第2課時(shí)對(duì)數(shù)函數(shù)性質(zhì)的應(yīng)用——(教學(xué)方式:拓展融通課習(xí)題講評(píng)式教學(xué))[課時(shí)目標(biāo)]進(jìn)一步掌握對(duì)數(shù)函數(shù)的圖象和性質(zhì),會(huì)解簡(jiǎn)單的對(duì)數(shù)不等式.會(huì)求對(duì)數(shù)型函數(shù)的單調(diào)性、值域等問(wèn)題.題型(一)解對(duì)數(shù)不等式[例1]解下列不等式:(1)log17x>log17(2)logx12>1(3)loga(2x5)>loga(x1).聽課記錄:|思|維|建|模|常見(jiàn)的對(duì)數(shù)不等式的3種類型(1)形如logax>logab的不等式,借助y=logax的單調(diào)性求解,如果a的取值不確定,需分a>1與0<a<1兩種情況討論;(2)形如logax>b的不等式,應(yīng)將b化為以a為底數(shù)的對(duì)數(shù)式的形式,再借助y=logax的單調(diào)性求解;(3)形如logax>logbx的不等式,可利用圖象求解.[針對(duì)訓(xùn)練]1.log3(x+2)>1的解集是()A.(2,+∞) B.(1,+∞)C.(0,1) D.(0,1)∪(1,+∞)2.不等式log12(2x+3)<log18(5x6)3.若實(shí)數(shù)x滿足不等式log2(x22x)>log2(x+4),則實(shí)數(shù)x的取值范圍是.

題型(二)對(duì)數(shù)型函數(shù)的單調(diào)性問(wèn)題[例2]求函數(shù)f(x)=log12(x22x3)聽課記錄:[變式拓展]1.若本例函數(shù)變?yōu)閒(x)=loga(x22x3),求f(x)的單調(diào)區(qū)間.2.若本例函數(shù)變?yōu)閥=loga(2ax),且在[0,1]上單調(diào)遞減,求a的取值范圍.|思|維|建|模|形如f(x)=logag(x)(a>0,且a≠1)的函數(shù)的單調(diào)區(qū)間的求法(1)先求g(x)>0的解集(也就是函數(shù)f(x)的定義域).(2)當(dāng)?shù)讛?shù)a>1時(shí),在g(x)>0這一前提下,g(x)的單調(diào)遞增區(qū)間是f(x)的單調(diào)遞增區(qū)間;g(x)的單調(diào)遞減區(qū)間是f(x)的單調(diào)遞減區(qū)間.(3)當(dāng)?shù)讛?shù)0<a<1時(shí),在g(x)>0這一前提下,g(x)的單調(diào)遞增區(qū)間是f(x)的單調(diào)遞減區(qū)間,g(x)的單調(diào)遞減區(qū)間是f(x)的單調(diào)遞增區(qū)間.[針對(duì)訓(xùn)練]4.函數(shù)f(x)=loga(ax3)在[1,3]上單調(diào)遞增,則a的取值范圍是()A.(1,+∞) B.(0,1)C.0,13 D.(3,+5.(1)求函數(shù)y=log12(4xx2)(2)已知函數(shù)f(x)=log2(x2+ax+3)2在(1,+∞)上單調(diào)遞增,求a的取值范圍.題型(三)對(duì)數(shù)函數(shù)性質(zhì)的綜合[例3]已知函數(shù)f(x)=log2(x+1)2.(1)若f(x)>0,求x的取值范圍;(2)若x∈(1,3],求f(x)的值域.聽課記錄:|思|維|建|模|(1)求對(duì)數(shù)型函數(shù)的值域一般是先求真數(shù)的范圍,然后利用對(duì)數(shù)函數(shù)的單調(diào)性求解;(2)判斷函數(shù)的奇偶性,一定要先求函數(shù)的定義域,再研究f(x)與f(x)的關(guān)系.[針對(duì)訓(xùn)練]6.已知函數(shù)f(x)=xlog2(4x+a)(a∈R且a≥0).(1)若函數(shù)f(x)為奇函數(shù),求實(shí)數(shù)a的值;(2)對(duì)任意的x∈12,1,不等式f(x)f(x)≤1恒成立,求實(shí)數(shù)a課下請(qǐng)完成課時(shí)跟蹤檢測(cè)(三十六)第2課時(shí)對(duì)數(shù)函數(shù)性質(zhì)的應(yīng)用[例1]解:(1)由題意可得x解得0<x<2.所以原不等式的解集為(0,2).