北師大版選修11 函數(shù)的極值與導(dǎo)數(shù) 課時(shí)作業(yè).doc

第6課時(shí)函數(shù)的極值與導(dǎo)數(shù)基礎(chǔ)達(dá)標(biāo)(水平一)1.函數(shù)f(x)=sin x+x2,x(0,)的極大值是().a.32+6b.-32+3c.32+3d.1+4【解析】f(x)=cos x+12,x(0,),由f(x)=0,即cos x=-12,得x=23,x0,23時(shí),f(x)0;x23,時(shí),f(x)0,x=23時(shí),f(x)有極大值f23=32+3.【答案】c2.已知函數(shù)f(x)=ax3+bx2+c,其導(dǎo)函數(shù)的圖象如圖所示,則函數(shù)f(x)的極小值是().a.a+b+cb.8a+4b+cc.3a+2bd.c【解析】由f(x)的圖象可知,當(dāng)x(-,0)(2,+)時(shí),f(x)0.f(x)在(-,0)和(2,+)上為減函數(shù),在(0,2)上為增函數(shù).當(dāng)x=0時(shí),f(x)取到極小值為f(0)=c.【答案】d3.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】f(x)=3ax2+b,由題意可知f(1)=3a+b=0,f(1)=a+b=-2,解得a=1,b=-3.【答案】a4.若a0,b0,且函數(shù)f(x)=4x3-ax2-2bx-2在x=1處有極值,則ab的最大值為().a.2 b.3 c.6 d.9【解析】f(x)=12x2-2ax-2b.由于函數(shù)f(x)在x=1處有極值,則有f(1)=0,即a+b=6(a,b0),由于a+b2ab,即aba+b22=9,當(dāng)且僅當(dāng)a=b=3時(shí)取最大值9.【答案】d5.直線y=a與函數(shù)y=13x3-x2的圖象有三個(gè)相異的交點(diǎn),則實(shí)數(shù)a的取值范圍是.【解析】f(x)=x2-2x,令f(x)=0,得x=0或x=2.f(0)=0,f(2)=-43,-43a0.【答案】-43a06.若a為函數(shù)f(x)=x3-12x的極小值點(diǎn),則a=.【解析】f(x)=3x2-12,當(dāng)x0;當(dāng)-2x2時(shí),f(x)2時(shí),f(x)0.x=2是f(x)的極小值點(diǎn).又a為f(x)的極小值點(diǎn),a=2.【答案】27.求函數(shù)f(x)=x2ex的極值.【解析】函數(shù)f(x)的定義域?yàn)閞,f(x)=2xex-x2ex(ex)2=x(2-x)ex,令f(x)=0,得x=0或x=2.當(dāng)x變化時(shí),f(x),f(x)的變化情況如下表:x(-,0)0(0,2)2(2,+)f(x)-0+0-f(x)04e-2由上表可以看出,當(dāng)x=0時(shí),函數(shù)取得極小值,且為f(0)=0;當(dāng)x=2時(shí),函數(shù)取得極大值,且為f(2)=4e-2.拓展提升(水平二)8.設(shè)函數(shù)f(x)=xex,則().a.x=1為f(x)的極大值點(diǎn)b.x=1為f(x)的極小值點(diǎn)c.x=-1為f(x)的極大值點(diǎn)d.x=-1為f(x)的極小值點(diǎn)【解析】f(x)=xex,f(x)=ex+xex=ex(1+x).當(dāng)f(x)0,即ex(1+x)0時(shí),解得x-1,當(dāng)x-1時(shí),函數(shù)f(x)為增函數(shù).同理可得,當(dāng)x0,f(-1)0,故不滿足.【答案】d10.已知函數(shù)y=xf(x)的圖象如圖所示(其中f(x)是函數(shù)f(x)的導(dǎo)函數(shù)),給出以下說(shuō)法:函數(shù)f(x)在區(qū)間(1,+)上是增函數(shù);函數(shù)f(x)在區(qū)間(-1,1)上無(wú)單調(diào)性;函數(shù)f(x)在x=-12處取得極大值;函數(shù)f(x)在x=1處取得極小值.其中正確的說(shuō)法有.【解析】從圖象上可以發(fā)現(xiàn),當(dāng)x(1,+)時(shí),xf(x)0,所以f(x)0,故f(x)在區(qū)間(1,+)上是增函數(shù),正確;當(dāng)x(-1,1)時(shí),f(x)0時(shí),令f(x)0,解得xm或xm3;令f(x)0,解得m3xm,函數(shù)f(x)在-,m3上單調(diào)遞增,在m3,m上單調(diào)遞減,在(m,+)上單調(diào)遞增,f(x)在x=m3處取得極大值fm3=12,解得m=32.當(dāng)m0,解得xm3;令f(x)xm,