4.4.2對(duì)數(shù)函數(shù)的圖象和性質(zhì)第1課時(shí)對(duì)數(shù)函數(shù)的圖象和性質(zhì)——(教學(xué)方式:深化學(xué)習(xí)課梯度進(jìn)階式教學(xué))[課時(shí)目標(biāo)]1.初步掌握對(duì)數(shù)函數(shù)的圖象和性質(zhì).2.會(huì)類(lèi)比指數(shù)函數(shù)研究對(duì)數(shù)函數(shù)的性質(zhì).3.掌握對(duì)數(shù)函數(shù)的圖象和性質(zhì)的簡(jiǎn)單應(yīng)用. 4.了解反函數(shù)的概念與圖象特點(diǎn).(一)對(duì)數(shù)函數(shù)的圖象和性質(zhì)項(xiàng)目0<a<1a>1圖象定義域

值域

過(guò)定點(diǎn),即x=1時(shí),y=0

函數(shù)值的變化當(dāng)0<x<1時(shí),,當(dāng)x>1時(shí),

當(dāng)0<x<1時(shí),,當(dāng)x>1時(shí),

單調(diào)性

|微|點(diǎn)|助|解|(1)注意點(diǎn):討論對(duì)數(shù)函數(shù)的性質(zhì)時(shí),若底數(shù)a的大小不確定,必須分a>1和0<a<1兩種情況進(jìn)行討論.(2)圖象的特點(diǎn):函數(shù)y=logax(a>0,且a≠1)的圖象無(wú)限靠近y軸,但永遠(yuǎn)不會(huì)與y軸相交;在同一坐標(biāo)系內(nèi),y=logax(a>0,且a≠1)的圖象與y=log1ax(a>0,且a≠1)的圖象關(guān)于x軸(即直線y=0)(3)底數(shù)對(duì)圖象的影響:比較圖象與y=1的交點(diǎn),此時(shí)y=1與對(duì)數(shù)函數(shù)圖象交點(diǎn)的坐標(biāo)為(a,1).交點(diǎn)的橫坐標(biāo)越大,對(duì)應(yīng)的對(duì)數(shù)函數(shù)的底數(shù)越大,即沿著直線y=1由左向右看,底數(shù)a增大(如圖).(二)反函數(shù)1.定義一般地,指數(shù)函數(shù)y=ax(a>0,且a≠1)與對(duì)數(shù)函數(shù)互為反函數(shù).它們的定義域與值域正好互換.

2.性質(zhì)(1)互為反函數(shù)的兩個(gè)函數(shù)的單調(diào)性相同,但單調(diào)區(qū)間不一定相同.(2)互為反函數(shù)的兩個(gè)函數(shù)的圖象關(guān)于直線y=x對(duì)稱(chēng).基礎(chǔ)落實(shí)訓(xùn)練1.判斷正誤(正確的畫(huà)“√”,錯(cuò)誤的畫(huà)“×”)(1)函數(shù)y=logax(a>0,且a≠1)的圖象過(guò)定點(diǎn)(1,0).()(2)函數(shù)y=logax(a>0,且a≠1)在(0,+∞)上具有單調(diào)性.()(3)由函數(shù)y=log2x的圖象向左平移1個(gè)單位長(zhǎng)度可得y=log2x+1的圖象.()2.已知函數(shù)f(x)=logax的圖象如圖所示,則a的取值可能是()A.10 B.1C.13 D.3.若函數(shù)y=f(x)是函數(shù)y=3x的反函數(shù),則f12的值為()A.log23 B.log32 C.19 D.4.若函數(shù)f(x)=log(a+1)x在(0,+∞)上單調(diào)遞減,則a的取值范圍為.

題型(一)對(duì)數(shù)函數(shù)的圖象及應(yīng)用[例1]函數(shù)y=|lg(x+1)|的圖象是()聽(tīng)課記錄:[例2]已知函數(shù)f(x)=loga(xm)+n的圖象恒過(guò)定點(diǎn)(3,5),則lgm+lgn的值是.

