Calculus Concepts and Applications 2nd Edition Solutions Manual

C o n c e p t s a n d A p p l i c a t i o n s Second Edition

P a u l A . F o e r s t e r

Solutions Manual

Calculus

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2005 by Key Curriculum Press. All rights reserved.

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10 9 8 7 6 5 4 12 11 10 09 08

ISBN: 978-1-55953-657-8

Contents

Chapter 1 Limits, Derivatives, Integrals, and Integrals .................................................... 1

Chapter 2 Properties of Limits ................................................................................................... 9

Chapter 3 Derivatives, Antiderivatives, and Indefinite Integrals .............................. 28

Chapter 4 Products, Quotients, and Parametric Functions ......................................... 51

Chapter 5 Definite and Indefinite Integrals ....................................................................... 82

Chapter 6 The Calculus of Exponential and Logarithmic Functions ..................... 118

Chapter 7 The Calculus of Growth and Decay ............................................................ 139

Chapter 8 The Calculus of Plane and Solid Figures ................................................... 168

Chapter 9 Algebraic Calculus Techniques for the Elementary Functions .......... 213

Chapter 10 The Calculus of MotionAverages, Extremes, and Vectors ............. 266

Chapter 11 The Calculus of Variable-Factor Products ................................................. 292

Chapter 12 The Calculus of Functions Defined by Power Series ............................. 313

iii

Overview

This Solutions Manual contains the answers to all problems in Calculus: Concepts andApplications. Solutions or key steps in the solutions are presented for all but the simplestproblems.

In most cases the solutions are presented in the form your students would be expected touse. For instance, decimal approximations are displayed as exact answers using ellipsisformat for a mathematical-world answer, then rounded to an appropriate number ofdecimal places with units of measurement applied for the corresponding real-worldanswer. An answer such as f(3) = 13.7569... 13.8 cm indicates that the precise answer,13.7569... , has been retained in memory in the students calculator without round-off forpossible use in subsequent computations. The ellipses indicate that the student chooses notto write all the digits on his or her paper.

Because the problems applying to the real world may be somewhat unfamiliar to both youand your students, fairly complete solutions are presented for these. Often commentary isincluded over and above what the student would be expected to write to further guide yourevaluation of students solutions, and in some cases reference is provided to later sectionsin which more sophisticated solutions appear. Later in the text, the details of computingdefinite integrals by the fundamental theorem are omitted because students are usuallyexpected to do these numerically. However, exact answers such as V = 8/3 are presentedwhere possible in case you choose to have your students do the algebraic integration.

Solutions are not presented for journal entries because these are highly individual for eachstudent. The prompts in most problems calling for journal entries should be sufficient toguide students in making their own responses.

Where programs are called for, you may use as a model the programs in the InstructorsResource Book. Check the publishers Web page (see the address on the copyright page ofthis manual) for further information on programs for specific models of the graphingcalculator.

If you or your students find any mistakes, please report them to Key Curriculum Press bysending in the Correction/Comment Form in the back of this book.

Paul A. Foerster

v

Calculus Solutions Manual Problem Set 1-2 1 2005 Key Curriculum Press

Chapter 1Limits, Derivatives, Integrals, and Integrals

Problem Set 1-11. a. 95 cm

b. From 5 to 5.1: average rate 26 34. cm/sFrom 5 to 5.01: average rate 27 12. cm/sFrom 5 to 5.001: average rate 27 20. cm/s So the instantaneous rate of change of d att = 5 is about 27.20 cm/s.

c. Instantaneous rate would involve divisionby zero.

d. For t = 1.5 to 1.501, rate 31.42 cm/s.The pendulum is approaching the wall: Therate of change is negative, so the distance isdecreasing.

e. The instantaneous rate of change is the limitof the average rates as the time intervalapproaches zero. It is called the derivative.

f. Before t = 0, the pendulum was not yetmoving. For large values of t, the pendulumsmotion will die out because of friction.

2. a. x = 5: y = 305, price is $3.05x = 10: y = 520, price is $5.20x = 20: y = 1280, price is $12.80

b. x = 5.1, rate 46 822. /ftx = 5.01, rate 46.9820 /ftx = 5.001, rate 46.9982 /ft

c. 47 /ft. It is called the derivative.d. x = 10: 44 /ft. x = 20: 128 /fte. The 20-ft board costs more per foot than the

10-ft board. The reason is that longer boardsrequire taller trees, which are harder to find.

Problem Set 1-2Q1. Power function, or polynomial functionQ2. f (2) = 8 Q3. Exponential functionQ4. g (2) = 9 Q5.

1

1x

h(x)

Q6. h (5) = 25 Q7. y = ax2 + bx + c, a 0Q8. y = x Q9. y = |x|

Q10. Derivative1. a. Increasing slowly b. Increasing fast2. a. Increasing fast b. Decreasing slowly

3. a. Decreasing fast b. Decreasing slowly4. a. Decreasing slowly b. Increasing slowly5. a. Increasing fast b. Increasing slowly

c. Decreasing slowly d. Increasing fast6. a. Decreasing fast b. Increasing slowly

c. Increasing fast d. Decreasing fast7. a. Increasing slowly b. Increasing slowly

c. Increasing slowly8. a. Decreasing fast b. Decreasing fast

c. Decreasing fast9. a. Increasing fast b. Neither increasing

nor decreasingc. Increasing fast d. Increasing slowly

10. a. Decreasing slowly b. Decreasing fastc. Decreasing fast d. Neither increasing

nor decreasing

11. a.

100 200

50

100

T(x) (C)

x (s)

x = 40: rate 1 1. /sx = 100: rate = 0/sx = 140: rate 0 8. /s

b. Between 0 and 80 s the water is warmingup, but at a decreasing rate.

Between 80 and 120 s the water is boiling,thus staying at a constant temperature.

Beyond 120 s the water is cooling down,rapidly at first, then more slowly.

12. a.

1 2 3 4 5 6 7 8

10

20

30

40

50

60

70v(x) (ft/s)

x (s)

2 Problem Set 1-2 Calculus Solutions Manual 2005 Key Curriculum Press

x = 2: rate 18 (ft/s)/s x = 5: rate = 0 (ft/s)/s x = 6: rate 11 (ft/s)/s

b. Units are (ft/s)/s, sometimes written as ft/s2. The physical quantity is acceleration.

13. a.

3 4 72

18

h(x)

x

Increasing at x = 3 Decreasing at x = 7

b. h (3) = 17, h(3.1) = 17.19 Average rate =

0.190.1

= 1.9 ft/s

c. From 3 to 3.01:

average rate =0.0199

0.01= 1.99 ft/s

From 3 to 3.001:

average rate =0.001999

0.001= 1.99 ft/s

The limit appears to be 2 ft/s.

d. h (7) = 9, h(7.001) = 8.993999 Average rate =

0.0060010.001

= 6.001 ft /s The derivative at x = 7 appears to be 6 ft/s.

The derivative is negative because h(x) is decreasing at x = 7.

14. a.

10

100

300

500f(t)

tDecrease Increase

Not much

b. Enter y2 =y1 (x ) y1(1)

x 1

t r(t) = y2 (foxes/year) 0.97 110.5684 0.98 109.7361 0.99 108.9001 1 undefined 1.01 107.2171 1.02 106.3703 1.03 105.5200

c. Substituting 1 for t causes division by zero, so r(1) is undefined. Estimate: r approaches the average of r(0.99) and r(1.01), 108.0586 foxes/year. (Actual is 108.0604 .) The instantaneous rate is called the derivative.

d.

f (4.01) f (4)0.01

= 129.9697

f (4) f (3.99)0.01 = 131.4833

Instantaneous rate = (129.9697 131.4833)/2 = 130.7265 foxes/year (actual: 130.7287) The answer is negative because the number of foxes is decreasing.

15. a. Average rate =a(2.1) a(2)

0.1=

52.9902 mm2/h

b. r (t) = 200(1.2t ) 200(1.22 )t 2

2

204060

r(t ) (mm2/hr)

t (mm)

r(2) is undefined. c. r(2.01) = 52.556504

52.556504 52.508608 = 0.04789 Use the solver to find t when r(t) = 52.508608 + 0.01 = 52.518608 . t = 2.002088 , so keep t within 0.002 unit of 2.

16. a. v(x ) = 43x 3 v(6) = 288

b. 6 to 6.1: average rate =43 (6.13 63 )

0.1=

146.4133 5.9 to 6: average rate =

43 (63 5.93)

0.1=

141.6133 Estimate of instantaneous rate is

(146.4133 + 141.6133)/2 = 144.0133 = 452.4312 cm3/cm.

c. r (x ) =43 x3 43 63

x 6

) (cm /cm)

6

r(x

144

48

3

x

r(6) is undefined.


d. r(6.1) = 146.4133 = 459.9710 r(6.1) is 7.5817 units from the derivative. Use the solver feature to find x if r(x) = 144 + 0.1. x = 6.001326 , so keep x within 0.00132 unit of 6.

17. a. i. 1.0 in./s ii. 0.0 in./s iii. 1.15 in./s b. 1.7 s, because y = 0 at that time

18. a. i. 0.395 in./min ii. 0.14 in./min iii. 0.105 in./min

b. The rate is negative, because y is decreasing as the tire goes down.

19. a. Quadratic (or polynomial) b. f(3) = 30 c. Increasing at about 11.0 (2.99 to 3.01)

20. a. Quadratic (or polynomial) b. f(1) = 12 c. Increasing at about 6.0 (0.99 to 1.01)

21. a. Exponential b. Increasing, because the rate of change from

1.99 to 2.01 is positive. 22. a. Exponential

b. Increasing, because the rate of change from 3.01 to 2.99 is positive.

23. a. Rational algebraic b. Decreasing, because the rate of change from

3.99 to 4.01 is negative. 24. a. Rational algebraic


25. a. Linear (or polynomial) b. Decreasing, because the rate of change from

4.99 to 5.01 is negative. 26. a. Linear (or polynomial)


27. a. Circular (or trigonometric) b. Decreasing, because the rate of change from

1.99 to 2.01 is negative. 28. a. Circular (or trigonometric)

b. Decreasing, because the rate of change from 0.99 to 1.01 is negative.

29. Physical meaning of a derivative: instantaneous rate of change

To estimate a derivative graphically: Draw a tangent line at the point on the graph and measure its slope.

To estimate a derivative numerically: Take a small change in x, find the corresponding

change in f(x), then divide. Repeat, using a smaller change in x. See what number these average rates approach as the change in x approaches zero.

The numerical method illustrates the fact that the derivative is a limit.

