Ch 2 Test Bank

1. Estimate the slope of the tangent line to the curve at x = 0.

A) B) C) D) Ans:

1 2 –2 0 B Difficulty: Easy

2. Estimate the slope of the tangent line to the curve at x = 2.

A) B)

2 –2 1 C) 4 1 D) 2 Ans: D Difficulty: Easy

3. List the points A, B, C, D, and E in order of increasing slope of the tangent line.

A) B) C) D) Ans:

B, C, E, D, A A, E, D, C, B E, A, D, B, C A, B, C, D, E B Difficulty: Easy

4. Compute the slope of the secant line between the points x = –3.1 and x = –3. Round your answer to the thousandths place. f ( x) = sin(2 x ) A) –0.995 B) 1.963 C) 5.963 D) –1.991 Ans: B Difficulty: Easy 5. Compute the slope of the secant line between the points x = –1 and x = –0.9. Round your answer to the thousandths place. f ( x ) = e 0.3 x A) B) C) D) Ans:

0.226 4.929 0.320 0.451 A Difficulty: Easy

6. Find an equation of the tangent line to y = f(x) at x = –2. f ( x ) = x3 + x2 + x A) B) C) D) Ans:

y = –2x + 4 y = 9x – 12 y = 2x + 4 y = 9x + 12 D Difficulty: Moderate

7. Find an equation of the tangent line to y = f(x) at x = 2. f ( x) = 2 x 3 + 5 A) y = 9x – 16 B) y = –24x – 27 C) y = 24x – 27 D) y = 24x + 27 Ans: C Difficulty: Moderate 8. Find the average velocity for an object between t = 4 sec and t = 4.1 sec if f(t) = –16t2 + 100t + 10 represents its position in feet. A) –29.6 ft/s B) –28 ft/s C) –31.2 ft/s D) 154 ft/s Ans: A Difficulty: Moderate 9. Find the average velocity for an object between t = –2 sec and t = –1.9 sec if f(t) = 5sin(t) + 5 represents its position in feet. (Round to the nearest thousandth.) A) –2.081 B) –1.616 C) –1.850 D) 1.850 Ans: C Difficulty: Moderate

10. The graph below gives distance in miles from a starting point as a function of time in hours for a car on a trip. Find the fastest speed (magnitude of velocity) during the trip. Describe how the speed during the first 2 hours compares to the speed during the last 2 hours. Describe what is happening between 2 and 3 hours.

Ans: The fastest speed occured during the last 2 hours of the trip when the car traveled at about 70 mph. The speed during the first 2 hours is 60 mph while the speed from 8 to 10 hours is about 70 mph. Between 2 and 3 hours the car was stopped. Difficulty: Moderate 11. Compute f′ (3) for the function f ( x ) = 2 3 6 B) 49 15 – C) 98 6 – D) 49 Ans: D Difficulty: Moderate

4 . x +5 2

A)

3 12. Compute f′ (2) for the function f ( x ) = 4 x − 5 x . A) 58 B) 43 C) 38 D) –43 Ans: B Difficulty: Moderate

13. Compute the derivative f′ (x) of f ( x) = 6 x 2 + 9 . −12 x f ′( x) = A) 6 x2 + 9 6x f ′( x) = B) 6 x2 + 9 −6 x f ′( x) = C) 6 x2 + 9 −6 x f ′( x) = D) 12 x + 9 Ans: B Difficulty: Moderate 14. Compute the derivative f′ (x) of f ( x) = −2 (2 x − 5) 2 −2 f ′( x) = B) (2 x − 5) 2 −1 f ′( x) = C) (2 x − 5) 2 2 f ′( x) = D) (2 x − 5) 2 Ans: A Difficulty: Moderate A)

f ′( x) =

1 . 2x − 5

15. Below is a graph of f ( x ) . Sketch a graph of f ′( x) .

Ans:

Difficulty: Moderate

16. Below is a graph of f ( x ) . Sketch a graph of f ′( x) .

Ans:

Difficulty: Difficult 17. Below is a graph of f ′( x) . Sketch a graph of a plausible f ( x ) .

Ans: Answers may vary. Below is one possible answer.