聽(tīng)課記錄:[變式拓展]若本例1中的函數(shù)y=|lg(x+1)|變?yōu)閥=f(x)=|log3x|,若f(a)>f(2),則a的取值范圍為.

|思|維|建|模|(1)對(duì)數(shù)函數(shù)y=logax(a>0且a≠1)的圖象恒過(guò)定點(diǎn)(1,0),即當(dāng)x=1時(shí),y=0,令其真數(shù)等于1,可得圖象過(guò)定點(diǎn)的坐標(biāo).(2)對(duì)數(shù)型函數(shù)圖象的變換方法①作y=f(|x|)的圖象時(shí),保留y=f(x)(x>0)的圖象不變,當(dāng)x<0時(shí),y=f(|x|)的圖象與y=f(x)(x>0)的圖象關(guān)于y軸對(duì)稱(chēng).②作y=|f(x)|的圖象時(shí),保留y=f(x)的x軸及上方圖象不變,把x軸下方圖象以x軸為對(duì)稱(chēng)軸翻折上去即可.③有關(guān)對(duì)數(shù)函數(shù)平移也符合“左加右減,上加下減”的規(guī)律.④y=f(x)與y=f(x)的圖象關(guān)于y軸對(duì)稱(chēng),y=f(x)與y=f(x)的圖象關(guān)于x軸對(duì)稱(chēng),y=f(x)與y=f(x)的圖象關(guān)于原點(diǎn)對(duì)稱(chēng).[針對(duì)訓(xùn)練]1.設(shè)a與b均為實(shí)數(shù),a>0且a≠1,已知函數(shù)y=loga(x+b)的圖象如圖所示,則a+2b的值為()A.6 B.8 C.10 D.122.函數(shù)f(x)=loga|x|+1(a>1)的圖象大致為()3.若函數(shù)y=loga(x+b)+c(a>0,且a≠1)的圖象恒過(guò)定點(diǎn)(3,2),則實(shí)數(shù)b=,c=.

