g=f(xx-xx-xx-x0(n+…，我們將g(x(稱為函數(shù)f(x(在點x=x0處的泰勒展開式.2.麥克勞林公式：f=fxn+Rn泰勒展開式x=0的特殊形式)1122有(n+1(階導數(shù)，那么對于?x∈(a,b(，有f(xx-xx-x0(2+… 稱該式為函數(shù)f(x(在x=0處的n階等函數(shù)轉(zhuǎn)化為多項式函數(shù)，通過計算多項式函數(shù)值近似求出原函數(shù)的值，如sinx=x cosx，則運用上面的想法求2cossin的近似值為 ()A.0.50B.-0.46C.-0.54D.0.56所以cos1-1≈-0.46. 14.(2025·陜西咸陽·三模)英國數(shù)學家泰勒發(fā)現(xiàn)的泰勒公式有如下特殊形式：當f(x)在x=0處的33n(n∈N*(階導數(shù)都存在時，f(x(=f(0(+f/(0(x+xxn+…．該公fx(=-sinx?fⅡ(0(=0；f(3((x(=-cosx?f(3((0(=-1；f(4((x(=sinx?f(4((0(=0；f(6((x(=-sinx?f(6((0(=0；f(7((x(=-cosx?f(7((0(=-1；求出sinx=x-+-+…，令x=2，5-7+…=2-+-+…≈0.91A.sin1<cos1B.sin1≈0.84(精確到小數(shù)點后兩位)C.cosD.當x>0時，sinx>x2>9當x>0時，令f(x(=sinx-x+，則f/(x(=cosx-1+x2，fⅡ(x(=x-sinx>0，所以f/(x(在(0,+∞(上為增函數(shù)，則f/(x(>f/(0(=0，所以f(x(在(0,+∞(上為增函數(shù)，則f(x(>f(0(=0，故當x>0時，sinx-x+>0恒成立，即sinx>x-．故D選項正確．44(n((x(為函數(shù)f(x(的n階導數(shù)，f(n((x(=[f(n-1((x([/(n≥2，n∈N*(，若f(n((x(存在，則稱f(x(n階可導.英國數(shù)學家泰勒發(fā)現(xiàn)：若f(x(在x0附近n+1階可導，則可構造x(=f(xx-xx-x0(2+●●●x-x0(n(稱其為f(x(在x0處的n次泰勒多項式)來逼近f(x(在x0附近的函數(shù)值.下列說法正確的是()A.若f(x(=sinx，則f(n((x(=sin(x+nπ(B.若f(x，則f(n((x(=(-1(n(n!(x-(n+1(【詳解】對于A，若f(x(=sinx，則f/(x(=cosx=sin(x+f(2((x(=-sinx=sin(x+πf(3((x(=-cosx=sin(x+f(4((x(=sinx=(x+2π(，所以f(n((x(=sin(x+，故A錯誤；對于B，若f(x(=，則f/(x(=-=-x-2,f(2((x(=(-x-2(/=2x-3=(-1(2(2!(x-3，f(3((x(=(2x-3(/=-6x-4=(-1(3(3!(x-4，f(4((x(=(-6x-4(/=24x-5=(-1(4(4!(x-5，觀察可知f(n((x(=(-1(n(n!(x-(n+1(，故B正確；對于C，f(x(=cosx，則f/(x(=-sinx,f(2((x(=-cosx,f(3((x(=sinx，因為f(0(=1,f/(0(=0,f(2((0(=-1,f(3((0(=0，所以f(x(=cosx在x0=0處的3次泰勒多項式T故C正確對于D，f(x(=ex的n階導數(shù)f(n((x(=ex，得T3(x(=f(0(+x-0(2=1+x，≈T7.(24-25高三上·重慶渝中·階段練習)英國數(shù)學家泰勒發(fā)現(xiàn)的泰勒公式有如下特殊形式:當f(x(在x=0處的n(n∈N*(階導數(shù)都存在時，f(x)=f(0)+f/(0)x+x2+….注:fⅡ(x(表示f(x(的2階導數(shù)，即為f/(x(的導數(shù)，f(n((x((n≥3(表示f(x(的n階導數(shù)，即為f(n+1((x((n≥3(的導數(shù).n!表示n的階乘，即n!=1×2×3×…×n.該公式也稱為麥克勞林公55【詳解】令f(x)=sinx，(x)=-sinx，f(3((x(=-cosx，f(4((x(=sinx，故f(0)=0,f/(0)=1,fⅡ(0)=0,f(3)(0)=-1,…，8.