微積分補(bǔ)充練習(xí)及解答自編外

A{x1x2解：由1x31x23或1x23， 3x5A{x1x2解：由1x31x23或1x23， 3x5，或1x1. 2.(2)|x1|(0)(1)0|2x1|解：(1)由0|2x1|222x12，而且2x10. x ，而且x .解集是( ,)(,)112 2(2)由|x1|x1，即1x1.解集是(113f(x1xxf(x解：令t x1，xt xt1)2f(t)(t1)22(t1)t21 f(x)x2（1）f(x)g(x)（2）f(xlnx2g(x)2lnx2x（3）f(x)g(xx（1）f(x)（2）f(xx0g(xx0相同同(1)y x2(2)yxx260，即x26 （1）的定義域?yàn)閧x|x同(1)y x2(2)yxx260，即x26 （1）的定義域?yàn)閧x|x 6或x6}.1x20x21x2x21x21，且x20x1x3x2。故函數(shù)的定義域?yàn)?3)1).y2log3(xy2log3x2)xlog3(x2)2y x32y2，因此函數(shù)的反函數(shù)為y32x2sin2 (sin 1(1)yxy解：(1)該函數(shù)是由函數(shù)yarctanu，u v，vx2復(fù)合而成的(2)y2uuv2vsintt11x8．f(x)fxf(0)及f(220，所以在函數(shù)的第一個(gè)表達(dá)式f(x1x2xf(2)12)1f(0)101；又因?yàn)?0f(x2x中，用2xf(2)224yu2 uev vx1，當(dāng)x x1時(shí)（2）yyu2 uev vx1，當(dāng)x x1時(shí)（2）y(ex11)21x1ye111)21e21)21x1ye111)21e01)2101exe yxarcsin ycos2x y（1）f(xx2arcsinx，則f(x(x2arcsinxx2arcsinxf(x)(x)2arcsin(x)x2arcsinxf(x)f(xf(x)f(xf(x（2）f(x)cos2x1f(x)cos2(x1cos2x1f(x)f(xf(xexexexexf(xf(x)f(xf(xyln(x2x211yln(x21ln10，所以函數(shù)有下界0xx12.f(x，求(1f(x)的定義域(2)f(2)及f(a2,(a為常數(shù)(1f(x)的定義域?yàn)?0)[0R(2)因?yàn)?02xf(2)2a20所以在第二個(gè)表達(dá)式中，用a2替代x，得到f(a2) 13.f(x是定義在(a,a上的函數(shù)(a0)(1)f(xf(x(2)f(xf(x(3)f(x（1）H(xf(xf(1)f(xf(x(2)f(xf(x(3)f(x（1）H(xf(xf(xH(x)f(x)f[(x)]f(x)f(x)H(x)f(xf(x（2）令G(x)f(xf(xG(x)f(x)f[(x)]f(x)f(x)G(x)f(xf(x（3）由（1）可知f(xf(x也是偶函數(shù)，再由（2）可知f(xf(x1[f(x)f(x)]1[f(x)f(x)]f(x)f(x14.求過(guò)點(diǎn)(31并且斜率為4yy0k(xx0，其中k4x0y(1)(4)(x3)y4xy0115.求過(guò)點(diǎn)(24)(49)y xy2 x2y4x(2)9 4，從而得5x6y34016．解：固定成本C02000000元，變動(dòng)成本為C1(Q4Q元，其中Q總成本函數(shù)為C(Q20000004Q 2000000平均成本函數(shù)為C(Q)R(QL(QR(Q16．解：固定成本C02000000元，變動(dòng)成本為C1(Q4Q元，其中Q總成本函數(shù)為C(Q20000004Q 2000000平均成本函數(shù)為C(Q)R(QL(QR(QC(Q8Q20000004Q4Q200000017．解：設(shè)總成本函數(shù)為C(xax2bx200xa102b10200a100b10020010ab100ab解這個(gè)方程組，得a0.03,b0，因此，總成本函數(shù)為C(x0.03x2200 0.03x2平均成本函數(shù)為C(x)0.03x 18．Rax2bxca02b0ca2b2c6a42b4cc即4a2bc6a0.5,b4,c0R0.5x24x1 1 , (2)0,1, , （1）數(shù)列的通項(xiàng)為y ynn(1)lim|an|0liman0(2)若數(shù)列{a2n}與{a2n1收斂且極限是相同的，那么{an}(2)0,1, , （1）數(shù)列的通項(xiàng)為y ynn(1)lim|an|0liman0(2)若數(shù)列{a2n}與{a2n1收斂且極限是相同的，那么{an}也收斂（1）|an||an0|1,3,5,7,9 (2)0,1,0,10,1,2n 2n(n)（1）（2）04．