INSTRUCTOR SOLUTIONS MANUAL -...

Typewritten TextINSTRUCTOR SOLUTIONS MANUAL

INSTRUCTORS SOLUTIONS MANUAL

to accompany

ADAMS / ESSEX

CALCULUS: A COMPLETE COURSE; CALCULUS: SINGLE VARIABLE; and CALCULUS: SEVERAL VARIABLES

Eighth Edition

Prepared by

Robert A. Adams University of British Columbia

Christopher Essex

University of Windsor Ontario

Toronto

Copyright 2014 Pearson Canada Inc., Toronto, Ontario. Pearson Canada. All rights reserved. This work is protected by Canadian copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the Internet) will destroy the integrity of the work and is not permitted. The copyright holder grants permission to instructors who have adopted Calculus: A Complete Course, Eighth Edition, by Adams/Essex to post this material online only if the use of the website is restricted by access codes to students in the instructors class that is using the textbook and provided the reproduced material bears this copyright notice.

Line

FOREWORD

These solutions are provided for the benefit of instructors using the textbooks:

Calculus: A Complete Course (8th Edition),

Single-Variable Calculus (8th Edition),and

Calculus of Several Variables (8th Edition)

by R. A. Adams and Chris Essex, published by Pearson Education Canada. For the most part,the solutions are detailed, especially in exercises on core material and techniques. Occasion-ally some details are omitted for example, in exercises on applications of integration, theevaluation of the integrals encountered is not always given with the same degree of detail asthe evaluation of integrals found in those exercises dealing specifically with techniques of in-tegration.

Instructors may wish to make these solutions available to their students. However, studentsshould use such solutions with caution. It is always more beneficial for them to attempt ex-ercises and problems on their own, before they look at solutions done by others. If they ex-amine solutions as study material prior to attempting the exercises, they can lose much ofthe benefit that follows from diligent attempts to develop their own analytical powers. Whenthey have tried unsuccessfully to solve a problem, then looking at a solution can give them ahint for a second attempt. SeparateStudent Solutions Manualsfor the books are availablefor students. They contain the solutions to the even-numbered exercises only.

Apr, 2012.

R. A. [email protected]

Chris [email protected]

Copyright 2014 Pearson Canada Inc.

Highlight

CONTENTS

Solutions for Chapter P 1Solutions for Chapter 1 23Solutions for Chapter 2 40Solutions for Chapter 3 82Solutions for Chapter 4 109Solutions for Chapter 5 179Solutions for Chapter 6 215Solutions for Chapter 7 270Solutions for Chapter 8 318Solutions for Chapter 9 353Solutions for Chapter 10 393Solutions for Chapter 11 421Solutions for Chapter 12 450Solutions for Chapter 13 494Solutions for Chapter 14 541Solutions for Chapter 15 583Solutions for Chapter 16 614Solutions for Chapter 17 641Solutions for Chapter 18 648Solutions for Chapter 18 extended 669

Solutions for Appendices 699

NOTE: Chapter 18 extended is only needed by users ofCalculus of Several Variables (Eighth Edition)


INSTRUCTORS SOLUTIONS MANUAL SECTION P.1 (PAGE 10)

CHAPTER P. PRELIMINARIES

Section P.1 Real Numbers and the Real Line(page 10)

1.2

9= 0.22222222 = 0.2

2.1

11= 0.09090909 = 0.09

3. If x = 0.121212 , then 100x = 12.121212 = 12+ x .Thus 99x = 12 andx = 12/99 = 4/33.

4. If x = 3.277777 , then 10x 32 = 0.77777 and100x 320 = 7 + (10x 32), or 90x = 295. Thusx = 295/90 = 59/18.

5. 1/7 = 0.142857142857 = 0.1428572/7 = 0.285714285714 = 0.2857143/7 = 0.428571428571 = 0.4285714/7 = 0.571428571428 = 0.571428note the same cyclic order of the repeating digits

5/7 = 0.714285714285 = 0.7142856/7 = 0.857142857142 = 0.857142

6. Two different decimal expansions can represent the samenumber. For instance, both 0.999999 = 0.9 and1.000000 = 1.0 represent the number 1.

7. x 0 andx 5 define the interval [0, 5].

8. x < 2 andx 3 define the interval [3, 2).

9. x > 5 or x < 6 defines the union(,6) (5, ).

10. x 1 defines the interval(,1].

11. x > 2 defines the interval(2, ).

12. x < 4 or x 2 defines the interval(,), that is, thewhole real line.

13. If 2x > 4, thenx < 2. Solution: (,2)

14. If 3x + 5 8, then 3x 8 5 3 andx 1. Solution:(, 1]

15. If 5x 3 7 3x , then 8x 10 andx 5/4. Solution:(, 5/4]

16. If6 x

4

3x 42

, then 6 x 6x 8. Thus 14 7xand x 2. Solution: (, 2]

17. If 3(2 x) < 2(3 + x), then 0< 5x and x > 0. Solution:(0, )

18. If x2 < 9, then|x | < 3 and3 < x < 3. Solution:(3, 3)

19. Given: 1/(2 x) < 3.CASE I. If x < 2, then 1< 3(2 x) = 6 3x , so 3x < 5and x < 5/3. This case has solutionsx < 5/3.CASE II. If x > 2, then 1> 3(2 x) = 63x , so 3x > 5and x > 5/3. This case has solutionsx > 2.Solution: (, 5/3) (2, ).

20. Given: (x + 1)/x 2.CASE I. If x > 0, thenx + 1 2x , so x 1.CASE II. If x < 0, thenx + 1 2x , so x 1. (notpossible)Solution: (0, 1].

21. Given: x2 2x 0. Thenx(x 2) 0. This is onlypossible ifx 0 andx 2. Solution: [0, 2].

22. Given 6x2 5x 1, then(2x 1)(3x 1) 0, soeither x 1/2 andx 1/3, or x 1/3 andx 1/2.The latter combination is not possible. The solution set is[1/3, 1/2].

23. Given x3 > 4x , we havex(x2 4) > 0. This is possibleif x < 0 andx2 < 4, or if x > 0 andx2 > 4. Thepossibilities are, therefore,2 < x < 0 or 2 < x < .Solution: (2, 0) (2,).

24. Given x2x 2, thenx2x2 0 so (x2)(x+1) 0.This is possible ifx 2 andx 1 or if x 2 andx 1. The latter situation is not possible. The solutionset is [1, 2].

25. Given:x

2 1 +

4

x.

CASE I. If x > 0, thenx2 2x + 8, so thatx2 2x 8 0, or (x 4)(x + 2) 0. This ispossible forx > 0 only if x 4.CASE II. If x < 0, then we must have(x 4)(x +2) 0,which is possible forx < 0 only if x 2.Solution: [2, 0) [4, ).

26. Given:3

x 1 1 then(x 1)(x + 1) > 0, so that3(x+1) < 2(x1). Thusx < 5. There are no solutionsin this case.CASE II. If 1 < x < 1, then(x 1)(x + 1) < 0, so3(x + 1) > 2(x 1). Thusx > 5. In this case allnumbers in(1, 1) are solutions.CASE III. If x < 1, then(x 1)(x + 1) > 0, so that3(x + 1) < 2(x 1). Thusx < 5. All numbersx < 5are solutions.Solutions: (,5) (1, 1).

27. If |x | = 3 thenx = 3.

28. If |x 3| = 7, thenx 3 = 7, sox = 4 or x = 10.

29. If |2t + 5| = 4, then 2t + 5 = 4, so t = 9/2 ort = 1/2.

30. If |1 t | = 1, then 1 t = 1, so t = 0 or t = 2.

1


SECTION P.1 (PAGE 10) ADAMS and ESSEX: CALCULUS 8

31. If |8 3s| = 9, then 8 3s = 9, so 3s = 1 or 17, ands = 1/3 or s = 17/3.

32. If

s

2 1

= 1, then

s

2 1 = 1, sos = 0 or s = 4.

33. If |x | < 2, thenx is in (2, 2).

34. If |x | 2, thenx is in [2, 2].

35. If |s 1| 2, then 1 2 s 1 + 2, sos is in [1, 3].

36. If |t + 2| < 1, then2 1 < t < 2 + 1, so t is in(3,1).

37. If |3x 7| < 2, then 7 2 < 3x < 7 + 2, so x is in(5/3, 3).

38. If |2x + 5| < 1, then5 1 < 2x < 5 + 1, sox is in(3,2).

39. If

x

2 1

1, then 1 1

x

2 1 + 1, so x is in [0, 4].

40. If

2

x

2

|x 3| says that the distancefrom x to 1 is greater than the distance fromx to 3, sox must be to the right of the point half-way between1and 3. Thusx > 1.

42. |x 3| < 2|x | x2 6x + 9 = (x 3)2 < 4x2 3x2 + 6x 9 > 0 3(x + 3)(x 1) > 0. Thisinequality holds ifx < 3 or x > 1.

43. |a| = a if and only if a 0. It is false if a < 0.

44. The equation|x 1| = 1 x holds if |x 1| = (x 1),that is, if x 1 < 0, or, equivalently, ifx < 1.

45. The triangle inequality|x + y| |x | + |y| implies that

|x | |x + y| |y|.

Apply this inequality withx = a b and y = b to get

|a b| |a| |b|.

Similarly, |a b| = |b a| |b| |a|. Since

|a| |b|

is equal to either|a| |b| or |b| |a|, depending on thesizes ofa and b, we have

|a b|

|a| |b|

.

Section P.2 Cartesian Coordinates in thePlane (page 16)

1. From A(0, 3) to B(4, 0), 1x = 4 0 = 4 and1y = 0 3 = 3. |AB| =

42 + (3)2 = 5.

2. From A(1, 2) to B(4,10), 1x = 4 (1) = 5 and1y = 10 2 = 12. |AB| =

52 + (12)2 = 13.

3. From A(3, 2) to B(1,2), 1x = 1 3 = 4 and1y = 2 2 = 4. |AB| =

(4)2 + (4)2 = 4

2.

4. From A(0.5, 3) to B(2, 3), 1x = 2 0.5 = 1.5 and1y = 3 3 = 0. |AB| = 1.5.

5. Starting point: (2, 3). Increments1x = 4, 1y = 7.New position is(2 + 4, 3 + (7)), that is,(2, 4).

6. Arrival point: (2,2). Increments1x = 5, 1y = 1.Starting point was(2 (5),2 1), that is,(3, 3).

7. x2 + y2 = 1 represents a circle of radius 1 centred at theorigin.

8. x2 + y2 = 2 represents a circle of radius

2 centred atthe origin.

9. x2 + y2 1 represents points inside and on the circle ofradius 1 centred at the origin.

10. x2 + y2 = 0 represents the origin.

11. y x2 represents all points lying on or above theparabolay = x2.

12. y < x2 represents all points lying below the parabolay = x2.

13. The vertical line through(2, 5/3) is x = 2; the hori-zontal line through that point isy = 5/3.

14. The vertical line through(

2,1.3) is x =

2; thehorizontal line through that point isy = 1.3.

15. Line through(1, 1) with slopem = 1 isy = 1 + 1(x + 1), or y = x + 2.

16. Line through(2, 2) with slopem = 1/2 isy = 2 + (1/2)(x + 2), or x 2y = 6.

17. Line through(0, b) with slopem = 2 is y = b + 2x .

18. Line through(a, 0) with slopem = 2 isy = 0 2(x a), or y = 2a 2x .

19. At x = 2, the height of the line 2x + 3y = 6 isy = (6 4)/3 = 2/3. Thus(2, 1) lies above the line.

20. At x = 3, the height of the linex 4y = 7 isy = (3 7)/4 = 1. Thus(3, 1) lies on the line.

21. The line through(0, 0) and (2, 3) has slopem = (3 0)/(2 0) = 3/2 and equationy = (3/2)x or3x 2y = 0.

22. The line through(2, 1) and (2, 2) has slopem = (2 1)/(2 + 2) = 3/4 and equationy = 1 (3/4)(x + 2) or 3x + 4y = 2.

23. The line through(4, 1) and (2, 3) has slopem = (3 1)/(2 4) = 1/3 and equationy = 1

1

3(x 4) or x + 3y = 7.

