Answer
$f^{-1}(x)=(\displaystyle \frac{3x}{x-1})^{2}$
Domain of $f^{-1}(x)=(-\infty,0]\cup(1,\infty)$
Range of $f^{-1}(x)=$ domain od f =$[0,9)\cup(9,\infty)$
Work Step by Step
Domain of f = $\{x\in \mathbb{R}|x\geq 0,\ x\neq 9\}.$
(1) Solve the equation for x, writing $y=f(x)$
$y=\displaystyle \frac{\sqrt{x}}{\sqrt{x}-3},\quad $... multiply with $(\sqrt{x}-3)$
$y\sqrt{x}-3y=\sqrt{x}$
$y\sqrt{x}-\sqrt{x}=3y$
$\sqrt{x}(y-1)=3y$
$\displaystyle \sqrt{x}=\frac{3y}{y-1}$
$x=(\displaystyle \frac{3y}{y-1})^{2},\ \displaystyle \quad\frac{3y}{y-1}\geq 0$
(2) interchange $x\leftrightarrow y, \quad y=f^{-1}(x)$
$y=f^{-1}(x)=(\displaystyle \frac{3x}{x-1})^{2},\ \displaystyle \quad\frac{3x}{x-1}\geq 0$
To find the domain, divide $\mathbb{R}$ into intervals and test the sign of $\displaystyle \frac{3x}{x-1}$
$\left[\begin{array}{llll}
& (-\infty,0] & [0,1) & (1,\infty) \\
test~point & -1 & 0.5 & 2\\
value & \frac{-3}{-4} & \frac{1.5}{-0.5} & \frac{6}{1}\\
sign & + & - & +
\end{array}\right]$
Domain of $f^{-1}(x)=(-\infty,0]\cup(1,\infty)$
Range of $f^{-1}(x)=$ domain od f =$[0,9)\cup(9,\infty)$
