Question 33: Let the function $f(x)=a x^{2}+b x+c$ ($a$, $b$, $c \in \mathbb{R}$), satisfying $\max _{x \in[0,1]}|f(x)| \leq 1$. Let $g(x)=$ $cx^{2}+bx+a$. Try to find the maximum possible value of $\max _{x \in[0,1]}|g(x)|$.

保留源文本的换行和格式如下：

Question 33: Let the function $f(x)=a x^{2}+b x+c$ ($a$, $b$, $c \in \mathbb{R}$), satisfying $\max _{x \in[0,1]}|f(x)| \leq 1$. Let $g(x)=$ $cx^{2}+bx+a$. Try to find the maximum possible value of $\max _{x \in[0,1]}|g(x)|$.
$$
\begin{array}{l}
g(x)=f(0) \cdot x^{2}+\left(4 f\left(\frac{1}{2}\right)-3 f(0)-f(1)\right) x+\left(2 f(1)-4 f\left(\frac{1}{2}\right)+2 f(0)\right) \\
=f(0) \cdot\left(x^{2}-3 x+2\right)+4 f\left(\frac{1}{2}\right)(x-1)-f(1) \cdot(x-2)
\end{array}
$$

Therefore, when $x \in [0,1]$, we have
$$
\begin{array}{l}
|g(x)|=\left|f(0) \cdot\left(x^{2}-3 x+2\right)+4 f\left(\frac{1}{2}\right)(x-1)-f(1) \cdot(x-2)\right| \\
\leq|f(0)| \cdot\left|x^{2}-3 x+2\right|+4\left|f\left(\frac{1}{2}\right)\right| \cdot|x-1|+|f(1)| \cdot|x-2| \\
\leq 2|f(0)|+4\left|f\left(\frac{1}{2}\right)\right|+2|f(1)| \leq 2+4+2=8
\end{array}
$$

On the other hand, if we take $f(0)=1, f\left(\frac{1}{2}\right)=-1, f(1)=1$, i.e., $a=8, b=-8, c=1$, then the conditions are satisfied, and $\max _{x \in[0,1]}|g(x)|=8$.
In conclusion, the maximum possible value of $\max _{x \in[0,1]}|g(x)|$ is 8.