Answer
$$
f(x)=\frac{\ln x}{ x^{2}} ; \quad\quad[1,4]
$$
The critical number in the given interval is: $e ^{\frac{1}{2}}$.
The absolute maximum, $0.18393 $ occurs when $x=e ^{\frac{1}{2}} $ and the absolute minimum, $0 $ occurs when $x=1.$
Work Step by Step
$$
f(x)=\frac{\ln x}{ x^{2}} ; \quad\quad[1,4]
$$
First look for critical numbers in the interval $(1,4)$.
find the points where the derivative is $0$
$$
\begin{aligned}
f^{\prime}(x) &=\frac{x^{2}.(\frac{1}{x}) -\ln x . (2x)}{ x^{4}} \\
&=\frac{x - 2x\ln x }{ x^{4}} \\
&=\frac{x(1- 2\ln x) }{ x^{4}}\\
&=\frac{(1- 2\ln x) }{ x^{3}}\\
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =\frac{(1- 2\ln x) }{ x^{3}}=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& (1- 2\ln x)=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& 2\ln x=1\\
\Rightarrow\quad\quad\quad\quad\quad\\
& \ln x=\frac{1}{2}\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x=e ^{\frac{1}{2}} \approx 1.64872\\
\end{aligned}
$$
So, the critical number in the given interval is: $e ^{\frac{1}{2}}$,
also $ f^{\prime}(x) $ is not exist when $x=0 $ but $0 \notin [1,4] $, so $x=0$ is not critical number in the given interval.
Now, evaluate the function at the critical numbers and the endpoints:
$$
\begin{aligned}
\text{when } x=1: \\
f(1) &=\frac{\ln (1)}{ (1)^{2}} \\
&=\frac{0}{ 1} \\
&=0
\end{aligned}
$$
$$
\begin{aligned}
\text{when } x=e ^{\frac{1}{2}}: \\
f(e ^{\frac{1}{2}}) &=\frac{\ln (e ^{\frac{1}{2}})}{ (e ^{\frac{1}{2}})^{2}} \\
&=\frac{1}{2e}\\
&\approx 0.18393
\end{aligned}
$$
and
$$
\begin{aligned}
\text{when } x=4: \\
f(4) &=\frac{\ln (4)}{ (4)^{2}} \\
&=\frac{\ln \left(2\right)}{8} \\
&\approx 0.08664
\end{aligned}
$$
We conclude that, the absolute maximum, $0.18393 $ occurs when $x=e ^{\frac{1}{2}} $ and the absolute minimum, $0 $ occurs when $x=1.$
