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Madoreo7
Answered

If f(x) is differentiable for the closed interval [-4, 0] such that f(-4) = 5 and f(0) = 9, then there exists a value c, -4 < c < 0 such that:

Af(c) = 0
Bf '(c) = 0
Cf (c) = 1
Df '(c) = 1

Verified Answer :

Benjamin16

Mean value theorem

For \(f:[a,b] \rightarrow \mathbb{R}\) exist \(c \in (a,b)\) such that 

\(f'(c)=\dfrac{f(b)-f(a)}{b-a}\)


\(f'(c)=\dfrac{f(-4)-f(0)}{-4-0}=\dfrac{5-9}{-4}=\dfrac{-4}{-4}=\boxed{1}\)

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