Mathematics for IB Diploma Applications and Interpretation HL 2 Worked Solutions [1 ed.] 9781925489873

Table of contents :
Chapter 0
Book Cover
Book Information
Table of Contents
Chapter 1 - Exponentials
Exercise 1A
Exercise 1B
Exercise 1C
Exercise 1D
Exercise 1E
Exercise 1F
Exercise 1G.1
Exercise 1G.2
Exercise 1H
Exercise 1I
Review Set 1A
Review Set 1B
Chapter 2 - Logarithms
Exercise 2A
Exercise 2B
Exercise 2C
Exercise 2D
Exercise 2E
Review Set 2A
Review Set 2B
Chapter 3 - Approximations and Error
Exercise 3A
Exercise 3B
Review Set 3A
Review Set 3B
Chapter 4 - Loans and Annuities
Exercise 4A
Exercise 4B
Review Set 4A
Review Set 4B
Chapter 5 - Modelling
Exercise 5A
Exercise 5B
Exercise 5C.1
Exercise 5C.2
Exercise 5C.3
Exercise 5D
Review Set 5A
Review Set 5B
Chapter 6 - Direct and Inverse Variation
Exercise 6A
Exercise 6B
Exercise 6C
Exercise 6D
Exercise 6E
Exercise 6F
Review Set 6A
Review Set 6B
Chapter 7 - Bivariate Statistics
Exercise 7A
Exercise 7B
Exercise 7C
Exercise 7D
Exercise 7E
Exercise 7F
Exercise 7G
Review Set 7A
Review Set 7B
Chapter 8 - Non-Linear Modelling
Exercise 8A
Exercise 8B
Exercise 8C
Exercise 8D
Exercise 8E
Review Set 8A
Review Set 8B
Chapter 9 - Vectors
Exercise 9A.1
Exercise 9A.2, 9B.1
Exercise 9B.2
Exercise 9B.3
Exercise 9C
Exercise 9D
Exercise 9E
Exercise 9F
Exercise 9G
Exercise 9H
Exercise 9I
Exercise 9J
Exercise 9K
Exercise 9L.1
Exercise 9L.2
Exercise 9L.3
Exercise 9M
Review Set 9A
Review Set 9B
Chapter 10 - Vector Applications
Exercise 10A
Exercise 10B
Exercise 10C
Exercise 10D
Exercise 10E
Exercise 10F.1
Exercise 10F.2
Review Set 10A
Review Set 10B
Chapter 11 - Complex Numbers
Exercise 11A
Exercise 11B, 11C
Exercise 11D
Exercise 11E
Exercise 11F
Exercise 11G
Exercise 11H.1
Exercise 11H.2
Exercise 11H.3
Exercise 11I
Exercise 11J
Review Set 11A
Review Set 11B
Chapter 12 - Matrices
Exercise 12A
Exercise 12B, 12C
Exercise 12D
Exercise 12E
Exercise 12F.1
Exercise 12F.2
Exercise 12F.3
Exercise 12G.1
Exercise 12G.2
Exercise 12H
Review Set 12A
Review Set 12B
Chapter 13 - Eigenvalues and Eigenvectors
Exercise 13A.1
Exercise 13A.2
Exercise 13B
Exercise 13C
Exercise 13D.1
Exercise 13D.2
Review Set 13A
Review Set 13B
Chapter 14 - Affine Transformations
Exercise 14A
Exercise 14B
Exercise 14C
Exercise 14D
Exercise 14E
Exercise 14F
Exercise 14G
Review Set 14A
Review Set 14B
Chapter 15 - Graph Theory
Exercise 15A
Exercise 15B
Exercise 15C
Exercise 15D.1
Exercise 15D.2
Exercise 15E
Exercise 15F
Exercise 15G.1
Exercise 15G.2
Exercise 15H
Exercise 15I
Exercise 15J
Exercise 15K.1
Exercise 15K.2
Review Set 15A
Review Set 15B
Chapter 16 - Voronoi Diagrams
Exercise 16A
Exercise 16B
Exercise 16C
Exercise 16D
Exercise 16E
Review Set 16A
Review Set 16B
Chapter 17 - Introduction to Differential Calculus
Exercise 17A.1
Exercise 17A.2
Exercise 17B
Exercise 17C.1
Exercise 17C.2
Exercise 17D
Exercise 17E
Exercise 17F
Review Set 17A
Review Set 17B
Chapter 18 - Rules of Differentiation
Exercise 18A
Exercise 18B.1
Exercise 18B.2
Exercise 18C
Exercise 18D
Exercise 18E
Exercise 18F
Exercise 18G
Exercise 18H
Review Set 18A
Review Set 18B
Chapter 19 - Properties of Curves
Exercise 19A
Exercise 19B
Exercise 19C
Exercise 19D
Exercise 19E
Exercise 19F
Review Set 19A
Review Set 19B
Chapter 20 - Applications of Differentiation
Exercise 20A
Exercise 20B
Exercise 20C
Exercise 20D
Review Set 20A
Review Set 20B
Chapter 21 - Introduction to Integration
Exercise 21A.1
Exercise 21A.2
Exercise 21B
Exercise 21C
Exercise 21D
Review Set 21A
Review Set 21B
Chapter 22 - Techniques for Integration
Exercise 22A
Exercise 22B
Exercise 22C
Exercise 22D
Exercise 22E
Review Set 22A
Review Set 22B
Chapter 23 - Definite Integrals
Exercise 23A
Exercise 23B
Exercise 23C
Exercise 23D
Exercise 23E
Exercise 23F.1
Exercise 23F.2
Exercise 23G
Review Set 23A
Review Set 23B
Chapter 24 - Kinematics
Exercise 24A
Exercise 24B.1
Exercise 24B.2
Exercise 24C
Exercise 24D
Exercise 24E
Exercise 24F
Exercise 24G
Review Set 24A
Review Set 24B
Chapter 25 - Differential Equations
Exercise 25A
Exercise 25B
Exercise 25C
Exercise 25D
Exercise 25E
Exercise 25F
Review Set 25A
Review Set 25B
Chapter 26 - Coupled Differential Equations
Exercise 26A
Exercise 26B
Exercise 26C
Exercise 26D
Review Set 26A
Review Set 26B
Chapter 27 - Discrete Random Variables
Exercise 27A
Exercise 27B
Exercise 27C.1
Exercise 27C.2
Exercise 27D
Exercise 27E
Exercise 27F
Exercise 27G
Exercise 27H
Exercise 27I
Review Set 27A
Review Set 27B
Chapter 28 - The Normal Distribution
Exercise 28A.1
Exercise 28A.2
Exercise 28B.1
Exercise 28B.2
Exercise 28C.1
Exercise 28C.2
Exercise 28D
Review Set 28A
Review Set 28B
Chapter 29 - Estimation and Confidence Intervals
Exercise 29A
Exercise 29B
Exercise 29C
Exercise 29D
Exercise 29E.1
Exercise 29E.2
Exercise 29E.3
Exercise 29F
Review Set 29A
Review Set 29B
Chapter 30 - Hypothesis Testing
Exercise 30A
Exercise 30B
Exercise 30C
Exercise 30D
Exercise 30E
Exercise 30F
Exercise 30G
Exercise 30H
Exercise 30I
Exercise 30J.1
Exercise 30J.2
Review Set 30A
Review Set 30B
Chapter 31 - x^2 Hypothesis Tests
Exercise 31A
Exercise 31B
Exercise 31C
Exercise 31D.1
Exercise 31D.2
Review Set 31A
Review Set 31B