(24-25高三上·重慶沙坪壩·階段練習)記f(n((x(為函數(shù)f(x(的n階導函數(shù)，f/(x(=f(1((x(且有f(n((x(=[f(n-1((x(['(n∈N*(，若f(n((x(存在，則稱f(x(n階可導.英國數(shù)學家泰勒發(fā)現(xiàn)：若f(x(在x0附近n+1階可導，則可構造Tn(x(=f(x0(+x-x0(+x-x0(2+…x-x0(n(稱為n次泰勒多項式)來逼近f(x(在x0附近的函數(shù)值，例如：f(x(=ex在x0=0 則f(x在x0=-1處的5次泰勒多項式中x3的系數(shù)為.【詳解】Tn(x(=f(x0(+(x-x0(+(x-x0(2+…+(x-x0(n，因為f(x(=-，f(-1(=1所以f/(x(=x-2f(2((x(=-2x-3,f(3((x(=6x-4，f(4((x(=-24x-5,f(5((x(=120x-6，又f/(-1(=1!,f(2((-1(=2!,f(3((-1(=6=3!，f(4((-1(=24=4!，f(5((-1(=120=5!. f(x)=f(x0(+(x-x0(+(x-x0(2+r(3)3(!x0((x-x0(3+…(z-x0(n+o(((x-x0(*((n∈N*(，我們稱上式為函數(shù)f(x)在x=x0處的泰勒展開式，其中o(x-x0(a(為(x-x0(n的高階無窮小量f(n)(x)=[f(n-1)(x)[/．特別地，當f(x)在x=0處n階連續(xù)可導，則稱f(x)=f(0)+f/(0)x+x2+x3+…+xn+a(xⅡ(為函數(shù)f(x)的麥克勞林公式．如f(x)=sinx的麥克勞林公式為sinx=x-+-…+(-1)n-1-+o(x2n-1(，66由sinx=x-+-…+(-1)n-1-+o(x2n-1(兩邊同時求導得cosx=1-+-所以cos=1-++…=+…≈0.88，所以cos≈0.881-=cosx+-1(x≥0)，則g/(x)=-sinx+x，設G(x)=-sinx+x(x≥0)，則G/(x)=-cosx+1≥0，ax-(sinx-cosx+2)≥0令h(x)=eax-sinx+cosx-2(x≥0)，則h/(x)=aeax-cosx-sinx，①當a≥1時，若x≥0,eax≥ex,h/(x)≥ex-cosx-sinx，所以ex≥x+1≥sinx+1，則h/(x)≥ex-cosx-sinx≥1-cosx≥0，ax≥sinx-cosx+2恒成立．(0)=a-1<0，77(3)(-∞,1]．于是e-x=1-x+-+-+…+(-1)n+…，②(3)F(x)=g(x)-a(1+=-a(1+，則F/ax，設G(x)=-ax,G/(x)=-a．(x)在(0,+∞(上為增函數(shù).88I(lna)=-a=a+-a=-a(<0，所以當-lna<x<0時，F(xiàn)I(x)>0；當0<x<lna時，F(xiàn)I(x)<0． x-xx-xx-xx-x0(n，我們將g(x)稱為函數(shù)f(x)在點x=x0處的n階泰勒展開式．例如，y=cosx在點x=0處的泰勒展開式為+ x>xa>logax．【答案】(1)g(x(=x-x2+x3ffg(x)=ln1+(x-0)+(x-0)2+(x-0)3=x-x2+x3．99令h(x)=f1(x)-g1(x)=cosx-1+，則h’(x)=-sinx+x．則不等式ax>xa今xlna>alnx今<今-<0．x>xa恒成立．a(chǎn)x=t>0，不等式xa>logax今aat>t今-alna<0．:m(t)max=m(e)=-alna.ta>logax恒成立．取M=max{x0,at{，對于任意x>M，ax>xa>logax成立，x>xa>logax情況為Dn種，可以用全排列n!減去有裝正確的情況種數(shù)，即Dn=n!1-+-…+f(0)+f/(0)x+x2+…+xn+…，其中f(n)(x)(n≥3)表示f(x)的克勞林公式.(1)求出D2,D3,D4的值；【詳解】(1)∵Dn=n!1-+-…+(-1)n，3!(1-+-=2,D4=4!(1-+-+=9,n≈.令h(x)=cosx-1+(x>0)，由h/(x)=-sinx+x>0，再令p(x)=-sinx+x,則p/(x)=-cosx所以n<2tan+4tan+…+2ntan<(2+-+(2+-+…命題得證.