, 2024

Solution:
Observe that $f(x) = (x+3)^2 - 3$. Now, we claim that

Claim 1. $f^{k}(x) = (x+3)^{2^{k}} - 3$ for all positive integers $k$.

Proof. We use induction. The base case $k=1$ is clear. To show the inductive step, note that $f^{k}(x) = (x+3)^{2^{k}} - 3$ implies

$$f^{k+1}(x) = f(f^{k}(x)) = f((x+3)^{2^{k}} - 3) = (((x+3)^{2^{k}} - 3) + 3)^2 - 3 = (x+3)^{2^{k+1}} - 3$$

Thus, if $r$ is a real root of $f^{6}$, then $(r+3)^{64} = 3$, so $r+3 = \pm \sqrt[64]{3}$, and hence $r = \pm \sqrt[64]{3} - 3$. The largest value of $r$ is thus $\sqrt[64]{3} - 3$.