Chapter 4 - Applications of the Derivative - Review Exercises - Page 330: 10

$$\eqalign{ & {\text{Absolute maximum: }}\left( { - 1,\frac{\pi }{2}} \right){\text{ and }}\left( {1,\frac{\pi }{2}} \right) \cr & {\text{Absolute minimum: }}\left( {0,0} \right) \cr} $$

$$\eqalign{ & g\left( x \right) = x{\sin ^{ - 1}}x,{\text{ on }}\left[ { - 1,1} \right] \cr & {\text{Differentiate}} \cr & f'\left( x \right) = x{\sin ^{ - 1}}x \cr & f'\left( x \right) = x\left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right) + {\sin ^{ - 1}}x\left( 1 \right) \cr & f'\left( x \right) = \frac{x}{{\sqrt {1 - {x^2}} }} + {\sin ^{ - 1}}x \cr & {\text{Calculate the critial points, set }}f'\left( x \right) = 0 \cr & \frac{x}{{\sqrt {1 - {x^2}} }} + {\sin ^{ - 1}}x = 0 \cr & {\text{For }}x = 0 \cr & \frac{0}{{\sqrt {1 - {x^2}} }} + {\sin ^{ - 1}}\left( 0 \right) = 0 \cr & {\text{Therefore}} \cr & {\text{We have the critical point}} \cr & x = 0 \cr & {\text{Evaluating }}f{\text{ at }}x = 0{\text{ and the endpoints, we obtain}} \cr & f\left( { - 1} \right) = \left( { - 1} \right){\sin ^{ - 1}}\left( { - 1} \right) = \frac{1}{2}\pi \cr & f\left( 0 \right) = \left( 0 \right){\sin ^{ - 1}}\left( 0 \right) = 0 \cr & f\left( 1 \right) = \left( 1 \right){\sin ^{ - 1}}\left( 1 \right) = \frac{1}{2}\pi \cr & {\text{The largest of those function values are:}} \cr & f\left( { - 1} \right) = f\left( 1 \right) = \frac{1}{2}\pi \cr & {\text{Which are absolute maximum on }}\left( { - 1,\frac{\pi }{2}} \right),{\text{ and }}\left( {1,\frac{\pi }{2}} \right) \cr & {\text{The smallest of those function values are:}} \cr & f\left( 0 \right) = 0 \cr & {\text{Which are absolute minimum on }}\left[ { - 1,1} \right] \cr & \cr & {\text{Absolute maximum: }}\left( { - 1,\frac{\pi }{2}} \right){\text{ and }}\left( {1,\frac{\pi }{2}} \right) \cr & {\text{Absolute minimum: }}\left( {0,0} \right) \cr & \cr & {\text{Graph}} \cr} $$