📝 题目
例 1 求证: 方程 ${x}^{3} + {px} + q = 0\left( {p > 0}\right)$ 有且仅有一个根.
💡 答案与解析
证 考虑 $f\left( x\right) = {x}^{3} + {px} + q = 0$ . 因为 $\mathop{\lim }\limits_{{x \rightarrow + \infty }}f\left( x\right) = + \infty$ ,所以 $\exists b > 0$ ,使得 $f\left( b\right) > 0$ . 又 $\mathop{\lim }\limits_{{x \rightarrow - \infty }}f\left( x\right) = - \infty$ ,所以 $\exists a < 0$ ,使得 $f\left( a\right)$ $< 0$ . 由介值定理, $\exists c \in \left( {a,b}\right)$ ,使得 $f\left( c\right) = 0$ ,即 ${c}^{3} + {pc} + q = 0$ . 由 $p > 0$ ,对 $\forall {x}_{2} > {x}_{1}$ ,有
$$ f\left( {x}_{2}\right) - f\left( {x}_{1}\right) = {x}_{2}^{3} - {x}_{1}^{3} + p\left( {{x}_{2} - {x}_{1}}\right) $$
$$ = \left( {{x}_{2} - {x}_{1}}\right) \left( {{x}_{2}^{2} + {x}_{1}{x}_{2} + {x}_{1}^{2}}\right) + p\left( {{x}_{2} - {x}_{1}}\right) $$
$$ \text{ (因为 }{x}_{1}{x}_{2} \geq - \frac{{x}_{1}^{2} + {x}_{2}^{2}}{2}\text{ ) } $$
$$ \geq \left( {{x}_{2} - {x}_{1}}\right) \left( {\frac{{x}_{1}^{2} + {x}_{2}^{2}}{2} + p}\right) > 0, $$
即函数 $f\left( x\right)$ 是单调递增的,因此只有一个根.