高等数学-一元微分学-可导函数的中值定理习题
考察中值定理的定义
考察罗尔定理的定义
例1
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image-20200419191114818考察拉格朗日中值定理定义
例1
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image-20200419195152638例3
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image-20200420195433746其实这道题就是求极限,没考察中值定理
例4
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image-20200421210034865考察泰勒定理
例1
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image-20200420110442108分母也可使用泰勒展开替换
例3
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image-20200420111931538其实此题不用泰勒展开,多次使用洛必达法则,也可求出a和b,而且更简单
例4
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image-20200421210647408求证存在\(\xi\)使等式成立
求证\(f^{(n)}(\xi)=0\)
一般会用到罗尔定理
找多个相同值点,多次使用罗尔定理
例1
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image-20200417134756621本题使用2次零点定理+罗尔定理证明
例2
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image-20200417141636783介值定理+罗尔定理得两个等值点,再用一次罗尔定理证明 函数值相加要想到用介值定理
例3
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image-20200420095349655求证仅含\(\xi\)的等式
即求证仅包含\(\xi\)不包含其他字母的等式
一般方法包括积分还原法和微分方程法
微分中值定理证明题中构造辅助函数的方法
积分还原法求证仅\(\xi\)的等式
左侧可化为\(\frac{f^{\prime}(x)}{f(x)}=[\ln f(x)]^{\prime}\)的式子
一般要求要证的式子包含一个函数及其导数(差一阶),有些可以变换到求证\(\frac{f^{\prime}(x)}{f(x)}=\cdot \cdot \cdot\)的形式,我们可以将左右都看作是函数的导数。
而\(\frac{f^{\prime}(x)}{f(x)}=[\ln f(x)]^{\prime}\),即证\([\ln f(x)]^{\prime}-[balabala]^{\prime}=0\), 即证\(\{\ln [f(x) \cdot balabala] \}^{\prime}=0\)。
则我们一般可以取辅助函数\(\phi (x) = \ln [f(x) \cdot balabala\)来求证。
例1
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image-20200422113734732找其他原函数
变为式子=0
的形式,式子左侧作为函数,试着积分,可解的的话,就可作为辅助函数。
例1
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image-20200421155825050例2
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image-20200421171619179微分方程法
将要证的方程看作微分方程,解出函数的形式,作为辅助函数。
例1
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image-20200420102659501例2
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image-20200421154745450求证含\(\xi\)和a与b的等式
这里a,b指区间端点,一般是在[a,b]上连续,在(a,b)上可导
a,b与\(\xi\)可分离
若a,b侧有如下形式,则使用对应的方法求解
批注 2020-04-17 201716例1
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image-20200417202354879例2
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image-20200417204213695a,b与\(\xi\)不可分离
\(\xi\)变为x,去分母,移项,变为式子=0
, 将其看作(?)'=0
,以?部分作辅助函数来证明
例1
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image-20200421163100435求证存在\(\xi\)和\(\eta\)使等式成立
仅有\(f^{\prime}(\xi), f^{\prime}(\eta)\)
方法是找3个点,使用2次拉格朗日中值定理
例1
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image-20200421191417002\(\xi,\eta\)复杂度不同
方法:留下偏复杂一部分,凑成某式的导数,用拉格朗日中值定理;或者凑成连个两个导数的比值,用柯西中值定理。
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例5
这题第二问要想到使用第一问的结论
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image-20200422115033884求证存在\(\epsilon,\xi,\eta\)使等式成立
例1
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image-20200421143943207拉格朗日中值定理使用
出现\(f(b)-f(a)\)想到使用拉格朗日中值定理
例1
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image-20200418152406592出现\(f(a), f(c), f(b)\) 想到使用2次拉格朗日中值定理
例1
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image-20200419185949907不等式的证明
例1
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image-20200421204128250其实设\(f(x)=x,g(x)=ln(x)\)用柯西中值定理来做也简单。