高考數(shù)學(xué)二輪復(fù)習(xí)專題1函數(shù)與導(dǎo)數(shù)專項(xiàng)突破1突破3利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)課件

突破3利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)在近幾年的高考中,函數(shù)與方程、不等式的交匯是考查的熱點(diǎn),常以指數(shù)函數(shù)、對數(shù)函數(shù)以及三角函數(shù)為載體考查函數(shù)的零點(diǎn)(方程的根)問題,難度較大,多以壓軸題出現(xiàn).利用導(dǎo)數(shù)研究函數(shù)零點(diǎn)的方法(1)通過最值(極值)判斷零點(diǎn)個(gè)數(shù)的方法借助導(dǎo)數(shù)研究函數(shù)的單調(diào)性、極值后,通過極值的正負(fù)、函數(shù)單調(diào)性判斷函數(shù)圖象走勢,從而判斷零點(diǎn)個(gè)數(shù)或者通過零點(diǎn)個(gè)數(shù)求參數(shù)范圍.(2)數(shù)形結(jié)合法求解零點(diǎn)對于方程解的個(gè)數(shù)(或函數(shù)零點(diǎn)個(gè)數(shù))問題,可利用函數(shù)的值域或最值,結(jié)合函數(shù)的單調(diào)性,畫出草圖,數(shù)形結(jié)合確定其中參數(shù)的范圍.(3)構(gòu)造函數(shù)法研究函數(shù)零點(diǎn)①根據(jù)條件構(gòu)造某個(gè)函數(shù),利用導(dǎo)數(shù)確定函數(shù)的單調(diào)區(qū)間及極值點(diǎn),根據(jù)函數(shù)零點(diǎn)的個(gè)數(shù)尋找函數(shù)在給定區(qū)間的極值以及區(qū)間端點(diǎn)的函數(shù)值與0的關(guān)系,從而求解.②解決此類問題的關(guān)鍵是將函數(shù)零點(diǎn)、方程的根、曲線交點(diǎn)相互轉(zhuǎn)化,突出導(dǎo)數(shù)的工具作用,體現(xiàn)轉(zhuǎn)化與化歸的思想方法.考點(diǎn)一判斷或證明函數(shù)零點(diǎn)的個(gè)數(shù)例1(2024江西南昌模擬)已知函數(shù)f(x)=x(x-3)+(x+2)ex.(1)求函數(shù)f(x)的最小值;(2)若g(x)=f(x)+x(3-x)+lnx+ex(x2-3x-1),求函數(shù)g(x)的零點(diǎn)個(gè)數(shù).解

(1)(方法一)因?yàn)閒(x)=(x+2)ex+x2-3x,所以f'(x)=(x+3)ex+2x-3.記m(x)=f'(x),則m'(x)=(x+4)ex+2,①當(dāng)x≤-3時(shí),(x+3)ex≤0,2x-3≤-9,可得f'(x)<0,故函數(shù)f(x)在區(qū)間(-∞,-3]上單調(diào)遞減.②當(dāng)x>-3時(shí),(x+4)ex>0,m'(x)>0,可知函數(shù)m(x)單調(diào)遞增,又m(0)=0,所以當(dāng)-3<x<0時(shí),m(x)<0;當(dāng)x>0時(shí),m(x)>0,所以函數(shù)f(x)在區(qū)間(-3,0)內(nèi)單調(diào)遞減,在區(qū)間(0,+∞)內(nèi)單調(diào)遞增.由①②知函數(shù)f(x)的單調(diào)遞減區(qū)間為(-∞,0),單調(diào)遞增區(qū)間為(0,+∞),故f(x)min=f(0)=2.(方法二)因?yàn)閒(x)=(x+2)ex+x2-3x,所以f'(x)=(x+3)ex+2x-3.令m(x)=f'(x),則m'(x)=(x+4)ex+2,令t(x)=m'(x),則t'(x)=(x+5)ex,當(dāng)x<-5時(shí),t'(x)<0,t(x)單調(diào)遞減,當(dāng)x>-5時(shí),t'(x)>0,t(x)單調(diào)遞增,故t(x)min=t(-5)=-e-5+2>0,所以m'(x)>0,故f'(x)在定義域上單調(diào)遞增.易知f'(0)=0,故當(dāng)x<0時(shí),f'(x)<0,f(x)單調(diào)遞減,當(dāng)x>0時(shí),f'(x)>0,f(x)單調(diào)遞增,故f(x)min=f(0)=2.當(dāng)0<x<x0時(shí),h(x)<0,g'(x)>0;當(dāng)x0<x<1時(shí),h(x)>0,g'(x)<0;當(dāng)x>1時(shí),h(x)>0,g'(x)>0,所以g(x)在區(qū)間(0,x0)內(nèi)單調(diào)遞增,在區(qū)間(x0,1)內(nèi)單調(diào)遞減,在區(qū)間(1,+∞)內(nèi)單調(diào)遞增,[對點(diǎn)訓(xùn)練1](2024福建寧德模擬)已知函數(shù)f(x)=ex-4sinx,其中e為自然對數(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,+∞)內(nèi)單調(diào)遞增.考點(diǎn)二根據(jù)函數(shù)零點(diǎn)的個(gè)數(shù)求參數(shù)范圍例2(2024江蘇常州模擬)已知函數(shù)f(x)=1+lnx+2ax(a∈R).(1)討論函數(shù)f(x)的單調(diào)性;(2)若f(x)存在兩個(gè)零點(diǎn),求a的取值范圍.[對點(diǎn)訓(xùn)練2](2024北京懷柔模擬)已知函數(shù)f(x)=2ax-ex2(a>0且a≠1).(1)當(dāng)a=e時(shí),求函數(shù)f(x)在點(diǎn)(0,f(0))處的切線方程;(2)若函數(shù)f(x)存在兩個(gè)極值點(diǎn),設(shè)x1是極小值點(diǎn),x2是極大值點(diǎn),若x1<x2,求實(shí)數(shù)a的取值范圍.解

