Question

How do you use the first and second derivatives to sketch `f(x)=e^(-x^2)`?

Answer 1

`f(x)` has a maximum in `x=0`, is strictly increasing for `x<0` and strictly decreasing for `x>0`, is concave down in the interval `(-1/sqrt(2), 1/sqrt(2))`, concave up outside the interval and has two inflection points for `x=+-1/sqrt(2)`

Explanation:

Using the chain rule we can calculate:

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

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

`f''(x) = (-2+4x^2)e^(-x^2)`

So we can see that `f(x)` is strictly increasing for `x<0` and strictly decreasing for `x>0` and hat in `x=0` it has a critical point.

The second derivative is positive for:

`(4x^2-2)>0` or `|x|>1/sqrt(2)`

Considering the sign of the second derivative this critical point is a maximum, and `f(x)` is concave down in the interval `(-1/sqrt(2), 1/sqrt(2))`, concave up outside the interval and has two inflection points for `x=+-1/sqrt(2)`

graph{e^-(x^2) [-1.293, 1.207, -0.15, 1.1]}