（北京專用）2019版高考數(shù)學(xué)一輪復(fù)習(xí)第三章導(dǎo)數(shù)及其應(yīng)用第二節(jié)導(dǎo)數(shù)與函數(shù)的單調(diào)性課件文

1、第二節(jié)導(dǎo)數(shù)與函數(shù)的單調(diào)性2總綱目錄教材研讀函數(shù)的導(dǎo)數(shù)與單調(diào)性的關(guān)系考點(diǎn)突破考點(diǎn)二利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間考點(diǎn)二利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間考點(diǎn)一利用導(dǎo)數(shù)判斷（證明）函數(shù)的單調(diào)性考點(diǎn)三已知函數(shù)的單調(diào)性求參數(shù)的范圍考點(diǎn)三已知函數(shù)的單調(diào)性求參數(shù)的范圍3函數(shù)的導(dǎo)數(shù)與單調(diào)性的關(guān)系函數(shù)的導(dǎo)數(shù)與單調(diào)性的關(guān)系函數(shù)y=f(x)在某個(gè)區(qū)間內(nèi)可導(dǎo),(1)若f (x)0,則f(x)在這個(gè)區(qū)間內(nèi)單調(diào)遞增 ;(2)若f (x)0,則f(x)在這個(gè)區(qū)間內(nèi)單調(diào)遞減 ;(3)若f (x)=0,則f(x)在這個(gè)區(qū)間內(nèi)是常數(shù)函數(shù) .教材研讀教材研讀41.已知函數(shù)f(x)的導(dǎo)函數(shù)f (x)=ax2+bx+c的圖象如圖所示,則f(x)的圖

2、象可能是() D5答案答案 D由題圖可知,當(dāng)xx1時(shí),由導(dǎo)函數(shù)f (x)=ax2+bx+c0,知相應(yīng)的函數(shù)f(x)在該區(qū)間上單調(diào)遞減;當(dāng)0 x0知相應(yīng)的函數(shù)f(x)在該區(qū)間上單調(diào)遞增.62.函數(shù)f(x)=(x-3)ex的單調(diào)遞增區(qū)間是()A.(-,2) B.(0,3) C.(1,4) D.(2,+)答案答案 D由f(x)=(x-3)ex,得f (x)=(x-2)ex,令f (x)0,得x2,故f(x)的單調(diào)遞增區(qū)間是(2,+).D73.下列函數(shù)中,在(0,+)上為增函數(shù)的是()A.f(x)=sin 2x B.f(x)=xexC.f(x)=x3-x D.f(x)=-x+ln x答案答案 B對(duì)于A

3、,易得f(x)=sin 2x的單調(diào)遞增區(qū)間為(kZ);對(duì)于B, f (x)=ex(x+1),當(dāng)x(0,+)時(shí), f (x)0,函數(shù)f(x)=xex在(0,+)上為增函數(shù);,44kkB對(duì)于C, f (x)=3x2-1,令f (x)0,得x或x0,得0 x0).令g(x)=x3+x2-x+1,則g(x)=3x2+2x-1=(3x-1)(x+1),令g(x)=(3x-1)(x+1)0,得x,令g(x)=(3x-1)(x+1)0,得0 x0,所以g(x)在(0,+)上恒大于零,于是,當(dāng)x(0,+)時(shí), f (x)=ex0恒成立,所以當(dāng)a=-1時(shí),函數(shù)f(x)在(0,+)上為增函數(shù).10,31,3131

4、322273221xxxx方法技巧方法技巧用導(dǎo)數(shù)法判斷函數(shù)f(x)在(a,b)內(nèi)的單調(diào)性的步驟求f (x).確定f (x)在(a,b)內(nèi)的符號(hào).作出結(jié)論:f (x)0時(shí)為增函數(shù);f (x)0時(shí)為減函數(shù).提醒提醒研究含參數(shù)函數(shù)的單調(diào)性時(shí),需注意依據(jù)參數(shù)取值對(duì)不等式解集的影響進(jìn)行分類討論.1-1已知函數(shù)f(x)=ax3+x2(xR)在x=-處取得極值.(1)確定a的值;(2)若函數(shù)g(x)=f(x)ex,討論g(x)的單調(diào)性.43解析解析(1)對(duì)f(x)求導(dǎo)得f (x)=3ax2+2x,因?yàn)閒(x)在x=-處取得極值,所以f =0,即3a+2=-=0,解得a=.(2)由(1)得g(x)=ex,故g

5、(x)=ex+ex=ex=x(x+1)(x+4)ex.434316943163a 83123212xx2322xx3212xx3215222xxx12令g(x)=0,解得x=0或x=-1或x=-4.當(dāng)x-4時(shí),g(x)0,故g(x)為減函數(shù);當(dāng)-4x0,故g(x)為增函數(shù);當(dāng)-1x0時(shí),g(x)0時(shí),g(x)0,故g(x)為增函數(shù).綜上知,g(x)在(-,-4)和(-1,0)內(nèi)為減函數(shù),在(-4,-1)和(0,+)內(nèi)為增函數(shù).典例典例2已知函數(shù)f(x)=xln x.(1)求曲線y=f(x)在點(diǎn)(1, f(1)處的切線方程;(2)求f(x)的單調(diào)區(qū)間;(3)若對(duì)于任意x,都有f(x)ax-1,求