Citation preview

MATHEMATICS

Applications and Interpretation HL

2

MichaellHaese

MarlqHumphries}

(ChrisiSangwin [NgodVo)

for use with

IB Diploma Programme Mark Humphries Ngoc Vo

Joseph Small Michael Mampusti

dINAOM

HAESE

SNOILNTOS

%

MATHEMATICS: APPLICATIONS AND INTERPRETATION Mark Humphries

B.Sc.(Hons.)

Ngoc Vo Michael Mampusti

B.Ma.Sc. B.Ma.Adv.(Hons.), Ph.D.

Joseph Small

HL WORKED

SOLUTIONS

B.Ma.Sc.

Published by:

Haese Mathematics

152 Richmond Road, Marleston, SA 5033, AUSTRALIA

Telephone:

Email: Web:

+61 8 8210 4666

[email protected] www.haesemathematics.com

National Library of Australia Card Number & ISBN

978-1-925489-87-3

© Haese & Harris Publications 2020

First Edition

2020

Artwork by Brian Houston and Bronson Mathews.

Typeset in Australia by Charlotte Frost and Deanne Gallasch. Typeset in Times Roman 10. This book is available on Snowflake only.

This book has been developed independently of the International Baccalaureate Organization (IBO). This book i in no way connected with, or endorsed by, the TBO. This book is copyright. Except as permitted by the Copyright Act (any fair dealing for the purposes of private study, rescarch, criticism or review), no part of this publication may be reproduced. stored in a retricval system,

or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise,

without the prior permission of the publisher. Enquiries to be made to Hacse Mathematics.

Copying for educational purposes: Where copies of part or the whole of the book are made under Part VB of the Copyright Act, the law requires that the educational institution or the body that administers it has given a

remuneration notice to Copyright Agency Limited (CAL). For information, contact the Copyright Agency Limited.

