6. Given $f(x)=x^{3}+b x^{2}+c x+d$ is an increasing function on $(-\infty, 0)$, and a decreasing function on $[0,2]$, and the equation $f(x)=0$ has three roots, which are $\alpha, 2, \beta$, then the range of $|\alpha-\beta|$ is
6. $[3,+\infty)$ Detailed Explanation: (1) $f^{\prime}(x)=3 x^{2}+2 b x+c, \because f(x)$ is an increasing function on $(-\infty, 0)$, and a decreasing function on $[0,2]$, $\therefore$ when $x=0$, $f(x)$ reaches a maximum value, $\therefore f^{\prime}(0)=0, \therefore c=0$.
(2) $\because f(2)=0, \therefore d=-4(b+2) \cdot f^{\prime}(0)=3 x^{2}+2 b x=0$ has two roots, $x_{1}=0, x_{2}=-\frac{2 b}{3}$,
$\therefore f(x)$ is a decreasing function on $[0,2]$, $\therefore x_{2}=-\frac{2 b}{3} \geqslant 2, \therefore b \leqslant-3$.
(3) $\alpha, 2, \beta$ are the roots of the equation $f(x)=0$, let $f(x)=(x-2)(x-\alpha)(x-\beta)$,
$$
\begin{array}{l}
\therefore f(x)=x^{3}-(2+\alpha+\beta) x^{2}+(2+2 \alpha+2 \beta) x-2 \alpha \beta, \\
\therefore|\alpha-\beta|=\sqrt{(\alpha+\beta)^{2}-4 \alpha \beta}=\sqrt{(b+2)^{2}+2 d}=\sqrt{(b+2)^{2}-8(b+2)}=\sqrt{(b-2)^{2}-16} . \\
\therefore b \leqslant-3, \therefore|\alpha-\beta| \geqslant 3 .
\end{array}
$$