考點規(guī)范練16導(dǎo)數(shù)的綜合應(yīng)用一、基礎(chǔ)鞏固1.已知函數(shù)f(x)=x3+ax2+bx+c在x=-23與x=1處都取得極值.(1)求a,b的值及函數(shù)f(x)的單調(diào)區(qū)間;(2)若對于x-1,2,不等式f(x)<c2恒成立,求c的取值范圍.解(1)f(x)=x3+ax2+bx+c,f'(x)=3x2+2ax+b.又f(x)在x=-23與x=1處都取得極值,f'-23=129-43a+b=0,f'(1)=3+2a+b=0,兩式聯(lián)立解得a=-12,b=-2,f(x)=x3-12x2-2x+c,f'(x)=3x2-x-2=(3x+2)(x-1),令f'(x)=0,得x1=-23,x2=1,當(dāng)x變化時,f'(x),f(x)的變化情況如下表:x-,-23-23-23,11(1,+)f'(x)+0-0+f(x)極大值極小值函數(shù)f(x)的遞增區(qū)間為-,-23與(1,+);遞減區(qū)間為-23,1.(2)f(x)=x3-12x2-2x+c,x-1,2,當(dāng)x=-23時,f-23=2227+c為極大值,而f(2)=2+c,則f(2)=2+c為最大值,要使f(x)<c2(x-1,2)恒成立,只需c2>f(2)=2+c,解得c<-1或c>2.c的取值范圍為(-,-1)(2,+).2.設(shè)函數(shù)f(x)=ax2-a-ln x,g(x)=1x-eex,其中aR,e=2.718為自然對數(shù)的底數(shù).(1)討論f(x)的單調(diào)性;(2)證明:當(dāng)x>1時,g(x)>0;(3)確定a的所有可能取值,使得f(x)>g(x)在區(qū)間(1,+)內(nèi)恒成立.(1)解f'(x)=2ax-1x=2ax2-1x(x>0).當(dāng)a0時,f'(x)<0,f(x)在(0,+)內(nèi)單調(diào)遞減.當(dāng)a>0時,由f'(x)=0有x=12a.當(dāng)x0,12a時,f'(x)<0,f(x)單調(diào)遞減;當(dāng)x12a,+時,f'(x)>0,f(x)單調(diào)遞增.(2)證明令s(x)=ex-1-x,則s'(x)=ex-1-1.當(dāng)x>1時,s'(x)>0,所以ex-1>x,從而g(x)=1x-1ex-1>0.(3)解由(2),當(dāng)x>1時,g(x)>0.當(dāng)a0,x>1時,f(x)=a(x2-1)-lnx<0.故當(dāng)f(x)>g(x)在區(qū)間(1,+)內(nèi)恒成立時,必有a>0.當(dāng)0<a<12時,12a>1.由(1)有f12a<f(1)=0,而g12a>0,所以此時f(x)>g(x)在區(qū)間(1,+)內(nèi)不恒成立.當(dāng)a12時,令h(x)=f(x)-g(x)(x1).當(dāng)x>1時,h'(x)=2ax-1x+1x2-e1-x>x-1x+1x2-1x=x3-2x+1x2>x2-2x+1x2>0.因此,h(x)在區(qū)間(1,+)內(nèi)單調(diào)遞增.又因為h(1)=0,所以當(dāng)x>1時,h(x)=f(x)-g(x)>0,即f(x)>g(x)恒成立.綜上,a12,+.3.已知函數(shù)f(x)=(x-k)ex+k,kZ.(1)當(dāng)k=0時,求函數(shù)f(x)的單調(diào)區(qū)間;(2)若當(dāng)x(0,+)時,不等式f(x)+5>0恒成立,求k的最大值.解(1)當(dāng)k=0時,f(x)=x·ex,f'(x)=ex+xex=ex(x+1),當(dāng)x(-,-1)時,f'(x)<0;當(dāng)x(-1,+)時,f'(x)>0;f(x)在(-,-1)內(nèi)是減函數(shù),在(-1,+)內(nèi)是增函數(shù).