30. Problems 13 and 14 involve estimating the value of a limit.

Problem Set 1-3 Q1. 72 ft2 Q2. y = cos x Q3. y = 2x Q4. y = 1/x Q5. y = x2 Q6. f(5) = 4 Q7. Q8.

x

y

x

y

Q9. Q10. x = 3

x

y

1. f(x) = 0.1x2 + 7 2. f(x) = 0.2x2 + 8 a. Approximately 30.8 a. Approximately 22.2 b. Approximately 41.8 b. Approximately 47.1

x

f (x)7

1 5 6

f(x)

x

2 3 5

8

3. h(x) = sin x 4. g( x) = 2 x + 5 a. Approximately 2.0 a. Approximately 7.9 b. Approximately 1.0 b. Approximately 12.2

1

3

x

h(x)

1 1 2

6

x

g(x)


5. There are approximately 6.8 squares between thecurve and the x-axis. Each square represents(5)(20) = 100 feet. So the distance is about(6.8)(100) = 680 feet.

6. There are approximately 53.3 squares between thecurve and the x-axis. Each square represents(0.5)(10) = 5 miles. So the distance is about(53.3)(5) = 266.5 miles.

7. Derivative =

tan . tan .. .

.

1 01 0 991 01 0 99

3 42K

8. Derivative = 7 (exactly, because that is theslope of the linear function)

9. a.

100

5 108.7t

v(t)

60

The range is 0 y 32.5660 .b. Using the solver, x = 8.6967 8.7 s.c. By counting squares, distance 150 ft.

The concept used is the definite integral.

d. Rate

=

v v( . ) ( . ). .

.

5 01 4 995 01 4 99

3 1107K

About 3.1 (ft/s)/sThe concept is the derivative.The rate of change of velocity is calledacceleration.

10. a.

2

10

5

1 3 4 5t

v(t)

b. v(4) = 9.3203 9.3 ft/sDomain: 0 t 4Range: 0 v(t) 9.3203

c. By counting squares, the integral from t = 0to t = 4 is about 21.3 ft. The units of theintegral are (ft/s) s = ft. The integral tellsthe length of the slide.

d. Rate

=

v v( . ) ( . ). .

.

3 01 2 993 01 2 99

1 8648K

About 1.86 (ft/s)/sThe derivative represents the acceleration.

11. From t = 0 to t = 5, the object travels about11.4 cm. From t = 5 to t = 9, the object travelsback about 4.3 cm. So the object is located about11.4 4.3 = 7.1 cm from its starting point.

12. See the text for the meaning of derivative.13. See the text for the meaning of definite integral.14. See the text for the meaning of limit.

Problem Set 1-4Q1. y changes at 30 Q2. Derivative 500Q3. Q4. f (3) = 9

x

y

Q5. 100 Q6. sin (/2) = 1Q7. 366 days Q8. DerivativeQ9. Definite integral Q10. f (x) = 0 at x = 4

1. a.

30

20,000

t

v(t)

b. Integral + + + +5 0 5 0 5 10 15( . ( ) ( ) ( ) ( )v v v vv(20) + v(25) + 0.5v(30)) = 5(56269.45) =281347.26 281 000, ftThe sum overestimates the integral becausethe trapezoids are circumscribed about theregion and thus include more area.

c. The units are (ft/s)(s), which equals feet, sothe integral represents the distance thespaceship has traveled.

d. Yes, it will be going fast enough, becausev(30) = 27,919.04 , which is greater than27,000.

2. a. v(t) = 4 + sin 1.4t

5

t3

v(t)


b. A definite integral has the units of thex-variable times the y-variable. Distance =rate time. Because v(t) is distance/time and t is time, their product is expressed inunits of distance.

c. See graph in part a.Distance + + +0 5 0 5 0 0 5 1. ( . ( ) ( . ) ( )v v vv(1.5) + v(2) + v(2.5) + 0.5v(3)) =0.5(26.041) = 13.02064 13.0 ft

d. v(3) = 3.128 3.1 mi/hMaximum speed was 5 mi/h at about 1.12 h.

3. Distance + + + + + =0 6 150 230 150 90 40 0. ( )396 ft

4. Volume 3(2500 + 8000 + 12000 + 13000 +11000 + 7000 + 4000 + 6000 + 4500) =204,000 ft3

5. Programs will vary depending on calculator. Seethe program TRAPRULE in the InstructorsResource Book for an example. The programgives T20 = 23.819625.

6. See the program TRAPDATA in the InstructorsResource Book for an example. The programgives T7 = 33, as in Example 2.

7. a.

1 4

7

f(x)

x

b. T10 = 18.8955T20 = 18.898875T50 = 18.89982These values underestimate the integral,because the trapezoids are inscribed in theregion.

c. T10: 0.0045 unit from the exact answerT20: 0.001125 unit from the exact answerT50: 0.00018 unit from the exact answerTn is first within 0.01 unit of 18.9 whenn = 7.T7 = 18.8908 , which is 0.0091 unitfrom 18.9.Because Tn is getting closer to 18.9 as nincreases, Tn is within 0.01 unit of 18.9 forall n 7.

8. a.

1 3

1

g(x)

x

b. T10 = 8.6700T20 = 8.659650 = 8.65672475These values overestimate the integral,because the trapezoids are circumscribedabout the region.

c. T10: 0.01385 unit from answerT20: 0.003465 unit from answerT50: 0.0005545 unit from answerTn is first within 0.01 of 8.65617024 whenn = 12.T12 = 8.665795 , which is 0.009624 unitfrom 8.65617024 .Because Tn is getting closer to the exactanswer as n increases, Tn is within 0.01 unitof the answer for all n 12.

9. From the given equation,y x= ( / ) .40 110 1102 2 Using the trapezoidalrule program on the positive branch with n = 100increments gives 6904.190 for the top half ofthe ellipse. Doubling this gives an area of13,808.38 cm2. The estimate is too lowbecause the trapezoids are inscribed within theellipse. The area of an ellipse is ab, where aand b are the x- and y-radii, respectively. Sothe exact area is (110)(40) = 4400 =13,823.007 cm2, which agrees both with theanswer and with the conclusion that thetrapezoidal rule underestimates the area.

10. Integral = 1(0.0 + 2.1 + 7.9 + 15.9 + 23.8 +29.7 + 31.8 + 29.7 + 23.8 + 15.9 + 7.9 +2.1 + 0) = 190.6The integral will have the units (in.2)(in.) = in.3,representing the volume of the football.

11. n = 10: integral 21.045n = 100: integral 21.00045n = 1000: integral 21.0000045Conjecture: integral = 21The word is limit.

12. The trapezoidal rule with n = 100 givesintegral 156.0096.Conjecture: integral = 156

13. If the trapezoids are inscribed (graph concavedown), the rule underestimates the integral.If the trapezoids are circumscribed (graph concaveup), the rule overestimates the integral.

Concave downInscribed trapezoids

Underestimates integral

Concave upCircumscribed trapezoids

Overestimates integral


Problem Set 1-51. Answers will vary.

Problem Set 1-6

Review Problems

R1. a. When t = 4, d = 90 80 sin [1, 2(4 3)] 15.4 ft.

b. From 3.9 to 4: average rate 40.1 ft/sFrom 4 to 4.1: average rate 29 3. ft/sInstantaneous rate 34.7 ft/sThe distance from water is decreasing, so he isgoing down.

c. Instantaneous rate d d( . ) ( . ).

.

5 01 4 990 02

70 8

d. Going up at about 70.8 ft/se. Derivative

R2. a. Physical meaning: instantaneous rate ofchange of a functionGraphical meaning: slope of a tangent line toa function at a given point

b. x = 4: decreasing fastx = 1: increasing slowlyx = 3: increasing fastx = 5: neither increasing nor decreasing

c. From 2 to 2.1:

average rate = =5 50 1

43 65472 1 2.

.

. K

From 2 to 2.01:

average rate = =5 50 01

402 01 2.

.

.5614K

From 2 to 2.001:

average rate =

=

5 50 001

40 26832 001 2.

.

. K

Differences between average rates andinstantaneous rates, respectively:43.6547 40.235947 = 3.418740.5617 40.235947 = 0.325540.2683 40.235947 = 0.03239The average rates are approaching theinstantaneous rate as x approaches 2.The concept is the derivative.The concept used is the limit.

d. t = 2: 3.25 m/st = 18: 8.75 m/st = 24: 11.5 m/sHer velocity stays constant, 7 m/s, from 6 sto 16 s. At t = 24, Mary is in her final sprinttoward the finish line.

R3. By counting squares, the integral isapproximately 23.2.Distance 23 2. ft (exact answer: 23.2422)Concept: definite integral

R4. a.

1 4x

5

f (x)

The graph agrees with Figure 1-6c.b. By counting squares, integral 15.0.

(Exact answer is 15.)c. T6 = 0.5(2.65 + 5.575 + 5.6 + 5.375 + 4.9 +

4.175 + 1.6) = 14.9375The trapezoidal sum underestimates theintegral because the trapezoids are inscribed inthe region.

d. T50 = 14.9991; Difference = 0.0009T100 = 14.999775; Difference = 0.000225The trapezoidal sums are getting closer to 15.Concept: limit

R5. Answers will vary.

Concept Problems

C1. a. f (3) = 32 7.3 + 11 1b. f (x) f (3) = x2 7x + 11 + 1 = x2 7x + 12c. f x f

x

x x

x

x x

x

( ) ( )

( )( )

33

7 123

4 33

2=

+= =

x 4, if x 3d. The limit is found by substituting 3 for x

in (x 4).Limit = exact rate = 3 4 = 1

C2. The line through (3, f (3)) with slope 1 isy = x + 2.

3

2 x

f(x)

The line is tangent to the graph. Zooming in bya factor of 10 on the point (3, 2) shows that thegraph becomes straighter and looks almost likethe tangent line. (Soon students will learn thatthis property is called local linearity.)


2

3

C3. a. f x x xx

x x

x( ) ( )( )= +

=

=

4 19 213

4 7 33

2

4x 7, x 3When x = 3, 4x 7 = 4 3 7 = 5.

b.

1 2 3 4 5 6

1

2

3

4

5

6f(x) (ft)

x (s)

5.8

4.2

2.8 3.2

c. 5.8 = 4(3 + ) 7 4.2 = 4(3 ) 75.8 = 12 + 4 7 4.2 = 12 4 74 = 0.8 4 = 0.8 = 0.2 = 0.2

d. 4(3 + ) 7 = 5 + 12 + 4 7 = 5 +

4 = = 14

There is a positive value of , namely

14 , for

each positive value of , no matter how small is.

e. L = 5, c = 3. . . . but not equal to 3 isneeded so that you can cancel the (x 3)factors without dividing by zero.

Chapter Test

T1. Limit, derivative, definite integral, indefiniteintegral

T2. See the text for the definition of limit.T3. Physical meaning: instantaneous rateT4.

2 5

3

6

x

y

T5. Concept: definite integralBy counting squares, distance 466.(Exact answer is 466.3496 .)

T6.

