Fractional Function which f(f(x)) = x

The condition $f(f(x)) = x$ tells us that the function is self-inverse , that is, $f^{-1}(x) = f(x)$ . Inverting $f(x) = \frac{ax+b}{cx+d}$ gives us $f^{-1}(x) = \frac{-dx+b}{cx-a}$ . Matching coefficients in these rational functions gives us the condition $a = -d$ .

The two fixed points of $f(x) = \frac{ax+b}{cx-a} \$ , x = 19 and x = 97, leads to $(a \cdot 19) + b = 19 [(c \cdot 19) - a]$ and $(a \cdot 97) + b = 97 [(c \cdot 97) - a]$ , produces the system of equations $38a = 19^2 c - b \ , \ 194a = 97^2c - b$ , from which we obtain the relations $a = 58c \ , \ b = -1843c$ .

Thus, our self-inverse function is

$f(x) = \frac{58cx \ - \ 1843c}{cx-58c} \ = \ \frac{58x \ - \ 1843}{x-58}$ .

Since 1843 is not a multiple of 58, there is no further simplification of this rational function (this is important, a point raised in one of the recent set of multiple-choice questions involving rational functions of linear polynomials). We see that the "limit at infinity" of this function is 58 , so the range of $f(x)$ does not include this value. It should be noted that the domain of the function also does not include 58, as is expected for a self-inverse function. (Such functions are symmetrical about the line $y = x$ . )

From the first two conditions, we know that the equation $f(x)=x$ has two distinct solutions. Rewriting the equation we get: $f(x)=\frac{ax+b}{cx+d}=x \hspace{5pt}\Leftrightarrow\hspace{5pt} cx^2+(d-a)x-b=0$ This equation has two solutions only if $c\neq0$ . Obsereve that if we denote $a'=a/c, b'=b/c$ and $d'=d/c$ we get $\frac{a'x+b'}{x+d'}=\frac{ax+b}{cx+d}=f(x)$ , so we may assume that $c=1$ .

Observe that the coefficients of the composition $f\circ f$ correspond to the coefficients of the matrix $\begin{pmatrix} a & b \\ 1 & d \end{pmatrix}\cdot \begin{pmatrix} a & b \\ 1 & d \end{pmatrix} = \begin{pmatrix} a^2+b & b(a+d) \\ a+d & d^2+b \end{pmatrix}$ So, for the last condition ( $f(f(x))=x$ ) to hold we need the last matrix to be diagonal. This implies that $d=-a$ . We have $f(x)=\frac{ax+b}{x-a}$ Now we will use the first two conditions to find $a,b$ : we know that $19,97$ are the solutions of the equation $x^2-2ax-b=0$ (from the equation above, substituting $c=1, d=-a$ ). So $x^2-2ax-b=(x-19)(x-97)=x^2-116x+1843$ So $a=58$ , $b=-1843$ and $f(x)=\frac{58x-1843}{x-58}$ It is easy to check that the only value $f(x)$ can never achieve is $a/c=\boxed{58}$

This is from 1997 AIME .