08 04 04 0. 8. The matrix 04 08 4 5 has determinant D 4 1 5 0 4. 9. The matrix 0 1 2 1 has determinant D 2 2 1 3 1. 10. The matrix 3 2 11. The matrix 2 5 does not have a determinant because it is not square.
SECTION 11.4 Determinants and Cramer’s Rule
12. The matrix
3 0
1 13. The matrix 2
14. The matrix
759
does not have a determinant because it is not square.
1 8 has determinant D 1 1 1 1 1 1 1 . 2 2 8 4 8 8 1 12
22 14 05
10
has determinant D 22 10 05 14 22 07 29. 1 0 12
In Solutions 15–20, A 3 5 2 . 0 0 4
15. M11 5 4 0 2 20, A11 12 M11 20
16. M33 1 5 3 0 5, A33 16 M33 5
17. M12 3 4 0 2 12, A12 13 M12 12
18. M13 3 0 0 5 0, A13 14 M13 0
20. M32 1 2 3 12 72 , A32 15 M32 72 19. M23 1 0 0 0 0, A23 15 M23 0 2 1 0 2 4 2 6 4 4. Since M 0, the . Therefore, expanding by the first column, M 2 21. M 0 2 4 1 3 0 1 3
matrix has an inverse. 1 2 5 22. M 2 3 2 . Therefore, 3 5 3 2 2 3 2 2 3 2 9 10 2 6 6 5 10 9 19 24 5 0, and so 5 M 1 3 3 5 3 3 5
the matrix does not have an inverse. 30 0 20 23. M 0 10 20 . Therefore, expanding by the first row, 40 0 10 0 10 10 20 20 30 100 0 20 0 400 3000 8000 5000, and so M 1 exists. M 30 0 10 0 40
2 32 12 2 32 2 4 1 1 4 2 8 3 1 5 4, and the 24. M 2 4 0 . Therefore, M 1 1 2 2 2 2 4 2 1 2 1 2 matrix has an inverse. 1 3 7 3 7 M 25. M . Therefore, expanding by the second row, 2 2 0 8 2 2 0 2 2 M 0, the matrix does not have an inverse.
1 3 8 0 2
2 6 14 16 0. Since
760
CHAPTER 11 Matrices and Determinants
2 4 M . Therefore, expanding by the first row, 1 6 4 1 3 0 3
0 1 0
26. M 2 1
has an inverse.
1 3 3 0
2 0 2
0 27. M 1 1 M 1
1
6 4 2. Since M 0, the matrix
2 0 1 . Therefore, expanding by the third row, 0 0 2 6 4 1 1 3 3 3 3 0 1 3 3 3 3 3 6 6 4 4, and so M 1 exists. 2 4 1 1 2 0 1 0 2 0 1 4 2 0 6 4 1 6 4 6 4 1
3 4 0 4 28. M 0 1 6 0 1 0 2 0 and so M 1 exists.
1 2 2 2 0 2 . Therefore, M 1 4 0 4 2 3 4 4 0 1 0 1 6 0
1 2 1 29. 2 2 1 6. The matrix has an inverse. 1 2 2
1 2 2 2 2 6 3 4 4 4
10 20 31 30. 10 11 45 20 40 50
6 16 2 2 92,
1080. The matrix has an inverse.
7 1 3 2 5 1 10 2 3 9 11 5 2 18 18 13 31. 0. The matrix has no inverse. 12. The matrix has an inverse. 32. 2 6 3 30 4 24 0 31 5 15 10 39 1 10 2 10 4 3 2 8 6 24 33. 20 15 3 12 9 6
0 2 35. M 2 3 0 M 1 2 3
10 1 0. The matrix has no inverse. 27 1
0 4 6 1 1 3 1 2 3 0 1 7 0 6 2 1 3 6 3 0 7
3 5 10 2 2 2 26 3 34. 6 9 16 45 8 12 20 36
8. The matrix has an inverse.
0 0 4 6 2 1 1 3 , by replacing R3 with R3 R2 . Then, expanding by the third row, 0 0 1 0 3 0 1 7 1 6 2 0 3 1 18. 0
SECTION 11.4 Determinants and Cramer’s Rule
36. M
2
3 1 7
6 2 3 . 7 7 0 5 3 12 4 0 4
761
2 0 1 7 2 3 1 7 4 0 2 3 4 6 2 3 , M Then 7 0 5 7 7 0 5 7 3 0 4 0 3 12 4 0
by replacing C2 with C2 3C3 . So expanding about the second column, 2 1 7 2 1 4 2 7 7 22 3 5 1183. 3 M 7 4 2 3 7 7 3 4 3 4 3 4 0 1 2 3 4 5 1 2 3 4 1 2 3 0 2 4 6 8 0 2 4 6 1 2 60 2 120. 37. M 0 0 3 6 9 , so M 5 5 4 0 2 4 20 3 0 0 3 6 0 2 0 0 0 4 8 0 0 3 0 0 0 4 0 0 0 0 5 2 1 6 4 2 1 6 4 0 1 6 4 7 2 2 5 11 2 2 5 7 2 2 5 38. M . Then, M , by replacing C1 with C1 2C2 . So 4 2 10 8 0 2 10 8 4 2 10 8 6 1 1 4 8 1 1 4 6 1 1 4 1 6 0 1 6 0 1 6 4 1 6 4 1 6 1 6 104 M 11 2 10 8 8 2 2 5 11 2 10 0 8 2 2 13 88 2 10 2 10 2 10 0 1 1 8 2 10 8 1 1 4 88 2 104 2 32 4 1 0 39. B 2 1 1 4 0 3 4 1 4 0 1 0 6 12 4 2 1 1 (a) B 2 4 0 4 3 0 3 4 1 4 1 4 6 2 3 (b) B 1 2 1 4 0 (c) Yes, as expected, the results agree.
40. If we expand along the first row of each submatrix, we see that the determinant is 210 1024. 2x y 9 2 9 9 1 2 1 25. 41. Then D 10, and D y 5, Dx x 2y 8 8 2 1 2 1 8 Dy Dx 10 25 Hence, x 2, y 5, and so the solution is 2 5. D D 5 5 6x 12y 33 6 33 33 12 6 12 9, and D y 6, Dx 12. 42. Then D 4x 7y 20 20 7 4 7 4 20 Dy Dx 3 9 12 Hence, x and y 2, and so the solution is 32 2 . D D 6 2 6
762
CHAPTER 11 Matrices and Determinants
1 3 3 6 1 6 8. Then, D 12, and D y 20, Dx 3 1 1 2 3 2 Dy Dx 8 12 Hence, x 04,and so the solution is 06 04. 20 06, y D D 20 1 1 1 1 1x 1y 1 2 1 1 1 1 2 3 2 3 3 44. Then, D 1 6 , Dx 3 3 , and D y 1 1 1 3 1x 1 y 3 4 2 2 6 4 6 4 6 2 1 Dy Dx 1 Hence, x 31 2, y 1 6, and so the solution is 2 6. D D x 6y 3 43. 3x 2y 1
6
1.
6
04 04 04 12 04 12 08. 32, and D y 08, Dx Then, D 12 32 32 16 12 16 Dy Dx 32 08 Hence, x 4, y 1,and so the solution is 4 1. D D 08 08 10x 17y 21 10 21 21 17 10 17 30. 12, and D y 30, Dx 46. Then, D 20x 31y 39 20 39 39 31 20 31 Dx 2 , y D y 30 1, and so the solution is 2 1 . Hence, x 12 30 5 5 D D 30 x y 2z 0 47. Then expanding by the second row, 3x z 11 x 2y 0 04x 12y 04 45. 12x 16y 32
1 1 D 3 0 1 2 1 0 D y 3 11 1 0
Therefore, x 44 11 5x 3y z 48. 4y 6z 7x 10y
2 1 1 3 2 0 2 1 1 11 1 0
0 1 2 1 1 1 2 2 12 1 11, Dx 11 0 1 11 1 0 1 2 2 0 0 2 0 1 1 0 1 1 2 11. 22, and Dz 3 0 11 11 0 1 2 1 2 0
44,
11 4, y 22 11 2, z 11 1, and so the solution is 4 2 1. 5 3 1 6 5 3 0 4 28 426 398, 6 Then D 0 4 6 1 22 7 10 7 10 7 10 0 13 5 6 3 1 6 1 22 4 0 22 6 3 272 126 398, D y 0 22 6 1 6 Dx 22 4 6 1 13 10 13 10 7 13 13 10 0 7 13 0 5 3 6 5 3 5 5 6 6 428 1562 1990. 6 154 642 796, and Dz 0 4 22 4 22 7 10 7 13 7 13 7 10 13
796 1990 Therefore, x 398 398 1, y 398 2, and z 398 5, and so the solution is 1 2 5.
SECTION 11.4 Determinants and Cramer’s Rule
2x1 3x2 5x3 1 49. x 1 x 2 x3 2 2x2 x3 8
Then, expanding by the third row, 2 3 5 2 5 D 1 1 1 2 1 1 0 2 1
2 3 1 1
6 1 7,
1 3 5 2 1 2 1 1 1 3 30 20 7, Dx 2 1 1 5 3 1 8 2 8 1 2 1 8 2 1
2 1 5 2 5 Dx 1 2 1 8 2 1 1 0 8 1
2 3 1 Dx 1 1 2 3 0 2 8
2 1 1 2
24 3 21, and
2 3 2 1 6 8 14. 8 2 1 1 1 2
21 14 Thus, x1 7 7 1, x2 7 3, x 3 7 2, and so the solution is 1 3 2.
c 2 2a 50. a 2b c 9 3a 5b 2c 22
2 0 1 1 2 2 1 18 1 19, 1 Then D 1 2 1 2 3 5 5 2 3 5 2 2 0 1 9 2 2 1 18 1 19, 1 Da 9 2 1 2 22 5 5 2 22 5 2
2 2 1 1 9 1 1 9 1 1 2 Db 1 9 1 2 3 22 3 2 22 2 3 22 2
2 0 2 Dc 1 2 9 3 5 22
2 9 2 5 22
1 2 2 3 5
2 2 0.
Hence, a 1, b 5, and c 0, and so the solution is 1 5 0.
80 10 5 95, and
763
764
CHAPTER 11 Matrices and Determinants
1x 1y 1z 3 5 2 51. 23 x 25 y 32 z x 45 y z
7 10 10x 6y 15z 21 11 20x 12y 45z 33 10 5x 4y 5z 9 9 5
Then
10 6 15 20 12 20 45 12 45 15 6 2400 1950 300 750, D 20 12 45 10 5 4 5 5 4 5 5 4 5 21 6 15 33 12 33 45 12 45 5040 1440 3600 0, 15 6 Dx 33 12 45 21 9 4 9 5 4 5 9 4 5 10 21 15 20 33 20 45 33 45 D y 20 33 45 10 2400 6825 5175 750, and 15 21 5 9 5 5 9 5 5 9 5 10 6 21 20 12 20 33 12 33 2400 2070 420 750. 21 6 Dz 20 12 33 10 5 4 5 9 4 9 5 4 9
Therefore, x 0, y 1, z 1, and so the solution is 0 1 1. 5 2 1 0 2x y 5 3 0 3 24 35 11, 1 52. 5x Then D 5 0 3 2 3z 19 0 7 4 7 0 4 7 4y 7z 17 5 1 0 19 3 0 3 60 82 22, 1 Dx 19 0 3 5 17 7 4 7 17 4 7 2 5 0 5 3 19 3 D y 5 19 3 2 164 175 11, and 5 0 7 17 7 0 17 7 2 1 5 2 1 2 5 52 85 33. Thus, x 2, y 1, and z 3, and so the 17 Dz 5 0 19 4 5 0 5 19 0 4 17
solution is 2 1 3. 0 3 5 3y 5z 4 2 0 2 1 12 70 58, 5 53. 2x Then D 2 0 1 3 z 10 4 7 4 0 4x 7y 4 7 0 0 0 4 5 4 3 5 4 5 4 5 216, and 378, D y 2 10 1 4 Dx 10 0 1 7 10 1 10 1 4 0 0 0 7 0 0 3 4 0 4 3 4 120 56 176. 7 Dz 2 0 10 4 2 10 0 10 4 7 0 108 88 189 108 88 Thus, x 189 29 , y 29 , and z 29 , and so the solution is 29 29 29 .
SECTION 11.4 Determinants and Cramer’s Rule
2 5 0 2x 5y 4 54. Then D 1 1 1 x y z8 3x 3 0 5 5z 0 2 4 5 0 4 5 220, D y 1 Dx 8 1 1 5 8 1 3 0 0 5 2 5 4 5 4 Dz 1 1 8 3 1 8 3 0 0
1 1 1 1 10 40 50, 5 2 3 5 0 5
0 1 1 8 1 80 32 48, and 4 8 1 2 3 5 0 5 0 5 4
132. Thus, x 22 , y 24 , and z 66 , and so the solution is 22 24 66 . 5 25 25 5 25 25
x y z0 2z 0 Then 55. y z 0 x 2z 1 1 1 1 1 1 1 1 2 0 1 2 0 0 1 1 0 D 1 0 1 0 1 2 0 1 2 0 1 1 0 2 0 1 2 0 1 2 0 1 0 2 0 2 0 1 1 1 1 2 1 4, 0 1 1 1 1 1 1 0 0 0 1 1 1 Dx 1 0 0 1 1 1 0 1 1 0 1 1 1 1 0 1 0 2 0
Dy
Dz
1 1 1 1 2 0 0 1 1 2 0 0 0 1 0 0 1 1 1 2 0 1 1 0 1 1 1 1 2 0 0 1 2 0 1 1 0 1 0 0 0 1 0 1 0 1 0
D y
1 0
Dy D
2 1 0 1 1 1 1 0 1 0
1 1 2 2 1
2,
1 1 1 0 1 1 2 1 1, 1 1 2 1 0 1 0 0
1 1 0 1 1 1 2 1 1, and 2 1 1 0 1 0
1 0 1 1 1 2 0 0 0 2 0 0 1 0 1 1 0 0 1 1 1 0 2 1 1 1
765
1 1 2 2 4. Hence, we have x Dx 2 1 , 2 D 4 2 1 1
Dz D 1 1 1 1 4 ,z , and 1, and the solution is 12 14 14 1 . D D 4 4 4 4 4
766
CHAPTER 11 Matrices and Determinants
xy1 1 1 0 0 0 1 0 1 1 0 yz 2 0 1 1 0 1 1 1 0 1 1 1 2, 56. Then D 1 0 1 1 1 0 1 1 1 0 1 1 1 0 0 1 1 z 3 1 0 1 0 0 1 x 4 1 0 0 1 1 1 0 0 2 1 0 1 1 0 2 1 1 0 3 1 1 1 1 1 1 3 2, Dx 1 0 1 1 1 3 1 1 1 2 3 0 1 1 4 1 0 1 0 1 4 0 1 0 0 1 4 0 0 1 1 1 0 0 0 1 0 2 1 0 0 2 1 0 1 0 3 1 1 1 3 1 4, Dy 1 2 1 1 1 0 1 1 3 1 1 0 3 1 1 1 1 4 1 0 1 1 0 1 4 0 1 1 4 0 1 1 1 1 0 1 1 0 1 2 0 0 1 2 0 1 1 3 1 1 1 0, 1 Dz 1 0 3 1 1 1 2 0 1 0 0 3 1 1 2 4 1 0 3 1 0 4 1 1 0 4 1 1 1 0 1 1 0 1 1 1 2 0 1 1 2 0 1 1 2 1 3 4 2 6. 1 1 D 1 0 1 3 1 1 1 2 0 0 1 3 1 3 1 3 0 4 0 1 3 0 0 4 1 0 0 4 Dy Dx Dz D 2 4 0 6 Hence, the solution is x 1, y 2, z 0, and 3. D D D D 2 2 2 2
0 0 1 1 57. Area 6 2 1 2 3 8 1
6 2 1 2 3 8
1 48 6 1 42 21 2 2
y
3 5 1 2 2
12 [5 2 6 10] 12 [3 16] 12 19 19 2
(6, 2)
1 (0, 0)
1 0 1 5 1 1 1 58. Area 3 5 1 1 2 2 2 1 2 2 1
(3, 8)
x
1
y (_2, 2)
(3, 5)
1 (1, 0)
x
SECTION 11.4 Determinants and Cramer’s Rule
1 3 1 2 1 9 1 1 1 3 59. Area 2 9 1 1 2 2 5 1 6 1 5 6 1 12 [1 9 6 3 2 5 1 12 45]
2 9 1 5 6
y (_1, 3)
2 5 1 7 1 2 1 1 1 5 60. Area 7 2 1 2 2 2 3 1 4 1 3 4 1
62.
a 0 0 0 0 0 b 0 0 0 0 0 c 0 0 a 0 0 0 d 0 0 0 0 0 e
a a a a a 0 a a a a 0 0 a a a a 0 0 0 a a 0 0 0 0 a
x 12 13 63. 0 x 1 23 0 0 x 2 x 1 1 64. 1 1 x x 1 x
a a a a a a a 0 a a a 2 a 0 a a 0 0 a a 0 0 a 0 0 0 a
1 x x x
1 0 0 1 0 1
a b x a 66. x x b x 0 1 1
x
7 2 1 3 4
(_2, 5)
d 0 abc 0 e
abcde
a a a3 0 a
a5
y (7, 2)
1
x
1 (3, _4)
x 12 0 x 2 x x 1 0 x 0, 1, or 2 0 x 2 0 x 1
1 x x 1 x
1 0 x 65. x 2 1 0 x 0 1
b 0 0 0 c 0 0 0 c 0 0 ab 0 d 0 0 0 d 0 0 0 e 0 0 0 e
2
(5, _6)
12 [2 2 4 5 7 3 28 6] 12 [2 6 5 4 34] 12 12 20 34 12 66 33
61.
(2, 9)
1
12 [15 3 3 57] 12 63 63 2
767
1 1 x 1
x 0 x x 2 1 x x 2 2x 1 0 x 12 0 x 1
2 x 1 x 0 1 x 2 0 x 2 1 x 1 x 0
a x a 1 x x
a b x x b
1 [ax x x a] a x b bx
ax x 2 ax ax ab bx x 2 ax bx ab x a x b 0 x a or x b
768
CHAPTER 11 Matrices and Determinants
1 x x2 67. 1 y y 2 1 z z2
68.
y y2 1 z z2
x x2 1 z z2
x x2 1 y y2
yz 2 y 2 z xz 2 x 2 z x y 2 x y 2
yz 2 y 2 z x z 2 x 2 z x y 2 x y 2 x yz x yz x yz xz 2 y 2 z yz 2 x 2 y x 2 z zy 2 x yz z x y xz y 2 yz x x y xz y 2 yz z x x y x z y 2 yz z x x y z y y z z x x y y z
x 2y 6z 5
3x 6y 5z 8 2x 6y 9z 7 (a) If x 1, y 0, and z 1, then x 2y 6z 12 06 1 5, 3x 6y 5z 3 16 05 1 8, and 2x 6y 9z 2 1 6 0 9 1 7. Therefore, x 1, y 0, z 1 is a solution of the system. 1 0 6 1 2 6 1 2 6 . Then, M 3 6 5 3 0 5 (replacing C2 with C2 2C1 ), so (b) M 3 6 5 2 2 9 2 6 9 2 6 9 1 6 2 5 18 46. M 2 3 5 1 2 6 x 5 (c) We can write the system as a matrix equation: 3 6 5 y 8 or M X B. Since M 0, M has 2 6 9 z 7 an inverse. If we multiply both sides of the matrix equation by M 1 , then we get a unique solution for X, given by X M 1 B. Thus, the equation has no other solution.
(d) Yes, since M 0.
1 1 69. (a) If three points lie on a line then the area of the “triangle” they determine is 0, that is, 2 Q 2
a1 b1 1 a2 b2 1 0 a3 b3 1
Q 0. If the points are not collinear, then the point form a triangle, and the area of the triangle determined by these
points is nonzero. If Q 0, then 12 Q 12 0 0, so the “triangle” has no area, and the points are collinear. 6 4 1 y 2 10 6 4 6 4 (6, 13) (b) (i) 2 10 1 6 13 6 13 2 10 6 13 1 (2, 10) 26 60 78 24 60 8 34 104 68 0
Thus, these points are collinear.
(_6, 4) 1 1
x
SECTION 11.4 Determinants and Cramer’s Rule
5 10 1 2 6 (ii) 2 6 1 15 2 15 2 1
5 10 15 2
5 10 2 6
769
y (_5, 10) (2, 6)
4 90 10 150 30 20 94 140 50 4
These points are not collinear. Note that this is difficult to determine from the diagram.
1 2
(15, _2)
x
x y 1 70. (a) Let M x1 y1 1 . Then, expanding by the third column, x2 y2 1 x1 y1 x y x y x1 y2 x2 y1 x y2 x2 y x y1 x1 y M x2 y2 x2 y2 x1 y1 x1 y2 x2 y1 x y2 x2 y x y1 x1 y x2 y x1 y x y2 x y1 x1 y2 x2 y1
x2 x1 y y2 y1 x x1 y2 x2 y1 So M 0 x2 x1 y y2 y1 x x1 y2 x2 y1 0 x2 x1 y y2 y1 x x1 y2 x2 y1 y y1 x y y1 y x x1 x 1 2 1 2 x2 x1 y y2 y1 x x1 y2 x1 y1 x1 y1 x2 y1 y 2 x2 x1 x2 x1 x2 x1 y y1 y y1 y 2 x x1 y1 y y1 2 x x1 , which is the “two-point” form of the equation for the line x2 x1 x2 x1 passing through the points x1 y1 and x2 y2 . (b) Using the result of part (a), the line has equation x y 1 20 50 x y 20 50 1 0 10 25 10 25 10 25 1
x y 20 50
0
500 500 25x 10y 50x 20y 0 25x 30y 1000 0 5x 6y 200 0.
71. (a) Let x be the amount of apples, y the amount of peaches, and z the amount of pears (in pounds). x y z 18 We get the model 075x 090y 060z 1380 075x 090y 060z 180
770
CHAPTER 11 Matrices and Determinants
1 1 1 075 090 075 060 090 060 0 090 135 045, 1 1 (b) D 075 090 060 1 075 090 075 060 090 060 075 090 060 18 1 1 1380 090 1380 060 090 060 0 72 108 36, 1 1 Dx 1380 090 060 18 180 090 180 060 090 060 180 090 060 1 18 1 075 1380 075 060 1380 060 1 18 D y 075 1380 060 1 075 180 075 060 180 060 075 180 060 72 162 117 27, and 1 1 18 075 090 075 1380 090 1380 18 1 Dz 075 090 1380 1 075 090 075 180 090 180 075 090 180
108 117 243 18. Dy Dx Dz 36 27 18 So x 8; y 6; and z 4. D D D 045 045 045 Thus, Muriel buys 8 pounds of apples, 6 pounds of peaches, and 4 pounds of pears.
72. (a) Using the points 10 25, 15 3375, and 40 40,we substitute for x and y and get the system 100a 10b c 25 225a 15b c 3375 1600a 40b c 40 100 10 1 (b) D 225 15 1 1600 40 1
225 15 1 1600 40
100 10 1 1600 40
100 10 1 225 15
9000 24,000 4000 16,000 1500 2250 15,000 12,000 750 3750, 25 10 1 25 25 10 3375 15 10 1 1 Da 3375 15 1 1 40 3375 15 40 40 40 40 40 1
1350 600 1000 400 375 3375 750 600 375 1875, 100 25 1 100 25 100 25 225 3375 1 1 Db 225 3375 1 1 225 3375 1600 40 1600 40 1600 40 1
9000 54,000 4000 40,000 3375 5625 45,000 36,000 2250 11,250, and
SECTION 11.4 Determinants and Cramer’s Rule
100 10 25 Dc 225 15 3375 1600 40 40
225 15 25 1600 40
100 10 3375 1600 40
100 10 40 225 15
25 9,000 24,000 3375 4,000 16,000 40 1,500 2,250
771
25 15,000 3375 12,000 40 750 375,000 405,000 30,000 0.
Thus, a
Db Dc Da 1875 11,250 0 005, b 3, and c 0. Thus, the model is D D D 3750 3,750 3,750
y 005x 2 3x.
73. Using the determinant formula for the area of a triangle, we have 1000 2000 1 1000 2000 5000 4000 1 1 1 Area 5000 4000 1 1 2 2 2000 6000 2000 6000 2000 6000 1
1000 2000 1 5000 4000
12 22,000,000 2,000,000 6,000,000 12 14,000,000 7,000,000
Thus, the area is 7,000,000 ft2 .
74. (a) The coordinates of the vertices of the surrounding rectangle are a1 b1 , a2 b1 , a2 b3 , and a1 b3 . The area of the surrounding rectangle is given by a2 a1 b3 b1 a2 b3 a1 b1 a2 b1 a1 b3 a1 b1 a2 b3 a1 b3 a2 b1 . (b) The area of the three blue triangles are as follows:
Area of a1 b1 a2 b1 a2 b2 : 12 a2 a1 b2 b1 12 a2 b2 a1 b1 a2 b1 a1 b2 Area of a2 b2 a2 b3 a3 b3 : 12 a2 a3 b3 b2 12 a2 b3 a3 b2 a2 b2 a3 b3 Area of a1 b a1 b3 a3 b3 : 12 a3 a1 b3 b1 12 a3 b3 a1 b1 a3 b1 a1 b3 . Thus the sum of the areas of the blue triangles, B, is
B 12 a2 b2 a1 b1 a2 b1 a1 b2 12 a2 b3 a3 b2 a2 b2 a3 b3 12 a3 b3 a1 b1 a3 b1 a1 b3 12 a1 b1 a1 b1 a2 b2 a2 b3 a3 b2 a3 b3 12 a1 b2 a1 b3 a2 b1 a2 b2 a3 b1 a3 b3
a1 b1 12 a2 b3 a3 b2 12 a1 b2 a1 b3 a2 b1 a3 b1 So the area of the red triangle A is the area of the rectangle minus the sum of the areas of the blue triangles, that is, A a1 b1 a2 b3 a1 b3 a2 b1 a1 b1 12 a2 b3 a3 b2 12 a1 b2 a1 b3 a2 b1 a3 b1 a1 b1 a2 b3 a1 b3 a2 b1 a1 b1 12 a2 b3 a3 b2 12 a1 b2 a1 b3 a2 b1 a3 b1
12 a1 b2 a2 b3 a3 b1 12 a1 b3 a2 b1 a3 b2 a b 1 1 1 (c) We first find Q a2 b2 1 by expanding about the third column. a3 b3 1 a1 b1 a1 b1 a2 b2 1 1 a2 b3 a3 b2 a1 b3 a3 b1 a1 b2 a2 b1 Q 1 a3 b3 a2 b2 a3 b3 a1 b2 a2 b3 a3 b1 a1 b3 a2 b1 a3 b2
So 12 Q 12 a1 b2 a2 b3 a3 b1 12 a1 b3 a2 b1 a3 b2 , the area of the red triangle. Since 12 Q is not always positive, the area is 12 Q.
75. (a) If A is a matrix with a row or column consisting entirely of zeros, then if we expand the determinant by this row or column, we get A 0 A1 j 0 A2 j 0 Anj 0.
772
CHAPTER 11 Matrices and Determinants
(b) Use the principle that if matrix B is a square matrix obtained from A by adding a multiple of one row to another, or a multiple of one column to another, then A B. If we let B be the matrix obtained by subtracting the two rows (or columns) that are the same, then matrix B will have a row or column that consists entirely of zeros. So B 0 A 0.
(c) Again use the principle that if matrix B is a square matrix obtained from A by adding a multiple of one row to another, or a multiple of one column to another, then A B. If we let B be the matrix obtained by subtracting the proper multiple of the row (or column) from the other similar row (or column), then matrix B will have a row or column that consists entirely of zeros. So B 0 A 0.
75. Gaussian elimination is superior, since it takes much longer to evaluate six 5 5 determinants than it does to perform one five-equation Gaussian elimination.
CHAPTER 11 REVIEW 1. (a) 2 3
2. (a) 2 3
(b) Yes, this matrix is in row-echelon form.
(b) Yes, this matrix is in row-echelon form.
(c) No, this matrix is not in reduced row-echelon form, since the leading 1 in the second row does not have a
(c) Yes, this matrix is in reduced row-echelon form. x 6 (d) y 0
0 above it. x 2y 5 (d) y 3
3. (a) 3 4
(b) Yes, this matrix is in row-echelon form. (c) Yes, this matrix is in reduced row-echelon form. 8z 0 x (d) y 5z 1 0 0
5. (a) 3 4 (b) No, this matrix is not in row-echelon form. The leading 1 in the second row is not to the left of the one above it. (c) No, this matrix is not in reduced row-echelon form. y 3z 4 (d) x y 7 x 2y z 2
4. (a) 3 4 (b) No, this matrix is not in row-echelon form, since the leading 1 in the second row is not to the left of the one above it. (c) Since this matrix is not in row-echelon form, it is not in reduced row-echelon form. x 3y 6z 2 (d) 2x y 5 z0
6. (a) 4 4
(b) No, this matrix is not in row-echelon form. The leading 1 in the fourth row is not to the left of the one above it. (c) No, this matrix is not in reduced row-echelon form. x 8y 6z 4 y 3z 5 (d) 2z 7 x y z 0
CHAPTER 11
1
2 2
6
1 1 0 1
R R 1 2 7. 1 1 0 1 1 2 1 3 7 2
2 2 1 3
1 1 0 1
R2 R1 R2 0 6 R3 2R1 R3 7 0
3 2 3 3
Review
773
1 1 0 1
R R R 3 2 3 7 0 9 0
3 2 0 1
Thus, z 2, 3y 2 2 7 3y 3 y 1, and x 1 1 x 0, and so the solution is 0 1 2.
1 1 1 2
8. 1 0
1 3 6 2 3 5
R2 R1 R2
1 1 1 2
0 0
2 2 4 2 3 5
1 1 1 2
0 0
R3 R2 R3
1 2
3 2
9. 2 1 1 2 2 7 11 9
1 2
3 2
0 3 5 6 0 3 5 5
R2 2R1 R2 R3 2R1 R3
2 2 4 . Thus, z 1; 2y 2 1 4 0 1 1
y 1; and x 1 1 2 x 2, and so the solution is 2 1 1.
1 2
0 0
R3 R2 R3
3 2
0
1 1 1
2
10. 1 1 3 6 3 1 5 10 R1 R2 R1
R2 R1 R2 R3 3R1 R3
1 0 2 4
1 1 1 2
0 0
2 2 4 2 2 4
R3 R2 R3
1 1 1 2
0
0 0
2 2 4 0 0 0
6 . 1
3 5
The last row corresponds to the equation 0 1, which is always false. Thus, there is no solution.
7 . 2
0 0
1 2 R2
1 1 1 2
1 1 2 0 0 0
0 1 1 2 . Let z t. Then y t 2 y 2 t and x 2t 4 x 4 2t, and so the 0 0 0 0
solutions are 4 2t 2 t t, where t is any real number.
1
1
1
1
0
1 1 4 1 1 11. 1 2 0 4 7 2 2 3 4 3 1 1 1 0 1 4 R3 3R2 R3 0 0 13 0 0 1
1
1
0 2 0 3 R4 2R1 R4 0 0 1 0 1 0 5 6 R3 R4 0 12 11 2 3 0 R2 R1 R2 R3 R1 R3
1
1
0
5 2 1 1 3 7 1 2 3 1
1
1
1
4
5
6 0 1 2 3 0 13 12 11
1
1
1
1
0
0 1 4 5 6 0 3 1 3 7 0 0 1 2 3 1 1 1 1 0 6 R4 13R3 R4 0 1 4 5 . 0 0 1 2 3 0 0 0 14 28
R3 R2 R3
0
Therefore, 14 28 2, z 2 2 3 z 1, y 4 1 5 2 6 y 0, and x 0 1 2 0 x 1. So the solution is 1 0 1 2.
774
CHAPTER 11 Matrices and Determinants
1 0 3
0 1
0 1 0 4 5 12. 0 2 1 1 0 2 1 5 4 4 1 0 0 1 R3 R4 0 0 0 0 1 0 3 0 1 0 1 0 4 5 0 0 1 0 1 0 0 0 1 1
1 1 3 2
13. 2 1 1 2 3 0 4 4 1 1 3 0 1 5 3 0 0 0
3
0
1
1 1
5 0 1 9 10
0 4
0 1
1 0 0 1 0 0 0 0
R2 4R4 R2
23 0
3
R4 3R3 R4
R1 3R3 R1
1 0
0 1 0 4 0 2 1 1 0 1 1 4
R4 2R1 R4
1 1
0 0 0 0 1 0 0 1 3
2
5 0 6
1 0
0 1
3
0
1
0 1 0 4 5 0 0 1 9 10 0 0 1 0 1
R3 2R2 R3 R4 R2 R4
3
1 0
0 1 0 4 5 R3 1 0 0 1 0 1 9 R4 0 0 0 9 9 2 1 . Therefore, the solution is 2 1 1 1. 1 1
0 3 5 2 0 3 5 2 4 4 1 0 3 3 R1 R2 R1 5 2 0 1 3 3 0 0 0 0
R2 2R1 R2 R3 3R1 R3
2
R3 R2 R3
1 1
0 0
3
2
3 5 2 0 0 0
1 3 R2
. The system is dependent, so let z t: y 5 t 2 3 3
y 53 t 23 and x 43 t 43 x 43 t 43 . So the solution is 43 t 43 53 t 23 t , where t is any real number.
1 1
0
1
14. 1 1 2 3 1 3 2 1 1 1 0 1 0 1 1 1 0 0 0 0
R2 R1 R2 R3 R1 R3
R1 R2 R1
1 1
0
1
1 1 0 1
1 3 R2 R3 0 2 2 2 R 2 R4 0 2 2 2 0 2 2 2 0 0 0 0 1 0 1 2 0 1 1 1 . Since the system is dependent, let z t. Then y t 1 0 0 0 0
y t 1 and x t 2 x t 2. So, the solution is t 2 t 1 t where t is any real number.
1 1
1 1 0
1 1
1 1 0
1 2 R4
1 1
1 1 0
2 3R1 R2 1 R2 R1 R R 0 2 4 2 2 0 1 2 1 1 3 1 1 1 2 1 0 1 0 1 . Since the system is dependent, Let z s and t. Then y 2s t 1 y 2s t 1 and 0 1 2 1 1
15.
x s 1 x s 1. So the solution is s 1 2s t 1 s t, where s and t are any real numbers.
1 1 3 16. 2 1 6 1 2 9
R2 2R1 R2 R3 R1 R3
1 1 3 0 3 0 0 1 6
R2 3R3 R2
1 1 3 0 0 18 . Since the second row corresponds 0 1 6
to the equation 0 18, which is always false, this system has no solution.
CHAPTER 11
1 1 1 0 17. 3 2 1 6 1 4 3 3
R2 3R1 R2 R3 R1 R3
1 1 1 0 0 5 4 6 0 5 4 3
R3 R2 R3
Review
775
1 1 1 0 0 5 4 6 . The last row of this 0 0 0 3
matrix corresponds to the equation 0 3, which is always false. Hence there is no solution. 1 2 3 2 1 2 3 2 1 2 3 2 R2 2R1 R2 3 R2 R3 0 5 11 3 R 18. 0 5 11 3 . Since the third 2 1 5 1 R3 4R1 R3 0 5 11 2 0 0 0 1 4 3 1 6
row corresponds to the equation 0 1, which is always false, this system has no solution. 1 0 0 1 1 1 0 0 1 1 1 1 1 1 2 R2 R1 R2 1 1 1 1 0 0 1 1 2 1 1 1 1 1 0 R1 12 R3 R3 R1 R3 19. 0 1 1 2 1 2 0 0 2 2 1 1 1 1 2 R4 2R1 R4 2 4 4 2 6 0 4 4 4 4 2 4 4 2 6 1 0 0 1 1 1 0 0 1 1 R1 R3 R1 0 1 1 2 1 0 1 1 2 1 14 R3 R3 R2 R3 R2 2R3 R2 R4 4R2 R4 0 0 0 4 0 121 R4 0 0 0 0 0 R4 R3 R4 0 0 0 12 0 0 0 0 1 0 1 0 0 0 1 0 1 1 0 1 . This system is dependent. Let z t, so y t 1 y t 1; x 1 x 1. So the 0 0 0 1 0 0 0 0 0 0 solution is 1 t 1 t 0, where t is any real number. 1 1 2 3 0 1 1 2 3 0 R 3R R 313 0 1 1 1 1 20. 0 1 1 1 1 3 2 7 10 2 0 1 1 1 2
R3 R2 R3
1 1 2 3 0 0 1 1 1 1 . Since the 0 0 0 0 1
third row corresponds to the equation 0 1, which is always false, this system has no solution.
21. A 3 3 and B 2 3 have different dimensions, so they are not equal. 25 1 5 1 5 e0 B, so A and B are equal. 22. A 0 21 log 1 12 0 12 In Solutions 23–34, the matrices A, B, C, D, E, F, and G are defined as follows:
A
1
2 0 1 4
D 0 1 2 0
B
E
2 1
12
1
1 2 4 2 1 0
4 0 2
C
F 1 1 0 7 5 0
23. A B is not defined because the matrix dimensions 1 3 and 2 3 are not compatible.
3
G
1 2
2 32 2 1 5
776
CHAPTER 11 Matrices and Determinants
24. C D
1 2
3
1
4
12 1
2 32 0 1 2 2 1 2 0 4 1 3 1 4 2 3 25. 2C 3D 2 2 2 3 0 1 2 1 2 0
5 2
1
1 6
3 12
4 18
4 3 0 3 4 4 2 6 0 2
0 2
26. 5B 2C is not defined because the matrix dimensions 2 3 and 3 2 are not compatible. 27. G A 5 2 0 1 10 0 5
28. AG is undefined because the matrix dimensions 1 3 and 1 1 are not compatible. 1 3 1 3 4 2 11 2 2 2 7 1 2 4 10 1 2 4 2 3 3 11 29. BC 30. C B 2 2 2 2 8 1 2 2 1 0 1 92 2 1 0 4 3 8 2 1 2 1 1 3 2 14 4 0 2 4 0 2 2 1 2 4 30 22 2 2 3 3 3 1 1 0 31. B F 32. FC 1 1 0 2 2 2 2 1 0 9 1 4 27 57 2 1 7 5 0 7 5 0 2 2 1 3 3 7 1 4 1 11 2 2 2 2 2 1 2 1 15 3 1 3 33. C D E 2 2 0 1 1 2 2 4 2 2 12 1 1 2 1 2 0 0 1 12 1 4 0 2 1 6 1 4 4 0 2 0 2 12 12 34. F 2C D 4 2 1 1 0 4 3 0 1 1 1 0 4 4 7 5 0 4 2 2 0 7 5 0 6 2 20 34 In Solutions 35–44, the matrices A and B are defined as follows: 1 4 1 3 0 3 B A 1 1 0 2 1 2 2 0 2 1 6 0 27 0 21 6 42 24 35. AB 2 36. A2 B 20 5 13 3 37 22 5 22 7 3 42 27 19 14 26 8 32 4 4 7 1 11 7 37. A1 B A 38. B AB 1 3 3 2 1 4 3 35 13 18 80 7 16 3 3 2 1 39. AB 12 40. B A 12 41. A1 3 1 1 1 1 44. A B A 4 43. A B A 4 42. A 3 2 5 3 52 1 0 3 52 2 5 1 0 and B A . 45. AB 2 6 1 1 0 1 1 1 2 6 0 1
CHAPTER 11
Review
777
2 1 3 3 2 52 2 52 1 0 0 3 2 1 3 1 0 0 2 2 1 1 2 0 1 0 and B A 1 1 2 2 2 1 0 1 0 . 46. AB 2 2 1 0 1 1 1 1 1 0 0 1 1 1 1 0 1 1 0 0 1
In Solutions 47–52, A
2 1 3 2
, B
1 2 2
4
, and C
0 1 3
.
2 4 0
47. A 3X B 3X B A X 13 B A. Thus, X 13
1 2 4
2
2 1 3 2
48. 12 X 2B A X 2B 2A X 2A 2B 2 A B. 2 1 1 2 3 1 6 2 2 . X 2 3 2 2 4 1 6 2 12
1 3
1 3 5
2
.
Thus,
49. 2 X A 3B X A 32 B X A 32 B. Thus, 3 3 7 2 1 2 2 1 2 1 2 . 3 2 X 2 3 6 0 8 2 4 3 2 3 2 50. 2X C 5A 2X 5A C, but the difference 5A C is not defined because the dimensions of 5A and C are not the same. 51. AX C A1 AX X A1 C. Now 2 1 2 1 0 1 3 2 1 2 2 6 1 . Thus, X A1 C . A1 4 3 3 2 3 2 2 4 0 3 2 4 5 9 52. AX B A1 AX X A1 B. From Exercise 65, 2 1 2 1 1 2 4 8 1 1 . Thus, X A B . A 3 2 3 2 2 4 7 14
1 4
2
53. D
2 9
. Then D 1 9 2 4 1, and so D 1
9 4 2
1
.
3 1 3 2 4 . 8 . Then D 2 3 1 2 8, and so D 1 1 54. D 8 1 1 1 2 1 3 8 4
55. D
2
4 12 2
6
. Then D 4 6 2 12 0, and so D has no inverse.
1 2 . Then D 2 56. D 1 1 2 3 2 0 3 2 2 4 0
1 2 4 0 2
2 2 6 4 2 0, and so D has no inverse.
778
57.
58.
59.
60.
CHAPTER 11 Matrices and Determinants
3 0 2 3 4 12 9 1. So D 1 exists. 1 . Then, D 1 D 2 3 0 2 3 4 2 4 2 1 1 3 1 1 1 0 1 3 1 1 1 0 3 0 1 1 0 0 R2 2R1 R2 R2 1 R2 R1 0 9 2 2 3 0 3 2 3 0 0 1 0 R 2 3 0 0 1 0 2R3 R3 4R1 R3 4 2 1 0 0 1 0 14 3 4 4 1 4 2 1 0 0 1 1 3 1 1 1 0 1 3 1 1 1 0 1 3 1 1 1 0 R3 R2 R3 9R2 R3 3 R2 R3 0 27 6 6 9 0 R 0 27 6 6 9 0 1 R3 0 1 0 2 1 2 R1 3R2 R1 3 0 28 6 8 8 2 0 1 0 2 1 2 0 9 2 2 3 0 3 2 3 1 0 1 5 4 6 1 0 0 3 2 3 1 2 R3 0 1 0 . Thus, D 1 2 1 2 2 1 2 0 1 0 2 1 2 . R1 R3 R1 8 6 9 0 0 2 16 12 18 0 0 1 8 6 9 1 2 3 1 0 0 1 2 3 2 4 2 5 4 5 1 4 6 1. So D 1 exists. 2 4 5 0 1 0 3 2 . Then, D 1 D 2 4 5 2 5 2 6 5 6 2 5 6 0 0 1 2 5 6 1 2 3 1 0 0 1 2 3 1 0 0 R R R 3R R R2 2R1 R2 2 3 1 3 1 0 0 1 2 1 0 0 1 0 2 0 1 R3 2R1 R3 0 1 0 2 0 1 0 0 1 2 1 0 1 3 2 1 2 0 5 3 0 1 0 0 1 3 2 1 1 2R2 R1 0 1 0 2 0 1 R 0 1 0 2 0 1 . Thus, D 2 0 1 . 2 1 0 0 0 1 2 1 0 0 0 1 2 1 0 1 0 0 1 1 0 0 1 1 0 0 0 1 2 R2 2 0 2 0 2 0 2 0 2 0 2 0 1 0 0 1 3 3 R3 1 D . Thus, D 0 3 3 2 3 24 and D exists. 0 0 3 3 1 0 4 0 0 3 3 0 0 1 0 4 R4 0 0 4 0 0 0 4 0 0 0 4 0 0 0 1 1 0 0 14 1 0 0 1 1 0 0 0 1 0 0 0 1 0 0 14 R1 R4 R1 0 1 0 1 0 1 0 1 0 1 0 0 0 1 0 0 0 1 0 1 R2 R4 R2 2 2 4 2 4 1 . Therefore, D . 0 0 1 1 0 0 1 1 0 0 1 0 0 0 1 0 0 0 1 1 R3 R4 R3 3 3 4 3 4 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 14 4 4 1 0 1 0 0 1 0 1 D . Thus, 1 1 1 2 1 2 1 2 1 0 1 0 1 1 1 2 1 1 1 2 1 1 0 1 0 1 0. D 1 1 2 1 1 2 1 2 1 2 1 2 2 1 2 1 2 1 2 2 3
0 1
Hence, D 1 does not exist.
CHAPTER 11
61.
12 5 x
5 2
x y
2
5
10 17 10
. If we let A
65
12 5 5 2
, then A1
Review
779
2 5 2 5 1 , and so 24 25 5 12 5 12
. Therefore, the solution is 65 154. 5 12 17 154 6 5 x 1 6 5 . , then 62. If we let A 8 7 y 1 8 7 7 5 7 5 1 7 5 7 5 x 6 1 2 , and so 2 2 . 2 1 A1 2 42 40 8 6 4 3 4 3 1 8 6 y 7 y
Therefore, the solution is 6 7. 1 2 1 5 2 1 5 1 0 0 1 2 2 0 1 0 2 1 5 x 3 R R 1 2 1 63. 1 2 2 y 4 . Let A 1 2 2 . Then 1 2 2 0 1 0 2 1 5 1 0 0 1 1 0 3 1 0 3 0 0 1 1 0 3 0 0 1 1 0 3 z 6 1 2 2 0 1 0 1 2 2 0 1 0 R 2R R R1 2R2 R1 R2 2R1 R2 2 3 2 0 3 1 1 2 0 0 1 1 1 0 2 R3 R1 R3 R3 R3 2R2 0 2 1 0 1 1 0 2 1 0 1 1 1 0 4 2 1 4 1 0 4 2 1 4 1 0 0 6 3 8 R1 4R3 R1 R3 0 1 1 1 0 2 0 1 0 1 1 1 . 0 1 1 1 0 2 R2 R3 R2 0 0 1 2 1 3 0 0 1 2 1 3 0 0 1 2 1 3 1 1 6 3 8 x 6 3 8 12 3 1 1 Hence, A1 1 1 1 and y 1 1 1 4 12 , and so the solution is 1 1 2 1 3 z 2 1 3 6 12 1 1 1 . 12 12 12 2 0 3 x 5 2 0 3 2 0 3 1 0 0 R R 1 2 64. 1 1 6 y 0 . Let A 1 1 6 . Then 1 1 6 0 1 0 3 1 1 z 5 3 1 1 3 1 1 0 0 1 1 1 6 0 1 0 1 1 6 0 1 0 1 1 6 0 1 0 2 2R1 R2 2 2R3 R2 2 0 3 1 0 0 R 0 2 9 1 2 0 R 0 2 9 1 2 0 R3 3R1 R3 3 1 1 0 0 1 0 4 17 0 3 1 0 0 1 2 1 1 1 1 0 12 5 6 1 1 0 12 5 6 1 R1 9R3 R1 1 R2 R1 0 2 0 17 7 9 2 R2 0 1 0 17 7 9 R 2 2 2 R2 6R3 2R2 0 0 1 2 1 1 0 0 1 2 1 1 7 3 3 7 3 3 x 10 1 0 0 72 32 32 2 2 2 2 2 2 5 0 1 0 17 7 9 . Hence, A1 17 7 9 and y 17 7 9 0 20 , and 2 2 2 2 2 2 2 2 2 z 5 0 0 1 2 1 1 2 1 1 2 1 1 5 so the solution is 10 20 5.
65. (a) The i jth entry of A represents how many pounds of vegetable j were sold on day i, and the ith entry of B represents the price of vegetable i.
780
CHAPTER 11 Matrices and Determinants
685 100 . The jth entry of AB represents the total revenue on day j. (b) AB 14 12 16 410 050 25 16 30
150
66. (a) We are given that x y 18, and since the total is $600, 20x 50y 600. 20 50 x 600 (b) 1 1 y 18 1 5 1 50 1 3 , so 30 (c) A1 1 2 20 1 50 1 1 20 30 3 1 1 5 18 5 20 30 10 600 600 3 3 . Brodie received 10 30 X A1 B 30 1 2 1 600 2 18 20 12 8 18 30 3 30 3 $20 bills and 8 $50 bills.
2 13 60 78 18. 208 210 2, and D y 6 30 2 1 and y 18 9 , and so the solution is 1 9 . Therefore, x 10 5 10 5 5 5
2 7 67. D 6 16
13 7 32 42 10, Dx 30 16
12 11 108 77 185, Dx 68. D 9 7 12 140 Dy 240 980 740. Therefore, x 7 20
140 11 1260 220 1480, and 9 20
740 1480 185 8 and y 185 4, and so the solution is 8 4.
2 1 5 2 1 1 7 195 39 156, 3 69. D 1 7 0 5 1 7 5 4 5 4 3 0 1 5 0 1 9 7 495 27 522, 3 Dx 9 7 0 5 9 7 9 4 9 4 3 2 0 5 2 0 1 9 D y 1 9 0 5 180 54 126, and 3 1 9 5 9 5 9 3 2 1 0 2 1 2 1 117 117 234. 9 Dz 1 7 9 9 1 7 5 4 5 4 9
522 87 , y 126 21 , and z 234 3 , and so the solution is 87 21 3 . Therefore, x 156 26 156 26 156 2 26 26 2
CHAPTER 11
Review
781
3 4 1 3 1 1 4 52 11 41, 1 70. D 1 0 4 4 1 4 2 5 2 1 5 10 4 1 10 1 20 4 880 20 860, 1 Dx 20 0 4 4 20 4 30 5 30 1 5 3 10 1 10 1 10 1 20 4 660 80 40 540, and D y 1 20 4 3 2 1 20 4 30 5 30 5 2 30 5 3 4 10 3 10 1 20 40 50 10. 1 Dz 1 0 20 4 1 20 2 30 2 1 30 860 540 10 860 540 10 Therefore, x 860 41 41 , y 41 , and z 41 , and so the solution is 41 41 41 .
1 3 1 3 1 1 3 1 3 1 1 1 4 8 10 11. 71. The area is 3 1 1 2 2 2 2 2 2 2 3 1 2 2 1 5 2 1 1 5 5 2 5 2 1 1 1 21 3 27 51 . 72. The area is 1 5 1 2 2 2 2 4 1 4 1 1 5 4 1 1
73. Let x be the amount invested in Bank A, y the amount invested in Bank B, and z the amount invested in Bank C. x y z 60,000 y z 60,000 x We get the following system: 002x 0025y 003z 1575 2x 25y 3z 157,500 which 2x 2x 2z y y 2z 0
1
1 1
60,000
has matrix representation 2 25 3 157,500 2 1 2 0 1 1 1 60,000 1 R1 R2 R1 0 1 0 40,000 0 R3 05R2 R3 0 05 1 37,500 0
1 0 40,000 0 1 17,500
$2500 in Bank A, $40,000 in Bank B, and $17,500 in Bank C.
1
1
1
60,000
R 1R 2 0 05 1 3 3 37,500 0 3 0 120,000 1 0 1 2500 R1 R3 R1 0 1 0 40,000 . Thus, she invests 0 0 1 17,500
R2 2R1 R2 R3 2R1 R3
0 1 20,000
782
CHAPTER 11 Matrices and Determinants
74. Let x be the amount of haddock, y the amount of sea bass, and z the amount of red snapper, in pounds. Our system 1 1 1 560 1 1 1 560 R R R R 125R R 2 3 2 3 1 3 0 has the matrix representation 075 0 255 125 075 200 575 125 0 200 320 125 0 200 320 1 1 1 560 1 1 1 560 1 1 1 560 4 R2 R 125R R 4 R3 3 2 3 0 075 0 3 3 255 0 340 340 0 1 0 1 0 0 125 075 380 0 125 075 380 0 0 075 45 1 1 1 560 1 1 0 500 1 0 0 160 R R R 1 R3 R1 1 2 1 0 1 0 340 R 0 1 0 340 0 1 0 340 . Thus, he caught 160 lb of 0 0 1 60 0 0 1 60 0 0 1 60 haddock, 340 lb of sea bass, and 60 lb of red snapper.
CHAPTER 11 TEST
1 8 0
0
4
1. 0 1 7 10 is in row-echelon form, but not reduced row-echelon form because the 1 in the second row does not have a 0 0 0 0 0 above it.
0 0
1 0 0
1 0 0
0
0 0 2 5 2. 0 1 2 7 1 0 3 0 3.
0 0 1
is in neither row-echelon nor reduced row-echelon form.
is in reduced row-echelon form. 3
4. 0 1 0 2 is in reduced row-echelon form. 0 0 1 32
1 1 2 0 1 1 2 0 x y 2z 0 R R R 2 1 2 5. 2x 4y 5z 5 has the matrix representation 2 4 5 5 0 2 1 5 0 2 3 5 2y 3z 5 0 2 3 5 1 1 2 0 1 1 2 0 1 R3 R3 R2 R3 5 2 0 2 1 5 0 2 1 5 . Thus z 0, 2y 0 5 y 2 , and 0 0 2 0 0 0 1 0 x 52 2 0 0 x 52 . Thus, the solution is 52 52 0 .
CHAPTER 11
2x 3y z 3 2 3 1 3 6. has the matrix representation x 2y 2z 1 1 2 2 1 4x y 5z 4 4 1 5 4 1 2 2 1 1 2 2 1 R2 2R1 R2 3 R2 R3 0 7 3 5 R 0 7 3 5 R3 4R1 R3 0 7 3 8 0 0 0 3
R1 R2
3R3 4R2 R3
1 2 0
3
0 3 1 2 0 0 7 7
1 7 R3
1 2 0
3
783
1 2 2 1 2 3 1 3 4 1 5 4
.
Since the last row corresponds to the equation 0 3, this system has no solution. 3 1 2 0 3 x 2y R R R 3 1 3 7. has the matrix representation 3y z 2 0 3 1 2 x 2y z 2 1 2 1 2
Test
1
2 0
3
0 3 1 2 0 4 1 5
0 3 1 2 . 0 0 1 1
Thus z 1, 3y 1 2 y 1, and x 2 1 3 x 1. Hence, the solution is 1 1 1. 1 3 1 0 1 3 1 0 x 3y z 0 R2 3R1 R2 0 5 1 1 8. has the matrix representation 3x 4y 2z 1 3 4 2 1 R3 R R 1 3 x 2y 0 5 1 1 1 1 2 0 1 1 3 1 0 1 3 1 0 1 0 25 35 1 R 3R R R3 R2 R3 1 1 1 1 . 1 2 1 0 5 1 1 5 R2 0 1 5 5 0 1 5 5 0 0 0 0 0 0 0 0 0 0 0 0
Since this system is dependent, let z t. Then y 15 t 15 y 15 t 15 and x 25 t 35 x 25 t 35 . Thus, the solution is 25 t 35 15 t 15 t .
