Given `f(x)=(2x)/((1+5x^2)^2)` for `-2<=x<=2` find the absolute max and min:

Since the function is continuous and on a closed interval, we are guaranteed both a maximum and a minimum. We must check any critical values as well as the endpoints:

The critical values are where the first derivative is zero or fails to exist.

We use the quotient rule to find the first derivative:

`f'(x)=(2(1+5x^2)^2-(2x)(2)(1+5x^2)(10x))/((1+5x^2)^4)`

`=((1+5x^2)[2(1+5x^2)-40x^2])/((1+5x^2)^4)`

`=(2-30x^2)/((1+5x^2)^3)`

The first derivative is continuous everywhere so we need only check its zeros:

`f'(x)=0==>2-30x^2=0==>x=+-sqrt(1/15)`

---------------------------------------------------------------

**The function's absolute maximum occurs at `x=sqrt(1/15)` while its absolute minimum occurs at `x=-sqrt(1/15)` regardless of the interval.**

----------------------------------------------------------------

The graph: