Deranged Function (1997 AIME Problem #12)

$f(x)=\dfrac {ax+b}{cx+d}$

$\implies f\left (f(x)\right ) =\dfrac {(a^2+bc)x+b(a+d)}{c(a+d)x+d^2+bc}$

$f\left (f(x)\right ) =x\implies$

$c(a+d)x^2+(d^2+bc)x=(a^2+bc)x+b(a+d)$

Comparing the coefficients of like terms we get $a+d=0$

So, $f(x)=\dfrac {ax+b}{cx-a}$

$f(19)=19\implies 38a+b=361c$

$f(97)=97\implies 194a+b=9409c$

Solving we get $a=58c,b=-1843c$

So, $f(x)=\dfrac {58x-1843}{x-58}$

So, the domain of $f(x)$ is $(-\infty, 58)\cup (58,\infty)$

For $x\neq 58,(x-58)f(x)=58x-1843$

If the value of $f(x)$ be $58$ , then the L. H. S. of the equation is

$58(x-58)=58x-3364$

Hence the above equation connecting $f(x)$ is not satisfied for any value of $x$

So, $58$ is not in the range of $f(x)$

So the only number out of the domain as well as the range of $f(x)$ is $\boxed {58}$ .

We can also compute the inverse of $f(x)$ since we have its inverse is the same as itself, $y=\frac{ax+b}{cx+d}\implies f^{-1}(x)=\frac{dx-b}{-cx+a}$ which means that $a, b, c$ and $d$ must be proportional to $d, -b, -c$ and $a$ . Thus, $a=-d$ and the rest follows.