Mark (√) against

Given f(x) = x².e^-x

⇒ f’(x) = 2x. e^-x – x² e^-x

⇒ Put f’(x) = 0

⇒ - (x²– 2x)e^-x = 0

⇒ x = 0 or x =2.

Now as there is a -ve sign before f’(x)

When x>2 the function is decreasing

x<0 function is decreasing

But in the interval (0,2) the function is increasing.