Single Variable Calculus, Chapter 4, 4.2, Section 4.2, Problem 24

Show that $18 \leq f(8) - f(2) \leq 30$. Suppose that $3 \leq f'(x) \leq 5$ for all values of $x$.
If $3 \leq f'(x) \leq 5$ for all values of $x$, then

By the definition of Mean Value Theorem...
$\displaystyle f'(c) = \frac{f(8) - f(2)}{8-2}$
For $c$ that is in the closed interval $[2,8]$.

We can say that $f$ is differentiable in the given range and continuous on $[2,8]$ because $f$ is differentiable everywhere. Since $\displaystyle f'(c) = \frac{f(8)-f(2)}{6}$, it means that the Mean Value Theorem is satisfied.

Notice that, $3 \leq f'(c) \leq 5$ implies $6 \times 3 \leq 6 f'(c) \leq 6 \times 5$ which equals $18 \leq f(8) - f(2) \leq 30$