數(shù)學(xué)上冊(cè)-課本答案第二章

習(xí)題 P8下列函數(shù)在給定區(qū)間上是否滿足羅爾定理的所有條件?如滿足就求出定理中的數(shù)值解

f(x)2x2x3,[1,1.5]; f(x) 1

,[2,2]14f(1)f(1.5)0.滿足.f(x)4x1, 14滿足.f(x) ,(1x2下列函數(shù)在給定區(qū)間上是否滿足拉格朗日中值定理的所有條件?如滿足就求出定理中的數(shù)值(1)f(x)x2,[0,a],a0; f(x)lnx,[1,2];(3)f(x)x35x2x2,[1,解滿足滿足滿足

a2022(a0), a2ln2ln11(21),1a2 ln2(9)(32101)[0(1)],

5 5 3

sinsin解由柯西中值公式

cos)(0cos

1sin 或 1sin,11sin2(

1

從而cos (0,1),而(0,)1( 2不用求出函數(shù)f(x)(x1)(x2)(x3)(x4)的導(dǎo)數(shù),說(shuō)明方程f(x)0有幾個(gè)實(shí)根,并它們f(1f(2)f(3)f(4);故有1122334,f(x0f(x在(ab內(nèi)具有二階導(dǎo)數(shù),f(x1f(x2f(x3,ax1x2x3b,證明:(x1x3內(nèi)至少有一點(diǎn),f(0x11x22x3,f(1

f(20,故存在12,f((1)|sinasinb||ab|; 1

ln(1x) (x0);(3)ex1 (x0)exexx1);(5)當(dāng)0x

時(shí),tanx2x2;(6)x0時(shí)

ln(1x)arctanx 證

tan

1在區(qū)間[0,xf(xln(1x運(yùn)用拉格朗日中值定理,x1

ln(1x) 1

ln(1x)ln1 1

(0x)在區(qū)間[0,xf(xexx1運(yùn)用拉格朗日中值定理,(exx1)0(e1)(x0) ((0,x))f(x)exex;f(x)exe0(x1tan令f(x) ,x

f(x)f(1)0(x1xsec2xtan 2xsinf(x) 0 2x2cos2tan 其中我們已經(jīng)知道2xsin2x0(x0).于是f(x) 在區(qū)間(0,)內(nèi)是嚴(yán)格單調(diào)增加的,從而 所欲證在區(qū)間[0,x](x0)f(x1xln(1xarctanx運(yùn)用拉格朗日中值定理,(1x)ln(1x)arctanx0ln(1) (x0) (0x)從而得證11

12證

arctanx ;(2)2arctanx

1

(x1)1f(xarctanx1

,1x 21 f(x) 1x 21 1x11x1f(x在(上是個(gè)常數(shù)函數(shù),f(x

f(0)0x,得所欲證f(x2arctanx

1x2,

2[1(1x2)x2x)]11f 1 (111f(xx1處是右連續(xù)的,f(x在[1上是個(gè)常數(shù)函數(shù),f(xf(1(x1(1)limlnx;(2)lim2x1;(3)limxxcosx;(4)limcotx;(5)limxtanx1x

x03

tan

xsin x0ln1

sin(6)lim ;(7)lim ;(8)limxsinx;(9)lim(ax1)x(a0) x

tan2(10)lime

;(11)

ln(1x)(12)lim2arctanx100 x0 xarccot x 1解(1)原式1x1

ln2x03xln ln1(cosxxsin sinx(sinxxcos原式=

1coscsc2

1

sin

111x原式=lim lim 1x x0sinxsin原式limxtanx(x)3lim1sec2xlim2secxtanxsec sin lim2sec3xsinx1 原式limex1

