Calculus, 10th Edition (Anton)

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

Chapter 3 - The Derivative In Graphing And Applications - 3.4 Absolute Maxima And Minima - Exercises Set 3.4 - Page 222: 14

Answer

$$\eqalign{ & {\text{max}} = \sqrt 2 {\text{ at }}x = \frac{{3\pi }}{4} \cr & {\text{min}} = - 1{\text{ at }}x = 0 \cr} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \sin x - \cos x;\,\,\left[ {0,\pi } \right] \cr & f\left( x \right){\text{ is }}\,{\text{continuous }}\,{\text{and}}\,{\text{differentiable everywhere}}{\text{, the }} \cr & {\text{absolute}}\,{\text{ extrema }}\,{\text{must }}\,{\text{occur}}\,{\text{ }}\,{\text{either }}\,{\text{at }}\,{\text{endpoints }}\,{\text{of }} \cr & {\text{the interval or at solutions to the equation }}f'\left( x \right) = 0{\text{ in}} \cr & {\text{the open interval }}\left( {0,\pi } \right){\text{.}} \cr & \cr & {\text{Differentiating }}f\left( x \right) \cr & f'\left( x \right) = \cos x + \sin x \cr & {\text{setting }}f'\left( x \right) = 0 \cr & \cos x + \sin x = 0 \cr & {\text{For the interval }}\left[ {0,\pi } \right]{\text{ we obtain }} \cr & x = \frac{{3\pi }}{4} \cr & {\text{Thus}}{\text{, there is a stationary points at }}x = \frac{{3\pi }}{4} \cr & {\text{Evaluating }}f\left( x \right){\text{ at the endpoints}}{\text{, }}x = - \frac{\pi }{4},{\text{ }}x = \frac{\pi }{2}{\text{ and the}} \cr & {\text{stationary point}}{\text{.}} \cr & f\left( 0 \right) = \sin \left( 0 \right) - \cos \left( 0 \right) = - 1 \cr & f\left( {\frac{{3\pi }}{4}} \right) = \sin \left( {\frac{{3\pi }}{4}} \right) - \cos \left( {\frac{{3\pi }}{4}} \right) = \sqrt 2 \cr & f\left( \pi \right) = \sin \left( \pi \right) - \cos \left( \pi \right) = 1 \cr & {\text{Then we can conclude that the absolute minimum of }}f\left( x \right){\text{ on}} \cr & {\text{the }}\,{\text{interval }}\,\,\left[ {0,\pi } \right]{\text{ is}}\,{\text{ }} - 1{\text{ at }}x = 0.{\text{ }}\,{\text{And}}\,{\text{ the }}\,{\text{absolute}} \cr & {\text{maximum is }}\sqrt 2 {\text{ at }}x = \frac{{3\pi }}{4} \cr & \cr & {\text{max}} = \sqrt 2 {\text{ at }}x = \frac{{3\pi }}{4} \cr & {\text{min}} = - 1{\text{ at }}x = 0 \cr} $$
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