Question

What are the points of inflection, if any, of `f(x)=2e^(-x^2)`?

Answer 1

`f(x) = 2e^(-x^2)` has two inflection points for `x=+-1/sqrt(2)`

Explanation:

A necessary condition for `f(x)` to have an inflection point in `x=barx` is that:

`f''(bar x)=0`

So we calculate:

`f'(x) =-4xe^(-x^2)`
`f''(x)=-4e^(-x^2)+8x^2e^(-x^2)=e^(-x^2)(8x^2-4)`

As the exponential is always positive, `f''(x)` is null when:

`8x^2-4=0`

That is:

`x=+-1/sqrt(2)`

To be sure that these are inflection points for `f(x)` we must check that `f''(x)` changes sign around its zeroes: this is certainly the case, because `8x^2-4=0` is a second degree polynomial and changes sign on the two sides of single order zeroes.

graph{2e^(-x^2) [-2.54, 2.46, -2.26, 0.24]}