Answer
$$
f(x)=x^{2} e^{-0.5x} ; \quad\quad[2,5]
$$
The critical number in the given interval is: $4$.
The absolute maximum, $2.16536 $ occurs when $x=4 $ and the absolute minimum, $1.47151$ occurs when $x=2.$
Work Step by Step
$$
f(x)=x^{2} e^{-0.5x} ; \quad\quad[2,5]
$$
First look for critical numbers in the interval $(2,5)$.
find the points where the derivative is $0$
$$
\begin{aligned} f^{\prime}(x) &=x^{2}\left(-\frac{1}{2} e^{-0.5 x}\right)+2 x e^{-0.5 x} \\
&=e^{-0.5 x}\left(2 x-\frac{1}{2} x^{2}\right)
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =e^{-0.5 x}\left(2 x-\frac{1}{2} x^{2}\right)=0\\
&=\frac{x}{2}e^{-0.5 x}\left(4 - x\right)=0
\Rightarrow\quad\quad\quad\quad\quad\\
& x=0 \quad \text {or} \quad(4-x)=0 \\
\Rightarrow\quad\quad\quad\quad\quad\\
& x=0 \quad \text {or} \quad x=4 \\
\end{aligned}
$$
We know that $x=4 \in [2,5]$ but $0 \notin [2,5]$
So, the critical number in the given interval is: $4$.
Now, evaluate the function at the critical numbers and the endpoints:
$$
\begin{aligned}
\text{when } x=2: \\
f(2) &=(2)^{2} e^{-0.5(2)} \\
&=\frac{4}{e} \\
&\approx 1.47151
\end{aligned}
$$
$$
\begin{aligned}
\text{when } x=4: \\
f(4) &=(4)^{2} e^{-0.5(4)} \\
&=\frac{16}{e^2} \\
&\approx 2.16536
\end{aligned}
$$
and
$$
\begin{aligned}
\text{when } x=5: \\
f(5) &=(5)^{2} e^{-0.5(5)} \\
&=5^2\cdot \frac{1}{e^{2.5}}\\
&\approx 2.05212
\end{aligned}
$$
We conclude that, the absolute maximum, $2.16536 $ occurs when $x=4 $ and the absolute minimum, $1.47151$ occurs when $x=2.$