vault backup: 2026-01-28 13:17:36

pull/6/head
王轲楠 2 months ago
parent 498f80eae3
commit ea569b2a32

@ -43,7 +43,7 @@ $\color{red} x^2 + \color{blue} y^2 = 1$
<span style='color:orange; font-weight:bold'>加粗橘色</span>
数学公式中的颜色文字(虽然大概用不到就是了)
\color{red} x^2 + \color{blue} y^2 = 1
$\color{red} x^2 + \color{blue} y^2 = 1$
```
**章节符号** §

@ -1,4 +1,5 @@
>[!note] 定理:
>如果函数$f(x)$在区间$[a,b]$上可导,则其导函数$f'(x)$在$[a,b]$上有介值性质,即若$f(x)$在$[a,b]$上的值域为$[m,M]$,则$\forall \xi\in[m,M]$,总$\exists \eta\in[a,b],$有$\xi=f'(\eta)$.
**证明**:若$m=M$,结论显然成立.若$m<M$,设$f'(x_1)=m,f'(x_2)=M$,不妨设$x_1<x_2$.任取$\xi\in(m,M)$,令$$g(x)=f(x)-\xi x,x\in[a,b].$$于是$g(x)$在$[a,b]$上可导,且$$g'(x_1)=f'(x_1)-\xi<0,g'(x_2)=f'(x_2)-\xi>0.$$由零值定理,$\exists \eta\in(x_1,x_2) \subset(a,b),g'(\eta)=0\implies f'(\eta)=\xi$,证毕.
**证明**:若$m=M$,结论显然成立.若$m<M$,设$f'(x_1)=m,f'(x_2)=M$,不妨设$x_1<x_2$.任取$\xi\in(m,M)$,令$$g(x)=f(x)-\xi x,x\in[a,b].$$于是$g(x)$在$[a,b]$上可导,且$$g'(x_1)=f'(x_1)-\xi<0,g'(x_2)=f'(x_2)-\xi>0.$$由零值定理,$\exists \eta\in(x_1,x_2) \subset(a,b),g'(\eta)=0\implies f'(\eta)=\xi$,证毕.

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