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@ -324,34 +324,6 @@ $$
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---
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>[!example] 例2
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设 $f(x)$ 在 $[a, b]$ 上连续,在 $(a, b)$ 内可导,证明存在不同的 $\xi, \eta \in (a, b)$,使得:
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$$f'(\xi) = \frac{a+b}{2\eta} f'(\eta)$$
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**解析**:
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1. 对 $f(x)$ 与 $g(x) = \frac{x^2}{2}$ 应用柯西中值定理,存在 $\eta \in (a, b)$ 使得:
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$$
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\frac{f(b)-f(a)}{(b^2 - a^2)/2} = \frac{f'(\eta)}{\eta}
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$$
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整理得:
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$$
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f(b)-f(a) = \frac{b^2 - a^2}{2\eta} f'(\eta)
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$$
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2. 对 $f(x)$ 应用拉格朗日中值定理,存在 $\xi \in (a, b)$ 使得:
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$$
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f(b)-f(a) = (b-a) f'(\xi)
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$$
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3. 联立两式,消去 $f(b)-f(a)$ 得:
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$$
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(b-a) f'(\xi) = \frac{(b-a)(a+b)}{2\eta} f'(\eta)
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$$
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由于 $b-a \neq 0$,约去后即得:
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$$
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f'(\xi) = \frac{a+b}{2\eta} f'(\eta)
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$$
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## 多次运用中值定理
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多次运用中值定理一般有如下特征:
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