高考總復(fù)習(xí)優(yōu)化設(shè)計(jì)GAOKAOZONGFUXIYOUHUASHEJI第3課時(shí)利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)高考解答題專項(xiàng)一2026考點(diǎn)一確定函數(shù)零點(diǎn)的個(gè)數(shù)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.已知函數(shù)f(x)=2lnx+mx2-(2m+1)x,其中m∈R.(1)若函數(shù)y=f(x)圖象僅有一條垂直于y軸的切線,求m的取值范圍;(2)討論函數(shù)f(x)的零點(diǎn)個(gè)數(shù).①當(dāng)m≤0時(shí),∵x>0,∴mx-1<0,∴當(dāng)x∈(0,2)時(shí),f'(x)>0,f(x)單調(diào)遞增;當(dāng)x∈(2,+∞)時(shí),f'(x)<0,f(x)單調(diào)遞減,此時(shí)f(x)max=f(2)=2ln

2+2m-2(2m+1)=2ln

2-2m-2.當(dāng)f(2)=0,m=ln

2-1時(shí),f(x)只有一個(gè)零點(diǎn)x=2;當(dāng)f(2)<0,ln

2-1<m≤0時(shí),f(x)沒有零點(diǎn);當(dāng)f(2)>0,m<ln

2-1<0時(shí),∵當(dāng)x>0且x→0或x→+∞時(shí),f(x)→-∞,∴f(x)分別在(0,2)和(2,+∞)上各有唯一零點(diǎn),此時(shí)f(x)有兩個(gè)零點(diǎn).∴f(x)在(0,+∞)上單調(diào)遞增.當(dāng)x>0且x→0時(shí),f(x)→-∞;當(dāng)x→+∞時(shí),f(x)→+∞,∴f(x)在(0,+∞)上有唯一零點(diǎn).又∵當(dāng)x→+∞時(shí),f(x)→+∞,∴函數(shù)f(x)在區(qū)間(2,+∞)上有唯一零點(diǎn).綜上所述,當(dāng)m<ln

2-1時(shí),函數(shù)f(x)有且僅有2個(gè)零點(diǎn);當(dāng)ln

2-1<m≤0時(shí),函數(shù)f(x)沒有零點(diǎn);當(dāng)m=ln

2-1或m>0時(shí),函數(shù)f(x)有且僅有1個(gè)零點(diǎn).方法點(diǎn)撥利用單調(diào)性和函數(shù)零點(diǎn)存在定理確定零點(diǎn)個(gè)數(shù)(1)討論函數(shù)的單調(diào)性,確定函數(shù)的單調(diào)區(qū)間;(2)在每個(gè)單調(diào)區(qū)間上,利用函數(shù)零點(diǎn)存在定理判斷零點(diǎn)的個(gè)數(shù);(3)注意區(qū)間端點(diǎn)的選取技巧;(4)含參數(shù)時(shí)注意分類討論.對(duì)點(diǎn)訓(xùn)練1已知函數(shù)f(x)=-lnx+2x-2.(1)求曲線y=f(x)的斜率等于1的切線方程;(2)求函數(shù)f(x)的極值;(3)設(shè)g(x)=x2f(x)-2f(x),判斷函數(shù)g(x)的零點(diǎn)個(gè)數(shù),并說(shuō)明理由.x0=1,所以y0=-ln

