What is the intuition of this problem about $|f'(x)|$? $\newcommand{\d}{\mathrm{d}}$Recently, I met a problem, which says that:

Given that $f(x)$ have continious derivatives on close interval $[0,2]$, $f(0)=f(2)=0$, $M=\max_{[0,2]} |f(x)|$. Prove that:

(1) there exists a number $\xi \in (0,2)$, s.t. $|f'(\xi)| \geq M$.




(2) If for every $x \in (0,2)$, $|f'(x)|\leq M$, then $M=0$.


I can prove the first problem. However, the second problem makes me mad. So I look up the solution. The proof is here:

We know that there exists a number $x_0$, s.t. $M=|f(x_0)|=|f(x_0)-f(0)|=|\int_{0}^{x_0}f'(x)dx|\leq\int_{0}^{x_0}|f'(x)|dx\leq\int_{0}^{x_0}Mdx\leq Mx_0$,


$M=|f(x_0)|=|f(x_0)-f(2)|=|\int_{x_0}^{2}f'(x)dx|\leq\int_{x_0}^{2}|f'(x)|dx\leq\int_{x_0}^{2}Mdx\leq M(2-x_0)$


So,$M(1-x_0) \leq 0 and  M(1-x_0)\geq 0$. If $x_0 \neq 1$, $M$ is apparently $0$, if $x_0=1$, then $M \leq \int_{0}^{1}|f'(x)|dx \leq M$ and $M \leq \int_{1}^{2}|f'(x)|dx \leq M$, if $M \neq 0$, then it contradicts.

I can understand this solution, but I don't know the intuition or motivation behind this. I don't believe that there exists a solution which has no motivation, but I can't find the motivation behind this problem.
In fact, everytime I meet this kind of problems, I often have no idea, I really think that this kind of problems are hard to solve, can anyone give some advice or some resource about this kind of problems?
 A: $|f'(x)| \le M \ne 0$ for $x \in (0,2)$ means that the function cannot change values too quickly (the graph of $f$ is never too steep). However, it does need to change quickly enough to go from $0$ to $M$ (or $-M$) and then back to zero. It turns out that the restriction on the steepness of the function is too strict to allow it to reach $M$ or $-M$.
The gist of the solution is that applying the mean value theorem shows that the location of the maximizer (or minimizer) $x_0$ has to satisfy $x_0 \ge 1$ (cannot reach $\pm M$ from $f(0)=0$ quickly enough due to constrained slope) and $x_0 \le 1$ (cannot reach $\pm M$ from $f(2)=0$ quickly enough). So the remaining case to deal with is when $x_0=1$. There is an edge case to be considered (the tent function in my comment) that gets ruled out only because of non-differentiability at $x=1$.
A: The intuition is that to go from $0$ up to $M$ (or vice versa) over less than a unit interval, the average slope has to be greater than $M$.
To make the previous paragraph into a proof, suppose without loss of generality that $f(x_0)=M$ and that $0<x_0\leq1$ (if $x_0>1$, we do a symmetric argument using the endpoint $2$ instead of $0$, and if $f(x_0)<0$ we work with $-f$).
If $x_0<1$, by the Mean Value Theorem,
$$
M=f(x_0)-f(0)=f'(\xi)\,x_0,
$$
so if $M>0$ we get
$$
f'(\xi)=\frac{M}{x_0}>M,
$$
a contradiction. Thus $M=0$.
When $x_0=1$, the only possibility for $f$ is to have $f'=M$ on $(0,1)$ and $f'=-M$ on $(1,2)$. That would contradict the differentiability at $1$.
