柯西中值定理

设x1x2>0x_1x_2>0x1​x2​>0，证明：在x1x_1x1​和x2x_2x2​之间存在一点ξ\xiξ，使得：

x1ex2−x2ex1=(1−ξ)eξ(x1−x2)x_1e^{x_2}-x^2e^{x_1}=(1-\xi)e^{\xi}(x_1-x_2)x1​ex2​−x2ex1​=(1−ξ)eξ(x1​−x2​)

解：
\qquad令f(x)=1xex,g(x)=1xf(x)=\dfrac 1x e^x,g(x)=\dfrac 1xf(x)=x1​ex,g(x)=x1​

∵x1x2>0\qquad \because x_1x_2>0∵x1​x2​>0

∴x1,x2\qquad \therefore x_1,x_2∴x1​,x2​同号，g′(x)=−1x2≠0g'(x)=-\dfrac{1}{x^2} \neq 0g′(x)=−x21​​=0

∵f(x),g(x)\qquad \because f(x),g(x)∵f(x),g(x)在x1,x2x_1,x_2x1​,x2​之间的闭区间内连续，在x1,x2x_1,x_2x1​,x2​之间的开区间内可导

\qquad且g′(x)≠0g'(x)\neq 0g′(x)​=0

∴\qquad \therefore∴在x1x_1x1​和x2x_2x2​之间存在一点ξ\xiξ，使得f(x2)−f(x1)g(x2)−g(x1)=f′(ξ)g′(ξ)\dfrac{f(x_2)-f(x_1)}{g(x_2)-g(x_1)}=\dfrac{f'(\xi)}{g'(\xi)}g(x2​)−g(x1​)f(x2​)−f(x1​)​=g′(ξ)f′(ξ)​

\qquad即1x2ex2−1x1ex11x2−1x1=−1ξ2eξ+1ξeξ−1ξ2\dfrac{\frac{1}{x_2}e^{x_2}-\frac{1}{x_1}e^{x_1}}{\frac{1}{x_2}-\frac{1}{x_1}}=\dfrac{-\frac{1}{\xi^2}e^\xi+\frac{1}{\xi}e^\xi}{-\frac{1}{\xi^2}}x2​1​−x1​1​x2​1​ex2​−x1​1​ex1​​=−ξ21​−ξ21​eξ+ξ1​eξ​

\qquad化简得x1ex2−x2ex1=(1−ξ)eξ(x1−x2)x_1e^{x_2}-x^2e^{x_1}=(1-\xi)e^{\xi}(x_1-x_2)x1​ex2​−x2ex1​=(1−ξ)eξ(x1​−x2​)

\qquad得证在x1x_1x1​和x2x_2x2​之间存在一点ξ\xiξ，使得x1ex2−x2ex1=(1−ξ)eξ(x1−x2)x_1e^{x_2}-x^2e^{x_1}=(1-\xi)e^{\xi}(x_1-x_2)x1​ex2​−x2ex1​=(1−ξ)eξ(x1​−x2​)