📝 题目
例 18 求方程 ${x}^{3} = {3px} + q\left( {p > 0}\right)$ 恰有三个实根的条件.
💡 答案与解析
解 令 $f\left( x\right) = {x}^{3} - {3px}$ ,如图 2.12 所示.
$$ {f}^{\prime }\left( x\right) = 3{x}^{2} - {3p}\overset{\text{ 令 }}{ \Rightarrow }0 \Rightarrow x = \pm \sqrt{p}, $$
$$ {f}^{\prime \prime }\left( x\right) = {6x}, $$
$$ f\left( \sqrt{p}\right) = - 2{p}^{\frac{3}{2}},\;f\left( {-\sqrt{p}}\right) = 2{p}^{\frac{3}{2}}. $$
由图可见,当 $\left| q\right| < {2p}\sqrt{p}$ 时,方程 ${x}^{3} = {3px} + q$ 恰有三个实根.
\begin{center} \includegraphics[max width=0.2\textwidth]{images/019.jpg} \end{center} \hspace*{3em}
图 2.12