Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.5 Linear Approximation and Differentials - 4.5 Exercises - Page 289: 55

Answer

$$\eqalign{ & \left( a \right)L\left( x \right) = 1 - x \cr & \left( b \right){\text{graph}} \cr & \left( c \right)0.9 \cr & \left( d \right)1\% {\text{ error}} \cr} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \frac{1}{{x + 1}};{\text{ }}a = 0;{\text{ }}\frac{1}{{1.1}} \cr & {\text{Differentiate }}f\left( x \right) \cr & f\left( x \right) = {\left( {x + 1} \right)^{ - 1}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\left( {x + 1} \right)}^{ - 1}}} \right] \cr & f'\left( x \right) = - {\left( {x + 1} \right)^{ - 2}} \cr & f'\left( x \right) = - \frac{1}{{{{\left( {x + 1} \right)}^2}}} \cr & {\text{Evaluate }}f\left( x \right){\text{ and }}f'\left( x \right){\text{ at }}a = 0 \cr & f\left( 0 \right) = \frac{1}{{0 + 1}} = 1 \cr & f'\left( 0 \right) = - \frac{1}{{{{\left( {0 + 1} \right)}^2}}} = - 1 \cr & \cr & \left( a \right){\text{Use the linear approximation formula }}\left( {{\text{See page 287}}} \right) \cr & f\left( x \right) = L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right){\text{ }}\left( {\bf{1}} \right) \cr & {\text{Substitute }}f\left( a \right){\text{ and }}f'\left( a \right){\text{ into }}\left( {\bf{1}} \right) \cr & L\left( x \right) = 1 - 1\left( {x - 0} \right) \cr & L\left( x \right) = 1 - x \cr & \cr & \left( b \right){\text{The graph of the function and the linear approximation }} \cr & {\text{at }}x = 0{\text{ is shown below}}{\text{.}} \cr & \cr & \left( c \right){\text{ Estimating the given value function at }}\frac{1}{{1.1}} \cr & f\left( x \right) = \frac{1}{{x + 1}} \to \frac{1}{{1.1}} = \frac{1}{{0.1 + 1}} \to x = 0.1 \cr & L\left( x \right) = 1 - x \cr & L\left( {0.1} \right) = 1 - 0.1 \cr & L\left( {0.1} \right) = 0.9 \cr & \cr & {\text{Therefore}}{\text{,}} \cr & \frac{1}{{1.1}} \approx L\left( {0.1} \right) \cr & \frac{1}{{1.1}} \approx 0.9 \cr & \cr & \left( d \right){\text{ The percent error is:}} \cr & \frac{{\left| {{\text{approximation}} - {\text{exact}}} \right|}}{{{\text{exact}}}} \times 100\% \cr & {\text{The exact value given by a calculator is }} \cr & \frac{1}{{1.1}} \approx 0.\overline {90} \cr & \cr & \frac{{\left| {0.9 - 0.\overline {90} } \right|}}{{0.\overline {90} }} \times 100\% \approx 1\% {\text{ error}} \cr} $$
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