## Calculus 8th Edition

$f(x) =xe^{\frac{-x^{2}}{8}}$ 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).