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
f(c) = 0
f '(c) = 0
f (c) = 1
f '(c) = 1

Answer 1

$\bf \textit{mean value theorem}\\\\ f'(c)=\cfrac{f(b)-f(a)}{b-a}\qquad \begin{cases} a=-4\\ b=0 \end{cases}\implies f'(c)=\cfrac{f(0)-f(-4)}{0-(-4)} \\\\\\ f'(c)=\cfrac{9-5}{0+4}\implies f'(c)=\cfrac{4}{4}\implies f'(c)=1$