Accelerated Calculus
Chapter 4a Sample Test (SOLUTIONS): Derivative Tests, Optimization, and Related Rates
1. Let h(x) be a function defined for all x ≠ 0 such that h(4) = –3 and the derivative of h is given
x2  2
by h / (x) 
for all x ≠ 0.
x
a. Find all values of x for which the graph of h(x) has a horizontal tangent, and determine
whether h(x) has a local maximum, a local minimum, or neither at each of these values.
Justify your answers.

x2  2
2
critical points: h / ( x ) 
 x   0  x   2 and x = 0 (not in domain)
x
x
2
h( x )  1  2  0 for all x.
x
So, x = 2 and x =  2 are both minimums of h(x) with horizontal tangents.
b. On what intervals, if any, is the graph of h(x) concave up? Justify your answer.
2
h( x )  1  2  0 for all x, so h(x) is concave up on its entire domain: ( ,0)  (0, ) .
x
c. Write an equation for the line tangent to the graph of h(x) at x = 4.
2
h(4)  3 and h(4)  4   3.5 , so y  3  3.5( x  4) is the tangent line.
4
d. Does the line tangent to the graph of h(x) at x = 4 lie above or below the graph of h for
x > 4? Why?
Since h(x) is concave up on its entire domain, tangent line will lie under the graph of h(x).
2. Consider the function f (x)  xekx , where k  0.
a. Find all critical points of f. Determine whether the graph of f has a local maximum,
minimum, or something else at each critical point.
1
f ( x)  e kx  kxe kx  e kx (1 kx)  0 when x  .
k
 kx
 k (1/ k )
f ( x )  ke (2  kx ) ; f (1 / k )  ke
(2  k (1 / k )  k / e , which is negative, so
1
x  is a maximum.
k
b. Find all inflection points of f.
f ( x )  ke kx (2  kx)  0  x  2 / k .
c. Use your results from parts a and b to identify any local extrema or inflection points of
the graph of f (x)  xe6x .
x = 1/6 is a minimum and 2/6 is an inflection point.

Accelerated Calculus
Chapter 4a Sample Test (SOLUTIONS): Derivative Tests, Optimization, and Related Rates
3. Consider the graph of g ( x ) , given at right. Determine the location of all relative extrema
and inflection points of g on the interval [−4, 4].
Local Maximum: x = 4
Local Minimum: x = −4
Inflection Points: x = −3, −1, 1, 3.
4. For what value(s) of x does the tangent line to the curve y  1  40 x3  3x5 have the largest
slope? All conclusions must be fully justified using calculus-based techniques.
The slope of the tangent line is given by m  120 x 2  15 x 4 . m achieves a maximum when
m  240 x  60 x 3  0  x  0 or x  4 .
m( x)  240  180 x 2 ; m(0)  240 (min) and m(4)  2640 (max).
So, at x  4 the slope of the tangent to y  1  40 x3  3x5 is greatest.
5. A 13-foot ladder is leaning against a wall when its base starts to slide away. By the time the
base is 12 feet from the house, the base is moving at the rate of 5 ft/sec.
a. How fast is the top of the ladder sliding down the wall at that moment when the base is
12 feet from the house?
x 2  y 2  132
dx
dy
dy
dy
2x  2 y
 0 ; 2(12)(5)  2(5)
0
 12
dt
dt
dt
dt
R
ft/sec.
Therefore, the top of the ladder is sliding down the wall at
a rate of 12 ft/sec when the base is 12 ft from the house.
Wall
Ladder
Ground
b. At what rate is the area of the triangular region R changing at that moment?
1
A  xy
2
dA 1 dy 1 dx dA 1
1
 x  y ;
 (12)( 12)  (5)(5)  59.5 ft 2 / sec
dt 2 dt 2 dt
dt 2
2
Therefore, the area is decreasing at a rate of 59.5 square ft/sec when the base is 12 ft from
the house.
Accelerated Calculus
Chapter 4a Sample Test (SOLUTIONS): Derivative Tests, Optimization, and Related Rates
c. Make an accurate graph of Area of R versus Distance from wall for distances between 0 and
13 feet.
1
A  x 169  x 2
2
d. Where is the area of R greatest? Justify your response.
x ( 2 x )
A  169  x 2 
0
2 169  x 2
84.5  x 2 , x  84.5  9.192
6. For smooth (i.e. continuous and differentiable on its domain) functions, determine whether
each of the following statements are True or False.
a. If a(2)  0 and a(2)  3 , then a( x) has a local maximum at x = 2.
False. It would be a minimum.
b. If b( x ) has an inflection point at x = 0, then b( x ) +2 also has an inflection point at x = 0.
True
c. If d ( x ) has a local minimum at x = −2, then d ( x)  0 .
False. d ( 2) could be 0 and still be a minimum.
Accelerated Calculus
Chapter 4a Sample Test (SOLUTIONS): Derivative Tests, Optimization, and Related Rates
7. Ship A is traveling due west toward Lighthouse Rock at a speed of 15 km/hr. Ship B is
traveling due north away from Lighthouse Rock at a speed of 10 km/hr. Let x be the distance
between Ship A and Lighthouse Rock at time t, and let y be the distance between Ship B and
Lighthouse Rock at time t, as shown in the figure below.
a. Find the distance, in km, between Ship A and Ship B when x = 4 and y = 3.
Distance between Ship A and Ship B is 5 km when x = 4 and y = 3.
b. Find the rate of change, in km/hr, of the distance
between the two ships when x = 4 and y = 3.
x2  y2  d 2
dx
dy
dd
2x  2 y
 2d
dt
dt
dt
dd
dd
 6 km/hr
;
dt
dt
Therefore, the distance is decreasing at a rate of 6km/hr when x = 4 and y = 3.
2(4)( 15)  2(3)(10)  2(5)
c. Let θ be the angle shown in the diagram below. Find the rate of change of θ, in radians
per hour, when x = 4 and y = 3.
y
dd
d dy
 sin   d cos  

, so d  sin( )  y and
;
d
dt
dt dt
3
4 d
d 17
( 6)   5  
 10 

radians/hr
5
5 dt
dt
5
Therefore, the angle shown is increasing at a rate of 17/5 radians/hr.
sin( ) 