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

1 Answer
May 1, 2017

#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#