Difficulty: Moderate 18. Below is a graph of f ′( x) . Sketch a graph of a plausible f ( x ) .

Ans: Answers may vary. Below is one possible answer.

Difficulty: Difficult

19. The table below gives the position s(t) for a car beginning at a point and returning 5 hours later. Estimate the velocity v(t) at two points around the third hour and use them to estimate the acceleration a(t). Interpret the results. t (hours) s(t) (miles)

0 0

1 15

2 50

3 80

4 70

5 0

Ans: The velocity is the change in distance traveled divided by the elapsed time. From hour 3 to 4 the average velocity is (70 − 80)/(4 − 3) = −10 mph. Likewise, the velocity between hour 2 and hour 3 is about 30 mph. The acceleration is (− 10 − 30)/(4 − 2) = −20 mph. The negative value of acceleration indicates a decrease in forward velocity or an increase in reverse velocity, or a combination of both. The car is in the process of slowing, turning around, and heading back to the starting point. Difficulty: Easy 20. Use the distances f(t) to estimate the velocity at t = 1.8. (Round to 2 decimal places.) t 1.2 f(t) 43 A) B) C) D) Ans:

1.4 48

1.6 53. 5

1.8 58

2 2.2 62. 67.5 5

2.4 73

–2250.00 32.22 22.50 25.00 C Difficulty: Easy

(1 + h)3 + (1 + h) − 2 equals f ′(a ) for some function f ( x) and some constant h →0 h a. Determine which of the following could be the function f ( x ) and the constant a. f ( x) = x 3 − x and a = −1 A) f ( x) = x 3 + x 2 and a = 0 B) f ( x) = x 3 + x − 20 and a = 0 C) f ( x) = x 3 + x and a = 1 D) Ans: D Difficulty: Moderate

21. lim

1 1 − 2 22. (h + 4) 16 equals f ′(a ) for some function f ( x) and some constant a. lim h →0 h Determine which of the following could be the function f ( x ) and the constant a. 1 f ( x) = 2 and a = 4 A) x 4 f ( x) = 2 and a = 4 B) x 1 f ( x) = − 2 and a = 5 C) x 1 f ( x) = − 2 and a = −4 D) x Ans: A Difficulty: Moderate 23. Find the derivative of f(x) = –x2 – 5x + 1. A) –x – 5 B) –2x2 + 1 C) –2x – 5 D) 2x + 5 Ans: C Difficulty: Easy 24. Find the derivative of f ( x) = – A)

f ′( x) = –

5 –4 x2

5 –4 x2 5 f ′( x) = – 4 C) x 5 f ′( x) = 2 – 8 x 2 D) x Ans: B Difficulty: Easy B)

f ′( x) =

5 – 4x + 4 . x

25. Find the derivative of f ( x) =

–5 x 2 – 5 x + 2 . 4x

10 x + 5 4 5x 5 f ′( x) = + B) 2 4 5 1 f ′( x) = – – 2 C) 4 2x 5x2 5x 1 D) f ′( x) = – – + 4 4 2x Ans: C Difficulty: Moderate A)

f ′( x) = –

26. Find the derivative of f ( x) =

–9 x 2 + x + 7 . x

27 x 1 7 + – 2 2 x 2 x3 36 x – 2 f ′( x) = – B) x 27 x 1 7 f ′( x) = – – + C) 2 2 x 2 x3 1 7 f ′( x) = –27 x + + D) x x3 Ans: A Difficulty: Moderate A)

f ′( x) = –

5 27. Find the third derivative of f ( x) = 3 x + 7 x –

12 x4

A)

f ′′′( x) = 180 x 2 –

B)

f ′′′( x) = 180 x 2 + 7 +

12 x4

4 x3 12 f ′′′( x) = 180 x 2 + 4 D) x Ans: D Difficulty: Moderate C)

f ′′′( x) = 60 x3 –

2 . x

28. Find the second derivative of y = x + A) B) C)

7 . x

d2y 21 = 1+ 2 dx 4 x5 d2y 21 = 2 dx 4 x5 d2y 21 =– 2 dx 4 x5

d2y 21 = 2 dx 4 x3 Ans: B Difficulty: Moderate D)