1+2-2=0,故切線方程為y=x-1.考向2.利用兩個(gè)函數(shù)圖象的交點(diǎn)確定零點(diǎn)個(gè)數(shù)例2.已知函數(shù)f(x)=axex,a為非零實(shí)數(shù).(1)求函數(shù)f(x)的單調(diào)遞減區(qū)間;(2)討論方程f(x)=(x+1)2的實(shí)數(shù)解的個(gè)數(shù).解(1)f'(x)=aex+axex=a(x+1)ex,令f'(x)=0,得x=-1.①當(dāng)a>0時(shí),令f'(x)<0,得x<-1,所以f(x)的單調(diào)遞減區(qū)間為(-∞,-1);②當(dāng)a<0時(shí),令f'(x)<0,得x>-1,所以f(x)的單調(diào)遞減區(qū)間為(-1,+∞).綜上所述,當(dāng)a>0時(shí),f(x)的單調(diào)遞減區(qū)間為(-∞,-1);當(dāng)a<0時(shí),f(x)的單調(diào)遞減區(qū)間為(-1,+∞).當(dāng)x∈(-∞,-1)時(shí),g'(x)>0,函數(shù)g(x)單調(diào)遞增;當(dāng)x∈(-1,0)∪(0,+∞)時(shí),g‘(x)<0,函數(shù)g(x)單調(diào)遞減.又因?yàn)間(-1)=0,所以當(dāng)x∈(-∞,-1)時(shí),g(x)∈(-∞,0);當(dāng)x∈(-1,0)時(shí),g(x)∈(-∞,0);當(dāng)x∈(0,+∞)時(shí),g(x)∈(0,+∞).作出函數(shù)的圖象(如圖).所以,當(dāng)a>0時(shí),原方程有且只有一個(gè)解;當(dāng)a<0時(shí),原方程有兩個(gè)解.誤區(qū)警示在借助函數(shù)圖象研究函數(shù)零點(diǎn)問(wèn)題時(shí),要準(zhǔn)確畫出函數(shù)的圖象,不僅要研究函數(shù)的單調(diào)性與極值的情況,還要關(guān)注函數(shù)值的正負(fù)以及變化趨勢(shì),把函數(shù)圖象與x軸有無(wú)交點(diǎn),哪一區(qū)間在x軸上方,哪一區(qū)間在x軸下方等情況分析清楚,這樣才能準(zhǔn)確地研究直線與圖象交點(diǎn)的個(gè)數(shù)情況.對(duì)點(diǎn)訓(xùn)練2已知函數(shù)f(x)=ax2+(a-2)x-xlnx.(1)設(shè)a=0.①求曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程;②試問(wèn)f(x)有極大值還是極小值?并求出該極值.(2)若f(x)在(0,e)上恰有兩個(gè)零點(diǎn),求a的取值范圍.解(1)∵a=0,∴f(x)=-2x-xln

x,定義域?yàn)?0,+∞),f'(x)=-3-ln

x.①由f'(1)=-3及f(1)=-2,得y=f(x)在點(diǎn)(1,f(1))處的切線方程為y+2=-3(x-1),即y=-3x+1.②令f'(x)>0,得0<x<e-3;令f'(x)<0,得x>e-3.∴f(x)在(0,e-3)內(nèi)單調(diào)遞增,在(e-3,+∞)上單調(diào)遞減,∴f(x)在x=e-3處取得極大值,且極大值為f(e-3)=e-3,f(x)沒有極小值.(2)(方法1)由f(x)=0,得ax-ln

x+a-2=0,則a(x+1)=ln

x+2.∴當(dāng)0<x<1時(shí),φ(x)>0,當(dāng)1<x<e時(shí),φ(x)<0,∴h(x)在(0,1)內(nèi)單調(diào)遞增,在(1,e)內(nèi)單調(diào)遞減,則h(x)max=h(1)=1.考點(diǎn)二已知函數(shù)零點(diǎn)個(gè)數(shù)求參數(shù)取值范圍例3.已知函數(shù)f(x)=ex-a(x+2).(1)當(dāng)a=1時(shí),討論f(x)的單調(diào)性;(2)若f(x)有兩個(gè)零點(diǎn),求a的取值范圍.解(1)當(dāng)a=1時(shí),f(x)=ex-x-2,則f'(x)=ex-1.當(dāng)x<0時(shí),f'(x)<0;當(dāng)x>0時(shí),f'(x)>0.所以f(x)在區(qū)間(-∞,0)上單調(diào)遞減,在區(qū)間(0,+∞)上單調(diào)遞增.(2)f'(x)=ex-a.當(dāng)a≤0時(shí),f'(x)>0,所以f(x)在區(qū)間(-∞,+∞)上單調(diào)遞增,故f(x)至多存在1個(gè)零點(diǎn),不符合題意.當(dāng)a>0時(shí),令f'(x)=0,得x=ln

