AP Calculus ABeasymcq1 pt

For which function does Rolle's Theorem guarantee a number cc in (2,2)(-2,2) with f(c)=0f'(c)=0?

A.f(x)=x3f(x)=x^3
B.f(x)=x2f(x)=x^2
C.f(x)=exf(x)=e^x
D.f(x)=1xf(x)=\dfrac{1}{x}

Explanation

Core Concept

f(x)=x2f(x)=x^2 is continuous on [2,2][-2,2] and differentiable on (2,2)(-2,2), and f(2)=f(2)=4f(-2)=f(2)=4. Rolle's Theorem therefore guarantees some cc in (2,2)(-2,2) with f(c)=0f'(c)=0.

Correct Answer

Bf(x)=x2f(x)=x^2

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