Calculus: Early Transcendentals (2nd Edition)

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

Chapter 4 - Applications of the Derivative - 4.1 Maxima and Minima - 4.1 Exercises - Page 243: 57

Answer

$$\eqalign{ & \left. a \right)x = - 0.9646,{\text{ }}x = 2.1769,{\text{ }}x = 5.3185 \cr & \left. b \right) \cr & {\text{Abs}}{\text{. max 3}}{\text{.7163 at }}x = 2.1769 \cr & {\text{Abs}}{\text{. min }} - {\text{32}}{\text{.7697 at }}x = 5.3185 \cr} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = {2^x}\sin x{\text{ on }}\left[ { - 2,6} \right] \cr & \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{2^x}\sin x} \right] \cr & f'\left( x \right) = {2^x}\left( {\cos x} \right) + \sin x\left( {{2^x}\ln 2} \right) \cr & f'\left( x \right) = {2^x}\left( {\cos x + \sin x\left( {\ln 2} \right)} \right) \cr & \cr & a.{\text{ Find the critical points}}{\text{, let }}f'\left( x \right) = 0 \cr & {2^x}\left( {\cos x + \sin x\left( {\ln 2} \right)} \right) = 0 \cr & \underbrace {{2^x} = 0}_{{\text{No solution}}}\,{\text{ or }}\cos x + \sin x\left( {\ln 2} \right) = 0 \cr & \cos x = - \sin x\left( {\ln 2} \right) \cr & {\text{Solve for }}x \cr & \frac{{\cos x}}{{\sin x}} = \ln \left( {\frac{1}{2}} \right) \cr & \cot x = \ln \left( {\frac{1}{2}} \right) \cr & x = {\cot ^{ - 1}}\left[ {\ln \left( {\frac{1}{2}} \right)} \right] \cr & {\text{Using a graphing utility we found that for the interval }}\left[ { - 2,6} \right] \cr & x = - 0.9646,{\text{ }}x = 2.1769,{\text{ }}x = 5.3185 \cr & \cr & \left( {\text{b}} \right){\text{Evaluate }}f\left( x \right){\text{ at the critical points and the endpoints }} \cr & f\left( { - 0.9646} \right) = {2^{ - 0.9646}}\sin \left( { - 0.9646} \right) \approx - 0.4211 \cr & f\left( {2.1769} \right) = {2^{2.1769}}\sin \left( {2.1769} \right) \approx 3.7163,{\text{ }}\left( {{\text{Largest}}} \right) \cr & f\left( {5.3185} \right) = {2^{5.3185}}\sin \left( {5.3185} \right) \approx - 32.7697,{\text{ }}\left( {{\text{Smallest}}} \right) \cr & {\text{Therefore}}{\text{,}} \cr & {\text{*The absolute maximum of }}f\left( x \right){\text{ is 3}}{\text{.7163 at }}x = 2.1769 \cr & {\text{*The absolute minimum of }}f\left( x \right){\text{ is }} - {\text{32}}{\text{.7697 at }}x = 5.3185 \cr & \cr & \left( {\text{c}} \right){\text{Graph}} \cr} $$
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