5 10 15 20 25 30 35 40

5

10

15

20

25Speed (ft/s)

Time (s)

T7 = 5(2.5 + 5 + 5 + 10 + 20 + 25 + 20 + 5) =462.5Trapezoidal rule probably underestimates theintegral, but some trapezoids are inscribed andsome circumscribed.

T7. Concept: derivative

5 10 15 20 25 30 35 40

5

10

15

20

25Speed (ft/s)

Time (s)

Slope 1.8 (ft/s)/s(Exact answer is 1.8137 .)Name: acceleration

T8. The roller coaster is at the bottom of the hill at25 s because thats where it is going the fastest.The graph is horizontal between 0 and 10 secondsbecause the velocity stays constant, 5 ft/s, as theroller coaster climbs the ramp.

T9. Distance = (rate)(time) = 5(10) = 50 ftT10. T5 = 412.5; T50 = 416.3118 ;

T100 = 416.340219

T11. The differences between the trapezoidal sum andthe exact sum are:For T5: difference = 3.8496For T50: difference = 0.03779For T100: difference = 0.009447The differences are getting smaller, so Tn isgetting closer to 416.349667 .


T12. From 30 to 31:

average rate = = y y( ) ( ) .31 30

11 9098K

From 30 to 30.1:

average rate = = y y( . ) ( )

.

.

30 1 300 1

1 8246K

From 30 to 30.01:

average rate =

=

y y( . ) ( ).

.

30 01 300 01

1 8148K

T13. The rates are negative because the roller coaster isslowing down.

T14. The differences between the average rates andinstantaneous rate are:For 30 to 31: difference = 0.096030For 30 to 31.1: difference = 0.010833For 30 to 30.01: difference = 0.001095

The differences are getting smaller, so the averagerates are getting closer to the instantaneous rate.

T15. Solve . gettingy x yx

( ) ( ),

= +30

301 81379936 1

x = 30.092220 . So keep x within 0.092unit of 30, on the positive side.

T16. Concept: derivative

T17.

=

=f f f( ) ( . ) ( . ). . .

4 4 3 3 74 3 3 7

35 290 6

10

T18. Answers will vary.


Chapter 2Properties of Limits

Problem Set 2-1

1. a. f ( )2 8 10 22 2

00

=

+

=

No value for f (2) because of division by zero.b.

x f (x)1.997 2.9941.998 2.9961.999 2.9982 undefined2.001 3.0022.002 3.0042.003 3.006

Yes, f (x) stays close to 3 when x is keptclose to 2, but not equal to 2.

c. To keep f (x) within 0.0001 unit of 3, keepx within 0.00005 unit of 2. To keep f (x)within 0.00001 unit of 3, keep x within0.000005 unit of 2. To keep f (x) arbitrarilyclose to 3, keep x within 12 that distanceof 2.

d. The discontinuity can be removed bydefining f (2) to equal 3.

2.

3

2

3

x

g(x)

3

2

g(x)

x

The limit seems to be 2.3.

3

2

x

h(x)

2

32.7

h(x)

x

There appears to be no limit, because the graphcycles infinitely as it approaches x = 3.

Problem Set 2-2Q1. Q2.

3

8

x

y

1

x

y

Q3. Q4.

6

4

x

y

22

4

x

y

Q5. Q6. Trapezoidal rule

1

4

x

y

Q7. Counting squaresQ8. Slope of the tangent lineQ9. Instantaneous rate of change

Q10. B1. See the text for the definition of limit.2. f (x) might be undefined at x = c, or might have a

value at x = c that is different from the limit.3. Has a limit, 3 4. Has a limit, 25. Has a limit, 3 6. Has a limit, 57. Has no limit 8. Has no limit9. Has a limit, 7 10. Has a limit, 20

11. Has no limit 12. Has no limit13. lim ( ) .

xf x

=

35 For = 0.5, 0.2 or 0.3.


14. lim ( ) .x

f x

=

23 For = 0.5, 0.8.

15. lim ( ) .x

f x

=

64 For = 0.7, 0.5 or 0.6.

(The right side is more restrictive.)16. lim ( ) .

xf x

=

42 For = 0.8, 0.7 or 0.8.

(The left side is more restrictive.)17. lim ( ) .

xf x

=

52 For = 0.3, 0.5 or 0.6.

(The right side is more restrictive.)18. lim ( ) .

xf x

=

36 For = 0.4, 0.1.

19. a. The graph should match Problem 13.b. lim ( )

xf x

=

35

c. Graph is symmetrical about x = 3.Let 5 2 sin (x 3) = 5 + 0.5 = 5.5. sin (x 3) = 0.25x = 3 + sin 1 (0.25)Max. = 3 [3 + sin 1 (0.25)] = 0.25268

d. Let 5 2 sin (x 3) = 5 + . sin (x 3) = /2x = 3 + sin 1 (/2)Max. = 3 [3 + sin 1 (/2)] =sin 1 (/2) = sin 1 (/2), which is positivefor any positive value of .


xf x

=

23

c. The graph is symmetrical about x = 2.Let (x 2)3 + 3 = 3 + 0.5 = 3.5. = + .x 2 0 53

Max. . . . = + = =2 0 5 2 0 5 0 79373 3 Kd. Let (x 2)3 + 3 = 3 + .

x = 2 + 1/3Max. = 2 + 1/3 2 = 1/3, which ispositive for any positive value of .


xf x

=

64

c. The right side is more restrictive.Let 1 + 3(7 x)1/3 = 4 0.7 = 3.3. x = 7 (2.3/3)3Max. = [7 (2.3/3)3] 6 = 0.5493

d. Because the right side is more restrictive, set1 3 7 43+ = x . x = 7 [(3 )/3]3Max. = 7 [(3 )/3)3] 6 = 1 [(3 )/3]3,which is positive for all positive values of .


xf x

=

42

c. The left side is more restrictive.Let 1 + 24 x = 2 + 0.8 = 2.8.

24 x = 1.8

x = 4 1 82

log .log

Max . log .

log. =

=4 4

1 82

0 84799K

d. Because the left side is more restrictive, set1 + 24 x = 2 + . 24 x = 1 +

x = +4 12

log( )log

Max. log(1 ) log(1 ) = + =

+4 42 2log log

,

which is positive for all > 0.23. a. The graph should match Problem 17.

b. lim ( )x

f x

=

52

c. The right side is more restrictive.Let (x 5)2 + 2 = 2 + 0.3 = 2.3. = + .x 5 0 3

Max = ( . ) .. 5 0 3 5 0 54772+ = Kd. Because the right side is more restrictive, set

(x 5)2 + 2 = 2 + . = + x 5 Max. ( ) , = + =5 5 which ispositive for all > 0.


xf x

=

36

c. The graph is symmetrical about x = 3.Let 6 2(x 3)2/3 = 6 0.4 = 5.6. x = 3 + 0.23/2Max. = (3 + 0.23/2) 3 = 0.08944

d. Let 6 2(x 3)2/3 = 6 . x = 3 + (/2)3/2Max. = [3 + (/2)3/2] 3 = (/2)3/2, whichis positive for all > 0.

25. a. f ( ) ( )( ) ( )( )2 5 6 5 13 5 25 2

5 00

00

2=

+

= =

The graph has a removable discontinuity atx = 2.Limit = 22 6(2) + 13 = 5

b. When f (x) = 5.1, x = 1.951191 .1 = 2 1.951191 = 0.048808When f (x) = 4.9, x = 2.051316 .2 = 2.051316 2 = 0.051316 max. = 0.048808


c.f(x)

x

c = 2

L = 5

26. a.

2

8

x

y

The graph is linear.There is a removable discontinuity at x = 2.The limit appears to be 9.

b. f ( ) ( ) ( )

2 4 2 7 2 22 2

00

2=

=

Indeterminate form

c. f x x xx

x x( ) ( )( ) ,= +

= + 4 1 2

24 1 2

Limit = 4(2) + 1 = 9If x 2, then (x 2) 0. Canceling is adivision process, but because (x 2) 0,you do not risk dividing by zero.

d. If f (x) = 9.001, x = 2.00025.If f (x) = 8.999, x = 1.99975.1 = 2.00025 2 = 0.000252 = 2 1.99975 = 0.00025Largest number is 0.00025.

e. L = 9, c = 2, = 0.001, = 0.00025

27. a. m t d t dt

t

t( ) ( ) ( ) =

=

44

3 484

2

b. Removable discontinuity at x = 4.

30

4

m(t)

t

c. Limit = 24 ft/s

d. m t t tt

t t( ) ( )( )

,=

+= +

3 4 44

3 12 4 if

3t + 12 = 24.12 t = 4.043t + 12 = 23.88 t = 3.96Keep t within 0.04 s of 4 s.

e. The limit of the average velocity is theinstantaneous velocity.

Problem Set 2-3Q1. 13Q2. Q3.

2

3

y

x

4

y

x

Q4. Q5.

x

y

x

y

1

1

Q6. (x 10)(x + 10) Q7. 75%Q8. Product of x and y, where x varies and y may

varyQ9.

3 1 8 22 21 3 15 21

1 5 7 0x2 5x + 7

Q10. D1.

10

2x

y

g h

g + h

lim ( ) , lim ( ) , lim ( )x x x

f x g x h x

= = =

2 2 210 4 6 and

= +

lim ( ) lim ( ) lim ( ),x x x

f x g x h x2 2 2

Q.E.D.

x f (x)1.96 9.96401.97 9.97221.98 9.98101.99 9.99022.00 102.01 10.01022.02 10.02092.03 10.03222.04 10.0439

All these f (x) values are close to 10.


2.

9

1.8 f

g

3x

y

lim ( ) . lim ( )x x

f x g x

= =

3 31 8 9 and

=

lim ( ) . lim ( ),x x

f x g x3 3

0 2 Q.E.D.

x f (x)2.96 1.752322.97 1.764182.98 1.776082.99 1.788023.00 1.83.01 1.812023.02 1.824083.03 1.836183.04 1.84832

All these f (x) values are close to 1.8.3.

3

7Limit = 7

x

f(x)

The limit is 7 because f (x) is always close to 7,no matter what value x takes on. (It shouldntbother you that f (x) = 7 for x 3 if you think ofthe definition of limit for a while.)

4.

x

f(x) = x

6

Limit = 6

lim ( ) .x

f x

=

66 The y-value equals the x-value.

5.

1

5y

1

y1

y2 y

2

y

x

lim , lim . , limx x x

y y y y

= = =

1 1 1 2 1 1 22 1 5 3 and

2 1 5 31 1 1 2 1 1 2

( . ) , lim lim lim= =

x x xy y y y

x y3 = f (x)0.997 2.97390.998 2.98250.999 2.99121 31.001 3.00871.002 3.01741.003 3.0262

All these f (x) values are close to 2(1.5) = 3.

6. 2 8 33 6

0 53 = = and sin.

