a.2. answers to exercises 419 · a.2. answers to exercises 419 a.2 answers to exercises section 1.1...

A.2. ANSWERS TO EXERCISES 419

A.2 Answers to Exercises

Section 1.1

Try This

1. a) −0.263158, b) −0.251256,c) −0.250125

Among the three slopes the best

approximation of the slope of the

tangent line is −0.250125

2.51128 , the total area of the eight rect-

angles is a better approximation

Check-It Out

1. 22 2.13

True or False:

T,T,F,T,F,

T ,T,T,T,T

Exercises Section 1.1

1. a) 4.1, b) 4.01, c) 4.001

2. a) 0.513167, b) 0.501256,

c) 0.500125

3. a)512 , b)

√48160 , c)

√10930

The arc length is the sum of parts

a), b), and c), which is about

1.13021.

4. a)

√1716 , b)

516 , c)

√4116 d)

√65

16

The arc length is the sum of parts

a), b), c), and d), which is about

1.47428.

5. a)1925 , b) same as part a)

6. a)296 , b)

643140 , c) part b)

7. a) 25, b) Approximate amount of

wetlands lost to erosion

8. a) −79, 207.9 gal/hr

b) −79, 920.1 gal/hr

9. a) 16.16 ft/sec b) 16.016 ft/sec

10. a) 12.6 ft/sec, b) 12.06 ft/sec

c) 6(2 + h) ft/sec

Section 1.2

Try This

1. 32 2. i) 4, ii) 4

3. i) does not exist, ii) 1

4. Note, f(x) = − 1x decreases

without bound as x approaches 0

and x �= 0.

Then f(x) does not approach a

single number as x approaches 0.

Thus, limx→0

−1

|x|fails to exist.

5. i) Note, cos

�π

[12k−2]+2

�= 1 and

cos

�π�

12k+1−2

�+2

�= −1 for all

integers k. Moreover,12k − 2

and1

2k+1 − 2 approach −2 as kincreases without bound. Then

cos

�π

x+2

�does not approach a

single number as x appoaches −2.

Thus, limx→−2

cos

�π

x+ 2

�does not

exist.

ii) limx→0

cos

�π

x+ 2

�= 0

6. Choose δ = �/2.

7. L = 1, choose δ = 0.03.

8. Choose δ = Min� �

3 , 1�.

Check-It Out

1. 3 2. 1 3. 4 4. no limit

5. no limit 6. hint δ =ε3

True or False:

F,F,T,F,F,

F,T,T,T,F


1. 0.25 2. 0.25 3. 5 4. 0.125

5. 4 6. −0.625 7. 9 8. 0.25

9. 2 10. 2 11. 4 12. −2

13. 5 14. 3 15. 7 16. 0

17. undefined 18. undefined

19. 1 20. 1

21. no limit 22. undefined

23. δ = .0025 24. δ =0.002

7

25. δ =0.023 26. δ = 0.005(2

√2 + 3)

27. L = 2, δ =118 28. L = 9, δ =

18

29. L = 9, δ =110 30. L = 2, δ =

18

31. L = 5, δ =ε2 32. L = 1, δ = ε

33. L = 10, δ = min{ ε8 , 1}

34. L = −3, δ = min{ ε5 , 1}

35. L =13 , δ = min{6ε, 1}

36. L = 1, δ = min{ε, 1}37. L = 2, δ = min{(2 +

√3)ε, 1}

38. L = 4, δ = min{ ε3 , 1}

39.√145− 12 ≈ 0.04159 in.

40. 2

�36 +

2π − 12 ≈ 0.1056 in.

π d2

4 = 36π ± 2

41. 4.85in. ≤ h ≤ 5.15in.

42. a) L = 2(1 + 2α) ≈ 2.0000288 m

b) Let A =0.0001

2(7.2×10−6).

T must satisfy

|T − 70| < A or 63.1 < T < 76.9

43. a) 0.0821(2)(293) = 48.1106 atm

b) Let V1 =48.110658.1106 ≈ 0.828.

Then 0.828 < V < 1.172 liter.

44. a)109 lux b) E is very large

45. a)p(4.05)−p(4)

0.05 ≈ $5.95/hr

b)p(4+h)−p(4)

4 = 6− h.If h ≈ 0, marginal profit is

$6/hr.

46. a)p(1.01)−p(1)

0.01 ≈ 0.151 m/hr.

b)p(1+h)−p(1)

h =2−49h

10 .

If h ≈ 0, instantaneous

velocity is 0.2 m/sec.

Section 1.3

Try It

1. 2, 2, −4

2. −2, −3/2, 3

3. 3, does not exist 4. 3/2, −1/2

5.

√3

3 ,

√3

3 6.14 ,

16

7. − 14 , −1 8. 1/2, 2

9. 1/3, 0 10. 2, 1

Check It Out

1. 9 2. −1 3.12

4. 7 5. −10

True or False:

T,F,F,F,F,

T,T,T,T,T


1. 4 2. 5 3. 2 4. π

5.38 6.

87 7. 3 8.

83

9.12 10.

√3

3

11. 6, 10, 37 ,

74 12.

16 ,

16 ,

12 , 3

13. 3, 1 14. 4, 4

15. −4 16. 3 17. 27 18. 48

19. 4 20. −6 21. 7 22. −7

23.14 24. 6 25. 1 26.

12

27. 12 28. 75 29.34 30.

83

31. − 19 32. −4 33. − 1

25 34. −1

35.16 36.

114 37.

13 38.

112

39. 1 40. −1 41.

√36 42. 2

√5

420 APPENDIX A. APPENDIX

43.45 44. − 8

17 45. 3 46.14

47. 1 48. 1 49. 4 50.12

51. 0 52. No limit

53. −1 54.13

55. −3 55. 5

59. a) 1, b) no limit 60. 4

61. 0 62. 0 63. 4 64. 5

Section 1.4

Try This

1. Nonremovable discontinuity

at x = −1

2. Removable discontinuity at x = 0.

If h(0) = 1, then h

becomes continuous at x = 0.

3. g is continuous on (−∞,∞)

4. y = cscx is continuous in

D = {x : x �= nπ, n an integer }.At each point not belonging to D,

y = cscx has a nonremovable

discontinuity.

5. Undefined, 0

6. 0, −1, 1

7. [−1/2, 1/2], (−1, 1)

8. Note,√1 + x2 is defined and

2 + cosx �= 0 for all x. Since a

quotient of continuous functions is

continuous, g(x) =√

1+x2

2+cos x is

continuous everywhere.

9. Continuous for all x in (−∞,∞)

10. Since g(x) = 4+x−x3is continuous,

g(1) = 4 and g(2) = −26, the

Intermediate Value Theorem

assures g(x0) = 0 for some number

x0 between 1 and 2.

Check-It Out

1. Removable discontinuity at x = −1.

Redefine f(−1) =12 in order to be

continuous at x = −1

2. 2, −2 3. (−∞,∞) 4. 2

5. If f(x) = x3 − x− 106, then

f(0) < 0 and f(107) > 0.

Intermediate Value Theorem as-

sures f(x0) = 0 for some xo in

[0, 107]

True or False

T,F, F,T,F

T,F,F,T,F


1. Yes, f(c) = L 2. No

3. Yes, f(c) = 4 4. Yes, f(c) = 10

5. No, unless f(1) = 2

6. No, unless limx→3−1

f(x) = 4

7. No, unless g(0) = 0

8. No, p(2) may or may not be defined

9. 2, 2, 2 10. 1, 1, 1

11. 0, 1, undefined

12. 3, 2, undefined

13. 0 14. Undefined

15. Undefined 16. Undefined

17. 1 18. −1 19. −1 20. 2

21. 0 22. Undefined

23. 0 24. 0 25. 4 26. 5

27. 1 28. −1 29. 1 30. −7

31. x = −1, 2, nonrem.

32. x = ±2, nonrem.

33. x = 3, nonremov;

x = −2, remov., f(−2) =45

34. x = 3, remov, f(3) = 227

35. x = 3, nonrem.

36. x = 4, nonrem.

37. x = 6, nonrem.

38. x = 2, remov. , f(2) = 1

39. x =π(2k+1)

4 , nonrem.

40. x = π(2k + 1),

nonrem.

43. ∅ 44. ∅45. x = −2, nonrem. 46. ∅47. x = 1, nonrem. 48. ∅49. ∅ 50. x = 3, nonrem.

51. x = 0, remov, f(0) = 1

52. x = 0, remov, f(0) = π

53. (−∞, 1) ∪ (1,∞)

54. (−∞, 0) ∪ (0,∞)

55. (−∞, 1) ∪ (1,∞)

56. (−∞,∞) 57. [−2, 2]

58. (−∞, 0) ∪ (0, 2) ∪ (2,∞)

59. [−4,∞) 60. (−∞, 1]

61. (−∞,∞) 62. (−∞,∞)

63. x =12 + k, integer k

64. x = kπ, integer k

65. x = 2k, integer k

66. x = k, integer k �= 0

67. a = 1 68. a = 1

69. a = 0, b = −4

70. a =

√3−12 , b = 1−

√3

2

Section 1.5

Try This

1. −∞, ∞ 2. −∞, ∞3. δ =

4�

1/M 4. δ = (1/M)2

5. a) x = 2 and x = −3, b) x = nπ

6. x = 2 7. −∞, ∞

Check-It Out

1. −∞, ∞, undefined

2. Line x = 5

True or False

F,T,F,T,F

T,T,T,F,F


1. ∞,−∞ undefined

2. −∞,∞ undefined

3. −∞,∞ undefined

4. ∞,∞,∞5. −∞,−∞ 6. ∞,−∞7. −∞,∞ 8. ∞,−∞9. Lines x = ±2

10. Lines x = −2, 3

11. Lines x = −4, 0, 2

12. Lines x = 0,±3

13. Lines x = −3,±2

14. Lines x = −3,±1, 2

15. Lines x = k, integer k

16. Lines x =π4 +

kπ2 , integer k

17. Lines x =2π3 + 2kπ,

x =4π3 + 2kπ, integer k

18. Lines x =π6 + 2kπ,

x =5π6 + 2kπ, integer k

19. Lines x =π2 + kπ

20. Lines x = kπ

21. None 22. None

23. None 24. None

25. ∞ 26. −∞ 27. −∞ 28. ∞29. Undefined 30. Undefined

31. −∞ 32. −∞ 33. ∞ 34. 0

35. ∞ 36. Undefined

37. ∞ 38. −∞ 39.14 40.

23

41.310 42.

13 43. −∞ 44. ∞

45. ∞ 46. ∞47. ∞, δ =

0.2M > 0;

If 0 < x < δ then0.2x > M

48. ∞, δ =6M > 0

If 0 < 2− x < δ then6

2−x > M

49. ∞, δ =1

M−1 , M > 1;

If 0 < −1− x < δ

thenx

1+x > M

50. ∞, δ =9

M−3 , M > 3;

If 0 < x− 3 < δthen

3xx−3 > M


51. −∞, δ =

�4− 1

N , N < 0;

If 0 < x− 2 < δ, then 14−x2 < N

52. −∞, δ = − 83

N3 , N < 0;

If 0 < −x < δ, then 83√x

< N

59. a)12 (π − α− sinα), c) 2

Chapter 1 Review Exercises

1. 3 +√2 +

√3 2.

51325

3. a) 2.1, b) 2.01, c) 2 + h

4. a) 0.249224, b) 0.249992,

c)1√

4+h+2

5. 1.05573,1.00505,1.0005,

0.9995,0.995049,0.954451

Approx. limit is 1

6. 1.7948,1.81188,1.81361

1.81399,1.81572,1.83318,

Approx. limit is 1.814

7. 0.0303, 0.0030, 0.0003,

0.9994, 0.9940, 0.9412

Approx. limits: 0,1,

and does not exist

8. 237, 241, 241, 121, 121, 123

Approx. limits: 241,121,

and does not exist

9. 1 10. 5

11. Does not exist

12. Does not exist

13. ∞, ∞ 14. ∞, −∞

15. −∞, ∞ 16. −∞, −∞

17. −5 18. −5 19. − 19 20.

59

21. − 12 22.

112 23. 0 24. ∞

25. 0 26. 0 27. 1 28. 2

29. 1 30. −1 31. − 15 32. 0

33. 2 34.12 35. 2 36.

14

37. 0 38. 0

39. Continuous on (−∞,∞)

40. Continuous on (−∞,∞)

41. Discontinuous at x = −6,−1

removable discontinuity at x = −6,

redefine f(−6) =95

42. Nonremovable discontinuity

at x = 0

43. In proof, let δ =ε3

Then |f(x)− 35| = 3|x− 12| < ε

for 0 < |x− 12| < δ

44. In proof, let δ = Min�1, ε

14

�

Then |f(x)− 10| =|x− 2||3x+ 5| < ε

for 0 < |x− 2| < δ

45. In proof, let

δ = Min

�1, (

√8 + 3)ε

�

Then |f(x)− (−1)| =|x−9|√x+3

< ε

for 0 < |x− 9| < δ

46. In proof, let

δ = Min

�1, (

√3 + 2)ε

�

Then |f(x)− 2| = |x−3|√x+1+2

< ε

for 0 < |x− 3| < δ

47. −∞ 48. −∞ 49. ∞ 50. ∞51. Let δ =

4√M

.

Then16x2 > M

whenever 0 < |x| < δ

52. Let M > 1 and δ =4

M−1 .

Thenx

x−4 > M whenever

4M−1 > x− 4 > 0.

53. Let N < 0, δ = Min�1,− 12

N

�.

Note12

x−1 < N if − 12N > 1− x.

Thus,12

x−1 < N whenever

0 < 1− x < δ.

54. Let δ = Min

�1,

� 8M

�2�.

Note if 8 < x < 10 then

x√9−x

> M if1√9−x

> M8 .

Thus,x√9−x

> M

whenever 0 < 9− x < δ.

55. Find the δ’s for 2 + x2

and 2 + x4.

Then choose the smaller of the δ�s,i.e., δ = Min

�√ε, 4

√ε�.

56. a =3√9− 3

57. [0.84375, 0.8125]

58. If f(x) ≥ x for all x then

f(1) ≥ 1. Thus, f(1) = 1.

If f(x) ≤ x for all x then

f(0) ≤ 0. Thus, f(0) = 0.

Else, f(x1) > 0 and f(x2) < 0

for some 0 ≤ x1, x2 ≤ 1.

Thus, by Intermediate Value

Theorem, f(x) = x has a solution.

59. a) Note BC =12

tan(α/2) .

