Evaluate the absolute maximum and minimum of `q(x)=xe^(-x^2/2)` on the interval [0,2].

The function is countinuous on the closed interval, and thus has both a maximum and a minimum. The extrema can only occur at critical points or the endpoints of the interval. The critical values will be when `q'(x)=0`

`q'(x)=e^(-x^2/2)+x(-x)e^(-x^2/2)`

`q'(x)=0` ==> `e^(-x^2/2)=x^2e^(-x^2/2)` Since `e^(-x^2/2)!=0` we can divide to get:

`x^2=1 ==>x=+-1` . Only x=1 is in the interval.

`q(0)=0`

`q(1)=e^(-1/2)~~.60653`

`q(2)=2e^(-2)~~.27067`

**So the minimum is 0 at x=0 and the maximum is `e^(-1/2)~~.60653` at x=1.**

The graph:

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