Number of real solutions of $f(f(x))$

Question

Given $f(x)=x^3-12x+3$, find number of real solutions of $f(f(x))=0$.

My Try: Since $f(x)$ is having three real toots say $\alpha$, $\beta$ and $\gamma$, we have $$f(f(\alpha))=f(f(\beta))=f(f(\gamma)) =f(3).$$

Hence by Rolles theorem $\exists$ atleast one $c $ such that $f'(c)=0$

But how to count number of roots?

Answer 1

Hint: What matters are the critical points and values. Thus you could first locate where $f'$ vanishes and the values at those points. The critical points cuts ${\Bbb R}$ into three intervals on which $f$ is monotone, now find the image of each such interval and convince yourself that each image contains the three roots of $f$, i.e. all in all there are 9 real solutions.

Answer 2

$f(x)=x^3 - 12 x + 3$ has $3$ real solutions, as you stated: $\alpha,\beta,\gamma\in\mathbb{R}$

$f(f(x))=0 $ means $\left(x^3-12 x+3\right)^3-12 \left(x^3-12 x+3\right)+3=0$

translates in three $3-$rd degree equations

$\alpha ^3-12 \alpha +3=\alpha;\;\beta ^3-12 \beta +3=\beta;\;\gamma^3-12 \gamma +3=\gamma$

that is

$\alpha ^3-13 \alpha +3=0;\;\beta ^3-13 \beta +3=0;\;\gamma^3-13\gamma +3=0$

Each of which you can easily prove has three real roots, so the global number of real roots is $9$

Hope it is useful