Inverse of rational function $y= \frac{3-x}{1+x^2}$ I have the function $$y= \frac{3-x}{1+x^2}$$ and I want to find the inverse of this function.
I know that 
$$x= \frac{1 \pm \sqrt{1-4y(3-y)}}{2y}$$ 
My question is how do I find the domain where the function is 
$$x= \frac{1- \sqrt{1-4y(3-y)}}{2y}$$ 
and 
$$x= \frac{1+ \sqrt{1-4y(3-y)}}{2y}$$ 
respectively?

I guess you got the idea, you just need to be more careful with the signs.
To find the inverse of:
$$inverse\:\frac{3-x}{1-x^2}$$
We'll need to find the inverse function in the form $y=g(x)$, given the above $f(x)$.
First, we need to get the quadratic equation solution general formula:
given:
$$Ax^2+Bx+C=0$$
$y_1(x)$ and $y_2(x)$ are roots such that:
$$y_1(x)=\frac{-B+\sqrt{B^2-4AC}}{2A}$$
and
$$y_2(x)=\frac{-B-\sqrt{B^2-4AC}}{2A}$$
To get the inverse of $f(x)$, we do two steps:
1-Interchange the variable name $x$ with $y$.
2-Solve the resulting equation for $y$.
1-Interchange the variable name $x$ with $y$ given:
$$f(x)=y=\frac{3-x}{1-x^2}$$
You get:
$$x=\frac{3-y}{1-y^2}$$
2-Solve for $y$:
$$ x(1-y^2) = 3-y$$
$$x-xy^2 = 3-y$$
$$x-xy^2-3+y=0$$
$$(-x)y^2+(1)y+(x-3)
Here $A=-x$, $B=1$, $C=x-3$
$$y_1(x)=\frac{-1+\sqrt{1^2-4(-x)(x-3)}}{2x}$$
Simplifying to get:
$$y_1\left(x\right)=\frac{-1+\sqrt{1+4\left(x\right)\left(x-3\right)}}{2x}$$
You can do the same for $y_2(x)$ 
$$y_2\left(x\right)=\frac{-1-\sqrt{1+4\left(x\right)\left(x-3\right)}}{2x}$$