4 3 29. Using the position function s (t ) = –3t – 4t –

A)

v(t ) = –12t 3 – 12t 2 +

B)

v(t ) = –9t 3 – 8t 2 +

4 , find the velocity function. t

4 t2

4 t2

4 t2 4 v(t ) = 12t 3 + 12t 2 + 2 D) t Ans: A Difficulty: Moderate C)

v(t ) = –12t 3 – 12t 2 –

3 30. Using the position function s (t ) = 4 t – , find the velocity function. t 4 3 v (t ) = – – 2 A) t t 2 3 v (t ) = + 2 B) t t 2 3 v (t ) = – + 2 C) t t 2 6 v (t ) = + 2 D) t t Ans: B Difficulty: Moderate

31. Using the position function s (t ) = –3t 3 – 4t – 4 , find the acceleration function. a (t ) = –9t A) a (t ) = –6t B) a (t ) = –18t C) a (t ) = –18t – 4 D) Ans: C Difficulty: Moderate 32. Using the position function s (t ) = A)

a (t ) = –

B)

a (t ) =

7 – 9 , find the acceleration function. t

21 4 t5 7

4 t5 7 a (t ) = – C) t3 21 a (t ) = D) 4 t5 Ans: D Difficulty: Moderate 33. The height of an object at time t is given by h(t ) = −16t 2 – 9t – 8 . Determine the object's velocity at t = 4. A) 137 B) –129 C) –137 D) –73 Ans: C Difficulty: Easy 34. The height of an object at time t is given by h(t ) = –8t 2 – 2t . Determine the object's acceleration at t = 4. A) –136 B) –16 C) –66 D) 16 Ans: B Difficulty: Easy 35. Find an equation of the line tangent to f ( x) = 4 x 2 – 6 x – 5 at x = –1. g ( x) = –14 x – 9 A) g ( x) = –8 x – 9 B) g ( x) = –14 x – 1 C) g ( x) = –8 x – 1 D) Ans: A Difficulty: Easy

36. Find an equation of the line tangent to f ( x) = 7 x – 5 x – 5 at x = 5.  –7 5 + 50  7 g ( x) =  5 +5  x – –10 2    7 5 – 10  7 g ( x) =  5 +5 B)  x + 5 2    7 5 – 25  7 g ( x) =  5 C)  x + 2  10   7 5 – 50  7 g ( x) =  5 –5 D)  x + 2  10  Ans: D Difficulty: Moderate A)

37. Determine the real value(s) of x for which the line tangent to f ( x) = 2 x 4 – 4 x 2 + 6 is horizontal. A) x = –1, x = 1 B) x = 0, x = –1, x = 1 C) x = 0 D) x = 0, x = 1 Ans: B Difficulty: Easy 38. Determine the real value(s) of x for which the line tangent to f ( x) = 4 x 2 – x – 6 is horizontal. 1 x = ,x =0 A) 8 1 ± 97 B) x= 8 1 x= C) 8 D) x = 0 Ans: C Difficulty: Easy

39. Find the second-degree polynomial (of the form ax2 + bx + c) such that f(0) = 0, f '(0) = 5, and f ''(0) = 1. x2 A) + 5x 2 x2 B) − + 5x 2 2 x C) − 5x + 1 2 x2 D) − + 5x + 1 2 Ans: A Difficulty: Moderate

(

)

9  x + 2 x  –5 x 2 +  . x  25 9 f ′( x) = –30 x 2 + x3/ 2 – 3/ 2 A) 2 2x 25 9 f ′( x) = –30 x 2 – x3/ 2 – 3/ 2 B) 2 2x 25 9 f ′( x) = 30 x 2 – x3/ 2 + 3/ 2 C) 2 2x 25 3/ 2 36 9 f ′( x ) = –30 x 2 – x + – 3/ 2 D) 2 x 2x Ans: B Difficulty: Moderate

40. Find the derivative of

(

)

3 41. Find the derivative of f ( x) = –4 x + 1 x .

