# How do you find the critical numbers of e^(-x^2)?

May 15, 2017

Domain: $\left(- \infty , + \infty\right)$
Range: $\left(0 , 1\right]$

#### Explanation:

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

${\lim}_{\text{x->+-oo}} f \left(x\right) = 0$

and

$f \left(x\right) \ge 0 \forall x \in \mathbb{R}$

$f ' \left(x\right) = {e}^{- {x}^{2}} \cdot \left(- 2 x\right)$ [Chain rule]

$\therefore f ' \left(x\right) = 0$ at $x = 0$

Hence: ${f}_{\max} = f \left(0\right) = {e}^{0} = 1$

Thus the domain of $f \left(x\right)$ is $\left(- \infty , + \infty\right)$ and
the range of $f \left(x\right)$ is $\left(0 , 1\right]$

This can be seen by the graph of $f \left(x\right)$ below:

graph{e^(-x^2) [-2.433, 2.435, -1.217, 1.216]}