资源预览内容
第1页 / 共13页
第2页 / 共13页
第3页 / 共13页
第4页 / 共13页
第5页 / 共13页
第6页 / 共13页
第7页 / 共13页
第8页 / 共13页
第9页 / 共13页
第10页 / 共13页
亲,该文档总共13页,到这儿已超出免费预览范围,如果喜欢就下载吧!
资源描述
1微分中值定理应用举例微分中值定理应用举例单调性与极值单调性与极值1.函数在上,比较的大小.)(xf0,1/( )0fx /(1),(0),(1)(0)ffff解:在上满足拉氏中值定理条件,存在,使得.)(xf0,10,1/(1)(0)( )fff由于,所以单调增加,而,所以,/( )0fx /( )fx01/(0)( )(1)fff即./(0)(1)(0)(1)ffff2.函数在0,1上,比较的大小.)(xf/( )0,(0)0fxf/(1),(0),(1)(0)ffff解:由于,所以单调增加,而,所以在上,/( )0fx /( )fx/(0)0f0,1/( )0fx 同上题讨论有/(0)(1)(0)(1)ffff3.在内,判断在,0内的( )()f xfx 0,/( )0,( )0fxfx/( ),( )fxfx符号.解:,所以在内为奇函数,为偶函数,为( )()f xfx )(xf, /( )fx/( )fx奇函数,在内,所以在,0内.0,/( )0,( )0fxfx/( )0,( )0fxfx4.已知函数在区间内具有二阶导数,且严格递增, )(xf1,1/( )fx,则:A.在内均有;B.在内均有/(1)(1)1ff1,1( )f xx 1,1 , 1,1;C. 在内均有,在内均有;( )f xx1,1( )f xx1,1( )f xxD. 在内均有,在内均有.1,1( )f xx1,1( )f xx解:令,则,( )( )F xf xx(1)(1) 10Ff /( )( ) 1Fxfxx1,1x1x 1,1x/( )( ) 1Fxfx/( )0Fx 0/( )0Fx ( )( )F xf xx减( )F x0增( )F x( )0F x 极小值( )0F x 选择 B.25 .设处处可导,则)(xfA.必;B. 必lim( ) xf x /lim( ) xfx /lim( ) xfx lim( ) xf x C. 必;D. 必lim( ) xf x /lim( ) xfx /lim( ) xfx lim( ) xf x 解:选择 D (A,C 的反例,B 的反例)yx2yx6.设函数在上有界且可导,则)(xf0,A. 必 ;B. 存在,必;lim( )0 xf x /lim( )0 xfx /lim( ) xfx /lim( )0 xfx C. 必; D. 存在,必; 0lim( )0 xf x /0lim( )0 xfx /0lim( ) xfx /0lim( )0 xfx 解:选择 A (B,C,D 的反例)( )f xx7. 设函数在的邻域内连续,且,则在处)(xf0x (0)0f 0( )lim21 cosxf x x0x A. 不可导; B.可导,且; C.取极大值; D.取极小值)(xf/(0)0f解:20000( )( )1( )(0)1( )(0)limlim2lim2lim21 cos00 2xxxxf xf xf xff xf xxxxxx所以 0000( )(0)1( )(0)1( )(0)limlimlimlim0000xxxxf xff xff xfxxxxxxx所以在可导,且.)(xf0x /(0)0f,而,所以在的某邻域内, 0( )lim21 cosxf x x1 cos0,20x0x ( )0f x 所以在处取极小值.(0)0f0x )(xf8. 为恒大于 0 的可导函数,且,则当时( ), ( )f x g x/( ) ( )( )( )0fx g xf x gxaxbA. ;B. ;( ) ( )( ) ( )f x g bf b g x( ) ( )( ) ( )f x g af a g xC. ; D. ( ) ( )( ) ( )f x g xf b g b( ) ( )( ) ( )f x g xf a g a解:,所以为减函数,/2( )( ) ( )( )( )0( )( )f xfx g xf x gx g xgx( ) ( )f x g x即当时,又为恒大于 0,axb( )( )( ) ( )( )( )f bf xf a g bg xg a( ), ( )f x g x3所以,选择 A( ) ( )( ) ( )f x g bf b g x9.设有二阶连续导数,且,)(xf/(0)0f/0( )lim1 xfx xA.(0)f是( )f x的极大值;B. (0)f是( )f x的极小值;C. 0,(0)f是曲线( )yf x的拐点;D. (0)f不是( )f x的极值;0,(0)f也不是曲线( )yf x的拐点.解:,所以在的邻域内,即曲线是凹的,又/0( )lim10 xfx x 0x /( )0fx ,所以(0)f是的极小值.选择 B/(0)0f)(xf10.