What are the points of inflection, if any, of f(x)=2e^(-x^2) f(x)=2ex2?

1 Answer
Dec 21, 2016

f(x) = 2e^(-x^2)f(x)=2ex2 has two inflection points for x=+-1/sqrt(2)x=±12

Explanation:

A necessary condition for f(x)f(x) to have an inflection point in x=barxx=¯x 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]}