多項式函數(shù)近似表示，具體形式為：f(x(=f(x0(+f/(x0((x-xx-x0(2+…x-x0(n+…(其中f(n)(x)表示f(x)的n次導數(shù))，以上公式我們稱為函數(shù)f(x)在x=x0處sinx>x+1．(參考數(shù)據(jù)ln【詳解】(1)因為函數(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x=1+x+x2+…+xn+…；cosx=1-x2+x4+…+x2n+….eix=1+(ix)+(ix)2+(ix)3+iπ令f(x)=sinx-x+x3,f’(x)=cosx-1+x2,fⅡ(x)=x-sinx，所以f//(x(在(0,上單調(diào)遞增，所以f(x)在(0,上單調(diào)遞增，所以f(x)>f(0)=0，而g(0)=0,g=-ln>0，則sinx>xx3>lnx-lnx-x≥1；即f/(x(=ex.x(<0，φ(x(在(-∞,0(上單調(diào)遞減，x(>0，φ(x(在(0,+∞(上單調(diào)遞增，則xex-lnx-x=elnx+x-(lnx+x(≥1，解法2：設g(x(=xex-lnx-x，則g/(x(=ex+xex--1=(x+1((ex-，時，g(x(>0，g(x(在(x0,+∞(上單調(diào)所以g(x(min=g(x0(=x0ex-lnx0-x0=1，所以xex-lnx-x≥1成立.設h(x(=xe2x再令M(x(=x2(1+2x(e2x-1+2lnx，則M(x(=2xe2x(2x2+4x+1(+>0，所以M(x(在x>0時單調(diào)遞增，又M=-2ln2-1<0，M(1(=3e2-1>0，所以M又因為M(x(=x2(1+2x(e2x-1+2lnx=(1+2x(e2x+2lnx-1+2lnx，且M(x0(=x(1+2x0(e2x-1+2lnx0=(1+2x0(e2x+2lnx-1+2lnx0=(1+2x0(e0-1+2lnx0=0，時，h(x(<0,h(x(在(0,x0(上單調(diào)遞減，牛頓法的基本思想是通過迭代來逼近方程的根.首先選擇一個接近函數(shù)零點的初始值x0，計算相應的函數(shù)值f(0)和切線斜率k=f′(0).然后計算穿過點(x0,f(x0((并且斜率為k=f′(0)的直線與x軸的交點的xxk(k∈N(是函數(shù)零點近似解的初始值，過點Pk(xk,f(xk((的切線為y=f/(xk((x-xk(+f(xk(，切線與2=()A.BC.D-2可得f/(x(=2x，則f/(x0(=f/(2(=4，同理f/(x1(=ff(x(的切線l:y-f(x0(=f/(x0((x-x0(，則l與x軸的交點的橫坐標x1=xf/(x0(≠0(，稱x1是r的第一次近似值；過點(x1,f(x1((作曲線y=f(x(的切線，則該切線與x軸的交點的橫坐標為A.若取初始近似值為1，則過點(1,f(1((作曲線y=f(x(的切線y=2x-3則y-(-2(=2(x-1(，即y=2x-4，則A錯誤；x0=1?x1=x0-=1-=2，x2=x1-=2-=，B錯誤；根據(jù)題意，可知x1=xx2=xx3=xxn+1=xn上述式子相加，得xn+1=x0----…-，所以x3=x0---，C不正確，則D正確.為函數(shù)f(x(的牛頓數(shù)列.已知數(shù)列{xn{為函數(shù)f(x(=x2-x-2的牛頓數(shù)列，an=ln，且a1=1，xn>2(n∈N+(，數(shù)列{an{的前n項和為Sn.則S2024=()A.22024-1B.22023-1C.2024-1D.2023-1A.1B.2C.3D.4所以曲線y=f(x(在點(2,2(處的切線方程為y-2=4(x-2(，所以曲線y=f(x(在點,處的切線方程為y-=3(x-，過點(x0,f(x0((作曲線f(x)在(x0,f(x0((處的切線，切線方程為l1，當f/(x0(≠0時，稱l1與x軸交點的①切線l1的方程為y=f/(x0((x-x0(+f(x0(；A.3B.2C.1D.0x0,f(x0((，則斜率k=f/(x0(，所以切線方程為y=f/(x0((x-x0(+f(x0(，故①正確；對于②,由①可知：y=f/(xn((x-xn(+f(xn(，f(x(的切線l,l與x軸的交點的橫坐標x1=xf/(x0(≠0(，稱x1是r的一次近似值；過點(x1,f(x1((作曲線y=f(x(的切線，則該切線與x軸的交點的橫坐標為x2=xf/(x1(≠0(，次函數(shù)f(x)有兩個不相等的實根b,c，其中c>b．