df/dx = - x e^-x + e^-x
= 0 for max or min within interval
BUT you did not say the interval
in general
e^-x = x e^x
1 = x is one place where the slope is zero
is the second derivative + or - ?
d^2f/dx^2 = -x(-e^-x) -e^-x - e^-x
= xe^-x -2 e^-x)
= (1/e^x)(x-2)
when x = 1
= (1/e) (-2) which is - so that is a max
now I do not know your interval so you better check the ends., I suspect the linmit of x/e^x ---> 0 as x ---> infinity so that would end up being a minimum
Find the absolute maximum and minimum values of each function on the given interval. f(x)=x(e^-x)
2 answers
Oh and by the way when x = 0 we have 0/1