Question 890728
Domain : All x, no restrictions.
Since both numerator and denominator are squared, there are no negative values. 
The minimum value occurs when {{{x=0}}} and equals {{{0}}}.
As {{{x}}} approaches {{{infinity}}}, 
{{{lim(x->infinity,f=1/(1/x^2+1)=1/(9+1)=1)}}}
.
.
.
{{{graph(300,300,-5,5,-5,5,x^2/(1+x^2))}}}
.
.
.
So the range is [0,1).