教材+真題特色微專(zhuān)題：三次函數(shù)圖象及性質(zhì) 解析版

于點(diǎn)P(a,b)成中心對(duì)稱(chēng)圖形的充要條件是函數(shù)y=f(x+a)-b為奇函數(shù).(1)求函數(shù)f(x)=x3-3x2圖象的對(duì)稱(chēng)中心；【解析】(1)∵f(x)=x3-3x2=(x-1)-x)=(-x)3-3(-x)=-x3+3x=-g(x).∴g(x)為奇函數(shù).∴f(x)=x3-3x2的圖象關(guān)于點(diǎn)(1,-2)對(duì)稱(chēng).即f(x)=x3-3x2的圖象的對(duì)稱(chēng)中心是點(diǎn)(1,-2).解：函數(shù)f(x)=x3-x2-2x+1的定義域?yàn)镽．對(duì)f(x)求導(dǎo)數(shù)得f/(x)=x2-x-2=(x+1)(x-2).x(-∞,-1)-1(-1,2)2f/(x)+0-0+f(x)11小值.f(x)=x3-4x+4有極小值，并且極小值 -．上述結(jié)論可以從函數(shù)f(x)=x3-4x+4在區(qū)間[0,3[上的圖象(圖5.3-16)得到直觀驗(yàn)證.5.人教B版2019年選擇性必修三P113復(fù)習(xí)題B組第4題：已知【答案】(-∞,-9[∪[3,+∞((1)你能歸納函數(shù)f(x(圖象的大致形狀嗎？它的圖象有什么222-3ac≤02-3ac>02-3ac≤02-3ac>0(1)求f(x(的極值；(2)求f(x(的單調(diào)區(qū)間；(3)求f(x(在區(qū)間[-5,1[上的最大值與最小值。2-33f/(x0(=-(x1-x2(2。A.x=3是f(x)的極小值點(diǎn)B.當(dāng)0<x<1時(shí)，f(x)<f(x2(C.當(dāng)1<x<2時(shí)，-4<f(2x-1)<0D.當(dāng)-1<x<0時(shí)，f(2-x)>f(x)【解析】對(duì)A，因?yàn)楹瘮?shù)f(x(的定義域?yàn)镽，而f/(x(=2(x-1((x-4(+(x-1(2=3(x-1((x-3(，易知當(dāng)xf/(x(<0f/(x(>0，函數(shù)f(x(在(-∞,1(上單調(diào)遞增，在44-x2=x(1-x(>0，所以1>x>x2>0，而由上可知，函數(shù)f(x(在(0,1(上單調(diào)遞增，所以f(x(>f(x2(，錯(cuò)f(2x-1(>f(3(，即-4<f(2x-1(<0，正確；對(duì)D，當(dāng)-1<x<0時(shí)，f(2-x)-f(x)=(1-x(2(-2-x(-(x-1(2(x-4(=(x-1(2(2-2x(>0，所以f(2-x)>f(x)，正確；故選ACD.D.存在a，使得點(diǎn)(1,f(1((為曲線y=f(x)的對(duì)稱(chēng)中心∈(-∞,0(∪(a,+∞(時(shí)f/(x)>0，故f(x)x=a處取到極小值，由f(0)=1>0，f(a)有一個(gè)零點(diǎn)，又f(-1)=-1-3a<0，f(2a)=4a3+1>0，則f(-1)f(0)<0,f(a)f(2a)<0，則f(x)在(-1,f(x)=f(2b-x)，即2x3-3ax2+1=2EQ\*jc3\*hps12\o\al(\s\up3(3),3)f(x)+f(2-x)=2x3-3ax2+1+2(2-x)3-3a(2-x)2+1=(12-6a)x2+(12a-24)x+18-12a，r12-6a=0(18-12a=6-6af(1))是f(x)的對(duì)稱(chēng)中心，D選項(xiàng)正確.A.f(x)有兩個(gè)極值點(diǎn)B.f(x)有三個(gè)零點(diǎn)C.點(diǎn)(0,1)是曲線y=f(x)的對(duì)稱(chēng)中心D.直線y=2x是曲線y=f(x)的切線x(=3x2-1，令f/(x(>0得x>或x<-，令f/(x)<0得-<x<，55()2a(x-a(2(x-b(的極大值點(diǎn)，∴在x=a左右附近都是小于零的.f(x(的圖象如下圖所示：2f(x(的圖象如下圖所示：22成立.=f(x)在點(diǎn)(0,0(處的切線方程為()A.y=-2xB.y=-xC.y=2xD.y=x故選D.66【解析】由f'(x(=6x2-2ax=0得x=0,x=，因?yàn)楹瘮?shù)f(x)在(0,+∞)上有且僅有一個(gè)零點(diǎn)且f1]上單調(diào)遞減，所以f(x)max=f(0)，f(x)min=min{f(-1),f(1){=f(-1)，最大值與最小值的和為f(0)+f(-1)=1-4=-3.7.