6、實(shí)數(shù)a的取值范圍.1,ee考點(diǎn)二利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間考點(diǎn)二利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間解析解析(1)因?yàn)楹瘮?shù)f(x)=xln x,所以f (x)=ln x+x=ln x+1,f (1)=ln 1+1=1.又因?yàn)閒(1)=0,所以曲線y=f(x)在點(diǎn)(1, f(1)處的切線方程為y=x-1.(2)函數(shù)f(x)=xln x的定義域?yàn)?0,+),由(1)可知, f (x)=ln x+1.令f (x)=0,解得x=.f(x)與f (x)在區(qū)間(0,+)上的變化情況如下:1x1exf (x)-0+f(x)單調(diào)遞減極小值單調(diào)遞增10,e1e1,e所以,f(x)的單調(diào)遞增區(qū)間是;f(x)的單調(diào)遞減區(qū)間是.(

7、3)當(dāng)xe時(shí),“f(x)ax-1”等價(jià)于“aln x+”.令g(x)=ln x+,x,則g(x)=-=,x.當(dāng)x時(shí),g(x)0,所以g(x)在區(qū)間(1,e)上單調(diào)遞增.而g=-ln e+e=e-11.5,1,e10,e1e1x1x1,ee1x21x21xx1,ee1,1e1,1e1eg(e)ln e+=1+0,解集在定義域內(nèi)的部分為單調(diào)遞增區(qū)間;(4)解不等式f (x)0,得x0,所以f(x)的單調(diào)增區(qū)間為(0,+).(2)f (x)=(x-a+1)ex.令f (x)=0,得x=a-1.所以當(dāng)a-11,即a2時(shí),在1,2上, f (x)0恒成立, f(x)單調(diào)遞增;當(dāng)a-12,即a3時(shí),在1,

8、2上, f (x)0恒成立, f(x)單調(diào)遞減;當(dāng)1a-12,即2a0, f(x)單調(diào)遞增.綜上,無論a為何值,當(dāng)x1,2時(shí), f(x)的最大值都為f(1)或f(2).f(1)=(1-a)e, f(2)=(2-a)e2,f(1)-f(2)=(1-a)e-(2-a)e2=(e2-e)a-(2e2-e).所以當(dāng)a=時(shí), f(1)-f(2)0, f(x)max=f(1)=(1-a)e.當(dāng)a=時(shí), f(1)-f(2)0, f(x)max=f(2)=(2-a)e2.222eeee2e 1e 1222eeee2e 1e 1考點(diǎn)三已知函數(shù)的單調(diào)性求參數(shù)的范圍考點(diǎn)三已知函數(shù)的單調(diào)性求參數(shù)的范圍典例典例3 (2

9、016北京海淀期中)已知函數(shù)f(x)=x3+x2+ax+1,曲線y=f(x)在點(diǎn)(0,1)處的切線為l.(1)若直線l的斜率為-3,求函數(shù)f(x)的單調(diào)區(qū)間;(2)若函數(shù)f(x)在區(qū)間-2,a上是單調(diào)函數(shù),求實(shí)數(shù)a的取值范圍.13解析解析(1)因?yàn)閒(0)=1,所以曲線y=f(x)經(jīng)過點(diǎn)(0,1),又f (x)=x2+2x+a,所以f (0)=a=-3,所以f (x)=x2+2x-3.令f (x)=0,解得x1=-3,x2=1.當(dāng)x變化時(shí), f (x), f(x)的變化情況如下表:x(-,-3)-3(-3,1)1(1,+)f (x)+0-0+f(x)極大值極小值所以函數(shù)f(x)的單調(diào)遞增區(qū)間為

10、(-,-3),(1,+),單調(diào)遞減區(qū)間為(-3,1).(2)當(dāng)函數(shù)f(x)在區(qū)間-2,a上單調(diào)遞減時(shí), f (x)0對(duì)x-2,a恒成立,即f (x)=x2+2x+a0對(duì)x-2,a恒成立,所以即解得-3a0.又-2a,所以-2a0.當(dāng)函數(shù)f(x)在區(qū)間-2,a上單調(diào)遞增時(shí), f (x)0對(duì)x-2,a恒成立,只要f (x)=x2+2x+a在-2,a上的最小值大于或等于0即可,易知函數(shù)f (x)=x2+2x+a圖象的對(duì)稱軸為x=-1, ( 2)0, ( )0,ffa22( 2)2 ( 2)0,20,aaaa 當(dāng)-2a-1時(shí), f (x)在-2,a上的最小值為f (a),即f (a)=a2+3a0,解

11、得a0或a-3,此種情形不成立.當(dāng)-1a時(shí), f (x)在-2,a上的最小值為f (-1),即f (-1)=1-2+a0,解得a1.綜上,實(shí)數(shù)a的取值范圍是-20(或f (x)0(或f (x)min0時(shí),若函數(shù)f(x)在區(qū)間(1,2)內(nèi)單調(diào)遞減,求k的取值范圍.1x解析解析(1)函數(shù)f(x)的定義域?yàn)閤|x0.f (x)=k-+=.當(dāng)k0時(shí),令f (x)0,解得0 x1,此時(shí)函數(shù)f(x)為單調(diào)遞增函數(shù);令f (x)1,此時(shí)函數(shù)f(x)為單調(diào)遞減函數(shù).當(dāng)k0時(shí),當(dāng)1時(shí),令f (x)0,解得0 x1,此時(shí)函數(shù)f(x)為單調(diào)遞增函數(shù);令f (x)0,解得x1,即0k0,解得0 x,此時(shí)函數(shù)f(x)為單調(diào)遞增函數(shù);令f (x)0,解得1x,此時(shí)函數(shù)f(x)為單調(diào)遞減函數(shù).綜上所述,當(dāng)k0時(shí),函數(shù)f(x)的單調(diào)遞增區(qū)間為(0,1),單調(diào)遞減區(qū)間