2



24. The line through(2, 0) and (0, 2) has slopem = (2 0)/(0 + 2) = 1 and equationy = 2 + x .

25. If m = 2 andb =

2, then the line has equationy = 2x +

2.

26. If m = 1/2 andb = 3, then the line has equationy = (1/2)x 3, or x + 2y = 6.

27. 3x + 4y = 12 hasx-intercepta = 12/3 = 4 and y-interceptb = 12/4 = 3. Its slope isb/a = 3/4.

y

x

3x + 4y = 12

Fig. P.2.27

28. x + 2y = 4 hasx-intercepta = 4 and y-interceptb = 4/2 = 2. Its slope isb/a = 2/(4) = 1/2.

y

x

x + 2y = 4

Fig. P.2.28

29.

2x

3y = 2 hasx-intercepta = 2/

2 =

2and y-interceptb = 2/

3. Its slope is

b/a = 2/

6 =

2/3.y

x2x

3y = 2

Fig. P.2.29

30. 1.5x 2y = 3 hasx-intercepta = 3/1.5 = 2 and y-interceptb = 3/(2) = 3/2. Its slope isb/a = 3/4.

y

x

1.5x 2y = 3

Fig. P.2.30

31. line through(2, 1) parallel to y = x + 2 is y = x 1; lineperpendicular toy = x + 2 is y = x + 3.

32. line through(2, 2) parallel to 2x + y = 4 is2x + y = 2; line perpendicular to 2x + y = 4 isx 2y = 6.

33. We have

3x + 4y = 62x 3y = 13

H 6x + 8y = 126x 9y = 39.

Subtracting these equations gives 17y = 51, soy = 3and x = (139)/2 = 2. The intersection point is(2, 3).

34. We have

2x + y = 85x 7y = 1

H 14x + 7y = 565x 7y = 1.

Adding these equations gives 19x = 57, sox = 3 andy = 8 2x = 2. The intersection point is(3, 2).

35. If a 6= 0 andb 6= 0, then(x/a) + (y/b) = 1 representsa straight line that is neither horizontal nor vertical, anddoes not pass through the origin. Puttingy = 0 we getx/a = 1, so thex-intercept of this line isx = a; puttingx = 0 gives y/b = 1, so they-intercept isy = b.

36. The line (x/2) (y/3) = 1 hasx-intercepta = 2, andy-interceptb = 3.

y

x

3

x

2

y

3= 1

2

Fig. P.2.36

37. The line through(2, 1) and (3, 1) has slopem = (1 1)/(3 2) = 2 and equationy = 1 2(x 2) = 5 2x . Its y-intercept is 5.

3



38. The line through(2, 5) and (k, 1) hasx-intercept 3, soalso passes through(3, 0). Its slopem satisfies

1 0k 3

= m =0 53 + 2

= 1.

Thus k 3 = 1, and sok = 2.

39. C = Ax + B. If C = 5, 000 whenx = 10, 000 andC = 6, 000 whenx = 15, 000, then

10, 000A + B = 5, 00015, 000A + B = 6, 000

Subtracting these equations gives 5, 000A = 1, 000, soA = 1/5. From the first equation, 2, 000+ B = 5, 000,so B = 3, 000. The cost of printing 100,000 pamphlets is$100, 000/5 + 3, 000= $23, 000.

40. 40 and40 is the same temperature on both theFahrenheit and Celsius scales.

C

-50

-40

-30

-20

-10

10

20

30

40

F-50 -40 -30 -20 -10 10 20 30 4050 60 70 80

(40,40)

C =5

9(F 32)

C = F

Fig. P.2.40

41. A = (2, 1), B = (6, 4), C = (5, 3)

|AB| =

(6 2)2 + (4 1)2 =

25 = 5

|AC | =

(5 2)2 + (3 1)2 =

25 = 5

|BC | =

(6 5)2 + (4 + 3)2 =

50 = 5

2.Since |AB| = |AC |, triangle ABC is isosceles.

42. A = (0, 0), B = (1,

3), C = (2, 0)

|AB| =

(1 0)2 + (

3 0)2 =

4 = 2

|AC | =

(2 0)2 + (0 0)2 =

4 = 2

|BC | =

(2 1)2 + (0

3)2 =

4 = 2.Since |AB| = |AC | = |BC |, triangle ABC is equilateral.

43. A = (2, 1), B = (1, 3), C = (3, 2)

|AB| =

(1 2)2 + (3 + 1)2 =

17

|AC | =

(3 2)2 + (2 + 1)2 =

34 =

2

17

|BC | =

(3 1)2 + (2 3)2 =

17.

Since |AB| = |BC | and |AC | =

2|AB|, triangle ABCis an isosceles right-angled triangle with right angle atB. Thus ABC D is a square ifD is displaced fromCby the same amountA is from B, that is, by increments1x = 2 1 = 1 and1y = 1 3 = 4. ThusD = (3 + 1, 2 + (4)) = (2,2).

44. If M = (xm , ym) is the midpoint ofP1P2, then the dis-placement ofM from P1 equals the displacement ofP2from M :

xm x1 = x2 xm , ym y1 = y2 ym.

Thus xm = (x1 + x2)/2 and ym = (y1 + y2)/2.

45. If Q = (xq , yq) is the point onP1P2 that is two thirds ofthe way fromP1 to P2, then the displacement ofQ fromP1 equals twice the displacement ofP2 from Q:

xq x1 = 2(x2 xq), yq y1 = 2(y2 yq).

Thus xq = (x1 + 2x2)/3 and yq = (y1 + 2y2)/3.

46. Let the coordinates ofP be (x, 0) and those ofQ be(X,2X). If the midpoint of P Q is (2, 1), then

(x + X)/2 = 2, (0 2X)/2 = 1.

The second equation implies thatX = 1, and the sec-ond then implies thatx = 5. Thus P is (5, 0).

47.

(x 2)2 + y2 = 4 says that the distance of(x, y) from(2, 0) is 4, so the equation represents a circle of radius 4centred at(2, 0).

48.

(x 2)2 + y2 =

x2 + (y 2)2 says that(x, y) isequidistant from(2, 0) and (0, 2). Thus(x, y) mustlie on the line that is the right bisector of the line from(2, 0) to (0, 2). A simpler equation for this line isx = y.

49. The line 2x + ky = 3 has slopem = 2/k. This lineis perpendicular to 4x + y = 1, which has slope4,providedm = 1/4, that is, providedk = 8. The line isparallel to 4x + y = 1 if m = 4, that is, if k = 1/2.

50. For any value ofk, the coordinates of the point of inter-section ofx + 2y = 3 and 2x 3y = 1 will also satisfythe equation

(x + 2y 3) + k(2x 3y + 1) = 0

4



because they cause both expressions in parentheses to be0. The equation above is linear inx and y, and so rep-resents a straight line for any choice ofk. This line willpass through(1, 2) provided 1+ 4 3+ k(2 6+ 1) = 0,that is, if k = 2/3. Therefore, the line through the pointof intersection of the two given lines and through thepoint (1, 2) has equation

x + 2y 3 +2

3(2x 3y + 1) = 0,

or, on simplification,x = 1.

Section P.3 Graphs of Quadratic Equations(page 22)

1. x2 + y2 = 16

2. x2 + (y 2)2 = 4, or x2 + y2 4y = 0

3. (x + 2)2 + y2 = 9, or x2 + y2 + 4y = 5

4. (x 3)2 + (y + 4)2 = 25, or x2 + y2 6x + 8y = 0.

5. x2 + y2 2x = 3x2 2x + 1 + y2 = 4(x 1)2 + y2 = 4centre: (1, 0); radius 2.

6. x2 + y2 + 4y = 0x2 + y2 + 4y + 4 = 4x2 + (y + 2)2 = 4centre: (0, 2); radius 2.

7. x2 + y2 2x + 4y = 4x2 2x + 1 + y2 + 4y + 4 = 9(x 1)2 + (y + 2)2 = 9centre: (1, 2); radius 3.

8. x2 + y2 2x y + 1 = 0x2 2x + 1 + y2 y + 14 =

14

(x 1)2 +(

y 12)2 = 14

centre: (1, 1/2); radius 1/2.

9. x2 + y2 > 1 represents all points lying outside the circleof radius 1 centred at the origin.

10. x2 + y2 < 4 represents the open disk consisting of allpoints lying inside the circle of radius 2 centred at theorigin.

11. (x + 1)2 + y2 4 represents the closed disk consisting ofall points lying inside or on the circle of radius 2 centredat the point(1, 0).

12. x2 + (y 2)2 4 represents the closed disk consisting ofall points lying inside or on the circle of radius 2 centredat the point(0, 2).

13. Together,x2 + y2 > 1 andx2 + y2 < 4 represent annulus(washer-shaped region) consisting of all points that areoutside the circle of radius 1 centred at the origin andinside the circle of radius 2 centred at the origin.

14. Together,x2 + y2 4 and(x + 2)2 + y2 4 represent theregion consisting of all points that are inside or on boththe circle of radius 2 centred at the origin and the circleof radius 2 centred at(2, 0).

15. Together,x2+ y2 < 2x and x2+ y2 < 2y (or, equivalently,(x 1)2 + y2 < 1 andx2 + (y 1)2 < 1) represent theregion consisting of all points that are inside both thecircle of radius 1 centred at(1, 0) and the circle of radius1 centred at(0, 1).

16. x2 + y2 4x + 2y > 4 can be rewritten(x 2)2+(y +1)2 > 9. This equation, taken together withx + y > 1, represents all points that lie both outside thecircle of radius 3 centred at(2, 1) and above the linex + y = 1.

17. The interior of the circle with centre(1, 2) and radius6 is given by(x + 1)2 + (y 2)2 < 6, or

x2 + y2 + 2x 4y < 1.

18. The exterior of the circle with centre(2,3) and ra-dius 4 is given by(x 2)2 + (y + 3)2 > 16, orx2 + y2 4x + 6y > 3.

19. x2 + y2 < 2, x 1

20. x2 + y2 > 4, (x 1)2 + (y 3)2 < 10

21. The parabola with focus(0, 4) and directrixy = 4 hasequationx2 = 16y.

22. The parabola with focus(0,1/2) and directrixy = 1/2has equationx2 = 2y.

23. The parabola with focus(2, 0) and directrixx = 2 hasequationy2 = 8x .

24. The parabola with focus(1, 0) and directrixx = 1 hasequationy2 = 4x .

25. y = x2/2 has focus(0, 1/2) and directrixy = 1/2.y

x

(0,1/2)

y=1/2

y=x2/2

Fig. P.3.25

26. y = x2 has focus(0,1/4) and directrixy = 1/4.

5



y

x

y=1/4

(0,1/4)

y=x2

Fig. P.3.26

27. x = y2/4 has focus(1, 0) and directrixx = 1.y

x

x=1

(1,0)

x=y2/4

Fig. P.3.27

28. x = y2/16 has focus(4, 0) and directrixx = 4.y

x

(4,0)

x=y2/16x=4

Fig. P.3.28

29.

y

x

(3, 3)

4

(4,2)

3

y = x2

Version (b)

Version (c)

Version (d)

Version (a)

Fig. P.3.29

a) has equationy = x2 3.

b) has equationy = (x 4)2 or y = x2 8x + 16.

c) has equationy = (x 3)2 + 3 or y = x2 6x + 12.

d) has equationy = (x 4)2 2, or y = x2 8x + 14.

30. a) If y = mx is shifted to the right by amountx1, theequationy = m(x x1) results. If(a, b) satisfies thisequation, thenb = m(ax1), and sox1 = a(b/m).Thus the shifted equation isy = m(x a + (b/m)) = m(x a) + b.

b) If y = mx is shifted vertically by amounty1,the equationy = mx + y1 results. If (a, b)satisfies this equation, thenb = ma + y1, andso y1 = b ma. Thus the shifted equation isy = mx + b ma = m(x a) + b, the sameequation obtained in part (a).

31. y =

(x/3) + 1

32. 4y =

x + 1

33. y =

(3x/2) + 1

34. (y/2) =

4x + 1

35. y = 1 x2 shifted down 1, left 1 givesy = (x + 1)2.

36. x2 + y2 = 5 shifted up 2, left 4 gives(x + 4)2 + (y 2)2 = 5.

