微积分补充练习及解答自编外

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）的定义域为{x|x同(1)y x2(2)yxx260，即x26 （1）的定义域为{x|x 6或x6}.1x20x21x2x21x21，且x20x1x3x2。故函数的定义域为(3)1).y2log3(xy2log3x2)xlog3(x2)2y x32y2，因此函数的反函数为y32x2sin2 (sin 1(1)yxy解：(1)该函数是由函数yarctanu，u v，vx2复合而成的(2)y2uuv2vsintt11x8．f(x)fxf(0)及f(220，所以在函数的第一个表达式f(x1x2xf(2)12)1f(0)101；又因为20f(x2x中，用2xf(2)224yu2 uev vx1，当x x1时（2）yyu2 uev vx1，当x x1时（2）y(ex11)21x1ye111)21e21)21x1ye111)21e01)2101exe yxarcsin ycos2x y（1）f(xx2arcsinx，则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，所以函数有下界0xx12.f(x，求(1f(x)的定义域(2)f(2)及f(a2,(a为常数(1f(x)的定义域为(0)[0R(2)因为202xf(2)2a20所以在第二个表达式中，用a2替代x，得到f(a2) 13.f(x是定义在(a,a上的函数(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也是偶函数，再由（2）可知f(xf(x1[f(x)f(x)]1[f(x)f(x)]f(x)f(x14.求过点(31并且斜率为4yy0k(xx0，其中k4x0y(1)(4)(x3)y4xy0115.求过点(24)(49)y xy2 x2y4x(2)9 4，从而得5x6y34016．解：固定成本C02000000元，变动成本为C1(Q4Q元，其中Q总成本函数为C(Q20000004Q 2000000平均成本函数为C(Q)R(QL(QR(Q16．解：固定成本C02000000元，变动成本为C1(Q4Q元，其中Q总成本函数为C(Q20000004Q 2000000平均成本函数为C(Q)R(QL(QR(QC(Q8Q20000004Q4Q200000017．解：设总成本函数为C(xax2bx200xa102b10200a100b10020010ab100ab解这个方程组，得a0.03,b0，因此，总成本函数为C(x0.03x2200 0.03x2平均成本函数为C(x)0.03x 18．Rax2bxca02b0ca2b2c6a42b4cc即4a2bc6a0.5,b4,c0R0.5x24x1 1 , (2)0,1, , （1）数列的通项为y ynn(1)lim|an|0liman0(2)若数列{a2n}与{a2n1收敛且极限是相同的，那么{an}(2)0,1, , （1）数列的通项为y ynn(1)lim|an|0liman0(2)若数列{a2n}与{a2n1收敛且极限是相同的，那么{an}也收敛（1）|an||an0|1,3,5,7,9 (2)0,1,0,10,1,2n 2n(n)（1）（2）04．讨论极限lim|x||x|x解：因为 lim11； lim(1)1，所以极 lim|x|x5f(x，讨论limf(x是否存在x解：因为当x0时，f(x)ex，而且 ，从而limex，所limf(x)lim 0ex因为当x0时，f(xsinx，所以limf(解：因为当x0时，f(x)ex，而且 ，从而limex，所limf(x)lim 0ex因为当x0时，f(xsinx，所以limf(xlimsinx0limf(xlimf(x0，因此极限limf(x06.limf(x)0lim|f(x|0limf(x0，所以对任意给定的正数0，存在0，使当0xf(x)|f(x)||f(x|lim|f(x|0f7x0时，哪些是无穷小量，哪些是无穷大量，哪些既不是无穷大量也xxxsin1x,ln|x|12x, x20.01xx1（1）xx， ，x20.01x，0.1ln|x|x ,1（2）x是无穷小量的有：sin1 1x,ln|x|x，x20.01x1 2x,x|x|x8.f(x)xlimf(x )x1x1xx1x f(xlimx10 limf(xlim(x1)0x1xx1x xxx1f(x9．以下数列当n1limf(x )x1x1xx1x f(xlimx10 limf(xlim(x1)0x1xx1x xxx1f(x9．以下数列当n1 }, 1lim[1(1)] lim(1)n 0解：因为 0, 0，所以当nn10．x0sinx2是比tanx高阶的无穷小.（本题超前！以后再做1:首先约定0 x00x2，0sin1（tan31，得：sinsinsin00，故0tantantantansinx2是比tanx2（可以利用等价无穷小关系）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：因为xx3xxx2x13 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：因为x2x33x(xh)2（） （1）lim ，（2）x3n12Lx2（4）（5）（6）5xnx1解：(1)lim1sin1lim1limsin10sin00xxx1，所以limsin10另解：因为lim10sinxx(xh)2(x22xhh2)lim(2xh)2x02x(2)1(3)lim(11L1)lim(21)1解：(1)lim1sin1lim1limsin10sin00xxx1，所以limsin10另解：因为lim10sinxx(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，确定a与b13．若xx2ax6lim(x2)0，所以lim(xaxb0解：因为x222ab0，从而得b2a4x2axx2ax2a(x24)a(xxxxlim(x2a)4aa2，b2a42(2)48x14f(x，问当a为何值时，极限limf(x2x xlimf(x)limexe01limf(x)lim(2xa0aalimf(xlimf(x，即a1时，极限limf(x15x0ex1x是等价无穷小.