2025高考數(shù)學(xué)二輪復(fù)習(xí)-專題1 函數(shù)與導(dǎo)數(shù) 第5講 利用導(dǎo)數(shù)研究函數(shù)的零點【課件】

第5講利用導(dǎo)數(shù)研究函數(shù)的零點2025基礎(chǔ)回扣?考教銜接以題梳點?核心突破目錄索引

基礎(chǔ)回扣?考教銜接1.(人B選必三6.2節(jié)習(xí)題)已知函數(shù)f(x)=x3-x2-x-1的圖象與直線y=c有3個不同的交點,求實數(shù)c的取值范圍.2.(人A選必二第五章例題)給定函數(shù)f(x)=(x+1)ex.(1)判斷函數(shù)f(x)的單調(diào)性,并求出f(x)的極值;(2)畫出函數(shù)f(x)的大致圖象;(3)求出方程f(x)=a(a∈R)的解的個數(shù).解

(1)函數(shù)的定義域為R.f'(x)=(x+1)'ex+(x+1)(ex)'=(x+2)ex.令f'(x)=0,解得x=-2.f'(x),f(x)的變化情況如表所示.x(-∞,-2)-2(-2,+∞)f'(x)-0+f(x)單調(diào)遞減單調(diào)遞增所以f(x)在區(qū)間(-∞,-2)上單調(diào)遞減,在區(qū)間(-2,+∞)上單調(diào)遞增.當(dāng)x=-2時,f(x)有極小值f(-2)=.(2)令f(x)=0,解得x=-1.當(dāng)x<-1時,f(x)<0;當(dāng)x>-1時,f(x)>0.真題體驗1.(2023·全國乙,文8)若函數(shù)f(x)=x3+ax+2存在3個零點,則a的取值范圍是(

)A.(-∞,-2)

B.(-∞,-3)C.(-4,-1) D.(-3,0)B(方法二)令f(x)=0,得-ax=x3+2,易知x≠0,所以-a=設(shè)g(x)=,則函數(shù)f(x)存在3個零點等價于函數(shù)g(x)=的圖象與直線y=-a有三個不同的交點.g'(x)=.當(dāng)x>1時,g'(x)>0,函數(shù)g(x)在(1,+∞)內(nèi)單調(diào)遞增,當(dāng)x<1且x≠0時,g'(x)<0,函數(shù)g(x)在(-∞,0),(0,1)內(nèi)單調(diào)遞減,且g(1)=3,當(dāng)x從左側(cè)趨近于0時,g(x)→-∞,當(dāng)x從右側(cè)趨近于0時,g(x)→+∞,當(dāng)x→+∞時,g(x)→+∞,由此可作出函數(shù)g(x)的大致圖象,如圖所示.根據(jù)以上信息,我們畫出f(x)的大致圖象如圖所示.3.(人A選必二第五章習(xí)題)已知函數(shù)f(x)=ae2x+(a-2)ex-x.(1)討論f(x)的單調(diào)性;(2)若f(x)有兩個零點,求實數(shù)a的取值范圍.由圖知,當(dāng)-a>3時,函數(shù)g(x)=的圖象與直線y=-a有三個交點,即函數(shù)f(x)有3個零點,所以a<-3.故選B.2.(2024·全國甲,文16)當(dāng)x>0時,曲線y=x3-3x與曲線y=-(x-1)2+a有兩個交點,則a的取值范圍是

.

(-2,1)解析

令x3-3x=-(x-1)2+a,得x3+(x-1)2-3x=a.令f(x)=x3+(x-1)2-3x,x>0,則f'(x)=3x2+2(x-1)-3=3x2+2x-5=(x-1)(3x+5).由f'(x)=0(x>0),得x=1.∴當(dāng)x∈(0,1)時,f'(x)<0,f(x)單調(diào)遞減;當(dāng)x∈(1,+∞)時,f'(x)>0,f(x)單調(diào)遞增.∴f(x)min=f(1)=-2.又f(0)=1,f(2)=3>1,曲線y=x3-3x與y=-(x-1)2+a在(0,+∞)上有兩個不同的交點等價于y=f(x)與y=a有兩個不同的交點,∴-2<a<1.3.(2021·全國甲,理21(2))已知a>0且a≠1,函數(shù)f(x)=(x>0).若曲線y=f(x)與直線y=1有且僅有兩個交點,求a的取值范圍.以題梳點?核心突破考點一探究零點個數(shù)例1(2024·河南鄭州三模)已知函數(shù)f(x)=eax-x.(1)若a=2,求曲線y=f(x)在(1,f(1))處的切線方程;(2)討論f(x)的零點個數(shù).解