討論極限lim|x||x|x解：因?yàn)? lim11； lim(1)1，所以極 lim|x|x5f(x，討論limf(x是否存在x解：因?yàn)楫?dāng)x0時(shí)，f(x)ex，而且 ，從而limex，所limf(x)lim 0ex因?yàn)楫?dāng)x0時(shí)，f(xsinx，所以limf(解：因?yàn)楫?dāng)x0時(shí)，f(x)ex，而且 ，從而limex，所limf(x)lim 0ex因?yàn)楫?dāng)x0時(shí)，f(xsinx，所以limf(xlimsinx0limf(xlimf(x0，因此極限limf(x06.limf(x)0lim|f(x|0limf(x0，所以對(duì)任意給定的正數(shù)0，存在0，使當(dāng)0xf(x)|f(x)||f(x|lim|f(x|0f7x0時(shí)，哪些是無(wú)窮小量，哪些是無(wú)窮大量，哪些既不是無(wú)窮大量也xxxsin1x,ln|x|12x, x20.01xx1（1）xx， ，x20.01x，0.1ln|x|x ,1（2）x是無(wú)窮小量的有：sin1 1x,ln|x|x，x20.01x1 2x,x|x|x8.f(x)xlimf(x )x1x1xx1x f(xlimx10 limf(xlim(x1)0x1xx1x xxx1f(x9．以下數(shù)列當(dāng)n1limf(x )x1x1xx1x f(xlimx10 limf(xlim(x1)0x1xx1x xxx1f(x9．以下數(shù)列當(dāng)n1 }, 1lim[1(1)] lim(1)n 0解：因?yàn)?0, 0，所以當(dāng)nn10．x0sinx2是比tanx高階的無(wú)窮小.（本題超前！以后再做1:首先約定0 x00x2，0sin1（tan31，得：sinsinsin00，故0tantantantansinx2是比tanx2（可以利用等價(jià)無(wú)窮小關(guān)系）x0sinx2:x2tanx:xsin limx0tanx0sinx2是比tanx（sinsinsinx)limx1100tantanx0tanx0sinx2是比tanxx3x3n2n（1）lim(x3x6)(2)(3)xx2x1nn34n22x2(6)lim(2x3x（4）xx25x1 1(1)lim(x23x6)223264x3x (2)（5）x0sinx2是比tanxx3x3n2n（1）lim(x3x6)(2)(3)xx2x1nn34n22x2(6)lim(2x3x（4）xx25x1 1(1)lim(x23x6)223264x3x (2)（5）xx2x11 1x2x1x3x0，所以解：因?yàn)閤x3xxx2x13 3nn0000 lim(3)14nn34n21022x2102(4)xx25x 1 1x )(5)1 1(1x)(1xx2 (1x)(1xx22x(1x)(2x1(1x)(1xx2 x1(1x)(1xx22x11x 1112(6)lim(2x33x0，所以lim(2x33x6)解：因?yàn)閤2x33x(xh)2（） （1）lim ，（2）x3n12Lx2（4）（5）（6）5xnx1解：(1)lim1sin1lim1limsin10sin00xxx1，所以limsin10另解：因?yàn)閘im10sinxx(xh)2(x22xhh2)lim(2xh)2x02x(2)1(3)lim(11L1)lim(21)1解：(1)lim1sin1lim1limsin10sin00xxx1，所以limsin10另解：因?yàn)閘im10sinxx(xh)2(x22xhh2)lim(2xh)2x02x(2)1(3)lim(11L1)lim(21)13n11 (3n3n1lim 3 3 (4)n3n11 1(n(1lim12L1 x2(x1)(xx 5x4lim(x1)(x4)limx4(6)x1x2ax6，確定a與b13．若xx2ax6lim(x2)0，所以lim(xaxb0解：因?yàn)閤222ab0，從而得b2a4x2axx2ax2a(x24)a(xxxxlim(x2a)4aa2，b2a42(2)48x14f(x，問(wèn)當(dāng)a為何值時(shí)，極限limf(x2x xlimf(x)limexe01limf(x)lim(2xa0aalimf(xlimf(x，即a1時(shí)，極限limf(x15x0ex1x是等價(jià)無(wú)窮小.