2014 AP” CALCULUS AB FREE-RESPONSE QUESTIONS
x-2-2<x<-1-1-1<x<11<x<33 f(x) 12 Positive 8 Positive 2 Positive 7 f'(x)-5 Negative 0 Negative 0 Positive 1 2 g(x) -1 Negative 0 Positive 3 Positive 1 3 Positive Positive 0 Negative-2 2 g'(x) 2 5 The twice-differentiable functions f and gare defined for all real numbers x. Values of f, f’, g, and g’ for various values of x are given in the table above. (a) Find the x-coordinate of each relative minimum of f on the interval [-2,3]. Justify your answers. (b) Explain why there must be a value c, for-1 <c< 1, such that f” (c) = 0. (c) The function h is defined by h(x) = ln(f(x)). Find h’ (3). Show the computations that lead to your answer. (d) Evaluate [f'(g(x))g'(x)dx
The Correct Answer and Explanation is :
Problem Breakdown and Solution:
(a) Finding the x-coordinate of each relative minimum of ( f(x) ) on the interval ([-2, 3]):
A relative minimum occurs where ( f'(x) ) changes from negative to positive, indicating a change in the direction of ( f(x) ).
- From the table, observe ( f'(x) ):
- At ( x = -2 ), ( f'(x) = -5 ) (negative).
- At ( x = 0 ), ( f'(x) = 0 ) (stationary).
- At ( x = 1 ), ( f'(x) = 1 ) (positive).
- At ( x = 3 ), ( f'(x) = 2 ) (positive).
Since ( f'(x) ) changes from negative to positive at ( x = 0 ), ( x = 0 ) is the location of a relative minimum.
Answer: The relative minimum of ( f(x) ) occurs at ( x = 0 ).
(b) Explaining why there must be a value ( c ) such that ( f”(c) = 0 ) for (-1 < c < 1):
The function ( f(x) ) is twice differentiable (and thus continuous). Between (-1 < x < 1), ( f'(x) ) changes from ( -5 ) at ( x = -1 ) to ( 1 ) at ( x = 1 ). By the Mean Value Theorem, ( f'(x) ) must have an intermediate slope value of ( 0 ) within this interval. Since ( f”(x) ) represents the rate of change of ( f'(x) ), ( f”(c) = 0 ) for some ( c ) in ((-1, 1)).
(c) Finding ( h'(3) ), where ( h(x) = \ln(f(x)) ):
By the chain rule:
[
h'(x) = \frac{f'(x)}{f(x)}.
]
At ( x = 3 ), the table gives ( f'(3) = 2 ) and ( f(3) = 7 ). Therefore:
[
h'(3) = \frac{f'(3)}{f(3)} = \frac{2}{7}.
]
Answer: ( h'(3) = \frac{2}{7} ).
(d) Evaluating (\int f'(g(x)) g'(x) \, dx):
Let ( u = g(x) ). Then ( du = g'(x) dx ), and the integral becomes:
[
\int f'(g(x)) g'(x) \, dx = \int f'(u) \, du.
]
The integral of ( f'(u) ) is ( f(u) + C ), where ( u = g(x) ). Thus:
[
\int f'(g(x)) g'(x) \, dx = f(g(x)) + C.
]
Answer: ( f(g(x)) + C ).
Explanation (300 words):
In (a), relative extrema occur where ( f'(x) ) changes sign. Observing the sign change at ( x = 0 ), it indicates a relative minimum. For (b), the continuity and differentiability of ( f(x) ) ensure the conditions of the Mean Value Theorem are met. Since ( f'(x) ) transitions from negative to positive, ( f”(x) = 0 ) must occur within the interval. For (c), the logarithmic derivative rule, ( h'(x) = \frac{f'(x)}{f(x)} ), simplifies the calculation at ( x = 3 ) using the provided table values. Finally, in (d), the substitution ( u = g(x) ) transforms the integral, leveraging the fundamental theorem of calculus to yield ( f(u) + C ).