37. y = (x 1)2 1 shifted down 1, right 1 givesy = (x 2)2 2.

38. y =

x shifted down 2, left 4 givesy =

x + 4 2.

6



39. y = x2 + 3, y = 3x + 1. Subtracting these equationsgivesx2 3x + 2 = 0, or (x 1)(x 2) = 0. Thusx = 1 orx = 2. The corresponding values ofy are 4 and 7. Theintersection points are(1, 4) and (2, 7).

40. y = x2 6, y = 4x x2. Subtracting these equationsgives2x2 4x 6 = 0, or 2(x 3)(x + 1) = 0. Thusx = 3or x = 1. The corresponding values ofy are 3 and5.The intersection points are(3, 3) and (1,5).

41. x2+ y2 = 25, 3x +4y = 0. The second equation says thaty = 3x/4. Substituting this into the first equation gives25x2/16 = 25, sox = 4. If x = 4, then the secondequation givesy = 3; if x = 4, theny = 3. Theintersection points are(4, 3) and (4, 3). Note thathaving found values forx , we substituted them into thelinear equation rather than the quadratic equation to findthe corresponding values ofy. Had we substituted intothe quadratic equation we would have got more solutions(four points in all), but two of them would have failed tosatisfy 3x + 4y = 12. When solving systems of nonlinearequations you should always verify that the solutions youfind do satisfy the given equations.

42. 2x2 + 2y2 = 5, xy = 1. The second equation says thaty = 1/x . Substituting this into the first equation gives2x2 + (2/x2) = 5, or 2x4 5x2 + 2 = 0. This equationfactors to(2x2 1)(x2 2) = 0, so its solutions arex = 1/

2 andx =

2. The corresponding values

of y are given byy = 1/x . Therefore, the intersectionpoints are(1/

2,

2), (1/

2,

2), (

2, 1/

2), and

(

2,1/

2).

43. (x2/4) + y2 = 1 is an ellipse with major axis between(2, 0) and (2, 0) and minor axis between(0,1) and(0, 1).

y

x

x24 +y

2=1

Fig. P.3.43

44. 9x2 + 16y2 = 144 is an ellipse with major axis between(4, 0) and (4, 0) and minor axis between(0,3) and(0, 3).

y

x

9x2+16y2=144

Fig. P.3.44

45.(x 3)2

9+

(y + 2)2

4= 1 is an ellipse with centre at

(3, 2), major axis between(0, 2) and (6, 2) andminor axis between(3, 4) and (3, 0).

y

x

(3,2)

(x3)29 +

(y+2)24 =1

Fig. P.3.45

46. (x 1)2 +(y + 1)2

4= 4 is an ellipse with centre at

(1, 1), major axis between(1, 5) and (1, 3) and minoraxis between(1,1) and (3, 1).

y

x

(x1)2+ (y+1)2

4 =4

(1,1)

Fig. P.3.46

47. (x2/4) y2 = 1 is a hyperbola with centre at the ori-gin and passing through(2, 0). Its asymptotes arey = x/2.

7



y

x

x24 y

2=1

y=x/2

y=x/2

Fig. P.3.47

48. x2 y2 = 1 is a rectangular hyperbola with centre atthe origin and passing through(0, 1). Its asymptotesare y = x .

y

x

x2y2=1

y=x

y=x

Fig. P.3.48

49. xy = 4 is a rectangular hyperbola with centre atthe origin and passing through(2, 2) and (2, 2). Itsasymptotes are the coordinate axes.

y

x

xy=4

Fig. P.3.49

50. (x 1)(y + 2) = 1 is a rectangular hyperbola with centreat (1, 2) and passing through(2,1) and (0, 3). Itsasymptotes arex = 1 and y = 2.

y

x

x = 1

y = 2

(x 1)(y + 2) = 1

Fig. P.3.50

51. a) Replacingx with x replaces a graph with its re-flection across they-axis.

b) Replacingy with y replaces a graph with its re-flection across thex-axis.

52. Replacingx with x and y with y reflects the graph inboth axes. This is equivalent to rotating the graph 180

about the origin.

53. |x | + |y| = 1.In the first quadrant the equation isx + y = 1.In the second quadrant the equation isx + y = 1.In the third quadrant the equation isx y = 1.In the fourth quadrant the equation isx y = 1.

y

x

1|x | + |y| = 1

11

1

Fig. P.3.53

Section P.4 Functions and Their Graphs(page 32)

1. f (x) = 1 + x2; domainR, range [1, )

2. f (x) = 1

x ; domain [0, ), range(, 1]

3. G(x) =

8 2x ; domain(, 4], range [0, )

4. F(x) = 1/(x 1); domain(, 1) (1, ), range(, 0) (0,)

8



5. h(t) =t

2 t

; domain(, 2), rangeR. (The equa-

tion y = h(t) can be squared and rewritten ast2 + y2t 2y2 = 0, a quadratic equation int having realsolutions for every real value ofy. Thus the range ofhcontains all real numbers.)

6. g(x) =1

1

x 2; domain(2, 3) (3, ), range

(, 0) (0, ). The equationy = g(x) can be solvedforx = 2 (1 (1/y))2 so has a real solution providedy 6= 0.

7.

y

x

y

x

y

x

y

x

graph (i)

graph (iii) graph (iv)

graph (ii)

Fig. P.4.7

Graph (ii) is the graph of a function because verticallines can meet the graph only once. Graphs (i), (iii),and (iv) do not have this property, so are not graphs offunctions.

8.

y

x

y

x

y

x

y

x

graph (a) graph (b)

graph (d)graph (c)

Fig. P.4.8

a) is the graph ofx(1x)2, which is positive forx > 0.

b) is the graph ofx2 x3 = x2(1 x), which is positiveif x < 1.

c) is the graph ofx x4, which is positive if 0< x < 1and behaves likex near 0.

d) is the graph ofx3 x4, which is positive if0 < x < 1 and behaves likex3 near 0.

9.x f (x) = x4

0 00.5 0.06251 1

1.5 5.06252 16

y

x

y = x4

Fig. P.4.9

9



10.x f (x) = x2/3

0 00.5 0.629961 1

1.5 1.31042 1.5874

y

x

y = x2/3

Fig. P.4.10

11. f (x) = x2 + 1 is even: f (x) = f (x)

12. f (x) = x3 + x is odd: f (x) = f (x)

13. f (x) =x

x2 1is odd: f (x) = f (x)

14. f (x) =1

x2 1is even: f (x) = f (x)

15. f (x) =1

x 2is odd about(2, 0): f (2 x) = f (2+ x)

16. f (x) =1

x + 4is odd about(4, 0):

f (4 x) = f (4 + x)

17. f (x) = x26x is even aboutx = 3: f (3 x) = f (3+ x)

18. f (x) = x3 2 is odd about(0, 2):f (x) + 2 = ( f (x) + 2)

19. f (x) = |x3| = |x |3 is even: f (x) = f (x)

20. f (x) = |x + 1| is even aboutx = 1:f (1 x) = f (1 + x)

21. f (x) =

2x has no symmetry.

22. f (x) =

(x 1)2 is even aboutx = 1:f (1 x) = f (1 + x)

23.y

xy=x2

24.y

x

y=1x2

25.y

x

y=(x1)2

26.y

x

y=(x1)2+1

27.y

x

y=1x3

28.y

x

y=(x+2)3

10



29.y

x

y=

x+1

30.y

x

y=

x+1

31.y

x

y=|x|

32.y

x

y=|x|1

33.y

x

y=|x2|

34.y

x

y=1+|x2|

35.y

x

y= 2x+2

x=2

36.y

x

x=2

y= 12x

37.y

x

y= xx+1

x=1

y=1

38.y

x

x=1

y=1

y= x1x

11



39.y

x

y= f (x)+2(1,3)

2 (2,2)

y

x

y= f (x)(1,1)

2

Fig. P.4.39(a) Fig. P.4.39(b)

40.y

x

y= f (x)+2(1,3)

2 (2,2)

y

x1

y= f (x)1(2,1)1

Fig. P.4.40(a) Fig. P.4.40(b)41.

y

x

y= f (x+2)(1,1)

2

42.y

x

(2,1)

1 3

y= f (x1)

43.y

x2

y= f (x)(1,1)

44.y

x

y= f (x)(1,1)

2

45.y

x

(3,1)

2 4

y= f (4x)

46.y

x

(1,1)

y=1 f (1x)

(1,1)

47. Range is approximately [0.18, 0.68].y

-1.0

-0.8

-0.6

-0.4

-0.2

0.2

0.4

0.6

0.8

x-5 -4 -3 -2 -1 1 2 3 4y = 0.18

y = 0.68y =

x + 2x2 + 2x + 3

Fig. P.4.47

48. Range is approximately(, 0.17].y

-7

-6

-5

-4

-3

-2

-1x-5 -4 -3 -2 -1 1 2 3 4

y = 0.17

y =x 1x2 + x

Fig. P.4.48

12



49.y

-1

1

2

3

4

5

x-5 -4 -3 -2 -1 1 2 3 4

y = x4 6x3 + 9x2 1Fig. P.4.49

Apparent symmetry aboutx = 1.5.This can be confirmed by calculatingf (3 x), whichturns out to be equal tof (x).

50.y

-1

1

2

x-5 -4 -3 -2 -1 1 2 3 4

y =3 2x + x2

2 2x + x2

Fig. P.4.50

Apparent symmetry aboutx = 1.This can be confirmed by calculatingf (2 x), whichturns out to be equal tof (x).

51.y

-2

-1

1

2

3

4

x-3 -2 -1 1 2 3 4 5 6

y =x 1x 2

y = x + 3

y = x 1

Fig. P.4.51

Apparent symmetry about(2, 1), and about the linesy = x 1 and y = 3 x .These can be confirmed by noting thatf (x) = 1+

1

x 2,

so the graph is that of 1/x shifted right 2 units and upone.

52.y

-2

-1

1

2

3

4

5

x-7 -6 -5 -4 -3 -2 -1 1 2

y =2x2 + 3x

x2 + 4x + 5

Fig. P.4.52

Apparent symmetry about(2, 2).This can be confirmed by calculating shifting the graphright by 2 (replacex with x 2) and then down 2 (sub-tract 2). The result is5x/(1 + x2), which is odd.

53. If f is both even and odd thef (x) = f (x) = f (x),so f (x) = 0 identically.

Section P.5 Combining Functions to MakeNew Functions (page 38)

1. f (x) = x , g(x) =

x 1.D( f ) = R, D(g) = [1,).D( f + g) = D( f g) = D( f g) = D(g/ f ) = [1,),D( f/g) = (1, ).( f + g)(x) = x +

x 1

( f g)(x) = x

x 1( f g)(x) = x

x 1

( f/g)(x) = x/

x 1(g/ f )(x) = (

1 x)/x

2. f (x) =

1 x , g(x) =

1 + x .D( f ) = (, 1], D(g) = [1,).D( f + g) = D( f g) = D( f g) = [1, 1],D( f/g) = (1, 1], D(g/ f ) = [1, 1).( f + g)(x) =

1 x +

1 + x

( f g)(x) =

1 x

1 + x

( f g)(x) =

1 x2

( f/g)(x) =

(1 x)/(1 + x)(g/ f )(x) =

(1 + x)/(1 x)

13



3.

y = x

y = x2

y = x x2

y

x

4.y

-2

-1

1

x-2 -1 1

y = x

y = x3

y = x3 x

5.y

x

y = x + |x |

y = |x |

y = x = |x |

y = x

6.y

-1

1

2

3

4

x-2 -1 1 2 3 4 5

y = |x |

y = |x 2|

y = |x | + |x 2|

7. f (x) = x + 5, g(x) = x2 3.f g(0) = f (3) = 2, g( f (0)) = g(5) = 22f (g(x)) = f (x2 3) = x2 + 2g f (x) = g( f (x)) = g(x + 5) = (x + 5)2 3f f (5) = f (0) = 5, g(g(2)) = g(1) = 2f ( f (x)) = f (x + 5) = x + 10g g(x) = g(g(x)) = (x2 3)2 3