a.當(dāng)x∈(-∞,ln

a)時(shí),f'(x)<0;當(dāng)x∈(ln

a,+∞)時(shí),f'(x)>0.所以f(x)在區(qū)間(-∞,ln

a)上單調(diào)遞減,在區(qū)間(ln

a,+∞)上單調(diào)遞增,故當(dāng)x=ln

a時(shí),f(x)取得最小值,最小值為f(ln

a)=-a(1+ln

a).因?yàn)閒(-2)=e-2>0,所以f(x)在區(qū)間(-∞,ln

a)上存在唯一零點(diǎn).由(1)得,當(dāng)x>2時(shí),ex-x-2>0,方法總結(jié)已知函數(shù)零點(diǎn)個(gè)數(shù)求參數(shù)取值范圍問(wèn)題的解法

對(duì)點(diǎn)訓(xùn)練3已知函數(shù)f(x)=(x-1)ex+ax2,a∈R.(1)若a<0,討論函數(shù)f(x)的單調(diào)性;(2)若函數(shù)f(x)有2個(gè)不同的零點(diǎn),求實(shí)數(shù)a的取值范圍.解(1)f'(x)=xex+2ax=x(ex+2a).已知a<0,令f'(x)=0,得x1=0,x2=ln(-2a).h'(x)<0,h(x)單調(diào)遞減;當(dāng)x>0時(shí),h'(x)>0,h(x)單調(diào)遞增.因?yàn)閔(1)=0,所以當(dāng)x<1時(shí),h(x)<0.若f(x)有2個(gè)不同的零點(diǎn),則函數(shù)h(x)的圖象與直線y=-a有兩個(gè)交點(diǎn),即-a<0,解得a>0,故實(shí)數(shù)a的取值范圍為(0,+∞).考點(diǎn)三:隱零點(diǎn)問(wèn)題例4.(2024江蘇宿遷模擬)已知函數(shù)

.(1)若a=0,求f(x)的極值;(2)若f(x)≥1恒成立,求實(shí)數(shù)a的取值范圍.(2)由題意可得axex+1-ln

x≥x,x>0.因?yàn)閔'(x)=aex·x+aex=aex(x+1)>0,所以h(x)在(0,+∞)上單調(diào)遞增,所以當(dāng)x∈(0,x0)時(shí),h(x)<0;當(dāng)x∈(x0,+∞)時(shí),h(x)>0,則當(dāng)x∈(0,x0)時(shí),g'(x)<0;當(dāng)x∈(x0,+∞)時(shí),g'(x)>0,則g(x)在(0,x0)內(nèi)單調(diào)遞減,在(x0,+∞)上單調(diào)遞增,方法總結(jié)隱零點(diǎn)的處理思路第一步,用零點(diǎn)存在定理判定導(dǎo)函數(shù)零點(diǎn)的存在性,其中難點(diǎn)是通過(guò)合理賦值,敏銳捕捉零點(diǎn)存在的區(qū)間,有時(shí)還需結(jié)合函數(shù)的單調(diào)性明確零點(diǎn)的個(gè)數(shù);第二步,虛設(shè)零點(diǎn)并確定取值范圍,抓住零點(diǎn)方程實(shí)施代換,如指數(shù)與對(duì)數(shù)互換,超越函數(shù)與簡(jiǎn)單函數(shù)的替換,利用同構(gòu)思想等解決,需要注意的是代換可能不止一次.對(duì)點(diǎn)訓(xùn)練4(2024河北邢臺(tái)高三統(tǒng)考期末)已知函數(shù)f(x)=sinx+x2.(2)證明由(