\(f(x)=xe^{-x^2/128}\)
\(\implies f'(x)=e^{-x^2/128}+x\left(-\dfrac{2x}{128}\right)e^{-x^2/128}=\left(1-\dfrac1{64}x^2\right)e^{-x^2/128}\)
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
\(f''(-8)=\dfrac1{4\sqrt e}>0\)
\(f''(8)=-\dfrac1{4\sqrt e}<0\)
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.