Acknowledgements: While cvery attempt has been made to trace and acknowledge copyright, the authors and

publishers apologise for any accidental infringement where copyright has proved uniraceable. They would be pleased to come to a suitable agreement with the rightful owner. Disclaimer: All the intemet addresses (URLS) given in this book were valid at the time of publication. While the authors and publisher regret any inconvenience that changes of address may cause readers, no responsibility for any such changes can be accepted by either the authors or the publisher.

FOREWORD This book gives you fully worked solutions for every question in Exercises, Review Sets, Activities, and Investigations (which do not involve student experimentation) in cach chapter of our textbook

Mathematics: Applications and Interpretation HL.

Correct answers can sometimes be obtained by different methods. In this book, where applicable, each

worked solution is modelled on the worked example in the textbook.

Be aware of the limitations of calculators and computer modelling packages. Understand that when your

calculator gives an answer that is different from the answer you find in the book. you have not necessarily

made a mistake, but the book may not be wrong either.

Please contact us if you notice any errors in this book.

MH

e-mail: web:

[email protected] www.haesemathematics.com

JS

NV

MM

TABLE OF CONTENTS

Chapter 1

EXPONENTIALS

Chapter 2

LOGARITHMS

Chapter 3

Chapter 4

68

APPROXIMATIONS

AND

ERROR

99

Chapter 5

LOANS AND ANNUITIES

114

MODELLING

141

Chapter 6

DIRECT AND INVERSE VARIATION

180

Chapter 7

BIVARIATE STATISTICS

207

NON-LINEAR MODELLING

260

VECTORS VECTOR APPLICATIONS

297

COMPLEX NUMBERS

434

Chapter 12

MATRICES

504

Chapter 13

EIGENVALUES AND EIGENVECTORS

555

Chapter 14

AFFINE TRANSFORMATIONS

616

Chapter 15

GRAPH THEORY

671

Chapter 16

VORONOI

735

Chapter 17

INTRODUCTION TO DIFFERENTIAL CALCULUS

777

Chapter 18

RULES OF DIFFERENTIATION

807

Chapter 19

PROPERTIES OF CURVES

Chapter 20

APPLICATIONS

875 964

Chapter 8

Chapter 9

Chapter 10 Chapter 11

Chapter 21

DIAGRAMS

INTRODUCTION

OF DIFFERENTIATION TO INTEGRATION

386

1015

Chapter 22

TECHNIQUES FOR INTEGRATION

1041

Chapter 23

DEFINITE INTEGRALS

1076

Chapter 24

KINEMATICS

1135

Chapter 25

DIFFERENTIAL EQUATIONS

1190

COUPLED

1232

Chapter 26 Chapter 27

DIFFERENTIAL EQUATIONS

DISCRETE RANDOM

VARIABLES

1282

Chapter 28

THE NORMAL DISTRIBUTION

1332

Chapter 29

ESTIMATION AND CONFIDENCE INTERVALS

1380

Chapter 30

HYPOTHESIS TESTING

1418

Chapter 31

x? HYPOTHESIS TESTS

1468

Chapter 1

EXPONENTIALS

2'x 2 —gtt3

X

—27%

=93

h (V2 3_= (27)(o518

B

72-5>

g

2

¢ 167 = (2)%

=2°

=2

=8

2 3 d 257 = (5%)7 =5

=

=k27

i

~1

= (273

=i 2

g 9 =372 =33

L4z

=271

=125 3

S

Chapter 1 (Exponentials)

Exercise IB_

7

EXERCISE 1

) 11

a

22xz? — ,T

2

a

3

b

1

zZxz2

=27 &

P +222+1) =2?x2®+a? x2?+a? x1

1o

2zt ?

b

or

VT

27(2"+1) =2"x2"4+2"x1

=27 42"

:

d

B B 2 ,z]xzu +z2xT =z

zxz ?

=l :z

0

=2°+22' +2* €

_3

¢

T(T7+2)

=T"xT+T" )

%2

+z

=z+1

e

=

3*(2-37)

—3Tx3

v:;:;z

33

*3

f

"

=z?

=2(3)~

£

27%(2°+5)

e B

h

3§

=145277) 1

i,

x’z

i

*XT

+T

=5

I

a

s 1 =z?+4+z7+1 1 1 oz %222 —z+527) : 1

=2

*x2?—z P

2XT

+5%)

x5% 4577 x 5%

3B +5+377)

=3 X3 +F X543 x3F :3h+5(3,)+30

=3 +5(3%) +1

'

*xzl4+z? x527

1

:

2227 3272

L e

— gt _g(g2e) 90 — 9% 397 1

=227 — 27 +52° 3

11

X222 +122 X3z

=5+1

1

+T

57%(5%

=27 427 +20

k

1

=5"+50

2 i@’ +a+a?)

_13

Xz? +x?

