例9 设f(x)在?a,b?上连续,在?a,b?内可导?ab?0?. 证明存在???a,b?,使得证明 方法一 令
bf?a??af?b?b?a?f?????f'???. f?a??ka?f?b??kbbf?a??af?b?b?a?k,则.构造函数F?x??f?x??kx,由条
件易知F(x)在?a,b?上连续,在?a,b?内可导且F?a??F?b?.由罗尔定理,存在???a,b?,使得
F'????0,即f?????f'????k,故
bf?a??af?b?b?a?f?????f'???. 1x,h?x??f?x?x方法二 利用柯西中值定理. 令g?x??22,因ab?0,知0??a,b?，从而
g?x?,h?x?在?a,b?内可导,且?g'?x????h'?x???0,g?a??g?b?.故g?x?,h?x?满足柯西中值
f?b?定理的条件, b1b??f?a?a1a??f'????f?????21?f?????f'???.
?2
例10 设f(x)在?a,b?上连续,在?a,b?内可导?0?a?b?. 证明存在?,???a,b?,使得f'????a?bf'???
2?证明 由拉格朗日中值定理,????a,b?, 使得
f'????f?b??f?a?b?a??b?a?.f?b??f?a?b2?a2 （1）.
令g?x??x2,则f(x),g?x?在?a,b?内可导,且?f'?x??2??g'?x??2?0?0?a?b?,g?a??g?b?.故g?x?,h?x?满足柯西中值定理的条件, ????a,b?, 使得f?b??f?a??f'??? （2）.由
b2?a22?(1)(2)立得f'????a?bf'???.
2?
例11 设f(x)定义在?0,c?,f'(x)存在且单调递减,f?0??0.试证对于0?a?b?a?b?c,恒有f?a?b??f?a??f?b? (*).
证明 (1) 当a?0时,(*)式显然成立.
(2) 当a?0时,即a?0时,对f(x)在?0,a?,?b,a?b?上分别使用拉格朗日中值定理，则
??1??0,a?与?2??b,a?b?使得f'??1??f?a??f?0?a?0?f?a?a (1),
f'??2??f?a?b??f?b?(a?b)?ba?f?a?b??f?b?a?a (2).又f'(x)单调递减,有f'??1??f'??2? (3).,即(*)式成立.
由(1),(2) ,(3)有,
f?a?f?a?b??f?b?
练习 设f(x)在?a,b?上满足f''(x)?0.试证明对任意x1,x2??a,b?,x1?x2,恒有
f(x1?x22)?f?x1??fx22??.
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提示 不妨设x1?x2,对f(x)在?x1,??x1?x2??x1?x2?,上分别使用拉格朗日中值定理. ,x2???22???
例12 设f(x)在?a,b?上连续,在?a,b?内可导.又f(x)不为形如Ax?B的函数, 证明存在???a,b?,使得f'????证明 令g(x)?f?b??f?a?b?af?b??f?a?b?a.
(x?a)?f?a?,则f(a)?g?a?,f(b)?g?b?.由题设条件,
f(x)?g?x?不成立.因此存在c??a,b?使得f(c)?g?c?.下面分两种情形.
（1）若f(c)?g?c?,则
f?c??f?a?c?a?g?c??g?a?c?ac?ab?cb?c?f?b??f?a?b?a?b?ab?a?b?a.在?a,c?上对f(x)应用拉格
.
朗日中值定理,存在???a,c?使得f'????（2）若f(c)?g?c?,则
f?b??f?c?b?c?f?c??f?a?f?b??f?a?g?b??g?c??f?b??f?a?.在?c,b?上对f(x)应用拉格
.
朗日中值定理,存在???c,b?使得f'????注 例12的结论可变更为f'????f?b??f?c?f?b??f?a?f?b??f?a?b?a.
例13 设f(x)是在?a,b?上可微的非常值函数,且f?a??f?b??0. 证明存在???a,b?,使得f'?????b?a??24baf?x?dx.
证明 设M?supf'?x?,只要证明M?a?x?b?b?a??24baf?x?dx.若M??,则上式显然成立.设
M有限,由拉格朗日中值定理,对任意x??a,b?,存在?1??a,x?,使得
x?a?f'??1??Mf?x??f?a?,所以
f?x??M?x?a?.同理存在?2??x,b?,使得
a?bf?b??f?x?b?xa?b?f'??2??M,所以f?x??M?b?x?.
?baf?x?dx??2af?x?dx??ba?b2f?x?dx?
?2aM?x?a?dx??ba?b2M?b?x?dx??b?a?24a?bM.因此M??b?a??24baf?x?dx.
a?b若其中等式成立,则?2af?x?dx??2aM?x?a?dx与
?ba?b2f?x?dx??ba?b2M?b?x?dx同时成立.
??M?x?a?,这推出f?x????M?b?x?,?a?x?a?b2a?b2.由于M?0,故f(x)在x?a?b2不可导,这与条件
?x?b 7
矛盾.
注 例13的结论可改进为f'?????b?a??24baf?x?dx.
例14 设f(x)在?a,b?上有二阶导数,且f'?a??f'?b??0. 证明存在???a,b?,使得f''????4f?b??f?a?.
