Chapter 3 - The Derivative In Graphing And Applications - 3.4 Absolute Maxima And Minima - Exercises Set 3.4 - Page 222: 12

$$\eqalign{ & {\text{max}} = \root 3 \of {121} {\text{ at }}x = 3 \cr & {\text{min}} = 0{\text{ at }}x = - 1{\text{ and }}x = 0 \cr} $$

$$\eqalign{ & f\left( x \right) = {\left( {{x^2} + x} \right)^{2/3}};\,\,\left[ { - 2,3} \right] \cr & f\left( x \right){\text{ is }}\,{\text{continuous }}{\text{, the absolute}}\,{\text{ extrema }}\,{\text{must }}\,{\text{occur}}\,{\text{ }} \cr & \,{\text{either }}\,{\text{at }}\,{\text{endpoints }}\,{\text{of the interval or at solutions to the }} \cr & {\text{equation }}f'\left( x \right) = 0{\text{ in the open interval }}\left( { - 2,3} \right){\text{.}} \cr & {\text{Differentiate }}f\left( x \right) \cr & f'\left( x \right) = \frac{2}{3}{\left( {{x^2} + x} \right)^{ - 13}}\left( {2x + 1} \right) \cr & f'\left( x \right) = \frac{{2\left( {2x + 1} \right)}}{{3\root 3 \of {{x^2} + x} }} \cr & {\text{Since }}\,f'\left( { - 1} \right){\text{ }}\,{\text{and }}\,{\text{ }}f'\left( 0 \right){\text{ }}\,{\text{is }}\,{\text{undefined }}\,{\text{and they are in the}} \cr & {\text{domain }}\,{\text{of}}\,{\text{ }}f\left( x \right),\,\,\,x = 0{\text{ }}\,{\text{and }}\,x = - 1{\text{ }}\,{\text{are }}\,{\text{critical }}\,{\text{points}}{\text{. }} \cr & {\text{In adition}}{\text{, }}f'\left( x \right) = 0{\text{ when:}} \cr & 2\left( {2x + 1} \right) = 0 \cr & x = - \frac{1}{2} \cr & {\text{Then evaluating }}f\left( x \right){\text{ the points}}{\text{, }}x = \left\{ { - 2, - 1, - \frac{1}{2},0,3} \right\} \cr & f\left( { - 2} \right) = {\left( {{{\left( { - 2} \right)}^2} - 2} \right)^{2/3}} = {2^{2/3}} = \root 3 \of 4 \cr & f\left( { - 1} \right) = {\left( {{{\left( { - 1} \right)}^2} - 1} \right)^{2/3}} = 0 \cr & f\left( { - \frac{1}{2}} \right) = {\left( {{{\left( { - \frac{1}{2}} \right)}^2} - \frac{1}{2}} \right)^{2/3}} = {\left( { - \frac{1}{4}} \right)^{2/3}} = \root 3 \of {\frac{1}{{16}}} \cr & f\left( 0 \right) = {\left( {{{\left( 0 \right)}^2} + 0} \right)^{2/3}} = 0 \cr & f\left( 3 \right) = {\left( {{{\left( 3 \right)}^2} + 3} \right)^{2/3}} = {11^{2/3}} = \root 3 \of {121} \cr & {\text{Then we can conclude that the absolute minimum of }}f\left( x \right){\text{ on}} \cr & {\text{the}}\,\,{\text{ interval }}\,\,\left[ { - 2,3} \right]{\text{ is }}\, - 0\,{\text{ at }}x = - 1,0.{\text{ }}\,{\text{And}}\,\,{\text{ the}}\,{\text{ absolute }} \cr & {\text{maximum is }}\root 3 \of {121} {\text{ at }}x = 3. \cr & \cr & {\text{max}} = \root 3 \of {121} {\text{ at }}x = 3 \cr & {\text{min}} = 0{\text{ at }}x = - 1{\text{ and }}x = 0 \cr & \cr} $$