第3課時(shí)利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)高考總復(fù)習(xí)優(yōu)化設(shè)計(jì)GAOKAOZONGFUXIYOUHUASHEJI2025考點(diǎn)一確定函數(shù)零點(diǎn)的個(gè)數(shù)(多考向探究預(yù)測(cè))考向1利用單調(diào)性和函數(shù)零點(diǎn)存在定理確定零點(diǎn)個(gè)數(shù)例1(2024·福建寧德模擬)已知函數(shù)f(x)=ex-4sin

x,其中e為自然對(duì)數(shù)的底數(shù).(1)求曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程;(2)證明:f(x)在[0,+∞)上有兩個(gè)零點(diǎn).(1)解

因?yàn)閒(x)=ex-4sin

x,所以f'(x)=ex-4cos

x,則f(0)=1,f'(0)=-3,故所求切線方程為y-1=-3(x-0),即3x+y-1=0.(2)證明

設(shè)g(x)=f'(x)=ex-4cos

x,則g'(x)=ex+4sin

x.顯然當(dāng)x∈(0,π]時(shí),g'(x)>0,當(dāng)x∈(π,+∞)時(shí),g'(x)>eπ-4>0,所以f'(x)在(0,+∞)上單調(diào)遞增.[對(duì)點(diǎn)訓(xùn)練1](2024·江蘇南京模擬)已知函數(shù)f(x)=ex+(a-e2)x,其中a∈R.(1)若a=e2-2,求函數(shù)f(x)在[0,2]上的最值;(2)當(dāng)a<0時(shí),證明:F(x)=f(x)-

ax2在(0,2)內(nèi)存在唯一零點(diǎn).(1)解

當(dāng)a=e2-2時(shí),f(x)=ex-2x,所以f'(x)=ex-2.令f'(x)>0,得x>ln

2;令f'(x)<0,得x<ln

2,所以f(x)在[0,ln

2]上單調(diào)遞減,在[ln

2,2]上單調(diào)遞增,又f(0)=1,f(ln

2)=2-2ln

2,f(2)=e2-4>1,所以f(x)在[0,2]上的最小值為2-2ln

2,最大值為e2-4.考向2數(shù)形結(jié)合確定零點(diǎn)個(gè)數(shù)例2(2024·江西贛州模擬)已知函數(shù)f(x)=.(1)求函數(shù)f(x)的最值;(2)討論函數(shù)g(x)=aex-ln

x-1的零點(diǎn)個(gè)數(shù).(2)函數(shù)g(x)=aex-ln

x-1的零點(diǎn)個(gè)數(shù)就是方程aex-ln

x-1=0的解的個(gè)數(shù),整理[對(duì)點(diǎn)訓(xùn)練2](2024·重慶巴蜀期末)已知函數(shù)f(x)=ax-ln

x-2.(1)當(dāng)a=1時(shí),求函數(shù)f(x)的極值;(2)討論函數(shù)f(x)的零點(diǎn)個(gè)數(shù).解

(1)當(dāng)a=1時(shí),f(x)=x-ln

x-2(x>0),f'(x)=1-(x>0),令f'(x)>0,則x>1;令f'(x)<0,則0<x<1.故函數(shù)f(x)的單調(diào)遞增區(qū)間是(1,+∞),單調(diào)遞減區(qū)間為(0,1),當(dāng)x=1時(shí),函數(shù)取極小值f(1)=1-ln

1-2=-1,無極大值.作出函數(shù)g(x)的圖象(如圖所示),因此當(dāng)a>e時(shí),直線y=a與y=g(x)的圖象沒有交點(diǎn);當(dāng)a=e或a≤0時(shí),直線y=a與y=g(x)的圖象有1個(gè)交點(diǎn);當(dāng)0<a<e時(shí),直線y=a與y=g(x)的圖象有2個(gè)交點(diǎn).綜上,當(dāng)a>e時(shí),函數(shù)f(x)沒有零點(diǎn);當(dāng)a=e或a≤0時(shí),函數(shù)f(x)有1個(gè)零點(diǎn);當(dāng)0<a<e時(shí),函數(shù)f(x)有2個(gè)零點(diǎn).考點(diǎn)二已知函數(shù)零點(diǎn)個(gè)數(shù)求參數(shù)的取值范圍例3(12分)(2024·江蘇常州模擬)已知函數(shù)f(x)=1+ln

x+2ax(a∈R).(1)討論函數(shù)f(x)的單調(diào)性;突破口:結(jié)合定義域?qū)分類討論.(2)若f(x)存在兩個(gè)零點(diǎn),求a的取值范圍.關(guān)鍵點(diǎn):先結(jié)合(1)中函數(shù)單調(diào)性得到最值,確定參數(shù)a的取值范圍,然后借助零點(diǎn)存在定理進(jìn)行推理驗(yàn)證.審題指導(dǎo):(1)確定函數(shù)定義域,求導(dǎo)后,從x>0入手,分類討論f'(x)>0解的情況,從而確定函數(shù)單調(diào)性;(2)首先由(1)中函數(shù)的單調(diào)性得到函數(shù)最值,確定參數(shù)a的取值范圍,然后再尋找恰當(dāng)?shù)膮^(qū)間端點(diǎn),根據(jù)零點(diǎn)存在定理確定參數(shù)的取值范圍.不能忽視函數(shù)的定義域

