Question

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

Find the following:

• domain; found: `(-\infty,\infty)`
• x-intercepts; found: `-x^2=-\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.

Answer 1

See below.

Explanation:

Symmetry

`f(-x) = e^(-(-x)^2) = e^(-x^2) = f(x)`

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

Increase/Decrease and Extrema

`f'(x) = -2xe^(-x^2)`

Since `e^"real number"` is always positive, the sign of `f'` is the opposite of the sign of `x`.

So `f` is increasing on `(-oo,0)`, has local maximum `f(0) = 1` and then `f` decreases on `(0,oo)`

Concavity and Inflection

`f''(x) = (-2)(e^(-x^2)) + (-2x)(-2xe^(-x^2))`

`= 2e^(-x^2)(2x^2-1)`

`f''` is defined everywhere and is `0` only at `+-sqrt2/2`.

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

on `(-oo,-sqrt2/2)`, `f''(x) >0` and the graph of `f` is concave up
on `(-sqrt2/2,sqrt2/2)`, `f''(x) < 0` and the graph of `f` is concave down
on `(sqrt2/2,oo)`, `f''(x) >0` and the graph of `f` is concave up

Therefore, there are two inflection points: `(-sqrt2/2,e^(-1/2))` and `(sqrt2/2,e^(-1/2))`.

(Note that `e^(-1/2) = 1/sqrte` if you prefer to write it that way.)