=2"+20+3

=20 45(277)

i

Loz

=72 +2z' +32°

=230=1 g

iz L + 32 _i7) 2% (22 +227

1

=222 —2? +5

3

a2

(2-DE'+3)

=2"x2"42"x3-1x2"-3

=22 4+ 2(2%) -3

oy

getl 3

b

(3 +2)(3"+5)

=3

x3"+3"x5+2x37+10

=3% +7(3%) + 10

_i

?

Chapter 1 (Exponentials)

Exercise 1B

(5" ~2)(5" ~ 1)

=5"%x5"—5"

d

(2°+3)?

=(2°)2 2"+2x x3+3%

x4-2x5"+8

=5% —6(5%) +8

=22 46(2°) +9 (1t +7)?

@ -1?

=)

=(3°)?-2x3"x1+12

=3 -2(3") +1

+2x 4" x T+17

=477 1 14(4%) + 49

(F +2)(T —2)

h

(2 +3)2-3)

=277 -3 =229

=4-9

i

(e =

k

=

77%)2

=P

=2 RTR T+

=7 _2x 7P +77% =7 _2477%

5% 45

=57 X 57+ 5 =5°(5"+1)

1

(777)?

b

s g —5x5"—5 5(5" — 1) x;m -

x

y

2 +4+I—24

-2 =52-2x5x27"+(277)? =25-10(27") +27%

=3"(3%+1)

™ =TT =T +7*

6" 6

4 16

3n+2 4 gn

“©

=343

=6x6"" -6

=66

=42x4"-16

=16x4"-16 =16(4" - 1)

—1)

gntl | gn—t

92 _ on+3

=

2

=10(3")

I

8

x 2P

=2"x2"-2"x8

=27(2" -8)

="t

—on 1y

4fl+1

— (@21 gmnmt

x 22 4ot gqon-l

=5(2"1)

9°—4

4725

—

= (3 +2)(3° - 2)

=(2°+5)(2° - 5)

(21)2

— o2

g1

—92n—1

3 93 4 g2n—1

=271 g4 221 =9(2*"1) 16— 9°


+6(2) +9

= (3 +27)(3" - 27)

= (5+2)(5-2) 97 +10(3%) +25

h

=(3%)2+10(3%) +25

4462749 =(2*+3)?

{a® +6a+9=(a+3)*}

47— 14(27) +49 =(2%) —14(2%) +49

=@-7? {a® - 14a+49 = (a—T7)%}

= (3 +5)* {a® +10a + 25 = (a+5)%}

257 —4(5%) +4

=(5%)2 - 4(5%) +4

=(5"-2)*

{a® —da+4=(a—2)%}

(252 —22 -2 =(2°+1)(2°-2)

b

£ -7(2%) +12

d

{a?—a-2=(a+1)(a—2)} = (22 -7(2%) +12 {a® - Ta+12=(a—3)(a—14)}

47 4+9(2%) +18 = (292 +9(2°) +18 =(2°+3)(2° +6)

{a®*+9a+18 = (a+3)(a+6)}

f

=(2°)2 -2 -20

=(3+3)(3*-2)

{a*+a—6=(a+3)(a—2)}

=(2"-3)(2"-4) 47 —2720

(3°)2+3°-6

97 +9(3%) +14

=(3°)2+9(3°) + 14

=(+2)(3+7)

=(2°+4)(2" - 5) {a® —a—20=(a+4)(a—5)}

{a®>+9a+14=(a+2)(a+7)}

h

=(3"+5)(3"-1)

25" 45 -2 =) +5" -2 =(5*+2)(5° - 1)

{a® +4a—5=(a+5)(a—1)}

{®+a-2=(a+2)(a-1)}

97 +4(3%) -5

=(3)+43) -5

497 — 7" 112 =(7)?-17(7") +12

= (7" —4)(7" —3)

{a® —Ta+12=(a—4)(a—3)} 12" 6"

e12"6"

_2"6"

_(12\"

—on

—on

=

=3

b

X

-

-

2910%

= 10"

4

2

2a

L@

=107

a

9

10

Chapter | (Exponentials) Exercise 1B 60

cC

3

6"

"

2bgh

3

= T

=

=3

=3

—3b

6\b

77

or

5°7%

g

7z

am\®

24k

T

(4"

('zn)

-0

_(1\"

F

6

or

8_“

_

=)

30

=3

(gye 3

sntl

h

_

=

=(%)(2\k

_ 5!

=)

=9

—

_

e s

W

=

_ 293¢

=5* or

"

=!

=)

9

w

4n

T anpn

35°

_

an

=

(7)

_gb

357

e

d

= =g

(8

i

8

6 4 am

a

—

om

2n 4 19n

—

b

on

o1

an

_ gmgm 4 om

_2n 4 onen

_amam

_2m@Em 1) T

_2M(146m) T

_2n(an 42 T

-

om

-

3" 1

4

19t

3¢

3¢

on

-

=1+6"

.

e

142n _6npenan

g

T T 1+om

—5 _ six5_5" =

B

h

q4n —

g

=

=2

on

g

n(m —1) o

=4

=271

omen

o 1 5% x5 — 5" 4

_en142m)

sn+l _gn

on

=442

T+

S

8


0, the graph lies above the asymptote. k0.

