设 x 1 x 2 > 0 x_1x_2>0 x1x2>0，证明：在 x 1 x_1 x1和 x 2 x_2 x2之间存在一点 ξ \xi ξ，使得：
x 1 e x 2 − x 2 e x 1 = ( 1 − ξ ) e ξ ( x 1 − x 2 ) x_1e^{x_2}-x^2e^{x_1}=(1-\xi)e^{\xi}(x_1-x_2) x1ex2−x2ex1=(1−ξ)eξ(x1−x2)
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f(x)=\dfrac 1x e^x,g(x)=\dfrac 1x
f(x)=x1ex,g(x)=x1
∵ x 1 x 2 > 0 \qquad \because x_1x_2>0 ∵x1x2>0
∴ x 1 , x 2 \qquad \therefore x_1,x_2 ∴x1,x2同号， g ′ ( x ) = − 1 x 2 ≠ 0 g'(x)=-\dfrac{1}{x^2} \neq 0 g′(x)=−x21=0
∵ f ( x ) , g ( x ) \qquad \because f(x),g(x) ∵f(x),g(x)在 x 1 , x 2 x_1,x_2 x1,x2之间的闭区间内连续，在 x 1 , x 2 x_1,x_2 x1,x2之间的开区间内可导
\qquad 且 g ′ ( x ) ≠ 0 g'(x)\neq 0 g′(x)=0
∴ \qquad \therefore ∴在 x 1 x_1 x1和 x 2 x_2 x2之间存在一点 ξ \xi ξ，使得 f ( x 2 ) − f ( x 1 ) g ( x 2 ) − g ( x 1 ) = 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 1 x 2 e x 2 − 1 x 1 e x 1 1 x 2 − 1 x 1 = − 1 ξ 2 e ξ + 1 ξ e ξ − 1 ξ 2 \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}} x21−x11x21ex2−x11ex1=−ξ21−ξ21eξ+ξ1eξ
\qquad 化简得 x 1 e x 2 − x 2 e x 1 = ( 1 − ξ ) e ξ ( x 1 − x 2 ) x_1e^{x_2}-x^2e^{x_1}=(1-\xi)e^{\xi}(x_1-x_2) x1ex2−x2ex1=(1−ξ)eξ(x1−x2)
\qquad 得证在 x 1 x_1 x1和 x 2 x_2 x2之间存在一点 ξ \xi ξ，使得 x 1 e x 2 − x 2 e x 1 = ( 1 − ξ ) e ξ ( x 1 − x 2 ) x_1e^{x_2}-x^2e^{x_1}=(1-\xi)e^{\xi}(x_1-x_2) x1ex2−x2ex1=(1−ξ)eξ(x1−x2)