在函數(shù)f(x)圖像上橫坐標為x1A.xB.a因為二次函數(shù)f(x(有兩個不等實數(shù)根b,c，所以不妨設f(x(=a(x-b((x-c(，因為f/(x(=a(2x-b-c(，所以f/(xn(=a(2xn-b-c(，所以在橫坐標為xn的點處的切線方程為：y-f(xn(=a(2xn-b-c((x-xn(，0,x1,…,xn-xn-1|足夠小時，我們可以把xn的值作為函數(shù)f(x(零A.切線l1的方程為6x-y-4=0B.x對于D，由題意得f(x(在(xn,f(xn((處的切線方程為y-f(xn(=f/(xn((x-xn)， 2 xn+1= 2 xn+1=由已知得f/(x(=3x2+3>0，則f(x(單調(diào)遞增，由零點存在性定理得存在r∈,1(作為f(x(零點，n-1知函數(shù)f(x(=(x-b((x-c(，其中c>b.在f(x(圖象上橫坐標為x1的點處作曲線y=f(x(的切線，切A.x(e是自然對數(shù)的底數(shù))B.數(shù)列{an{是遞增數(shù)列故A正確；由f(x(=(x-b((x-c(，得f/(x(=(2x-b-c(，所以f/(xn(=(2xn-b-c(，所以在橫坐標為xn的點處的切線方程為y-f(xn(=(2xn-b-c((x-xn(，所以ln【詳解】由f(x(=x3-x+1，f/(x可得f/(x1(=f/(-1.5(=3×(-1.5)2-1=5.75，f(x1(=f(-1.5(=-0.875，令y=0，得x2=-≈-1.35.f(x1)時，r的n+1次近似值xn+1與n次近似值xn的關系為：xn+1=．由此推理得r的n+1次近似值與r的n次近似值的關系式為xn+1=xnxn+1=xn-=xn-=xn+，=x0+=1+=，x2=x1+=×+×=，數(shù)列{xn{稱為函數(shù)f(x(的“牛頓數(shù)列”.已知數(shù)列{xn{為函數(shù)f(x(=x2-x的牛頓數(shù)列，且數(shù)列 范圍.n=2n(3)-9≤t≤由a1=2,an=ln則2=ln=則有x2==，所以a2=ln=2ln=4；所以an+1=lnlnlnann-1=2n；x(<0，g(x(是減函數(shù)，x(>0，g(x(是增函數(shù)，當n為奇數(shù)時，原式化簡為-t≤Sn+，所以當n=1時，-t≤9，所以t≥-9；綜上可知，-9≤t分．其中牛頓在《流數(shù)法與無窮級數(shù)》(TheMethodofFluxionsandInifiniteSeries)n．一般地，作點(xn,f(xn((處曲線y=f(x(的切線ln+1【詳解】(1)函數(shù)f(x(=x3-x2+2x-5，求導得f/(則f/(1(=3，而f(1(=-3，y=f(x(在x=1處的切線l1方程為y-(-3(=3(x-1(，即3x-y-6=0.y=f(x(在x=2處的切線l2為y-3=10(x-2(，令y=0，得可得f(xn(=x+axn+b,f/(xn(=2xn+a，過點(xn,f(xn((作曲線y=f(x(的切線ln+1:y-f(xn(=f/(xn((x-xn(，nn+1=an+(n+2-1(dn，假設數(shù)列{dn{中存在4項di,dj,dk,dl(其中i,j,k,l成等差數(shù)列)成等比數(shù)列，所以在數(shù)列{dn{中不存在4項di,dj,dk,dl成等比數(shù)列.=f(x(在點(x3,f(x3((處的切線，以此類推，直到求得滿足精度的方程f(x(=0的近似解xn(n≥2(為止.已知函數(shù)f(x(=x-lnx-2，在橫坐標為x1的點處作f(x(的切線，切線與x軸交點的橫坐標為(1)求f(x(的單調(diào)區(qū)間；【詳解】(1)函數(shù)f(x(=x-lnx-2的定由f/(x(<0(2)設f(x(在點(2,f(2((處的切線為l：y-f(2(=f/(2((x-2(， 所以方程x-lnx-2=0的一個近似根x2為3.39.