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: 4

Answer

The function is continuous on $[0,8]$ $f(0) = 1$ $f(8) = 4$ However, the function does not satisfy the conclusion of the Mean Value Theorem on $[0,8]$

Work Step by Step

The function is continuous on $[0,8]$ $f(0) = 1$ $f(8) = 4$ However, there is no number $c$ in the interval $(0,8)$ such that $f'(c) = \frac{f(8)-f(0)}{8-0} = \frac{4-1}{8-0} = \frac{3}{8}$ Therefore, the function does not satisfy the conclusion of the Mean Value Theorem on $[0,8]$
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