16 3 x +1 3 4 f ′( x) = – 3 x – 1 B) 3 16 3 f ′( x) = – x +1 C) 3 8 f ′( x) = – 3 x + 2 D) 3 Ans: C Difficulty: Moderate A)

f ′( x) =

42. Find the derivative of f ( x) =

–9 x – 4 . 6x + 1

–15 (6 x + 1) 2 3 – B) 2 3 C) 2 15 D) (6 x + 1) 2 Ans: D Difficulty: Moderate A)

43. Find the derivative of f ( x) =

2x . 4 x2 – 9

–8 x 2 – 18 (4 x 2 – 9)2 1 – 2 B) 2x 8 x 2 + 18 C) (4 x 2 – 9)2 1 D) 2x 2 Ans: A Difficulty: Moderate A)

44. A small company sold 1000 widgets this year at a price of $12 each. The price will increase by $1.75 per year and each year 200 more widgets will be sold than the previous year. Determine the annual rate of revenue increase in the third year. A) $3100/year B) $5550/year C) $8950/year D) $3750/year Ans: B Difficulty: Moderate 45. Find an equation of the line tangent to h( x) = f ( x ) g ( x ) at x = 3 if f (3) = –3 , f ′(3) = 3 , g (3) = 3 , and g ′(3) = –1 . y = –6 x – 27 A) y = –6 x – 45 B) y = 12 x – 45 C) y = 12 x + 27 D) Ans: C Difficulty: Moderate

46. Find an equation of the line tangent to h( x) =

f ( x) at x = 1 if f (1) = –3 , g ( x)

f ′(1) = –1 , g (1) = 1 , and g ′(1) = 2 . y = –7 x – 8 A) y = 5x – 8 B) y = –7 x + 4 C) y = 5x + 2 D) Ans: B Difficulty: Moderate 47. The Dieterici equation of state, Pe an / VRT (V − nb) = nRT , gives the relationship between pressure P, volume V, and temperature T for a liquid or gas. At the critical point, P′(V ) = 0 and P′′(V ) = 0 with T constant. Using the result of the first derivative and substituting it into the second derivative, find the critical volume Vc in terms of the constants n, a, b, and R.  an 2 nRT   1  − an / VRT ′ P ( V ) = = 0 gives the result that Ans:  2 −  e (V − nb)   V − nb  V an(V − nb) RT = . V2  −2an 2  − an / VRT 2nRT 2an2 a2 n3 P′′(V ) =  3 + − + =0 e 3 2 2 4 2  V (V − nb) (V − nb) V (V − nb) V (V − nb) RT  . When the result of the first derivative is substituted for RT in the parentheses, the result is that Vc = 2nb. Difficulty: Difficult 48. Find the derivative of f ( x) = x 2 – 1 . 2x f ′( x) = A) x2 – 1 4x f ′( x) = B) x2 – 1 –x f ′( x) = C) x2 – 1 x f ′( x) = D) x2 – 1 Ans: D Difficulty: Moderate

49. Find the derivative of f ( x) = A)

f ′( x) =

B)

f ′( x) =

C)

f ′( x) =

D)

f ′( x) =

Ans: A

–8 5x2 + 3

40 x (5 x 2 + 3)3 –80 x (5 x 2 + 3)3 –40 x (5 x 2 + 3)3 –16 x

(5 x 2 + 3)3 Difficulty: Moderate x . x +6   x 3 x2 + 6  

50. Find the derivative of f ( x) =

2

A)

 1 1 −  2  x x2 + 6 

B)

 x  −  2   x +6 2 x x2 + 6

C)

 1 1 − x x2 + 6  2  x x2 + 6 

(

(

D) Ans: B

) (

1

x2 + 6

3

)

(

1

)

(

)

−

(

2 x2 x2 + 6

)

2

Difficulty: Moderate

)

    

.