題型(二)比較大小問(wèn)題[例3]比較下列各組數(shù)的大小.(1)log1245(2)log123與log(3)loga2與loga3.聽(tīng)課記錄:|思|維|建|模|比較對(duì)數(shù)值大小的常用方法(1)同底數(shù)的利用對(duì)數(shù)函數(shù)的單調(diào)性.(2)同真數(shù)的利用對(duì)數(shù)函數(shù)的圖象或用換底公式轉(zhuǎn)化.(3)底數(shù)和真數(shù)都不同,找中間量.[提醒]比較數(shù)的大小時(shí)先利用性質(zhì)比較出與0或1的大小.[針對(duì)訓(xùn)練]4.已知a=log2e,b=ln2,c=log1213,則a,b,c的大小關(guān)系為(A.a>b>c B.b>a>cC.c>b>a D.c>a>b5.已知log12m<log12n<0,則(A.n<m<1 B.m<n<1C.1<m<n D.1<n<m題型(三)反函數(shù)[例4]若函數(shù)f(x)與g(x)的圖象關(guān)于直線y=x對(duì)稱(chēng),函數(shù)f(x)=12(?x),則f(2)+g(4)=(A.3 B.4 C.5 D.6聽(tīng)課記錄:|思|維|建|模|互為反函數(shù)的函數(shù)的性質(zhì)(1)同底數(shù)的指數(shù)函數(shù)與對(duì)數(shù)函數(shù)互為反函數(shù).(2)互為反函數(shù)的定義域與值域互換.(3)互為反函數(shù)的兩個(gè)函數(shù)的圖象關(guān)于直線y=x對(duì)稱(chēng).[針對(duì)訓(xùn)練]6.函數(shù)y=ax(a>0,且a≠1)的反函數(shù)的圖象過(guò)點(diǎn)(a,a),則a的值為()A.2 B.1C.2或12 D.7.函數(shù)y=log3x13≤x≤81的反函數(shù)的定義域?yàn)?A.(0,+∞) B.1C.(1,4) D.[1,4]課下請(qǐng)完成課時(shí)跟蹤檢測(cè)(三十五)4.4.2對(duì)數(shù)函數(shù)的圖象和性質(zhì)第1課時(shí)對(duì)數(shù)函數(shù)的圖象和性質(zhì)?課前預(yù)知教材(一)(0,+∞)R(1,0)y>0y<0y<0y>0減函數(shù)增函數(shù)(二)1.y=logax(a>0,且a≠1)[基礎(chǔ)落實(shí)訓(xùn)練]1.(1)√(2)√(3)×2.A3.B4.(1,0)?課堂題點(diǎn)研究[例1]選A由于函數(shù)y=lg(x+1)的圖象可由函數(shù)y=lgx的圖象左移一個(gè)單位長(zhǎng)度而得到,函數(shù)y=lgx的圖象與x軸的交點(diǎn)是(1,0),故函數(shù)y=lg(x+1)的圖象與x軸的交點(diǎn)是(0,0),即函數(shù)y=|lg(x+1)|的圖象與x軸的公共點(diǎn)是(0,0),顯然四個(gè)選項(xiàng)只有A選項(xiàng)滿足.[例2]解析:因?yàn)楹瘮?shù)f(x)=loga(xm)+n的圖象恒過(guò)定點(diǎn)(3,5),故3m=1,且n=5,則m=2,n=5.所以lgm+lgn=lg2+lg5=lg10=1.答案:1[變式拓展]解析:作出函數(shù)f(x)的圖象,如圖所示.由于f(2)=f12,故結(jié)合圖象可知,當(dāng)f(a)>f(2)時(shí),a的取值范圍為0,12∪(2,答案:0,12∪(2,[針對(duì)訓(xùn)練]1.選C令f(x)=y=loga(x+b),由題圖可知,f(0)=logab=2,f(3)=loga(3+b)=0,即a2=b,?3+b=1,a>0,a≠1,2.選C∵函數(shù)f(x)=loga|x|+1(a>1)是偶函數(shù),∴f(x)的圖象關(guān)于y軸對(duì)稱(chēng),當(dāng)x>0時(shí),f(x)=logax+1單調(diào)遞增;當(dāng)x<0時(shí),f(x)=loga(x)+1單調(diào)遞減,又∵圖象過(guò)(1,1),(1,1)兩點(diǎn),結(jié)合選項(xiàng)可知選C.3.解析:∵函數(shù)的圖象恒過(guò)定點(diǎn)(3,2),∴將(3,2)代入y=loga(x+b)+c,得2=loga(3+b)+c.又當(dāng)a>0,且a≠1時(shí),loga1=0恒成立,∴c=2,3+b=1.∴b=2,c=2.答案:22[例3]解:(1)y=log12x在(0,+∞)上單調(diào)遞減,因?yàn)?5<67,所以lo(2)法一log123lo=lg3lg12lg3因?yàn)閥=lgx是增函數(shù),所以lg15<lg12<0<lg所以log123log153<0.所以log法二因?yàn)樵趚∈(1,+∞)上,y=log15x的圖象在y=log12x圖象的上方,所以log(3)當(dāng)a>1時(shí),y=logax為增函數(shù),所以loga2<loga3;當(dāng)0<a<1時(shí),y=logax為減函數(shù),所以loga2>loga3.[針對(duì)訓(xùn)練]4.選Dlog1213=log23>log2e,則c>a.因?yàn)閎=ln2=1log2e<1<log2e=a,所以a>b.因此c5.選D因?yàn)?<12<1,log12m<log12n<0,所以m>n[例4]選D法一∵函數(shù)f(x)與g(x)的圖象關(guān)于直線y=x對(duì)稱(chēng),f(x)=12(?x)=2x,∴g(x)=log2x,∴f(2)+g(4)=22法二∵f(x)=12(?x),∴f(2)=4,故函數(shù)f(x)的圖象經(jīng)過(guò)點(diǎn)(2,4).∵函數(shù)f(x)與g(x)的圖象關(guān)于直線y=x對(duì)稱(chēng),∴函數(shù)g(x)的圖象經(jīng)過(guò)點(diǎn)(4,2).∴g(4)=2,∴f(2)+f([針對(duì)訓(xùn)練]6.選B法一函