Question

What are the absolute extrema of `f(x) =x/(x^2-x+1) in[0,3]`?

Answer 1

Absolute minimum is `0` (at `x=0`) and absolute maximum is `1` (at `x=1`).

Explanation:

`f'(x) = ((1)(x^2-x+1)-(x)(2x-1))/(x^2-x+1)^2 = (1-x^2)/(x^2-x+1)^2`

`f'(x)` is never undefined and is `0` at `x=-1` (which is not in `[0,3]`) and at `x=1`.

Testing the endpoints of the intevral and the critical number in the interval, we find:

`f(0) = 0`
`f(1) = 1`
`f(3) = 3/7`

So, absolute minimum is `0` (at `x=0`) and absolute maximum is `1` (at `x=1`).