<s>[INST] An expression for $f(2x)$ as a rational function of $f(x)$ A real-valued function $f$ defined on the $\mathbb{R}\setminus\{1\}$ by
\begin{equation*}
f(x) = \frac{3x + 1}{x - 1} = 3 + \frac{4}{x - 1}
\end{equation*}
is invertible, and
\begin{equation*}
f^{-1}(x) = \frac{x + 1}{x - 3} = 1 + \frac{4}{x - 3} .
\end{equation*}
Evaluate constants $A$, $B$, and $C$ such that
\begin{equation*}
f(2x) = \frac{Af(x) + B}{f(x) + C}
\end{equation*}
for every real number $x$ distinct from $1/2$ and 1.
This is a problem from a past high school competition given in a county - Monroe County - in New York.  (I will post a solution to the problem.)  Does every rational function that is equal to the quotient of two linear functions have such a property?  Can $f(3x)$ be expressed analogously?  How about $f(kx)$ for any positive integer $k$?  Is there any advantage to such an expression over substituting "$2x$" in for "$x$"?
 [/INST] Here is the intended solution to the problem.  This is a lot of work for a problem on a ten-minute competition.  (There are six topics at each competition, and in each topic, there are three problems. The problem in the post comes from the topic called "Functions." I doubt anyone solved this problem during the competition.)
Solution
If $x > 1$, for example, there is exactly one real number $y > 3$ such that $f^{-1}(y) = x$.
\begin{equation*}
f(2x) = \frac{Ay + B}{y + C} .
\end{equation*}
$2x > 1$, $f(2x) > 3$, and $f(2x)$ is in the domain of $f^{-1}$.
\begin{equation*}
\frac{2y + 2}{y - 3} = 2x = f^{-1}\bigl(f(2x)\bigr)
= f^{-1}\left(\frac{Ay + B}{y + C}\right)
= \frac{\dfrac{Ay + B}{y + C} + 1}{\dfrac{Ay + B}{y + C} - 3}
= \frac{(A + 1)y + B + C}{(A - 3)y + B - 3C} .
\end{equation*}
\begin{equation*}
(2y + 2)\bigl[ (A - 3)y + B - 3C \bigr] = (2A - 6)y^{2} + (2A + 2B - 6C - 6)y + 2B - 6C ,
\end{equation*}
and
\begin{equation*}
(y - 3)((A + 1)y + B + C) = (A + 1)y^{2} + [-3A + B + C - 3]y - 3B - 3C .
\end{equation*}
For every real number $y > 3$,
\begin{equation*}
(2A - 6)y^{2} + (2A + 2B - 6C - 6)y + 2B - 6C = (A + 1)y^{2} + (-3A + B + C - 3)y - 3B - 3C .
\end{equation*}
So, $(A, \, B, \, C)$ is the solution to the system of linear equations
\begin{equation*}
\begin{cases}
2x - 6 = x + 1 \\
2x + 2y - 6z - 6 = -3x + y + z - 3 \\
2y - 6z = -3y - 3z
\end{cases}
\end{equation*}
in the variables $x$, $y$, and $z$. The only solution to this system of equations is $(7, \, 3, \, 5)$. So, $A = 7$, $B = 3$, and $C = 5$.
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