.

r( ).

3 80 5

16= =

x r (x)2.9997 15.98942.9998 15.99292.9999 15.99643 163.0001 16.00353.0002 16.00703.0003 16.0105

All these r (x) values are close to 16.lim ( )

.

.

xf x

3 6

3 620

, so the limit of a quotient

cannot be applied because of division by zero.

7. lim ( ) limx x

f x x x

= +3 3

2 9 5

= +

lim lim limx x x

x x3

2

3 39 5 Limit of a sum

(or difference)= +

lim lim limx x x

x x x3 3 3

9 5

Limit of a product,limit of a constant

= (3)(3) 9(3) + 5 Limit of x= 9 27 + 5 = 13

8. lim ( ) limx x

f x x x

= + 1 1

2 3 6

= lim lim limx x x

x x

+ 1

2

1 13 6

Limit of a sum= lim lim lim

x x xx x x

+

1 1 13 6


= (1)(1) + 3(1) 6 Limit of x= 1 3 6 = 8


9.

8

2 x

r(x)

r( ) =2 2 4 2 122 2

4 8 120

00

2( ) ( )( )

+=

+ =

r xx x

xx x( ) ( )( ) ,= +

+=

6 22

6 2

lim ( )x

r x

= =

22 6 8

Proof:lim ( ) lim ( )x x

r x x

=

2 26 Because x 2

= + lim lim

2 2x xx ( )6 Limit of a sum

= 2 6 = 8, Q.E.D. Limit of x, limit of aconstant

10.13

5x

f(x)

f ( ) ( )5 5 3 5 405 5

25 15 400

00

2=

+

=

+ =

f x x xx

x x( ) ( )( )

,=

+= +

8 55

8 5

lim ( )x

f x

= + =5

5 8 13


f x x

= +5 5

8 Because x 5= lim lim

x xx

+

5 58 Limit of a sum

= 5 + 8 = 13, Q.E.D. Limit of x, limit of aconstant

11.41

10

5

x

f(x)

f ( ) ( ) ( ) =

=

=

5 5 3 5 4 5 305 5

125 75 20 300

00

3 2

f x x x xx

x x x( ) ( )( )

,=

+ += + +

222 6 5

52 6 5

lim ( ) ( )x

f x

= + + =5

25 2 5 6 41


f x x x

= + +5 5

2 2 6 Because x 5

= lim lim ( ) limx x x

x x

+ +5

2


= lim lim limx x x

x x x

+ +5 5 5

2 6 Limit of a product,limit of a constanttimes a function,limit of a constant

= 5 5 + 2 5 + 6 = 41, Q.E.D.Limit of x

12.

28

3x

f(x)

f ( ) ( )3 3 3 5 3 213 3

27 9 15 210

00

3 2=

+

=

+ =

f x x x xx

x x x( ) ( )( )

,=

+ += + +

224 7 3

34 7 3

lim ( ) ( )x

f x

= + + =3

23 4 3 7 28

Proof:lim ( ) lim )x x

f x x x

= + +3 3

2 4 7 (Because x 3

= + +

lim lim limx x x

x x3

2


= + +

lim lim limx x x

x x x3 3 3

4 7 Limit of a product,limit of a constanttimes a function,limit of a constant

= 3 3 + 4 3 + 7 = 28, Q.E.D.Limit of x

13.

9

f(x)

x

1

f ( ) ( ) ( ) ( )( ) = +

+1 1 4 1 2 1 3

1 1

3 2

=

+ +=

1 4 2 30

00


f x x x xx

x x x( ) ( )( ) ,= + ++

= + 2

25 3 11

5 3 1

lim ( ) ( ) ( )x

f x

= + =1

21 5 1 3 9


f x x x

= +1 1

2 5 3Because x 1

= + + lim lim ( ) lim

x x xx x

1

2

1 15 3

Limit of a sum= +

lim lim lim

x x xx x x

1 1 15 3

Limit of a product,limit of a constanttimes a function,limit of a constant

= (1)(1) + (5)(1) + 3 = 9, Q.E.D.Limit of x

14.

17

2 xf(x)

f ( ) ( ) ( )2 2 11 2 21 2 2 102 2

16 88 84 2 100

00

4 3 2=

+

=

+ =

f x x x x xx

( ) ( )( )

=

+ +3 29 3 5 22

= + + x x x x3 29 3 5 2, limx

f x

= + + = 2

3 22 9 2 3 2 5 17( ) ( ) ( )


f x x x x

= + +2 2

3 29 3 5Because x 2

= + + +

lim lim ( ) lim limx x x x

x x x2

32

2

2 29 3 5

Limit of a sum= +

lim lim lim ( ) lim limx x x x x

x x x x x2 2 2 2 2

9+ +

3 5

2limx

x


= 2 2 2 + (9)(2 2) + 3 2 + 5 = 17,Q.E.D. Limit of x

15.x f (x)

4.990 40.88014.991 40.89214.992 40.90404.993 40.91604.994 40.92804.995 40.94004.996 40.95204.997 40.96404.998 40.97604.999 40.98805 undefined5.001 41.01205.002 41.02405.003 41.03605.004 41.04805.005 41.06005.006 41.07205.007 41.08405.008 41.09605.009 41.1080

The table shows that f (x) will be within 0.1 unitof lim ( )

xf x

=

541 if we keep x within 0.008 unit

of 5.16.

f (x)

x

9

1

When x is close to 1, f (x) is close to 9.

17. f x x xx x

x x

x x

x

x( ) ( )( )( )( )=

+

+=

=

2

25 66 9

2 33 3

23

You cannot find the limit by substituting intothe simplified form because the denominator stillbecomes zero.

18. f x xx x

x x x

x x

x x

x

( ) ( )( )( )( )=

+=

+ +

=

+ +

3

2

2

2

84 4

2 2 42 2

2 42

You cannot find the limit by substituting intothe simplified form because the denominator stillgoes to zero.


19. a. 5(0)1/2 = 0 = v(0)5(1)1/2 = 5 = v(1)5(4)1/2 = 10 = v(4)5(9)1/2 = 15 = v(9)5(16)1/2 = 20 = v(16)

b. a

v v( ) .9 9 001 99 001 9

0 8333101 =( . ) ( ).

K

Conjecture: a( ) . /9 0 83 5 6= =Units of a(t): (mi/h)/s

c. av t v

t

t

tt t( ) lim ( ) ( )

lim

/9 9

95 15

99 91 2

= =

=

+

=

+

lim ( )( )( )lim

/

/ /

/

t

t

t

t t

t

9

1 2

1 2 1 2

9 1 2

5 33 3

53

=

56

, which agrees with the conjecture.d. Distance = integral of v(t) from 1 to 9. By the

trapezoidal rule with n = 100 increments,integral 86.6657 . The units are(mi/h) s. To convert to ft, multiply by 5280and divide by 3600, getting 127.1111(exact: 127 19 ) . The truck went about 127 ft.

20. a. Derivative =2 1 22 1 2

12 613 3

.

.

.

b. xx

x x x

x

3 282

2 2 42

( )( )

=

+ +=

x x2 2 4+ + , provided x 2. This expressionapproaches 12 as x approaches 2.

Proof:

lim

lim ( )x x

x

xx x

= + +

2

3

2

282

2 4

Because x 2= + +

lim lim limx x x

x x2

2

2 22 4

Limit of a sum= + +

lim lim limx x x

x x x2 2 2

2 4


= 2 2 + 2 2 + 4 = 12, Q.E.D.Limit of x

c. The line through point (2, 8) with slope 12 isy = 12x 16. The line appears to be tangentto the graph of f at point (2, 8).

12

1

2x

f(x)

8

21. By the symmetric difference quotient,

derivative =

0 7 0 72 0 01

0 059945 01 4 99

. .

( . ) . .. .

K

22. By the trapezoidal rule with n = 100,integral 11 8235. K .

23. Prove that limx c

n nx c

= for any positive integer n.

Proof:Anchor:

If n = 1, limx c

x c c

=1 1

= by the limit of x.

Induction Hypothesis:Assume that the property is true for n = k. =

limx c

k kx c

Verification for n = k + 1:lim lim ( )x c

kx c

kx x x

+

=

1

= =

lim limx c

kx c

kx x c c By the inductionhypothesis

=+ck 1

Conclusion:

=

limx c

n nx c for all integers n 1, Q.E.D.24. Answers will vary.

Problem Set 2-4Q1. Instantaneous rate of changeQ2. Product of x and y, where x varies and y

can varyQ3. 0.0005Q4.

Q5. Exponential functionQ6.

y = cos xx

Q7. (x + 6)(x 1) Q8. 53Q9. 120 Q10. 103

1. a. Has left and right limitsb. Has no limitc. Discontinuous. Has no limit


2. a. Has left and right limitsb. Has a limitc. Discontinuous. No f (3)

3. a. Has left and right limitsb. Has a limitc. Continuous

4. a. Has left and right limitsb. Has a limitc. Continuous

5. a. Has no left or right limitb. Has no limitc. Discontinuous. No limit or f (2)

6. a. Has left and right limitsb. Has a limitc. Continuous (Note that the x-value 5 is not at

the discontinuity.)7. a. Has left and right limits

b. Has a limitc. Discontinuous. f (1) limit

8. a. Has left and right limitsb. Has no limitc. Discontinuous. No limit

9. a. Has left and right limitsb. Has a limitc. Discontinuous. No f (c)

10. a. Has left and right limitsb. Has no limitc. Discontinuous. No limit, no f (c)

11. Answers may vary. 12. Answers may vary.

3

x

f(x)

4

f(x)

x


5

x

f(x)

2

x

f(x)

f(2)


x

f(x)

6

2

x

f(x)


10

f(x)

x

2

2

5 x

f(x)


x

f(x)6

4

1

x

f(x)

5

3

21. Discontinuous at x = 322. Discontinuous at x = 1123. Discontinuous at x = /2 + n, where n is an

integer24. Nowhere discontinuous25.

x

f(x)

2

1

23

Discontinuous because limx

f x

=

22( ) and f (2) = 3

26.

2

12

x

g(x)

Discontinuous because g(x) has no limit as xapproaches 2


27.

2

3

x

s(x)

Discontinuous because s(x) has no limit as xapproaches 2 from the left (no real functionvalues to the left of x = 2)

28.

2x

p(x)

1

Discontinuous because p(x) has no limit as xapproaches 2

29.

2

1

x

h(x)

Discontinuous because there is no value of h(2)30.

2

3

x

f(x)

Discontinuous because f (x) has no limit as xapproaches 2

31.

c f c( ) lim ( )x c f x lim ( )x c f x + lim ( )x c f x Continuous?1 4 2 2 2 removable2 1 1 1 1 continuous4 5 5 2 none step5 none none none none infinite

32.

c f c( ) lim ( )x c f x lim ( )x c f x + lim ( )x c f x Continuous?1 3 2 3 none step2 1 4 4 4 removable3 5 5 5 5 continuous5 5 5 none none infinite

33. a.

x

d(x)

2

3

b. lim ( ) , lim ( ) .x x

d x d x +

= =

2 23 3 Limit = 3.

