解答題專項(xiàng)一　第3課時(shí)　利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)

第3課時(shí)利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)第四章解答題專項(xiàng)一考點(diǎn)一確定函數(shù)零點(diǎn)的個(gè)數(shù)(多考向探究預(yù)測)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)例題(2022·山東棗莊一模)已知函數(shù)f(x)=xex-asinx(a∈R).(1)若?x∈[0,π],f(x)≥0,求a的取值范圍;(2)當(dāng)a≥-59時(shí),試討論f(x)在(0,2π)內(nèi)零點(diǎn)的個(gè)數(shù),并說明理由.解

(1)f'(x)=(x+1)ex-acos

x,①若a≤0,當(dāng)x∈[0,π]時(shí),-a≥0,sin

x≥0,f(x)=xex+(-a)sin

x≥0,當(dāng)且僅當(dāng)x=0時(shí)取等號,可見,a≤0符合題意.可見,當(dāng)x∈[0,π]時(shí),f'(x)≥0,當(dāng)且僅當(dāng)a=1,且x=0時(shí)取等號.所以f(x)在[0,π]上單調(diào)遞增,所以f(x)≥f(0)=0.所以0<a≤1符合題意.③若a>1,因?yàn)閥=(x+1)ex在[0,π]上單調(diào)遞增,y=-acos

x在[0,π]上單調(diào)遞增,所以f'(x)=(x+1)ex-acos

x在[0,π]上單調(diào)遞增,當(dāng)x∈(0,x0)時(shí),f'(x)<f'(x0)=0,f(x)單調(diào)遞減,所以f(x)<f(0)=0.可見,a>1不符合題意.綜上,a的取值范圍是(-∞,1].(2)①若-59≤a≤0,由(1)知,當(dāng)x∈(0,π]時(shí),f(x)>0,f(x)在(0,π]內(nèi)無零點(diǎn).當(dāng)x∈(π,2π)時(shí),-1≤sin

x<0,0<-sin

x≤1,-asin

x≥a,又由y=xex單調(diào)遞增,則f(x)=xex-asin

x>πeπ+a>3e3-59>3×2.73-59=0.049>0.可見,若-59≤a≤0,f(x)在(0,2π)內(nèi)無零點(diǎn).②若0<a≤1,由(1)知,當(dāng)x∈(0,π]時(shí),f(x)>0,f(x)在(0,π]內(nèi)無零點(diǎn).當(dāng)x∈(π,2π)時(shí),-sin

x>0,f(x)=xex+a(-sin

x)>xex>0.可見,若0<a≤1,f(x)在(0,2π)內(nèi)無零點(diǎn).當(dāng)x∈(0,x0)時(shí),f'(x)<f'(x0)=0,f(x)單調(diào)遞減;當(dāng)x∈(x0,π)時(shí),f'(x)>f'(x0)=0,f(x)單調(diào)遞增.又f(0)=0,所以f(x0)<f(0)=0.又f(π)=πeπ>0,由零點(diǎn)存在定理及f(x)的單調(diào)性可知,存在唯一的x1∈(x0,π),使得f(x1)=0.可見,f(x)在(0,π]內(nèi)存在唯一的零點(diǎn).當(dāng)x∈(π,2π)時(shí),sin

x<0,-asin

x>0,所以f(x)=xex-asin

x>xex>0,所以f(x)在(π,2π)內(nèi)沒有零點(diǎn).可見,f(x)在(0,2π)有且僅有1個(gè)零點(diǎn).綜上所述,若-59≤a≤1,f(x)在(0,2π)內(nèi)無零點(diǎn);若a>1,f(x)在(0,2π)內(nèi)有且僅有1個(gè)零點(diǎn).規(guī)律方法

