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?
 A: 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.
A: 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}$$
A: The graph of an inverse function is the graph of the original function reflected about the line $y=x,$ and, to be a function, a graph must pass the vertical line test. Another way to say this is that it fails to be a function when the original function fails a horizontal line test. The function is monotonic decreasing up to its minimum then monotonic increasing up to its maximum and finally monotonic decreasing at the end. You can now use calculus or some other means to calculate the minimum $m$, the maximum $M$ and you'll have three regions where the inverses are well defined - $(-\infty,m], [m,M]$ and $[M,\infty)$. Note that every inflection point should be checked this way as locally that's where the function violates injectivity unless it's constant. This function just happens to have a unique maximum and minimum.
