Question

Xe^(-x^2/128) absolute max and absolute min

Answer 1

$f(x)=xe^{-x^2/128}$
$\implies f'(x)=e^{-x^2/128}+x\left(-\dfrac{2x}{128}\right)e^{-x^2/128}=\left(1-\dfrac1{64}x^2\right)e^{-x^2/128}$

Extrema can occur when the derivative is zero or undefined.

$\left(1-\dfrac1{64}x^2\right)e^{-x^2/128}=0\implies 1-\dfrac1{64}x^2=0\implies x^2=64\implies x=\pm8$

Maxima occur where the first derivative is zero and the second derivative is negative; minima where the second derivative is positive. You have

$f''(x)=-\dfrac1{32}xe^{-x^2/128}+\left(1-\dfrac1{64}x^2\right)\left(-\dfrac{2x}{128}\right)e^{-x^2/128}=\left(-\dfrac3{64}x+\dfrac1{4096}x^3\right)e^{-x^2/128}$

At the critical points, you get

$f''(-8)=\dfrac1{4\sqrt e}>0$
$f''(8)=-\dfrac1{4\sqrt e}$

So you have a minimum at $\left(-8,-\dfrac8{\sqrt e}\right)$ and a maximum at $\left(8,\dfrac8{\sqrt e}\right)$.

Meanwhile, as $x\to\pm\infty$, it's clear that $f(x)\to0$, so these extrema are absolute on the function's domain.