# How do you use the first and second derivatives to sketch f(x)=e^(-x^2)?

Dec 12, 2016

$f \left(x\right)$ has a maximum in $x = 0$, is strictly increasing for $x < 0$ and strictly decreasing for $x > 0$, is concave down in the interval $\left(- \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}\right)$, concave up outside the interval and has two inflection points for $x = \pm \frac{1}{\sqrt{2}}$

#### Explanation:

Using the chain rule we can calculate:

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

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

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

So we can see that $f \left(x\right)$ is strictly increasing for $x < 0$ and strictly decreasing for $x > 0$ and hat in $x = 0$ it has a critical point.

The second derivative is positive for:

$\left(4 {x}^{2} - 2\right) > 0$ or $| x | > \frac{1}{\sqrt{2}}$

Considering the sign of the second derivative this critical point is a maximum, and $f \left(x\right)$ is concave down in the interval $\left(- \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}\right)$, concave up outside the interval and has two inflection points for $x = \pm \frac{1}{\sqrt{2}}$

graph{e^-(x^2) [-1.293, 1.207, -0.15, 1.1]}