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

$\mathrm{df} \frac{x}{\mathrm{dx}} = \frac{x ' \cdot \left({x}^{2} + 1\right) - x \cdot \left({x}^{2} + 1\right) '}{{x}^{2} + 1} ^ 2 = \frac{1 - {x}^{2}}{1 + {x}^{2}} ^ 2$
Hence the first derivative nullifies at $x = \pm 1$ but because x ε (0,2)
We have that at $x = 1$ we got a maximum which is
$f \left(1\right) = \frac{1}{2}$