Madoreo7
contestada

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:
a. f(c) = 0
b. f '(c) = 0
c. f (c) = 1
d. f '(c) = 1

Respuesta :

Benjamin16
Mean value theorem

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

[tex]f'(c)=\dfrac{f(b)-f(a)}{b-a}[/tex]


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

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