（本題超前！以后再做證：令uex1，則exu1xln(u1x0u0ex u0ln(uuux0ex1x（1）limtanxsinx（3）（4）x0sinxsinx2sin2x1x，（7）lim（5）（6）x3sin(xsintanxsinsin解(1)證：令uex1，則exu1xln(u1x0u0ex u0ln(uuux0ex1x（1）limtanxsinx（3）（4）x0sinxsinx2sin2x1x，（7）lim（5）（6）x3sin(xsintanxsinsin解(1)lim1tanx(1cosx)11(11)0x0tantanxsintansinsin) 10 55 ) 11 x0sin sin5x5 x0sin 2sin2sin2212xsinxsin令u，則x ，當(dāng)x時(shí)，u0，從limxsin1 sinulimsinu1.u02x1(2x11)(2x1 sin(2x11)sin2x11)sinx012x1011x0sinx2sinlimxsin1100lim xx )sinx2xsin x0sin(x3)(x(x(x2)1(32)5x3sin(xsin(x x3sin(x22)，（2） （3）xx（4）lim(1x)x （5）lim1xx02222lim解：(1)limxxx 11xxlim xlim e4x1 3x1x x2)x2x22)，（2） （3）xx（4）lim(1x)x （5）lim1xx02222lim解：(1)limxxx 11xxlim xlim e4x1 3x1x x2)x2xe2 1)x2lim(3))x2)x2x21(4)lim(1x)xlim13((5)lim1x lim1x lim1xx e2x02x02x02L2n2 2n22n2解 2n2 2n2 2n212n2 2n2 2n22n2 2n2 2n2L2n2 2n2 2n22n2 2n212n2 n(2n2 2n22L2n2n)2n2 又因?yàn)閚2n2 n2n21 )1L2n2 2n22n2 1cos12x （1）（）（3）（）sintanx03arcsinx0arctanx(1)解：因?yàn)楫?dāng)x0，sinx～x，1cosx ，所1)1L2n2 2n22n2 1cos12x （1）（）（3）（）sintanx03arcsinx0arctanx(1)解：因?yàn)楫?dāng)x0，sinx～x，1cosx ，所1lim1cosx1sin2(2)x0tanx～x，12x1～2x112x11tan(3)x0arcsinx～xln(15x～5x5x03arcsin sin(4)因?yàn)楫?dāng)x0，arctanx～x， sinlimsin x0arctan 20.x0xsinxxyxx0f(0)011f(0limxsin10f(0lim(x11f(0f(0，極限limf(xx021.xxxxx25x（1）y（2） （3） xxxx25x解(1)yxx由x2xxxxx25x（1）y（2） （3） xxxx25x解(1)yxx由x2x20，即(x2)(x1)0，得x x1.因?yàn)閤2和x1時(shí)，x2x1x25x(x2)(xx2 x2因?yàn)閤2x x2(x2)(xx 2 x25x(x2)(xxx1因?yàn)閤2x x1(x2)(xx1xyex20x2x2x2limex2x2xxxx(3)f(x)xx0f(0)0f(0lim(x11f(0lim(x11f(0f(0，極限limf(xx022.ln(1x21（3） x0（1）（2）x0sin(11ln(1x2（1）x0sin(1sin(1f(x)x0f(0)0 ln(1x2f(0)0所以x0sin(11（2）lim111limcos1xsin(1f(x)x0f(0)0 ln(1x2f(0)0所以x0sin(11（2）lim111limcos1xlim)cos01cos(1)cos(11110（3） x0x0ln(1x～x ln(1x) x1.x0x03x2xx23f(x，求bf(xx1x1f(14，極限limf(xlim(3x2b312b3b當(dāng)limf(xf(1，即3b4，從而b1f(xx124.求常數(shù)a和bxsin1xxxaxxyx（2）xsinaxxxyxx0f(xax2x0f(xxbx0f(0)2f(0lim(ax2)2f(0lim(xbbf(0f(0f(0)，即b2f(xxf(0lim(xbbf(0f(0f(0)，即b2f(xx0續(xù)．因此，當(dāng)b2axsin1xxxf(x)sinsinx0f(x當(dāng)x0時(shí)，f(x)x x0f(0)af(0limsinx1f(0lim(xsin1bbf(0f(0f(0)ab1f(xx0處連續(xù)．因此，當(dāng)ab125.證明方程sinxx1至少有一個(gè)介于2和2f(xsinxx1f(x22f(2)sin(2)(2)11sin20f(2)sin2213sin20由根的存在定理可知，在(22)x0f(x0sinx0x010方程sinxx1至少有一個(gè)介于2和2x026.yxexx21x0x1xf(xxexx210，1f(0)0e002110f(1)1e121e20由零點(diǎn)定理可知，在(01xf(x0yxexx21x0x1x1f(xxxxf(x)f(1)limx1f(1xf(x)fxx22x(xf(1)xx1f(xxxxf(x)f(1)limx1f(1xf(x)fxx22x(xf(1)xxxlim(x1)(11)y3xyf(xxf(x3(xx23x2)limyx0 yxxx3．f(xx0f(0)0limf(xlimx2sin10，所以limf(xf(0)f(xx0x2sin1f(x)f(0)limxsin10f(xx0又因?yàn)閤f(0)0（1）y （2） （3）xx（4）ytan（1）yy(x1)x11 （2）yxy(x2 x ．2(3)y xxx4解：因?yàn)閥 xx xx2x4yx4) (4)ytanytanu)sec2u（(3)y xxx4解：因?yàn)閥 xx xx2x4yx4) (4)ytanytanu)sec2u（2）（3）(1)ylnxylnx)1(2)y2x1解：y(2x)2xln2， 20ln2ln2(3)ysinxcos1ysinx)cosx 6.ysinx1 cosysinx)cosx，切線的斜率為ky1 3(x) 法線的斜率為k1k 3，法線方程為y2 3(x6)（1）y4x3 （2）（3）yx22xlogx，（6）ylnsin（4）y xxex（5）y1cos(1)y4x3 x2ln2解：y(4x3 x2lnx)(4x3)(x)(2lnx)12x22(2)yy2解：y(4x3 x2lnx)(4x3)(x)(2lnx)12x22(2)yy(x2ex)(x2exx2(ex)2xexx2ex(3)yx22xlogxyx22xlogx22x22x)logx)222x2xln2(4)y x解：y(xxex)(x)(xex) (x)exx(ex) exxex2 2sin(5)y1cossinx)(sinx)(1cosx)sinx(1cosy1coscosx(1cosx)sinx(sinx)cosxcos2xsin2 1cos(6)y1xlnln (lnx) ln lnx(x) 1解：y )（1）ysinxcosxtanxcotxcscysinxcosxtanxcotxcsc(sinx)(cosx)(tanx)(cotx)(csccosxsinxsec2xcsc2xcscxcotx（2）yln3xarcsinxtanyln3xarcsinxtanx)xln3)(arcsinx)tansec2x1(3)yarctanxarccosxyyln3xarcsinxtanx)xln3)(arcsinx)tansec2x1(3)yarctanxarccosxy(arctanxarccosx2x)(arctanx)arccosx)2x2xln211（4）yxlnxlnxyxlnxlnxex)xlnx)lnx)ex(x)lnxx(lnx)(lnx)xlnx(x)lnxx11lnxexlnx11lnxex（5）y1sintsint)(tsint)(1sint)tsint(1siny1sin(sinttcost)(1sint)tsint(cossintsin2ttcosttcostsinttsintcos(1sinsintsin2ttcos(6)yexsinxlogyexsinxlogx)exsinx)logx)ex)sinxex(sinx)exsinxexcosxxln(7)ysecxlnx x解：y(secxlnx x2x)(secxlnx)(x2x(secx)lnxsecx(lnx)[(x)2x x(2xsecxtanxlnx x2ln22y xx34sin1，求x1解：y[x(x34(secx)lnxsecx(lnx)[(x)2x x(2xsecxtanxlnx x2ln22y xx34sin1，求x1解：y[x(x34sin1)](x)(x34sin1) x(x34x34(x4sin1) x(3x00) x223x113210x單位的總成本是C(x)20003x50x（元MC(x)C(x)20003x50x)03503 2 32.55.5．