Then A1 =144 cos3(α/2)

sin(α/2) .

b) The area of a sector isθr2

2 .

ThenA2 = 144

�1

tan(α/2) − π−α2

�.

c)A1A2

→ 3 as α → π2

60. a) v(2) = −64 ft/sec

b) 3 sec at the ground for s(3) = 0

61. a) R(100) = $150,

MR(100) = 0.5 dollars/brush

b) R(125) = $156.25,

MR(125) = 0 dollars/brush

62. ∞

Chapter 1 Multiple Choice Test

1. A 6. A2. B 7. B3. C 8. D4. A 9. C5. A 10. C

11. D 16. C12. A 17. B13. B 18. B14. C 19. A15. C 20. D

Section 2.1

Try This

1. −3 2. 6, −12 3. −9, −1

4. lim∆x→0

f(0 +∆x)− f(0)

∆x=

lim∆x→0

3√∆x

∆x=

lim∆x→0

1

3�

(∆x)2= ∞

5. 4, 2x 6. m =14 , y =

14x+ 2

7. g�(x) = 2/x2; Since the tangent lines

to the graph of g(x) = −2/x have

positive slopes and g�(x) is the

slope of the tangent line at x, we

obtain g�(x) > 0. Consequently,

the points (x, g�(x)) on the graph

of y = g�(x) lie above the x-axis.

8. If f(x) = 2|x− 1|, then

limx→1+

f(x)− f(1)

x− 1=

limx→1+

2|x− 1|x− 1

= 2 and

limx→1−

f(x)− f(1)

x− 1=

limx→1−

2|x− 1|x− 1

= −2.

Since the one-sided limits are not

equal, f �(x) does not exist.

If g(x) = [[x+ 2]], then

limx→0+

g(x)− g(0)

x− 0=

limx→0+

[[x+ 2]]− 2

x=

limx→0+

2− 2

x= 0 and

limx→0−

g(x)− g(0)

x− 0=

limx→0−

1− 2

x= lim

x→0−

−1

x, which

does not exist. Thus, g�(0) does

not exist.

9. f �(−1) = lim

x→−1

f(x)− f(−1)

x+ 1

= limx→−1

3√x+ 1

x+ 1


= limx→−1

1

3�

(x+ 1)2= ∞.

Then f �(−1) does not exist.

Check-It Out

1. 6 2. 10x

True or False:

T,T,T,T,T,

T,F,T,F,F


1. 10 2.37 3. −8 4. 6

5. 0 6. 0 7.14 8.

14

9.23 10. 2 11. −10 12. 9

13. −3 14. − 114 15.

14 16.

112

17. 0 18. −2 19. 3 20. −2

21. 1 22. 5 23. 3 24.23

25. 6x 26. 2x− 6

27.−1

(t+4)228.

−2(t+1)2

29.1

2√x+4

30.1√x−1

31.

√3x

2x 32.2√2x

x

33. 6x2 34. 4x3

35. 6x− 1 36. −2x+ 2

37. − 4x2 38.

−8(x−4)2

39.2√x

40.1

2√x+9

41. limx→1+

f(x)− f(1)

x− 1= 1

limx→1−

f(x)− f(1)

x− 1= −1

1x

2

y

42. limx→1+

f(x)− f(1)

x− 1= 2

limx→1−

f(x)− f(1)

x− 1= −2

1x

2

y

43. limx→0+

f(x)− f(0)

x= 0

limx→0−

f(x)− f(0)

x= 0

1x

23

y

44. limx→0+

f(x)− f(0)

x= 0

limx→0−

f(x)− f(0)

x= 1

1x

23

y

45. limx→−2

f(x)− f(−2)

x+ 2= ∞

�1�3x

�2

23

y

46. limx→1

f(x)− f(1)

x− 1= −∞

2 3x

�2

23

y

47. x = 3 48. x = ±2

49. x = −2 50. x = 3

51. x = 1 52. x = ±3

53. a) f(1) = 3, f(3) = −2;

b) 5; c) The line through

(1, 3) and (3,−2)

54. a)f(1)−f(0)

1−0 ; b)f(2)−f(3)

2−3

55. a)

x

y

b)

1x

y

56. a)

x

y

b)

x

y

57. a) D, E; b) less; c) see graph below

B

A

C

D

E

x

y

59. Yes 60. No

61. a =32 , b = − 1

2 , c = 0


62. f �(0) = lim

x→0

f(x)− f(0)

x=

limx→0

1− cosx

x2=

1

2

limx→0

�sinx

x

�21

1 + cosx=

1

2

Section 2.2

Try This

1. p�(x) = 0, a�(t) = 0

2. f �(x) = 5x4

, g�(x) = − 2x3 ,

y� = − 12√x3

, p�(t) = 1

3. 6,−2, 112 4. y = 12x− 16

5. f �(x) = 12x3

, g�(x) = − 12x4 ,

y� = 6√x3

6. f �(x) = 42x5 − 32,

g�(x) = − 28x8 +

16x3 − 5

7. y� = −π sinx,

y� = 2 cos(x)− π,

y� = 12 (cosx−

√3 sinx)

8. y = −√2x+

√2(4+π)

4

9. −94.4 ft/sec, −95.84 ft/sec

10. −64 ft/sec

Check-It Out

1. 0 2. 1 3. 48x3 4. − sinx

5. −33.6 ft/sec 6. −32 ft/sec

True or False

T,F, F,F,T,

F,F,T,T,T


1. 6x2 − 5 2. −15x4+ π

3. 3x2+ 2x 4. 16x+ 10

5. 4 cosx 6. −π sin t

7. 1 + 3 sinx 8.1√t− cos t

9. − 1x2 10.

1

33√x2

− 2

55√x3

11.1

33√t2

− 12t

√t

12. x− 32

13. 6√x− 3

2√x+

12x

√x

14. −15√x− 2

x√

x+

4√x

15. − 2x2 +

8x3 − 15

x4 16. − 12x2 − 3π

8x4

17. 32 18. 7 19. − 116 20.

114

21. − 227 22. − 1

64 23. −2 24. 2

25. −√22 26. −10

√2

27. −2 28. −1

29.815 30. 0 31. 0 32. − 17

4

33. y = 6x− 9 34. y = −4x+ 5

35. y = 8x− 12 36. y = −2x+ 9

37. y =52x+ 10 38. y = − x

16 − 1

39. y =√3x+ 1−

√36 π

40. y = − 2√3

3 x+4√3π−69

41. ave. rate = 30.3, f �(5) = 30

42. ave. rate ≈ 47.12, f �(4) = 47

43. ave. rate =6−6

√2

π , f �(π/4) = −

√2

2

44. ave. rate =24

√2−24π ,

f �(π/6) = 2

√3

45. −12.32 ftsec

, s�(1) = −12ftsec

,

|s�(1)| = 12ftsec

46. 14.16 ftsec

, s�(0.49) = 14.32 ftsec

,

|s�(0.49)| = 14.32 ftsec

47. −33.751 msec

s�(5.99) ≈ −33.70 msec

,

|s�(5.99)| ≈ 33.70 msec

48. ≈ 7.50 msec

,

s�(0.25) = 7.55 msec

,

|s�(0.25)| = 7.55 msec

49. 2.5 sec, v(2.5) = −56 ft/sec

50. −128√2 ft/sec

51. −64 ft/sec 52. −90.2 m/sec

53. 64 ft/sec 54. 19.6 m/sec

55. (±2, 4) 56. (5, 1/5)

57. y = 7x± 4

58. y = 2x, y = 2x+ 4/3

59. a = 3, b = −2, c = 1

60. a = 1, b = −3, c = 2,

and y = −x+ 1

61. y = −x2 +

32 62.

�12 ,

√2

2

�

63. (2, 1) 64. y = −x2 + 2

65. f(x) = x2 66. f(x) = x3+ 2x

Section 2.3

Try This

1.dydx = 36x2

+ 14x+ 1

2.dydx = x cosx+ sinx,

dydx = cos

2 x− sin2 x = cos 2x

3. h�(x) = 40

�x3

cosx+ 3x2sinx

�,

h�(x) = 40x3

cosx+ 120x2sinx

4.dydx = − 11

(3x−4)25

dydx =

203(10+x)2

6. y� = 32x

7, y� = − 1

2x3 , y�= 2

7.ddx

� 1xn

�=

xn·0−1·nxn−1

x2n =

−nxn−1

x2n = − nxn+1

8.dydx =

2(tan(x)+1)tan(x)−1

9.dydx =

1−cos xsin2 x

= cscx(cscx− cotx)

10. Average acceleration and accelera-

tion are both −32 ft/sec per sec

11.10

27x2 3√x2

Check-It Out

1. x4cosx+ 4x3

sinx

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

3. − 52x4 4. −32 ft/sec

2

True or False

F,T,F,T,F

T,T,T,F,T


1. y� = 3− 4x

2. y� = 8x3 − 24x2 − 1

3. f �(x) = 3x2 − a2 − 1

4. f �(x) = 4x3 − 2(a2 + 1)x+ a

5. y� = x sec2 x+ tanx

6. x�x sec

2 x+ 2 tanx�

7. x secx (2 + x tanx)

8. p�(x) = cos2 x− sin

2 x

9. f �(x) = 4(2x− 5)

10. y� = 24x2+ 24x+ 6

11. f �(x) = −

1

(x− 1)2

12. f �(x) =

12

(3− 2x)2

13. g�(x) =x(2a3 − x3

)

(x3 + a3)2

14. g�(x) =−4c2x

(x2 − c2)2

15. −x sinx+ cosx

x2

16.sinx− x cosx

sin2 x

17.8x

(x2 + 5)218.

2x3+ 3x2

+ 1

(x+ 1)2

19. 1 20. − 13x2 − 4

3x3

21.3x2 22.

12x2

5 23. − 43x3 24.

6x5

25.2 cos x

5 − 13 26.

sec2x2

27 −4 28. − 12 29. −17 30. −3

31. −7 32. 0 33. − 49900 34. − 4

π2

35. −1 36. −1−π

2

37. −2 38. −2

39. y = − 12x+ 1 40. y = x

41. y = − 425x+

1225 42. y = − 4

3x+73

43. y =43x− 2π

9 +

√33

44. y = 2√3x− 2

√3π3 + 2


45. y = −√2x+

√2π4 +

√2

46. y = −4x+8π3 −

√3

47. 3a+ b 48.3a−b

9

49. b− 3a 50. 2(3a+ b)

51. 2ax− 3x2 52. 9x2+ 22x+ 5

53. a) a(t) = 6t− 12; b) t = 2 sec

54. a) ≈ −0.51 in./sec2

b) −0.5 in./sec2

55. 121 ft 56. 60 ft

57. 6 58. 7! or 5040

59. − 24x5 60. (−1)

n n!xn+1

61. −5 cosx 62. −2 cosx

63. R(x) = x− 0.002x2

R�(x) = 1− 0.004x

64. 3.276t2 + 6.3t+ 5.2

Section 2.4

Try This

1. (f ◦ g)(x) = 3(5x− 7)4,

(f ◦ g)�(x) = 60(5x− 7)3

2. a) f(x) = x3, g(x) = 5x− 2,

b) f(x) =√x, g(x) = 1− x2

,

c) f(x) = tanx, g(x) = 2πx

3. a) 2 sec 2x tan 2x,

b) 15(5x− 2)2, c)

3x2

2√

1+x3

4. a) 2(sin2 x− cos

2 x),

b) 12 sinx(1− cosx)2

5. a) − 12x(x2−1)3

, b) − 4x(1+x2)3/2

6. Vertical tangent lines at (3, 0)

and (−2, 0),

f �(x) = 0 at the (1/2,− 3

�25/4)

7. y� = 3(3x− 1)3(x+ 4)

2(7x+ 15)

8. f �(x) = cos2 x+2 sin x(1+sin x)

2 cos2 x√1+sin x

or f �(x) = (sin x+1)3/2

2 cos2 x

9. h�(x) = −1

(1+x)2

�1+x1−x

10. a) −x sin(2x2),

b) 2 sec(2x+ 1) tan(2x+ 1),

c) − csc2 3√x

33√x2

11. f �(x) = 12 sin 6x

Check-It Out

1. 24(4x+ 1)2

2. f(x) = 10 cscx, g(x) = 12x

3. 24x(x2 − 1)3 4. 2x cos(x2

)

True or False

F,F,F,T,F,

T,T,T,T,T


1. (f ◦ g)(x) = 2(4x− 5)3

(f ◦ g)�(x) = 24(4x− 5)2

2. (f ◦ g)(x) = 12x2+ 1

(f ◦ g)�(x) = 24x

3. (f ◦ g)(x) =√1− 2x

(f ◦ g)�(x) = −1√1−2x

4. (f ◦ g)(x) = 3√9− x

(f ◦ g)�(x) = −1

3 3√

(9−x)2

5. (f ◦ g)(x) = 2 sin(12x)(f ◦ g)�(x) = 24 cos(12x)

6. (f ◦ g)(x) = −3 cos(8x)(f ◦ g)�(x) = 24 sin(8x)

7. (f ◦ g)(x) = tan(x3 − 10)

(f ◦ g)�(x) =3x2

sec( x3 − 10)

8. (f ◦ g)(x) = csc(x2)2

(f ◦ g)�(x) =−x csc(x2

) cot(x2)

9. a) y = u4, u = 3x+ 1

b) y = tanu, u = x2

c) y = 3 cos2 u, u = 5x

10. a) y =√u, u = 2x+ 1

b) y = sinu, u =x

x+1

c) y = u2, u = 2 sinx+ 1

11. 15(3x+ 1)4 12. 8x

�x2 − 3

�3

13.7

2√7x+4

14.x2

(x3+1)2/3

15. − 4(4x+1)

3(2x2+x−7)7/3

16. − 4x

3(x2+4)5/3

17. 6x2sec

2�2x3

�

18. −6x2csc

2�x3

�

19. 6 cos(3x) sin(3x)

20. −8 cos(4x) sin(4x)

21. 10 sec2 x tanx

22. 6 cot 3x csc23x

23.12 (cos(2x)− 2x) 24.

12 (x+ sin 3x)

25.sec

�x

x+1

�tan

�x

x+1

�

(x+1)2

26. −cot

�2x

x+1

�csc

�2x

x+1

�

(x+1)2

27. −3 cot2(sinx) csc2(sinx) cosx

28. −8 sec2(cos(2x)) sin 2x

29.3(x−4)x

2(x−3)3/230.