Continuous.34. a.

32

1x

h(x)

b. lim ( ) , lim ( ) .x x

h x h x +

= =

1 13 2 No limit.

Not continuous.35. a.

2x

m(x)9

7

b. lim ( ) , lim ( ) .x x

m x m x +

= =

2 29 7 No limit.

Not continuous.36. a.

1

2

q(x)

x

b. lim ( ) , lim ( ) .x x

q x q x +

= =

1 12 2 Limit = 2.

Continuous.37. 9 22 = 2k

k = 2.5g(x)

x

2

5


38. 0.4(1) + 1 = k(1) + 2 k = 0.6

x

f(x)

1

1.4

39. (32)k = 3k 3 k = 1/2.

3

x

u(x)1

40. k + 5 = (1)2k k = 5/2

x

v(x)

1

5

41. a. b 1 = a(1 2)2 b 1 = ab. a = 1 b = 0. Continuous at x = 1.

x

f(x)

1 1

a = 1, b = 0

c. For example, a = 1 b = 2. Continuousat x = 1.

1

1 x

f(x)

e.g., a = 1, b = 2

42. lim ( ) limx x

f x k x k

= =

2 2

2 2 2 4lim ( ) lim . . ( )x x

f x kx k k + +

= = =

2 21 5 1 5 2 3

For f (x) to be continuous at x = 2, these twolimits must be equal, so find k such thatk2 4 = 3kk2 3k 4 = 0(k 4)(k + 1) = 0so k = 4 and k = 1 are the two values of k thatwill make f (x) continuous at x = 2.

43. Let T( ) = the number of seconds it takes tocross.

T( )if

40sin

, if 0 or 90

=

=

< < < <

24 90

90 180,

90

40

44. a.

x

f(x)

4

1

b. f (x) seems to approach 4 as x approaches 1.c. f (1.0000001) = 1.0000001 + 3 + 10 13

4.0000001, which is close to 4.d. There is a vertical asymptote at x = 0. You

must get x much closer to 1 than x =1.0000001 for the discontinuity to show up.

45. For any value of c, P(c) is determined by additionand multiplication. Because the set of realnumbers is closed under multiplication andaddition, P(c) will be a unique, real number forany real value x = c. P(c) is the limit of P(x) asx approaches c by the properties of the limit of aproduct of functions (for powers of x), the limitof a constant times a function (for multiplicationby the coefficients), and the limit of a sum (forthe individual terms). Therefore, P is continuousfor all values of x.

46. a. limx0

|sgn x| = 1 but f (0) = 0lim ( ) ( ),x

f x f

0

0 so discontinuous

b.

x

g(x)

2

3

c.

x

h(x)

1

1 1


d. For x > 0, a(x) = x/x = 1 = sgn x.For x < 0, a(x) = (x)/x = 1 = sgn x.For x = 0, a(0) is not defined. a(x) = sgn x for all x 0, Q.E.D.

e.

2

2

2x

f(x)

Problem Set 2-5Q1. No limit Q2. 3Q3. 4 Q4. 3Q5. 2 Q6. NoQ7. No Q8. YesQ9. No Q10. Yes

1. limx

f x

( ) = limx

f x 3

4( ) = lim

xf x

+=

33( ) lim

xf x

=

1( )

limx

f x

=

21( ) lim

xf x

=

3( )

limx

f x +

=

32( ) lim

xf x

( ) does not

exist.2. lim

xg x

=( ) 2 lim

xg x

=

24( )

limx

g x +

=

23( ) lim

xg x

=

1( )

limx

g x

=

23( ) lim

xg x

=

34( )

lim ( )x

g x

= 2


x

f(x)

2

2x

f(x)


7

5

x

f(x)

x

f(x)

7. a.

3

2

x

f(x)

b. limx

f x +

3

( ) = , limx

f x

3

( ) = ,lim ( ),x

f x3

none, limx

f x

( ) = ,2limx

f x

( ) = 2

c. 2 13

100+

=

x

13

98x

=

x 3 198

=

x = =3 1

983 0102. K

x f (x)3.01 1023.001 10023.0001 10002

All of these f (x) values are greater than 100.lim ( )x

f x +

= 3

means that f (x) can be keptarbitrarily far from zero just by keeping xclose enough to 3 on the positive side.There is a vertical asymptote at x = 3.

d. 21

32 001+

=

x.

13

0 001

3 10001003

x

x

x

=

=

=

.

x f (x)1004 2.000991005 2.000991006 2.00099All of these f (x) values are within 0.001 unitof 2. lim

xf x

=( ) 2 means that you can keep

f (x) arbitrarily close to 2 by making the valueof x arbitrarily large. y = 2 is a horizontalasymptote.

8. a.

/2

1x

g(x)

b. lim ( ) , lim ( )/ /x x

g x g x +

= = 2 2

The limit is infinite because |g(x)| can be keptarbitrarily far from zero. You cant saylim ( )

/xg x

=

2 because the left and right


limits are not the same (one is positive and theother is negative).

c. sec x = 1000 cos x = 0.001x = arccos (0.001) = 1.57179

x g(x)1.5717 1106.51.5716 1244.21.5715 1421.1

All of these f (x) values are less than 1000.lim ( )

/xg x

+=

2 means that arbitrarily far

g(x) can be kept arbitrarily far from zero in thenegative direction by keeping x close enoughto 2 on the positive side.The line x = 2 is a vertical asymptote.

9. a.

5

3r(x)

x

2

b. lim ( )x

r x

= 2 because (sin x)/x approaches zero.

c. r( ) sin( ) . ,28 2 28

282 00967= + = K which is

within 0.01 unit of 2.

r( ) sin( )32 2 32

32= + = 2.01723 ,K which is

more than 0.01 unit away from 2.y

x

1.992.01

28 32

r

Keeping r(x) within 0.01 unit of 2 means youwant to keep sin . ,xx < 0 01 or |sin x| 1, orx > 100. So D = 100.

d. The line y = 2 is an asymptote. Even thoughr (x) oscillates back and forth across this line,the limit of r(x) is 2 as x approaches infinity,satisfying the definition of asymptote.

e. The graph suggests that lim ( ) .x

r x

=

03

(The exact value is e, 2.7182 .)10. a. h x x x( ) ( / )= +1 1

10

1

2

3

x

h(x)

b. There is a compromise number (bigger than1, but finite) that wins. (The exact limit is e.)

11. The limit is infinite. y is unbounded as xapproaches infinity. If there were a number Esuch that log x < E for all x > 0, then you couldlet x = 102E so that log x = log 102E = 2E, whichis greater than E, which was assumed to be anupper bound.

12. Wanda, heres what happens to a fraction whenthe denominator gets close to zero: 10 1 10. = ,

10.0001

10.00001, , ,= =10 000 100 000. The answers

just keep getting bigger and bigger. When thedenominators get bigger and bigger, the fractiongets closer and closer to zero, like this:

110

1100

110000 1 01 001= = =. , 0. , 0. .

13. a. The definite integral is the product of theindependent and dependent variables. Becausedistance = (rate)(time), the integral representsdistance in this case.

b. T9 = 17.8060052 T45 = 17.9819616 T90 = 17.9935649T450 = 17.9994175

c. The exact answer is 18. It is a limit becausethe sums can be made as close to it as youlike, just by making the number of trapezoidslarge enough (and thus keeping their widthsclose to zero). The sums are smaller than theintegral because each trapezoid is inscribedunder the graph and thus leaves out a part ofits respective strip of the region.

d. Tn is 0.01 unit from 18 when it equals 17.99.From part b, this occurs between n = 45 and n= 90. By experimentation,T66 = 17.9897900 and T67 = 17.9900158 .Therefore, the approximation is within 0.01unit of 18 for any value of n 67.An alternative solution is to plot the graph ofthe difference between 18 and Tn as a functionof the number of increments, n, or to do aregression analysis to find an equation. Thebest-fitting elementary function is an inversepower variation function, y = (5.01004)


(x 1.48482). The graph of this function and threeof the four data points are shown here. UseTRACE or the solver feature of your grapher tofind n 67.

0.01

0.1

y

n

67

90

14. a. Work = force distance. Because a definiteintegral measures the y-variable times thex-variable, it represents work in this case.

b. By the trapezoidal rule, T10 = 24.147775and T100 = 24.004889 . The units arefoot-pounds.

c. The integer is 24.d. By experimentation, T289 = 24.001003 and

T290 = 24.000998 . D = 290

15. Length = 100 sec x = 100/cos xLength > 1000 100/cos x > 1000cos x < 0.1 (because cos x is positive)x > cos1 0.1 (because cos is decreasing)x > 1.4706289/2 1.4706289 = 0.100167x must be within 0.100167 radian of /2.The limit is (positive) infinity.

16. a. f (2) = 5 2 0 (1/0), which has theform 0 .g(2) = 5 2 0 (1/0)2, which has theform 0 .h(2) = 5 2 02 (1/0), which has theform 0 .

b. f x x xx

x x( ) ( )= = 5 2 12

5 2

,

=

( )limx

f x2

10

g x x xx

x

xx( ) ( )= = 5 2 1

25

222( ) ,

( )limx

g x2

is infinite.

h x x xx

x x x( ) ( ) = ( ), = 5 2 12

5 2 22

( ) =limx

h x2

0

c. The indeterminate form 0 could approachzero, infinity, or some finite number.

Problem Set 2-6

Q1. 53 Q2. 53Q3. Undefined Q4. 5Q5. Undefined Q6. Does not exist

Q7. 1 Q8. +Q9. Indeterminate Q10. C

1. IVT applies on [1, 4] because f is a polynomialfunction, and polynomial functions arecontinuous for all x.f (1) = 18, f (4) = 3 There is a value x = c in (1, 4) for whichf (c) = 8.Using the intersect or solver feature,c = 1.4349 , which is between 1 and 4.

f (x)

x

f(1)

f(4)8

1 4c

2. IVT applies on [0, 6] because f is a polynomialfunction, and polynomial functions arecontinuous for all x.f (0) = 8, f (6) = 0.224 There is a value x = c in (0, 6) for whichf (c) = 1.Using the intersect or solver feature,c = 5.8751 , which is between 0 and 6.

0.224

8

0 6c xf(x)

3. a. For 1 y < 2 or for 5 < y 8, the conclusionwould be true. But for 2 y 5, it would befalse because there are no values of x in [1, 5]that give these values for f (x).

b. The conclusion of the theorem is true becauseevery number y in [4, 6] is a value of g(x) forsome value of x in [1, 5].

