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: 10


$L(x)=x+1$ at $x=x_{0}=8$

Work Step by Step

$f(x)=x+1 $, $L(x)=f(a)+f'(a)(x-a)$ $f'(x)=1$ at $x_{0}=8$ the function and its derivative is close to the value 8.1 so $x_{0}=8$ $f(8)=8+1=9$ $f'(8)=1=1$ then $L(x)=9+(x-8)=x+1$ $L(x)=x+1$ at $x=x_{0}$ Thus, the final answer is: $L(x)=x+1$ at $x=x_{0}$
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