1.泰勒公式有如下特殊形式：當(dāng)f(x(在x=0處的n(n∈N*(階導(dǎo)數(shù)都存在時(shí)，f(x(=f(0(+f/(0(x+ xn+…．注：f”(x(表示f(x(的2階導(dǎo)數(shù)，即為f/(x(的導(dǎo)數(shù)，f(n((x((n≥3(表示f(x(的n階導(dǎo)數(shù)，該公式也稱麥克勞林公式．2.【極值點(diǎn)第二充分條件】已知函數(shù)f((1)若f”(x0(>0，則f(x)在x0處取得極小值；(2)若f”(x0(<0，則f(x)在x0處取得極大值.3.帕德近似是法國(guó)數(shù)學(xué)家亨利．帕德發(fā)明的用有理多項(xiàng)式近似特定函數(shù)的方法．給定兩個(gè)正整數(shù)m，n，函數(shù)f(x)在x=0處的[m,n]階帕德近似定義為：R，且滿足：f(0)=R(0)，f/,f(m+n)(0)=R(m+n)(0)．(注：f”(x)=[f/(x)[/，f川(x)=[f”(x)[/，f(4)(x)=[f川(x)[/，f(5)(x)=[f(4)(x)[/，…；f(n)(x)為f(n-1)(x)的導(dǎo)數(shù)).4.拉格朗日中值定理又稱拉氏定理：如果函數(shù)f(x(在[a,b[上連續(xù)，且在(a,b(上可導(dǎo)，則必有ξ∈(a,b(，使得f/(ξ((b-a(=f(b(-f(a(.知識(shí)卡片1：一般地，如果函數(shù)f(x(在區(qū)間[a,b[上連續(xù)，用分點(diǎn)a=x0<x1<…<xi-1<xi<…<xn=b將區(qū)間[a,b[等分成n個(gè)小區(qū)間，在每個(gè)小區(qū)間[xi-1,xi[上任取一點(diǎn)ξi(i=1,2,…,n(，作和式if(ξ1(Δx f(x(在區(qū)間[a,b[上的定積分，記作bfdx即dxf(ξi(.這里，a與b分別叫做積分式.從幾何上看，如果在區(qū)間[a,b[上函數(shù)f(x(連續(xù)且恒有f(x(≥0，那么定積分bfdx表示由直線x=a,x=b(a≠b(,y=0和曲線y=f(x(所圍成的曲邊梯形的面積.知識(shí)卡片2：一般地；如果f(x(是區(qū)間[a,b[上的連續(xù)函數(shù)，并且F/(x(=f(x(，那么dx=FF(b)-F(a).這個(gè)結(jié)論叫做微積分基本定理，又叫做牛頓-萊布尼茨公式. ,1(時(shí)，有不等式(1+x(n≤1+nx成則稱f(x(為[a,b[上的凹函數(shù)；若f，則稱f(x(為凸函數(shù).若f(x(是區(qū)間[a,b[恒成立(當(dāng)且僅當(dāng)x1=x2=…=xn時(shí)等號(hào)成立).(2)設(shè)f(x(=-2sinx，若區(qū)間[a,b[滿足：當(dāng)f(x(定義域?yàn)閇a,b[時(shí)，值域也為[a,b[，則稱區(qū)間[a,b[為(1)由題意，得sinx>x-，所以>1->1-=1->，(2)對(duì)于函數(shù)f(x(=-2sinx，有-2≤f(x(≤2,-2≤a<b≤2，,b[，故f(x(最小值為-2，于是a=-2，所以-,?[a,b[，此時(shí)f(x(的定義域?yàn)閇-2,2[，值域?yàn)閇-2,2[，符合題意．當(dāng)a>-時(shí)，f(x(在[a,b[上單調(diào)遞減，故sinb+sina>0,-(sina+sinb(<0，與a+b=-2(sina+sinb(矛盾；綜上所述，f(x(有唯一的“和諧區(qū)間”[-2,2[．①若函數(shù)f(x(可導(dǎo)，我們通常把導(dǎo)函數(shù)f/(x(的導(dǎo)數(shù)叫做f(x(的二階導(dǎo)數(shù)，記作f”(x(.類似的，函數(shù)f(x(的二階導(dǎo)數(shù)的導(dǎo)數(shù)叫做函數(shù)f(x(的三階導(dǎo)數(shù)，記作f川(x(，函數(shù)f(x(的三階導(dǎo)數(shù)的導(dǎo)數(shù)叫做函數(shù)f(x(的四階導(dǎo)數(shù)……，一般地，函數(shù)f(x(的n-1階導(dǎo)數(shù)的導(dǎo)數(shù)叫做函數(shù)f(x(的n階導(dǎo)數(shù)，記作f(n((x(=[fn-1(x([/，n≥4；③若函數(shù)f(x(在包含x0的某個(gè)開區(qū)間(a,b)上具有任意階的導(dǎo)數(shù)，那么對(duì)于任意x∈(a,b(有g(shù)(x)=f(xx-xx-xx-x0(n+…，我們將g(x(稱為函數(shù)f(x(在點(diǎn)x=x0處的泰勒展開式.例如f1(x)=ex在點(diǎn)x=0處的(1)求出f(x)=cosx在點(diǎn)x=0處的泰勒展開式g(x(；(1)f(x)=cosx，f/(x)=-sinx，f''(x(=-cosx，…,所以f(0)=cos0=1，f/(0)=-sin0=0，f''(x(=-cos0=-1，…,由cosxx-0(2n+…(3)因?yàn)?(1-1+1-1+1-1(1-1-… 兩邊求導(dǎo)可得：-sinx所以sinxf(x(<xg(x(；>0.所以φ(x(在(-∞,0[上遞減，在[0,+∞(上遞增，從而ex-1-x=φ(x(≥φ(0(=0，即ex≥1+x.(2)設(shè)h(x(=xg(x(-f(x(=x則h/(x故對(duì)x>0有h/(x(=x.，所以h(x(在[0,+∞(上遞增.從而對(duì)x∈(0,+∞(有h(x(>h(0(=0，即xg(x(-f(x(>0，故f(x(<xg(x(.(3)據(jù)已知有F(x(=g(x(-a故F/(xax，F(xiàn)”(xa(我們約定F”(x(是F/(x(的導(dǎo)數(shù)).據(jù)e≠e-”(xa>1-a≥0.所以F/(x(在(-∞,0[和[0,+∞(上遞增，從而在(-∞,+∞(上遞增.故對(duì)x<0有F/(x(<0，對(duì)x>0有F/(x(>0.所以F(x(在(-∞,0[上遞減，在[0,+∞(上遞增，從而x=0是F(x(的極小值點(diǎn)，滿足條件；xx從而e2-e-2>e0-e-0=0，xa-2(+1-a=0.這就說明F/(x(在ln上遞減，從而對(duì)0<x<2ln有F/(x(<F/(0(故F(x(在ln上遞減，從而x=0不可能是F(x(的極小值點(diǎn)，不滿足條件.4.18世紀(jì)早期英國(guó)牛頓學(xué)派最優(yōu)秀代表人物之一的數(shù)學(xué)家泰勒(BrookTaylor克勞林公式)有如下特殊形式：當(dāng)f(x(在x=0處的n(n∈N*(階導(dǎo)數(shù)都存在時(shí)，f(x(=f(0(+f/(0(.x xn+…．其中，f″(x(表示f(x(的二階導(dǎo)數(shù)，即為f/(x(的導(dǎo)數(shù)，f(n((x((n≥3(表示f(x(的n階導(dǎo)數(shù)．(1)記f(x(=cosx，則f/(x(=-sinx,fⅡ(x(=-cosx,f(3((x(=sinx,f(4((x(=cosx，令g(x(=sinx-(x，則g/(x(=cosxx2,gⅡ(x(=-sinx+x,g川(x(=1-cosx，∵g川(x(≥0恒成立，∴gⅡ(x(在(0,+∞(遞增，∴gⅡ(x(>gⅡ(0(=0,∴g/(x(在(0,+∞(遞增，∴g/(x(>g/(0(=0,∴g(x(在(0,+∞(遞增，g(x(=sinx-(xg(0(=0，即sinx>x 令φ(x(=ln(x+1(-x,(x易得φ(x(在(-1,0(上遞增，在(0,+∞(上遞減，從而φ(x(≤φ(0(=0，即ln(x+1(≤x(當(dāng)且僅當(dāng)x=0時(shí)取等號(hào))，即n+k>0，到e其中e為自然對(duì)數(shù)的底數(shù)，0<θ<1,n!=n×(n-1(×(x)(n+1)(此處ε介于0和x之間)稱作拉格朗日余項(xiàng).