Calculus: Early Transcendentals, Chapter 4, Review, Section Review, Problem 6

f(x)=x^2e^-x
Now to find the absolute extrema of the function , that is continuous on a closed interval, we have to find the critical numbers that are in the interval and evaluate the function at the endpoints and at the critical numbers.
f'(x)=x^2(e^-x(-1))+e^-x(2x)
f'(x)=e^-x(2x-x^2)
f'(x)=(x(2-x))/e^x
Now to find the critical numbers, solve for x for f'(x)=0
(x(2-x))/e^x=0
x(2-x)=0
x=0 , x=2
f(-1)=(-1)^2e=e
f(0)=0
f(2)=2^2e^-2=4/e^2=0.541
f(3)=3^2e^-3=9/e^3=0.448
Absolute Maximum=e at x=-1
Absolute Minimum=0 at x=0
Now to find Local extrema let's check the sign of f'(x) at test points in the intervals (-1,0) , (0,2) and (2,3)
f'(-0.5)=-0.5e^-0.5(2+0.5)=-0.758
f'(1)=e^-1(2-1)=0.3678
f'(2.5)=2.5e^-2.5(2-2.5)=-0.102
Since the sign of f'(x) is changing from positive to negative from the intervals(0,2) and (2,3) ,
So there is Local Maximum at x=2
Local Maximum f(2)=4/e^2