Chapter 3 - The Derivative In Graphing And Applications - 3.4 Absolute Maxima And Minima - Exercises Set 3.4 - Page 222: 13

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

$$\eqalign{ & f\left( x \right) = x - 2\sin x;\,\,\left[ { - \frac{\pi }{4},\frac{\pi }{2}} \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( { - \frac{\pi }{4},\frac{\pi }{2}} \right){\text{.}} \cr & \cr & {\text{Differentiating }}f\left( x \right) \cr & f'\left( x \right) = 1 - 2\cos x \cr & {\text{setting }}f'\left( x \right) = 0 \cr & 1 - 2\cos x = 0 \cr & \cos x = \frac{1}{2}\,\,\, \to \,\,\,x = \frac{\pi }{3} \cr & {\text{Thus}}{\text{, there are stationary points at }}x = \frac{\pi }{6}{\text{ and }}x = - \frac{\pi }{6} \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 points}}{\text{.}} \cr & f\left( { - \frac{\pi }{4}} \right) = - \frac{\pi }{4} - 2\sin \left( { - \frac{\pi }{4}} \right) = \sqrt 2 - \frac{\pi }{4} \cr & f\left( {\frac{\pi }{3}} \right) = \frac{\pi }{3} - 2\sin \left( {\frac{\pi }{3}} \right) = \frac{\pi }{3} - \sqrt 3 \cr & f\left( {\frac{\pi }{2}} \right) = \frac{\pi }{2} - 2\sin \left( {\frac{\pi }{2}} \right) = \frac{\pi }{2} - 2 \cr & {\text{Then we can conclude that the absolute minimum of }}f\left( x \right){\text{ on}} \cr & {\text{the }}\,{\text{interval }}\,\,\left[ { - \frac{\pi }{4},\frac{\pi }{2}} \right]{\text{ is}}\,{\text{ }}\frac{\pi }{3} - \sqrt 3 {\text{ at }}x = \frac{\pi }{3}.{\text{ }}\,{\text{And}}\,{\text{ the }}\,{\text{absolute}} \cr & {\text{maximum is }}\sqrt 2 - \frac{\pi }{4}{\text{ at }}x = - \frac{\pi }{4} \cr & \cr & {\text{max}} = \sqrt 2 - \frac{\pi }{4}{\text{ at }}x = - \frac{\pi }{4} \cr & {\text{min}} = \frac{\pi }{3} - \sqrt 3 {\text{ at }}x = \frac{\pi }{3} \cr} $$