此時(shí)稱該式為函數(shù)f(x)在x=0處的n階泰勒公式，也稱作f(x)的n階麥克勞林公式.A.5B.6C.7D.8所以n的最小值為7.7.(2024·高二·陜西咸陽·階段練習(xí))給出定義：設(shè)f/(x(是函數(shù)y=f(x(的導(dǎo)函數(shù)，f”(x(是函數(shù)f/(x(的導(dǎo)函數(shù)，若方程f”(x(=0有實(shí)數(shù)解x=x0，則稱(x0,f(x0((為函數(shù)y=f(x(的“拐點(diǎn)”.經(jīng)研究發(fā)現(xiàn)所有的(1)若函數(shù)f(x(=x3+3x2-9x-1，求函數(shù)f(x(圖象的對(duì)稱中心；(2)已知函數(shù)g(x(=2mx3+[6ln(mx(-15[x2+，其中m>0.(ⅱ)若g(x1(+g(x2(=2(0<x1<x2(，求證：0<x1<<x2.(1)因?yàn)閒(x(=x3+3x2-9x-1，所以f/(x(=3x2+6x-9，所以f”(x(=6x+6.令f”(x(=0，解得x=-1，又f(-1(=-1+3+9-1=10，所以函數(shù)f(x(圖象的對(duì)稱中心為(-1,10(.(2)(i)因?yàn)間(x(=2mx3+[6ln(mx(-15[x2+x-+1，m>0，所以g/(x(=6mx2+6x+2[6ln(mx(-15[x+，”(x(=12mx+12ln(mx(-12=0，且mx-1+ln(mx(=0，令h(x(=x+lnx-1(x>0(，則h/(x(=1+>0在(0,+∞(上恒成立，所以h(x(=x+lnx-1在(0,+∞(上單調(diào)遞增，又h(1(=0，由零點(diǎn)存在性定理知，h(x(=x+lnx-1有唯一的零點(diǎn)所以g(x(的拐點(diǎn)為,1(. x(=12mx+12ln(mx(-12在(0,+∞(上單調(diào)遞增，gⅡ=0，x(>0，/(x(≥0在(0,+∞(上恒成立，∴g(x(在(0,+∞(上單調(diào)遞增，(x1(+g(x2(=2(0<x1<x2(，所以0<x1<<x2.8.(2024·高二·廣東東莞·階段練習(xí))記fⅡ(x(=(f/(x((/，f/(x(為f(x(的導(dǎo)函數(shù).若對(duì)?x∈D，fⅡ(x(>0，則稱函數(shù)y=f(x(為D上的“凸函數(shù)”.已知函數(shù)f(x(=exx3-ax2-1，a∈R.(2)若函數(shù)y=f(x(在(1,+∞(上有極值，求a的取值范圍.(1)由f(x(=ex-x3-ax2-1，得f/(x(=ex-x2-2ax，fⅡ(x(=ex-2x-2a，由于函數(shù)f(x(為R上的凸函數(shù)，故fⅡ(x(=ex-2x-2a>0，即2a<ex-2x，令g(x)=ex-2x，則g/(x)>0；故g(x)=ex-2x在(-∞,ln2)上單調(diào)遞減，在(ln2,+∞)上單調(diào)遞增，故g(x)min=g(ln2)=2-2ln2，故2a<2-2ln2,∴a<1-ln2，故a的取值范圍為(-∞,1-ln2)；(2)由f(x(=ex-x3-ax2-1，得f/(x(=ex-x2-2ax，函數(shù)y=f(x(在(1,+∞(上有極值，即f/(x(=ex-x2-2ax在(1,+∞(上有變號(hào)零點(diǎn)，即=2a在(1,+∞(上有解，令hx∈(1,+∞(,∴h/(x令m(x)=(x-1(ex-x2，則m/(x)=x(ex-2(>0,x∈(1,+∞(，即m(x)=(x-1(ex-x2在(1,+∞(上單調(diào)遞增，且1<x<x0時(shí)，m(x)<0，x>x0時(shí)，m(x)>0，故h(x)在(1,x0(上單調(diào)遞減，在(x0,+∞(上單調(diào)遞增，故hmin=h由于m=(x0-1(ex0-xex故h(x)>0,∴2a>0,a>0，即a的取值范圍為(0,+∞).9.(2024·上海普陀·一模)若函數(shù)y=f(x((x∈D(同時(shí)滿足下列兩個(gè)條件，則稱y=f(x(在D上具有性質(zhì)M．①y=f(x(在D上的導(dǎo)數(shù)f/(x(存在；②y=f/(x(在D上的導(dǎo)數(shù)fⅡ(x(存在，且fⅡ(x(>0(其中fⅡ(x(=[f/(x([/)恒成立．(1)判斷函數(shù)y=lg在區(qū)間(0,+∞(上是否具有性質(zhì)M？并說明理由．(1)令y=f(x(=lg，x∈(0,+∞(，則f/(xx∈(0,+∞(，fx(=>0恒成立，:函數(shù)y=lg在區(qū)間(0,+∞(上具有性質(zhì)M；“g(x(在x=1處取得極值，且g(x(為奇函數(shù)，:g(x(在x=-1處也取得極值，(g/(-1(=6-2a-b=0(b=6，:{g(g/(-1(=6-2a-b=0(b=6，:g(x(=2x3+，g/(x(=6x2-=6x2-6x-2，故g(x(在(0,1(單調(diào)遞減，在(1,+∞(單調(diào)遞增，滿足g(x(在x=1處取得極值，:存在實(shí)數(shù)c∈(0,+∞(，使gⅡ(x(>0在區(qū)間[c,+∞(上恒成立，:存在實(shí)數(shù)c，使得y=g(x(在區(qū)間[c,+∞(上具有性質(zhì)M，c的取值范圍是(0,+∞(；(3)“x∈(0,+∞(，(x+1([1+ln(x+1([:1+ln(x+1(>(x+1([1+ln(x+1([令F(x(=x，則F/(x令G(x(=x-ln(x+1(-1， x(>0，G(x(在區(qū)間(0,+∞(上單調(diào)遞增，:存在x0∈(2,3(，使G(x0(=x0-ln(x0+1(-1=0，:當(dāng)x∈(0,x0(時(shí)，G(x(<0，F(xiàn)/(x(<0，F(xiàn)(x(在區(qū)間(0,x0(上單調(diào)遞減，，F(xiàn)(x(在區(qū)間(x0,+∞(上單調(diào)遞增， (x0+1([1+ln(x0+1([:當(dāng)x∈(0,+∞(時(shí)，F(xiàn)(x(的最小值為F(x0(=x0，由G(x0(=x0-ln(x0+1(-1= (x0+1([1+ln(x0+1([:F(x0(==x0+1，“x0,3(，:F(x0(∈(3,4(，又“kF(x(恒成立，:k<F(x0(，:k的最大值為3.數(shù)f(x(在x=0處的[m,n[階帕德近似定義為：R(x，且滿足：f(0(=R(0(，,f(m+n((0(=R(m+n((0(.(注：fⅡ(x(=[f/(x([/，f川=[fⅡ(x([/，f(4((x(=[f川(x([/，f(5((x(=[f(4((x([/，…；f(n((x(為f(n-1((x(的導(dǎo)數(shù)).(1)求函數(shù)f(x(=ln(x+1(在x=0處的[1,1[階帕德近似函數(shù)R(x(；(2)在(1)的條件下，試比較f(x(與R(x(的大小；(3)在(1)的條件下，若h(xm(f(x(在(0,+∞(上存在極值，求m的取值范圍.(1)由f(x(=ln(x+1(得f/fⅡ則函數(shù)f(x(=ln(x+1(在x=0處的[1,1[階帕德近似函數(shù)R(x)=，故R(x)===-所以R(2)令F(x)=f(x)-R(x)，(x>-1)，則F,(x)=f,(x)-R,(x)=-=(x()-(21)=≥0恒成立，所以F(x)在(-1,+∞)上單調(diào)遞增，且F(0)=0，所以當(dāng)-1<x<0時(shí)，F(xiàn)(x)<0，即f(x)<R(x)，當(dāng)x>0時(shí)，F(xiàn)(x)>0，即f(x)>R(x)．