Answer
See the solution step.
Work Step by Step
Using the mean value theorem, there is $c \in (2,8)$ such that:
$$f'(c)=\frac{f(8)-f(2)}{8-2}$$
Since $3\leq f'(c) \leq 5$ it follows:
$$3\leq \frac{f(8)-f(2)}{8-2} \leq 5$$
$$3\leq \frac{f(8)-f(2)}{6} \leq 5$$
$$6\cdot 3\leq 6\cdot \frac{f(8)-f(2)}{6} \leq 6\cdot 5$$
$$18\leq f(8)-f(2) \leq 30$$
