Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.9 Local Linear Approximation; Differentials - Exercises Set 2.9 - Page 181: 8

Answer

$${\text{ }}\frac{1}{{1 + x}} \approx 1 - x$$

Work Step by Step

$$\eqalign{ & \frac{1}{{1 + x}} \approx 1 - x,{\text{ }}{x_0} = 0 \cr & {\text{Let }}f\left( x \right) = \frac{1}{{1 + x}} \cr & {\text{ }}f\left( {{x_0}} \right) = f\left( 0 \right) = \frac{1}{{1 + 0}} = 1 \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{{1 + x}}} \right] \cr & f'\left( x \right) = - \frac{1}{{{{\left( {1 + x} \right)}^2}}} \cr & f'\left( {{x_0}} \right) = - \frac{1}{{{{\left( {1 + 0} \right)}^2}}} = - 1 \cr & {\text{We can approximate values of }}f\left( x \right){\text{ by}} \cr & {\text{ }}f\left( x \right) \approx f\left( {{x_0}} \right) + f'\left( {{x_0}} \right)\left( {x - {x_0}} \right) \cr & {\text{then}} \cr & {\text{ }}\frac{1}{{1 + x}} \approx 1 + \left( { - 1} \right)\left( {x - 0} \right) \cr & {\text{ }}\frac{1}{{1 + x}} \approx 1 - x \cr} $$
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