51. Find the derivative of f ( x) =

( x 2 + 7)4 . 7

4 x( x 2 + 7)3 7 2 f ′( x ) = x( x 2 + 7)3 B) 7 8 f ′( x ) = x( x 2 + 7)3 C) 7 1 f ′( x ) = x( x 2 + 7)3 D) 7 Ans: C Difficulty: Moderate A)

f ′( x ) =

52. Find an equation of the line tangent to f ( x) = A) B) C) D) Ans:

1 2

x – 24

at x = 5.

y = –5x + 24 y = –5x y = 5x + 6 y = –5x + 26 D Difficulty: Moderate

53. Use the position function s (t ) = t 2 + 45 meters to find the velocity at t = 2 seconds. A) 7 m/s 2 B) m/s 7 1 C) m/s 7 1 D) m/s 7 Ans: B Difficulty: Moderate 54. Use the table of values to estimate the derivative of h( x) = f ( g ( x ) ) at x = 1. x f(x) g(x)

–3 –4 8

–2 –3 6

–1 –2 4

0 –3 4

1 –4 6

2 –5 8

3 –4 6

4 –2 4

5 0 3

A) B) C) D) Ans:

h′(1) ≈ –1 h′(1) ≈ 2 h′(1) ≈ 3 h′(1) ≈ –3 A Difficulty: Moderate

55. Compute the derivative of h( x) = f ( g ( x ) ) at x = –1 where f (–1) = 4 , g (–1) = –3 , f ′(–1) = 9 , f ′(–3) = –4 , g ′(–1) = 1 , and g ′(–3) = 6 . h′(–1) = 9 A) h′(–1) = 4 B) h′(–1) = –4 C) h′(–1) = –12 D) Ans: C Difficulty: Moderate 56. f ( x) = –3 x 5 + 2 x3 – x has an inverse g(x). Compute g ′(–2) . 1 g ′(–2) = – A) 217 1 g ′(–2) = – B) 10 1 g ′(–2) = C) 2 1 g ′(–2) = – D) 2 Ans: B Difficulty: Moderate 57. f ( x) = – x 3 + 5 x + 9 has an inverse g(x). Compute g ′(5) . 1 g ′(5) = – A) 2 1 g ′(5) = – B) 8 1 g ′(5) = C) 2 1 g ′(5) = D) 8 Ans: C Difficulty: Moderate

58. Find the derivative of f ( x) = –6sin( x) + 7 cos(3 x) − x . f ′( x) = –6 cos x – 21sin 3 x − 1 A) f ′( x) = –6 cos x – 7 sin 3 x − 1 B) f ′( x) = 6 cos x + 21sin 3 x − 1 C) f ′( x) = cos x – 3sin 3 x − 1 D) Ans: A Difficulty: Easy 59. Find the derivative of f ( x) = –2sin 2 x + 8 x 2 . f ′( x) = 4sin x cos x + 16 x A) f ′( x) = –4sin x cos x + 8 x B) f ′( x) = –4sin x + 16 x C) f ′( x) = –4sin x cos x + 16 x D) Ans: D Difficulty: Easy 60. Find the derivative of f ( x) = – sin x sec x . sec x f ′( x) = – A) 2 – tan x sec 2 x f ′( x) = – B) 2 – tan x sec 2 x f ′( x) = – C) – tan x sec x tan x f ′( x) = – D) 2 – tan x Ans: B Difficulty: Moderate –8cos x 2 61. Find the derivative of f ( x) = . x2 –16( x 2 sin x 2 + cos x2 ) A) f ′( x) = x3 16( x sin x 2 + cos x 2 ) B) f ′( x) = x3 16( x 2 sin x 2 + cos x2 ) C) f ′( x) = x3 16( x 2 sin x 2 + cos x2 ) D) f ′( x) = x4 Ans: C Difficulty: Moderate