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.對點(diǎn)訓(xùn)練(2022·廣東汕頭三模)已知函數(shù)f(x)=x-2sinx.(1)求f(x)在(0,π)的極值;(2)證明:函數(shù)g(x)=lnx-f(x)在(0,π)有且只有兩個(gè)零點(diǎn).使得φ(x0)=g'(x0)=0.當(dāng)x變化時(shí),g(x),g'(x)變化如下表:x(0,x0)x0(x0,π)g'(x)+0-g(x)單調(diào)遞增極大值g(x0)單調(diào)遞減考向2數(shù)形結(jié)合法確定零點(diǎn)個(gè)數(shù)例題(2022·山東日照三模)已知函數(shù)f(x)=(x-2)ex-ax+alnx(a∈R).(1)當(dāng)a=-1時(shí),求函數(shù)f(x)的單調(diào)區(qū)間;(2)當(dāng)a<e時(shí),討論f(x)的零點(diǎn)個(gè)數(shù).解

(1)當(dāng)a=-1時(shí),f(x)=(x-2)ex+x-ln

x,所以當(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,即f(x)的單調(diào)遞減區(qū)間是(0,1),單調(diào)遞增區(qū)間是(1,+∞).(2)由題意,函數(shù)f(x)=(x-2)ex-ax+aln

x=(x-2)ex-a(x-ln

x),x>0,當(dāng)x∈(0,1)時(shí),m'(x)<0,m(x)單調(diào)遞減;當(dāng)x∈(1,+∞)時(shí),m'(x)>0,m(x)單調(diào)遞增,又由m(1)=1,所以m(x)≥1.令f(x)=0,可得(x-2)ex-ax+aln

x=0,當(dāng)x>2時(shí),h'(x)>0,h(x)單調(diào)遞增.所以h(x)min=h(2)=2-ln

2>0,即x>0時(shí),h(x)>0恒成立.故當(dāng)0<x<1時(shí),g'(x)<0,g(x)單調(diào)遞減;當(dāng)x>1時(shí),g'(x)>0,g(x)單調(diào)遞增.所以g(x)min=g(1)=-e﹐又由x→0時(shí),g(x)→0,當(dāng)x→+∞時(shí),g(x)→+∞,畫出函數(shù)g(x)的示意圖象,如圖所示,結(jié)合圖象可得:當(dāng)a<-e時(shí),無零點(diǎn);當(dāng)a=-e或0≤a<e時(shí),有一個(gè)零點(diǎn);當(dāng)-e<a<0時(shí),有兩個(gè)零點(diǎn).規(guī)律方法

含參數(shù)的函數(shù)零點(diǎn)個(gè)數(shù),可轉(zhuǎn)化為方程解的個(gè)數(shù).若能分離參數(shù),可將參數(shù)分離出來后,用x表示參數(shù)的函數(shù),作出該函數(shù)的圖象,根據(jù)圖象特征求參數(shù)的范圍.當(dāng)x∈(1,+∞)時(shí),h(x)<0.∴h(x)的大致圖象如下圖:∴當(dāng)0<a<e時(shí),f(x)零點(diǎn)個(gè)數(shù)為0;當(dāng)a=e時(shí),f(x)零點(diǎn)個(gè)數(shù)為1;當(dāng)a>e時(shí),f(x)零點(diǎn)個(gè)數(shù)為2.考點(diǎn)二已知函數(shù)的零點(diǎn)個(gè)數(shù)求參數(shù)的取值范圍例題(12分)(2022·全國乙,理21)已知函數(shù)f(x)=ln(1+x)+axe-x.(1)當(dāng)a=1時(shí),求曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程;(2)若f(x)在區(qū)間(-1,0),(0,+∞)各恰有一個(gè)零點(diǎn),求a的取值范圍.【教師講評】

(1)第(1)問先算出切點(diǎn),再求導(dǎo)算出斜率,根據(jù)點(diǎn)斜式寫方程;(2)第(2)問先求導(dǎo),再對a分類討論,并將x分成(-1,0),(0,+∞)兩部分研究.本題的關(guān)鍵是對a的范圍進(jìn)行合理分類,否定和肯定并用,否定只需要找出一個(gè)特例即可,肯定則各方面都說明.規(guī)律方法

