2022年秋高中數(shù)學(xué)第六章導(dǎo)數(shù)及其應(yīng)用6.2利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)6.2.2導(dǎo)數(shù)與函數(shù)的極值最值課后習(xí)題新人教B版選擇性必修第三冊(cè)

歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助！6.2.2導(dǎo)數(shù)與函數(shù)的極值、最值必備知識(shí)基礎(chǔ)練1.函數(shù)f(x)的定義域?yàn)殚_(kāi)區(qū)間(a,b),其導(dǎo)函數(shù)f'(x)在(a,b)內(nèi)的圖象如圖所示,則函數(shù)f(x)在開(kāi)區(qū)間(a,b)內(nèi)的極大值點(diǎn)的個(gè)數(shù)為()A.1 B.2 C.3 D.42.函數(shù)f(x)的導(dǎo)函數(shù)為f'(x)=-x(x+2),則函數(shù)f(x)有()A.最小值f(0) B.最小值f(-2)C.極大值f(0) D.極大值f(-2)3.函數(shù)f(x)=x2·ex+1,x∈[-2,1]的最大值為()A.4e-1 B.1 C.e2 D.3e24.當(dāng)x=1時(shí),三次函數(shù)有極大值4,當(dāng)x=3時(shí)有極小值0,且函數(shù)的圖象過(guò)原點(diǎn),則此函數(shù)是()A.f(x)=x3+6x2+9x B.f(x)=x3-6x2+9xC.f(x)=x3-6x2-9x D.f(x)=x3+6x2-9x5.設(shè)函數(shù)f(x)=exx+a,若f(x)的極小值為e,則A.-12 B.12 C.36.已知函數(shù)f(x)=x3+ax2+bx+1,曲線y=f(x)在點(diǎn)(1,f(1))處的切線斜率為3,且x=23是y=f(x)的極值點(diǎn),則a+b=7.若函數(shù)f(x)=12x2-x+alnx有兩個(gè)不同的極值點(diǎn),則實(shí)數(shù)a的取值范圍是8.函數(shù)f(x)=x3-3x2-9x+k在區(qū)間[-4,4]上的最大值為10,則其最小值為.

9.定義在0,π2的函數(shù)f(x)=8sinx-tanx的最大值為.

10.已知函數(shù)f(x)=13x3+a2x2+ax+b,當(dāng)x=-1時(shí),函數(shù)f(x)的極值為-712,則f(2)=11.設(shè)函數(shù)f(x)=ln(2x+3)+x2.(1)討論f(x)的單調(diào)性;(2)求f(x)在區(qū)間-34關(guān)鍵能力提升練12.(多選題)關(guān)于函數(shù)f(x)=ex-2,下列結(jié)論不正確的是()A.f(x)沒(méi)有零點(diǎn) B.f(x)沒(méi)有極值點(diǎn)C.f(x)有極大值點(diǎn) D.f(x)有極小值點(diǎn)13.已知函數(shù)y=f(x)在R上可導(dǎo)且f(0)=2,其導(dǎo)函數(shù)f'(x)滿足f'(x)-f(x)x-2>0,若函數(shù)g(x)滿足exA.函數(shù)g(x)在(2,+∞)上單調(diào)遞增B.x=2是函數(shù)g(x)的極小值點(diǎn)C.當(dāng)x≤0時(shí),不等式f(x)≤2ex恒成立D.函數(shù)g(x)至多有兩個(gè)零點(diǎn)14.函數(shù)f(x)=4x-lnx的最小值為()A.1+2ln2 B.1-2ln2C.1+ln2 D.1-ln215.已知函數(shù)f(x)=ax2+2lnx,若當(dāng)a>0時(shí),f(x)≥2恒成立,則實(shí)數(shù)a的取值范圍是16.設(shè)f(x)=alnx+12x+32x+1,其中a∈R,曲線y=f(x)在點(diǎn)(1,(1)求a的值;(2)求函數(shù)f(x)的極值.17.已知函數(shù)f(x)=x3+klnx(k∈R),f'(x)為f(x)的導(dǎo)函數(shù).(1)當(dāng)k=6時(shí),①求曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程;②求函數(shù)g(x)=f(x)-f'(x)+9x(2)當(dāng)k≥-3時(shí),求證:對(duì)任意的x1,x2∈[1,+∞),且x1>x2,有f'(學(xué)科素養(yǎng)創(chuàng)新練18.已知函數(shù)f(x)=ex+ax2-x.(1)當(dāng)a=1時(shí),討論f(x)的單調(diào)性;(2)當(dāng)x≥0時(shí),f(x)≥12x3+1,求a的取值范圍19.已知函數(shù)g(x)=xlnx,f(x)=g(x(1)若函數(shù)f(x)在(1,+∞)上是減函數(shù),求實(shí)數(shù)a的最小值;(2)若存在x1,x2∈[e,e2],使f(x1)≤f'(x2)+a(a>0)成立,求實(shí)數(shù)a的取值范圍.