In Solutions 9–16, A
2 4
1 0 4
, B 1 1 , and C 1 1 2 . 2 4 3 0 0 1 3 2 3
9. A B is undefined because A is 2 2 and B is 3 2, so they have incompatible dimensions.
10. AB is undefined because A is 2 2 and B is 3 2, so they have incompatible dimensions. 2 4 2 4 12 22 6 12 6 10 2 3 11. B A 3B 1 1 2 4 3 1 1 0 1 3 3 3 2 3 0 3 0 6 9 9 0 3 9 1 0 4 2 4 14 4 36 58 2 3 2 3 1 1 3 3 0 3 12. C B A 1 1 2 2 4 2 4 0 1 3 3 0 8 1 18 28 3 2 3 4 3 2 1 2 A1 13. A 8 6 2 2 2 4 1 1
14. B 1 does not exist because B is not a square matrix.
15. det B is not defined because B is not a square matrix.
784
CHAPTER 11 Matrices and Determinants
1 0 4 1 1 1 2 4 16. det C 1 1 2 1 0 1 1 3 0 1 3 4x 3y 10 17. (a) The system 3x 2y 30 4 3 (b) We have D 3 2
1 4 3
is equivalent to the matrix equation
4 3 3 2
x y
10 30
.
2 3 x 2 3 10 70 4 2 3 3 1. So D 1 and . 3 4 y 3 4 30 90
Therefore, x 70 and y 90.
1 4 1 18. A 0 2 0 1 0 1
1 1 2 1 1
1 4 0 1 0 0, B 0 2 0 2 2. Since A 0, A does not have an inverse, and 3 1 3 0 1
since B 0, B does have an inverse.
1 0 0 1 2 0
0 2 0 0 1 0 0 0 1 3 6 1 z 14 2x 19. 3x y 5z 0 4x 2y 3z 2
1 2 R2
1 4 0 1 0 0
0 2 0 0 1 0 3 0 1 0 0 1
1 0 0 1 2 0
4 0 1 0 0
1 2 0
R1 2R2 R1 R3 6R2 R3
0 1 0 0 1 0 . Therefore, B 1 0 1 0 . 2 2 0 0 1 3 6 1 3 6 1
2 0 1 1 5 Then D 3 1 5 2 2 3 4 2 3
14 0 1 1 5 Dx 0 1 5 14 2 3 2 2 3
3 1 26 10 36, 1 4 2
0 1 182 2 180, 1 4 2
2 14 1 2 14 14 1 D y 3 0 5 3 120 300 180, and 5 4 2 2 3 4 2 3 2 0 14 Dz 3 1 0 4 2 2
0 2 0 0 1 0 0 12 1 3 0 1
R3 3R1 R3
1
1 0 2 2 2
3 1 4 140 144. 14 2 2
180 144 Therefore, x 180 36 5, y 36 5, z 36 4, and so the solution is 5 5 4.
Computer Graphics
785
20. Let x and y represent the number of pounds of almonds and walnuts respectively. Then the problem is modeled by the system
of equations
x
1 1 Then D , so det D 475 345 475 345
y 3
475x 345y 1191
3 1 det Dx 1191 345
1
1
1 3 1035 1191 156, and det D y 475 1191
345 475 13,
1191 1425 234. Then
D y 234 Dx 156 x 12 and y 18, so she bought 12 pounds of almonds and 18 pounds of walnuts. D D 13 13
FOCUS ON MODELING Computer Graphics
1. The data matrix D
0 1 1 0 0 0 1 1
square.
represents the gray
Reflection using T
TD
y
2 1
0
1
0
0 1
1
0
0 1
:
0 1 1 0 0 0 1 1
y
0 1
1
0 0 1 1
2 1
x
2
_1
1
0
1 2
x
_1
Expansion with c 2 using T
2 0
: 0 1 2 0 0 1 1 0 0 2 2 0 TD 0 1 0 0 1 1 0 0 1 1 y
2
_1
Shearing with c 1 using T
1 1
: 0 1 1 1 0 1 1 0 0 1 2 1 TD 0 1 0 0 1 1 0 0 1 1 y
2
1
0
1
1
2
x
0 _1
1
2
0
x
786
FOCUS ON MODELING
2. The data matrix D
0 1 1 0 0 0 1 1
square.
represents the
Reflection in y-axis using T1
T1 D
y
2
1
0
0 1
1
0 1 1 0 0 0 1 1 y
1
_1
Expansion in y-direction with c 3 using T2 T2 D
0 3
0 1 1 0 0 0 1 1
_1
3. (a) T
1 15
0 1 1 0 0
0
1 1
1
:
2
x
2
_1
1 0
0 1
3
0
1 0
y
0 1 1 0 0 0 3 3
1 0
0 3
:
0
Shear in y-direction with c 1 using T3
y
2
2
1
1
x
1 0
: 1 1 1 0 0 1 1 0 0 1 1 0 T3 D 1 1 0 0 1 1 0 1 2 1 3
1
x
3
0
1
_1
0
1
x
is a shear in the x-direction. 0 1 1 15 1 15 1 (b) T 1 1 0 1 0 1
(c) T 1 is a leftward shear in the x-direction.
(d) The result is the original matrix. Algebraically, T 1 T D T 1 T D I D D where I is the 2 2 identity 1 15 1 15 0 1 1 0 1 15 0 1 25 15 0 1 1 0 matrix: 0 1 0 1 0 0 1 1 0 1 0 0 1 1 0 0 1 1
Computer Graphics
4. (a) T
1 15 0 1
x-direction.
is an expansion by a factor of 3 in the
(b) S
1 0 0 2
y-direction.
is an expansion by a factor of 2 in the
y
y
2
2
1
1
0
1
3
2
x
0
_1
3 0
0 3 3 0
(c) T S D
0 1
1
1 0
0 2
0 1 1 0 0 0 1 1
3 0 0 1
0 1 1 0 0 0 2 2
y
2 1
0
This corresponds to expansion by a factor of 3 in the x-direction and a factor of 2 in the y-direction. 3 0 1 0 3 0 (d) W T S 0 1 0 2 0 2 3 0 0 1 1 0 0 3 3 0 . It is the same as T S D. (e) W D 0 2 0 0 1 1 0 0 2 2
5. (a) D
0 1 1 4 4 1 1 6 6 0 0
0 0 4 4 5 5 075 0 , (b) T 0 1 075 0 0 TD 0 1 0 1 025 , (c) S 0 1 1 025 0 SD 0 1 0
x
3
2
_1
0 0 2 2
787
7 7 8 8 0
1
3
2
_1
1 1 4 4 1 1 6 6 0 0 0 4 4 5 5 7 7 8 8 0
1 1 4 4 1 1 6 6 0 0 0 4 4 5 5 7 7 8 8 0
0 0
0 1 2 5 525 225 275 775 8 2 0
0 075 075 3 3 075 075 45 45 0 0 4
0 0 4 4 5
4 5 5
5
7
7
7
7
8
8 8 0
8 0
x
788
FOCUS ON MODELING
6. (a) The data matrix D
0 1 2 1 0 0 0 0 2 4 4 0
figure at right.
represents the
y
4 3 2 1
(b) T
1
1
, 0 1 1 1 0 1 2 1 0 0 TD 0 1 0 0 2 4 4 0 0 1 4 5 4 0 0 0 2 4 4 0
The transformation is a reflection about the x-axis and a shear in the x-direction. 1 1 1 0 1 1 (c) T 0 1 0 1 0 1
0
1
2
3
4
5
x
1
2
3
4
5
x
y 0 _1 _2 _3 _4
12
CONIC SECTIONS
12.1 PARABOLAS 1. A parabola is the set of all points in the plane equidistant from a fixed point called the focus and a fixed line called the directrix of the parabola. 2. The graph of the equation x 2 4 py is a parabola with focus F 0 p and directrix y p. So the graph of x 2 12y is a parabola with focus F 0 3 and directrix y 3.
3. The graph of the equation y 2 4 px is a parabola with focus F p 0 and directrix x p. So the graph of y 2 12x is a parabola with focus F 3 0 and directrix x 3. 4. (a) From top to bottom: focus 0 3, vertex 0 0, directrix y 3. (b) From left to right: directrix x 3, vertex 0 0, focus 3 0.
5. y 2 2x is Graph III, which opens to the right and is not as wide as the graph for Exercise 5.
6. y 2 14 x is Graph V, the only graph that opens downward.
7. x 2 6y is Graph II, which opens downward and is narrower than the graph for Exercise 6. 8. 2x 2 y is Graph I, the only graph that opens to the right.
9. y 2 8x 0 is Graph VI, which opens to the right and is wider than the graph for Exercise 1.
10. 12y x 2 0 is Graph IV, which opens downward and is wider than the graph for Exercise 3. 11. (a) x 2 8y, so 4 p 8 p 2. The focus is 0 2,
the directrix is y 2, and the focal diameter is 8.
12. (a) x 2 4y, so 4 p 4 p 1. The focus is
0 1, the directrix is y 1, and the focal diameter is 4.
y
(b)
y
(b)
1 1 1
13. (a) y 2 24x, so 4 p 24 p 6. The focus is
6 0, the directrix is x 6, and the focal diameter is 24.
(b)
1
x
x
14. (a) y 2 16x, so 4 p 16 p 4. The focus is 4 0, the directrix is x 4, and the focal diameter is 16.
(b)
y
y 2 2 1
1
x
x
789
790
CHAPTER 12 Conic Sections
15. (a) y 18 x 2 x 2 8y, so 4 p 8 p 2. The focus is 0 2, the directrix is y 2, and the focal diameter is 8.
16. (a) x 2y 2 y 2 12 x, so 4 p 12 p 18 . The focus is 18 0 , the directrix is x 18 , and the focal diameter is 12 .
y
(b)
y
(b)
1
1 x
1
1
17. (a) x 2y 2 y 2 12 x, so 4 p 12 p 18 . The focus is 18 0 , the directrix is x 18 , and the focal diameter is 12 .
18. (a) y 14 x 2 x 2 4y, so 4 p 4 p 1. The focus is 0 1, the directrix is y 1, and the focal diameter is 4.
y
(b) y
(b)
x
1
1
x
1
19. (a) 5y x 2 , so 4 p 5 p 54 . The focus is 0 54 , the directrix is y 54 , and the focal diameter is 5.
(b)
1
y
20. (a) 9x y 2 , so 4 p 9 p 94 . The focus is 94 0 , the directrix is x 94 , and the focal diameter is 9.
(b)
y
2 1 1
x
x
1
x
SECTION 12.1 Parabolas
21. (a) x 2 12y 0 x 2 12y, so 4 p 12
p 3. The focus is 0 3, the directrix is y 3,
and the focal diameter is 12.
22. (a) x 15 y 2 0 y 2 5x, so 4 p 5 p 54 . The focus is 54 0 , the directrix is x 54 , and the focal diameter is 5.
y
(b)
y
(b) 1 x
2
1 1
23. (a) 5x 3y 2 0 y 2 53 x. Then 4 p 53 5 . The focus is 5 0 , the directrix is p 12 12 5 , and the focal diameter is 5 . x 12 3
(b)
24. (a) 8x 2 12y 0 x 2 32 y. Then 4 p 32 p 38 . The focus is 0 38 , the directrix is y 38 , and the focal diameter is 32 . y
(b)
y
1 1
1 1
x
x
x
25. x 2 16y
26. x 2 8y 1 -5
-5
-2
5
27. y 2 13 x
5
28. 8y 2 x 1
2
5
-2 -1
-2
10
791
792
CHAPTER 12 Conic Sections
29. 4x y 2 0
30. x 2y 2 0 2
-2
-1
1
2
4
-2
31. Since the focus is 0 6, p 6 4 p 24. Hence, an equation of the parabola is x 2 24y. 32. Since the focus is 0 14 , p 14 4 p 1. So an equation of the parabola is x 2 y.
33. Since the focus is 8 0, p 8 4 p 32. Hence, an equation of the parabola is y 2 32x.
34. Since the focus is 5 0, p 5 4 p 20. Hence, an equation of the parabola is y 2 20x. 35. Since the focus is 0 34 , p 34 4 p 3. Hence, an equation of the parabola is x 2 3y. 1 0 , p 1 4 p 1 . Hence, an equation of the parabola is y 2 1 x. 36. Since the focus is 12 12 3 3
37. Since the directrix is x 4, p 4 4 p 16. Hence, an equation of the parabola is y 2 16x.
38. Since the directrix is y 12 , p 12 4 p 2. Hence, an equation of the parabola is x 2 2y.
1 , p 1 4 p 2 . Hence, an equation of the parabola is x 2 2 y. 39. Since the directrix is y 10 10 5 5
40. Since the directrix is x 18 , p 18 4 p 12 . Hence, an equation of the parabola is y 2 12 x.
1 , p 1 4 p 1 . Hence, an equation of the parabola is y 2 1 x. 41. Since the directrix is x 20 20 5 5
42. Since the directrix is y 5, p 5 4 p 20. Hence, an equation of the parabola is x 2 20y.
43. The focus is on the positive x-axis, so the parabola opens horizontally with 2 p 2 4 p 4. So an equation of the parabola is y 2 4x.
44. The focus is on the negative y-axis, so the parabola opens vertically with 2 p 6 4 p 12. So an equation of the parabola is x 2 12y.
45. The parabola opens downward with focus 10 units from 0 0, so p 10 4 p 40 and an equation of the parabola is x 2 40y.
46. Since the parabola opens upward with focus 5 units from the vertex, the focus is 5 0. So p 5 4 p 20. Thus an equation of the parabola is x 2 20y.
47. The directrix has y-intercept 6, and so p 6 4 p 24. Therefore, an equation of the parabola is x 2 24y. 48. Since the focal diameter is 8 and the focus is on the negative y-axis, 4 p 8. So the equation is x 2 8y.
49. p 2 4 p 8. Since the parabola opens upward, its equation is x 2 8y.
50. The directrix is x 2, and so p 2 4 p 8. Since the parabola opens to the left, its equation is y 2 8x. 51. p 4 4 p 16. Since the parabola opens to the left, its equation is y 2 16x.
52. p 3 4 p 12. Since the parabola opens downward, its equation is x 2 12y.
53. The focal diameter is 4 p 32 32 3. Since the parabola opens to the left, its equation is y 2 3x. 54. The focal diameter is 4 p 2 5 10. Since the parabola opens upward, its equation is x 2 10y.
55. The equation of the parabola has the form y 2 4 px. Since the parabola passes through the point 4 2, 22 4 p 4 4 p 1, and so an equation is y 2 x.
56. Since the directrix is x p, we have p2 16, so p 4, and an equation is y 2 4 px or y 2 16x.
SECTION 12.1 Parabolas
793
57. The area of the shaded region is width height 4 p p 8, and so p2 2 p 2 (because the parabola opens downward). Therefore, an equation is x 2 4 py 4 2y x 2 4 2y. 58. The focus is 0 p. Since the line has slope 12 , an equation of the line is y 12 x p. Therefore, the point where the line intersects the parabola has y-coordinate 12 2 p p 1. The parabola’s equation is of the form x 2 4 py, 1 5 (since p 0). Hence, an equation of the parabola is so 22 4 p p 1 p2 p 1 0 p 2 5 1 y. x2 2
59. (a) A parabola with directrix y p has equation x 2 4 py. If the directrix is y 12 , then p 12 , so an equation is x 2 4 12 y x 2 2y.
(b)
1
-4
-2
4 x@=_32y x@=_16y
-3
x 2 4y. If the directrix is y 4, then p 4, so an equation is
-4 x@=_4y x@=_2y
-5
x 2 4 4 y x 2 16y. If the directrix is y 8, then p 8, so
an equation is x 2 4 8 y x 2 32y.
As the directrix moves further from the vertex, the parabolas get flatter. (b)
10
focal diameter is 4 p 1, an equation is x 2 y. If the focal diameter is
y=x@
2y=x@
8
4y=x@
6
4 p 2, an equation is x 2 2y. If the focal diameter is 4 p 4, an
4
equation is x 2 4y. If the focal diameter is 4 p 8, an equation is x 2 8y.
2
-2
If the directrix is y 1, then p 1, so an equation is x 2 4 1 y
60. (a) If the focal diameter of a parabola is 4 p, it has equation x 2 4 py. If the
0 -1
8y=x@
2 -4
-2
0
2
4
As the focal diameter increases, the parabolas get flatter.
61. (a) Since the focal diameter is 12 cm, 4 p 12. Hence, the parabola has equation y 2 12x.
(b) At a point 20 cm horizontally from the vertex, the parabola passes through the point 20 y, and hence from part (a), y 2 12 20 y 2 240 y 4 15. Thus, C D 8 15 31 cm.
62. The equation of the parabola has the form x 2 4 py. From the diagram, the parabola passes through the point 10 1, and so 102 4 p 1 4 p 100 and p 25. Therefore, the receiver is 25 ft from the vertex.
63. With the vertex at the origin, the top of one tower will be at the point 300 150. Inserting this point into the equation x 2 4 py gives 3002 4 p 150 90000 600 p p 150. So an equation of the parabolic part of the cables is
x 2 4 150 y x 2 600y.
64. The equation of the parabola has the form x 2 4 py. From the diagram, the parabola passes through the point 100 379, and so 1002 4 p 379 1516 p 10000 and p 65963. Therefore, the receiver is about 65963 inches 55 feet from the vertex.
65. Many answers are possible: satellite dish TV antennas, sound surveillance equipment, solar collectors for hot water heating or electricity generation, bridge pillars, etc. 66. Yes. If a cone intersects a plane that is parallel to a line on the cone, the resulting curve is a parabola, as shown in the text.
794
CHAPTER 12 Conic Sections
12.2 ELLIPSES 1. An ellipse is the set of all points in the plane for which the sum of the distances from two fixed points F1 and F2 is constant. The points F1 and F2 are called the foci of the ellipse. y2 x2 2. The graph of the equation 2 2 a b where c a 2 b2 . So the graph of
1 with a b 0 is an ellipse with vertices a 0 and a 0 and foci c 0, x2 y2 2 1 is an ellipse with vertices 5 0 and 5 0 and foci 3 0 and 2 5 4
3 0.
x2 y2 3. The graph of the equation 2 2 1 with a b 0 is an ellipse with vertices 0 a and 0 a and foci 0 c, b a x2 y2 where c a 2 b2 . So the graph of 2 2 1 is an ellipse with vertices 0 5 and 0 5 and foci 0 3 and 4 5 0 3. 4. (a) From left to right: vertex 5 0, focus 3 0, focus 3 0, vertex 5 0. (b) From top to bottom: vertex 0 5, focus 0 3, focus 0 3, vertex 0 5. 5.
y2 x2 1 is Graph II. The major axis is horizontal and the vertices are 4 0. 16 4
y2 6. x 2 1 is Graph IV. The major axis is vertical and the vertices are 0 3. 9 7. 4x 2 y 2 4 is Graph I. The major axis is vertical and the vertices are 0 2. 8. 16x 2 25y 2 400 is Graph III. The major axis is horizontal and the vertices are 5 0. 9.
y2 x2 1. 25 9
(c)
(a) This ellipse has a 5, b 3, and so c2 a 2 b2 16 c 4. The vertices c are 5 0, the foci are 4 0, and the eccentricity is e 45 08. a
y
1 1
x
1
x
(b) The length of the major axis is 2a 10, and the length of the minor axis is 2b 6.
10.
y2 x2 1 16 25 (a) This ellipse has a 5, b 4, and so c2 25 16 9 c 3. The vertices are 0 5, the foci are 0 3, and the eccentricity is e ac 35 06.
(b) The length of the major axis is 2a 10, and the length of the minor axis is 2b 8.
(c)
y
1
SECTION 12.2 Ellipses
11.
y2 x2 1 36 81
(c)
(a) This ellipse has a 9, b 6, and so c2 81 36 45 c 3 5. The vertices are 0 9, the foci are 0 3 5 , and the eccentricity is
795
y
2 x
2
e ac 35 .
(b) The length of the major axis is 2a 18 and the length of the minor axis is 2b 12.
12.
x2 y2 1 4
(c)
(a) This ellipse has a 2, b 1, and so c2 4 1 3 c 3. The vertices are 2 0, the foci are 3 0 , and the eccentricity is e ac 23 .
y
1 1
x
(b) The length of the major axis is 2a 4 and the length of the minor axis is 2b 2.
13.
y2 x2 1 49 25
(c)
(a) This ellipse has a 7, b 5, and so c2 49 25 24 c 2 6. The vertices are 7 0, the foci are 2 6 0 , and the eccentricity is
y
2 2
x
2
x
e ac 2 7 6 .
(b) The length of the major axis is 2a 14 and the length of the minor axis is 2b 10.
14.
x2 y2 1 9 64
(c)
(a) This ellipse has a 8, b 3, and so c2 64 9 55 c 55. The vertices are 0 8, the foci are 0 55 , and the eccentricity is
y
2
e ac 855 .
(b) The length of the major axis is 2a 16 and the length of the minor axis is 2b 6.
15. 9x 2 4y 2 36
y2 x2 1 4 9
(a) This ellipse has a 3, b 2, and so c2 9 4 5 c 5. The vertices are 0 3, the foci are 0 5 , and the eccentricity is e ac 35 .
(b) The length of the major axis is 2a 6, and the length of the minor axis is 2b 4.
(c)
y
1 1
x
796
CHAPTER 12 Conic Sections
16. 4x 2 25y 2 100
y2 x2 1 25 4
(c)
(a) This ellipse has a 5, b 2, and so c2 25 4 21 c 21. The vertices are 5 0, the foci are 21 0 , and the eccentricity is
y
1 1
x
1
x
1
x
2
x
e ac 521 .
(b) The length of the major axis is 2a 10, and the length of the minor axis is 2b 4.
y2 x2 1 17. x 2 4y 2 16 16 4
(c)
(a) This ellipse has a 4, b 2, and so c2 16 4 12 c 2 3.The vertices are 4 0, the foci are 2 3 0 , and the eccentricity is
y
1
e ac 2 4 3 23 .
(b) The length of the major axis is 2a 8, and the length of the minor axis is 2b 4.
18. 4x 2 y 2 16
y2 x2 1 4 16
(c)
(a) This ellipse has a 4, b 2, and so c2 16 4 12 c 2 3. The vertices are 0 4, the foci are 0 2 3 , and the eccentricity is
y
1
e ac 2 4 3 23 .
(b) The length of the major axis is 2a 8, and the length of the minor axis is 2b 4.
19. 16x 2 25y 2 1600
y2 x2 1 100 64
(c)
(a) This ellipse has a 10, b 8, and so c2 100 64 36 c 6. The
y
2
vertices are 10 0, the foci are 6 0, and the eccentricity is e ac 35 .
(b) The length of the major axis is 2a 20 and the length of the minor axis is 2b 16.
y2 x2 1 49 2 (a) This ellipse has a 7, b 2, and so c2 49 2 47 c 47. The vertices are 7 0, the foci are 47 0 , and the eccentricity is
20. 2x 2 49y 2 98
e ac 747 .
(b) The length of the major axis is 2a 14 and the length of the minor axis is 2b 2 2.
(c)
y
1 1
x
SECTION 12.2 Ellipses
y2 x2 1 3 9 (a) This ellipse has a 3, b 3, and so c2 9 3 6 c 6. The vertices are 0 3, the foci are 0 6 , and the eccentricity is
21. 3x 2 y 2 9
(c)
797
y
1 1
x
1
x
1
x
1
x
e ac 36 .
(b) The length of the major axis is 2a 6 and the length of the minor axis is 2b 2 3. x2 y2 22. x 2 3y 2 9 1 9 3 (a) This ellipse has a 3, b 3, and so c2 9 3 6 c 6. The vertices are 3 0, the foci are 6 0 , and the eccentricity is
(c)
y
1
e ac 36 .
(b) The length of the major axis is 2a 6 and the length of the minor axis is 2b 2 3. y2 x2 1 2 4 (a) This ellipse has a 2, b 2, and so c2 4 2 2 c 2. The vertices are 0 2, the foci are 0 2 , and the eccentricity is
23. 2x 2 y 2 4
(c)
y
1
e ac 22 .
(b) The length of the major axis is 2a 4 and the length of the minor axis is 2b 2 2. y2 x2 1 4 3 (a) This ellipse has a 2, b 3, and so c2 4 3 c 1. The vertices are
24. 3x 2 4y 2 12
(c)
y
1
2 0, the foci are 1 0, and the eccentricity is e ac 12 .
(b) The length of the major axis is 2a 4 and the length of the minor axis is 2b 2 3. y2 x2 1 1 25. x 2 4y 2 1 1 4
(a) This ellipse has a 1, b 12 , and so c2 1 14 34 c 23 . The vertices are 1 0, the foci are 23 0 , and the eccentricity is 3 e ac 32 1 2 .
(b) The length of the major axis is 2a 2, and the length of the minor axis is 2b 1.
(c)
y 1
1
x
798
CHAPTER 12 Conic Sections
26. 9x 2 4y 2 1
y2 x2 1 19 14
(c)
5 c 5 . The vertices are (a) This ellipse has a 12 , and so c2 14 19 36 6 56 5 1 0 2 , the foci are 0 6 , and the eccentricity is e ac 12 35 .
y 0.5
0.5 x
(b) The length of the major axis is 2a 1, and the length of the minor axis is 2b 23 .
27. x 2 4 2y 2 x 2 2y 2 4
y2 x2 1 4 2
(c)
2, and so c2 4 2 2 c 2. The vertices are 2 0, the foci are 2 0 , and the eccentricity is
y
1
(a) This ellipse has a 2, b
x
1
e ac 22 .
(b) The length of the major axis is 2a 4, and the length of the minor axis is 2b 2 2. x2 y2 1 28. y 2 1 2x 2 2x 2 y 2 1 1 1
(c)
y 1
2 2 (a) This ellipse has a 1, b 2 , and so c2 1 12 12 c 22 . The vertices are 0 1, the foci are 0 22 , and the eccentricity is e ac 11 2 22 .
1
x
(b) The length of the major axis is 2a 2, and the length of the minor axis is 2b 2. 29. This ellipse has a horizontal major axis with a 5 and b 4, so an equation is
x2 52
y2 42
1
y2 x2 1. 25 16
y2 x2 y2 x2 1. 30. This ellipse has a vertical major axis with a 5 and b 2. Thus an equation is 2 2 1 4 25 2 5 31. This ellipse has a vertical major axis with c 2 and b 2. So a 2 c2 b2 22 22 8 a 2 2. So an equation is
x2
y2 y2 x2 1. 1 2 4 8 22 2 2
32. This ellipse has a vertical major axis with a 4 and c 3. So c2 a 2 b2 9 16 b2 b2 7. Thus an equation is
x2 y2 y2 x2 2 1 1. 7 7 16 4
x2 y2 2 1. Substituting 2 16 b 36 36 36 3 4 36 64 1 2 48. Thus, an the point 8 6 into the equation, we get 256 2 1 2 1 4 2 b 4 3 b b b y2 x2 1. equation of the ellipse is 256 48
33. This ellipse has a horizontal major axis with a 16, so an equation of the ellipse is of the form
SECTION 12.2 Ellipses
799
x2 y2 34. This ellipse has a vertical major axis with b 2, so an equation of the ellipse is of the form 2 2 1. Substituting the 2 a 4 16 4 4 1 3 4 4 . Thus, an equation of point 1 2 into the equation, we get 14 2 1 2 1 2 a 2 4 4 3 3 a a a x2 y2 x2 3y 2 the ellipse is 1 1. 4 163 4 16 35.
x2 y2 y2 x2 4x 2 1 1 y 2 20 25 20 20 25 5 4x 2 . y 20 5
y2 y2 1 1 x 2 y 2 12 12x 2 12 12 y 12 12x 2 .
36. x 2
4 2
5 -2 -5
-1
-2
1
2
5 -4 -5
37. 6x 2 y 2 36 y 2 36 6x 2 y 36 6x 2 .
x2 38. x 2 2y 2 8 2y 2 8 x 2 y 2 4 2 x2 y 4 . 2
5
-10
10
2
-5 -2
2 -2
39. The foci are 4 0, and the vertices are 5 0. Thus, c 4 and a 5, and so b2 25 16 9. Therefore, an equation of the ellipse is
y2 x2 1. 25 9
40. The foci are 0 3 and the vertices are 0 5. Thus, c 3 and a 5, and so c2 a 2 b2 9 25 b2 b2 25 9 16. Therefore, an equation of the ellipse is
y2 x2 1. 16 25
41. The foci are 1 0 and the vertices are 2 0. Thus, c 1 and a 2, so c2 a 2 b2 1 4 b2 b2 4 1 3. Therefore, an equation of the ellipse is
y2 x2 1. 4 3
42. The foci are 0 2 and the vertices are 0 3. Thus, c 2 and a 3, so c2 a 2 b2 4 9 b2 b2 9 4 5. Therefore, an equation of the ellipse is
x2 y2 1. 5 9
43. The foci are 0 10 and the vertices are 0 7. Thus, c 10 and a 7, so c2 a 2 b2 10 49 b2 b2 49 10 39. Therefore, an equation of the ellipse is
y2 x2 1. 39 49
800
CHAPTER 12 Conic Sections
44. The foci are 15 0 and the vertices are 6 0. Thus, c 15 and a 6, so c2 a 2 b2 15 36 b2
y2 x2 1. 36 21 45. The length of the major axis is 2a 4 a 2, the length of the minor axis is 2b 2 b 1, and the foci are on the b2 36 15 21. Therefore, an equation of the ellipse is
y2 1. 4 46. The length of the major axis is 2a 6 a 3, the length of the minor axis is 2b 4 b 2, and the foci are on the y-axis. Therefore, an equation of the ellipse is x 2
x-axis. Therefore, an equation of the ellipse is
x2 y2 1. 9 4
47. The foci are 0 2, and the length of the minor axis is 2b 6 b 3. Thus, a 2 4 9 13. Since the foci are on the y-axis, an equation is
x2 y2 1. 9 13
48. The foci are 5 0, and the length of the major axis is 2a 12 a 6. Thus, c2 a 2 b2 25 36 b2
y2 x2 1. 36 11 49. The endpoints of the major axis are 10 0 a 10, and the distance between the foci is 2c 6 c 3. Therefore, b2 36 25 11. Since the foci are on the x-axis, an equation is
x2 y2 1. 100 91 50. Since the endpoints of the minor axis are 0 3, we have b 3. The distance between the foci is 2c 8, so c 4. Thus, b2 100 9 91, and so an equation of the ellipse is
y2 x2 1. 25 9 51. The length of the major axis is 10, so 2a 10 a 5, and the foci are on the x-axis, so the form of the equation is 2 5 y2 4 x2 5 4 4 22 2 1. Since the ellipse passes through 2 1 5 2 , we have 1 2 25 25 25 b2 5 b b b y2 x2 1. b2 5, and so an equation is 25 5 52. The length of the minor axis is 10, so 2b 10 b 5, and the foci are on the y-axis, so the form of the equation is 2 2 5 40 x 2 y2 5 40 4 40 5 40 , we have 1 1 2 2 1. Since the ellipse passes through 25 a 25 25 a 2 5 a2 a 2 2 y x 1. a 2 50, and so an equation is 25 50 a 2 b2 c2 9 16 25, and an equation of the ellipse is
2 6 and 53. The eccentricity is 13 , so e 13 , and the foci are 0 2, so c 2. Thus, e ac a ce 13
y2 x2 1. b2 a 2 c2 36 4 32. The major axis lies on the y-axis, so an equation is 32 36 54. The eccentricity is e 075 34 and the foci are 15 0 32 0 , so c 32 . Thus, a ce 32 34 2 and y2 x2 7 1. b2 a 2 c2 4 94 74 . The major axis lies on the x-axis, so an equation is 4
4 3 55. Since the length of the major axis is 2a 4, we have a 2. The eccentricity is 2 ac 2c , so c 3. Then y2 b2 a 2 c2 4 3 1, and since the foci are on the y-axis, an equation of the ellipse is x 2 1.
4
56. The eccentricity is e 35 and the major axis has length 2a 12, so a 6. Thus, c ae 2 5 and y2 x2 1. b2 a 2 c2 36 20 16. The foci are on the x-axis, so an equation of the ellipse is
36
16
SECTION 12.2 Ellipses
4x 2 y 2 4 57. 4x 2 9y 2 36
y (0, 2)
Subtracting the first equation from the second gives 1
8y 2 32 y 2 4 y 2. Substituting y 2 in the first equation gives
(0, _2)
y2 1 9x 2 16y 2 144 144x 2 256y 2 2304 9 2 2 16x 9y 144 144x 2 81y 2 1296 y2 1 16
y
(_ 125 , 125)
Dividing the first equation by 100 gives x 2 y2
( 125 , 125)
1
x
1
Adding gives 175y 2 1008 y 12 5 . Substituting for y gives 2 1296 12 144 9x 2 144 2304 9x 2 16 12 5 25 25 x 5 , and so the 12 four points of intersection are 12 5 5 .
100x 2 25y 2 100 59. y2 x 2 1 9
x
1
4x 2 22 4 x 0, and so the points of intersection are 0 2.
2 x 16 58. 2 x 9
801
(
12 12 _5 ,_5
)
(
12 12 5,_5
)
y
y2 1. 4
1
(_1, 0)
y2
(1, 0) 1
0 Subtracting this equation from the second equation gives 9 4 1 1 y 2 0 y 0. Substituting y 0 in the second equation gives 9 4
x
x 2 02 1 x 1, and so the points of intersection are 1 0.
25x 2 144y 2 3600 3600x 2 20,736y 2 518,400 60. 144x 2 25y 2 3600 3600x 2 625y 2 90,000
y
Subtracting
3600 the second equation from the first, we have 20,111y 2 428,400, so y 2 169 2 60 2 y 60 3600 x 60 13 . Substituting for y gives 25x 144 13 13 , and so 60 the four points of intersection are 60 13 13 .
10
(_ 6013 , 6013 )
( 6013 , 6013 ) 10 x
(
_ 60 , _ 60 13 13
(
)
60 , _ 60 13 13
)
y2 x2 1 has a 4 and b 2. Thus, an equation of the ancillary circle is 61. (a) The ellipse x 2 4y 2 16 16 4 x 2 y 2 4.
(b) If s t is a point on the ancillary circle, then s 2 t 2 4 4s 2 4t 2 16 2s2 4 t2 16, which implies that 2s t is a point on the ellipse.
1 62. (a) x 2 ky 2 100 ky 2 100 x 2 y 100 x 2 . For the top k 1 half, we graph y 100 x 2 for k 4, 10, 25, and 50. k (b) This family of ellipses have common major axes and vertices, and the eccentricity increases as k increases.
k=4 2 1 k=50 -10-8 -6 -4 -2 0
k=10 k=25 2 4 6 8 10
802
63.
CHAPTER 12 Conic Sections
y2 x2 1 is an ellipse for k 0. Then a 2 4 k, b2 k, and so c2 4 k k 4 c 2. Therefore, all of k 4k the ellipses’ foci are 0 2 regardless of the value of k.
64. The foci are c 0, where c2 a 2 b2 . The endpoints of one latus rectum are the points c k, and the length is 2k. 2 a 2 c2 2 2 2 2 2 2 b k k c a c c k2 . Since Substituting this point into the equation, we get 2 2 1 2 1 2 2 a b b a a2 a 2b2 b4 b2 b2 b2 a 2 c2 , the last equation becomes k 2 2 k . Thus, the length of the latus rectum is 2k 2 . a a a a 65. Using the perihelion, a c 147,000,000, while using the aphelion, a c 153,000,000. Adding, we have 2 2 2a 300,000,000 a 150,000,000. So b2 a 2 c2 150 106 3 106 22,4911012 224911016 .
y2 x2 1. 16 22500 10 22491 1016 c 66. Using the eccentricity, e 025 c 025a. Using the length of the minor axis, 2b 10,000,000,000 a 18 b 5 109 . Since a 2 c2 b2 , a 2 025a2 25 1018 15 a 2 25 1018 a 2 80 3 10 16 5 5 9 9 9 5 109 . Since the Sun is at one focus of the ellipse, a 80 3 10 4 3 10 . Then c 025 4 3 10 3 the distance from Pluto to the Sun at perihelion is a c 4 53 109 53 109 3 53 109 387 109 km; the distance from Pluto to the Sun at aphelion is a c 4 53 109 53 109 5 53 109 645 109 km. Thus, an equation of the orbit is
67. Using the perilune, a c 1075 68 1143, and using the apolune, a c 1075 195 1270. Adding, we get
2a 2413 a 12065. So c 1270 12065 c 635. Therefore, b2 120652 6352 1,451,610. Since
y2 x2 1. 1,455,642 1,451,610 68. Placing the origin at the center of the sheet of plywood and letting the x-axis be the long central axis, we have 2a 8, so that a 4, and 2b 4, so that b 2. So c2 a 2 b2 42 22 12 c 2 3 346. So the tacks should be located 2 346 692 feet apart and the string should be 2a 8 feet long. a 2 1,455,642, an equation of Apollo 11’s orbit is
69. From the diagram, a 40 and b 20, and so an equation of the ellipse whose top half is the window is
x2 y2 1. 1600 400
252 h2 Since the ellipse passes through the point 25 h, by substituting, we have 1 625 4y 2 1600 1600 400 5 39 975 1561 in. Therefore, the window is approximately 156 inches high at the specified point. y 2 2 70. Have each friend hold one end of the string on the blackboard. These fixed points will be the foci. Then, keeping the string taut with the chalk, draw the ellipse. 71. We start with the flashlight perpendicular to the wall; this shape is a circle. As the angle of elevation increases, the shape of the light changes to an ellipse. When the flashlight is angled so that the outer edge of the light cone is parallel to the wall, the shape of the light is a parabola. Finally, as the angle of elevation increases further, the shape of the light is hyperbolic. 72. The shape drawn on the paper is almost, but not quite, an ellipse. For example, when the bottle has radius 1 unit and the compass legs are set 1 unit apart, then it can be shown that an equation of the resulting curve is 1 y 2 2 cos x. The graph of this curve differs very slightly from the ellipse with the same major and minor axis. This example shows that in mathematics, things are not always as they appear to be.
SECTION 12.3 Hyperbolas
803
12.3 HYPERBOLAS 1. A hyperbola is the set of all points in the plane for which the difference of the distances from two fixed point F1 and F2 is constant. The points F1 and F2 are called the foci of the hyperbola. y2 x2 2. The graph of the equation 2 2 1 with a 0, b 0 is a hyperbola with horizontal transverse axis, vertices a 0 a b x2 y2 and a 0 and foci c 0, where c a 2 b2 . So the graph of 2 2 1 is a hyperbola with vertices 4 0 and 4 3 4 0 and foci 5 0 and 5 0. y2 x2 3. The graph of the equation 2 2 1 with a 0, b 0 is a hyperbola with vertical transverse axis, vertices 0 a a b y2 x2 and 0 a and foci 0 c, where c a 2 b2 . So the graph of 2 2 1 is a hyperbola with vertices 0 4 and 4 3 0 4 and foci 0 5 and 0 5. 4. (a) From left to right: focus 5 0, vertex 4 0, asymptote y 34 x, asymptote y 34 x, vertex 4 0, focus 5 0.
(b) From top to bottom: focus 0 5, vertex 0 4, asymptote y 43 x, asymptote y 43 x, vertex 0 4, focus 0 5. x2 y 2 1 is Graph III, which opens horizontally and has vertices at 2 0. 5. 4 x2 6. y 2 1 is Graph IV, which opens vertically and has vertices at 0 1. 9 7. 16y 2 x 2 144 is Graph II, which pens vertically and has vertices at 0 3. 8. 9x 2 25y 2 225 is Graph I, which opens horizontally and has vertices at 5 0.
9.
y2 x2 1 4 16
(c)
(a) The hyperbola has a 2, b 4, and c2 16 4 c 2 5. The vertices are 2 0, the foci are 2 5 0 , and the asymptotes are y 42 x
y
1 1
x
1
x
1
x
y 2x.
(b) The transverse axis has length 2a 4. 10.
x2 y2 1 9 16
(c)
(a) The hyperbola has a 3, b 4, and c2 9 16 25 c 5. The vertices
y
1
are 0 3, the foci are 0 5, and the asymptotes are y 34 x.
(b) The transverse axis has length 2a 6.
11.
x2 y2 1 36 4 (a) The hyperbola has a 6, b 2, and c2 36 4 40 c 2 10. The vertices are 0 6, the foci are 0 2 10 , and the asymptotes are y 3x.
(b) The transverse axis has length 2a 12.
(c)
y
2
804
12.
CHAPTER 12 Conic Sections
y2 x2 1 9 64
(c)
(a) The hyperbola has a 3, b 8, and c2 9 64 73 c 73. The vertices are 3 0, the foci are 73 0 , and the asymptotes are y 83 x.
y
5 2
x
1
x
(b) The transverse axis has length 2a 6.
13.
x2 y2 1 1 25
(c)
y 1
(a) The hyperbola has a 1, b 5, and c2 1 25 26 c 26. The vertices are 0 1, the foci are 0 26 , and the asymptotes are y 15 x.
(b) The transverse axis has length 2a 2.
14.
y2 x2 1 2 1
(c)
(a) The hyperbola has a 2, b 1, and c2 2 1 3 c 3. The vertices are 2 0 , the foci are 3 0 , and the asymptotes are
y
1 1
x
1
x
y 22 x.
(b) The transverse axis has length 2a 2 2. 15. x 2 y 2 1
(c)
(a) The hyperbola has a 1, b 1, and c2 1 1 2 c 2. The vertices are 1 0, the foci are 2 0 , and the asymptotes are y x.
y
1
(b) The transverse axis has length 2a 2.
16.
y2 x2 1 16 12
(c)
(a) The hyperbola has a 4, b 2 3, and c2 16 12 28 c 28 2 7. The vertices are 4 0, the foci are 2 7 0 , and the
y
2 2
x
1
x
asymptotes are y 23 x.
(b) The transverse axis has length 2a 8. 17. 9x 2 4y 2 36
y2 x2 1 4 9
(a) The hyperbola has a 2, b 3, and c2 4 9 13 c 13. The vertices are 2 0, the foci are 13 0 , and the asymptotes are y 32 x.
(b) The transverse axis has length 2a 4.
(c)
y
1
SECTION 12.3 Hyperbolas
18. 25y 2 9x 2 225
x2 y2 1 9 25
(c)
(a) The hyperbola has a 3, b 5, and c2 25 9 34 c 34. The vertices are 0 3, the foci are 0 34 , and the asymptotes are y 35 x.
805
y
2 2
x
2
x
1
x
1
x
1
x
(b) The transverse axis has length 2a 6.
19. 4y 2 9x 2 144
x2 y2 1 36 16
(c)
(a) The hyperbola has a 6, b 4, and c2 a 2 b2 52 c 2 13. The vertices are 0 6, the foci are 0 2 13 , and the asymptotes are
y
2
y 32 x.
(b) The transverse axis has length 2a 12.
x2 y2 1 20. y 2 25x 2 100 100 4
(c)
(a) The hyperbola has a 10, b 2, and c2 100 4 104 c 2 26. The vertices are 0 10, the foci are 0 2 26 , and the asymptotes are
y
5
y 5x.
(b) The transverse axis has length 2a 20.
y2 x2 1 8 2 (a) The hyperbola has a 2 2, b 2, and c2 8 2 10 c 10. The vertices are 2 2 0 , the foci are 10 0 , and the asymptotes are
21. x 2 4y 2 8 0
(c)
y
1
y 2 x 12 x. 8
(b) The transverse axis has length 2a 4 2.
y2 x2 1 3 9 (a) The hyperbola has a 3, b 3, and c2 3 9 12 c 2 3. The vertices are 0 3 , the foci are 0 2 3 , and the asymptotes are
22. 3y 2 x 2 9 0
y 33 x.
(b) The transverse axis has length 2a 2 3.
(c)
y
1
806
CHAPTER 12 Conic Sections
x2 y2 1 23. x 2 y 2 4 0 y 2 x 2 4 4 4
y
(c)
(a) The hyperbola has a 2, b 2, and c2 4 4 8 2 2. The vertices are 0 2, the foci are 0 2 2 , and the asymptotes are y x.
1 1
x
1
x
(b) The transverse axis has length 2a 4.
x2 y2 1 24. x 2 3y 2 12 0 4 12 (a) The hyperbola has a 2, b 2 3, and c2 4 12 16 c 4. The
y
(c)
1
vertices are 0 2, the foci are 0 4, and the asymptotes are 2 x 3 x. y 3 2 3
(b) The transverse axis has length 2a 4. y2 25. 4y 2 x 2 1 1 x 2 1
y
(c)
1
4
(a) The hyperbola has a 12 , b 1, and c2 14 1 54 c 25 . The vertices are 0 12 , the foci are 0 25 , and the asymptotes are
1
x
1
x
1 y 12 1 x 2 x.
(b) The transverse axis has length 2a 1. 26. 9x 2 16y 2 1
y2 x2 1 19 116
y
(c)
1
1 25 c 5 . The (a) The hyperbola has a 13 , b 14 , and c2 19 16 144 12 1 5 vertices are 3 0 , the foci are 12 0 , and the asymptotes are 3 y 14 13 x 4 x.
(b) The transverse axis has length 2a 23 . 27. From the graph, the foci are 4 0, and the vertices are 2 0, so c 4 and a 2. Thus, b2 16 4 12, and since the vertices are on the x-axis, an equation of the hyperbola is
x2 y2 1. 4 12
28. From the graph, the foci are 0 13 and the vertices are 0 12, so c 13 and a 12. Then b2 c2 a 2 169 144 25, and since the vertices are on the y-axis, an equation of the hyperbola is
y2 x2 1. 144 25
29. From the graph, the vertices are 0 4, the foci are on the y-axis, and the hyperbola passes through the point 3 5. So
y2 x2 9 25 32 52 2 1. Substituting the point 3 5, we have 2 1 1 2 16 16 16 b b b 9 x2 y2 9 2 b2 16. Thus, an equation of the hyperbola is 1. 16 16 16 b
the equation is of the form
SECTION 12.3 Hyperbolas
807
x2 y2 30. The vertices are 2 3 0 , so a 2 3, so an equation of the hyperbola is of the form 2 2 1. Substituting b 2 3 the point 4 4 into the equation, we get hyperbola is
16 16 16 16 16 4 1 2 b2 48. Thus, an equation of the 1 2 12 b2 12 12 b b
y2 x2 1. 12 48
a 3 31. From the graph, the vertices are 0 3, so a 3. Since the asymptotes are y 3x x, we have 3 b 1. b b y2 x2 y2 Since the vertices are on the x-axis, an equation is 2 2 1 x 2 1. 9 3 1 b 1 3 b 32. The vertices are 3 0, so a 3. Since the asymptotes are y 12 x x, we have b . Since the a 3 2 2 4y 2 y2 x2 x2 1. 1 vertices are on the x-axis, an equation is 2 9 9 3 322 33. x 2 2y 2 8 2y 2 x 2 8 y 2 12 x 2 4 y 12 x 2 4
34. 3y 2 4x 2 24 3y 2 4x 2 24 y 2 43 x 2 8 y 43 x 2 8 10
5
-5
5
-10
10
-5
35.
-10
x2 y2 x2 y2 x2 1 1 y2 2 2 6 2 6 3 x2 2 y 3 5
-5
36.
y2 x2 1 16x 2 25y 2 1600 100 64 2 25y 2 16x 2 1600 y 2 16 25 x 64 2 y 16 25 x 64
20
5 -20
20
-5 -20
37. The foci are 5 0 and the vertices are 3 0, so c 5 and a 3. Then b2 25 9 16, and since the vertices are on the x-axis, an equation of the hyperbola is
x2 y2 1. 9 16
38. The foci are 0 10 and the vertices are 0 8, so c 10 and a 8. Then b2 c2 a 2 100 64 36, and since the vertices are on the y-axis, an equation of the hyperbola is
x2 y2 1. 64 36
39. The foci are 0 2 and the vertices are 0 1, so c 2 and a 1. Then b2 4 1 3, and since the vertices are on x2 1. the y-axis, an equation is y 2 3
808
CHAPTER 12 Conic Sections
40. The foci are 6 0 and the vertices are 2 0, so c 6 and a 2. Then b2 c2 a 2 36 4 32, and since the vertices are on the x-axis, an equation is
y2 x2 1. 4 32
b b 41. The vertices are 1 0 and the asymptotes are y 5x, so a 1. The asymptotes are y x, so 5 b 5. a 1 2 y Therefore, an equation of the hyperbola is x 2 1. 25 6 a 42. The vertices are 0 6, so a 6. The asymptotes are y 13 x x 13 b 18. Since the vertices are on b b y2 x2 the y-axis, an equation of the hyperbola is 1. 36 324 43. The vertices are 0 6, so a 6. Since the vertices are on the y-axis, the hyperbola has an equation of the form
x2 25 81 25 45 y2 2 1. Since the hyperbola passes through the point 5 9, we have 2 1 2 b2 20. 36 36 36 b b b x2 y2 1. Thus, an equation is 36 20
44. The vertices are 2 0, so a 2. Since the vertices are on the x-axis, the hyperbola has an equation of the form y2 30 30 9 5 x2 2 1. Since the hyperbola passes through the point 3 30 , we have 2 1 2 b2 24. 4 4 4 b b b y2 x2 1. Thus, an equation is 4 24 45. The asymptotes of the hyperbola are y x, so b a. Since the hyperbola passes through the point 5 3, its foci are on x2 y2 25 9 the x-axis, and its equation has the form, 2 2 1, so it follows that 2 2 1 a 2 16 b2 . Therefore, an a a a a x2 y2 equation of the hyperbola is 1. 16 16
46. The asymptotes of the hyperbola are y x, so b a. Since the hyperbola passes through the point 1 2, its foci are on
y2 x2 4 1 the y-axis, and its equation has the form 2 2 1, so it follows that 2 2 1 a 2 3. Therefore, an equation a a a a y2 x2 of the hyperbola is 1. 3 3
x2 42 12 y2 47. The foci are 0 3, so c 3 and an equation is 2 2 1. The hyperbola passes through 1 4, so 2 2 1 a b b a 2 2 2 2 2 2 2 2 4 2 2 2 16b a a b 16b 9 b 9 b b b 8b 9 0 b 1 b 9 0, Thus, b2 1, a 2 8, and an equation is
y2 x 2 1. 8
x2 y2 1. The hyperbola passes through 4 18 , 48. The foci are 10 0 , so c 10 and an equation is 2 b a 18 16 2 2 2 2 2 so 2 2 1 16b 18a a b 16b 18 10 b2 10 b2 b2 b4 24b2 180 0 a b y2 x2 b2 30 b2 6 0. Thus, b2 6, a 2 4, and an equation is 1. 4 6
49. The foci are 5 0, and the length of the transverse axis is 6, so c 5 and 2a 6 a 3. Thus, b2 25 9 16, and an equation is
y2 x2 1. 9 16
SECTION 12.3 Hyperbolas
809
50. The foci are 0 1, and the length of the transverse axis is 1, so c 1 and 2a 1 a 12 . Then y2 4x 2 x2 b2 c2 a 2 1 14 34 , and since the foci are on the y-axis, an equation is 1 4y 2 1. 14 34 3 y2 x2 1 has a 5 and b 5. Thus, the asymptotes are y x, and their 5 5 slopes are m 1 1 and m 2 1. Since m 1 m 2 1, the asymptotes are perpendicular.
51. (a) The hyperbola x 2 y 2 5
(b) Since the asymptotes are perpendicular, they must have slopes 1, so a b. Therefore, c2 2a 2 a 2
c2 , and 2
x2 y2 c2 since the vertices are on the x-axis, an equation is 1 1 1 x 2 y 2 . 2 2 2 2c 2c y2 x2 52. The hyperbolas 2 2 1 and a b
x2 y2 1 are conjugate to each other. a2 b2
y2 x2 (a) x 2 4y 2 16 0 1 and 4y 2 x 2 16 0 16 4 y2 x2 1. So the hyperbolas are conjugate to each other. 16 4 (b) They both have the same asymptotes, y 12 x.
y
1 1
x
(c) The two general conjugate hyperbolas both have asymptotes b y x. a 53.
x c2 y 2 x c2 y 2 2a. Let us consider the positive case only. Then x c2 y 2 2a x c2 y 2 , and squaring both sides gives x 2 2cx c2 y 2 4a 2 4a x c2 y 2 x 2 2cx c2 y 2 4a x c2 y 2 4cx 4a 2 . Dividing by 4 and squaring both sides gives a 2 x 2 2cx c2 y 2 c2 x 2 2a 2 cx a 4 a 2 x 2 2a 2 cx a 2 c2 a 2 y 2 c2 x 2 2a 2 cx a 4
a 2 x 2 a 2 c2 a 2 y 2 c2 x 2 a 4 . Rearranging the order, we have c2 x 2 a 2 x 2 a 2 y 2 a 2 c2 a 4 c2 a 2 x 2 a 2 y 2 a 2 c2 a 2 . The negative case gives the same result. 54. (a) The hyperbola
x2 y2 1 has a 3, b 4, so c2 9 16 25, and c 5. Therefore, the foci are F1 5 0 and 9 16
F2 5 0.
25 2569 25 16 (b) Substituting P 5 16 3 into an equation of the hyperbola, we get 9 16 9 9 1, so P lies on the hyperbola. 1 34 2 2 (c) d P F1 5 52 163 02 16 3 , and d P F2 5 5 163 0 3 900 256 3 .
16 (d) d P F2 d P F1 34 3 3 6 2 3 2a.
55. (a) From the equation, we have a 2 k and b2 16 k. Thus, c2 a 2 b2 k 16 k 16 c 4. Thus the foci of the family of hyperbolas are 0 4.
810
CHAPTER 12 Conic Sections
2 y2 x kx 2 x2 (b) 1 y2 k 1 . For y k k 16 k 16 k 16 k kx 2 , k 1, 4, 8, 12. As k the top branch, we graph y k 16 k
k=12 8
k=8
6
increases, the asymptotes get steeper and the vertices move further apart.
-10
4
k=4
2
k=1
0
10
56. d AB 500 2c c 250. (a) Since t 2640 and 980, we have d d P A d P B t 980 fts 2640 s 2,587,200 ft 490mi.
(b) c 250, 2a 490 a 245 and the foci are on the y-axis. Then b2 2502 2452 2475. Hence, an equation is x2 y2 1. 60,025 2475
2502 x2 (c) Since P is due east of A, c 250 is the y-coordinate of P. Therefore, P is at x 250, and so 1 2 2475 245 2502 x 2 2475 1 10205. Then x 101, and so P is approximately 101 miles from A. 2452 57. Since the asymptotes are perpendicular, a b. Also, since the sun is a focus and the closest distance is 2 109 , it follows 2 109 and that c a 2 109 . Now c2 a 2 b2 2a 2 , and so c 2a. Thus, 2a a 2 109 a 21 4 1018 x2 y2 a 2 b2 1 23 1019 . Therefore, an equation of the hyperbola is 23 1019 23 1019 32 2 x 2 y 2 23 1019 .
