Calculus early transcendental functions 4th edition smith test bank

Chapter Find the equation of the tangent line to y  x – x at x  A) y  –9 B) y  C) y  –9 x D) y  x Ans: A Difficulty: Moderate Section: 2.1 Find an equation of the tangent line to y = f(x) at x = f  x   x3  x  x A) y = –12x – 36 B) y = 34x + 63 C) y = 12x – 36 Ans: D Difficulty: Moderate Section: 2.1 D) y = 34x – 63 Find an equation of the tangent line to y = f(x) at x = f ( x)  x  A) y = 9x – 16 B) y = –24x – 27 C) y = 24x – 27 Ans: C Difficulty: Moderate Section: 2.1 Find the equation of the tangent line to y  A) 16 x 25 25 B) 16 y  – x 25 25 Ans: C Difficulty: Moderate D) y = 24x + 27 at x  x2 C) y D) 16 x 25 25 16 y x 25 25 y– Section: 2.1 Find the equation of the tangent line to y  x – at x  A) y  x – B) y  3x – C) y  x – 18 Ans: B Difficulty: Moderate Section: 2.1 D) y  3x – 18 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 Section: 2.1 Page 131 Chapter Compute the slope of the secant line between the points x = and x = 1.1 Round your answer to the thousandths place f  x   e 0.5 x A) 0.845 B) 5.529 C) 0.780 D) 1.691 Ans: A Difficulty: Easy Section: 2.1 List the points A, B, C, D, and E in order of increasing slope of the tangent line A) B, C, E, D, A B) A, E, D, C, B C) E, A, D, B, C Ans: B Difficulty: Easy Section: 2.1 D) A, B, C, D, E Use the position function s(t )  4.9t  meters to find the velocity at time t  seconds A) –43.1 m/sec B) –29.4 m/sec C) –28.4 m/sec D) –44.1 m/sec Ans: B Difficulty: Moderate Section: 2.1 10 Use the position function s(t )  t + meters to find the velocity at time t  –1 seconds 1 m/sec D) m/sec Difficulty: Moderate Section: 2.1 A) m/sec B) m/sec C) Ans: D 11 Find the average velocity for an object between t = sec and t = 3.1 sec if f(t) = –16t2 + 100t + 10 represents its position in feet A) 2.4 ft/s B) ft/s C) 0.8 ft/s D) 166 ft/s Ans: A Difficulty: Moderate Section: 2.1 Page 132 Chapter 12 Find the average velocity for an object between t = sec and t = 1.1 sec if f(t) = 5sin(t) + represents its position in feet (Round to the nearest thousandth.) A) 2.702 B) 2.268 C) 2.487 D) –2.487 Ans: C Difficulty: Moderate Section: 2.1 13 Estimate the slope of the tangent line to the curve at x = –2 A) –1 B) –2 C) D) Ans: B Difficulty: Easy Section: 2.1 14 Estimate the slope of the tangent line to the curve at x = A) 1 D) Difficulty: Easy Section: 2.1 B) –3 Ans: D C) Page 133 Chapter 15 The table shows the temperature in degrees Celsius at various distances, d in feet, from a specified point Estimate the slope of the tangent line at d  and interpret the result d 13 20 14 C m  4.67; The temperature is increasing 4.67 C per foot at the point feet from the specified point B) m  –0.33; The temperature is decreasing 0.33 C per foot at the point feet from the specified point C) m  –3; The temperature is decreasing C per foot at the point feet from the specified point D) m  20; The temperature is increasing 20 C per foot at the point feet from the specified point Ans: C Difficulty: Moderate Section: 2.1 A) 16 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 hours compares to the speed during the last hours Describe what is happening between and hours Ans: The fastest speed occurred during the last hours of the trip when the car traveled at about 70 mph The