
Problem 4. Let $f(x)=x^{3}-3 x+1$. Find the number of different real solutions of the equation $f(f(x))=0$.


Solution: Since $f^{\prime}(x)=3(x-1)(x+1)$ it follows that $f$ is strictly monotone non-decreasing in the intervals $(-\infty,-1]$ and $[1, \infty)$ and strictly monotone non-increasing in the interval $[-1,1]$. Moreover $\lim _{x \rightarrow \pm \infty} f(x)= \pm \infty, f(-1)=3, f(1)=-1, f(3)=19>0$ and so the equation $f(x)=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$ such that $x_{1}<-1<x_{2}<1<x_{3}<3$. Therefore $f(x)=x_{1}$ has only one real root (which is less than -1 ) and $f(x)=x_{2}$ and $f(x)=x_{3}$ have three distinct real roots each (one in each of the intervals $(-\infty,-1),(-1,1)$ and $(1, \infty))$. Since the roots of $f(f(x))=0$ are exactly the roots of these three equations, we conclude that it has seven distinct real roots.