(2)當(dāng)x>1時(shí),logx12>1=logxx解得x<12,此時(shí)不等式無(wú)解當(dāng)0<x<1時(shí),logx12>1=logxx,解得x>12,所以12<綜上所述,原不等式的解集為12(3)當(dāng)a>1時(shí),原不等式等價(jià)于2解得x>4.當(dāng)0<a<1時(shí),原不等式等價(jià)于2解得52<x<4綜上所述,當(dāng)a>1時(shí),原不等式的解集為(4,+∞);當(dāng)0<a<1時(shí),原不等式的解集為52[針對(duì)訓(xùn)練]1.選Blog3(x+2)>1?log3(x+2)>log33,又函數(shù)y=log3x在(0,+∞)上單調(diào)遞增,所以x+2>3,解得x>1,故不等式的解集為(1,+∞).2.解析:易知log18(5x6)3=log123(5x6)3=log由log12(2x+3)<log18(5x可得log12(2x+3)<log12(5又函數(shù)y=log12x在(0,+∞)所以可得2x+3>0,5x?6>0,2答案:63.解析:∵log2(x22x)>log2(x+4),∴x2?2x>x+4,x2?2答案:(4,1)∪(4,+∞)[例2]解:由題意,函數(shù)f(x)=log12(x22x設(shè)t=x22x3,令t>0,即x22x3>0,解得x>3或x<1,又由t=(x1)24在區(qū)間(3,+∞)上單調(diào)遞增,在區(qū)間(∞,1)上單調(diào)遞減,又由函數(shù)y=log12t結(jié)合復(fù)合函數(shù)單調(diào)性的判定方法,可得函數(shù)f(x)的單調(diào)遞增區(qū)間為(∞,1),單調(diào)遞減區(qū)間為(3,+∞).[變式拓展]1.解:由(1)知t=x22x3在(∞,1)上單調(diào)遞減,在(3,+∞)上單調(diào)遞增,若0<a<1,則y=logat單調(diào)遞減,所以f(x)=loga(x22x3)在(∞,1)上單調(diào)遞增,在(3,+∞)上單調(diào)遞減.若a>1,則y=logat單調(diào)遞增,所以f(x)=loga(x22x3)在(∞,1)上單調(diào)遞減,在(3,+∞)上單調(diào)遞增.2.解:令y=logat,t=2ax,當(dāng)0<a<1時(shí),y=logat為減函數(shù),t=2ax為減函數(shù),不合題意;當(dāng)a>1時(shí),y=logat為增函數(shù),t=2ax為減函數(shù),符合題意,需要2ax>0在[0,1]上恒成立,即(2ax)min>0,所以2a>0,解得a<2,從而1<a<2.綜上,a的取值范圍為(1,2).[針對(duì)訓(xùn)練]4.選D∵a>0,且a≠1,∴u=ax3為增函數(shù).∴若函數(shù)f(x)為增函數(shù),則f(x)=logau必為增函數(shù).∴a>1.又u=ax3在[1,3]上恒為正,∴a3>0,即a>3.5.解:(1)由4xx2>0,得函數(shù)的定義域是(0,4).令t=4xx2(0<x<4),則y=log1∵t=4xx2=(x2)2+4,∴t=4xx2(0<x<4)的單調(diào)遞減區(qū)間是[2,4),單調(diào)遞增區(qū)間是(0,2).又y=log12t在(0,+∞)上是減函數(shù),∴函數(shù)y=log12(4xx2)的單調(diào)遞減區(qū)間是(0,2),單調(diào)遞增區(qū)間是[2(2)因?yàn)楹瘮?shù)f(x)在(1,+∞)上單調(diào)遞增,又函數(shù)y=log2x在定義域上單調(diào)遞增,所以u(píng)=x2+ax+3在(1,+∞)上單調(diào)遞增,且u>0在(1,+∞)上恒成立,所以a2≤1且12+1×a+3≥0,解得a≥2,即a的取值范圍為[2,+∞)[例3]解:(1)∵x+1>0,∴x>1.函數(shù)f(x)的定義域?yàn)?1,+∞).∵f(x)>0,即log2(x+1)2>0,∴l(xiāng)og2(x+1)>2.∴x+1>4.∴x>3.∴x的取值范圍是(3,+∞).(2)∵x∈(1,3],∴x+1∈(0,4].∴l(xiāng)og2(x+1)∈(∞,2].∴l(xiāng)og2(x+1)2∈(∞,0].∴f(x)的值域?yàn)?∞,0].[針對(duì)訓(xùn)練]6.解:(1)因?yàn)楹瘮?shù)f(x)為奇函數(shù),所以f(x)+f(x)=0對(duì)定義域內(nèi)每一個(gè)元素x恒成立.