(3)f(x(在(xn,f(xn((處切線方程為y-f(xn(=f/(xn((x-xn(，xnxnf(3)=1-ln3<0,f(4)=2-ln4=2(1-ln2)>0，-lnx0-2=0，即lnx0=x0-2，x(>0，程近似解的方法，其過程如下：如圖，設r是f(x(=0的根，選取x0作為r的初始近似值，過點(x0,f(x0((作曲線y=f(x(的切線L，L的方程為y=f(x0(+f/(x0((x-x0(．如果f/(x0(≠0，則L與x軸的交點的橫坐標記為x1，稱x1為n，根據(jù)已有精確度ε,當|xn-r|<ε時，(2)函數(shù)h(x(=x(lnx-1(-lnx+ex-e-x．①試寫出函數(shù)h(x(的最小值m與r的關系式；(2)①m=-r2+r+e-r,-1<r<-，②證明見解析=-，所以x2=；(2)①因為h(x(=x(lnx-1(-lnx+ex-e-x，所以h/(x)=lnx-+ex+e-x，令φ(x)=lnx-+ex+e-x,x>0，由f(r)=0，得-1<r<-，且er=-r，則e-r=-，r=ln(-r(，所以h/=ln(-re-r+er=0，(x)>0，于是函數(shù)h(x)在(0,-r)上單調(diào)遞減，在(-r,+∞)上單調(diào)遞增，函數(shù)h(x)在x=-r處取得最小值m=h(-r)=-rln(-r(+r-ln(-r(+e-r-er=-r2+r+e-r，即m=-r2+2r,-1<r.②由①知，m=-r2+r+e-r,-1<r，令u(x)=-x2+x+e-x,x<0，求導得u/(x)=-e-x-2x+1，令y=-e-x-2x+1,x<0，求導得y/=e-x-2，則函數(shù)u/(x)在(-ln2,0)上單調(diào)遞減(0)=0,u/(-1)=-e+3>0，而-1<r，因此u(r)>u(-1(=e-2，所以m>e-2.方法--牛頓切線法：如圖，設r是y=f(x(的一個零點，任意選取x0作為r的初始近似值，在點(x0,f(x0((處作曲線y=f(x(的切線l1，l1與x軸的交點橫坐標為x1，稱x1為r的1次近似值；在點在點(xn,f(xn((處作曲線y=f(x(的切線ln+1，ln+1與x軸的交點橫坐標為xn+1，稱xn+1為r的n+1次(2)已知g(x(=(x-b((x-c((c>b(，數(shù)列{xn{為g(x(的牛頓數(shù)列.所以f(x(在(0,1(處切線l1：y-1=2x，令y=0，則得x所以r的2次近似值為-；(2)(i)g(x(=(x-b((x-c(=x2-(b+c(x+bc，所以g/(x(=2x-(b+c(，所以g/(xn(=2xn-所以在橫坐標為xn的點處的切線方程為：y-g(xn(=[2xn-(b+c([(x-xn(，所以h(x(在(1,+∞(上單調(diào)遞減，所以h(x(<h(1(=0，所以lnx<x-1(x>1(，r的初始近似值，過點(x0,f(x0((作曲線y=f(x(的切線L，L的方程為y=f(x0(+f/(x0((x-x0(.如果f/(x0(≠0，則L與x軸的交點的橫1為r的一階近似值.再過點(x1,f(x1((作曲線y=f(x(的切線，并求出切線與x軸的n有精確度ε,當|xn-xn-1|<ε時，給出近似解xn.已知函數(shù)f(x(=xn-nx,x∈R，其中n∈N,n≥2.f(x(與g(x(的大小；n<e)(2)(i)f(x(≥g(x(；(ii)證明見解析則h/(x(=3x2-1,k1=h/(-1(=2,h(-1(=1，曲線h(x(在x0=-1處的切線為y-1=2(x+1(，-x0|=0.5.k2=h/(-1.5(=，h(-1.5(=-，曲線h(x(在x1=-1.5處的切線為y2-x1|<0.5.f/(x(=nxn-1-n，則f/(x0(=n2-n曲線y=f(x(在點P處的切線方程為y=f/(x0((x-x0(，即g(x(=f/(x0((x-x0(，令F(x(=f(x(-g(x(，即F(x(=f(x(-f/(x0((x-x0(，則F/(x(=f/(x(-f/(x0(.