5x+23(x+1)2/3

√2x−1

31. − tan�cos

x2

�sec

2�cos

x2

�sin

�x2

�

32.12 cos

�x4

�sec

2�sin

x4

�

33. vertical (2, 0); horizontal (0,−8)

34. vertical none; horizontal�0, 1

4

�

35. vertical (−5, 0), (4, 0);

horizontal

�− 1

2 ,−35�

814

�

36. vertical (±2, 0);�0, 6 3

√2

�

37. vertical (±nπ, 0); horizontal at�(4n+1)π

2 , 1�and

�(4n+3)π

2 ,−1

�

for integer n

38. vertical none; horizontal at

(nπ, 4(−1)n) for integer n

39. y =32x− 5

2 40. y = −7x+ 15

41. y = 3√3x+ 1−

√36 π

42. y = −6√3x+ 2(1 +

√3π)

43. y = 6√3x+ 3−

√3π

44. y = −12√3x+ 6 + 2

√3π

45. 12 46. 6 47.32 48. 98

49. Apply chain rule to (f ◦ g)(x) = x

50. g��(x) = 3x2since g�(x) = x3

by

Exercise 49

51. f(1) = 2, g�(2) = 1/8

52. f(1) = −1,

(f−1)�(−1) =

1f �(1) = − 1

8

53. A(w) = 2w√1− w2,

A�(w) =

2−4w2√1−w2

54. A(w) = w(1 +√1− w2),

A�(w) = 1 +

1−2w2√1−w2

55. d =√x2 − 18x+ 121

d� = x−9√x2−18x+121

56. d =

�(x− 1)2 +

� 1x − 1

�2

d� = x4−x3+x−1x3A(x)

Section 2.5

Try This

1. a) 3dydx , b) 6y2 dy

dx ,

c) x dydx + y, d)

y2

3√

1+y3

dydx

2. a)1

3y2−1, b)

y2

1−2xy

3. − 12 4. y = x

5. a)−6y

(1+3y2)2, b)

−1y3

Check It Out

1. a) 2dydx + 3 b) 8y dy

dx + 2x

c) 12(x+ y)11�1 +

dydx

�

2.32 3.

−6y(3y2+1)3

True or False

F,F,T,T,F,

T,T,T,T,T



1. −xy 2.

xy

3.−2x−4y3y2+4x

4.y−2x+52y−x

5. −�

yx 6.

32

�yx

7. − 9y+2x9(2y+x) 8.

−y−2y2√xyx(1+4y

√xy)

9. −1 10. −1

11.x

cos(2y)+y cos(y2)

12.−x

sin(2y)−y sin(y2)

13.1

2 cos(2y)−1 14. − y+sin(x−y)x−sin(x−y)

15. − 4y3 16.

1y3

17.x4

12y3 18.6y3−8x2

9y5

19.2(y2−1)2−8x2y

(y2−1)3

20.6x(2y+1)2−18x4

(2y+1)3

21. − 67 22. 0 23. 1 24.

18

25. 2 26. − 12

27. y = −x+ 1 28. y =√3x− 2

29. y =−3√5x− 4√

530. y =

−5x√3

+ 10

31. y = − 5x4 +

94 32. y =

4x27 +

1927

33. y = 2x− 1 34. y = 3x− 2

35. y =−2x+14

5 36. y = −x+ 3

37. (2,±2√2) 38. (1,

√2)

43. Horizontal at (0, 0) and (33√2, 3 3

√4)

Vertical at (0, 0) and (33√4, 3 3

√2)

�3 23

, 3 43 ��3 4

3, 3 2

3 �3 43

x

3 43

y

44. Horizontal at (1, 2) and (−1,−2)

Vertical at (2, 1), (−2,−1)

1�2 �1 2x

�1

2

�2

1

y

Section 2.6

Try This

1. 1.75 2. −2.5

3. −80 in.2/min 4.

14π ft/hr

5. 1 ft/sec 6. 5/π ft3/hr

Check-It Out

1. 6 2. 6π

3. − 23 4. 4 in.

2/min

True or False

F,T,T,F,T,

T,F,T,F,T


1. 12 2. 6 3.136 4. 40

5. 11 6. 21 7. 4.5 8.316

9. 16 in.2/min 10. 2 ft/hr.

11. 32π ft3/hr 12. 8π in.

2/min

13. 216 in.3/min 14. 120 in.

2/hr

15.3

100 ft/min 16. −270 in.3/hr

17. − 2√3

9 radians/min,

dydt = − 2√

3ft/min, and

area increases at 2√3 sq. ft/min

18. −2.5 radians/hr, use

cotα = −m where α

is the angle in the problem

19. 72π ft3/hr 20. 22π ft

3/hr

21.209π ft/min 22.

3√2π

m/hr

23.11513 mph 24. 10

√7 mph

25.33√5units/sec 26. −4 units/sec

27.2513 ft/sec 28. 2

√2 miles/min.

29.dSdt =

dVdt

2r

30.drdt = k where V =

43πr

3,

dVdt = k(4πr2),

31.1013 ft/sec 32.

3613 ft/sec

33.π5 yd/sec

34. −2400 lb/ft3per min

35. 4 in.

36. 2.5 ft/sec =dydt ,

use 82+ y2 = x2

37. decreasing at

81√3−2162 cm

2/sec

38. 2.5 ft/min

39.1

30π cm3/min 40. −4 cm

2/min


1. A 6. B 11. A 16. C2. D 7. B 12. D 17. B3. C 8. C 13. D 18. C4. C 9. B 14. C 19. B5. C 10. B 15. C 20. B

Investigation Projects

1. P �(2) = −2000 bacteria/minute,

decreasing

2. a) N �(30) = 2700 people/day

b) N(31)−N(30) = 2699 people

c) Yes, if t = 10 and t = 50 days

3. a) C�(2000) = $2.40/item

b) C(2001)− C(2000) = $2.3998

c) C�(x) is the approximate cost

of the (x+ 1)st item

4. a) R(x) = 3x2+ 200x− 0.02x3

b) R�(100) = $200/item

c) C(101)− C(100) = $196.98

5. b) 8 cm/min

6. (h, k) = (−4, 3.5), r = 5√5/2

7. The tangent lines are

y = −x+15

4

y =1

8

�4

�17 +

√285

�x−

17

�15 +

√285

��

y =1

8

�−4

�√285− 17

�x+

17

�√285− 15

��

Section 3.1

Try This

1. a) max 2, min 1; b) no max , min 1;

c) no max , no min;

d) max 2, no min;

2. Max g(1) = 8, no min

3. a) f �(2) = 0, b) f �

(0) is undefined

4. a) x = 2,−4, b) x = 0, c) None

5. Max = 18, min = −10

6. Max = 4− 3√4, min = −2

√3/9

7. Max f(π/6) = 1.5,

min f(0) = f(π/2) = f(π) = 1

Check-It Out

1. 0, 1 2. π/3, 5π/3

3. Max f(2) = 8, min f(1) = −3

True or False

T,F,F,F,T,

T,F,T,T,T



1. max 5 at x = 3, min 1 x = 1

2. max 5 at x = 3, no min

3. no max, min 1 at x = 0

4. max 4 at x = 3, no min

5. f �(2) = f �

(−2) = 0

6. f �(1) = f �

(2) = 0

7. f �(0) undefined,

vertical tangent line

8. f �(0) undefined,

vertical tangent line

9. x = ±4 10. x = −6, 4

11. x = 0, 25 12. x = 0, 12

7

13. x = −2, 0 14. x = −2, 4

15. x =π6 ,

5π6 16. x =

π3 ,

5π3

17. x =π2 ,

3π2 , 7π

6 , 11π6

18. x =2π3 , 4π

3

19. Max f(4) = 0, min f(2) = −8

20. Max f(−2.5) = 12.5, min f(0) = 0

21. Max f(−3) = 59, min f(3) = −49

22. Max f(1) = f(−0.5) = 3,

min f(−1) = f(0.5) = 1

23. Max g(−5) = 110, min g(1) = 2

24. Max g(−1) = 2,

min g( 13 ) =2227 ≈ 0.8

25. Max g(2) = 9, min g(1) = 0

26. Max g(−1) = 0,

min g( 18 ) = − 21872048 ≈ −1.1

27. Max f(0) = 0,

min f( 23 ) = − 4√6

9 ≈ −1.1

28. Max f( 23 ) =2√

39 ,

min f(−1) = −√2

29. Max f(0) = f(2π) = 1,

min f(π) = −1

30. Max f(− 14 ) = 2, min f( 14 ) = 0

31. Max f(1) = f(−1) = 0,

min f(0) = −1

32. Max f(2) = 14 , min f(0) = 0

33. Max k(π3 ) = 3√3− π,

min k(π) = −3π

34. Max M(π6 ) = 1.5, min M(0) = 1

35. Let x ∈ [30, 50] be the diagonal.

The cost function is

C(x) = 12x+ 4[40−�

x2 − 900].

Minimum C(45√2) ≈ $499.41

36. Let x be the distance between Pand the shorter antenna. The

length is L(x) =√x2 + 502 +�

(100− x)2 + 752.

The minimum is L(40) = 25√41

≈ 160 ft. Thus, P must be 40 ft

from the shorter antenna.

37. Let x and y be the dimensions of

the rectangle. Assume x is the ra-

dius of the cylinder and y is height.

Since V (x) = πx2y = πx2(3 − x)

and V �= π3x(2−x), the maximum

is V (2) = 4π.

38. The minimum of

f(x) = 2

�1 + x2 +

�1 + (2− x)2

is f(0.461736) ≈ 4.03764

Section 3.2

Try This

1. a) c = 1, b) c = 3π/4

2. Since f(0) = 4 and f(1) = −6,

f has a zero in (0, 1). Since

f �(x) = −9− 3x2

has no zero,

f has exactly one zero by Rolle’s

Theorem. Then f has only one

x-intercept.

3. c = 1/4

4. Let s(t) be the distance driven in thours. Suppose s(t) is differen-

tiable. By the Mean Value The-

orem, there exists c in (0, 4) where

s�(c) =s(4)− s(0)

4− 0=

300− 0

4= 75 mph.

Thus, the instantaneous velocity is

75 mph at some time c.

Check It Out

1. c = 1 2. c = 1

3. By the Intermediate Value Theorem,

f has a zero for f(1) > 0 and

f(−1) < 0. If f(a) = f(b) = 0

and a �= b, then f �(c) = 0 for

some c by Rolle’s theorem. But

f �(x) = − sinx + 2 has no zero.

Then f has exactly one zero.

True or False

T,F,F,F,T,

T,T,F,F,T


1. c = 3 2. c =16

3. c =23 4. c =

6−√

33

5. c =π3 6. c =

π4

7. c = 0,±�

π2 8. c = 2

9. c =23 10. c = − 5

3

11. c = 0,±√2

2 12. c =2(3−

√3)

3

13. c = 1 +√2 14. c =

√10− 2

15. c = 1 16. c =14

21.f(6)−f(1)

5 = f �(c) by the Mean

Value Theorem. Then f(6) ≤ 14.

22.p(12)−p(10)

2 = p�(c) by the Mean

Value Theorem. Then p(12) ≥ 8.

Section 3.3

Try This

1. a) increasing on (−∞, 3),

decreasing on (3,∞);

b) increasing on (−∞, 0)

and (4,∞), decreasing on (0, 4)

2. Rel. max g(1) = 1/20,

rel. min g(0) = 0

3. Rel. max k(5π/3) = 2+√3

2 ,

rel. min k(4π/3) = − 2+√3

2

4. Rel. max f(0) = 6.25,

rel. min f(±1) = 6

5. 8 yards

Check It Out

1. inc (−2, 2), dec (−∞,−2) and (2,∞)

2. max f(−2) = 27, min f(1) = 0

3. max f(π4 ) =√2, min f( 5π4 ) = −

√2

4.14

True or False

T,F,T,T,F,

F,F,T,T,T


1. inc (−∞, 1), dec (1,∞)

2. inc (−∞,−1), dec (−1,∞)

3. inc (2,∞), dec (−∞, 2)

4. inc (3,∞), dec (−∞, 3)

5. inc (−∞,− 13 ) and (

12 ,∞)

dec (− 13 ,

12 )

6. inc (−∞,− 12 ) and

(34 ,∞), dec (− 1

2 ,34 )

7. inc (− 12 ,

13 ),

dec (−∞,− 12 ) and (

13 ,∞)

8. inc (13 ,

32 ),

dec (−∞, 13 ) and (

32 ,∞)

9. inc (−∞,∞) 10. dec (−∞,∞)

11. dec (−∞, 32 ) and (

32 ,∞)

12. inc (−∞,−4) and (−4,∞)

13. inc (−∞,−3) and (−3, 0)

dec (0, 3) and (3,∞)

14. inc (0, 2) and (2,∞)

dec (−∞,−2) and (−2, 0)


15. inc (0, π2 ) and (π, 3π

2 ),

dec (π2 ,π) and (

3π2 , 2π)

16. inc (π2 ,π) and (

3π2 , 2π)

dec (0, π2 ) and (π, 3π

2 )

17. inc (0, 3π4 ) and (

7π4 , 2π)

dec (3π4 , 7π

4 )

18. inc (0, π4 ) and (

5π4 , 2π)

dec (π4 ,

5π4 )

19. inc (0, π2 ), (

7π6 , 3π

2 ), (11π6 , 2π),

dec (π2 ,

7π6 ) and (

3π2 , 11π

6 )

20. inc (7π6 , 11π

6 ),

dec (0, 7π6 ) and (

11π6 , 2π)

21. min f(2) = −16, no max

22. min g(−1) = −12, no max

23. rel max h(−3) = 42,

rel min h(1) = 10

24. rel max y(−1) = 0,

rel min h( 53 ) = − 25627

25. rel max y(− 59 ) =

25681 ,

rel min y( 13 ) = 0

26. rel max y( 32 ) = 20,

rel min y(− 32 ) = −34

27. rel max f( 12 ) =4316 ,

rel min f(3) = −52

and f(0) = 2

28. rel max f(0) = − 14

29. min f(0) = −9

30. rel max f( 3√2) =

3√26

31. rel min k(− 3√4) =

− 3√412

32. rel max f(−1) = −2,

rel min f(1) = 2

33. rel max f( 25 ) =64

√10

25 ,

min f(0) = f(2) = 0

34. rel max f(0) = 0,

min f(±3) = −81

35. max f(4) = 6, no rel min

36. min f(0) = 1

37. rel max f(− 2627 ) =

5527 ,

rel min f(−1) = 2

38. rel max f(−2) = − 323 ,

rel min f(2) = 323

39. rel max f(2) = 3√4,

rel min f(0) = f(3) = 0

40. rel min f( 17 ) = − 216343

√7

41. max g(π2 ) = g( 3π2 ) = 5

min g(π) = 4

42. max g(0) = −2

min g(±π2 ) = −3

43. max v(π6 ) = 2, min v( 7π6 ) = −2

44. max A(2π3 ) =

√3 +

π3

min A(− 2π3 ) = −

√3− π

45. rel max f( 7π6 ) = f( 11π6 ) =54 ,

rel min f(π2 ) = −1, f( 3π2 ) = 1

46. rel max f(π3 ) = f( 5π3 ) =54 ,

rel min f(π) = −1

47. − 14 48. 2

49. 9 ft2,

p2

16 ft2 50. 16 ft, 4

√A ft

51. a) inc (−�

52 , 0) and (

�52 ,∞);

dec on (−∞,−�

52 ) and (0,

�52 )

b) (±�

52 ,

32 )

52. a) inc (12 ,∞), dec (0, 1

2 )

b) (12 ,

�12 )

53. x = 2, y = 3 54. x = 2, y = 4

55. a) Possibly p(x) = (x− 2)2+ 1

2x

1

y

b) Possibly p(x) = (x− 1)3 − 3

1x

�3

y

56. a) Possibly v(t) = − 516 t

3+

154 t

�2 2x

�5

5

y

b) Possibly v(t) = −t4 + 2t2

�1 1x

�1

1

y

63. y =14 (x

2 − 3x+ 6)

64. y =−4x3+30x2−48x+103

27

Section 3.4

Try This

1. Concave up (−∞,−1) and (0,∞),

concave downward on (−1, 0)

2. Concave up (−∞, 1) and (1,∞),

concave down on (−1, 1)

3. Concave up on (−∞, 1),

concave down on (1,∞)

4. Concave up on (0, 1),

concave down (−∞, 0) and (1,∞),

pts of inflection (0,−2) and (1,−1)

5. Since y�� = 10(x−1)

9x4/3 , the graph of y is

concave down on (−∞, 1), concaveupward on (1,∞), and the point of

inflection is (1,−5).