8. f (x) = 2/x , g(x) = x/(1 x).f f (x) = 2/(2/x) = x; D( f f ) = {x : x 6= 0}f g(x) = 2/(x/(1 x)) = 2(1 x)/x;

D( f g) = {x : x 6= 0, 1}g f (x) = (2/x)/(1 (2/x)) = 2/(x 2);

D(g f ) = {x : x 6= 0, 2}g g(x) = (x/(1 x))/(1 (x/(1 x))) = x/(1 2x);

D(g g) = {x : x 6= 1/2, 1}

9. f (x) = 1/(1 x), g(x) =

x 1.f f (x) = 1/(1 (1/(1 x))) = (x 1)/x;

D( f f ) = {x : x 6= 0, 1}f g(x) = 1/(1

x 1);

D( f g) = {x : x 1, x 6= 2}g f (x) =

(1/(1 x)) 1 =

x/(1 x);D(g f ) = [0, 1)

g g(x) =

x 1 1; D(g g) = [2, )

10. f (x) = (x + 1)/(x 1) = 1 + 2/(x 1), g(x) = sgn(x).f f (x) = 1 + 2/(1 + (2/(x 1) 1)) = x;D( f f ) = {x : x 6= 1}

f g(x) =sgnx + 1sgnx 1

= 0; D( f g) = (, 0)

g f (x) = sgn(

x + 1x 1

)

={ 1 if x < 1 or x > 1

1 if 1 < x < 1 ;

D(g f ) = {x : x 6= 1, 1}g g(x) = sgn(sgn(x)) = sgn(x); D(g g) = {x : x 6= 0}

f (x) g(x) f g(x)

11. x2 x + 1 (x + 1)212. x 4 x + 4 x13.

x x2 |x |

14. 2x3 + 3 x1/3 2x + 315. (x + 1)/x 1/(x 1) x16. 1/(x + 1)2 x 1 1/x2

17. y =

x .y = 2 +

x : previous graph is raised 2 units.

y = 2 +

3 + x : previous graph is shiftend left 3 units.y = 1/(2 +

3 + x): previous graph turned upside down

and shrunk vertically.

14



y

x

y =

x

y = 2 +

x

y = 2 +

x + 3

y = 1/(2 +

x + 3)

Fig. P.5.17

18.y

x

y = 2xy = 2x 1

y = 1 2x

y =

1 2x

y =1

1 2x

y =1

1 2x

1

Fig. P.5.18

19.y

x82

(1,2)

y=2 f (x)

20.y

x

2

y=(1/2) f (x)

21.y

x

y= f (2x)(1/2,1)

1

22.y

x

y= f (x/3)

63

23.y

x

y=1+ f (x/2)

(2,2)

24.y

x

y=2 f ((x1)/2)

1 5

25.y

x

y = f (x)(1, 1)

2

26.y

x

y = g(x)(1, 1)

-2

15



27. F(x) = Ax + B(a) F F(x) = F(x) A(Ax + B) + B = Ax + B A[(A 1)x + B] = 0Thus, eitherA = 0 or A = 1 and B = 0.(b) F F(x) = x A(Ax + B) + B = x (A2 1)x + (A + 1)B = 0Thus, eitherA = 1 or A = 1 and B = 0

28. x = 0 for 0 x < 1; x = 0 for 1 x < 0.

29. x = x for all integersx .

30. x = x is true for all realx ; if x = n + y wherenis an integer and 0 y < 1, thenx = n y, so thatx = n and x = n.

31.y

x

y = x x

32. f (x) is called the integer part ofx because| f (x)|is the largest integer that does not exceedx ; i.e.|x | = | f (x)| + y, where 0 y < 1.

y

x

y = f (x)

Fig. P.5.32

33. If f is even andg is odd, then: f 2, g2, f g, g f ,and f f are all even. f g, f/g, g/ f , and g g are odd,and f + g is neither even nor odd. Here are two typicalverifications:

f g(x) = f (g(x)) = f (g(x)) = f (g(x)) = f g(x)( f g)(x) = f (x)g(x) = f (x)[g(x)]= f (x)g(x) = ( f g)(x).

The others are similar.

34. f even f (x) = f (x)f odd f (x) = f (x)f even and odd f (x) = f (x) 2 f (x) = 0 f (x) = 0

35. a) Let E(x) = 12 [ f (x) + f (x)].Then E(x) = 12 [ f (x) + f (x)] = E(x). Hence,E(x) is even.Let O(x) = 12 [ f (x) f (x)].Then O(x) = 12 [ f (x) f (x)] = O(x) andO(x) is odd.

E(x) + O(x)= 12 [ f (x) + f (x)] +

12 [ f (x) f (x)]

= f (x).

Hence, f (x) is the sum of an even function and anodd function.

b) If f (x) = E1(x) + O1(x) where E1 is even andO1is odd, then

E1(x) + O1(x) = f (x) = E(x) + O(x).

Thus E1(x) E(x) = O(x) O1(x). The left side ofthis equation is an even function and the right sideis an odd function. Hence both sides are both evenand odd, and are therefore identically 0 by Exercise36. HenceE1 = E and O1 = O. This shows thatf can be written in only one way as the sum of aneven function and an odd function.

Section P.6 Polynomials and Rational Func-tions (page 45)

1. x2 7x + 10 = (x + 5)(x + 2)The roots are5 and2.

2. x2 3x 10 = (x 5)(x + 2)The roots are 5 and2.

3. If x2 + 2x + 2 = 0, thenx =2

4 8

2= 1 i .

The roots are1+ i and1 i .x2 + 2x + 2 = (x + 1 i )(x + 1 + i ).

4. Rather than use the quadratic formula this time, let uscomplete the square.

x2 6x + 13 = x2 6x + 9 + 4= (x 3)2 + 22

= (x 3 2i )(x 3 + 2i ).

The roots are 3+ 2i and 3 2i .

5. 16x4 8x2 + 1 = (4x2 1)2 = (2x 1)2(2x + 1)2. Thereare two double roots: 1/2 and1/2.

6. x4 + 6x3 + 9x2 = x2(x2 + 6x + 9) = x2(x + 3)2. Thereare two double roots, 0 and3.

16



7. x3 + 1 = (x + 1)(x2 x + 1). One root is1. The othertwo are the solutions ofx2 x + 1 = 0, namely

x =1

1 4

2=

1

2

3

2i.

We have

x3 + 1 = (x + 1)

(

x 1

2

3

2i

)(

x 1

2+

3

2i

)

.

8. x4 1 = (x2 1)(x2 + 1) = (x 1)(x + 1)(x i )(x + i ).The roots are 1,1, i , andi .

9. x6 3x4 + 3x2 1 = (x2 1)3 = (x 1)3(x + 1)3. Theroots are 1 and1, each with multiplicity 3.

10. x5 x4 16x + 16 = (x 1)(x4 16)= (x 1)(x2 4)(x4 + 4)= (x 1)(x 2)(x + 2)(x 2i )(x + 2i ).

The roots are 1, 2,2, 2i , and2i .

11. x5 + x3 + 8x2 + 8 = (x2 + 1)(x3 + 8)= (x + 2)(x i )(x + i )(x2 2x + 4)

Three of the five roots are2, i andi . The remain-ing two are solutions ofx2 2x + 4 = 0, namely

x =2

4 162

= 1

3 i . We have

x5+x3+8x2+8 = (x+2)(xi )(x+i )(xa+

3 i )(xa

3 i ).

12. x9 4x7 x6 + 4x4 = x4(x5 x2 4x3 + 4)= x4(x3 1)(x2 4)= x4(x 1)(x 2)(x + 2)(x2 + x + 1).

Seven of the nine roots are: 0 (with multiplicity 4),1, 2, and2. The other two roots are solutions ofx2 + x + 1 = 0, namely

x =1

1 4

2=

1

2

3

2i.

The required factorization ofx9 4x7 x6 + 4x4 is

x4(x1)(x2)(x+2)

(

x 1

2+

3

2i

)(

x 1

2

3

2i

)

.

13. The denominator isx2 + 2x + 2 = (x + 1)2 + 1 which isnever 0. Thus the rational function is defined for all realnumbers.

14. The denominator isx3 x = x(x 1)(x + 1) whichis zero if x = 0, 1, or1. Thus the rational function isdefined for all real numbers except 0, 1, and1.

15. The denominator isx3 + x2 = x2(x + 1) which is zeroonly if x = 0 or x = 1. Thus the rational function isdefined for all real numbers except 0 and1.

16. The denominator isx2 + x 1, which is a quadraticpolynomial whose roots can be found with the quadraticformula. They arex = (1

1 + 4)/2. Hence the

given rational function is defined for all real numbersexcept(1

5)/2 and(1 +

5)/2.

17.x3 1x2 2

=x3 2x + 2x 1

x2 2

=x(x2 2) + 2x 1

x2 2

= x +2x 1x2 2

.

18.x2

x2 + 5x + 3=

x2 + 5x + 3 5x 3x2 + 5x + 3

= 1 +5x 3

x2 + 5x + 3.

19.x3

x2 + 2x + 3=

x3 + 2x2 + 3x 2x2 3xx2 + 2x + 3

=x(x2 + 2x + 3) 2x2 3x

x2 + 2x + 3

= x 2(x2 + 2x + 3) 4x 6 + 3x

x2 + 2x + 3

= x 2 +x + 6

x2 + 2x + 3.

20.x4 + x2

x3 + x2 + 1=

x(x3 + x2 + 1) x3 x + x2

x3 + x2 + 1

= x +(x3 + x2 + 1) + x2 + 1 x + x2

x3 + x2 + 1

= x 1 +2x2 x + 1x3 + x2 + 1

.

21. As in Example 6, we wanta4 = 4, soa2 = 2and a =

2, b =

2a = 2. Thus

P(x) = (x2 2x + 2)(x2 + 2x + 2).

22. Following the method of Example 6, we calculate

(x2bx+a2)(x2+bx+a2) = x4+a4+(2a2b2)x2 = x2+x2+1

provideda = 1 andb2 = 1 + 2a2 = 3, sob =

3. ThusP(x) = (x2

3x + 1)(x2 +

3x + 1).

23. Let P(x) = an xn + an1xn1 + + a1x + a0, wheren 1. By the Factor Theorem,x 1 is a factor ofP(x) if and only if P(1) = 0, that is, if and only ifan + an1 + + a1 + a0 = 0.

24. Let P(x) = an xn + an1xn1 + + a1x + a0, wheren 1. By the Factor Theorem,x + 1 is a factor ofP(x) if and only if P(1) = 0, that is, if and only ifa0 a1+a2 a3 + + (1)nan = 0. This condition saysthat the sum of the coefficients of even powers is equalto the sum of coefficients of odd powers.

17



25. Let P(x) = an xn + an1xn1 + + a1x + a0, where thecoefficientsak , 0 k n are all real numbers, so thatak = ak . Using the facts about conjugates of sums andproducts mentioned in the statement of the problem, wesee that ifz = x + i y, wherex and y are real, then

P(z) = anzn + an1zn1 + + a1z + a0= an zn + an1zn1 + + a1z + a0= P(z).

If z is a root of P, then P(z) = P(z) = 0 = 0, and z isalso a root ofP.

26. By the previous exercise,z = u iv is also a root ofP. ThereforeP(x) has two linear factorsx u ivand x u + iv. The product of these factors is the realquadratic factor(x u)2 i2v2 = x2 2ux + u2 + v2,which must also be a factor ofP(x).

27. By the previous exercise

P(x)

x2 2ux + u2 + v2=

P(x)

(x u iv)(x u + iv)= Q1(x),

where Q1, being a quotient of two polynomials with realcoefficients, must also have real coefficients. Ifz = u+ivis a root of P having multiplicity m > 1, then it mustalso be a root ofQ1 (of multiplicity m 1), and so,therefore,z must be a root ofQ1, as must be the realquadraticx2 2ux + u2 + v2. Thus

P(x)

(x2 2ux + u2 + v2)2=

Q1(x)

x2 2ux + u2 + v2= Q2(x),

where Q2 is a polynomial with real coefficients. We cancontinue in this way until we get

P(x)

(x2 2ux + u2 + v2)m= Qm(x),

where Qm no longer hasz (or z) as a root. Thusz and zmust have the same multiplicity as roots ofP.

Section P.7 The Trigonometric Functions(page 57)

1. cos(

3

4

)

= cos(

4

)

= cos

4=

1

2

2. tan3

4= tan

3

4= 1

3. sin2

3= sin

(

3

)

= sin

3=

3

2

4. sin(

7

12

)

= sin(

4+

3

)

= sin

4cos

3+ cos

4sin

3

=1

2

1

2+

1

2

3

2=

1 +

3

2

2

5. cos5

12= cos

(

2

3

4

)

= cos2

3cos

4+ sin

2

3sin

4

= (

1

2

)(

1

2

)

+

(3

2

)

(

1

2

)

=

3 12

2

6. sin11

12= sin

12

= sin(

3

4

)

= sin

3cos

4 cos

3sin

4

=

(3

2

)

(

1

2

)

(

1

2

)(

1

2

)

=

3 12

2

7. cos( + x) = cos(

2 ( x))

= cos(

( x))

= cos( x) = cosx

8. sin(2 x) = sinx

9. sin(

3

2 x

)

= sin(

(

x

2

))

= sin(

x

2

)

= sin(

2 x

)

= cosx

10. cos(

3

2+ x

)

= cos3

2cosx sin

3

2sinx

= (1)( sinx) = sinx

11. tanx + cotx =sinx

cosx+

cosx

sinx

=sin2 x + cos2 x

cosx sinx

=1

cosx sinx

18



12.tanx cotxtanx + cotx

=

( sinx

cosx

cosx

sinx

)

( sinx

cosx+

cosx

sinx

)

=

(

sin2 x cos2 xcosx sinx

)

(

sin2 x + cos2 xcosx sinx

)

= sin2 x cos2 x

13. cos4 x sin4 x = (cos2 x sin2 x)(cos2 x + sin2 x)= cos2 x sin2 x = cos(2x)

14. (1 cosx)(1 + cosx) = 1 cos2 x = sin2 x implies1 cosx

sinx=

sinx

1 + cosx. Now

1 cosxsinx

=1 cos 2

( x

2

)

sin 2( x

2

)

=1

(

1 2 sin2( x

2

))

2 sinx

2cos

x

2

=sin

x

2

cosx

2

= tanx

2

15.1 cosx1 + cosx

=2 sin2

( x

2

)

2 cos2( x

2

) = tan2( x

2

)

16.cosx sinxcosx + sinx

=(cosx sinx)2

(cosx + sinx)(cosx sinx)

=cos2 x 2 sinx cosx + sin2 x

cos2 x sin2 x

=1 sin(2x)

cos(2x)= sec(2x) tan(2x)

17. sin 3x = sin(2x + x)= sin 2x cosx + cos 2x sinx= 2 sinx cos2 x + sinx(1 2 sin2 x)= 2 sinx(1 sin2 x) + sinx 2 sin3 x= 3 sinx 4 sin3 x

18. cos 3x = cos(2x + x)= cos 2x cosx sin 2x sinx= (2 cos2 x 1) cosx 2 sin2 x cosx= 2 cos3 x cosx 2(1 cos2 x) cosx= 4 cos3 x 3 cosx

19. cos 2x has period .

y

x2/2

1y = cos(2x)

Fig. P.7.19

20. sinx

2has period 4 .

y

x 2

1

1

Fig. P.7.20

21. sinx has period 2.y

x2 431

1

1

y = sin(x)

Fig. P.7.21

22. cosx

2has period 4.

y

x1

35

1

1

Fig. P.7.22

23.y

-3

-2

-1

1

2

x

y = 2 cos(

x

3

)

19



24.y

-1

1

2

x

y = 1 + sin(

4

)

25. sinx =3

5,

2< x <

cosx = 4

5, tanx =

3

4

x

53

4

Fig. P.7.25

26. tanx = 2 wherex is in [0,

2]. Then

sec2 x = 1 + tan2 x = 1 + 4 = 5. Hence,secx =

5 and cosx =

1

secx=

1

5,

sinx = tanx cosx =2

5

.

27. cosx =1

3,

2< x < 0

sinx =

8

3=

2

3

2

tanx =

8

1= 2

2

x

8

1

3

Fig. P.7.27

28. cosx = 5

13wherex is in

[

2, ]

. Hence,

sinx =

1 cos2 x =

1 25

169=

12

13,

tanx = 12

5.

29. sinx = 1

2, < x 0 so the particle ismoving to the right.

8. Average velocity over [t k, t + k] is

3(t + k)2 12(t + k) + 1 [3(t k)2 12(t k) + 1](t + k) (t k)

=1

2k

(

3t2 + 6tk + 3k2 12t 12k + 1 3t2 + 6tk 3k2

+ 12t 12k + 1)

=12tk 24k

2k= 6t 12 m/s,

which is the velocity at timet from Exercise 7.

9.y

1

2

t1 2 3 4 5

y = 2 +1

sin(t)

Fig. 1.1.9

At t = 1 the height isy = 2 ft and the weight ismoving downward.

10. Average velocity over [1, 1 + h] is

2 +1

sin(1 + h)

(

2 +1

sin

)

h

=sin( + h)

h=

sin cos(h) + cos sin(h)h

= sin(h)

h.

h Avg. vel. on [1,1 + h]1.0000 00.1000 -0.9836316430.0100 -0.9998355150.0010 -0.999998355

11. The velocity att = 1 is aboutv = 1 ft/s. The indicates that the weight is moving downward.

23


SECTION 1.1 (PAGE 63) ADAMS and ESSEX: CALCULUS 8

12. We sketched a tangent line to the graph on page 55 inthe text att = 20. The line appeared to pass throughthe points(10, 0) and (50, 1). On day 20 the biomass isgrowing at about(1 0)/(50 10) = 0.025 mm2/d.

13. The curve is steepest, and therefore the biomass is grow-ing most rapidly, at about day 45.

14. a)profit

255075

100125150175

year2008 2009 2010 2011 2012

Fig. 1.1.14

b) Average rate of increase in profits between 2010 and2012 is

174 622012 2010

=112

2= 56 (thousand$/yr).

c) Drawing a tangent line to the graph in (a) att = 2010 and measuring its slope, we find thatthe rate of increase of profits in 2010 is about 43thousand$/year.

Section 1.2 Limits of Functions (page 71)

1. From inspecting the graphy

x1 1

1

y = f (x)

Fig. 1.2.1

we see that

limx1

f (x) = 1, limx0

f (x) = 0, limx1

f (x) = 1.

2. From inspecting the graph

y

x1 2 3

1y = g(x)

Fig. 1.2.2

we see that

limx1

g(x) does not exist

(left limit is 1, right limit is 0)

limx2

g(x) = 1, limx3

g(x) = 0.

3. limx1

g(x) = 1

4. limx1+

g(x) = 0

5. limx3+

g(x) = 0

6. limx3

g(x) = 0

7. limx4

(x2 4x + 1) = 42 4(4) + 1 = 1

8. limx2

3(1 x)(2 x) = 3(1)(2 2) = 0

9. limx3

x + 3x + 6

=3 + 33 + 6

=2

3

10. limt4

t2

4 t=

(4)2

4 + 4= 2

11. limx1

x2 1x + 1

=12 11 + 1

=0

2= 0

12. limx1

x2 1x + 1

= limx1

(x 1) = 2

13. limx3

x2 6x + 9x2 9

= limx3

(x 3)2

(x 3)(x + 3)= lim

x3

x 3x + 3

=0

6= 0

14. limx2

x2 + 2xx2 4

= limx2

x

x 2=

24

=1

2

15. limh21

4 h2does not exist; denominator approaches 0

but numerator does not approach 0.

16. limh03h + 4h2

h2 h3= lim

h0

3 + 4hh h2

does not exist; denomi-

nator approaches 0 but numerator does not approach 0.

24


INSTRUCTORS SOLUTIONS MANUAL SECTION 1.2 (PAGE 71)

17. limx9

x 3

x 9= lim

x9

(

x 3)(

x + 3)(x 9)(

x + 3)

= limx9

x 9(x 9)(

x + 3)

= limx9

1

x + 3=

1

6

18. limh0

4 + h 2

h

= limh0

4 + h 4h(

4 + h + 2)

= limh0

1

4 + h + 2=

1

4

19. limx

(x )2

x=

02

2= 0

20. limx2

|x 2| = | 4| = 4

21. limx0

|x 2|x 2

=| 2|2

= 1

22. limx2

|x 2|x 2

= limx2

{

1, if x > 21, if x < 2.

Hence, limx2

|x 2|x 2

does not exist.

23. limt1

t2 1t2 2t + 1

limt1

(t 1)(t + 1)(t 1)2

= limt1

t + 1t 1

does not exist

(denominator 0, numerator 2.)

24. limx2

4 4x + x2

x 2= lim

x2

|x 2|x 2

does not exist.

25. limt0

t

4 + t

4 t= lim

t0

t (

4 + t +

4 t)(4 + t) (4 t)

= limt0

4 + t +

4 t

2= 2

26. limx1

x2 1

x + 3 2= lim

x1

(x 1)(x + 1)(

x + 3 + 2)(x + 3) 4

= limx1

(x + 1)(

x + 3 + 2) = (2)(

4 + 2) = 8

27. limt0

t2 + 3t(t + 2)2 (t 2)2

= limt0

t (t + 3)t2 + 4t + 4 (t2 4t + 4)

= limt0

t + 38

=3

8

28. lims0

(s + 1)2 (s 1)2

s= lim

s0

4s

s= 4

29. limy1

y 4y + 3y2 1

= limy1

(

y 1)(y 3)(

y 1)(y + 1)(y + 1)=

24

=12

30. limx1

x3 + 1x + 1

= limx1

(x + 1)(x2 x + 1)x + 1

= 3

31. limx2

x4 16x3 8

= limx2

(x 2)(x + 2)(x2 + 4)(x 2)(x2 + 2x + 4)

=(4)(8)

4 + 4 + 4=

8

3

32. limx8

x2/3 4x1/3 2

= limx8

(x1/3 2)(x1/3 + 2)(x1/3 2)

= limx8

(x1/3 + 2) = 4

33. limx2

(

1

x 2

4

x2 4

)

= limx2

x + 2 4(x 2)(x + 2)

= limx2

1

x + 2=

1

4

34. limx2

(

1

x 2

1

x2 4

)

= limx2

x + 2 1(x 2)(x + 2)

= limx2

x + 1(x 2)(x + 2)

does not exist.

35. limx0

2 + x2

2 x2

x2

= limx0

(2 + x2) (2 x2)x2(

2 + x2 +

2 x2)

= limx0

2x2

x2(

2 + x2) +

2 x2)

=2

2 +

2

=1

2

36. limx0

|3x 1| |3x + 1|x

= limx0

(3x 1)2 (3x + 1)2

x (|3x 1| + |3x + 1|)= lim

x0

12xx (|3x 1| + |3x + 1|)

=121 + 1

= 6

37. f (x) = x2

limh0

f (x + h) f (x)h

= limh0

(x + h)2 x2

h

= limh0

2hx + h2

h= lim

h02x + h = 2x

25



38. f (x) = x3

limh0

f (x + h) f (x)h

= limh0

(x + h)3 x3

h

= limh0

3x2h + 3xh2 + h3

h= lim

h03x2 + 3xh + h2 = 3x2

39. f (x) = 1/x

limh0

f (x + h) f (x)h

= limh0

1

x + h

1

xh

= limh0

x (x + h)h(x + h)x

= limh0

1

(x + h)x=

1

x2

40. f (x) = 1/x2

limh0

f (x + h) f (x)h

= limh0

1

(x + h)2

1

x2

h

= limh0

x2 (x2 + 2xh + h2)h(x + h)2x2

= limh0

2x + h

(x + h)2x2=

2x

x4=

2

x3

41. f (x) =

x

limh0

f (x + h) f (x)h

= limh0

x + h

x

h

= limh0

x + h xh(

x + h +

x)

= limh0

1

x + h +

x=

1

2

x

42. f (x) = 1/

x

limh0

f (x + h) f (x)h

= limh0

1

x + h

1

xh

= limh0

x

x + h

h

x

x + h

= limh0

x (x + h)h

x

x + h(

x +

x + h)

= limh0

1

x

x + h(

x +

x + h)

=1

2x3/2

43. limx/2

sinx = sin/2 = 1

44. limx/4

cosx = cos/4 = 1/

2

45. limx/3

cosx = cos/3 = 1/2

46. limx2/3

sinx = sin 2/3 =

3/2

47.x (sinx)/x

1.0 0.841470980.1 0.998334170.01 0.999983330.001 0.999999830.0001 1.00000000

It appears that limx0

sinx

x= 1.

