关于微分中值定理的专题讨论

$\bf命题：$设$f\left( x \right)$在$\left[ {0,1} \right]$上二次可微，且$f\left( 0 \right) = f\left( 1 \right) = f'\left( 0 \right) = f'\left( 1 \right) = 0$，则存在$\xi \in \left( {0,1} \right)$，使得$f''\left( \xi \right){\rm{ = }}f\left( \xi \right)$

1

$\bf命题：$设$f\left( x \right)$在$\left[ {0,1} \right]$上连续，在$\left( {0,1} \right)$上可微，且$f\left( 0 \right) = 0,f\left( 1 \right) = \frac{1}{2}$，则存在$\xi ,\eta \in \left( {0,1} \right)$，使得$f'\left( \xi \right){\rm{ + }}f'\left( \eta \right){\rm{ = }}\xi {\rm{ + }}\eta $

1

$\bf命题：$设$f(x)$在$[0,1]$上可导，且$f\left( 0 \right) = 0,f\left( 1 \right) = 1$，证明：存在$\xi,\eta \in(0,1)$，使得$\frac{1}{{f'\left( \xi \right)}} + \frac{1}{{f'\left( \eta \right)}} = 2$

$\bf命题：$设$f\left( x \right)$在$\left[ {a,b} \right]$上连续且在$(a,b)$上可微，若$f\left( a \right) = 0,f\left( x \right) \ne 0,\forall x \in \left( {a,b} \right)$，则对任意的自然数$m,n$，存在$\xi ,\eta \in \left( {a,b} \right)$，使得$\frac{{nf'\left( \xi \right)}}{{f\left( \xi \right)}} = \frac{{mf'\left( \eta \right)}}{{f\left( \eta \right)}}$

1

$\bf命题：$设设$f\left( x \right)$在$\left[ {0,1} \right]$上连续，在$(0,1)$上可微，且$f(0)=0,f(1)=1$，则存在不同的点$\xi ,\eta \in \left( {0,1} \right)$，使得$f'\left( \xi \right)f'\left( \eta \right) = 1$

1

$\bf命题：$设$f(x),g(x)$在$\left[ {a,b} \right]$上连续，在$(a,b)$内具有二阶导数且存在相等最大值，$f\left( a \right) = g\left( a \right),f\left( b \right) = g\left( b \right)$，证明：存在$\xi \in(a,b)$，使得$f''\left( \xi \right) = g''\left( \xi \right)$

$\bf命题：$设$f\left( x \right)$在$\left( {a, + \infty } \right)$可导且导函数有界，则存在常数$c,\xi $，使得当$x \ge \xi $时，有$f\left( x \right) < cx$

1

$\bf命题：$设$f(x)$在$\left[ {a,b} \right]$上二次可微，若存在$c \in \left( {a,b} \right)$，使得$f\left( c \right) > 0$，且$f\left( a \right) = f\left( b \right) = 0$，则存在$\xi \in \left( {a,b} \right)$，使得$f''\left( \xi \right) < 0$

1

$\bf命题：$$（1）$设$f\left( x \right)$在$\left( {a, + \infty } \right)$上可微，且$\lim \limits_{x \to \begin{array}{*{20}{c}}{{a^ + }}\end{array}} f\left( x \right) = \lim \limits_{x \to \begin{array}{*{20}{c}}{ + \infty }\end{array}} f\left( x \right)$，则存在$\xi \in \left( {a, + \infty } \right)$，使得$f'\left( \xi \right) = 0$

$（2）$若$f\left( x \right)$在$\left( {a, + \infty } \right)$上二次可微，则存在$\eta \in \left( {a, + \infty } \right)$，使得$f''\left( \eta \right) = 0$

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$\bf命题：$设$f(x)$是$\left[ {a, + \infty } \right)$上有界的可微函数，且$\left| {f'\left( x \right)} \right|$单调，则$\lim \limits_{x \to \begin{array}{*{20}{c}}{ + \infty }\end{array}} xf'\left( x \right) = 0$