參考答案6.2.2導(dǎo)數(shù)與函數(shù)的極值、最值1.B依題意,記函數(shù)y=f'(x)的圖象與x軸的交點(diǎn)的橫坐標(biāo)自左向右依次為x1,x2,x3,x4,當(dāng)a<x<x1時(shí),f'(x)>0;當(dāng)x1<x<x2時(shí),f'(x)<0;當(dāng)x2<x<x4時(shí),f'(x)≥0;當(dāng)x4<x<b時(shí),f'(x)<0.因此,函數(shù)f(x)分別在x=x1,x=x4處取得極大值,選B.2.C由f'(x)=-x(x+2),令f'(x)=-x(x+2)>0,解得-2<x<0,即函數(shù)的單調(diào)遞增區(qū)間為(-2,0),令f'(x)=-x(x+2)<0,解得x>0或x<-2,即函數(shù)的單調(diào)遞減區(qū)間為(-∞,-2),(0,+∞),所以函數(shù)有極大值f(0),極小值f(-2).故選C.3.C∵f'(x)=(x2+2x)ex+1=x(x+2)ex+1,∴令f'(x)=0,解得x=-2或x=0.又∵當(dāng)x∈[-2,1]時(shí),ex+1>0,∴當(dāng)-2<x<0時(shí),f'(x)<0;當(dāng)0<x<1時(shí),f'(x)>0.∴f(x)在(-2,0)上單調(diào)遞減,在(0,1)上單調(diào)遞增.又f(-2)=4e-1,f(1)=e2,∴f(x)的最大值為e2.4.B∵三次函數(shù)的圖象過(guò)原點(diǎn),故可設(shè)為f(x)=x3+bx2+cx,∴f'(x)=3x2+2bx+c.又x=1,3是f'(x)=0的兩個(gè)根,∴1+3=解得b∴f(x)=x3-6x2+9x.又y'=3x2-12x+9=3(x-1)(x-3),∴當(dāng)x=1時(shí),f(x)極大值=4,當(dāng)x=3時(shí),f(x)極小值=0,滿足條件,故選B.5.B由已知得f'(x)=ex(x+a-1)(x+a)2(x≠-a),令f'(x)=0,有x=1-a,且f(x)在(-∞,1-a)上單調(diào)遞減,在(1-a,+∞)上單調(diào)遞增,∴f(x)的極小值為f(1-a)=e1-a=6.-2∵f'(x)=3x2+2ax+b,∴f解得a=2,b=-4,∴a+b=2-4=-2.7.0,14因?yàn)楹瘮?shù)f(x)=12x2-x+alnx有兩個(gè)不同的極值點(diǎn),所以f'(x)=x-1+ax=x2-x+ax=所以方程x2-x+a=0在(0,+∞)上有兩個(gè)不同的實(shí)數(shù)根,所以Δ=1-4a>0,a>0,解得08.-71f'(x)=3x2-6x-9=3(x-3)(x+1).令f'(x)=0,得x=3或x=-1.又f(-4)=k-76,f(3)=k-27,f(-1)=k+5,f(4)=k-20,則f(x)max=k+5=10,得k=5,∴f(x)min=k-76=-71.9.33已知函數(shù)f(x)=8sinx-tanx,那么f'(x)=8cosx-1co令f'(x)=0,得cosx=12∵x∈0,π2,∴x=π3.當(dāng)x∈0,π3時(shí),f'(x)>0,函數(shù)f(x)在區(qū)間0,π3上單調(diào)遞增;當(dāng)x∈π3,π2時(shí),f'(x)<0,函數(shù)f(x)在區(qū)間π3,∴當(dāng)x=π3時(shí),函數(shù)f(x)取得最大值fπ3=33.10.53已知函數(shù)f(x)=13x3+a2x2所以f'(x)=x2+2a2x+a.