（1）ysin2x3，（2）ylnsin4x，(3)yexx(4)y(3x4(5)yeaxsinbx，（6）y(1)ysin2ysin2x32sinx3(sinx32sinx3cosx3(x36x2sinx3cos(2)ylnsincos4x(4x)4cosx4cotylnsin4x)(sin4x)(3)yexyexxx)(ex)(xx2)ex()(x2)ex )x ex(4)y(3x4y3x41)100100(3x41)99(3x41)100(3x41)99(12x3100(3x41)9912x31200x3(3x4x ex(4)y(3x4y3x41)100100(3x41)99(3x41)100(3x41)99(12x3100(3x41)9912x31200x3(3x4(5)yeaxsinyeaxsinbx)eax)sinbxeax(sinbx)aeaxsinbxeaxbcos(6)ylnsin解：y xln3(ln xln3 xlnsinlnsin xln3 () xln )sin (1)ysin2xe2xy(sin2xe2x)(sin2x)e2x)cos2x(2x)e2x(2x)cos2x2e2x2(cos2xe2x(2)y(5x24)31y5x2431x](5x24)31x(5x2431x(10x0)31x(5x24)[(1x)310x31x(5x24)(1x)3(110x31x (5x24)(1(3)y1cossin2x)(sin2x)(1cos2x)sin2x(1cosy1cos(2cos2x)(1cos2x)sin2x(02sin2cos2x2cos22x2sin22x2cos2x(3)y1cossin2x)(sin2x)(1cos2x)sin2x(1cosy1cos(2cos2x)(1cos2x)sin2x(02sin2cos2x2cos22x2sin22x2cos2x 1cos(4)設(shè)y 1x2，求y(1x2)(1x2)(02x)212111（1）ysin2xcosysin2xcos2x)(sin2x)cos2xsin2x(cos2 )cos2x (2sin 2sinxcosx(cos2x2sin2xsin 2sinxcosx1cos2x2sin2xsin2（2）y3xcosxtanx3xcosxtanx2) 3x) )(tanx2y(3x(3x)sinx()sec2x2(x2 cos3x1sinx2xsec2 2（3)y 4x(4x23ln34x3 (3x)4x23x(4x24解：y( ) 4(4x2(4x23xln34x23x(4x23ln34x3 (3x)4x23x(4x24解：y( ) 4(4x2(4x23xln34x23x(4x2)3xln324(4x2(4x21t(4)yy 11(3t)(1t)(3t)(1t) sec(2t)sec(2t)(1t2（5）y(xa2x2 a2解：y[(x a2x2)1](xa2x2)2(x a2x2a2x2)2[(x)a2x2(x(xa2x2)2(a2x22a2(xa2x2)22a2(xa2x2)2a214.yef(xyyef(x2]ef(x2)f(x2ef(x2)f(x2(x22xef(x2)f(x2（1）y(11lnyln(11)xxln(11)x[ln(x1)lnx]xln(x1)xln(lny)[xln(x1)xlnx][xln(x1)](x（1）y(11lnyln(11)xxln(11)x[ln(x1)lnx]xln(x1)xln(lny)[xln(x1)xlnx][xln(x1)](xln1yln(x1)xx(lnxx1lnx1]y[ln(11) yy[ln(x1) x(x（2）y(x2)(xx(xx(xlnyln [lnxln(x1)ln(x2)ln(x(x2)(x (x2)(x (lny)1[lnxln(x1)ln(x2)ln(x1y1[1(x1)(x2)(x3 xxx1(13 y1y(1 xyexey(xyexey)(xy)(ex)(ey)0yxyexeyy0(xey)yexyexyxey17．x22xyy22xx2(x22xyy2)2x2y2x(xey)yexyexyxey17．x22xyy22xx2(x22xyy2)2x2y2xy2yyy1xyxx2x22xyy22x，得44yy24y0y4x21201或1245k 所求切線方程為y0 (x2)或y4 (x2)，即y x1或y xxx218．