34. Answer: 4
Let $a, b, c$ be fixed. Set $f(x)=a x^{2}+b x+c$. Then
$$
f(-1)=a-b+c, \quad f(0)=c, \quad f(1)=a+b+c
$$

Solve the system to get
$$
a=\frac{1}{2} f(-1)-f(0)+\frac{1}{2} f(1), \quad b=-\frac{1}{2} f(-1)+\frac{1}{2} f(1) .
$$

Suppose $|f(x)| \leq 1$ for all $|x| \leq 1$. Then
$$
\begin{aligned}
|2 a x+b| & =\left|\left(x-\frac{1}{2}\right) f(-1)-2 f(0) x+\left(x+\frac{1}{2}\right) f(1)\right| \\
& \leq\left|x-\frac{1}{2}\right|+2|x|+\left|x+\frac{1}{2}\right| \\
& \leq\left|x-\frac{1}{2}\right|+\left|x+\frac{1}{2}\right|+.2 \leq 4 .
\end{aligned}
$$

Moreover, $\left|2 x^{2}-1\right| \leq 1$ whenever $|x| \leq 1$, and $|2 x|=4$ is achieved at $x= \pm 1$.
34. Let $M$ be a positive integer. It is known that whenever $\left|a x^{2}+b x+c\right| \leq 1$ for all $|x| \leq 1$, then $|2 a x+b| \leq M$ for all $|x| \leq 1$. Find the smallest possible value of $M$.