高中數(shù)學(xué) 第二章 基本初等函數(shù)（Ⅰ）2.2 對(duì)數(shù)函數(shù) 2.2.2 對(duì)數(shù)函數(shù)及其性質(zhì) 第1課時(shí) 對(duì)數(shù)函數(shù)的圖象及性質(zhì)課件 新人教版必修1

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.初步理解對(duì)數(shù)函數(shù)的概念初步理解對(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.了了 解反函數(shù)的概念解反函數(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.通過類比思想通過類比思想, ,利利 用指數(shù)函數(shù)探索對(duì)數(shù)函數(shù)的圖象及性質(zhì)用指數(shù)函數(shù)探索對(duì)數(shù)函數(shù)的圖象及性質(zhì), ,學(xué)會(huì)研究函數(shù)的方法學(xué)會(huì)研究函數(shù)的方法. . 自主學(xué)習(xí)自主學(xué)習(xí) 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)

2、叫做對(duì)數(shù)函數(shù), ,其中其中x x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)探究知識(shí)探究 y=logy=loga ax(a0,x(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,x(a0,且且a1)a1)和指數(shù)函數(shù)和指數(shù)函數(shù)y=ay=ax x(a0,(a0,且且a1)a1)互為互為 . . 反函數(shù)反函數(shù) 自我檢測(cè)自我檢測(cè) D D B B C C 5.5.函數(shù)函數(shù)y=2+logy=2+log2 2x(x1)x(x1)的值域?yàn)榈闹涤驗(yàn)? . 解析解析: :當(dāng)當(dāng)x1x1時(shí)時(shí),log,log2 2x0,x0,

3、所以所以y=2+logy=2+log2 2x2.x2. 答案答案: :2,+)2,+) 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 課堂探究課堂探究 【例例1 1】 (1) (1)下列函數(shù)中下列函數(shù)中, ,是對(duì)數(shù)函數(shù)的是是對(duì)數(shù)函數(shù)的是. . y=logy=loga ax x2 2(a0,(a0,且且a1);a1);y=logy=log2 2x-1;x-1;y=2logy=2log8 8x;x;y=logy=logx xa(x0,a(x0,且且 x1);x1);y=logy=log5 5x.x. 解析解析: :(1)(1)中真數(shù)不是自變量中真數(shù)不是自變量x,x,不是對(duì)數(shù)函數(shù)不是對(duì)數(shù)函數(shù); ; 中對(duì)

4、數(shù)式后減中對(duì)數(shù)式后減1,1,故不是對(duì)數(shù)函數(shù)故不是對(duì)數(shù)函數(shù); ; 中中l(wèi)oglog8 8x x前的系數(shù)是前的系數(shù)是2,2,而不是而不是1,1,故不是對(duì)數(shù)函數(shù)故不是對(duì)數(shù)函數(shù); ; 中底數(shù)是自變量中底數(shù)是自變量x,x,而非常數(shù)而非常數(shù)a,a,故不是對(duì)數(shù)函數(shù)故不是對(duì)數(shù)函數(shù); ; 是對(duì)數(shù)函數(shù)是對(duì)數(shù)函數(shù). . 答案答案: :(1)(1) 答案答案: :(2)f(x)=log(2)f(x)=log3 3x x 方法技巧方法技巧 (1)(1)判斷一個(gè)函數(shù)是對(duì)數(shù)函數(shù)必須是形如判斷一個(gè)函數(shù)是對(duì)數(shù)函數(shù)必須是形如y=logy=loga ax(a0 x(a0且且a1)a1)的的 形式形式, ,即必須滿足以下條件即必須滿

5、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)若已知對(duì)數(shù)函數(shù)過定點(diǎn)求解析式時(shí)若已知對(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)=logf(x)=loga ax(a0 x(a0 且且a1),a1),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. 解析解析: :(1)a(1)a2 2-a+1=1,-a+1=1,解得解得a=0a=0或或a=1.a=1. 又又a+10,a+10,且且a+11

6、,a+11,所以所以a=1.a=1. (2)(2)因?yàn)楹瘮?shù)因?yàn)楹瘮?shù)y=logy=loga ax(a0 x(a0且且a1)a1)的圖象恒過的圖象恒過(1,0),(1,0), 則令則令x+1=1,x+1=1,得得x=0,x=0, 此時(shí)此時(shí)y=logy=loga a(0+1)-2=-2,(0+1)-2=-2, 所以函數(shù)圖象恒過定點(diǎn)所以函數(shù)圖象恒過定點(diǎn)(0,-2).(0,-2). 答案答案: :(1)1(1)1(2)(0,-2)(2)(0,-2) 即時(shí)訓(xùn)練即時(shí)訓(xùn)練1 1- -1:1:(1)(1)函數(shù)函數(shù)f(x)=(af(x)=(a2 2-a+1)log-a+1)loga+1 a+1x x是對(duì)數(shù)函數(shù) 是

7、對(duì)數(shù)函數(shù), ,則實(shí)數(shù)則實(shí)數(shù)a=a=; ; (2)(2)函數(shù)函數(shù)y=logy=loga a(x+1)-2(a0(x+1)-2(a0且且a1)a1)的圖象恒過點(diǎn)的圖象恒過點(diǎn). . 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)(2)函數(shù)函數(shù)y=logy=loga a|x|+1(0a1)|x|+1(0a0a0且且a1,a1,函數(shù)函數(shù)y=ay=ax x與與y=logy=loga a(-x)(-x)的圖象可能是的圖象可能是( () ) 解析解析: :由由y=logy=loga a(-x)(-x)的定義域?yàn)榈亩x域?yàn)?-,0)(-,0)知知, ,圖象應(yīng)在圖象應(yīng)在y y軸左側(cè)軸左側(cè), ,可排除可排除A,DA,D 選項(xiàng)選項(xiàng). . 當(dāng)當(dāng)a1a1時(shí)時(shí),y=a,y=ax x應(yīng)為增函數(shù)應(yīng)為增函數(shù),y=log,y=loga a(-x)(-x)應(yīng)為減函數(shù)應(yīng)為減函數(shù), ,可知可知B B項(xiàng)正確項(xiàng)正確. . 而對(duì)而對(duì)C C項(xiàng)項(xiàng), ,由圖象知由圖象知y=ay=ax x遞減遞減0a10a1y=logy=loga a(-x)(-x)應(yīng)為增函數(shù)應(yīng)為增函數(shù). .與與C C圖不符圖不符. . 故選故選B.B. 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)的定義域問題題型三題型三 方法技巧方法技巧 求對(duì)數(shù)型復(fù)合函數(shù)的定義域時(shí)