AP Calculus ABhardmcq1 pt

Let ff be a function continuous on the closed interval [2,5][-2, 5] and differentiable on the open interval (2,5)(-2, 5). If f(2)=1f(-2) = 1 and f(5)=22f(5) = 22, which of the following statements must be true?

A.There exists a value cc in (2,5)(-2, 5) such that f(c)=3f'(c) = 3.
B.There exists a value cc in (2,5)(-2, 5) such that f(c)=0f'(c) = 0.
C.The function ff must have a relative extremum on the open interval (2,5)(-2, 5).
D.There exists a value cc in (2,5)(-2, 5) such that f(c)=0f(c) = 0.

Explanation

Core Concept

Correct. By the Mean Value Theorem, since ff is continuous on [2,5][-2, 5] and differentiable on (2,5)(-2, 5), there must exist a cc in (2,5)(-2, 5) such that f(c)=f(5)f(2)5(2)=2217=3f'(c) = \frac{f(5) - f(-2)}{5 - (-2)} = \frac{22 - 1}{7} = 3.

Correct Answer

AThere exists a value cc in (2,5)(-2, 5) such that f(c)=3f'(c) = 3.

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