因為f/(x(=nxn-1-n在(0,+∞(上單調(diào)遞增，所以F/(x(在(0,+∞(上單調(diào)遞增.x0(<0所以F(x(在(0,x0(上單調(diào)遞減，在(x0,+∞(上單調(diào)遞增，所以對任意的正實數(shù)x都有F(x(≥F(x0(=0，(ii)證明：因為f(x(=nxn-1-n在(0,+∞(上單調(diào)遞增，f(1(=0，所以f(x(在(0,1(上單調(diào)遞減，在(1,+∞(上單調(diào)遞增，所以x=1是f(x(在(0,+∞(的極小值點，也是f(x(在(0,+∞(的最小值點，即f(x(min=f(1(=1-n<0.又f(0(=f(x0(=0，所以當方程f(x(=a有兩個根時，1-n<a<0，曲線y=f(x(過點(1,1-n(和點(0,0(的割線方程為y=(1-n(x.下面證明：f(x(<(1-n(x(0<x<1(.設u(x(=f(x(-(1-n(x=xn-x(0<x<1)，所以u(x(在(0上單調(diào)遞減，u(x(<u(0(所以當0<x<1時，u(x(<0，即f(x(<(1-n(x.曲線y=f(x(過點(1,1-n(和點(x0,0(的割線方程為y=(x-x0(.下面證明：f(x(<(x-x0((1<x<x0(.設v(x(=f(x(-(x-x0(=xn-nx-(x-x0((1<x<x0(，則v(x(=nxn-1-nx<x0(，即v(x(在(1,x0(上單調(diào)遞增，因為f(1+=(1+n-1-n<e-1-n<0，所以n->0，即v(x0(>0.所以v(x(在(1,x3(上單調(diào)遞減，v(x(<v(1(=0，在(x3,x0(上單調(diào)遞增，v(x(<v(x0(=0，f(x(x-x0(.f(x2(<(x2-x0(，解得x2>+x0②.由②-①,得x2-xx0=x即證得x2-x1>n則在(a,b(內(nèi)至少存在一點ξ,使得f如圖所示，在滿足定理條件的曲線y=f(x)上至少存在一點P(ξ,f(ξ)(，該曲線在該點處的切線平行于曲如f(x)=x3在x=0處的切線斜率為(1)f(b)-f(a)=f’(ξ)(b-a)，(a<ξ<b).(2)f(b)-f(a)=f’(a+θ(b-a)((b-a)，(0<θ<1).(3)f(a+h)-f(a)=f’(a+θh(h，(0<θ<1).至少存在一點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A.3B.2C.1D.0則f(4(-f(1(=f/(c((4-3(， 1t(=4t3-8t-3=0在[1,2[上的零點個數(shù)，t(=12t2-8>0，所以g(t(在[1,2[上單調(diào)上至少存在一個實數(shù)t，使得f(b(-f(a(=f/(t((b-a(，其中t稱為“拉格朗日中值”.函數(shù)g(x(=A.B.C.2D.即g/(t(=t+1=，所以t=.那么在開區(qū)間(a,b(上至少存在一個實數(shù)t，使得f(b(-f(a(=f/(t((b-a(，其中t被稱為拉格朗日中值.函數(shù)f(x(=2x+x2在區(qū)間[0，4[上(a,Ob(內(nèi)至少存在一點c，使得f(b(-f(a(=f/(c((b-a(成立，其中c叫做f(x(在[a,Ob[上的“拉 ()A.1B.2C.3D.4即=-sinx0，則在區(qū)間(a,b(上至少存在一個數(shù)ξ,使得f(b)-f(a)=f/(ξ)(b-a)，其中ξ稱為拉格朗日中值.則g(x)=lnx在區(qū)間[1,e[上的拉格朗日中值ξ=.朗日中值ξ滿足，g(e(-g(1(=g/(ξ((e-1(所以g區(qū)間(a,b(上至少存在一個實數(shù)t，使得f(b(-f(a(=f/(t((b-a(，其中t稱為“拉格朗日”中值.函數(shù)38.(24-25高二下·河北邢臺·階段練習)如果函數(shù)f(x(滿足在b-a①函數(shù)f(x(在區(qū)間[a,b[上連續(xù)(函數(shù)圖象沒有間斷)；②函數(shù)f(x(在開區(qū)間(a,b(內(nèi)可導(導數(shù)存在)．則在區(qū)間(a,b(內(nèi)至少存在一點ξ,使得f/(ξ(=f(b(-f(b(-f(a(b-at2-cost1|≤|t2-t1|；(3)已知函數(shù)h(x(=xex-x2-alnx(a≥0(在區(qū)間(0,+∞(上滿足拉格朗日中值定理的兩個條件，當x(=-sinx，t2-cost1|=|t2-t1|，可得|cost2-cost1|≤|t2-t1|，t2-cost1|≤|t2-t1|成立.