Find smallest $A\in \mathbb{R}$ such that $|f'(0)|\leq A.$ Problem
Find smallest $A\in \mathbb{R}$ such that for 3-degree polynomial satisfying the condition $\forall x\in[0;1],\ |f(x)|\le 1$ holds $|f'(0)|\le A$.
Solution
Assume that the 3-degree polynomial is $$ax^3+bx^2+cx+d,(a \neq 0).$$Let $x=0,\dfrac{1}{3},\dfrac{2}{3},1$ respectively. Then $$\begin{cases}d=f(0),\\\dfrac{a}{27}+\dfrac{b}{9}+\dfrac{c}{3}+d=f\left(\dfrac{1}{3}\right),\\\dfrac{8a}{27}+\dfrac{4b}{9}+\dfrac{2c}{3}+d=f\left(\dfrac{2}{3}\right),\\a+b+c+d=f\left(1\right).\end{cases}$$
By solving the linear equation system, we obtain $$\begin{align*}f'(0)&=c=-\frac{11}{2}f(0)+9f\left(\frac{1}{3}\right)-\frac{9}{2}f\left(\frac{2}{3}\right)+f(1)\\&\leq\frac{11}{2}|f(0)|+9\left|f\left(\frac{1}{3}\right)\right|+\frac{9}{2}\left|f\left(\frac{2}{3}\right)\right|+|f(1)|\\&\leq\frac{11}{2}+9+\frac{9}{2}+1\\&=20.\end{align*}$$
Hence, $$A=20.$$
This is right? I'm not sure for that.

I think an approach from interval arithmetic could be appropriate. Here the equation
$$
f([0,1])=[0,1]
$$
must be satisfied for your requirements.
$$
f([0, 1]) = a[0,1]^3+b[0, 1]^2+c[0,1]+d
$$
$$
= a[0,1] + b[0,1] +c[0,1] + d
$$
$$
= (a +b + c) [0,1] + d = [0,1]
$$
In order for the last equation to be valid statement $d = [d_1, d_2]$ must be an interval, too. From the the rule of addition of intervals we obtain $d_1=0$. So we can look at only the upper limits of the intervals.
$$
a + b + c + d_2 = 1
$$
Any combination of the coefficients satisfying this equation will fulfill the requirement. Thus you can choose and let $c = A = 0$ leaving
$$
a+b+d = 1,
$$
where $d\geq 0$.
