Meaning of $\left| (f\circ f)(x) \right|$ is not differentiable only on single point, when $f(x)$ is a cubic function?

Question

• $f(x)$ is a cubic function, and its leading coefficient is $1$.
• $f(x)$ intersects with $x$-axis only on $(1,0)$.
• $f'(1) \ne 0$.
• $\left| (f\circ f)(x) \right|$ is not differentiable only on $x=2$.

What is the number of all $f(4)$s, when $f(4) \in \mathbb{Z}$?

My approach:

1. $\left| (f\circ f)(x) \right|$ is not differentiable only on $x=2$.
2. It means that the graph of $(f\circ f)(x)$ has to pass $x=2$, but not tangent to $x$-axis.
3. So, the only number that satisfies $(f\circ f)(x) = 0$ and $f'(f(x))f'(x) \ne 0$ is $2$.
4. But, since $f(x)=0$ only on $x=1$, step 3 is equivalent to:
5. The only number satisfies $f(x)=1$ and $f'(1)f'(x) \ne 0$ is $2$.
6. Applying $f'(1) \ne 0$,
7. $2$ is the only number that satisfies $f(x)=1$ and $f'(x) \ne 0$.

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8. $f(x) = (x-1)(x^2+px+q)$.
9. $x^2+px+q$ cannot have any roots, so $p^2-4q < 0$.

And I'm stuck here. I have thought about getting $f'(x)$ on step 8, and applying step 7 on it, but I didn't do it because it would give me some meaningless equations which cannot be solved.

Are these steps correct, without any leap of logic?