又由h/(x(=(x+1(ex-2x-，設λ(x(=(x+1(ex-2x-，有λ/(x(=(x+2(ex-2+≥(x+2(ex-2=xex+2(ex-1(>0，可得函數(shù)λ(x(單調(diào)遞增，使得h使得hg(x(=f(x(+ae-x-x2.(2)0<a<所以f(-x(=-f(x(，即ae-x+bex=-aex-be-x，所以(a+b((ex+e-x(=0，所以a+b=0．所以當a=1時，f(x(=ex-e-x，所以f/(x(=ex+e-x， (e-e-1(-(e-1-e(1-(-1(由題意知即求方程ex+ (e-e-1(-(e-1-e(1-(-1(令m(x(=ex+e-x-(e-e-1(，則m/(x(=ex-e-x所以當-1<x<0時，m/(x(<0，所以m(x(單調(diào)遞減；所以m(x(min=m(0(=2+e-1-e<0，又m(-1(=2e-1>0，m(1(=2e-1>0，則h/(x(=-1+=，x>0．令h/(x(=0=，因為h(x(在定義域內(nèi)有三個不同的極值點，令n(x(=aex-2x-(ae-a-1(，則n(0(=2a+1-ae，n(1(=a-1，n/(x(=aex-2．，所以n(x(單調(diào)遞減，因為g(x(在區(qū)間[0,1[上有且只有一個“拉格朗日中值點”，x(≥0，n(x(在[0,1[上單調(diào)遞增，x(≤0，n(x(在[0,1[上單調(diào)遞減，，所以n(x(單調(diào)遞增，所以n(x(min=n(lnlnae-a-1(=3-2lnae+a， a(<0，μ(a(單調(diào)遞減，則u(a(max=u-=1-2ln(e-1((((2)若函數(shù)f(x(在區(qū)間[1,2[上單調(diào)遞減，(2)a≤-2所以f/(x(≥0，f(x(在區(qū)間x∈[1,2[上為增函數(shù)，所以函數(shù)f(x(的最小值為f(1(=-2；(2)由題意可得f/(x(=2x-3+=，即a≤-2x2+3x在x∈[1,2[恒成立，只需a≤(-2x2+3x(min即可，設A(x1,y1(，B(x2,y2(是f(x(上不同的兩點，且0<x1<x2，由題意可得f(x1(=x-3x1+alnx1，f(x2(=x-3x2+alnx2，函數(shù)f(x(在拉格朗日平均值點處的切線斜率k=f/=x1+x2-3+，令h(t(=lnt則h/(t所以在(1,+∞(上不存在t使得lnt+=2，即不存在這樣的A,B兩點使得f/=則在區(qū)間(a,b(上至少存在一個實數(shù)ξ,使得f/(ξ(成立．已知函數(shù)f(x(=ex-x2+mx．【答案】(1)m≥-1+2ln2x(=ex-2x+m≥1對所有x>0成立.整理得m≥-ex+2x+1.設函數(shù)g(x(=-ex+2x+1(x>0)，則g/(x(=-ex+2.g(x(max=g(ln2(=-eln2+2ln2+1=-2+2ln2+1=-1+2ln2.故m≥-1+2ln2；(2)當m=-時，函數(shù)f(x(=ex-xx.驗證是否存在區(qū)間[a,b](a>0)使得>f.方法一：不等式>f等價于>.=f/(t)，由于函數(shù)f/(x(=ex-2x-，[f/(x([/=ex-2在x>ln2時大于零，x-2>0(即x>ln2)時，[f/(x([/=ex-2>0，在f(x)為凸函數(shù)；當x<ln2時，[f/(x([/=ex-2<0，f(x(為凹函數(shù).(3)f/(x(=ex-2x+m由于m<-1，設g(x(=ex-2x+m，g/(x(=ex-2，令g/(x)=0得x=ln2.由于m<-1，且2-2ln2+m<1-2ln2=1-ln4<0，所以g(ln2)<0.g(x(=+∞和g(x(=+∞,所以g(x)=0有兩個解x1和x2，x1<ln2<x2.因為f/(0(=1+m<0，所以x1<0.x=2x1-m和ex=2x2-m，相減得：ex-ex=2(x2-x1(①,需要證明ex-ex<-4m，即：2(x2-x1(<-4m?x2-x1<-2m，設d=x2-x1，需證明d<4ex-8x1②.x(ed-1(=2(x2-x1(，ex=2x1-m，所以ed-1=③,設p(x(=ex-1-x-x2，令q(x(=p/(x(=ex-1-x，q所以ed>1+d+d2對于d>0恒成立，構造函數(shù)h(xex+8x-2，只需證明在x=x1<0的條件下h(x(<0成立.