6. Since s�(t) = 4t(4− t2) and

s��(t) = 16 − 12t2, the rel. max.

value is s(±2) = 21 and the rel.

min. value is s(0) = 5.

7. rel max f(π) = 3, rel min f(π3 ) = − 32

Check It Out

1. concave up (−∞,−2) and (0,∞)

concave down (−2, 0)

2. (0, 2)

3. rel max f(0) = 0, rel min f(±2) = −4

True or False

F,T,F,T,T,

F,T,F,T,T


1. concave up (−∞,∞)

2. concave down (−∞,∞)

3. concave up (1,∞),

concave down (−∞, 1),

inflection pt (1, 2)

4. concave up (π, 2π),

concave down (0,π),

inflection pt (π, 0)

5. concave up (−∞, 1) and (3,∞);

concave down (1, 3);

inflection pts (1, 14) and (3, 30)

6. concave up (0,∞);

concave down (−∞, 0);


7. concave up (−∞,−2) and (0,∞);

concave down (−2, 0); inflection pt

(0, 0) and (−2, 6 3√2)

8. concave up (−∞, 0) and (1,∞);

concave down (0, 1);

inflection pt (0, 0) and (1, 3)



inflection point (8,−48)



inflection point (2, 24 3√2)

11. concave up (π4 ,

3π4 ); concave down

(0, π4 ) and (

3π4 ,π); inflection pts

(π4 ,

32 ) and (

3π4 , 3

2 )



(0, π4 ) and (

3π4 ,π); inflection pts

(π4 ,

2+π4 ) and (

3π4 , 2+3π

4 )


13. concave up (3π4 ,π);

concave down (0, 3π4 );

inflection pts (3π4 , 0)



(0, π4 ) and (

5π4 , 2π); inflection pts

(π4 , 0), (

5π4 , 0)

15. concave up (−∞,−2) and (2,∞);

concave down (−2, 2); inflection

points (±2, 4)

16. concave up (−∞,−5) and (5,∞);

concave down (−5, 5);

inflection points (±5, 12 )

17. concave up (−4, 0) and (4,∞);

concave down (−∞,−4) and (0, 4);

inflection points (−4,−2), (0, 0),(4, 2)

18. concave up (−7, 0) and (7,∞);

concave down (−∞,−7) and (0, 7);

inflection points (−7,−1), (0, 0),(7, 1)

19. concave up (−∞, 0); concave down

(0,∞); inflection pts (0, 0)

20. concave down (−1,∞);

no inflection pt

21. concave up (−∞, 1); no inflection pt

22. concave up (0, 1); concave down

(−1, 0); inflection pt (0, 0)

23. Rel. max. value: f(0) = 5

rel. min. value: f(2) = 1

24. Rel. max. value: f(1) = 5

rel. min. value: f(0) = 4

25. rel max g(−2) = 32,

rel min g( 43 ) =36427

26. rel max f(−2) = 27,

rel min f( 13 ) =38627

27. rel max C(±3√2) = 6724;

rel min C(0) = 6400

28. rel max A(±6√2) = 9409;

rel min A(0) = 4225

29. rel max g(π4 ) = 3,

rel min g( 3π4 ) = −3

30. rel max f( 2π3 ) = 5,

rel min f(π3 ) = −3

31. rel max L(−π3 ) = −

√3 +

4π3 ,

rel min L(π3 ) =√3− 4π

3

32. rel max R(2π3 ) = −

√3 +

8π3 ,

rel min R(π3 ) =

√3 +

4π3

33. rel max y(4) = 1,

rel min y(−4) = −1

34. rel max Q(−5) = 1,

rel min Q(5) = −1

35. rel max f(0) = 1, rel min f(4) = 9

36. rel max g(−3√3) =

9√3

2 ,

rel min g(3√3) = − 9

√3

2

37. max K(1) = 3, no rel min

38. max M(±8) = 16,

rel min g(0) = 0

39. f(x) = (x− 1)2+ 2

40. f(x) = −(x− 3)2 − 1

41. f(x) = 16 (x

3 − 3x2+ 12)

42. f(x) = − 52x

3+

152 x

43. (1, 1), (−1,−1) 44.

√3 3√42

45.√5 46. (

π2 , 1)

47.8 4√33 48. 2

√2

49.16 4√3

3 50. (2, 1/8), (−2,−1/8)

51. False; f(x) = x4

52. False; f(x) = −√x

53. Show d2y/dx2 > 0. The converse

is false, try f(x) = 1x for x > 0

54. For instance, f , g, f �, g�, f ��

, and

g�� are all positive functions

Section 3.5

Try This

1. a) limit is 0

x 10 100 1000

f(x) 0.198 0.02 0.0002

b) limit is 0

x −10 −100 −1000

f(x) −0.198 −0.02 −0.0002

2. If x > M =1ε , then

1x < ε

3. a) 0, b) 0, c) undefined

4. a) − 12 , b) 0

5. a) 1, b) − 13

6. a) 2, b) −2

7. N(24) ≈ 44, 083,

N(t) → 50, 000 as t → ∞

8. a) −∞, b) ∞

9. a) ∞, b) −∞

Check It Out

1. a) 0, b) − 12 , c) −∞, d) −2

2. y = −1 3. −10400

True or False

F,T,T,F,T,

T,T,F,T,F


1. a) −2, b) −2

2. a) 3, b) 3

3. a) 0, b) 0

4. a) 0, b) 0

5. a) 2, b) −2

6. a) 1, b) 1

7. a) 0, b) 0

8. a) 0, b) 0

9. 2 10. 1

11. 1 12. 1 13.32 14.

12

15. ∞ 16. 0 17. 0 18. ∞19. 2 20.

13 21. −4 22. 2

23.23 24.

43

25. 0 26. 0 27. 0 28. 3

29. 0 30. ∞ 31. ∞ 32. ∞33. 0 34. 0 35.

12 36. 3

37. −2 38. 0 39.83 40.

23

41. y = −3 42. y = ±1

43. y = 0 44. y = 2

45. Possibly f(x) = 3x2

x2+1

�1 1x

3

y

46.

�1 1x

2

3

y

47. Possibly f(x) = arctanx

x

Π

2

�Π2

y

48.

�1 1x

2

4

y

49. P (t) → 14 as t → ∞.

About 25% of adults will have

sinus problems towards the end

of spring

50.12

51. a) C(x) = 40, 000 + 5x

b) As x → ∞,C(x)x → $5/item

52. a) A(11) ≈ 17.96 PSI; b) 25 PSI


Section 3.6

Try This

1. F �(w) = 5w3

(w − 4),

F ��(w) = 20w2

(w − 3),

rel max F (0) = 0,

rel min F (4) = −256,

point of inflection (3,−162)

3 4x

�256

�162

y

2. R�(x) = x−1

2x3/2 , R��(x) = 3−x

4x5/2 ,

relative minimum R(1) = 2,

point of inflection (3, 4/√3)

1 3x

2

4

3

y

3. g�(t) = 6t(t2−1)2

, g��(t) = − 6(3t2+1)(t2−1)3

relative minimum g(0) = 4

2 3x

23

y

4. H�(t) = 2(1− 3√t)

t2/3,

H��(t) = 2( 3√t−2)

3t5/3,

rel max H(1) = 3,

point of inflections (0, 0), (8, 0)

1 8x

23

y

5. N �(x) = − 2 sin x+1

(2+sin x)2,

N ��(x) = 2 cos x(sin x−1)

(2+sin x)3,

rel max T (11π/6 + 2kπ) =√3/3,

rel min T (7π/6 + 2kπ) = −√3/3,

points of inflection (π/2 + kπ, 0)

7 Π6

11 Π6

Π

23 Π2

x

33

�33

y

6. g�(t) = t(t−6)(t−3)2

, g��(t) = − 18(t−3)3

rel max g(0) = 0,

rel min g(6) = 12

6x

12

y

Check It Out

1. D = {x : x �= ±1}asymp. lines x = ±1

symmetric about the origin

2. D = (−∞,∞)

R = (−∞,−4] ∪ [0,∞)

inc (−∞,−2) and (0,∞)

dec (−2,−1) and (−1, 0)

conc up (−1,∞)

conc down (−∞,−1)

3. critical num t = π4 ,

5π4

inc (0, π4 ) and (

5π4 , 2π)

dec (π4 ,

5π4 )

True or False

F,T,F,F,F

T,F,F,T,F


1. Max f(±√3) = 11,

rel.min f(0) = 2,

inc on (−∞,−√3) and (0,

√3),

dec on (−√3, 0) and (

√3,∞),

limx→∞

f(x) = −∞,

limx→−∞

f(x) = −∞,

range (−∞,√3]

2. Rel max f(−√3) = − 3

√3

2 ,

rel.min f(√3) =

3√3

2 ,

inc on (−∞,−√3) and (

√3,∞),

dec on (−√3,−1), (−1, 1), and

(1,√3), lim

x→∞f(x) = ∞,

limx→−∞

f(x) = −∞,

range (−∞,∞)

3. Rel max f(0) = 0,

rel.min f(2) = −23√4,

inc on (−∞, 0) and (2,∞),

dec on (0, 2), limx→∞

f(x) = ∞,

limx→−∞

f(x) = −∞,

range (−∞,∞)

4. Max f(π/4) =√2,

min f(5π/4) = −√2,

inc on (0, π4 ) and (

5π4 , 2π),

dec on (π4 ,

5π4 ),

limx→∞

f(x) and

limx→−∞

f(x) do not exist,

range [−√2,

√2]

5. Intercept (0, 0), max f(0) = 0,

asymptotes x = ±1, y = 1

range (−∞, 0] ∪ (1,∞)

�3 �2 �1 1 2 3

�4

�2

2

4

6. Intercept (0, 0),

inflection pt (0, 0),

asymptotes x = ±2, y = 0,

range (−∞,∞)

�10 �5 5 10

�10

�5

5

10


rel max f(3√3) = −9

√3,

rel min f(−3√3) = 9

√3,


asymptotes x = ±3

and y = 2x, range (−∞,∞)

�10 �5 5 10

�30

�20

�10

10

20

30


rel max f(0) = 0,

rel min f(±√2) = 4,

asymptotes x = ±1,

range (−∞, 0] ∪ [4,∞)

�4 �2 2 4

�10

�5

5

10

9. Min f(4) = 2√2,

inflection pt (8, 4√6

3 ),

asymp x = 2, range [2√2,∞)

4 8

2 2

4


max f(1) =√2

2 ,

min f(−1) = −√2

2 ,

inflection pts (0, 0), ( 4√5,

4√5√6),

and (− 4√5,−

4√5√6)

asymp y = 0, range [−√2

2 ,√

22 ]

�1 1 54� 54

1


11. Intercepts (0, 0), (8, 0),

p�(x) = − 43

x−2x2/3 ,

p��(x) = − 49

x+4x5/3 ,

max p(2) = 63√2,

inflection pts (0, 0)

and (−4,−123√4),

range (−∞, 6 3√2]

2�4 8

�12 22�3

6 23

12. Intercepts (0, 0), (36, 0),

F �(x) = − 4(x−9)

27x2/3 ,

F ��(x) = − 4(x+18)

81x5/3 ,

max f(9) = 33√9,

inflection pts (0, 0) and

(−18,−63√18), range (−∞, 3 3

√9]

9�18

3 93

�6 183

13. Intercepts (0, 0), (±2√6, 0),

rel max f(0) = 0,

rel min f(±2√3) = −6,

inflection pts (−2,− 103 )

and (2,− 103 ), range [−6,∞)

�2 3 2 32�2

�6

6

� 103

14. Intercepts (0, 0), (± 2√30

3 , 0),

rel max f(−2√2) =

64√2

15 ,

rel min f(2√2) = − 64

√2

15 ,

inflection pts (0, 0), (2,− 5615 ),

(−2, 5615 ), range (−∞,∞)

�2 2 2

� 5615

64 215

15. Max (π3 ,

4√3

3 ), min (5π3 ,− 4

√3

3 ),


Π

35 Π3

2 ΠΠ

4

3

� 4

3

16. Max (π6 ,

5√3

3 ), min (5π6 ,− 5

√3

3 ),

inflection pt (π2 , 0), (

3π2 , 0),

Π

65 Π6

2 Π3 Π2

Π

2

5 33

�5 33

17. Max (π3 ,

4π3 −

√3),

min (−π3 ,

√3− 4π

3 ),


Π

3�Π3

Π

2�Π2

4 Π3� 3

3 � 4 Π3

18. Rel. max (− π12 ,

2π3 −

√3),

rel. min (π12 ,

√3− 2π

3 ),


� Π12

Π

12Π

4�Π4

3 � 2 Π3

2 Π3� 3

19. y =x

x−3 20. y =x3

x2−1

21. y =1x + x+ 1 22. y =

1x+1 +

x|x|x

23. Possibly f(x) = x2

1

1

3


1 3

1

2


1

1

2

26. Possibly f(x) = x3 − 3x2

1 3�1

�3

27. s0 = 25 ft, max ht s(3) = 169 ft

28. Use F �= − [2 cos(2θ) + 1]×

[sin(θ + sin 2θ) + 1]

Max F (0) = 1,

Min F (π) = −1

rel max F (π3 ) = − sin(

√3

2 − π6 ),

rel min F (2π3 ) = sin(

√32 − π

6 ),

ΠΠ

32 Π3

1

�1

29. S�= (cos 3θ + cos θ)/2

Max S(π4 ) = S( 3π4 ) =

√23 ,

Min S(0) = S(π) = 0,

rel min S(π2 ) =13

Π

4Π

23 Π4

1

30. y� = −3(sin 3t+ sin 4t) =

2 sin(7x2 ) cos(

x2 ),

critical nos x = 0, 2π7 , 4π

7 , 6π7

Π2 Π7

4 Π7

6 Π7

1

74

32. Inflection pt (− k3 ,

2k3

27 )

If k = 0, no local extrema.