16

Chapter 1 (Exponentials)

b

i

Exercise 1E

Each point will become 1 times its previous distance from the y-axis.

q

If 0 < g 1, g =1, the graph remains unchanged. il

the graph becomes steeper, and if

The graph is stretched horizontally with invariant y-axis and scale factor l.

q

iii The horizontal asymptote y = 0 will remain the same.
2.

y =27 y = 10"

a#1,

have

above

horizontal asymptote and are increasing.

as

a>0,

y =

the

0

y = 2%

corresponds to €, and corresponds to B.

the graphs of y =a "

and

y = (—1~)

a

are

Chapter 1 (Exponentials) y=-5%

has

k

1.

the graph lies below the horizontal asymptote y =0 and is decreasing. y = —5"

d y=(4)"

corresponds to E.

has k>0

and 0—3}.

Chapter 1 (Exponentials) e

Exercise 1E

y=2-2°

i When z=0,

y=2-2°

il The horizontal asymptote is y = 2.

—2-1

=1

the y-intercept is 1.

il

When

When

z=2,

y=2-2% =2-4 ==2 z=-2, y=2-2"2 =2-= 1

v

=8_1

i~3

=1 1 v f

The domainis

{z |z €R}.

Therangeis

{y|y3}.

21

22

Chapter 1 (Exponentials)

g

Exercise 1E

y=3-277 i

When

=0,

y=3-2°

il

=3-1

The horizontal asymptote is y = 3.

=3

the y-intercept is 2.

iiil When

z=2,

y=3-272 =35

iv

1

=12 _1

i1 _u 1

When z=-2,

y=3-2"02 =322

=3-4 =-1 v

The domainis

{x |z € R}.

Therangeis

{y|y < 3}.

ix3T41

hy

i When

o=0,

y=—-§x3"+1 -3 x1+1

i The horizontal asymptote is y = 1.

2

the y-intercept is . ili

v

When

z =2,

When

z=-2,

y=-1x3"?+1

The domain is {x | x € R}.

The rangeis

{y|y < 1}.

Chapter 1 (Exponentials) Exercise IE a

y=kx2+c Substituting

(0, —5)

into the equation gives

—5=kx2"+c o k+e=—5

Substituting

.. (1)

(2, 10)

into the equation gives

10=kx2+c ©.

dk+c=10

Now

.. (2

(2)— (1) gives

3k=15 . k=5 andso c=-10.

b y=5x27-10 When

10

a

z=6,

y=5x2°—10

=5x64—10 =310

f(0)=35-ad" 35-1 =25 the y-intercept is 2.5.

Pis (0,25). b

The point

(—1,2)

lies on the graph.

o f(-1)=2

© 35-a'=2

- a=15

¢

1

f(z)=3.5—1.5"%

has horizontal asymptote

a y=kx2+c Substituting

(0, 4)

into the equation gives

4=kx2+c

kt+e=4 Substituting

s Now

.. (1)

(2, 7)

into the equation gives

T=kx2+c

dk+e=T7

(2)—(1)

.2

gives

3k=3

. k=1

andso

the exponential model is y = 2% + 3.

c=3.

y =3.5.

23

24

Chapter| (Exponentials) b

Exercise 1E

y=kx3+c Substituting

(—1, 3)

into the equation gives

3=kx3'+¢

tkte=3 Substituting

.. ()

(1, —13)

into the equation gives

-13=kx3"+c © 3k4+e=-13 ... (2) Now

(2)—(1)

gives

$k=-16 . k=-6

andso

c=5.

the exponential model is y = —6 x 37 + 5.

¢

y=kx2"%+c Substituting

(-2, 9)

into the equation gives

9=kx2"4¢ =kx2¥+ec L dkte=9 .. (1) Substituting

(1, —5)

into the equation gives

—5=kx2"'+c

lkte=-5 Now

(1)—(2)

..

gives

k=14 .

k=4

the exponential model is

@

|

The

graph

lies

c=-T.

y =4 x 277 — 7.

above

the

horizontal

asymptote.

k must be positive.

il

The horizontal asymptote lies below the T-axis.

¢ must be negative.

y=kx4"

+c

Substituting (1, —4)

into the equation gives

—4

=k x4' 4+¢

s dkte=—4 .. (D) Substituting (2, 2) into the equation gives 2=k x 4%+ ¢ 16k+c=2 ... (2) gives

12k=6

- k=

i &

(2)— (1)

&

Now

£

b

m_

12

andso

the exponential function is y = 4 x 47 — 6.

