Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.10 - Linear Approximations and Differentials. - 3.10 Exercises: 1

Answer

$L(x) = 16x + 23$

Work Step by Step

$f(x) = x^3 - x^2 + 3$ , $a =-2$ $f(a) = (-2)^3 - (-2)^2 + 3$ $f(a) = -9$ $f'(x) = 3x^2 - 2x$ $f'(-2) = 3(-2)^2 - 2(-2)$ $f'(-2) = 16$ $L(x) = f(a) + f'(a)(x-a)$ $L(x) = -9 + 16(x-(-2))$ $L(x) = -9 + 16(x+2)$ $L(x) = -9 + 16x+32$ $L(x) = 16x + 23$
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