（本题超前！以后再做证：令uex1，则exu1xln(u1x0u0ex u0ln(uuux0ex1x（1）limtanxsinx（3）（4）x0sinxsinx2sin2x1x，（7）lim（5）（6）x3sin(xsintanxsinsin解(1)证：令uex1，则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，则x ，当x时，u0，从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 又因为n2n2 n2n21 )1L2n2 2n22n2 1cos12x （1）（）（3）（）sintanx03arcsinx0arctanx(1)解：因为当x0，sinx～x，1cosx ，所1)1L2n2 2n22n2 1cos12x （1）（）（3）（）sintanx03arcsinx0arctanx(1)解：因为当x0，sinx～x，1cosx ，所1lim1cosx1sin2(2)x0tanx～x，12x1～2x112x11tan(3)x0arcsinx～xln(15x～5x5x03arcsin sin(4)因为当x0，arctanx～x， sinlimsin x0arctan 20.x0xsinxxyxx0f(0)011f(0limxsin10f(0lim(x11f(0f(0，极限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.因为x2和x1时，x2x1x25x(x2)(xx2 x2因为x2x x2(x2)(xx 2 x25x(x2)(xxx1因为x2x x1(x2)(xx1xyex20x2x2x2limex2x2xxxx(3)f(x)xx0f(0)0f(0lim(x11f(0lim(x11f(0f(0，极限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，极限limf(xlim(3x2b312b3b当limf(xf(1，即3b4，从而b1f(xx124.求常数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续．因此，当b2axsin1xxxf(x)sinsinx0f(x当x0时，f(x)x x0f(0)af(0limsinx1f(0lim(xsin1bbf(0f(0f(0)ab1f(xx0处连续．因此，当ab125.证明方程sinxx1至少有一个介于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至少有一个介于2和2x026.yxexx21x0x1xf(xxexx210，1f(0)0e002110f(1)1e121e20由零点定理可知，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又因为xf(0)0（1）y （2） （3）xx（4）ytan（1）yy(x1)x11 （2）yxy(x2 x ．2(3)y xxx4解：因为y xx xx2x4yx4) (4)ytanytanu)sec2u（(3)y xxx4解：因为y xx xx2x4yx4) (4)ytanytanu)sec2u（2）（3）(1)ylnxylnx)1(2)y2x1解：y(2x)2xln2， 20ln2ln2(3)ysinxcos1ysinx)cosx 6.ysinx1 cosysinx)cosx，切线的斜率为ky1 3(x) 法线的斜率为k1k 3，法线方程为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单位的总成本是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)设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 所求切线方程为y0 (x2)或y4 (x2)，即y x1或y xxx218．f(xx1处可导，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：因为y1t33t2x1t22t 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．求下列函数的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：数值9.8(2)d(arttanx)1(3)d(1e3xC) 解法一：数值9.8fxxx9.81取x9，x0.8，则f(9) 93，f(x)，f(9)22f(9.8f(90.8f(9)f310.83.13339.890.89(10.8)310.0889解法二因为10.0889 9.831.04453.133529.设某商品的需求函数为Q80010P（P为价格，Q为需求量），成本函数为（２）解：由Q80010PP800.1QR(Q)QPQ(800.1Q)80Q0.1Q2L(QR(QC(Q80Q0.1Q2500020Q60Q0.1Q25000（1）L(Q60Q0.1Q25000)600.2Q经济意义：当需求量Q154时，再增加销售（或生产）30L(400)600.240020经济意义：当需求量Q400时，再增加销售（或生产）201f(xsinx在区间[0，2]上满足罗尔定理的条件，并求出此时定理中的数值.f(xcosx在区间（02）内有定义，所以在区间（02）f(0)f(21f(xsinx在区间[0，2]上满足罗尔定理的条件，并求出此时定理中的数值.f(xcosx在区间（02）内有定义，所以在区间（02）f(0)f(20f(x)在区间[02]上满足罗尔定理的条件.由f(xcosx0x2k 值 值.f(xlnx是初等函数，且在区间[1，2]上有定义，所以它在区间[1，2]f(x1在区间（1，2）f(x在区间（1，2）f(x在区间[1，2]上满足拉格朗日中值定理的条件.