62. Find an equation of the line tangent to f ( x) = x cos x at x = 2 . (Round coefficients to 3 decimal places.) y = –1.402 x + 3.637 A) y = –2.235 x + 3.637 B) y = –2.235 x – 3.637 C) y = –1.402 x – 3.637 D) Ans: B Difficulty: Moderate 63. Find an equation of the line tangent to f ( x) = tan 4 x at x = 3 . (Round coefficients to 3 decimal places.) y = 4.74 x – 17.488 A) y = –5.617 x + 14.308 B) y = 5.617 x – 17.488 C) y = 5.617 x – 16.764 D) Ans: C Difficulty: Moderate 64. Use the position function s (t ) = 9 cos t + 7t 2 feet to find the velocity at t = 4 seconds. (Round answer to 2 decimal places.) A) v(4) = 62.81 ft/s B) v(4) = 49.19 ft/s C) v(4) = –49.19 ft/s D) v(4) = 61.88 ft/s Ans: A Difficulty: Moderate 65. Use the position function s (t ) = –3sin(2t ) – 3 meters to find the velocity at t = 1 seconds. (Round answer to 2 decimal places.) A) v(1) = –5.46 m/s B) v(1) = –3.24 m/s C) v(1) = 1.25 m/s D) v(1) = 2.5 m/s Ans: D Difficulty: Moderate 66. A weight hanging by a spring from the ceiling vibrates up and down. Its vertical position is given by s (t ) = 8sin(7t ) . Find the maximum speed of the weight and its position when it reaches maximum speed. A) speed = 8, position = 56 B) speed = 56, position = 0 C) speed = 7, position = 8 D) speed = 56, position = 7 Ans: B Difficulty: Moderate

67. The total charge in an electrical circuit is given by Q(t ) = 2sin(2t ) + t + 5 . The dQ current is the rate of change of the charge, i (t ) = . Determine the current at dt t = 0 (Round answer to 2 decimal places.) A) i (0) = 3 i (0) = 5 B) i (0) = 10 C) D) i (0) = 1 Ans: B Difficulty: Moderate 68. For f ( x) = sin x , find f (158) ( x) . A) cos x B) –cos x C) sin x D) –sin x Ans: D Difficulty: Easy 69. Given that lim x→0

sin x sin(–7t ) = 1 , find lim . t → 0 x 5t

1 5 –35 B) 7 – C) 5 1 – D) 7 Ans: C Difficulty: Easy A)

cos x − 1 –7 cos t − 1 = 0 , find lim . x→0 t →0 x 2t

70. Given that lim A)

0

7 2 C) –14 2 – D) 7 Ans: A Difficulty: Easy B)

–

5t sin x = 1 , find lim . t →0 sin(–7t ) x→0 x

71. Given that lim

–35 1 B) 5 7 – C) 5 5 – D) 7 Ans: D Difficulty: Easy A)

sin x tan(–7t ) = 1 , find lim . x→0 t →0 x 6t

72. Given that lim

1 7 7 – B) 6 6 – C) 7 1 D) 6 Ans: B Difficulty: Moderate A)

–

73. Find the derivative of f ( x) = x –8e –8 x . f ′( x) = ( –8 x –7 + 8 x –8 ) e–8 x A) B) C)

f ′( x) = –8 x –9 e –8 – 8 x –8 e–8 x −1 f ′( x) = ( –8 x –9 – 8 x –8 ) e–8 x

f ′( x) = –8 x –9 – 8e –8 x D) Ans: C Difficulty: Easy 74. Find the derivative of f ( x) = 6 –6 x + 4 . f ′( x) = 6 –6 x + 4 (–6) ln 6 A) f ′( x) = (–6)6 –6 x + 4 B) f ′( x) = 6 –6 x + 4 ln 6 C) f ′( x) = 6 –6 x + 4 (–6 x + 4) ln 6 D) Ans: A Difficulty: Easy

75. Find the derivative of f ( x) = ln ( 8 x ) . 1 1 f ′( x) = + A) x 8 8 f ′( x) = B) x 1 f ′( x) = C) 8x 1 f ′( x) = D) x Ans: D Difficulty: Easy 76. Find the derivative of f ( x) = ln

(

)

2x .

