11 Let $f(x)=a x^{3}+b x+c$ (where $a, b, c$ are real numbers), when $0 \leqslant x \leqslant 1$, $0 \leqslant f(x) \leqslant 1$. Find the maximum possible value of $b$.
10 From $\left\{\begin{array}{l}f(0)=c, \\ f(1)=a+b+c, \\ f\left(\sqrt{\frac{1}{3}}\right)=\frac{a}{3 \sqrt{3}}+\frac{b}{\sqrt{3}}+c\end{array}\right.$ we know
$$
2 b=3 \sqrt{3} f\left(\sqrt{\frac{1}{3}}\right)-f(1)-(3 \sqrt{3}-1) f(0) \leqslant 3 \sqrt{3} \text {. }
$$
$f(x)=\frac{3 \sqrt{3}}{2}\left(x-x^{3}\right)$ satisfies the conditions, the maximum possible value of $b$ is $\frac{3 \sqrt{3}}{2}$.