58. (a) These equally spaced concentric circles can be used as a kind of measure where we count the number of rings. In the case of the red dots, the sum of the number of wave crests from each center is a constant, in this case 17. As you move out one wave crest from the left stone you move in one wave crest from the right stone. Therefore this satisfies the geometric definition of an ellipse. (b) Similarly, in the case of the blue dots, the difference of the number of wave crests from each center is a constant. As you move out one wave crest from the left center you also move out one wave crest from the right stone. Therefore this satisfies the geometric definition of a hyperbola. 59. Some possible answers are: as cross-sections of nuclear power plant cooling towers, or as reflectors for camouflaging the location of secret installations. 60. The wall is parallel to the axis of the cone of light coming from the top of the shade, so the intersection of the wall and the cone of light is a hyperbola. In the case of the flashlight, hold it parallel to the ground to form a hyperbola.
12.4 SHIFTED CONICS 1. (a) If we replace x by x 3 the graph of the equation is shifted to the right by 3 units. If we replace x by x 3 the graph is shifted to the left by 3 units. (b) If we replace y by y 1 the graph of the equation is shifted upward by 1 unit. If we replace y by y 1 the graph is shifted downward by 1 unit. 2. x 2 12y, from top to bottom: focus 0 3, vertex 0 0, directrix y 3. x 32 12 y 1, from top to bottom: focus 3 4, vertex 3 1, directrix y 2.
SECTION 12.4 Shifted Conics
3.
4.
5.
811
x2 y2 x 32 y 12 1, from left to right: vertex 0, focus 0, focus 0, vertex 0. 1, from 5 3 3 5 52 42 52 42 left to right: vertex 2 1, focus 0 1, focus 6 1, vertex 8 1.
x2 y2 2 1, from left to right: focus 5 0, vertex 4 0, asymptote y 34 x, asymptote y 34 x, vertex 4 0, 2 4 3 x 32 y 12 focus 5 0. 1, from left to right: focus 2 1, vertex 1 1, asymptote y 34 x 13 4 , 42 32 asymptote y 34 x 54 , vertex 7 1, focus 8 1.
y 12 x 22 1 9 4
y
(c)
x2 y2 1 by shifting it 2 units to 9 4 the right and 1 unit upward. So a 3, b 2, and c 9 4 5. The
(a) This ellipse is obtained from the ellipse
1 x
1
center is 2 1, the vertices are 2 3 1 1 1 and 5 1, and the foci are 2 5 1 .
(b) The length of the major axis is 2a 6 and the length of the minor axis is 2b 4.
6.
x 32 y 32 1 16
(c)
x2 y 2 1 by shifting to the right (a) This ellipse is obtained from the ellipse 16 3 units and downward 3 units. So a 4, b 1, and c 16 1 15. The center is 3 3, the vertices are 3 4 3 1 3 and 7 3, and the foci are 3 15 3 .
y
1 x
1
(b) The length of the major axis is 2a 8 and the length of the minor axis is 2b 2.
7.
x2 y 52 1 9 25 x2 y2 (a) This ellipse is obtained from the ellipse 1 by shifting it 5 units 9 25 downward. So a 5, b 3, and c 25 9 4. The center is 0 5, the vertices are 0 5 5 0 10 and 0 0, and the foci are 0 5 4 0 9 and 0 1.
(b) The length of the major axis is 2a 10 and the length of the minor axis is 2b 6.
(c)
y
1 1
x
812
CHAPTER 12 Conic Sections
y 22 1 8. x 2 4
(c)
y
1
y2 1 by shifting it 2 units (a) This ellipse is obtained from the ellipse x 2 4 downward. So a 2, b 1, and c 4 1 3. The center is 0 2, the
x
1
vertices are 0 2 2 0 4 and 0 0, and the foci are 0 2 3 0 2 3 and 0 2 3 .
(b) The length of the major axis is 2a 4 and the length of the minor axis is 2b 2.
9.
y 12 x 52 1 16 4
y
(c)
y2 x2 1 by shifting it 5 units to 16 4 the left and 1 units upward. So a 4, b 2, and c 16 4 2 3. The
(a) This ellipse is obtained from the ellipse
1 1
x
center is 5 1, the vertices are 5 4 1 9 1 and 1 1, and the foci are 5 2 3 1 5 2 3 1 and 5 2 3 1 .
(b) The length of the major axis is 2a 8 and the length of the minor axis is 2b 4.
10.
y 1 x 12 1 36 64
(c)
y2 x2 1 by shifting it 1 unit to the 36 64 left and 1 unit downward. So a 8, b 6, and c 64 36 2 7. The center is 1 1, the vertices are 1 1 8 1 9 and 1 7, and the foci are 1 1 2 7 1 1 2 7 and 1 2 7 1 .
(a) This ellipse is obtained from the ellipse
y
2 2
x
(b) The length of the major axis is 2a 16 and the length of the minor axis is 2b 12.
11. 4x 2 25y 2 50y 75 4x 2 25 y 12 25 75
x2 y 12 1 25 4
y2 x2 1 by shifting it 1 unit 25 4 upward. So a 5, b 2, and c 25 4 21. The center is 0 1, the
(c)
y
(a) This ellipse is obtained from the ellipse
vertices are 5 1 5 1 and 5 1, and the foci are 21 1 . 21 1 21 1 and
(b) The length of the major axis is 2a 10 and the length of the minor axis is 2b 4.
1 1
x
SECTION 12.4 Shifted Conics
12. 9x 2 54x y 2 2y 46 0 9 x 32 81 y 12 1 46 0
(c)
x 32 y 12 1 4 36
813
y
1
x2
x
2
y2
1 by shifting it 3 units to 4 36 the right and 1 unit downward. So a 6, b 2, and c 36 4 4 2. The
(a) This ellipse is obtained from the ellipse
center is 3 1, the vertices are 3 1 6 3 7 and 3 5, and the foci are 3 1 4 2 3 1 4 2 and 3 4 2 1 .
(b) The length of the major axis is 2a 12 and the length of the minor axis is 2b 4.
13. x 32 8 y 1
(b)
(a) This parabola is obtained from the parabola x 2 8y by shifting it 3 units to the
y
1
right and 1 unit down. So 4 p 8 p 2. The vertex is 3 1, the focus is 3 1 2 3 1, and the directrix is y 1 2 3.
14. y 12 16 x 3
(b)
(a) This parabola is obtained from the parabola y 2 16x by shifting it 3 units to
x
1
x
y
2
the right and 1 unit down. So 4 p 16 p 4. The vertex is 3 1, the focus is 3 4 1 7 1, and the directrix is x 3 4 1.
15. y 52 6x 12 6 x 2
1
y
(b)
1 1
(a) This parabola is obtained from the parabola y 2 6x by shifting to the right
x
2 units and down 5 units. So 4 p 6 p 32 . The vertex is 2 5, the focus is 2 32 5 12 5 , and the directrix is x 2 32 72 .
16. y 2 16x 8 16 x 12
(a) This parabola is obtained from the parabola y 2 16x by shifting to the right 12 unit. So 4 p 16 p 4. The vertex is 12 0 , the focus is 1 4 0 9 0 , and the directrix is x 1 4 7 . 2 2 2 2
(b)
y
2 1
x
814
CHAPTER 12 Conic Sections
17. 2 x 12 y x 12 12 y
y
(b)
(a) This parabola is obtained from the parabola x 2 12 y by shifting it 1 unit to the right. So 4 p 12 p 18 . The vertex is 1 0, the focus is 1 18 , and the directrix is y 18 .
1 x
1
2 2 18. 4 x 12 y x 12 14 y
y
(b)
1
(a) This parabola is obtained from the parabola x 2 14 y by shifting it 12 unit to 1 . The vertex is 1 0 , the focus is the left, so 4 p 14 p 16 2 1 1 1 1 1 1 . 2 0 16 2 16 , and the directrix is y 0 16 16
19. y 2 6y 12x 33 0 y 32 9 12x 33 0 y 32 12 x 2
1
(b)
x
y
(a) This parabola is obtained from the parabola y 2 12x by shifting it 2 units to the right and 3 units upward. So 4 p 12 p 3. The vertex is 2 3, the
2
focus is 2 3 3 5 3, and the directrix is x 2 3 1.
20. x 2 2x 20y 41 0 x 12 1 20y 41 0 x 12 20 y 2
x
1
(b)
y
(a) This parabola is obtained from the parabola x 2 20y by shifting it 1 unit to the left and 2 units upward, so 4 p 20 p 5. The vertex is 1 2, the focus
1
is 1 2 5 1 7, and the directrix is y 2 5 3.
21.
y 32 x 12 1 9 16 x2 y2 1 by shifting it 9 16 1 unit to the left and 3 units up. So a 3, b 4, and c 9 16 5. The
(b)
1
x
1
x
y
(a) This hyperbola is obtained from the hyperbola
center is 1 3, the vertices are 1 3 3 4 3 and 2 3, the foci are 1 5 3 6 3 and 4 3, and the asymptotes are
y 3 43 x 1 y 43 x 1 3 y 43 x 13 3 and y 43 x 53 .
1
SECTION 12.4 Shifted Conics
22. x 82 y 62 1
(b)
y
(a) This hyperbola is obtained from the hyperbola x 2 y 2 1 by shifting to the right 8 units and downward 6 units. So a 1, b 1, and c 1 1 2.
10
The center is 8 6, the vertices are 8 1 6 7 6 and 9 6, the foci are 8 2 6 , and the asymptotes are y 6 x 8
20
x
_10
_20
y x 14 and y x 2.
x 12 1 23. y 2 4
815
y
(b)
x2 (a) This hyperbola is obtained from the hyperbola y 2 1 by shifting it 1 unit 4 to the left. So a 1, b 2, and c 1 4 5. The center is 1 0, the
1 x
1
vertices are 1 1 1 1 and 1 1, the foci are 1 5 1 5 and 1 5 , and the asymptotes are y 12 x 1 y 12 x 12 and y 12 x 12 .
24.
y 12 x 32 1 25
y
(b)
y2 x 2 1 by shifting to the 25 left 3 units and upward 1 unit. So a 5, b 1, and c 25 1 26. The
(a) This hyperbola is obtained from the hyperbola
5 1
x
center is 3 1, the vertices are 3 1 5 3 4 and 3 6, the foci are 3 1 26 , and the asymptotes are y 1 5 x 3 y 5x 16 and y 5x 14.
25.
y 12 x 12 1 9 4
y
(b)
x 2 y2 1 by shifting it 1 unit 9 4 to the left and 1 unit downward, so a 3, b 2, and c 9 4 13. The center is 1 1, the vertices are 1 3 1 4 1 and 2 1, the foci are 1 13 1 1 13 1 and 13 1 1 , and the
1
(a) This hyperbola is obtained from the hyperbola
1
x
5
x
asymptotes are y 1 23 x 1 y 23 x 13 and y 23 x 53 .
26.
x2 y 22 1 36 64 x2 y2 1 by shifting it (a) This hyperbola is obtained from the hyperbola 36 64 2 unit downward. So a 6, b 8, and c 36 64 10. The center is 0 2, the vertices are 0 2 6 0 8 and 0 4, the foci are
0 2 10 0 12 and 0 8, and the asymptotes are y 2 34 x y 34 x 2 and y 34 x 2.
(b)
y 5
816
CHAPTER 12 Conic Sections
27. 36x 2 72x 4y 2 32y 116 0 36 x 12 36 4 y 42 64 116 0
y
(b) x 12 y 42 1 36 4
y2 x 2 1 by shifting it 1 unit 36 4 to the left and 4 units upward. So a 6, b 2, and c 36 4 2 10.
2
(a) This hyperbola is obtained from the hyperbola
1
x
The center is 1 4, the vertices are 1 4 6 1 2 and 1 10, the foci are 1 4 2 10 1 2 10 4 and 1 4 2 10 , and
the asymptotes are y 4 3 x 1 y 3x 7 and y 3x 1. 28. 25x 2 9y 2 54y 306 25x 2 9 y 32 81 306
x 2 y 32 1 9 25
(b)
x2 y2 1 by shifting it 9 25 3 units downward. So a 3, b 5, and c 9 25 34. The center is 0 3, the vertices are 3 3 3 3 and 3 3, the foci are 34 3 , and the asymptotes are 34 3 34 3 and
y
2
(a) This hyperbola is obtained from the hyperbola
1
x
y 3 53 x y 53 x 3 and y 53 x 3.
29. This is a parabola that opens down with its vertex at 0 4, so its equation is of the form x 2 a y 4. Since 1 0 is a point on this parabola, we have 12 a 0 4 1 4a a 14 . Thus, an equation is x 2 14 y 4.
30. This is a parabola that opens to the right with its vertex at 6 0, so its equation is of the form y 2 4 p x 6, with
p 0. Since the distance from the vertex to the directrix is p 6 12 6, an equation is y 2 4 6 x 6
y 2 24 x 6
31. This is an ellipse with the major axis parallel to the x-axis, with one vertex at 0 0, the other vertex at 10 0, and one focus at 8 0. The center is at 010 2 0 5 0, a 5, and c 3 (the distance from one focus to the center). So b2 a 2 c2 25 9 16. Thus, an equation is
y2 x 52 1. 25 16
32. This is an ellipse with the major axis parallel to the y-axis. From the graph the center is 2 3, with a 3 and b 2. Thus, an equation is
x 22 y 32 1. 4 9
33. This is a hyperbola with center 0 1 and vertices 0 0 and 0 2. Since a is the distance form the center to a vertex, a we have a 1. The slope of the given asymptote is 1, so 1 b 1. Thus, an equation of the hyperbola is b 2 2 1 x 1. y 34. From the graph, the vertices are 2 0 and 6 0. The center is the midpoint between the vertices, so the center is 26 00 4 0. Since a is the distance from the center to a vertex, a 2. Since the vertices are on the x-axis, the 2 2 y2 x 42 42 0 42 2 1. Since the point 0 4 lies on the hyperbola, we have 2 1 4 4 b b 2 2 2 y 3y 2 16 16 16 16 4 4 x x 2 1 3 2 b2 . Thus, an equation of the hyperbola is 16 1 1. 4 3 4 4 16 b b
equation is of the form
3
SECTION 12.4 Shifted Conics
817
35. The ellipse with center C 2 3, vertices V1 8 3 and V2 12 3, and foci F1 4 3 and F2 8 3 has
x 22 y 32 1. The distance between the vertices 2 a b2 is 2a 12 8 20, so a 10. Also, the distance from the center to each focus is c 2 4 6, so a horizontal major axis, so its equation has the form
b2 a 2 c2 100 36 64. Thus, an equation is
y 32 x 22 1. 100 64
36. The ellipse with vertices V1 1 4 and V2 1 6 and foci F1 1 3 and F2 1 5 is centered midway between the vertices; that is, it has C 1 12 4 6 1 1. The distance between the vertices is 2a 6 4 10, so a 5.
Also, the distance from the center to each focus is c 1 3 4, so b2 a 2 c2 25 16 9. Thus, an equation is x 12 y 12 1. 9 25
37. The hyperbola with center C 1 4, vertices V1 1 3 and V2 1 11, and foci F1 1 5 and F2 1 13 has y 42 x 12 1. The distance between the vertices a2 b2 is 2a 11 3 14, so a 7. Also, the distance from the center to each focus is c 4 5 9, so a vertical transverse axis, so its equation has the form
b2 c2 a 2 81 49 32. Thus, an equation is
x 12 y 42 1. 49 32
38. The hyperbola with vertices V1 1 1 and V2 5 1, and foci F1 4 1 and F2 8 1 is centered midway between the vertices; that is, it has C 12 1 5 1 2 1. It has a horizontal transverse axis, so its equation has the form
y 12 x 22 1. The distance between the vertices is 2a 5 1 6, so a 3. Also, the distance from the 2 a b2 x 22 y 12 1. center to each focus is c 2 4 6, so b2 c2 a 2 36 9 27. Thus, an equation is 9 27
39. The parabola with vertex V 3 5 and directrix y 2 has an equation of the form x 32 4 p y 5. The distance from the vertex to the directrix is p 5 2 3, so an equation is x 32 12 y 5.
40. The parabola with focus F 1 3 and directrix x 3 has vertex midway between the focus and the directrix; that is, it has V 12 3 1 3 2 3. The distance from the vertex to the directrix is p 2 3 1. (Since the directrix is to the right of the focus, p is negative.) Thus, an equation is y 32 4 1 x 2 4 x 2.
41. The hyperbola with foci F1 1 5 and F2 1 5 that passes through the point 1 4 is centered midway between the foci; y2 x 12 that is, it has C 1 12 5 5 1 0. It has a vertical transverse axis, so its equation has the form 2 1. a b2 The point 1 4 lies on the transverse axis, so it is a vertex and we have a 4 0 4. Also, the distance from the center to each focus is c 5 0 5, so b2 c2 a 2 25 16 9. Thus, an equation is
y2 x 12 1. 16 9
42. The hyperbola with foci F1 2 2 and F2 4 2 that passes through the point 3 2 is centered midway between the foci; that is, it has C 12 2 4 2 1 2. It has a horizontal transverse axis, so its equation has the form
y 22 x 12 1. The point 3 2 lies on the transverse axis, so it is a vertex and we have a 3 1 2. 2 a b2 Also, the distance from the center to each focus is c 4 1 3, so b2 c2 a 2 9 4 5. Thus, an equation is x 12 y 22 1. 4 5
818
CHAPTER 12 Conic Sections
43. The ellipse with foci F1 1 4 and F2 5 4 that passes through the point 3 1 is centered midway between the foci; y 42 x 32 that is, it has C 12 1 5 4 3 4, and so its equation has the form 1. The distance from 2 a b2 the center to each focus is c 3 1 2, so c2 a 2 b2 a 2 b2 4. Substituting the point x y 3 1 into the
3 32 1 42 1 b2 25. From above, we have a 2 b2 4 29. Thus, an 2 a b2 x 32 y 42 equation of the ellipse is 1. 29 25 44. The ellipse with foci F1 3 4 and F2 3 4 and x-intercepts 0 and 6 is centered midway between the foci; that is, it has x 32 y 2 C 3 12 4 4 3 0, and so its equation has the form 2 1. Because 0 is an x-intercept, we substitute b2 a 2 2 0 3 2 1 b 3. We also have c 4 0 4, so a 2 b2 c2 25 and an 0 0 into this equation, obtaining b2 a y2 x 32 equation is 1. 9 25 45. The parabola that passes through the point 6 1, with vertex V 1 2 and horizontal axis of symmetry has an equation of equation of the ellipse, we have
1 . the form y 22 4 p x 1. Substituting the point 6 1 into this equation, we have 1 22 4 p 6 1 p 28 Thus, an equation is y 22 17 x 1.
46. Because 6 2 is lower than V 4 1, the parabola opens downward, so an equation is x 42 4 p y 1.
Substituting the point 6 2, we have 6 42 4 p 2 1 p 1, so an equation is x 42 4 y 1. y
47. y 2 4 x 2y y 2 8y 4x y 2 8y 16 4x 16
y 42 4 x 4. This is a parabola with 4 p 4 p 1. The vertex is
4 4, the focus is 4 1 4 3 4, and the directrix is x 4 1 5. 1 x
1
48. 9x 2 36x 4y 2 0 9 x 2 4x 4 36 4y 2 0
y
y2 x 22 1. This is an ellipse with a 3, 4 9 b 2, and c 9 4 5. The center is 2 0, the foci are 2 5 , the
9 x 22 4y 2 36
1 x
1
vertices are 2 3, the length of the major axis is 2a 6, and the length of the minor axis is 2b 4.
49. x 2 5y 2 2x 20y 44 x 2 2x 1 5 y 2 4y 4 44 1 20 y 22 x 12 x 12 5 y 22 25 1. This is a hyperbola 25 5 with a 5, b 5, and c 25 5 30. The center is 1 2, the foci are 1 30 2 , the vertices are 1 5 2 4 2 and 6 2, and the asymptotes
are y 2 55 x 1 y 55 x 1 2 y 55 x 2 55 and
y 55 x 2 55 .
y
1 2
x
SECTION 12.4 Shifted Conics y
50. x 2 6x 12y 9 0 x 2 6x 9 12y x 32 12y. This is a
parabola with 4 p 12 p 3. The vertex is 3 0, the focus is 3 3,
_1
1
x
and the directrix is y 3.
y
51. 4x 2 25y 2 24x 250y 561 0 4 x 2 6x 9 25 y 2 10y 25 561 36 625 4 x 32 25 y 52 100
x 32
1 x
1
y 52
1. This is an ellipse 25 4 with a 5, b 2, and c 25 4 21. The center is 3 5, the foci are 3 21 5 , the vertices are 3 5 5 2 5 and 8 5, the length
of the major axis is 2a 10, and the length of the minor axis is 2b 4. 52. 2x 2 y 2 2y 1 2x 2 y 2 2y 1 1 1 2x 2 y 12 2
y
y 12 1. This is an ellipse with a 2, b 1, and c 2 1 1. 2 The center is 0 1, the foci are 0 1 1 0 0 and 0 2, the vertices are 0 1 2 , the length of the major axis is 2a 2 2, and the length of the minor
1
x2
1
x
axis is 2b 2.
53. 16x 2 9y 2 96x 288 0 16 x 2 6x 9y 2 288 0 16 x 2 6x 9 9y 2 144 288 16 x 32 9y 2 144
y
2 x
1
y2 x 32 1. This is a hyperbola with a 4, b 3, and 16 9 c 16 9 5. The center is 3 0, the foci are 3 5, the vertices are
3 4, and the asymptotes are y 43 x 3 y 43 x 4 and y 4 43 x. 54. 4x 2 4x 8y 9 0 4 x 2 x 8y 9 0
y
2 4 x 2 x 14 8y 9 1 4 x 12 8 y 1
2 x 12 2 y 1. This is a parabola with 4 p 2 p 12 . The vertex is 1 1 , the focus is 1 3 , and the directrix is y 1 1 1 . 2 2 2 2 2 55. x 2 16 4 y 2 2x x 2 8x 4y 2 16 0 x 2 8x 16 4y 2 16 16 4y 2 x 42 y 12 x 4.
Thus, the conic is degenerate, and its graph is the pair of lines y 12 x 4 and y 12 x 4.
1 1
x
y
1 1
x
819
820
CHAPTER 12 Conic Sections y
56. x 2 y 2 10 x y 1 x 2 10x y 2 10y 1 x 2 10x 25 y 2 10y 25 1 25 25 x 52 y 52 1. This is a hyperbola with a 1, b 1, and c 1 1 2. The center is 5 5, the foci are 5 2 5 , the vertices are 5 1 5 4 5 and 6 5, and the
1 x
1
asymptotes are y 5 x 5 y x and y x 10. 57. 3x 2 4y 2 6x 24y 39 0 3 x 2 2x 4 y 2 6y 39 3 x 2 2x 1 4 y 2 6y 9 39 3 36
y
(1, 3)
1 x
1
3 x 12 4 y 32 0 x 1 and y 3. This is a degenerate conic whose
graph is the point 1 3.
58. x 2 4y 2 20x 40y 300 0 x 2 20x 4 y 2 10y 300 x 2 20x 100 4 y 2 10y 25 300 100 100 x 102 4 y 52 100. Since u 2 2 0 for all u, R, there is no x y such that x 102 4 y 52 100. So there is no solution, and the graph is empty.
59. 2x 2 4x y 5 0 y 2x 2 4x 5.
-2
2
4
-5 -10
60. 4x 2 9y 2 36y 0 4x 2 9 y 2 4y 0 4x 2 9 y 2 4y 4 36 9 y 2 4y 4 36 4x 2 y 2 4y 4 4 49 x 2 y 2 4 49 x 2 y 2 4 49 x 2 61. 9x 2 36 y 2 36x 6y x 2 36x 36 y 2 6y 9x 2 36x 45 y 2 6y 9 9 x 2 4x 5 y 32 y 3 9 x 2 4x 5 y 3 3 x 2 4x 5
5
-4
-2
2
2
2 y 1 x 2 y 1 12 x 2 y 1 12 x 2.So
conic.
-5
4
4
-10
62. x 2 4y 2 4x 8y 0 x 2 4x 4 y 2 2y 0 x 2 4x 4 4 y 2 2y 1 0 4 y 12 x 22
y 1 12 x 2 12 x 2 and y 1 12 x 2 12 x. This is a degenerate
2
SECTION 12.4 Shifted Conics
63. 4x 2 y 2 4 x 2y F 0 4 x 2 x y 2 8y 4 x 2 x 14 y 2 8y 16 16 1 F 4 x (a) For an ellipse, 17 F 0 F 17.
821
F 1 2 y 12 17 F 2
(b) For a single point, 17 F 0 F 17. (c) For the empty set, 17 F 0 F 17.
64. The parabola x 2 y 100 x 2 y 100 has 4 p 1 p 14 . The vertex is 0 100 and so the focus is 399 0 100 14 0 399 4 . Thus, one vertex of the ellipse is 0 100, and one focus is 0 4 . Since the second focus of 401 the ellipse is 0 0, the second vertex is 0 14 . So 2a 100 14 401 4 a 8 . Since 2c is the distance between 399 4012 3992 2 2 2 the foci of the ellipse, 2c 399 4 , c 8 , and then b a c 64 64 25. The center of the ellipse is 2 2 y 399 x x2 8y 3992 8 and so its equation is 0 399 1. 2 1, which simplifies to 8 25 25 160,801 401 8
65. (a) x 2 4 p y p, for
p 2, 32 , 1, 12 , 12 , 1, 32 , 2. 3
1 2
p= 2
p= 1
2
0
4
8
-4 3
1
_2
(c) The parabolas become narrower as the vertex moves toward the origin.
-8 p= _ 2
focus of x 2 4 py is at 0 p, so this point is also shifted p units at the origin.
4 -4
x 2 4 py vertically p units so that the vertex is at 0 p. The
vertically to the point 0 p p 0 0. Thus, the focus is located
8
-8
(b) The graph of x 2 4 p y p is obtained by shifting the graph of
p= _1 _2
66. Since 0 0 and 1600 0 are both points on the parabola, the x-coordinate of the vertex is 800. And since the highest point it reaches is 3200, the y-coordinate of the vertex is 3200. Thus the vertex is 800 3200, and the equation is of the form x 8002 4 p y 3200. Substituting the point 0 0, we get 0 8002 4 p 0 3200 640,000 12,800 p
p 50. So an equation is x 8002 4 50 y 3200 x 8002 200 y 3200. y
67. Since the height of the satellite above the earth varies between 140 and 440, the length of the major axis is 2a 140 2 3960 440 8500 a 4250. Since the center of the earth is at one focus, we have a c earth radius 140 3960 140 4100
c a 4100 4250 4100 150. Thus, the center of the ellipse is 150 0. So b2 a 2 c2 42502 1502 18,062,500 22500 18,040,000. Hence,
an equation is
410
Path of satellite Earth
3960 Center of ellipse is (_150, 0)
140 x
y2 x 1502 1. 18,062,500 18,040,000
68. (a) We assume that 0 1 is the focus closer to the vertex 0 0, as shown in the figure in the text. Then the center of the ellipse is 0 a and 1 a c. So c a 1 and a 12 a 2 b2 a 2 2a 1 a 2 b2 b2 2a 1.
x2 x2 y a2 y 22 1. If we choose 1. If we choose a 2, then we get 2a 1 3 4 a2 x2 y 52 1. (Answers will vary, depending on your choices of a 1.) a 5, then we get 9 25 Thus, the equation is
822
CHAPTER 12 Conic Sections
(b) Since a vertex is at 0 0 and a focus is at 0 1, we must have c a 1 (a 0), and the center of the hyperbola is 0 a. So c a 1 and a 12 a 2 b2 a 2 2a 1 a 2 b2 b2 2a 1. Thus the equation x2 x2 y a2 y 22 1. If we let a 2, then we get 1. If we let a 5, then we get 2a 1 4 5 a2 x2 y 52 1. (Answers will vary, depending on your choices of a.) 25 11
is
(c) Since the vertex is at 0 0 and the focus is at 0 1, we must have
y
p 1. So x 02 4 1 y 0 x 2 4y, and there is no other
possible parabola.
(d) Graphs will vary, depending on the choices of a in parts (a) and (b). (e) The ellipses are inside the parabola and the hyperbolas are outside the
1 1
parabola. All touch at the origin.
x
12.5 ROTATION OF AXES 1. If the x- and y-axes are rotated through an acute angle to produce the new X- and Y -axes, then the x y-coordinates x y and the XY -coordinates X Y of a point P in the plane are related by the formulas x X cos Y sin , y X sin Y cos , X x cos y sin , and Y x sin y cos . 2. (a) In general, the graph of the equation Ax 2 Bx y C y 2 Dx E y F 0 is a conic section.
AC . B (c) The discriminant of this equation is B 2 4AC. If the discriminant is 0 the graph is a parabola, if it is positive the graph is an ellipse, and if it is negative the graph is a hyperbola. 2 and 3. x y 1 1, 45 . Then X x cos y sin 1 1 1 1 2 2 Y x sin y cos 1 1 1 1 0. Therefore, the XY -coordinates of the given point are X Y 2 0 . (b) To eliminate the x y-term from this equation we rotate the axes through an angle that satisfies cot 2
2
2
4. x y 2 1, 30 . Then X x cos y sin 2 23 1 12 12 1 2 3 and Y x sin y cos 2 12 1 23 12 2 3 . Therefore, the XY -coordinates of the given point are X Y 12 1 2 3 12 2 3 . 3 3 , 60 . Then X x cos y sin 3 12 3 23 0 and Y x sin y cos 3 23 3 12 2 3. Therefore, the XY -coordinates of the given point are X Y 0 2 3 .
5. x y
6. x y 2 0, 15 . Then X x cos y sin 2 cos 15 0 sin 15 19319 and Y x sin y cos 2 sin 15 0 cos 15 05176. Therefore, the XY -coordinates of the given point are approximately X Y 19319 05176. 7. x y 0 2, 55 . Then X x cos y sin 0 cos 55 2 sin 55 16383 and Y x sin y cos 0 sin 55 2 cos 55 11472. Therefore, the XY -coordinates of the given point are approximately X Y 16383 11472.
SECTION 12.5 Rotation of Axes
823
8. x y 2 4 2 , 45 . Then X x cos y sin 2 1 4 2 1 5 and 2 2 1 1 Y x sin y cos 2 4 2 3. Therefore, the XY -coordinates of the given point are 2
2
X Y 5 3.
9. x 2 3y 2 4, 60 . Then x X cos 60 Y sin 60 12 X 23 Y and y X sin 60 Y cos 60 23 X 12 Y . 2 2 3 23 X 12 Y 4 Substituting these values into the equation, we get 12 X 23 Y X2 3XY 3XY 3XY 3Y 2 3X 2 Y2 X2 9 2 3Y 2 3Y 2 3 3XY 3 4 X 4 4 2 4 4 2 4 4 4 4 4 2 2 2X 2 2 3XY 4 X 2 3XY 2.
10. y x 12 , 45 . Then x X cos 45 Y sin 45 22 X 22 Y and y X sin 45 Y cos 45 22 X 22 Y . Substituting these values into the equation, we get 2 2 X2 2 2 2 2 2 22 X 22 Y 1 22 Y 1 2 X 2 Y 2 X 2 Y 1 2 Y2 X2 2 22 X 22 Y 1 2Y 1 2 2 2 2 Y X 22 X 2X 22 Y 1 2Y 1 22 Y 0 X 2 Y 2 2XY 3 2X 2Y 2 0. 2 2 11. x 2 y 2 2y, cos1 35 . So cos 35 and sin 45 . Then
2 2 X cos Y sin 2 X sin Y cos 2 2 X sin Y cos 35 X 45 Y 45 X 35 Y 2 45 X 35 Y
24XY 16Y 2 16X 2 24XY 9Y 2 8X 6Y 7X 2 48XY 7Y 2 8X 6Y 9X 2 0 25 25 25 25 25 25 5 5 25 25 25 5 5 7Y 2 48XY 7X 2 40X 30Y 0.
12. x 2 2y 2 16, sin1 35 . So sin 35 and cos 45 . Then x 45 X 35 Y and y 35 X 45 Y . Substituting these values into the equation, we get 4 X 3 Y 2 2 3 X 4 Y 2 16 16 X 2 24 XY 9 Y 2 2 9 X 2 24 XY 16 Y 2 16 5 5 5 5 25 25 25 25 25 25 16 X 2 18 X 2 9 Y 2 32 Y 2 24 XY 48 XY 16 34 X 2 41 Y 2 24 XY 16 34X 2 41Y 2 24XY 400. 25 25 25 25 25 25 25 25 25
13. x 2 2 3x y y 2 4, 30 . Then x X cos 30 Y sin 30 23 X 12 Y 12 3X Y and y X sin 30 Y cos 30 12 X 23 Y 12 X 3Y . Substituting these values into the 2 1 2 1 X 3Y 3X Y 2 3 12 3X Y 2 X 3Y 4 equation, we get 12 2 2 2 3X Y 2 3 3X Y X 3Y X 3Y 16 3X 2 2 3XY Y 2 6X 2 4 3XY 6Y 2 X 2 2 3XY 3Y 2 16 8X 2 8Y 2 16
X2 Y2 1. This is a hyperbola. 2 2
824
CHAPTER 12 Conic Sections
1 X 1 Y 1 X Y 14. x y x y, 4 . Therefore, x X cos 4 Y sin 4 2 2 2 1 X 1 Y 1 X Y . Substituting into the equation gives and y X sin 4 Y cos 4 2 2 2 1 1 1 1 X Y X Y X Y X Y 1 X 2 Y 2 2 X X 2 Y 2 2 2X 2 2 2 2 2 2 2 X 2 Y2 1. This is a hyperbola. 2 2
15. (a) x y 8 0x 2 x y 0y 2 8. So A 0, B 1, and C 0, and so the
(c)
discriminant is B 2 4AC 12 4 0 0 1. Since the discriminant is
1
positive, the equation represents a hyperbola. (b) cot 2
y
x
1
AC 0 2 90 45 . Therefore, B 2 2 2 X 2 Y . After substitution, the original 2 2 2 2 Y 2 X 2 Y 8 X2 Y2
x 22 X 22 Y and y equation becomes 22 X
X Y X Y 8 1. This is a hyperbola with a 4, b 4, and c 4 2. Hence, the vertices 2 16 16 are V 4 0 and the foci are F 4 2 0 .
16. (a) x y 4 0 0x 2 x y 0y 2 4 0. So A 0, B 1, and C 0,
y
(c)
and so the discriminant is B 2 4AC 12 4 0 0 1. Since the
1
discriminant is positive, the equation represents a hyperbola.
x 22 X 22 Y and y equation becomes 22 X
x
1
AC 0 2 90 45 . Therefore, (b) cot 2 B 2 2 2 X 2 Y . After substitution, the original 2 2 2 2 Y 2 X 2 Y 40 Y2 X2
X Y X Y 4 1. This is a hyperbola with a 2 2, b 2 2, and c 4. Hence, the 2 8 8 vertices are V 0 2 2 and the foci are F 0 4.
17. (a) x 2 2 3x y y 2 2 0. So A 1, B 2 3, and C 1, and so the 2 discriminant is B 2 4AC 2 3 4 1 1 0. Since the discriminant is positive, the equation represents a hyperbola.
(c)
y
1 1
11 AC 1 2 60 30 . Therefore, (b) cot 2 B 2 3 3
x 23 X 12 Y and y 12 X 23 Y . After substitution, the original equation becomes 2 2 3 1 1 X 3Y 1 X 3Y 2 3 23 X 12 Y 20 2 X 2Y 2 2 2 2 3 X 2 3 XY 1 Y 2 3 3X 2 2XY 3Y 2 14 X 2 23 XY 34 Y 2 2 0 4 2 4 2 X 2 34 32 14 XY 23 3 23 Y 2 14 32 34 2 2X 2 2Y 2 2 Y 2 X 2 1.
x
SECTION 12.5 Rotation of Axes
18. (a) 13x 2 6 3x y 7y 2 16. Then A 13, B 6 3, and C 7, and so 2 the discriminant is B 2 4AC 6 3 4 13 7 256. Since the
(c)
825
y
1
discriminant is negative, the equation represents an ellipse.
1
x
13 7 AC 1 2 60 30 . Therefore, (b) cot 2 B 6 3 2
x 23 X 12 Y and y 12 X 23 Y . After substitution, the original equation becomes 2 2 1 X 3Y 7 1 X 3Y 13 23 X 12 Y 6 3 23 X 12 Y 16 2 2 2 2 13 3X 2 2 3XY Y 2 3 3 3X 2 2XY 3Y 2 74 X 2 2 3XY 3Y 2 16 4 2 9 7 XY 13 3 6 3 7 3 Y 2 13 9 21 16 16X 2 4Y 2 16 X 2 39 4 2 4 2 2 2 4 2 4
Y2 1. This is an ellipse with a 2, b 1, and c 4 1 3. Thus, the vertices are V 0 2 and the 4 foci are F 0 3 `. X2
19. (a) 11x 2 24x y 4y 2 20 0. So A 11, B 24, and C 4, and so the discriminant is B 2 4AC 242 4 11 4 0. Since the
discriminant is positive, the equation represents a hyperbola.
AC 11 4 7 7 . Therefore, cos 2 25 B 24 24 cos 1725 35 and sin 1725 45 . Hence, 2 2
(c)
y
1 1
(b) cot 2
3X 45 Y and y 45 X 5 equation becomes 2 11 35 X 45 Y 24 35 X 11 9X 2 24XY 16Y 2 25 x
3 Y . After substitution, the original 5
7 , we have Since cos 2 25
2 10626 , so 53 .
2 4 45 X 35 Y 20 0 24 12X 2 7XY 12Y 2 4 16X 2 24XY 9Y 2 20 0 25 25
4Y 5
4 X 3Y 5 5
x
X 2 99 288 64 XY 264 168 96 Y 2 176 288 36 500 125X 2 500Y 2 500 1 X 2 Y 2 1. 4
826
CHAPTER 12 Conic Sections
20. (a) 21x 2 10 3x y 31y 2 144. Then A 21, B 10 3, and C 31, 2 and so the discriminant is B 2 4AC 10 3 4 21 31 2304.
(c)
y
1
Since the discriminant is negative, the equation represents an ellipse.
1
x
21 31 AC 1 (b) cot 2 2 120 60 . B 10 3 3
Therefore, x 12 X 23 Y and y 23 X 12 Y . After substitution, the original equation becomes 2 2 3 3 1 1 21 12 X 23 Y 10 3 12 X 23 Y 144 2 X 2 Y 31 2 X 2 Y 21 X 2 2 3XY 3Y 2 5 3 2 2 144 3X 2 2XY 3Y 2 31 4 2 4 3X 2 3XY Y 15 93 XY 21 3 10 3 31 3 Y 2 63 15 31 144 36X 2 16Y 2 144 X 2 21 4 2 4 2 2 2 4 2 4 1 X 2 1 Y 2 1. This is an ellipse with a 3, b 2, and c 9 4 5. 4 9 21. (a)
2 3x 3x y 3. So A 3, B 3, and C 0, and so the discriminant 3 0 9. Since the discriminant is positive, is B 2 4AC 32 4
(c)
y
1
the equation represents a hyperbola.
x
1
1 AC 2 60 30 . Therefore, (b) cot 2 B 3
x 23 X 12 Y and y 12 X 23 Y . After substitution, the equation 2 1 X 3Y 3 becomes 3 23 X 12 Y 3 23 X 12 Y 2 2 3 3 4 3X 2 2 3XY Y 2 4 3X 2 2XY 3Y 2 3 3 3 3 3 3 3 X2 3 Y 2 3 3 X2 6 Y2 1 Y 2 1. This X 2 3 4 3 3 4 3 XY 6 4 4 4 4 2 2 2 2 3 is a hyperbola with a 2 and b 2 3. 3
22. (a) 153x 2 192x y 97y 2 225. Then A 153, B 192, and C 97, and so the discriminant is B 2 4AC 1922 4 153 97 22500.
Since the discriminant is negative, the equation represents an ellipse.
AC 153 97 56 56 (b) cot 2 cos 2 . Therefore, B 192 192 200 156200 4 3 16 12 cos 156200 2 20 5 and sin 2 20 5
(c)
y
1
cos1 54 369 . Substituting gives 2 3 X 4 Y 97 3 X 4 Y 2 225 153 45 X 35 Y 192 45 X 35 Y 5 5 5 5 153 16X 2 24XY 9Y 2 192 12X 2 7XY 12Y 2 97 9X 2 24XY 16Y 2 225 25 25 25
1
X 2 2448 2304 873 XY 3672 1344 2328 Y 2 1377 2304 1552 5625 5625X 2 625Y 2 5625 X 2 19 Y 2 1. This is an ellipse with a 3, b 1, and c 9 1 2 2.
x
SECTION 12.5 Rotation of Axes
23. (a) x 2 2x y y 2 x y 0. So A 1, B 2, and C 1, and so the
y
(c)
discriminant is B 2 4AC 22 4 1 1 0. Since the discriminant is
1
zero, the equation represents a parabola.
1
AC (b) cot 2 0 2 90 45 . Therefore, B
827
x
x 22 X 22 Y and y 22 X 22 Y . After substitution, the original equation becomes 2 2 X 2Y 2 X 2Y 2 X 2Y 2 2 2 2 2 2 2
2 2 2 22 X 22 Y 22 X 22 Y 0 2 X 2 Y 1 X 2 XY 1 Y 2 X 2 Y 2 1 X 2 XY Y 2 2Y 0 2X 2 2Y 0 X 2 2 Y . This is a 2 2 2 2
1 . parabola with 4 p 1 and hence the focus is F 0 2
4 2
24. (a) 25x 2 120x y 144y 2 156x 65y 0. Then A 25, B 120, and
(c)
y
C 144, and so the discriminant is
B 2 4AC 1202 4 25 144 0. Since the discriminant is zero,
the equation represents a parabola.
AC 25 144 119 cos 2 119 169 . Therefore, B 120 120 1119169 5 . Hence, cos 1119169 12 13 2 13 and sin 2
(b) cot 2
2 2
x
Since cos 2 119 169 , we have
12X 5Y 5X 12Y 2 452 , so 23 . x and y . Substituting gives 13 13 13 13 5Y 2 5Y 12Y 12Y 2 5Y 5X 5X 12X 12X 12X 144 120 156 25 13 13 13 13 13 13 13 13 13 13 Y2 12Y X2 5X 65 0 25 122 120 12 5 144 52 25 52 120 5 12 144 122 13 13 169 169 Y X 156 12 65 5 156 5 65 12 0 169Y 2 169X 0 X Y 2 . This is a parabola with 13 13 4 p 1.
828
CHAPTER 12 Conic Sections
25. (a) 2 3x 2 6x y 3x 3y 0. So A 2 3, B 6, and C 0, and so the discriminant is B 2 4AC 62 4 2 3 0 36. Since the
y
(c)
2
discriminant is positive, the equation represents a hyperbola. AC 2 3 1 (b) cot 2 2 120 60 . Therefore, B 6 3
2
x
x 12 X 23 Y and y 23 X 12 Y , and substituting gives 2 3 3 3 1 1 1 2 3 12 X 23 Y 6 12 X 23 Y 2 X 2Y 3 2 X 2 Y 3 2 X 2Y 0 3 X 2 23XY 3Y 2 3 3X 2 2XY 3Y 2 3 X 3Y 3 3X Y 0 2 2 2 2 X 2 23 3 2 3 X 23 3 2 3 XY 3 3 Y 2 3 2 3 3 2 3 Y 32 32 0 3X 2 2 3X 3 3Y 2 0 X 2 2X 3Y 2 0 3Y 2 X 2 2X 1 1 X 12 3Y 2 1. This is a hyperbola with a 1, b 33 , c 1 13 2 , and C 1 0. 3
26. (a) 9x 2 24x y 16y 2 100 x y 1. Then A 9, B 24, and
C 16, and so the discriminant is B 2 4AC 242 4 9 16 0. Since the discriminant is zero, the equation represents a parabola.
(c)
y 1 1
x
AC 9 16 7 7 369 . Now cos 2 25 B 24 24 4 and sin 1725 3 , and so cos 1725 2 5 2 5
(b) cot 2
x 45 X 35 Y , y 35 X 45 Y . By substitution, 2 3 X 4 Y 16 3 X 4 Y 2 100 4 X 3 Y 3 X 4 Y 1 9 45 X 35 Y 24 45 X 35 Y 5 5 5 5 5 5 5 5 9 2 2 24 12X 2 7XY 12Y 2 16 9X 2 24XY 16Y 2 100 1 X 7 Y 1 25 16X 24XY 9Y 25 25 5 5 625Y 2 500X 3500Y 2500 5Y 2 28Y 4X 20 196 196 96 24 14 2 4 X 24 . This is a 5 Y 2 28 5 Y 25 4X 20 5 4X 5 4 X 5 Y 5 5 5 14 . parabola with 4 p 45 and V 24 5 5
27. (a) 52x 2 72x y 73y 2 40x 30y 75. So A 52, B 72, and C 73, and so the discriminant is
B 2 4AC 722 4 52 73 10,000. Since the discriminant is decidedly negative, the equation represents an ellipse.
SECTION 12.5 Rotation of Axes
829
52 73 7 AC . Therefore, as in Exercise 19(b), we get cos 35 , sin 45 , and B 72 24 x 35 X 45 Y , y 45 X 35 Y . By substitution, 2 4 X 3 Y 73 4 X 3 Y 2 52 35 X 45 Y 72 35 X 45 Y 5 5 5 5 40 35 X 45 Y 30 45 X 35 Y 75 52 9X 2 24XY 16Y 2 72 12X 2 7XY 12Y 2 73 16X 2 24XY 9Y 2 25 25 25
(b) cot 2
24X 32Y 24X 18Y 75 468X 2 832Y 2 864X 2 864Y 2 1168X 2 657Y 2 1250Y 1875 2500X 2 625Y 2 1250Y 1875 100X 2 25Y 2 50Y 75 X 2 14 Y 12 1. This is an ellipse with a 2, b 1, c 4 1 3, and center C 0 1.
(c)
y
1 x
1
7 , we have 2 cos1 7 10626 and so 53 . Since cos 2 25 25
28. (a) 7x 24y2 49x 2 336x y 576y 2 600x 175y 25. So A 49, B 336, and C 576, and so the
discriminant is B 2 4AC 3362 4 49 576 0. Since the discriminant is zero, the equation represents a parabola. AC 49 576 527 1527625 7 and (b) cot 2 . Therefore, cos 25 cos 2 527 625 2 B 336 336 7 24 24 7 24 sin 1527625 2 25 . Substituting x 25 X 25 Y and y 25 X 25 Y gives 7 X 24 Y 2 336 7 X 24 Y 24 X 7 Y 576 24 X 7 Y 2 49 25 25 25 25 25 25 25 25 7 X 24 Y 175 24 X 7 Y 25 600 25 25 25 25 49 2 2 336 168X 2 527XY 168Y 2 576 576X 2 336XY 49Y 2 625 49X 336XY 576Y 625 625
168X 576Y 168X 49Y 25 X 2 2401 56,448 331,776 XY 16,464 177,072 193,536 Y 2 28,224 56,448 28,224
6252 Y 15,625 1 Y 1 . This is a parabola with 390,625X 2 390,625Y 15,625 25X 2 25Y 1 X 2 Y 25 25 1 4 p 1 and vertex 0 25 . (c)
y
1 1
x
1 527 14748 and so 74 . Since cos 2 527 625 , we have 2 cos 625
830
CHAPTER 12 Conic Sections
29. (a) The discriminant is B 2 4AC 42 4 2 2 0. Since the discriminant is 0, the equation represents a parabola. (b) 2x 2 4x y 2y 2 5x 5 0 2y 2 4x y 2x 2 5x 5 2 y 2 2x y 2x 2 5x 5 2 y 2 2x y x 2 2x 2 5x 5 2x 2 2 y x2 5x 5 y x2 52 x 52 y x 52 x 52 y x 52 x 52
5
5
30. (a) The discriminant is B 2 4AC 22 4 1 3 8 0. Since the discriminant is negative, the equation represents an ellipse. (b) x 2 2x y 3y 2 8 3y 2 2x y 8 x 2 3 y 2 23 x y 8 x 2
3 y 2 23 x y 19 x 2 8 x 2 13 x 2 3 y 13 x 2 y 13 x 83 29 x 2 y 13 x 83 29 x 2 y 13 x 83 29 x 2
2
5
8 23 x 2
-5
5 -5
31. (a) The discriminant is B 2 4AC 102 4 6 3 28 0. Since the discriminant is positive, the equation represents a hyperbola. (b) 6x 2 10x y 3y 2 6y 36 3y 2 10x y 6y 36 6x 2 3y 2 2 5x 3 y 36 6x 2 y 2 2 53 x 1 y 12 2x 2 2 2 y 2 2 53 x 1 y 53 x 1 53 x 1 12 2x 2 2 10 2 2 y 53 x 1 25 9 x 3 x 1 12 2x
10
-10
10 -10
2 y 53 x 1 79 x 2 10 3 x 13 y 53 x 1 79 x 2 10 3 x 13 y 53 x 1 79 x 2 10 3 x 13
32. (a) The discriminant is B 2 4AC 62 4 9 1 0. Since the discriminant is 0, the equation represents a parabola. (b) 9x 2 6x y y 2 6x 2y 0 y 2 6x y 2y 9x 2 6x
5
y 2 2 3x 1 y 9x 2 6x
y 2 2 3x 1 y 3x 12 3x 12 9x 2 6x 2 y 3x 1 1 y 3x 1 1 y 3x 1 1. So y 3x or y 3x 2. This is a degenerate conic.
-2
2 -5
SECTION 12.5 Rotation of Axes
831
33. (a) 7x 2 48x y 7y 2 200x 150y 600 0. Then A 7, B 48, and C 7, and so the discriminant is
B 2 4AC 482 4 7 7 0. Since the discriminant is positive, the equation represents a hyperbola. We 7 AC cos 45 and sin 35 . now find the equation in terms of XY -coordinates. We have cot 2 B 24 Therefore, x 45 X 35 Y and y 35 X 45 Y , and substitution gives
2 3 X 4 Y 7 3 X 4 Y 2 200 4 X 3 Y 150 3 X 4 Y 600 0 7 45 X 35 Y 48 45 X 35 Y 5 5 5 5 5 5 5 5 7 16X 2 24XY 9Y 2 48 12X 2 7XY 12Y 2 7 9X 2 24XY 16Y 2 25 25 25
160X 120Y 90X 120Y 600 0 112X 2 168XY 63Y 2 576X 2 336XY 576Y 2 63X 2 168XY 112Y 2 6250X 15,000 0 25X 2 25Y 2 250X 600 0 25 X 2 10X 25 25Y 2 600 625 X 52 Y 2 1. This is a hyperbola with a 1, b 1, c 1 1 2, and center C 5 0.
(b) In the XY -plane, the center is C 5 0, the vertices are V 5 1 0 V1 4 0 and V2 6 0, and the foci are F 5 2 0 . In the x y-plane, the center is C 45 5 35 0 35 5 45 0 C 4 3, the vertices are 12 4 3 3 4 24 18 V1 45 4 35 0 35 4 45 0 V1 16 5 5 and V2 5 6 5 0 5 6 5 0 V2 5 5 , and the foci are F1 4 45 2 3 35 2 and F2 4 45 2 3 35 2 . (c) In the XY -plane, the equations of the asymptotes are Y X 5 and Y X 5. In the x y-plane, these equations
become x 35 y 45 x 45 y 35 5 7x y 25 0. Similarly, x 35 y 45 x 45 y 35 5 x 7y 25 0.
34. (a) 2 2 x y2 7x 9y 2 2x 2 4 2x y 2 2y 2 7x 9y. Therefore, A 2 2, B 4 2, and C 2 2, X Y X Y AC x y 2X. Thus the 0 2 90 45 . Thus x , y and so cot 2 B 2 2 2 16X 2Y 8X 2 16X 2Y 4X 2 8X Y 4 X 2 2X 1 Y 4 equation becomes 2 2 2X 2 X 12 14 Y 4. This is a parabola.
1 . Thus, in XY -coordinates the vertex is V 1 4 and the focus is F 1 63 . In x y-coordinates, (b) 4 p 14 p 16 16 5 2 3 2 79 2 47 2 V and F 32 32 . 2 2 (c) The directrix is Y 65 16 . Thus
65 x y or y x 6516 2 . 16 2
35. We use the hint and eliminate Y by adding: x X cos Y sin x cos X cos2 Y sin cos and
y X sin Y cos y sin X sin2 Y sin cos , and adding these two equations gives x cos y sin X cos2 sin2 x cos y sin X. In a similar manner, we eliminate X by subtracting:
x X cos Y sin x sin X cos sin Y sin2 and y X sin Y cos y cos X sin cos Y cos2 , so x sin y cos Y cos2 sin2 x sin y cos Y . Thus, X x cos y sin and Y x sin y cos .
832
36.
CHAPTER 12 Conic Sections
x
y 1. Squaring both sides gives x 2 x y y 1 2 x y 1 x y, and squaring both sides again
gives 4x y 1 x y2 1 x y x x 2 x y y x y y 2 4x y x 2 y 2 2x y 2x 2y 1 AC x 2 y 2 2x y 2x 2y 1 0. Then A 1, B 2, and C 1, and so cot 2 0 2 90 B
45 . Therefore, x 22 X 22 Y and y 22 X 22 Y , and substituting gives 2 2 X 2Y 2 X 2Y 2 X 2Y 2 2 2 2 2 2 2 2 22 X 22 Y 2 22 X 22 Y 2 22 X 22 Y 1 0 1 X 2 2XY Y 2 X 2 Y 2 1 X 2 2XY Y 2 2 2X 1 0 X 2 1 1 1 XY 1 1 2 2 2 2 1 . This is a parabola with Y 2 12 1 12 2 2X 1 0 2Y 2 2 2X 1 Y 2 2X 12 2 X 2 2 1 4 p 2 and vertex V 0 . However, in the original equation we must have x 0 and y 0, so we get only the 2 2 part of the parabola that lies in the first quadrant.
