11. Let $f(x)=\max \left|x^{3}-a x^{2}-b x-c\right|(1 \leqslant x \leqslant 3)$, find the minimum value of $f(x)$ when $a, b, c$ take all real numbers.
11. Solution: Let $x=x^{\prime}+2$,
$$
\begin{array}{l}
f(x)=\max \left|x^{3}-a x^{2}-b x-c\right|(1 \leqslant x \leqslant 3) \Leftrightarrow f\left(x^{\prime}\right)=\max \left|x^{\prime 3}-a_{1} x^{\prime 2}-b_{1} x^{\prime}-c_{1}\right|\left(-1 \leqslant x^{\prime} \leqslant 1\right), \\
\text { Let } g\left(x^{\prime}\right)=x^{\prime 3}-a_{1} x^{\prime 2}-b_{1} x^{\prime}-c_{1}, x^{\prime} \in[-1,1],
\end{array}
$$

Then $4 g(1)-4 g(-1)=8-8 b_{1}, 8 g\left(\frac{1}{2}\right)-8 g\left(-\frac{1}{2}\right)=2-8 b_{1}$,
$$
\begin{array}{r}
\therefore 24 f\left(x^{\prime}\right) \geqslant 4|g(1)|+4|g(-1)|+8\left|g\left(\frac{1}{2}\right)\right|+8\left|g\left(-\frac{1}{2}\right)\right| \\
\geqslant\left|4 g(1)-4 g(-1)-8 g\left(\frac{1}{2}\right)+8 g\left(-\frac{1}{2}\right)\right|=6 .
\end{array}
$$

Therefore, $f\left(x^{\prime}\right) \geqslant \frac{1}{4} \Rightarrow \max \left|x^{\prime 3}-a_{1} x^{\prime 2}-b_{1} x^{\prime}-c_{1}\right| \geqslant \frac{1}{4}\left(-1 \leqslant x^{\prime} \leqslant 1\right)$.
At this point, $a_{1}=0, b_{1}=\frac{3}{4}, c_{1}=0, \because-1 \leqslant x^{\prime} \leqslant 1, \therefore$ let $x^{\prime}=\cos \theta$,
then $\left|g\left(x^{\prime}\right)\right|=\left|\cos ^{3} \theta-\frac{3}{4} \cos \theta\right| \leqslant \frac{1}{4}|\cos \theta| \leqslant \frac{1}{4}$.
$\therefore$ When $\cos \theta= \pm 1$, $\left|g\left(x^{\prime}\right)\right|=\frac{1}{4}, \therefore f\left(x^{\prime}\right)_{\text {min }}=\frac{1}{4} . \therefore f(x)=\max \left|x^{3}-a x^{2}-b x-c\right|(1 \leqslant x \leqslant 3)$ $\Leftrightarrow \max \left|\left(x^{\prime}+2\right)^{3}-a\left(x^{\prime}+2\right)^{2}-b\left(x^{\prime}+2\right)-c\right|\left(-1 \leqslant x^{\prime} \leqslant 1\right)=\max \mid x^{\prime 3}-(a-6) x^{\prime 2}-(4 a+b-12) x^{\prime}$ $-(4 a+2 b+c-8) \left\lvert\, \geqslant \frac{1}{4} . \therefore\right.$ When $\left\{\begin{array}{l}a^{\prime}=a-6=0 \\ b^{\prime}=4 a+b-12=\frac{3}{4} \\ c^{\prime}=4 a+2 b+c-8=0\end{array} \Rightarrow\left\{\begin{array}{l}a=6 \\ b=-\frac{45}{4} \\ c=\frac{13}{2}\end{array}\right.\right.$, $f(x)=\frac{1}{4}$.