Question

If `y=e^(−x^2)`, what are the points of inflection, concavity and critical points?

Answer 1

`f(x) = e^(-x^2)` has a relative maximum for `x=0`, two inflection points in `x= +-1/sqrt(2)`, is concave down in the interval between the two inflection points and concave up outside

Explanation:

Given:

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

we can calculate the first and second order derivatives:

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

`f''(x) = -2e^(-x^2) +4x^2e^(-x^2)= 2(2x^2-1)e^(-x^2)`

we can therefore determine that:

(1) By solving the equation:

`f'(x) = 0 => -2xe^(-x^2) = 0`

we can see that `f(x)` has a single critical point for `x=0`, this point is a relative maximum since `f''(0) = -2 < 0`

Looking at the second derivative, we can see that `2e^(-x^2)` is always positive and non null, so that inflection points and concavity are determined by the factor `(2x^2-1)`, so:

(2) `f(x)` has two inflection points for:

`2x^2-1 = 0 => x=+-1/sqrt(2)`

(3) As `(2x^2-1)` is a second order polynomial with leading positive coefficient, we know that is is negative in the inteval between the roots, and positive outside, so:

`f(x)` is concave up in `(-oo,-1/sqrt(2))` and in `(1/sqrt(2),+oo)`

`f(x)` is concave down in `(-1/sqrt(2),1/sqrt(2))`