22n?2n?1?0.令f?n??n?2n?1.则f?0???1,f?1???2,f?2???1,f?3??2.因此当
n?3时,所要证的不等式成立.
例9 在数列 1,2,3344,?,nn,?中求出最大的数.
1?解 设y?x,则lny?lnx.两边求导得???lnx???2?,所以y'?xxyxx?x?x1111y'111x?1x2(1?lnx).
当0?x?e时, y'?0.故y?xx递减;当e?x???时, y'?0,故y?xx递增.所以1?32,
3?44???nn??.因为23?32,故2?33,因此所求的最大数为33.
2n?xx??x例10讨论函数f?x???1?x??e的极值. ????2!n!???2n?12nx???x?xx??x?x解 求导得f?x???1?x?x???x?e?1?x??e??e.由??????n!?n?1?!?2!2!n!?????nf'(x)?0解出x?0.分两种情况讨论:
(1)若n为偶数,则对一切x,f'(x)?0.故f(x)不存在极值点.
(1)若n为偶数,则当x?0时f'(x)?0;当x?0时f'(x)?0.故f(x)在x?0处有极值点(且为最值点) f(0)?1.
例11设f(x)在?a,b?上有二阶导数,且满足f''(x)??1?x2?f'?x??f?x??0与
f(a)?f?b??0.证明f(x)在?a,b?上恒为0.
证明 若f(x)在?a,b?上不恒为0.设f(x)在?a,b?上的最大者与最小值分别为M与m.则
M?m若
M且M?0?m.
?0,则存在???a,b?使得f????M?0.由Fermat定理,
f'????0.由题设条件
f''(?)?1???2?f'????f????0.因此f''(?)?f????M?m?0?0.这推出f(x)在x??取最小值,
这是不可能的.
同理可证m?0也是不可能的.因此必有M
例12证明
121,这说明f(x)在?a,b?上恒为0.
??2dx2x220?147.
?x?1证明 设f(x)?2x?x?1,则f'(x)?4x?12x2.因此函数f(x)的稳定点为x?14.
?x?1当0?x?14时, f'(x)?0;当14?x?121?时, f'(x)?0.又f(0)?f????1.
?4? 16
1?因此f(x)在?0,1?上的最小值为f??????78,最大值为1.
?2??4?因此当0?x?12时,
78?f?x??1,1?2x12?78.
?x?1由于被积函数在积分区间上连续且不恒为常数, 所以
1211??20dx??2dx2x210??x?1?2870dx?1287?147.
例13设f(x)在???,???上具有二阶导数,且满足 (1)f''(x)?0;
(2)limf'(x)???0,limf'(x)???0;
x???x???(3)存在一点x0使得f(x0)?0.
证明f(x)?0在???,???上有且仅有两个根.
分析 由条件(1)(2)知f'(x)在???,???上严格递增,且由负变正.因此f(x)在???,???上由递减变为递增,故在某点达到最小值,由(3)知最小值为负的.
证明 由条件(1)(2)知f(x)在某点?达到最小值,且f????f(x0)?0.在??,???上f'(x)?0且单调递增.取a??使得f(a)?0.在?a,???上, f'(x)?f'?a??0.由例2知方程f(x)?0在?a,???上有且只有一个根,从而在??,???上有且仅有一个根.
同理可证f(x)?0在???,??上有且仅有一个根.所以f(x)?0在???,???上有且仅有两个根.
例14设F(x)在???,???上具有二阶导数,且恒有F''(x)?0.证明: (1)对任意实数x1,x2,有
F?x1??F?x2?2?x1?x2??F??;
2??(2)对任意实数x1,x2,?,xn与任意正实数p1,p2,?,pn,有
p1F?x1??p2F?x2????pnF?xn?p1?p2???pn?p1x1?p2x2???pnxn?F??p1?p2???pn????. ???(3)对?0,?2?????2上的任意连续函数f(x)有?F?f?x??sinxdx?F?2f?x?sindx?. ?0?0???证明 由于(1)是(2)的特殊情形,故先证(2).记a?由泰勒公式
p1x1?p2x2???pnxnp1?p2???pn,
17
F?xi??F?a??F'?a??xi?a??F''??i?2?xi2?a?
?F?a??F'?a??xi?a?(i?1,2,?,n),
乘以pi后相加得
p1F?x1??p2F?x2????pnF?xn???p1?p2???pn?F?a??F'?a??p1?x1?a??p2?x2?a????pn?xn?a????p1?p2???pn?F?a?,即(2)成立.
i?i?（2）将?0,??分为n等分,令pi?sin,xi?f(),其中i?1,2,?,n.由(2),
???2?2n2n?ni???i??fsin?????2n?2n?i?1??F?ni???sin?2n??i?1??n?F?f?2n??sini?1??i?????ni?2n.
???i?1sini?2n令n???两边取极限,注意到F(x)的连续性,
i?2n??n??i??imf?sin?n?????2n??i?1F?ni??limsin?2n?n???i?1???2n??????nn???lim?F?f?2n??sini?1??i?????ni?2n??2n,
???2nn???lim?i?1sini?2n??2n??2??0f?x?sindx即 F???2sinxdx??0????????20F?f?x??sinxdx?.