37. (a) Z
x
X cos sin , Z , and R .
cos x cos sin X X cos Y sin Y . Equating the entries in this Thus Z R Z y sin cos Y X sin Y cos y
Y
sin
matrix equation gives the first pair of rotation of axes formulas. Now cos sin cos sin 1 1 and so Z R 1 Z R cos2 sin2 sin cos sin cos X cos sin x x cos y sin . Equating the entries in this matrix equation gives Y sin cos y x sin y cos
the second pair of rotation of axes formulas. cos 1 sin 1 cos 2 sin 2 (b) R1 R2 sin 1 cos 1 sin 2 cos 2 cos 1 cos 2 sin 1 sin 2 cos 1 sin 2 sin 1 cos 2 cos 1 2 sin 1 2 sin 1 2 cos 1 2 sin 1 cos 2 cos 1 sin 2 sin 1 sin 2 cos 1 cos 2 38. (a) Using A A cos2 B sin cos C sin2 , B 2 C a sin cos B cos2 sin2 , and C A sin2 B sin cos C cos2 , we first expand the terms. 2 2 B 2 C a sin cos B cos2 sin2
2 4 C A2 sin2 cos2 4B C A sin cos cos2 sin2 B 2 cos2 sin2 4 C A2 sin2 cos2 4B C A sin cos3 4B C A sin3 cos
B 2 cos4 2B 2 sin2 cos2 B 2 sin4
SECTION 12.5 Rotation of Axes
833
4A C 4 A cos2 B sin cos C sin2 A sin2 B sin cos C cos2
4A2 sin2 cos2 4AB sin cos3 4AC cos4 4AB sin3 cos 4B 2 sin2 cos2
4BC sin cos3 4AC sin4 4BC sin3 cos 4C 2 sin2 cos2
Since there are many terms in this expansion, we find the coefficients of like trignometric terms. 2 B Term 4A C Sum cos4
B2
sin cos3
4B C A
sin2 cos2 sin3 cos sin4 So 2
B 4A C
4 C A2 2B 2 4B C A B2
4AC
B 2 4AC
4AB 4BC
4BC 4AB 4AB 4BC 0 8AC 2B 2 2 B 2 4AC
4A2 4B 2 4C 2 4AB 4BC 4AC
4BC AB 4AB 4BC 0 B 2 4AC
B 2 4AC cos4 2 B 2 4AC sin2 cos2 B 2 4AC sin4
2 cos4 2 sin2 cos2 sin4 B 2 4AC cos2 sin2 B 2 4AC 12 B 2 4AC
B 2 4AC
So B 2 4AC B 2 4A C .
(b) Using A A cos2 B sin cos C sin2 and C A sin2 B sin cos C cos2 , we have A C A cos2 B sin cos C sin2 A sin2 B sin cos C cos2 A sin2 cos2 C sin2 cos2 AC
(c) Since F F, F is also invariant under rotation. 39. Let P be the point x1 y1 and Q be the point x2 y2 and let P X 1 Y1 and Q X 2 Y2 be the images of P and Q under the rotation of . So X 1 x1 cos y1 sin Y1 x1 sin y1 cos , X 2 x2 cos y2 sin and Y2 x2 sin y2 cos . Thus d P Q X 2 X 1 2 Y2 Y1 2 , where X 2 X 1 2
2 2 x2 cos y2 sin x1 cos y1 sin x2 x1 cos y2 y1 sin
x2 x1 2 cos2 x2 x1 y2 y1 sin cos y2 y1 2 sin2 and Y2 Y1 2
2 2 x2 sin y2 cos x1 sin y1 cos x2 x1 sin y2 y1 cos
x2 x1 2 sin2 x2 x1 y2 y1 sin cos y2 y1 2 cos2 So
X 2 X 1 2 Y2 Y1 2 x2 x1 2 cos2 x2 x1 y2 y1 sin cos y2 y1 2 sin2 x2 x1 2 sin2 x2 x1 y2 y1 sin cos y2 y1 2 cos2
x2 x1 2 cos2 y2 y1 2 sin2 x2 x1 2 sin2 y2 y1 2 cos2 x2 x1 2 cos2 sin2 y2 y1 2 sin2 cos2 x2 x1 2 y2 y1 2
834
CHAPTER 12 Conic Sections
Putting these equations together gives d P Q X 2 X 1 2 Y2 Y1 2 x2 x1 2 y2 y1 2 d P Q.
12.6 POLAR EQUATIONS OF CONICS 1. All conics can be described geometrically using a fixed point F called the focus and a fixed line called the directrix. For a distance from P to F fixed positive number e the set of all points P satisfying e is a conic section. If e 1 the conic is a distance from P to parabola, if e 1 the conic is an ellipse, and if e 1 the conic is a hyperbola. The number e is called the eccentricity of the conic. ed ed 2. The polar equation of a conic with eccentricity e has one of the following forms: r or r . 1 e cos 1 e sin
2 3 3 3. Substituting e 23 and d 3 into the general equation of a conic with vertical directrix, we get r 1 23 cos
r
6 . 3 2 cos
4 3 3 4. Substituting e 43 and d 3 into the general equation of a conic with vertical directrix, we get r 1 43 cos
r
12 . 3 4 cos
5. Substituting e 1 and d 2 into the general equation of a conic with horizontal directrix, we get r r
2 . 1 sin
12 1 sin
1 4 2 6. Substituting e 12 and d 4 into the general equation of a conic with horizontal directrix, we get r 1 12 sin 4 . r 2 sin 20 45 r . 7. r 5 sec r cos 5 x 5. So d 5 and e 4 gives r 1 4 cos 1 4 cos 12 06 2 r . 8. r 2 csc r sin 2 y 2. So d 2 and e 06 gives r 1 06 sin 1 06 sin 9. Since this is a parabola whose focus is at the origin and vertex at 5 2, the directrix must be y 10. So d 10 and 10 1 10 . e 1 gives r 1 sin 1 sin d P F 2 2 10. Since the vertex is at 2 0 we have d P F 2. Now, since e we get 04 d P 5. d P d P 04 04 7 28 The directrix is x 7, so substituting e 04 and d 7 we get r r . 1 04 cos 1 04 cos 6 is Graph II. The eccentricity is 1, so this is a parabola. When 0, we have r 3 and when 11. r 2 , we 1 cos have r 6. 2 1 12. r is Graph III. r , so e 12 and this is an ellipse. When 0, r 2, and when , r 23 . 2 cos 1 1 cos 2
3 13. r is Graph VI. e 2, so this is a hyperbola. When 0, r 3, and when , r 3. 1 2 sin 5
14. r
5 3 is Graph I. r , so e 1 and this is a parabola. When 0, r 53 and when , r 53 . 3 3 sin 1 sin
SECTION 12.6 Polar Equations of Conics
835
4 12 , so e 23 and this is an ellipse. When 0, r 4, and when , r 4. is Graph IV. r 2 3 2 sin 1 3 sin
15. r
6 12 is Graph V. r , so e 32 and this is a hyperbola. When 0, r 12 5 , and when , 2 3 cos 1 32 cos we have r 12.
16. r
4 has e 1 and d 4, so 1 sin it represents a parabola.
17. (a) The equation r
3
18. (a) The equation r
3 2 has e 1 2 2 sin 1 sin
and d 32 , so it represents a parabola. V(3/4, ¹/2) _4
_8
_4 y=_4
O
4 V(2, 3¹/2)
ed , 1 e sin the directrix is parallel to the polar axis and has equation y 4. The vertex is 2 32 .
e 1 and d 53 , so it represents a parabola. x=5/3
O
O
4
ed , 1 e sin the directrix is parallel to the polar axis and has equation y d 32 . The vertex is 34 2 .
(b) Because the equation is of the form r
5
5 3 has 3 3 cos 1 cos
1
y=3/2 2
8
(b) Because the equation is of the form r
19. (a) The equation r
_2
2
V(5/6, 0)
3
ed , 1 e cos the directrix is parallel to the polar axis and has equation x d 53 . The vertex is 56 0 .
(b) Because the equation is of the form r
2
20. (a) The equation r
2 5 has 5 5 cos 1 cos
e 1 and d 25 , so it represents a parabola. x=_2/5
1 V(1/5, ¹) O
1
ed , 1 e cos the directrix is parallel to the polar axis and has equation x d 25 . The vertex is 15 .
(b) Because the equation is of the form r
836
CHAPTER 12 Conic Sections
21. (a) The equation r
2 4 has e 12 1, so it represents an 1 2 cos 1 2 cos
x=_4
ellipse.
1
ed (b) Because the equation is of the form r with d 4, the directrix is 1 e cos vertical and has equation x 4. Thus, the vertices are V1 4 0 and V2 43 .
Vª(4/3, ¹)
1
O
VÁ(4, 0)
(c) The length of the major axis is 2a V1 V2 4 43 16 3 and the center is at the midpoint of V1 V2 , 43 0 . The minor axis has length 2b where 2 2 b2 a 2 c2 a 2 ae2 83 83 12 16 3 , so 8 3 2b 2 16 3 3 462.
22. (a) The equation r
6 2 has e 23 1, so it represents an 2 3 2 sin 1 3 sin
VÁ(6, ¹/2)
ellipse. ed with d 3, the directrix is 1 e sin horizontal and has equation y 3. Thus, the vertices are V1 6 2 and V2 65 32 .
(b) Because the equation is of the form r
1 O Vª(6/5, 3¹/2)
1 y=_3
(c) The length of the major axis is 2a V1 V2 6 65 36 5 and the center is at the midpoint of V1 V2 , 12 5 2 . The minor axis has length 2b where 2 18 2 2 36 , so b2 a 2 c2 a 2 ae2 18 5 5 3 5 12 5 2b 2 36 5 5 537.
23. (a) The equation r
3 12 has e 34 1, so it represents an 4 3 sin 1 34 sin
ellipse. ed (b) Because the equation is of the form r with d 4, the directrix is 1 e sin horizontal and has equation y 4. Thus, the vertices are V1 12 7 2 and V2 12 32 .
96 (c) The length of the major axis is 2a V1 V2 12 7 12 7 and the center is at 3 the midpoint of V1 V2 , 36 7 2 . The minor axis has length 2b where
2 48 3 2 144 , so b2 a 2 c2 a 2 ae2 48 7 7 4 7 24 7 2b 2 144 7 7 907.
VÁ(12/7, ¹/2) O 1 4 8 Vª(12, 3¹/2)
y=4
SECTION 12.6 Polar Equations of Conics 9
24. (a) The equation r
18 2 has e 34 1, so it represents an 4 3 cos 1 34 cos
x=6
ellipse.
VÁ(18/7, 0)
ed (b) Because the equation is of the form r with d 6, the directrix is 1 e cos 0 and V2 18 . vertical and has equation x 6. Thus, the vertices are V1 18 7
Vª(18, ¹)
2 O
5
144 (c) The length of the major axis is 2a V1 V2 18 7 18 7 and the center is at the midpoint of V1 V2 , 54 7 . The minor axis has length 2b where
2 72 3 2 324 , so b2 a 2 c2 a 2 ae2 72 7 7 4 7 36 7 2b 2 324 7 7 1361.
8 has e 2 1, so it represents a hyperbola. 1 2 cos ed with d 4, the transverse axis is (b) Because the equation has the form r 1 cos horizontal and the directrix has equation x 4. The vertices are V1 83 0 and
x=4
25. (a) The equation r
VÁ(8/3, 0)
Vª(8, 0)
O
V2 8 8 0.
(c) The center is the midpoint of V1 V2 , 16 3 0 . To sketch the central box and the
asymptotes, we find a and b. The length of the transverse axis is 2a 16 3 , and so 2 2 a 83 , and b2 c2 a 2 ae2 a 2 83 2 83 64 3 , so 8 3 b 64 3 3 462.
10 has e 4 1, so it represents a hyperbola. 1 4 sin ed with d 52 , the transverse axis (b) Because the equation has the form r 1 cos
26. (a) The equation r
VÁ(2, 3¹/2)
is vertical and the directrix has equation y 52 . The vertices are
10 3 3 V1 10 3 2 3 2 and V2 2 2 . (c) The center is the midpoint of V1 V2 , 83 32 . To sketch the central box and the
asymptotes, we find a and b. The length of the transverse axis is 2a 43 , and so 2 2 a 23 , and b2 c2 a 2 ae2 a 2 23 4 23 20 3 , so b 20 3 258.
y=_5/2 Vª(10/3, 3¹/2)
837
838
CHAPTER 12 Conic Sections
27. (a) The equation r
10 20 has e 32 1, so it represents a 2 3 sin 1 32 sin
VÁ(4, 3¹/2) O
hyperbola. (b) Because the equation has the form r
y=_20/3
ed with d 20 3 , the transverse axis 1 cos
is vertical and the directrix has equation y 20 3 . The vertices are 3 3 V1 20 2 20 2 and V2 4 2 . (c) The center is the midpoint of V1 V2 , 12 32 . To sketch the central box and the
Vª(20, 3¹/2)
asymptotes, we find a and b. The length of the transverse axis is 2a 16, and so 2 a 8, and b2 c2 a 2 ae2 a 2 8 32 82 80, so b 80 4 5 894.
28. (a) The equation r
3 6 has e 72 1, so it represents a 7 2 7 cos 1 2 cos
hyperbola. ed with d 67 , the transverse axis (b) Because the equation has the form r 1 cos is horizontal and the directrix has equation x 67 . The vertices are V1 23 0 and V2 65 65 0 . (c) The center is the midpoint of V1 V2 , 14 15 0 . To sketch the central box and the asymptotes, we find a and b. The length of the transverse axis is 2a 4 , and b2 c2 a 2 ae2 a 2 4 7 2 4 2 a 15 15 2 15 b 45 2 5 5 089.
29. (a) r
4 e 3, so the conic is a hyperbola. 1 3 cos
(b) The vertices occur where 0 and . Now 0 r
x=6/7 1 O VÁ(2/3, 0)
Vª(6/5, 0)
8 , and so 15 4 , so 5
4 1, 1 3 cos 0
4 4 and r 2. Thus the vertices are 1 0 and 1 3 cos 2 2 .
(_2, ¹)
(1, 0)
8
8 3 r e 1, so the conic is a parabola. 3 3 cos 1 cos 8 8 4 (b) Substituting 0, we have r . Thus the vertex is 43 0 . 3 3 cos 0 6 3
30. (a) r
1
( 43 , 0)
SECTION 12.6 Polar Equations of Conics
2 e 1, so the conic is a parabola. 1 cos 2 (b) Substituting , we have r 22 1. Thus the vertex is 1 . 1 cos
31. (a) r
(1, ¹)
1
10
32. (a) r
10 3 r e 23 , so the conic is an ellipse. 3 2 sin 1 23 sin
(10, ¹2 )
3 (b) The vertices occur where 2 and 2 . Now 2 10 10 10 10 10 and 32 r 2. Thus, r 3 3 2 sin 1 5 3 2 sin 2 2 3 . the vertices are 10 and 2 2 2
33. (a) r
1
(2, 3¹ 2 )
1 6 6 2 r e 12 , so the conic is an ellipse. 2 sin 1 12 sin
(2, ¹2)
3 (b) The vertices occur where 2 and 2 . Now 2
1
6 6 6 6 2 and 32 r 6. Thus, the 3 2 sin 2 3 1 2 sin 2 vertices are 2 2 and 6 32 .
r
(6, 3¹ 2 )
5
34. (a) r
5 2 e 32 , so the conic is a hyperbola. r 2 3 sin 1 32 sin
1
3 (b) The vertices occur where 2 and 2 . Now 2 5 5 5 5 and 32 r 55 1. Thus, r 3 2 3 sin 1 2 3 sin 2 2 3 the vertices are 5 2 and 1 2 .
(
¹ _5, 2
)
(1, 3¹ 2)
7
7 2 r e 52 , so the conic is a hyperbola. 2 5 sin 1 52 sin 7 3 7 7 (b) The vertices occur where 2 and 2 . r 2 5 sin 3 3 and 2 7 7 1. Thus, the vertices are 7 and 32 r 7 3 2 2 5 sin 32 1 32 .
35. (a) r
(_ 73 , ¹2 )
1
(1, 3¹ 2)
8
36. (a) r
8 3 r e 13 , so the conic is an ellipse. 3 cos 1 13 cos
(b) The vertices occur where 0 and . Now 0 8 8 r 84 2 and r 82 4. Thus, the vertices 3 cos 0 3 cos are 2 0 and 4 .
(4, ¹)
(2, 0) 1
839
840
CHAPTER 12 Conic Sections 1
37. (a) r
1 4 e 34 , so the 4 3 cos 1 34 cos
conic is an ellipse. The vertices occur where 0 and . Now 0 r
(b) If the ellipse is rotated through 3 , the equation of the resulting conic is r
1 1 and 4 3 cos 0
1 . 4 3 cos 3
1.0
1 17 . Thus, the vertices r 4 3 cos are 1 0 and 17 . We have d 13 , so the
0.5
directrix is x 13 .
x=-1/3
0.5
O
1
2
38. (a) r
2 5 e 35 , so the 5 3 sin 1 35 sin
conic is an ellipse. The vertices occur where 2 and 32 . Now 2 r and 32 r
2 1 5 3 sin 2
(b) If the ellipse is rotated through 23 , the equation 2 . of the resulting conic is r 5 3 sin 23 0.5
2
14 . Thus, the 5 3 sin 32
1 3 2 vertices are 1 2 and 4 2 . We have d 3 , so the directrix is y 23 .
1
O y=_2/3
-1.0
-0.5 -0.5
SECTION 12.6 Polar Equations of Conics
2 e 1, so the conic is a parabola. 1 sin 2 Substituting 2 , we have r t 1 sin 1, 2 so the vertex is 1 . Because d 2, the directrix 2
39. (a) r
is y 2.
(b) If the ellipse is rotated through 4 , the equation of the resulting conic is r
2 . 1 sin 4 2
y=2 O
-8
-6
-4
-2
2 -2 -4 -6 -8
9 9 2 e 1, so the conic 2 2 cos 1 cos is a parabola. Substituting 0, we have 9 9 , so the vertex is 94 0 . r 2 2 cos 0 4
40. (a) r
Because d 92 , the directrix is x 92 .
(b) If the ellipse is rotated through 56 , the equation of the resulting conic is r
9 . 2 2 cos 56
x=9/2
O
10
5 1
5 -5
41. The ellipse is nearly circular when e is close to 0 and becomes more elongated as
4
e 1 . At e 1, the curve becomes a parabola.
e=0.4 e=0.6
2
2 -2
4 e=0.8
e=1
-4
d where d 12 , d 2, and 1 sin d 10. As d increases, the parabolas get flatter while the vertex moves further
42. (a) Shown are the graphs of the conics r
10 d=10
from the focus at the origin.
-10
10 d=2
-10
1
d=2
841
842
CHAPTER 12 Conic Sections
e where e 05, e 1, and 1 e sin e 10. As e increases, the conic changes from an ellipse to a parabola, and
(b) Shown are the graphs of the conics r
finally to a hyperbola. The vertex gets closer to the directrix (shown as a
2
e=10 -2
2
e=0.5 e=1 -2
dashed line in the figure).
ed we need to show that ed a 1 e2 . 1 e cos e2 d 2 From the proof of the Equivalent Description of Conics we have a 2 2 . Since the conic is an ellipse, e 1 1 e2 and so the quantities a, d, and 1 e2 are all positive. Thus we can take the square roots of both sides and maintain 2d 2 a 1 e2 ed e ed equality. Thus a 2 r . ed a 1 e2 . As a result, r 2 a 1 e cos 1 e cos 1 e2 1 e2
43. (a) Since the polar form of an ellipse with directrix x d is r
(b) Since 2a 299 108 we have a 1495 108 , so a polar equation for the earth’s orbit (using e 0017) is 1495 108 1 00172 149 108 . r 1 0017 cos 1 0017 cos
44. (a) Using the form of the equation from Exercise 27, the perihelion distance occurs when and the aphelion distance occurs when 0. Since the focus is at the origin, the perihelion distance is 2 a 1 e a 1 e2 a 1 e 1 e a 1 e 1 e a 1 e and the aphelion distance is a 1 e. 1e 1e 1e 1e (b) Given e 0017 and 2a 299 108 we have a 1495 108 . Thus the perihelion distance is
1495 108 [1 0017] 1468 108 km and the aphelion distance is 1495 108 1 0017 1520 108 km.
45. From Exercise 44, we know that at perihelion r 443 109 a 1 e and at aphelion r 737 109 a 1 e. 737 109 1e a 1 e 1664 1664 1 e 1 e 9 a 1 e 1e 443 10 0664 1664 1 e 1664 0664 2664e e 025. 2664 Dividing these equations gives
46. Since the focus is at the origin, the distance from the focus to any point on the conic is the absolute value of the r-coordinate of that point, which we can obtain from the polar equation. 47. The r-coordinate of the satellite will be its distance from the focus (the center of the earth). From the r-coordinate we can easily calculate the height of the satellite.
CHAPTER 12 REVIEW 1. (a) y 2 4x. This is a parabola with 4 p 4 p 1. The vertex is 0 0, the
(b)
y
focus is 1 0, and the directrix is x 1.
1 1
x
CHAPTER 12 1 y 2 y 2 12x. This is a parabola with 4 p 12 p 3. The vertex 2. (a) x 12 is 0 0, the focus is 3 0, and the directrix is x 3.
(b)
Review
843
y
2 x
1
3. (a) 18 x 2 y x 2 8y. This is a parabola with 4 p 8 p 2. The vertex is 0 0, the focus is 0 2, and the directrix is y 2.
y
(b)
1
4. (a) x 2 8y. This is a parabola with 4 p 8 p 2. The vertex is 0 0,
_1
7. (a) y 22 4 x 2. This is a parabola with 4 p 4 p 1. The vertex is
(b)
x
1
x
y
vertex is 0 0, the focus is 0 2, and the directrix is y 2.
(b)
1
1
(b)
6. (a) 2x y 2 0 y 2 2x. This is a parabola with 4 p 2 p 12 . The vertex is 0 0, the focus is 12 0 , and the directrix is x 12 .
x
y
(b)
the focus is 0 2, and the directrix is y 2.
5. (a) x 2 8y 0 x 2 8y. This is a parabola with 4 p 8 p 2. The
1
y
1 x
1
y
2 2, the focus is 1 2, and the directrix is x 3.
1 1
x
844
CHAPTER 12 Conic Sections
8. (a) x 32 20 y 2. This is a parabola with 4 p 20 p 5. The
(b)
9. (a) 12 y 32 x 0 y 32 2x. This is a parabola with 4 p 2 p 12 . The vertex is 0 3, the focus is 12 3 , and the directrix is x 12 .
(b)
y
1
vertex is 3 2, the focus is 3 7, and the directrix is y 3.
x
5
y
1 1
10. (a) 2 x 12 y x 12 12 y. This is a parabola with 4 p 12 p 18 . The vertex is 1 0, the focus is 1 18 , and the directrix is y 18 .
x
y
(b)
1 1 x
11. (a) 12 x 2 2x 2y 4 x 2 4x 4y 8 x 2 4x 4 4y 8
y
(b)
x 22 4 y 3. This is a parabola with 4 p 4 p 1. The vertex is 2 3, the focus is 2 3 1 2 2, and the directrix is y 3 1 4.
12. (a) x 2 3 x y x 2 3x 3y x 2 3x 94 3y 94 2 x 32 3 y 34 . This is a parabola with 4 p 3 p 34 . The vertex is 32 34 , the focus is 32 34 34 32 0 , and the directrix is
1 1
(b)
y 34 34 32 .
13. (a)
y2 x2 1. This is an ellipse with a 5, b 3, and c 25 9 4. The 9 25 center is 0 0, the vertices are 0 5, and the foci are 0 4.
(b) The length of the major axis is 2a 10 and the length of the minor axis is 2b 6.
y
1
(c)
x
1
x
y
1 1
x
CHAPTER 12
14. (a)
y2 x2 1. This is an ellipse with a 7, b 3, and c 49 9 2 10. 49 9 The center is 0 0, the vertices are 7 0, and the foci are 2 10 0 .
Review
845
y
(c)
1 x
1
(b) The length of the major axis is 2a 14 and the length of the minor axis is 2b 6.
15. (a)
y2 x2 1. This is an ellipse with a 7, b 2, and c 49 4 3 5. 49 4 The center is 0 0, the vertices are 7 0, and the foci are 3 5 0 .
y
(c)
1 x
1
(b) The length of the major axis is 2a 14 and the length of the minor axis is 2b 4.
16. (a)
y2 x2 1. This is an ellipse with a 6, b 2, and c 36 4 4 2. 4 36 The center is 0 0, the vertices are 0 6, and the foci are 0 4 2 .
y
(c)
1 x
1
(b) The length of the major axis is 2a 12 and the length of the minor axis is 2b 4.
y2 x2 1. This is an ellipse with a 4, b 2, and 17. (a) x 2 4y 2 16 16 4 c 16 4 2 3. The center is 0 0, the vertices are 4 0, and the foci are 2 3 0 .
y
(c)
1 x
1
(b) The length of the major axis is 2a 8 and the length of the minor axis is 2b 4.
y2 x2 1. This is an ellipse with a 12 , b 13 , and 19 14 c 14 19 16 5. The center is 0 0, the vertices are 0 12 , and the foci are 0 16 5 .
18. (a) 9x 2 4y 2 1
y 0.5
(c)
0.5 x
(b) The length of the major axis is 2a 1 and the length of the minor axis is 2b 23 .
19. (a)
y2 x 32 1This is an ellipse with a 4, b 3, and 9 16 c 16 9 7. The center is 3 0, the vertices are 3 4, and the foci are 3 7 .
(b) The length of the major axis is 2a 8 and the length of the minor axis is 2b 6.
(c)
y
1 1
x
846
CHAPTER 12 Conic Sections
20. (a)
y 32 x 22 1. This is an ellipse with a 5, b 4, and 25 16 c 25 16 3. The center is 2 3, the vertices are
(c)
y 1 x
1
2 5 3 3 3 and 7 3, and the foci are 2 3 3 1 3 and 5 3.
(b) The length of the major axis is 2a 10 and the length of the minor axis is 2b 8.
21. (a)
x 22 y 32 1. This is an ellipse with a 6, b 3, and 9 36 c 36 9 3 3. The center is 2 3, the vertices are 2 3 6 2 9 and 2 3, and the foci are 2 3 3 3 .
y
(c)
1 x
1
(b) The length of the major axis is 2a 12 and the length of the minor axis is 2b 6.
22. (a)
x2 y 52 1. This is an ellipse with a 5, b 3, and 3 25 c 25 3 22. The center is 0 5, the vertices are 0 5 5 0 10 and 0 0, and the foci are 0 5 22 .
y 1
(c)
1
x
(b) The length of the major axis is 2a 10 and the length of the minor axis is 2b 2 3. 23. (a) 4x 2 9y 2 36y 4x 2 9 y 2 4y 4 36 4x 2 9 y 22 36
y
(c)
x2 y 22 1. This is an ellipse with a 3, b 2, and 9 4 c 9 4 5. The center is 0 2, the vertices are 3 2, and the foci are 5 2 .
1 1
x
(b) The length of the major axis is 2a 6 and the length of the minor axis is 2b 4.
24. (a) 2x 2 y 2 2 4 x y 2x 2 4x y 2 4y 2 2 x 2 2x 1 y 2 4y 4 2 2 4 2 x 12 y 22 8 x 12 y 22 1. This is an ellipse with a 2 2, b 2, and 4 8 c 8 4 2. The center is 1 2, the vertices are 1 2 2 2 , and
the foci are 1 2 2 1 0 and 1 4. (b) The length of the major axis is 2a 4 2 and the length of the minor axis is 2b 4.
(c)
y
1 1
x
CHAPTER 12
y2 y2 x2 x2 1 0. This is a hyperbola with a 4, b 3, and 9 16 16 9 c 16 9 25 5. The center is 0 0, the vertices are 0 4, the foci
25. (a)
(b)
y2 x2 1. This is a hyperbola with a 7, b 4 2, and 49 32 c 49 32 9. The center is 0 0, the vertices are 7 0, the foci are
(b)
28. (a)
y2 x2 1. This is a hyperbola with a 2, b 7, and 4 49 c 4 49 53. The center is 0 0, the vertices are 2 0, the foci are 53 0 , and the asymptotes are y 72 x.
x2 y2 1. This is a hyperbola with a 5, b 2, and 25 4 c 25 4 29. The center is 0 0, the vertices are 0 5, the foci are 0 29 , and the asymptotes are y 52 x.
y2 x2 1. This is a hyperbola with a 4, b 2 2, 29. (a) x 2 2y 2 16 16 8 and c 16 8 24 2 6. The center is 0 0, the vertices are 4 0, the foci are 2 6 0 , and the asymptotes are y 2 4 2 x y 1 x.
y
2
y 12 x.
x
2
x
1
x
1
x
1
x
1
x
y
(b)
y
2
(b)
y
2
(b)
y
1
2
2
x y 1. This is a hyperbola 30. (a) x 2 4y 2 16 0 4y 2 x 2 16 4 16 with a 2, b 4, and c 4 16 2 5. The center is 0 0, the vertices are 0 2, the foci are 0 2 5 , and the asymptotes are y 24 x
1
2
9 0, and the asymptotes are y 4 7 2 x.
27. (a)
847
1
are 0 5, and the asymptotes are y 43 x.
26. (a)
Review
(b)
y
1
848
31. (a)
CHAPTER 12 Conic Sections
y2 x 42 1. This is a hyperbola with a 4, b 4 and 16 16 c 16 16 4 2. The center is 4 0, the vertices are 4 4 0 which are 8 0 and 0 0, the foci are 4 4 2 0 , and the asymptotes
y
(b)
1
1 x
are y x 4.
32. (a)
y 22 x 22 1. This is a hyperbola with a 2 2, b 2 2 and 8 8 c 8 8 4. The center is 2 2, the vertices are 2 2 2 2 , the
(b)
y 1
x
1
foci are 2 4 2 2 2 and 6 2, and the asymptotes are y 2 x 2 y x and y x 4.
33. (a)
x 12 y 32 1. This is a hyperbola with a 2, b 6, and 4 36 c 4 36 2 10. The center is 1 3, the vertices are 1 3 2 1 1 and 1 5, the foci are 1 3 2 10 , and the
(b)
1
1 8 asymptotes are y 3 13 x 1 y 13 x 10 3 and y 3 x 3 .
34. (a)
x2 y 32 1. This is a hyperbola with a 3, b 4, and 3 16 c 3 16 19. The center is 0 3, the vertices are 0 3 3 , the foci are 0 3 19 , and the asymptotes are y 3 43 x
(b)
2
x
2
x
y
1
y 43 x 3 and y 43 x 3.
35. (a) 9y 2 18y x 2 6x 18 9 y 2 2y 1 x 2 6x 9 9 9 18
y
y
(b)
x 32 y 12 9 y 12 x 32 18 1. This is a 2 18 hyperbola with a 2, b 3 2, and c 2 18 2 5.The center is 3 1, the vertices are 3 1 2 , the foci are 3 1 2 5 , and
the asymptotes are y 1 13 x 3 y 13 x and y 13 x 2. 36. (a) y 2 x 2 6y y 2 6y 9 x 2 9 y 32 x 2 9
x2 y 32 1. This is a hyperbola with a 3, b 3, and 9 9 c 9 9 3 2. The center is 0 3, the vertices are 0 3 3 0 6 and 0 0, the foci are 0 3 3 2 , and the asymptotes are y 3 x y x 3 and y x 3.
1 1
(b)
x
y
1 1
x
CHAPTER 12
Review
849
37. This is a parabola that opens to the right with its vertex at 0 0 and the focus at 2 0. So p 2, and the equation is y 2 4 2 x y 2 8x.
38. This is an ellipse with the center at 0 0, a 12, and b 5. The equation is then
x2 y2 y2 x2 1. 2 1 2 144 25 12 5
39. From the graph, the center is 0 0, and the vertices are 0 4 and 0 4. Since a is the distance from the center to a
vertex, we have a 4. Because one focus is 0 5, we have c 5, and since c2 a 2 b2 , we have 25 16 b2
b2 9. Thus an equation of the hyperbola is
x2 y2 1. 16 9
40. This is a parabola that opens to the left with its vertex at 4 4, so its equation is of the form y 42 4 p x 4 with
p 0. Since 0 0 is a point on this hyperbola, we must have 0 42 4 p 0 4 16 16 p p 1. Thus the
equation is y 42 4 x 4.
41. From the graph, the center of the ellipse is 4 2, and so a 4 and b 2. The equation is
y 22 x 42 1 42 22
x 42 y 22 1. 16 4 42. From the graph, the center is at 1 0, and the vertices are 0 0 and 2 0. Since a is the distance form the center to a b vertex, a 1. From the graph, the slope of one of the asymptotes is 1 b 1. Thus an equation of the hyperbola is a x 12 y 2 1. 43.
x2 x2 y 1 y 1 x 2 12 y 1. This is a parabola with 12 12 4 p 12 p 3. The vertex is 0 1, the focus is 0 1 3 0 2, and
y 1
the directrix is y 1 3 4.
y2 y x2 12x 2 y 2 12y 0 12x 2 y 2 12y 36 36 12 144 12 2 2 x y 6 1. This is an ellipse with a 6, b 3, and 3 36 c 36 3 33. The center is 0 6, the foci are 0 6 33 , and the
y
x2 y2 1. This is a hyperbola with a 12, b 12, 45. x 2 y 2 144 0 144 144 and c 144 144 12 2. The center is 0 0, the foci are 0 12 2 , the
y
44.
vertices are 0 6 6 0 0 and 0 12.
vertices are 0 12, and the asymptotes are y x.
x
1
1 1
x
4 4
x
850
CHAPTER 12 Conic Sections
x 32 y 2 1. This is a 46. x 2 6x 9y 2 x 2 6x 9 9y 2 9 9 hyperbola with a 3, b 1, and c 9 1 10. The center is 3 0, the foci are 3 10 0 , the vertices are 3 3 0 6 0 and 0 0, and the
y
1 1
x
asymptotes are y 13 x 3 y 13 x 1 and y 13 x 1.
47. 4x 2 y 2 8 x y 4 x 2 2x y 2 8y 0 4 x 2 2x 1 y 2 8y 16 4 16 4 x 12 y 42 20
y
x 12 y 42 1. This is an ellipse with a 2 5, b 5, and 5 20 c 20 5 15. The center is 1 4, the foci are 1 4 15 , and the vertices are 1 4 2 5 .
3 3
48. 3x 2 6 x y 10 3x 2 6x 6y 10 3 x 2 2x 1 6y 10 3 2 2 y 13 . 1 3 x 12 6y 13 3 x 12 6 y 13 x 6 6 1 13 This is a parabola with 4 p 2 p 2 . The vertex is 1 6 the focus is 1 1 10 1 5 , and the directrix is y 13 1 8 . 1 13 6 2 6 3 6 2 3
y
1 x
1
y
49. x y 2 16y x 64 y 2 16y 64 y 82 x 64. This is a parabola with 4 p 1 p 14 . The vertex is 64 8, the focus is 64 14 8 255 8 , and the directrix is x 64 14 257 4 4 .
2 10
50. 2x 2 4 4x y 2 y 2 2x 2 4x 4 y 2 2 x 2 2x 1 4 2 y2 x 12 1. This is a hyperbola with a 2, 2 b 1, and c 2 1 3. The center is 1 0, the foci are 1 3 , the vertices are 1 2 , and the asymptotes are y 2 x 1 y 2 x 2 and y 2 x 2. y 2 2 x 12 2
x
x
y
1 1
x
CHAPTER 12
51. 2x 2 12x y 2 6y 26 0 2 x 2 6x y 2 6y 26 2 x 2 6x 9 y 2 6y 9 26 18 9 2 x 32 y 32 1
x 32 1 2
c
Review
851
y x
1
_1
y 32 1. This is an ellipse with a 1, b 22 , and
1 12 22 . The center is 3 3, the foci are 3 3 22 , and the
vertices are 3 3 1 3 4 and 3 2.
52. 36x 2 4y 2 36x 8y 31 36 x 2 x 4 y 2 2y 31 36 x 2 x 14 4 y 2 2y 1 31 9 4
y
1
2 2 y 12 36 x 12 4 y 12 36 x 12 1. This is a 9
1
x
hyperbola with a 1, b 3, and c 1 9 10. The center is 12 1 , the foci are 12 10 1 , the vertices are 12 1 1 12 1 and 32 1 , and the asymptotes are y 1 3 x 12 y 3x 52 and y 3x 12 .
1 25 5 2 1 2 75 . 2 53. 9x 2 8y 2 15x8y27 0 9 x 2 53 x 25 36 8 y y 4 27 4 2 9 x 6 8 y 2 4 However, since the left-hand side of the equation is greater than or equal to 0, there is no point that satisfies this equation. The graph is empty. y 54. x 2 4y 2 4x 8 x 2 4x 4 4y 2 8 4 x 22 4y 2 12 y2 x 22 1. This is an ellipse with a 2 3, b 3, and 12 3 c 12 3 3. The center is 2 0, the foci are 2 3 0 1 0 and 5 0, and the vertices are 2 2 3 0 .
1 1
x
55. The parabola has focus 0 1 and directrix y 1. Therefore, p 1 and so 4 p 4. Since the focus is on the y-axis and the vertex is 0 0, an equation of the parabola is x 2 4y.
56. The parabola with vertex at the origin and focus F 5 0 has p 5, so 4 p 20. The focus is on the x-axis, so an equation is y 2 20x.
57. The ellipse with center at the origin and with x-intercepts 2 and y-intercepts 5 has a vertical major axis, a 5, and b 2, so an equation is
x2 y2 1. 4 25
58. The hyperbola has vertices 0 2 and asymptotes y 12 x. Therefore, a 2, and the foci are on the y-axis. Since the slopes of the asymptotes are 12
a y2 x2 b 2a 4, an equation of the hyperbola is 1. b 4 16
59. The ellipse has center C 0 4, foci F1 0 0 and F2 0 8, and major axis of length 10. Then 2c 8 0 c 4. Also,
since the length of the major axis is 10, 2a 10 a 5. Therefore, b2 a 2 c2 25 16 9. Since the foci are on
the y-axis, the vertices are on the y-axis, and an equation of the ellipse is
x2 y 42 1. 9 25
852
CHAPTER 12 Conic Sections
60. The hyperbola has center C 2 4, foci F1 2 7 and F2 2 1, and vertices V1 2 6 and V2 2 2. Thus, 2a 6 2 4 a 2. Also, 2c 7 1 6 c 3. So b2 9 4 5. Since the hyperbola has center C 2 4, its equation is y 42 x 22 1. 4 5
61. The ellipse has foci F1 1 1 and F2 1 3, and one vertex is on the x-axis. Thus, 2c 3 1 2 c 1, and so the
center of the ellipse is C 1 2. Also, since one vertex is on the x-axis, a 2 0 2, and thus b2 4 1 3. So an
equation of the ellipse is
x 12 y 22 1. 3 4
62. The parabola has vertex V 5 5 and directrix the y-axis. Therefore, p 0 5 p 5 4 p 20. Since the parabola opens to the right, its equation is y 52 20 x 5.
63. The ellipse has vertices V1 7 12 and V2 7 8 and passes through the point P 1 8. Thus, 2a 12 8 20 8 12 x 72 y 22 7 2. Thus an equation of the ellipse has the form 1. a 10, and the center is 7 2 2 100 b
1 72 8 22 1 3600 36b2 100b2 64b2 3600 b2 225 4 . 100 b2 4 x 72 y 22 y 22 x 72 1 1. Therefore, an equation of the ellipse is 2254 100 225 100 Since the point P 1 8 is on the ellipse,
64. The parabola has vertex V 1 0, horizontal axis of symmetry, and crosses the y-axis where y 2. Since the parabola has a horizontal axis of symmetry and V 1 0, its equation is of the form y 2 4 p x 1. Also, since the parabola crosses
the y-axis where y 2, it passes through the point 0 2. Substituting this point gives 22 4 p 0 1 4 p 4.
Therefore, an equation of the parabola is y 2 4 x 1.
65. The length of the major axis is 2a 186,000,000 a 93,000,000. The eccentricity is e ca 0017, and so c 0017 93,000,000 1,581,000. (a) The earth is closest to the sun when the distance is a c 93,000,000 1,581,000 91,419,000.
(b) The earth is furthest from the sun when the distance is a c 93,000,000 1,581,000 94,581,000. y
66. We sketch the LORAN station on the y-axis and place the x-axis halfway between them as
A
x2 y2 suggested in the exercise. This gives us the general form 2 2 1. Since the ship is 80 miles a b closer to A than to B we have 2a 80 a 40Since the foci are 0 150, we have c 150.
150
x
Thus b2 c2 a 2 1502 402 20900. So this places the ship on the hyperbola given by the
150
x2 y2 1600 y2 225 y2 1. When x 40, we get 1 1600 20,900 1600 20,900 1600 209 y 415. (Note that y 0, since A is on the positive y-axis.) Thus, the ship’s position is
equation
B
approximately 40 415.
67. (a) The graphs of
x2 y2 1 for k 1, 2, 4, and 8 are shown in 16 k 2 k2
the figure. (b) c2 16 k 2 k 2 16
c 4. Since the center is 0 0,
the foci of each of the ellipses are 4 0.
y k=8
4
40
2 1 1 1
x
CHAPTER 12
68. (a) The graphs of y kx 2 for k 12 , 1, 2, and 4 are shown in the figure. 1 1 1 y. Thus the foci are 0 . (b) y kx 2 x 2 y 4 k 4k 4k
10
k=4
Review
853
k=2
8
k=1
6 4
(c) As k increases, the focus gets closer to the vertex.
1
k=2
2 0
-2
2
69. (a) x 2 4x y y 2 1. Then A 1, B 4, and C 1, so the discriminant is 42 4 1 1 12. Since the discriminant is positive, the equation represents a hyperbola. 11 AC 0 2 90 45 . Therefore, x 22 X 22 Y and y 22 X 22 Y . (b) cot 2 B 4 Substituting into the original equation gives 2 2 2 X 2Y 2 X 2Y 2 X 2Y 2 X 2Y 4 1 2 2 2 2 2 2 2 2 y 1 X 2 2XY Y 2 2 X 2 XY XY Y 2 1 X 2 2XY Y 2 1 (c) 2 2 3X 2 Y 2 1 3X 2 Y 2 1. This is a hyperbola with a 1 , b 1, and 3 1 2 1 c 3 1 . Therefore, the hyperbola has vertices V 0 and foci 3 3 2 F 0 , in XY -coordinates.
1
x
1
3
70. (a) 5x 2 6x y 5y 2 8x 8y 8 0. Then A 5, B 6, and C 5, so the discriminant is 62 4 5 5 64. Since the discriminant is negative, the equation represents an ellipse. AC (b) cot 2 0 2 90 45 . Therefore, x 22 X 22 Y and y 22 X 22 Y . Substituting B into the original equation gives 2 2 X 2Y 5 22 X 22 Y 6 22 X 22 Y 2 2 2 5 22 X 22 Y 8 22 X 22 Y 8 22 X 22 Y 8 0 5 X 2 2XY Y 2 3 X 2 Y 2 5 X 2 2XY Y 2 4 2X 4 2Y 4 2X 4 2Y 8 0 2 2 5X 2 3X 2 3Y 2 5Y 2 8 2Y 8 0 X2 Y 12 1. This ellipse has 2X 2 8 Y 2 2Y 12 8 4 6 32 a 6, b 32, and c 6 32 3 2 2 . Therefore, the vertices are V 6 1 and the foci are F 3 2 2 1 .
(c)
y
1 1
71. (a) 7x 2 6 3x y 13y 2 4 3x 4y 0. Then A 7, B 6 3, and C 13, so the discriminant is 2 6 3 4 7 13 256. Since the discriminant is negative, the equation represents an ellipse.
x
854
CHAPTER 12 Conic Sections 7 13 AC 1 2 60 30 . Therefore, x 23 X 12 Y and y 12 X 23 Y . B 6 3 3 Substituting into the original equation gives 2 1 X 3Y 7 23 X 12 Y 6 3 23 X 12 Y 2 2 2 13 12 X 23 Y 4 3 23 X 12 Y 4 12 X 23 Y 0 7 3X 2 2 3XY Y 2 3 3 3X 2 3XY XY 3Y 2 4 2 2 2 6X 2 3Y 2X 2 3Y 0 13 4 X 2 3XY 3Y y 9 13 8X Y 2 7 9 39 0 4X 2 8X 16Y 2 0 (c) X 2 21 4 2 4 4 2 4 1 4 X 2 2X 1 16Y 2 4 X 12 4Y 2 1. This ellipse has a 1, x 1 b 12 , and c 1 14 12 3. Therefore, the vertices are V 1 1 0 V1 0 0 and V2 2 0 and the foci are F 1 12 3 0 .
(b) cot 2
72. (a) 9x 2 24x y 16y 2 25. Then A 9, B 24, and C 16, so the discriminant is 242 4 9 16 0. Since the discriminant is zero, the equation represents a parabola.
(b) cot 2
AC 9 16 7 7 cos 2 , so cos B 24 24 25
1725 35 , sin 2
53 , and thus x 35 X 45 Y and y 45 X 35 Y . Substituting,
1725 45 2 y
(c)
9x 2 24x y 16y 2 25 2 4 X 3 Y 16 4 X 3 Y 2 25 9 35 X 45 Y 24 35 X 45 Y 5 5 5 5
1 x
1
25X 2 25 X 2 1 X 1. Thus the graph is a degenerate conic that consists of two lines. Converting back to x y-coordinates, we see that X 35 x 45 y, so 35 x 45 y 1 3x 4y 5.
73. 5x 2 3y 2 60 3y 2 60 5x 2 y 2 20 53 x 2 . This conic is an ellipse.
5
-5
5 -5
74. 9x 2 12y 2 36 0 12y 2 9x 2 36 y 2 34 x 2 3 y 34 x 2 3. This conic is a hyperbola.
5
-5
5 -5
CHAPTER 12
75. 6x y 2 12y 30 y 2 12y 30 6x y 2 12y 36 66 6x y 62 66 6x y 6 66 6x y 6 66 6x. This conic is
Review
20
a parabola.
10
-10
10
76. 52x 2 72x y 73y 2 100 73y 2 72x y 52x 2 100 0. Using the quadratic formula,
1
72x 72x2 473 52x 2 100 y 146
72x
10,000x 2 29,200 146
10 2 36 73 x 73 73 25x
-2
2
-1
7325x 2
72x20146
This conic is an ellipse.
1 e 1. Therefore, this is a 1 cos parabola.
2 e 23 . Therefore, this is an 3 2 sin ellipse.
77. (a) r
78. (a) r
(b)
(b)
( 25 , ¹2) 1
( 21 , ¹)
_1
1
(2, 3¹ 2 ) 4 e 2. Therefore, this is a 1 2 sin hyperbola.
12 e 4. Therefore, this is a 1 4 cos hyperbola.
79. (a) r
80. (a) r
(b)
(b)
(_4, 3¹ 2) ( 43 , ¹2) 1
(_4, 0)
( 125 , ¹)
1
855
856
CHAPTER 12 Conic Sections
CHAPTER 12 TEST 1. x 2 12y. This is a parabola with 4 p 12 p 3. The focus is 0 3 and the directrix is y 3.
2.
y _1
y2 x2 1. This is an ellipse with a 4, b 2, and c 16 4 2 3. The 16 4 vertices are 4 0, the foci are 2 3 0 , the length of the major axis is
y
x2 y2 1. This is a hyperbola with a 3, b 4, and c 9 16 5. The 9 16
y
vertices are 0 3, the foci are 0 5, and the asymptotes are y 34 x.
x
1
x
1
x
1
2a 8, and the length of the minor axis is 2b 4.
3.
1
1
4. The parabola with vertex 0 0 and focus 4 0 has p 4, so 4 p 16. The focus lies to the right of the vertex, so an equation is y 2 16x.
5. The ellipse with foci 3 0 and vertices 4 0 has a 4 and c 3, so c2 a 2 b2 9 16 b2 b2 16 9 7. Thus, an equation is
x2 y2 1. 16 7
6. The hyperbola has foci 0 5 and asymptotes y 34 x. Since the foci are 0 5, c 5, the foci are on a 3 a a 34 b. Then the y-axis, and the center is 0 0. Also, since y 34 x x, it follows that b b 4 2 3 2 2 c2 52 25 a 2 b2 34 b b2 25 16 b b 16, and by substitution, a 4 4 3. Therefore, an equation of the hyperbola is
x2 y2 1. 9 16
7. This is a parabola that opens to the left with its vertex at 0 0. So its equation is of the form y 2 4 px with p 0. Substituting the point 4 2, we have 22 4 p 4 4 16 p p 14 . So an equation is y 2 4 14 x y 2 x.
8. This is an ellipse tangent to the x-axis at 0 0 and with one vertex at the point 4 3. The center is 0 3, and a 4 and b 3. Thus the equation is
x2 y 32 1. 16 9
CHAPTER 12
Test
857
9. This a hyperbola with a horizontal transverse axis, vertices at 1 0 and 3 0, and foci at 0 0 and 4 0. Thus the center x 22 y2 1 is 2 0, and a 3 2 1 and c 4 2 2. Thus b2 22 12 3. So an equation is 2 3 1 y2 1. x 22 3 y 10. 16x 2 36y 2 96x 36y 9 0 16 x 2 6x 36 y 2 y 9 16 x 2 6x 9 36 y 2 y 14 9 144 9 1 2 2 y 12 3 x 16 x 32 36 y 12 144 1. This is an 9 4 ellipse with a 3, b 2, and c 9 4 5. The center is 3 12 , the vertices are 3 3 12 0 12 and 6 12 , and the foci are h c k 3 5 12 3 5 12 and 3 5 12 .
2
x
1
11. 9x 2 8y 2 36x 64y 164 9 x 2 4x 8 y 2 8y 164 9 x 2 4x 4 8 y 2 8y 16 164 36 128
y
y 42 x 22 1. This is a hyperbola 9 x 22 8 y 42 72 8 9 with a 2 2, b 3, and c2 a 2 b2 17. The center is 2 4, the foci are 2 17 4 , the vertices are 2 2 2 4 , and the asymptotes are
1 1
x
y 4 3 4 2 x 2 y 3 4 2 x 4 3 2 2 and y 3 4 2 x 4 3 2 2 .
12. 2x y 2 8y 8 0 y 2 8y 16 2x 8 16 y 42 2 x 4. This is a parabola with 4 p 2 p 12 . The vertex is 4 4, the focus is 4 12 4 72 4 , and the directrix is x 4 12 92 .
y 1 1
x
13. The ellipse with center 2 0, foci 2 3 and major axis of length 8 has a horizontal major axis with 2a 8 a 4.
y2 x 22 1. 7 16 14. The parabola has focus 2 4 and directrix the x-axis (y 0). Therefore, 2 p 4 0 4 p 2 4 p 8, and the Also, c 3, so b2 a 2 c2 16 9 7. Thus, an equation is
vertex is 2 4 p 2 2. Hence, an equation of the parabola is x 22 8 y 2 x 2 4x 4 8y 16 x 2 4x 8y 20 0.
15. We place the vertex of the parabola at the origin, so the parabola contains the points 3 3, and the equation is of the form y 2 4 px. Substituting the point 3 3, we get 32 4 p 3 9 12 p p 34 . So the focus is 34 0 , and we should place the light bulb 34 inch from the vertex.
16. (a) 5x 2 4x y 2y 2 18. Then A 5, B 4, and C 2, so the discriminant is 42 4 5 2 24. Since the discriminant is negative, the equation represents an ellipse.
858
FOCUS ON MODELING
52 3 AC . Thus, cos 2 35 and so cos 135 (b) cot 2 255, 2 B 4 4 135 5 sin 5 . It follows that x 2 5 5 X 55 Y and y 55 X 2 5 5 Y . By 2 2 2 5 X 2 5Y 2 5 X 2 5Y 18 substitution, 5 2 5 5 X 55 Y 4 2 5 5 X 55 Y 5 5 5 5 4X 2 4XY Y 2 45 2X 2 4XY XY 2Y 2 25 X 2 4XY 4Y 2 18
2 2 8 2 1 8 4 18 6X 2 Y 2 18 X Y 1. This is an X 2 4 85 25 XY 4 12 5 5 Y 5 5 3 18 ellipse with a 3 2 and b 3. y (c) (d) In XY -coordinates, the vertices are V 0 3 2 . Therefore, in
1 x
1
x y-coordinates, the vertices are x 3 2 and y 6 2 5 5 3 2 6 2 3 2 6 2 V1 , and x and y 5 5 5 5 2 . V2 3 2 6 5
Since cos 2 35 we have
5
2 cos1 35 5313 , so 27 . 17. (a) Since the focus of this conic is the origin and the directrix is x 2, the equation has the form
ed . Subsituting e 12 and d 2 we 1 e cos 1 2 get r r . 1 2 cos 1 2 cos
(b) r
3 r 2 sin 1
3 2 . So e 12 and the 1 sin 2
conic is an ellipse.
r
O
O
1
1
FOCUS ON MODELING Conics in Architecture 1. Answers will vary. 2. (a) The difference P F2 P A c is a constant (the length of the string). Also, P F1 P A d is constant. Subtracting these two equations, we find that P F2 P F1 c d is a constant. This is the definition of a hyperbola. (b) Reflect the entire apparatus through a line perpendicular to F1 F2 .
4. As the lids are twisted more, the vertices of the hyperbolic cross sections get closer together. 5. (a) The tangent line passes though the point a a 2 , so an equation is y a 2 m x a.
Conics in Architecture
(b) Because the tangent line intersects the parabola at only the one point a a 2 , the system
y a 2 m x a y x2
859
has
only one solution, namely x a, y a 2 . y a 2 m x a y a 2 m x a a 2 m x a x 2 x 2 mx am a 2 0. This quadratic (c) 2 yx y x2 has discriminant m2 4 1 am a 2 m 2 4am 4a 2 m 2a2 . Setting this equal to 0, we find m 2a. (d) An equation of the tangent line is y a 2 2a x a y a 2 2ax 2a 2 y 2ax a 2 .
6. (a) F1 and F2 are points of tangency of the spheres to the plane, and the vertical line through Q 1 and Q 2 is also tangent to each sphere. Since P lies outside of both spheres, we see that P F1 P Q 1 and P F2 P Q 2 . (b) Each sphere is tangent to the enclosing cylinder along a circle in a vertical plane. The distance between these two planes is constant, so P Q 1 P Q 2 Q 1 Q 2 is constant for any choice of P.
(c) Combining the results of parts (a) and (b), we see that P F1 P F2 P Q 1 P Q 2 is constant.
(d) By definition, an ellipse is a curve consisting of all points whose distances to two fixed points F1 and F2 add up to a constant. That is the case here.
13
SEQUENCES AND SERIES
13.1 SEQUENCES AND SUMMATION NOTATION 1. A sequence is a function whose domain is the natural numbers. 2. The nth partial sum of a sequence is the sum of the first n terms of the sequence. So for the sequence an n 2 the fourth partial sum is S4 12 22 32 42 30.
3. an n 3. Then a1 1 3 2, a2 2 3 1, a3 3 3 0, a4 4 3 1, and a100 100 3 97.
