📝 题目
例 8 设 $f\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上连续,在(0,1)内可导, $f\left( 0\right) = f\left( 1\right) =$ 0. 求证: 对于 $\forall {x}_{0} \in \left( {0,1}\right) ,\exists \xi \in \left( {0,1}\right)$ ,使得 ${f}^{\prime }\left( \xi \right) = f\left( {x}_{0}\right)$ .
💡 答案与解析
证 令 $F\left( x\right) = f\left( x\right) - {xf}\left( {x}_{0}\right)$ ,则 $F\left( 0\right) = 0$ ,且
$$ F\left( {x}_{0}\right) = f\left( {x}_{0}\right) - {x}_{0}f\left( {x}_{0}\right) = \left( {1 - {x}_{0}}\right) f\left( {x}_{0}\right) , $$
$$ F\left( 1\right) = f\left( 1\right) - f\left( {x}_{0}\right) = - f\left( {x}_{0}\right) , $$
由此推出 $F\left( {x}_{0}\right) \cdot F\left( 1\right) = - \left( {1 - {x}_{0}}\right) {\left( f\left( {x}_{0}\right) \right) }^{2}$ .
下面分两种情况讨论:
第一种情况, $f\left( {x}_{0}\right) = 0$ . 根据罗尔定理,有
$$ \exists \xi \in \left( {0,{x}_{0}}\right) \subset \left( {0,1}\right) \text{ , } $$
使得 ${F}^{\prime }\left( \xi \right) = 0$ ,即得 ${f}^{\prime }\left( \xi \right) = f\left( {x}_{0}\right)$ ,从而本题得证.
第二种情况, $f\left( {x}_{0}\right) \neq 0$ . 则 $F\left( {x}_{0}\right)$ 与 $F\left( 1\right)$ 异号,于是根据连续函数的中间值定理, $\exists \eta \in \left( {{x}_{0},1}\right)$ ,使得 $F\left( \eta \right) = 0$ . 现在对 $F\left( x\right)$ 在 $\left\lbrack {0,\eta }\right\rbrack$ 上用罗尔定理,我们有
$$ \exists \xi \in \left( {0,\eta }\right) \subset \left( {0,1}\right) \text{ , } $$
使得 ${F}^{\prime }\left( \xi \right) = 0$ ,即得 ${f}^{\prime }\left( \xi \right) = f\left( {x}_{0}\right)$ ,从而本题也得证.