(2)不等式f(x)+5>0恒成立(x-k)ex+k+5>0在x(0,+)時恒成立,令F(x)=(x-k)ex+k+5,F'(x)=ex(x-k+1)(xR),當(dāng)x(-,k-1)時,f'(x)<0;當(dāng)x(k-1,+)時,f'(x)>0;f(x)在(-,k-1)內(nèi)是減函數(shù),在(k-1,+)內(nèi)是增函數(shù).若k-10,即k1,當(dāng)x(0,+)時,F(x)>F(0)0.而F(0)=5>0恒成立,k1符合題意.若k-1>0,即k>1,當(dāng)x(0,+)時,只需F(x)min=F(k-1)=-ek-1+5+k>0即可.令h(k)=-ek-1+5+k,h'(k)=1-ek-1<0恒成立,即h(k)=-ek-1+5+k單調(diào)遞減.h(2)=-e+7>0,h(3)=-e2+8>0,h(4)=-e3+9<0,1<k3,綜上,k的最大值為3.4.已知函數(shù)f(x)=ln x+ax2+(2a+1)x.(1)討論f(x)的單調(diào)性;(2)當(dāng)a<0時,證明f(x)-34a-2.(1)解f(x)的定義域為(0,+),f'(x)=1x+2ax+2a+1=(x+1)(2ax+1)x.若a0,則當(dāng)x(0,+)時,f'(x)>0,故f(x)在(0,+)單調(diào)遞增.若a<0,則當(dāng)x0,-12a時,f'(x)>0;當(dāng)x-12a,+時,f'(x)<0.故f(x)在區(qū)間0,-12a內(nèi)單調(diào)遞增,在區(qū)間-12a,+內(nèi)單調(diào)遞減.(2)證明由(1)知,當(dāng)a<0時,f(x)在x=-12a取得最大值,最大值為f-12a=ln-12a-1-14a.所以f(x)-34a-2等價于ln-12a-1-14a-34a-2,即ln-12a+12a+10.設(shè)g(x)=lnx-x+1,則g'(x)=1x-1.當(dāng)x(0,1)時,g'(x)>0;當(dāng)x(1,+)時,g'(x)<0.所以g(x)在區(qū)間(0,1)內(nèi)單調(diào)遞增,在區(qū)間(1,+)內(nèi)單調(diào)遞減.故當(dāng)x=1時,g(x)取得最大值,最大值為g(1)=0.所以當(dāng)x>0時,g(x)0.從而當(dāng)a<0時,ln-12a+12a+10,即f(x)-34a-2.二、能力提升5.設(shè)函數(shù)f(x)=(1-x2)ex.(1)討論f(x)的單調(diào)性;(2)當(dāng)x0時,f(x)ax+1,求a的取值范圍.解(1)f'(x)=(1-2x-x2)ex.令f'(x)=0得x=-1-2,x=-1+2.當(dāng)x(-,-1-2)時,f'(x)<0;當(dāng)x(-1-2,-1+2)時,f'(x)>0;當(dāng)x(-1+2,+)時,f'(x)<0.所以f(x)在(-,-1-2),(-1+2,+)內(nèi)單調(diào)遞減,在(-1-2,-1+2)內(nèi)單調(diào)遞增.(2)f(x)=(1+x)(1-x)ex.當(dāng)a1時,設(shè)函數(shù)h(x)=(1-x)ex,h'(x)=-xex<0(x>0),因此h(x)在0,+)內(nèi)單調(diào)遞減,而h(0)=1,故h(x)1,所以f(x)=(x+1)h(x)x+1ax+1.當(dāng)0<a<1時,設(shè)函數(shù)g(x)=ex-x-1,g'(x)=ex-1>0(x>0),所以g(x)在0,+)內(nèi)單調(diào)遞增,而g(0)=0,故exx+1.當(dāng)0<x<1時,f(x)>(1-x)(1+x)2,(1-x)(1+x)2-ax-1=x(1-a-x-x2),取x0=5-4a-12,則x0(0,1),(1-x0)(1+x0)2-ax0-1=0,故f(x0)>ax0+1.當(dāng)a0時,取x0=5-12,則x0(0,1),f(x0)>(1-x0)(1+x0)2=1ax0+1.綜上,a的取值范圍是1,+).6.(2018全國,文21)已知函數(shù)f(x)=13x3-a(x2+x+1).