=+m(ln(x+1)，(x>0)，h,(x)=-++m(×=，因?yàn)閔(x)=-m(f(x)在(0,+∞)上存在極值，所以h,(x)在(0,+∞)上存在變號(hào)零點(diǎn)，令g(x)=mx2+x-(x+1)ln(x+1)，x>0，g,(x)=2mx+1-1-ln(x+1)=2mx-ln(x+1)，記m(x(=g,(x(=2mx-ln(x+1)，則m,(x(=2m②當(dāng)m≥時(shí)，由于m≥,所以2m(x+1(-1≥x+1-1=x>0，故m,(x)>0，故g,(x)在(0,+∞)上單,(x)>0，所以g,min=glnm+ln令H(x)=1-x+lnx，0<x<1，H,(x)=-1+>0，所以H(x)在(0,1)上單調(diào)遞增，H(x)<H(1(=0，所以g-1(=1-2m+ln(2m)<0，故(mx2-1(-(mx-1((x+1(=x(1-m(>0，進(jìn)而>mx-1，所以1+-ln(x+1)>mx-ln(x+1)；令Glnln>mx-ln=m-ln 設(shè)k(x)=lnx-x+1，則當(dāng)x>1時(shí)kl(x)=-1<0,k(x(單調(diào)遞減，當(dāng)0<x<1時(shí)，kl(x)>0,k(x(單調(diào)遞增，故當(dāng)k(x)≤k(1(=0，故lnx≤x-1,故lnx≤x，所以G(x)>m(x+1)-2、x+1-m=(x+1)-m+(x+1)-2、x+1，(x+1)-m=-m>0，所以m的取值范圍為(0,．?dāng)?shù)f(x(在x=0處的[m,n[階帕德近似定義為：R(x(=，且滿足：f(0(=R(0(,fl(0(=Rl(0(,fⅡ(0(=RⅡ(0(,...,f(m+n((0(=R(m+n((0(.(注：fⅡ(x(=[fl(x([l，f川(x(=[fⅡ(x([l,f(4((x(=[fⅡ(x([l,f(5((x(=[f(4((x([l,...,f(n((x(為f(n-1((x(的導(dǎo)數(shù))已知f(x(=ln(x+1(在x=0處的[1,1[階帕德近似為g(x(1)求實(shí)數(shù)m,n的值；(3)設(shè)a為實(shí)數(shù)，討論方程f(x(-g(x(=0的(1)由f(x(=ln(x+1(,g(x(=，有f(0(=g(0(，(x令φ(x(=f(x(-g(x(=ln(x+1(-(x≥0(，則(x所以φ(x(在其定義域(-1,+∞(內(nèi)為增函數(shù)，又φ(0(=f(0(-g(0(=0，:x≥0時(shí)，φ(x(=f(x(-g(x(≥φ(0(=0，得證.(3)h(x(=f(x(-g(x(=ln(x+1(-的定義域是(-1,+∞(，l(x(≥0，所以h(x(在(-1,+∞(上單調(diào)遞增，且h(0(=0，所以h(x(在(-1,+∞(上存在1個(gè)零點(diǎn)；②當(dāng)a>2時(shí)，令t(x(=x2+(4-2a((x+1(=x2+(4-2a(x+(4-2a(，a-2(-√a2-2a〈0,x2=(a-2(+va2-2a〈0.x(-1,x1(x1(x1,x2(x2(x2,+∞(h/(x(+0-0+h(x(極大值h(x1(極小值h(x2(，所以h(x(在(x1,x2(上存在1個(gè)零點(diǎn)，且h(x1(>h(0(=0,h(x2(<h(0(=0；當(dāng)x∈(-1,x1(時(shí)，因?yàn)閔(e-a-1(=lne-a-1<e-a-1<0，而h(x(在(-1,x1(單調(diào)遞增，且h/(x1(=0，而h(e-a-1(<0，故-1<e-a-1<x1，所以h(x(在(-1,x1(上存在1個(gè)零點(diǎn)；a-1>0，而h(x(在(x2,+∞(單調(diào)遞增，且h/(x2(=0，而h(ea-1(>0，所以ea-1>x2，所以h(x(在(x2,+∞(上存在1個(gè)零點(diǎn).從而h(x(在(-1,+∞(上存在3個(gè)零點(diǎn).當(dāng)a>2時(shí)，方程f(x(-g(x(=12.給定兩個(gè)正整數(shù)m，n，函數(shù)f(x(在x=0處的[m,n[階帕德近似定義為：R(x且滿足：f(0(=R(0(，f/(0(=R/(0(，fⅡ(0(=RⅡ(0(…f(m+n((0(=R(m+n((0(.已知f(x(=ln(x+1(在x=0處的[1,1[階帕德近似R(x(=注：fⅡ(x(=[f/(x([/，f川(x(=[fⅡ(x([/，f(4((x(=[f川(x([/，f(5((x(=[f(4((x([/，…(1)因?yàn)閒(x(=ln(x+1(，R(x(=，f，-1=-2(b-ac(c，c=， (2)(x+c(f=(x+ln+1(，令+1=t，則(x+c(f=lnt，t∈(0,1(U(1,+∞(，令h(t(=lntt∈(0,1(U(1,+∞(，所以h(t(在(0,1(單調(diào)遞增，在(1,+∞(單調(diào)遞增， t∈(1,+∞(,h(t(>h(1(=0,lnt所以lnt>1，綜上，(x+c(f>1.x+時(shí)，即ln(1+x+>1,(x+ln(1+>1，由(2)知上式成立，所以x∈(-∞,-1(U(0,+∞(，x<e等價(jià)于xln(1+<1，所以解集為(0,+∞(.數(shù)f(x(在x=0處的[m,n[階帕德近似定義為：R(x且滿足：f(0(=R(0(，f/(0(=R/(0(，fⅡ(0(=R/(0(…，f(m+n((0(=R(m+n((0((注：fⅡ(x(=[f/(x([/，f川(x(=[fⅡ(x([/，f(4((x(=[f川(x([/，f(5((x(=[f(4((x([/，…,fn(x(為f(n-1((x(的導(dǎo)數(shù))已知f(x(=ln(x+1(在x=0處的[1,1[階帕德近似為R(x(=．(2)求證：(x+b(f因?yàn)閒(x(=ln(x+1(，所以f/(x(=，fⅡ(x(=-(0(=fⅡ(0(ln1.1=f(0.1(≈R(0.1(==≈0.095所以φ(t(在(0,1(上單調(diào)遞增，在(1,+∞(上單調(diào)遞增t(<φ(1(=0，可得lntlnt>1成立，t(>φ(1(=0，x<e<(1+x+成立，所以使e<(1+x+成立的x的取值范圍為x>0或x<-1不妨令h(x(=lnx-x+1，函數(shù)定義域?yàn)?0,1(∪(1,+∞(．x(>0，h(x(單調(diào)遞增；/(x(<0，h(x(單調(diào)遞減，所以h(x(<h(1(=0，則當(dāng)x∈(-∞,-1(∪(0,+∞(時(shí)，有l(wèi)n(1+<(所以使(1+x<e成立的x的取值范圍為(0,+∞綜上可得不等式(1+x<e<(1+x-的解集為(0,+∞(． (2)已知函數(shù)f(x(=ex-ax2-(e-a-1(x-1,a∈R在區(qū)間(0,1(內(nèi)有零點(diǎn)，求a的取值范圍．(1)設(shè)F(x(=ax4+bx3+cx2-(a+b+c(x,x∈[0,1[，則F/(x(=4ax3+3bx2+2cx-(a+b+c(，所以函數(shù)F(x(在[0,1[上連續(xù)，在區(qū)間(0,1(上可導(dǎo)，又F(0(=0,F(1(=a+b+c-a-b-c=0，故F(0(=F(1(，(2)因?yàn)楹瘮?shù)f(x(=ex-ax2-(e-a-1(x-1,a∈R在區(qū)間(0,1(內(nèi)有零點(diǎn)，,1(，由f(x(=ex-ax2-(e-a-1(x-1可得f/(x(=ex-2ax-(e-a-1(，所以函數(shù)f(x(在[0,x1[上連續(xù)，在(0,x1(上可導(dǎo)，又f(0(=e0-0-0-1=0，f(x1(=0，因?yàn)楹瘮?shù)f(x(在[x1,1[上連續(xù)，在(x1,1(上可導(dǎo)，又f(1(=e-a-e+a+1-1=0，f(x1(=0，所以方程ex-2ax-(e-a-1(=0在(0,1(上至少有兩個(gè)不等的實(shí)數(shù)根，設(shè)g(x(=ex-2ax-(e-a-1(,x∈(0,1(，則g/(x(=ex-2a，x(>0，函數(shù)g(x(在(0,1(上單調(diào)遞增，所以方程ex-2ax-(e-a-1(=0在(0,1(上至多有一個(gè)根，矛盾，x(<0，函數(shù)g(x(在(0,1(上單調(diào)遞減，所以方程ex-2ax-(e-a-1(=0在(0,1(上至多有一個(gè)根，矛盾，，函數(shù)g(x(在(0,ln2a(上單調(diào)遞減，，函數(shù)g(x(在(ln2a,1(上單調(diào)遞增，所以g(ln2a(<0，又g(0(=1-(e-a-1(=2+a-e，g(1(=e-2a-(e-a-1(=1-a，所以a的取值范圍(e-2,1(．