$9 \cdot 58$ If the quadratic function $f(x)=a x^{2}+b x+c$ has values whose absolute values do not exceed 1 on $[0,1]$, what is the maximum possible value of $|a|+|b|+|c|$?
[Solution] Without loss of generality, let $a>0$. From the assumption, we have
$$
|c| \leqslant 1,|a+b+c| \leqslant 1 .
$$

If $b \geqslant 0$, then $|a|+|b|+|c|=|a+b|+|c|$
$$
\leqslant|a+b+c|+2|c| \leqslant 3 .
$$

If $b \leqslant-2 a$, then $a+b \leqslant-a$, thus $|a| \leqslant|a+b| \leqslant 2$.
From this, we get $\quad|a|+|b|+|c| \leqslant 2+4+1=7$.
If $-2 a < b < 0$, from (1) we derive
$$
-a-1-c \leqslant b \leqslant-a+1-c<0 .
$$

From (3) and (4), we get
$$
(-a+1-c)^{2} \leqslant b^{2} \leqslant 4 a(1+c),
$$

which means
$$
a^{2}-2(3+c) a+(1-c)^{2} \leqslant 0 .
$$

From this, we know
$$
2<a \leqslant 3+c+\sqrt{(3+c)^{2}-(1-c)^{2}}=3+c+\sqrt{8(1+c)},
$$

and using $|c| \leqslant 1$, so $\quad|a|=a \leqslant 8$.
From (3), we derive $|b| \leqslant \sqrt{4 a(1+c)} \leqslant \sqrt{8 a} \leqslant 8$.
Thus, $|a|+|b|+|c| \leqslant 17$.
On the other hand, it is easy to see that $\max _{0 \leqslant x \leqslant 1}\left|8 x^{2}-8 x+1\right|=1$. Therefore, the maximum value of $|a|+|b|+|c|$ is 17.