AP Calculus
AP problem #6
All work must be completed on a separate sheet of paper. Calculators are not allowed.
1. Let f  x  be the function defined by f  x   k  12 x  3x2  2 x3 where k is a constant.
a)
b)
c)
d)
On what interval is the function increasing. Justify your answer.
If the relative maximum value of f is 4, what is the value of k?
Find the interval where the function is concave up. Justify your answer.
Find the relative minimum value of the function.
2. The figure below shows the graph of f  , the derivative of f. The domain of f is 4  x  4 . The
derivative of f is an even function and f   3  f   3  0 .
a) For what value of x in the interval 4  x  4 does f have a relative maximum? Justify your
answer.
b) For what value of x in the interval 4  x  4 does f have a relative minimum? Justify your
answer.
c) For what values of x is the graph of f concave downward? Use f  to justify your answer.
3. Consider the relation defined by the equation tan y  x  y for x in the open interval 2  x  2 .
dy
2
  tan y  .
dx
b) Find the x and y coordinate of each point where the tangent line to the graph is vertical.
d2y
c) Find
in terms of y.
dx 2
a) Show that
AP #6 Key (25 points)
1. a) f   x  =12 + 6x – 6x2 = -6(x – 2)(x + 1) critical values at 2 and -1.
f  x  is increasing from (-1, 2) b/c f   x  > 0 (4 pts.)
b) max is at x = 2, so 4 = k + 12(2) + 3(2)2 – 2(2)3
k = -16 (2 pts.)

c) f  x  =6 – 12x, inflection point when x = 1/2.
f  x  is concave up on  ,1/ 2)  b/c f   x  >0 (4 pts.)
d) min is at x = -1, so -16 – 12 + 3(-1)2 – 2(-1)3 = -23 minimum value is -23 (2 pts.)
2. a) f  x  has a relative maximum at x = 3 b/c f   x  changes from positive to negative. (2 pts.)
b) f  x  has a relative minimum at x = -3 b/c f   x  changes from negative to positive. (2 pts.)
c) (-2, 0) and (2, 4) b/c f   x  < 0 and/or f   x  is decreasing. (2 pts.)
3. a) sec2 y

b) tan y  0 at 0,  , and   , points:
2
c)

dy
dy
dy dy
dy
dy
1
 1
1,
sec 2 y  1  1 ,
, sec2 y 

 tan 2 y (2 pts.)
2
dx
dx
dx dx
dx
dx
sec y  1
d2y
dy
3
,
 2  tan y   sec2 y
2
dx
dx

 0, 0  ,  ,   , and   ,  
 2  tan y   sec 2 y  tan y  =
3
2

(3 pts.)
2sec2 y
(2 pts.)
tan 5 y
AP #6 Key (25 points)
1. a) f   x  =12 + 6x – 6x2 = -6(x – 2)(x + 1) critical values at 2 and -1.
f  x  is increasing from (-1, 2) b/c f   x  > 0 (4 pts.)
b) max is at x = 2, so 4 = k + 12(2) + 3(2)2 – 2(2)3
k = -16 (2 pts.)
c) f   x  =6 – 12x, inflection point when x = 1/2.
f  x  is concave up on  ,1/ 2)  b/c f   x  >0 (4 pts.)
d) min is at x = -1, so -16 – 12 + 3(-1)2 – 2(-1)3 = -23 minimum value is -23 (2 pts.)
2. a) f  x  has a relative maximum at x = 3 b/c f   x  changes from positive to negative. (2 pts.)
b) f  x  has a relative minimum at x = -3 b/c f   x  changes from negative to positive. (2 pts.)
c) (-2, 0) and (2, 4) b/c f   x  < 0 and/or f   x  is decreasing. (2 pts.)
3. a) sec2 y

b) tan 2 y  0 at 0,  , and   , points:
c)

dy
dy
dy dy
dy
dy
1
 1
1,
sec 2 y  1  1 ,
, sec2 y 

 tan 2 y (2 pts.)
2
dx
dx
dx dx
dx
dx
sec y  1
d2y
dy
3
 2  tan y   sec2 y
,
2
dx
dx

 0, 0  ,  ,   , and   ,  
 2  tan y   sec 2 y  tan y  =
3
2

(3 pts.)
2sec2 y
(2 pts.)
tan 5 y