If k > 0, local max f(− 2k3 ) =

4k3

27 ,

local min f(0) = 0;

If k < 0, local min f(− 2k3 ) =

4k3

27 ,

local max f(0) = 0

33. G(t) = t3−3t2−9t+198

�1 1 3

3

1

�1

Section 3.7

Try This

1. 25 ft by 25 ft 2. (3.5,√3.5)

3. Length 4√2, width 2

√2

4. x = 6 feet from the 3-ft post

5. r = h =103√π

=10

3√π2

π cm

Check It Out

1. x = 200 ft, y = 600 ft

2. p(4) = 16, minimum value of p

True or False

F, T,F,F,F,

T,F,F,T,F



1. D = 1.25 when x = 1

2. D = 1 when x = 0

3. A = 25 when w = 5

4. A = 9/√2 when b = 3

5. S = π + 2 when r = 1

6. V = 16π when r = 2

7. a) (172 ,

√342 ), b) (0, 0)

8. a) (2, 2√2), b) (0, 0)

9.12 10.

√33

11. x = 100√6 ft, y = 50

√6 ft

12. r =20

4+π 13. 2 in.

14.4√3

9+4√3or

48√3−6411

15. (0, 0), (8, 0), (0, 4); not the mini-

mum perimeter

16. 3√3 17. r =

3√15π2

π , h =2

3√15π2

π

18. radius100π m, length of one

side of rectangle is 100 m

19.187 miles from the point on

the highway closest to Town A

20. BP =√3 miles

21.x2

8 square feet.

22. base b, perimeter p, two sides

of the same lengthp−b2

23. x = 4 workers 24. 220 students

25. 165 ft by 300 ft

26. y = 20 ft, x = 7.5 ft

27.|mp−q+b|√

m2+1

28.2π(3−

√6)

3 radians ≈ 33◦

29. θ =π3 30.

32πr3

81

31. c =√27 when x = 3

32.427 33. 2 radians

34. C = 54 35. K = −3

36. max A+B, min A−B

37. max A+B, min −B2+8A2

8A

38. Draw a right triangle through the

centers of the first and second

circles. By the Pythagorean

Theorem, the center of the second

circle is (2√6, 3).

Let (h, k) be the center, and let rbe the radius of the third circle.

Draw a right triangle that joins the

centers of the first and third circles.

h2+ (k − 2)

2= (r + 2)

2.

Similarly, we obtain

(2

√6− h)2 + (k − 3)

2= (r + 3)

2.

Subtract the two equations above

to obtain

r = 12− 2

√6h− k.

Rewrite the second equation as

(2

√6−h)2+(k−3)

2= ((r+2)+1)

2.

Then substitute the first equation

to obtain

(2

√6− h)2 + (k − 3)

2=

(

�h2 + (k − 2)2 + 1)

2.

Solving for k, we find

k =192

√6− 336h+ 23

√6h2

24(√6− h)

.

If h = 0 then k = 8 and r = 4.

Thus, the radius of the third circle

is r = 4 if its center lies in the

y-axis.

We skip the details in evaluating

the limit of the points of contact of

the first and third circles. As the

radius of the third circle increases

to ∞, the limit is�−8√6

25,96

25

�.

Section 3.8

Try This

1. x4 ≈ 1.25992

2. x4 ≈ 0.739085, |x4 − x3| ≈ 0.00003

3. Let g(x) =�� f(x)f

��(x)(f �(x))2

��.The graph of g is dashed below:

f �x� � x2 � 60

g�x� 87x

0.2

y

Then g(x) < 0.2 for x in (7, 8).

By Theorem 3.11, we can apply

Newton’s Method with x1 = 8.

Thus, limn→∞

xn =

√60.

4. Let g(x) =�� f(x)f

��(x)(f �(x))2

��.The graph of g is dashed below:

f �x� � sin�x� � x � 2g�x�

541

x

0.2

y

Then g(x) < 0.2 for x in (1, 54 ).

By Theorem 3.11, we can apply

Newton’s Method with x1 = 1.

Thus, limn→∞

xn = c where f(c) = 0.

Also, x2 ≈ 1.10292

Check It Out

1. x2 =34 2. x3 =

1712

True or False

T,T,F,T,F,T


1. x2 = 3.00, x3 = 2.800

2. x2 = −2.750, x3 ≈ −2.342

3. x2 = 1.500, x3 ≈ 1.296

4. x2 = 1.500, x3 ≈ 1.153

5. x2 = 1.000, x3 = −1.000

6. x2 ≈ 2.759, x3 ≈ 2.755

7. x2 ≈ 3.134, x3 ≈ 3.142

8. x2 ≈ 0.167, x3 = 0.164

9. x1 = −1, xn = (−1)n

10. x1 = 1, xn = (−1)n2n−1

11. x1 = 1, xn = (−1)n+1

12. x2 is undefined for f �(−3) = 0

13. x1 = 5; as n → ∞ then

xn = 2x1 − x21 → −∞

14. x1 = 2.5, as n → ∞|xn| =

�� 1−3x2

2

�� → ∞

15. x4 ≈ 1.7100

16. x5 ≈ 3.9330

17. x3 ≈ 3.1416

18. x4 ≈ 3.1416

19. Let f(x) = x2 − 10 and

g(x) =�� f(x)f

��(x)(f �(x))2

��.

From the graphs, we see g(x) ≤ 0.3

when 3 ≤ x ≤ 3.5.

By Theorem 3.11, xn converges

to a zero of f(x) if x1 = 3.

Then x4 ≈ 3.162.

f �x� g�x�3 3.5

x

0.3

y

20. The graphs of f(x) = x3+ 5x+ 10

and g(x) =�� f(x)f

��(x)(f �(x))2

�� are shown.

Note g(x) ≤ 0.6 for x in [−2,−1].

By Theorem 3.11, xn converges to

a zero of f(x) if x1 = −1. Then

x4 ≈ −1.423.

f �x�g�x�

�2 �1x

0.6

y


21. Let f(x) = cos(πx2) and

g(x) =�� f(x)f

��(x)(f �(x))2

��. From the

graphs, if 0.35 ≤ x ≤ 0.45 then

g(x) ≤ 0.7. By Theorem 3.11,

xn converges to a zero of f(x) if

x1 = 0.35. Then x4 ≈ 0.399.

f �x�g�x�

0.35 0.45x

0.7y

22. Let f(x) = x2sin(x)− 2x and

g(x) =�� f(x)f

��(x)(f �(x))2

��.In [6, 7], we find g(x) ≤ 0.9.By Theorem 3.11, xn converges

to a zero of f(x) if x1 = 7.

Thus, x4 ≈ 6.591.

f �x�g�x�6 7

x

0.9y

25. Let f(x) = ax2+ bx+ c, a > 0,

and g(x) = f(x)f ��(x)(f �(x))2 .

Then g(x) → 12 as x → ∞.

Note, g�(x) = 2a(b2−4ac)(2ax+b)2

> 0

if x �= − b2a . Thus, g(x) is increas-

ing on (− b2a ,∞) and (−∞,− b

2a ).

In particular, |g(x)| < 12 if

x �= − b2a . Hence, Newton’s

method converges to a zero of f(x)if x �= − b

2a .

Section 3.9

Try This

1. Linearization L(x) = x;tan

π24 ≈ L( π

24 ) =π24

2. dy = −0.05,�y =

1√1.1

− 1 ≈ −0.0465

3. Let f(x) =√x. Then√

15 ≈ f(16) + f �(16)�x =

318 = 3.875 where �x = −1.

4. sin 29◦ ≈ sin

π6 + cos

π6

�− π

180

�=

12 +

√3

2

�− π

180

�≈ 0.4849

5. If x is the height, then

f(x) = 16x is the volume.

The propagated error is

|�f | = 16 |�x| ≈ 1.6

since |�x| ≤ 0.1.The percentage relative error

is

�� ff(6)

�� · 100% or

��16�x

16(6)

�� · 100% ≈ 1.7%.

Check-It Out

1. L(x) = x4 + 1;

√4.1 ≈ L(4.1) = 2.025

2. Let y = g(x) = tanx and

dx = 0.005. Then

dy = g�(π4 )dx = 0.01

3. Let f(x) = x3be the volume

of a cube with side x, and

let �x be the error in the

measurement of the side

where |�x| ≤ 116 .

The propagated error is

|f(12 +�x)− f(12)| ≈f �(12) · 1

16 ≈ 13 cu. in.

True or False

T,T,F,F,T

F,T,T,F,T


1. L(x) = 12x− 6;

(2.1)3 ≈ L(2.1) = 9.2

2. L(x) = −x4 +

34 ;

1(1.9)2

≈ L(1.9) = 0.275

3. L(x) = x34 +

172 ;

√285 ≈ L(285) ≈ 16.88

4. L(x) = x48 +

83 ;

3√70 ≈ L(70) ≈ 4.125

5. L(x) = 2x+ 1− π2 ;

tan(π4 + 0.1) ≈

L(π4 + 0.1) = 1.2

6. L(x) = 2x3 +

2√3− π

9 ;

sec(π6 + 0.2) ≈ L(π6 + 0.2) =

2√3+

215 ≈ 1.288

7. L(x) = − x12 + 2;

3√7 ≈ L(1) = 23

12 ≈ 1.917

8. L(x) = − x64 +

12 ;

2√17

≈ L(1) = 3164 ≈ 0.484

9. L(x) = −√3

2 x+3+

√3π

6 ;

cos(61◦) ≈ L(π3 +

π180 ) =

12 − π

120√

3≈ 0.48

10. L(x) = x√2+

4−π4√2;

sin(44◦) ≈ L(π4 − π

180 ) =

180−π180

√2≈ 0.69

11. �y = 1.01; dy = 1

12. �y =

�1041 − 1

2 ≈ −0.006;

dy = − 1160 ≈ −0.006

13. �y = cot� 23π

90

�− 1 ≈ −0.034;

dy = − π90 ≈ −0.035

14. �y = csc� 8π45

�− 2 ≈ −0.11;

dy = −π√3

45 ≈ −0.12

15. �y ≈ −0.238; dy = −0.24

16. �y ≈ −0.0175; dy ≈ −0.0177

17. 5.1 18. 1.92

19.

√3

2 − π360 ≈ 0.857

20. 2−√3π

180 ≈ 1.97

21. 32.8 22. 9.11

23. 3 sq. in. 24. 0.785 sq. in.

25. 50 cubic inch 26. 3.75 sq. in.

27. 349 ft 28. 0.013 inch

29. 0.08 inch 30.140 radian ≈ 1.4◦


1. A 6. C 11. A 16. B2. A 7. B 12. D 17. C3. C 8. B 13. C 18. A4. C 9. C 14. C 19. B5. B 10. C 15. C 20. B


3. 2000 books

Section 4.1

Try This

1. a) πx+ C b) y = x3+ C

2. a) 3x5+ x2

+ C b) 2 sinx+ C

3. a)�x−3dx =

x−2

−2 + C =

− 12x2 + C

b)�x1/3dx =

x4/3

4/3 + C =

34x

4/3+ C

c)�4x−1/2dx =

4x1/2

1/2 + C =

8√x+ C

4. 2x4 − 8x2+K 5. 4x+

x3

3 +K

6. sin t+ C

7. y = 2t+ sec t + 1 8. 5 sec

Check-It Out

1. a) 4x+ C, b) 5x2 − x+ C,

c) −2 cos t+K, d) −3

x3+ C

2. y = x2+ x+ 3

True or False

F,T,T,T,F,

T,F,F,T,F



1. 10x+ C 2. πx+ C

3. 20t2 + C 4. 6t2 + C

5. 5 sin t+ C 6. 3 tan t+ C

7. sec t+ C 8. − 13 cos t+ C

9. − 1w3 + C 10.

−13w3 + C

11. − 116w2 + C 12.

17w

7+ C

13.310m

10/3+D

14.23 (m

3/2 −√m) +D

15.59m

9/5+m+K

16.7211m

11/6+K

17.13 s

3 − 12 s

2 − 2s+K

18.12 s

2 − 25 s

5/2+K

19.112 s

3 − 12 s

2+ s+ C

20. − 25 s

5/2+ 3s2 − 8s3/2 + 8s+ C

21. −3 cos θ − tanθ + C

22. θ − sec θ + C

23.13x

3+ tanx− x+ C

24. 2− cscα+G

25. − cscx+ C

26. sinα− tanα+G

27. − cotβ + P

28. − cscβ + cotβ + P

29. y = 2x+ x2

30. y =14x

4 − 13x

3+

1−6√2

12

31. y = 2√t

32. y = − 611 t

11/6+

37 t

7/3+

977

33. y = cosx

34. y = − cotx+

√33

35. a) v(t) = −32t+ 16

b) s(t) = −16t2 + 16t+ 96

c) s(0.5) = 100, highest point

36. a) s(6) = 102 ft, highest point

b) v(6 +

√5864 ) = −8

√586

≈ −194 ft/sec

37. 1 +

√224 ≈ 2.2 sec 38. 144 ft

39. y =x2

2 − x3

6 + x+53 40.