15.(2024·高三·陜西安康·開學(xué)考試)定理：如果函(1)設(shè)f(x)=x(x-1)(x-2)(x-4)，記f(x)的導(dǎo)數(shù)為f/(x)，試用上述定理，說明方程f/(x)=0根的個(gè)a+b(1)函數(shù)f(x)=x(x-1)(x-2)(x-4)的定義域?yàn)镽，函數(shù)f(x)在R上的圖象連續(xù)不斷，求導(dǎo)得f/(x)=4x3-21x2+28x-8，顯然f(0)=f(1)=f(2)=f(4)=a+b在閉區(qū)間[0,1]上f(x)的圖象是連續(xù)不斷的，在開區(qū)間(0,1)內(nèi)每一點(diǎn)存在導(dǎo)數(shù)，且f(0)=f(1)，，使得f/所以方程f/(x)=0有3個(gè)根，分別在區(qū)間(0,1),(1,2),(2,4)內(nèi).(2)令函數(shù)h(x)=f(x)-x，依題意，h(x)在閉區(qū)間[a,b]上的圖象是連續(xù)不斷的曲線，求導(dǎo)得h/(x)=f/(x)-，則h(x)在開區(qū)間(a,b)內(nèi)每一點(diǎn)存在導(dǎo)數(shù)，且h(a)=f(a)-b=h即f，整理得f(b)-f(a)=f/(c)(b-a)，所以在開區(qū)間(a,b)內(nèi)至少存在一點(diǎn)c，使得f(b)-f(a)=f/(c)(b-a).<x2， f，把F(x)的二次項(xiàng)系數(shù)表示成關(guān)于f的函數(shù)G(f)，并求G(f)的值(1)由題意G(f)=++=++=f-e+2又f>e，所以G(f)>e-e+2=-e+2．即G(f)的值域是(-e+2,+∞(；(a+b)[d(b-c)+e(c-a)+f(a-b)]=d(b-c)(a+b)+e(c-a)(a+b)+f(a2-b2)=d(b-c)([(b+c)+(a-c)]+e(c-a)[(c+a)+(b-c)]+f(a2-b2)=d(b2-c2)+e(c2-a2)+f(a2-b2)+d(b-c)(a-c)+e(c-a)(b-c)d(b-c)+e(c-a)+f(a-b)所以(a+b)[d(b-c)+e(c-a)+f(a-b)]>d(b2-c2)+e(c2-a2)+f(a2-b2)，所以a+b<d(b2-c2(+e(c2d(b-c)+e(c-a)+f(a-b)(b+c)[d(b-c)+e(c-a)+f(a-b)]=d(b2-c2)+e(c-a)(b+c)+f(a-b)(b+c)=d(b2-c2)+e(c-a)(c-a+b-a)+f(a-b)(a+b+c-a)=d(b2-c2)+e(c2-a2)+f(a2-b2)+e(c-a)(b-a)+f(a-b)(c-a)d(b-c)+e(c-a)+f(a-b)，所以(b+c)[d(b-c)+e(c-a)+f(a-b)]<d(b2-c2)+e(c2-a2)+f(a2-b2)，所以b+c>d(b2-c2(+ed(b-c)+e(c-a)+f(a-b)，17.(2024·江蘇鹽城·模擬預(yù)測(cè))根據(jù)多元微分求條件極值理論，要求二元函數(shù)z=f(x,y)在約束條件g(x,y)(Lx(x,y,λ)=fx(x,y)+λgx(x,y)=0Ly(x,y,λ)=fy(x,y)+λgy(x,y)=0，解此方程組，得出解(x,y)，就是二元函數(shù)z=f(x,y)在約束條件g(x,(Lλ(x,y,λ)=g(x,y)=0y)的可能極值點(diǎn)．x,y的值代入到f(x,y)中即為極值．(1)求函數(shù)f(x,y)=x2y2+2xy+xy2關(guān)于變量y的導(dǎo)數(shù)并求當(dāng)x=1處的導(dǎo)數(shù)值．(2)利用拉格朗日乘數(shù)法求：設(shè)實(shí)數(shù)x,y滿足g(x,y)=4x2+y2+xy-1=0，求f(x,y)=2x+y的最大z2≥.(1)函數(shù)f(x,y)=x2y2+2xy+xy2，對(duì)變量y求導(dǎo)得：fy(x,y)=2x2y+2x+2xy，(2)令L(x,y,λ)=2x+y+λ(4x2+y2+xy-1)，(x=y=|或{|55Lλ(x,(x=y=|或{|55Lλ(x,y,λ)=4x2+y2+xy-1=0λ=1055(λ=-55于是函數(shù)f(x,y)在約束條件g(x,y)=0的可能極值點(diǎn)是(-,-,(，當(dāng)x=-,y=-時(shí)，函數(shù)f(x,y)的一個(gè)極值為函數(shù)f(-,-=-，當(dāng)x=,y=時(shí)，函數(shù)f(x,y)的一個(gè)極值為函數(shù)f,(=，4所以f(x,y)的最大值為210.2+-(λ1+λ2)2=+λ+λ+(λ1+λ21b(a-b)≥a2+1b(a-b)≥a2+ sh(x2(-x2-1[.[ch(x1(+sh(x1([>sin(x1+x2(-sinx1-x2cosx1.(1)平方關(guān)系：ch2(x(-sh2(x(=1；倍角公式：sh(2x(=2sh(x(ch((sh(x)(,=ch(x)(sh(x)(,=ch(x)x(-sh2(x(=22=-=1；倍角公式：sh(2x(===2sh(x(c導(dǎo)數(shù)：(sh(x)(,===chx，(ch(x)(,==sh(2)構(gòu)造函數(shù)F(x(=sh(x(-kx，x∈(0,+∞(，由(1)可知F,(x(=ch(x(-k，所以F,(x(=ch(x(-k>0，故F(x)為嚴(yán)格增函數(shù)，②當(dāng)k>1時(shí)，令G(x(=F,(x(,x∈(0則G,(x(=sh(x(>0，可知G(x(是嚴(yán)格增函數(shù)，,(x)=G(x)<G(x0)=0，則F(x)在(0,x0)上為嚴(yán)格減函數(shù)，所以原式變?yōu)?ex-x2-1(.ex>sin(x1+x2(-sinx1-x2cosx1，即證ex+x-sin(x1+x2(>ex-sinx1+x2(ex-cosx1(，設(shè)函數(shù)f(x(=ex-sinx，即證f(x1+x2(>f(x1(+x2f,(x1(，f,(x(=ex-cosx，設(shè)t(x(=f,(x(=ex-cosx，t,(x(=ex+sinx，x>0時(shí)t,(x(>0，t(x(在(0,+∞(上單調(diào)遞增，即f,(x(=ex-cosx在(0,+∞(上單調(diào)遞增，設(shè)h(x(=f(x+x1(-f(x1(-xf,(x1(,(x>0(，則h,(x(=f,(x+x1(-f,(x1(，由于f/(x(=ex-cosx在(0,+∞(上單調(diào)遞增，x+x1>x1，所以f/(x+x1(>f/(x1(，即h/(x(>0，故h(x(在(0,+∞(上單調(diào)遞增，所以f(x+x1(-f(x1(-xf/(x1(>0，即f(x+x1(>f(x1(+xf/(x1(，因此f(x1+x2(>f(x1(+x2f/(x1(恒成立，所以原不等式成立，得證.數(shù)的單調(diào)性以及在其他不等式的證明等方面都有著極其廣泛的應(yīng)用．伯努利不等式的一種常見形式,n∈N*．n<、2n+1-、2n-1，2+…+bn n<3(n*(；(2)已知函數(shù)f(x(=ax-elogax-e(a>1(.②討論f(x(的極值點(diǎn)的個(gè)數(shù).n<3.(2)①當(dāng)a=e時(shí)，f(x(=ex-elnx-e，f/(x(=e由y=ex,y=-均在(0,+∞(單調(diào)遞增，所以f/(x(在(0,+∞(上單調(diào)遞增，，f(x(單調(diào)遞減；，f(x(單調(diào)遞增，從而f(x(≥f(1(=0，得證.