f(xx1處可導(dǎo)，求abaxf(xx1f(xx1limf(x)limf(x)flimx2lim(axb)aab1，即b1f(x)fx2f(1)lim(x1)xxf(x)faxbax(1a)f(1)xxxxxx因此，得a2b1a1219f(x)ln(1xf(xsinxf(xxxf(x)x0f(x)f(0)limsinxln(10)limsinxf(0)xf(xf(xxxf(x)x0f(x)f(0)limsinxln(10)limsinxf(0)xf(x)fln(1x)ln(1f(0)xxxlimln(1x)limx1f(0)f(0)1f(0)1xxf(x)y1解：因?yàn)閥1t33t2x1t22t t t yytet)t)ett(et)ettetx(et)et(t)et ette e(1t) (1)y xln1 ln1 y(xlnlnxx(ln lnxx1y 1 ln1 y(xlnlnxx(ln lnxx1y lnx ln ln (ln222 x2lnx 2 x2ln(2)ysinxcos ysinxcosx)cosxsiny(cosxsinx)sinxcosx(sinxcos(1)yxsinxy()y(xsinx)(x)sinxx(sinx)sinxxcosy(sinxxcosx)(sinx)(xcosx)cosx(x)cosxx(coscosxcosxxsinx2cosxxsin y()2 20 (2)y3x32x2x1y3x32x2x1)9x24x1y(9x24x1)18x23．求下列函數(shù)的n(1)yeayeax)eax(ax)aeay(aeax)aeax(ax)a2eay(a2eax)a2eax(ax)a3eay(n)(2)yxeyxex)(x)exx(ex)exxex(1x)ey[(1x)ex](1x)ex(1x)(ex)ex(1x)ex(2x)ey[(2x)ex](2x)ex(2x)(ex)ex(2x)ex(3y[(2x)ex](2x)ex(2x)(ex)ex(2x)ex(3x)ey(n(nx)e24yf(xy(1)x2y2(x2y2a22x2yyyyx(xx(x)yxyyxyyy y (2)y1y1xeyy0(x)eyx(ey)eyxeyeyey12(ey)(2y)ey(2(2eyy(2y)ey(y)(2y)2ey(3y)(2eyy 2e(3y)ye(3 (2(2（1）y2xy2xx2ex)2x)x2ex)22xexdyydx(22xexx2ex（2）yy(xex2)(x)ex2x(ex2)ex2xex2(2x)(12x2dyydx(12x2)ex21（3）y1 (1ex)cosx(1ex)(cos 2xexcosx1（3）y1 (1ex)cosx(1ex)(cos 2xexcosx(1ex)(sin)cos2cos2 2xexcosx(1ex)(sindyydxcos2（4）ysiny(sin2x)cos2x(2x)2cosdyydx2cos(1)xyexyydyd(xydexdxdyexyd(xy)exydxex(1exy)dy(exyxdy 11ex(2)x2y2R2（Rydyd(x2y2dx2dy22xdx2ydyydydyx (1)d C) (2)d(arttanx)1(3)d(1e3xC) 解法一：數(shù)值9.8(2)d(arttanx)1(3)d(1e3xC) 解法一：數(shù)值9.8fxxx9.81取x9，x0.8，則f(9) 93，f(x)，f(9)22f(9.8f(90.8f(9)f310.83.13339.890.89(10.8)310.0889解法二因?yàn)?0.0889 9.831.04453.133529.設(shè)某商品的需求函數(shù)為Q80010P（P為價(jià)格，Q為需求量），成本函數(shù)為（２）解：由Q80010PP800.1QR(Q)QPQ(800.1Q)80Q0.1Q2L(QR(QC(Q80Q0.1Q2500020Q60Q0.1Q25000（1）L(Q60Q0.1Q25000)600.2Q經(jīng)濟(jì)意義：當(dāng)需求量Q154時(shí)，再增加銷售（或生產(chǎn)）30L(400)600.240020經(jīng)濟(jì)意義：當(dāng)需求量Q400時(shí)，再增加銷售（或生產(chǎn)）201f(xsinx在區(qū)間[0，2]上滿足羅爾定理的條件，并求出此時(shí)定理中的數(shù)值.f(xcosx在區(qū)間（02）內(nèi)有定義，所以在區(qū)間（02）f(0)f(21f(xsinx在區(qū)間[0，2]上滿足羅爾定理的條件，并求出此時(shí)定理中的數(shù)值.f(xcosx在區(qū)間（02）內(nèi)有定義，所以在區(qū)間（02）f(0)f(20f(x)在區(qū)間[02]上滿足羅爾定理的條件.