`f(x) = (x^2)(e^-x), [-1,3]` Find the local and absolute extreme values of the function on the given interval.
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.
Now to find the critical numbers, solve for x for f'(x)=0
`x=0 , x=2`
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)
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