②函數(shù)f(x(的定義域?yàn)?0,+∞(，f/(x(=axlna-=，設(shè)g(x(=xaxln2a-e，a>1，顯然函數(shù)g(x(在(0,+∞(上單調(diào)遞增，g(x(與f/(x(同號(hào)，當(dāng)a>e時(shí)，g(0(=-e<0，g(1(=ag(x(>0，故f(x(在(0,x0(單調(diào)遞減，在(x0,+∞(單調(diào)遞增；所以函數(shù)f(x(在(0,+∞(上有且僅有一個(gè)極值點(diǎn)；當(dāng)a=e時(shí)，由①知，函數(shù)f(x(在(0,+∞(上有且僅有一個(gè)極值點(diǎn)；又g(1(=aln2a-e<0，所以函數(shù)g(x(在(1,內(nèi)有一個(gè)零點(diǎn)x1，(x(>0故f(x(在(0,x1(單調(diào)遞減，在(x1,+∞(單調(diào)遞增；所以函數(shù)f(x(在(0,+∞(上有且僅有一個(gè)極值點(diǎn)；綜上所述，函數(shù)f(x(在(0,+∞(上有且僅有一個(gè)極值點(diǎn).f(x)的定義域?yàn)閇a,b[，如果對(duì)于[a,b[內(nèi)任意兩數(shù)x1,x2，都有f≤,則稱f(x(為 小組成員又了解到了琴生不等式(Jensen不等式)：若f(x)是區(qū)間[a,b[上的凹函數(shù)，則對(duì)任意的x1=x2=…=xn∈[a,b[,有不等式f≤恒成立(當(dāng)且僅當(dāng)x1=x2=…不等式求多元最值問題，關(guān)鍵是構(gòu)造函數(shù).小組成員選擇了反比例型函數(shù)f(x(=-和對(duì)數(shù)函數(shù)g(x(ri=ex),1[時(shí)，不等式g(mx2+x(≤0恒成立，求實(shí)數(shù)m的取值范圍.(1)記函數(shù)f(x(=-，首先證明其凹凸性：?x1,x2∈(0,1(,則-f=+=1-x1+1-x2-2=[(1-x1(+(1-x2([2-4(1-x1((1-x2(= 2(1-x1((1-x2(1-x1+1-x22(1-x1((1-x2((1-x1+1-x2(所以f(x在(0,1(為凹函數(shù).(2)設(shè)ri=ex,因?yàn)閞i≥1,故xi≥0(i=1,2,3…n(要證++...+≥.只需證++…+≥由琴生不等式，只需證h(x(=在[0,+∞)為凹函數(shù).設(shè)x1,x2≥0,h 即證+1((ex1+1+ex2+1(≥2(ex1+1((ex2+1(,x1+ex2(≥0.即證-e2≥0，e≥1+x(≤0恒成立，即loga(mx2+x(≤0，因?yàn)閍>1,即0<mx2+x≤1恒成可得-<m≤在x∈(0,1[時(shí)恒成立．≥1，-∈(-∞,-1[，所以m>-1．由可得所以m≤0．故-1<m≤0．則稱f(x(為[a,b[上的凹函數(shù)；若f，則稱f(x(為凸函數(shù).若f(x(是區(qū)間[a,b[f恒成立(當(dāng)且僅當(dāng)x1=x2=…=xn時(shí)等號(hào)成立).設(shè)ri=ex)則-f=-=1-x1+1-x2-2=[(1-x1(+(1-x2([2-4(1-x1((1-x2(2(1-x1((1-x2(1-x1+1-x22(1-x1((1-x2((1-x1+1-x2(2(1-x1((1-x2((1-x1+1-x2(=[2(1-x1((1-x2((1-x1+1-x2(上也為凹函數(shù).(3)設(shè)ri=ex，因?yàn)閞i≥1，所以xi≥0(i=1,2,3,…,n(，由琴生不等式，只需證h(x(=在[0,+∞(上為凹函數(shù). 即證+1((ex1+1+exi+1(≥2(exi+1((ex2+1(，即證-e2(≥0(*(,又≥0,e≥1,(*(式顯然成立，所以≥h成立，h(x(在[0,+∞(上為凹函數(shù)，則得證.l=0，則稱函數(shù)f(x(為區(qū)間[0,a[上的k階無窮遞降函數(shù).(1)證明f(x(=x3-3x不是區(qū)間[0,3[上的2階無窮遞降函數(shù)； 1(2)計(jì)算：l+x(x； 1(1)記H(x(=f(x(-f=x3-x， 因?yàn)镠所以f(x(≥f在區(qū)間[0,3[不恒成立，所以，f(x(=x3-3x不是區(qū)間[0,3[上的2階無窮遞降函數(shù).11所以llng(x(=1，所以l+x(x=e.所以f(t(>f>...>f，t(>1.=f(x(在點(diǎn)Q1(x1,f(x1((處切線，切線與x軸交于點(diǎn)P2(x2,0(，再作y=f(x(在點(diǎn)Q2(x2,f(x2((處切線，切線與x軸交于點(diǎn)P3(x3,0(，再作y=f(x(在點(diǎn)Q3(x3,f(x3((處切線，以此類推，直到求得滿足精度的零點(diǎn)近似解xn(n≥2(為止．(3)設(shè)函數(shù)f(x(=x2，令xn+1=g(xn(=xn，且b-a=1，若函數(shù)h(x(=ex-ax2，φ(x(=xlnx-x2+(1)由函數(shù)f(x(=x3-6，則f/(x(=3x2，那么在Q點(diǎn)處的切線方程為y+5=3(x-1(，(2)設(shè)Pn(xn,0(，則Qn(xn,f(xn((，因?yàn)閒(x(=2x，所以f/(x(=2xln2，則Qn(xn,f(xn((處切線為y=2xln2.(x-xn(+2x，切線與x軸相交得Pn+1(xn+1,0(，PPf(xn(=2-=(2-log2e(n-1=，(3)曲線y=f(x(在(xn,f(xn((處的切線為y-f(xn(=f/(xn((x-xn(，所以切線與x軸交點(diǎn)橫坐標(biāo)為xn+1=xn，當(dāng)函數(shù)f(x(=x2時(shí)，即xn+1=xn-=xn-=xn-xn=xn=g(xn(，故h(x(=ex-x2，曲線y=h(x(的一條切線方程為y=(e-2(x+1，x-x2≥xlnx-x2+(e-1(x+1，x+(1-e(x-xlnx-1≥0，又h(1(=e-1，故曲線y=h(x(在x=1處的切線方程為y=( 證明：設(shè)m(x(=h(x(-(e-2(x-1=ex-x2-(e-2(x-1，x>0，則m/(x(=ex-2x-(e-2(，令F(x(=m/(x(，則F/(x(=ex-2且在(0,+∞(上單增，當(dāng)x∈(0,ln2(時(shí)，F(xiàn)/(x(<0，故m/(x(單調(diào)遞減；當(dāng)x∈(ln2,+∞(時(shí)，F(xiàn)/(x(>0，故m/(x(單調(diào)遞0那么m(x(在(0,x0(上單調(diào)遞增，在(x0,1(上單調(diào)遞減，在(1,+∞(上單調(diào)遞增，又m(0(=m(1(=0，∴m(x(=ex-x2-(e-2(x-1≥0(x>0(，當(dāng)且僅當(dāng)x=1時(shí)取等號(hào)，所以x≥lnx+1，即≥lnx+1．所以，ex+(2-e(x-1≥xlnx+x，x+(1-e(x-xlnx-1≥0成立，當(dāng)x=1時(shí)等號(hào)成立．另要證h(x(≥φ(x(，即ex-x2≥xlnx-x2+(e-1(x+1，等價(jià)于xlnx+(e-1(x+1-ex≤0，令G(x(=lnx+e-1G/(x，，G(x(單調(diào)遞增；,+∞(時(shí)，G/(x(<0，G(x(單調(diào)遞減；∴G(x(有最大值F(1(=0，即G(x(≤0恒成立，即當(dāng)x>0時(shí)，h(x(≥φ(x(.l=0，則稱函數(shù)f(x(為區(qū)間[0,a[上的k階無窮遞降函數(shù).(1)試判斷f(x(=x3-3x是否為區(qū)間[0,3[上的2階無窮遞降函數(shù)； 1 1(1)設(shè)F(x(=f(x(-f=x3-x，由于F所以f(x(≥f不成立，故f(x(=x3-3x不是區(qū)間[0,3[上的2階無窮遞降函數(shù).