Find a function within a function

Question

Let $$f(x)=4x(1-x).$$ Find no. of solutions of $$f(f(f(f(x)))))=x/3.$$ I tried it by plotting the graph but I'm not able tho carry of further transformation for $f(f(x))$. Is there any easier way to solve it

Answer 1

Draw the graph of $f$! It starts at $(0,0)$, climbs monotonically to $\bigl({1\over2},1\bigr)$, and then descends again monotonically to $(1,0)$. In other words, $f$ maps each of the intervals $\bigl[0,{1\over2}\bigr]$ and $\bigl[{1\over2},1\bigl]$ bijectively onto the interval $[0,1]$.

When $x$ runs from $0$ to ${1\over2}$ then $y=f(x)$ runs monotonically from $0$ to $1$. It follows that $f\bigl(f(x)\bigr)=f(y)$ runs first from $0$ to $1$ and then back to $0$. Similarly, when $x$ runs from ${1\over2}$ to $1$ then $y=f(x)$ runs monotonically from $1$ to $0$. It follows that $f\bigl(f(x)\bigr)=f(y)$ runs first from $0$ to $1$ and then back to $0$.

Looking at the whole interval $0\leq x\leq 1$ we see that $f\circ f$ creates four images $[0,1]$ of the original $x$-interval $[0,1]$. As a consequence $f\circ f\circ f$ creates eight images, and $g:=f^{\circ 4}$ sixteen such images. Therefore we can say that the graph of $$g:\>[0,1]\to[0,1],\qquad x\mapsto f^{\circ4}(x)$$ starts at $(0,0)$, makes eight smooth waves up to $y=1$ and back to $y=0$, then ends at $(1,0)$. The line $y={x\over3}$ then intersects this graph in totally $16$ points, one of them being $(0,0)$.

The equation $g(x)={x\over3}$ is a polynomial equation of degree $16$. Since we have established $16$ real roots we can be sure that there are no other real intersections of the two graphs in question.