由題意知f'(-1)=0,f(-1)=-712即1解得a當(dāng)a=1,b=-14時(shí),f'(x)=x2+2x+此時(shí)函數(shù)是R上的增函數(shù),函數(shù)f(x)沒(méi)有極值,不合題意;當(dāng)a=-12,b=-1時(shí),f'(x)=x2+12x-12=12(x+1)·(2x-當(dāng)x<-1或x>12時(shí),f'(x)>當(dāng)-1<x<12時(shí),f'(x)<所以函數(shù)f(x)在(-∞,-1)和12,+∞上單調(diào)遞增,函數(shù)f(x)在-1,12上單調(diào)遞減.當(dāng)x=-1時(shí),f(x)取得極大值,符合題意,所以a所以f(x)=13x3+14x2-12所以f(2)=5311.解易知f(x)的定義域?yàn)?3(1)f'(x)=22x+3+2當(dāng)-32<x<-1時(shí),f'(x)>當(dāng)-1<x<-12時(shí),f'(x)<當(dāng)x>-12時(shí),f'(x)>從而f(x)在區(qū)間-32,-1,-(2)由(1)知,f(x)在區(qū)間-34,14上的最小值為f-又因?yàn)閒-34-f14=ln32+916-所以f(x)在區(qū)間-34,14上的最大值為f12.ACD令f(x)=0,解得x=ln2,所以f(x)有零點(diǎn),所以A選項(xiàng)不正確.f'(x)=ex>0,所以f(x)在R上遞增,沒(méi)有極值點(diǎn),所以B選項(xiàng)正確,C,D選項(xiàng)不正確.故選ACD.13.C∵exg(x)=f(x),∴g(x)=f(x)ex,則g'(x)=f'(x)-f(x)ex,由題意得當(dāng)x>2時(shí),f'(x)-f(x)>0,故y=g(x)在(2,+∞)上單調(diào)遞增,選項(xiàng)A正確;當(dāng)x<2時(shí),f'(x)-f(x)<0,故y=g(x)在(-∞,2)上單調(diào)遞減,故x=2是函數(shù)y=g(x)的極小值點(diǎn),故選項(xiàng)B正確;由y=g(x)在(-∞,2)上單調(diào)遞減,則y=g(x)在(-∞,0)上單調(diào)遞減,由g(0)=f(0)e0=2,得當(dāng)x≤0時(shí),g(x)≥g(0),∴f(x)ex≥2,故f(x)≥2ex,故選項(xiàng)C錯(cuò)誤;若g(2)<0,則y=g(x)有2個(gè)零點(diǎn),若g(2)=0,則函數(shù)y=g(x14.Af'(x)=4-1x=4x令f'(x)>0,得x>14;令f'(x)<0,得0<x<1所以當(dāng)x=14時(shí),函數(shù)有最小值為f14=4×14-ln14=1+ln4=1+2ln2.故選15.[e,+∞)由f(x)=ax2+2lnx,得f'(x)=2(x2-a)x3,又函數(shù)f令f'(x)=0,得x=-a(舍去)或x=a.當(dāng)0<x<a時(shí),f'(x)<0;當(dāng)x>a時(shí),f'(x)>0.故x=a是函數(shù)f(x)的極小值點(diǎn),也是最小值點(diǎn),且f(a)=lna+1.要使f(x)≥2恒成立,需lna+1≥2恒成立,則a≥e.16.解(1)因?yàn)閒(x)=alnx+12x故f'(x)=ax由于曲線y=f(x)在點(diǎn)(1,f(1))處的切線垂直于y軸,故該切線斜率為0,即f'(1)=0,從而a-12+32=0,(2)由(1),知f(x)=-lnx+12x+3f'(x)=-1x?12x2+32=3x2-2x-1因?yàn)閤2=-13不在定義域內(nèi),舍去當(dāng)x∈(0,1)時(shí),f'(x)<0,故f(x)在(0,1)上單調(diào)遞減;當(dāng)x∈(1,+∞)時(shí),f'(x)>0,故f(x)在(1,+∞)上單調(diào)遞增.故f(x)在x=1處取得極小值f(1)=3.17.(1)解①當(dāng)k=6時(shí),f(x)=x3+6lnx(x>0),故f'(x)=3x2+6x可得f(1)=1,f'(1)=9,所以曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程為y-1=9(x-1),即y=9x-8.