# Consider the function f(x)=e^(-x^(2))?

## Find the following: domain; found: $\left(- \setminus \infty , \setminus \infty\right)$ x-intercepts; found: $- {x}^{2} = - \setminus \infty$ thus there is NO x-intercept y-intercept; found: $y = 1$ symmetry vertical asymptotes; found: NONE horizontal asymptotes; found: $x = 0$ (exponential function rule?) intervals of increase and decrease local maxima and minima intervals of concavity inflection points If any characteristics are not present in the function, state "NONE". Then graph the function I apologize for the confusion regarding the parts I can understand/answer! I have worked out which ones I actually need help with now.

Nov 23, 2016

See below.

#### Explanation:

Symmetry

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

So the function is even and the graph of the function is symmetric with respect to the $y$-axis.

Increase/Decrease and Extrema

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

Since ${e}^{\text{real number}}$ is always positive, the sign of $f '$ is the opposite of the sign of $x$.

So $f$ is increasing on $\left(- \infty , 0\right)$, has local maximum $f \left(0\right) = 1$ and then $f$ decreases on $\left(0 , \infty\right)$

Concavity and Inflection

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

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

$f ' '$ is defined everywhere and is $0$ only at $\pm \frac{\sqrt{2}}{2}$.

Analysis of the sign of $f ' '$ shows that the sign of $f ' '$ agrees with that of $2 {x}^{2} - 1$, so:

on $\left(- \infty , - \frac{\sqrt{2}}{2}\right)$, $f ' ' \left(x\right) > 0$ and the graph of $f$ is concave up
on $\left(- \frac{\sqrt{2}}{2} , \frac{\sqrt{2}}{2}\right)$, $f ' ' \left(x\right) < 0$ and the graph of $f$ is concave down
on $\left(\frac{\sqrt{2}}{2} , \infty\right)$, $f ' ' \left(x\right) > 0$ and the graph of $f$ is concave up

Therefore, there are two inflection points: $\left(- \frac{\sqrt{2}}{2} , {e}^{- \frac{1}{2}}\right)$ and $\left(\frac{\sqrt{2}}{2} , {e}^{- \frac{1}{2}}\right)$.

(Note that ${e}^{- \frac{1}{2}} = \frac{1}{\sqrt{e}}$ if you prefer to write it that way.)