关于微分中值定理的专题讨论

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

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

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

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

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

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

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

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

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

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

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

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

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

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

1

$f命题：$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$frac{{fleft( xi   ight)}}{{int_a^xi  {fleft( x ight)dx} }} - frac{{gleft( xi   ight)}}{{int_xi ^b {gleft( x ight)dx} }} = 1$

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

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

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

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

$f(04华师三)$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$f'left( xi   ight) = frac{{{eta ^2}f'left( eta   ight)}}{{ab}}$

附录

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

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

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

方法一   方法二   方法三   方法四   方法五