②依題意,g(x)=x3-3x2+6lnx+3x,x∈(0,+∞).從而可得g'(x)=3x2-6x+6x?3x2,整理可得g'(x)=3(x-1)3當(dāng)x變化時(shí),g'(x),g(x)的變化情況如下表:x(0,1)1(1,+∞)g'(x)-0+g(x)↘極小值↗所以,函數(shù)g(x)的單調(diào)遞減區(qū)間為(0,1),單調(diào)遞增區(qū)間為(1,+∞);g(x)的極小值為g(1)=1,無(wú)極大值.(2)證明由f(x)=x3+klnx(x>0),得f'(x)=3x2+kx對(duì)任意的x1,x2∈[1,+∞),且x1>x2,令x1x2=t則(x1-x2)[f'(x1)+f'(x2)]-2[f(x1)-f(x2)]=(x1-x2)3x12+kx1+3x22+kx2-2x13?x23+klnx1x2=x13?x23-3x12x2+3x1x22+kx1x令h(x)=x-1x-2lnx,x∈[1,+∞)當(dāng)x>1時(shí),h'(x)=1+1x2由此可得h(x)在[1,+∞)上單調(diào)遞增,所以當(dāng)t>1時(shí),h(t)>h(1),即t-1t-2lnt>0因?yàn)閤2≥1,t3-3t2+3t-1=(t-1)3>0,k≥-3,所以,x23(t3-3t2+3t-1)+kt-1t-2lnt≥(t3-3t2+3t-1)-3t-1t-2lnt=t3-3t2+6lnt+3t-1.②由(1)②可知,當(dāng)t>1時(shí),g(t)>g(1),即t3-3t2+6lnt+3t>1,故t3-3t2+6lnt+3t-1>0.由①②③可得(x1-x2)[f'(x1)+f'(x2)]-2[f(x1)-f(x2)]>0.所以,當(dāng)k≥-3時(shí),對(duì)任意的x1,x2∈[1,+∞),且x1>x2,有f'(18.解(1)當(dāng)a=1時(shí),f(x)=ex+x2-x,f'(x)=ex+2x-1.故當(dāng)x∈(-∞,0)時(shí),f'(x)<0;當(dāng)x∈(0,+∞)時(shí),f'(x)>0.所以f(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增.(2)f(x)≥12x3+1等價(jià)于12x3設(shè)函數(shù)g(x)=12x3-a則g'(x)=-12x3-ax2+x+1-32x2+2ax-1e-x=-12x[x2-(2a+3)x+4a+2]e-x=-12x(x-2a-1)·(x-2)e①若2a+1≤0,即a≤-12,則當(dāng)x∈(0,2)時(shí),g'(x)>0所以g(x)在(0,2)上單調(diào)遞增,而g(0)=1,故當(dāng)x∈(0,2)時(shí),g(x)>1,不合題意.②若0<2a+1<2,即-12<a<12,則當(dāng)x∈(0,2a+1)∪(2,+∞)時(shí),g'(x)<0;當(dāng)x∈(2a+1,2)時(shí),g'(x)>所以g(x)在(0,2a+1),(2,+∞)上單調(diào)遞減,在(2a+1,2)上單調(diào)遞增.由于g(0)=1,所以g(x)≤1當(dāng)且僅當(dāng)g(2)=(7-4a)e-2≤1,即a≥7-所以當(dāng)7-e24≤a<12時(shí),③若2a+1≥2,即a≥12,則g(x)≤12x3+x+1e-x.由于0∈7-e24,12,故由②可得12x3+x+1e-x≤1.故當(dāng)a≥12時(shí),綜上,實(shí)數(shù)a的取值范圍是7-19.解由已知函數(shù)g(x),f(x)的定義域均為(0,1)∪(1,+∞),且f(x)=xlnx-ax(a>(1)函數(shù)g'(x)=lnx因?yàn)閒(x)在/