Chapter 4 - Applications of the Derivative - 4.2 What Derivatives Tell Us - 4.2 Exercises - Page 257: 43

$$\eqalign{ & \left( a \right){\text{Critical points }}x = - \sqrt 3 {\text{ and }}x = \sqrt 3 \cr & \left( b \right){\text{Local min at }}x = - \sqrt 3 ;\,\,\,\,\,\,{\text{Local max at }}x = \sqrt 3 \cr & \left( c \right){\text{Absolute min:}}\, - \,6\sqrt 3 {\text{ at }}x = - \sqrt 3 \cr & \,\,\,\,\,\,{\text{Absolute max:}}\,\,28{\text{ at }}x = - 4 \cr} $$

$$\eqalign{ & f\left( x \right) = - {x^3} + 9x{\text{ on the interval }}\left[ { - 4,3} \right] \cr & {\text{Derivative}} \cr & f'\left( x \right) = - {x^3} + 9x \cr & f'\left( x \right) = - 3{x^2} + 9 \cr & {\text{Set the derivative to 0}} \cr & - 3{x^2} + 9 = 0 \cr & {x^2} = 3 \cr & {\text{Solving the equation we obtain}} \cr & x = - \sqrt 3 {\text{ and }}x = \sqrt 3 \cr & \cr & {\text{Evaluating that point and the endpoints we obtain}} \cr & f\left( { - 4} \right) = - {\left( { - 4} \right)^3} + 9\left( { - 4} \right) = 28 \cr & f\left( { - \sqrt 3 } \right) = - {\left( { - \sqrt 3 } \right)^3} + 9\left( { - \sqrt 3 } \right) = - 6\sqrt 3 \cr & f\left( {\sqrt 3 } \right) = - {\left( {\sqrt 3 } \right)^3} + 9\left( {\sqrt 3 } \right) = 6\sqrt 3 \cr & f\left( 3 \right) = - {\left( 3 \right)^3} + 9\left( 3 \right) = 0 \cr & {\text{Then}}{\text{, }} \cr & {\text{The largest result is 28 at }}x = - 4\,\,\,\left( {{\text{Absolute maximum}}} \right) \cr & {\text{The smallest result is }} - 6\sqrt 3 {\text{ at }}x = - \sqrt 3 \,\,\left( {{\text{Absolute and local}}} \right){\text{ minimum}} \cr & {\text{ and local maximum at }}x = \sqrt 3 \cr & \cr & {\text{We can conclude that}} \cr & \left( a \right){\text{Critical points }}x = - \sqrt 3 {\text{ and }}x = \sqrt 3 \cr & \left( b \right){\text{Local min at }}x = - \sqrt 3 ;\,\,\,\,\,\,{\text{Local max at }}x = \sqrt 3 \cr & \left( c \right){\text{Absolute min:}}\, - \,6\sqrt 3 {\text{ at }}x = - \sqrt 3 \cr & \,\,\,\,\,\,{\text{Absolute max:}}\,\,28{\text{ at }}x = - 4 \cr} $$