## Calculus: Early Transcendentals 8th Edition

$18 \leq f(8) - f(2) \leq 30$
Using the Mean Value Theorem, we have: $f'(c) = \frac{f(8) - f(2)}{8 - 2}$ $f'(c) = \frac{f(8) - f(2)}{6}$ $3 \leq f'(c) \leq 5$ $3 \leq \frac{f(8) - f(2)}{6} \leq 5$ Multiply each side by $6$: $(6)3 \leq (6) \frac{f(8) - f(2)}{6} \leq 5 (6)$ $(6)3 \leq f(8) - f(2) \leq 5 (6)$ $18 \leq f(8) - f(2) \leq 30$