Question 39, Let $a, b$ be real numbers, and the function $f(x)=x^{3}+a x^{2}+b x$. If there exist three real numbers $x_{1}, x_{2}, x_{3}$ satisfying $x_{1}+1 \leq x_{2} \leq x_{3}-1$, and $f\left(x_{1}\right)=f\left(x_{2}\right)=f\left(x_{3}\right)$, and the minimum value of $|a|+2|b|$.
Question 39, Answer: Let $f\left(x_{1}\right)=f\left(x_{2}\right)=f\left(x_{3}\right)=-c$, then $x_{1}, x_{2}, x_{3}$ are the three roots of the polynomial $g(x)=x^{3}+ax^{2}+bx+c=0$. According to Vieta's formulas, we have: $\left\{\begin{array}{c}x_{1}+x_{2}+x_{3}=-a \\ x_{1} x_{2}+x_{2} x_{3}+x_{1} x_{3}=b\end{array}\right.$ Therefore, we know:
$$
\begin{array}{l}
a^{2}-3 b=\left(x_{1}+x_{2}+x_{3}\right)^{2}-3\left(x_{1} x_{2}+x_{2} x_{3}+x_{1} x_{3}\right)=\frac{\left(x_{2}-x_{1}\right)^{2}+\left(x_{3}-x_{2}\right)^{2}+\left(x_{3}-x_{1}\right)^{2}}{2} \\
\geq \frac{1^{2}+1^{2}+2^{2}}{2}=3
\end{array}
$$

Thus, we know
$$
b \leq \frac{a^{2}-3}{3}
$$
(1) If $|a| \leq \sqrt{3}$, then $|b| \geq \frac{3-a^{2}}{3}$. In this case, $|a|+2|b| \geq |a|+2 \cdot \frac{3-a^{2}}{3}=\frac{19}{8}-\frac{2\left(|a|-\frac{3}{4}\right)^{2}}{3} \geq \frac{19}{8}-$ $\frac{2\left(\sqrt{3}-\frac{3}{4}\right)^{2}}{3}=\sqrt{3}$.
(2) If $|a|>\sqrt{3}$, then $|a|+2|b| \geq |a|>\sqrt{3}$.
In either case, we have $|a|+2|b| \geq \sqrt{3}$. When $a=\sqrt{3}, b=0, x_{1}=-1-\frac{\sqrt{3}}{3}, x_{2}=-\frac{\sqrt{3}}{3}$, $x_{3}=1-\frac{\sqrt{3}}{3}, f\left(x_{1}\right)=f\left(x_{2}\right)=f\left(x_{3}\right)=\frac{2 \sqrt{3}}{9}$, the equality holds.