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

##### 1 Answer

The absolute extrema of a function in a closed interval

So, let's find the local extrema:

if

So our function is decresing in

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

So the *candidates* are:

and it's easy to understand that the absolute extrema are

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