Question

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

Answer 1

`f(x)` has one point of inflection, which is at `x approx .161`

Explanation:

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

To find points of inflection, find where `f''(x)` changes sign.

Use the product rule and chain rule:
`f'(x)=(2)(e^(-x))+(2x)(-e^(-x))+(2x)(e^x)+(x^2)(e^x)`

Rewrite/simplify:
`f'(x)=2e^(-x)-2xe^(-x)+2xe^x+x^2e^x`

More differentiation...
`f''(x)=-2e^(-x)-(2)(e^(-x))-(2x)(-e^(-x))+(2)(e^x)+(2x)(e^x)+(2x)(e^x)+(x^2)(e^x)`

More simplification...
`f''(x)=-4e^(-x)+2xe^(-x)+2e^x+4xe^x+x^2e^x`

Set `f''(x)` equal to zero using graphing calculator:
`0=f''(x)`

`x approx .16059649`