speed during the first hours is 60 mph while the speed from to 10 hours is about 70 mph Between and hours the car was stopped Difficulty: Moderate Section: 2.1 17 Compute f(3) for the function f ( x)  x3  x A) 150 B) 130 C) 120 D) –130 Ans: B Difficulty: Moderate Section: 2.2 Page 134 Chapter 18 Compute f(4) for the function f ( x)  x 4 1 B) C) – D) – 25 25 25 Ans: D Difficulty: Moderate Section: 2.2 A) 19 Compute the derivative function f(x) of f ( x)  21 (3 x  1) B) 3 f ( x)  (3 x  1) Ans: A Difficulty: Moderate A) C) f ( x)  D) 3x  7 (3 x  1) 21 f ( x)  (3 x  1) f ( x)  Section: 2.2 20 Compute the derivative function f(x) of f ( x)  x  A) B) f ( x)  f ( x)  Ans: B 8 x C) 4x  4x 4x  Difficulty: Moderate D) Section: 2.2 Page 135 f ( x)  4 x 4x2  4 x f ( x )  8x  Chapter 21 Below is a graph of f ( x) Sketch a plausible graph of a continuous function f ( x ) Ans: Answers may vary Below is one possible answer Difficulty: Moderate Section: 2.2 Page 136 Chapter 22 Below is a graph of f ( x ) Sketch a graph of f ( x) Ans: Difficulty: Moderate 9+ Section: 2.2 Page 137 Chapter 23 Below is a graph of f ( x ) Sketch a graph of f ( x) Ans: Difficulty: Difficult Section: 2.2 Page 138 Chapter 24 Below is a graph of f ( x) Sketch a plausible graph of a continuous function f ( x ) Ans: Answers may vary Below is one possible answer Difficulty: Difficult 25 Section: 2.2 Compute the right-hand derivative D f (0)  lim h 0 derivative D f (0)  lim h 0 f (h)  f (0) h if x  f (h)  f (0) and the left-hand h  4x + f ( x)    –8 x + if x  A) D f (0)  –8 , D f (0)  B) D f (0)  , D f (0)  –8 Ans: A Difficulty: Moderate C) D) Section: 2.2 Page 139 D f (0)  , D f (0)  D f (0)  –2 , D f (0)  –2 Chapter 26 Numerically estimate the derivative f (0) for f ( x)  xe3 x A) B) C) D) Ans: D Difficulty: Moderate Section: 2.2 27 The table below gives the position s(t) for a car beginning at a point and returning hours later Estimate the velocity v(t) at two points around the third hour t (hours) s(t) (miles) 0 15 50 80 70 Ans: The velocity is the change in distance traveled divided by the elapsed time From hour to the average velocity is (70 − 80)/(4 − 3) = −10 mph Likewise, the velocity between hour and hour is about 30 mph Difficulty: Easy Section: 2.2 28 Use the distances f(t) to estimate the velocity at t = 2.2 (Round to decimal places.) t 1.6 f(t) 49 1.8 2.2 2.4 2.6 54 59.5 64 68.5 73.5 79 A) –2250.00 B) 29.09 C) 22.50 D) 25.00 Ans: C Difficulty: Easy Section: 2.2 29 5 x – x if x  For f ( x)   find all real numbers a and b such that f (0) exists  ax  b if x  A) a  10, b any real number B) a  4, b  Ans: D Difficulty: Moderate C) D) Section: 2.2 Page 140 a  –6, b any real number a  –6, b  Chapter 92 Use the position function s(t )  cos 2t – t feet to find the velocity at t = seconds (Round answer to decimal places.) A) v(3) = –5.44 ft/s B) v(3) = –6.56 ft/s Ans: A Difficulty: Moderate C) D) Section: 2.6 v(3) = 6.56 ft/s v(3) = –7.92 ft/s 93 Use the position function s (t )  sin(2t ) + meters to find the velocity at t = seconds (Round answer to decimal places.) A) v(4) = 13.85 m/s B) v(4) = –9.15 m/s Ans: D Difficulty: Moderate C) D) Section: 2.6 v(4) = –1.02 m/s v(4) = –2.04 m/s 94 Use the position function to find the velocity at time t  t0 Assume units of feet and seconds sin10t s(t )  , t  