Single Variable Calculus, Chapter 7, 7.2-1, Section 7.2-1, Problem 64

Determine the absolute maximum and absolute minimum values of $\displaystyle f(x) = x^2 e^{\frac{-x}{2}}$ on the interval $[-1,6]$

To determine the critical numbers, we set $f'(x) = 0$, so..


$
\begin{equation}
\begin{aligned}

\text{if } f(x) =& x^2 e^{\frac{-x}{2}} \text{ then by using Product Rule..}
\\
\\
f'(x) =& \left[ x^2 e^{\frac{-x}{2}} \left( \frac{-1}{2} \right) + 2x e^{\frac{-x}{2}} \right]
\\
\\
f'(x) =& xe^{\frac{-x}{2}} \left( 2 - \frac{x}{2} \right)

\end{aligned}
\end{equation}
$


When $f'(x) = 0$, then..

$\displaystyle 0 = x e^{\frac{-x}{2}} \left( 2 - \frac{x}{2} \right)$

We have,

$xe^{\frac{-x}{2}} = 0 $ and $ \displaystyle 2 - \frac{x}{2} = 0$

The real solution and the critical number is..


$
\begin{equation}
\begin{aligned}

2 - \frac{x}{2} =& 0
\\
\\
\frac{x}{2} =& 2
\\
\\
x =& 4

\end{aligned}
\end{equation}
$



So we have either a maximum or a minimum at $x = 4$, if we evaluate $f(x)$ with $x = 4$, the intervals $x = \pm 1$ and $x = 6$ and $x = 0$,

$
\begin{array}{ccc}
\text{when } x = 4, & & \text{when }x = -1, \\
f(4) = 4^2 (e^{\frac{-4}{2}}) & & f(-1) = (-1)^2 (e^{\frac{-4(-1)}{2}} ) \\
f(4) = 2.1654 & & f(-1) = 1.6487 \\
\text{when } x = 6,& & \text{when } x = 0,\\
f(6) = 6^2 (e^{\frac{-6}{2}}) & & f(0) = (0)^2 (e^{\frac{-0}{2}}) \\
f(6) = 1.7923 & & f(0) = 0
\end{array}
$

Therefore, the absolute maximum is $f(4) = 2.1654$ and the absolute minimum is $f(0) = 0$.