ex

ex

1x0x(ex x0(ex1) x0(1x)ex x0ex(1 原式=

sin3xcos

1

sin

1

1xsinxcos x3sin yxsinx,則lnysinxlnxlncsclimlny

1 limsinxtanx0 x0cscxcot 于是limxsinxlimylimelnye0

tt

at1t

tt

atlnaln.1. 1x2 原式=

=…=

t1原式=limln(1t)lim1t原式

t10arctanx

1t2令y arctanx

,則lnyx(lnlnarctanx,ln2lnarctan1xlimlny 1x

arctan

1x2

1 2 2

xarctanx1 從而limarctanxlimylimelnyex (1)f(x)x5x32x1,x01;(2)f(x)ex,x0a(略11(1)f(x)xex;(2)f(x)11(略y2x33x212x;(2)yxex;(3)yxln(1x);

y1x解

yln(x1x2);(6)yx|sin2x|x(,1(1)y6x26x126(x2)(x1)00y (, (0, (, (0, y 1

(, (, (1, (0,

yx211x

1,y 1 1 (1 (, (2, (1, (0, yln(x1x2) ) x111221yx11122112 kx(k12y

)

.y

kx(k

)12 (k12

)x(k1)

12 (k )x(k1)1212當(dāng)kx(k )且12cos2x0即當(dāng)x5k時(shí)y0,在該點(diǎn)左側(cè)附近y0,在右側(cè)附1212y0.單調(diào)增加區(qū)間(k,5k),單調(diào)減少區(qū)間(5k,(k )12 12)當(dāng)(k )x(k1)且12cos2x0即當(dāng)x(k5時(shí)y0,在該點(diǎn)左側(cè)附近y0,在12)6側(cè)附近y0.單調(diào)增加區(qū)間(k ),(k5,單調(diào)減少區(qū)間(k5,(k1)12 12

f(x)

x 1

x

ln(1x)

e2 xln(1x)解當(dāng)x0時(shí),lnf(x) ,

limlnf(x)limln(1x)xlim1 1

x02(1

f(x)limelnf(x)e122

f(0)

f(x,

0處連續(xù)設(shè)limf(xk,求lim[f(xaf(x 解lim[f(xa)f lim[f()a]alimf() (x,xa) x(1)yx33x27yx4yxsin(1)yx33x27yx4yxsinx(4)yx2lnxyx11yxx(7)y2exex2y2(x1)31y32(x1)3解x(,0(0,2(2,00ymaxminy1

(x2)(x2)x(,(2,0(0,2(2,00ymaxminy1cosx0,x2k.在每一個(gè)駐點(diǎn)的兩側(cè),y符號(hào)不變(都是正的),故這些駐點(diǎn)都 (,)1 (,)1 e0min12 1

y 11 (,4)y340max54(3411xlnx yxx xx2(1lnx,xe.經(jīng)過(guò)駐點(diǎn)時(shí),y由正變負(fù),yxx1yeea為何值時(shí),yasinx

sin3x在x 處取得極值?它是極大值還是極小值?并求此值 解yacosxcos3x.

0a110a2x xy(asinx3sin3x)由 (2

30)0x處取得極大值3x 318.求下列函數(shù)在給定區(qū)間上的最大值與最小值x12(3)yx ,0xx1212解y ,沒(méi)有使導(dǎo)數(shù)為零的點(diǎn)或使導(dǎo)數(shù)不存在的點(diǎn).12x0處取得它在[0,4]0,x4

0,

6,故y 氧氣的恢復(fù).設(shè)f(t)是某 氧氣的衡量標(biāo)準(zhǔn).當(dāng)f(t)1時(shí)是標(biāo)準(zhǔn)水平,時(shí)間t以周計(jì)算.t0時(shí) ,一些有機(jī)物開(kāi)始氧化,里氧氣的數(shù)量是由模t2tf(t)

t2

,0t確定的.試問(wèn)何時(shí)里氧氣水平最低?何時(shí)氧氣水平最高ft 1(t21)t2tt2 ( t21,f t2 t21,得駐點(diǎn)y 11012 1氧氣水平當(dāng)t=1時(shí)最低(為),t=0時(shí)最高(2.(

yx

;

y

arctan解(2)yx2ex)2xexx2ex1x22x)exy(x24x2)ex(x2 2)(x2 (,2 20拐 (22,220拐 (2y(5)yearctanx,y

earctan