# What are the values and types of the critical points, if any, of f(x)=3e^(-2x^2))?

Jan 8, 2017

$x = 0$ is a local maximum.

#### Explanation:

We can find critical points by equating the first derivative to zero:

$\frac{d}{\mathrm{dx}} \left(3 {e}^{- 2 {x}^{2}}\right) = - 12 x {e}^{- 2 {x}^{2}}$

So $f ' \left(x\right) = 0 \implies x = 0$ and $x = 0$is the only critical point.

We can now look at the sign of $f ' \left(x\right)$ and see:

$x < 0 \implies f ' \left(x\right) > 0$
$x > 0 \implies f ' \left(x\right) < 0$

So the point $x = 0$ is a local maximum.

graph{3e^(-2x^2) [-10, 10, -5, 5]}