方企勤 第二章 一元函数微分学 第15题
📝 题目
例 15 设 $f\left( x\right)$ 是非负函数,在 $\left\lbrack {a,b}\right\rbrack$ 上二阶可导,且 ${f}^{\prime \prime }\left( x\right) \neq 0$ , 求证: 方程 $f\left( x\right) = 0$ 在(a, b)内如果有根,就只能有一个根.
💡 答案解析
证 设 ${x}_{0} \in \left( {a,b}\right)$ ,使得 $f\left( {x}_{0}\right) = 0$ . 首先,我们有 ${f}^{\prime }\left( {x}_{0}\right) = 0$ . 事实上,由假设 $f\left( x\right) \geq 0$ ,
$$ \left. \begin{array}{l} {f}^{\prime }\left( {x}_{0}\right) = {f}_{ - }^{\prime }\left( {x}_{0}\right) = \mathop{\lim }\limits_{{x \rightarrow {x}_{0} - }}\frac{f\left( x\right) - f\left( {x}_{0}\right) }{x - {x}_{0}} \leq 0 \\ {f}^{\prime }\left( {x}_{0}\right) = {f}_{ + }^{\prime }\left( {x}_{0}\right) = \mathop{\lim }\limits_{{x \rightarrow {x}_{0} + }}\frac{f\left( x\right) - f\left( {x}_{0}\right) }{x - {x}_{0}} \geq 0 \end{array}\right\} \Rightarrow {f}^{\prime }\left( {x}_{0}\right) = 0. $$
其次,我们假定存在 ${x}_{1},{x}_{2} \in \left( {a,b}\right) ,{x}_{1} \neq {x}_{2}$ (不妨设 ${x}_{1} < {x}_{2}$ ),使得 $f\left( {x}_{1}\right) = f\left( {x}_{2}\right) = 0$ . 那么根据上述证明,我们有 ${f}^{\prime }\left( {x}_{1}\right) = {f}^{\prime }\left( {x}_{2}\right) = 0$ . 再在 $\left\lbrack {{x}_{1},{x}_{2}}\right\rbrack$ 上对 ${f}^{\prime }\left( x\right)$ 用罗尔中值定理,则存在 $\xi \in \left( {{x}_{1},{x}_{2}}\right)$ ,使得 ${f}^{\prime \prime }\left( \xi \right) = 0$ . 这与 ${f}^{\prime \prime }\left( x\right) \neq 0$ 的假定矛盾.