Question

What are the absolute extrema of `f(x)= xsqrt(25-x^2) in [-4,5]`?

Answer 1

The absolute minimum is `-25/2` (at `x=-sqrt(25/2)`). The absolute maximum is `25/2` (at `x=sqrt(25/2)`).

Explanation:

`f(-4) = -12` and `f(5)=0`

`f'(x) = sqrt(25-x^2)+x/(cancel(2)sqrt(25-x^2))*-cancel(2)x`

`= (25-x^2-x^2)/sqrt(25-x^2) = (25-2x^2)/sqrt(25-x^2)`

The critical numbers of `f` are `x=+-sqrt(25/2)` Both of these are in `[-4,5]`..

`f(-sqrt(25/2)) = -sqrt(25/2)sqrt(25-25/2)`

`= -sqrt(25/2)sqrt(25/2) = -25/2`

By symmetry (`f` is odd), `f(sqrt(25/2)) = 25/2`

Summary:
`f(-4) = -12`
`f(-sqrt(25/2)) = -25/2`
`f(sqrt(25/2)) = 25/2`
`f(5)=0`

The absolute minimum is `-25/2` (at `x=-sqrt(25/2)`).
The absolute maximum is `25/2` (at `x=sqrt(25/2)`).