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 "$x$"?
 A: 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 $f(x)$.
A: This question is easily answered, IF you know the relation between fractional-linear transformations and $2$-by-$2$ matrices.
A fractional-linear transformation is, as you’ve identified, quotient of two linear polynomials that doesn’t reduce to a constant, so representable as $x\mapsto\frac{ax+b}{cx+d}$. The wonderful fact, verifiable with a minimum of work, is that these transformations are well representable by matrices, $\begin{pmatrix}a&b\\c&d\end{pmatrix}=M$ in this case. If $\,f$ and $g$ are frac. lin. tfs., represented by matrices $M$ and $N$ as above, then the wonderfulness is that $f\circ g$ is represented by $MN$. You must be aware, however, that if $\lambda$ is a nonzero constant, $\lambda M$ and $M$ represent the same fractional-linear transformation.
You may not recognize $x\mapsto2x$ as a fractional-linear transformation, but it is, and its matrix representation is $P=\begin{pmatrix}2&0\\0&1\end{pmatrix}$. Your original frac. lin. tf. is $f$, with matrix $Q=\begin{pmatrix}3&1\\1&-1\end{pmatrix}$. Looking at the problem finally, you see that you are being asked to find $R$ such that $RQ=QP$. Solve for $R$ in noncommutative matrix-land, and you find that $R=QPQ^{-1}$. Fortunately, they’ve given you $Q^{-1}$. Thus
$$
R=\begin{pmatrix}3&1\\1&-1\end{pmatrix}
\begin{pmatrix}2&0\\0&1\end{pmatrix}
\begin{pmatrix}1&1\\1&-3\end{pmatrix}=
\begin{pmatrix}7&3\\1&5\end{pmatrix}\,,
$$
from which you read off your coefficients.
To a student armed with the theory, it should take about two minutes to finish this question off.
A: 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 $C = 5$.
A: Let $\displaystyle f(x)=\frac{Ax+B}{Cx+D}$, $x\in\Bbb{R}\setminus\{-\frac DC\}$. For every $k\in\Bbb{N}$, we assume there exists $P,Q,T\in\Bbb{R}$ satisfy
$$f(kx)=\frac{Pf(x)+Q}{f(x)+T}$$
Then
$$f(kx)[f(x)+T]=Pf(x)+Q$$
$$\frac{Akx+B}{Ckx+D}\cdot\frac{(A+CT)x+(B+DT)}{Cx+D}=\frac{(AP+CQ)x+(BP+DQ)}{Cx+D}$$
$$(Akx+B)\left[(A+CT)x+(B+DT)\right]=(Ckx+D)[(AP+CQ)x+(BP+DQ)]$$
Comparing both sides we get
\begin{equation*}
\begin{cases}
A(A+CT)=C(AP+CQ) \\
B(A+CT)+Ak(B+DT)=D(AP+CQ)+Ck(BP+DQ) \\
B(B+DT)=D(BP+DQ)
\end{cases}
\end{equation*}
We have to solve the system of $P,Q,T$:
$$ \left[
    \begin{array}{ccc|c}
      AC&C^2&-AC&A^2\\
      (AD+kBC)&CD(k+1)&-(BC+kAD)&AB(k+1)\\
      BD&D^2&-BD&B^2\\
    \end{array}
\right] $$
Finally we get 
$$Q=-\frac{AB}{CD},T=\frac{(\frac AC+\frac BDk)\beta-2AB(k+1)}{(k-1)\alpha},P=T+\frac{\beta}{CD}$$
where $\alpha=AD-BC$ and $\beta=AD+BC$.
If the solution exists, the conditions are 
$$CD\neq0,\alpha\neq0,k\neq1$$
