University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.11 - Linearization and Differentials - Exercises - Page 198: 13



Work Step by Step

$f(x)=e^{-x}$, $L(x)=f(a)+f'(a)(x-a)$ $f'(x)=-e^{-x}$ $x_{0}=0$ $f(0)={e^0}=1$ $f'(0)=-e^{-0}=-1$ then $L(x)=1-1(x-0)=-x+1$ $L(x)=-x+1$ Thus, the final answer is: $L(x)=-x+1$
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