1 4x 1 f ′( x) = B) x 1 f ′( x) = C) 2x 1 1 1 f ′( x) =  +  D) 2  x 2 Ans: C Difficulty: Easy A)

f ′( x) =

77. Find an equation of the line tangent to f ( x) = 2 x at x = 3. A) 8 ( x ln 2 − (1 + 3ln 2) ) B) C)

( x ln 2 + (1 − 3ln 2) ) 8 ( x ln 2 + (1 − 3ln 2) ) ( x ln 2 + (3ln 2 − 1) )

D) Ans: C

Difficulty: Moderate

78. Find an equation of the line tangent to f ( x) = 3ln( x 4 ) at x = 3. x y = + (ln 3 − 1) A) 3 x  y = 12  + (ln 3 − 1)  B) 3  x  y = 12  + (1 − ln 3)  C) 3  x y = + (1 − ln 3) D) 3 Ans: B Difficulty: Moderate 79. The value of an investment is given by v(t ) = (900)4t . Find the instantaneous percentage rate of change. (Round to 2 decimal places.) A) 1.39 % per year B) 49.91 % per year C) 138.63 % per year D) 25.97 % per year Ans: C Difficulty: Moderate 80. A bacterial population starts at 200 and quadruples every day. Calculate the percent rate of change rounded to 2 decimal places. A) 160.94 % B) 138.63 % C) 1.39 % D) 88.63 % Ans: B Difficulty: Moderate 81. An investment compounded continuously will be worth f (t ) = Ae rt , where A is the investment in dollars, r is the annual interest rate, and t is the time in years. APY can be defined as ( f (1) − A) / A , the relative increase of worth in one year. Find the APY for an interest rate of 11%. Express the APY as a percent rounded to 2 decimal places. APY = 111.63% A) APY = 10.63% B) APY = 11.63% C) APY = 12.63% D) Ans: C Difficulty: Moderate

82. Find the derivative of f ( x) = x cos5 x .  cos 5 x  f ′( x) = x cos5 x  − 5(sin 5 x) ln x  A)  x  cos5 x f ′( x ) = (−5sin 5 x) x B) f ′( x) = (cos 5 x) x cos5 x −1 C) f ′( x) = x cos5 x (ln x − 5sin 5 x) D) Ans: A Difficulty: Moderate 83. Find the derivative of f ( x) = ( x 4 )3 x . f ′( x) = x12 x (ln x + 12) A) f ′( x) = 12 x12 x −1 B) f ′( x) = 12 x12 x C) f ′( x) = 12 x12 x (ln x + 1) D) Ans: D Difficulty: Easy

84. The position of a weight attached to a spring is described by s (t ) = e −2t sin 3t . Determine and graph the velocity function for positive values of t and find the approximate first time when the velocity is zero. Find the approximate position of the weight the first time the velocity is zero. Round answers to tenths.

Ans: v(t ) = e −2t (3cos 3t − 2sin 3t ) . The velocity is first zero at about 0.3 and its position is about 0.4.

Difficulty: Moderate

85. Compute the slope of the line tangent to 4 x 2 + 5 xy + 7 y 2 = 1 at (7, 7) . 3 A) slope = – 19 13 slope = – B) 19 19 slope = – C) 13 9 D) slope = – 7 Ans: B Difficulty: Moderate 86. Find the derivative y ′( x ) implicitly if – y 2 + 7 xy = –9 . 7y y′ ( x ) = – A) 2 y xy – 7 x 7 y xy 4 y + 7x 7y y′ ( x ) = C) 4 y xy + 7 x 7y y′ ( x ) = D) 4 y + 7 x xy Ans: C Difficulty: Moderate B)

y′ ( x ) =

87. Find the derivative y ′( x ) implicitly if –4sin xy + 2 x = –5 . 1 y y ′( x ) = – + A) 2 x cos xy x 1 y y ′( x ) = − B) 2 x x cos xy cos xy y y ′( x ) = − C) 2x x 1 y y ′( x ) = − D) 2 x cos xy x Ans: D Difficulty: Moderate 88. Find the location of all horizontal and vertical tangents for x 2 − xy 2 = 16 . A) horizontal: none; vertical: (–4, 0), (4, 0) B) horizontal: (4, 0); vertical: (–4, 0), (4, 0) C) horizontal: (–4, 0), (4, 0); vertical: none D) horizontal: none; vertical: (4, 0) Ans: A Difficulty: Moderate

