2024-2025學(xué)年高中數(shù)學(xué)第一章導(dǎo)數(shù)及其應(yīng)用1.3.2函數(shù)的極值與導(dǎo)數(shù)課時(shí)素養(yǎng)評(píng)價(jià)含解析新人教A版選修2-2

PAGE課時(shí)素養(yǎng)評(píng)價(jià)六函數(shù)的極值與導(dǎo)數(shù)(15分鐘30分)1.函數(shù)f(x)=ax3+x+1有極值的充要條件是 ()A.a>0 B.a≥0 C.a<0 D.a≤0【解析】選C.因?yàn)閒′(x)=3ax2+1,所以f′(x)=3ax2+1=0?3a=-QUOTE<0,即a<0,反之a(chǎn)<0,f′(x)=3ax2+1=0肯定有根.2.設(shè)函數(shù)f(x)的定義域?yàn)镽,x0QUOTE是f(x)的極大值點(diǎn),以下結(jié)論肯定正確的是 ()A.?x∈R,f(x)≤fQUOTEB.-x0是fQUOTE的微小值點(diǎn)C.-x0是-f(x)的微小值點(diǎn)D.-x0是-fQUOTE的微小值點(diǎn)【解析】選D.對(duì)于A選項(xiàng),函數(shù)的極大值不肯定是函數(shù)的最大值,所以錯(cuò);對(duì)于B中的f(-x)是將f(x)的圖象關(guān)于y軸對(duì)稱,所以-x0是其極大值點(diǎn),錯(cuò)誤;對(duì)于C中的-f(x)是將f(x)的圖象關(guān)于x軸對(duì)稱,所以x0才是其微小值點(diǎn),錯(cuò)誤;而對(duì)于D中的-f(-x)是將f(x)的圖象關(guān)于原點(diǎn)對(duì)稱,故-x0是其微小值點(diǎn),正確.3.如圖是函數(shù)y=f(x)的導(dǎo)函數(shù)y=f′(x)的圖象,給出下列說法:①-3是函數(shù)y=f(x)的極值點(diǎn);②-1是函數(shù)y=f(x)的最小值點(diǎn);③y=f(x)在x=0處切線的斜率小于零;④y=f(x)在區(qū)間(-3,1)上單調(diào)遞增.則正確說法的序號(hào)是________.

【解析】依據(jù)導(dǎo)函數(shù)圖象可知當(dāng)x∈(-∞,-3)時(shí),f′(x)<0,在x∈(-3,1)時(shí),f′(x)≥0,所以函數(shù)y=f(x)在(-∞,-3)上單調(diào)遞減,在(-3,1)上單調(diào)遞增,故④正確;則-3是函數(shù)y=f(x)的微小值點(diǎn),故①正確;因?yàn)樵?-3,1)上單調(diào)遞增,所以-1不是函數(shù)y=f(x)的最小值點(diǎn),故②不正確;因?yàn)楹瘮?shù)y=f(x)在x=0處的導(dǎo)數(shù)大于0,所以切線的斜率大于零,故③不正確.答案:①④4.f(x)=x(x-c)2在x=2處有極大值,則常數(shù)c的值為________.