256

Section 4.2

Try This

1.152 2. 8 3.

15150 , 3

4. 2 +√2 +

√3 < A < 3 +

√2 +

√3

5. S =(n+1)(2n+1)

6n2 , s =(n−1)(2n−1)

6n2

6. 8 7. 16/3

Check-It Out

1. 2 +5n 2.

116

3.4(n+1)(2n+1)

3n2

True or False

T,T,F,F,F,

F,T,T,T,T


1. 14 2.212 3.

4936 4. 3 + 6a

5. 68 6.2920 7. 60 8.

118

9.

20�

i=3

i− 1

i+ 110.

n�

i=1

3

n

�1 +

i

n

11.

n�

i=1

��3i

n

�2

−3i

n

�·3

n

12.

n�

k=1

sin

�kπ

n

�π

n

13. 670 14. 1225

15. 2640 16. 14,760

17.n+1n , limit 1

18.3n+12n , limit

32

19.n−1n , limit 1

20.14n2+9n+1

2n2 , limit 7

21.(n+1)(2n+1)

6n2 , limit13

22.8n2+3n+1

2n2 , limit 4

23.3(8n2+9n+3)

n2 , limit 24

24.(n+2)(n+1)(n−1)

4n3 , limit14

25.3625 ,

3125 26.

533840 ,

319420

27.(9+

√3)π

12 ,(7+

√3)π

12

28.18 (3 + 2

√3 +

√10 +

√11),

18 (3 + 2

√2 +

√10 +

√11)

29.92 ,

72 30.

44627 ,

37427

31.(1+

√3)π

12 ,(3+

√3)π

36

32.(9+

√3)π

12 ,(7+

√3)π

12

33. S(n) = 9�1 +

1n

�,

s(n) = 9�1− 1

n

�

34. S(n) = 6 +4n , s(n) = 6− 4

n

35. S(n) = 1− 16

�2− 1

n

� �1− 1

n

�

s(n) = 1− 16

�2 +

1n

� �1 +

1n

�

36. S(n) = 14n2+3n+16n2 ,

s(n) = 14n2−3n+16n2

37. S(n) = 4(n+1)2

n2 , s(n) = 4(n−1)2

n2

38. S(n) = (7n−1)(25n+7)4n2 ,

s(n) = (7n+1)(25n−7)4n2

39.13 40.

143 41. 12 42.

56

43.72 44. 16 45. 39 46.

56

47. 16 48.203 49.

52 50.

352

51.23 52.

56

Section 4.3

Try This

1. 1 2.π(1+2

√3)

6

3.a+b2 4. 4, 10

5. 2, 1.5 6. 0,−4

7. 2(b3 − a3), (b−a)−(b3−a3)3

Check-It Out

1. 182 2. 3 3. 9.5

True or False

T,F,T,T,T,

F,F,T,F,F


1. 12 2. 12 3. −3 4. 3

5. 10 6. 0

7.25π4 8. 2− π

4

9.9π2 10. 1 11. −4 12. −12

13. −6 14. −4 15.92 16. −4

17. a) −4, b) 0 c) 8, d) 5

18. a) 31, b) −10 c) 20, d) 8

19. 25 20. 8 21. 1 22. 6

23.52 24. 6 25. 18 26. 4

27.1123 28.

163 29. 16 30. 6

31.� 5−2(2x− 5)dx

32.� 40 3x(4− x)2dx

33.� 30

√5 + x2dx 30.

� 52

2x2 dx

35. 8 36. 20 37.32 38. 2

39.π + 2

440. π + 4

41.π4 42.

32

43.π4 +

32 44.

52

45. 1 +π4 46.

π2

47.� 10 x2dx =

13

48.� 10

√x dx =

23

49.12 50.

√2− 1

55. S(n) = 5π6 , s(n) = π

3

56.(2

√3−1)π24


Section 4.4

Try This

1. 130 and 1 2. 2.5

3. a/2 4. 1/4

5. −10.875 meters/sec

6. 3, 2, 1.5

7.x

x4+1, f(x) + x sin

2 x

8. − sin(x)√cosx

Check-It Out

1.103 2.

2π 3.

√2

True or False

F,T,F,T,T,

F,T,T,F,T


1. 9 2. 15 3. 9 4. 2

5. − 23 6. − 74

3 7.72 8. 5

9. 3 10. −2√3

11.16 12. 4− 2

√2

13.π + 2

414.

π − 2

4

15.78 16.

43 17. 7 18.

23

19. 2 20. ] 1

21. 1 22. 3√2− 2

√3

23. 16 24.4 + π2

2

25. 12 26.163 27. 20 28.

34

29.323 30. 4 31.

43 32.

13

33.3112 34.

43

35. 10 36.12 (2b+ dm+ cm)

37.83 38.

c3 (c

2+ cd+ d2)

39.43 40.

32 41.

2π 42. 0

43.1

1 + x244.

√x2 + 1

45. − tanx 46. −1 + x2

x

47. 4√1− tan3 x sec

2 x

48.sinx cosx

1 + sin5 x

49. 65x3 50. 2x sin(x6)− sin(x3

)

51.32 52. ±

√3

3

53. ±1, 0 54.√3

55. f(q(x))q�(x)− f(p(x))p�(x)

56. p�� x

1 f(t)dt�· f(x)

57. 1 58. f(x) = xn+ C

59. Hint:

�1t2

�= 1 if

�2/3 ≤ t ≤ 1.

62.� 10 x2dx =

13 63.

� 10 x4dx =

15

64.� 2π0 sinx dx = 0 65. − cos x

x2

66.1−(

� x0 sin(t2)dt)

2

�1+(

� x0 sin(t2)dt)

2�2 · sin(x2

)

Section 4.5

Try This

1.23 (x

2+ 4)

3/2+ C 2. − cosx2

+ C

3.−1

3(3x2+1)+C 4.

13 cos(x3−3x2

)+C

5.15 (x

2+ 1)

5+ C 6. 2

√tan t+ C

7.149 8.

2815 9.

13 10. 0, 46

15

Check-It Out

1.1

π + 1(5 + 3x2

)π+1

+ C

2.−1

3(x3

+ 3x) + C

3.1

3

�2

√2− 1

�+ C 4. 0

True or False

F,T,F,T,T,

T,T,F,F,T


1.(2t2+3)4

16 + C 2.1

6(9−x3)2+ C

3. 2√sin t+ C 4. − sin

� 1t

�+ C

5.215 (w + 4)

3/2(3w − 8) + C

6. −2

3

√1− w(w + 2) + C

7.112

�x2

+ 4�6

+C 8. − (6−x3)5

15 +C

9.13 sin(3t)+C 10. −2 cos

�√t�+C

11. − (1+ 1w )

5

5 + C

12.12

�1− 1

w2

�4+ C

13.14 tan(4θ) + C

14. −2 cot

�θ2

�− θ + C

15. 2(x4+ x2

+ 1)4+ C

16. − (y5−5y3+4)9

5 + C

17.−2√x+1

+C 18.110 (

√x− 1)

5+C

19.−2

1+sin θ + C 20.2

2 cosα+1 + C

21.14 (1 + tanα)4 + C

22. − 13 csc(3α) + C

23.25 (2y − 1)

5/2+

23 (2y − 1)

3/2+ C

24.23 (1 + 3y)3/2 − 2(1 + 3y)1/2 + C

25.110 26.

25532 27. 2 28.

645

29. 1 30.13 31.

√26 32.

12

33.13 34. 1

35. 3 + 2√3 36. 2

√2/3

37. 0 38. 0 39. 100 40. 2

41.32 42.

13 43.

23 44. 2

45.12A 46. W

47.1a − 1

b 48.12B

Section 4.6

Try This

1. 0.088333

2.110

�f(1)+2

�f(1.2)+f(1.4)+f(1.6)+

f(1.8)

�+ f(2)

�≈ −0.418021

where f(x) = sin(πx)x

3.13 (

1√2+

4√5+

2√10

+4√17

+

2√26

+4√37

+1√50

) ≈ 1.76327

4. Since

|f ��(x)| = |− 4x2

cosx2 − 2 sinx2|we find |f ��

(x)| ≤ 6 = K on [0, 1].

Then

|ET | ≤K(b− a)3

12n2=

6

12(62)=

1

72.

An upper bound is172 .

5. Since

|f ��(x)| = |− 4x2

sinx2+ 2 cosx2|

we obtain |f ��(x)| ≤ 6 on [0, 1].

Choose n ≥�

105/2 or n ≥ 224.

True or False

T, T, T, T, F, F, T, T, T, T


1. a) 1.68333, b) 1.62222

2. a) 1.11667, b) 1.10000

3. a) 0.765496, b) 0.77753

4. a) 3.06198, b) 3.11013

5. a) 3.06198, b) 3.14876

6. a) 2.97842, b) 0.26247

7. a) 2.62331, b) 2.60391

8. a) 0.35233, b) 0.35698

9. a) 8.15121, b) 8.14594

10. a) 3.10675, b) 3.09894

11. ET ≈ 0.00116, (b) ES ≈ 0.00002

using |f ��(x)| ≤ 2

9 and

|f (4)(x)| ≤ 56

81

12. ET ≈ 0.16667, (b) ES ≈ 0.00417

using |f ��(x)| ≤ 4 and

|f (4)(x)| ≤ 6

13. ET ≈ 0.01563, (b) ES ≈ 0.00011

using |f ��(x)| ≤ 3 and

|f (4)(x)| ≤ 5


14. ET ≈ 0.00926, (b) ES ≈ 0.00069

using |f ��(x)| ≤ 2

9 and

|f (4)(x)| ≤ 80

81

15. a) n ≥ 366, b) n ≥ 26

16. a) n ≥ 130, b) n ≥ 11

17. a) n ≥ 238, b) n ≥ 19

18. a) n ≥ 255, b) n ≥ 9

19. 14 both by TR and SR

20. 132

21. TR 2712 ft, SR 26

23 ft

22.� 100 60D(x)dx ≈ 2910 ft

3by TR,

2900 ft3by SR

23.� 300 f(t)π( 12 )

2dt ≈ 138 ft3by TR,

137 ft3by SR

24.� 120 w(x)dx ≈ 126 ft

2by TR,

12623 ft

2by SR


1. B 6. A 11. B 16. D2. C 7. A 12. A 17. B3. B 8. B 13. B 18. B4. A 9. B 14. B 19. B5. C 10.B 15. B 20. C21. B 22.D


1.π4 2. 8x4f(x4

) + 2� x4

0 f(s)ds

3.� 10 (16x3 − 4x)dx = 2 4.

sin 12

5. If f(x) =� x2

√t3 + 1 dt, then the

limit is f �(2) = 3

6.� 10

dx(1+x)4

=724

7. If f(x) =� 2xx t cos t dt, then the limit

is f �(x) = 3x cosx

8. a) 0 b) does not exist

Section 5.1

Try This

1. ln 3+ln(x+1), lnx− ln 4, ln√9− x

2. 20.086, 0.247, 1.386, −0.693

3. secx cscx, 33x−1 , x+ 2x lnx

4.3e2

− 1 5. y = x

6. y� = 2x − 2

2x+1 +12 tanx

7. y� = x2/3 sin2 x√x+1

·�

23x + 2 cotx− 1

2(x+1)

�

8. y� = 2−2x2x−3−x2

9. y = 4 ln |x|− 12x

2+ C

10. y = ln | lnx|+ 1

Check-It Out

1. − tanx 2. 1 3. y =x

2e4+

32

4. y = 5x− 2 ln |x|− 5

5. 3e3 − 2e2 − e3 + e2

True or False

F,T,T,T,T,

T,T,T,T,T


1. a) lnx+ 2 ln y + ln z

b) 2 lnx+ ln y − ln z

c) lnx+ ln y − 13 ln z

2. a) ln

�3�

x2(x+1)x2+3

�

b) ln

�8√

x3+1

�c) ln

�x2−4x2+1

3.3x 4.

3(ln x)2

x

5. − tanx 6.cos(ln x)

x

7.x

x2+18.

tt2+1

− 1t

9.2−ln x2x

√x

10.sin x(1+ln | sec x|)

cos2 x

11.2 cos(2x)

x − 2 sin(2x) ln(x2)

12. 2(ln t) cos(2t) + sin(2t)t

13.1

x ln x 14.

√4+x2+x

4+x2+x√

4+x2

15.1

x2−116.

1−ln t2t3/2

√ln t

17. − cscx 18. sec t 19.(ln x)2+2

x

20.1

x2+1− 1

x((ln x)2+1)

21. y = x+ 1 22. y =32x− 3

2

23. y = 8x− 5 24. y = − 32x+ 4

25. x2√x2 − 1

�2x +

xx2−1

�

26.

�x2

+ 1

x2 − 1

�x

x2+1+

xx2−1

�

27.(x− 2)(x+ 3)

(x+ 2)(x− 3)×

�1

x−2 +1

x+3 − 1x+2 − 1

x−3

�

28.

√3x− 2

x(x+ 1)2

�3

2(3x− 2)−

1

x−

2

x+ 1

�

29.x(x − 1)3/2

√x + 1

�1

x−

3

2(x − 1)−

1

2(x + 1)

�

30.�

(2x− 1)(x+ 2)(x− 3) ×12

�2

2x−1 +1

x+2 +1

x−3

�

31. min f(e−1) = −e−1

,

concave up (0,∞)

��1 1x

1

2

y

32. min f(e−1/2) = − 1

2e

concave down (0, 1e3/2

)

concave up (1

e3/2,∞)

��1�2 1x

1

y

33. max f(e) = e, concave down (0,∞)

�1x

�

y

34. min f(e−4) = −e−4

,

concave up (0,∞)

��4x

0.5

y


35. min f(1) = −1,

conc. down (0, e−1),

conc. up (e−1,∞),

��1 1x

�1

y

36. min f(1) = 1, concave up (0,∞),

1x

1

y

39. y = 3 + lnx 40. y =12 (ln t)2

41. y =12 (t

2 − (ln t)2 + 1)

42. y = x lnx− e

43. y = x+ lnx− 1

44. y = 4 lnx+1x − 4

45.� 21 ln(x)dx = ln(4)− 1

1 2x

y

46. −� 11/2 ln(x)dx =

12 (1− ln 2)

112

x

y

47.� e2

1ln xx dx = 2

1 �2x

0.5

y

48.� ee−1

2xdx = 4

��1x

4

y

49. If (a, ln a) is the point of tangency,

then the y-intercept of the tangent

line is (0, ln(a)− 1). Also, (0, ln a)is the y intercept of the horizon-

tal line through the point of tan-

gency. The distance between the

y-intercepts is 1 unit. See figure

below.

