第21練三次函數(shù)的性質(zhì)(分值：40分)三次函數(shù)f(x)=ax3+bx2+cx+d(a≠0)的性質(zhì)(1)奇偶性：f(x)不可能為偶函數(shù)；當(dāng)且僅當(dāng)b=d=0時(shí),f(x)是奇函數(shù).(2)對(duì)稱性：中心對(duì)稱,且對(duì)稱中心是-b3a,f-b3a.引申性質(zhì)：若y=f(x)是可導(dǎo)函數(shù).①若圖象關(guān)于點(diǎn)(m,n)對(duì)稱,則y=f'(x)的圖象關(guān)于直線x=m對(duì)稱；②若圖象關(guān)于直線x=m對(duì)稱,則y=f'(x)的圖象關(guān)于點(diǎn)(m,0)對(duì)稱；③奇函數(shù)的導(dǎo)數(shù)是偶函數(shù),偶函數(shù)的導(dǎo)數(shù)是奇函數(shù),周期函數(shù)的導(dǎo)數(shù)還是周期函數(shù).(3)根與系數(shù)的關(guān)系設(shè)ax3+bx2+cx+d=0(a≠0)的三個(gè)根分別為x1,x2,x3,則ax3+bx2+cx+d=a(x-x1)(x-x2)(x-x3)=ax3-a(x1+x2+x3)x2+a(x1x2+x2x3+x3x1)x-ax1x2x3.比較系數(shù)可得x1+x2+x3=-ba,x1x2+x2x3+x3x1=ca,x1x2x(4)三次函數(shù)切線條數(shù)設(shè)f(x)=ax3+bx2+cx+d(a≠0)在對(duì)稱中心N-b3a,f-b3a處的切線為l,函數(shù)f(x)及l(fā)把坐標(biāo)平面分成Ⅰ,Ⅱ,Ⅲ,Ⅳ四個(gè)區(qū)域,如圖1(a<0①當(dāng)定點(diǎn)P在對(duì)稱中心N或在Ⅰ和Ⅲ區(qū)域時(shí),過點(diǎn)P的切線有且僅有一條；②當(dāng)定點(diǎn)P在函數(shù)f(x)的圖象上(除去對(duì)稱中心)或在切線l上時(shí),過點(diǎn)P的切線有兩條；③當(dāng)定點(diǎn)P在Ⅱ和Ⅳ區(qū)域時(shí),過點(diǎn)P的切線有三條.一、單項(xiàng)選擇題(每小題5分,共10分)1.已知函數(shù)f(x)=-x3+ax2-x-1在R上單調(diào),則實(shí)數(shù)a的取值范圍是()A.(-∞,-3]∪[3,+∞B.[-3,C.(-∞,-3)∪(3,+∞D(zhuǎn).(-3,答案B解析由f(x)=-x3+ax2-x-1,得f'(x)=-3x2+2ax-1,因?yàn)楹瘮?shù)在R上單調(diào),所以f'(x)=-3x2+2ax-1≤0在R上恒成立,則Δ=4a2-12≤0?-3≤a≤3所以實(shí)數(shù)a的取值范圍是[-3,32.已知函數(shù)f(x)=2x3-3x2+9x-72,則f

12026+f

22026+f

32026+A.2024 B.2025C.2025 D.2026答案B解析由f(x)=2x3-3x2+9x-72,可得f'(x)=6x2-6x+9,令g(x)=6x2-6x+9,g'(x)=12x令g'(x)=12x-6=0,得x=12,又f

12=2×123-3×122+9×12-72=12,所以對(duì)稱中心為12,12,f

12026+f

20252026=1,f

22026+f

20242026=1,…,f

10122026+f

10142026二、多項(xiàng)選擇題(每小題6分,共12分)3.(2024·合肥模擬)已知三次函數(shù)f(x)=x3+ax+2,下列敘述正確的是()A.當(dāng)a=1時(shí),函數(shù)f(x)無極值點(diǎn)B.函數(shù)f(x)的圖象關(guān)于點(diǎn)(0,2)中心對(duì)稱C.過點(diǎn)(0,2)的切線有兩條D.當(dāng)a<-3時(shí),函數(shù)f(x)有3個(gè)零點(diǎn)答案ABD解析對(duì)于A,當(dāng)a=1時(shí),f(x)=x3+x+2,f'(x)=3x2+1>0,f(x)單調(diào)遞增,無極值點(diǎn),故A正確；對(duì)于B,因?yàn)閒(x)+f(-x)=4,所以函數(shù)f(x)的圖象關(guān)于點(diǎn)(0,2)中心對(duì)稱,故B正確；對(duì)于C,設(shè)切點(diǎn)坐標(biāo)為(x1,f(x1)),則切線方程為y-f(x1)=f'(x1)(x-x1),因?yàn)榍芯€過點(diǎn)(0,2),所以2-f(x1)=f'(x1)(-x1),即2-x13-ax1-2=-3x13-ax1,解得即只有一個(gè)切點(diǎn),即只有一條切線,故C錯(cuò)誤；對(duì)于D,f'(x)=3x2+a,當(dāng)a<-3時(shí),令f'(x)=0,得x=±-當(dāng)x∈-∞,--a3時(shí),f'(x)>0,當(dāng)x∈--a3,-a3時(shí),f'(x)當(dāng)x∈-a3,+∞時(shí),f'(x)>0,f(又f(x)有極大值為f