t A) B) v( )  ft/sec v( )  – Ans: C 10  C) D) ft/sec Difficulty: Moderate v( )  10 v( )   2 ft/sec ft/sec Section: 2.6 95 A weight hanging by a spring from the ceiling vibrates up and down Its vertical position is given by s (t )  9sin(7t ) Find the maximum speed of the weight and its position when it reaches maximum speed A) speed = 9, position = 63 B) speed = 63, position = Ans: B Difficulty: Moderate 96 C) D) Section: 2.6 sin x sin(7t )  , find lim x 0 t 0 x –8t Given that lim Ans: C A) – D) Difficulty: Easy Section: 2.6 B) –56 C) – Page 157 speed = 7, position = speed = 63, position = Chapter 97 cos x  cos t   , find lim x 0 t 0 x 2t Given that lim A) B) Ans: A 98 6t sin x  , find lim t  sin(7t ) x 0 x Given that lim A) 42 C) D) 6 Difficulty: Easy Section: 2.6 B) Ans: D 99 1 C) D) 2 Difficulty: Easy Section: 2.6 Given that lim x 0 sin x tan(7t )  , find lim t  x 8t B) C) D) 8 Ans: B Difficulty: Moderate A) Section: 2.6 100 For f ( x)  sin x , find f (22) ( x) A) cos x B) –cos x C) sin x D) –sin x Ans: D Difficulty: Easy Section: 2.6 101 The total charge in an electrical circuit is given by Q(t )  3sin(3t )  t + The current is dQ the rate of change of the charge, i (t )  Determine the current at t  (Round dt answer to decimal places.) A) i (0)  B) i (0)  10 C) i (0)  12 D) i (0)  Ans: B Difficulty: Moderate Section: 2.6 102 Find the derivative of f ( x)  x –9e –2 x A) f ( x)   –9 x –8 + x –9  e –2 x C) B) D) f ( x)  –9 x –10e –2 – x –9e –2 x 1 Ans: C Difficulty: Easy Section: 2.7 Page 158 f ( x)   –9 x –10 – x –9  e –2 x f ( x)  –9 x –10 – 2e –2 x Chapter 103 Differentiate the function f ( x)  e3 x cos x A) C) f ( x)  –12e3 x sin x x x B) D) f ( x)  3e cos x + 4e sin x Ans: D Difficulty: Moderate Section: 2.7 f ( x)  12e3 x sin x f ( x)  3e3 x cos x – 4e3 x sin x 104 Find the derivative of f ( x)  93 x + f ( x)  93 x + (3ln 9) f ( x)  (3)93 x + Ans: A Difficulty: Easy A) B) C) D) f ( x)  93 x + ln f ( x)  93 x + (3x + 8) ln Section: 2.7 105 Differentiate the function w f ( w)  3w e – 3w B) f ( w)  3w C) f ( w)  3w 3w e 3e e Difficulty: Moderate Section: 2.7 A) f ( w)  Ans: A D) f ( w)  3w  e3 w 106 Find the derivative of f ( x)  ln  x  1 B) f ( x)  C) f ( x)   x x 2x Difficulty: Easy Section: 2.7 A) f ( x)  Ans: D 107 Find the derivative of f ( x)  ln  x  3x B) f ( x)  C) f ( x)  6x 3x 2x Difficulty: Easy Section: 2.7 A) f ( x)  Ans: C D) f ( x)  Page 159 D) f ( x)  1 1   x  Chapter 108 Differentiate the function f (t )  ln(t + 8t ) A) t + 8t B) f (t )  5t + Ans: D Difficulty: Moderate f (t )  C) f (t )  (5t + 8) ln(t + 8t ) D) f (t )  5t + t + 8t Section: 2.7 109 Differentiate the function g ( x)  sin x ln( x5  3) A) C) x sin x x 3 B) D) sin x g ( x)   cos x ln( x5  3)  x 3 Ans: A Difficulty: Moderate Section: 2.7 g ( x)  cos x ln( x  3)  x cos x x5  cos x g ( x)  x 3 g ( x )  110 Differentiate the function x h( x )  e A) h( x)  e B) h( x)  e ln C) h( x)  e x e ln Ans: C Difficulty: Moderate Section: 2.7 x x x D) h( x )  e x e 111 Find an equation of the line tangent to f ( x)  3x at x = y  27  x ln  (1  3ln 3)  C) B) y  x ln  (1  3ln 3) Ans: C Difficulty: Moderate D) Section: 2.7 A) y  27  x ln  (1  3ln 3)  y  x ln  (3ln  1) 112 Find an equation of the line tangent to f ( x)  3ln( x ) at x = A) x  (ln  1) B) x  y  12   (ln  1)  2  Ans: B Difficulty: Moderate C) y D) Section: 2.7 Page 160 x  y  12   (1  ln 2)  2  x y   (1  ln 2) x Chapter 113 Find all values of x for which the tangent line to f ( x)  x 2e –4 x is horizontal A) x  Ans: D B) x  0, x  –4 C) x  0, x  Difficulty: Moderate D) x  0, x  Section: 2.7 114 The value of an investment is given by v(t )  (600)4t Find the instantaneous percentage rate of change (Round to decimal places.) A) 1.39 % per year B) 33.27 % per year Ans: C Difficulty: Moderate C) D) Section: 2.7 138.63 % per year 17.31 % per year 115 A bacterial population starts at 300 and quadruples every day Calculate the percent rate of change rounded to decimal places A) 160.94 % B) 138.63 % C) 1.39 % D) 88.63 % Ans: B Difficulty: Moderate Section: 2.7 116 Use logarithmic differentiation to find the derivative of f ( x)  x cos x A)  cos x  f ( x)  x cos x   2(sin x) ln x   x  cos x f ( x)  (2sin x) x f ( x)  (cos x) x cos x 1 B) C) D) f ( x)  xcos x (ln x  2sin x) Ans: A Difficulty: Moderate Section: 2.7 117 Find the derivative of f ( x)  ( x3 )3 x A) f ( x)  x9 x (ln x  9) B) f ( x)  x9 x1 Ans: D Difficulty: Easy C) D) Section: 2.7 Page 161 f ( x)  x9 x f ( x)  x9 x (ln x  1) Chapter 118 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 (3cos3t  2sin 3t ) The velocity is first zero at about 0.3 and its position is about 0.4 Difficulty: Moderate Section: 2.7 Page 162 Chapter 119 An investment compounded continuously will be worth f (t )  Aert , 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 5% Express the APY as a percent rounded to decimal places A) APY  105.13% B) APY  4.13% Ans: C Difficulty: Moderate C) D) Section: 2.7 APY  5.13% APY  6.13% 120 Compute the slope of the line tangent to 3x + 3xy + y  34 at (2, –1) 15 B) slope = 8 Difficulty: Moderate A) slope = Ans: B Section: 2.8 C) slope = D) slope = 15 14 121 Find the derivative y ( x ) implicitly x2 y – y  5x A) y ( x )  B) y( x)  Ans: B C) xy +  xy 2 x2 y – Difficulty: Moderate D) xy – 2 xy + 12 y( x)  2x2 y y( x)  – Section: 2.8 122 Find the derivative y ( x ) implicitly if y – xy  –6 A) y  x   – B) y  x   Ans: C y y xy + x C) y xy 8y – x Difficulty: Moderate D) y y xy – x y y  x   y – x xy y  x   Section: 2.8 123 Find the derivative y ( x ) implicitly if 4sin xy + x  –5 A) y  x cos xy x B) y y( x)  –  x x cos xy Ans: D Difficulty: Moderate C) y( x)  D) Section: 2.8 Page 163 5cos xy y  4x x y y( x)  –  x cos xy x y( x)  – Chapter 124 Find the derivative y ( x ) implicitly xe y  y cos x  A) ey 9sin x  xe y B) ey y( x)   9sin x Ans: D Difficulty: Difficult C) y( x )   D) 9sin x ey e y  y sin x y( x)  cos x  xe y y( x)   Section: 2.8 125 Find the derivative y ( x ) implicitly e5 y  ln( y – 1)  3x A) 3( y – 1) 5( y – 1)e5 y  y B) (3  5e5 x )( y – 1) y( x)  2y Ans: A Difficulty: Difficult C) y( x)  D) 3( y – 1) 5( y – 1)e5 y  3( y – 1)  y y( x)  5( y – 1)e5 y y( x)  Section: 2.8 126 Find an equation of the tangent line at the given point x  16 y  at (4, 1) 4 1 A) y  – x  B) y  – x  C) y  x  12 Ans: C Difficulty: Moderate Section: 2.8 D) y  1 x 12 127 Find an equation of the tangent line at the given point x y  y  at (2, 1) Ans: 13 y  – x 5 Difficulty: Moderate Section: 2.8 128 Find the second derivative, y( x) , of –2 x3 + y  –3 A) y( x)  y  xy y  y  y( x)   2y xy Ans: B Difficulty: Moderate B) C)  y  y( x)  –  2y