# If  y=e^(−x^2), what are the points of inflection, concavity and critical points?

Jan 13, 2017

$f \left(x\right) = {e}^{- {x}^{2}}$ has a relative maximum for $x = 0$, two inflection points in $x = \pm \frac{1}{\sqrt{2}}$, is concave down in the interval between the two inflection points and concave up outside

#### Explanation:

Given:

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

we can calculate the first and second order derivatives:

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

$f ' ' \left(x\right) = - 2 {e}^{- {x}^{2}} + 4 {x}^{2} {e}^{- {x}^{2}} = 2 \left(2 {x}^{2} - 1\right) {e}^{- {x}^{2}}$

we can therefore determine that:

(1) By solving the equation:

$f ' \left(x\right) = 0 \implies - 2 x {e}^{- {x}^{2}} = 0$

we can see that $f \left(x\right)$ has a single critical point for $x = 0$, this point is a relative maximum since $f ' ' \left(0\right) = - 2 < 0$

Looking at the second derivative, we can see that $2 {e}^{- {x}^{2}}$ is always positive and non null, so that inflection points and concavity are determined by the factor $\left(2 {x}^{2} - 1\right)$, so:

(2) $f \left(x\right)$ has two inflection points for:

$2 {x}^{2} - 1 = 0 \implies x = \pm \frac{1}{\sqrt{2}}$

(3) As $\left(2 {x}^{2} - 1\right)$ is a second order polynomial with leading positive coefficient, we know that is is negative in the inteval between the roots, and positive outside, so:

$f \left(x\right)$ is concave up in $\left(- \infty , - \frac{1}{\sqrt{2}}\right)$ and in $\left(\frac{1}{\sqrt{2}} , + \infty\right)$

$f \left(x\right)$ is concave down in $\left(- \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}\right)$