Find first and second derivatives, critical number, inflection points.
f(x) = xe^x
f'(x) = ('x) (e^x) + x(e^x)('x)
f'(x) = e^x+xe^x
f'(x) = e^x(1+x)
Critical number x = -1
f"(x) = e^x(1) + (1+x)(e^x)
f"(x) = e^x +e^x(1+x)
Critical Number x=-1
I hope I am close to the correct answer.
first of all, we do not find critical points from the second derivative, we use the second derivative to find concavity and possible inflection points. second of all,
Originally Posted by confusedagain
e^x(2 + x) \:=\:0\quad\Rightarrow\quad x = -2
If you prefer, use e^x(2 + x)=0\implies x=-2, this yields