--a3>f(0)=2>0,所以若函數(shù)f(x)有則f(x)有極小值為f

-a3=2a3-a3+2<0,4.(2025·長沙模擬)已知三次函數(shù)f(x)=ax3+bx2+cx+d(a≠0)有三個(gè)不同的零點(diǎn)x1,x2,x3(x1<x2<x3),函數(shù)g(x)=f(x)-1也有三個(gè)零點(diǎn)t1,t2,t3(t1<t2<t3),則()A.b2>3acB.若x1,x2,x3成等差數(shù)列,則x2=-bC.x1+x3<t1+t3D.x12+x22+x32答案ABD解析因?yàn)閒(x)=ax3+bx2+cx+d,則f'(x)=3ax2+2bx+c,a≠0,對(duì)稱中心為-對(duì)于A,因?yàn)閒(x)有三個(gè)不同的零點(diǎn),所以f(x)必有兩個(gè)極值點(diǎn),即f'(x)=3ax2+2bx+c=0有兩個(gè)不同的實(shí)根,所以Δ=4b2-12ac>0,即b2>3ac,故A正確；對(duì)于B,由x1,x2,x3成等差數(shù)列,及三次函數(shù)的中心對(duì)稱性,可知(x2,f(x2))為f(x)的對(duì)稱中心,所以x2=-b3a,對(duì)于C,函數(shù)g(x)=f(x)-1,當(dāng)g(x)=0時(shí),f(x)=1,則直線y=1與y=f(x)圖象的交點(diǎn)的橫坐標(biāo)即為t1,t2,t3,當(dāng)a>0時(shí),畫出f(x)與y=1的圖象,由圖可知,x1<t1,x3<t3,則x1+x3<t1+t3,當(dāng)a<0時(shí),則x1+x3>t1+t3,故C錯(cuò)誤；對(duì)于D,由題意得a整理得x得(x1+x2+x3)2-2(x1x2+x2x3+x3x1)=(t1+t2+t3即x12+x22+x32=t12三、填空題(共5分)5.已知三次函數(shù)y=f(x)有三個(gè)零點(diǎn)x1,x2,x3,且在點(diǎn)(xi,f(xi))處的切線的斜率為ki(i=1,2,3),則1k1+1k2+1答案0解析設(shè)f(x)=a(x-x1)(x-x2)(x-x3)(a≠0),則f'(x)=a(x-x1)(x-x2)+a(x-x1)(x-x3)+a(x-x2)(x-x3),由導(dǎo)數(shù)的幾何意義可得k1=f'(x1)=a(x1-x2)(x1-x3),k2=f'(x2)=a(x2-x1)(x2-x3),k3=f'(x3)=a(x3-x1)(x3-x2),所以1k1+1k2+1k3=1a(x四、解答題(共13分)6.(13分)對(duì)于三次函數(shù)f(x)=ax3+bx2+cx+d(a≠0),給出定義：設(shè)f'(x)是函數(shù)y=f(x)的導(dǎo)數(shù),f″(x)是f'(x)的導(dǎo)數(shù),若方程f″(x)=0有實(shí)數(shù)解x0,則稱點(diǎn)(x0,f(x0))為函數(shù)y=f(x)的“拐點(diǎn)”.經(jīng)過探究發(fā)現(xiàn)：任何一個(gè)三次函數(shù)都有“拐點(diǎn)”；任何一個(gè)三次函數(shù)都有對(duì)稱中心,且“拐點(diǎn)”就是對(duì)稱中心.設(shè)函數(shù)f(x)=13x3-m2x2+(1)當(dāng)m=1時(shí),求f

1100+f

2100+…+f

98100+f(2)若不等式2xlnx+f'(x)+3≥0恒成立,求實(shí)數(shù)m的取值范圍.(7分)解(1)函數(shù)f(x)=13x3-m2x2當(dāng)m=1時(shí),f(x)=13x3-12x2∵f'(x)=x2-x,∴f″(x)=2x-1,令f″(x)=2x-1=0,解得x=1則對(duì)稱中心的縱坐標(biāo)為f

12=1,∴f(x)+f(1-x)=2,∴f

1100+f

99100=2,f

2100+f

98100=2,…,f

49100+f

51100則f

1100+f

2100+…+f

98100+f

99100(2)∵2xlnx+f'(x)+3≥0,f'(x)=x2-mx,即mx≤2xlnx+x2+3,又x>0,∴m≤2xlnx+x2+3x在x∈令t(x)=2xlnx+x2+3∴m≤t(x)