微積分第4章習(xí)題選解

(1)f(x)x3(x1),[0,1]fx在[0,1]上連續(xù)，在(0,1)f(0f(10f(x)4x33x2x2(4x3),3(0,1)4fxgx在[abgx0f(a)f(b0(a,bf()g()f()g(證作輔助函數(shù)hx)fx),則hx)fx)gxfx)gx)g( 顯然hx在[a,b(a,b使h()0即f()g(f()g(cosxcosyxy證xyf(t)cost，在x,yfxx,yfxfysinxysin1cosxcosyxyx1時,exex證令f(t)etet,f(tetex1在[1,xf(t(1xexex(eex1而1x1exex(eex10fxx2x4x8)fx0有幾個實根,指出所在區(qū)間,并進一步f(x)0有幾個實根. f(x)有三個零點2,4,8,由羅爾定理知,f(x)在區(qū)間(2,4),(4,8)內(nèi)分別至少有一個零點.而f(x)是二次多項式,故只有兩個零點.進一步地,fxfx(2,8)fxarctanx12

1x2

(3)在區(qū)間(1,內(nèi),fx2(1)f(3)2f(0)0 2(1x2)4xf(x) 1x

21

2x1x

(1x2 1x

12

1x(1x2

0f(3) 1arctan(3)1()2x(1,fx2

8.設(shè)fx在[a,b上連續(xù),在(a,b內(nèi)可導(dǎo),證明存在(a,bbf(b)af(a)b

證作輔助函數(shù)gxxfx則gx)fxxfxgx在[ab(a,b),g(b) bf(b)af

f()f() b b

g()limx1lim 1x1ln x11/6lim2x53x22x1lim10x46x26

x23x

2xlimxsinxlim1cosxlimx2/21 x 3x x03x 1 limxarctanx 1x x 1xx0xarcsin x011x

1x x

2limcos(sinx)1limsin(sinx)cosx1 3x limcos(sinx1limsin2x21 3xtanx

3xsec2x

tan2

x0xsinln

1

1cos1

lim 2x01x2sin2lim lim lim 0x0cot

x0xcsc2

limlntan7xlim7tan3xsec27xlim73x1x0lntan x03tan7xsec2 x037lim(2x)tanxlim2x 2x x cot xcsc2

lim(1

)limexx1limexx1limex

1

x0

ex

x(ex

x

2 limln(exx limex

e2elim(ex

x x

x x0elim(sinx)xlimexlnsinxlim

lnsin1/xlim

cosxxsinxe01

ln 1/limlnxln(1x)limlnxxlim lim

limx0x

x01/ x01/x limx(ln

ln2lnarctanlnarctanx limx

arctanx)xex ex 1/ ex(1x)arctanxeg(x)e3.設(shè)f(x)

,xxg(01g(01fxx(g(x)ex)(g(x)ex解當(dāng)x0時，f(x) xx0

(0)

f(x)fx

g(x)exxlimg(x)e

limg(x)e

g(x)e

g(01 x

2

解解

yx1xy11,駐點x1x0xx(,(1,(0,(1,y+--+y↗↘↘↗yxlnxylnx1駐點x1ex0x(0,1/ (1/e,yy 解

y2xxy2xx2xxx

,(0,

x1定義域0x2(1,2)y

xln(1x)xx22fxxln(1xfx

f(x)1 1

1

0x0fxf(00xln(1xx2x再證右式.令g(x)ln(1x)x ,2g(x)

1

1x

x1

0x0gx)g(0)0即ln(1xxx22x2x綜上所述,xln(1x)x 2解

yxeyex(1x),yex(x2)x(2-02e+2拐點 )y(x1)3x2 yx3x3,y x3

3 3

3(5x2)y9

x3(5x2)

1 x3 3

4 x3(5x215x) 9

4 x3(5x1) 9拐點可疑點:x0x15x155(150(0,f-0++f1拐點 )5設(shè)點(1,3yx3ax2bx14a,b之值,并求出該曲線 y3x22axb,y6x2a,y(1)62a0,a3又y(11ab143b9y3x26x93x3x1駐點x13x(,(1,(3,yy+↗-↘+↗2f(x)32(x1)3 f(x)

