Answer
The absolute maximum is $f(1) = \sqrt{e}$
The absolute minimum is $f(-2) = -\frac{2}{e}$
Work Step by Step
$f(x) = xe^{x/2}$
We can find the points when $f'(x) = 0$:
$f(x) = xe^{x/2}$
$f'(x) = e^{x/2}+\frac{xe^{x/2}}{2} = 0$
$\frac{xe^{x/2}}{2} = -e^{x/2}$
$x = -2$
We can find the value of $f(-2)$ and the values of the function at the endpoints of the interval $[-3, 1]$:
$f(-2) = (-2)e^{-2/2} = -\frac{2}{e}$
$f(-3) = (-3)e^{-3/2} = -\frac{3}{e^{3/2}} \gt -\frac{2}{e}$
$f(1) = (1)e^{1/2} = \sqrt{e}$
The absolute maximum is $f(1) = \sqrt{e}$
The absolute minimum is $f(-2) = -\frac{2}{e}$