由f(xcosx0x2k 值 值.f(xlnx是初等函數(shù)，且在區(qū)間[1，2]上有定義，所以它在區(qū)間[1，2]f(x1在區(qū)間（1，2）f(x在區(qū)間（1，2）f(x在區(qū)間[1，2]上滿足拉格朗日中值定理的條件.從而存在一點(diǎn)(12)f(2ln2f(1ln10f(1ln201解得 (1,2)，因此，此時(shí)定理中的數(shù)值 3．sinx2sinx2f(xsinxx1x2x1x2sinx2sin0x2x1x2x1x2f(xx1x2]上滿足拉格朗日中值定理的f(x2)f(x1)f()(x2x1)，(x1,x2)sinx2sinx1cos(x2x1)1x2x2cosx2sinx2sinx2 f(xarctanxarccotxf(x)(arctanxarccotx)(arctanx)(arccotx)01 1 所以f(x)c.又因?yàn)閏f(1)arctan1arccot ，所以，得f(x) arctanxarccotx f(xarctanxarccotxf(x)(arctanxarccotx)(arctanx)(arccotx)01 1 所以f(x)c.又因?yàn)閏f(1)arctan1arccot ，所以，得f(x) arctanxarccotx5ex(1)x0(exex x0(x2 x02x 20 x0x2(2)limln(12x)1x01 12exe(3)sin(exex)(sinx)exexexesinexee0(n(4)x(xnn(n解： x x(2x x2xlnx(2xlnL(5)limx2lnlnlimxlnx x0 (6) ex(ex1[x(exex1解： ) exx(exexx0ex1 x0exlnlimxlnx x0 (6) ex(ex1[x(exex1解： ) exx(exexx0ex1 x0exexx02(7)limxcos1cos1sin11解：limx 1limsin1sin0xsin6.驗(yàn)證極限xxcos11sinxsin11xxcos 1 1(xsin1cos(xcos x1sin7.f(xf(xf(x)x(x解：令f(x)x(x2)0，得x1 x22列表判別如下：因此，區(qū)間(0)、(2f(x(,0(0,2(2,ff解：因?yàn)楫?dāng)x(02yxsinx)1cosx0解：因?yàn)楫?dāng)x(02yxsinx)1cosx0，所以函數(shù)yxsinx證明：當(dāng)x0ex1x證：f(xexx1f(xexx1)ex1．x0f(x)0f(x在(0x0f(x)f(0)0，即exx10，因此 1(x0)10.(1)yx3解：函數(shù)的定義域?yàn)?y(x3x)3x2xx(3xyx(3x10x1x0列表判別如下：(、(0f(x(,0)f(x(2)yxln(x解：函數(shù)的定義域?yàn)?1y[xln(x1)] 1y0x01x0f(x0，因此，區(qū)間(0f(x當(dāng)1x0f(x0，因此，區(qū)間(10f(x(3)y2x33x2解：函數(shù)的定義域?yàn)?y(2x33x212x)6x26x126(x2)(x(1,ff令y6(x2)(x1)0，得x1 x21．列表判別如下：因此，區(qū)間(2)、(1f(x的單調(diào)增區(qū)間；區(qū)間(2,1f(x(4)y1x1x)1xxy1令y6(x2)(x1)0，得x1 x21．列表判別如下：因此，區(qū)間(2)、(1f(x的單調(diào)增區(qū)間；區(qū)間(2,1f(x(4)y1x1x)1xxy1 (1 (1因此，區(qū)間(1、(1f(x(5)f(x) x3 x3解：函數(shù)的定義域?yàn)?xxx f(x)(x3 x31) 0x3 x3(x1) f(xx10x1x0f(x列表判別如下：因此，區(qū)間(0)、(1f(x的單調(diào)增區(qū)間；區(qū)間(0,1f(x11.(1)yxex解：函數(shù)的定義域?yàn)?y(xex)(x)exx(ex)exxex(1x)ex(,ff(,ffy1x)ex]1x)ex1x(ex)ex1x)ex2x)ex．yy1x)ex]1x)ex1x(ex)ex1x)ex2x)ex．y0，即(2x)ex0x2．x2f(x0，因此區(qū)間2x2f(x0，因此區(qū)間(2，+)為凹區(qū)間．點(diǎn)(22e2為曲線的拐點(diǎn)．(2)yln(1x2解：函數(shù)的定義域?yàn)?(1x2)11(2x)(1x2)(2x)(1x2(1x22(1x2)2(1x2y(1x2 (1x2 1令y0，解得x1 x21．