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

Question

4424 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-15T14:15:34

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

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

$lim_"x->+-oo" f(x) =0$

and

$f(x)>=0 forall x in RR$

$f'(x) = e^(-x^2) * (-2x)$ [Chain rule]

$:. f'(x) = 0$ at $x=0$

Hence: $f_max = f(0) = e^0 = 1$

Thus the domain of $f(x)$ is $(-oo,+oo)$ and
the range of $f(x)$ is $(0,1]$

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

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

How do you find the critical numbers of e^(-x^2)e−x2e^(-x^2)?