(2014高考數(shù)學(xué)課標(biāo)1理科A.(2,+∞)B.(-∞,-2)C.(1,+∞)D.(-∞,-1)(x)<0;x∈,+∞(,f/(x)>0;且f(0)=1>0,f(x)有小于零的零點(diǎn),不符合題意．當(dāng) 且x0>0,只需f>0,即a2>4,a<-2．選B令t=,則問(wèn)題又等價(jià)于a=-t3+3t有唯一的正零根,即y=a與y=-t3+3t有唯一的交點(diǎn)且交點(diǎn)在在y軸右側(cè)記f(t)=-t3+3t,f/(t)=-3t2+3,由f/(t)=0,t=±1,t∈(-∞,-1(,f/(t)<0;t∈(-1,1(,f/(t)>0;,t∈(1,+∞(,f/(t)<0,要使a=-需a<f(-1)=-2,選BA.[-5,-3]B.-6,-L|ΓC.[-6,-2]D.[-4,-3]773-x2+4x+3≥0可化為a≥--+，令f(x(=--+，則-6；當(dāng)-2≤x<0時(shí)，ax3-x2+4x+3≥0可化是-6≤a≤-2，即實(shí)數(shù)a的取值范圍是[-6，-2].1.已知三次函數(shù)f(x)=x3-(4m-1)x2+(15m2-2m-7)x+2在定義域R上無(wú)極值點(diǎn)，則m的取值A(chǔ).m<2或m>4B.m≥2或m≤4C.2≤m≤4D.2<m<4f/(x(=x2-2(4m-1(x+15m2-2m-7，由題意得導(dǎo)函數(shù)f/(x(=x2-2(4m-1(x+15m2-2m-2-2(4m-1(x+15m2-2m-7≥0恒成立，∴Δ=4(4m-1(2-4(15m2-2m-7(-f/(-2)3.三次函數(shù)f(x)=mx3-x在(-∞,+∞)上是減函數(shù)，則m的取值范圍是()A.m<0B.m<1C.m≤0D.m≤1882-1≤0minA.0B.1C.2D.3所以f(0)>0>f(1)，又x趨向-∞時(shí)f(x)趨向-∞,x趨向+∞時(shí)f(x5.已知三次函數(shù)f(x(=ax3+x2+cx+有三個(gè)不同的零點(diǎn)x1,x2,x3(x1<x2<x3(，函數(shù)g(x(=f(x(-1.則()A.3ac<1C.若g(x(恰有兩個(gè)不同的零點(diǎn)m,n(m<n)，則2m+n=-D.若g(x(有三個(gè)不同的零點(diǎn)t1,t2,t3(t1<t2<t3(，則xEQ\*jc3\*hps13\o\al(\s\up3(2),1)+xEQ\*jc3\*hps13\o\al(\s\up3(2),2)+xEQ\*jc3\*hps13\o\al(\s\up3(2),3)=tEQ\*jc3\*hps13\o\al(\s\up3(2),1)+tEQ\*jc3\*hps13\o\al(\s\up3(2),2)+tEQ\*jc3\*hps13\o\al(\s\up3(2),3),x2,x3成2<-EQ\*jc3\*hps12\o\al(\s\up3(2),1)EQ\*jc3\*hps12\o\al(\s\up3(2),2)EQ\*jc3\*hps12\o\al(\s\up3(2),3)EQ\*jc3\*hps12\o\al(\s\up3(2),1)EQ\*jc3\*hps12\o\al(\s\up3(2),2)EQ\*jc3\*hps12\o\al(\s\up3(2),3)A.f(x(的值域?yàn)镽B.f(x(圖象的對(duì)稱(chēng)中心為(0,2(C.當(dāng)b-3a>0時(shí)，f(x(在區(qū)間(-1,1(內(nèi)單調(diào)遞減99D.當(dāng)ab>0時(shí)，f(x(有兩個(gè)極值點(diǎn)f(x(-2=ax3-bx滿足g(-x(=-ax3+bx=-g(x(，可知g(x(為奇函數(shù)，BD.A.f(x)有三個(gè)零點(diǎn)B.x=0是f(x)的極小值點(diǎn)C.f(x)的圖象關(guān)于點(diǎn)(-1,-2)對(duì)稱(chēng)D.