# What are the points of inflection, if any, of f(x)=2e^(-x^2) ?

Dec 21, 2016

$f \left(x\right) = 2 {e}^{- {x}^{2}}$ has two inflection points for $x = \pm \frac{1}{\sqrt{2}}$

#### Explanation:

A necessary condition for $f \left(x\right)$ to have an inflection point in $x = \overline{x}$ is that:

$f ' ' \left(\overline{x}\right) = 0$

So we calculate:

$f ' \left(x\right) = - 4 x {e}^{- {x}^{2}}$
$f ' ' \left(x\right) = - 4 {e}^{- {x}^{2}} + 8 {x}^{2} {e}^{- {x}^{2}} = {e}^{- {x}^{2}} \left(8 {x}^{2} - 4\right)$

As the exponential is always positive, $f ' ' \left(x\right)$ is null when:

$8 {x}^{2} - 4 = 0$

That is:

$x = \pm \frac{1}{\sqrt{2}}$

To be sure that these are inflection points for $f \left(x\right)$ we must check that $f ' ' \left(x\right)$ changes sign around its zeroes: this is certainly the case, because $8 {x}^{2} - 4 = 0$ is a second degree polynomial and changes sign on the two sides of single order zeroes.

graph{2e^(-x^2) [-2.54, 2.46, -2.26, 0.24]}