1

$\bf命题：$设$f(x)$在$[a,b]$上连续，在$(a,b)$上可微，且满足条件\[f\left( a \right)f\left( b \right) > 0,f\left( a \right)f\left( {\frac{{a + b}}{2}} \right) < 0\]证明：对每个实数$k$，存在$\xi \in \left( {a,b} \right)$，使得$f'\left( \xi \right) - kf\left( \xi \right) = 0$

1

$\bf命题：$设$f\left( x \right) \in {C^1}\left[ {a,b} \right]$，且存在$c \in \left( {a,b} \right)$，使得$f'\left( c \right) = 0$，则存在$\xi \in \left( {a,b} \right)$，使得$f'\left( \xi \right) = \frac{{f\left( \xi \right) - f\left( a \right)}}{{b - a}}$

1

$\bf命题：$设$f(x)$在$[a,b]$上可导，且$f'\left( a \right) = f'\left( b \right)$，则存在$\xi \in \left( {a,b} \right)$，使得$f'\left( \xi \right)\left( {\xi - a} \right) = f\left( \xi \right) - f\left( a \right)$

1

$\bf命题：$设$f(x),g(x)$为$[a,b]$上的正值连续函数，则存在$\xi \in (a,b)$，使得$\frac{{f\left( \xi \right)}}{{\int_a^\xi {f\left( x \right)dx} }} - \frac{{g\left( \xi \right)}}{{\int_\xi ^b {g\left( x \right)dx} }} = 1$

1

$\bf命题：$设$a>1$，函数$f:\left( {0, + \infty } \right) \to \left( {0, + \infty } \right)$可微，证明：存在趋于无穷的正数列${x_n}$，使得$f'\left( {{x_n}} \right) < f\left( {a{x_n}} \right),n = 1,2, \cdots $

1

$\bf命题：$设$f(x)$在$[-2,2]$上二阶可导，且$\left| {f\left( x \right)} \right| \leqslant 1\left( { - 2 \leqslant x \leqslant 2} \right)$，\[\frac{1}{2}{\left[ {f'\left( 0 \right)} \right]^2} + {f^3}\left( 0 \right) > \frac{3}{2}\]证明：存在${x_0} \in \left( { - 2,2} \right)$，使得$f''\left( {{x_0}} \right) + 3{f^2}\left( {{x_0}} \right) = 0$

1

$\bf命题：$设$f(x)$在$[-2,2]$上二阶可导，且$\left| {f\left( x \right)} \right| \leqslant 1$,${f^2}\left( 0 \right){\text{ + }}{\left( {f'\left( 0 \right)} \right)^2}{\text{ = }}4$，证明：存在$\xi \in \left( { - 2,2} \right)$，使得$f''\left( \xi \right) + f\left( \xi \right) = 0$

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$\bf(03中科院五)$设$f(x)$在$[a,b]$上连续，在$(a,b)$内可导，且$f\left( a \right) = 0,f\left( x \right) > 0\left( {a < x < b} \right)$，证明：不存在常数$M>0$，使得$$0 \leqslant f'\left( x \right) \leqslant Mf\left( x \right),x \in \left( {a,b} \right]$$

$\bf(04华师三)$设$f(x)$在$[a,b]$上连续，在$(a,b)$内可导，则存在$\xi,\eta \in(a,b)$，使得$f'\left( \xi \right) = \frac{{{\eta ^2}f'\left( \eta \right)}}{{ab}}$

[b]附录[/b]

$\bf命题：$$\bf(Gronwall-Bellman不等式微分形式)$设$f(x)$在$\left[ {0, + \infty } \right)$上可微，$f(0)=0$，且存在$A>0$，使得\[\left| {f'\left( x \right)} \right| \le A\left| {f\left( x \right)} \right|,\forall x \in \left[ {0, + \infty } \right)\]证明：$f\left( x \right) \equiv 0$

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$\bf命题：$设${f\left( x \right)}$在$\left[ {a,b} \right]$上连续，在$\left( {a,b} \right)$上二次可微，则存在$\xi \in \left( {a,b} \right)$，使得

\[f\left( a \right) + f\left( b \right) - 2f\left( {\frac{{a + b}}{2}} \right) = \frac{{{{\left( {b - a} \right)}^2}}}{4}f''\left( \xi \right)\]

方法一 方法二 方法三 方法四 方法五