4. a. f f f( ) , ( ) , ( . ) . ,2 4 3 8 0 5 2 1 414= = = = K f ( )5 8=

b. f is continuous at x = 3 because it has a limitand a function value and they both equal 8.

c. f is continuous nowhere else. Because thesets of rational and irrational numbers aredense, there is a rational number between anytwo irrational numbers, and vice versa. Sothere is no limit of f (x) as x approaches anynumber other than 3.

d. The conclusion is not true for all values of ybetween 1 and 4. For instance, if y = 3, thenc would have to equal log2 3. But log2 3 isirrational, so f (c) = 8, which is not between1 and 4.


5. Let f (x) = x2. f is a polynomial function, so it iscontinuous and thus the intermediate valuetheorem applies. f (1) = 1 and f (2) = 4, so thereis a number c between 1 and 2 such that f (c) = 3.By the definition of square root, c = 3 , Q.E.D.

6. Prove that if f is continuous, and if f (a) ispositive and f (b) is negative, then there is at leastone zero of f (x) between x = a and x = b.Proof:f is continuous, so the intermediate valuetheorem applies. f (a) is positive and f (b) isnegative, so there is a number x = c between aand b for which f (c) = 0. Therefore, f has at leastone zero between x = a and x = b, Q.E.D.

7. The intermediate value theorem is called anexistence theorem because it tells you that anumber such as 3 exists. It does not tell youhow to calculate that number.

8. Telephone your sweethearts house. An answer tothe call tells you the existence of thesweetheart at home. The call doesnt tell suchthings as how to get there, and so on. Also,getting no answer does not necessarily mean thatyour sweetheart is out.

9. Let f (t) = Jesses speed Kays speed. f (1) =20 15 = 5, which is positive. f (3) = 17 19 =2, which is negative. The speeds are assumed tobe continuous (because of laws of physics), so fis also continuous and the intermediate valuetheorem applies.So there is a value of t between 1 and 3 forwhich f (t) = 0, meaning that Jesse and Kay aregoing at exactly the same speed at that time.The existence of the time tells you neither whatthat time is nor what the speed is. An existencetheorem, such as the intermediate value theorem,does not tell these things.

10. Let f (x) = number of dollars for x-ounce letter.f does not meet the hypothesis of the IVT on theinterval [1, 9] because there is a stepdiscontinuity at each integer value of x. There isno value of c for which f (c) = 2 because f (x)jumps from 1.98 to 2.21 at x = 8.

11. You must assume that the cosine is functioncontinuous. Techniques: c = cos 1 0.6 = 0.9272 Using the solver feature, c = 0.9272... Using the intersect feature, c = 0.9272...

12. You must assume that 2x is continuous.f (0) = 20 = 1, because any positive number to the0 power equals 1.

c = = =log loglog2

3 32

1 5849. ...

Using the solver feature, c = 1.5849... Using the intersect feature, c = 1.5849...

13. This means that a function graph has a highpoint and a low point on any interval in whichthe function is continuous.

x

f(x)

a bc1 c2

If the function is not continuous, there may be apoint missing where the maximum or minimumwould have been.

x

f(x)

a b

Another possibility would be a graph with avertical asymptote somewhere between a and b.

14. Prove that if f is continuous on [a, b], the imageof [a, b] under f is all real numbers between theminimum and maximum values of f (x),inclusive.

Proof:By the extreme value theorem, there are numbersx1 and x2 in [a, b] such that f (x1) and f (x2) are theminimum and maximum values of f (x) on [a, b].Because x1 and x2 are in [a, b], f is continuous onthe interval whose endpoints are x1 and x2. Thus,the intermediate value theorem applies on thelatter interval. Thus, for any number y betweenf (x1) and f (x2), there is a number x = c betweenx1 and x2 for which f (c) = y, implying that theimage of [a, b] under f is all real numbersbetween the minimum and maximum values off (x), inclusive, Q.E.D.

Problem Set 2-7Review Problems

R0. Answers will vary.

R1. a. f ( )3 36 51 153 3

00

=

+

=

Indeterminate form


b. f x x x( ) = ,4 5 3

99

9

9

y

x

At x = 3 there is a removable discontinuity.c. For 0.01, keep x within 0.0025 unit of 3. For

0.0001, keep x within 0.000025 unit of 3. Tokeep f (x) within unit of 7, keep x within14 unit of 3.

R2. a. L f xx c

=

lim ( ) if and only if for any number

> 0, no matter how small,there is a number > 0 such that if x iswithin units of c, but x c, then f (x) iswithin units of L.

b. limx

f x

=

12( )

limx

f x2

( ) does not exist.limx

f x

=

34( )

limx

f x4

( ) does not exist.limx

f x5

( ) = 3c. lim

xf x

=

23( )

Maximum : 0.6 or 0.7d. The left side of x = 2 is the more restrictive.

Let 2 + x 1 = 3 0.4 = 2.6. x = 1 + 0.62 = 1.36 maximum value of is 2 1.36 = 0.64.

e. Let f x( ) = 3 .2 1 3+ = x x = (1 )2 + 1Let = 2 [(1 )2 + 1] = 1 (1 )2,which is positive for all positive < 1. If 1, simply take = 1. Then will bepositive for all > 0.

R3. a. See the limit property statements in the text.b.

20

3x

g(x)

19

The limit of a quotient property does notapply because the limit of the denominatoris zero.

g x x x xx

( ) = +

( )( )3 10 23

2

g(x) = x2 10x + 2, x 3

You can cancel the (x 3) because thedefinition of limit says but not equal to 3.

lim ( )x

g x3

= + +

lim lim ( ) limx x x

x x3

2

3 310 2

Limit of a sum= +

lim lim limx x x

x x x3 3 3

10 2


= 3 3 10(3) + 2 Limit of x= 19, which agrees with the graph.

c. f (x) = 2x,g x x x

x

x x

x( ) = +

=

2 8 153

3 53

( )( )

= x + 5, x 3lim ( ) , limx x

f x g x

= =

3 38 2 ( )

p(x) = f (x) g(x)limx

p x

= =

38 2 16( )

x p(x)2.997 15.99072.998 15.99382.999 15.99693 undefined3.001 16.00303.002 16.00613.003 16.0092

All these p(x) values are close to 16.

r xf xg x

( ) = ( )( )

lim ( )x

r x

= =

3

82

4

3

5

10

15

20y

x

f

g

r

d. For 5 to 5.1 s: average velocity = 15.5 m/s.

Average velocity = =f t ft

( ) ( )

55

35 5 505

5 2 55

2t t

t

t t

t

( )( )

= =


5(t 2), for t 5. Instantaneous velocity =limit = 5(5 2) = 15 m/s.The rate is negative, so the distance above thestarting point is getting smaller, whichmeans the rock is going down.Instantaneous velocity is a derivative.

R4. a. f is continuous at x = c if and only if1. f (c) exists2. lim ( )

x cf x

exists

3. limx c

f x f c

=( ) ( ) f is continuous on [a, b] if and only if f is

continuous at every point in (a, b), andlim ( ) ( )x a

f x f a +

= and limx b

f x f b

=( ) ( ).b.

c f c( ) limx c f x ( ) lim ( )x c f x + lim ( )x c f x Continuous?1 none none none none infinite

2 1 3 3 3 removable

3 5 2 5 none step

4 3 3 3 3 continuous

5 1 1 1 1 continuous

c. i. ii.

x

y

1

x

y

2

iii. iv.

x

y

3

x

y

4

v. vi.

5x

y

L

f(6)

x

y

6

vii.

1

5

2

x

y

d.

2

4

x

f(x)

2

The left limit is 4 and the right limit is 2, sof is discontinuous at x = 2, Q.E.D.Let 22 = 22 6(2) + k. k = 12

R5. a. limx

f x

= 4

( ) means that f (x) can be keptarbitrarily far from 0 on the positive side justby keeping x close enough to 4, but not equalto 4.limx

f x

=( ) 5 means that f (x) can be made tostay arbitrarily close to 5 just by keeping xlarge enough in the positive direction.

b. lim ( )x

f x

does not exist.

limx

f x

=

21( )

limx

f x

= 2

( ) lim

xf x

+=

2( )

limx

f x

=( ) 2c. f (x) = 6 2 x

limx

f x

=( ) 6f (x) = 5.999 = 6 2 x2 x = 0.001

x = log

log0 001

2.

x = 9.965...

x f (x)10 5.99902320 5.99999904630 5.99999999907

All of these f (x) values are within 0.001 of 6.d. g(x) = x 2

lim ( )x

g x

= 0

g(x) = 106 = x 2x2 = 10 6x = 10 3

x g(x)0.0009 1.2345 106

0.0005 4,000,0000.0001 1 108


All of these g(x) values are larger than1,000,000.

e. v(t) = 40 + 6 tn Trapezoidal Rule

50 467.9074100 467.9669200 467.9882400 467.9958

The limit of these sums seems to be 468.By exploration,T222 = 467.98995T223 = 467.99002 D = 223

R6. a. See the text statement of the intermediatevalue theorem.The basis is the completeness axiom.See the text statement of the extreme valuetheorem.The word is corollary.

b. f (x) = x3 + 5x2 10x + 20f (3) = 8, f (4) = 4So f (x) = 0 for some x between 3 and 4 bythe intermediate value theorem.The property is continuity.The value of x is approximately 3.7553.

c.

4

3x

f(x)

f (6) = 1 and f (2) = 5 by tracing on thegraph or by simplifying the fraction to getf (x) = x + 7, then substituting. You will notalways get a value of x if y is between 1 and5. If you pick y = 3, there is no value of x.This fact does not contradict the intermediatevalue theorem. Function f does not meet thecontinuity hypothesis of the theorem.

Concept Problems

C1.

hh

f

g

y

x

7

4

Conjecture: lim ( )x

f x

=

47

C2. f (1) = 12 6 1 + 9 = 4As x 1 from the left, f (x) 12 + 3 = 4.As x 1 from the right, f (x) 12 6 + 9 = 4. lim

xf x f

14 1( ) = = ( )

f is continuous at x = 4, Q.E.D.For the derivative, from the left side,f x f

x

x

x

x x

x

( ) ( )

( )( )

11

3 41

1 11

2=

+=

+=

x + 1, x 1 = + =

limx

f x1

1 1 2( )For the derivative, from the right side,f x f

x

x x

x

x x

x

( ) ( )

( )( )

11

6 9 41

1 51

2=

+= =

x 5, x 1 lim

xf x

+

11 5 4( ) = =

So f is continuous at x = 1, but does not have avalue for the derivative there because the rate ofchange jumps abruptly from 2 to 4 at x = 1. Ingeneral, if a function has a cusp at a point, thenthe derivative does not exist, but the function isstill continuous.

