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

May 27, 2016

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

#### Explanation:

$f ' \left(x\right) = \frac{\left(1\right) \left({x}^{2} - x + 1\right) - \left(x\right) \left(2 x - 1\right)}{{x}^{2} - x + 1} ^ 2 = \frac{1 - {x}^{2}}{{x}^{2} - x + 1} ^ 2$

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

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

$f \left(0\right) = 0$
$f \left(1\right) = 1$
$f \left(3\right) = \frac{3}{7}$

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