Problems mainly based on a composite function $f(f(x))={f(x)}^3$ 
For $f(x)$ which is differentiable over all $\mathbb{R}$, $f(x)$ meets four conditions
(a) $f(4) < 1 $
(b) For all $\mathbb{R}$, $f(f(x))={f(x)}^3$
(c) if $p < q$, $0\le f(p) \le f(q)$
(d).  if $p < q$, $f'(p) + \cfrac{3}{4}p \le f'(q) + \cfrac{3}{4}q$
for all $f(x)$ that meets above conditions, $M(x)$ is the function where $f(4)$ become largest number. (maximum)
Find the value of $$\int_{-2}^{2} M(x) dx$$
Source: A Korean hs problem

I was struggling to understand what (b) means, and I couldn't figure it out from the start.
(detailed questions after some comments)
I was mainly curious about the condition (b).
Simply thought, $f(x)$ is $0$, $1$ or $x^3$
But it is not likely, because of the condition (a)
I think the problem wanted me to think about the domain and range of $f(x)$
But at that point, I wasn't able to move forward.
I don't know how to connect that thought with other conditions.
 A: Some observations:

*

*Rearranging condition 'd' , we find: $ \frac{f'(p) - f'(q)}{p-q} \leq \frac{3}{4}$, take the limit as $ p \to q$ with $p>q$ on LHS. Take the upper side limit to avoid problems:
$$ \lim_{ p \to q^+} \frac{ f'(p) - f'(q)}{q-p} \leq \frac34$$
We find that: $ -f''(x) \leq \frac34$(*)


*Equation b, we find that if $f(x)=0$ at any $ x \in \mathbb{R}$ , then it must be that $x=0$ is a zero for the function as well.


*The c. condition can be written as: $f'(x) \geq 0$


*If $f(4)<1$, then it must be that $f(f(4))<f(4)$, by the facts that $f(4)<4$ and $f(x)$ is an increasing function.
We can also prove by induction on the above fact that upon setting the number of compositions to infinity, the function goes to zero. Since $0 \leq f(x)$, we find the lower bound is zero.


*(From Lutz Lehman) , if a function is continuous for all $x \in \mathbb{R}$, then it must be that the range can be written as an open interval $(0,b)$. Further, we can identify a polynomial equation which the upper bound must satisfy.

Let $b$ be the upper bound, if $b>1$ then $f(b) \geq b$, then $b^3 \geq b$, this is only without contradiction if $b^3 = b$.
If $b<1$, then we find that $1 \leq b$
Hence, we arrive at the conclusion of Lutz Lehman on the bounds:
$$0 \leq b <1$$

Example function:
As noted by Lutz Lehman, $f(x)=0$ seems to be a function satisfying the criteria in the question + the ones derived above.

*: I assumed existence of second derivative to make this work, otherwise there is no reason for upper limit and lower limit to agree.
**: I assumed $f'(x) \neq 0$