设函数在xa的某个邻域内连续, ( )f a为的极大值,则存在0,当)(xf)(xf时,必有:,xaaA. ; B. ;( )( )0xaf xf a( )( )0xaf xf aC.; D. .2( )( )lim0()()taf tf xxatx2( )( )lim0()()taf tf xxatx解:( )f a为的极大值,则存在0,和时,)(xf,xaa,xa a都有,所以时, ,所以 A,B 都不正确.( )( )f xf a,xaa( )( )0f xf a,由于,所以.22( )( )( )( )lim()()taf tf xf af x txax( )( )0f af x2( )( )0()f af x ax选择 C11.设函数在内有定义, 是函数的极大值点,则)(xf, 00x )(xfA. 必是的驻点;B.必是的极小值点0x)(xf0x()fxC. 必是的极小值点; D.对一切都有0x( )f xx0( )()f xf x解:选择 B12. ,则在处2( )( )lim1()xaf xf a xa xa4A. 导数存在,且; B.取极大值; C.取极小值; D . 导数不存在)(xf/( )0fa )(xf解:,所以在的某去心邻域内有,2( )( )lim1()xaf xf a xa xa( )( )0f xf a所以在处,取极大值.xa)(xf9 .证明:的最大值(1,2,)nn n L证明:令,1 ( )xf xx(1)x 1ln( )xxf xe,所以时,11 / 222111( )ln1 lnxxfxxxxxxxxxe/( )0fx 且时,时,所以时的唯一极大值,也是xe/( )0fx xe/( )0fx ( )f e1 ( )xf xx最大值.而的最大值必是中的一个,而,所以是(1,2,)nn n L32, 332333的最大值.(1,2,)nn n L不等式的证明不等式的证明1.当时,证明:;0x 1arctan2xx证明:令1( )arctan2f xxx,所以时单调减,/ 2211( )01fxxx0x 1( )arctan2f xxx而,1lim( )limarctan02xxf xxx所以时,即.0x 1( )arctan02f xxx1arctan2xx2. 当时,证明:;0x 11ln(1)1xx证明:时,令,0x 11( )ln(1)1f xxx, 单调减,/ 222111( )01(1)1xfxxxx xx ( )f x而,11lim( )limln(1)01xxf xxx5所以时,即.0x 11( )ln(1)01f xxx11ln(1)1xx方法二,时, ,令,0x 1ln(1)ln(1)lnxxx( )lnf xx则在区间上用拉格朗日中值定理有:,1x x/11ln(1)ln(1)ln( )xxfx其中,所以,即有.1xx111 1xx11ln(1)1xx3.证明:;221ln(1)1xxxx证明:设22( )1ln(1)1f xxxxx 则/2222122( )ln(1)(1) 12 12 1xxfxxxx xxxx ,令,得唯一驻点2ln(1)xx/( )0fx 0x ,所以是的极小值点,所以又/21( )0 1fx x 0x ( )f x( )(0),f xf(0)0f所以,即.( )0f x 221ln(1)1xxxx4.当,证明 ;1x ln(1) ln1xx xx证明:因为,所以,所证等价于1x ln ,10xx1ln(1)lnxxxx零,则,所以时单调增加,( )lnf xxx/( )ln10fxx 1x ( )lnf xxx而,所以,即,即.11xx(1)( )f xf x1ln(1)lnxxxxln(1) ln1xx xx5.,证明:;1,xae()()aa xaxa证明:只需证ln()()lnaaxaxa令,则,( )ln()()lnf xaaxaxa/( )lnafxaax/ 2( )0afx ax 所以单调减少,而,所以时/( )fx/(0)1 ln0fa 10x /( )(0)0fxf即单调减少,而,所以时,( )f x(0)0f10x ( )(0)0f xf即,即.ln()()lnaaxaxa()()aa xaxa66.设,证明:baebaab证明:只需证明,设,lnlnbaab( )lnlnf xxaax,所以单调增加,/ 2( )ln,( )0aafxafxxx/( )lnafxax又,所以时,/( )ln10faa bxae/( )ln0afxax故单调增加.( )lnlnf xxaax因此,时,而,bxe( )lnln( )f xbxxbf a( )0f a 所以,即时,.( )lnln0f bbaabbaelnlnbaab所以.baab7设在上可导,且单调递减,( )f x0,/( )fx证明:对任意正数,都有, a b1()(2 )(2 )2f abfafb证明:不妨设,令0ab1( )()(2 )(2 )2F xf axfafx则,当时有,由于单调递减/( )()(2 )Fxfaxfxxa2axx/( )fx所以,即,所以单调增,即时/()(2 )faxfx/( )0Fx ( )F xxa( )( )F xF
网站客服QQ:2055934822
金锄头文库版权所有
经营许可证:蜀ICP备13022795号 | 川公网安备 51140202000112号