<s>[INST] 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?
 [/INST] Domain of $x$ is the range of $y$, and vice versa.
You have a square root for $x=f(y)$, so you should solve the condition for the expression within that square root is bigger than or equal $0$. Or:
$1-12y+4y^2\ge 0$
Solving $1-12y+4y^2=0$ yields $(1/2)(3+2\sqrt{2})$ and $(1/2)(3-2\sqrt{2})$. These two are real values, the coefficient a of the original quadratic expression is $4$, or $4>0$. Hence, the expression will have a negative value between the 2 roots.
Therefore, for this, $y<(1/2)(3+2\sqrt{2})$ and $y>(1/2)\sqrt{3-2\sqrt{2}}$ will make the square root of $f(y)$ "meaningful", thus, that is the domain of $x$. And yes, the two $x=f(y)$ share the same domain
However, based on the comments, I suggest you to double-check on your math again, just to make sure the $x=f(y)$ is written correctly. The way to process after that is the same though.
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