# What are the critical points and also inflection points of 3e^(-2x(^2))?

Aug 7, 2015

Assuming that the function is: $f \left(x\right) = 3 {e}^{- 2 {x}^{2}}$, the only critical number is: $0$, and the inflection points are: $\left(\frac{1}{\sqrt{12}} , 3 {e}^{- \frac{1}{6}}\right)$ and $\left(- \frac{1}{\sqrt{12}} , 3 {e}^{- \frac{1}{6}}\right)$

#### Explanation:

$f \left(x\right) = 3 {e}^{- 2 {x}^{2}}$

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

$f ' \left(x\right)$ is never undefined and is $0$ when $x = 0$.

The only critical number is $0$.

$f ' ' \left(x\right) = - 12 {e}^{- 2 {x}^{2}} - 12 x \left(- 12 x {e}^{- 2 {x}^{2}}\right)$

$= - 12 {e}^{- 2 {x}^{2}} \left(1 - 12 {x}^{2}\right)$

#f''(x) is never undefined, so the only chance it has to change signs is when it is zero.

Solving: $- 12 {e}^{- 2 {x}^{2}} \left(1 - 12 {x}^{2}\right) = 0$, gives us

$x = \pm \frac{1}{\sqrt{12}}$

Observe that the sign of $f ' ' \left(x\right)$ is the opposite of the sign of $1 - 12 {x}^{2}$ which does, indeed, change sign at both $+ \frac{1}{\sqrt{12}}$ and $- \frac{1}{\sqrt{12}}$

$f$ is an even function so we calculate: $f \left(\pm \frac{1}{\sqrt{12}}\right) = 3 {e}^{- 2 \left(\frac{1}{12}\right)} = 3 {e}^{- \frac{1}{6}}$

The inflection points are: $\left(\frac{1}{\sqrt{12}} , 3 {e}^{- \frac{1}{6}}\right)$ and $\left(- \frac{1}{\sqrt{12}} , 3 {e}^{- \frac{1}{6}}\right)$

(Simplify and rationalize if you prefer.)