📝 题目
例 2 设 $f\left( x\right)$ 在 $\left\lbrack {a,b}\right\rbrack$ 上连续,在(a, b)内除仅有的一个点外都可导. 求证: $\exists c \in \left( {a,b}\right)$ ,使得 $\left| {f\left( b\right) - f\left( a\right) }\right| \leq \left( {b - a}\right) \left| {{f}^{\prime }\left( c\right) }\right|$ .
💡 答案与解析
证 设函数 $f\left( x\right)$ 在点 $d \in \left( {a,b}\right)$ 处不可导. 分别在(a, d)上和在 (d, b)上对 $f\left( x\right)$ 用微分中值定理,我们得
$f\left( d\right) - f\left( a\right) = \left( {d - a}\right) {f}^{\prime }\left( {c}_{1}\right)$ 和 $f\left( b\right) - f\left( d\right) = \left( {b - d}\right) {f}^{\prime }\left( {c}_{2}\right) ,$
其中 ${c}_{1} \in \left( {a,d}\right)$ 和 ${c}_{2} \in \left( {d,b}\right)$ . 将以上两个等式相加,我们得
$$ f\left( b\right) - f\left( a\right) = \left( {d - a}\right) {f}^{\prime }\left( {c}_{1}\right) + \left( {b - d}\right) {f}^{\prime }\left( {c}_{2}\right) . $$
由此我们得到
$$ \left| {f\left( b\right) - f\left( a\right) }\right| \leq \left( {d - a}\right) \left| {{f}^{\prime }\left( {c}_{1}\right) }\right| + \left( {b - d}\right) \left| {{f}^{\prime }\left( {c}_{2}\right) }\right| $$
$$ \leq \left( {d - a}\right) \left| {{f}^{\prime }\left( c\right) }\right| + \left( {b - d}\right) \left| {{f}^{\prime }\left( c\right) }\right| $$
$$ = \left( {b - a}\right) \left| {{f}^{\prime }\left( c\right) }\right| \text{ , } $$
其中 $\left| {{f}^{\prime }\left( c\right) }\right| = \max \left\{ {\left| {{f}^{\prime }\left( {c}_{1}\right) }\right| ,\left| {{f}^{\prime }\left( {c}_{2}\right) }\right| }\right\}$ ;
$$ c = \left\{ \begin{array}{ll} {c}_{1}, & \text{ 当 }\left| {{f}^{\prime }\left( {c}_{1}\right) }\right| \geq \left| {{f}^{\prime }\left( {c}_{2}\right) }\right| , \\ {c}_{2}, & \text{ 当 }\left| {{f}^{\prime }\left( {c}_{1}\right) }\right| < \left| {{f}^{\prime }\left( {c}_{2}\right) }\right| . \end{array}\right. $$