# How do you find the critical points for f(x)=3e^(-2x(^2))?

Jun 10, 2018

$\left(0 , 3\right)$ is a maximum critical point

#### Explanation:

First, you need to differentiate your equation

$f \left(x\right) = 3 {e}^{- 2 {x}^{2}}$
$f ' \left(x\right) = - 4 x \times 3 {e}^{- 2 {x}^{2}}$
$f ' \left(x\right) = - 12 x {e}^{- 2 {x}^{2}}$
$f ' ' \left(x\right) = - 12 x \times - 4 x {e}^{- 2 {x}^{2}} + {e}^{- 2 {x}^{2}} \times - 12$
$f ' ' \left(x\right) = 48 {x}^{2} {e}^{- 2 {x}^{2}} - 12 {e}^{- 2 {x}^{2}}$

For stationary points/critical points, $f ' \left(x\right) = 0$

$- 12 x {e}^{- 2 {x}^{2}} = 0$

$- 12 x = 0$ or ${e}^{- 2 {x}^{2}} = 0$

There is no solution for ${e}^{- 2 {x}^{2}} = 0$ since the graph NEVER goes to 0. It only APPROACHES 0.

$- 12 x = 0$
$x = 0$

To find whether it is maximum or minimum, you sub $x = 0$ into $f ' ' \left(x\right)$. If the answer is greater than zero ie $> 0$, then it is a minimum. If the answer is smaller than zero ie $< 0$, then it is a maximum.

$f ' ' \left(0\right) = - 12 {e}^{0} = - 12 < 0$
Therefore, at $x = 0$, it is a maximum
To find the coordinate, sub $x = 0$ back into $f \left(x\right)$ and you will get $y = 3$ --> $\left(0 , 3\right)$ is a maximum