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
`</span>c \in (a,b)` such that


`f'(c)=(f(b)-f(a))/(b-a)`



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