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 $f'(x)$?
 A: 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.
A: We know that $|{f(x)}| \leqq |{e^x-e^2}|$, putting $x=2$ we get
$f(2)=0$
Hence $|{f(x)-f(2)}| \leqq |{e^x-e^2}|$.
Now, we replace $x$ by $2+h$ where $h$ is a positive quantity.
Dividing both sides by $h$ and imposing limit gives us
$$ \lim_{h \to 0} |\frac{f(2+h)-f(2)}{h}| \leqq \lim_{h \to 0} 
|\frac{e^{(2+h)}-e^2}{h}| $$.
After evaluating the limit, we get |$f'(2)$| $\leqq$ $e^2$.
Hence, the maximum value of $f'(2)=e^2$.
If a function $f(x)$ whose value of $a$ is positive and whose $c=5$ can be shown to satisfy $|f(x)| \leqq |e^x-e^2|$, the maximum value of $f'(2)$ will be $e^2$ and will be attained.
This is because in that case, $f'(x)$ will be an upward facing parabola attaining all the values from a certain minimum to infinity.
One such function would be $$f(x)=\frac{x^3}{8}-x^2-x+5$$.
Now, we show that this function satisfies $|f(x)| \leqq |e^x-e^2|$.
For positive $x$, we get only 2 roots of $f(x)$ which are $2$ and $3+√29$.
Firstly, we consider $x$  in the interval $[0,2]$.
In this interval, $f(x)$ is negative and $e^x-e^2$ is also negative.
We consider $g(x)=e^x-e^2-f(x)$.
We can see that $g"(x)=e^x-3x/4+2$ is always positive in this interval.
Hence $g'(x)$ is increasing. Also, as $g'(0)$ is positive, $g(x)$ is increasing. And $g(2)=0$ tells us that $g(x)$ is always negative in [0,2]. Hence for interval $[0,2]$ we are done.
Now, we consider the interval $[2, 3+√29]$.
In this interval, both $f(x)$ and $e^x-e^2$ both are positive.
Considering $h(x)=e^x-e^2-f(x)$, we see that $h'''(x)$ is positive.
This means that $h"(x)$ is increasing. Also, as $h"(2)$ is positive, $h"(x)$ is always positive. This means that $h'(x)$ is increasing.
It is evident that $h(x)$ is always positive. Hence for this interval also, we are done.
Now it's only left to prove for the interval $[3+√29, \infty]$.
One can proceed in similar fashion to show that.