不能漏掉這種情形

分類討論后對(duì)結(jié)果進(jìn)行綜述

借助(1)問的結(jié)果

尋找恰當(dāng)合理的區(qū)間端點(diǎn)值

構(gòu)造函數(shù)

根據(jù)零點(diǎn)存在定理判定

[對(duì)點(diǎn)訓(xùn)練3](12分)(2024·遼寧錦州模擬)已知函數(shù)f(x)=x3-aln

x.(1)當(dāng)a=1時(shí),求函數(shù)f(x)的極值;(2)函數(shù)f(x)在區(qū)間(1,e]上存在兩個(gè)不同的零點(diǎn),求實(shí)數(shù)a的取值范圍.考點(diǎn)三可化為函數(shù)零點(diǎn)的參數(shù)問題例4(2023·北京,20)設(shè)函數(shù)f(x)=x-x3eax+b,曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程為y=-x+1.(1)求a,b的值;(2)設(shè)函數(shù)g(x)=f'(x),求g(x)的單調(diào)區(qū)間;(3)求f(x)的極值點(diǎn)個(gè)數(shù).解

(1)因?yàn)閒(x)=x-x3eax+b,所以f'(x)=1-(3x2+ax3)eax+b,因?yàn)榍€y=f(x)在(1,f(1))處的切線方程為y=-x+1,所以f(1)=-1+1=0,f'(1)=-1,則(3)由(1)得f(x)=x-x3e-x+1(x∈R),f'(x)=1-(3x2-x3)e-x+1,由(2)知f'(x)在(0,x1),(x2,+∞)上單調(diào)遞減,在(-∞,0),(x1,x2)上單調(diào)遞增,當(dāng)x<0時(shí),f'(-1)=1-4e2<0,f'(0)=1>0,即f'(-1)f'(0)<0,所以f'(x)在(-∞,0)上存在唯一零點(diǎn),不妨設(shè)為x3,則-1<x3<0,此時(shí)當(dāng)x<x3時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)x3<x<0時(shí),f'(x)>0,f(x)單調(diào)遞增;所以f(x)在(-∞,0)上有一個(gè)極小值點(diǎn).當(dāng)x∈(0,x1)時(shí),f'(x)在(0,x1)上單調(diào)遞減,則f'(x1)=f'(3-)<f'(1)=1-2<0,故f'(0)f'(x1)<0,所以f'(x)在(0,x1)內(nèi)存在唯一零點(diǎn),不妨設(shè)為x4,則0<x4<x1,此時(shí)當(dāng)0<x<x4時(shí),f'(x)>0,f(x)單調(diào)遞增;當(dāng)x4<x<x1時(shí),f'(x)<0,f(x)單調(diào)遞減,所以f(x)在(0,x1)上有一個(gè)極大值點(diǎn).當(dāng)x∈(x1,x2)時(shí),f'(x)在(x1,x2)上單調(diào)遞增,則f'(x2)=f'(3+)>f'(3)=1>0,故f'(x1)f'(x2)<0,所以f'(x)在(x1,x2)上存在唯一零點(diǎn),不妨設(shè)為x5,則x1<x5<x2,此時(shí)當(dāng)x1<x<x5時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)x5<x<x2時(shí),f'(x)>0,f(x)單調(diào)遞增,所以f(x)在(x1,x2)上有一個(gè)極小值點(diǎn);當(dāng)x>x2=3+>3時(shí),3x2-x3=x2(3-x)<0,所以f'(x)=1-(3x2-x3)e-x+1>0,則f(x)單調(diào)遞增,所以f(x)在(x2,+∞)上無極值點(diǎn).綜上可知,f(x)在(-∞,0)和(x1,x2)上各有一個(gè)極小值點(diǎn),在(0,x1)上有一個(gè)極大值點(diǎn),共有3個(gè)極值點(diǎn).[對(duì)點(diǎn)訓(xùn)練4](2024·廣東深圳模擬)已知函數(shù)f(x)=xex+ax2(a∈R).(1)當(dāng)a=時(shí),求曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程;(2)若函數(shù)g(x)=xln

x+xex-f(x)有兩個(gè)極值