4. an 2n 1. Then a1 2 1 1 1, a2 2 2 1 3, a3 2 3 1 5, a4 2 4 1 7, and a100 2 100 1 199. 1 1 1 1 1 1 1 1 1 . Then a1 , a2 , a3 , a4 , and 5. an 2n 1 2 1 1 3 2 2 1 5 2 3 1 7 2 4 1 9 1 1 a100 . 2 100 1 201
6. an n 2 1. Then a1 12 1 0, a2 22 1 3, a3 32 1 8, a4 42 1 15, and a100 1002 1 9999.
7. an 5n . Then a1 51 5, a2 52 25, a3 53 125, a4 54 625, and a100 5100 79 1069 . n 1 2 3 1 , a 1 4 1 , and . Then a1 1 13 , a2 1 19 , a3 1 27 8. an 1 4 3 3 3 3 3 81 100 3100 19 1048 . a100 1 3
1 1 1 1n 11 12 13 14 . Then a1 1, a2 , a3 , a4 , and 2 2 2 2 4 9 16 n 1 2 3 42 1 1100 . a100 2 10,000 100 1 1 1 1 1 1 1 1 1 1 , and a100 . 1, a2 , a3 , a4 10. an 2 . Then a1 2 2 2 2 2 4 9 16 10,000 n 1 2 3 4 100 9. an
11. an 1 1n . Then a1 1 11 0, a2 1 12 2, a3 1 13 0, a4 1 14 2, and a100 1 1100 2.
1 2 3 4 1n1 n 12 1 13 2 14 3 15 4 . Then a1 , a2 , a3 , a4 , and n1 11 2 21 3 31 4 41 5 100 1101 100 . a100 101 101 13. an n n . Then a1 11 1, a2 22 4, a3 33 27, a4 44 256, and a100 100100 10200 . 12. an
14. an 3. Then a1 3, a2 3, a3 3, a4 3, and a100 3. 15. an 2 an1 3 and a1 4. Then a2 2 4 3 14, a3 2 14 3 34, a4 2 34 3 74, and a5 2 74 3 154.
1 2 2 1 1 a 24 4 4, a3 , a4 3 , and a5 9 . 16. an n1 and a1 24. Then a2 6 6 6 3 6 9 6 54 17. an 2an1 1 and a1 1. Then a2 2 1 1 3, a3 2 3 1 7, a4 2 7 1 15, and a5 2 15 1 31. 1 1 1 1 2 1 3 1 5 , a3 and a1 1. Then a2 , a4 , and a5 . 18. an 1 an1 11 2 3 5 8 1 1 1 2 1 3 2
3
5
861
862
CHAPTER 13 Sequences and Series
19. an an1 an2 , a1 1, and a2 2. Then a3 2 1 3, a4 3 2 5, and a5 5 3 8.
20. an an1 an2 an3 and a1 1, a2 1, and a3 1. Then a4 1 1 1 3 and a5 3 1 1 5. 21. (a) a1 7, a2 11, a3 15, a4 19, a5 23, a6 27, a7 31, a8 35, a9 39, a10 43 (b)
22. (a) a1 2, a2 6, a3 12, a4 20, a5 30,
a6 42, a7 56, a8 72, a9 90, a10 110
(b) 40
100
20
50
0
0 0
5
10
12 12 23. (a) a1 12 1 12, a2 2 6, a3 3 4, 12 12 12 a4 12 4 3, a5 5 , a6 6 2, a7 7 , 3 12 4 12 6 a8 12 8 2 , a9 9 3 , a10 10 5
0
5
10
24. (a) a1 6, a2 2, a3 6, a4 2, a5 6, a6 2, a7 6, a8 2, a9 6, a10 2
(b)
(b)
5 10 0
5
0
5
10
0 0
5
10
25. (a) a1 2, a2 05, a3 2, a4 05, a5 2,
a6 05, a7 2, a8 05, a9 2, a10 05
(b)
26. (a) a1 1, a2 3, a3 2, a4 1, a5 3,
a6 2, a7 1, a8 3, a9 2, a10 1
(b) 2 1 0 0
5
10
4 2 0 -2 -4
5
10
27. 2, 4, 6, 8, . All are multiples of 2, so a1 2, a2 2 2, a3 3 2, a4 4 2, . Thus an 2n.
28. 1, 3, 5, 7, . All are odd numbers, so a1 2 1, a2 2 2 1, a3 3 2 1, a4 4 2 1, . Thus an 2n 1. 29. 2, 4, 8, 16, . All are powers of 2, so a1 2, a2 22 , a3 23 , a4 24 , . Thus an 2n .
1 2 1 , 1 , . The denominators are all powers of 3, and the terms alternate in sign. Thus a 1 , a 1 , 30. 13 , 19 , 27 1 2 81 31 32 3 4 n 1 1 1 a3 , a4 , . So an . 3n 33 34 31. 2, 3, 8, 13, . The difference between any two consecutive terms is 5, so a1 5 1 7, a2 5 2 7, a3 5 3 7, a4 5 4 7, . Thus, an 5n 7.
32. 7, 4, 1, 2, . The difference between any two consecutive terms is 3, so a1 3 1 10, a2 3 2 10, a3 3 3 10, a4 3 4 10, . Thus, an 3n 10.
33. 5, 25, 125, 625, . These terms are powers of 5, and the terms alternate in sign. So a1 12 51 , a2 13 52 , a3 14 53 , a4 15 54 , . Thus an 1n1 5n .
34. 3, 03, 003, 0003, . The ratio of any two consecutive terms is 10, so a1 3 100 , a2 3 101 , a3 3 102 , n1 1 a4 3 103 , . Thus an 3 10 .
SECTION 13.1 Sequences and Summation Notation
863
7 , 9 , . We consider the numerator separately from the denominator. The numerators of the terms differ 35. 1, 34 , 59 , 16 25 2 1 1 2 2 1 2 3 1 2 4 1 by 2, and the denominators are perfect squares. So a1 , a2 , a3 , a4 , 2 2 2 1 2 3 42 2 5 1 2n 1 , . Thus an . a5 2 5 n2
12 22 32 36. 34 , 45 , 56 , 67 , . Both the numerator and the denominator increase by 1, so a1 , a2 , a3 , 13 23 33 42 n2 a4 , . Thus an . 43 n3 37. 0, 2, 0, 2, 0, 2, . These terms alternate between 0 and 2. So a1 1 1, a2 1 1, a3 1 1, a4 1 1, a5 1 1, a6 1 1, Thus an 1 1n . n1 38. 1, 12 , 3, 14 , 5, 16 , . So a1 1, a2 21 , a3 31 , a4 41 , . Thus an n 1 .
39. a1 1, a2 3, a3 5, a4 7, . Therefore, an 2n 1. So S1 1, S2 1 3 4, S3 1 3 5 9, S4 1 3 5 7 16, S5 1 3 5 7 9 25, and S6 1 3 5 7 9 11 36. 40. a1 12 , a2 22 , a3 32 , a4 42 , . Therefore, an n 2 . So S1 12 1, S2 122 5, S3 532 59 14, S4 14 42 14 16 30, S5 30 52 30 25 55, and S6 55 62 55 36 91.
1 1 1 1 1 4 1 1 41. a1 13 , a2 2 , a3 3 , a4 4 , . Therefore, an n . So S1 13 , S2 13 2 , S3 13 2 3 13 27 , 3 9 3 3 3 3 3 3 1 1 1 1 1 1 1 1 1 1 1 1 , and S5 13 2 3 4 5 121 , S6 13 2 3 4 5 6 364 S4 13 2 3 4 40 81 243 729 . 3 3 3 3 3 3 3 3 3 3 3 3 42. a1 1, a2 1, a3 1, a4 1, . Therefore, an 1n . So S1 1, S2 1 1 0, S3 0 1 1, S4 1 1 0, S5 0 1 1, and S6 1 1 0. 2 2 2 2 2 2 2 2 43. an n . So S1 23 , S2 2 89 , S3 23 2 3 26 , and S4 23 2 3 4 80 27 81 . Therefore, 3 3 3 3 3 3 3 3 n 3 1 Sn . 3n 1 1 . So S1 12 13 , S2 12 13 13 14 12 13 13 14 12 14 , n1 n2 1 1 S3 2 3 13 14 14 15 12 13 13 14 14 15 12 15 , and S4 12 13 13 14 14 15 15 16 12 13 13 14 14 15 15 16 12 16 . Therefore, 1 1 1 1 1 1 1 1 1 1 Sn 12 13 13 14 n1 n2 2 3 3 n1 n1 n2 2 n2 .
44. an
n n 1. So S1 1 2 1 2, S2 1 2 2 3 1 2 2 3 1 3, 1 2 2 3 3 4 1 2 2 3 3 4 1 4, S3 S4 1 2 2 3 3 4 4 5 1 2 2 3 3 4 4 51 5
45. an
Therefore, 1 2 2 3 n n 1 Sn 1 2 2 3 3 n n n 1 1 n 1
864
CHAPTER 13 Sequences and Series
k log k log k 1. So S1 log 1 log 2 log 2, S2 log 2 log 2 log 3 log 3, k 1 S3 log 2 log 2 log 3 log 3 log 4 log 2 log 2 log 3 log 3 log 4 log 4, and
46. an log
S4 log 2 log 2 log 3 log 3 log 4 log 4 log 5 log 2 log 2 log 3 log 3 log 4 log 4 log 5 log 5
Therefore, Sn log n 1. 4 47. k1 k 1 2 3 4 10 4 48. k1 k 2 1 22 32 42 1 4 9 16 30 3
1 1 1 6 3 2 11 k1 k 1 2 3 6 6 6 6 j 1 2 3 4 99 100 1 1 1 1 1 1 1 0 50. 100 j1 1 1 1 1 1 1 1 49.
51.
52.
8
i i 1 1 1
12
02020202 8
i 4 10 10 10 10 10 10 10 10 10 10 90 53. k1 2k1 20 21 22 23 24 1 2 4 8 16 31 54. i31 i2i 1 21 2 22 3 23 2 8 24 34
5
55. 385 56. 15,550 57. 46,438 58. 0153146 59. 22 60. 0688172 4 3 3 3 3 3 61. k1 k 1 2 3 4 1 8 27 64 4 j 1 0 1 2 3 3 2 15 62. 0 j1 j 1 2 3 4 5 3 2 5 63. 6k0 k 4 4 5 6 7 8 9 10 64. 9k6 k k 3 6 9 7 10 8 11 9 12 54 70 88 108 k 3 4 5 100 65. 100 k3 x x x x x n j1 x j 12 x 13 x 2 14 x 3 1n1 x n x x 2 x 3 1n1 x n 66. j1 1 67. 2 4 6 50 25 k1 2k 10 68. 2 5 8 29 k1 3k 1 2 69. 12 22 32 102 10 k1 k 70.
71. 72. 73. 74. 75.
1 1 1 1 1 1k 100 k2 k ln k 2 ln 2 3 ln 3 4 ln 4 5 ln 5 100 ln 100 999 1 1 1 1 1 k1 12 23 34 999 1000 k k 1 1 2 3 n k 2 2 2 nk1 2 12 2 3 n k k 1 x x 2 x 3 x 100 100 k0 x k1 k x k1 1 2x 3x 2 4x 3 5x 4 100x 99 100 k1 1 2, 2 2, 2 2 2, 2 2 2 2, . We simplify each term in an attempt to determine a formula for an . So a1 212 , a2 2 212 232 234 , a3 2 234 274 278 , a4 2 278 2158 21516 , . Thus
n n an 22 12 .
76. G 1 1, G 2 1, G 3 2, G 4 3, G 5 5, G 6 8, G 7 13, G 8 21, G 9 34, G 10 55 77. (a) A1 $2004, A2 $200801, A3 $201202, A4 $201605, A5 $202008, A6 $202412
SECTION 13.1 Sequences and Summation Notation
865
(b) Since 3 years is 36 months, we get A36 $214916. 78. (a) I1 0, I2 $050, I3 $150, I4 $301, I5 $503, I6 $755 (b) Since 5 years is 60 months, we have I60 $97700.
79. (a) P1 35,700, P2 36,414, P3 37,142, P4 37,885, P5 38,643 (b) Since 2014 is 10 years after 2004, P10 42,665.
80. (a) The amount she owes at the end of the month, An , is the amount she owes at the beginning of the month, An1 , plus the interest, 0005An1 , minus the $200 she repay her uncle. Thus An An1 0005An1 200 An 1005An1 200.
(b) A1 9850, A2 969925, A3 954774, A4 939548, A5 924246, A6 908867. Thus she owes her uncle $908867 after six months.
81. (a) The number of catfish at the end of the month, Pn , is the population at the start of the month, Pn1 , plus the increase in population, 008Pn1 , minus the 300 catfish harvested. Thus Pn Pn1 008Pn1 300 Pn 108Pn1 300. (b) P1 5100, P2 5208, P3 5325, P4 5451, P5 5587, P6 5734, P7 5892, P8 6064, P9 6249, P10 6449, P11 6665, P12 6898. Thus there should be 6898 catfish in the pond at the end of 12 months.
82. (a) Let n be the years since 2002. So P0 $240,000. Each month the median price of a house in Orange County increases to 106 the previous months. Thus Pn 106n P0 . (b) Since 2010 2002 8. Thus P8 382,52353. Thus in 2010, the median price of a house in Orange County should be $382,524.
83. (a) Let Sn be his salary in the nth year. Then S1 $30,000. Since his salary increase by 2000 each year, Sn Sn1 2000. Thus S1 $30,000 and Sn Sn1 2000.
(b) S5 S4 2000 S3 2000 2000 S2 2000 4000 S1 2000 6000 $38,000. 84. (a) Let n be the days after she starts, so C0 4. And each day the concentration increases by 10%. Thus after n days the brine solution is Cn 110Cn1 with C0 4. (b) C8 110C7 110 110C6 1108 C0 1108 4 86. Thus the brine solution is 86 g/L of salt.
85. Let Fn be the number of pairs of rabbits in the nth month. Clearly F1 F2 1. In the nth month each pair that is two or more months old (that is, Fn2 pairs) will add a pair of offspring to the Fn1 pairs already present. Thus Fn Fn1 Fn2 . So Fn is the Fibonacci sequence. 86. (a) an n 2 . Then a1 12 1, a2 22 4, a3 32 9, a4 42 16. (b) an n 2 n 1 n 2 n 3 n 4, a1 12 1 1 1 2 1 3 1 4 1 0 1 2 3 1, a2 22 2 1 2 2 2 3 2 4 4 1 0 1 2 4, a3 32 3 1 3 2 3 3 3 4 9 2 1 0 1 9,
a4 42 4 1 4 2 4 3 4 4 16 3 2 1 0 16.
Hence, the sequences agree in the first four terms. However, for the second sequence, a5 52 5 1 5 2 5 3 5 4 25 4 3 2 1 49, and for the first sequence, a5 52 25, and thus the sequences disagree from the fifth term on. (c) an n 2 n 1 n 2 n 3 n 4 n 5 n 6 agrees with an n 2 in the first six terms only. (d) an 2n and bn 2n n 1 n 2 n 3 n 4.
866
CHAPTER 13 Sequences and Series
a n if an is even 2 87. an1 3an 1if an is odd
With a1 11, we have a2 34, a3 17, a4 52, a5 26, a6 13, a7 40,
a8 20, a9 10, a10 5, a11 16, a12 8, a13 4, a14 2, a15 1, a16 4, a17 2, a18 1, (with 4, 2, 1 repeating). So a3n1 4, a3n2 2, and a3n 1, for n 5. With a1 25, we have a2 76, a3 38, a4 19, a5 58, a6 29, a7 88, a8 44, a9 22, a10 11, a11 34, a12 17, a13 52, a14 26, a15 13, a16 40, a17 20, a18 10, a19 5, a20 16, a21 8, a22 4, a23 2, a24 1, a25 4, a26 2, a27 1, (with 4, 2, 1 repeating). So a3n1 4, a3n2 2, and a3n3 1 for n 7. We conjecture that the sequence will always return to the numbers 4, 2, 1 repeating.
88. an anan1 anan2 , a1 1, and a2 1. So a3 a31 a31 a2 a2 1 1 2, a4 a42 a41 a2 a3 1 2 3, a5 a53 a52 a2 a3 1 2 3, a6 a63 a63 a3 a3 2 2 4, a7 a74 a73 a3 a4 2 3 5, a8 a85 a84 a3 a4 2 3 5, a9 a95 a95 a4 a4 3 3 6, and a10 a106 a105 a4 a5 3 3 6. The definition of an depends on the values of certain preceding terms. So an is the sum of two preceding terms whose choice depends on the values of an1 and an2 (not on n 1 and n 2).
13.2 ARITHMETIC SEQUENCES 1. An arithmetic sequence is sequence where the difference between successive terms is constant. 2. The sequence an a n 1 d is an arithmetic sequence where a is the first term and d is the common difference. So, for the arithmetic sequence an 2 5 n 1 the first term is 2 and the common difference is 5. 3. True. The nth partial sum of an arithmetic sequence is the average of the first and last terms times n.
4. True. If we know the first and second terms of an arithmetic sequence then we can find any other term.
5. (a) a1 7 3 1 1 7, a2 7 3 2 1 10, a3 7 3 3 1 13, a4 7 3 4 1 16,
6. (a) a1 10 20 1 1 10, a2 10 20 2 1 10, a3 10 20 3 1 30,
a5 7 3 5 1 19
a4 10 20 4 1 50,
(b) The common difference is 3. (c)
a5 10 20 5 1 70
an 20
(b) The common difference is 20.
15
(c)
10
60
5 0
an
40 1
2
3
4
5 n
20 0
1
2
3
4
5 n
SECTION 13.2 Arithmetic Sequences
7. (a) a1 6 4 1 1 6,
8. (a) a1 10 4 1 1 10,
a2 6 4 2 1 10,
a2 10 4 2 1 6,
a3 6 4 3 1 14,
a3 10 4 3 1 2,
a4 6 4 4 1 18, a5 6 4 5 1 22 (b) The common difference is 4. (c)
a4 10 4 4 1 2, a5 10 4 5 1 6
(b) The common difference is 4.
an 0
(c) 1
2
867
3
4
5 n
an 10
_10
5
_20
0 _5
3 1
2
4
5 n
_10
9. (a) a1 52 1 1 52 , a2 52 2 1 32 , a3 52 3 1 12 , a4 52 4 1 12 ,
a3 12 3 1 1, a4 12 4 1 32 ,
a5 52 5 1 32
a5 12 5 1 2 (b) The common difference is 12 .
(b) The common difference is 1. (c)
10. (a) a1 12 1 1 0, a2 12 2 1 12 ,
an
(c)
2
2 4 0 _2
an
1
2
3
1 5 n 0
1
2
3
4
5 n
11. a 9, d 4, an a d n 1 9 4 n 1. So a10 9 4 10 1 45. 12. a 5, d 4, an a d n 1 5 4 n 1. So a10 5 4 10 1 31.
13. a 07, d 02, an a d n 1 07 02 n 1. So a10 07 02 10 1 25. 14. a 14, d 32 , an a d n 1 14 32 n 1. So a10 14 32 10 1 12 .
15. a 52 , d 12 , an a d n 1 52 12 n 1. So a10 52 12 10 1 2. 16. a 3, d 3, an a d n 1 3 3 n 1. So a10 3 3 10 1 10 3.
17. a4 a3 a3 a2 a2 a1 6. The sequence is arithmetic with common difference 6.
18. a4 a3 a3 a2 a2 a1 12. The sequence is arithmetic with common difference 12. 19. Since a3 a2 7 and a4 a3 6, the terms of the sequence do not have a common difference. This sequence is not arithmetic. 20. a4 a3 a3 a2 a2 a1 32. The sequence is arithmetic with common difference 32.
21. Since a2 a1 4 2 2 and a4 a3 16 8 8, the terms of the sequence do not have a common difference. This sequence is not arithmetic. 22. a4 a3 a3 a2 a2 a1 2. This sequence is arithmetic with common difference 2. 3 23. a4 a3 32 0 32 , a3 a2 0 32 , a2 a1 32 3 32 . This sequence is arithmetic with common 2 difference 32 . 8 4 24. a4 a3 ln 16 ln 8 ln 16 8 ln 2, a3 a2 ln 8 ln 4 ln 4 ln 2, a2 a1 ln 4 ln 2 ln 2 ln 2. This sequence is arithmetic with common difference ln 2.
868
CHAPTER 13 Sequences and Series
25. a4 a3 77 60 17, a3 a2 60 43 17, 4 a1 43 26 17. This sequence is arithmetic with common difference 17. 1 and a a 1 1 1 , the terms of the sequence do not have a common difference. 26. Since a4 a3 15 14 20 3 2 4 3 12 This sequence is not arithmetic.
27. a1 4 7 1 11, a2 4 7 2 18, a3 4 7 3 25, a4 4 7 4 32, a5 4 7 5 39. This sequence is arithmetic, the common difference is d 7 and an 4 7n 4 7n 7 7 11 7 n 1. 28. a1 4 21 6, a2 4 22 8, a3 4 23 12, a4 4 24 20, a5 4 25 36. Since a4 a3 8 and a3 a2 4, the terms of the sequence do not have a common difference. This sequence is not arithmetic. 1 1 1 1 1 1 1 1 1 1 , a2 , a3 , a4 , a5 . Since 29. a1 1 2 1 3 1 2 2 5 1 2 3 7 1 2 4 9 1 2 5 11
2 and a a 1 1 2 , the terms of the sequence do not have a common difference. This a4 a3 19 17 63 3 2 7 3 21 sequence is not arithmetic.
30. a1 1 12 32 , a2 1 22 2, a3 1 32 52 , a4 1 42 3, a5 1 52 72 . This sequence is arithmetic, the common difference is d 12 and an 1 n2 1 12 n 12 12 32 12 n 1. 31. a1 6 1 10 4, a2 6 2 10 2, a3 6 3 10 8, a4 6 4 10 14, a5 6 5 10 20. This sequence is arithmetic, the common difference is d 6 and an 6n 10 6n 6 6 10 4 6 n 1. 32. a1 3 11 1 2, a2 3 12 2 5, a3 3 13 3 0, a4 3 14 4 7,
a5 3 15 5 2. Since a4 a3 7 and a3 a2 5, the terms of the sequence do not have a common difference. This sequence is not arithmetic.
33. 4, 10, 16, 22, . Then d a2 a1 10 4 6, a5 a4 6 22 6 28, an 4 6 n 1, and a100 4 6 99 598.
34. 1, 11, 23, 35, . Then d a2 a1 11 1 12, a5 a4 12 35 12 47, an 1 12 n 1, and a100 1 12 99 1187.
35. 29, 11, 7, 25, . Then d a2 a1 11 29 18, a5 a4 18 25 18 43, an 29 18 n 1, and a100 29 18 99 1753. 36. 64, 49, 34, 19, . Then d a2 a1 49 64 15, a5 a4 15 19 15 4, an 64 15 n 1, and a100 64 15 99 1421. 37. 4, 9, 14, 19, . Then d a2 a1 9 4 5, a5 a4 5 19 5 24, an 4 5 n 1, and a100 4 5 99 499.
38. 11, 8, 5, 2, . Then d a2 a1 8 11 3, a5 a4 3 2 3 1, an 11 3 n 1, and a100 11 3 99 286.
39. 12, 8, 4, 0, . Then d a2 a1 8 12 4, a5 a4 4 0 4 4, an 12 4 n 1, and a100 12 4 99 384.
8 10 7 1 1 8 1 19 7 1 40. 76 , 53 , 13 6 , 3 , . Then d a2 a1 6 6 2 , a5 a4 2 3 2 6 , an 6 2 n 1, and 152 a100 76 12 99 7297 6 3 .
41. 25, 265, 28, 295, . Then d a2 a1 265 25 15, a5 a4 15 295 15 31, an 25 15 n 1, and a100 25 15 99 1735.
42. 15, 123, 96, 69, . Then d a2 a1 123 15 27, a5 a4 27 69 27 42, an 15 27 n 1, and a100 15 27 99 2523.
43. 2, 2 s, 2 2s, 2 3s, . Then d a2 a1 2 s 2 s, a5 a4 s 2 3s s 2 4s, an 2 n 1 s, and a100 2 99s.
44. t, t 3, t 6, t 9, . Then d a2 a1 t 3 t 3, a5 a4 3 t 9 3 t 12, an t 3 n 1, and a100 t 3 99 t 297.
SECTION 13.2 Arithmetic Sequences
869
45. a50 1000 and d 6. Thus, a50 a1 d 50 1 1000 a1 6 50 1 a1 1000 294 706 and a2 706 6 712. 46. a100 750 and d 20. Thus, 750 a1 20 99 a1 1230 and a5 1230 20 4 1150.
1 . Thus, a 1 8 1 5 and 47. a14 23 and a9 14 , so 23 a1 13d and 14 a1 8d 23 14 5d d 12 1 4 12 12
5 1 n 1. an 12 12 48. a12 118 and a8 146, so 118 a1 11d and 146 a1 7d 118 146 4d d 7. Thus, a1 146 7 7 195 and an 195 7 n 1.
49. a1 25 and d 18, so an 601 25 18 n 1 601 33. Thus, 601 is the 33rd term of the sequence.
50. a1 3500 and d 15, so an 2795 2795 3500 15 n 1 n 48. Thus, 2795 is the 48th term of the sequence.
51. a 3, d 5, n 20. Then S20 20 2 2 3 19 5 1010. 52. a 10, d 8, n 30. Then S30 30 2 [2 10 29 8] 3180.
53. a 40, d 14, n 15. Then S15 15 2 [2 40 14 14] 870. 25 54. a 2, d 23, n 25. Then S25 2 2 2 24 23 6850.
55. a1 55, d 12, n 10. Then S10 10 2 2 55 9 12 1090. 56. a2 8, a5 95, n 15. Thus a2 a d 8 and a5 a 4d 95. Subtracting the first equation from the second gives 3d 15 d 05. Substituting for d in the first equation gives a 05 8 a 75. Thus S15 15 2 2 75 14 05 165.
57. 1 5 9 401 is a partial sum of an arithmetic series with a 1 and d 5 1 4. The last term is
401 an 1 4 n 1, so n 1 100 n 101. So the partial sum is S101 101 2 1 401 101 201 20,301. 3 3 58. 3 2 0 2 3 30 is a partial sum of an arithmetic sequence with a 3 and d 32 3 32 . The
621 last term is 30 an 3 32 n 1, so 22 n 1 n 23. So the partial sum is S23 23 2 3 30 2 3105. 59. 250 233 216 97 is a partial sum of an arithmetic sequence with a 250 and d 233 250 17. The last
10 term is 97 an 250 17 n 1, so n 1 97250 17 9 n 10. So the partial sum is S10 2 250 97 1735. 60. 89 85 81 13 is a partial sum of an arithmetic sequence with a 89 and d 85 89 4. The last term is 20 13 an 89 4 n 1, so n 1 1389 4 19 n 20. So the partial sum is S20 2 89 13 1020. 61. 07 27 47 567 is a partial sum of an arithmetic sequence with a 07 and d 27 07 2. The last term
is 567 an 07 2 n 1 28 n 1 n 29. So the partial sum is S29 29 2 07 567 8323. 62. 10 99 98 01 is a partial sum of an arithmetic sequence with a 10 and d 01. The last term
is 01 an 10 01 n 1, so 99 n 1 n 100. So the partial sum is S100 100 2 10 01 505. 10 63. k0 3 025k is a partial sum of an arithmetic sequence with a 3 025 0 3 and d 025. The last term is a11 3 025 10 55. So the partial sum is S11 11 2 3 55 4675. 20 64. n0 1 2n is a partial sum of an arithmetic sequence where a 1 2 0 1, d 2, and the last term is a21 1 2 20 39. So the partial sum is S21 21 2 1 39 399.
65. We have an arithmetic sequence with a 5 and d 2. We seek n such that 2700 Sn
n [2a n 1 d]. Solving for 2
n [10 2 n 1] 5400 10n 2n 2 2n n 2 4n 2700 0 n 50 n 54 0 2 n 50 or n 54. Since n is a positive integer, 50 terms of the sequence must be added to get 2700. n 66. a 12 and d 8, so Sn 2700 2700 [2 12 n 1 8] 4n 2 8n 2700 0 4 n 27 n 25 0. 2 Thus, 2700 is the sum of the first 25 terms of the sequence. n, we have 2700
870
CHAPTER 13 Sequences and Series
67. Let x denote the length of the side between the length of the other two sides. Then the lengths of the three sides of the triangle are x a, x, and x a, for some a 0. Since x a is the longest side, it is the hypotenuse, and by the Pythagorean Theorem, we know that x a2 x 2 x a2 x 2 2ax a 2 x 2 x 2 2ax a 2 x 2 4ax 0 x x 4a 0 x 4a (x 0 is not a possible solution). Thus, the lengths of the three sides are x a 4a a 3a, x 4a, and x a 4a a 5a. The lengths 3a, 4a, 5a are proportional to 3, 4, 5, and so the triangle is similar to a 3-4-5 triangle.
68. P 10110 10210 10310 101910 101231910 . Now, 1 2 3 19 is an arithmetic series with 1 19 a 1, d 1, and n 19. Thus, 1 2 3 19 S19 19 190, and so P 1019010 1019 . 2 69. The sequence 1, 35 , 37 , 13 , is harmonic if 1, 53 , 73 , 3, forms an arithmetic sequence. Since 53 1 73 53 3 73 23 , the sequence of reciprocals is arithmetic and thus the original sequence is harmonic. 70. The two original numbers are 3 and 5. Thus, the reciprocals are 13 and 15 , and their average is 1 1 1 1 5 3 4 . Therefore, the harmonic mean is 15 . 2 3 5 2 15 15 15 4
71. The diminishing values of the computer form an arithmetic sequence with a1 12,500 and common difference d 1875. Thus the value of the computer after 6 years is a7 12,500 7 1 1875 $1250. 72. The number of poles in a layer can be viewed as an arithmetic sequence, where a1 25 and the common difference is 1. The number of poles in the first 12 layers is S12 12 2 [2 25 11 1] 6 39 234.
73. The increasing values of the man’s salary form an arithmetic sequence with a1 30,000 and common difference d 2300. Then his total earnings for a ten-year period are S10 10 2 [2 30,000 9 2300] 403,500. Thus his total earnings for the 10 year period are $403,500.
74. The number of cars that can park in a row can be viewed as an arithmetic sequence, where a1 20 and the common difference is 2. Thus the number of cars that can park in the 21 rows is S21 21 2 [2 20 20 2] 105 80 840. 75. The number of seats in the nth row is given by the nth term of an arithmetic sequence with a1 15 and common n difference d 3. We need to find n such that Sn 870. So we solve 870 Sn [2 15 n 1 3] for n. We have 2 n 2 2 2 870 27 3n 1740 3n 27n 3n 27n 1740 0 n 9n 580 0 x 20 x 29 0 2 n 20 or n 29. Since the number of rows is positive, the theater must have 20 rows. 76. The sequence is 16, 48, 80, . This is an arithmetic sequence with a 16 and d 48 16 32. (a) The total distance after 6 seconds is S6 62 32 5 32 3 192 576 ft. (b) The total distance after n seconds is Sn n2 [32 32 n 1] 16n 2 ft.
77. The number of gifts on the 12th day is 1 2 3 4 12. Since a2 a1 a3 a2 a4 a3 1, the number of gifts on the 12th day is the partial sum of an arithmetic sequence with a 1 and d 1. So the sum is 1 12 6 13 78. S12 12 2 78. (a) We want an arithmetic sequence with4 terms, so let a1 10 and a4 18. Since the sequence is arithmetic, 8 46 are the a4 a1 3d 18 10 8 3d 8 d 83 . Therefore, a2 10 83 38 and a 10 2 3 3 3 3 two arithmetic means between 10 and 18.
(b) We want an arithmetic sequence with5 terms, so let a1 10 and a5 18. Since the sequence is arithmetic, a5 a1 4d 18 10 8 4d 8 d 2. Therefore, a2 10 2 12, a3 10 2 2 14, and a4 10 3 2 16 are the three arithmetic means between 10 and 18.
SECTION 13.3 Geometric Sequences
871
(c) We want an arithmetic sequence with6 terms, with the starting dosage a1 100 and the final dosage a6 300. Since the sequence is arithmetic, a6 a1 5d 300 100 200 5d 200 d 40. Therefore, a2 140, a3 180, a4 220, a5 260, and a6 300. The patient should take 140 mg, then 180 mg, then 220 mg, then 260 mg, and finally arrive at 300 mg.
13.3 GEOMETRIC SEQUENCES 1. A geometric sequence is a sequence where the ratio between successive terms is constant. 2. The sequence an ar n1 is a geometric sequence where a is the first term and r is the common ratio. So, for the geometric sequence an 2 5n1 the first term is 2 and the common ratio is 5. 3. True. If we know the first and second terms of a geometric sequence then we can find all other terms. 4. (a) The nth partial sum of a geometric sequence an ar n1 is given by Sn a
1 rn . 1r
k1 a ar ar 2 ar 3 is is an infinite geometric series. If r 1, then this series (b) The series k1 ar converges and its sum is S a 1 r. If r 1 the series diverges.
5. (a) a1 7 30 7, a2 7 31 21,
6. (a) a1 6 050 6, a2 6 051 3,
a3 7 32 63, a4 7 33 189,
a3 6 052 15, a4 6 053 075,
a5 7 34 567
a5 6 054 0375
(b) The common ratio is 3. (c)
(b) The common ratio is 05.
an
an
(c)
600
6 4
400
2
200
0 _2
500 300
2
3
4
5 n
_4
100
0
1
2
3
4
5 n
0 1 7. (a) a1 52 12 52 , a2 52 12 54 ,
2 3 5 , a3 52 12 58 , a4 52 12 16 4 5 a5 52 12 32
8. (a) a1 30 1, a2 31 3, a3 32 9, a4 33 27, a5 34 81
(b) The common ratio is 3. (c)
60
an
40
2
0
an 80
(b) The common ratio is 12 . (c)
1
20
1
2
3
4
5 n
_2
9. a 7, r 4. So an ar n1 7 4n1 and a4 7 43 448. 10. a 3, r 2. So an ar n1 3 2n1 and a4 3 23 24. n1 3 5 . 11. a 52 , r 12 . So an ar n1 52 12 and a4 52 12 16
0
1
2
3
4
5 n
872
CHAPTER 13 Sequences and Series
12. a
n1 n 3 3, r 3. So an ar n1 3 3 3 and a4 3 3 9.
a2 a a 6 12 24 2, and 4 2. Since these ratios are the same, the sequence is geometric with the 2, 3 a1 3 a2 6 a3 12 common ratio 2. a 31 48 93 a 16 and 3 . Since these ratios are not the same, this is not a geometric sequence. 14. 2 a1 3 a2 48 16 a 1536 768 384 1 a 1 a 1 15. 2 , 3 , and 4 . Since these ratios are the same, the sequence is geometric with a1 3072 2 a2 1536 2 a3 768 2 13.
the common ratio 12 . 16.
1 144 a2 , a1 432 3
a3 1 a 1 48 16 , and 4 . Since these ratios are the same, the sequence is a2 144 3 a3 48 3
geometric with the common ratio 13 . 17.
1 a 1 a 1 32 34 38 a2 , 3 , and 4 . Since these ratios are the same, the sequence is geometric with the a1 3 2 a2 32 2 a3 34 2
common ratio 12 . 18.
1 9 a2 , a1 27 3 common ratio 13 .
a3 1 3 , and a2 9 3
a4 1 . Since these ratios are the same, the sequence is geometric with the a3 3
19.
2 a 4 a2 13 15 and 4 . Since these ratios are not the same, this is not a geometric sequence. a1 12 3 a3 14 5
20.
a2 e4 a e6 a e8 2 e2 , 3 4 e2 , and 4 6 e2 . Since these ratios are the same, the sequence is geometric with the a1 a2 a3 e e e
common ratio e2 . a 11 121 a 11, 3 11, and 21. 2 a1 10 a2 11 with the common ratio 11.
a4 1331 11. Since these ratios are the same, the sequence is geometric a3 121
1
22.
1 a2 a 1 3 41 and 4 81 . Since these ratios are not the same, this is not a geometric sequence. a1 2 a3 4 2
6
23. a1 2 31 6, a2 2 32 18, a3 2 33 54, a4 2 34 162, and a5 2 35 486. This sequence is geometric, the common ratio is r 3, and an a1 r n1 6 3n1 .
24. a1 4 31 7, a2 4 32 13, a3 4 33 31, a4 4 34 85, and a5 4 35 247. Since a2 a3 31 13 7 and a 13 are different ratios, this is not a geometric sequence. a1 2 1 1 1 1 1 1 1 1 1 25. a1 , a2 2 , a3 3 , a4 4 , and a5 5 . This sequence is geometric, the 4 16 64 256 1024 4 4 4 4 n1 common ratio is r 14 and an a1r n1 14 14 . 26. a1 11 21 2, a2 12 22 4, a3 13 23 8, a4 14 24 16, and
a5 15 25 32. This sequence is geometric, the common ratio is r 2, and an a1r n1 2 2n1 . 27. Since ln a b b ln a, we have a1 ln 50 ln 1 0, a2 ln 51 ln 5, a3 ln 52 2 ln 5, a4 ln 53 3 ln 5, a5 ln 54 4 ln 5. Since a1 0 and a2 0, this sequence is not geometric. a a 4 27 28. a1 11 1, a2 22 4, a3 33 27, a4 44 256, and a5 55 3125. Since 2 4 and 3 are a1 1 a2 4 different, this is not a geometric sequence.
SECTION 13.3 Geometric Sequences
873
a 29. 2, 6, 18, 54, . Then r 2 62 3, a5 a4 3 54 3 162, and an 2 3n1 . a1 14 n1 a2 3 28 56 2 2 56 2 112 2 30. 7, 14 . 3 , 9 , 27 , . Then r a 7 3 , a5 a4 3 27 3 81 , and an 7 3 1 009 a 03, a5 a4 03 00081 03 000243, and 31. 03, 009, 0027, 00081, . Then r 2 a1 03 an 03 03n1 .
n1 a 2 2, a5 a4 2 2 2 2 4, and an 2, 2, 2 2, . Then r 2 2 . a1 1 1 , . Then r a2 12 1 , a a 1 1 1 1 , a 144 1 n1 . 33. 144, 12, 1, 12 n 5 4 12 12 12 12 144 12 a1 144 a 2 1 2 1 , and a 8 1 n1 . 34. 8, 2, 12 , 18 , . Then r 14 , a5 a4 18 14 32 n 4 a1 8 4 a 353 35. 3, 353 , 373 , 27, . Then r 2 323 , a5 a4 323 27 323 3113 , and a1 3 n1 3 32n23 32n13 . an 3 323
32. 1,
t2 n1 a2 t t4 t t5 t2 t3 t4 t t , and an t 36. t, , , , . Then r 2 , a5 a4 . 2 4 8 a1 t 2 2 8 2 16 2
n1 a s 27 37. 1, s 27 , s 47 , s 67 , . Then r 2 s 27 , a5 a4 s 27 s 67 s 27 s 87 , and an s 27 s 2n27 . a1 1 a 5c1 38. 5, 5c1 , 52c1 , 53c1 , . Then r 2 5c , a5 a4 a1 5 n1 5 5cnc 5cnc1 . an 5 5c 3 2 a 6 39. a1 15, a2 6. Thus r 2 and a4 a1r 41 15 25 a1 15 5 12 1 1 1 40. a1 , a . Thus r 6, so a6 65 648. 12 2 2 112 12
41. a3
5c 53c1 5c 54c1 , and
24 . 25
1 a a r5 9 and a6 9. Thus, 6 1 2 r 3 r 3 27, so r 3. Therefore, a1 32 13 3 a3 13 a1r
1 and a 1 3 1 . a1 27 2 27 9
a7 8 2 81 32 329 a 12 . Thus, , so r . Therefore, a4 r 3 a1 a1 34 r3 r3 9 a4 12 27 3 827 2 r n1 81 2 and the nth term is a1 r n . 2 3
42. a4 12 and a7
9 a 9216 a 18 512 r 8. Therefore, a1 23 and 43. a3 18 and a6 9216. Thus, r 3 6 a3 18 64 32 r 9 an 8n1 . 32 a6 729256 a 54 27 3 3 44. a3 54 and a6 729 384 and r . Thus, a1 23 256 . Thus, r a 54 512 8 r 382 3 a2 384 38 144. a 729 1728, a2 1728 075 1296, and a3 1296 075 972. 45. r 075 and a4 729, so a1 34 r 0753
874
CHAPTER 13 Sequences and Series
6 a 18 1 . 46. r 16 and a3 18, so a1 23 648 and a7 648 16 2 72 r 16 n1 47. a 1536 and r 12 , so an ar n1 6 1536 12 1536 21n log2 6 log2 1536 1 n n 1 log2 1536 6 1 log2 256 9. Thus, 6 is the ninth term.
n5 468,750 48. a2 30 and a5 3750, so r 3 3750 30 125 and r 5. Thus, an 468,750 3750 5
,750 n 5 log5 468 3750 log5 125 n 8. Therefore, the eighth term of the sequence is 468,750.
1 26 5 63 315. 49. a 5, r 2, n 6. Then S6 5 12 1 1 4 80 2 3 2 81 80 . 50. a 23 , r 13 , n 4. Then S4 3 81 2 3 1 1 3
3
a ar 5 a 2 51. a3 28, a6 224, n 6. So 6 2 r 3 . So we have r 3 6 224 28 8, and hence r 2. Since a3 a r , we a3 a3 ar a 28 1 26 7 63 441. get a 23 2 7. So S6 7 12 r 2 012 a 000096 a 52. a2 012, a5 000096, n 4. So r 3 5 0008 r 02, and thus a1 2 06. a2 012 r 02 Therefore, S4 06
1 024 07488. 1 02
a 53. 1 3 9 2187 is a partial sum of a geometric sequence, where a 1 and r 2 31 3. Then the last term is a1 2187 an 1 3n1 n 1 log3 2187 7 n 8. So the partial sum is S8 1
1 38 3280. 13
1 1 is a partial sum of a geometric sequence for which a 1 and r a2 2 1 . The last 54. 1 12 14 18 512 2 a1 1 10 n1 1 12 1 1 341 term is an , where an 1 2 , so n 10. So the partial sum is S10 1 512 . 512 1 1 2
30 a 2. The 55. 15 30 60 960 is a partial sum of a geometric sequence for which a 15 and r 2 a1 15 last term is an 960 15 2n1 n 7, so the partial sum is S7 15
1 27 645. 1 2
a 1 2560 56. 5120 2560 1280 20 is a partial sum of a geometric sequence for which a 5120 and r 2 . a1 5120 2 9 n1 1 12 n 9, so the partial sum is S9 5120 10,220. The last term is an 20 5120 12 1 12 a 57. 125 125 125 12,500,000 is a partial sum of a geometric sequence for which a 125 and r 2 10. The a1 1 108 13,888,88875. 1 10 a 1 58. 10,800 1080 108 0000108 is a partial sum of a geometric sequence for which a 10,800 and r 2 . a1 10 9 1 n1 1 10 1 The last term is an 0000108 10,800 10 n 9, so the partial sum is S9 10,800 11,999999988. 1 1 10 last term is an 12,500,000 125 10n1 n 8, so the partial sum is S8 125
SECTION 13.3 Geometric Sequences
59.
5
k1
k1
3 12
3
5 1 12 1 12
93 16
1 26 105 1 2 k1 5 k1 1 23 5 211 63. 3 23 3 2 27 1 3 k1
61.
6
5 2k1 5
875
5 k1 1 32 55 60. 8 32 8 3 2 k1 1 2 5
1 56 39,060 15 k1 6 k1 1 32 6 1330 64. 64 32 64 k1 1 32
62.
6
10 5k1 10
1 is an infinite geometric series with a 1 and r 1 . Therefore, it is convergent with sum 65. 1 13 19 27 3 3 a 1 . S 1r 2 1 1 3
66. 1 12 14 18 is an infinite geometric series with a 1 and r 12 . Therefore, it is convergent with sum S
2 1 1 . 3 1 3 1 2 2
1 is an infinite geometric series with a 1 and r 1 . Therefore, it is convergent with sum 67. 1 13 19 27 3
S
3 a 1 . 1 1r 4 1 3
4 8 is an infinite geometric series with a 2 and r 2 . Therefore, it is convergent with sum 68. 25 25 125 5 5 2 5
2
2 53 . 2 3 1 5 5 2 3 69. 1 32 32 32 is an infinite geometric series with a 1 and r 32 1 1 1 1 1 70. 6 8 10 12 is an infinite geometric series with a 6 and r 3 3 3 3 3 S
1. Therefore, the series diverges. 1 1 . Therefore, it is convergent with 2 9 3
1
sum S
1 9 1 a 6 3 . 6 1 1r 648 3 8 1 9
71. 3 32 34 38 is an infinite geometric series with a 3 and r 12 . Therefore, it is convergent with sum 3 2. S 1 12
72. 1 1 1 1 is an infinite geometric series with a 1 and r 1. Because r 1 1 1, the series diverges. 73. 3 3 11 3 112 3 113 is an infinite geometric series with a 3 and r 11 1. Therefore, the series diverges. 10
100 3 3 10 3 . Therefore, it is convergent 74. 100 9 3 1 10 is an infinite geometric series with a 9 and r 100 10 9 100 100 a 9 10 1000 9 with sum S 100 9 13 117 . 13 3 1r 1 10 10 1 1 is an infinite geometric series with a 1 and r 1 . Therefore, the sum of the series is 75. 1 12 4 2 2 2 2 2
S
1 2
1 1 2
1 2 1. 21
876
CHAPTER 13 Sequences and Series
76. 1 2 2 2 2 4 is an infinite geometric series with a 1 and r 2. Because r 2 1, the series diverges. 7 7 7 is an infinite geometric series with a 7 and r 1 . Thus 77. 0777 10 100 1000 10 10 7 7 a 10 1 . 0777 1r 9 1 10
53 53 53 53 78. 02535353 02 1000 100,000 10,000,000 is an infinite geometric series (after the first term) with a 1000 53
1 . Thus 005353 1000 53 100 53 , and so 02535353 2 53 2 99 53 251 . and r 100 1000 99 990 1 10 990 990 990 1 100
3 3 3 3 1 79. 0030303 100 10,000 1,000,000 is an infinite geometric series with a 100 and r 100 . Thus 3
0030303
1 3 a 1001 . 1r 99 33 1 100
25 25 80. 211252525 211 1025 ,000 1,000,000 100,000,000 is an infinite geometric series (after 25
25 10,000 1 , and so the first term) with a 1025 ,000 and r 100 . Thus 000252525 1 9900 1 100 211252525
25 211 99 25 20,914 10,457 211 . 100 9900 9900 9900 4950
112 112 112 112 81. 0112 0112112112 1000 1,000,000 1,000,000,000 is an infinite geometric series with a 1000 and 112
112 1 . Thus 0112112112 a 10001 . r 1000 1r 999 1 1000
123 123 123 123 1 82. 0123123123 1000 1,000,000 1,000,000,000 is an infinite geometric series with a 1000 and r 1000 .
Thus 0123123123
123 41 123 1000 . 1 999 333 1 1000
a 83. Since we have 5 terms, let us denote a1 5 and a5 80. Also, 5 r 4 because the sequence is geometric, and so a1 r 4 80 5 16 r 2 . If r 2, the three geometric means are a2 10, a3 20, and a4 40. (If r 2, the three geometric means are a2 10, a3 20, and a4 40, but these are not between 5 and 80.)
84. The sum is given by a b a 2 2b a 3 3b a 10 10b a a 2 a 3 a 10 b 2b 3b 10b a
1 a 10 10 1 a 10 b 10b a 55b 1a 2 1a
a a3 85. (a) 5 3 5 3 is neither arithmetic nor geometric because a2 a1 a3 a2 and 2 . a1 a2 (b) 13 1 53 73 is arithmetic with d 23 , so the next term is a5 73 23 3. (c) 3 3 3 3 9 is geometric with r 3, so the next term is a5 9 3.
(d) 3 32 0 32 is arithmetic with d 32 , so the next term is a5 32 32 3. 86. (a) 1 1 1 1 is geometric with r 1, so the next term is a5 1. 1 (b) 5 3 5 6 5 1 is geometric with r 516 , so a5 1 516 . 6 5 a (c) 2 1 12 2 is neither arithmetic nor geometric because a2 a1 a3 a2 and 2 a1
a3 . a2
SECTION 13.3 Geometric Sequences
877
(d) x 1 x x 1 x 2 is arithmetic with d 1, so a5 x 2 1 x 3. 87. (a) The value at the end of the year is equal to the value at beginning less the depreciation, so Vn Vn1 02Vn1 08Vn1 with V1 160,000. Thus Vn 160,000 08n1 .
(b) Vn 100,000 08n1 160,000 100,000 08n1 0625 n 1 log 08 log 0625 log 0625 211. Thus it will depreciate to below $100,000 during the fourth year. n1 log 08 88. Let an denote the number of ancestors a person has n generations back. Then a1 2, a2 4, a3 8, . Since 4 8 2, this is a geometric sequence with r 2. Therefore, a 2 214 215 32,768. 15 2 4 89. Since the ball is dropped from a height of 80 feet, a 80 . Also since the ball rebounds three-fourths of the distance n fallen, r 34 . So on the nth bounce, the ball attains a height of an 80 34 . Hence, on the fifth bounce, the ball goes 5 a5 80 34 80243 1024 19 ft high.
90. a 5000, r 108. After 1 hour, there are 5000 108 5400, after 2 hours, 5400 108 5832, after 3 hours, 5832 108 629856, after 4 hours, 629856 108 68024448, and after 5 hours, 68024448 108 7347 bacteria. After n hours, the number of bacteria is an 5000 108n .
91. Let an be the amount of water remaining at the nth stage. We start with 5 gallons, so a 5. When 1 gallon (that is, 15 of the mixture) is removed, 45 of the mixture (and hence 45 of the water in the mixture) remains. Thus, a1 5 45 a2 5 45 45 , n 3 and in general, an 5 45 . The amount of water remaining after 3 repetitions is a3 5 45 64 25 , and after 5 5 repetitions it is a5 5 45 1024 625 .
92. Let aC denote the term of the geometric series that is the frequency of middle C. Then aC 256 and aC1 512. Since aC 256 this is a geometric sequence, r 512 256 2, and so aC2 r 2 22 64.
93. Let an be the height the ball reaches on the nth bounce. From the given information, an is the geometric sequence n an 9 13 . (Notice that the ball hits the ground for the fifth time after the fourth bounce.) 2 3 4 (a) a0 9, a1 9 13 3, a2 9 13 1, a3 9 13 13 , and a4 9 13 19 . The total distance traveled is 8 a0 2a1 2a2 2a3 2a4 9 2 3 2 1 2 13 2 19 161 9 17 9 ft. (b) The total distance traveled at the instant the ball hits the ground for the nth time is 2 3 4 n1 Dn 9 2 9 13 2 9 13 2 9 13 2 9 13 2 9 13 2 3 4 n1 1 1 1 1 9 2 9 9 3 9 3 9 3 9 3 9 13
n n3 n 1 13 1 1 2 9 9 18 9 27 1 3 3 1 13
1 2n 2n 1. At the end of 30 days, she 12 will have S30 230 1 1,073,741,823 cents $10,737,41823. To become a billionaire, we want 2n 1 1011 or 11 approximately 2n 1011 . So log 2n log 1011 n 365. Thus it will take 37 days. log 2
94. We have a geometric sequence with a 1 and r 2. Then Sn 1
878
CHAPTER 13 Sequences and Series
95. Let a1 1 be the man with 7 wives. Also, let a2 7 (the wives), a3 7a2 72 (the sacks), a4 7a3 73 (the cats), and a5 7a4 74 (the kits). The total is a1 a2 a3 a4 a5 1 7 72 73 74 , which is a partial sum of a geometric sequence with a 1 and r 7. Thus, the number in the party is S5 1
96. (a) (b)
10
k1
1 k1 50 2
50
k1 1 k1 50 2
10 1 12 1 12
1 75 2801. 17
100 1 00009765 999023 mg
50 100 mg 1 12
97. Let an be the height the ball reaches on the nth bounce. We have a0 1 and an 12 an1 . Since the total distance d traveled includes the bounce up as well and the distance down, we have 2 3 4 d a0 2 a1 2 a2 1 2 12 2 12 2 12 2 12 i 2 3 1 1 1 1 12 12 12 1 1 3 2 1 1 i0 2
Thus the total distance traveled is about 3 m. 2 98. The time required for the ball to stop bouncing is t 1 1 1 which is an infinite geometric series with 2 2 2 2 1 2 2 21 2 2. Thus the a 1 and r 1 . The sum of this series is t 2 21 21 21 21 1 1 2 time required for the ball to stop is 2 2 341 s. 99. (a) If a square has side x, then by the Pythagorean Theorem the length of the side of the square formed by x2 x x2 x 2 x 2 joining the midpoints is, . In our case, x 1 and the side of the 2 2 4 4 2 2 1 first inscribed square is , the side of the second inscribed square is 1 1 1 , the side of the 2 2 2 2 3 third inscribed square is 1 , and so on. Since this pattern continues, the total area of all the squares is 2 2 4 6 2 3 1 1 1 1 12 12 12 2. A 12 1 2 2 2 1 12 2 3 (b) As in part (a), the sides of the squares are 1, 1 , 1 , 1 , . Thus the sum of the perimeters is 2 2 2 2 3 1 1 1 4 , which is an infinite geometric series with a 4 and r 1 . Thus S 414 4 2 2 2 2 21 4 2 424 2 4 2 4 8 4 2. the sum of the perimeters is S 21 21 21 21 1 1 2
2 2 100. Let An be the area of the disks of paper placed at the nth stage. Then A1 R 2 , A2 2 12 R 2R , 2 1 1 2 2 2 2 A3 4 14 R 4 R , . We see from this pattern that the total area is A R 2 R 4 R . Thus, R2 2 R 2 . the total area, A, is an infinite geometric series with a1 R 2 and r 12 . So A 1 12
SECTION 13.4 Mathematics of Finance
879
101. Let an denote the area colored blue at nth stage. Since only the middle squares are colored blue, an 19 area remaining yellow at the n 1 th stage. Also, the area remaining yellow at the nth stage is 89 of the area 2 3 remaining yellow at the preceding stage. So a1 19 , a2 19 89 , a3 19 89 , a4 19 89 , . Thus the total area 2 3 colored blue A 19 19 89 19 89 19 89 is an infinite geometric series with a 19 and r 89 . So the total area is A
1 9
1. 1 89
102. Let a1 , a2 , a3 , be a geometric sequence with common ratio r. Thus a2 a1 r, a3 a1 r 2 , , an a1 r n1 . Hence, 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 n1 1 1 , , , and so a2 a1 r a1 r a3 a1 r 2 a1 r an a1 r n1 a1 r a1 r 2 a1 r n1 1 1 1 1 , , , is a geometric sequence with common ratio . a1 a2 a3 r 103. a1 a2 a3 is a geometric sequence with common ratio r . Thus a2 a1 r, a3 a1 r 2 , , an a1 r n1 . Hence log a2 log a1r log a1 log r, log a3 log a1 r 2 log a1 log r 2 log a1 2 log r, , log an log a1 r n1 log a1 log r n1 log a1 n 1 log r , and so log a1 log a2 log a3 is an arithmetic
sequence with common difference log r. 104. Since a1 a2 a3 is an arithmetic sequence with common difference d, the terms can be expressed as a2 a1 d, 2 a3 a1 2d, , an a1 n 1 d. So 10a2 10a1 d 10a1 10d , 10a3 10a1 2d 10a1 10d , , n1 , and so 10a1 10a2 10a3 is a geometric sequence with common ratio r 10d . 10an 10a1 n1d 10a1 10d
13.4 MATHEMATICS OF FINANCE 1. An annuity is a sum of money that is paid in regular equal payments. The amount of an annuity is the sum of all the individual payments together with all the interest. 2. The present value of an annuity is the amount that must be invested now at interest rate i per time period in order to provide n payments each of amount R. 1 00610 1 1 in 1 1000 $13,18079. i 006 008 1006666724 1 1 in 1 R 500, n 24, i 00066667. So A f R 500 $12,96659. 12 i 00066667 1 01220 1 1 in 1 n 20, R $5000, i 012. So A f R 5000 $360,26221. i 012 006 1 in 1 10320 1 R 500, n 20, i 003. So A f R 500 $13,43519. 2 i 003 008 1 00216 1 1 in 1 n 16, R $300, i 002. So A f R 300 $5,59179. 4 i 002 1 00540 1 1 in 1 n 40, R 2000, i 005. So A f R 2000 $241,59955. i 005 iAf 010 0025 5000 A f 5000, n 4 2 8, i 0025. So R $57234. n 4 1 i 1 10258 1 iAf 006 0005 2000 0005, n 8. Then R $24566. A f $2000, i 12 1 in 1 1 00058 1
3. n 10, R $1000, i 006. So A f R 4. 5. 6. 7. 8. 9. 10.