�a, ln�a��a x

ln�a�ln�a��1

y

50.dydx =

y(3x−2)x(2y2+1)

51. a) ∞ b) 0

Section 5.2

Try This

1. a) 5 ln |x+ 1|+ C;

b) 2 ln |x3+ 1|+ C

2. 4 ln 17 3. 4 + ln 16

4. y = 3− 3ln x

7.ln 24 8.

6(√3−

√2)

π

Check-It Out

1. ln(t2 + 9) + C

2.14 ln |1 + 4 secx|+ C

3.14 ln 2

4. 2x− 43 ln |3x+ 2|+ C s

5. y = − 2ln x + 3

6. y = − ln | lnx|+ e

True or False

T,F,T,F,F,

F,T,F,T,T


1. π ln |t|+ C 2.25 ln |x|+ C

3. ln |x+ 5|+ C 4.14 ln |4x+ 1|+ C

5.23 ln

52 6. ln 1331

7.116 (8t+ 3− 3 ln |8t+ 3|) + C

8. t2 +13 ln |3t+ 2|− 4

9 + C

9.1π ln | sec(πx)|+ C

10.12π ln | sin(2πθ)|+ C

11. ln32 12.

12 ln

53

13.ln 3

414.

ln 3

2π

15.13 ln |2 + 3 sec t|+ C

16. − ln(1 + cos2 t) + C

17.12 ln | tan 2x|+ C

18. ln | tanx|+ C

19.14 20. (ln 2)

2

21. 2(√2− 1) 22. 8

23. 2 ln |1 +√x|+ C

24. x− 2√x+ 2 ln |1 +

√x|+ C

25. ln | cos θ + θ sin θ|+ C

26. − ln |α cosα− sinα|+ C

27. ln | sinx+ cosx|+ C

28. ln | sinx+ 1|+ C

29. ln 2 30. 5 ln52

31.12 ln(2 +

√3) 32. ln

43

33. y = −3 ln |2− x|34. y = t− ln |t+ 1|+ 1

35. y =12 ln | sec 2x|

36. y = ln |1− sin t|37. s(t) = −10 ln |t|+ 18

38. s(t) = − 19 ln |3t+ 1|+ 1

3 t

39.−1

1 + x240.

2√1− 4x2

41.1

x√4 + lnx

42.√9 + x2 −

√9+(ln x)2

x

43.ln(3)3 44.

ln 22

45.3π ln

√3+1√3−1

46.2 ln(2)

π

47. (1, 0) 48. − 14 < m < 0


Section 5.3

Try This

1. yes, yes 2. yes, no, yes

3. p−1(x) = x+1

3 with domain

and range (−∞,∞);

f−1(x) = 1

x − 1 with domain

{x : x �= 0} and range

{y : y �= −1}

4. p(1) = 2, p−1(2) = 1,

(p−1)�(2) = −1/6

Check-It Out

1. f−1(x) =

1

1− x,

D = {x : x �= 1},R = {y : y �= 0}

2. Since g�(x) =−4

(x− 4)2< 0,

g is decreasing and one-to-one.

3. p−1(3) = 0, (p−1

)�(3) =

1

2

4.2

5

True or False

F, F, T, T, T,

F, T, T, T, F


1. yes 2. yes 3. no 4. no

5. yes 6. yes

7. yes

1x

1

y

8. yes

1x

1

y

9. yes

1x

1

y

10. yes

1x

1

y

11. f−1(x) = 5x

7−2x

D = {x|x �= 72}

R = {y|y �= − 52}

12. g−1(t) = t+4

1−t

D = {t|t �= 1}R = {y|y �= −1}

13. M−1(x) = −

√1− x

D = {x|x ≤ 1}R = {y|y ≤ 0}

14. N−1(x) = 2− x2

D = {x|x ≤ −1}R = {y|y ≤ 1}

15. C−1(t) = t2 + 3

D = {t|t ≥ 0}R = {y|y ≥ 3}

16. R−1(t) =

√t− 2

D = {t|t ≥ 3}R = {y|y ≥ 1}

17. If g(x1) = g(x2) then

πx1 − 3 = πx2 − 3.

Thus, x1 = x2.

Hence g is one-to-one.

18. If f(x1) = f(x2) then

37x1 + ln 2 =

37x+ ln 2.

Thus,37x1 =

37x2 and x1 = x2.

Hence f is one-to-one.

19. If s(t1) = s(t2) then

t12t1+3 =

t22t2+3 .

Cross-multiply:

2t1t2 + 3t1 = 2t1t2 + 3t2

Thus, t1 = t2.

Hence s is one-to-one.

20. If R(t1) = R(t2) then

4t14−3t1

=4t2

4−3t2.

Cross-multiply:

16t1 − 12t1t2 = 16t2 − 12t1t2

Thus, t1 = t2.Hence R is one-to-one.

21. If C(w1) = C(w2) then

w1 − 1

w1 + 1=

w2 − 1

w2 + 1.

Cross-multiply:

w1w2 + w1 − w2 − 1 =

w1w2 − w1 + w2 − 1

Thus, 2w1 = 2w2 or w1 = w2.

Hence C is one-to-one.

22. If f(t1) = f(t2) then

1− 2t11 + 2t1

=1− 2t21 + 2t2

.

Cross-multiply:

1 + 2t2 − 2t1 − 4t1t2 =

1− 2t2 + 2t1 − 4t1t2

Thus, 4t2 = 4t1 or t1 = t2Hence f is one-to-one.

23.1

1124. −1

25. −16

326.

1

4

27. −1

228.

4

3

29. y is increasing for

dy

dx=

4x(x+ 1)

(2x+ 1)2> 0

when x > 0. Then y has an

inverse function.

30. −9

31. Differentiate (F ◦ F )(x) = x to

obtain f(F (x))f(x) = 1.

Integrate to show F (x) = x.

32.23

34. f(x) = x if 0 ≤ x < 1, and

f(x) = 3− x if 1 ≤ x < 2

Section 5.4

Try This

1. t = A−1005 2. w = e4 − 10

3.dydx = ex cos x

(cosx− x sinx),dydx = esin xx cosx+ esin x

4. Rel. min. point�1, e−1/2

�,

relative max. point�−1,−e−1/2

�

5. T (5) ≈ 87◦F ; since T �

(5) ≈ −3.3◦F ,

the soup cools at the rate of 3.3◦Fper minute.

6. 2ex/2 + C,13 e

x3+ C

7.12 (1− e−1

) 8. y = 1 + t+ e2t

Check It Out

1. x =12 ln(y − 1)

2. f �(x) = e2x + e−2x


3.13 (e− 1)

4. y = − sin t+ e−3t+ 1

True or False:

F,T,T,T,T,

F,T,F,T,T


1. x =12 ln y 2. x = ln

�y+13

�

3. t = 10 ln

�5

A+4

�

4. t = 5 ln

�20

30−A

�

5. h = e6P+4 6. w = e(5−10V )/2

7. y� = 2xex2+2 8. f �

(t) =t (2− t)

et

9. g�(t) =1− 2t

e2t

10. y� = 10ex sin x(x cosx+ sinx)

11. R�(x) = e3x (5− 3x)

12. P �(w) =

1− w lnw

wew

13. y� = −2ecos 2x sin 2x

14. y� = −1

15. L�(x) =

4e2x

(ex − 1)(1 + ex)(1 + e2x)

16. f �(x) = 20

�e20x − e−20x

�

17. r�(x) =24e4x

√1 + e4x

18. A�(t) = esin t

(cos t cos 3t− 3 sin 3t)

19.1

2e2x + C 20. −ecos x + C

21. −1

2e−t2

+ C

22. 2et +e2t

2+ t+ C

23.2√1− et

�et − 1

�

3+ C

24.3

2x2 − x+ C

25. 2e√w+ C

26. ln√e2w + e−2w + C

27.e− 1

228. 3(e− 1)

29. e√3 − e 30. ln

3

�1 + e

2

31. y = 3x+ 1 32. y = −2x+ 2

33. y =1

2x+ 2 34. y = 8e3x− 3e3

35. y =

√3e

2x+

√e−

π

12

√3e

36. y = −1

37. e− e−1 38. 3(e− 1)

39.19e6−1

2e640. 10 ln(5)− 8

41.12 (e− e−1

) 42.12

�1− 1

e16

�

43. y = − cos t+ 14 e

−2t − 14

44. y = ex + e−x

45. y = ln

�2

1+e−2t

46. y =12 ln[(e2t + 1)(e−2t

+ 1)]− ln 2

47.dy

dx=

1

1 + 2e2xy(x+ 2)

48.dy

dx=

6x+ yexy

4y − xexy

49.dy

dx=

5 + e2y

4− 2xye2y

50.dy

dx=

ex + y

3− x+ e−y

51.dy

dx=

2xyey

1− x2yey

52.dy

dx=

2x− yexy

xexy − 2y53. y = 4x− 4 54. y = x

55. min. (0, 1)

56. max (1, 1e ), inflec. pt (2, 2

e2)

57. max (1, 1√e), min (−1,− 1√

e),

inflec. pts (0, 0), (√3,

√3

e3/2),

and (−√3,−

√3

e3/2)

58. max (π4 ,

e−π/4√2

),

min (5π4 ,− e−5π/4

√2

),

inflec. pt (π2 , e

−π/2),

and (3π2 ,−e−3π/2

)

59. min (e−1/2,− 12e ),

inflec. pt (e−3/2,− 32e3

)

60. min. pt. (e, e−1),

inflec. pt (e3/2, 32e3/2

)

61. a = ln

�A+

√A2+42

�

63.13 64. x = µ; x = µ± σ

Section 5.5

Try This

1. r = 4 ln 1.02 ≈ 0.079

2.0.3 ln 200

ln 2 ≈ 2.29 mm

3.13 (log2(10) + 1) ≈ 1.4, 3

25 ≈ 0.1

4. y� = sec2(x)2tan x

ln 2,

p�(t) = 12√

t ln 122√t

,

M �(x) = 1

2 ln(10)(x+√x)

5. xsin x� sin x

x + cos(x) lnx�

6.2

ln 10 ,3−

√3

ln 3

7. a) $1,051,161.90 b) $1,051,267.50

8. $12,460.77

9. P (15) ≈ 27, 167

P (t) → 40, 000 as t → ∞

Check It Out

1. 2(e2) ≈ 167.6

2. a) 3π3xlnπ, b) 2

(2t−4) ln 3

3. a)π5s

5 lnπ , b)3

4 ln 2

True or False:

F,T,F,F,T,

T,T,F,T,F


1. −3 2. −1 3.1

24.

2

3

5.log(7)−1

3 ≈ −0.1

6.log3(8/5)

2 ≈ 0.2

7.ln 5

365 ln(1+0.12/365) ≈ 13.4

8.323 ≈ 10.7

9.√101− 1 ≈ 9.0

10. e(1

ln 2− 1ln 3 )

−1≈ 6.5

11.

y�2 x

1 2x

1

2

y

12.

y�log3x

1 3x

1

y

13.

y�log ��x�

�10 1x

1

y


14.

y�10 x�3

1 2x

10

3

y

15.

y�log2�x�1x

1

2

y

16.

y�4 x �1

1 2x

1

4

y

17.

y�log1�2�x�4��4 4x

1

�2

y

18.

y��1�3�x�1�21 2

x

3

�2

y

19. 2xln 2 20. 5 · 3t ln 3

21.3

√tln 3

2√t

22. 3x2

24x

(ln 16 + x ln 9)

23. 3t10

3tln(3) ln(10)

24.� 12ln 12

�xln

�12

ln(12)

�

25.1−ln x

x2 26.t+2

(t2+t) ln 4

27. 10s�ln(s) + 1

s ln 10

�

28.2x(x2 ln(2)−2x+ln(2))

(x2+1)2

29.1

2(x−1) ln a 30.1

x ln(10) ln(x)

31.cot(x)ln 10 32. − 2 tan(2w)

ln 3

33.2x cot(x)−1

x ln 4 34.θ tan(θ)−1

θ ln 8

35. y = 10x ln 4 + 2− ln 4

36. y = 28 ln(7)x+ 14(1− ln 7)

37. y = 8√3 ln(2)x+ 4− 8π ln 2√

3

38. y =

√30 ln 10

2 x+√10−

√30π ln 10

12

39. y =x

25 ln 5 + 2− 1ln 5

40. y =2

ln 27x+ 2− 2ln 3

41. y = 2�1 +

1ln 10

�x− 20

ln 10

42. y =x

ln 4 + 1− 1ln 2

43.2x

ln 2 + C 44.10t

ln 10 + C

45.5x

2

2 ln 5 + C 46.2√

t+1

ln 2 + C

47.63x

3 ln 6 + C 48.23x

3 ln 2 − 52x

2 ln 5 + C

49.ln(1+52x)

2 ln 5 +C 50.ln(9+32x)

2 ln 3 +C

51.3sin t

ln 3 + C 52.5(2−x)2

−2 ln 5 + C

53.−2

√8cos t+1ln 8 + C

54.2

3 ln 3

�3tan t

+ 1�3/2

+ C

55.18ln 3 − 4

ln 2 56.3

ln 4

57.15ln 4 58.

−2π ln 10

59.e−12 60.

1ln 9

61.4−2

√2

ln 2 62. log4(1 + 4x)

63. xx(ln(x)+ 1) 64. x1/x

�1−ln x

x2

�

65. xln x�

ln x2

x

�

66. (2x+4)3x+6

�3x+6x+2 + 3 log(2x+ 4)

�

67. (log2 x)log3 x · 1+ln(log2 x)

x ln 3

68. (log x)10x

×�10

x(ln 10) ln(log x) + 10x

x(log x) ln 10

�

69.y(x ln y−y)x(ln x−x) 70.

