Question 1209607
<pre>
{{{r(x) = (x^2 + 1)/(1 - x)^2}}}}

{{{r(x) = (x^2 + 1)/(1 -2x+ x^2)}}}}

So r(x) has two horizontal asymptotes, the x-axis, and

y = quotient of leading coefficients
or {{{y=1/1}}} or y=1.  

That might make one think that 1 would be eliminated from the range
but it wouldn't because (0,1) is a point on the graph.

The numerator and denominator are both always positive.

So the range is all positive numbers.

{{{(matrix(1,3,0,",",infinity))}}}

Edwin</pre>