# How do you find the critical points of: f(x)=(x)(e^(-x^2))?

Oct 21, 2015

See the explanation section, below.

#### Explanation:

$f \left(x\right) = \left(x\right) \left({e}^{- {x}^{2}}\right)$

$f ' \left(x\right) = \left(1\right) \left({e}^{- {x}^{2}}\right) + \left(x\right) \left({e}^{- {x}^{2}} \left(- 2 x\right)\right)$ (product and chain rules)

$= {e}^{- {x}^{2}} \left(1 - 2 {x}^{2}\right)$

$f ' \left(x\right)$ is never undefined and is $0$ at $x = \pm \frac{1}{\sqrt{2}}$.

Both $\frac{1}{\sqrt{2}}$ ans $- \frac{1}{\sqrt{2}}$ are in the domain of $f$ (Domain is $\mathbb{R}$), so both are critical numbers for $f$.

Note

Some treatments have "critical points" being points in 2-space, so we also find $y = f \left(x\right)$ at these critical numbers. I do not use that terminology, but if your teacher does, then you should find $\left(\frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2} \sqrt[4]{e}}\right)$ and $\left(- \frac{1}{\sqrt{2}} , - \frac{1}{\sqrt{2} \sqrt[4]{e}}\right)$ (or whatever form is expected).