?b?a?2证明1 令F?x??f'(x),则F(x)在?a,b?上可微,F?a??F?b??0.若F(x)为常值函数,则F(x)?0,因此f?a??f?b?.故结论显然成立. 若F(x)不是常值函数,由例13, 存在
???a,b?,使得F'?????b?a??24baF?x?dx?4?b?a?2f?b??f?a?.
从而f''????4?b?a?2f?b??f?a?.
证明2 利用泰勒公式
f(x)?f?x0??f'?x0??x?x0??12f''????x?x0?.
2得
f(a?b2a?b2)?f?a??f'?a?b?a2b?a2?b?a?,
?f''??1???22???b?a?, ?f''??2???2?2?1212f()?f?b??f'?b?a?b2其中, a??1?1??2?b,所以
2?b?a?f?a??f''??1???2?2??b?a?,
f?b??f''??2???2?2?12f?b??f?a??12?b?a?24.f''??1??f''??2?24,
f''??1??f''??2???b?a?2f?b??f?a? (*).
若f''??1??4?b?a?2f?b??f?a?且f''??2??4?b?a?42f?b??f?a?,
则
12f''??1??f''??2??12?f''????1f''??2????b?a?2f?b??f?a?
这与(*)式矛盾,因此?可取?1或?2.
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例15 设f(x)在?a,??上可微,且limf'(x)?A?0. 证明limf(x)??.
x??x??证明 由limf'(x)?A?0,?M?0,当x?M时, f'(x)?x??A2.在区间?M,x?上应用拉格朗日
A2中值定理得,
f?x??f?Mx?M??f'????A2,所以f(x)?f?M???x?M?.固定M,令x??,
得limf(x)??.
x??推论 设f(x)在?a,??上可微,且limf(x)与limf'(x)均存在且有限,则limf'(x)?0.
x??x??x??证明 假设limf'(x)?A?0.若A?0,由例15, limf(x)??.若A?0,类似可证
x??x??limf(x)???,产生矛盾.
x??
练习 设f(x)在???,???上二次可导,且有界.试证存在x0????,???,使得f''?x0??0. 证明 若f''?x?变号,则由导数的介值性,存在x0????,???使得f''?x0??0.若f''?x?不变号,例如f''?x??0(f''?x??0类似证明),则f'?x?严格递增.取x0????,???使f'?x0??0.若f'?x0??0,则当x?x0时, 对f(x)在[x,x0]上应用拉格朗日中值定理,存在???x,x0?,使得 f(x)?f?x0??f'????x?x0??f?x0??f'?x0??x?x0?,令x???,则f(x)???.若
f'?x0??0,则当x?x0时, 对f(x)在[x0,x]上应用拉格朗日中值定理,存在???x0,x?,使得 f(x)?f?x0??f'????x?x0??f?x0??f'?x0??x?x0?,令x???,则f(x)???,这与f(x)有界性矛盾.
例16 在区间?0,x?（或?x,0?）上对f(x)?ln?1?x?应用拉格朗日中值定理得,
ln?1?x??0?x?11?x??0??1?1?.求证lim??x?012.
证明 由ln?1?x??0?x?1?x?得, ??x?ln?1?x?.
xln?1?x?x?ln?1?x?所以lim??limx?ln?1?x??lim(等价无穷小代换) 2x?0x?0xln?1?x?x?0x1??limx?01111?x(L’Hospital法则)?. ?limx?02(1?x)22x
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例17 由拉格朗日中值定理,当x?0时x?1?证明 (1)
14???12x?1212x??,0???1.
; (2) lim??x?0?14,lim??x??.
证明 (1)由2x???????141414(x?1?(2x?1?2x)2x?1?x得
?x
x(x?1)?4x)
[1?2(x(x?1)?x)]. (*)
其中
0?x(x?1)?x?xx?x?1??x?12.
所以
14???1214.
(2) 由(*)式得
x?0lim???lim?x?0[1?2(x(x?1)?x)]?14,
lim??limx??x??1??4???1?1??.
x?1?x??x??22x
例18 设f(x)在?0,???上可导,f?0??0,并设A?0使f'?x??Af?x?(x??0,???).试证明在?0,???上f(x)?0.
证明1 因f(x)在?0,???上可导,f?0??0, 故当x?0时,由拉格朗日中值定理,存在
1??时有?1??0,x?,使得f?x??f?0??f'??1??x?0??f'??1?x?Af??1??x.当限制x?0,?2A???f?x??f?x??1212f??1?(?1??0,x?).重复使用此式可得 f??1??122f??2????12nf??n?,这里0??n??n?1????1?x?12A.由f(x)的连续性,存在M?0使得f?x??M(x??0,??M1?).故0?f?x??n?0(n???).从而
22A??1?i?1i??).利用数学归纳法可证在一切?(i?1,2,?)上恒有f(x)?0,f(x)?0(x?0,?2A??2A,2A?????故在?0,???上f(x)?0.
证明2 因f?x?在?0,??1?1?的连续,故存在x1??0,使f?x1??maxf?x??M.所以???12A??2A?0?x?2A 10