Question:

Let $g(x) = \frac{x+4}{x-2}$. We need $f(x) = f(g(x))$, meaning $x$ and $g(x)$ must map to the same value under $f$. First, find fixed points: $g(x) = x \Rightarrow \frac{x+4}{x-2} = x \Rightarrow x+4 = x^2-2x \Rightarrow x^2-3x-4 = 0 \Rightarrow x = 4$ or $x = -1$. At fixed points, $f(x) = f(g(x))$ is automatically satisfied. Next, check if $g(g(x)) = x$ (2-cycles): $g(g(x)) = g\left(\frac{x+4}{x-2}\right) = \frac{\frac{x+4}{x-2}+4}{\frac{x+4}{x-2}-2} = \frac{x+4+4(x-2)}{x+4-2(x-2)} = \frac{5x-4}{-x+8}$. Setting this equal to $x$ gives $5x-4 = -x^2+8x \Rightarrow x^2-3x-4 = 0$, yielding the same solutions. Therefore, any orbit under $g$ either reaches a fixed point or cycles back. The minimum configuration is when $f$ has exactly 2 roots at the fixed points $x = 4$ and $x = -1$.