从而存在一点(12)f(2ln2f(1ln10f(1ln201解得 (1,2)，因此，此时定理中的数值 3．sinx2sinx2f(xsinxx1x2x1x2sinx2sin0x2x1x2x1x2f(xx1x2]上满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.又因为cf(1)arctan1arccot ，所以，得f(x) arctanxarccotx f(xarctanxarccotxf(x)(arctanxarccotx)(arctanx)(arccotx)01 1 所以f(x)c.又因为cf(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.验证极限xxcos11sinxsin11xxcos 1 1(xsin1cos(xcos x1sin7.f(xf(xf(x)x(x解：令f(x)x(x2)0，得x1 x22列表判别如下：因此，区间(0)、(2f(x(,0(0,2(2,ff解：因为当x(02yxsinx)1cosx0解：因为当x(02yxsinx)1cosx0，所以函数yxsinx证明：当x0ex1x证：f(xexx1f(xexx1)ex1．x0f(x)0f(x在(0x0f(x)f(0)0，即exx10，因此 1(x0)10.(1)yx3解：函数的定义域为(y(x3x)3x2xx(3xyx(3x10x1x0列表判别如下：(、(0f(x(,0)f(x(2)yxln(x解：函数的定义域为(1y[xln(x1)] 1y0x01x0f(x0，因此，区间(0f(x当1x0f(x0，因此，区间(10f(x(3)y2x33x2解：函数的定义域为(y(2x33x212x)6x26x126(x2)(x(1,ff令y6(x2)(x1)0，得x1 x21．列表判别如下：因此，区间(2)、(1f(x的单调增区间；区间(2,1f(x(4)y1x1x)1xxy1令y6(x2)(x1)0，得x1 x21．列表判别如下：因此，区间(2)、(1f(x的单调增区间；区间(2,1f(x(4)y1x1x)1xxy1 (1 (1因此，区间(1、(1f(x(5)f(x) x3 x3解：函数的定义域为(xxx f(x)(x3 x31) 0x3 x3(x1) f(xx10x1x0f(x列表判别如下：因此，区间(0)、(1f(x的单调增区间；区间(0,1f(x11.(1)yxex解：函数的定义域为(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，因此区间2x2f(x0，因此区间(2，+)为凹区间．点(22e2为曲线的拐点．(2)yln(1x2解：函数的定义域为((1x2)11(2x)(1x2)(2x)(1x2(1x22(1x2)2(1x2y(1x2 (1x2 1令y0，解得x1 x21．这两个点将定义域分成三个区间：(1、(11及(1所以点(1ln2)，(1,ln2)是曲线的拐点；区间(1,1是凹区间，区间(1(3)yx42x31解：函数的定义域为(y(x42x31)4x36x2y(4x36x2)12x212x12x(x1)令y0，即12x(x1)0，解得x1 x2这两个点将定义域分成三个区间：(0)、(01及(1所以点(01，(1,0)(0)(1(0,1(,0(0,1(1,yy(,(1,yy(1)f(x) (x 解：函数的定义域为( f(x) (x1)3]0 (x (x ．x1 (1)f(x) (x 解：函数的定义域为( f(x) (x1)3]0 (x (x ．x1 x1f(xx1f(x0x1时，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是函数的极大因为f(2)x2因为f(2)240f(2)32126因为f(2)240f(2)32126612是函数的极小值，x2(4)f(x)x39x215x1解：函数的定义域为(f(x)(x39x215x1)3x218x153(x26x5)3(x1)(xf(x0，即3(x1)(x50x11x25f(x)(3x218x15)6x因为f(16118120f(11391215118是函数的极大x1是极大值点．又因为f(56518120f(553952155124x5(5)f(x)2x2解：函数的定义域为(f(x)(2x2x4)4x4x34x(1x2)4x(1x)(1f(x0，即4x(1x)(1x0x11x20x31f(x)(4x4x3)4因为f(14121)280f(12(1)21)41x1又因为f(0)4120240f(0)202040是函数的极小值，xf(14121280f(1212141x113．f(x1解：函数的定义域为((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．这两个特殊点将定义域分四个区间：(1、(11及(1列表判别如下：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上的最小值为f(0)0f(1x(2)f(x)x2x 解：f(x)(x x3)1 1 1 由f(x)0，即1 0，解得驻点x1；而x0时，f(x)不存在，即x0是可导点．比较f(0) f(1)1f(1)5f(8)2f(x在[1上的最小值为f(1) ；在x8处取得最大值2(,(1,(1,f(xf(x (3)f(x)sin2x[ f(xsin2xx)2cos2x1由f(x)2cos2x10即cos2x1当 x时解得驻点x，x 比较f() ，f ) (3)f(x)sin2x[ f(xsin2xx)2cos2x1由f(x)2cos2x10即cos2x1当 x时解得驻点x，x 比较f() ，f ) ) f()sin()()0 ，f()sin 0 的大小，得到f(x)在 ]上的最小值为f() ；最大值为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在所给区间[0(2)f(x)f(x在所给区间(0f(xf(0)ln(102)0解：设长方体容器的底面边长为xhVx2h，所以容器