The domain of #f(x)# is #RR#
To determine the points of inflexion, we calculte #f''(x)# and find out when it is #=0#
We start by calculating the first derivative
#u(x)=x# #=># #u'(x)=1#
#v(x)=1+x^2# #=># #v'(x)=2x#
#f'(x)=(u'v-uv')/v^2= (1*(1+x^2)-x*2x)/(1+x^2)^2#
#=(1+x^2-2x^2)/(1+x^2)^2=(1-x^2)/(1+x^2)^2#
#f(x)=0# when #1-x^2=0# #=># #x=+-1#
from #x=oo# to #x=-1#
#f'(x)# is negative
from #x=-1# to #x=1#
#f'(x)# is positive
from #x=1# to #x=oo#
#f'(x)# is negative
So we have a minimum at #(-1,-1/2)#
and a maximum at # (1,1/2)#
Let's calculate #f''(x)#
#u=1-x^2# , #u'=-2x#
#v=(1+x^2)^2# so #v'=4x(1+x^2)#
So #f''(x)=(-2x(1+x^2)^2-(1-x^2)(4x(1+x^2)))/(1+x^2)^4#
#f''(x)=0# when #x=0# and tis correspond to the point #(0,0)#
#f''(x)=(-2x(1+x^2)(1+x^2+2-2x^2))/(1+x^2)^4#
This is #=0# for #x=0#
and #3-x^2=0# # => # #x=+-sqrt3#
graph{x/(1+x^2) [-5, 5, -2.5, 2.5]}