Chapter 1 (Exponentials)

=0,

y=3x4"—6

1-6

=

x

ol

¢ When

=

5 T the y-intercept is —4%.

d The horizontal asymptote of y =4 x 4% —6 is y=—6. 13

f(r)=kxa"+c The graph has horizontal asymptote The y-intercept is 10, so

.

y =2,

so

c¢=2.

f(0) = 10

kxa®+2=10 . k=8

The graph passes through

(5, 258),

so

f(5) = 258

. 8xa°+2=258

. 8a° =256 s G082

;. a=2

So, the exponential function is f(z) =8 x 27 +2.

14 fz)=kxa*+c The graph has horizontal asymptote The y-interceptis 1, so

y =4,

so

c=4.

f(0) =1

L kxa®+4=1 . k=-3

The graph passes through (L. 2), so f(1) =2 . —3xa'+4=2

a . a=}

)

So, the exponential function is f(z) = —3 x ()" +4.

15

f(z) =ka"+c f)=11 - kate=11 f2)=17 f3)=29 Now

(2)—(1)

.

ka’+c=17 ka®+c=29

gives

.. (1) .. (2

3)

ka®>—ka=6 o kala—1)=6

and (3)—(2) gives ka® —ka® =12

.. (4)

. ka*(a—1)=12

. alka(a—1)] =12 6a=12 s a=2

{using (4)}

Exercise IE

25

26

Chapter 1 (Exponentials)

Exercise

Substituting a = 2 into (4) gives Substituting @ =2 and k=3

1E

k(2)(1

k=3

into (1) gives

3(2)+c= 11 .

..

c=5

the exponential function is f(x) = 3 x 2% + 5.

f0)=3x2045=8 the y-intercept is 8.

16

f(z)=ka " +c f-2)=21 ka®+e=21 f(Hy=0 ©okal4e=0 f@)=-3 L ka?+e=-% Now

(1)—(2)

gives

.. (1) .. (2) ..(Q3)

ka® —ka ' =21

ka® — k= 21a . k(a®—1)=21a 2la 2 ln—'L:,_] and

(2)—(3)

gives

ka™'

- (4)

—

.. (5)

.

+

2la

Equating (4) and (5) gives

3a?

Fo1-

Using technology,

26D

a =2

{a>0}

Substituting a =2 into (4) gives k= 2‘3152)1 S k=6 Substituting

a =2

and

k =6

2

into (1) gives

6(2)? +c¢ =21 . 24+c¢=21 Loe=-3

So, the exponential function is f(z) =6 x 27% — 3.

a

f(0)=6x20-3

=3

the y-intercept is 3. b

The horizontal asymptote is y = —3.

1

a

Exercise 1F

il

and

ii

|

when

2" =06

when

Wegraph

Y,

z~16. z~—0.7.

=2%

and

Yy

=3

on the same set of axes, and find their

point of intersection.

[EXE]:Show coordinates

¥1=2°(x)

v2-3

"

We graph

%=1.584902501

Wegraph

Y; =2%

B [EXETshon coordinates ¥1=2%(x)

[T 7309055042 ¥=0.0. v

Yo =20

B Texcishon coordimates y2=20" = T

e it amiopaoms_vzo

on

The solution is z ~ 4.32.

We graph

Y; = 4%

and

v

Yz =100

on

the same set of axes, and find their point of intersection. @

wisecr|

msect

The solution is 2 ~ —0.737.

the same set of axes, and find their point

of intersection.

m

[

_1¥=3

and

Yy = 0.6

point of intersection.

/

The solution is z ~ 1.58.

Y, = 2%

on the same set of axes, and find their

44‘_”““

a

27

From the graph:

i 2" =3

b

Chapter 1 (Exponentials)

[EXE]:Show coords

=3.321926005_

¥=100

The solution is 2 ~ 3.32.

Chapter | (Exponentials)

¢

Wegraph

Y; =3%

and

Y, =30

We graph Y; = (1.2)% and Y2 =3 on

on

the same set of axes, and find their point of intersection.

= o [x=3. 085003274

e

Exercise 1F

the same set of axes, and find their point

of intersection.

——wTSECT 5

o0

The solution is z ~ 3.10.

The solution is z ~ 6.03.

Wegraph Y; = (1.04)% and Y» =4.238

We graph Y; = (0.9)% and Y =0.5 on

on the same set of axes, and find their

the same set of axes, and find their point

The solution is z ~ 36.8.

The solution is z =~ 6.58.

point of intersection.

a Wegraph Y; =3x2X

and Y, =93 on

the same set of axes, and find their point

of intersection.

The solution is = ~ 4.95.

¢ We graph

Y; =8x 3%

and

Yo =120

point of intersection.

v-120

The solution is z ~ 2.46.

b

We graph Y = 40x(0.8)X and Y5 = 10

on the same set of axes, and find their point of intersection.