即f(x)+f(x)=log2(4x+a)xlog2(4x+a)+x=log2[(4x+a)(4x+a)]=log2[1+a(4x+4x)+a2]=0,則1+a(4x+4x)+a2=1,即a(4x+4x+a)=0.又因?yàn)閍≥0,所以4x+4x+a>0,故a=0.(2)因?yàn)閒(x)=log22x4x+a,所以f(x)=log2由f(x)f(x)=log21+a·4得0<1+a·4x4x+a≤2,又a≥0,故只需要1+a·4x≤2·4x+2a,即a(4x2)≤2·4x1對(duì)任意的x∈12,1恒成立.因?yàn)閤∈12,1,所以4x2>0,故因?yàn)閥=2+34x?2在所以2+34x?2min=7綜上所述,實(shí)數(shù)a的取值范圍為0,7板塊綜合指、對(duì)函數(shù)圖象與性質(zhì)的綜合[例1]選C由題意,根據(jù)函數(shù)f(x)=loga(x+b)的圖象,可得0<a<1,0<b<1,根據(jù)指數(shù)函數(shù)y=ax(0<a<1)的圖象與性質(zhì),結(jié)合圖象變換向下移動(dòng)b個(gè)單位長(zhǎng)度,可得函數(shù)g(x)=axb的圖象只有選項(xiàng)C符合.故選C.[例2]選A由題意知關(guān)于x的不等式4x32≤logax在x∈0,12恒成立,所以當(dāng)x∈0,12時(shí),函數(shù)y=4x32的圖象不在由圖可知0<a<1,loga12≥1[針對(duì)訓(xùn)練]1.選C函數(shù)g(x)的定義域是(∞,0),排除A、B;若0<a<1,則f(x)=ax是減函數(shù),此時(shí)g(x)=loga?1x是減函數(shù),C、D都不滿足;若a>1,則f(x)=ax是增函數(shù),此時(shí)g(x)=loga?1x是增函數(shù)2.解析:根據(jù)指數(shù)函數(shù)和對(duì)數(shù)函數(shù)的圖象,畫出f(x)的圖象如圖所示,數(shù)形結(jié)合可知,要滿足題意,只需a∈(0,1].答案:(0,1][例3]選B因?yàn)閥=0.2x在R上單調(diào)遞減,所以0<0.20.3<0.20=1.所以0<a<1.又2b=0.3且y=log2x在(0,+∞)上單調(diào)遞增,所以b=log20.3<log21=0.所以b<0.又y=log0.3x在(0,+∞)上單調(diào)遞減,所以log0.30.2>log0.30.3=1.所以c>1.綜上可知,c>a>b.故選B.[例4]選C由已知得f(x)=loga(a2x2ax2)<0=loga1.又當(dāng)0<a<1時(shí),函數(shù)y=logax在(0,+∞)上單調(diào)遞減,得a2x2ax2>1,即a2x2ax3>0,(ax3)·(ax+1)>0.因?yàn)閍x+1>0,所以ax>3.又0<a<1,所以x<loga3.故選C.[針對(duì)訓(xùn)練]3.選C對(duì)于A,函數(shù)y=3x在R上單調(diào)遞增,因?yàn)?<x<y<1,故3x<3y,A錯(cuò)誤;對(duì)于B,根據(jù)底數(shù)a對(duì)對(duì)數(shù)函數(shù)y=logax的影響:當(dāng)0<a<1時(shí),在x∈(1,+∞)上“底小圖高”.因?yàn)?<x<y<1,所以logx3>logy3,B錯(cuò)誤;對(duì)于C,函數(shù)y=log4x在(0,+∞)上單調(diào)遞增,故log4x<log4y,C正確;對(duì)于D,函數(shù)y=14x在R上單調(diào)遞減,故14x>14.解析:因?yàn)橐筬(x)=lg(2xb)在x∈[1,+∞)時(shí),恒有f(x)≥0,所以有2xb≥1在x∈[1,+∞)時(shí)恒成立,即2x≥b+1在x∈[1,+∞)上恒成立.又因?yàn)橹笖?shù)函數(shù)g(x)=2x在定義域上是增函數(shù).所以只需2≥b+1成立即可,解得b≤1.答案:(∞,1][例5]選A依題意,函數(shù)g(x)=loga(a2x+t)的定義域?yàn)镽,令u=a2x+t,顯然u>0,函數(shù)y=logau在(0,+∞)上單調(diào)性與u=a2x+t在R上單調(diào)性相同,則函數(shù)g(x)在R上單調(diào)遞增,顯然t≥0,而當(dāng)t=0時(shí),函數(shù)g(x)=2x不滿足條件②,因此t>0,由于函數(shù)g(x)在[m,n]上的值域?yàn)閇m,n],則loga(a2m+t)=m,loga(a2n+