It is bijective so it should have one. I was solving some of my homework when I came upon this. The answer key had a question mark for the solution. I looked it up on Wolfram Alpha and there is a solution: $f^{-1}(x)=\frac{2x^2+1}{2x^2}\pm\frac{\sqrt{4x^2+1}}{2x^2}$
I tried to backtrack from the solution but I had no luck. Could someone point me on the right path?
On
The inverse function of $y=\frac{\sqrt{x}}{x-1}$ is $x=\frac{\sqrt{y}}{y-1}$. What we need to do is to express this implicit-function form into the explicit form.
We can consider this as solving the equation $x = \frac{\sqrt{y}}{y-1}$ with the variable $y$. Note that this equation can be observed as $$x(\sqrt{y})^2 - \sqrt{y} - x=0,$$ the quadratic equation with the variable $\sqrt{y}$. Apply the quadratic formula, and just square it!
First of all, we need to properly set the domain and co-domain. The domain is $$ D=\{x\in \mathbb{R}: x \ge 0, x \ne 1\} = [0,1[\cup]1,+\infty[ $$
and the co-domain is $D^* =\mathbb{R}$. As you have mentioned, $f:D \to D^*$ is one-to-one, and therefore invertible. An expression for the inverse can be obtained using the method mentioned by @SteveKim: $$ \frac{\sqrt{x}}{x-1} = y \Leftrightarrow y x -\sqrt{x} - y = 0 \Leftrightarrow y (\sqrt{x})^2-\sqrt{x}-y = 0, $$
yielding $$ \sqrt{x} = \dfrac{1\pm \sqrt{1+4y^2}}{2y}. $$
When $y > 0$, we must choose the + sign and get, after squaring, $$ x = 1+\dfrac{1+\sqrt{1+4y^2}}{2y^2} $$
when $y < 0$, we must choose the - sign and get $$ x = 1+\dfrac{1-\sqrt{1+4y^2}}{2y^2}.$$
Finally, the inverse is given by $$ f^{-1}(y)=\begin{cases} 1+\dfrac{1-\sqrt{1+4y^2}}{2y^2},& y <0\\ 0, & y=0\\ 1+\dfrac{1+\sqrt{1+4y^2}}{2y^2}, & y >0 \end{cases} $$