880
CHAPTER 13 Sequences and Series
1 104520 009 1 1 in 0045. So A p R 1000 $13,00794. 2 i 0045 30 1 1 008 008 1 1 in 12 12. R 300, n 30, i . So A p R 300 $813265. 008 12 i 11. R 1000, n 20, i
12
13. R $200, n 20, i
1 1 004520 009 1 1 in 0045. So A p R 200 $260159. 2 i 0045
14. A p 50,000, n 10 2 20, i 15. A p $12 000, i
i Ap 008 004 50,000 004. So R $367909. n 2 1 1 i 1 10420
i Ap 0105 000875 12000 000875, n 48. Then R $30724. n 12 1 1 i 1 1 00087548
009 00075. Over a 30 year period, n 30 12 360, and the monthly payment is 12 i Ap 00075 80,000 R $64370. Over a 15 year period, n 15 12 180, and so the monthly 1 1 in 1 10075360 80,000 00075 $81141. payment is R 1 10075180
16. A p 80,000, i
i Ap 008 0006667 100,000 0006667, n 360. Then R $73376. n 12 1 1 i 1 1 0006667360 Therefore, the total amount paid on this loan over the 30 year period is 360 73376 $264,15360.
17. A p $100,000, i
i Ap 006 0005 200,000 0005, n 180. Then R $168771. Therefore, 12 1 1 in 1 1 0005180 the total amount paid on this loan over the 15 year period is 180 168771 $303,78780.
18. A p $200,000, i
1 1005360 1 1 in 006 0005. So A p R 3500 $583,770.65. 12 i 0005 Therefore, Dr. Gupta can afford a loan of $583,770.65.
19. R 3500, n 12 30 360, i
009 1 10075360 1 1 in 00075. So A p R 650 $80,78321. 12 i 00075 Therefore, the couple can afford a loan of $80,78321.
20. R 650, n 12 30 360, i
008 000667. The amount borrowed is 12 1 10066736 1 1 in Ap R 220 $7,02060. So she purchased the car for i 000667 $7,02060 $2000 $902060.
21. R 220, n 12 3 36, i
22. R $30, i
1 1 000833312 1 1 in 010 0008333, n 12. Then A p R 30 $34124. 12 i 0008333
00975 0008125. 23. A p 100,000, n 360, i 12 i Ap 0008125 100,000 (a) R $85915. 1 1 in 1 1 0008125360
(b) The total amount that will be paid over the 30 year period is 360 85915 $309,29400. (c) R $85915, i
00975 1 0008125360 1 0008125, n 360. So A f 85915 $1,841,51929. 12 0008125
SECTION 13.4 Mathematics of Finance
24. (a) For the 30-year mortgage, A p 300,000, n 12 30 360, and i 0065 12 , so R30
i Ap 1 1 in
0065 300,000 12 360 $189620. 1 1 0065 12
For the 15-year mortgage, A p 300,000, n 12 15 180, and i 00575 12 , so R15
i Ap 1 1 in
monthly payment.
00575 300,000 12 180 $249123. The 15-year mortgage has the larger 1 1 00575 12
(b) The total amount of the 30-year mortgage is 360 189620 $682,63200, and the total amount of the 15-year mortgage is 180 249123 $448,42140. The 15-year mortgage has the lower total payment.
25. A p $640, R $32, n 24. We want to solve the equation R
i Ap 1 1 in
x . So we can express R for the interest rate i. Let x be the interest rate, then i 12 x 640 12 as a function of x by R x . We graph R x and y 32 in x 24 1 1 12 the rectangle [012 022] [30 34]. The x-coordinate of the intersection is about
34 32 30 0.15
0.20
01816, which corresponds to an interest rate of 1816%.
26. A p $12,500, R $420, n 36. We want to solve for the interest rate using the
i AP x . So we can . Let x be the interest rate, then i n 12 1 1 i x 12,500 12 . We graph express R as a function of x as follows: R x x 36 1 1 12 R x and y 420 in the rectangle [012 013] by [415 425]. The x-coordinate of equation R
425 420 415 0.120
0.125
the intersection is about 01280, which corresponds to an interest rate of 1280%.
27. A p $18999, R $1050, n 20. We want to solve the equation
i Ap x . So for the interest rate i. Let x be the interest rate, then i 1 1 in 12 x 18999 12 we can express R as a function of x by R x . We graph x 20 1 1 12 R x and y 1050 in the rectangle [010 018] [10 11]. The x-coordinate of R
the intersection is about 01168, which corresponds to an interest rate of 1168%.
11.0 10.5 10.0 0.10
0.15
0.130
881
882
CHAPTER 13 Sequences and Series
28. A p $2000 $200 $1800, R $88, n 24. We want to solve for the interest
89
i AP x . . Let x be the interest rate, then i 12 1 1 in x 1800 12 . So we can express R as a function of x as follows: R x x 24 1 1 12 [014 [87 We graph R x and y 88 in the viewing rectangle 016] by 89]. The rate using the equation R
88 87 0.14
0.15
0.16
x-coordinate of the intersection is about 01584, which corresponds to an interest
rate of 1584%.
R . The present value of an annuity is the sum of 1 ik the present values of each of the payments of R dollars, as shown in the time line.
29. (a) The present value of the kth payment is PV R 1 ik Time
1
2
3
4
Payment
R
R
R
R
¤¤¤
n-1
n
R
R
Present value R/(1+i) R/(1+i)@ R/(1+i)# R/(1+i)$ R/(1+i)n-1 R/(1+i)n
(b)
R R R R 2 3 1i 1 in 1 i 1 i 2 n1 R 1 R 1 1 R R 1i 1i 1i 1i 1i 1i 1i 1 R 1 rn and r . Since Sn a , we have This is a geometric series with a 1i 1i 1r n 1 1 1 1 in 1 1 in 1 1 in R 1 i R R Ap . R 1 1 1i i 1 i 1 1 1 i 1 1i 1i Ap
R . The amount of money to be invested now (A p ) to ensure an 1 ik annuity in perpetuity is the (infinite) sum of the present values of each of the payments, as shown in the time line.
30. (a) The present value of the kth payment is P V Time
1
2
3
4
Payment
R
R
R
R
Present value R/(1+i) R/(1+i)@ R/(1+i)# R/(1+i)$ R/(1+i)n-1 R/(1+i)n
¤¤¤
n-1
n
R
R
¤¤¤
SECTION 13.5 Mathematical Induction
883
R R R R R and . This is an infinite geometric series with a n 2 3 1i 1i 1 i 1 i 1 i R 1 1 i R 1i R. . Therefore, A p r 1 1i 1i i i 1 1i 5000 R (c) Using the result from part (b), we have R 5000 and i 010. Then A p $50,000. i 010 (d) We are given two different time periods: the interest is compounded quarterly, while the annuity is paid yearly. In order to use the formula in part (b), we need to find the effective annual interest rate produced by 8% interest compounded quarterly, which is 1 inn 1. Now i 8% and it is compounded quarterly, n 4, so the effective
(b) A p
annual yield is 1 0024 1 008243216. Thus by the formula in part (b), the amount that must be invested is 3000 A p 008243216 $36,39356.
31. (a) Using the hint, we calculate the present value of the remaining 240 payments with R 72417, i 00075, and n 240. Since A p R
1 10075240 1 1 in 72417 80,48784, they still owe $80,48784 on their i 00075
mortgage. (b) On their next payment, 00075 80,48784 $60366 is interest and $72417 60366 $12051 goes toward the principal.
13.5 MATHEMATICAL INDUCTION 1. Mathematical induction is a method of proving that a statement P n is true for all natural numbers n. In Step 1 we prove that P 1 is true. 2. (b) We prove “if P kis true then P k 1 is true.” 3. Let P n denote the statement 2 4 6 2n n n 1. Step 1: P 1 is the statement that 2 1 1 1, which is true. Step 2: Assume that P k is true; that is, 2 4 6 2k k k 1. We want to use this to show that P k 1 is true. Now 2 4 6 2k 2 k 1 k k 1 2 k 1
induction hypothesis
k 1 k 2 k 1 [k 1 1] Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 4. Let P n denote the statement 1 4 7 10 3n 2
n 3n 1 . 2
1 [3 1 1] 12 , which is true. 2 2 k 3k 1 Step 2: Assume that P k is true; that is. 1 4 7 3k 2 . We want to use this to show that 2 P k 1 is true. Now k 3k 1 3k 1 induction hypothesis 1 4 7 10 3k 2 [3 k 1 2] 2 2 2 3k k 6k 2 3k 5k 2 k 3k 1 6k 2 k 1 3k 2 k 1 [3 k 1 1] 2 2 2 2 2 2 Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. Step 1: P 1 is the statement that 1
884
CHAPTER 13 Sequences and Series
n 3n 7 . 2 1 3 1 7 Step 1: We need to show that P 1 is true. But P 1 says that 5 , which is true. 2 k 3k 7 Step 2: Assume that P k is true; that is, 5 8 11 3k 2 . We want to use this to show that 2 P k 1 is true. Now
5. Let P n denote the statement 5 8 11 3n 2
5 8 11 3k 2 [3 k 1 2]
k 3k 7 3k 5 2
induction hypothesis
3k 2 7k 6k 10 3k 2 13k 10 2 2 2 3k 10 k 1 k 1 [3 k 1 7] 2 2
Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 6. Let P n denote the statement 12 22 32 n 2 Step 1: P 1 is the statement that 12
n n 1 2n 1 . 6
123 , which is true. 6
Step 2: Assume that P k is true; that is, 12 22 32 k 2
k k 1 2k 1 . We want to use this to show that 6
P k 1 is true. Now, k k 1 2k 1 k 12 induction hypothesis 6 2k 2 k 6k 6 k 2k 1 6 k 1 k 1 k 1 6 6 2k 2 7k 6 k 1 k 2 2k 3 k 1 6 6
12 22 32 k 2 k 12
k 1 [k 1 1] [2 k 1 1] 6
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 7. Let P n denote the statement 1 2 2 3 3 4 n n 1 Step 1: P 1 is the statement that 1 2
n n 1 n 2 . 3
1 1 1 1 2 , which is true. 3
Step 2: Assume that P k is true; that is, 1 2 2 3 3 4 k k 1
k k 1 k 2 . We want to use this to 3
show that P k 1 is true. Now 1 2 2 3 3 4 k k 1 k 1 [k 1 1] k k 1 k 2 k 1 k 2 3 k k 1 k 2 3 k 1 k 2 k 1 k 2 k 3 3 3 3
Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
induction hypothesis
SECTION 13.5 Mathematical Induction
8. Let P n denote the statement 1 3 2 4 3 5 n n 2 Step 1: P 1 is the statement that 1 3
885
n n 1 2n 7 . 6
129 , which is true. 6
Step 2: Assume that P k is true; that is, 1 3 2 4 3 5 k k 2
k k 1 2k 7 . We want to use this to 6
show that P k 1 is true. Now 1 3 2 4 3 5 k k 2 k 1 [k 1 2] k k 1 2k 7 k 1 k 3 6 2k 2 7k 6k 18 k 2k 7 6 k 3 k 1 k 1 6 6 6
induction hypothesis
k 1 [k 1 1] [2 k 1 7] 6
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
9. Let P n denote the statement 13 23 33 n 3
n 2 n 12 . 4
12 1 12 , which is clearly true. 4 k 2 k 12 Step 2: Assume that P k is true; that is, 13 23 33 k 3 . We want to use this to show that 4 P k 1 is true. Now
Step 1: P 1 is the statement that 13
13 23 33 k 3 k 13
k 2 k 12 k 13 4 k 12 k 2 4 k 1 4
k 12 k 22
induction hypothesis
k 12 k 2 4k 4 4
k 12 [k 1 1]2
4 4 Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
10. Let P n denote the statement 13 33 53 2n 13 n 2 2n 2 1 . Step 1: P 1 is the statement that 13 12 2 12 1 , which is clearly true. Step 2: Assume that P k is true; that is, 13 33 53 2k 13 k 2 2k 2 1 . We want to use this to show that P k 1 is true. Now
13 33 53 2k 13 2k 13 k 2 2k 2 1 2k 13
induction hypothesis
2k 4 k 2 8k 3 12k 2 6k 1 2k 4 8k 3 11k 2 6k 1 k 2 2k 1 2k 2 4k 1 k 12 2 k 12 1
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
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CHAPTER 13 Sequences and Series
11. Let P n denote the statement 23 43 63 2n3 2n 2 n 12 . Step 1: P 1 is true since 23 2 12 1 12 2 4 8. Step 2: Assume that P k is true; that is, 23 43 63 2k3 2k 2 k 12 . We want to use this to show that P k 1 is true. Now 23 43 63 2k3 [2 k 1]3 2k 2 k 12 [2 k 1]3
induction hypothesis
2k 2 k 12 8 k 1 k 12 k 12 2k 2 8k 8 2 k 12 k 22 2 k 12 [k 1 1]2
Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
1 1 1 n 1 . 12 23 34 n n 1 n1 1 12 , which is clearly true. Step 1: P 1 is the statement that 12 1 1 1 k 1 . We want to use this to show that Step 2: Assume that P k is true; that is 12 23 34 k k 1 k 1 P k 1 is true. Now
12. Let P n denote the statement
1 1 1 1 k 1 12 23 k k 1 k 1 k 2 k 1 k 1 k 2
induction hypothesis
k k 2 1 k 2 2k 1 k 12 k 1 k 2 k 1 k 2 k 1 k 2 k 1 k 1 k 2 k 1 1
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 13. Let P n denote the statement 1 2 2 22 3 23 4 24 n 2n 2 1 n 1 2n . Step 1: P 1 is the statement that 1 2 2 [1 0], which is clearly true.
Step 2: Assume that P k is true; that is, 1 2 2 22 3 23 4 24 k 2k 2 1 k 1 2k . We want to use
this to show that P k 1 is true. Now
1 2 2 22 3 23 4 24 k 2k k 1 2k1 2 1 k 1 2k k 1 2k1
induction hypothesis
2 1 k 1 2k k 1 2k 2 1 2k 2k 2 1 k 2k1 2 1 [k 1 1] 2k1
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 14. Let P n denote the statement 1 2 22 2n1 2n 1.
Step 1: P 1 is the statement that 1 21 1, which is clearly true.
Step 2: Assume that P k is true; that is, 1 2 22 2k1 2k 1. We want to use this to show that P k 1 is true. Now
1 2 22 2k1 2k 2k 1 2k 2 2k 1 2k1 1 Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
induction hypothesis
SECTION 13.5 Mathematical Induction
887
15. Let P n denote the statement n 2 n is divisible by 2.
Step 1: P 1 is the statement that 12 1 2 is divisible by 2, which is clearly true.
Step 2: Assume that P k is true; that is, k 2 k is divisible by 2. Now k 12 k 1 k 2 2k 1 k 1 k 2 k 2k 2 k 2 k 2 k 1. By the induction hypothesis, k 2 k is divisible by 2, and clearly 2 k 1 is divisible by 2. Thus, the sum is divisible by 2, so P k 1 is true. Therefore, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
16. Let P n denote the statement that 5n 1 is divisible by 4.
Step 1: P 1 is the statement that 51 1 4 is divisible by 4, which is clearly true.
Step 2: Assume that P k is true; that is, 5k 1 is divisible by 4. We want to use this to show that P k 1 is true. Now, 5k1 1 5 5k 1 5 5k 5 4 5 5k 1 4 which is divisible by 4 since 5 5k 1 is divisible by 4 by the induction hypothesis. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
17. Let P n denote the statement that n 2 n 41 is odd.
Step 1: P 1 is the statement that 12 1 41 41 is odd, which is clearly true.
Step 2: Assume that P k is true; that is, k 2 k 41 is odd. We want to use this to show that P k 1 is true. Now, k 12 k 1 41 k 2 2k 1 k 1 41 k 2 k 41 2k, which is also odd because k 2 k 41 is odd by the induction hypothesis, 2k is always even, and an odd number plus an even number is always odd. Therefore, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 18. Let P n denote the statement that n 3 n 3 is divisible by 3.
Step 1: P 1 is the statement that 13 1 3 3 is divisible by 3, which is true.
Step 2: Assume that P k is true; that is, k 3 k 3 is divisible by 3. We want to use this to show that P k 1 is true. Now, k 13 k 1 3 k 3 3k 2 3k 1 k 1 3 k 3 k 3 3k 2 3k k 3 k 3 3 k 2 k , which is divisible by 3, since k 3 k 3 is divisible by 3 by the induction hypothesis, and 3 k 2 k is divisible by 3. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
19. Let P n denote the statement that 8n 3n is divisible by 5.
Step 1: P 1 is the statement that 81 31 5 is divisible by 5, which is clearly true.
Step 2: Assume that P k is true; that is, 8k 3k is divisible by 5. We want to use this to show that P k 1 is true. Now, 8k1 3k1 8 8k 3 3k 8 8k 8 5 3k 8 8k 3k 5 3k , which is divisible by 5 because 8k 3k
is divisible by 5 by our induction hypothesis, and 5 3k is divisible by 5. Thus P k 1 follows from P k. So by the
Principle of Mathematical Induction, P n is true for all n.
20. Let P n denote the statement that 32n 1 is divisible by 8.
Step 1: P 1 is the statement that 32 1 8 is divisible by 8, which is clearly true.
Step 2: Assume that P k is true; that is, 32k 1 is divisible by 8. We want to use this to show that P k 1 is true. Now, 32k1 1 9 32k 1 9 32k 9 8 9 32k 1 8, which is divisible by 8, since 32k 1 is divisible by 8 by the induction hypothesis. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true
for all n.
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CHAPTER 13 Sequences and Series
21. Let P n denote the statement n 2n . Step 1: P 1 is the statement that 1 21 2, which is clearly true.
Step 2: Assume that P k is true; that is, k 2k . We want to use this to show that P k 1 is true. Adding 1 to both sides
of P k we have k 1 2k 1. Since 1 2k for k 1, we have 2k 1 2k 2k 2 2k 2k1 . Thus k 1 2k1 ,
which is exactly P k 1. Therefore, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
22. Let P n denote the statement n 12 2n 2 , for all n 3.
Step 1: P 3 is the statement that 3 12 2 32 or 16 18, which is true.
Step 2: Assume that P k is true; that is, k 12 2k 2 , k 3. We want to use this to show that P k 1 is true. Now k 22 k 2 4k 4 k 2 2k 1 2k 3 k 12 2k 3 2k 2 2k 3
induction hypothesis
2k 2 2k 3 2k 1
because 2k 1 0 for k 3
2k 2 4k 2 2 k 12 Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n 3. 23. Let P n denote the statement 1 xn 1 nx, if x 1.
Step 1: P 1 is the statement that 1 x1 1 1x, which is clearly true.
Step 2: Assume that P k is true; that is, 1 xk 1 kx. Now, 1 xk1 1 x 1 xk 1 x 1 kx,
by the induction hypothesis. Since 1 x 1 kx 1 k 1 x kx 2 1 k 1 x (since kx 2 0), we have
1 xk1 1 k 1 x, which is P k 1. Thus P k 1 follows from P k. So the Principle of Mathematical Induction, P n is true for all n. 24. Let P n denote the statement 100n n 2 , for all n 100.
Step 1: P 100 is the statement that 100 100 1002 , which is true.
Step 2: Assume that P k is true; that is, 100k k 2 . We want to use this to show that P k 1 is true. Now 100 k 1 100k 100 k 2 100 k 2 2k 1 k 12
induction hypothesis because 2k 1 100 for k 100
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n 100. 25. Let P n be the statement that an 5 3n1 .
Step 1: P 1 is the statement that a1 5 30 5, which is true.
Step 2: Assume that P k is true; that is, ak 5 3k1 . We want to use this to show that P k 1 is true. Now, ak1 3ak 3 5 3k1 , by the induction hypothesis. Therefore, ak1 3 5 3k1 5 3k , which is exactly P k 1. Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
26. an1 3an 8 and a1 4. Then a2 3 4 8 4, a3 3 4 8 4, a4 3 4 8 4, , and the conjecture is that an 4. Let P n denote the statement that an 4. Step 1: P 1 is the statement that a1 4, which is true. Step 2: Assume that P k is true; that is, ak 4. We want to use this to show that P k 1 is true. Now, ak1 3 ak 8 3 4 8 4, by the induction hypothesis. This is exactly P k 1, so by the Principle of Mathematical Induction, P n is true for all n.
SECTION 13.5 Mathematical Induction
889
27. Let P n be the statement that x y is a factor of x n y n for all natural numbers n. Step 1: P 1 is the statement that x y is a factor of x 1 y 1 , which is clearly true.
Step 2: Assume that P k is true; that is, x y is a factor of x k y k . We want to use this to show that P k 1 is true. Now, x k1 y k1 x k1 x k y x k y y k1 x k x y x k y k y, for which x y is a factor because x y is a factor of x k x y, and x y is a factor of x k y k y, by the induction hypothesis. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
28. Let P n be the statement that x y is a factor of x 2n1 y 2n1 .
Step 1: P 1 is the statement that x y is a factor of x 1 y 1 , which is clearly true.
Step 2: Assume that P k is true; that is, x y is a factor of x 2k1 y 2k1 . We want to use this to show that P k 1 is true. Now x 2k11 y 2k11 x 2k1 y 2k1 x 2k1 x 2k1 y 2 x 2k1 y 2 y 2k1 x 2k1 x 2 y 2 x 2k1 y 2k1 y 2
for which x y is a factor. This is because x y is a factor of x 2 y 2 x y x y and x y is a factor of
x 2k1 y 2k1 by our induction hypothesis. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n .
29. Let P n denote the statement that F3n is even for all natural numbers n. Step 1: P 1 is the statement that F3 is even. Since F3 F2 F1 1 1 2, this statement is true. Step 2: Assume that P k is true; that is, F3k is even. We want to use this to show that P k 1 is true. Now, F3k1 F3k3 F3k2 F3k1 F3k1 F3k F3k1 F3k 2 F3k1 , which is even because F3k is even by the induction hypothesis, and 2 F3k1 is even. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 30. Let P n denote the statement that F1 F2 F3 Fn Fn2 1. Step 1: P 1 is the statement that F1 F3 1. But F1 1 2 1 F3 1, which is true. Step 2: Assume that P k is true; that is, F1 F2 F3 Fk Fk2 1. We want to use this to show that P k 1 is true. Now Fk12 1 Fk3 1 Fk2 Fk1 1 Fk2 1 Fk1 F1 F2 F3 Fk Fk1 by the induction hypothesis. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 31. Let P n denote the statement that F12 F22 F32 Fn2 Fn Fn1 . Step 1: P 1 is the statement that F12 F1 F2 or 12 1 1, which is true. Step 2: Assume that P k is true, that is, F12 F22 F32 Fk2 Fk Fk1 . We want to use this to show that P k 1 is true. Now 2 2 Fk Fk1 Fk1 F12 F22 F32 Fk2 Fk1
Fk1 Fk Fk1 Fk1 Fk2
induction hypothesis
by definition of the Fibonacci sequence
890
CHAPTER 13 Sequences and Series
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
32. Let P n denote the statement that F1 F3 F2n1 F2n . Step 1: P 1 is the statement that F1 F2 , which is true since F1 1 and F2 1. Step 2: Assume that P k is true; that is, F1 F3 F2k1 F2k , for some k 1. We want to use this to show that P k 1 is true; that is, F1 F3 F2k11 F2k1 . Now F1 F3 F2k1 F2k1 F2k F2k1
induction hypothesis
F2k2
definition of F2k2
F2k1 Therefore, P k 1 is true. So by the Principle of Mathematical Induction, P n is true for all n.
33. Let P n denote the statement
Step 1: Since
1 1 1 0
2
1 1 1 0
1 1 1 0
n
1 1
1 0
k1
Fn1
Fn
Fn
Fn1
2 1
k
1 1 1 0
Step 2: Assume that P k is true; that is,
1 1 1 0
1 1
.
F2 F1
Fk1
Fk
Fk Fk1
1 1
Fk1 Fk Fk1
1 1 1 0
Fk Fk1
F3 F2
1 0
k
Fk
, it follows that P 2 is true.
. We show that P k 1 follows from this. Now,
Fk1
Fk
Fk Fk1
Fk2 Fk1 Fk1
Fk
1 1 1 0
induction hypothesis
by definition of the Fibonacci sequence
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n 2.
34. Let a1 1 and an1
1 Fn , for n 1 . Let P n be the statement that an , for all n 1. 1 an Fn1
Step 1: P 1 is the statement that a1
F1 F , which is true since a1 1 and 1 11 1. F2 F2
Step 2: Assume that P k is true; that is, ak ak1
1 (definition of ak1 ) 1 ak
Fk . We want to use this to show that P k 1 is true. Now Fk1
1 Fk 1 Fk1
(induction hypothesis)
Fk1 F k1 (definition of Fk2 ). Fk Fk1 Fk2
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n 100.
SECTION 13.5 Mathematical Induction
891
35. Since F1 1, F2 1, F3 2, F4 3, F5 5, F6 8, F7 13, our conjecture is that Fn n, for all n 5. Let P n denote the statement that Fn n.
Step 1: P 5 is the statement that F5 5 5, which is clearly true.
Step 2: Assume that P k is true; that is, Fk k, for some k 5. We want to use this to show that P k 1 is true. Now, Fk1 Fk Fk1 k Fk1 (by the induction hypothesis) k 1 (because Fk1 1). Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n 5.
36. Since 100 10 103 , 100 11 113 , 100 12 123 , , our conjecture is that 100n n 3 , for all natural numbers n 10. Let P n denote the statement that 100n n 3 , for n 10.
Step 1: P 10 is the statement that 100 10 1,000 103 1,000, which is true.
Step 2: Assume that P k is true; that is, 100k k 3 . We want to use this to show that P k 1 is true. Now 100 k 1 100k 100 k 3 100
induction hypothesis
k3 k2
because k 10
k 3 3k 2 3k 1 k 13 Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n 10. 37. (a) P n n 2 n 11 is prime for all n. This is false as the case for n 11 demonstrates: P 11 112 11 11 121, which is not prime since 112 121.
(b) n 2 n, for all n 2. This is true. Let P n denote the statement that n 2 n. Step 1: P 2 is the statement that 22 4 2, which is clearly true.
Step 2: Assume that P k is true; that is, k 2 k. We want to use this to show that P k 1
is true. Now k 12 k 2 2k 1. Using the induction hypothesis (to replace k 2 ), we have
k 2 2k 1 k 2k 1 3k 1 k 1, since k 2. Therefore, k 12 k 1, which is exactly P k 1. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
(c) 22n1 1 is divisible by 3, for all n 1 . This is true. Let P n denote the statement that 22n1 1 is divisible by 3. Step 1: P 1 is the statement that 23 1 9 is divisible by 3, which is clearly true.
Step 2: Assume that P k is true; that is, 22k1 1 is divisible by 3. We want to use this to show that P k 1 is true. Now, 22k11 1 22k3 1 4 22k1 1 3 1 22k1 1 3 22k1 22k1 1 , which is
divisible by 3 since 22k1 1 is divisible by 3 by the induction hypothesis, and 3 22k1 is clearly divisible by 3. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
(d) The statement n 3 n 12 for all n 2 is false. The statement fails when n 2: 23 8 2 12 9.
(e) n 3 n is divisible by 3, for all n 2. This is true. Let P n denote the statement that n 3 n is divisible by 3. Step 1: P 2 is the statement that 23 2 6 is divisible by 3, which is clearly true.
Step 2: Assume that P k is true; that is, k 3 k is divisible by 3. We want to use this to show that P k 1 is true. Now k 13 k 1 k 3 3k 2 3k 1 k 1 k 3 3k 2 2k k 3 k 3k 2 2k k k 3 k 3 k 2 k . The term k 3 k is divisible by 3 by our induction hypothesis, and the term 3 k 2 k is clearly divisible by 3. Thus k 13 k 1 is divisible by 3, which is exactly P k 1. So by the Principle of Mathematical Induction, P n is true for all n.
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CHAPTER 13 Sequences and Series
(f) n 3 6n 2 11n is divisible by 6, for all n 1. This is true. Let P n denote the statement that n 3 6n 2 11n is divisible by 6. Step 1: P 1 is the statement that 13 6 12 11 1 6 is divisible by 6, which is clearly true.
Step 2: Assume that P k is true; that is, k 3 6k 2 11k is divisible by 6. We show that P k 1 is then also true. Now
k 13 6 k 12 11 k 1 k 3 3k 2 3k 1 6k 2 12k 6 11k 11 k 3 3k 2 2k 6 k 3 6k 2 11k 3k 2 9k 6 k 3 6k 2 11k 3 k 2 3k 2 k 3 6k 2 11k 3 k 1 k 2
In this last expression, the first term is divisible by 6 by our induction hypothesis. The second term is also divisible by 6. To see this, notice that k 1 and k 2 are consecutive natural numbers, and so one of them must be even (divisible by 2). Since 3 also appears in this second term, it follows that this term is divisible by 2 and 3 and so is divisible by 6. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
38. The induction step fails when k 2, that is, P 2 does not follow from P 1. If there are only two cats, Midnight and
Sparky, and we remove Sparky, then only Midnight remains. So at this point, we still know only that Midnight is black. Now removing Midnight and putting Sparky back leaves Midnight alone. So the induction hypothesis does not allow us to
conclude that Sparky is black.
13.6 THE BINOMIAL THEOREM 1. An algebraic expression of the form a b, which consists of a sum of two terms, is called a binomial. 2. We can find the coefficients in the expansion of a bn from the nth row of Pascal’s Triangle. So a b4 1a 4 4a 3 b 6a 2 b2 4ab3 1b4 .
n n! 4 4! . So 4. k k! n k! 3 3!1! 4 4 4 3 4. To expand a bn we can use the Binomial Theorem. Using this theorem we find a b4 a a b 0 1 4 4 4 4 2 2 ab3 b . a b 3 4 2
3. The binomial coefficients can be calculated directly using the formula
5. x y6 x 6 6x 5 y 15x 4 y 2 20x 3 y 3 15x 2 y 4 6x y 5 y 6
6. 2x 14 2x4 4 2x3 6 2x2 4 2x 1 16x 4 32x 3 24x 2 8x 1 2 3 4 1 1 1 1 4 1 1 4 4 3 2 x 4x 6x 4x x 4 4x 2 6 2 4 7. x x x x x x x x 8. x y5 x 5 5x 4 y 10x 3 y 2 10x 2 y 3 5x y 4 y 5
9. x 15 x 5 5x 4 10x 3 10x 2 5x 1 6 a b a 3 6a 2 a b 15a 2 b 20a ab b 15ab2 6 ab2 b b3 10. a 3 6a 2 ab 15a 2 b 20ab ab 15ab2 6b2 ab b3
or a 3 6a 52 b12 15a 2 b 20a 32 b32 15ab2 6a 12 b52 b3 . 5 5 4 3 2 11. x 2 y 1 x 2 y 5 x 2 y 10 x 2 y 10 x 2 y 5x 2 y 1 x 10 y 5 5x 8 y 4 10x 6 y 3 10x 4 y 2 5x 2 y 1 6 12. 1 2 16 6 15 2 15 14 2 20 13 2 2 15 12 4 6 1 4 2 23 1 6 2 30 40 2 60 24 2 8 99 70 2
SECTION 13.6 The Binomial Theorem
893
13. 2x 3y3 2x3 3 2x2 3y 3 2x 3y2 3y3 8x 3 36x 2 y 54x y 2 27y 3 3 2 3 14. 1 x 3 13 3 12 x 3 3 1 x 3 x 3 1 3x 3 3x 6 x 9 5 4 3 2 1 5 1 1 1 1 1 x x2 x2 x 15. 5 x 10 x 10 x x 5 x x x x x x 1 5 10 10 5 72 2 12 5x x 52 x x x x x 2 x 3 x 4 x 5 x 5 x 1 x5 16. 2 25 5 24 10 23 10 22 5 2 3240x 20x 2 5x 3 58 x 4 32 2 2 2 2 2 2 6 5 4! 6! 6 15 17. 4! 2! 2 1 4! 4 8 7 6 5! 8 8! 8 7 56 18. 3! 5! 3 2 1 5! 3 100 99 98! 100 100! 19. 4950 98! 2! 98! 2 1 98 10 9 8 7 6 5! 10 10! 20. 3 2 7 6 252 5! 5! 5 4 3 2 1 5! 5 3 2! 4 3 2! 3! 4! 3 4 18 21. 1! 2! 2! 2! 1 2! 2 1 2! 1 2 5! 5 4 3! 5 4 3! 5 5 5! 22. 10 10 100 2! 3! 3! 2! 2 1 3! 3! 2 1 2 3 23. 50 51 52 53 54 55 1 15 25 32 5 5! 5! 5! 5 5 5 5 5 5! 24. 1 0. Notice that the first and sixth 1 1! 4! 2! 3! 3! 2! 4! 1! 0 1 2 3 4 5 terms cancel, as do the second and fifth terms and the third and fourth terms. 25. x 2y4 40 x 4 41 x 3 2y 42 x 2 4y 2 43 x 8y 3 44 16y 4 x 4 8x 3 y 24x 2 y 2 32x y 3 16y 4 26. 1 x5 50 15 51 14 x 52 13 x 2 53 12 x 3 54 1 x 4 55 x 5 1 5x 10x 2 10x 3 5x 4 x 5 2 3 4 5 6 1 6 6 4 1 6 6 6 5 1 6 3 1 6 2 1 6 6 1 1 27. 1 1 1 1 1 1 1 x x x x x x x 2 0 1 3 4 5 6 15 6 20 15 6 1 1 2 3 4 5 6 x x x x x x 4 2 3 4 4 4 4 2 4 3 2 28. 2A B 0 2A 1 2A B 2 2A2 B 2 43 2A B 2 44 B 2 29.
30.
31. 32.
16A4 32A3 B 2 24A2 B 4 8AB 6 B 8 20 19 The first three terms in the expansion of x 2y20 are 20 x 20 , 20 2y 40x 19 y, and 0 x 1 x 20 18 2 18 2 2 x 2y 760x y . 30 12 30 12 29 The first four terms in the expansion of x 12 1 are 30 x 15 , 30 x x 1 30x 292 , 0 1 12 27 3 30 12 28 2 x x 1 435x 14 , and 30 1 4060x 272 . 2 3 25 23 13 24 253 The last two terms in the expansion of a 23 a 13 are 25 a 25a 263 , and 25 a 253 . 24 a 25 a 40 1 40 40 , 40 x 39 1 The first three terms in the expansion of x x are 40 x 40x 38 , and 0 1 x x 40 38 1 2 780x 36 . x 2 x
894
CHAPTER 13 Sequences and Series
18 33. The middle term in the expansion of x 2 1 occurs when both terms are raised to the 9th power. So this term is 18 2 9 9 x 1 48,620x 18 . 9 16 4 16 16 34. The fifth term in the expansion of ab 120 is 20 4 ab 1 4845a b . 2 23 2 23 35. The 24th term in the expansion of a b25 is 25 23 a b 300a b . 3 27 3 27 36. The 28th term in the expansion of A B30 is 30 27 A B 4060A B . 1 99 99 37. The 100th term in the expansion of 1 y100 is 100 99 1 y 100y . 25 2 24 1 1 25x 47 . x is 25 38. The second term in the expansion of x 2 1 x x
39. The term that contains x 4 in the expansion of x 2y10 has exponent r 4. So this term is 10 4 104 13,440x 4 y 6 . 4 x 2y 12 r 40. The rth term in the expansion of 2y is 12 2 y 12r . The term that contains y 3 occurs when 12 r 3 r 9 3 2 y 3520 2y 3 . r 9 . Therefore, the term is 12 9 r 2 12r 12 r 242r 41. The rth term is 12 r a b . Thus the term that contains b8 occurs where 24 2r 8 r 8. So r a b 8 8 8 8 the term is 12 8 a b 495a b . 8r 8r 8r xr 1 r8 8r 8r r8 8r x 2r8 . So the term that does not contain x occurs 42. The rth term is r8 8xr 2x 2 x 2 4 1 when 2r 8 0 r 4. Thus, the term is 84 8x4 17,920. 2x 43. x 4 4x 3 y 6x 2 y 2 4x y 3 y 4 x y4
44. x 15 5 x 14 10 x 13 10 x 12 5 x 1 1 [x 1 1]5 x 5 45. 8a 3 12a 2 b 6ab2 b3 30 2a3 31 2a2 b 32 2ab2 33 b3 2a b3 4 4 3 2 46. x 8 4x 6 y 6x 4 y 2 4x 2 y 3 y 4 40 x 2 41 x 2 y 42 x 2 y 2 43 x 2 y 3 44 y 4 x 2 y h 3x 2 3xh h 2 x 3 3x 2 h 3xh 2 h 3 x 3 3x 2 h 3xh 2 h 3 x h3 x 3 3x 2 3xh h 2 47. h h h h 4 4 4 3 x 1 x h 42 x 2 h 2 43 xh 3 44 h 4 x 4 x 4 4x 3 h 6x 2 h 2 4xh 3 h 4 x 4 x h4 x 4 0 48. h h h 3 6x 2 h 4xh 2 h 3 3 2 2 3 4 h 4x 4x h 6x h 4xh h 4x 3 6x 2 h 4xh 2 h 3 h h 100 49. 101100 1 001100 . Now the first term in the expansion is 100 1, the second term is 0 1 100 98 100 99 2 1 1 001 1, and the third term is 2 1 001 0495. Now each term is nonnegative, so 101100 1 001100 1 1 0495 2. Thus 101100 2. n n n n n! n! n! n! 1. 1. Therefore, 50. . 0!n! 1 n! n!0! n! 1 n 0 0 n n n n 1! n n n! n n 1! n! n. n. Therefore, 51. 1! n 1! 1 n 1! 1 n1 1 n 1! 1! n 1! 1 n n n. 1 n1
SECTION 13.6 The Binomial Theorem
52.
n r
n! n! r! n r ! n r! r!
n
n n r
895
for 0 r n.
n! n! . r 1! [n r 1]! r! n r! n! n! r n! n r 1 n! (b) r 1! [n r 1]! r! n r ! r r 1! n r 1! r! n r 1 n r ! r n! n r 1 n! r! n r 1! r! n r 1! Thus a common denominator is r! n r 1!.
53. (a)
n r 1
r
(c) Therefore, using the results of parts (a) and (b), r n! n! n n n! n r 1 n! r! n r 1! r! n r 1! r 1 r r 1! [n r 1]! r! n r!
n! r n r 1 n! n 1 r n! n r 1 n! n1 n 1! r! n r 1! r! n r 1! r! n 1 r ! r ! n 1 r! r n 54. Let P n be the proposition that is an integer for the number n, 0 r n. r n 0 Step 1: Suppose n 0. If 0 r n, then r 0, and so 1, which is obviously an integer. Therefore, P 0 r 0 is true. k1 Step 2: Suppose that P k is true. We want to use this to show that P k 1 must also be true; that is, is an r k k k 1 by the key property of binomial coefficients integer for 0 r k 1. But we know that r 1 r r k k (see Exercise 49). Furthermore, and are both integers by the induction hypothesis. Since the sum of two r 1 r k 1 integers is always an integer, must be an integer. Thus, P k 1 is true if P k is true. So by the Principal of r n Mathematical induction, is an integer for all n 0, 0 r n. r
55. By the Binomial Theorem, the volume of a cube of side x 2 inches is 3 3 3 2 3 3 3 x x 2 x 22 2 x 3 3 2x 2 3 4x 8 x 3 6x 2 12x 8. The volume x 23 0 1 2 3 of a cube of side x inches is x 3 , so the difference in volumes is x 3 6x 2 12x 8 x 3 6x 2 12x 8 cubic inches. n pr q nr . In this case, n r p 09, q 01, n 5, and r 3, so the probability that the archer hits the target exactly 3 times in 5 attempts is 5 5 3 53 P 093 012 00729. 09 01 2 53
56. By the Binomial Theorem, the coefficient of pr in the expansion of p qn is
57. Notice that 100!101 100!100 100! and 101!100 101 100!100 101100 100!100 . Now
100! 1 2 3 4 99 100 and 101100 101 101 101 101. Thus each of these last two expressions consists of 100 factors multiplied together, and since each factor in the product for 101100 is larger than each factor in the product for
100!, it follows that 100! 101100 . Thus 100!100 100! 100!100 101100 . So 100!101 101!100 .
896
CHAPTER 13 Sequences and Series
58.
11 2 1214 13318 1 4 6 4 1 16 1 5 10 10 5 1 32
Conjecture: The sum is 2n . Proof: 2n 1 1n
n
n
11 1n1
n
12 1n2
n
1n 10 1 2 n n n n n 0 1 2 n n n n n n n 0 n 1 59. 0 0 1 1 1 1 1 1n1 12 1n2 1n 10 0 1 2 n n n n n n 1n 1k k n 0 1 2 0
10 1n
CHAPTER 13 REVIEW 1 4 9 16 100 n2 12 22 32 42 102 . Then a1 , a2 , a3 , a4 , and a10 . n1 11 2 21 3 31 4 41 5 10 1 11 8 2n 21 22 23 24 . Then a1 11 2, a2 12 2, a3 13 , a4 14 4, and an 1n n 1 2 3 3 4 210 1024 512 a10 110 . 10 10 5 1 2 1n 1 11 1 12 1 13 1 an , a3 . Then a1 0, a2 0, 3 3 3 8 4 n 1 2 33 1 2 1 14 1 110 1 , and a . a4 10 3 3 64 32 500 4 10 n n 1 1 1 1 2 2 1 3 3 1 4 4 1 . Then a1 1, a2 3, a3 6, a4 10, and an 2 2 2 2 2 10 10 1 a10 55. 2 654 2n! 2 1! 2 2! 2 3! an n . Then a1 1 1, a2 2 3, a3 3 15, 2 n! 8 2 1! 2 2! 2 3! 8765 2 4! 2 10! 105, and a10 10 654,729,075. a4 4 16 2 4! 2 10! n1 11 21 3! 31 4! an 3, a3 6, . Then a1 1, a2 2! 1! 2! 2! 2 2 2 2 41 5! 10 1 11! 10, and a10 55. a4 2! 3! 2! 9! 2 2 an an1 2n 1 and a1 1. Then a2 a1 4 1 4, a3 a2 6 1 9, a4 a3 8 1 16, a5 a4 10 1 25, a6 a5 12 1 36, and a7 a6 14 1 49. 1 1 1 1 1 a a a a a a ,a 4 ,a 5 , and an n1 and a1 1. Then a2 1 , a3 2 , a4 3 n 2 2 3 6 4 24 5 5 120 6 6 720 1 a . a7 6 7 5040 an an1 2an2 , a1 1 and a2 3. Then a3 a2 2a1 5, a4 a3 2a2 11, a5 a4 2a3 21, a6 a5 2a4 43, and a7 a6 2a5 85.
1. an 2.
3.
4.
5.
6.
7.
8.
9.
CHAPTER 13
Review
897
12 3an1 and a1 3 312 . Then a2 3a1 3 3 3 312 334 , 3a2 3 334 374 378 , a4 3a3 3 378 3158 31516 , 3a4 3 31516 33116 33132 , a6 3a5 3 33132 36332 36364 , 3a6 3 36364 312764 3127128 .
10. an a3 a5 a7
11. (a) a1 2 1 5 7, a2 2 2 5 9,
a3 2 3 5 11, a4 2 4 5 13,
a5 2 5 5 15 an
(b)
5 5 5 5 5 5 12. (a) a1 1 , a2 2 , a3 3 , 2 4 8 2 2 2 5 5 5 5 , a5 5 a4 4 16 32 2 2 an
(b)
14 12 10 8 6 4 2 0
2 1
1
2
3
4
0
5 n
1
2
3
4
5 n
(c) S5 7 9 11 13 15 55
5 5 5 5 5 155 (c) S5 2 4 8 16 32 32
(d) This sequence is arithmetic with common difference
(d) This sequence is geometric with common ratio 12 .
2. 31 3 32 9 33 27 , 13. (a) a1 2 , a2 3 , a3 4 4 8 16 2 2 2
7 1 2 , a2 4 3, 2 2 2 5 3 3 4 5 a3 4 , a4 4 2, a5 4 2 2 2 2 2
14. (a) a1 4
34 81 35 243 , a5 6 a4 5 32 64 2 2 (b)
(b)
an
3
4
2
2
0
(c) S5
an
1 1
2
3
4
5 n
633 3 9 27 81 243 4 8 16 32 64 64
(d) This sequence is geometric with common ratio 32 .
0
1
2
3
4
5 n
5 3 25 7 (c) S5 3 2 2 2 2 2 (d) This sequence is arithmetic with common difference 12 .
15. 5 55 6 65 . Since 55 5 6 55 65 6 05, this is an arithmetic sequence with a1 5 and d 05. Then a5 a4 05 7. 16. 2 2 2 3 2 4 2 . Since 2 2 2 3 2 2 2 4 2 3 2 2, this is an arithmetic sequence with a1 2 and d 2. Then a5 a4 2 4 2 2 5 2. 17. t 3 t 2 t 1 t . Since t 2 t 3 t 1 t 2 t t 1 1, this is an arithmetic sequence with a1 t 3 and d 1. Then a5 a4 1 t 1. 4 2, this is a geometric sequence with a 2 and r 2. Then 18. 2 2 2 2 4 . Since 2 2 2 2 1 2 2 2 a5 a4 r 4 2 4 2. 1 t2 t 1 1 1 19. t 3 t 2 t 1 . Since 3 2 , this is a geometric sequence with a1 t 3 and r . Then a5 a4 r 1 . t t t t t t
898
CHAPTER 13 Sequences and Series
3 2 20. 1 32 2 52 . Since 32 1 2 32 , and 2 , the series is neither arithmetic nor geometric. 1 32 1
1
2
21. 34 12 13 29 . Since 23 31 91 23 , this is a geometric sequence with a1 34 and r 23 . Then 4 2 2 4 a5 a4 r 9 3 27 .
2
3
1 1 2 1 1 1 a a and 22. a 1 2 . Since 1 a a a 1 a 1 1 1 a5 a4 r 2 3 . a a a 6i 12 23. 3 6i 12 24i . Since 2i, 3 6i ratio r 2i.
1 1 , this is a geometric sequence with a1 a and r . Then a a
2 2i 24i 2 2i, 2i, this is a geometric sequence with common i 12 i
a 2 2i 24. The sequence 2, 2 2i, 4i, 4 4i, 8, is geometric (where i 2 1), since 2 1 i, a1 2 4i 2 2i 8 8i a 1 1 i i 4i 4 4i a3 1 i, 4 1 1 2 1 1 i, a2 2 2i 2 2i 2 2i 8 a3 4i i i i i 8 4 4i 32 32i 8 a5 1 i. Thus the common ratio is i 1 and the first term is 2. So the a4 4 4i 4 4i 4 4i 32 nth term is an a1r n1 2 1 in1 .
25. a6 17 a 5d and a4 11 a 3d. Then, a6 a4 17 11 a 5d a 3d 6 6 2d d 3. Substituting into 11 a 3d gives 11 a 3 3, and so a 2. Thus a2 a 2 1 d 2 3 5. 26. a20 96 and d 5. Then 96 a20 a 19 5 a 95 a 96 95 1. Therefore, an 1 5 n 1. 2 27. a3 9 and r 32 . Then a5 a3 r 2 9 32 81 4.
a5 3 28. a2 10 and a5 1250 27 . Then r a 2 n1 an ar n1 6 53 .
1250 27 125 r 5 and a a a2 10 6. Therefore, 1 27 3 5 10 r 3
29. (a) An 32,000 105n1
(b) A1 $32,000, A2 32,000 1051 $33,600, A3 32,000 1052 $35,280, A4 32,000 1053 $37,044, A5 32,000 1054 $38,89620, A6 32,000 1055 $40,84101, A7 32,000 1056 $42,88306,
A8 32,000 1057 $45,02721
30. (a) An 35,000 1,200 n 1 (b) A8 35,000 1,200 7 $43,400. The salary for this teacher is higher for the first 6 years is higher, but from the 7th year on, the salary of the teacher in Exercise 29 is higher. 31. Let an be the number of bacteria in the dish at the end of 5n seconds. So a0 3, a1 3 2, a2 3 22 , a3 3 23 , . Then, clearly, an is a geometric sequence with r 2 and a 3. Thus at the end of 60 5 12 seconds, the number of bacteria is a12 3 212 12,288.
32. Let d be the common difference in the arithmetic sequence a1 , a2 , a3 , , so that an a1 n 1 d, n 1, 2, 3, , and let e be the common difference for b1 , b2 , b3 , , so that bn b1 n 1 e. Then an bn a1 n 1 a b1 n 1 e a1 b1 n 1 d e, n 1, 2, 3, . Thus a1 b1 a2 b2 is an arithmetic sequence with first term a1 b1 and common difference d e.
CHAPTER 13
Review
899
33. Suppose that the common ratio in the sequence a1 a2 a3 is r . Also, suppose that the common ratio in the sequence b1 b2 b3 is s. Then an a1r n1 and bn b1 s n1 , n 1 2 3 . Thus an bn a1r n1 b1 s n1 a1 b1 rsn1 . So the sequence a1 b1 a2 b2 a3 b3 is geometric with first term a1 b1 and common ratio rs. 34. (a) Yes. If the common difference is d, then an a1 n 1 d. So an 2 a1 2 n 1 d, and thus the sequence a1 2 a2 2 a3 2 is an arithmetic sequence with the common difference d, but with the first term a1 2. (b) Yes. If the common ratio is r, then an a1 r n1 . So 5an 5a1 r n1 , and the sequence 5a1 5a2 5a3 is also geometric, with common ratio r, but with the first term 5a1 .
35. (a) 6 x 12 is arithmetic if x 6 12 x 2x 18 x 9. 12 x x 2 72 x 6 2. (b) 6 x 12 is geometric if 6 x 36. (a) 2 x y 17 is arithmetic. Therefore, 15 17 2 a4 a1 a 3d a 3d. So d 5, and hence, x a d 2 5 7 and y a 2d 2 2 5 12. 13 a4 ar 3 3 r 17 r . So (b) 2 x y 17 is geometric. Therefore, 17 2 2 a1 a 2 13 23 17 13 17 23 1713 and y a ar 2 2 2 2 213 1723 . x a2 ar 2 17 3 2 2 2 37.
38.
6
2 2 2 2 2 k3 k 1 3 1 4 1 5 1 6 1 16 25 36 49 126
4
2i
21
22
23
24
4
6
8
210 140 126 120
596
i 1 2i 1 2 1 1 2 2 1 2 3 1 2 4 1 2 3 5 7 105 105 6 k1 0 1 2 3 4 5 39. k1 k 1 2 2 2 3 2 4 2 5 2 6 2 7 2 2 6 16 40 96 224 384 5 40. m1 3m2 31 30 31 32 33 13 1 3 9 27 121 3 10 2 2 2 2 2 2 2 2 2 2 2 41. k1 k 1 0 1 2 3 4 5 6 7 8 9 42. 43. 44.
100
1 1 1 1 1 1 1 1 j2 j 1 1 2 3 4 5 98 99
50
k1
10
3k
2k1
3 32 33 34 349 350 2 3 4 5 50 51 2 2 2 2 2 2
2 n 2 1 2 2 2 3 2 9 2 10 n1 n 2 1 2 2 2 3 2 9 2 10 2
45. 3 6 9 12 99 3 1 3 2 3 3 3 33 2 46. 12 22 33 1002 100 k1 k
33
k1 3k
47. 1 23 2 24 3 25 4 26 100 2102
48.
1 212 2 222 3 232 4 242 100 21002 k2 100 k1 k 2
1 1 1 1 1 999 k1 k k 1 12 23 34 999 1000
49. 1 09 092 095 is a geometric series with a 1 and r
09 09. Thus, the sum of the series is 1
1 096 1 0531441 468559. 1 09 01 50. 3 37 44 10 is an arithmetic series with a 3 and d 07. Then 10 an 3 07 n 1 07 n 1 7 S6
143 n 11. So the sum of the series is S11 11 2 3 10 2 715. 51. 5 2 5 3 5 100 5 is an arithmetic series with a 5 and d 5. Then 100 5 an 5 5 n 1 n 100. So the sum is S100 100 5 100 5 50 101 5 5050 5. 2
900
CHAPTER 13 Sequences and Series
52. 13 23 1 43 33 is an arithmetic series with a 13 and d 13 . Then an 33 13 13 n 1 n 99. So the 2 99 1 99 100 1650. sum is S99 99 2 3 2 3 3 6 53. n0 3 4n is a geometric series with a 3, r 4, and n 7. Therefore, the sum of the series is S7 3
54.
