parmenides51 wrote: The function $g : [0, \infty) \to [0, \infty)$ satisfies the functional equation: $g(g(x)) = \frac{3x}{x + 3}$, for all $x \ge 0$. You are also told that for $2 \le x \le 3$: $g(x) = \frac{x + 1}{2}$. (a) Find $g(2021)$. (b) Find $g(1/2021)$. For $x>0$, let $h(x)=\frac 1{g(\frac 1x)}$. So that : $h(h(x))=x+\frac 13$ and $h(x)=\frac {2x}{x+1}$ $\forall x\in\left[\frac 13,\frac 12\right]$ We have $h(x+\frac 13)=h(x)+\frac 13$ and $h(\frac 13)=\frac 12$ and so ; $h(2021)=h(\frac 13+6062\frac 13)=h(\frac 13)+6062\frac 13=\frac{12127}6$ $h(\frac 1{2021})=h(\frac 1{2021}+\frac 13)-\frac 13$ $=\frac{2(\frac 1{2021}+\frac 13)}{(\frac 1{2021}+\frac 13)+1}-\frac 13$ $=\frac{4057}{24261}$ And so $\boxed{g(2021)=\frac{24261}{4057}}$ and $\boxed{g(\frac 1{2021})=\frac 6{12127}}$