The solution is = ~ 6.21.

on the same set of axes, and find their

— 5 [1=2.400073821

of intersection.

We graph Y2

=

34

Y; = 21 x (1.05)%

and

on the same set of axes, and

find their point of intersection.

et The solution is z ~ 9.88.

Chapter1 (Exponentials) Exercise IF e

We graph Y; =500 x (0.95)% and

We graph

Y2 = 350

Y> = 470 on the same set of axes, and find their point of intersection.

on the same set of axes, and

find their point of intersection.

Y; = 250 x (1.125)

[=5.350815043 The solution is z ~ 6.95.

g

Wegraph

Y; =3%+5

and

Y =50

point of intersection.

[B_[ExEl:Show coordinates

and

Y2 = 30 on the same set of axes, and find their point of intersection.

[x=7. 228289519 o, Oy-s0 The solution is x ~ 7.23.

When z=-2,

y=5 5=4xa2-7 12=4a"2 a® = La=m

{a >0}

and

Y> = 80 on the same set of axes, and find their point of intersection.

The solution is =~ 1.71.

Y; =20+ 80 x (0.75)%

a>0

Y, = 60+ 10 x (1.5)X

P

v=50

y=4xa"-7,

We graph

— 5Tt x=1.700511201 "¥=80

The solution is z ~ 3.46.

i We graph

y=470

[8_[EXE]:Showa coordinates B < B oo vo=so

——= o —+—+—tmeect|

[x=3.464973821

and

The solution is z =~ 5.36.

on the same set of axes, and find their

a

29

1 :

Chapter 1 (Exponentials) Exercise 1F b

,1/:4x(fi)’—7 The z-intercept occurs where

y =0

8 /l>0,

d>0

As the intensity of light diminishes as the depth increases, we would expect that

0 < a < 1.

40

Chapter | (Exponentials)

b

Exercise 1G.2

When d=1, L=95 © 10xa' =95 .


0 So, the domainis {z | > 0} and the rangeis il

As

z— 0%,

f(z) — —oc,

As

z—

f(x)— oc,

f(0) When

o0,

so the vertical asymptote is x = 0. so there is no horizontal asymptote.

is undefined, so there is no y-intercept. f(z) =0,

logz=2

v @=10

oz =100 So, the z-intercept is 100.

{y |y € R}.

Iny=3-=z

- oy=etT @) =T

79

80

Chapter 2 (Logarithms)

b

Exercise 2D

f(z) =log(z+1) i

log(x+1)

is defined when

z+1>

So, the domain is {x |z > —1} il

As z — —1%, As

x — o0,

f(z) — —oc, f(x)— oo,

f(0) =log1 =0, When

f(z)=0,

0, thatis, when

= > —1

and the range is {y |y € R}. so the vertical asymptote is = = —1.

so there is no horizontal asymptote.

so the y-intercept is 0. log(z+1)=0

L z+1=10° . z=0

So, the z-intercept is 0.

¢

f(z)=2logz+1 i il

logz is defined when

= >0

So, the domain is {x | z > 0} and the range is {y | y € R}. As

= — 0%,

f(z)— —oc,

so the vertical asymptote is = = 0.

As & — o0, f(z) — oo, so there is no horizontal asymptote. f(0) is undefined, so there is no y-intercept.

When f(z) =0,

2logz+1=0

- 2logr=—1 - logz=-%

s

LT=10T7 = s So, the z-intercept is V‘E'

Chapter 2 (Logarithms)

d

Exercise 2D

f(z)=1-logz i

logx is defined when

z > 0

So, the domain is {x | 2 >0} ii

As

2 — 0%,

f(xr) — oc,

As

x—

f(x)— —o00,

00,

and the range is {y |y € R}.

so the vertical asymptote is = = 0. so there is no horizontal asymptote.

£(0) is undefined, so there is no y-intercept.

When

f(z)=0,

1—logz=0

logz =1

So, the a-intercept is 10. i

x =10

y

y=1-loga

3

a

For

y=Inz,

when

y =0,

Inx

z=1 Ais

y=Inz

asits z-intercept is 1,

so B must be y = In(z — 2).

b

y =In(z —2)

is a horizontal translation of

y=Inz, 2 units to the right. y = In(z +2) is a horizontal translation of y=Inz,

2 units to the left.

y=In(z+2)

81

82

Chapter 2 (Logarithms) Exercise 2D ¢

y=Inz As

has domain

x — 0%,

y—

y=In(z —2)

{x |z >0}

—o0,

so

has domain

As

x— 2T,

y — —o0,

As

¢ — 2%,

y=Inz

has vertical asymptote

{z |z >2}. so

y=In(z—2)

has vertical asymptote

y=1In(z +2) has domain {z|a > —2}. 4

y=In(z?),

=2z

y—

z = 0.