8
1 47 35 1 47 9831. 1 4
k2 is a geometric series with a 7, r 512 , and n 9. Thus, the sum of the series is k0 7 5 1 592 1 5 1 592 5 55 1 592 7 S9 7 7
15 1 5 1 5 1 5 74 1 625 5 5 3125 5467 1092 5 79088
4 8 is a geometric series with a 1 and r 2 . Therefore, it is convergent with sum 55. 1 25 25 125 5 5 1 a . S 1r 7 1 2 5
56. 01 001 0001 00001 is an infinite geometric series with a 01 and r 01. Therefore, it is convergent with 1 01 . sum S 1 01 9
57. 5 5 101 5 1012 5 1013 is an infinite geometric series with a 5 and r 101. Because r 101 1, the series diverges. 1 1 1 58. 1 12 13 32 is an infinite geometric series with a 1 and r . Thus, it is convergent with sum 3 3 3 1 3 S 12 3 3 . 31 1 1 3
2
59. 1 98 98 diverges.
3 98 is an infinite geometric series with a 1 and r 98 . Because r 98 1, the series
60. a ab2 ab4 ab6 is an infinite geometric series with first term a and common ratio b2 . Because b 1, b2 1, a . 1 b2 61. We have an arithmetic sequence with a 7 and d 3. Then n n n Sn 325 [2a n 1 d] [14 3 n 1] 11 3n 650 3n 2 11n 3n 50 n 13 0 2 2 2 is inadmissible). Thus, 13 terms must be added. n 13 (because n 50 3 and so the series is convergent with sum S
62. We have a geometric series with S3 52 and r 3. Then 52 S3 a 3a 9a 13a a 4, and so the first term is 4. 1 215 2 215 1 65,534, and so the total 63. This is a geometric sequence with a 2 and r 2. Then S15 2 12 number of ancestors is 65,534. 10816 1 12,500 10816 1 $30,32428. 64. R 1000, i 008, and n 16. Thus, A 1000 108 1 65. A 10,000, i 003, and n 4. Thus, 10,000 R
10,000 003 1034 1 R $239027. 003 1034 1
009 00075. 12 60,000 00075 (a) If the period is 30 years, n 360 and R $48277. 1 10075360
66. A 60,000 and i
CHAPTER 13
(b) If the period is 15 years, n 180 and R
Review
901
60,000 00075 $60856. 1 10075180
67. Let P n denote the statement that 1 4 7 3n 2
n 3n 1 . 2
1 [3 1 1] 12 , which is true. 2 2 k 3k 1 Step 2: Assume that P k is true; that is, 1 4 7 3k 2 . We want to use this to show that 2 P k 1 is true. Now Step 1: P 1 is the statement that 1
1 4 7 10 3k 2 [3 k 1 2]
k 3k 1 3k 1 2
induction hypothesis
3k 2 k 6k 2 k 3k 1 6k 2 2 2 2
3k 2 5k 2 k 1 3k 2 2 2 k 1 [3 k 1 1] 2
Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
1 1 1 n 1 . 13 35 57 2n 1 2n 1 2n 1 1 1 Step 1: P 1 is the statement that , which is true. 13 211 1 1 1 1 k Step 2: Assume that P k is true; that is, . We want to use this to 13 35 57 2k 1 2k 1 2k 1 show that P k 1 is true. Now 1 1 1 1 1 13 35 57 2k 1 2k 1 2k 1 2k 3
68. Let P n denote the statement that
k 1 2k 1 2k 1 2k 3
k 2k 3 1 2k 2 3k 1 2k 1 2k 3 2k 1 2k 3
k 1 k 1 k 1 2k 1 2k 3 2 k 1 1 2k 1 2k 3
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
induction hypothesis
902
CHAPTER 13 Sequences and Series
1 1 1 1 1 1 1 n 1. 69. Let P n denote the statement that 1 1 2 3 n
Step 1: P 1 is the statement that 1 11 1 1, which is clearly true. 1 1 1 1 1 1 1 k 1. We want to use this to Step 2: Assume that P k is true; that is, 1 1 2 3 k show that P k 1 is true. Now 1 1 1 1 1 1 1 1 1 1 1 2 3 k k 1 1 1 1 1 1 1 1 1 1 1 1 2 3 k k 1 1 k 1 1 induction hypothesis k1 k 1 1
Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
70. Let P n denote the statement that 7n 1 is divisible by 6.
Step 1: P 1 is the statement that 71 1 6 is divisible by 6, which is clearly true.
Step 2: Assume that P k is true; that is, 7k 1 is divisible by 6. We want to use this to show that P k 1 is true. Now 7k1 1 7 7k 1 7 7k 7 6 7 7k 1 6, which is divisible by 6. This is because 7k 1 is divisible by
6 by the induction hypothesis, and clearly 6 is divisible by 6. Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 71. an1 3an 4 and a1 4. Let P n denote the statement that an 2 3n 2. Step 1: P 1 is the statement that a1 2 31 2 4, which is clearly true.
Step 2: Assume that P k is true; that is, ak 2 3k 2. We want to use this to show that P k 1 is true. Now ak1 3ak 4 3 2 3k 2 4
definition of ak1 induction hypothesis
2 3k1 6 4 2 3k1 2
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.
72. Let P n denote the statement that F4n is divisible by 3. Step 1: Show that P 1 is true, but P 1 is true since F4 3 is divisible by 3. Step 2: Assume that P k is true; that is, F4k is divisible by 3. We want to use this to show that P k 1 is true. Now, F4k1 F4k4 F4k2 F4k3 F4k F4k1 F4k1 F4k2 F4k F4k1 F4k1 F4k F4k1 2 F4k 3 F4k1 , which is divisible by 3 because F4k is divisible by 3 by our induction hypothesis, and 3 F4k1 is clearly divisible by 3. Thus, P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 5! 54 54 5 5 5! 73. 10 10 100 2! 3! 3! 2! 2 2 2 3 10! 10 9 10 9 8 7 10 10 10! 74. 45 210 255 2! 8! 6! 4! 2 432 2 6 75. 5k0 5k 50 51 52 53 54 55 2 0!5!5! 1!5!4! 2!5!3! 2 1 5 10 32 8 2 80 88 2 81 87 2 82 86 2 83 85 84 84 2 2 82 2 282 2 562 702 12,870 76. 8k0 8k 8k 77. A B3 30 A3 31 A2 B 32 AB 2 33 B 3 A3 3A2 B 3AB 2 B 3
CHAPTER 13
Test
903
78. x 25 50 x 5 51 x 4 2 52 x 3 22 53 x 2 23 54 x 24 55 25
x 5 10x 4 40x 3 80x 2 80x 32 6 79. 1 x 2 60 16 61 15 x 2 62 14 x 4 63 13 x 6 64 12 x 8 65 x 10 66 x 12 80. 81. 82.
83.
1 6x 2 15x 4 20x 6 15x 8 6x 10 x 12 2x y4 40 2x4 41 2x3 y 42 2x2 y 2 43 2x y 3 44 y 4 16x 4 32x 3 y 24x 2 y 2 8x y 3 y 4 3 19 3 19 The 20th term is 22 19 a b 1540a b . 20 23 20 23 19 The first three terms in the expansion of b23 b13 b b are 20 b403 , 20 0 1 18 2 b23 b13 20b373 , and 20 b13 190b343 . 2 r 10r . The term that contains A6 occurs when r 6. Thus, the The rth term in the expansion of A 3B10 is 10 r A 3B 10 6 term is 6 A 3B4 210A6 81B 4 17,010A6 B 4 .
CHAPTER 13 TEST 1. an 2n 2 n a1 1, a2 6, a3 15, a4 28, a5 45, a6 66, and S6 1 6 15 28 45 66 161.
2. an1 3an n, a1 2 a2 3 2 1 5, a3 3 5 2 13, a4 3 13 3 36, a5 3 36 4 104, and a6 3 104 5 307.
3. (a) The common difference is d 5 2 3. (b) an 2 n 1 3
(c) a35 2 3 35 1 104
3 1. 4. (a) The common ratio is r 12 4 n1 (b) an a1r n1 12 14 101 (c) a10 12 14 38 65,3536 4
1
1 1. r 15 , so a5 ra4 25 5. (a) a1 25, a4 15 . Then r 3 5 25 125 8 1 15 97,656 58 1 (b) S8 25 1 12,500 3125 1 5
6. (a) a1 10 and a10 2, so 9d 8 d 89 and a100 a1 99d 10 88 78. 10 2 10 9 8 [2a (b) S10 10 60 1 d] 10 2 2 9
7. Let the common ratio for the geometric series a1 a2 a3 be r, so that an a1r n1 , n 1, 2, 3, . Then 2 n1 an2 a1r n1 a12 r 2 . Therefore, the sequence a12 , a22 , a32 , is geometric with common ratio r 2 . 5 2 1 12 1 22 1 32 1 42 1 52 0 3 8 15 24 50 8. (a) n1 1 n (b) 6n3 1n 2n2 13 232 14 242 15 252 16 262 2 4 8 16 10 2
3
9
9. (a) The geometric sum 13 22 23 24 210 has a 13 , r 23 , and n 10. So 3 3 3 3 10 1 3 1 1024 58,025 . S10 13 123 123 3 59,049 59,049
904
FOCUS ON MODELING 1 1 1 has a 1 and r 212 1 . Thus, (b) The infinite geometric series 1 12 2 2 232 2 2 2 21 1 S 2 2. 11 2
21
21
21
n n 1 2n 1 . 6 123 , which is true. Step 1: Show that P 1 is true. But P 1 says that 12 6 k k 1 2k 1 . We want to use this to show that Step 2: Assume that P k is true; that is, 12 22 32 k 2 6 P k 1 is true. Now
10. Let P n denote the statement that 12 22 32 n 2
12 22 32 k 2 k 12
k k 1 2k 1 k 12 6
induction hypothesis
k 1 2k 2 k 6k 6 k 1
k k 1 2k 1 6 k 12 6 6 2 2 k 1 2k 7k 6 k 1 2k k 6k 6 6 6 k 1 k 2 2k 3 k 1 [k 1 1] [2 k 1 1] 6 6
Thus P k 1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 5 2 3 4 5 11. 2x y 2 50 2x5 51 2x4 y 2 52 2x3 y 2 53 2x2 y 2 54 2x y 2 55 y 2 32x 5 80x 4 y 2 80x 3 y 4 40x 2 y 6 10x y 8 y 10
3 7 3 3 12. 10 3 3x 2 120 27x 128 414,720x
13. (a) Each week he gains 24% in weight, that is, 024an . Thus, an1 an 024an 124an for n 1.
a0 is given to be 085 lb. Then a0 085, a1 124 085, a2 124 124 085 1242 085, a3 124 1242 085 1243 085, and so on. So we can see that an 085 124n .
(b) a6 124a5 124 124a4 1246 a0 1246 085 31 lb (c) The sequence a1 a2 a3 is geometric with common ratio 124.
FOCUS ON MODELING Modeling with Recursive Sequences 00365 00001. Thus the amount in the account at 365 the end of the nth day is An 10001An1 with A0 $275,000. (b) A0 $275,000, A1 10001A0 10001 275,000 $275,02750,
1. (a) Since there are 365 days in a year, the interest earned per day is
A2 10001A1 10001 10001A0 100012 A0 $275,05500,
A3 10001A2 100013 A0 $275,08251, A4 100014 A0 $275,11002, A5 100015 A0 $275,13753,
A6 100016 A0 $275,16504, A7 100017 A0 $275,19256 (c) An 10001n 275,000 2. (a) Tn Tn1 15 with T1 5.
(b) T1 5, T2 T1 15 5 15 65, T3 T2 15 5 15 15 5 2 15 80, T4 T3 15 5 2 15 15 5 3 15 95, T5 T4 15 5 3 15 15 5 4 15 110, T6 T5 15 5 4 15 15 5 5 15 125
Modeling with Recursive Sequences
905
(c) This is an arithmetic sequence with Tn 5 15 n 1.
(d) Tn 65 5 15n 15 615 15n n 41So Sheila swims 65 minutes on the 41st day.
(e) Using the partial sum of an arithmetic sequence, she swims for 30 2 [2 5 30 1 15] 15 535 8025 minutes 13 hours 225 minutes. 003 00025. Thus the amount in the account at the 3. (a) Since there are 12 months in a year, the interest earned per day is 12 end of the nth month is An 10025An1 100 with A0 $100. (b) A0 $100, A1 10025A0 100 10025 100 100 $20025, A2 10025A1 100 10025 10025 100 100 100 100252 100 10025 100 100 $30075, A3 10025A2 100 10025 100252 100 10025 100 100 100 100253 100 100252 100 10025 100 100 $40150, A4 10025A3 100 10025 100253 100 100252 100 10025 100 100 100 100254 100 100253 100 100252 100 10025 100 100 $50251
(c) An 10025n 100 100252 100 10025 100 100, the partial sum of a geometric series, so 10025n1 1 1 10025n1 An 100 100 . 1 10025 00025
1002561 1 $658083. 00025 4. (a) The amount An of pollutants in the lake in the nth year is 30% of the amount from the preceding year (030An1 ) plus the amount discharged that year (2400 tons). Thus An 030An1 2400. (d) Since 5 years is 60 months, we have A60 100
(b) A0 2400, A1 030 2400 2400 3120,
A2 030 [030 2400 2400] 2400 0302 2400 2400 2400 2400 3336, A3 030 0302 2400 2400 2400 2400 2400 0033 2400 0302 2400 2400 2400 2400 34008, A4 030 0033 2400 0302 2400 2400 2400 2400 2400
0034 2400 0033 2400 0302 2400 2400 2400 2400 34202
(c) An is the partial sum of a geometric series, so An 2400
1 030n1
2400
1 030 34286 1 030n1
1 030n1 070
(e) 4000 2000
1 0307 0 34278 tons. The sum of a geometric 0 070 1 34286 tons. series, is A 2400 070 5. (a) Un 105Un1 010 105Un1 110 105 Un1 1155Un1 with U0 5000. (b) U0 $5000, U1 1155U0 $5775, (d) A6 2400
U2 1155U1 1155 1155 5000 11552 5000 $667013, U3 1155U2 1155 11552 5000 11553 5000 $770399, U4 1155U3 1155 11553 5000 11554 5000 $889811.
(c) Using the pattern we found in part (b), we have Un 1155n 5000.
10
20
906
FOCUS ON MODELING
(d) U10 115510 5000 $21,12467. 6. (a) In the nth year since Victoria’s initial deposit the amount Vn in her CD is the amount from the preceding year (Vn1 ), plus the 5% interest earned on that amount (005Vn1 ), plus $500 times the number of years since her initial deposit (500n). Thus Vn 105Vn1 500n. (b) Using the sequence mode on a calculator and scrolling down, we see that U31 435,52342 and U32 503,02955, while V 31 455,67849 and V 32 494,46242. Thus, Ursula’s savings surpass Victoria’s savings in the 32nd year.
14
COUNTING AND PROBABILITY
14.1 COUNTING 1. The Fundamental Counting Principle says that if one event can occur in m ways and a second event can occurs in n ways, then the two events can occur in order in m n ways. So if you have two choices for shoes and three choices for hats, then the number of different shoe-hat combinations you can wear is 2 3 6. 2. The number of ways of arranging r objects from n objects in order is called the number of permutations of n objects taken r n! . at a time, and is given by the formula P n r n r! 3. The number of ways of choosing r objects from n objects is called the number of combinations of n objects taken r at a n! . time, and is given by the formula Cn r r! n r! 4. (a) False. When counting combinations order does not matter. (b) True. When counting permutations order matters. (c) False. For a set of n distinct objects, the number of different combinations of these objects is less than the number of different permutations. (d) True. If we have a set with 5 distinct objects then the number of different ways of choosing 2 members of this set is the same as the number of ways of choosing 3 members. 9! 9! 9 8 72 7! 9 2!
5. P 8 3
8! 8! 8 7 6 336 5! 8 3!
6. P 9 2
7. P 11 4
11! 11! 11 10 9 8 7920 7! 11 4!
8. P 10 5
10! 10! 30,240 5! 10 5!
10. P 99 3
99! 99! 941,094 99 3! 96!
9. P 100 1 11. C 8 3
100! 100! 100 100 1! 99!
8! 876 8! 56 3! 8 3! 3! 5! 321
13. C 11 4
11 10 9 8 11! 330 4! 7! 4321
15. C 100 1
100! 100 1 100 1! 99!
12. C 9 2
9! 9! 98 36 2! 9 2! 2! 7! 21
14. C 10 5
10! 10! 252 5! 10 5! 5! 5!
16. C 99 3
99! 99 98 97 156,849 3! 96! 321
17. By the Fundamental Counting Principle, the number of possible single-scoop ice cream cones is number of ways to number of ways to 4 3 12. choose the flavor choose the type of cone 18. By the Fundamental Counting Principle, the possible number of 3-letter words is number of ways to number of ways to number of ways to . choose the 1st letter choose the 2nd letter choose the 3rd letter
(a) Since repetitions are allowed, we have 26 choices for each letter. Thus, there are 26 26 26 17,576 words.
(b) Since repetitions are not allowed, we have 26 choices for the 1st letter, 25 choices for the 2nd letter, and 24 choices for the 3rd letter. Thus there are 26 25 24 15,600 words.
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CHAPTER 14 Counting and Probability
19. (a) By the Fundamental Counting Principle, the possible number of ways 8 horses can complete a race, assuming no ties in any position, is number of ways to number of ways to number of ways to 87654321 choose the 1st finisher choose the 2nd finisher choose the 8th finisher 8! 40,320
20. 21.
22.
23.
24.
25.
(b) By the Fundamental Counting Principle, the possible number of ways the first, second, and third place can be decided, number of ways to number of ways to number of ways to assuming no ties, is 8 7 6 336. choose the 1st finisher choose the 2nd finisher choose the 3th finisher Since there are four choices for each of the five questions, by the Fundamental Counting Principle there are 4 4 4 4 4 1024 different ways the test can be completed. The number of possible seven-digit phone numbers is number of ways to number of ways to number of ways to . Since the first digit cannot be a 0 or a 1, there choose the 2nd digit choose the 7th digit choose the 1st digit are only 8 digits to choose from, while there are 10 digits to choose from for the other six digits in the phone number. Thus the number of possible seven-digit phone numbers is 8 10 10 10 10 10 10 8,000,000. Since a runner can only finish once, there are no repetitions. And since we are assuming that there is no tie, the number of number of ways to number of ways to number of ways to different finishes is 5 4 3 2 1 120. choose the 1st runner choose the 2nd runner choose the 5th runner Since there are 4 main courses, there are 6 ways to choose a main course. Likewise, there are 5 drinks and 3 desserts so there are 5 ways to choose a drink and 3 ways to choose a dessert. So the number of different meals consisting of a main course, a number of ways to number of ways to number of ways to drink, and a dessert is 4 5 3 60. choose the main course choose a drink choose a dessert By the Fundamental Counting Principle, the number of different routes from town A to town D via towns B and C is number of routes number of routes number of routes 4 5 6 120. from C to D from A to B from B to C The number of possible sequences of heads and tails when a coin is flipped 5 times is number of possible number of possible number of possible 2 2 2 2 2 outcomes on the 1st flip outcomes on the 2nd flip outcomes on the 5th flip 25 32
Here there are only two choices, heads or tails, for each flip. 26. Since each die has six different faces, the number of different outcomes when rolling a red and a white die is 6 6 36. 27. Since there are six different faces on each die, the number of possible outcomes when a red die and a blue die and a white die are rolled is number of possible number of possible number of possible 6 6 6 63 216. outcomes on the red die outcomes on the blue die outcomes on the white die 28. The number of possible skirt-blouse-shoe outfits is number of ways number of ways number of ways 5 8 12 480. to choose a skirt to choose a blouse to choose shoes 29. The number of different California license plates possible is number of ways to number of ways number of ways 9 263 103 158,184,000. choose a nonzero digit to choose 3 letters to choose 3 digits 30. The number of possible ID numbers consisting of one letter followed by 3 digits is 26 10 10 10 26,000. 31. Since successive numbers cannot be the same, the number of possible choices for the second number in the combination is only 59. The third number in the combination cannot be the same as the second in the combination, but it can be the same as the first number, so the number of possible choices for the third number in the combination is also 59. So the number of possible combinations consisting of a number in the clockwise direction, a number in the counterclockwise direction, and then a number in the clockwise direction is 60 59 59 208,860.
SECTION 14.1 Counting
909
32. The number of possible license plates of two letters followed by three digits is number of ways to number of ways 262 103 676,000. Since 676,000 8,000,000, there will not be choose 2 letters to choose 3 digits enough different license plates for the state’s 8 million registered cars. 33. Since a student can hold only one office, the number of ways that a president, a vice-president and a secretary can be chosen from a class of 30 students is number of ways number of ways to number of ways 30 29 28 24,360. to choose a president choose a vice-president to choose a secretary 34. The number of ways a chairman, vice-chairman, and a secretary can be chosen if the chairman must be a Democrat and the vice-chairman must be a Republican is number of ways to choose
a Democratic chairman
from the 10 Democrats
number of ways to choose
a Republican vice-chairman from the 10 Republicans
number of ways to choose a secretary from
10 7 15 1050.
the remaining 15 members
35. We have seven choices for the first digit and 10 choices for each of the other 8 digits. Thus, the number of Social Security numbers is 7 108 700,000,000. 36. The number of possible ways to arrange three girls and four boys is number of ways to number of ways 3! 4! 144. arrange 3 girls to arrange 4 boys 37. (a) The number of ways to select 5 of the 8 objects is C 8 5
8! 56. 5! 3!
(b) A set with 8 elements has 28 256 subsets.
38. We may choose any subset of the 8 available brochures. There are 28 256 ways to do this. 39. Each subset of toppings constitutes a different way a hamburger can be ordered. Since a set with 10 elements has 210 1024 subsets, there are 1024 different ways to order a hamburger. 40. We consider a set of 20 objects (the shoppers in the mall) and a subset that corresponds to those shoppers that enter the store. Since a set of 20 objects has 220 1,048,576 subsets, there are 1,048,576 outcomes to their decisions. 41. (a) The number of ways to seat ten people in a row of ten chairs is 10! 3,628,800. (b) The number of ways to choose six out of ten people and seat them in six chairs is C 10 6 6!
10! 6! 4!
10! 151,200. 4! 42. The number of ways of selecting 3 objects in order (a 3-letter word) from 6 distinct objects (the 6 letters) assuming that the letters cannot be repeated is P 6 3 6 5 4 120. 6!
43. In selecting these officers, order is important and repetition is not allowed, so the number of ways of choosing 3 officers from 15 students is P 15 3 2730. 44. The number of ways of selecting 3 objects in order (a 3-digit number) from 4 distinct objects (the 4 digits) with no repetition of the digits is P 4 3 4 3 2 24. 45. Since the order of finish is important, we want the number of permutations of 8 objects (the contestants) taken three at a 8! 8! 8 7 6 336. time, which is P 8 3 5! 8 3!
46. The number of ways of ordering 8 pieces in order (without repeats) is P 8 8 8! 40,320.
47. The number of ways of ordering 9 distinct objects (the contestants) is P 9 9 9! 362,880. Here a runner cannot finish more than once, so no repetitions are allowed, and order is important. 48. The number of ways of ordering three of the five distinct flags is P 5 3 60.
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CHAPTER 14 Counting and Probability
49. The number of ways of ordering 1000 distinct objects (the contestants) taking 3 at a time is P 1000 3 1000 999 998 997,002,000. We are assuming that a person cannot win more than once, that is, there are no repetitions. 50. In selecting these officers, order is important and repetition is not allowed, so the number of ways of choosing 4 officers from 30 students is P 30 4 657,720. 51. We first place Jack in the first seat, and then seat the remaining four students. Thus the number of these arrangements is number of ways to number of ways to seat P 1 1 P 4 4 1! 4! 24. seat Jack in the first seat the remaining four students 52. We start by first placing Jack in the middle seat, and then we place the remaining four students in the remaining four seats. number of ways number of ways to seat Thus the number of a these arrangements is 1 P 4 4 1! 4! 24. the remaining four students to seat Jack 53. Here we have 6 objects, of which 2 are blue marbles and 4 are red marbles. Thus the number of distinguishable permutations 6 5 4! 6! 15. is 2! 4! 2 4! 54. Here we have 14 objects (the 14 balls) of which 5 are red balls, 2 are white balls, and 7 are blue balls. So the number of 14! 72,072. distinguishable permutations is 5! 2! 7! 55. The number of distinguishable permutations of 12 objects (the 12 coins), from like groups of size 4 (the pennies), of size 3 12! 277,200. (the nickels), of size 2 (the dimes) and of size 3 (the quarters) is 4! 3! 2! 3! 56. The word ELEEMOSYNARY has 12 letters of which 3 are E, 2 are Y, and the remaining letters are distinct. So we wish to find the number of distinguishable permutations of 12 objects (the 12 letters) from like groups of sizes 3 and 2 and seven 12! like groups of size 1. We get 39,916,800. 3! 2! 1! 1! 1! 1! 1! 1! 57. The number of distinguishable permutations of 12 objects (the 12 ice cream cones) from like groups of size 3 (the vanilla cones), of size 2 (the chocolate cones), of size 4 (the strawberry cones), and of size 5 (the butterscotch cones) is 14! 2,522,520. 3! 2! 4! 5! 58. This is the number of distinguishable permutations of 7 objects (the students) from like groups of size 3 (the ones who stay in the 3-person room), size 2 (the ones who stay in the 2-person room), size 1 (the one who stays in the 1-person room), and 7! size 1 (the one who sleeps in the car). This number is 420. 3! 2! 1! 1! 59. The number of distinguishable permutations of 8 objects (the 8 cleaning tasks) from like groups of size 5, 2, and 1 workers, 8! respectively is 168. 5! 2! 1! 60. The number of distinguishable permutations of 30 objects (the students) from like groups of sizes 8, 11, and 11 is 30! 4,128 ,840,588,600. 8! 11! 11! 61. Here we are interested in the number of ways of choosing three objects (the three members of the committee) from a set of 25! 2300. 25 objects (the 25 members). The number of combinations of 25 objects taken three at a time is C 25 3 3! 22! 6! 62. We want the number of ways of choosing a group of three from a group of six. This number is C 6 3 20. 3! 3! 12! 63. We want the number of ways of choosing a group of three from a group of 12. This number is C 12 3 220. 3! 9! 64. We want the number of ways of choosing a group of six people from a group of ten people. The number of combinations of 10! 210. ten objects (people) taken six at a time is C 10 6 6! 4!
SECTION 14.1 Counting
911
65. We want the number of ways of choosing a group (the 5-card hand) where order of selection is not important. The number 52! 2,598,960. of combinations of 52 objects (the 52 cards) taken 5 at a time is C 52 5 5! 47! 66. Since order is not important in a 7-card hand, the number of combinations of 52 objects (the 52 cards) taken 7 at a time is 52! 133,784,560. C 52 7 7! 45! 67. The order of selection is not important, hence we must calculate the number of combinations of 10 objects (the 10 questions) 10! taken 7 at a time, this gives C 10 7 120. 7! 3! 68. In this exercise, we assume that the pizza toppings cannot be repeated, so we are interested in the number of ways to select a 16! 560. subset of 3 toppings from a set of 16 toppings. The number of ways this can occur is C 16 3 3! 13! 69. We assume that the order in which he plays the pieces in the recital is not important, so the number of combinations of 12! 495. 12 objects (the 12 pieces) taken 8 at a time is C 12 8 8! 4! 70. The order in which the skirts are selected is not important and no skirt is repeated. So the number of combinations of 8! 56. eight skirts taken five at a time is C 8 5 5! 3! 71. The order in which the pants are selected is not important and no pair of pants is repeated, so the number of combinations of 10! ten pairs of pants taken three at a time is C 10 3 120. 3! 7! 72. (a) Since Jack must go on the field trip, we first pick Jack to go on the field trip, and then select the six other students from 29! the remaining 29 students. Since C 29 6 475,020, there are 475,020ways to select the students to go on 6! 23! the field trip with Jack. (b) We first take Jack out of the class of 30 students and select the 7 students from the remaining 29 students. Thus there 29! are C 29 7 1,560,780 ways to pick the 7 students for the field trip. 7! 22! (c) We are interested only in the group of seven students taken from the class of 30 students, not the order in which they are 30! 2,035,800. picked. Thus the number is C 30 7 7! 23! 73. Since the order in which the numbers are selected is not important, the number of combinations of 49 numbers taken 6 at a 49! 13,983,816. time is C 49 6 6! 43! 74. The number of distinguishable permutations of 13 objects (the total number of blocks he must travel) which can be 13! partitioned into like groups of size 8 (the east blocks) and of size 5 (the north blocks) is 1,287. 8! 5! 20! 15,504. 75. (a) The number of ways of choosing 5 students from the 20 students is C 20,5 5! 15! 12! (b) The number of ways of choosing 5 students for the committee from the 12 females is C 12 5 792. 5! 7! (c) We use the Fundamental Counting Principle to count the number of possible committees with 3 females and 2 males. Thus, we get number of ways to choose the number of ways to choose the C 12 3 C 8 2 220 28 6160. 3 females from the 12 females 2 males from the 8 males 76. We pick the two men from the group of ten men, and we pick the two women from the group of ten women. So the number of ways 2 men and 2 women can be chosen is number of ways to pick number of ways to pick C 10 2 C 10 2 45 45 2025. 2 of the 10 men 2 of the 10 women
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CHAPTER 14 Counting and Probability
77. The number of ways the committee can be chosen is number of ways to number of ways to number of ways to C 20 1 C 19 1 C 18 4 choose 1 president choose 1 vice-president choose 4 other members 20 19 3060 1,162,800 78. (a) We choose 2 girls from the girls on the camping trip and the 4 boys from the boys on the camping trip. Thus the number of ways to pick the 6 to gather firewood is number of ways to number of ways to C 9 2 C 16 4 36 1820 65,520. choose 2 of 9 girls choose 4 of 16 boys (b) Method 1: We consider the number of ways of selecting the group of 6 from the 25 campers and subtract off the groups that contain no girls and those that contain one girls. Thus the number groups that contain at least two girls is C 25 6 C 9 0 C 16 6 C 9 1 C 16 5 177,100 1 8008 9 4368 129,780. Method 2: In the method we construct all the groups that are possible. Thus the number of groups is groups with 2 girls and 4 boys
groups with 3 girls and 3 boys
groups with 4 girls and 2 boys
groups with 5 girls and 1 boy
groups with 6 girls
C 9 2 C 16 4 C 9 3 C 16 3 C 9 4 C 16 2 C 9 5 C 16 1 C 9 6 C 16 0 36 1820 84 560 126 120 126 16 84 1
65,520 47,040 15,120 2,016 84 129,780. 79. The number of ways the committee can be chosen is number of ways to number of ways to number of ways to number of ways to choose 2 of 6 freshmen choose 3 of 8 sophomores choose 4 of 12 juniors choose 5 of 10 seniors C 6 2 C 8 3 C 12 4 C 10 5 15 56 495 252 104,781,600 80. The leading and supporting roles are different (order counts), while the extra roles are not (order doesn’t count). Also, the male roles must be filled by the male actors and the female roles filled by the female actresses. Thus, number of ways the actors and actresses can be chosen is number of ways the number of ways the number of ways to number of ways to leading and the supporting leading and the supporting choose 5 of 8 male extras choose 3 of 10 female extras actors can be chosen
actresses can be chosen
P 10 2 P 12 2 C 8 5 C 10 3 90 132 56 120 79,833,600.
81. We choose 3 forwards from the forwards, 2 defensemen from the defensemen, and the goalie from the two goalies. Thus the number of ways to pick the 6 starting players is number of ways to number of ways to number of ways to C 12 3 C 6 2 C 2 1 pick 3 of 12 forwards pick 2 of 6 defensemen pick 1 of 2 goalies 220 15 2 6600 82. To order a pizza, we must make several choices. First the size (4 choices), the type of crust (2 choices), and then the toppings. Since there are 14 toppings, the number of possible choices is the number of subsets of the 14 toppings, that is 214 choices. So by the Fundamental Counting Principle, the number of possible pizzas is 4 2 214 131,072. 83. We count the total number of committees and subtract the number that contain both Barry and Harry. The total number of committees possible is C 10 4 and the number that contain both Barry and Harry is C 8 2, so the number of possible committees is C 10 4 C 8 2 210 28 182.
SECTION 14.1 Counting
913
84. Method 1: We consider the number of 5-member committees that can be formed from the 26 interested people and subtract the numbers of those that contain no teacher and those that contain no students. Thus the number of committees is C 26 5 C 12 5 C 14 5 65,780 792 2002 62,986. Method 2: Here we construct all the committees that are possible. Thus the number of committees is committees with
1 student and 4 teachers
committees with
2 students and 3 teachers
committees with
3 students and 2 teachers
committees with
4 students and 1 teacher
C 14 1 C 12 4 C 14 2 C 12 3 C 14 3 C 12 2 C 14 4 C 12 1 14 495 91 220 364 66 1001 12 6,930 20,020 24,024 12,012 62,986 85. Since the two algebra books must be next to each other, we first consider them as one object. So we now have four objects to arrange and there are 4! ways to arrange these four objects. Now there are 2 ways to arrange the two algebra books. Thus the number of ways that 5 mathematics books may be placed on a shelf if the two algebra books are to be next to each other is 2 4! 48.
86. We treat John and Jane as one object and Mike and Molly as one object, so we need the number of ways of permuting 8 objects. We then multiply this by the number of ways of arranging John and Jane within their group and arranging Mike and Molly within their group. Thus the number of possible arrangements is P 8 8 P 2 2 P 2 2 8! 2! 2! 161,280. 87. (a) To find the number of ways the men and women can be seated we first select and place a man in the first seat and then select 1 of arrange the arrange the other 7 people. Thus we get C 4 1 P 7 7 4 7! 20,160. the 4 men remaining 7 people (b) To find the number of ways the men and women can be seated we first select and place a woman in the first and last seats and then arrange the other 6 people. Thus we get arrange 2 of arrange the P 4 2 P 6 6 12 6! 8,640. the 4 women remaining 6 people 88. (a) We treat the women as one object and find the number of ways of permuting 5 objects. We then permute the 4 women. Thus the number of arrangements is P 5 5 P 4 4 5! 4! 2880. (b) There are two ways to solve this exercise. Method 1: The number of ways the men and women can be seated is number of ways the number of ways the number of ways to select P 4 4 P 4 4 C 2 1 men can be arranged women can be arranged the gender of the first seat 4! 4! 2 1152
Method 2: There are 8 choices for the first seat, 4 for the second, 3 for the third, 3 for the fourth, 2 for the fifth, 2 for the sixth, 1 for the seventh, and 1 for the eighth. Thus the number of ways of seating the men and women in the required fashion is 8 4 3 3 2 2 1 1 1152.
89. The number of ways the top finalist can be chosen is
number of ways to choose the 6
semifinalists from the 30
number of ways to choose the 2
number of ways
to choose the winner C 30 6 C 6 2 C 2 1
finalists from the 6
from the 2 finalists
593,775 15 2 17,813,250
90. There are many different possibilities here, so we consider the complement where no professor is chosen for the delegates and subtract this number from the way to select 3 people from the group of 8 people, which is C 8 3. If the professor cannot to be selected, then we must select 3 people from a group of 5 and this can be done in C 5 3 ways. Thus the number of delegations that contain a professor is C 8 3 C 5 3 56 10 46. 91. Since there are 26 letters, the possible number of combinations of the first and the last initials is 26 26 676. Since 677 676, there must be at least two people that have the same first and last initials in any group of 677 people.
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CHAPTER 14 Counting and Probability
92. When two objects are chosen from ten objects, it determines a unique set of eight objects, those not chosen. So choosing two objects from ten objects is the same a choosing eight objects from ten objects. In general, every subset of r objects chosen from a set of n objects determines a corresponding set of n r objects, namely, those not chosen. Therefore, the total number of combinations for each type are equal. 93. We are only interested in selecting a set of three marbles to give to Luke and a set of two marbles to give to Mark, not the order in which we hand out the marbles. Since both C 10 3 C 7 2 and C 10 2 C 8 3 count the number of ways this can be done, these numbers must be equal. (Calculating these values shows that they are indeed equal.) In general, if we wish to find two distinct sets of k and r objects selected from n objects (k r n), then we can either first select the k objects from the n objects and then select the r objects from the n k remaining objects, or we can first select the r objects n n r n nk from the n objects and then the k objects from the n r remaining objects. Thus . r k k r 94. (a) x y5 x y x y x y x y x y
x y x y x y x x x y yx yy
x y x y x x x x x y x yx x yy yx x yx y yyx yyy
x y x x x x x x x y x x yx x x yy x yx x x yx y x yyx x yyy yx x x yx x y yx yx yx yy yyx x yyx y yyyx yyyy
x x x x x x x x x y x x x yx x x x yy x x yx x x x yx y x x yyx x x yyy x yx x x x yx x y x yx yx x yx yy x yyx x x yyx y x yyyx x yyyy yx x x x yx x x y yx x yx yx x yy yx yx x yx yx y yx yyx yx yyy yyx x x yyx x y yyx yx yyx yy yyyx x yyyx y yyyyx yyyyy (b) There are ten terms that contain two x’s and three y’s. Their sum is x x yyy x yx yy x yyx y x yyyx yx x yy yx yx y yx yyx yyx x y yyx yx yyyx x (c) To count the number of terms with two x’s, we must count the number of ways to pick two of the five positions to contain an x. This number is C 5 2. n (d) In the Binomial Theorem, the coefficient is the number of ways of picking r positions in a term with n factors to r contain an x. By definition, this is C n r .
14.2 PROBABILITY 1. The set of all possible outcomes of an experiment is called the sample space. A subset of the sample space is called an event. The sample space for the experiment of tossing two coins is S H H H T T H T T , and the event “getting at least 3 n E . one head” is E H H H T T H. The probability of getting at least one head is P E n S 4 2. (a) The probability of E or F occurring is P E F P E P F P E F.
(b) If the events E and F have no outcome in common (that is, the intersection of E and F is empty), then the events are called mutually exclusive. So in drawing a card from a deck, the event E of “getting a heart” and the event F of “getting a spade” are mutually exclusive.
(c) If E and F are mutually exclusive, then the probability of E or F is PE F P E P F. n E F . So in tossing a die, the conditional 3. The conditional probability of E given that F occurs is P E F n F probability of the event E “getting a six” given that that the event F “getting an even number” has occurred is 1 P E F . 3
SECTION 14.2 Probability
915
4. (a) The probability of E and F occurring is PE F P E P F E.
(b) If the occurrence of E does not affect the probability of the occurrence F, then the events are called independent. So in tossing a coin twice, the event E of “getting heads on the first toss” and the event F of “getting heads on the second toss” are independent.
(c) If E and F are independent events, then the probability of E and F is PE F P E P F. 5. (a) S 1 2 3 4 5 6 (b) E 2 4 6 (c) E 5 6
6. (a) There are two possible outcomes of the coin toss and 52 possible outcomes of drawing a card, so n S 2 52 104. (b) H A H A H A H A
(c) T J T Q T K T J T Q T K T J T Q T K T J T Q T K
(d) H 2 H 3 H4 H5 H6 H7 H8 H9 H10 H J H Q H K H A
7. Let H stand for head and T for tails. (a) The sample space is S H H H T T H T T .
1 (b) Let E be the event of getting exactly two heads, so E H H. Then P E nE nS 4 .
3 (c) Let F be the event of getting at least one head. Then F H H H T T H, and P F nF nS 4 . 2 1 (d) Let G be the event of getting exactly one head, that is, G H T T H . Then P G nG nS 4 2 . 8. Let H stand for heads and T for tails; the numbers 1, 2, , 6 are the faces of the die. (a) S H 1 H2 H3 H 4 H5 H6 T 1 T 2 T 3 T 4 T 5 T 6
3 1 (b) Let E be the event of getting heads and rolling an even number. So E H2 H4 H6, and P E nE nS 12 4 .
(c) Let F be the event of getting heads and rolling a number greater than 4. So F H5 H6, and 2 1 P F nF nS 12 6 .
3 1 (d) Let G be the event of getting tails and rolling an odd number. So G T 1 T 3 T 5, and P G nG nS 12 4 . 1 9. (a) Let E be the event of rolling a six. Then P E nE nS 6 . 3 1 (b) Let F be the event of rolling an even number. Then F 2 4 6. So P F nF nS 6 2 . 1 (c) Let G be the event of rolling a number greater than 5. Since 6 is the only face greater than 5, P G nG nS 6 . 2 1 10. (a) Let E be the event of rolling a two or a three. Then P E nE nS 6 3 . 3 1 (b) Let F be the event of rolling an odd number. So F 1 3 5, and P F nF nS 6 2 . 1 (c) Let G be the event of rolling a number divisible by 3. So G 3 6, and P G nG nS 3 . 4 1 11. (a) Let E be the event of choosing a king. Since a deck has four kings, P E nE nS 52 13 . (b) Let F be the event of choosing a face card. Since there are three face cards per suit and four suits, 12 3 P F nF nS 52 13 .
3 10 . (c) Let F be the event of choosing a face card. Then P F 1 P F 1 13 13
13 1 12. (a) Let E be the event of choosing a heart. Since there are 13 hearts, P E nE nS 52 4 . 26 1 (b) Let F be the event of choosing a heart or a spade. Since there are 13 hearts and 13 spades, P F nE nS 52 2 . (c) Let G be the event of choosing a heart, a diamond or a spade. Since there are 13 cards in each suit, 39 3 P G nG nS 52 4 .
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CHAPTER 14 Counting and Probability
5 13. (a) Let E be the event of selecting a red ball. Since the jar contains five red balls, P E nE nS 8 . (b) Let F be the event of selecting a yellow ball. Since there is only one yellow ball, 1 7 P F 1 P F 1 nF nS 1 8 8 .
0 (c) Let G be the event of selecting a black ball. Since there are no black balls in the jar, P G nG nS 8 0.
14. (a) Let E be the event of selecting a white or a yellow ball. Since there are two white balls and one yellow ball, 3 5 P E 1 P E 1 nE nS 1 8 8 .
(b) Let F be the event of selecting a red, a white or a yellow ball. Since all the types of balls are in the jar, P E 1. (c) Let G be the event of selecting a white ball. Since there are 2 white balls, 2 6 3 P E 1 P E 1 nE nS 1 8 8 4 .
1287 C 13 5 0000495. C 52 5 2,598,960 (b) Let E be the event of choosing five cards of the same suit. Since there are four suits and 13 cards in each suit, 5,148 4 C 13 5 000198. n E 4 C 13 5. Also, n S C 52 5. Therefore, P E C 52 5 2,598,960 (c) Let E be the event of dealing five face cards. Since there are 3 face cards for each suit and 4 suits, 792 C 12 5 0000305. P E C 52 5 2,598,960 (d) Let E be the event of dealing a royal flush (ace, king, queen, jack, and 10 of the same suit). Since there is only one such 4 4 153908 106 . sequence for each suit, there are only 4 royal flushes, so P E C 52 5 2,598,960
15. (a) Let E be the event of dealing five hearts. Since there are 13 hearts, P E
16. (a) Let E be the event of choosing three defective CDs. Since there are four defective CDs, n E C 4 3. Also, C 4 3 4 1 n E 0018. n S C 12 3. Therefore, P E n S C 12 3 220 55 (b) Let E be the event of choosing three functioning CDs. Since there are eight functioning CDs, n E C 8 3, so C 8 3 56 14 n E 0255. P E n S C 12 3 220 55 17. (a) Let E be the event of choosing two red balls. Since there are three red balls, n E C 3 2. Also, n S C 8 2. C 3 2 3 n E 0107. Therefore, P E n S C 8 2 28 (b) Let E be the event of choosing two white balls. Since there are five white balls, n E C 5 2, so C 5 2 10 5 n E 0357. P E n S C 8 2 28 14 3 . 18. (a) Let E be the event of choosing a T. Since 3 of the 16 letters are Ts, P E 16
(b) Let F be the event of choosing a vowel. Since there are 6 vowels, (c) Let F be the event of choosing a vowel. Then P F 1 38
6 3. P F 16 8 5. 8
19. (a) Let E be the probability that at least one card is a spade. The number of hands that do not contain a spade is the number 7411 C 39 5 0778. of possible 5-card hands using the other three suits, that is, C 39 5. Thus, P E 1 C 52 5 9520 (b) Let E be the probability that at least one card is a face card. The number of hands that do not contain a face card is the number of possible 5-card hands using the cards of the deck that are not face cards, that is, C 40 5. Thus, 6221 C 40 5 0747. P E 1 C 52 5 8330
SECTION 14.2 Probability
917
20. (a) S 1 1 1 2 1 3 1 4 1 5 1 6 2 1 2 2 2 3 2 4 2 5 2 6 3 1 3 2 3 3 3 4 3 5 3 6 4 1 4 2 4 3 4 4 4 5 4 6 5 1 5 2 5 3 5 4 5 5 5 6 6 1 6 2 6 3 6 4 6 5 6 6 6 1. (b) Let E be the event of getting a sum of 7. Then E 1 6 2 5 3 4 4 3 5 2 6 1, and P E 36 6
4 1. (c) Let F be the event of getting a sum of 9. Then F 3 6 4 5 5 4 6 3, and P F 36 9
6 1. (d) Let E be the event that the two dice show the same number. Then P E 36 6 (e) Let E be the event that the two dice show different numbers. Then E is the event that the two dice show the same number. Thus, P E 1 P E 1 16 56 . 5 (f) Let E be the event of getting a sum of 9 or higher. Then P E 10 36 18 .
3 21. (a) Let E be the event that the spinner stops on red. Since 12 of the regions are red, P E 12 16 4 .
(b) Let F be the event that the spinner stops on an even number. Since 8 of the regions are even-numbered, 8 1. P F 16 2
4 1. (c) Since 4 of the even-numbered regions are red, P E F P E P F P E F 34 12 16 4 1. 22. (a) Let E be the event that the spinner stops on blue. Since only 4 of the regions are blue, P E 16 4
8 1. (b) Let F be the event that the spinner stops on an odd number. Since 8 of the regions are odd-numbered, P F 16 2
(c) Since none of the odd-numbered regions are blue, P E F P E P F 14 12 34 .
23. (a) Yes, the events are mutually exclusive since the number cannot be both even and odd. So P E F P E P F 36 36 1. (b) No, the events are not mutually exclusive since 6 is both even and greater than 4. So P E F P E P F P E F 36 26 16 23 . 24. (a) No, the events are not mutually exclusive since 4 is greater than 3 and also less than 5. So P E F P E P F P E F 36 46 16 1. (b) Yes, the events are mutually exclusive since there are only 2 numbers less than 3, namely 1 and 2, but they are not divisible by 3. So P E F P E P F 26 26 23 . 25. (a) No, the events E and F are not mutually exclusive since the jack, queen, and king of spades are both face cards and 12 3 11 spades. So P E F P E P F P E F 13 52 52 52 26 . (b) Yes, the events E and F are mutually exclusive since the card cannot be both a heart and a spade. So 13 1 P E F P E P F 13 52 52 2 .
26. (a) No, events E and F are not mutually exclusive since a king can be a club. So 4 1 4 P E F P E P F P E F 13 52 52 52 13 .
(b) No, events E and F are not mutually exclusive since an ace can be a spade. So 4 13 1 4 . P E F P E P F P E F 52 52 52 13
27. (a) Let E denote a roll of five and F a roll greater than three. Using the formula for conditional probability, we have 1 n E F . P E F n F 3 (b) Let E denote a roll of three and F an odd roll. Using the formula for conditional probability, we have 1 n E F . P E F n F 3
918
CHAPTER 14 Counting and Probability
4 1 n E F . n F 12 3 n E F 1 (b) Let E denote drawing a king and F drawing a spade. Then P E F . n F 13 n E F 1 (c) Let E denote drawing a spade and F drawing a king. Then P E F . n F 4
28. (a) Let E denote drawing a queen and F drawing a face card. Then P E F
29. Let E denote the spinner stopping on an even number and F the spinner stopping on red. Then n E F 4 1 P E F . n F 12 3 30. Let E denote the spinner stopping on a number divisible by 3 and F the spinner stopping on blue. Then n E F 1 P E F . n F 4 31. (a) There is only one red ball numbered 3 and one green ball numbered 3. If the ball drawn is numbered 3, then the probability it is red is 12 . (b) There is only one ball numbered 7 and it is green, so if the ball drawn is numbered 7, then the probability it is green is 1. (c) There are two even-numbered red balls and three even-numbered green balls, so if the ball is even-numbered, then the probability it is a red ball is 25 . (d) There are five red balls, and two are even-numbered, so if the ball drawn is red, then the probability it is even-numbered is 25 . 32. (a) If the first ball drawn is red, there are four red balls and seven green balls remaining, so the probability that the second 4 . ball is red is 11 5. (b) If the first ball drawn is green, there are five red balls and six green balls remaining, so the required probability is 11
(c) If the first ball drawn is odd-numbered, then there are six odd-numbered and five even-numbered balls remaining, so the 5 . required probability is 11
(d) If the first ball drawn is even-numbered, then there are seven odd-numbered and four even-numbered balls remaining, 4 . so the required probability is 11
33. (a) Let E be the event of drawing a black ball first. Because the jar contains seven black balls and three white balls, the 7 . The probability of the second ball being white is 3 1 , so the probability of the first ball being black is P E 10 9 3 7 1 7 probability of the intersection is 10 3 30 .
7 2 7 . (b) Here the probability of the second ball being black is 69 23 , so the probability of the intersection is 10 3 15
34. (a) Let E be the event of drawing a red sock. Since three pairs are red, the drawer contains six red socks, and so 6 1 P E nE nS 18 3 . 5 (b) Let F be the event of drawing another red sock. Since there are 17 socks left of which 5 are red, P F nF nS 17 . 5 5 . (c) In this case, the probability is P E F P E P F 13 17 51 4 and the probability of the second being a king is 4 , so the 35. (a) The probability of the first card being an ace is 52 51 4 4 4 . probability of the intersection is 52 51 663
4 and the probability of the second being an ace is 3 , so the (b) The probability of the first card being an ace is 52 51 4 3 1 . probability of the intersection is 52 51 221
36. (a) Yes, the first roll does not influence the outcome of the second roll.
1 1 . (b) The probability of getting a six on both rolls is P E F P E P F 16 6 36
SECTION 14.2 Probability
919
37. Let E be the event of getting a 1 on the first roll, and let F be the event of getting an even number on the second roll. Since 1 . these events are independent, P E F P E P F 16 36 12 38. (a) Yes, they are independent. The toss of the coin does not influence the roll of the die.
(b) The probability of getting a tail is 12 and the probability of getting an even number is 36 12 , so the probability of the intersection is 12 12 14 . 39. (a) Yes. What happens on spinner A does not influence what happens on spinner B.
2 1. (b) The probability that A stops on red and B stops on yellow is P E F P E P F 24 8 8
40. (a) Let E A and E B be the event that the respective spinners stop on a purple region. Since these events are independent, 1 . P E A E B P E A P E B 14 28 16 (b) Let FA and FB be the event that the respective spinners stop on a blue region. Since these events are independent, 1. P FA FB P FA P FB 14 18 32
41. (a) Let B and G stand for “boy”and “girl”. Then S B B B B G B B B BG B B B BG B B B BG GG B B G BG B G B BG BGG B BG BG B BGG BGGG G BGG GG BG GGG B GGGG 1. (b) Let E be the event that the couple has only boys. Then E B B B B and P E 16
(c) Let F be the event that the couple has 2 boys and 2 girls. Then
6 3. F GG B B G BG B G B BG BGG B BG BG B BGG, so P F 16 8
2 1. (d) Let G be the event that the couple has 4 children of the same sex. Then G B B B B GGGG, and P G 16 8 (e) Let H be the event that the couple has at least 2 girls. Then H is the event that the couple has fewer than two girls. 5 11 . Thus, H B B B B G B B B BG B B B BG B B B BG, so n H 5, and P H 1 P H 1 16 16
42. Let E be the event that a 13-card bridge hand consists of all cards from the same suit. Since there are exactly 4 such hands 4 63 1012 . (one for each suit), P E C 52 13 43. Let E be the event that the ball lands in an odd numbered slot. Since there are 18 odd numbers between 1 and 36
9 P E 18 38 19 . 44. (a) Let E be the event that the toddler arranges the word FRENCH. Since the letters are distinct, there are P 6 6 ways of 1 1 00014. arranging the blocks of which only one spells the word FRENCH. Thus P E P 6 6 720 (b) Let E be the event that the toddler arranges the letters in alphabetical order. Since there are P 6 6 ways of arranging 1 1 00014. the blocks of which only one is in alphabetical order, P E P 6 6 720 45. Let E be the event of picking the 6 winning numbers. Since there is only one way to pick these, 1 1 715 108 . P E C 49 6 13,983,816 46. Let E be the event that no women are chosen. The number of ways that no women are chosen is the same as the number of C 11 6 000078. ways that only men are chosen, which is C 11 6. Thus P E C 30 6
47. The sample space consist of all possible True-False combinations, so n S 210 . Let E be the event that the student 1 1 . answers all 10 questions correctly. Since there is only one way to answer all 10 questions correctly, P E 10 1024 2 48. Let E be the event that the batch will be discarded. Thus, E is the event that at least one defective bulb is found. It is easier to find E , the event that no defective bulbs are found. Since there are 10 bulbs in the batch of which 8 are non-defective, C 8 3 C 8 3 . Thus P E 1 P E 1 1 04667 05333. P E C 10 3 C 10 3
920
CHAPTER 14 Counting and Probability
49. (a) Let E be the event that the monkey types “Hamlet” as his first word. Since “Hamlet” contains 6 letters and there are 48 1 typewriter keys, P E 6 818 1011 . 48 (b) Let F be the event that the monkey types “to be or not to be” as his first words. Since this phrase has 18 characters 1 (including the blanks), P F 18 547 1031 . 48 1 1 00014. 50. (a) Let E be the event that the monkey arranges the 6 blocks to spell HAMLET. Then P E 720 6! (b) The probability that the monkey arranges the 6 blocks to spell HAMLET three consecutive times is the probability of 1 3 268 109 . three independent events E, and hence is equal to [P E]3 720
51. Let E be the event that the toddler will arrange the 8 blocks to spell TRIANGLE or INTEGRAL. The number of ways of arranging these blocks is the number of distinguishable permutations of 8 blocks. Since no two blocks are the same, the 2 496 105 number of distinguishable permutations is 8!. Two of these arrangements result in event E, so P E 8! or 00000496. 52. Let E be the event that you predict the correct order for the horses to finish the race. Since there are eight horses, there are P 8 8 8! ways that the horses could finish, with only one being the correct order. Thus, 1 1 248 105 . P E P 8 8 40,320 53. (a) Let E be the event that the pea is tall. Since tall is dominant, E T T T t t T . So P E 34 . (b) E is the event that the pea is short. So P E 1 P E 1 34 14 .
