Question

What are the absolute extrema of `f(x)=x/(x^2+1) in(0,2)`?

Answer 1

The first derivative is

`df(x)/dx=(x'*(x^2+1)-x*(x^2+1)')/(x^2+1)^2=(1-x^2)/(1+x^2)^2`

Hence the first derivative nullifies at `x=+-1` but because `x ε (0,2)`

We have that at `x=1` we got a maximum which is

`f(1)=1/2`