48.x (1 cosx)/x2

1.0 0.459697690.1 0.499583470.01 0.499995830.001 0.499999960.0001 0.50000000

It appears that limx0

1 cosxx2

=1

2.

49. limx2

2 x = 0

50. limx2+

2 x does not exist.

51. limx2

2 x = 2

52. limx2+

2 x = 2

53. limx0

x3 x does not exist.

(x3 x < 0 if 0 < x < 1)

54. limx0

x3 x = 0

55. limx0+

x3 x does not exist. (See # 9.)

56. limx0+

x2 x4 = 0

57. limxa

|x a|x2 a2

= limxa

|x a|(x a)(x + a)

= 1

2a(a 6= 0)

58. limxa+

|x a|x2 a2

= limxa+

x ax2 a2

=1

2a

59. limx2

x2 4|x + 2|

=0

4= 0

60. limx2+

x2 4|x + 2|

=0

4= 0

26



61. f (x) ={ x 1 if x 1

x2 + 1 if 1 < x 0(x + )2 if x > 0

limx1

f (x) = limx1

x 1 = 1 1 = 2

62. limx1+

f (x) = limx1+

x2 + 1 = 1 + 1 = 2

63. limx0+

f (x) = limx0+

(x + )2 = 2

64. limx0

f (x) = limx0

x2 + 1 = 1

65. If limx4

f (x) = 2 and limx4

g(x) = 3, then

a) limx4

(

g(x) + 3)

= 3 + 3 = 0

b) limx4

x f (x) = 4 2 = 8

c) limx4

(

g(x))2

= (3)2 = 9

d) limx4

g(x)

f (x) 1=

32 1

= 3

66. If lim x a f (x) = 4 and limxa

g(x) = 2, then

a) limxa

(

f (x) + g(x))

= 4 + (2) = 2

b) limxa

f (x) g(x) = 4 (2) = 8

c) limxa

4g(x) = 4(2) = 8

d) limxa

f (x)

g(x)=

4

2= 2

67. If limx2

f (x) 5x 2

= 3, then

limx2

(

f (x) 5)

= limx2

f (x) 5x 2

(x 2) = 3(2 2) = 0.

Thus limx2 f (x) = 5.

68. If limx0

f (x)

x2= 2 then

limx0 f (x) = limx0 x2f (x)

x2= 0 (2) = 0,

and similarly,

limx0f (x)

x= lim

x0x

f (x)

x2= 0 (2) = 0.

69.y

-0.4

-0.2

0.2

0.4

0.6

0.8

1.0

1.2

x-3 -2 -1 1 2

y =sinx

x

Fig. 1.2.69

limx0

sinx

x= 1

70.y

-0.4

-0.2

0.2

0.4

0.6

0.8

x-0.08 -0.04 0.04 0.08

y =sin(2x)

sin(3x)

Fig. 1.2.70

limx0 sin(2x)/ sin(3x) = 2/3

71.y

-0.1

0.1

0.20.30.40.5

0.60.70.8

x0.2 0.4 0.6 0.8 1.0

y =sin

1 x

1 x2

Fig. 1.2.71

limx1

sin

1 x

1 x2 0.7071

27



72.y

-1.2

-1.0

-0.8

-0.6

-0.4

-0.2

x0.2 0.4 0.6 0.8

y =x

x

sinx

Fig. 1.2.72

limx0+

x

x

sinx= 1

73.y

-0.2

-0.1

0.1

x-0.2 -0.1 0.1

y = xy = x sin(1/x)

y = x

Fig. 1.2.73

f (x) = x sin(1/x) oscillates infinitely often asx ap-proaches 0, but the amplitude of the oscillations decreasesand, in fact, limx0 f (x) = 0. This is predictable be-cause|x sin(1/x)| |x |. (See Exercise 95 below.)

74. Since

5 2x2 f (x)

5 x2 for 1 x 1, andlimx0

5 2x2 = limx0

5 x2 =

5, we have

limx0 f (x) =

5 by the squeeze theorem.

75. Since 2 x2 g(x) 2 cosx for all x , and sincelimx0(2 x2) = limx0 2 cosx = 2, we havelimx0 g(x) = 2 by the squeeze theorem.

76. a)y

1

2

3

x-2 -1 1

y = x2

y = x4

(1, 1) (1, 1)

Fig. 1.2.76

b) Since the graph off lies between those ofx2 andx4, and since these latter graphs come together at(1, 1) and at (0, 0), we have limx1 f (x) = 1and limx0 f (x) = 0 by the squeeze theorem.

77. x1/3 < x3 on (1, 0) and (1,). x1/3 > x3 on(, 1) and (0, 1). The graphs ofx1/3 and x3 inter-sect at(1, 1), (0, 0), and(1, 1). If the graph ofh(x)lies between those ofx1/3 and x3, then we can determinelimxa h(x) for a = 1, a = 0, anda = 1 by thesqueeze theorem. In fact

limx1

h(x) = 1, limx0

h(x) = 0, limx1

h(x) = 1.

78. f (x) = s sin1

xis defined for allx 6= 0; its domain is

(, 0) (0, ). Since| sin t| 1 for all t , we have| f (x)| |x | and |x | f (x) |x | for all x 6= 0.Since limx0 = (|x |) = 0 = limx0 |x |, we havelimx0 f (x) = 0 by the squeeze theorem.

79. | f (x)| g(x) g(x) f (x) g(x)Since lim

xag(x) = 0, therefore 0 lim

xaf (x) 0.

Hence, limxa

f (x) = 0.If lim

xag(x) = 3, then either3 lim

xaf (x) 3 or

limxa f (x) does not exist.

Section 1.3 Limits at Infinity and InfiniteLimits (page 78)

1. limx

x

2x 3= lim

x

1

2 (3/x)=

1

2

2. limx

x

x2 4= lim

x

1/x

1 (4/x2)=

0

1= 0

28



3. limx

3x3 5x2 + 78 + 2x 5x3

= limx

3 5

x+

7

x38

x3+

2

x2 5

= 3

5

4. limx

x2 2x x2

= limx

1 2

x21

x 1

=1

1= 1

5. limx

x2 + 3x3 + 2

= limx

1

x+

3

x3

1 +2

x3

= 0

6. limx

x2 + sinxx2 + cosx

= limx

1 +sinx

x2

1 +cosx

x2

=1

1= 1

We have used the fact that limxsinx

x2= 0 (and simi-

larly for cosine) because the numerator is bounded whilethe denominator grows large.

7. limx

3x + 2

x

1 x

= limx

3 +2

x

1

x 1

= 3

8. limx

2x 1

3x2 + x + 1

= limx

x

(

2 1

x

)

|x |

3 +1

x+

1

x2

(but |x | = x as x )

= limx

2 1

x

3 +1

x+

1

x2

=2

3

9. limx

2x 1

3x2 + x + 1

= limx

2 1

x

3 +1

x+

1

x2

= 2

3

,

becausex implies thatx < 0 and so

x2 = x .

10. limx

2x 5|3x + 2|

= limx

2x 5(3x + 2)

= 2

3

11. limx3

1

3 xdoes not exist.

12. limx3

1

(3 x)2=

13. limx3

1

3 x=

14. limx3+

1

3 x=

15. limx5/2

2x + 55x + 2

=0

252

+ 2= 0

16. limx2/5

2x + 55x + 2

does not exist.

17. limx(2/5)

2x + 55x + 2

=

18. limx2/5+

2x + 55x + 2

=

19. limx2+

x

(2 x)3=

20. limx1

x

1 x2=

21. limx1+

1

|x 1|=

22. limx1

1

|x 1|=

23. limx2

x 3x2 4x + 4

= limx2

x 3(x 2)2

=

24. limx1+

x2 x

x x2= lim

x1+

1

x2 x=

25. limx

x + x3 + x5

1 + x2 + x3

= limx

1

x2+ 1 + x2

1

x3+

1

x+ 1

=

26. limx

x3 + 3x2 + 2

= limx

x +3

x2

1 +2

x2

=

27. limx

x

x + 1(

1

2x + 3)

7 6x + 4x2

= limx

x2(

1 +1

x

)(

1

x

2 +3

x

)

x2(

7

x2

6

x+ 4

)

=1(

2)

4=

1

4

2

28. limx

(

x2

x + 1

x2

x 1

)

= limx

2x2

x2 1= 2

29



29. limx

(

x2 + 2x

x2 2x)

= limx

(x2 + 2x) (x2 2x)

x2 + 2x +

x2 2x= lim

x

4x

(x)(

1 +2

x+

1 2

x

)

= 4

1 + 1= 2

30. limx

(

x2 + 2x

x2 2x)

= limx

x2 + 2x x2 + 2x

x2 + 2x +

x2 2x= lim

x

4x

x

1 +2

x+ x

1 2

x

= limx

4

1 +2

x+

1 2

x

=4

2= 2

31. limx

1

x2 2x x

= limx

x2 2x + x

(

x2 2x + x)(

x2 2x x)

= limx

x2 2x + x

x2 2x x2

= limx

x(

1 (2/x) + 1)2x

=2

2= 1

32. limx

1

x2 + 2x x= lim

x

1

|x |(

1 + (2/x) + 1= 0

33. By Exercise 35,y = 1 is a horizontal asymptote (at theright) of y =

1

x2 2x x. Since

limx

1

x2 2x x= lim

x

1

|x |(

1 (2/x) + 1= 0,

y = 0 is also a horizontal asymptote (at the left).Now

x2 2x x = 0 if and only if x2 2x = x2, that

is, if and only if x = 0. The given function is undefinedat x = 0, and wherex2 2x < 0, that is, on the interval[0, 2]. Its only vertical asymptote is atx = 0, where

limx01

x2 2x x

= .

34. Since limx

2x 5|3x + 2|

=2

3and lim

x

2x 5|3x + 2|

= 2

3,

y = (2/3) are horizontal asymptotes ofy = (2x 5)/|3x + 2|. The only vertical asymptoteis x = 2/3, which makes the denominator zero.

35. limx0+

f (x) = 1

36. limx1

f (x) =

37.y

-1

1

2

3

x1 2 3 4 5 6

y = f (x)

Fig. 1.3.37

limx2+ f (x) = 1

38. limx2

f (x) = 2

39. limx3

f (x) =

40. limx3+

f (x) =

41. limx4+

f (x) = 2

42. limx4

f (x) = 0

43. limx5

f (x) = 1

44. limx5+

f (x) = 0

45. limx

f (x) = 1

46. horizontal: y = 1; vertical: x = 1, x = 3.

47. limx3+

x = 3

48. limx3

x = 2

49. limx3

x does not exist

50. limx2.5

x = 2

51. limx0+

2 x = limx2

x = 1

52. limx3

x = 4

53. limtt0

C(t) = C(t0) except at integerst0lim

tt0C(t) = C(t0) everywhere

limtt0+

C(t) = C(t0) if t0 6= an integerlim

tt0+C(t) = C(t0) + 1.5 if t0 is an integer

30



y

x

6.00

4.50

3.00

1.50

1 2 3 4

y = C(t)

Fig. 1.3.53

54. limx0+

f (x) = L(a) If f is even, then f (x) = f (x).Hence, lim

x0f (x) = L.

(b) If f is odd, then f (x) = f (x).Therefore, lim

x0f (x) = L.

55. limx0+

f (x) = A, limx0

f (x) = B

a) limx0+

f (x3 x) = B (sincex3 x < 0 if 0 < x < 1)

b) limx0

f (x3 x) = A (becausex3 x > 0 if1 < x < 0)

c) limx0

f (x2 x4) = A

d) limx0+

f (x2 x4) = A (sincex2 x4 > 0 for0 < |x | < 1)

Section 1.4 Continuity (page 87)

1. g is continuous atx = 2, discontinuous atx = 1, 0, 1, and 2. It is left continuous atx = 0and right continuous atx = 1.

y

1

2

x-2 -1 1 2

(1, 2)

(1, 1)y = g(x)

Fig. 1.4.1

2. g has removable discontinuities atx = 1 andx = 2.Redefineg(1) = 1 and g(2) = 0 to makeg continuousat those points.

