Chapter 4 - Applications of the Derivative - 4.1 Maxima and Minima - 4.1 Exercises - Page 243: 46

$$\eqalign{ & \left( {\text{a}} \right)x = - 1{\text{ and }}x = 1 \cr & \left( {\text{b}} \right){\text{absolute maximum of }}1{\text{ at }}x = 1 \cr & {\text{absolute minimum of }} - 1{\text{ at }}x = 1 \cr & \left( {\text{c}} \right){\text{ graph}} \cr} $$

$$\eqalign{ & f\left( x \right) = x\sqrt {2 - {x^2}} {\text{ on }}\left[ { - \sqrt 2 ,\sqrt 2 } \right] \cr & \cr & {\text{Differentiate}} \cr & f'\left( x \right) = \underbrace {\frac{d}{{dx}}\left[ {x\sqrt {2 - {x^2}} } \right]}_{{\text{use product rule}}} \cr & f'\left( x \right) = x\left( {\frac{{ - 2x}}{{2\sqrt {2 - {x^2}} }}} \right) + \sqrt {2 - {x^2}} \cr & f'\left( x \right) = \sqrt {2 - {x^2}} - \frac{{{x^2}}}{{\sqrt {2 - {x^2}} }} \cr & \cr & \left( a \right){\text{ Set the derivative equals to 0}} \cr & f'\left( x \right) = 0 \cr & \sqrt {2 - {x^2}} - \frac{{{x^2}}}{{\sqrt {2 - {x^2}} }} = 0 \cr & \sqrt {2 - {x^2}} = \frac{{{x^2}}}{{\sqrt {2 - {x^2}} }} \cr & {\text{Square both sides}} \cr & 2 - {x^2} = \frac{{{x^4}}}{{2 - {x^2}}} \cr & 4 - 4{x^2} + {x^4} = {x^4} \cr & 4 - 4{x^2} = 0 \cr & {x^2} = 1 \cr & {\text{Factoring}} \cr & x = - 1{\text{ and }}x = 1 \cr & {\text{And the derivative is not defined at }}x = \pm \sqrt 2 ,{\text{ but they are}} \cr & {\text{the endpoints}}{\text{, then the critical points are:}} \cr & x = - 1{\text{ and }}x = 1 \cr & \cr & \left( {\text{b}} \right){\text{Evaluate }}f\left( x \right){\text{ at the critical points and the endpoints }} \cr & f\left( { - \sqrt 2 } \right) = \left( { - \sqrt 2 } \right)\sqrt {2 - {{\left( { - \sqrt 2 } \right)}^2}} = 0 \cr & f\left( { - 1} \right) = \left( { - 1} \right)\sqrt {2 - {{\left( { - 1} \right)}^2}} = - 1,{\text{ }}\left( {{\text{Smallest}}} \right) \cr & f\left( 1 \right) = \left( 1 \right)\sqrt {2 - {{\left( 1 \right)}^2}} = 1,{\text{ }}\left( {{\text{Largest}}} \right) \cr & f\left( {\sqrt 2 } \right) = \left( {\sqrt 2 } \right)\sqrt {2 - {{\left( {\sqrt 2 } \right)}^2}} = 0 \cr & {\text{*The largest of these function values is }}f\left( 1 \right) = 1,{\text{ then}} \cr & {\text{the absolute maximum of }}f\left( x \right){\text{ on }}\left[ { - \sqrt 2 ,\sqrt 2 } \right]{\text{ is 1}} \cr & {\text{*The smallest of these function values is }}f\left( { - 1} \right) = - 1,{\text{ then}} \cr & {\text{the absolute minimum of }}f\left( x \right){\text{ on }}\left[ { - \sqrt 2 ,\sqrt 2 } \right]{\text{ is }} - {\text{1}} \cr & {\text{*From the graph we can see that the endpoints corresponds}} \cr & {\text{ to neither a local maximum nor a local minimum}}{\text{.}} \cr & \cr & \left( {\text{c}} \right){\text{Graph}} \cr} $$