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: 4

Answer

$L\left( x \right) = - \sqrt 3 x + \frac{{\sqrt 3 \pi }}{6} + \frac{1}{2}$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \cos 2x,{\text{ }}a = \frac{\pi }{6} \cr & {\text{Evaluate }}f\left( x \right){\text{ at }}a = \frac{\pi }{6} \cr & f\left( {\frac{\pi }{6}} \right) = \cos 2\left( {\frac{\pi }{6}} \right) \cr & f\left( {\frac{\pi }{6}} \right) = \cos \left( {\frac{\pi }{3}} \right) \cr & f\left( {\frac{\pi }{6}} \right) = \frac{1}{2} \cr & {\text{Differentiating the given function}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\cos 2x} \right] \cr & f'\left( x \right) = - 2\sin 2x \cr & {\text{Evaluate }}f'\left( x \right){\text{ at }}a = \frac{\pi }{6} \cr & f'\left( {\frac{\pi }{6}} \right) = - 2\sin 2\left( {\frac{\pi }{6}} \right) \cr & f'\left( {\frac{\pi }{6}} \right) = - 2\sin \left( {\frac{\pi }{3}} \right) \cr & f'\left( {\frac{\pi }{6}} \right) = - \sqrt 3 \cr & {\text{The linear approximation at }}f\left( a \right){\text{ is given by: }} \cr & L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right){\text{ }}\left( {\bf{1}} \right) \cr & {\text{Substituting }}f\left( {\frac{\pi }{6}} \right) = \frac{1}{2},{\text{ }}f'\left( {\frac{\pi }{6}} \right) = - \sqrt 3 {\text{ and }}a = \frac{\pi }{6}{\text{ into }}\left( {\bf{1}} \right) \cr & L\left( x \right) = \frac{1}{2} - \sqrt 3 \left( {x - \frac{\pi }{6}} \right) \cr & {\text{Simplifying}} \cr & L\left( x \right) = \frac{1}{2} - \sqrt 3 x + \frac{{\sqrt 3 \pi }}{6} \cr & L\left( x \right) = - \sqrt 3 x + \frac{{\sqrt 3 \pi }}{6} + \frac{1}{2} \cr} $$
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