當(dāng)0<x<1時(shí)，f(x)>f(x2(極小值點(diǎn)，故B正確；對(duì)于C，因?yàn)閒(-1+x)+f(-1-x)=(-1+x+2)2(-1+x-1)+(-1-x+2)2(-1-x-1)=(x+1(2(x-2(+(1-x(2(-2-x(=-4，即f(-1+x)+f(-1-x)=2×(-2(，則f(x)的圖象關(guān)于點(diǎn)(-1,2A.f(x(在(0,+∞(上單調(diào)遞增B.當(dāng)f(x(有且僅有3個(gè)零點(diǎn)時(shí)，b的取以令f/(x(>0解得x<0或x>2，令f/(x(<0解得0<x<2，故f(x(在區(qū)間(-∞,0(或(2,+∞(上單調(diào)(0(>0(0(>0EQ\*jc3\*hps21\o\al(\s\up7(b),8)EQ\*jc3\*hps21\o\al(\s\up7(>),a)x(=6ax-6a，x(=2-6ax,axEQ\*jc3\*hps12\o\al(\s\up3(3),0)-3axEQ\*jc3\*hps12\o\al(\s\up3(2),0)+b(，所以在點(diǎn)C處的切線方程為：y-(axEQ\*jc3\*hps12\o\al(\s\up3(3),0)-3axEQ\*jc3\*hps12\o\al(\s\up3(2),0)+b(=(3axEQ\*jc3\*hps12\o\al(\s\up3(2),0)-6ax0((x-x0(，又因?yàn)榍芯€過(guò)點(diǎn)P(2,a(，所以a-(axEQ\*jc3\*hps12\o\al(\s\up3(3),0)-3axEQ\*jc3\*hps12\o\al(\s\up3(2),0)+b(=(3axEQ\*jc3\*hps12\o\al(\s\up3(2),0)-6ax0((2-x0(，解得2axEQ\*jc3\*hps12\o\al(\s\up3(3),0)EQ\*jc3\*hps12\o\al(\s\up3(2),0)數(shù)可轉(zhuǎn)化為y=g(x(與y=b圖象交點(diǎn)個(gè)數(shù)，g/(x(=6ax2-18ax+12a=6a(x-1((x-2(，9.已知三次函數(shù)ax3+bx2+x+c無(wú)極值，且滿足a+≤8，則a2-b2=.2-4a≤0，即a≥b2>0，所以a+≥b2+≥2=4=16-4=12.10.已知三次函數(shù)f(x)=mx3+nx2+px+2q的圖象如圖所示，則EQ\*jc3\*hps21\o\al(\s\up7(f),f)EQ\*jc3\*hps15\o\al(\s\up11(/),/)EQ\*jc3\*hps21\o\al(\s\up7(p),2)EQ\*jc3\*hps21\o\al(\s\up7(m),3m)f(0(=-1，故f(x(=x3+2x-1的對(duì)稱(chēng)中心為(0,-1(.故當(dāng)x1+x2=0時(shí)，f(x1(+f(x2(=2×(-1(=-2.=0．記f(x(的極大值為M，則M的取值范圍是.【詳解】因?yàn)閒(2-t(=f(t(=f(t(=0，所以t是f(x(的零點(diǎn)也是極值點(diǎn)，2-t也是f(x(的零點(diǎn)，不妨設(shè)f(x(=(x+t-2((x-t)2，故f(x(=(x-t)2+2(x+t-2((x-t)=(x-t((x-t+2x+2t-4(=f(x(單調(diào)遞減，可得f(x(的極大值M=f=(t-1)3，因?yàn)閠∈(2)若x=1是函數(shù)f(x(的極值點(diǎn)，證明：函數(shù)f(x(的圖象關(guān)于點(diǎn)M,f成中心對(duì)稱(chēng).當(dāng)4a<x<2時(shí)，f(x(<0,f(x(單調(diào)遞減.(4a,2(時(shí)，f(x(單調(diào)遞減.故f(x(=x3-x2+2x+2，所以M,，因?yàn)閒(3-x(+f(x(=(3-x)3-(3-x)2+2(3-x(+2+x3-x2+2x+2=[(3-x)3+x3[-[(3-x)2+x2[+10=(9x2-27x+27(-(2x2-6x+9(+10=，所以函數(shù)f(x(的圖象關(guān)于點(diǎn)M,成中心對(duì)稱(chēng).14.已知函數(shù)f(x(=x(x-c)2.(2)若函數(shù)y=f(x(的圖象與直線y=-x+c相切，求實(shí)數(shù)c的值.x(=(x-c)2+2(x-c(?x=(x-c((3x-c(，當(dāng)c=2時(shí)，則f'(x(=(x-2((3x-2(，令f'(x(>0，解得x<或x>2；令f'