xy D)  y  y( x)  –  2y xy Section: 2.8 Page 164 Chapter 129 Find the second derivative, y( x) , of –3 y  –2 x3 + x – cos y C) –4 x  (– cos y – 3)( y) y( x)  –3 y + sin y B) D) –2 x  (cos y – 6) y y( x)  –6 y – cos y Ans: D Difficulty: Moderate Section: 2.8 A) –12 x  (cos y – 3) y y( x)  –6 y – sin y –12 x  (cos y + 6)( y) y( x)  –6 y – sin y 130 Find the derivative of f ( x)  cos 1 ( x5 – 2) A) B) f ( x)  x sin( x5 – 2) cos ( x5 – 2) C) f ( x)  5x4 cos ( x5 – 2) D) Ans: D Difficulty: Moderate f ( x)  f ( x)   5x4  ( x5 – 2) 5x4  ( x5 – 2) Section: 2.8 131 Find the derivative of f ( x)  tan 1 (3/ x) A)  x2 B) f ( x)    x2 Ans: A Difficulty: Moderate C) f ( x)   D)  9x2 f ( x)    3x f ( x)   Section: 2.8 132 Find the derivative of f ( x)  5e3tan 1 x A) 30 3tan 1 x e  x2 B) 3tan 1 x f ( x)  e  x2 Ans: C Difficulty: Moderate C) f ( x)  D) Section: 2.8 Page 165 15 3tan 1 x e  x2 3tan 1 x f ( x)  e  x2 f ( x)  Chapter 133 Find the derivative of f ( x)  4sec1 ( x5 ) A) B) f ( x)  x5 x10  x D) x2 1 Difficulty: Moderate 4x4 f ( x)  x –20 x5 f ( x)  Ans: A C) 20 x f ( x)  x2  5x4 x4 x8  Section: 2.8 134 Find the location of all horizontal and vertical tangents for x  xy  49 A) B) C) D) Ans: horizontal: none; vertical: (–7, 0), (7, 0) horizontal: (7, 0); vertical: (–7, 0), (7, 0) horizontal: (–7, 0), (7, 0); vertical: none horizontal: none; vertical: (7, 0) A Difficulty: Moderate Section: 2.8 135 Find the location of all horizontal and vertical tangents for x  xy  81  A) B) C) D)    horizontal:  –9, –3  ,  –9,3  ; vertical: (0, 0) horizontal:  –9, –3  ,  –9,3  ; vertical: none horizontal:  9, –3  ,  9,3  ; vertical: (–81, 0) horizontal: –9, –3 , –9,3 ; vertical: (–81, 0) Ans: C Difficulty: Moderate Section: 2.8 Page 166 Chapter 136 Sketch the graph of the function f ( x)  cosh( x / 8) A) y x -10 -8 -6 -4 -2 -2 10 -4 B) y x -10 -8 -6 -4 -2 -2 10 -4 C) y x -10 -8 -6 -4 -2 -2 10 -4 D) y x -10 -8 -6 -4 -2 -2 10 -4 Ans: B Difficulty: Moderate Section: 2.9 Page 167 Chapter 137 Find the derivative of f ( x)  cosh x A) B) f ( x)   C) cosh x x cosh x x Difficulty: Moderate D) f ( x)  Ans: D f ( x)   f ( x)  sinh x x sinh x x Section: 2.9 138 Find the derivative of f ( x)   x 3 A) f ( x)   x  f ( x)   x  sech x Ans: B Difficulty: Moderate B) C) f ( x)  sech x D) f ( x)  3sech x Section: 2.9 139 Find the derivative of f ( x)  sech4 x A) f ( x)  –4sech4 x x B) f ( x)  4sech4 x x Ans: A Difficulty: Moderate C) D) Section: 2.9 f ( x)  4sech x f ( x)  sech x 140 Find the derivative of f ( x)  x sinh10 x A) B) C) D) Ans: 141 f ( x)  40 x3 cosh10 x f ( x)  x3 cosh10 x f ( x)  x3 sinh10 x  10 x cosh10 x f ( x)  x3 sinh10 x  x cosh10 x C Difficulty: Moderate Section: 2.9 Find the derivative of f ( x)  cosh x x–2 C) 4( x – 2) sinh x  cosh x ( x – 2) B) D) ( x – 2) sinh x  cosh x f ( x)  ( x – 2) Ans: A Difficulty: Moderate Section: 2.9 A) f ( x)  Page 168 4sinh x x–2 4sinh x f ( x )  ( x – 2) f ( x)  Chapter 142 Find the derivative of f ( x)  cosh 1 x A) B) f ( x)  f ( x)  Ans: C C) 64  x x  64 Difficulty: Moderate D) f ( x)  f ( x)  64 x   64 x Section: 2.9 143 A general equation for a catenary is y  a cosh( x / b) Find a and b to match the following characteristics of a hanging cable The ends are 20 m apart and have a height of y  20 m The height in the middle is y  10 m  ln(  2)  10 , y  10 cosh  x  10 ln(  2)   Difficulty: Moderate Section: 2.9 Ans: a  10, b  144 Suppose that the vertical velocity v(t ) of a falling object of mass m  30 kg subject to gravity and air drag is given by  9.8k  9.8m v(t )    t  k  m  for some positive constant k Suppose k  0.5 and find the terminal velocity vT by computing lim v(t ) t  A) vT  –96.8 m/sec B) vT  –48.4 m/sec Ans: C Difficulty: Moderate C) D) Section: 2.9 vT  –24.2 m/sec vT  –12.1 m/sec 145 Determine if the function satisfies Rolle's Theorem on the given interval If so, find all values of c that make the conclusion of the theorem true f ( x)  36  x ,  –9, 9 A) x  B) x  36 C) x  –6, x  D) Rolle's Theorem not satisfied Ans: A Difficulty: Easy Section: 2.10 146 Using the Mean Value Theorem, find a value of c that makes the conclusion true for f ( x)  x3  5x , in the interval [1,1] A) c  1.129 B) One or more hypotheses fail Ans: C Difficulty: Easy Section: 2.10 Page 169 C) c  0.295 D) c  Chapter 147 Using the Mean Value Theorem, find a value of c that makes the conclusion true for    f ( x)  cos x,   ,   2 A) One or more hypotheses fail Ans: B Difficulty: Easy C) c  B) c   D) c  881 Section: 2.10 148 Prove that x3 + x –  has exactly one solution Ans: Let f  x   x + x – The function f(x) is continuous and differentiable everywhere Since f(0) < and f(1) > 0, f(x) must have at least one zero The derivative of f ( x)  x3 + x – is f ( x)  27 x + , which is always greater than zero Therefore f(x) can only have one zero Difficulty: Moderate Section: 2.10 149 Find all functions g such that g ( x)  f ( x) f ( x)  x g ( x)  24 x3 g ( x)  x 5 C) g ( x)  24 x3  C, for some constant C D) g ( x)  x5  C , for some constant C Ans: D Difficulty: Easy Section: 2.10 A) B) 150 Find all the functions g ( x) such that g ( x)  A) x8 B) g ( x)  – 10  c 5x Ans: D Difficulty: Moderate x9 C) g ( x)  – D) 12 25 x8 g ( x)  –  c 4x g ( x)  Section: 2.10 151 Find all the functions g ( x) such that g ( x)  – sin x g ( x)  – cos x  c A) B) g ( x)  cos x  c Ans: B Difficulty: Moderate g ( x)  cos x C) D) g ( x)  sin x  c Section: 2.10 Page 170 Chapter 152 Determine if the function f ( x)  x3 + x + is increasing, decreasing, or neither A) Increasing B) Decreasing C) Neither Ans: A Difficulty: Easy Section: 2.10 153 Determine if the function f ( x)  –5x – x + is increasing, decreasing, or neither A) Increasing B) Decreasing C) Neither Ans: C Difficulty: Easy Section: 2.10 154 Explain why it is not valid to use the Mean Value Theorem for the given function on the specified interval Show that there is no value of c that makes the conclusion of the theorem true ,  3, 5 f ( x)  x–4 Ans: The function is not continuous on the specified interval, so the Mean Value Theorem does not apply Note that f (3)  1 and f (5)  , so that f (5)  f (3)  ( 1)   (5)  (3) Also, f ( x )   ( x – 4) Since f ( x)  for all x in the domain of f, there is no value of c such that f (5)  f (3) f (c)  That is, there is no value of c such that f (c)  (5)  (3) Difficulty: Moderate Section: 2.10 Page 171 ... 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 hours... compares to the speed during the last hours Describe what is happening between and hours Ans: The fastest speed occurred during the last hours of the trip when the car traveled at about 70 mph The