Number of solutions for composite function

Question

If $f(x)=4x(1-x)$

The function is defined over $\Bbb{R}$.

Find the number of real solutions of $$f\circ f\circ f(x)=\frac x3.$$

Answer 1

Writing out the left hand side yields the polynomial $$f\circ f\circ f=-16384x^8+65536x^7-106496x^6+90112x^5-42240x^4+10752x^3-1344x^2+64x.$$ This is clearly divisible by $-64x$, and dividing this out leaves the remainder $$256x^7-1024x^6+1664x^5-1408x^4+660x^3-168x^2+21x-1.$$ The rational root test yields factors $x-1$ and $x-\frac{1}{2}$, the latter with multiplicity $2$. Dividing these out leaves $$64x^4+128x^3+48x^2-16x+1=(8x^2+8x-1)^2,$$ which has the two roots $\frac{1}{4}(2\pm\sqrt{2})$, each with multiplicity $2$. Note that both are in the interval $(0,1$).

This shows that the function $(f\circ f\circ f)(x)$ on the reals equals $0$ precisely at these five points, and that it is tangent to the $x$-axis at the three zeros between $0$ and $1$. This allows you to sketch the graph of $(f\circ f\circ f)(x)$ on some interval containing $[0,1]$, and from this it should be clear how many solutions to $(f\circ f\circ f)(x)=\frac{x}{3}$ there are.

I leave it to you to find the heights of the peaks of $(f\circ f\circ f)(x)$ between its zeros.