Answer
The absolute maximum value at (2, 1.213) and the absolute minimum value at (-1, -0.882).
Work Step by Step
$f(x) =xe^{\frac{-x^{2}}{8}}$
Apply product rule of differentiation.
$f'(x) =e^{\frac{-x^{2}}{8}}(1-\frac{x^{2}}{4})$
Now put, $f'(x) =0$
$e^{\frac{-x^{2}}{8}}(1-\frac{x^{2}}{4})=0$
Since, $e^{\frac{-x^{2}}{8}}\ne0$
Thus, $(1-\frac{x^{2}}{4})=0$
This implies $x=2$
Now at $x=-1, 2, 4$, we have
$f(-1)=-1 e^{\frac{-1}{8}}=-0.882$
$f(2) =2e^{\frac{-2^{2}}{8}}=\frac{2}{\sqrt e}=1.213$
$f(4)=4e^{-2}= 0.54134$
Therefore, the absolute maximum value at (2, 1.213) and the absolute minimum value at (-1, -0.882).