(1)若a=2,則f(x)=e2x-x,f'(x)=2e2x-1.又f(1)=e2-1,切點為(1,e2-1),曲線y=f(x)在(1,f(1))處的切線斜率k=f'(1)=2e2×1-1=2e2-1,故所求切線方程為y-(e2-1)=(2e2-1)(x-1),即y=(2e2-1)x-e2.(方法二

分離參數(shù))[對點訓(xùn)練1](2024·陜西安康模擬預(yù)測)已知函數(shù)f(x)=aexsinx+x-cosx,f'(x)為f(x)的導(dǎo)函數(shù),f'(0)=2.(1)求a的值;(2)求f'(x)在(0,π)上的零點個數(shù).解

(1)由f(x)=aexsin

x+x-cos

x,則f'(x)=aexsin

x+aexcos

x+1+sin

x=aex(sin

x+cos

x)+1+sin

x,又f'(0)=2,所以a+1=2,即a=1.(2)由(1)可知f'(x)=ex(sin

x+cos

x)+1+sin

x,設(shè)g(x)=f'(x)=ex(sin

x+cos

x)+1+sin

x,則g'(x)=ex(sin

x+cos

x)+ex(cos

x-sin

x)+cos

x=cos

x(2ex+1),考點二根據(jù)零點個數(shù)求參數(shù)取值范圍例2(2024·河南一模)已知函數(shù)f(x)=alnx-x2+a(a∈R).(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)若f(x)有兩個零點,求實數(shù)a的取值范圍.規(guī)律方法

已知函數(shù)零點個數(shù)求參數(shù)取值范圍問題的解法[對點訓(xùn)練2]已知函數(shù)f(x)=ex-ax2.(1)若函數(shù)f(x)的圖象與直線y=x-1相切,求實數(shù)a的值;(2)若函數(shù)g(x)=f(x)-x+1有且只有一個零點,求實數(shù)a的取值范圍.當(dāng)x∈(-∞,0)時,h'(x)>0,函數(shù)h(x)單調(diào)遞增;當(dāng)x∈(0,2)時,h'(x)<0,函數(shù)h(x)單調(diào)遞減;當(dāng)x∈(2,+∞)時,h'(x)>0,函數(shù)h(x)單調(diào)遞增.設(shè)t(x)=ex-x+1,則t'(x)=ex-1,當(dāng)x∈(-∞,0)時,t'(x)<0,函數(shù)t(x)單調(diào)遞減;當(dāng)x∈(0,+∞)時,t'(x)>0,函數(shù)t(x)單調(diào)遞增.畫出函數(shù)h(x)的圖象如圖所示.考點三隱零點問題*例3(2024·江西贛州一模)已知函數(shù)f(x)=ex-1-lnx.(1)求f(x)的單調(diào)區(qū)間;(2)已知m>0,若函數(shù)g(x)=f(x)-m(x-1)有唯一的零點x0.求證:1<x0<2.∴當(dāng)x>0時,l(x)即f'(x)單調(diào)遞增,又f'(1)=0,∴當(dāng)x∈(0,1)時,f'(x)<0,f(x)單調(diào)遞減;當(dāng)x∈(1,+∞)時,f'(x)>0,f(x)單調(diào)遞增.∴f(x)的減區(qū)間為(0,1),增區(qū)間為(1,+∞).[對點訓(xùn)練3](2024·山東淄博一模)已知函數(shù)f(x)=ex-sinx-