這兩個(gè)點(diǎn)將定義域分成三個(gè)區(qū)間：(1、(11及(1所以點(diǎn)(1ln2)，(1,ln2)是曲線的拐點(diǎn)；區(qū)間(1,1是凹區(qū)間，區(qū)間(1(3)yx42x31解：函數(shù)的定義域?yàn)?y(x42x31)4x36x2y(4x36x2)12x212x12x(x1)令y0，即12x(x1)0，解得x1 x2這兩個(gè)點(diǎn)將定義域分成三個(gè)區(qū)間：(0)、(01及(1所以點(diǎn)(01，(1,0)(0)(1(0,1(,0(0,1(1,yy(,(1,yy(1)f(x) (x 解：函數(shù)的定義域?yàn)? f(x) (x1)3]0 (x (x ．x1 (1)f(x) (x 解：函數(shù)的定義域?yàn)? f(x) (x1)3]0 (x (x ．x1 x1f(xx1f(x0x1時(shí)，f(x)0因此f(1) x1(2)f(x)xln(1x1f(x)[xln(1x)]11 1f(x00x011x0f(x0x0f(x0f(0)0ln(10)0x0(3)f(x)3xx0 3(x2f(x)(3x )3 f(x0x240x2x2f(x)(312) 0f(2)32)126612是函數(shù)的極大因?yàn)閒(2)x2因?yàn)閒(2)240f(2)32126因?yàn)閒(2)240f(2)32126612是函數(shù)的極小值，x2(4)f(x)x39x215x1解：函數(shù)的定義域?yàn)?f(x)(x39x215x1)3x218x153(x26x5)3(x1)(xf(x0，即3(x1)(x50x11x25f(x)(3x218x15)6x因?yàn)閒(16118120f(11391215118是函數(shù)的極大x1是極大值點(diǎn)．又因?yàn)閒(56518120f(553952155124x5(5)f(x)2x2解：函數(shù)的定義域?yàn)?f(x)(2x2x4)4x4x34x(1x2)4x(1x)(1f(x0，即4x(1x)(1x0x11x20x31f(x)(4x4x3)4因?yàn)閒(14121)280f(12(1)21)41x1又因?yàn)閒(0)4120240f(0)202040是函數(shù)的極小值，xf(14121280f(1212141x113．f(x1解：函數(shù)的定義域?yàn)?(2x)(1x2)(2x)(1x2)(1x2)22(1x2)(1x22(1x2(1x2f(x))1令f(x)0，即(2x)(1x2)(2x)(1x2)(1x2)22(1x2)(1x22(1x2(1x2f(x))1令f(x)0，即2(1x2)0，解得x x1．這兩個(gè)特殊點(diǎn)將定義域分四個(gè)區(qū)間：(1、(11及(1列表判別如下：f(11x121f(1)1x1(1)f(x)8x2f(x)8x2x416x4x34x(2x)(2xf(x0，即4x(2x)(2x0x0x2（舍去x2（舍去．f(0)0f(1)f(1)817f(x在[11上的最小值為f(0)0f(1x(2)f(x)x2x 解：f(x)(x x3)1 1 1 由f(x)0，即1 0，解得駐點(diǎn)x1；而x0時(shí)，f(x)不存在，即x0是可導(dǎo)點(diǎn)．比較f(0) f(1)1f(1)5f(8)2f(x在[1上的最小值為f(1) ；在x8處取得最大值2(,(1,(1,f(xf(x (3)f(x)sin2x[ f(xsin2xx)2cos2x1由f(x)2cos2x10即cos2x1當(dāng) x時(shí)解得駐點(diǎn)x，x 比較f() ，f ) (3)f(x)sin2x[ f(xsin2xx)2cos2x1由f(x)2cos2x10即cos2x1當(dāng) x時(shí)解得駐點(diǎn)x，x 比較f() ，f ) ) f()sin()()0 ，f()sin 0 的大小，得到f(x)在 ]上的最小值為f() ；最大值為f ) (4)f(x)x(x[2,解：f(x)[x(x1)3](x)(x1)3x[(x1)3](x1)3 (x (x1)3[3(x1)x] (4x3)33(xf(x0，即4x30x3x1f(xx0f(3)33f(1)f(2)23f(2)2的大小，得到f(x) [22]f3)332f(2)23215.f(xln(1x2，x[0(1)f(x)f(xln(1x21x0f(x0f(x在所給區(qū)間[0(2)f(x)f(x在所給區(qū)間(0f(xf(0)ln(102)0解：設(shè)長(zhǎng)方體容器的底面邊長(zhǎng)為xhVx2h，所以容器