# What is the absolute extrema of the function: 2x/(x^2 +1) on closed interval [-2,2]?

Apr 29, 2015

The absolute extrema of a function in a closed interval $\left[a , b\right]$ can be or local extrema in that interval, or the points whose ascissae are $a \mathmr{and} b$.

So, let's find the local extrema:

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

$y ' \ge 0$

if

$- {x}^{2} + 1 \ge 0 \Rightarrow {x}^{2} \le 1 \Rightarrow - 1 \le x \le 1$.

So our function is decresing in $\left[- 2 , - 1\right)$ and in $\left(1 , 2\right]$ and it is growing in $\left(- 1 , 1\right)$, and so the point $A \left(- 1 - 1\right)$ is a local minimum and the point $B \left(1 , 1\right)$ is a local maximum.

Now let's find the ordinate of the points at the extrema of the interval:

$y \left(- 2\right) = - \frac{4}{5} \Rightarrow C \left(- 2 , - \frac{4}{5}\right)$

$y \left(2\right) = \frac{4}{5} \Rightarrow D \left(2 , \frac{4}{5}\right)$.

So the candidates are:

$A \left(- 1 - 1\right)$
$B \left(1 , 1\right)$
$C \left(- 2 , - \frac{4}{5}\right)$
$D \left(2 , \frac{4}{5}\right)$

and it's easy to understand that the absolute extrema are $A$ and $B$, as you can see:

graph{2x/(x^2 +1) [-2, 2, -5, 5]}