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: 16

Answer

On the graph, we can see that the tangent line at the point $x=0.8386$ is parallel to the secant line.

Work Step by Step

$f(x) = e^{-x}$ $f'(x) = -e^{-x}$ 1. $f$ is continuous on the interval $[0,2]$ 2. $f$ is differentiable on the interval $(0,2)$ $f$ satisfies the hypotheses of the Mean Value Theorem on the interval $[0,2]$ According to the Mean Value Theorem Theorem, there is a number $c$ in the interval $(0,2)$ such that $f'(c) = \frac{f(2)-f(0)}{2-0}$ Then: $f'(c) = \frac{f(2)-f(0)}{2-0}$ $-e^{-c} = \frac{e^{-2}-e^{-0}}{2-0}$ $e^{-c} = \frac{1-(1/e^2)}{2}$ $-c = ln\Big(\frac{1-(1/e^2)}{2}\Big)$ $c = -ln\Big(\frac{1-(1/e^2)}{2}\Big)$ $c \approx 0.8386$ On the graph, we can see that the tangent line at the point $x=0.8386$ is parallel to the secant line.
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