TITLE: Number of distinct solutions of $f(f(x))=0$
QUESTION [6 upvotes]: Let $f(x)=x^3-3x+1.$ Then what is the number of different real solutions of the
equation $f(f(x))=0$?
$f(x)$ has three roots and $f(f(x))$ will be 0 when value of $f(x)$ is equal to its root. But this approach is turning out to be tedious as finding exact values of roots of $f(x)$ is not possible Could someone suggest a better approach?

REPLY [2 votes]: $$f(x)=x^3 -3x+1$$
$$f'(x)=3x^2-3$$
So the maximum is at $(-1,3)$ , minimum at $(1,-1)$
Thus $f(x)$ has $3$ real root.
Now in order for $f(f(x))=0$, $f(x)$ must be equal to one of its roots.
So it is really asking how many roots does each $f(x)=k$ have, where $k$ is a root of $f(x)$.
As observed from the calculation above, the minimum is at $y=-1$ and the maximum is at $y=3$. So if $f(x)>3$ or $f(x)<-1$, there is only one root. Whereas if $f(x)$ equals $-1$ or $3$, there are two distinct roots. If $-1<f(x)<3$, there are three distinct roots.