方企勤 第一章 分析基础 第1题
📝 题目
例 1 设函数 $f\left( x\right) ,g\left( x\right)$ 在(a, b)上严格单调增加,求证: 函数
$$ \varphi \left( x\right) \overset{\text{ 定义 }}{ = }\max \{ f\left( x\right) ,g\left( x\right) \} ,\;\psi \left( x\right) \overset{\text{ 定义 }}{ = }\min \{ f\left( x\right) ,g\left( x\right) \} $$
也在(a, b)上严格单调增加.
💡 答案解析
证 $\forall {x}_{1},{x}_{2} \in \left( {a,b}\right)$ 且设 ${x}_{2} > {x}_{1}$ ,因为 $f\left( x\right) ,g\left( x\right)$ 在(a, b)上严格单调增加,所以 $f\left( {x}_{2}\right) > f\left( {x}_{1}\right) ,g\left( {x}_{2}\right) > g\left( {x}_{1}\right)$ . 于是
$$ \left. \begin{array}{l} \varphi \left( {x}_{2}\right) = \max \left\{ {f\left( {x}_{2}\right) ,g\left( {x}_{2}\right) }\right\} > f\left( {x}_{2}\right) > f\left( {x}_{1}\right) \\ \varphi \left( {x}_{2}\right) = \max \left\{ {f\left( {x}_{2}\right) ,g\left( {x}_{2}\right) }\right\} > g\left( {x}_{2}\right) > g\left( {x}_{1}\right) \end{array}\right\} \Rightarrow \varphi \left( {x}_{2}\right) > \varphi \left( {x}_{1}\right) . $$
同理可证 $\psi \left( x\right)$ 在(a, b)上严格单调增加.