3. g has no absolute maximum value on [2, 2]. It takeson every positive real value less than 2, but does not takethe value 2. It has absolute minimum value 0 on thatinterval, assuming this value at the three pointsx = 2,x = 1, andx = 1.

4. Function f is discontinuous atx = 1, 2, 3, 4, and 5. fis left continuous atx = 4 and right continuous atx = 2and x = 5.y

-1

1

2

3

x1 2 3 4 5 6

y = f (x)

Fig. 1.4.4

5. f cannot be redefined atx = 1 to become continuousthere because limx1 f (x) (= ) does not exist. (isnot a real number.)

6. sgnx is not defined atx = 0, so cannot be either continu-ous or discontinuous there. (Functions can be continuousor discontinuous only at points in their domains!)

7. f (x) ={

x if x < 0x2 if x 0 is continuous everywhere on the

real line, even atx = 0 where its left and right limits areboth 0, which is f (0).

8. f (x) ={

x if x < 1x2 if x 1 is continuous everywhere on the

real line except atx = 1 where it is right continuous,but not left continuous.

limx1

f (x) = limx1

x = 1 6= 1

= f (1) = limx1+

x2 = limx1+

f (x).

9. f (x) ={

1/x2 if x 6= 00 if x = 0

is continuous everywhere ex-

cept atx = 0, where it is neither left nor right continuoussince it does not have a real limit there.

10. f (x) ={

x2 if x 10.987 if x > 1

is continuous everywhere

except atx = 1, where it is left continuous but not rightcontinuous because 0.987 6= 1. Close, as they say, but nocigar.

31



11. The least integer functionx is continuous everywhereon R except at the integers, where it is left continuousbut not right continuous.

12. C(t) is discontinuous only at the integers. It is continu-ous on the left at the integers, but not on the right.

13. Sincex2 4x 2

= x + 2 for x 6= 2, we can define thefunction to be 2+ 2 = 4 at x = 2 to make it continuousthere. The continuous extension isx + 2.

14. Since1 + t3

1 t2=

(1 + t)(1 t + t2)(1 + t)(1 t)

=1 t + t2

1 tfor

t 6= 1, we can define the function to be 3/2 at t = 1to make it continuous there. The continuous extension is1 t + t2

1 t.

15. Sincet2 5t + 6t2 t 6

=(t 2)(t 3)(t + 2)(t 3)

=t 2t + 2

for t 6= 3,we can define the function to be 1/5 at t = 3 to make itcontinuous there. The continuous extension is

t 2t + 2

.

16. Sincex2 2x4 4

=(x

2)(x +

2)

(x

2)(x +

2)(x2 + 2)=

x +

2

(x +

2)(x2 + 2)for x 6=

2, we can define the function to be 1/4 at

x =

2 to make it continuous there. The continuous

extension isx +

2

(x +

2)(x2 + 2). (Note: cancelling the

x +

2 factors provides a further continuous extension tox =

2.

17. limx2+ f (x) = k 4 and limx2 f (x) = 4 = f (2).Thus f will be continuous atx = 2 if k 4 = 4, that is,if k = 8.

18. limx3 g(x) = 3 m andlimx3+ g(x) = 1 3m = g(3). Thus g will be con-tinuous atx = 3 if 3 m = 1 3m, that is, if m = 1.

19. x2 has no maximum value on1 < x < 1; it takes allpositive real values less than 1, but it does not take thevalue 1. It does have a minimum value, namely 0 takenon at x = 0.

20. The Max-Min Theorem says that a continuous functiondefined on a closed, finite interval must have maximumand minimum values. It does not say that other functionscannot have such values. The Heaviside function is notcontinuous on [1, 1] (because it is discontinuous atx = 0), but it still has maximum and minimum values.Do not confuse a theorem with its converse.

21. Let the numbers bex and y, wherex 0, y 0, andx + y = 8. If P is the product of the numbers, then

P = xy = x(8 x) = 8x x2 = 16 (x 4)2.

ThereforeP 16, so P is bounded. ClearlyP = 16 ifx = y = 4, so the largest value ofP is 16.

22. Let the numbers bex and y, wherex 0, y 0, andx + y = 8. If S is the sum of their squares then

S = x2 + y2 = x2 + (8 x)2

= 2x2 16x + 64 = 2(x 4)2 + 32.

Since 0 x 8, the maximum value ofS occurs atx = 0 or x = 8, and is 64. The minimum value occurs atx = 4 and is 32.

23. SinceT = 100 30x + 3x2 = 3(x 5)2 + 25, T willbe minimum whenx = 5. Five programmers should beassigned, and the project will be completed in 25 days.

24. If x desks are shipped, the shipping cost per desk is

C =245x 30x2 + x3

x= x2 30x + 245

= (x 15)2 + 20.

This cost is minimized ifx = 15. The manufacturershould send 15 desks in each shipment, and the shippingcost will then be $20 per desk.

25. f (x) =x2 1

x=

(x 1)(x + 1)x

f = 0 at x = 1. f is not defined at 0.f (x) > 0 on (1, 0) and (1,).f (x) < 0 on (, 1) and (0, 1).

26. f (x) = x2 + 4x + 3 = (x + 1)(x + 3)f (x) > 0 on (, 3) and (1,)f (x) < 0 on (3, 1).

27. f (x) =x2 1x2 4

=(x 1)(x + 1)(x 2)(x + 2)

f = 0 at x = 1.f is not defined atx = 2.f (x) > 0 on (, 2), (1, 1), and(2, ).f (x) < 0 on (2, 1) and (1, 2).

28. f (x) =x2 + x 2

x3=

(x + 2)(x 1)x3

f (x) > 0 on (2, 0) and (1,)f (x) < 0 on (, 2) and (0, 1).

29. f (x) = x3 + x 1, f (0) = 1, f (1) = 1.Since f is continuous and changes sign between 0 and 1,it must be zero at some point between 0 and 1 by IVT.

30. f (x) = x3 15x + 1 is continuous everywhere.f (4) = 3, f (3) = 19, f (1) = 13, f (4) = 5.Because of the sign changesf has a zero between4and3, another zero between3 and 1, and anotherbetween 1 and 4.

32



31. F(x) = (x a)2(x b)2 + x . Without loss of generality,we can assume thata < b. Being a polynomial,F iscontinuous on [a, b]. Also F(a) = a and F(b) = b.Since a < 12(a + b) < b, the Intermediate-Value Theoremguarantees that there is anx in (a, b) such thatF(x) = (a + b)/2.

32. Let g(x) = f (x) x . Since 0 f (x) 1 if 0 x 1,therefore,g(0) 0 and g(1) 0. If g(0) = 0 let c = 0,or if g(1) = 0 let c = 1. (In either casef (c) = c.)Otherwise,g(0) > 0 andg(1) < 0, and, by IVT, thereexistsc in (0, 1) such thatg(c) = 0, i.e., f (c) = c.

33. The domain of an even function is symmetric about they-axis. Since f is continuous on the right atx = 0,therefore it must be defined on an interval [0, h] forsomeh > 0. Being even,f must therefore be definedon [h, h]. If x = y, then

limx0

f (x) = limy0+

f (y) = limy0+

f (y) = f (0).

Thus, f is continuous on the left atx = 0. Being contin-uous on both sides, it is therefore continuous.

34. f odd f (x) = f (x)f continuous on the right lim

x0+f (x) = f (0)

Therefore, lettingt = x , we obtain

limx0

f (x) = limt0+

f (t) = limt0+

f (t)

= f (0) = f (0) = f (0).

Therefore f is continuous at 0 andf (0) = 0.

35. max 1.593 at0.831, min0.756 at 0.629

36. max 0.133 atx = 1.437; min0.232 atx = 1.805

37. max 10.333 atx = 3; min 4.762 atx = 1.260

38. max 1.510 atx = 0.465; min 0 atx = 0 and x = 1

39. root x = 0.682

40. root x = 0.739

41. roots x = 0.637 andx = 1.410

42. roots x = 0.7244919590 andx = 1.220744085

43. fsolve gives an approximation to the single real root to10 significant figures; solve gives the three roots (includ-ing a complex conjugate pair) in exact form involving the

quantity(

108+ 12

69)1/3

; evalf(solve) gives approxi-

mations to the three roots using 10 significant figures forthe real and imaginary parts.

Section 1.5 The Formal Definition of Limit(page 92)

1. We require 39.9 L 40.1. Thus

39.9 39.6 + 0.025T 40.10.3 0.025T 0.512 T 20.

The temperature should be kept between 12C and 20C.

2. Since 1.2% of 8,000 is 96, we require the edge lengthxof the cube to satisfy 7904 x3 8096. It is sufficientthat 19.920 x 20.079. The edge of the cube must bewithin 0.079 cm of 20 cm.

3. 3 0.02 2x 1 3 + 0.023.98 2x 4.021.99 x 2.01

4. 4 0.1 x2 4 + 0.11.9749 x 2.0024

5. 1 0.1

x 1.10.81 x 1.21

6. 2 0.01 1

x 2 + 0.01

1

2.01 x

1

1.990.5025 x 0.4975

7. We need0.03 (3x +1)7 0.03, which is equivalentto 0.01 x 2 0.01 Thus = 0.01 will do.

8. We need0.01

2x + 3 3 0.01. Thus

2.99

2x + 3 3.018.9401 2x + 3 9.0601

2.97005 x 3.030053 0.02995 x 3 0.03005.

Here = 0.02995 will do.

9. We need 8 0.2 x3 8.2, or 1.9832 x 2.0165.Thus, we need0.0168 x 2 0.0165. Here = 0.0165 will do.

33



10. We need 1 0.05 1/(x + 1) 1 + 0.05,or 1.0526 x + 1 0.9524. This will occur if0.0476 x 0.0526. In this case we can take = 0.0476.

11. To be proved: limx1

(3x + 1) = 4.Proof: Let > 0 be given. Then|(3x + 1) 4| < holdsif 3|x 1| < , and so if|x 1| < = /3. This confirmsthe limit.


(5 2x) = 1.Proof: Let > 0 be given. Then|(5 2x) 1| < holdsif |2x 4| < , and so if|x 2| < = /2. This confirmsthe limit.


x2 = 0.

Let > 0 be given. Then|x2 0| < holds if|x 0| = |x | < =

.


x 21 + x2

= 0.Proof: Let > 0 be given. Then

x 21 + x2

0

=|x 2|1 + x2

|x 2| <

provided|x 2| < = .

15. To be proved: limx1/2

1 4x2

1 2x= 2.

Proof: Let > 0 be given. Then ifx 6= 1/2 we have

1 4x2

1 2x 2

= |(1+2x)2| = |2x 1| = 2

x 1

2

<

provided|x 12 | < = /2.


x2 + 2xx + 2

= 2.Proof: Let > 0 be given. Forx 6= 2 we have

x2 + 2xx + 2

(2)

= |x + 2| <

provided|x + 2| < = . This completes the proof.


1

x + 1=

1

2.

Proof: Let > 0 be given. We have

1

x + 1

1

2

=

1 x2(x + 1)

=|x 1|2|x + 1|

.

If |x 1| < 1, then 0< x < 2 and 1< x + 1 < 3, so that|x + 1| > 1. Let = min(1, 2). If |x 1| < , then

1

x + 1

1

2

=|x 1|2|x + 1|

0 be given. Ifx 6= 1, we have

x + 1x2 1

(

1

2

)

=

1

x 1(

1

2

)

=|x + 1|2|x 1|

.

