Answer
$9 \leq f(4) \leq 21$
Work Step by Step
1. The function $f$ is continuous on the interval on the closed interval $[0,4]$
2. $f'(x)$ exists for all numbers in the interval $(0,4)$.
Thus $f$ is differentiable on the open interval $(0,4)$
$f$ satisfies the hypotheses of the Mean Value Theorem on the interval $[0,4]$
Therefore, according to the Mean Value Theorem Theorem, there is a number $c$ in the interval $(0,4)$ such that $f'(c) = \frac{f(4)-f(0)}{4-0}$
It is given that $2 \leq f'(c) \leq 5$
Then:
$2 \leq \frac{f(4)-f(0)}{4-0} \leq 5$
$2 \leq \frac{f(4)-f(0)}{4} \leq 5$
$8 \leq [f(4)-f(0)] \leq 20$
$8 \leq [f(4)-1] \leq 20$
$9 \leq f(4) \leq 21$