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

1 Answer
Krishna V.
Jun 1, 2017

The points of inflection are at #-1.2247, 0, and 1.2247#.

Explanation:

To find points of inflection for a function, you take the second derivative of the function and set the second derivative to equal 0. The product rule is used below:

#f(x) = xe^(-x^2)#

#f'(x) = (x*-2xe^(-x^2))+(e^(-x^2)*1)=-2x^2e^(-x^2)+e^(-x^2)#

#f^((2))(x) = (-4x*e^(-x^2))+(-2x^2*-2xe^(-x^2))-2xe^(-x^2)#

#f^((2))(x) = -4xe^(-x^2)+4x^3e^(-x^2)-2xe^(-x^2)#

#f^((2))(x) = -6xe^(-x^2)+4x^3e^(-x^2)#

Now we set it equal to 0:

#0 = -6xe^(-x^2)+4x^3e^(-x^2)#

If we graph this function, the roots are approximately at #-1.2247, 0, and 1.2247#:
graph{-6xe^(-x^2)+4x^3e^(-x^2) [-5, 5, -2.5, 2.5]}

All of these values go from positive to negative or vice versa in an interval test, so they are all points of inflection.

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