Answer
$$
f(x)=\frac{x-1}{x^{2}+1}; \quad\quad[1,5]
$$
Critical number in the given interval is: $1+\sqrt{2}$ .
The absolute maximum, $0.2071 $ occurs when $x=1+\sqrt{2} $ and the absolute minimum, $0 $ occurs when $x=1.$
Work Step by Step
$$
f(x)=\frac{x-1}{x^{2}+1}; \quad\quad[1,5]
$$
First look for critical numbers in the interval $(1,5)$.
find the points where the derivative is $0$
$$
\begin{aligned}
f^{\prime}(x) &=\frac{(x^{2}+1)(1)-(x-1)(2x)}{(x^{2}+1)^{2}}\\
&=\frac{-x^{2}+2x+1}{(x^{2}+1)^{2}}\\
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =\frac{-x^{2}+2x+1}{(x^{2}+1)^{2}}=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& -x^{2}+2x+1=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& \:x_{1,\:2}=\frac{-2\pm \sqrt{2^2-4\left(-1\right)1}}{2\left(-1\right)}\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x=1-\sqrt{2} \quad \text {or} \quad x=1+\sqrt{2} \\
\end{aligned}
$$
We know thar $x=1+\sqrt{2} \in [1,5]$ but $1-\sqrt{2} \notin [1,5]$
So, critical numbers in the given interval is: $1+\sqrt{2}$ .
Now, evaluate the function at the critical numbers and the endpoints:
$$
\begin{aligned}
\text{when } x=1+\sqrt{2}: \\
f(1+\sqrt{2}) &=\frac{(1+\sqrt{2})-1}{(1+\sqrt{2})^{2}+1} \\
&=\frac{\sqrt{2}-1}{2} \\
&\approx 0.2071
\end{aligned}
$$
$$
\begin{aligned}
\text{when } x=1: \\
f(1) &=\frac{(1)-1}{(1)^{2}+1} \\
&=\frac{0}{2} \\
&=0
\end{aligned}
$$
and
$$
\begin{aligned}
\text{when } x=5: \\
f(5) &=\frac{(5)-1}{(5)^{2}+1} \\
&=\frac{2}{13} \\
&\approx 0.15384
\end{aligned}
$$
We conclude that, the absolute maximum, $0.2071 $ occurs when $x=1+\sqrt{2} $ and the absolute minimum, $0 $ occurs when $x=1.$
