Number of real zeroes of iterated polynomial: $x^3-2x+1$

Solution 1:

A sketch immediately reveals what is going on.

The polynomial $$ y = x^{\,3} - 2x + 1 $$ is a depressed cubic, with three real zeros at $x=\{-\phi, 1/\phi, 1\}$.

It has a local maximum $(x_{max},y_{max})=( - \sqrt{2/3}, \; 1+4*\sqrt{6}/9)$ and a local minimum $(x_{min},y_{min})=( \sqrt{2/3}, \; 1-4*\sqrt{6}/9)$.

The range $[y_{min},\,y_{max}]$ includes the two zeros $1/\phi$ and $1$, but not the lower at $-\phi$.

Then the sketch shows that at the first iteration $$ y_{\,2} (y_{\,1} (x)) $$ we will have that:
- the lower zero will remain, while the upper two will be replicated $3$ times;
- the maximum will remain, while the minimum will be replicated $3$ times;
- in between each triple $zero,min,zero$ a new maximum will appear;

However it is not easy to predict what the value of the two new maxima will be, and thus how many of the zeros $1/\phi$ and $1$ they will encompass in the following iteration, specially in the long run.
And that confirms @Jirky comment.