【解析】因?yàn)閤=2是f(x)的極大值點(diǎn),又f(x)=x(x2-2cx+c2),所以f′(x)=x(2x-2c)+x2-2cx+c2=3x2-4cx+c2.所以f′(2)=c2-8c+12=0.得c=2或c=6.當(dāng)c=2時(shí),不能取極大值,故c=6.答案:65.設(shè)函數(shù)f(x)=x2-2x-4lnx,求f(x)的極值.【解析】由已知,f(x)的定義域?yàn)镼UOTE,f′(x)=2x-2-QUOTE=QUOTE,令f′(x)=0,得2x2-2x-4=0.又x>0,所以x=2,當(dāng)0<x<2時(shí),f′(x)<0;當(dāng)x>2時(shí),f′(x)>0.因此,當(dāng)x=2時(shí),f(x)有微小值,微小值為fQUOTE=-4ln2,f(x)無極大值.(30分鐘60分)一、選擇題(每小題5分,共25分)1.函數(shù)f(x)=ax3+bx在x=1處有極值-2,則a,b的值分別為 ()A.1,-3B.1,3C.-1,3D.-1,-3【解析】選A.f′(x)=3ax2+b,由題意可知QUOTE解得QUOTE2.如圖是定義在QUOTE上的函數(shù)f(x)的導(dǎo)函數(shù)的圖象,則函數(shù)f(x)的極值點(diǎn)的個(gè)數(shù)為 ()A.2 B.3 C.4 D.5【解析】選B.設(shè)導(dǎo)函數(shù)的零點(diǎn)分別為x1,x2,x3,x4.則函數(shù)f(x)在(a,x1)上單調(diào)遞增,(x1,x2)上單調(diào)遞減,(x2,x3)上單調(diào)遞增,(x3,x4)上單調(diào)遞增,(x4,b)上單調(diào)遞減,故函數(shù)f(x)在x1取極大值,在x2取微小值,在x4取極大值.3.如圖是函數(shù)f(x)=x3+bx2+cx+d的大致圖象,則QUOTE+QUOTE等于 ()A.QUOTE B.QUOTE C.QUOTE D.QUOTE【解析】選C.函數(shù)f(x)=x3+bx2+cx+d的圖象過點(diǎn)(0,0),(1,0),(2,0),得d=0,b+c+1=0,4b+2c+8=0,則b=-3,c=2,f′(x)=3x2+2bx+c=3x2-6x+2,且x1,x2是函數(shù)f(x)=x3+bx2+cx+d的兩個(gè)極值點(diǎn),即x1,x2是方程3x2-6x+2=0的實(shí)根,所以QUOTE+QUOTE=(x1+x2)2-2x1x2=4-QUOTE=QUOTE.4.函數(shù)f(x)=aex-sinx在x=0處有極值,則a的值為 ()A.-1 B.0 C.1 D.e【解析】選C.由題意得:f′(x)=aex-cosx,因?yàn)閒(x)在x=0處有極值,所以f′(0)=a-cos0=a-1=0,解得:a=1,經(jīng)檢驗(yàn)滿意題意.5.若x=-2是函數(shù)f(x)=QUOTEx3-ax2-2x+1的一個(gè)極值點(diǎn),則函數(shù)f(x)的微小值為 ()A.-QUOTE B.-QUOTE C.QUOTE D.QUOTE【解析】選B.因?yàn)閒(x)=QUOTEx3-ax2-2x+1,所以f′(x)=x2-2ax-2,由題意得f′QUOTE=2+4a=0,解得a=-QUOTE,所以f(x)=QUOTEx3+QUOTEx2-2x+1,所以f′(x)=x2+x-2=QUOTE.當(dāng)x<-2或x>1時(shí),f′(x)>0;當(dāng)-2<x<1時(shí),f′(x)<0.所以,函數(shù)y=f(x)的單調(diào)遞增區(qū)間為QUOTE和QUOTE,單調(diào)遞減區(qū)間為QUOTE,當(dāng)x=1時(shí),函數(shù)y=f(x)取得微小值fQUOTE=QUOTE+QUOTE-2+1=-QUOTE.二、填空題(每小題5分,共15分)6.假如函數(shù)y=f(x)的導(dǎo)函數(shù)的圖象如圖所示,給出下列推斷:①函數(shù)y=f(x)在區(qū)間QUOTE內(nèi)單調(diào)遞增;②函數(shù)y=f(x)在區(qū)間QUOTE內(nèi)單調(diào)遞減;③函數(shù)y=f(x)在區(qū)間(4,5)內(nèi)單調(diào)遞增;④當(dāng)x=2時(shí),函數(shù)y=f(x)有微小值;⑤當(dāng)x=-QUOTE時(shí),函數(shù)y=f(x)有極大值.則上述推斷正確的是________.(填序號(hào))

【解析】函數(shù)的單調(diào)性由導(dǎo)數(shù)的符號(hào)確定,當(dāng)x∈(-∞,-2)時(shí),f′(x)<0,所以f(x)在(-∞,-2)上為減函數(shù),同理f(x)在(2,4)上為減函數(shù),在(-2,2)上是增函數(shù),在(4,+∞)上為增函數(shù),所以可解除①和②,③正確.由于函數(shù)在x=2的左側(cè)遞增,右側(cè)遞減,所以x=2時(shí),函數(shù)有極大值;而在x=-QUOTE的左右兩側(cè),函數(shù)的導(dǎo)數(shù)都是正數(shù),故函數(shù)在x=-QUOTE的左右兩側(cè)均為增函數(shù),所以x=-QUOTE不是函數(shù)的極值點(diǎn).解除④和⑤.答案:③7.若函數(shù)f(x)=QUOTEx2+(a-1)x-alnx沒有極值,則a=________.

【解析】f′(x)=(x-1)QUOTE,x>0,當(dāng)a≥0時(shí),QUOTE+1>0.令f′(x)<0,得0<x<1;令f′(x)>0,得x>1.f(x)在x=1處取微小值.當(dāng)a<0時(shí),方程QUOTE+1=0必有一個(gè)正數(shù)解x=-a,(1)若a=-1,此正數(shù)解為x=1,此時(shí)f′(x)=QUOTE≥0,f(x)在(0,+∞)上單調(diào)遞增,無極值.(2)若a≠-1,此正數(shù)解為x≠1,f′(x)=0必有2個(gè)不同的正數(shù)解,f(x)存在2個(gè)極值.綜上,a=-1.答案:-18.已知函數(shù)f(x)=lnx-QUOTE的微小值大于0,則實(shí)數(shù)m的取值范圍為________.

