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

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Answer 1 · 2015-11-04T05:18:44

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

$f(1)=1/2$

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