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