f(x)= 2x/(x^2+1)

First we will determine the derivative f'(x)

Let f(x)= u/v

u= 2x ==> u'=2

v= x^2+1 ==> v'= 2x

f'(x)= (u'v-uv') /v^2

= [2(x^2+1) -(2x)(2x)]/( x^2+1)^2

= 2x^2+2 -4x^2 / (x^2+1)^2

= -2(x^2-1)/ (x^2+1)^2

= -2(x-1)(x+1)/ (x^2+1)^2

Reduce similar:

= -2(x-1)/(x^2+1)

Now we sill find critical values:

f'(x)=0 ==> x=1 is a critical value

The extreme value is:

f(1)= 2/2 =1