??20sinxdx???所以 F?2f?x?sindx???0???20F?f?x??sinxdx.
f?x?h??f?x?h?2例15设f(x)在???,???上有二阶连续导数,且对任意x与h成立f(x)?证明f''(x)?0恒成立. 注 逆命题由例14（1）知成立.
.
证明 用反证法. 设命题不真,即存在x0使得f''(x0)?0.由于二阶导数连续,存在h??(x0?h,x0?h)时f''(?)?0.由泰勒公式, f?x0?h??f?x0??f'?x0?h??0,当
f''??1?22h,
f?x0?h??f?x0??f'?x0?h?f''??2?2h22h,其中x0??1?x0?h, x0?h??2?x0.于是
f?x0?h??f?x0?h??2f?x0??2?f''??1??f''??2???2f(x0),
18
即f(x0)?f?x0?h??f?x0?h?2,这与条件矛盾.
例16设f(x)在?0,1?上存在二阶导数,且f(0)?f?1??0，又maxf(x)?2,证明minf''(x)??16.
x??0,1?x??0,1?证明 因f(x)在?0,1?上连续,从而有最大最小值.又maxf(x)?2,f(0)?f?1??0,故最大值
x??0,1?在?0,1?内部达到.所以存在x0??0,1?使得f?x0??maxf(x)?2.因f?x0?为极大值,由Fermat
x??0,1?定理有f'?x0??0.在
x0处分别对0,1按泰勒公式展开,存在?,???0,1?使
f''??2f''???20?f?0??f?x0??f'?x0??0?x0??0?f?1??f?x0??f'?x0??1?x0????0?x0??2?22122f''???x0,
?1?x0??2?122f''????1?x0?，
其中?介于0与x0之间, ?介于0与1?x0之间.
?4所以minf''(x)?min{f''(?),f''???}?min???2,?x??0,1???x04??.
????1?x0?2而当x0?1???0,?时, min?2??4?4??2,???1?x0?2?x0?4????2??16;
x0??当x0?1???,1?时, min?2??4?4??2,???1?x0?2?x0?4???16. ???2??1?x0??故minf''(x)??16.
x??0,1?例17 设f(x)在?0,1?上存在二阶导数,且f(0)?f?1??0，又minf(x)??1,证明maxf''(x)?8.
x??0,1?x??0,1?证明 设f(x)在?0,1?上的最小值为f(a)??1,其中a??0,1?.则f'(a)?0.由泰勒公式得,
f?x??f?a??f'?a??x?a??f''??2??x?a?2 , 其中?介于x与a之间.令x分别为0,1,则
0??1?f''??1?22a2a2,
0??1?2f''??2?2?1?a?2,其中
?120??1?a??2?112.所以
f''(?1)?,f''(?2)??1?a?2.容易看出,当a时, f''(?1)?8;当a?时, f''(?2)?8,
因此maxf''(x)?8.
x??0,1?例18 证明:当x?0时,不等式x?x36?sinx?x?x36?xx5成立.
f''?x??x?sinx,f'''?x??1?cosx.所以
1202证明 (1)令f?x??sinx?x?x3,则
6f'?x??cosx?1?,
2f?0??f'?0??f''?0??0,f'''?x??0(x??0,2??).
19
f''?0?2由泰勒公式得f?x??f?0??f'?0?x?所以对任意x??0,2??，x?(2)再令F?x??F''?x??0x2?13!f'''???xx33?0(x??0,2??).其中???0,x?.
x,所以x?x3x36?sinx.而当x?6时,x?6??1?sinx,
x36?sinx(x?0）.
x?x36?x5120?sinx,则
F'?x??1?x22?x424?cosF''?x???x?6?sinx.由(1),
,于是F'?x?单调递增.当xx5?0时, F'?x??F'?0??0,F?x?单调递增.故F?x??F?0??0即x?x36?120?sinx?0,即sinx?x?x36?x5.从而结论得证.
120a?b?例19 设f(x)是在?a,b?上二次可微,且f????0.
?2?证明
?f?x?dxab?M?b?a?243,其中M?maxf''?x?.
a?x?b证明 令x0?a?b2,则f?x0??0.在x?x0处利用泰勒展开式,
f''??2f?x??f?x0??f'?x0??x?x0???b?x?2x0?,其中?介于x与x0之间.于是有
??baf?x?dx?f'?x0??ba?x?x0?dx?M2?131?2af''????x?x0?dx
3212?baf''????x?x0?dx?2?x?x0??M24?b?a?3,
其中M?maxf''?x?.
a?x?b例20设f?x?h??f?x??hf'?x????证明lim??1n?1hnn!f?n??x??h? (1) (0???1)且f?n?1??x??0.
.
h?0证明 由f(x)在x处的泰勒展开式
f?x?h??f?x??hf'?x????hnn!f?n??x??hn?1?n?1?!hn?1f?n?1??x??o?hn?1? (2).
由(1)-(2)得
hnn!1n!f?n?(f?n??x??h???n?f?n??x?)??n?1?!?n?1?f?n?1??x??o?hn?1?.
所以
??x??h??hf?x??1?n?1?!1f?x??o?1?,
即
1n!???f?n??x??h???hf?n??x?1n!1??n?1?!?n?1?f?n?1??x??o?1?.
1f?n?1?对上式两端令h又f?n?1??0,则limh?0???f?x???n?1?!?x?.
?x??0,故limh?0??.
n?1 20