Answer :

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.

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