Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 3 - Applications of Differentiation - 3.3 Exercises - Page 183: 43

Answer

$$\eqalign{ & f\left( x \right){\text{ has a relative minimum at }}x = \frac{{5\pi }}{4} \cr & f\left( x \right){\text{ has a relative maximum at }}x = \frac{\pi }{4}{\text{ }} \cr} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \sin x + \cos x \cr & \left( {\text{a}} \right){\text{Calculating the first derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\sin x + \cos x} \right] \cr & f'\left( x \right) = \cos x - \sin x \cr & {\text{Calculating the critical points, set }}f'\left( x \right) = 0 \cr & f'\left( x \right) = \cos 2x \cr & \cos x - \sin x = 0 \cr & \cos x = \sin x \cr & {\text{On the interval }}\left( {0,2\pi } \right){\text{ }}\cos x = \sin x{\text{ for }}x = \frac{\pi }{4},\frac{{5\pi }}{4} \cr & \cr & \left( {\text{b}} \right){\text{Set the intervals: }} \cr & \left( {0,\frac{\pi }{4}} \right),\left( {\frac{\pi }{4},\frac{{5\pi }}{4}} \right),\left( {\frac{{5\pi }}{4},2\pi } \right) \cr & {\text{Making a table of values }}\left( {{\text{See examples on page 180 }}} \right) \cr} $$ \[\begin{array}{*{20}{c}} {{\text{Interval}}}&{{\text{Test Value}}}&{{\text{Sign of }}f'\left( x \right)}&{{\text{Conclusion}}} \\ {\left( {0,\frac{\pi }{4}} \right)}&{x = \frac{\pi }{6}}&{\frac{{ - 1 + \sqrt 3 }}{2} > 0}&{{\text{Increasing}}} \\ {\left( {\frac{\pi }{4},\frac{{5\pi }}{4}} \right)}&{x = \frac{\pi }{2}}&{ - 1 < 0}&{{\text{Decreasing}}} \\ {\left( {\frac{{5\pi }}{4},2\pi } \right)}&{x = \frac{{3\pi }}{2}}&{1 > 0}&{{\text{Increasing}}} \end{array}\] $$\eqalign{ & *f'\left( x \right){\text{ changes from negative to positive at }}x = \frac{{5\pi }}{4},{\text{ so }} \cr & f\left( x \right){\text{ has a relative minimum at }}x = \frac{{5\pi }}{4}{\text{ }} \cr & *f'\left( x \right){\text{ changes from positive to negative at }}x = \frac{\pi }{4},{\text{ so }} \cr & f\left( x \right){\text{ has a relative maximum at }}x = \frac{\pi }{4}{\text{ }} \cr & \cr & \left( {\text{c}} \right){\text{Graph}} \cr} $$
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