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 258: 3

Answer

$L\left( x \right) = \frac{1}{{12}}x + \frac{4}{3}$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \root 3 \of x ,{\text{ }}a = 8 \cr & {\text{Evaluate }}f\left( x \right){\text{ at }}a = 8 \cr & f\left( 8 \right) = \root 3 \of 8 \cr & f\left( 8 \right) = 2 \cr & {\text{Differentiating the given function}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\root 3 \of x } \right] \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^{1/3}}} \right] \cr & f'\left( x \right) = \frac{1}{3}{x^{1/3 - 1}} \cr & f'\left( x \right) = \frac{1}{3}{x^{ - 2/3}} \cr & f'\left( x \right) = \frac{1}{{3{x^{2/3}}}} \cr & {\text{Evaluate }}f'\left( x \right){\text{ at }}a = 8 \cr & f'\left( 8 \right) = \frac{1}{{3{{\left( 8 \right)}^{2/3}}}} \cr & f'\left( 8 \right) = \frac{1}{{12}} \cr & {\text{The linear approximation at }}f\left( a \right){\text{ is given by: }} \cr & L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right){\text{ }}\left( {\bf{1}} \right) \cr & {\text{Substituting }}f\left( 8 \right) = 2,{\text{ }}f'\left( 8 \right) = \frac{1}{{12}}{\text{ and }}a = 8{\text{ into }}\left( {\bf{1}} \right) \cr & L\left( x \right) = 2 + \frac{1}{{12}}\left( {x - 8} \right) \cr & {\text{Simplifying}} \cr & L\left( x \right) = 2 + \frac{1}{{12}}x - \frac{2}{3} \cr & L\left( x \right) = \frac{1}{{12}}x + \frac{4}{3} \cr} $$
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