`f(x) = (x^2)(e^-x), [-1,3]` Find the local and absolute extreme values of the function on the given interval.

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Chapter 4, Review - Problem 6 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`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 

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