1980 AHSME Problems/Problem 14

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As $f(x)=cx/2x+3$, we can plug that into $f(f(x))$ and simplify to get $c^2x/2cx+6x+9 = x$ . However, we have a restriction on x such that if $x=-3/2$ we have an undefined function. We can use this to our advantage. Plugging that value for x into $c^2x/2cx+6x+9 = x$ yields $c/2 = -3/2$, as the left hand side simplifies and the right hand side is simply the value we have chosen. This means that $c=-3$, which is answer choice A.

Alternatively, after simplifying the function to $c^2x/2cx+6x+9 = x$, multiply both sides by $2cx+6x+9$ and divide by $x$ to yield $c^2=2cx+6x+9$. This can be factored to $x(2c+6) + (3+c)(3-c) = 0$. This means that both $2c+6$ and either one of $3+c$ or $3-c$ are equal to 0. $2c+6=0$ yields $c=-3$ and the other two yield $c=3,-3$. The clear solution is $c=-3$.