(2)設(shè)g(x(=則lng(xln(1+x設(shè)h(x(3)令x-π=t，則原不等式等價(jià)于tant.sin2t≥t3,t，記f(t則f即有對(duì)任意t，均有f(t(>f所以f(t(>f所以f(t(>1,t，證畢!新時(shí)期，為研究變量和函數(shù)提供了重要的方法和手段．對(duì)于函數(shù)f(x,f(x(在區(qū)間[a,b[上域(稱為曲邊梯形ABQP)的面積，根據(jù)微積分基本定理可得dx=lnb-lna，因?yàn)榍吿菪蜛BQP的面積小于梯形ABQP的面積，即S曲邊梯形ABQP<S梯形A ①證明：對(duì)任意兩個(gè)不相等的正數(shù)x1,x2，曲線y=f(x(在(x1,f(x1((和(x2,f(x2((處的切線均不重合；②當(dāng)b=-1時(shí)，若不等式f(x(≥2sin(x-1(恒成立，求實(shí)數(shù)a的取值范圍．(1)在曲線y取一點(diǎn)M過點(diǎn)M,作f(x(的切線分別交AP,BQ于M1,M2，因?yàn)镾曲邊梯形ABQP>S梯形ABMM，可得lnb-lna>.(|AM1|+|BM2|(.|AB|=.2..(b-a(，即<．(2)①由函數(shù)f(x(=ax2+bx+xlnx，可得f/(x(=2ax+lnx+b+1，不妨設(shè)0<x1<x2，曲線y=f(x(在(x1,f(x1((處的切線方程為l1:y-f(x1(=f/(x1((x-x1(，即y=f/(x1(x+f(x1(-x1f/(x1(同理曲線y=f(x(在(x2,f(x2((處的切線方程為l2:y=f/(x2(x+f(x2(-x2f/(x2(，ff(x2(-x2f/(x2(，即對(duì)任意實(shí)數(shù)a,b及任意不相等的正數(shù)x1,x2,l1與l2均不重合．②當(dāng)b=-1時(shí)，不等式f(x(≥2sin(x-1(恒成立，所以h(x(=ax2-x+xlnx-2sin(x-1(≥0在(0,+∞(恒成立，所以h(1(≥0→a≥1，因?yàn)閍≥1，所以h(x(≥x2-x+xlnx-2sin(x-1(設(shè)H(x(=x2-x+xlnx-2sin(x-1(,H/(x(=2x+lnx-2cos(x-1((i)當(dāng)x∈[1,+∞(時(shí)，由2x≥2,,lnx≥0,-2cos(x-1(≥-2知H/(x(≥0恒成立，即H(x(在[1,+∞(為增函數(shù)，所以H(x(≥H(1(=0成立；(ii)當(dāng)x∈(0,1(時(shí)，設(shè)G(x(=2x+lnx-2cos(x-1(，可得G/(x(=2由2sin(x-1(≥-2,>0知G/(x(≥0恒成立，即G(x(=H/(x(在(0,1(為增函數(shù)．所以H/(x(<H/(1(=0，即H(x(在(0,1(為減函數(shù)，所以H(x(>H(1(=0成立，27.拉格朗日中值定理是微分學(xué)的基本定理之一，內(nèi)容為：如果函數(shù)f(x(在閉區(qū)間[a,b[上的圖象連續(xù)不間斷，在開區(qū)間(a,b(內(nèi)的導(dǎo)數(shù)為f/(x(，那么在區(qū)間(a,b(內(nèi)至少存在一點(diǎn)c，使得f(b(-f(a(=f/(c((b-a(成立，其中c叫做f(x(在[a,b[上的“拉格朗日中值點(diǎn)”．根據(jù)這個(gè)定理，可得函數(shù)f(x(=A.1B.eC.e-1D.【解析】由f(x(=lnx可得f,(x(=，令x0為函數(shù)f(x(=lnx在[1,e[上的“解得x0=e-1.法則，即在一定條件下通過對(duì)分子、分母分別求導(dǎo)再求極限來確定未定式值的方法，如A.B.C.1D.2法則，即在一定條件下通過對(duì)分子、分母分別求導(dǎo)再求極限來確定未定式值的方法，如A.B.C.1D.230.拉格朗日中值定理是微分學(xué)的基本定理之一，定理內(nèi)容為：如果函數(shù)f(x)在區(qū)間[a,b]上的圖像連續(xù)不間斷，在開區(qū)間(a,b)內(nèi)的導(dǎo)數(shù)為f,(x)，那么在區(qū)間(a,b)內(nèi)至少存在一點(diǎn)c，使得f(b)-f(a)=f,(c)(b在,|上的拉格朗日中值點(diǎn)的個(gè)數(shù)為．【解析】f,(x(=2cos(2x-0為函數(shù)f(x(在,π|上的拉格朗日中值點(diǎn)， 少存在一個(gè)數(shù)ξ,使得f(b(-f(a(=f/(ξ((b-a(，其中ξ稱為拉格朗日中值.函數(shù)g(x(=lnx+x在區(qū)間[1,2[上的拉格朗日中值ξ=.由拉格朗日中值的定義可知，函數(shù)g(x(=lnx+x在區(qū)間[1,2[上的拉格朗日中值ξ滿足，g(2(-g(1(=/(ξ((2-1(所以g/(ξ(=g(2(-g(1(=ln2+2-1=ln2+1(1)在閉區(qū)間[a,b[上是連續(xù)不斷的；(2)在區(qū)間(a,b(上都有導(dǎo)數(shù).則在區(qū)間(a,b(上至少存在一個(gè)實(shí)數(shù)t，使得f(b(-f(a(=f'(t((b-a(，其中t稱為“拉格朗日中值”.函方面留下了很多寶貴的成果．設(shè)函數(shù)f(x(在(a,b(上的導(dǎo)函數(shù)為f/(x(，f/(x(在(a,b(上的導(dǎo)函數(shù)為f(x(，若在(a,b(上fⅡ(x(<0恒成立，則稱函數(shù)f(x(在(a,b(上為“凸函數(shù)”，已知f(x(=ex-xlnx-【答案】m≥e2-【解析】函數(shù)f(x)=ex-xlnx-x2，求導(dǎo)得f/(x)=ex-1-lnx-mx，fⅡ(x)=ex--m，x)<0?m>ex-恒成立，而函數(shù)g(x)=ex-在(1,2)上單調(diào)遞增，g(x)<g(2)=e2-，則m≥e2-，所以實(shí)數(shù)m的取值范圍是m≥e2-.故答案為：m≥e2-了很多寶貴的成果.定義：函數(shù)f(x)在(a，b(上的導(dǎo)函數(shù)為f/(x)，f/(x)在(a，b(上的導(dǎo)函數(shù)為fⅡ(x(，若③函數(shù)f(x)=ex-xlnx-x2在(1，4(為“嚴(yán)格凸函數(shù)”，則m的取值范圍為[e-1，+∞(.【解析】f(x)=-x3+3x2+2的導(dǎo)函數(shù)f/(x)=-3x2+6x，fⅡ(x)=-6x+6，fⅡ(x)<0在(1，+∞(上恒成立，f(x)=ex-xlnx-x2的導(dǎo)函數(shù)f/(x)=ex-lnx-1-mx，fⅡ(x)=ex--m，函數(shù)f(x)=ex-xlnx-x2在(1，4(為“嚴(yán)格凸函數(shù)”可得ex--m<0在(1，4(上恒成立，即m>ex設(shè)g=ex所以g(x)在(1,4(單調(diào)遞增，所以g(x)<g(4)=e4-，所以m≥e4-，所以③不正確； f(x)=ex-xlnx-x2的導(dǎo)函數(shù)f,(x)=ex-lnx-1-mx，fⅡ(x)=ex--m，fⅡ(x)<0，所以ex--即m>ex-，設(shè)g(x)=ex-，則g(x)在(1,4(單調(diào)遞增，所以g(x)<e4-，所以m≥e4-，所以③不正確；37.英國(guó)數(shù)學(xué)家泰勒發(fā)現(xiàn)了如下公式：sinx=x-+-+…，c精確性.(1)cos0.3≈1-+=1-+==.2，故tanx>x,x∈(0,.故c>b>a而cos=1-+××-××-×-…<1-+×=+，且+=-=>0，故c>b>a.38.英國(guó)數(shù)學(xué)家泰勒發(fā)現(xiàn)了如下公式：ex=1+x+++…++…其中n!=1×2×3×4×…×n為自然對(duì)數(shù)的底數(shù)，e=2.71828……．以上公式稱為泰勒公式．