Question

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

Answer 1

At `x~=0.8`

Explanation:

Point of inflection occur where `(d^2f)/(dx^2)=0`

As we have `f(x)=xe^(x^2)-x^2e^x`

`(df)/(dx)=e^(x^2)+xe^(x^2)xx2x-2xe^x-x^2e^x`

= `e^(x^2)+2x^2e^(x^2)-2xe^x-x^2e^x`

and `(d^2f)/(dx^2)=2xe^(x^2)+2(2xe^(x^2)+x^2xxe^(x^2)xx2x)-2(e^x+xe^x)-(2xe^x+x^2e^x)`

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

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

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

at `x~=0.8` (from the graph given below) and this is the point of inflection, as may be seen from graph of `f(x)=xe^(x^2)-x^2e^x`

graph{6xe^(x^2)+4x^3e^(x^2)-2e^x-4xe^x-x^2e^x [-0.553, 1.947, -0.545, 0.705]}

graph{xe^(x^2)-x^2e^x [-0.553, 1.947, -0.545, 0.705]}