已知函數(shù)零點(diǎn)個(gè)數(shù)求參數(shù)取值范圍問題的解法

對點(diǎn)訓(xùn)練(12分)(2022·全國乙,文20)已知函數(shù)f(x)=ax--(a+1)lnx.(1)當(dāng)a=0時(shí),求f(x)的最大值;(2)若f(x)恰有一個(gè)零點(diǎn),求a的取值范圍.考點(diǎn)三可轉(zhuǎn)化為函數(shù)零點(diǎn)的函數(shù)問題例題已知函數(shù)f(x)=2ex-x2-2(x-1)(其中e為自然對數(shù)的底數(shù)).(1)求f(x)的單調(diào)區(qū)間;(2)若g(x)=f(x)+(a-2)ex有兩個(gè)極值點(diǎn),求實(shí)數(shù)a的取值范圍.解

(1)f(x)=2ex-x2-2(x-1),f'(x)=2(ex-x-1),令p(x)=f'(x),p'(x)=2(ex-1),令p'(x)>0,解得x>0,令p'(x)<0,解得x<0,所以p(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增,所以f'(x)=p(x)≥p(0)=0,所以f(x)的單調(diào)遞增區(qū)間為R,無單調(diào)遞減區(qū)間.(2)若g(x)=f(x)+(a-2)ex=aex-x2-2(x-1)有兩個(gè)極值點(diǎn),即g'(x)=aex-2x-2有兩個(gè)變號零點(diǎn).令h(x)=g'(x)=aex-2x-2,(ⅰ)當(dāng)a≤0時(shí),h(x)=aex-2x-2在R上單調(diào)遞減,最多只有一個(gè)零點(diǎn),不合題意;(ⅱ)當(dāng)a≥2時(shí),h(x)=aex-2x-2≥2(ex-x-1),由(1)可得ex≥x+1,知h(x)=aex-2x-2≥2(ex-x-1)≥0,最多只有一個(gè)零點(diǎn),不合題意.規(guī)律方法

若f'(x)為可導(dǎo)函數(shù)f(x)的導(dǎo)函數(shù),x0為f(x)的極值點(diǎn),則必有f'(x0)=0.曲線的切線條數(shù)、兩曲線的交點(diǎn)個(gè)數(shù)、極值點(diǎn)問題、方程根的個(gè)數(shù)等問題解決的關(guān)鍵是轉(zhuǎn)化為對應(yīng)函數(shù)的零點(diǎn)個(gè)數(shù)問題,通過數(shù)形結(jié)合等方式,研究函數(shù)的零點(diǎn)以確定相應(yīng)的數(shù)量關(guān)系.對點(diǎn)訓(xùn)練(2023·江蘇南京師大附中模擬)已知f(x)=ln(x+1)-ax(a∈R),g(x)=-sinx.(1)討論f(x)的單調(diào)性;(2)若函數(shù)f(x)與g(x)的圖象恰有一個(gè)交點(diǎn),求a的取值范圍.(2)函數(shù)f(x)與g(x)的圖象恰有一個(gè)交點(diǎn),等價(jià)于h(x)=f(x)-g(x)有一個(gè)零點(diǎn),h(x)=f(x)-g(x)=ln(x+1)-ax+sin

x,當(dāng)a≤0時(shí),當(dāng)-1<x<0時(shí),ln(x+1)<0,sin

x<0,則h(x)=ln(x+1)-ax+sin

x<0,當(dāng)0<x<π時(shí),ln(x+1)>0,sin

x>0,則h(x)=ln(x+1)-ax+sin

x>0,當(dāng)x≥π時(shí),ln(x+1)>1,ln(x+1)+sin

x>0,則h(x)=ln(x+1)-ax+sin

x>0,即h(x)除0之外無其他零點(diǎn),符合題意;則存在x0∈(0,π)使h'(x0)=0,即h(x)在(0,x0)上單調(diào)遞增,在