第一　對(duì)數(shù)函數(shù)的圖象及性質(zhì)ppt課件

1、2.2.22.2.2對(duì)數(shù)函數(shù)及其性質(zhì)對(duì)數(shù)函數(shù)及其性質(zhì)第一課時(shí)對(duì)數(shù)函數(shù)的圖象及性質(zhì)第一課時(shí)對(duì)數(shù)函數(shù)的圖象及性質(zhì)課標(biāo)要求課標(biāo)要求:1.:1.初步了解對(duì)數(shù)函數(shù)的概念初步了解對(duì)數(shù)函數(shù)的概念.2.2.掌握對(duì)數(shù)函數(shù)的圖象和性質(zhì)掌握對(duì)數(shù)函數(shù)的圖象和性質(zhì).3.3.了解反函數(shù)的概念了解反函數(shù)的概念, ,知道指數(shù)函數(shù)與對(duì)數(shù)函數(shù)互為反函數(shù)知道指數(shù)函數(shù)與對(duì)數(shù)函數(shù)互為反函數(shù).4.4.經(jīng)過類比思經(jīng)過類比思想想, ,利用指數(shù)函數(shù)探求對(duì)數(shù)函數(shù)的圖象及性質(zhì)利用指數(shù)函數(shù)探求對(duì)數(shù)函數(shù)的圖象及性質(zhì), ,學(xué)會(huì)研討函數(shù)的方法學(xué)會(huì)研討函數(shù)的方法. .自主學(xué)習(xí)自主學(xué)習(xí)新知建構(gòu)新知建構(gòu)自我整合自我整合【情境導(dǎo)學(xué)】【情境導(dǎo)學(xué)】導(dǎo)入某種細(xì)胞分裂時(shí)

2、導(dǎo)入某種細(xì)胞分裂時(shí), ,得到分裂個(gè)數(shù)得到分裂個(gè)數(shù)t t是分裂次數(shù)是分裂次數(shù)n n的函數(shù)的函數(shù), ,可以用指數(shù)可以用指數(shù)函數(shù)表示為函數(shù)表示為t=2n,t=2n,反過來(lái)反過來(lái), ,假設(shè)知道分裂后的細(xì)胞個(gè)數(shù)也可求出分裂的次假設(shè)知道分裂后的細(xì)胞個(gè)數(shù)也可求出分裂的次數(shù)數(shù)n,n,即即n=log2t,n=log2t,而且對(duì)于每一個(gè)細(xì)胞個(gè)數(shù)而且對(duì)于每一個(gè)細(xì)胞個(gè)數(shù)t,t,有獨(dú)一的分裂次數(shù)有獨(dú)一的分裂次數(shù)n n與之相對(duì)與之相對(duì)應(yīng)應(yīng), ,因此因此n n是關(guān)于是關(guān)于t t的函數(shù)的函數(shù). .習(xí)慣上仍用習(xí)慣上仍用x x表示自變量表示自變量,y,y表示它的函數(shù)表示它的函數(shù), ,即即y=log2x.y=log2x.這就是本節(jié)

3、我們要研討的對(duì)數(shù)函數(shù)這就是本節(jié)我們要研討的對(duì)數(shù)函數(shù). .1.1.對(duì)數(shù)函數(shù)的概念對(duì)數(shù)函數(shù)的概念函數(shù)函數(shù) 叫做對(duì)數(shù)函數(shù)叫做對(duì)數(shù)函數(shù), ,其中其中 是自變量是自變量, ,函數(shù)的定函數(shù)的定義域是義域是 . .知識(shí)探求知識(shí)探求y=logax(a0,y=logax(a0,且且a1)a1)(0,+)(0,+)2.2.對(duì)數(shù)函數(shù)的圖象與性質(zhì)對(duì)數(shù)函數(shù)的圖象與性質(zhì)a1a10a10a0,y=logax(a0,且且a1)a1)和指數(shù)函數(shù)和指數(shù)函數(shù)y=ax(a0,y=ax(a0,且且a1)a1)互為互為 . .探求探求1:1:同底數(shù)的指數(shù)、對(duì)數(shù)函數(shù)的定義域、值域有何關(guān)系同底數(shù)的指數(shù)、對(duì)數(shù)函數(shù)的定義域、值域有何關(guān)系? ?答

4、案答案: :同底數(shù)的指數(shù)函數(shù)的定義域是同底數(shù)對(duì)數(shù)函數(shù)的值域同底數(shù)的指數(shù)函數(shù)的定義域是同底數(shù)對(duì)數(shù)函數(shù)的值域, ,指數(shù)函數(shù)的指數(shù)函數(shù)的值域是對(duì)數(shù)函數(shù)的定義域值域是對(duì)數(shù)函數(shù)的定義域. .探求探求2:2:互為反函數(shù)的兩個(gè)函數(shù)圖象有何特征互為反函數(shù)的兩個(gè)函數(shù)圖象有何特征? ?答案答案: :關(guān)于直線關(guān)于直線y=xy=x對(duì)稱對(duì)稱. .反函數(shù)反函數(shù)自我檢測(cè)自我檢測(cè)1.(1.(概念概念) )以下函數(shù)是對(duì)數(shù)函數(shù)的是以下函數(shù)是對(duì)數(shù)函數(shù)的是( ( ) )(A)y=loga(2x)(A)y=loga(2x)(B)y=log22x(B)y=log22x(C)y=log2x+1(C)y=log2x+1(D)y=lg x(D