If |x +1| < 1, then2 < x < 0, so3 < x 1 < 1 and|x 1| > 1. Ler = min(1, 2). If 0 < |x (1)| < then |x 1| > 1 and|x + 1| < 2. Thus

x + 1x2 1

(

1

2

)

=|x + 1|2|x 1|

0 be given. We have

|

x 1| =

x 1

x + 1

|x 1| <

provided|x 1| < = . This completes the proof.


x3 = 8.Proof: Let > 0 be given. We have|x3 8| = |x 2||x2 + 2x + 4|. If |x 2| < 1,then 1 < x < 3 andx2 < 9. Therefore|x2 + 2x + 4| 9 + 2 3 + 4 = 19. If|x 2| < = min(1, /19), then

|x3 8| = |x 2||x2 + 2x + 4| 0 there exists a number > 0, depending on, such that

a < x < a implies | f (x) L| < .

22. We say that limx f (x) = L if the following condi-tion holds: for every number > 0 there exists a numberR > 0, depending on, such that

x < R implies | f (x) L| < .

23. We say that limxa f (x) = if the following con-dition holds: for every numberB > 0 there exists anumber > 0, depending onB, such that

0 < |x a| < implies f (x) < B.

34



24. We say that limx f (x) = if the following conditionholds: for every numberB > 0 there exists a numberR > 0, depending onB, such that

x > R implies f (x) > B.

25. We say that limxa+ f (x) = if the following con-dition holds: for every numberB > 0 there exists anumber > 0, depending onR, such that

a < x < a + implies f (x) < B.

26. We say that limxa f (x) = if the following con-dition holds: for every numberB > 0 there exists anumber > 0, depending onB, such that

a < x < a implies f (x) > B.

27. To be proved: limx1+1

x 1= . Proof: Let B > 0

be given. We have1

x 1> B if 0 < x 1 < 1/B, that

is, if 1 < x < 1 + , where = 1/B. This completes theproof.


x 1= . Proof: Let B > 0

be given. We have1

x 1< B if 0 > x 1 > 1/B,

that is, if 1 < x < 1, where = 1/B.. This completesthe proof.


x2 + 1

= 0. Proof: Let > 0

be given. We have

1

x2 + 1

=1

x2 + 1

R, whereR = 1/. This completes theproof.

30. To be proved: limx

x = . Proof: Let B > 0 begiven. We have

x > B if x > R where R = B2. This

completes the proof.

31. To be proved: if limxa

f (x) = L and limxa

f (x) = M, thenL = M.Proof: SupposeL 6= M. Let = |L M|/3. Then > 0. Since lim

xaf (x) = L, there exists1 > 0 such that

| f (x)L| < if |xa| < 1. Since limxa

f (x) = M, thereexists2 > 0 such that| f (x) M| < if |x a| < 2.Let = min(1, 2). If |x a| < , then

3 = |L M| = |( f (x) M) + (L f (x)| | f (x) M| + | f (x) L| < + = 2.

This implies that 3< 2, a contradiction. Thus the origi-nal assumption thatL 6= M must be incorrect. ThereforeL = M.


g(x) = M, then there exists > 0such that if 0< |x a| < , then |g(x)| < 1 + |M|.Proof: Taking = 1 in the definition of limit, we obtaina number > 0 such that if 0< |x a| < , then|g(x) M| < 1. It follows from this latter inequality that

|g(x)| = |(g(x) M)+ M| |G(x) M|+|M| < 1+|M|.


f (x) = L and limxa

g(x) = M, thenlimxa

f (x)g(x) = L M.Proof: Let > 0 be given. Since lim

xaf (x) = L, there

exists1 > 0 such that| f (x) L| < /(2(1 + |M|))if 0 < |x a| < 1. Since lim

xag(x) = M, there ex-

ists 2 > 0 such that|g(x) M| < /(2(1 + |L|)) if0 < |x a| < 2. By Exercise 32, there exists3 > 0such that|g(x)| < 1 + |M| if 0 < |x a| < 3. Let = min(1, 2, 3). If |x a| < , then

| f (x)g(x) L M = | f (x)g(x) Lg(x) + Lg(x) L M|= |( f (x) L)g(x) + L(g(x) M)| |( f (x) L)g(x)| + |L(g(x) M)|= | f (x) L||g(x)| + |L||g(x) M|

0 such that if 0< |x a| < , then|g(x)| > |M|/2.Proof: By the definition of limit, there exists > 0 suchthat if 0 < |x a| < , then |g(x) M| < |M|/2(since|M|/2 is a positive number). This latter inequalityimplies that

|M| = |g(x)+(Mg(x))| |g(x)|+|g(x)M| < |g(x)|+|M|2

.

It follows that |g(x)| > |M| (|M|/2) = |M|/2, asrequired.

35




g(x) = M where M 6= 0, then

limxa

1

g(x)=

1

M.

Proof: Let > 0 be given. Since limxa

g(x) = M 6= 0,

there exists1 > 0 such that|g(x) M| < |M|2/2 if0 < |x a| < 1. By Exercise 34, there exists2 > 0such that|g(x)| > |M|/2 if 0 < |x a| < 3. Let = min(1, 2). If 0 < |x a| < , then

1

g(x)

1

M

=|M g(x)||M||g(x)|

0 be given. Sincef is continuous atL,there exists a number > 0 such that if|yL| < , then| f (y) f (L)| < . Since limxc g(x) = L, there exists > 0 such that if 0< |x c| < , then |g(x) L| < .Taking y = g(x), it follows that if 0 < |x c| < , then| f (g(x)) f (L)| < , so that limxc f (g(x)) = f (L).

38. To be proved: if f (x) g(x) h(x) in an open intervalcontainingx = a (say, fora 1 < x < a + 1, where1 > 0), and if limxa f (x) = limxa h(x) = L, thenalso limxa g(x) = L.Proof: Let > 0 be given. Since limxa f (x) = L,there exists2 > 0 such that if 0< |x a| < 2,then | f (x) L| < /3. Since limxa h(x) = L,there exists3 > 0 such that if 0< |x a| < 3,then |h(x) L| < /3. Let = min(1, 2, 3). If0 < |x a| < , then

|g(x) L| = |g(x) f (x) + f (x) L| |g(x) f (x)| + | f (x) L| |h(x) f (x)| + | f (x) L|= |h(x) L + L f (x)| + | f (x) L| |h(x) L| + | f (x) L| + | f (x) L|

1.

18. limx1

x x2 = 0

19. limx

1 x2

3x2 x 1= lim

x

(1/x2) 13 (1/x) (1/x2)

= 1

3

20. limx

2x + 100x2 + 3

= limx

(2/x) + (100/x2)1 + (3/x2)

= 0

21. limx

x3 1x2 + 4

= limx

x (1/x2)1 + (4/x2)

=

22. limx

x4

x2 4= lim

x

x2

1 (4/x2)=

23. limx0+

1

x x2=

24. limx1/2

1

x x2=

1

1/4= 2

25. limx

sinx does not exist; sinx takes the values1 and 1in any interval(R, ), and limits, if they exist, must beunique.

26. limx

cosx

x= 0 by the squeeze theorem, since

1

x

cosx

x

1

xfor all x > 0

and limx(1/x) = limx(1/x) = 0.

27. limx0

x sin1

x= 0 by the squeeze theorem, since

|x | x sin1

x |x | for all x 6= 0

and limx0(|x |) = limx0 |x | = 0.

28. limx0

sin1

x2does not exist; sin(1/x2) takes the values1

and 1 in any interval(, ), where > 0, and limits, ifthey exist, must be unique.

29. limx

[x +

x2 4x + 1]

= limx

x2 (x2 4x + 1)x

x2 4x + 1

= limx

4x 1x |x |

1 (4/x) + (1/x2)

= limx

x [4 (1/x)]x + x

1 (4/x) + (1/x2)

= limx

4 (1/x)1 +

1 (4/x) + (1/x2)= 2.

Note how we have used|x | = x (in the second lastline), becausex .

30. limx

[x +

x2 4x + 1] = + =

31. f (x) = x3 4x2 + 1 is continuous on the whole real lineand so is discontinuous nowhere.

32. f (x) =x

x + 1is continuous everywhere on its domain,

which consists of all real numbers exceptx = 1. It isdiscontinuous nowhere.

33. f (x) ={

x2 if x > 2x if x 2

is defined everywhere and dis-

continuous atx = 2, where it is, however, left continuoussince limx2 f (x) = 2 = f (2).

34. f (x) ={

x2 if x > 1x if x 1

is defined and continuous ev-

erywhere, and so discontinuous nowhere. Observe thatlimx1 f (x) = 1 = limx1+ f (x).

35. f (x) = H(x 1) ={

1 if x 10 if x < 1

is defined everywhere

and discontinuous atx = 1 where it is, however, rightcontinuous.

36. f (x) = H(9 x2) ={ 1 if 3 x 3

0 if x < 3 or x > 3 is definedeverywhere and discontinuous atx = 3. It is rightcontinuous at3 and left continuous at 3.

37. f (x) = |x |+|x +1| is defined and continuous everywhere.It is discontinuous nowhere.

38. f (x) ={ |x |/|x + 1| if x 6= 1

1 if x = 1 is defined everywhereand discontinuous atx = 1 where it is neither left norright continuous since limx1 f (x) = , whilef (1) = 1.

37


CHALLENGING PROBLEMS 1 (PAGE 94) ADAMS and ESSEX: CALCULUS 8

Challenging Problems 1 (page 94)

1. Let 0 < a < b. The average rate of change ofx3 over[a, b] is

b3 a3

b a= b2 + ab + a2.

The instantaneous rate of change ofx3 at x = c is

limh0

(c + h)3 c3

h= lim

h0

3c2h + 3ch2 + h3

h= 3c2.

If c =

(a2 + ab + b2)/3, then 3c2 = a2 + ab + b2, sothe average rate of change over [a, b] is the instantaneousrate of change at

(a2 + ab + b2)/3.Claim:

(a2 + ab + b2)/3 > (a + b)/2.Proof: Sincea2 2ab + b2 = (a b)2 > 0, we have

4a2 + 4ab + 4b2 > 3a2 + 6ab + 3b2

a2 + ab + b2

3>

a2 + 2ab + b2

4=(

a + b2

)2

a2 + ab + b2

3>

a + b2

.

2. For x near 0 we have|x 1| = 1 x and |x +1| = x +1.Thus

limx0

x

|x 1| |x + 1|= lim

x0

x

(1 x) (x + 1)=

1

2.

3. For x near 3 we have|5 2x | = 2x 5, |x 2| = x 2,|x 5| = 5 x , and |3x 7| = 3x 7. Thus

limx3

|5 2x | |x 2||x 5| |3x 7|

= limx3

2x 5 (x 2)5 x (3x 7)

= limx3

x 34(3 x)

= 1

4.

4. Let y = x1/6. Then we have

limx64

x1/3 4x1/2 8

= limy2

y2 4y3 8

= limy2

(y 2)(y + 2)(y 2)(y2 + 2y + 4)

= limy2

y + 2y2 + 2y + 4

=4

12=

1

3.

5. Usea b =a3 b3

a2 + ab + b2to handle the denominator.

We have

limx1

3 + x 2

3

7 + x 2

= limx1

3 + x 4

3 + x + 2

(7 + x)2/3 + 2(7 + x)1/3 + 4(7 + x) 8

= limx1

(7 + x)2/3 + 2(7 + x)1/3 + 4

3 + x + 2=

4 + 4 + 42 + 2

= 3.

6. r+(a) =1 +

1 + a

a, r(a) =

1

1 + aa

.

a) lima0 r(a) does not exist. Observe that the rightlimit is and the left limit is.

b) From the following table it appears thatlima0 r+(a) = 1/2, the solution of the linear equa-tion 2x 1 = 0 which results from settinga = 0 inthe quadratic equationax2 + 2x 1 = 0.

a r+(a)

1 0.414210.1 0.48810

0.1 0.513170.01 0.49876

0.01 0.501260.001 0.49988

0.001 0.50013

c) lima0

r+(a) = lima0

1 + a 1

a

= lima0

(1 + a) 1a(

1 + a + 1)

= lima0

1

1 + a + 1=

1

2.

7. TRUE or FALSE

a) If limxa f (x) exists and limxa g(x) does not

exist, then limxa(

f (x) + g(x))

does not exist.

TRUE, because if limxa(

f (x) + g(x))

were to

exist then

limxa

g(x) = limxa

(

f (x) + g(x) f (x))

= limxa

(

f (x) + g(x))

limxa

f (x)

would also exist.

b) If ne

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