Answer
$$
f(x)=\frac{x}{x^{2}+2}; \quad\quad[0,4]
$$
Critical number in the given interval is: $\sqrt{2}$ .
The absolute maximum, $0.35355 $ occurs when $x=\sqrt{2} $ and the absolute minimum, $0 $ occurs when $x=0.$
Work Step by Step
$$
f(x)=\frac{x}{x^{2}+2}; \quad\quad[0,4]
$$
First look for critical numbers in the interval $(0,4)$.
find the points where the derivative is $0$
$$
\begin{aligned}
f^{\prime}(x) &=\frac{(x^{2}+2)(1)-x(2x)}{(x^{2}+2)^{2}}\\
&=\frac{-x^{2}+2}{(x^{2}+2)^{2}}\\
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =\frac{-x^{2}+2}{(x^{2}+2)^{2}}=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& -x^{2}+2=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x=-\sqrt{2} \quad \text {or} \quad x=\sqrt{2} \\
\end{aligned}
$$
We know thar $x=\sqrt{2} \in [0,4]$ but $-\sqrt{2} \notin [0,4]$
So, critical number in the given interval is: $\sqrt{2}$ .
Now, evaluate the function at the critical numbers and the endpoints:
$$
\begin{aligned}
\text{when } x=\sqrt{2}: \\
f(\sqrt{2}) &=\frac{(\sqrt{2})}{(\sqrt{2})^{2}+2} \\
&=\frac{\sqrt{2}}{4}\\
&\approx 0.35355
\end{aligned}
$$
$$
\begin{aligned}
\text{when } x=0: \\
f(0) &=\frac{(0)}{(0)^{2}+2} \\
&=\frac{0}{4}\\
&=0
\end{aligned}
$$
and
$$
\begin{aligned}
\text{when } x=4: \\
f(4) &=\frac{(4)}{(4)^{2}+2} \\
&=\frac{2}{9}\\
&\approx 0.2222
\end{aligned}
$$
We conclude that, the absolute maximum, $0.35355 $ occurs when $x=\sqrt{2} $ and the absolute minimum, $0 $ occurs when $x=0.$
