# How do you find all the extrema for f(x) = x/(x^2+x+1) on [-2,2]?

Sep 10, 2015

Relative Max: $\frac{1}{3}$ (at $1$), Relative min $- 1$ (at$- 1$)
Absolute max and min are the same as the relative.

#### Explanation:

$f \left(x\right) = \frac{x}{{x}^{2} + x + 1}$ on [-2,2]

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

$f ' \left(x\right)$ is never undefined and is $0$ at $x = \pm 1$

On $\left[- 2 , - 1\right)$, we have $f ' \left(x\right) < 0$
on $\left(- 1 , 1\right)$, we have $f ' \left(x\right) > 0$

So $f \left(- 1\right) = - 1$ is a relative minumum.

On $\left(- 1 , 1\right)$, we have $f ' \left(x\right) > 0$
on $\left(1 , 2\right]$, we have $f ' \left(x\right) < 0$

So $f \left(1\right) = \frac{1}{3}$ is a relative minumum.

$f \left(- 2\right) = - \frac{2}{3} > - 1$ and $f \left(2\right) = \frac{2}{7} < \frac{1}{3}$ so the relative extrema are also absolute on the interval.