設(shè)f(xg(x根x≥1+x；(1)令h(x(=ex-x-1，則h/(x(=ex-1，/(x(>0/(x(<0，故h(x(在(-∞,0(上單調(diào)遞減，在(0,+∞(上單調(diào)遞增，故h(x(在x=0處取得極小值，也是最x≥x+1；(2)<g(x(?ex-e-x<xex+xe-x?(1-x(ex-(1+x(e-x<0，x∈(0,+∞(， 令q(x(=(1-x(ex-(1+x(e-x，x∈(0,+∞(，則q/(x(=-xex+xe-x=-x(ex-e-x(，x(=-x(ex-e-x(<0，故q(x(=(1-x(ex-(1+x(e-x在x∈(0,+∞(上單調(diào)遞減故q(x(<q(0(=0，即(1-x(ex-(1+x(e-x<0，結(jié)論得證；(3)F(x(=g(x(-a(1+=-a(1+，則F/(x(=-ax，令G(x(=F/(x(，x(=-a≥-a≥1-1=0，故G(x(=F/(x(在R上單調(diào)遞增，故F(x(在x∈(0,+∞(上單調(diào)遞增，在x∈(-∞,0(上單調(diào)遞減，函數(shù)近似表示，具體形式為：f=f(x0(+f/(x0((x-xx-xx-x0(n+…(其中f(n)(x)表示f(x)的n次導(dǎo)數(shù))，以上公式我們稱為函數(shù)f(x)在x=x0處的秦勒展開式．(1)因?yàn)楹瘮?shù)f(x(在x=x0處的泰勒展開式為f(x(=f(x0(+f’(x0((x-x0(+(x-x0(2+…+(x-x0(n+…(其中f(n((x(表示f(x(的n次導(dǎo)數(shù))，ex=1+x+x2+…+xn+…；cosx=1-x2+x4+…+x2n+….eix=1+(ix)+(ix)2+(ix)3+(ix)4+…+(ix)n+…令f(x)=sinx-x+x3,f’(x)=cosx-1+x2,fⅡ(x)=x-sinx，所以fⅡ(x)>fⅡ(0)=0，所以f/(x)在(0,上單調(diào)遞增，所以f/(x)>f/(0)=0，所以f上單調(diào)遞增，所以f>f再令g=xx3-ln，x而g(0)=0,g=-ln>0，則sinx>x-x3>ln(x+1)，(1)設(shè)m(x(=ex-x-1，則m/(x)=ex-1，x)<0，所以m(x)在(-∞,0)上單調(diào)遞減，于是e-x=1-x+-+-+…+(-1)n+…②,由①②得f(x)==x+++…+-+…，g(x)==1+++…++…，(3)由知f 令h=f-kxx3，則h/(x解得x1=ln(k或x2=ln(k(x)<0，得φ(x)在(0,x2(上單調(diào)遞減，從而t(x(在(0,x2(上單調(diào)遞減，故t(x(<t(0(=1-k<0，即h/(x(<h/(0(=1-k<0，因此h(x)在(0,x2(上單調(diào)遞減，所以h(x)<h(0)=0，矛盾，41.(2024·高三·四川成都·開學(xué)考試)麥克勞林展開式是泰勒展開式的一種特殊形式，f(x(的麥克勞林展開式為：f(x(=f(0(+f/(0(xxnxn，其中f(n((0(表示f(x(的n階(1)因?yàn)閒/(x(=cosx,f(2((x(=-sinx,f(3((x(=-cosx,(2)設(shè)g(x(=ln(1+x(-x因?yàn)閤>0,所以g/(xg(x(單調(diào)遞增，所以g(x(>g(0(=ln1-0+0=0,所以ln(1+x(>x當(dāng)m=3時(shí)，設(shè)h(x(=ex+lnx+--3x=1+x+++lnx+--3xh(x(=+lnx+-2x,h/(x(=x+-2≥2-2=0當(dāng)x>1,h/(x(>0,h(x(單調(diào)遞增，則h(x(>h(1(=ln1+-2+=0，數(shù)f(x(在x=0處的[m,n[階帕德近似定義為：R(x且滿足：f(0(=R(0(，f/(0(=R/(0(，fⅡ(0(=RⅡ(0(，...，f(m+n((0(=R(m+n((0(.(注：fⅡ(x(=[f/(x([/，f川(x(=[fⅡ(x([/，f(4((x(=[f川(x([/，f(5((x(=[f(4((x([/，f(n((x(為f(n-1((x(的導(dǎo)數(shù))已知f(x(=ln(x+1(在x=0處的[1,1[階帕德近似為g(x(1)求實(shí)數(shù)m,n的值；(3)設(shè)a為實(shí)數(shù)，討論函數(shù)h(x(=f(x(-ax的單調(diào)性.(1)由f(x(=ln(x+1(，g(x(=知：f(0(=g(0(=0；令φ(x(=f(x(-g(x(=ln(x+1(-(x≥0(，則(xφ(x(在[0,+∞(上單調(diào)遞增，又φ(0(=f(0(-g(0(=0，∴φ(x(≥φ(0(=0，即當(dāng)x≥0時(shí)，f(x(≥g(x(.(3)由題意知：h(x(=f(x(-ax=ln(x+1(-ax(x>-1(，∴h(x(在(-1,+∞(上單調(diào)遞增； ∴h(x(在(-1,-1(上單調(diào)遞增，在-1,+∞(上單調(diào)遞減；綜上所述：當(dāng)a≤0時(shí)，h(x(在(-1,+∞(上單調(diào)遞增；當(dāng)a>0時(shí)，h(x(在(-1,-1(上單調(diào)遞增，在-1,+∞(上單調(diào)遞減.算機(jī)數(shù)學(xué)中有著廣泛的應(yīng)用.已知函數(shù)f(x)在x=0處的[m,n[階帕德近似定義為：R(x)=,且滿足：f(0)=R(0)，f/(0)=R/(0)，f(2)(0)=R(2)(0)，…,f(m+n)(0)=R(m+n)(0).其中f(2)(x)=[f/(x)[/，f(3)(x)=[f(2)(x)[/，…,f(m+n)(x)=[f(m+n-1)(x)[/.已知f(x)=ln(x+1)在x=0處的[2,2[階帕德近似為R(2)設(shè)h(x(=f(x(-R(x(，證明：xh(x)≥0；-1.(1)依題意可知，f(0)=0,R(0)=a，因?yàn)閒(0)=R(0)，所以a=0，(2)依題意，h(x)=f(x)-R(x)=ln(1+x)-，故h(x)在(-1,+∞)單調(diào)遞增，綜上，?x>-1，xh(x)≥0；(3)不妨設(shè)x1<x2<x3，令t(x)=lnx-λ(x-，/(x)>0當(dāng)λ>0時(shí)，令s(x)=-λx2+x-λ,其判別式Δ=1-4λ2，/(x)≤0，此時(shí)t(x)單調(diào)遞減，t(x)=0不存在三個(gè)不等實(shí)根；(x)=0存在兩個(gè)不等正實(shí)根r1,r2(r1<r2(，r1,r2(時(shí)，t/(x)>0，t(x)單調(diào)遞增，r2,+∞(時(shí)，t/(x)<0，t(x)單調(diào)遞減，(1)=1-2λ>0，故t(r1(<0,t(r2(>0，因?yàn)閘nx<x-1(x≠1)，所以所以t(λ4(=lnλ4-λ(λ4->2--λ5+=(2-λ5(+-2(>0，4,r1(，滿足t(x1(=0，又因?yàn)閠=ln-λ-x(=-lnx+λ(x-=-t(x)，數(shù)f(x(在x=0處的[m,n[階帕德近似定義為R(x，且滿足：f(0(=R(0(，f/(0(=R/(0(,fⅡ(0(=RⅡ(0(..f(m+n((0(=R(m+n((0(.已知f(x(=ln(x+1(在x=0處的[1,1[階帕德近似為R(x(=.注：fⅡ(x(=[f/(x([/,f川(x(=[fⅡ(x([/，f(4(x=[f川(x([/,f(5(x=[f(4(x[/...(2)求證：(x+b(f(1)∵R(x(=∴R/(x(=,RⅡ(x(=∵f(x(=ln(x+1(，則f/(xfⅡ(x 則(t:(x+ln(1+>1成立，即(x+b(f>1成立.，即x>0或x<-1又由(2)知(x+ln(1+>1成立，記h(x(=lnx-x+1,x∈(0,1(U(1,+∞(，則h/(x(=-1=，:當(dāng)0<x<1時(shí)，h/(x(>0,x>1:h(x(<h(1(=0，即lnx<x-1，x∈(0,1(U(1,+∞(:ln(1+<，x∈(-∞,-1(U(0,+∞(當(dāng)x∈(-∞,-1(時(shí)由ln(1+<，可知xln(1+<1不成立；x<e<(1+x+的解集為(0,+∞(.45.