(a) Let E be the event that the offspring will be tall. Since only offspring
54.
Parent 2
Parent 1
t
t
T
Tt
Tt
t
tt
tt
with genotype Tt will be tall, P E 24 12 . (b) E is the event that the offspring will not be tall (thus, the offspring is short). So P E 1 P E 1 12 12 .
55. Let E be the event that the player wins on spin 1, and let F be the event that the player wins on spin 2. What happens on the first spin does not influence what happens on the second spin, so the events are independent. Thus,
1 1 1 . P E F P E P F 38 38 1444 56. Let E be the event that the committee is all male and F the event it is all female. The sample space is the set of all ways that 5 people can be chosen from the group of 14. These events are mutually exclusive, so C 8 5 6 56 31 C 6 5 . P E F P E P F C 14 5 C 14 5 2002 1001 57. Let E, F and G denote the events of rolling two ones on the first, second, and third rolls, respectively, of a pair of dice. The 1 1 1 1 214 105 . events are independent, so P E F G P E P F P G 36 36 36 363 58. Let E be the event that a player has exactly 5 winning numbers and F be the event that a player has all 6 winning numbers. These events are mutually exclusive. For a players to have exactly 5 winning numbers means that the player has 5 of the 6 winning numbers and 1 number that was not selected in the lottery. So n E C 6 5 C 43 1. Thus, 1 C 6 5 C 43 1 00000185. P at least 5 winning numbers P E F P E P F C 49 6 C 49 6
59. Let E be the event that the marble is red and F be the event that the number is odd-numbered. Then E is the event that the marble is blue, and F is the event that the marble is even-numbered. 6 3 (a) P E 16 8 8 1 (b) P F 16 2
SECTION 14.2 Probability
921
6 8 3 11 (c) P E F P E P F P E F 16 16 16 16 8 5 13 . (d) P E F P E P F P E F 10 16 16 16 16
60. The number of ways a set of six numbers can be selected from a group of 49 numbers is C 49 6. Since the games are 2 1 independent, the probability of winning the lottery two times in a row is 511 1015 . C 49 6 61. The probability of getting 2 red balls by picking from jar B is 57 46 10 21 . The probability of getting 2 red balls by 5 15 . The probability of getting 2 red balls after putting all balls in one jar is picking one ball from each jar is 37 7 49 8 7 4 . Hence, picking both balls from jar B gives the greatest probability. 14 13 13 1 , and is the same for the second and third wheels. The events are 62. (a) The probability that the first wheel has a bar is 11 1 1 1 1 . independent, and so the probability of getting 3 bars is 11 11 11 1331
(b) The probability of getting a number on the first wheel is 10 11 , the probability of getting the same number on the second 1 , and the probability of getting the same number on the third wheel is 1 . Thus, the probability of getting wheel is 11 11 1 1 10 the same number on each wheel is 10 11 11 11 1331 . (c) We use the complement, no bar, to determine the probability of at least one bar. The probability that the first wheel does
not have a bar is 10 11 , and is the same for the second and third wheels. Since the events are independent, the probability 10 10 103 1000 1000 331 of getting no bar is 10 11 11 11 113 1331 , and so P at least one bar 1 1331 1331 .
63. Let E be the event that she opens the lock within an hour. The number of combinations she can try in one hour is 10 60 600. The number of possible combinations is P 40 3 assuming that no number can be repeated. Thus 600 5 600 0010 . P E P 40 3 59,280 494 64. (a) Let E be the event that the curriculum committee consists of 2 women and 4 men. So number of committees with 2 women and 4 men C 8 2 C 10 4 28 210 490 P E . number of ways to select 6 member committee C 18 6 18564 1547 (b) Let F be the event that curriculum committee consists of two or fewer women. Then F is the event that the committee has no woman or one woman or two women. Then P F
1 210 8 252 28 210 C 8 0 C 10 6 C 8 1 C 10 5 C 8 2 C 10 4 C 18 6 C 18 6 C 18 6 18,564 193 8106 18,564 442
249 (c) So F is the event the at curriculum committee has more that two women. The P F 1 P F 1 193 442 442 .
65. Let E be the event that Paul stands next to Phyllis. To find n E we treat Paul and Phyllis as one object and find the number of ways to arrange the 19 objects and then multiply the result by the number of ways to arrange Paul and Phyllis. 2 19! 2! 010. So n E 19! 2!. The sample space is all the ways that 20 people can be arranged. Thus P E 20! 20 66. Let E be the event that the monkey arranges the 6 blocks to spell BUBBLE. The number of ways of arranging these blocks is the number of distinguishable permutations of 6 blocks. Since there are three blocks labeled B, the number of distinguishable 1 1 6! 3! . permutations is . Only one of these arrangements spells the word BUBBLE. Thus P E 6! 3! 6! 120 3!
922
CHAPTER 14 Counting and Probability
67. Let E be the event that the monkey arranges the 11 blocks to spell PROBABILITY. The number of ways of arranging these blocks is the number of distinguishable permutations of 11 blocks. Since there are two blocks labeled B and two 11! . Only one of these arrangements spells the word blocks labeled I, the number of distinguishable permutations is 2! 2! 1 1 2! 2! PROBABILITY. Thus P E . 11! 11! 9,979,200 2! 2! 68. (a) Because the events are independent, the probability that the family has two boys given that the oldest child is a boy is 12 . (b) There are four equally likely possibilities: boy-boy, boy-girl, girl-boy, and girl-girl. In three cases, at least one of the children is a boy, and in one of those cases both children are boys. Thus, the probability that the family has two boys given that one of the children is a boy is 13 .
14.3 BINOMIAL PROBABILITY 1. A binomial experiment is one in which there are exactly two outcomes. One outcome is called success and the other is called failure. 2. If a binomial experiment has probability p of success then the probability of failure is 1 p. The probability of getting 3. 4. 5. 6. 7. 8.
exactly r successes in n trials of this experiment is C n r pr 1 pnr . P 2 successes in 5 C 5 2 072 033 013230 P 3 successes in 5 C 5 3 073 032 030870 P 0 success in 5 C 5 0 070 035 000243 P 5 successes in 5 C 5 5 075 030 016807 P 1 success in 5 C 5 1 071 034 002835 P 1 failure in 5 C 5 1 031 074 036015. Note that exactly one failure is the same as exactly 4 successes, and P 4 successes in 5 C 5 4 074 031 036015.
9. P at least 4 successes P 4 successes P 5 successes C 5 4 074 031 C 5 5 075 030 036015 016807 052822 10. P at least 3 successes P 3 successes P 4 successes P 5 successes
C 5 3 073 032 C 5 4 074 031 C 5 5 075 030 030870 036015 016807 083692
11. P at most 1 failure P 0 failure P 1 failure P 5 successes P 4 successes C 5 5 075 030 C 5 4 074 031 016807 036015 052822 12. P at most 2 failures P 0 failure P 1 failure P 2 failure
P 5 successes P 4 successes P 3 successes
C 5 5 075 030 C 5 4 074 031 C 5 3 073 032 016807 036015 030870 083692
SECTION 14.3 Binomial Probability
13. P at least 2 successes P 2 successes P 3 successes P 4 successes P 5 successes 013230 030870 036015 016807 096922 14. P at most 3 failures P 0 failure P 1 failure P 2 failures P 3 failures
P 5 successes P 4 successes P 3 successes P 2 successes 016807 036015 030870 013230 096922 (b)
15. (a) Outcome
Probability
1
02
2
02
3
02
4
02
5
02
Outcome
Probability
1
05
2
03
3
01
4
01
5
0
16. (a)
0.2
0
1
2
3
4
5
(b)
17. (a)
0.1
0
1
2
3
4
3
4
(b) r
Probability
0
4
1 16 1 4 3 8 1 4 1 16
r
Probability
0
00776
1
02592
2
03456
3
02304
4
00768
5
00102
1 2 3
18. (a)
1 16
0
(b)
1
2
0.5
0
1
2
3
4
5
923
924
CHAPTER 14 Counting and Probability
19. (a)
(b) r
Probability
0
02097
1
03670
2
02753
3
01147
4
00287
5
00043
6
000036
7
0000013
r
Probability
0
0000001
1
0000054
2
0001215
3
0014580
4
0098415
5
0354294
6
0531441
20. (a)
0.3
(b)
0
1
0
1
2
3
4
5
6
7
0.5
2
3
4
5
6
2 4 5 21. Here “success” is “face is 4” and P face is 4 16 . Then P 2 successes in 6 C 6 2 16 020094. 6 22. Here P success 08 and P failure 02. (a) P 0 success in 7 C 7 0 080 027 00000128 (b) P 7 successes in 7 C 7 7 087 020 0209715
(c) P he hits target more than once 1 [P 0 success in 7 P 1 success in 7] 1 C 7 0 080 027 C 7 1 081 026 099963
(d) P at least 5 successes P 5 successes P 6 successes P 7 successes C 7 5 085 022 C 7 6 086 021 C 7 7 087 020 085197
23. P 4 successes in 10 C 10 4 044 066 025082
24. The complement is that none of the raccoons had rabies.
P at least 1 had rabies 1 P none had rabies 1 C 4 0 010 094 1 06561 03439
25. (a) P 5 in 10 C 10 5 0455 0555 023403
(b) P at least 3 1 P at most 2 P 0 in 10 P 1 in 10 P 2 in 10 1 C 10 0 0450 05510 C 10 1 0451 0559 C 10 2 0452 0558 090044
26. (a) P 12 in 15 C 15 12 0112 093 331695 1010
SECTION 14.3 Binomial Probability
925
(b) P at least 12 P 12 in 15 P 13 in 15 P 14 in 15 P 15 in 15 C 15 12 0112 093 C 15 13 0113 092 C 15 14 0114 091 C 15 15 0115 090 340336 1010
27. (a) The complement of at least 1 germinating is no seed germinating, so P at least 1 germinates 1 P 0 germinates 1 C 4 0 0750 0254 099609.
(b) P at least 2 germinate P 2 germinates P 3 germinates P 4 germinates C 4 2 0752 0252 C 4 3 0753 0251 C 4 4 0754 0250
094922 (c) P 4 germinates C 4 4 0754 0250 031641
28. (a) P at least 3 boys P 3 boys P 4 boys P 5 boys C 5 3 053 052 C 5 4 054 051 C 5 5 055 050 05
(b) P at least 4 girls P 4 girls P 5 girls P 6 girls P 7 girls C 7 4 054 053 C 7 5 055 052 C 7 6 056 051 C 7 7 057 050 05
29. (a) P all 10 are boys C 10 10 05210 0480 00014456. (b) P all 10 are girls C 10 0 0520 04810 000064925. (c) P 5 in 10 are boys C 10 5 0525 0485 024413. 30. (a) P 2 in 12 C 12 2 022 0810 028347. (b) The complement of “at least 3” is “at most 2”, so
P at least 3 in 12 1 P at most 2 in 12 1 [P 0 in 12 P 1 in 12 P 2 in 12] 1 C 12 0 020 0812 C 12 1 021 0811 C 12 2 022 0810
044165 31. (a) P 3 in 3 C 3 3 00053 09950 0000000125.
(b) The complement of “one or more bulbs is defective” is “none of the bulbs is defective.” So P at least 1 defective 1 P none is defective 1 C 3 0 00050 09953 0014925.
32. P at least 1 in 10 1 P 0 in 10 1 C 10 0 0050 09510 040126
33. The complement of “2 or more workers call in sick” is “0 or 1 worker calls in sick.” So P 2 or more 1 [P 0 in 8 P 1 in 8] 1 C 8 0 0040 0968 C 8 1 0041 0967 0038147
34. P 3 in 5 favor C 5 3 063 042 03456
35. (a) P 6 in 6 C 6 6 0756 0250 017798 (b) P 0 in 6 C 6 0 0750 0256 000024414 (c) P 3 in 6 C 6 3 0753 0253 013184
926
CHAPTER 14 Counting and Probability
(d) P at least 2 seasick 1 P at most 1 seasick 1 [P 6 in 6 OK P 5 in 6 OK] 1 C 6 6 0756 0250 C 6 5 0755 0251 046606 36. (a) P machine breaks P at least 1 component fails 1 P 0 component fails 1 C 4 0 0010 0994 0039404 (b) P 0 component fails C 4 0 0010 0994 0960596 (c) P 3 components fail C 4 3 0013 0991 000000396 37. (a) The complement of “at least one gets the disease” is “none gets the disease.” Then P at least 1 gets the disease 1 P 0 gets the disease 1 C 4 0 0250 0754 068359. (b) P at least 3 get the disease P 3 get the disease P 4 get the disease C 4 3 0253 0751 C 4 4 0254 0750 005078
38. There are 52 cards in the deck of which 13 belong to any one suit, so P heart P spade P diamond 025. P club
(a) P 3 in 3 are hearts C 3 3 0253 0750 0015625 (b) P 2 in 3 are spades C 3 2 0252 0751 0140625 (c) P 0 in 3 are diamonds C 3 0 0250 0753 0421875 (d) P at least 1 is a club 1 P none is a club 1 C 3 0 0250 0753 0578125
39. Fred (a nonsmoker) is already in the room, concerns the remaining 4 participants assigned to the room. exercise sothis (a) P 1 in 4 is a smoker C 4 1 031
073 04116
(b) P at least 1 smoker 1 P 0 in 4 is a smoker 1 C 4 0 030 074 07599
40. (a) P 2 or more 1 [P 0 in 100 P 1 in 100] 1 C 100 0 0020 098100 C 100 1 0021 09899 059673
(b) Since P at least 1 interested 1 P 0 interested and 09835 0507, the telephone consultant needs to make at least 35 calls to ensure at least a 05 probability of reaching one or more interested parties.
41. (a) P 8 or more recover P 0 dies P 1 dies P 2 die C 10 0 060 0410 C 10 1 061 049 C 10 2 062 048 00123 (b) Yes, the drug appears to be effective. 42. (a) P 5 or more hits P 5 hits P 6 hits P 7 hits P 8 hits C 8 5 065 043 C 8 6 066 042 C 8 7 067 041 C 8 8 068 040 0594 (b) No, the coaching does not appear to have made any difference.
SECTION 14.4 Expected Value
927
0.3
43. (a) Number of heads
Probability
0
0003906
1
003125
2
0109375
3
021875
4
0273475
5
021875
6
0109375
7
003125
8
0003906
0.25 0.2 Probability
0.15 0.1 0.05 0
2
4
6
8
Number of heads
If n 8, then 4 heads has the greatest probability of occurring. If the coin is flipped 100 times, then 50 heads has the greatest probability of occurring. 0.3
(b) Number of heads
Probability
0
0001953
1
0017578
2
0070313
3
0164063
4
0246094
5
0246094
6
0164063
7
0070313
8
0017578
9
0001953
0.25 0.2 Probability
0.15 0.1 0.05 0
2
4
6
8
Number of heads
If n 9, then 4 and 5 heads are the most likely outcomes. If the coin is flipped 101 times, then 50 and 51 heads are the most likely outcomes.
14.4 EXPECTED VALUE 1. If a game gives payoffs of $10 and $100 with probabilities 09 and 01, respectively, then the expected value of this game is E 10 09 100 01 $19. 2. If you played the game in Exercise 1 many times then you would expect your average payoff per game to be about $19. 3. Mike gets $2 with probability 12 and $1 with probability 12 . Thus, E 2 12 1 12 15, and so his expected winnings are $150 per game.
4. The probability that Jane gets $10 is 16 , and the probability that she loses $1 is 56 . Thus E 10 16 1 56 0833, and so her expectation is $0833.
1 , the expected value of this game is 5. Since the probability of drawing the ace of spades is 52 1 1 51 49 094. So your expected winnings are $094 per game. E 100 52 52 52 6. The expected value of this game is E 3 12 2 12 52 25. So Tim’s expected winnings are $250 per game.
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CHAPTER 14 Counting and Probability
7. Since the probability that Carol rolls a six is 16 , the expected value of this game is E 3 16 050 56 55 6 09167. So Carol expects to win $092 per game. 2 14 , the probability that Albert gets one tail and one head is 8. The probability that Albert gets two tails is 12 2 2 C 2 1 12 12 , and the probability that Albert gets two heads is 12 14 . If Albert gets two heads, he will receive
$4, if he get one head and one tail, he will get $2 $1 $1, and if he get two tails, he will lose $2. Thus the expected value 1 of this game is E 4 14 1 2 14 1. So Albert’s expected winnings are $1 per game. 2
9. Since the probability that the die shows an even number equals the probability that that die shows an odd number, the expected value of this game is E 2 12 2 12 0. So Tom should expect to break even after playing this game many times.
10. Since there are 4 aces, 12 face cards and only one 8 of clubs, the expected value of this game is 4 26 12 13 1 $1425. E 104 52 52 52
11. Since it costs $050 to play, if you get a silver dollar, you win only 1 050 $050. Thus the expected value of this game 2 050 8 030. So your expected winnings are $030 per game. In other words, you is E 050 10 10 should expect to lose $030 per game.
8 7 56 , and the probability of not choosing 2 white 12. The probability of choosing 2 white balls (that is, no black ball) is 10 9 90 56 34 34 3111. 0 balls (at least one black) is 1 90 90 . Therefore, the expected value of this game is E 5 56 90 90
Thus, John’s expected winnings are $311 per game.
1 1 37 2 00526. 13. You can either win $35 or lose $1, so the expected value of this game is E 35 38 38 38 Thus the expected value is $00526 per game. 1 . After the first prize winner is selected, then 2 106 1 1 P winning the second prize . Similarly, P winning the third prize . So the expected 2 106 1 2 106 2 1 1 1 5 10 104 $0555. value of this game is E 106 2 106 2 106 1 2 106 2 (b) Since we expect to win $0555 on the average per game, if we pay $100, then our net outcome is a loss of $0445 per game. Hence, it is not worth playing, because on average you will lose $0445 per game.
14. (a) We have P winning the first prize
15. By the rules of the game, a player can win $10 or $5, break even, or lose $100. Thus the expected value of this game is 10 5 10 100 2 78 050. So the expected winnings per game are $050. E 10 100 0 100 100 100
16. Since the safe has a six digit combination, there are 106 possible combinations to the safe, of which only one is correct. The 1 106 1 0. expected value of this game is E 106 1 1 106 106 17. If the stock goes up to $20, she expects to make $20 $5 $15. And if the stock falls to $1, then she has lost $5 $1 $4. So the expected value of her profit is E 15 01 4 09 21. Thus, her expected profit per share is $210, that is, she should expect to lose $210 per share. She did not make a wise investment. 3 1 18. Since the wheels of the slot machine are independent, the probability that you get three watermelons is 11 . So the expected value of this game is E 475 13 025 1 13 $0246. 11
11
CHAPTER 14
Review
929
19. There are C 49 6 ways to select a group of six numbers from the group of 49 numbers, of which only one is a winning 1 1 set. Thus the expected value of this game is E 106 1 1 1 $093. C 49 6 C 49 6 20. (a) Since the life insurance policy costs $25 per year, we have the expected value E 7500 25 00003 25 09997 2275. (b) The expected yearly income is 450,000 2275 $10,237,500.
21. The expected number is E 2 015 3 045 4 030 5 010 335 hours of TV. 22. The expected number is 005 3 015 2 045 1 035 0 09 foreign languages.
23. The expected number is 3 030 2 045 1 015 0 010 195 times in any given week. 24. (a) Number of girls
Probability
0
1 8 3 8 3 8 1 8
1 2 3
(b) The expected number of girls is 0 18 1 38 2 38 3 18 15.
1 12 51 1 27 , and so the game is not fair. 25. (a) The expected value is 52 52 2 104 1 x 51 1 0 x 51 $2550. (b) The game is fair with payout x, where 52 52 2 2
26. (a) The expected value is 13 20 23 10 0, and so the game is fair.
3070 10 27. (a) The expected value is 16 16 30 35 36 2 36 9 , and so the game is not fair. 1 x 35 2 0 x $70. (b) The game is fair with payout x, where 36 36
1 28. (a) The expected value is 16 12 10 11 12 1 12 , and so the game is not fair.
(b) The game is fair with payout x, where 16 12 x 11 12 1 0 x $11. 1 1 1 600 1 1 1 1 1 23 , and so the game is not fair. 29. (a) The expected value is 52 6 2 52 6 2 624 1 1 1 1 1 (b) The game is fair with payout x, where 52 6 2 x 1 52 6 12 0 x $623. 30. (a) The expected value is 28 1 68 12 18 , and so the game is not fair. (b) The game is fair with payout x, where 28 x 68 12 0 28 x 38 x $150.
31. If you win, you win $1 million minus the price of the stamp. If you lose, you lose only the price of the stamp (currently 44 cents). So the expected value of this game is 999,99956 expect to lose 39 cents on each entry, and so it’s not worth it.
1 20 106 1 039. Thus, you 044 20 106 20 106
CHAPTER 14 REVIEW 1. The number of possible outcomes is number of outcomes number of outcomes number of ways 2 6 52 624. when a coin is tossed a die is rolled to draw a card 2. (a) The number of 3-digit numbers that can be formed using the digits 1–6 if a digit can be used any number of times is 6 6 6 216.
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CHAPTER 14 Counting and Probability
(b) The number of 3-digit numbers that can be formed using the digits 1–6 if a digit can be used only once is 6 5 4 120. 3. (a) Order is not important, and there are no repetitions, so the number of different two-element subsets is 54 5! 10. C 5 2 2! 3! 2 5! 20. 3! 4. Since the order in which the people are chosen is not important and a person cannot be bumped more than once (no (b) Order is important, and there are no repetitions, so the number of different two-letter words is P 5 2 repetitions), the number of ways that 7 passengers can be bumped is C 120 7 59488 1010 .
5. You earn a score of 70% by answering exactly 7 of the 10 questions correctly. The number of different ways to answer the 10! questions correctly is C 10 7 120. 7! 3! 6. There 2 ways to answer each of the 10 true-false questions and 4 ways to answer each of the 5 multiple choice questions. So the number of ways that this test can be completed is 210 45 1,048,576. 7. You must choose two of the ten questions to omit, and the number of ways of choosing these two questions is 10! 45. C 10 2 2! 8! 8. Since the order of the scoops of ice cream is not important and the scoops cannot be repeated, the number of ways to have a banana split is C 15 4 1365.
9. The maximum number of employees using this security system is number of choices number of choices number of choices 26 26 26 17,576. for the first letter for the second letter for the third letter 10. Since there are n! ways to arrange a group of size n and 5! 120, there are 5 students in this class.
11. We could count the number of ways of choosing 7 of the flips to be heads; equivalently we could count the number of ways of 10! 120. choosing 3 of the flips to be tails. Thus, the number of different ways this can occur is C 10 7 C 10 3 3! 7! 12. The number ways to form a license plate consisting of 2 letters followed by 3 numbers is 26 26 10 10 10 676,000. Since there are fewer possible license plates than 700,000, there must be fewer than 700,000 licensed cars in the Yukon. x! x! 10 20 x x 1 20 13. Let x be the number of people in the group. Then C x 2 10 2! x 2! x 2! x 2 x 20 0 x 5 x 4 0 x 5 or x 4. So there are 5 people in this group.
14. Each topping corresponds to a subset of a set with n elements. Since a set with n elements has 2n subsets and 211 2048, there are 11 toppings that the pizza parlor offers. 15. A letter can be represented by a sequence of length 1, a sequence of length 2, or a sequence of length 3. Since each symbol is either a dot or a dash, the possible number of letters is number of letters number of letters number of letters 23 22 2 14. using 3 symbols using 2 symbols using 1 symbol
16. Since the nucleotides can be repeated, the number of possible words of length n is 4n . Since 42 16 20 and 43 64, the minimum length of word needed is 3.
17. (a) Since we cannot choose a major and a minor in the same subject, the number of ways a student can select a major and a minor is P 16 2 16 15 240. (b) Again, since we cannot have repetitions and the order of selection is important, the number of ways to select a major, a first minor, and a second minor is P 16 3 16 15 14 3360.
(c) When we select a major and 2 minors, the order in which we choose the minors is not important. Thus the number of number of ways number of ways to ways to select a major and 2 minors is 16 C 15 2 16 105 1680. to select a major select two minors
CHAPTER 14
Review
931
18. (a) Solution 1: Since the leftmost digit of a three-digit number cannot be zero, there are 9 choices for this first digit and 10 choices for each of the other two digits. Thus, the number of three-digit numbers is 9 10 10 900. Solution 2: Since there are 999 numbers between 1 and 999, of which the numbers between 1 and 99 do not have three digits, there are 999 99 900 three-digit numbers. (b) There are 1001 numbers from 0–1000. From part (a), there are 900 three-digit numbers. Therefore the probability that 900 0899. the number chosen is a three-digit number is P E 1001
19. Because the letters are distinct, the number of anagrams of the word RANDOM is 6! 720. 20. Because two letters are the same, the number of anagrams of the word BLOB is
4! 12. 2!
21. Because three letters are the same, the number of anagrams of the word BUBBLE is
6! 120. 3!
22. Because there are two sets of four indistinguishable letters (I, S) and one set of two indistinguishable letters (P), the number 11! 34,650. of anagrams of the word MISSISSIPPI is 4! 4! 2! 23. (a) The possible number of committees is C 18 7 31,824.
(b) Since we must select the 4 men from the group of 10 men and the 3 women from the group of 8 women, the possible number of ways to number of ways to number of committees is C 10 4 C 8 3 210 56 11,760. choose 4 of 10 men choose 3 of 8 women (c) We remove Susie from the group of 18so the possible number of committees is C 17 7 19,448.
(d) The possible number of committees is possible number of possible number of possible number of committees with 5 women committees with 6 women committees with 7 women C 8 5 C 10 2 C 8 6 C 10 1 C 8 7 C 10 0 56 45 28 10 8 1 2808
(e) Since the committee is to have 7 members, “at most two men” is the same as “at least five women,” which we found in part (d). So the number is also 2808. (f) We select the specific offices first, then complete the committee from the remaining members of the group. So the number of possible committees is number of ways to choose number of ways to choose P 18 3 C 15 4 4896 1365 6,683,040. a chairman, a vice-chairman, 4 other members and a secretary
24. Method 1: We choose the 5 states first and then one of the two senators from each state. Thus the number of committees is C 50 5 25 67,800,320. Method 2: We choose one of 100 senators, then choose one of the remaining 98 senators (deleting the chosen senator and the other senator form that state), then choose one of the remaining 96 senators, continuing this way until the 5 senators are chosen. Finally, we need to divide by the number of ways to arrange the 5 senators. Thus the number of committees is 10098969492 67,800,320. 5! 2 25. (a) The probability that the ball is red is 10 15 3 .
8 . (b) The probability that the ball is even numbered is 15 2 . (c) The probability that the ball is white and an odd number is 15 7 5 12 4 (d) The probability that the ball is red or odd numbered is P red P odd P red odd 10 15 15 15 15 5 .
26. Let Rn denote the event that the nth ball is red and let Wn denote the event that the nth ball is white. 9 3 (a) P both balls are red P R1 R2 P R1 P R2 R1 10 15 14 7 .
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CHAPTER 14 Counting and Probability
(b) Solution 1: The probability that one is white and that the other is red is C 10 1 C 5 1 10 number of ways to select one white and one red . number of ways to select two balls C 15 2 21 Solution 2: P one white and one red P W1 R2 P R1 W2 P W1 P R2 W1 P R1 P W2 R1
5 10 10 5 10 15 14 15 14 21 (c) Solution 1: Let E be the event “at least one is red”. Then E is the event “both are white”. 5 4 2 . Thus P E 1 2 19 . P E P W1 W2 P W1 P W2 W1 15 14 21 21 21
9 19 Solution 2: P at least one is red P one red and one white P both red 10 21 21 21 (from (a) and (b)). (d) Since 5 of the 15 balls are both red and even-numbered, the probability that both balls are red and even-numbered is 5 4 2 15 14 21 .
2 1 1 . (e) Since 2 of the 15 balls are both white and odd-numbered, the probability that both are white is 15 14 105
27. (a) S H H H H H T H T H H T T T H H T H T T T H T T T . (b) P H H H 18
(c) P 2 or more heads P exactly 2 heads P 3 heads 38 18 48 12 (d) P tails on the first toss 48 12
28. The probability that you select a mathematics book is
4 2 number of ways to select a mathematics book 04. number of ways to select a book 10 5
29. Since rolling a die and selecting a card is independent, 4 1. P both show a six P die shows a six P card is a six 16 52 78 4 1 30. (a) P ace 52 13 (b) Let E be the event the card chosen is an ace, and let F be the event the card chosen is a jack. Then 4 4 2. P E F P E P F 52 52 13 (c) Let E be the event the card chosen is an ace, and let F be the event the card chosen is a spade. Then 4 13 1 4 . P E F P E P F P E F 52 52 52 13 (d) Let E be the event the card chosen is an ace, and let F be the event the card chosen is a red card. Then n E F 2 1. P E F 52 26 n S 1 1 1 1 . 31. (a) Since these events are independent, the probability of getting the ace of spades, a six, and a head is 52 6 2 624 1 1 1 (b) The probability of getting a spade, a six, and a head is 13 52 6 2 48 .
3 1 3 (c) The probability of getting a face card, a number greater than 3, and a head is 12 52 6 2 52 .
32. (a) The probability the first die shows some number is 1, and the probability the second die shows the same number is 16 . So the probability each die shows the same number is 1 16 16 .
(b) By part (a), the event of showing the same number has probability of 16 , and the complement of this event is that the dice show different numbers. Thus the probability that the dice show different numbers is 1 16 56 . 33. (a) Since there are four kings in a standard deck, P 4 kings
C 4 4 1 1 52515049 369 106 . C 52 4 270,725 4321 13121110
C 13 4 4321 11 000264. (b) Since there are 13 spades in a standard deck, P 4 spades 52515049 4165 C 52 4 4321
CHAPTER 14
(c) Since there are 26 red cards and 26 black cards, P all same color
Review
933
2 26252423 2 C 26 4 4321 92 011044. 52515049 833 C 52 4 4321
34. In the numbers game lottery, there are 1000 possible “winning” numbers. 1 . (a) The probability that John wins $500 is 1000 (b) There are P 3 3 6 ways to arrange the digits 1, 5, 9. However, if John wins only $50, it means that his number 159 5 1 . was not the winning number. Thus the probability is 1000 200
35. She knows the first digit and must arrange the other four digits. Since only one of the P 4 4 24 arrangements is correct, 1 . the probability that she guesses correctly is 24
36. The number of different pizzas is the number of subsets of the set of 12 toppings, that is, 212 4096. The number of pizzas with anchovies is the number of ways of choosing anchovies and then choosing a subset of the 11 remaining toppings, that 1 is, 1 211 2048. Thus, P getting anchovies 2048 4096 2 . Note that this makes intuitive sense: for each pizza combination without anchovies there is a corresponding one with anchovies, so half will have anchovies and half will not.
37. (a) Since there are only two colors of socks, any 3 socks must contain a matching pair. (b) Method 1: If the two socks drawn form a matching pair then they are either both red or both blue. So C 20 2 C 30 2 choosing a both red or 051. P P P both red P both blue C 50 2 C 50 2 matching pair both blue Method 2: The complement of choosing a matching pair is choosing one sock of each color. So C 20 1 C 30 1 1 049 051. P choosing a matching pair 1 P different colors 1 C 50 2 38. (a) number of codes choices for 1st digit choices for 2nd digit choices for 5th digit 10 10 10 10 10 105 100,000
(b) Since there are five numbers (0, 1, 6, 8 and 9) that can be read upside down, we have number of codes choices for 1st digit choices for 2nd digit choices for 5th digit 55 3125.
55 n E 1 5 . n S 32 10 (d) Suppose a zip code is turned upside down. Then the middle digit remains the middle digit, so it must be a digit that reads the same when turned upside down, that is, a 0, 1 or 8. Also, the last digit becomes the first digit, and the next to last digit becomes the second digit. Thus, once the first two digits are chosen, the last two are determined. Therefore, the number of zip codes that read the same upside down as right side up is number of codes choices for 1st digit choices for 2nd digit choices for 5th digit 5 5 3 1 1 75. (c) Let E be the event that a zip code can be read upside down. Then by parts (a) and (b), P E
39. (a) Order is important, and repeats are possible. Thus there are 10 choices for each digit. So the number of different Zip+4 codes is 10 10 10 109 .
(b) If a Zip+4 code is to be a palindrome, the first 5 digits can be chosen arbitrarily. But once chosen, the last 4 digits are determined. Since there are 10 ways to choose each of the first 5 digits, there are 105 palindromes. 5 (c) By parts (a) and (b), the probability that a randomly chosen Zip+4 code is a palindrome is 109 104 .
10
40. (a) Using the rule for the number of distinguishable combinations, the number of divisors of N is 7 1 2 1 5 1 144.
(b) An even divisor of N must contain 2 as a factor. Thus we place a 2 as one of the factors and count the number of distinguishable combinations of M 26 32 55 . So using the rule for the number of distinguishable combinations, the number of even divisors of N is 6 1 2 1 5 1 126.
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CHAPTER 14 Counting and Probability
(c) A divisor is a multiple of 6 if 2 is a factor and 3 is a factor. Thus we place a 2 as one of the factors and a 3 as one of the factors. Then we count the number of distinguishable combinations of K 26 31 55 . So using the rule for the number of distinguishable combinations, the number of even divisors of N is 6 1 1 1 5 1 84.
126 7 (d) Let E be the event that the divisor is even. Then using parts (a) and (b), P E nE nS 144 8 . 4 1 41. (a) P king 52 13
8 2 (b) P king or ace 52 13
4 1 number of kings . number of face cards 12 3 number of kings 4 1 (d) The probability that the card is a king given that it is not an ace is . number of non-aces 48 12 (c) The probability that the card is a king given that it is a face card is
42. (a)
(b) (c) (d)
1 3 12 0 1 Because each card is replaced, the probability that all three cards are kings is C 3 3 . 13 13 2197 4 4 4 1 1 Another method:We can calculate the probability as 3 . 52 52 52 2197 13 2 1 1 36 12 The probability that exactly two cards are jacks is C 3 2 . 13 13 2197 0 3 10 1000 3 . The probability that none of the cards is a face card is C 3 0 13 13 2197 “At least one of the cards is a face card” is the complement of the event in part (c), so its probability is 1197 1000 . 1 2197 2197
4 4 5 43. (a) P 4 sixes in 8 rolls C 8 4 16 0026048. 6
(b) There are three even numbers on a die and three odds numbers, so P even P odd 05. Thus P 2 or more evens in 8 rolls 1 P fewer than 2 evens 1 [P 0 evens P 1 even] 1 C 8 0 050 058 C 8 1 051 057 096484
44. (a) P 5 in 5 are white flesh C 5 5 035 070 000243 (b) P 0 in 5 are white flesh C 5 0 030 075 016807 (c) P 2 in 5 are white flesh C 5 2 032 073 03087
(d) P 3 or more are red flesh P 3 in 5 are red flesh P 4 in 5 are red flesh P 5 in 5 are red flesh C 5 3 032 073 C 5 4 031 074 C 5 5 030 075 083692
45. (a) The probability that nine or more patients would have recovered without the drug is C 12 9 0659 0353 C 12 10 06510 0352 C 12 11 06511 0351 C 12 12 06512 0350 0347.
(b) No, the drug does not appear to be effective.
CHAPTER 14
Test
935
46. The probabilities are as follows: Zero heads: C 4 0 070 034 00081 One heads: C 4 1 071 033 00756
Two heads: C 4 2 072 032 02646
Three heads: C 4 3 073 031 04116
Four heads: C 4 4 074 030 02401
Outcome (heads)
Probability
0
00081
1
00756
2
02646
3
04116
4
02401
47. There are 36 possible outcomes in rolling two dice and 6 ways in which both dice show the same numbers, namely, 1 1, 6 1 30 0. 2 2, 3 3, 4 4, 5 5, and 6 6. So the expected value of this game is E 5 36 36 1 , 48. Using the same logic as in Exercise 32(a), the probability that all three dice show the same number is 1 16 16 36 1 35 . Thus, the expected value of this game is while the probability they are not all the same is 1 36 36 1 35 30 E 5 36 1 36 36 083. So John’s expected winnings per game are $083, that is, he expects to
lose $083 per game.
49. Since Mary makes a guess as to the order of ratification of the 13 original states, the number of such guesses 1 is P 13 13 13!, while the probability that she guesses the correct order is . Thus the expected value is 13! 13! 1 1 0 000016. So Mary’s expected winnings are $000016. E 1,000,000 13! 13! 50. The expected number of times Liam goes jogging in any given week is 04 3 01 2 02 1 03 0 16.
CHAPTER 14 TEST 1. The order is fixed, but for each grandchild they have three choices of pictures. Thus, the number of possibilities is 3 3 3 3 81. 2. There are 4 main courses, 3 desserts, and 6 drinks to choose from, so the total number of possibilities is 4 3 6 72. 3. (a) If repetition is allowed, then each letter can be chosen in 26 ways and each digit in 10 ways, so the number of possible passwords is 264 103 456,976,000.
(b) If repetition is not allowed, then the first letter can be chosen in 26 ways, the second in 25 ways, the third in 24 ways, and the fourth in 23 ways. The first digit can be chosen in 10 ways, the second in 9 ways, and the third in 8 ways. Thus, in this case the total number of possible passwords is 26 25 24 23 10 9 8 258,336,000. 4. (a) Order is important in the arrangement, therefore the number of ways to arrange P 30 4 657,720.
(b) Here we are interested in the group of books to be taken on vacation so order is not important, therefore the number of ways to choose these books is C 30 4 27,405.
5. There are two choices to be made: choose a road to travel from Ajax to Barrie, and then choose a different road from Barrie to Ajax. Since there are 4 roads joining the two cities, we need the number of permutations of 4 objects (the roads) taken 2 at a time (the road there and the road back). This number is P 4 2 4 3 12. 6. A customer must choose a size of pizza and must make a choice of toppings. There are 4 sizes of pizza, and each choice of toppings from the 14 available corresponds to a subset of the 14 objects. Since a set with 14 objects has 214 subsets, the number of different pizzas this parlor offers is 4 214 65,536. 7. (a) We want the number of ways of arranging 4 distinct objects (the letters L, O, V, E). This is the number of permutations of 4 objects taken 4 at a time. Therefore, the number of anagrams of the word LOVE is P 4 4 4! 24.
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FOCUS ON MODELING
(b) We want the number of distinguishable permutations of 6 objects (the letters K, I, S, S, E, S) consisting of three like groups of size 1 and a like group of size 3 (the S’s). Therefore, the number of different anagrams of the word KISSES is 6! 6! 120. 1! 1! 1! 3! 3! 8. We choose the officers first. Here order is important, because the officers are different. Thus there are P 30 3 ways to do this. Next we choose the other 5 members from the remaining 27 members. Here order is not important, so there are C 27 5 ways to do this. Therefore the number of ways that the board of directors can be chosen is 27! 1,966,582,800. P 30 3 C 27 5 30 29 28 5! 22! 9. One card is drawn from a deck. 1 (a) Since there are 26 red cards, the probability that the card is red is 26 52 2 . 4 1 . (b) Since there are 4 kings, the probability that the card is a king is 52 13
2 1. (c) Since there are 2 red kings, the probability that the card is a red king is 52 26
10. Let R be the event that the ball chosen is red. Let E be the event that the ball chosen is even-numbered. 5 03846. (a) Since 5 of the 13 balls are red, P R 13 6 4615. (b) Since 6 of the 13 balls are even-numbered, P E 13
5 6 2 9 06923. (c) P R or E P R P E P R E 13 13 13 13
11. Let E be the event of choosing 3 men. Then P E
number of ways to choose 3 men C 5 3 n E 0022. n S number of ways to choose 3 people C 15 3
12. Two dice are rolled. Let E be the event of getting doubles. Since a double may occur in 6 ways, P E
n E 6 1 . n S 36 6
13. There are 4 students and 12 astrological signs. Let E be the event that at least 2 have the same astrological sign. Then E is the event that no 2 have the same astrological sign. It is easier to find E . So number of ways to assign 4 different astrological signs 12 11 10 9 P 12 4 55 . P E number of ways to assign 4 astrological signs 12 12 12 12 96 124 41 Therefore, P E 1 P E 1 55 96 96 0427. 14. (a) P 6 heads in 10 tosses C 10 6 0556 0454 023837. (b) “Fewer than 3 heads” is the same as “0, 1, or 2 heads.” So
P fewer than 3 heads P 0 head in 10 P 1 head in 10 P 2 heads in 10 C 10 0 0550 04510 C 10 1 0551 0459 C 10 2 0552 0458 002739
4 1 , the probability 15. A deck of cards contains 4 aces, 12 face cards, and 36 other cards. So the probability of an ace is 52 13 3 , and the probability of a non-ace, non-face card is 36 9 . Thus the expected value of this game of a face card is 12 13 52 13 52 1 1 3 5 9 85 0654, that is, about $065. is E 10 13 13 13 13
FOCUS ON MODELING The Monte Carlo Method 1. (a) You should find that with the switching strategy, you win about 90% of the time. The more games you play, the closer to 90% your winning ratio will be.
The Monte Carlo Method
937
1 , since there are ten doors and only (b) The probability that the contestant has selected the winning door to begin with is 10 9 . If the contestant switches, he exchanges a one is a winner. So the probability that he has selected a losing door is 10 1 , and the probability that he losing door for a winning door (and vice versa), so the probability that he loses is now 10 9 . wins is now 10
2. (a) You should find that you get a combination consisting of one head and one tail about 50% of the time. (b) The possible gender combinations are B B BG G B GG. Thus, the probability of having one child of each sex is 2 1. 4 2
3. (a) You should find that player A wins about 78 of the time. That is, if you play this game 80 times, player A should win approximately 70 times. (b) The game will end when either player A gets one more head or player B gets three more tails. Each toss is independent, and both heads and tails have probability 12 , so we obtain the following probabilities. Outcome
Probability
H
1 2 1 1 1 2 2 4 1 1 1 1 2 2 2 8 1 1 1 1 2 2 2 8
TH TTH TTT
Since Player A wins for any outcome that ends in heads, the probability that he wins is 12 14 18 78 . 4. (a) If you simulate 80 World Series with coin tosses, you should expect the series to end in 4 games about 10 times, in 5 games about 20 times, in 6 games about 25 times, and in 7 games about 25 times. (b) We first calculate the number of ways that the series can end with team A winning. (Note that a team must win the final game plus three of the preceding games to win the series.) To win in 4 games, team A must win 4 games right off the bat, and there is only 1 way this can happen. To win in 5 games, team A must win the final game plus 3 of the first 4 games, so this can happen in C 4 3 4 ways. To win in 6 games, team A must win the final game plus 3 of the first 54 10 ways. To win in 7 games, team A must win the final game plus 3 5 games, so this can happen in C 5 3 21 654 of the first 6 games, so this can happen in C 6 3 20 ways. By symmetry, it is also true for team B that 321 they can win in 4 games 1 way, in 5 games 4 ways, in 6 games 10 ways, and in 7 games 20 ways. The probability that any particular team wins a given game is 12 ; this fact, together with the assumption that the games are independent allows us to calculate the probabilities in the following table. Series
Number of ways
4 games
2
5 games
8
6 games
20
7 games
40
Probability 2 12 12 12 8 12 12 12 12 20 12 12 12 12 12 40 12 12 12 12 12 12
12 12 12 12
1 8 1 4 5 16 5 16
5 6 5 7 5 13 58. Thus, on average, we expect a World Series to end in (c) The expected value is 18 4 14 5 16 16 16
about 58 games.
5. With 1000 trials, you are likely to obtain an estimate for that is between 31 and 32.
938
FOCUS ON MODELING
6. Modify the TI-83 program in Problem 5 to the following: PROGRAM:PROB6 :0P :For(N,1,1000) :randX:randY :P+(X2 Y)P :End :Disp “PROBABILITY IS APPROX”,P/1000 You should find that the probability is very close to 13 . 7. (a) We can use the following TI-83 program to model this experiment. It is a minor modification of the one given in Problem 5. PROGRAM:PROB7 :0P :For(N,1,1000) :randX:randY :P+((X+Y)1)P :End :Disp “PROBABILITY IS APPROX”,P/1000 You should find that the probability is very close to 12 . (b) Following the hint, the points in the square for which x y 1 are the ones that lie below the line x y 1. This triangle has area 12 (it takes up half the square), so the probability that x y 1 is 12 .
APPENDIXES A
GEOMETRY REVIEW
1. Congruent by ASA 2. Congruent by SSS 3. Not necessarily congruent 4. Congruent by SAS 5. Similar 6. Similar 7. Similar 8. Not similar x 5 x 125 9. 6 150 36 y 10. y 30 y 25 y 21 x 92 7 y , so x 6. 11. 2 32 4 7 214 2 x 12. x 4 x 8 c ac x x 13. a ab ab b c ac 14. x a a x a b
C
15. From the diagram, we see that because AC E F and BC E D, a and b . Thus, ABC AE D E B F.
F
D
A
16.
b»
º
E
Œ
b
B
(a) Because AB F G, a , and so ADG GC F.
C G Œ
a
º
(b) Because a , b 180 a 90 b, and similarly a a.
F
Thus, ADG F E B.
a»
FE AD , so AD E B DG F E. DG EB FE AD , so AD E B DG F E D E2 , and (d) By similarity, DG EB the result follows. (c) By similarity,
A
a
D
82 62 10 18. x 732 552 48 19. x 22 12 3
E
b
B
17. x
939
940
APPENDIXES
20. x 2 3x2 202 10x 2 400 x 2 40 x 2 10
21. x 2 x 22 582 2x 2 4x 4 3364 2x 2 4x 3360 2 x 42 x 40 0 x 40 22. x 2 172 x 12 x 2 289 x 2 2x 1 2x 288 x 144
23. 52 122 169 132 , so the triangle is a right triangle.
24. 152 202 625 252 , so the triangle is a right triangle.
25. 82 102 164 122 , so the triangle is not a right triangle. 26. 62 172 325 182 , so the triangle is not a right triangle. 27. 482 552 5329 732 , so the triangle is a right triangle.
28. 132 842 7225 852 , so the triangle is a right triangle.
29. Let the other leg have length x. Then 112 x 2 x 12 121 x 2 x 2 2x 1 2x 122 x 61 cm. 30. Let the width of the rectangle be x. Then x 2 x 12 1692 2x 2 2x 1 28,561
2x 2 2x 28,560 2 x 120 x 119 0 x 119 ft. Then the length is x 1 120 ft, and the dimensions of the rectangles are 119 ft by 120 ft.
31. If the quadrilateral were a rectangle, we would have 172 212 272 . But this is false, so it is not a rectangle. 32. AB 202 152 25. The area of the triangle can be written as 12 BC C A 12 15 20 150, or as 1 AB h 1 25 h 150 1 25 h h 12. 2 2 2
33. The diagonal of the left face of the box is 32 42 5, so the length of the desired diagonal is 52 122 13. 2 34. (a) a 2 b2 m 2 n 2 2mn2 m 4 n 4 2m 2 n 2 4m 2 n 2 (b) m n a b c 2 m 4 n 4 2m 2 n 2 m 2 n 2 c2 2 1 3 4 5 so a b c is a Pythagorean triple.
35.
8
h a
B
d
b
CALCULATIONS AND SIGNIFICANT FIGURES
1. 327 01834 309
1
8 6 10
3
2
5 12 13
4
1
15 8 17
4
2
12 16 20
4
3
7 24 25
5
1
24 10 26
5
2
21 20 29
5
3
16 30 34
5
4
9 40 41
24 h 8 h and . Thus, 24a 8b b 3a a d b d h 24 4a d, and so h 6. a 4a
By similarity, 24
3
2. 10268 267 1294
APPENDIX C Graphing with a Graphing Calculator
3. 2836 501375 14,220
4.
201,186 3841 5238
5. 1363 252
6.
4273 2067
7. 33 64275 66787 33 70954 2300
701 8. 127105 759
9. 510 103 124 107 6007 106 380
10.
1361107 47717105 5066 1281876
11. The circumference is 2r 2 527 331 ft and the area is r 2 5272 873 ft2 . 12. The volume is 13 r 2 h 13 42672 523 997 cm3 . 2 m m 13. The force is F G 1 2 2 667428 1011 11,4262 266 1012 N. 57,200 r
14. The Sun and the Earth are 150 1011 m apart, with masses 19891 1030 kg and 5972 1024 kg respectively. m m 198911030 59721024 (a) The force is F G 1 2 2 667428 1011 352 1022 N. 2 1501011 r lb 793 1021 lb. (b) The force is 352 1022 N 352 1022 N 0225 1N
C
GRAPHING WITH A GRAPHING CALCULATOR
1. y x 4 2
(b) [0 4] by [0 4]
(a) [2 2] by [2 2] 2
4 3
1 -2
-1
-1
1
2
2 1 0 0
-2
(c) [8 8] by [4 40]
1
2
3
4
20
40
(d) [40 40] by [80 800] 40
800
30
600
20
400
10
200
-8 -6 -4 -2
2 4 6 8
-40
The viewing rectangle in part (c) produces the most appropriate graph of the equation.
-20
941
942
APPENDIXES
2. y x 2 7x 6
(b) [0 10] by [20 100]
(a) [5 5] by [5 5]
100
4 2
50 -4
-2
2
-2
4 0
-4
5
(c) [15 8] by [20 100]
10
(d) [10 3] by [100 20] 100 -10 -8 -6 -4 -2 50
2
-50
-14-12-10-8 -6 -4 -2
2 4 6 8
-100
The viewing rectangle in part (c) produces the most appropriate graph of the equation.
3. y 100 x 2
(a) [4 4] by [4 4]
(b) [10 10] by [10 10] 4
10
2 -4
-2
2
-2
4
-10
-5
-4
10
2
4
-10
(c) [15 15] by [30 110]
(d) [4 4] by [30 110]
100
100
50
50
-15 -10 -5
5
5
10 15
-4
The viewing rectangle in part (c) produces the most appropriate graph of the equation.
-2
APPENDIX C Graphing with a Graphing Calculator
4. y 2x 2 1000
(b) [10 10] by [100 100]
(a) [10 10] by [10 10] 10
-10
-5
100
5
10
-10
-5
-10
5
10
-100
(c) [10 10] by [1000 1000]
(d) [25 25] by [1200 200]
1000 -20
-10
10
20
-500 -10
-5
5
10 -1000
-1000
The viewing rectangle in part (d) produces the most appropriate graph of the equation. 5. y 10 25x x 3
(a) [4 4] by [4 4]
(b) [10 10] by [10 10] 4
10
2 -4
-2
-2
2
4
-10
-5
5
-4
10
-10
(c) [20 20] by [100 100]
(d) [100 100] by [200 200]
100
200 100
-20
-10
10
20
-100
100
-100
-100
-200
The viewing rectangle in part (c) produces the most appropriate graph of the equation. 6. y 8x x 2 (a) [4 4] by [4 4]
(b) [5 5] by [0 100]
4
100
2 50 -4
-2
-2 -4
2
4 -4
-2
0
2
4
943
944
APPENDIXES
(c) [10 10] by [10 40]
(d) [2 10] by [2 6]
40
6 4
20
-10
2
-5
5
-2 -2
10
2
4
6
8
10
From the graphs we see that the viewing rectangle in (d) produces the most appropriate graph of the equation. Note: Squaring both sides yields the equation y 2 8x x 2 x 42 y 2 16. Since this gives a circle, the original equation represents the top half of a circle. 7. y 100x 2 , [2 2] by [10 400]
8. y 100x 2 , [2 2] by [400 10] -2
400
-1
0
1
2
300 200
-100
100
-200 -300
-2
-1
0
1
2
9. y 4 6x x 2 , [4 10] by [10 20]
-400
10. y 03x 2 17x 3, [15 10] by [10 20]
20
20
10
10
-4 -2 -10
2
4
6
8 10
4 256 x 2 . We require that 256 x 2 0 4 16 x 16, so we graph y 256 x 2 in the
11. y
-15 -10
12. y
-5 -10
5
10
12x 17, [0 10] by [0 20] 20
viewing rectangle [20 20] by [1 5].
10 4 0 0
2
-20
-10
10
10
20
13. y 001x 3 x 2 5, [50 150] by [2000 2000]
14. y x x 6 x 9, [10 10] by [250 150]
2000
100 -10 100
-2000
5
-5 -100 -200
5
10
APPENDIX C Graphing with a Graphing Calculator
1 , [2 4] by [8 8] 15. y 2 x 2x
945
x , [10 10] by [02 02] 16. y 2 x 25 0.2
5
0.1 -2
-1
1
2
3
4
-10
-5
-5
-0.1
5
10
-0.2
18. y 2x x 2 5, [10 10] by [10 10]
17. y 1 x 1, [3 5] by [1 5]
10
4 2
-10 -2
2
-5
4
rectangle [4 4] by [1 3], there is no point of intersection. You can verify this by zooming in.
20. Although the graphs of y
49 x 2 and
y 15 41 3x appear to intersect in the viewing rectangle [8 8] by [1 8], there is no point of intersection. You can verify this by zooming in. 8
3
6
2
4
1 -2
10
-10
19. Although the graphs of y 3x 2 6x 12 and 7 x 2 appear to intersect in the viewing y 7 12
-4
5
2 2
-1
4
-8 -6 -4 -2
2 4 6 8
21. The graphs of y 6 4x x 2 and y 3x 18 appear to 22. The graphs of y x 3 4x and y x 5 appear to have have two points of intersection in the viewing rectangle
one point of intersection in the viewing rectangle [4 4]
[6 2] by [5 20]. You can verify that x 4 and
by [15 15]. The solution is x 2627.
x 3 are exact solutions.
10
20 -4
10
-2
2 -10
-6
-4
-2
2
4
946
APPENDIXES
23. x 2 y 2 9 y 2 9 x 2 y 9 x 2 . So we 24. y 12 x 2 1 y 12 1 x 2 graph the functions y1 9 x 2 and y2 9 x 2 in y 1 1 x 2 y 1 1 x 2 . So we graph the viewing rectangle [6 6] by [4 4]. the functions y1 1 1 x 2 and y2 1 1 x 2 in the viewing rectangle [3 3] by [1 3].
4 2
-6
-4
-2 -2
3 2
4
2
6
1
-4 -3
1 4x 2 25. 4x 2 2y 2 1 2y 2 1 4x 2 y 2 2 1 4x 2 y . So we graph the functions 2 1 4x 2 1 4x 2 and y2 in the viewing y1 2 2
-2
-1 -1
[5 5].
4
-4
-2
-2 -4
-0.5
3
2
0.5
0.5
2
26. y 2 9x 2 1 y 2 1 9x 2 y 1 9x 2 . So we graph the functions y1 1 9x 2 and y2 1 9x 2 in the viewing rectangle [5 5] by
rectangle [12 12] by [08 08].
-1.0 -0.5
1
1.0
2
4