calculate the critical points of f(x)=xe^-x

Question

i plugged in zero in the derivative, and i got:

e^-0*0e^-0 = 0
1*0= 0

so is the critical point zero then???

Answer 1

Your derivative was probably wrong
I got
f'(x) = x(-e^-x) + e^-x)
= e^-x(-x+1)

setting this = 0 gives x=1
if x=1
f(1) = 1(e^-1) = 1/e
so we have a critical point at (1,1/e)