AMC 12 Mock Exam 4 Problem 22

Let $g(x)=\frac1{1-x}$. Then $g(3)=-\frac12$, $g\left(-\frac12\right)=\frac23$, and $g\left(\frac23\right)=3$; none of these is $0$ or $1$. Writing $a=f(3)$, $b=f\left(-\frac12\right)$, $c=f\left(\frac23\right)$, the equation at $x=3,-\frac12,\frac23$ gives $$a+b=3,\qquad b+c=-\tfrac12,\qquad c+a=\tfrac23.$$ Adding, $a+b+c=\frac12\left(3-\frac12+\frac23\right)=\frac{19}{12}$, so $a=\frac{19}{12}-(b+c)=\frac{19}{12}+\frac12=\boxed{\tfrac{25}{12}}$.

(Such an $f$ exists: in general, solving the same $3\times3$ system at each orbit $\{x,g(x),g(g(x))\}$ defines $f$ consistently.)