(1)當(dāng)a=e時(shí),f(x)=2ex-ex2,f(0)=2,f'(x)=2ex-2ex,f'(0)=2,所以函數(shù)f(x)在點(diǎn)(0,f(0))處的切線方程為y-2=2(x-0),即y=2x+2.(2)f(x)=2ax-ex2,f'(x)=2(ax·ln

a-ex),設(shè)g(x)=ax·ln

a-ex,由題設(shè)g(x)有兩個(gè)變號零點(diǎn)x1,x2,且在x1的左側(cè)附近,g(x)<0,在x1的右側(cè)附近,g(x)>0,在x2的左側(cè)附近,g(x)>0,在x2的右側(cè)附近,g(x)<0.又g'(x)=ax·(ln

a)2-e,若a>1,則g'(x)為R上的增函數(shù),當(dāng)x∈(-∞,u)時(shí),g'(x)<0,當(dāng)x∈(u,+∞)時(shí),g'(x)>0,故g(x)在(-∞,u)內(nèi)單調(diào)遞減,在(u,+∞)內(nèi)單調(diào)遞增.因?yàn)間(x)有兩個(gè)變號零點(diǎn)x1,x2(x1<x2),故g(u)<0,且x1<u<x2,g(x1)=0,此時(shí)在x1的左側(cè)附近,g(x)>0,在x1的右側(cè)附近,g(x)<0,這與題設(shè)矛盾,故0<a<1,g'(x)為R上的減函數(shù),此時(shí)當(dāng)x∈(-∞,u)時(shí),g'(x)>0,當(dāng)x∈(u,+∞)時(shí),g'(x)<0,故g(x)在(-∞,u)內(nèi)單調(diào)遞增,在(u,+∞)內(nèi)單調(diào)遞減,故g(u)>0,考點(diǎn)三函數(shù)零點(diǎn)(方程根)的證明問題例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(diǎn)x0,證明:1<x0<2.(1)解

f(x)的定義域是(0,+∞),∴當(dāng)x>0時(shí),l(x)即f'(x)為增函數(shù),又f'(1)=0,∴當(dāng)x∈(0,1)時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)x∈(1,+∞)時(shí),f'(x)>0,f(x)單調(diào)遞增.∴f(x)的單調(diào)遞減區(qū)間為(0,1),單調(diào)遞增區(qū)間為(1,+∞).(2)證明

∵g(x)=f(x)-m(x-1)=ex-1-ln

x-mx+m(m>0),∴g'(x)=ex-1--m(x>0,m>0).由(1)可知g'(x)在(0,+∞)內(nèi)單調(diào)遞增,且g'(1)=-m<0,又g'(1+m)=em--m>em-(m+1).設(shè)φ(x)=ex-(x+1),則φ'(x)=ex-1,當(dāng)x>0時(shí),φ'(x)>0,φ(x)單調(diào)遞增,當(dāng)x<0時(shí),φ'(x)<0,φ(x)單調(diào)遞減,∴φ(x)≥φ(0)=0,當(dāng)且僅當(dāng)x=0時(shí)等號成立,∴g'(1+m)>em-(m+1)>0.∴存在唯一的t∈(1,1+m)?(1,+∞),使得g'(t)=0.∴當(dāng)x∈(0,t)時(shí),g'(t)<0,g(x)單調(diào)遞減;當(dāng)x∈(t,+∞)時(shí),g'(t)>0,g(x)單調(diào)遞增,∴g(x)min=g(t)=et-1-ln

t-mt+m.設(shè)p(x)=ex-x2,則p'(x)=ex-2x,設(shè)q(x)=ex-2x,則q'(x)=ex-2,由ex-2<0,得x<ln

2,此時(shí)q(x)單調(diào)遞減,由ex-2>0,得x>ln

2,此時(shí)q(x)單調(diào)遞增.則q(x)min=q(ln

2)=2-2ln

2>0,則p'(x)>0,∴p(x)=ex-x2單調(diào)遞增,當(dāng)x>0時(shí),p(x)>p(0)=1,∴ex-x2>0,即ex>x2.設(shè)k(x)=ln

x-x+1,則k'(x)=由k'(x)<0,得x>1,此時(shí)k(x)單調(diào)遞減,由k'(x)>0,得0<x<1,此時(shí)k(x)單調(diào)遞增,則k(x)max=k(1)=0,∴k(x)=ln

x-x+1≤0,即ln

x≤x-1,當(dāng)且僅當(dāng)x=1時(shí)等號成立.[對點(diǎn)訓(xùn)練3](2024江蘇徐州模擬)已知函數(shù)f(x)=ex+x2-x(lnx+a-1),其中a∈R,e為自然對數(shù)的底數(shù).當(dāng)x>1時(shí),g'(x)>0,g(x)單調(diào)遞增,當(dāng)0<x<1時(shí),g'(x)<0,g(x)單調(diào)遞減,所以g(x)的最小值為g(1)=e+2-a,故φ(a)=g(1)=e+2-a.(2)證明

由于x1,x2(x1<x2)為函數(shù)f(x)的兩個(gè)零點(diǎn),所以x1,x2(x1<x2)也是g(x)的兩個(gè)零點(diǎn),故φ(a)=g(1)=e+2-