Chapter 4 - Applications of the Derivative - Review Exercises - Page 330: 8

\[\begin{align} & \text{Absolute maximum on }\left( \frac{2}{\sqrt{5}},\frac{16\sqrt{2}}{5\sqrt[4]{5}} \right) \\ & \text{Asolute minimum on }\left( 4,-24 \right) \\ \end{align}\]

\[\begin{align} & f\left( x \right)=4{{x}^{1/2}}-{{x}^{5/2}}\text{ on }\left[ 0,4 \right] \\ & \text{Differentiate} \\ & f'\left( x \right)=4\left( \frac{1}{2}{{x}^{-1/2}} \right)-\frac{5}{2}\left( {{x}^{3/2}} \right) \\ & f'\left( x \right)=2{{x}^{-1/2}}-\frac{5}{2}{{x}^{3/2}} \\ & \text{Calculate the critical points, set }f'\left( x \right)=0 \\ & 2{{x}^{-1/2}}-\frac{5}{2}{{x}^{3/2}}=0 \\ & \text{Factoring} \\ & {{x}^{-1/2}}\left( 2-\frac{5}{2}{{x}^{2}} \right)=0 \\ & 2-\frac{5}{2}{{x}^{2}}=0 \\ & {{x}^{2}}=\frac{4}{5} \\ & x=\pm \frac{2}{\sqrt{5}} \\ & \text{only }\frac{2}{\sqrt{5}}\text{ is on the interval }\left[ 0,4 \right] \\ & \text{Therefore} \\ & \text{We have the critical points} \\ & x=\frac{2}{\sqrt{5}} \\ & \text{Evaluating }f\text{ at each of these points, and endpoints we have} \\ & f\left( 0 \right)=0 \\ & f\left( \frac{2}{\sqrt{5}} \right)=4{{\left( \frac{2}{{{5}^{1/2}}} \right)}^{1/2}}-{{\left( \frac{2}{{{5}^{1/2}}} \right)}^{5/2}}=\frac{4\sqrt{2}}{\sqrt[4]{5}}-\frac{4\sqrt{2}}{5\sqrt[4]{5}}=\frac{16\sqrt{2}}{5\sqrt[4]{5}}\approx 3.02 \\ & f\left( 4 \right)=-24 \\ & \text{The largest of these function values is:} \\ & f\left( \frac{2}{\sqrt{5}} \right)=\frac{16\sqrt{2}}{5\sqrt[4]{5}} \\ & \text{Which is the absolute maximum on }\left( \frac{2}{\sqrt{5}},\frac{16\sqrt{2}}{5\sqrt[4]{5}} \right) \\ & \text{The smallest of those function values are:} \\ & f\left( 4 \right)=-24 \\ & \text{Which is the absolute minimum on }\left( 4,-24 \right) \\ & \\ & \text{Critical points: }x=\frac{2}{\sqrt{5}} \\ & \text{ Absolute maximum on }\left( \frac{2}{\sqrt{5}},\frac{16\sqrt{2}}{5\sqrt[4]{5}} \right) \\ & \text{Asolute minimum on }\left( 4,-24 \right) \\ & \text{Graph} \\ \end{align}\]