Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 3 - Section 3.10 - Linear Approximations and Differentials. - 3.10 Exercises - Page 259: 8

Answer

$(1+x)^{-3} \approx 1-3x$ On the graph, we can see that the linear approximation is accurate to within 0.1 when $-0.1 \leq x \leq 0.1$

Work Step by Step

$f(x) = (1+x)^{-3}$ $f'(x) = -3~(1+x)^{-4}$ When $x = 0$: $f'(x) = -3~(1+x)^{-4}$ $f'(0) = -3~(1+0)^{-4}$ $f'(0) = -3$ We can find the linear approximation at $a=0$: $f(x) \approx f(a)+f'(a)(x-a)$ $f(x) \approx f(0)+f'(0)(x-0)$ $f(x) \approx 1+(-3)(x)$ $f(x) \approx 1-3x$ $(1+x)^{-3} \approx 1-3x$ On the graph, we can see that the linear approximation is accurate to within 0.1 when $-0.1 \leq x \leq 0.1$ We can verify this: When $x = -0.1$: $1-3x = 1-(3)(-0.1) = 1.3$ $(1+x)^{-3} = (1-0.1)^{-3} = 1.37$ When $x = 0.1$: $1-3x = 1-(3)(0.1) = 0.7$ $(1+x)^{-3} = (1+0.1)^{-3} = 0.75$
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