Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.2 - The Mean Value Theorem - 4.2 Exercises - Page 292: 26

Answer

$18 \leq f(8) - f(2) \leq 30$

Work Step by Step

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 $6$: $(6)3 \leq (6) \frac{f(8) - f(2)}{6} \leq 5 (6)$ $(6)3 \leq f(8) - f(2) \leq 5 (6)$ $18 \leq f(8) - f(2) \leq 30$
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