Question

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

Answer 1

The absolute extrema of a function in a closed interval `[a,b]` can be or local extrema in that interval, or the points whose ascissae are `a or b`.

So, let's find the local extrema:

`y'=2* (1*(x^2+1)-x*2x)/(x^2+1)^2=2*(-x^2+1)/(x^2+1)^2`.

`y'>=0`

if

`-x^2+1>=0rArrx^2<=1rArr-1<=x<=1`.

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

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

`y(-2)=-4/5rArrC(-2,-4/5)`

`y(2)=4/5rArrD(2,4/5)`.

So the candidates are:

`A(-1-1)`
`B(1,1)`
`C(-2,-4/5)`
`D(2,4/5)`

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]}