Calculus: Early Transcendentals 8th Edition

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

Chapter 4 - Review - Exercises - Page 360: 46

Answer

$9 \leq f(4) \leq 21$

Work Step by Step

1. The function $f$ is continuous on the interval on the closed interval $[0,4]$ 2. $f'(x)$ exists for all numbers in the interval $(0,4)$. Thus $f$ is differentiable on the open interval $(0,4)$ $f$ satisfies the hypotheses of the Mean Value Theorem on the interval $[0,4]$ Therefore, according to the Mean Value Theorem Theorem, there is a number $c$ in the interval $(0,4)$ such that $f'(c) = \frac{f(4)-f(0)}{4-0}$ It is given that $2 \leq f'(c) \leq 5$ Then: $2 \leq \frac{f(4)-f(0)}{4-0} \leq 5$ $2 \leq \frac{f(4)-f(0)}{4} \leq 5$ $8 \leq [f(4)-f(0)] \leq 20$ $8 \leq [f(4)-1] \leq 20$ $9 \leq f(4) \leq 21$
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