(1)若a=3,求f(x)的單調(diào)區(qū)間;(2)證明:f(x)只有一個零點.(1)解當(dāng)a=3時,f(x)=13x3-3x2-3x-3,f'(x)=x2-6x-3.令f'(x)=0,解得x=3-23或x=3+23.當(dāng)x(-,3-23)(3+23,+)時,f'(x)>0;當(dāng)x(3-23,3+23)時,f'(x)<0.故f(x)在(-,3-23),(3+23,+)單調(diào)遞增,在(3-23,3+23)單調(diào)遞減.(2)證明因為x2+x+1>0,所以f(x)=0等價于x3x2+x+1-3a=0.設(shè)g(x)=x3x2+x+1-3a,則g'(x)=x2(x2+2x+3)(x2+x+1)20,僅當(dāng)x=0時g'(x)=0,所以g(x)在(-,+)單調(diào)遞增,故g(x)至多有一個零點,從而f(x)至多有一個零點.又f(3a-1)=-6a2+2a-13=-6a-162-16<0,f(3a+1)=13>0,故f(x)有一個零點.綜上,f(x)只有一個零點.7.(2018廣東茂名二模)已知函數(shù)f(x)=ln x+12(x-1)2.(1)判斷f(x)的零點個數(shù);(2)若函數(shù)g(x)=ax-a,當(dāng)x>1時,g(x)的圖象總在f(x)的圖象的下方,求a的取值范圍.解(1)f(x)=lnx+12(x-1)2的定義域為(0,+),f'(x)=1x+x-1,1x+x2,f'(x)1>0,f(x)在(0,+)上為增函數(shù),又f(1)=0,f(x)在(0,+)上只有一個零點.(2)由題意,當(dāng)x>1時,12(x-1)2+lnx-ax+a>0恒成立.令h(x)=12(x-1)2+lnx-ax+a,則h'(x)=x+1x-1-a.當(dāng)a1時,h'(x)=x+1x-1-a>1-a0,h(x)在(1,+)上為增函數(shù).又h(1)=0,h(x)>0恒成立.當(dāng)a>1時,h'(x)=x2-(1+a)x+1x,令(x)=x2-(1+a)x+1,則=(1+a)2-4=(a+3)(a-1)>0.令(x)=0的兩根分別為x1,x2且x1<x2,x1+x2=1+a>0,x1·x2=1>0,0<x1<1<x2,當(dāng)x(1,x2)時,(x)<0,h'(x)<0,h(x)在(1,x2)上為減函數(shù),又h(1)=0,當(dāng)x(1,x2)時,h(x)<0.故a的取值范圍為(-,1.三、高考預(yù)測8.(2018河北石家莊一模)已知函數(shù)f(x)=x(ln x-ax)(aR).(1)若a=1,求函數(shù)f(x)的圖象在點(1,f(1)處的切線方程;(2)若函數(shù)f(x)有兩個極值點x1,x2,且x1<x2,求證:f(x2)>-12.(1)解由已知,f(x)=x(lnx-x),當(dāng)x=1時,f(x)=-1,f'(x)=lnx+1-2x,當(dāng)x=1時,f'(x)=-1,所以所求切線方程為x+y=0.(2)證明由已知可得f'(x)=lnx+1-2ax=0有兩個相異實根x1,x2,令h(x)=f'(x),則h'(x)=1x-2a,若a0,則h'(x)>0,h(x)單調(diào)遞增,f'(x)=0不可能有兩根;若a>0,令h'(x)=0得x=12a,可知h(x)在0,12a上單調(diào)遞增,在12a,+上單調(diào)遞減,令f'12a>0,解得0<a<12,由1e<12a有f'1e=-2ae<0,由1a2>12a有f'1a2=-2lna+1-2a<0,從而當(dāng)0<a<12時,函數(shù)f(x)有兩個極值點,當(dāng)x變化時,f'(x),f(x)的變化情況如下表:x(0,x1)x1(x1,x2)x2(x2,+)f'(x)-0+0-f(x)單調(diào)遞減f(x1)單調(diào)遞增f(x2)單調(diào)遞減因為f'(1)=1-2a>0,所以x1<1<x2,f(x)在區(qū)間1,x2上單調(diào)遞增,所以f(x2)>f(1)=-a>-12.8