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 "$x$"?

Rational functions of the form $x\mapsto \frac{ax+b}{cx+d}$ (with $ad\ne bc$) form a group under composition (Möbius transformations). You are given such an $f$ (and its inverse $f^{-1}$) and want some $g$ such that $g\circ f=f\circ 2$ (where the duplication function $2(x)=\frac{2x+0}{0x+1}$ is a Möbius transformation as well). We apply $\circ f^{-1}$ from the right to find $$ g=f\circ 2\circ f^{-1}.$$
In this particular case, 
$$g(x)=f\left(2\cdot \frac{x+1}{x-3}\right) =\frac{3\cdot2\cdot \frac{x+1}{x-3}+1 }{2\cdot \frac{x+1}{x-3}-1}=\ldots =\frac{7x+3}{x+5}$$
As long as we stay within Möbius transformations , there is nothing special about the $2$ in this problem. It would not be more complicated to ask how to express $f(\frac{x+42}{2-x})$ in terms of $f(x)$.