【解析】由f(x)=lnx-QUOTE,得f′(x)=QUOTE(x>0),令f′(x)=0,則x=-m,因?yàn)閒(x)=lnx-QUOTE的微小值大于0,必有微小值點(diǎn)-m>0,故m<0,所以當(dāng)x>-m時(shí),f′(x)>0,當(dāng)0<x<-m時(shí),f′(x)<0,所以f(x)在(0,-m)上單調(diào)遞減,在(-m,+∞)上單調(diào)遞增,所以f(x)的微小值=f(-m)=ln(-m)+1>0,所以m<-QUOTE.綜上m的取值范圍為QUOTE.答案:QUOTE三、解答題(每小題10分,共20分)9.已知函數(shù)f(x)=ax2+blnx在x=1處有極值QUOTE.(1)求a,b的值;(2)求f(x)的單調(diào)區(qū)間.【解析】(1)因?yàn)閒′(x)=2ax+QUOTE.又f(x)在x=1處有極值QUOTE,所以QUOTE即QUOTE解得a=QUOTE,b=-1.(2)由(1)可知f(x)=QUOTEx2-lnx,其定義域是QUOTE,f′(x)=x-QUOTE=QUOTE.由f′(x)<0,得0<x<1;由f′(x)>0,得x>1.所以函數(shù)y=f(x)的單調(diào)減區(qū)間是QUOTE,單調(diào)增區(qū)間是QUOTE.10.設(shè)f(x)=xlnx-ax2+(2a-1)x,a∈R.(1)令g(x)=f′(x),求g(x)的單調(diào)區(qū)間.(2)已知f(x)在x=1處取得極大值,求實(shí)數(shù)a的取值范圍.【解析】(1)由f′(x)=lnx-2ax+2a,可得g(x)=lnx-2ax+2a,x∈(0,+∞),則g′(x)=QUOTE-2a=QUOTE,當(dāng)a≤0時(shí),x∈(0,+∞)時(shí),g′(x)>0,函數(shù)g(x)單調(diào)遞增;當(dāng)a>0時(shí),x∈QUOTE時(shí),g′(x)>0,函數(shù)g(x)單調(diào)遞增,x∈QUOTE時(shí),g′(x)<0,函數(shù)g(x)單調(diào)遞減.所以當(dāng)a≤0時(shí),g(x)的單調(diào)遞增區(qū)間為(0,+∞);當(dāng)a>0時(shí),g(x)的單調(diào)遞增區(qū)間為QUOTE,單調(diào)遞減區(qū)間為QUOTE.(2)由(1)知,f′(1)=0.①當(dāng)a≤0時(shí),f′(x)單調(diào)遞增.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.所以f(x)在x=1處取得微小值,不合題意.②當(dāng)0<a<QUOTE時(shí),QUOTE>1,由(1)知f′(x)在QUOTE內(nèi)單調(diào)遞增,可得當(dāng)x∈(0,1)時(shí),f′(x)<0,x∈QUOTE時(shí),f′(x)>0,所以f(x)在(0,1)內(nèi)單調(diào)遞減,在QUOTE內(nèi)單調(diào)遞增,所以f(x)在x=1處取得微小值,不合題意.③當(dāng)a=QUOTE時(shí),QUOTE=1,f′(x)在(0,1)內(nèi)單調(diào)遞增,在(1,+∞)內(nèi)單調(diào)遞減,所以當(dāng)x∈(0,+∞)時(shí),f′(x)≤0,f(x)單調(diào)遞減,不合題意.④當(dāng)a>QUOTE時(shí),0<QUOTE<1,當(dāng)x∈QUOTE時(shí),f′(x)>0,f(x)單調(diào)遞增,當(dāng)x∈(1,+∞)時(shí),f′(x)<0,f(x)單調(diào)遞減,所以f(x)在x=1處取得極大值,符合題意.綜上可知,實(shí)數(shù)a的取值范圍為QUOTE.1.函數(shù)f(x)=QUOTEex(其中e=2.718…是自然對(duì)數(shù)的底數(shù))的極值點(diǎn)是________;極大值=________.

【解析】由已知得f′(x)=QUOTEex=QUOTEex=(x+2)(x-1)ex,因?yàn)閑x>0,令f′(x)=0,可得x=-2或x=1,當(dāng)x<-2時(shí)f′(x)>0,即函數(shù)f(x)在(-∞,-2)上單調(diào)遞增;當(dāng)-2<x<1時(shí),f′(x)<0,即函數(shù)f(x)在區(qū)間(-2,1)上單調(diào)遞減;當(dāng)x>1時(shí),f′(x)>0,即函數(shù)f(x)在區(qū)間(1,+∞)上單調(diào)遞增.故f(x)的極值點(diǎn)為-2,