89. Find the location of all horizontal and vertical tangents for x 2 + xy 2 + 1 = 0 . A) B) C) D)

( )( ) horizontal: ( –1, – 2 ) , ( –1, 2 ) ; vertical: (0, 0) horizontal: ( –1, – 2 ) , ( –1, 2 ) ; vertical: none horizontal: ( 1, – 2 ) , ( 1, 2 ) ; vertical: (–1, 0)

horizontal: –1, – 2 , –1, 2 ; vertical: (–1, 0)

Ans: C

Difficulty: Moderate

90. Find the second derivative, y ′′( x ) , of –4 x 3 – 3 y 3 = –4 . 2 y′ y ′′( x ) = – − A) 3 xy 2 y B)

( y′ ) 2 y ′′( x ) = – − 2y 3 xy

C)

( y′ ) 4 y ′′( x ) = − 2y 3 xy

2

2

( y′) 2 y ′′( x ) = + D) 2y 3 xy Ans: B Difficulty: Moderate 2

91. Find the second derivative, y ′′( x ) , of –3 y 2 = 2 x3 – 2 x – cos y . 4 x + (– cos y – 3)( y′) 2 y ′′( x ) = A) –3 y + sin y 2 x + (cos y – 6) y′ y ′′( x ) = B) –6 y – cos y 12 x + (cos y – 3) y 2 y ′′( x ) = C) –6 y 2 – sin y 12 x + (cos y + 6)( y′) 2 ′′ y ( x) = D) –6 y – sin y Ans: D Difficulty: Moderate

92. Find the derivative of f ( x) = –6sec −1 ( x5 ) . A)

f ′( x ) =

B)

f ′( x) =

C)

f ′( x) =

D)

f ′( x) =

Ans: A

–30 x 4 x10 − 1

x5

30 x 5 x2 −1

x

–6 x 4 x

x2 + 1 5x4

x4

x8 − 1

Difficulty: Moderate

93. Find the derivative of f ( x) = 7e3tan 42 3tan −1 x f ′( x) = e A) 1 − x2 7 3tan −1 x f ′( x) = e B) 1 + x2 21 3tan −1 x f ′( x) = e C) 1 + x2 3 3tan −1 x f ′( x) = e D) 1 − x2 Ans: C Difficulty: Moderate

−1

x

.

94. Using the Mean Value Theorem, find a value of c that makes the conclusion true for f ( x) = 4 x 3 + 5 x 2 , in the interval [−1,1]. c ≈ −1.129 A) B) One or more hypotheses fail c ≈ 0.295 C) c=0 D) Ans: C Difficulty: Easy 95. Using the Mean Value Theorem, find a value of c that makes the conclusion true  π π for f ( x) = cos x,  − ,   2 2 A) One or more hypotheses fail c=0 B) π c= C) 4 c ≈ .881 D) Ans: B Difficulty: Easy

96. Determine if the function f ( x) = 5 x3 + 4 x + 3 is increasing, decreasing, or neither. A) Increasing B) Decreasing C) Neither Ans: A Difficulty: Easy 97. Determine if the function f ( x) = –2 x 4 – 4 x 2 – 4 is increasing, decreasing, or neither. A) Increasing B) Decreasing C) Neither Ans: C Difficulty: Easy 98. Prove that 3 x 3 + 7 x – 1 = 0 has exactly one solution. 3 Ans: Let f ( x ) = 3 x + 7 x – 1 . The function f(x) is continuous and differentiable everywhere. Since f(0) < 0 and f(1) > 0, f(x) must have at least one zero. The derivative of f ( x) = 3 x 3 + 7 x – 1 is f ′( x) = 9 x 2 + 7 , which is always greater than zero. Therefore f(x) can only have one zero. Difficulty: Moderate 99. Find all the functions g ( x) such that g ′( x) =

1 . x9

1 8 x8 1 g ( x) = – +c B) 10 x10 2 g ( x) = C) 25 x8 1 g ( x) = – 8 + c D) 8x Ans: D Difficulty: Moderate A)

g ( x) = –

100. Find all the functions g ( x) such that g ′( x) = –7 sin 2 x. g ( x) = –14 cos 2 x A) 7 g ( x) = cos 2 x + c B) 2 g ( x ) = 7 cos 2 x C) 7 g ( x) = sin 2 x + c D) 2 Ans: B Difficulty: Moderate