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: 7

Answer

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

Work Step by Step

$f(x)={x}^2+2x $, $L(x)=f(a)+f'(a)(x-a)$ $f'(x)=2{x}+2$ $x_{0}=0$ $f(0)=0+0=0$ $f'(0)=0+2=2$ then $L(x)=0+2(x-0)=2x$ $L(x)=2x$ at $x=x_{0}$ Thus, the final answer is: $L(x)=2x$ at $x=x_{0}$
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