令u(x(=h/(x(=--4ex+8，uxex對于x<0恒成立，h所以h(x(單調(diào)遞減，又因為h(0(=-2<0，所以x<0時h(x(<0成立.b-a(2)已知函數(shù)g(x(=xln+aex-x(a>ii)證明見解析-f(b(<a-b，即f(a(-a<f(b(-令H(x(=f(x(-x=x++mlnx，可知H(x(在(1,3(上單調(diào)遞增，令h(x(=-x，顯然h(x(在(1,3(上單調(diào)遞減，故問題轉(zhuǎn)化為f/(x0(>1恒成立.故令P(x(=-x，顯然P(x(在(1,3(上單調(diào)遞減，即證f(b(-f>2f，又b<3，f(b(-f=f/(ξ2(.，b-a(則f(ξ2(>f(ξ1(，故f(ξ2(.3>f(ξ1(b-a(函數(shù)g(x(=xlnaex-x有兩個零點，即方程xln+aex-x=0有兩個根，即方程ln有2個根.令φ(x(=lnx+x-1(x>0(,φ/(x(=+1>0，-1=0的根，1-x2=lnt，所以tx2-x2=lnt，則x2=-,x1=tx2=-，要證ex+2x>32，即證x1+2x2>5ln2，又x1+2x令m(tm/(t令u(t(=-3lnt+t-+1,u/(t(=-+1+=，t(<uln即x1+2x2>5ln2，矩陣的運算條件：③矩陣相乘的條件為第一個矩陣的列數(shù)等于第二個矩陣的行數(shù).A,B的和(差),記作：A+B(A-B)①矩陣與實數(shù)的積αA++=①矩陣的乘積：第i行第j列元素Cij是矩陣A第i個行向量與矩陣B的第j個列向量的數(shù)量積，那么C矩陣叫做A與B的結(jié)合律：γ(AB(=(γA(B=A(γB(，(AB(C=A(BC(①互換矩陣的兩行；12記DDxDy可用二階行列式表示為n．(1)三階行列式的展開方法：①對角線方式展開1c2-a1b3c2-a2b1c3-a3b2c1這就是說，一個三階行列式可以表示為它的第一行的元素分別與它們的代數(shù)余子式乘積的和。l((2-1(((2-1(=()A.0B.C.D.2l50(．sixf(x1(f(x2(=，則a的取值范圍是(A.(-,B.(-,C.(-,-D.(-1,-所以(2a+2(2<<(2a+4(2，解得-1≤a<-；所以<(2a+4(2，解得-<a<-1，A.f(x(的圖像關于點,0(中心對稱B.f(x(的圖像關于y軸對稱C.f(x(在區(qū)間0,|上單調(diào)遞增D.f(x(最小正周期為π13241324a4-a2a3，若函數(shù)f(x(=A.f(x(的圖象關于點(π,0(中心對稱B.f(x(的圖象關于直線x=對稱C.f(x(在區(qū)間-,0|上單調(diào)遞增D.f(x(的最小正周期為πf(x)=cos2x-sin2x-3cos(2x+=cos2x+3sin2xf(x(=cosx-sinx-=cos(x+-．【答案】f(x(=-x3+(6+3a(x2+ax-3,-則f(x(=[x2+5x-2x2+x3x2+x[L|L|5x2-2x3+x2+3ax2+ax=-x3+(6+3a(x2+ax；若f(x)是R上的單調(diào)遞增函數(shù)，則f/(x)=-3x2+2(6+3a(若f(x)是R上的單調(diào)遞減函數(shù)，故答案為：①f(x)=-x3+(6+3a(x2+ax；②-3,-.的符號稱二階行列式，并規(guī)定二階的行列式計算如下：f(x)=2sinx2sin2x(2)sinC=+，tan∠BAD=3【詳解】(1)f(x)=2sinxcos(x+2x=2sinxcosx-sinx(-2sin2x=sin2x-3sin2x=sin2x-(1-cos2x)=3sin(2x+-，所以f(x)的對稱軸為x=+π(k∈Z 因為A=,A+B+C=π,所以C=-B，所以sinC=sin-B(=sincosB-cossinB=×+×=+，設∠BAD=θ,則∠CAD=-θ,BD=AD=3ADsinθsinB6,CD=AD=6ADBD=AD=3ADsinθ