Chapter 4 - Applications of the Derivative - 4.2 What Derivatives Tell Us - 4.2 Exercises - Page 257: 41

$$\eqalign{ & \left( a \right){\text{Critical points }}x = - \sqrt 2 {\text{ and }}x = \sqrt 2 \cr & \left( b \right){\text{Local min at }}x = - \sqrt 2 ;\,\,\,\,\,\,{\text{Local max at }}x = \sqrt 2 \cr & \left( c \right){\text{Absolute min:}}\, - \,2{\text{ at }}x = - \sqrt 2 \cr & \,\,\,\,\,\,{\text{Absolute max:}}\,\,2{\text{ at }}x = \sqrt 2 \cr} $$

$$\eqalign{ & f\left( x \right) = x\sqrt {4 - {x^2}} {\text{ on the interval }}\left[ { - 2,2} \right] \cr & {\text{Derivative}} \cr & f'\left( x \right) = x\left( {\frac{{ - 2x}}{{2\sqrt {4 - {x^2}} }}} \right) + \sqrt {4 - {x^2}} \left( 1 \right) \cr & f'\left( x \right) = \frac{{ - {x^2}}}{{\sqrt {4 - {x^2}} }} + \sqrt {4 - {x^2}} \cr & {\text{Set the derivative to 0}} \cr & \frac{{ - {x^2}}}{{\sqrt {4 - {x^2}} }} + \sqrt {4 - {x^2}} = 0 \cr & \frac{{ - {x^2} + 4 - {x^2}}}{{\sqrt {4 - {x^2}} }} = 0 \cr & \frac{{4 - 2{x^2}}}{{\sqrt {4 - {x^2}} }} = 0 \cr & 4 - 2{x^2} = 0 \cr & {x^2} = 2 \cr & {\text{Solving the equation we obtain}} \cr & x = - \sqrt 2 {\text{ and }}x = \sqrt 2 \cr & \cr & {\text{Evaluating that point and the endpoints we obtain}} \cr & f\left( { - 2} \right) = \left( { - 2} \right)\sqrt {4 - {{\left( { - 2} \right)}^2}} = 0 \cr & f\left( { - \sqrt 2 } \right) = \left( { - \sqrt 2 } \right)\sqrt {4 - {{\left( { - \sqrt 2 } \right)}^2}} = - 2 \cr & f\left( {\sqrt 2 } \right) = \left( {\sqrt 2 } \right)\sqrt {4 - {{\left( {\sqrt 2 } \right)}^2}} = 2 \cr & f\left( 2 \right) = \left( 2 \right)\sqrt {4 - {{\left( 2 \right)}^2}} = 0 \cr & {\text{Then}}{\text{, }} \cr & {\text{The largest result is }}2{\text{ at }}x = \sqrt 2 \,\,\,\left( {{\text{Absolute maximum}}} \right){\text{ and local}} \cr & {\text{The smallest result is }} - {\text{2 at }}x = - \sqrt 2 \,\,\left( {{\text{Absolute minimum}}} \right){\text{ and local}} \cr & \cr & {\text{We can conclude that}} \cr & \left( a \right){\text{Critical points }}x = - \sqrt 2 {\text{ and }}x = \sqrt 2 \cr & \left( b \right){\text{Local min at }}x = - \sqrt 2 ;\,\,\,\,\,\,{\text{Local max at }}x = \sqrt 2 \cr & \left( c \right){\text{Absolute min:}}\, - \,2{\text{ at }}x = - \sqrt 2 \cr & \,\,\,\,\,\,{\text{Absolute max:}}\,\,2{\text{ at }}x = \sqrt 2 \cr} $$