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: 8


$L(x)=-x+2$ at $x=x_{0}$

Work Step by Step

$f(x)=\frac{1}{x} $, $L(x)=f(a)+f'(a)(x-a)$ $f'(x)=-\frac{1}{x^2}$ at $x_{0}=1$ the function and its derivative is close to the value 0.9 $x_{0}=1$ $f(1)=1$ $f'(1)=-1$ then $L(x)=1-1(x-1)=-x+2$ $L(x)=-x+2$ at $x=x_{0}$ Thus, the final answer is: $L(x)=-x+2$ at $x=x_{0}$
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