(2024·江蘇揚(yáng)州·模擬預(yù)測(cè))帕德近似是法國(guó)數(shù)學(xué)家帕德發(fā)明的用多項(xiàng)式近似特定函數(shù)的方法．給定兩個(gè)正整數(shù)m，n，函數(shù)f(x(在x=0處的[m,n[階帕德近似定義為：R(x且滿足：f(0(=R(0(，f/(0(=R/(0(，fⅡ(0(=RⅡ(0(，…,f(m+n((0(=R(m+=[fⅡ(x([/，f(4((x(=[f川(x([/，f(5((x(=[f(4((x)[/，…已知f(x(=ex在x=0處的[1,1[階帕德近似為(1)由題意得R/(xRⅡ(xf(0(=f/(0(=fⅡ(0(=1，故R/(0(=a-b=1，RⅡ(0(=-2b(a-b(=1，-x的大小，令g(xe-x，g/(xe-xe-x，所以g/(x(>0，從而可得g(x(在(0,1(上單調(diào)遞增，所以f(x(<R(x(.(3)設(shè)u(x(=ex-x-1，u/(x(=ex-1，/(x(<0，u(x(在(-∞,0(上單調(diào)遞減，/(x(>0，u(x(在(0,+∞(上單調(diào)遞增，ean令y=(1-x(ex，y/=-xex，故該函數(shù)在(0,+∞(上遞減，n+1<ann≤;故-1>2n-1-1(=2n-1，aaa-e-2-an≥0令g(x(=e-e-，g/(x(=e+e--1,故g/(x(=e+e-≥0，所以g(x(在(0,單調(diào)遞增，所以g(x(≥g(0(=0，得證. 數(shù)f(x(在x=0處的[m,n[階帕德近似定義為：R(x，且滿足：f(0(=R(0(，f/(0(=R/(0(，fⅡ(0(=RⅡ(0(，……,f(m+n((0(=R(m+n((0(，注：fⅡ(x(=[f/(x([/，f川(x(=[fⅡ(x([/，f(4((x(=[f川(x([/，f(5((x(=[f(4((x([/，……已知函數(shù)f(x(=ln(x+1(.(1)求函數(shù)f(x(=ln(x+1(在x=0處的[1,1[階帕德近似R(x(.②若f(x(-m+1(R(x(≤1-cosx恒成立，求實(shí)數(shù)m的取值范圍.(1)由題可知函數(shù)f(x(=ln(x+1(在x=0處的[1,1[階帕德近似，則R(x(=，f/(x(=，fⅡ(x(=-，由f(0(=R(0(得a0=0，所以R(x(=，則R/(x(=，所以R(x(2)①令F(x(=-ln(x+1(，x∈(-1,0(∪(0,+∞(，因?yàn)镕/(x所以F(x(在x∈(-1,0(及(0,+∞(上均單調(diào)遞減.當(dāng)x∈(-1,0(，F(xiàn)(x(>F(0(=0，即>ln(x+1(，②由f(x(-m+1(R(x(≤1-cosx得ln(x+1(-mx+cosx-1≤0在(-1,+∞(上恒成立，令h(x(=ln(x+1(-mx+cosx-1，且h(0(=0，所以x=0是h(x(的極大值點(diǎn)，又h/(x(=-m-sinx，故h/(0(=1-m=0，則m=1，當(dāng)m=1時(shí)，h(x(=ln(x+1(-x+cosx-1，所以h/(xsinx-1=-sinx故h(x(在(-1,0(上單調(diào)遞增，所以當(dāng)x∈(-1,0(時(shí)，h(x(<h(0(=0，當(dāng)x∈(0,+∞(時(shí)，h(x(=[ln(x+1(-x[+(cosx-1(，令φ(x(=ln(x+1(-x，因?yàn)棣?(x(=-1<0，所以φ(x(在(0,+∞(上單調(diào)遞減，所以φ(x(<φ(0(=0，故當(dāng)x∈(0,+∞(時(shí)，h(x(=[ln(x+1(-x[+(cosx-1(<0，綜上，當(dāng)m=1時(shí)，f(x(-m+1(R(x(≤1-cosx恒成立.柯西(Cauchy(中值定理，它們被稱為微分學(xué)的三大中值定理.羅爾中值定理的描述如下：如果函數(shù)f(x(滿足三個(gè)條件①在閉區(qū)間[a,b[上的圖象是連續(xù)不斷的，②在開區(qū)間(a,b(內(nèi)是可導(dǎo)函數(shù)，③f(a(=f(b(，那么在(a,b(內(nèi)至少存在一點(diǎn)ξ(a<ξ<b)，使得等式f/(ξ(=0成立.(1)設(shè)方程a0xn+a1xn-1+…+an-1x=0有一個(gè)正根x=x0，證明：方程a0nxn-1+a1(n-1(xn-2+…+an-1(2)設(shè)函數(shù)f(x(是定義在R上的連續(xù)且可導(dǎo)函數(shù)，且f(-x(+f(x(=0.證明：對(duì)于m>0，方程f/(x(-f(m(m=0在(-m,f(m((3)設(shè)函數(shù)f(x(=ex-ax2-(e-a-1(x-1.證明：函數(shù)g(x(=2f(x(+f/(x(在區(qū)間(0，1(內(nèi)至少存在一個(gè)零點(diǎn).(1)證明：令函數(shù)f(x(=a0xn+a1xn-1+…+an-1x，顯然f(x(在[0,x0[上連續(xù)，在(0,x0(內(nèi)可導(dǎo)，則f/(x(=a0nxn-1+a1(n-1(xn-2+…+an-1，由條件知f(0(=f(x0(=0，即方程a0nxn-1+a1(n-1(xn-2+…+an-1=0必有一個(gè)小于x0的正根.由f(-x(+f(x(=0，得f(0(=0，所以g(0(=f(0(-0=0.因?yàn)間(-m(=f(-m(+f(m(=0，所以g(-m(=g(0(，即f/(x同理，因?yàn)間(m(=f(m(-f(m(=0，由羅爾中值定理知，x2(=f/(x2(-=0.故方程f/(x(-=0在(-m,m(內(nèi)至少存在兩個(gè)不同的解.(3)證明：令F(x(=e2xf(x(，則F/(x(=e2x[2f(x(+f/(x([=e2xg(x(.由f(x(=ex-ax2-(e-a-1(x-1，得f(0(=f(1(=0，則F(0(=F(1(=0，又因?yàn)镕(x(是連續(xù)且可導(dǎo)函數(shù)，g(ξ(=0.故函數(shù)g(x(=2f(x(+f/(x(區(qū)間(0，1(內(nèi)至少存在一個(gè)零點(diǎn). f(a)=f(b)，那么在區(qū)間(a,b)內(nèi)至少存在一點(diǎn)m，使得fI(m)=0．(2)已知函數(shù)f(x)=xlnx,g(x)=x2-bx+1，若對(duì)于區(qū)間(1,2)|f(x1)-f(x2)|>|g(x1)-g(x2)|成立，求實(shí)數(shù)b的取值范圍．【解析】t，則f-bt=f-at，令函數(shù)F(x)=f(x)-tx，則F(a)=F(b),FI(x)=fI(x)-t，即fI-t=0，:fI(2)依題意，fI(x)=lnx+1,gI(x)=x-b，由(1)得|fI(x)|>|gI(x)|,x∈(1,2)，于是lnx+1>|x-b|，即-1-lnx<b-x<lnx+1，因此x-lnx-1<b<x+lnx+1，令φ(x)=x-lnx-1(1<x<2)，:實(shí)數(shù)b的取值范圍是1-ln2≤b≤2．49.已知函數(shù)f(x(=x2-3x+alnx，a∈R.(1)由題意可知當(dāng)a=1時(shí)，f(x(=x2-3x+lnx，f(1(=1-3+0=-2，fI(x(=2x-3+，所以函數(shù)f(x(的在點(diǎn)(1,-2(處切線的斜率k=fI(1(=2-3+1=0，所以函數(shù)f(x(的在點(diǎn)(1,-2(處的切線為y=-2.(2)由題意可得fI(x(=2x-3若函數(shù)f(x(在區(qū)間[1，2[上單調(diào)遞減，則2x2-3x+a≤0在x∈[1,2[恒成立，即a≤-2x2+3x在x∈[1,2[恒成立，只需a≤(-2x2+3x(min即可，,2[時(shí)y