方企勤 第二章 一元函数微分学 第13题

解(1)依题意,存在 ${x}_{1},{x}_{2}$ 满足 $a < {x}_{1} < {x}_{2} < b$ ,使得

$$ \left. \begin{array}{l} \frac{f\left( {x}_{1}\right) }{{x}_{1} - a}{f}^{\prime }\left( a\right) > 0 \\ \frac{f\left( {x}_{2}\right) }{{x}_{2} - b}{f}^{\prime }\left( b\right) > 0 \end{array}\right\} \Rightarrow \frac{f\left( {x}_{1}\right) f\left( {x}_{2}\right) }{\left( {{x}_{1} - a}\right) \left( {{x}_{2} - b}\right) } > 0 $$

$$ \Rightarrow f\left( {x}_{1}\right) f\left( {x}_{2}\right) < 0\text{ . } $$

故存在 $\xi \in \left( {{x}_{1},{x}_{2}}\right) \subset \left( {a,b}\right)$ ,使 $f\left( \xi \right) = 0$ .

(2)令 $F\left( x\right) \overset{\text{ 定义 }}{ = }{\mathrm{e}}^{x}f\left( x\right)$ ,注意到 ${F}^{\prime }\left( x\right) = {\mathrm{e}}^{x}\left( {{f}^{\prime }\left( x\right) - f\left( x\right) }\right)$ . 因为 $F\left( a\right) = F\left( \xi \right) = F\left( b\right) = 0$ ,所以根据罗尔定理,存在 ${\xi }_{1} \in \left( {a,\xi }\right)$ , ${\xi }_{2} \in \left( {\xi ,b}\right)$ ,使得

$$ {F}^{\prime }\left( {\xi }_{1}\right) = {F}^{\prime }\left( {\xi }_{2}\right) = 0 \Rightarrow {f}^{\prime }\left( {\xi }_{1}\right) = f\left( {\xi }_{1}\right) ,{f}^{\prime }\left( {\xi }_{2}\right) = f\left( {\xi }_{2}\right) . $$

再令 $G\left( x\right) = {f}^{\prime }\left( x\right) - f\left( x\right)$ ,并改写 ${f}^{\prime \prime }\left( x\right) - f\left( x\right) = {G}^{\prime }\left( x\right) + G\left( x\right)$ ,则因为

$$ G\left( {\xi }_{1}\right) = G\left( {\xi }_{2}\right) = 0\overset{\text{ 罗尔定理 }}{ \rightarrow }\exists \eta \in \left( {{\xi }_{1},{\xi }_{2}}\right) , $$

使得 ${\left. {\left\lbrack {\mathrm{e}}^{x}G\left( x\right) \right\rbrack }^{\prime }\right| }_{x = \eta } = 0$ . 即得 ${f}^{\prime \prime }\left( \eta \right) - f\left( \eta \right) = {G}^{\prime }\left( \eta \right) + G\left( \eta \right) = 0$ .

\subsubsection{四、用反证法}