—o0,

so

y=In(z+2)

z = 2.

has vertical asymptote

z =

>0

{mlnb=1In(®™)}

So, the y-values are twice as large for y = In(z?) as they are for y = Inx. Therefore, yes, Kelly is correct.

5

a y=In@®) = 3lna isa . vertical stretch ; of y=Inz ¥

b y — 1..(1)z — ()

with scale factor 3.

reflection of y =Inz ¥

y=In(z%

= ~Ing isa in the z-axis.

y=Inz

¢

y=In(z+e) of y=Inz,

y=Inz

is a horizontal translation

d

e units to the left. y=In(z +e)

l-ex~-172

e

For £>0,

y=In@a®) +2=2z+2

upwards. For z 0,

in

y=In(z—2)—3

is a

y=Inxz

2 3 )

through

translation of

Chapter 2 (Logarithms)

6

a ln(es?) =lne+In(z?) —1+42lne {z>0}

7

a When

=2,

y=b

Y

Exercise 2D

y=a+2logz

b=>5—1log2

=log(10°) —log2

= log(1232) = log 50 000

b When =2, y=b=1og50000 {froma} *. a+2log2 = log50000 . a=1log50000 — 2log2 = log 50 000 — log(2?)

= log((20) =log 12500

8 fizx—e¥ a

gizm2-1

f is defined by - flisdefinedby

y = e** z=e%

© 2y=Inz

b

(9o f)(x) =g(f(z)) =g(e*)

=2e* —1 (go f)(x) is definedby

= (gof)"\(z)

isdefined by

y=2e*"—1

z=2¢* —1

oz 4+1=2e%

y_T+1 2v_zt

=1m@2z-1)

83

84

Chapter 2 (Logarithms)

Investigation 2

Logarithmic scales

|

INVESTIGATION 2 1

Ay 100

70

/L]

S

oo}—— e

’/

i

:

(]

=Y

20

2

As the y-values approach a power of 10, the minor tick marks become closer together. The spacing of the y-values 1, 2, ..... 9, and the y-values 10, 20, ...., 90, matches the spacing of the numbers on the slide rules in the Opening Problem.

3

From the graph, each distance measures ~ 4.7 cm. log 10 — log 1

log 20 — log2

log 50 — log 5

—10

= log(%)

=1

=log 10

=log10

=

=1

= log(%)

Chapter 2 (Logarithms)

&

log(5x 10%) = log5+log(10°)

and

Exercise 2E

log(10%%) = 2.5

=log5+2 =270

Since log(5 x 10%) > log(10*®),

1

a

5 x 10

is further to the right on the logarithmic scale.

P corresponds to the value

107" i = 5ok

Q corresponds to the value

10~! = 5

b

i log(k=)

=1

=3

*(W):f("’ 2

R

il

log(fi)

=log R

P

)

3

R 2

Qi -1

i 0

R 1

2

—log10

=logR—1
Cy,

vinegar is more acidic.

s Cv=10724

~0.00398

88

Chapter 2 (Logarithms)

Exercise 2E

il If C=200Cy,

iii

pH = —log(200Cy) = —1log 200 — log Cy = —1log 200 + 6.6 ~4.30

The H30™ concentration in vinegar ga is about

10724~ (=430) ~ 10190 x~ 79.2

times

greater than in the substance in ii.

d

If C>1,

then

logC >0

. pH=—logC

=h Fy

.

FA . log (F_o> =g =249

Ep _ 24

y

. FA=10*F, Now,

the brightness of an object varies inversely with the square of its distance from

the observer, so if star B is 2 times further away than star A but has the same apparent

magnitude, then star B must be 22 =4 times brighter than star A. Fy = 4F), 2.4, . the absolute magnitude of star Bis M = —2.5log (%’”’) 0

= —2.5log(4 x 10*4) = —2.5(log 4 + log(10*4))

= —2.5(log 4+ 2.4) ~ -7.51

b

From 2 d i, the observed flux density of the Sun is 10696 F;, If the Sun was 10 parsecs away from the Earth, then the observed flux density would be

T

—6

.

P

. the Sun has absolute magnitude ~

2

—2.5log (M)

14

0

~ —2.5log(2.35 x 1072:304)

~ —2.5(log 2.35 — 2.304)

~4.83

1

a

logvi0

b

=— log(10? 3

=loxtllF) a

log 27

3

a 32=10%*

~1.431

~ 1015051

¢

~log(107%)

—1 -3 b

log(0.58)




b

= i“(fs) B 91“(3 ) :1n(144x )

iln9-In2 1

S -2 —In3-In2

== 5

a

=log

a

log16 +2log3

b

16 +log(3?)

log36

b

= log(4 . x 9) = log(2?

= 1“(525) -u(Z)

log16 —2log 3

= log 16 — log(3?)

loghd