Let $f(x)=ax^3+bx^2+cx+5$. If $|f(x)|\le|e^x-e^2|$ for all $x\ge0$ and if the maximum value of $|12a+4b+c|$ is $m$, then find $[m]$ 
Let $f(x)=ax^3+bx^2+cx+5$. If $|f(x)|\le|e^x-e^2|$ for all $x\ge0$ and if the maximum value of $|12a+4b+c|$ is $m$, then find $[m]$ (where $[.]$ represents the greatest integer function.)

I first thought $|f(x)|\le|e^x-e^2|$ represents the boundedness of $f(x)$ but then realized a cubic function is unbounded. Also, the RHS of the inequality is not constant.
On observing $|12a+4b+c|$, I figured it is $f'(2)$. I don't think we can find its maximum value by finding the critical points of $f''(x)=0$ because that would give the extremum of $f'(x)$, not $f'(2)$.
If $|f(x)|\le|e^x-e^2|$, can we say anything about $f'(x)$?

We have $f(2) = 0$.
Then, we have, for all $x \ge 0, ~ x \ne 2$,
$$\left|\frac{f(x) - f(2)}{x - 2}\right|
\le \left|\frac{\mathrm{e}^x - \mathrm{e}^2}{x - 2}\right|$$
which results in $|f'(2)| \le \mathrm{e}^2$,
i.e. $|12a + 4b + c| \le \mathrm{e}^2$.
On the other hand,
when $a = - \frac58, ~ b = \frac{15}{4} + \frac12 \mathrm{e}^2, ~ c = -\frac{15}{2} - \mathrm{e}^2$ with $|12a + 4b + c| = \mathrm{e}^2$, it is not difficult to prove that $|f(x)| \le |\mathrm{e}^x - \mathrm{e}^2|$ for all $x\ge 0$.
Thus, the maximum value of $|12a + 4b + c|$ is $\mathrm{e}^2$.
Thus, $\lfloor m \rfloor = \lfloor \mathrm{e}^2 \rfloor = 7$.
We are done.

Hint:
We split into two cases:

*

*$0\le x \le 2$:

We need to prove that
$f(x) \le \mathrm{e}^2 - \mathrm{e}^x$ and $-(\mathrm{e}^2 - \mathrm{e}^x) \le f(x)$.
I only prove the former.
Let $g(x) = \mathrm{e}^2 - \mathrm{e}^x - f(x)$. We have
$g''(x) = \frac{15}{4}(x - 2) - \mathrm{e}^x - \mathrm{e}^2 < 0$. Thus, $g(x)$ is concave on $[0, 2]$.
Also, $g(0) > 0$ and $g(2) = 0$.
Thus, $g(x) \ge 0$ for all $x \in [0, 2]$.


*$2 \le x$:

Omitted.

Hint for the value of $a, b, c$:
Assume that $12a + 4b + c = \mathrm{e}^2$.
We have $b = -4a + \frac54 + \frac12 \mathrm{e}^2$ and $c = 4a - 5 - \mathrm{e}^2$.
Then consider $f(x) \le \mathrm{e}^2 - \mathrm{e}^x$ etc. to determine $a$.
I leave them for the author of the OP.
