If $|x|\leq 1$, $|ax^2+bx+c|\leq 1$, find the max possible value of $|2ax+b|$ 
Given $a,b,c\in \mathbb{R}$ such that  $|x|\leq 1$ and $|ax^2+bx+c|\leq 1$, find the maximum value of
$$
|2ax+b|
$$

My Attempt:
Set $x=1$ in $|ax^2+bc+c|\leq 1$ to get
$$
\tag{$\star$}|a+b+c|\leq 1
$$
Similarly, set $x=-1$ in $|ax^2+bc+c|\leq 1$ to get
$$
\tag{$\star\star$}|a-b+c|\leq 1
$$
Similarly, set $x=0$ in $|ax^2+bx+c|\leq 1$ to get
$$
\tag{$\star\star\star$}|c|\leq 1
$$
Now, adding  $(\star)$ and $(\star\star)$ gives
$$
|a+c|\leq 1
$$
Subtracting $(\star)$ from $(\star\star)$ gives
$$
|b|\leq 1
$$
Now, we have
$$
-1\leq (a+c)\leq 1\\
-1-c\leq a \leq 1-c
$$
Since $-1\leq c\leq 1$, we get
$$
-2 \leq a \leq 2,\; -1 \leq b\leq 1
$$
How can I complete the solution from this point?
 A: By the Markov brothers' inequality, for a polynomial $p$ of degree $n$ we have
$$ \max_{-1\le x\le 1} |p'(x)| \le n^2\max_{-1\le x\le 1} |p(x)| $$
with equality for Chebyshev polynomials.  Taking $p(x)=ax^2+bx+c$ we get $|2ax+b|\le 4$.
A: Without loss of generality we can assume that $b = 0$, otherwise graph of the function $x \mapsto ax^2+bx+c$ could be shifted ($x \mapsto x-b/(2a)$).
So we know that for $x$ such that $|x| \le 1$ we have $|ax^2+c| \le 1$ and want to maximalize $|2ax|$. Assume that $a > 0$ (if $a<0$, just multiply everything by $-1$). So $|2ax| = 2a |x|$. Due to the symmetry ($b = 0$) we can assume even that $x > 0$, so we are maximalizing $2ax$ for $x \in (0,1)$. Thus we are interested in how big $2a$ can get.
Setting $x \in \{0, 1\}$ we notice that $|c+a| \le 1$ and $|c| \le 1$. It means that $c$ lies in the interval $[-1,1]$ and $a$ lies in $[-2,2]$. Can $a$ be equal to $2$? Yes, if $c=-1$.
Conclusion: for $a = 2$, $b = 0$, $c = -1$, $x = 1$ we have $|2ax + b| = 4$ and $4$ can't be replaced by any larger number.