5、)y=lg xD D 2.(2.(解析式解析式) )假設(shè)對(duì)數(shù)函數(shù)過點(diǎn)假設(shè)對(duì)數(shù)函數(shù)過點(diǎn)(4,2),(4,2),那么其解析式為那么其解析式為( ( ) )(A)y=(A)y= (B)y=2x (B)y=2x(C)y=log4x(C)y=log4x (D)y=log2x (D)y=log2xD D 12x3.(3.(定義域定義域) )函數(shù)函數(shù)y=log3(x-4)y=log3(x-4)的定義域?yàn)榈亩x域?yàn)? ( ) )(A)R (B)(-,4)(4,+)(A)R (B)(-,4)(4,+)(C)(-,4) (D)(4,+)(C)(-,4) (D)(4,+)D D 4.(4.(單調(diào)性單調(diào)性) )函數(shù)函

6、數(shù)y=ln xy=ln x的單調(diào)遞增區(qū)間是的單調(diào)遞增區(qū)間是( ( ) )(A)e,+)(A)e,+) (B)(0,+) (B)(0,+)(C)(-,+)(C)(-,+)(D)1,+)(D)1,+)5.(5.(圖象圖象) )函數(shù)函數(shù)y=loga(x-2)+3(a0y=loga(x-2)+3(a0且且a1)a1)的圖象恒過定點(diǎn)的圖象恒過定點(diǎn).答案答案:(3,3):(3,3)B B題型一題型一 對(duì)數(shù)函數(shù)的概念對(duì)數(shù)函數(shù)的概念課堂探求課堂探求典例分析典例分析舉一反三舉一反三解析解析:(1):(1)由對(duì)數(shù)函數(shù)定義知由對(duì)數(shù)函數(shù)定義知, ,是對(duì)數(shù)函數(shù)是對(duì)數(shù)函數(shù). .應(yīng)選應(yīng)選D.D.答案答案:(1)D:(1)D

7、答案答案:(2)4:(2)4(3)2(3)2方法技巧方法技巧 (1) (1)判別一個(gè)函數(shù)是對(duì)數(shù)函數(shù)必需是形如判別一個(gè)函數(shù)是對(duì)數(shù)函數(shù)必需是形如y=logax(a0y=logax(a0且且a1)a1)的方式的方式, ,即必需滿足以下條件即必需滿足以下條件: :系數(shù)為系數(shù)為1;1;底數(shù)為大于底數(shù)為大于0 0且不等于且不等于1 1的常數(shù)的常數(shù); ;對(duì)數(shù)的真數(shù)僅有自變量對(duì)數(shù)的真數(shù)僅有自變量x.x.(2)(2)假設(shè)知對(duì)數(shù)函數(shù)過定點(diǎn)求解析式時(shí)假設(shè)知對(duì)數(shù)函數(shù)過定點(diǎn)求解析式時(shí), ,常用待定系數(shù)法常用待定系數(shù)法, ,設(shè)設(shè)f(x)=logax f(x)=logax (a0(a0且且a1),a1),將定點(diǎn)代入后利用指

8、對(duì)數(shù)式互化或指數(shù)冪的運(yùn)算性質(zhì)求將定點(diǎn)代入后利用指對(duì)數(shù)式互化或指數(shù)冪的運(yùn)算性質(zhì)求a.a.題型二題型二 對(duì)數(shù)函數(shù)的圖象特征對(duì)數(shù)函數(shù)的圖象特征(2)(2021(2)(2021河南高一期末河南高一期末) )函數(shù)函數(shù)y=lg|x-1|y=lg|x-1|的圖象是的圖象是( () )解析解析:(2):(2)當(dāng)當(dāng)x1x1時(shí)時(shí),y=lg|x-1|=lg(x-1),y=lg|x-1|=lg(x-1),當(dāng)當(dāng)x1x0y=loga(x+b)(a0且且a1)a1)的圖象過點(diǎn)的圖象過點(diǎn)(-(-1,0),(0,1),1,0),(0,1),那么那么lg a+lg b=lg a+lg b=. . 解析解析: :由題意可得由題意可

9、得0=loga(-1+b),1=logab,0=loga(-1+b),1=logab,解得解得a=b=2,a=b=2,所以所以lg a+lg b= lg a+lg b= 2lg 2.2lg 2.答案答案:2lg 2:2lg 2題型三題型三 與對(duì)數(shù)函數(shù)有關(guān)的定義域問題與對(duì)數(shù)函數(shù)有關(guān)的定義域問題【例【例3 3】 求以下函數(shù)的定義域求以下函數(shù)的定義域: :(1)f(x)=lg(x-2)+ ;(1)f(x)=lg(x-2)+ ;規(guī)范解答規(guī)范解答:(1):(1)要使函數(shù)有意義要使函數(shù)有意義, ,需滿足需滿足 2 2分分解得解得x2x2且且x3, 3x3, 3分分所以函數(shù)定義域?yàn)樗院瘮?shù)定義域?yàn)?2,3)(3,+). 4(2,3)(3,+). 4分分13x 20,30,xx(2)f(x)=logx+1(16-4x).(2)f(x)=logx+1(16-4x).規(guī)范解答規(guī)范解答:(2):(2)要使函數(shù)有意義要使函數(shù)有意義, ,需滿足需滿足 7 7分分解得解得-1x0-1x0或或0 x4, 90 x4, 9分分所以函數(shù)定義域?yàn)樗院瘮?shù)定義域?yàn)?-1,0)(0,4). 10(-1,0)(0,4). 10分分1641