bluecarneal wrote: For any function $f(x)$, define $f^1(x) = f(x)$ and $f^n (x) = f(f^{n-1}(x))$ for any integer $n \ge 2$. Does there exist a quadratic polynomial $f(x)$ such that the equation $f^n(x) = 0$ has exactly $2^n$ distinct real roots for every positive integer $n$? (6 points) Yes. Choose for example $f(x)=2x^2-1$. $f^n(x)$ has roots $\cos\left(\frac{\pi}{2^{n+1}}+\frac{k\pi}{2^n}\right)$ with $k=0,1,2,...,2^n-1$

That polynomial is not at all unexpected; in fact it is $T_2(x) = 2x^2-1$, the Chebyshev polynomial of the first kind of degree $2$. Since $T_k\circ T_\ell = T_{k\ell}$, it follows $T_2^n = T_{2^n}$, and that is all, since it is known $T_k$ has precisely $k$ real roots (all belonging to $(-1,1)$).