Question

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

Answer 1

Mean value theorem

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

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


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