Find the intervals of increase and decrease for the function f(x) = x^2e^(-x)

Question

anonymous · Answer

did you find the derivative?
it is 
$$-x(x-2)e^{-x}$$

anonymous · Answer

and you want to know where it is positive and where it is negative, because that will answer your question 
now $e^{-x}>0$ so we can ignore that term and just look at $-x(x-2)$

anonymous · Answer

this is 0 at 0 and at 2, negative on the interval $(-\infty,0)\cup(2,\infty)$ and positive on $0,2$

anonymous · Answer

In order to find the interval of increasing and decreasing you have to find the derivative (same apply if you want to find the concavity you have then to find the 2nd derivative)
if f(x)=x^2e^-x then f'(x)=2xe^-x + -x^2.e^-x  =e^-x(2x-x^2) now when the first derivative is negative => decreasing , if Positive => increasing we know that e^g(x) is always positive no matter what therefore,  e^(-x) is positive so now we only have to care about 2x-x^2 (when positive and when negative) 
2x-x^2 = x(2-x) so now you can do it in addition to the final answer that @satellite73 stated  |dw:1334113830854:dw|