C3. The graph is a y = x2 parabola with a stepdiscontinuity at x = 1. (Use the rise-runproperty. Start at the vertex. Then run 1, rise 1;run 1, rise 3; run 1, rise 5; . . . . Ignore thediscontinuity at first.) To create thediscontinuity, use the signum function withargument (x 1). Because there is no value forf (1), the absolute value form of the signumfunction can be used.

y x xx

=2 2 1

1+

| |

C4. The quantity | ( ) |f x L is the distance betweenf (x) and L. If this distance is less than , thenf (x) is within units of L. The quantity |x c|is the distance between x and c. The right part ofthe inequality, |x c| < , says that x is within units of c. The left part, 0 < |x c|, says that xdoes not equal c. Thus, this definition of limit isequivalent to the other definition.

Chapter Test

T1. f is continuous at x = c if and only if1. f (c) exists2. lim ( )

x cf x

exists

3. limx c

f x f c

=( ) ( )f is continuous on [a, b] if and only if f iscontinuous at all points in (a, b), andlim ( ) ( )x a

f x f a +

= and lim ( ) ( )x b

f x f b

= .


T2. a. limx

f x

=

23( ) lim

xf x

+=

24( )

lim ( )x

f x2

does not exist. lim ( )x

f x

=

62

lim ( )x

f x +

=

62 lim

xf x

=

62( )

b. f is continuous on [2, 6] because it iscontinuous for all values in (2, 6) andlim ( ) ( )x

f x f +

=

22 and lim ( ) ( )

xf x f

=

66 .

T3. See the text statement of the quotient property.T4. a. Left: 4; right: 4

b. Limit: 4c. Discontinuous

T5. a. Left: none; right: noneb. Limit: nonec. Discontinuous

T6. a. Left: 6; right: 6b. Limit: 6c. Continuous

T7. a. Left: 2; right: 3b. Limit: nonec. Discontinuous

T8.

4

4

x

y

T9. a. b.

x

f(x)

x

g(x)

1

c. d.

1

1

h(x)

x

1

1

x

s(x)

T10. a. f ( ) ( )( ) ,3 0 5 0 8 0 30 3

00

2=

+

=

an indeterminate formb. lim ( ) lim ( ),

x xf x x x x

= +

3 32 5 8 3

Definition of limitx c

= + +


x x3

2

3 35 8

Limit of a sum= + +


x x x3 3 3

5 8


= 3 3 + (5) 3 + 8 Limit of x= 2, Q.E.D.

T11. If k f x x xx x

= =

+ >

1 2

1 2

2

, ( ) ,,

lim limx x

f x f x +

= =

2 24 3( ) , ( )

f is discontinuous at x = 3.

2

5f(x)

x

T12. limx

f x k

=

2

22( )limx

f x k +

+2

2 ( ) = 4k = 2 + kk = 2/3

T13. See graph in T11.T14. a. lim

xT x

=( ) 20

From the graph, it appears that if x > 63 ft,then T(x) is within 1 of the limit.The graph of T has a horizontal asymptote atT = 20.

b. T = 20 + 8(0.97x) cos 0.5x. The amplitude ofthe cosine factor is 8 0 97( . ).x Make thisamplitude < 0.1.8(0.97c) = 0.1

0.97c = 0.0125

c =log .log .

0 01250 97

c = 143.8654 T is within 0.1 unit of 20 wheneverx > 143.8654 .

c. The time of day would be mid-afternoon,when the temperature of the surface ishighest.

T15. a. Use either TRACE or TABLE to show:d(0) = 0, d(10) = 6, d(20) = 14, d(30) = 24,d(40) = 36, and d(50) = 50.

b. Average rate =

=

d d( . ) ( ).

20 1 2020 1 20

14 0901 1420 1 20

0 901. .

= . cm/day


c. Average rate =

=

d t dt

( ) ( )2020

0 01 0 5 1420

0 01 70 2020

2. .

. ( )( )

t t

t

t t

t

+=

+=

0.01t + 0.7, t 20. The limit as t approaches20 is 0.01(20) + 0.7, which equals0.9 cm/day. This instantaneous rate is calledthe derivative.

d. The glacier seems to be speeding up becauseeach 10-day period it moved farther than it hadin the preceding 10-day period.

T16. c(0) = p(0) = 10, so each has the same speed att = 0. lim ( ) . lim ( ) .

t tc t p t

= = 16 Surprise for

Phoebe!

T17. f x kx xkx x

( ) , ,

=

>

2 2

10 2if if

limx

f x k k

= =

2

22 4( )limx

f x k +

=

210 2( )

Make 4k = 10 2k k = 5/3. There is a cuspat x = 2.

2

10f(x)

x

T18. h(x) = x3. h(1) = 1 and h(2) = 8, so 7 is betweenh(1) and h(2). The intermediate value theoremallows you to conclude that there is a realnumber between 1 and 8 equal to the cube rootof 7.

T19. Answers will vary.


Chapter 3Derivatives, Antiderivatives,and Indefinite Integrals

Problem Set 3-11. The graph is correct.

2. Average rate = = =f f( . ) ( ).

.

.

5 1 50 1

3 21 30 1

2.1 km/min3.

5

2

x

y

4. The graph of r has a removable discontinuityat x = 5.

rf f( ) ( ) ( )5 5 5

5 500

=

=

5. r x f x fx

x x

x( ) ( ) ( )=

=

+

55

8 18 35

2

=

= ( )( )

,

x x

xx x

5 35

3 5

=

f r xx

( ) lim ( )55

=

= =

lim ( )( ) lim( )x x

x x

xx

5 5

5 35

3 2

The derivative is the velocity of the spaceship,in km/min.

6. Find the equation of the line through (5, f (5)),or (5, 3), with slope 2.y 3 = 2(x 5) y = 2x 7

5

3

x

y

This line is tangent to the graph of f (x) at (5, 3).7. As you zoom in, the line and the graph appear to

be the same.8. Answers will vary.

Problem Set 3-2Q1. Instantaneous rate of changeQ2. x + 9 Q3. 18

Q4.

x

y

1

Q5. 9x2 42x + 49Q6. signQ7. Q8.

x

y

3

x

y

5

2

Q9. Newton and LeibnizQ10. D

1. See the text for the definition of derivative.2. Physical: Instantaneous rate of change of the

dependent variable with respect to theindependent variableGraphical: Slope of the tangent line to the graphof the function at that point

3. a. f xxx

( ) =3 0 6 5 433

2lim . .

= lim . ( )( )

.

x

x x

x

+=

3

0 6 3 33

3 6

b. Graph of the difference quotient m(x)

3

3.6

x

m(x)

c., d. Tangent line: y = 3.6x 5.4

31 x

f(x)

4. a. f x x f xxx

( ) = . , ( ) =0 2 6 0 2 7 26

2

6

2

+

lim . .

=

+=

lim . ( )( )

.

x

x x

x6

0 2 6 66

2 4


b.

6

2.4

x

m(x)

c., d. Tangent line: y = 2.4x + 7.2

6

7.2

xf(x)

5. f x xxx

+ + +

+( ) =2 5 1 5

22

2lim

= lim ( )( )x

x x

x

+ +

+=

2

2 32

1

6. ( ) =f x xxx

+ +

+4 6 2 10

44

2lim

=

+ +

+lim ( )( )x

x x

x4

4 24

2=

7. ( ) =f x x xxx

+ +

1 4 8 6

11

3 2lim

= = lim ( )( )

x

x x x

x1

21 3 21

4

8. f x x xxx

+

+( ) =1 4 6 8

11

3 2lim

=

+

+=

lim ( )( )x

x x x

x1

21 2 21

1

9. f xxx

+ +

( ) =3 0 7 2 0 1

33lim . .

= =

lim . ( )

.

x

x

x3

0 7 33

0 7

10. f xxx

( ) =4 1 3 3 2 244

lim . .

= =

lim . ( )

.

x

x

x4

1 3 44

1 3

11. fxx

+=

( ) =1 5 5

10

1lim

12. fxx

+

=

( ) =3 2 2

30

3lim

13. The derivative of a linear function equals theslope. The tangent line coincides with the graphof a linear function.

14. The derivative of a constant function is zero.Constant functions are horizontal and dontchange! The tangent line coincides with thegraph.

15. a. Find f ( 1) = 2, then plot a line through point(1, f (1)) using f ( 1) as the slope. The line isy = 2x 1.

b. Near the point (1, 1), the tangent line and thecurve appear nearly the same.

c. The curve appears to get closer and closer tothe line.

d. Near point (1, 1) the curve looks linear.e. If a graph has local linearity, the graph near

that point looks like the tangent line.Therefore, the derivative at that point could besaid to equal the slope of the graph at thatpoint.

16. a. f ( x) = x2 + 0.1 (x 1)2/3f ( 1) = 12 + 0.1(1 1)2/3 = 1 + 0 = 1, Q.E.D.The graph appears to be locally linear at(1, 1), because it looks smooth there.

b. Zoom in by a factor of 10,000.

1

1

c. The graph has a cusp at x = 1. It changesdirection abruptly, not smoothly.

d. If you draw a secant line through (1, 1) from apoint just to the left of x = 1, it has a largenegative slope. If you draw one from a pointjust to the right, it has a large positive slope.In both cases, the secant line becomes verticalas x approaches 1 and a vertical line hasinfinite slope. So there is no real numberequal to the derivative.

17. a.

3

5

x

f(x)7

b. First simplify the equation.

f x x xx

( ) ,,

=

+

=

2 3

7 3if if


The difference quotient is

m xx

x

x

x( ) = ( )

+=

2 73

53

3

5

x

m(x)

c.

x f (x)2.997 667.662.998 10012.999 20013.000 undefined3.001 19993.002 9993.003 665.66The difference quotients are all large positivenumbers on the left side of 3. On the rightside, they are large negative numbers. For aderivative to exist, the difference quotientmust approach the same number as x getscloser to 3.

18. a.

1

2

x

s(x)

b. m x xx

( ) = | sin ( ) |

11

1

1

x

m(x)

c. As x approaches 1 from the left, m(x)approaches 1. As x approaches 1 from theright, m(x) approaches 1. Because the left andright limits are unequal, there is no derivativeat x = 1.

19. a. +

f x x xxx

( ) = lim . . .

3

20 25 2 5 7 25 23

=

lim . ( )( )

x

x x

x3

0 25 3 73

=

lim ,x

x3

0 25 7 1 . ( ) = Q.E.D.The tangent line on the graph has slope 1.

b.

Draw secant linesfrom here.

f (x)

3

2

x

As the x-distance between the point and 3decreases, the secant lines (solid) approach thetangent line (dashed).

c. The same thing happens with secant linesfrom the left of x = 3. See the graph in part b.

d.

