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: 1



Work Step by Step

Given: $f(x)=x^3-2x+3$ L(x)=f(a)+f'(a)(x-a) $f'(x)=3x^2-2$ a=2 $f(2)=2^3-2\times2+3=7$ $f'(2)=3\times2^2-2=10$ then $L(x)=7+10(x-2)$ $L(x)=7+10x-20=10x-13$ $L(x)=10x-13$ Thus, the final answer: $L(x)=10x-13$
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