Answer
\[\begin{align}
& \text{Critical point: }x=\frac{1}{e} \\
& \text{ Absolute maximum on }\left( \frac{1}{e},10-\frac{2}{e} \right) \\
\end{align}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)=2x\ln x+10\text{ on }\left( 0,4 \right) \\
& \text{Differentiate} \\
& f'\left( x \right)=2x\left( \frac{1}{x} \right)+\ln x\left( 2 \right)+0 \\
& f'\left( x \right)=2+2\ln x \\
& \text{Calculate the critical points, set }f'\left( x \right)=0 \\
& 2+2\ln x=0 \\
& \ln x=-1 \\
& {{e}^{\ln x}}={{e}^{-1}} \\
& x=\frac{1}{e} \\
& \text{Evaluating }f\text{ at }x=\frac{1}{e}\text{, and }x=0.0001,\text{ }x=4 \\
& f\left( 0.0001 \right)\approx 10 \\
& f\left( \frac{1}{e} \right)=2\left( \frac{1}{e} \right)\ln \left( \frac{1}{e} \right)+10=10-\frac{2}{e}\approx 9.26 \\
& f\left( 4 \right)=2\left( 4 \right)\ln \left( 4 \right)+10=10+\ln {{4}^{8}}\approx 21.09 \\
& \text{Therefore,} \\
& \text{The smallest of those function values are:} \\
& f\left( \frac{1}{e} \right)=10-\frac{2}{e} \\
& \text{Which is the absolute minimum on }\left( \frac{1}{e},10-\frac{2}{e} \right) \\
& \\
& \text{Critical points: }x=\frac{1}{e} \\
& \text{ Absolute maximum on }\left( \frac{1}{e},10-\frac{2}{e} \right) \\
& \text{Graph} \\
\end{align}\]
