Chapter 3 - The Derivative In Graphing And Applications - 3.4 Absolute Maxima And Minima - Exercises Set 3.4 - Page 222: 9

$$\eqalign{ & {\text{max}} = 8{\text{ at }}x = 4 \cr & {\text{min}} = - 1{\text{ at }}x = 1 \cr} $$

$$\eqalign{ & f\left( x \right) = {\left( {x - 2} \right)^3};\,\,\left[ {1,4} \right] \cr & f\left( x \right){\text{ is }}\,{\text{continuous }}\,{\text{and}}\,{\text{differentiable everywhere}}{\text{, the }} \cr & {\text{absolute}}\,{\text{ extrema }}\,{\text{must }}\,{\text{occur}}\,{\text{ }}\,{\text{either }}\,{\text{at }}\,{\text{endpoints }}\,{\text{of }} \cr & {\text{the interval or at solutions to the equation }}f'\left( x \right) = 0{\text{ in}} \cr & {\text{the open interval }}\left( {1,4} \right){\text{.}} \cr & \cr & {\text{Differentiating }}f\left( x \right) \cr & f'\left( x \right) = 3{\left( {x - 2} \right)^2}\left( 1 \right) \cr & f'\left( x \right) = 3{\left( {x - 2} \right)^2} \cr & {\text{setting }}f'\left( x \right) = 0 \cr & 3{\left( {x - 2} \right)^2} = 0 \cr & x = 2 \cr & {\text{Thus}}{\text{, there is a stationary point at }}x = 2 \cr & {\text{Evaluating }}f\left( x \right){\text{ at the endpoints}}{\text{, }}x = 1,{\text{ }}x = 4{\text{ and }}x = 2 \cr & f\left( 1 \right) = {\left( {1 - 2} \right)^3} = - 1 \cr & f\left( 2 \right) = {\left( {2 - 2} \right)^3} = 0 \cr & f\left( 4 \right) = {\left( {4 - 2} \right)^3} = 8 \cr & {\text{Then we can conclude that the absolute minimum of }}f\left( x \right){\text{ on}} \cr & {\text{the interval }}\left[ {1,4} \right]{\text{ is }} - {\text{1 at }}x = 1.{\text{ And the absolute maximum is}} \cr & {\text{8 at }}x = 4. \cr & \cr & {\text{max}} = 8{\text{ at }}x = 4 \cr & {\text{min}} = - 1{\text{ at }}x = 1 \cr} $$