4(x1)333

不可導(dǎo)點x1x(,ff+↗-↘

f(1)3解

f(x)xlnxfxlnx1駐點x1ex0x(0,1/1/(1/e,yy-↘0+↗((

f1)1. 解

f(x)xefxex(1x駐點x1x(,1(1,ff+↗0-↘極大值f(1)1e100100x

f(x)

,[6,8]100x f(x)100x

0，x0f(0)10,f(6)8,f(8)6

f(010f(86f(x)xex2,(,)令 f(x)ex22x2ex2(12x2)e 0，x 1f

x)

2 同理,最大值為f ) Q6.Q1255PQP1004(自產(chǎn)自銷,產(chǎn)銷平衡,試問如何定價,才能使工廠獲得利潤最大?最大利潤是多Q解P255

QL(Q)RCQPCQ(25)1004Q 21Q100 L(Q)2Q5

0，Q52.5此時價格為P14.5L(Q)205

故當(dāng)Q52.5時利潤最大,最大利潤為

451.25或解LP(1255P4(1255P100145P5P2600令L145 0，P14.510.34每件售價多少時,方可獲最大利潤?最大利潤是多少?解設(shè)售價為xL(1204x20)(x3)1520x200x22760,

x[3,L1520

0，x3.8,160L4000x3.8Lmax12813.yexx0)解設(shè)切點(ueu則切線方程為yeueuxu它與兩條坐標(biāo)軸的交點分別為(0,1u)eu),(u1,0,截距分別為(1u)eu,u1,S(u)1(u1)2eu2

u0S(u)1(u1)(12

0，u12故當(dāng)u1時面積取到最大值,Smaxe

ylnxy x25x解lim 0,故有水平漸近線y0xx25x , 故有鉛直漸近線x1,x4x1x25x

x4x25x解

y4lim4

(x(x

4,故有水平漸近線y4故有鉛直漸近線x1(5)yxex解

limxex0故有水平漸近線y0解

y

2xx25x2x

., 2x x1x25x x4x25x

x1,x4a

2x

x2xx25x2b 2x25x

2x)

10x225x

10x故有斜漸近線y2x10y3xx3

xy

excosx excosxexsin cosx

2cos (cosx1)(ex2 x xcosx

ex211

lim

1 x x lim(cotx1)limxcosxsinxlimxcosxsinxlimxsinx

xsinxx

xx

xxe

e2(

e2(xlim x

limxx

1e

e

e

1/ x1/2e，a0，b0 1x1n

lim

ax2

)xlimln(axbx)ln2limaxlnabxlnblnalnbln(，原極限axbx

ax ax1bx1或解：lim( )xlim(1 ax1bx ax1bx

)limax1bx

lnalnlim(1 1

)ax1bx ex ex

limxln(ex1)，limlny

1 x0ln(ex 原極限y1x2xyxy(1)3

2x,y22x

0，x1拐點(1,0)y3(x1，即3xy30ey

1

y0y0

yxxx0) 解：yxx(1lnx)0,x ,極小值f

eeyy(xx33xy22y332確定，試討論該函數(shù)有無極值x3x23y22xyy6y2y0y0x2y2xx

2x4yy2xy2yy2(2yy2y2y)0x2y2y0y(210x24xxp60

（10000x60000）解：（1）Cx6000020x,Cx20（2）R(x)xp60x

x

,R(x)60

x（3）L(x)R(x)C(x)40x

x

60000L(x)40

L(20000)340000（4）x1000(60p),x1000,xp . 60（5）R(x)60

0,x30000,R(x)

0fxxlnxfxfxxln

,x f(x)1lnx,x xlnx,0xfx)0x1e

(1lnx),0xxx e1e1 1f+0-+f↗極大值1/↘↗,1f(x)1

x

(0,1)(1,(1,0)

,0x證明：當(dāng)02

cos2

tantancos2使tantan cos2

(，而cos2cos2cos21故cos2

cos2

cos2

cos2

tantan

。cos2利用中值定理證明：方程5ax44bx33cx22dxabcd在(0f(x)ax5bx4cx3dx2abcd)xf(x)5ax44bx33cx22dx