📝 题目
例 24 设 $f\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上连续,在(0,1)内有二阶导数,且
$$ f\left( 0\right) \cdot f\left( 1\right) > 0,\;{f}^{\prime \prime }\left( x\right) > 0\left( {\forall x \in \left( {0,1}\right) }\right) , $$
$$ {\int }_{0}^{1}f\left( x\right) \mathrm{d}x = 0. $$
求证:
( 1 )函数 $f\left( x\right)$ 在(0,1)内恰有两个零点;
(2)至少存在一点 $\xi \in \left( {0,1}\right)$ ,使得 ${f}^{\prime }\left( \xi \right) = {\int }_{0}^{\xi }f\left( t\right) \mathrm{d}t$ .
💡 答案与解析
证(1)函数 $f\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上有惟一的最小值点 ${x}_{m} \in \left( {0,1}\right)$ (见图 3.3),显然 $f\left( {x}_{m}\right) < 0$ ,否则 $f\left( x\right) \neq 0$ ,这与 $\displaystyle{\int }_{0}^{1}f\left( x\right) \mathrm{d}x = 0$ 矛盾. 又因为
$$ f\left( 0\right) \cdot f\left( 1\right) > 0,{f}^{\prime \prime }\left( x\right) > 0 \Rightarrow f\left( 0\right) > 0,f\left( 1\right) > 0. $$
\begin{center} \includegraphics[max width=0.2\textwidth]{images/024.jpg} \end{center} \hspace*{3em}
图 3.3
否则由凹函数的最大值在端点达到,导致 $f\left( x\right) < 0\left( {\forall x \in \left( {0,1}\right) }\right)$ , 这又与 $\displaystyle{\int }_{0}^{1}f\left( x\right) \mathrm{d}x = 0$ 矛盾. 于是有 $f\left( 0\right) \cdot f\left( {x}_{m}\right) < 0,f\left( {x}_{m}\right) \cdot f\left( 1\right)$ $< 0$ . 又因为 $f\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上连续,所以 $\exists {x}_{1} \in \left( {0,{x}_{m}}\right)$ ,使得 $f\left( {x}_{1}\right) =$ $0;\exists {x}_{2} \in \left( {{x}_{m},1}\right)$ ,使得 $f\left( {x}_{2}\right) = 0$ .
如果 $f\left( x\right)$ 在(0,1)内有三个零点,由罗尔定理,函数 ${f}^{\prime }\left( x\right)$ 在 (0,1)内有两个零点,导致 ${f}^{\prime \prime }\left( x\right)$ 有一个零点,这与 ${f}^{\prime \prime }\left( x\right) > 0$ 矛盾.
(2)令 $F\left( x\right) = {f}^{\prime }\left( x\right) - {\int }_{0}^{x}f\left( t\right) \mathrm{d}t$ ,注意到由 ${f}^{\prime }\left( {x}_{m}\right) = 0,{f}^{\prime }\left( x\right) \uparrow$ 推出 ${f}^{\prime }\left( {x}_{1}\right) < 0,{f}^{\prime }\left( {x}_{2}\right) > 0$ .
又根据第 (1) 小题, $f\left( x\right) > 0\left( {\forall x \in \left( {0,{x}_{1}}\right) \cup \left( {{x}_{2},1}\right) }\right)$ ,所以
$$ {\int }_{0}^{{x}_{1}}f\left( x\right) \mathrm{d}x > 0,\;{\int }_{{x}_{2}}^{1}f\left( x\right) \mathrm{d}x > 0, $$
于是
$$ {\int }_{0}^{{x}_{2}}f\left( x\right) \mathrm{d}x = {\int }_{0}^{1}f\left( x\right) \mathrm{d}x - {\int }_{{x}_{2}}^{1}f\left( x\right) \mathrm{d}x $$
$$ = 0 - {\int }_{{x}_{2}}^{1}f\left( x\right) \mathrm{d}x < 0. $$
故有
$$ {f}^{\prime }\left( {x}_{1}\right) - {\int }_{0}^{{x}_{1}}f\left( t\right) \mathrm{d}t < 0,\;{f}^{\prime }\left( {x}_{2}\right) - {\int }_{0}^{{x}_{2}}f\left( t\right) \mathrm{d}t > 0, $$
即 $F\left( {x}_{1}\right) < 0,F\left( {x}_{2}\right) > 0$ . 再由 $F\left( x\right)$ 的连续性,存在 $\xi \in \left( {0,1}\right)$ ,使得 $F\left( \xi \right) = 0$ ,即 ${f}^{\prime }\left( \xi \right) = {\int }_{0}^{\xi }f\left( t\right) \mathrm{d}t$ .
提问 第 (2) 小题这样证明对吗? 令 $F\left( x\right) = {f}^{\prime }\left( x\right) - {\int }_{0}^{x}f\left( t\right) \mathrm{d}t$ . 因为
$$ {f}^{\prime }\left( {x}_{m}\right) = 0,{f}^{\prime }\left( x\right) \uparrow \Rightarrow {f}^{\prime }\left( 0\right) < 0, $$
$$ {f}^{\prime }\left( 1\right) > 0 \Rightarrow F\left( 0\right) < 0,F\left( 1\right) > 0, $$
所以 $\exists \xi \in \left( {0,1}\right)$ ,使得 $F\left( \xi \right) = 0$ ,即得结论.
解答 这样证明不对. 因为 ${f}^{\prime }\left( 0\right) ,{f}^{\prime }\left( 1\right)$ 未必存在. 例如,设
$$ f\left( x\right) \overset{\text{ 定义 }}{ = }\frac{4}{3} - \sqrt{x} - \sqrt{1 - x}. $$
显然 $f\left( x\right)$ 在 $\left\lbrack {0,1}\right\rbrack$ 上连续,在(0,1)内有二阶导数,
$$ f\left( 0\right) \cdot f\left( 1\right) > 0,\;{\int }_{0}^{1}f\left( x\right) \mathrm{d}x = 0, $$
$$ {f}^{\prime \prime }\left( x\right) > 0\;\left( {\forall x \in \left( {0,1}\right) }\right) . $$
但这个函数在点 $x = 0,x = 1$ 处的导数不存在 (见图 3.4).
\begin{center} \includegraphics[max width=0.2\textwidth]{images/025.jpg} \end{center} \hspace*{3em}
图 3.4
\subsubsection{五、从定积分的信息提取被积函数的信息}