<s>[INST] If $|ax^2+bx+c|\leq 2\ \ \forall x\in[-1,1]$ then find the maximum value of $|cx^2+2bx+4a|\ \ \forall x\in [-2,2]$. 
If
$$\left|ax^2+bx+c\right|\leq 2\quad \forall x\in[-1,1]$$
then find the maximum value of
$$\left|cx^2+2bx+4a\right|\quad \forall x\in [-2,2].$$

My Attempt
Let $f(x)=ax^2+bx+c$, then
$$|cx^2+2bx+4a|=x^2|4\frac{a}{x^2}+2\frac{b}{x}+c|=x^2\left|f\left(\frac{2}{x}\right)\right|$$
But then I couldn't make any headway.
I also tried to express $a,b,c$ in terms of $f(-1),f(0),f(1)$ i.e $$c=f(0);b=\frac{f(1)-f(-1)}{2};a=\frac{f(-1)-2f(0)+f(1)}{2}$$ but again couldn't go further
 [/INST] We have
\begin{align*}
 |a - b + c| &\le 2, \tag{1}\\
 |a + b + c| &\le 2, \tag{2}\\
 |c| &\le 2. \tag{3}
\end{align*}
Using (1) and (2), we have
$$|a + c| + |b| \le 2. \tag{4}$$
(Note: If $(a + c)b \ge 0$, then $|a + c| + |b| = |a + c + b| \le 2$. If $(a + c)b < 0$, then $|a + c| + |b| = |a + c - b| \le 2$.)
Using (1) and (3), we have
$$|a - b| \le |a - b + c| + |-c| \le 4. \tag{5}$$
Using (2) and (3), we have
$$|a + b| \le |a + b + c| + |-c| \le 4. \tag{6}$$
Using (5) and (6), we have
$$|a| + |b| \le 4. \tag{7}$$
(Note: If $ab \ge 0$, then $|a| + |b| = |a + b| \le 4$.
If $ab < 0$, then $|a| + |b| = |a - b| \le 4$.)
Using (4) and (7), we have, for all $x\in [-2, 2]$,
\begin{align*}
 |cx^2 + 2bx + 4a|
 &\le |cx^2 + 4a| + |2bx| \\
 &\le \max(|4c + 4a|, |4a|) + 4|b| \tag{8}\\
 &\le 16
\end{align*}
where in (8) we have used
$$\max_{x\in [-2, 2]} |cx^2 + 4a|
= \max(|c\cdot (-2)^2 + 4a|, |c \cdot 2^2 + 4a|, |c \cdot 0^2 + 4a|).$$
On the other hand, when $a = -4, b = 0, c = 2$, we have $|ax^2 + bx + c| = |-4x^2 + 2|\le 2$ for all $x\in [-1, 1]$, and
the maximum of $|cx^2 + 2bx + 4a|= |2x^2 - 16|$ on $[-2, 2]$ is $16$.
Thus, the maximum of $|cx^2 + 2bx + 4a|$ on $[-2, 2]$ is $16$.
</s>