Chapter 4 - Applications of the Derivative - 4.1 Maxima and Minima - 4.1 Exercises - Page 243: 32

$$\eqalign{ & \left( a \right){\text{Critical points }}x = \left\{ {\frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4},\frac{{7\pi }}{4}} \right\} \cr & \left( b \right)f\left( {\frac{\pi }{4}} \right) = f\left( {\frac{{5\pi }}{4}} \right) = \frac{1}{2}{\text{ Is the absolute maximum}}\,\left( {\,{\text{and local}}} \right) \cr & \,\,\,\,\,\,\,f\left( {\frac{{3\pi }}{4}} \right) = f\left( {\frac{{7\pi }}{4}} \right) = \frac{1}{2}{\text{ Is the absolute minimum}}\,\left( {\,{\text{and local}}} \right) \cr} $$

$$\eqalign{ & f\left( x \right) = \sin x\cos x,{\text{ on the interval }}\left[ {0,2\pi } \right] \cr & \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \sin x\left( { - \sin x} \right) + \cos x\left( {\cos x} \right) \cr & f'\left( x \right) = {\cos ^2}x - {\sin ^2}x \cr & {\text{Setting the derivative equal to zero}}{\text{, we have}} \cr & {\cos ^2}x - {\sin ^2}x = 0 \cr & \left( {\cos x + \sin x} \right)\left( {\cos x - \sin x} \right) = 0 \cr & \cos x = - \sin x,\,\,\,\,\cos x = \sin x \cr & {\text{Solving this equation for the interval }}\left[ {0,2\pi } \right]{\text{ gives the critical points}} \cr & x = \frac{\pi }{4},\,x = \frac{{3\pi }}{4}\,,\,\,\,x = \frac{{5\pi }}{4}\,,\,{\text{ and }}x = \frac{{7\pi }}{4} \cr & {\text{All of them lies in the given interval }}\left[ {0,2\pi } \right]. \cr & {\text{These points and the endpoints are candidates for the location}} \cr & {\text{of absolute extrema:}} \cr & {\text{Points }}x = \left\{ {0,\frac{\pi }{4},\frac{{3\pi }}{4},\frac{{5\pi }}{4},\frac{{7\pi }}{4},2\pi } \right\} \cr & \cr & {\text{Evaluating these points}} \cr & f\left( 0 \right) = \sin \left( 0 \right)\cos \left( 0 \right) = 0 \cr & f\left( {\frac{\pi }{4}} \right) = \sin \left( {\frac{\pi }{4}} \right)\cos \left( {\frac{\pi }{4}} \right) = \frac{1}{2} \cr & f\left( {\frac{{3\pi }}{4}} \right) = \sin \left( {\frac{{3\pi }}{4}} \right)\cos \left( {\frac{{3\pi }}{4}} \right) = - \frac{1}{2} \cr & f\left( {\frac{{5\pi }}{4}} \right) = \sin \left( {\frac{{5\pi }}{4}} \right)\cos \left( {\frac{{5\pi }}{4}} \right) = \frac{1}{2} \cr & f\left( {\frac{{7\pi }}{4}} \right) = \sin \left( {\frac{{7\pi }}{4}} \right)\cos \left( {\frac{{7\pi }}{4}} \right) = - \frac{1}{2} \cr & f\left( {2\pi } \right) = \sin \left( {2\pi } \right)\cos \left( {2\pi } \right) = 0 \cr & \cr & {\text{The largest of these function values is }}f\left( {\frac{\pi }{4}} \right) = f\left( {\frac{{5\pi }}{4}} \right) = \frac{1}{2},{\text{ }} \cr & {\text{which is the absolute maximum }}\left( {{\text{and local}}} \right){\text{ on the interval }}\left[ {0,2\pi } \right]. \cr & \cr & {\text{The smallest of these function values is }}f\left( {\frac{{3\pi }}{4}} \right) = f\left( {\frac{{7\pi }}{4}} \right) = - \frac{1}{2},{\text{ }} \cr & {\text{which is the absolute minimum}}\,\left( {{\text{and local}}} \right){\text{ on the interval }}\left[ {0,2\pi } \right]. \cr & \cr & {\text{The next graph shows that the critical points corresponds to neither}} \cr & {\text{a local maximum nor a local minimum}}{\text{.}} \cr} $$