1+ln x1+ln y

71. $182.20; $109.33 per year at the end

of 10 years

72. $818.84; $97.77 per year at the end

of 6 years

73. P (0) = 1000

P �(0) = 200 units/minute

74. a) 200 b) C =ln 3100

75.(ln 4)2

2 x2 − x ln 2 + 1

76. a = 1, b = ln√2 = −c

77. Hint: odd function

78. Hint: f(±1) ≤ a, f �(0) = 0.

Section 5.6

Try This

1. y = Cex3−x2

2. y =

�4√t+ 1

3. 15871

�17,63915,871

�2≈ 19, 604

4.10005.3

� 12

�1/5.3ln(1/2) ≈

−114.7 units/year

5. t ≈ 3.83 hr, or 3 hr and 50 min

6. 48 minutes after it was taken from the

oven

Check It Out

1. a) y = Ce2x/ ln 2

, b) y2 = x2+ 3

2. 36,744,344

3. −58 mg/hr 4. −4◦F/min

True or False:

F, F, T, T, T,

T, F, F, F, T


5. y =2

1− 2x6. y =

3

1− x3

7. y = t 8. y = ln(2− e−t)

9. y = −2et4/4 10. y =

x

1 + x

11. y = 2 sec t 12. y =1

1− ln(sinx)

13. y =

�1 + 2 sin

2 t

14. y =2

1 + cos t

15.100 ln 0.9ln 0.96 ≈ 258 years

16. about 14.4 hours

17. a) 13.2 mg b) 50 years

18. 15,683 years

19. H = 280 (2/7)t/20 + 70

H(60) ≈ 76.5◦F

20. H = 325− 265 (175/265)t/2.5

H(3.4287) ≈ 175◦F

21. b) $182.21 22. b) $64.82/year

23. P (t) = 1256 (t+ 6400)

2

24. 71.7%

25. a) v(t) = 201+6t ft/sec

b) a(1) = − 12049 ≈ 2.4 ft

2/sec

26. Yes since s(3) = 99 where

s(t) = −11t2 + 66t

27. P �(2) = 101 people/year

since P (t) = (t+ 200)2/4

28. a) 6 miles since P (h) = 30(3√0.5)h

b) −1.7 inches of mercury per mile

29. 4π(0.15)2√e ≈ 0.47 cm

2/sec


Section 5.7

Try This

1. y = x+Cx

2. y = 1− e−x2

3. Amount of sugar in tank is

y(t) = t3−150t2+4375t+312506250

4. y(4) ≈ 143.8 mg, since

y(t) = 60ln 2

�4 + 2

−t/24(ln 2− 4)

�

5. I(t) = 6− 6e−5tamperes, where t

is in seconds

6. I(t) = 6 − 6et/2 amperes, where tis in seconds

7.

2�2x

2

�2

y

8. C,A,B

9. (0.1, 0.2), (0.2, 0.42), (0.3, 0.662)

Check It Out

1. y = e1−cos x − 1

2. y(10) ≈ 2.6 lb of salt

where y = 3− 3e−0.2t

3. Q(t) = 625

�1− e−5t

�

4. ye(1) = 3, ye(2) = 7

True or False:

F, F, T, T, T,

T, T, F, T, T


1. y = 3− 3e−x 2. y =1

2(e2x − 1)

3. y = (x+ 1)e−x2/2

4. y = (3x+ 2)e−x2

5. y = 5

�t−

1

t

�6. y = 4

�t2 − t

�

7. y =1− cos t2

t

8. y =10π + sin t5

5t4

9. y = sin t+ 2 cos t

10. y =1 + sin

2 x

2 sinx

11. y =x4

+ 3x

3

12. y =t3

2π

�1− π ln

�cos t2

��

13. y =2t3 + 3t2 − 6

6(1 + t)

14. y =x2

+ x+ 4√2x+ 1

15. y = xe−2x(1 + lnx)

16. y =e−3x

+ 1

3x

17. A, y = −1

4−

x

2+ Ce2x

18. B, y2 = 2(x+ C)

19. C, y = ln

�−1

x+ C

�

20. D, not separable or linear

21. y = 2ex−0.5 − x− 1

�2 2x

�2

2

0.5

y

22. y =2

5e1−x − 1 + x

�2 21x

�2

2y

23. y = x

24.1

y=

1

x+ 2

25. ye(1) = 1.5, ye(2) = 2.75,

ye(3) = 5.125

26. ye(1) = 2, ye(2) = 2 +√2/2,

ye(3) = 3 +√2/2,

ye(4) = 3 +√2

27. y = 40�1− e−t/20

�

sincedy

dt= 2−

5y

100,

y(0) = 0

28. y = 2.5�5− 3e−2t/25

�

sincedy

dt= 1−

2y

25, y(0) = 5

29. t ≈ 7.6 sec,

where y =2(625+100t+t2)

5(50+t)

sincedydt = 0.8− y

50+t , y(0) = 5

30. y(50) = 22.5 kg of sugar,

where y =400t+ t2

4 (200 + t)since

dy

dt= 0.5−

y

200 + t, y(0) = 0

31. 37.5 sec, where

y =2500 + 75t− t2

250since

dy

dt= 0.5−

2y

100− t, y(0) = 10

32. 1.5 lb

33. Q(t) = 11100

�1− e

−100t3

�

34. Q(t) = 57 e

−25t/3�e

7t3 − 1

�

max Q(t) ≈ .086 coulombs

when t = 37 ln

2518

35. I(t) = 52

�1− e−10t

�

36. I(t) = 94 e

−3t�e8t/3 − 1

�, max is

I(t) = 2/ 4√3 ≈ 1.5 amperes

if t = 34 ln 3

37. y(t) = 15ln 2

�32− 32

−t/12�

38. y(24) = 600ln 2 ≈ 866 mg

since y(t) = 1200ln 2

�1− 2

−t/24�

39. y(t) = $1, 000, 0000

sincedydt = 0.05y − 5(10

4),

y(0) = 106

40. y(4) = $22, 662.80

sincedydt = 0.2y − 4(10

4),

y(0) = 105, and

y(t) = − 100,0003

�−4 + e3t/10

�

41. y = ex − x− 1

Section 5.8

Try This

1.π3 ,

π4 ,

45 2. −π

3 ,

√74

3. 0 4.52

5.5

1+25x2 ,3√

e6x−1

6. −1 ≤ x ≤ 1, 0 < x < 1

7. x = 2, θ = tan−1

2− tan−1

0.5 ≈0.64 radians ≈ 37

◦

Check It Out

1. a)2√10

7 b) −π6

2. x =

√3

3

3. a)2x

x4+1b)

2 tan−1(t)t2+1

4.

√3

3 +π6

True or False:

F, T, T, T, F,

F T, T, T, T



1.2√13

13 2. −2√2

3.x√

1−x24.

x√1+x2

5.3√3−410 6.

169√3−24069

7.14 8. 1−

√32

9. −π6 10.

53

11.

√3

1−√3

12.

√34

13.45 14. − 1

3

15. y� =5

√1− 25x2

16. y� =−3x2

√1− x6

17. y� =3

9 + x2

18. y� =2

√x(1 + x)

19. y� =2

x√x4 − 1

20. y� =−1

√1− x2

21. y� =−1

|x|√4x2 − 1

22. y� =−1

|x| (arccsc x)√x2 − 1

23. y� =x−

√1− x2 sin−1 x

x2√1− x2

24. y� =−ecot

−1 x

1 + x2

25. y =x

2+

π − 2

4

26. y = −x+√3 +

π

6

27. y = x−√3 +

π

3

28. y =

√3

2x+ π −

√3

29. y� =1

2

�1− (x+ y)2 − 1

30. y� =

�1− x2y2 + y

�1− x2y2 − x

31. y� = ey�ex − 1

1+x2

�

32. y� = y

�sec2 y

| tan y|√

tan2 y−1− x

�−1

33. 10√7 ≈ 26.5 feet

34.√10/2 ≈ 1.6 yards

35. y = tan 2 ≈ 1.107, when x = 1

36. rel max pt≈�√

2+2√5

2 ,−0.60578

�,

rel min pt≈�

−√

2+2√5

2 , 3.74737

�

37.52π3 ≈ 54 miles/min

38. − 39610 ≈ −0.0639 radian/sec

39. θ�(t) = 3

4x√

x2−9

40. θ = x

�12

�1−

�25−x2

24

�2�−1

θ�(√13) =

√39

18 for x =√13

if θ =π3

46. No, if x < 0

47. a)

y�sin�sin�1�x��1 1x

�1

1

y

b)

y�sin�1�sin �x��Π2

�Π2

3 Π2

�3 Π2

x

Π2

�Π2

y

48.ddx

�tan

−1(cotx)

�= −1

domain is

{x : x �= nπ, n is an integer}

y�tan�1�cot x�Π�Π x

Π2

�Π2

y

Section 5.9

Try This

1. sin−1

�x4

�+ C,

15 arctan

� t5

�+ C,

17 sec

−1�w

7

�+ C

2. π 3.π2 − sin

−1�e−3

�≈ 1.5

4.13 tan

−1�x3

�+ ln(9 + x2

) + C

5.15 tan

−1�x−3

5

�+ C 6. π/3

7.12 tan

−1�e2x

�+ C

14 ln

�1 + e4x

�+ C

8.13 , −(ln 2)

2/4

Check It Out

1. a) sin−1

�x5

�+ C

b)12 tan

−1� t2

�+ C

c)1√2sec

−1�

s√2

�+ C

2. a)π6 b)

√3π6

3.π4 + ln

√2

True or False:

F,F,F,T,T,

T,T,T,T,F


1. sin−1

�x2

�+ C

2. sin−1

�s√3

�+ C

3.1√5tan

−1�

t√5

�+ C

4.13 tan

−1(3t) + C

5.1√2sec

−1�

|x|√2

�+ C

6. arcsec (4|x|) + C

7.12 arcsin(2t+ 1) + C

8. sin−1

(tanx) + C

9.12 sec

−1(e2x) + C

10. sec−1

(2ex) + C

11.π15

12.π3

13.π6

14.π4

15. ln� 32

�16.

√2

17.16 tan

−1�

3 sin θ2

�+ C

18. − 1√2tan

−1�√

2 cos θ�+ C

19.−1√2arctan

�√2 cotx

�+ C

20. Rewrite integral:� sec2 θdθ

1+2 tan2 θ=

√2

2 tan−1

(√2 tan θ) + C

21.13 tan

−1� t+1

3

�+ C

22.14 tan

−1� s−3

4

�+ C

23. sin−1

�x+12

�+ C

24. sin−1

�w−32

�+ C

25. arcsin�x+3

4

�+ C

26.13 arctan

�x−23

�+ C

27.π20 28.

π2

29.π3 30.

π48

31.12 sec

−1�

|x+3|2

�+ C

32.�sec

−1 x�2

+ C

33. − 14

�cos

−1(2x)

�2+ C

34.12 ln

�tan

−1(2s)

�+ C


35. y = 1− sin−1 x

36. y = 8 tan−1

� t2

�− 2π

37. y = 2esec−1 t

38. y =

�2 tan−1 x+

π2

39. y = t2e− tan−1 t

40. y =x cos−1 x−

√1−x2

x

41.π6 42.

π8 43.

π6 44.

π12

45. a = tan(1) 46. b = 3

Section 5.10

Try This

1. ln 19 ≈ 2.94, ln 199 ≈ 5.29,

ln 1999 ≈ 7.60;

d → ∞ as x → 1−

2.sinh x2√x, −2xcsch2(x2

),sinh x

2√cosh x

3. 0 4. ln(5/4)

5. (ln(2+√3), 4

√3), (ln(2−

√3),−4

√3)

7. a =2e

e2+1

8.15

�cosh

−1 12181 − cosh

−1 10081

�

9.12 cosh

−1� sin 2s

3

�+ C

10. Since tanh−1 x =

12 ln

1+x1−x ,

we find d = ln1+r1−r =

2

�12 ln

1+r1−r

�= 2 tanh

−1 r

Check It Out

1.

a) 2 cosh 3x sinh 3x

b) − [2sech 2x coth 3x] esech 2x

c)4√

16x2+1

d)10x

1−25x4

2.

a) − 19 sech

−1�x9

�+ C

b) cosh−1

� t9

�+ C

True or False:

F, T, T, T, T,

T, T, F, T,T


1.54 2. − 4

3 3.1213 4. 0

5. ln(√2− 1) 6.

ln 32

7. y� = 3 sinh(3x+ 1)

8. y� = 10 cosh 5x

9. y� = 1− 2xsech2x2

10. y� = csch2(1/x)x2

11. y� = sech (x2)−

2x2sech (x2

) tanh(x2)

12. y� = −csch ( 3√x) coth( 3√x)

33√x2

13. y� = tanhx

14. y� = e−x(coshx− sinhx)

15. y� = sech t 16. y� = sech x

17. y� = xcosh x�

cosh xx + sinh(x) lnx

�

18. y� = xsinh x�

sinh xx + cosh(x) lnx

�

19. y� =x

√x2 + 1

sinh

�x2 + 1

20. y� =1

xcosh (lnx)

21. y� = −2xex2�sech ex

2��

tanh ex2�

22. y� = 0

23. y� = 2 coth 2x

24. y� = 6x cosh2(x2 − 2) sinh(x2 − 2)

25. y� = coshx− sinhx

26. f �(t) = e2t

�2 coth t− csch

2t�

27. y� =6

√4x2 + 1

28. y� =2

4− x2

29. y� = | secx|

30. y� =1

2√x2 − x

31. y� = −2csch

−1x

|x|√1 + x2

32. y� = −2 sec 2x

33. y� = 2 sec 2x

34. y� = sinh−1 x

35. y� = −1

x√

1−x2

36. y� = −4x3√x8−1

37.14 38.

209

39.13 cosh(3x) + C

40. 2 sinh√x+ C

41.14 sinh

2(2x) + C

42.13 cosh

3(x− 1) + C

43. sinhx+13 sinh

3 x+ C

44.112 cosh

34x− 1

4 cosh 4x+ C

45. −2 cothx2 + C

46.√2 sinhx+ C

47. 2√2 sinh

x2 + C 48. tanh 1

49. ln 2− 35 50.

4+coth 4−coth 84

51.825 52.

1

3cosh

22− cosh 2 +

2

3

53. ln(cosh 1)

54. 2�tan

−1 e− π4

�

55. x = −1

2ln 3

56. rel max (0,−1),

rel min (1,− sinh 1)

57. 1− e−1

58. 8 tan−1

�e2

�− 2π

59.

a) f−1(x) = sinh

2�x

2

�

b) g−1(x) = 2 tanh

−1�x

2

�

68. y = ±1


1. C 6. A 11. B 16. A2. A 7. A 12. C 17. D3. B 8. C 13. A 18. A4. C 9. B 14. C 19. B5. B 10.B 15. A 20. D


2. hint: x ≈ 4.965

3.

a) T ≈ 5, 796 K

b) T ≈ 11, 146 K

a.2. answers to exercises 419 · a.2. answers to exercises 419 a.2 answers to exercises section 1.1...

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