Answer
Absolute maximum: $(2,\displaystyle \frac{2\sqrt{5}}{5})$
Absolute minimum: $(0,0)$
Work Step by Step
$ f(x)=\displaystyle \frac{x}{\sqrt{x^{2}+1}}=x(x^{2}+1)^{-1/2} \quad$ ... defined on $[0,2].$
$ f^{\prime}(x)=1\displaystyle \cdot(x^{2}+1)^{-1/2}+x[-\frac{1}{2}(x^{2}+1)^{-3/2}\cdot 2x]\quad$ ... product rule, chain rule
$=\displaystyle \frac{1}{\sqrt{x^{2}+1}}-\frac{x^{2}}{(x^{2}+1)\sqrt{x^{2}+1}}$
$=\displaystyle \frac{(x^{2}+1)}{\sqrt{x^{2}+1}}-\frac{x^{2}}{(x^{2}+1)\sqrt{x^{2}+1}}$
$=\displaystyle \frac{1}{(x^{2}+1)\sqrt{x^{2}+1}}\quad$ ... defined on $(0,2).$
$f^{\prime}(x)=0 \quad$ for no x$\in(0,2)$
( no critical numbers)
$\left[\begin{array}{lcccc}
x, \text{ interval} & 0 & (0,4) & 2\\
t=\text{ test number} & & 1 & \\
\text{ sign of }f^{\prime}(t) & & + & \\
f(x) & 0 & \nearrow & 2/\sqrt{5}\\
& min & & max
\end{array}\right]$
$\displaystyle \frac{2}{\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$
Absolute maximum: $(2,\displaystyle \frac{2\sqrt{5}}{5})$
Absolute minimum: $(0,0)$