3

4

x

Draw secant linesfrom here.

g(x)

e. A derivative is a limit. Because the left andright limits are unequal, there is no derivativeat x = 3.

f. m xx

x( ) =

cos

66

3

. By table,

x m(x)2.9 3.14012.99 3.14153 undefined3.01 3.14153.1 3.1401

Conjecture: The numbers are and .20. From Problem 19, parts b and c, the tangent line

is the limit of the secant lines as x approaches c.Because the slope of the secant line is the average


rate of change of f (x) for the interval from x to c(or from c to x) and the derivative, f (c), is thelimit of this average rate, the slope of the tangentline equals f (c).

Problem Set 3-3Q1. 3Q2.

2

y

x

5

Q3.y

x

Q4. 20%Q5. 3x2 2x 8Q6. 25x2 70x + 49Q7. log 6Q8.

Q9.

Q10. limx c

is missing.

1. a.40

f

2 2

f

x

y

b. f (x) is positive for 2 < x < 2. The graph off is increasing for these x-values.

c. f ( x) is decreasing for x satisfying |x| > 2.f (x) < 0 for these values of x.

d. Where the f graph crosses the x-axis, thef graph has a high point or a low point.

e. See the graph in part a.f. Conjecture: f is quadratic.

2.15

3x

y

g

g'

The graph does not have the high and lowpoints that are typical of a cubic function. As xincreases, the graph starts to roll over and forma high point, but it starts going back up againbefore that happens. This behavior is revealed bythe fact that the derivative is positive everywhere.Between x = 0 and x = 1, the derivative reachesa low point, indicating that the slope is aminimum, but the slope is still positive andthe graph of g is still going up.

3. a.200

h

h

x

y

2 1 2.5

b. The h graph looks like a cubic functiongraph. Conjecture: Seventh-degree functionhas a sixth-degree function for its derivative.

c. By plotting the graph using a friendly window,then tracing, the zeros of h are 2, 1, 2.5.

d. If h(x) = 0, the h graph has a high point ora low point. This is reasonable because ifh(x) = 0, the rate of change of h(x) is zero,which would happen when the graph stopsgoing up and starts going down, or viceversa.

e. See the graph in part a.4.

102 x

y

q

q'

The graph does not have the expected shape for aquartic function. The two high points and the lowpoint all appear to occur as a high point atx = 2. The derivative graph crosses the x-axis justonce, at x = 2, indicating that this is the onlyplace where the function graph is horizontal.


5. a.

4

5x

yf

f

b. Amplitude = 1, period = 2 = 6.283 c. The graph of f also has amplitude 1 and

period 2.d.

fg

f and g5

4

x

y

The graphs of f and g are the same shape,spaced 1 unit apart vertically. The graphs off and g are identical! This is to be expectedbecause the shapes of the f and g graphs arethe same.

6.

4

5x

yf

f

The function available on the grapher is y = cos x.The amplitude is 1, the period is 2, and theshape is sinusoidal. cos 0 = 1, and the graph is ata high point, y = 1, when x = 0.

7. 8.

x

y

4

3f

f

x

y

4

3f'

f

9. 10.

x

f

f

y

1

5

x

y

2

3f'

f

11. The derivative for f ( x) = 2x is consistently belowthat of the function itself. This fact implies thatf ( x) does not increase rapidly enough to make the

derivative equal the function value. So the basemust be greater than 2. By experimenting, 3 istoo large, but not by much. You can use trialand error with bases between 2 and 3, checkingthe results either by plotting the graph and thenumerical derivative or by constructing tables.An ingenious method that some students comeup with uses the numerical derivative andnumerical solver features to solvenDeriv(bx, x, 1) = b1 at x = 1. The answer isabout 2.718281 . (In Section 3-9, students willlearn that this number is e, the base of naturallogarithms.) The graph of f ( x) = 2.781xand its numerical derivative are shown here.

f and f '

x

y

1

1

12. Answer will vary depending on calculator.

13. a. Maximum area = (12.01)2 = 144.2401 in.2Minimum area = (11.99)2 = 143.7601 in.2Range is 143.7601 area 144.2401.Area is within 0.2401 in.2 of the ideal.

b. Let x be the number of inches.Area = x2.The right side of 12 is more restrictive, so setx2 = 144.02. x = 144.021/2 = 12.000833Keep the tile dimensions within 0.0008 in. of12 in.

c. The 0.02 in part b corresponds to , and the0.0008 corresponds to .

14. The average of the forward and backwardsdifference quotients equals

12

f x h f xh

f x f x hh

( ) ( ) ( ) ( )++

=

+

12

f x h f x hh

( ) ( )

=

+f x h f x hh

( ) ( )2

, Q.E.D.

15. a. f ( x) = x3 x + 1 f ( 1) = 1f x x

xx

+

( ) =1 1 111

3lim ( )

= =

+

lim

lim ( )( )

x x

x x

x

x x x

x1

3

111 1

1=

limx

x x1

1 2( + ) =


b. Forward: f f( . ) ( ).

.

.

.

1 1 10 1

1 231 10 1

2 31= =

Backwards: f f( ) ( . ).

.

.

.

1 0 90 1

1 0 8290 1

1 71= =

Symmetric:f f( . ) ( . )

( . ). .

.

.

1 1 0 92 0 1

1 231 0 8290 2

2 01= =

The symmetric difference quotient is closer tothe actual derivative because it is the averageof the other two, and the other two span theactual derivative.

c. f ( 0) = 1f x x

xx

+

( ) =0 1 100

3lim ( )

= lim lim( )x x

x x

xx

= =

0

3

02 1 1

d. Forward: f f( . ) ( ).

.

.

0 1 00 1

0 901 10 1

= = 0.99

Backwards: f f( . ) ( ).

.

.

0 1 00 1

1 099 10 1

=

= 0.99

Symmetric:f f( . ) ( . )

( . ). .

.

.

0 1 0 12 0 1

0 901 1 0990 2

0 99= =

All three difference quotients are equal becausef ( x) changes just as much from 0.1 to 0 as itdoes from 0 to 0.1.

16.h Backwards Forward Symmetric

0.1 1.1544 3.1544 10.01 3.6415 5.6415 10.001 9 11 1The backwards difference quotients are becominglarge and negative, while the forward differencequotients are becoming large and positive. Theiraverage, the symmetric difference quotient, isalways equal to 1.

17. Answers will vary.

Problem Set 3-4Q1. 9x2 24x + 16Q2. a3 + 3a2b + 3ab2 + b3Q3. See the text definition of derivative.Q4. f x h f x h

h( ) ( )+

2Q5. No limit (infinite) Q6. log 73Q7. 3 Q8. Pythagorean theoremQ9. 10 Q10. C

1. f ( x) = 5x4 f ( x) = 20x3

2. y = 11x8 dy/dx = 88x7

3. v = 0.007t 83 dv/dt = 0.581t 84

4. v x x v x x( ) = ( ) = 9

10

1812

5. M(x) = 1215 M (x) = 0 (Derivative of aconstant)

6. f (x) = 4.7723 f (x) = 0 (Derivative of aconstant)

7. y = 0.3x2 8x + 4 dy/dx = 0.6x 88. r = 0.2x2 + 6x 1 dr /dx = 0.4x + 6

9. ddx

x( )13 1=

10. f (x) = 4.5x2 x f (x) = 9x 111. y = x2.3 + 5x 2 100x + 4

dy/dx = 2.3x1.3 10x 3 100

12. ddx

x x x x x x( )/ 2 5 2 1 3 5 24 3 14 25

8 3+ = + /

13. v = (3x 4)2 = 9x2 24x + 16 dv/dx = 18x 2414. u = (5x 7)2 = 25x2 70x + 49

du/dx = 50x 7015. f (x) = (2x + 5)3 = 8x3 + 60x2 + 150x + 125

f (x) = 24x2 + 120x + 15016. f (x) = (4x 1)3 = 64x3 48x2 + 12x 1

f (x) = 192x2 96x + 1217. P x x x P x x( ) ( ) = + =

2

24 1

18. Q x x x x Q x x x( ) = + ( ) = + 3 2

2

3 21 1+

19. f (x) = 7x4

=+

f x x h x

hh( ) lim ( )

0

4 47 7

= + + +

lim ( )h

x x h xh h x0

3 2 2 3 328 42 28 7 28=

By formula, f (x) = 7 4x3 = 28x3, which checks.20. g(x) = 5x3

g x x h xhh

+

( ) = lim

0

3 35 5( )

= + + =

limh

x xh h x0

2 2 215 15 5 15( )By formula, g (x) = 5 3x2 = 15x2, which checks.

21. v(t) = 10t2 5t + 7

v tt h t h t t

hh

+ + + +

( ) = lim [ ( ) ( ) ] ( )0

2 210 5 7 10 5 7

=

+

lim h

th h hh0

220 10 5

= +

lim ( )h

t h t0

20 10 5 20 5=

By formula, v (t) = 10 2t 5 = 20t 5, whichchecks.


22. s t t t( ) = + .4 26 3 7 =

+ + + +

s tt h t h t t

hh( ) lim [( ) ( ) . ] [ . ]

0

4 2 4 26 3 7 6 3 7

=

+ + +

lim h

t h t h th h th hh0

3 2 2 3 4 24 6 4 12 6

= + + +

lim ( )h

t t h th h t h0

3 2 2 34 6 4 12 6

= 4 123t t By formula, s t t t t t ( ) = = 4 6 2 4 123 3 , which checks.

23. Mae should realize that you differentiatefunctions, not values of functions. If yousubstitute a value for x into f (x) = x4, you getf (3) = 34 = 81, which is a new function, g(x) =81. It is the derivative of g that equals zero.Moral: Differentiate before you substitute for x.

24. a. v(x) = h(x) = 10x + 20b. The book was going down at 10 m/s.

The velocity is 10, so h(x) is decreasing.c. The book was 15 m above where he threw it.d. 2 s. The book is at its highest point when the

velocity is zero. v(x) = 0 if and only if x = 2.25.

x

f

f

y

9

7

26.

6

x

g

g'

y3

27. a.

f

f

y

x

10

1

b. The graph of f is shown dashed in part a.c. There appear to be only two graphs because

the exact and the numerical derivative graphsalmost coincide.

d. f (3) = 6.2f (3) = 3.8 (by formula)f (3) 3.8000004 (depending on grapher)The two values of f ( )3 are almost identical!

28. a. g(x) = x 1. Conjecture: g(x) = 1 x 2.Conjecture is confirmed.

y1

y 2 and y3

1

1

x

y

b. h(x) = x1/ 2. Conjecture: g(x) = 0.5x 1/ 2.Conjecture is confirmed.

y1

y2 and y31

2

x

y

c. e(x) = 2x. Conjecture: e (x) = x 2x 1.Conjecture is refuted!

y1

y2

y3

111

x

y

29. f x x x( ) = /1 2 2 13+f x x f =( ) = + , ( )

Calculus Concepts and Applications 2nd Edition Solutions Manual

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