高考數(shù)學(xué)第二章函數(shù)2.6對(duì)數(shù)與對(duì)數(shù)函數(shù)課件文新人教A版.ppt

2.6 對(duì)數(shù)與對(duì)數(shù)函數(shù),知識(shí)梳理,考點(diǎn)自測(cè),1.對(duì)數(shù)的概念 (1)根據(jù)下圖的提示填寫與對(duì)數(shù)有關(guān)的概念: (2)a的取值范圍 . 2.對(duì)數(shù)的性質(zhì)與運(yùn)算法則 (1)對(duì)數(shù)的運(yùn)算法則 如果a0,且a1,M0,N0,那么 loga(MN)= ;,指數(shù),對(duì)數(shù),冪,真數(shù),底數(shù),a0,且a1,logaM+logaN,logaM-logaN,知識(shí)梳理,考點(diǎn)自測(cè),知識(shí)梳理,考點(diǎn)自測(cè),4.對(duì)數(shù)函數(shù)的圖象與性質(zhì),(0,+),(1,0),增函數(shù),減函數(shù),知識(shí)梳理,考點(diǎn)自測(cè),5.反函數(shù) 指數(shù)函數(shù)y=ax(a0,且a1)與對(duì)數(shù)函數(shù) (a0,且a1)互為反函數(shù),它們的圖象關(guān)于直線 對(duì)稱.,y=logax,y=x,知識(shí)梳理,考點(diǎn)自測(cè),1.對(duì)數(shù)的性質(zhì)(a0,且a1,M0,b0) (1)loga1=0; (2)logaa=1; (3)logaMn=nlogaM(nR);,2.換底公式的推論 (1)logab·logba=1,即logab= (2)logab·logbc·logcd=logad. 3.對(duì)數(shù)函數(shù)的圖象與底數(shù)大小的比較 如圖,直線y=1與四個(gè)函數(shù)圖象交點(diǎn)的橫坐標(biāo)即為相應(yīng)的底數(shù).,知識(shí)梳理,考點(diǎn)自測(cè),知識(shí)梳理,考點(diǎn)自測(cè),×,×,×,×,知識(shí)梳理,考點(diǎn)自測(cè),A.abc B.acb C.cab D.cba,B,知識(shí)梳理,考點(diǎn)自測(cè),3.(2017河南焦作模擬)若函數(shù)y=a|x|(a0,且a1)的值域?yàn)閥|0y1,則函數(shù)y=loga|x|的圖象大致是 ( ),A,解析:若函數(shù)y=a|x|(a0,且a1)的值域?yàn)閥|0y1,則0a1,由此可知y=loga|x|的圖象大致是選項(xiàng)A中的圖象.,知識(shí)梳理,考點(diǎn)自測(cè),4.(教材習(xí)題改編P68練習(xí)T3)下列運(yùn)算結(jié)果正確的序號(hào)是 . log212-log23=2; +log315=1; 4log23=9;logsin 45°2=-2.,5.(教材例題改編P71例7(2)函數(shù)y=loga(4-x)+1(a0,且a1)的圖象恒過點(diǎn) .,(3,1),解析:當(dāng)4-x=1,即x=3時(shí),y=loga1+1=1. 所以函數(shù)的圖象恒過點(diǎn)(3,1).,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,對(duì)數(shù)式的化簡(jiǎn)與求值 例1化簡(jiǎn)下列各式:,思考對(duì)數(shù)運(yùn)算的一般思路是什么?,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,解題心得對(duì)數(shù)運(yùn)算的一般思路: (1)首先利用冪的運(yùn)算把底數(shù)或真數(shù)進(jìn)行變形,化成分?jǐn)?shù)指數(shù)冪的形式,使冪的底數(shù)最簡(jiǎn),然后正用對(duì)數(shù)運(yùn)算性質(zhì)化簡(jiǎn)合并. (2)將對(duì)數(shù)式化為同底數(shù)對(duì)數(shù)的和、差、倍數(shù)運(yùn)算,然后逆用對(duì)數(shù)的運(yùn)算性質(zhì),轉(zhuǎn)化為同底對(duì)數(shù)真數(shù)的積、商、冪的運(yùn)算.,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,D,4,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,對(duì)數(shù)函數(shù)的圖象及其應(yīng)用,C,B,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,思考應(yīng)用對(duì)數(shù)型函數(shù)的圖象主要解決哪些問題? 解題心得應(yīng)用對(duì)數(shù)型函數(shù)的圖象可求解的問題: (1)對(duì)一些可通過平移、對(duì)稱變換作出其圖象的對(duì)數(shù)型函數(shù),在求解其單調(diào)性(單調(diào)區(qū)間)、值域(最值)、零點(diǎn)時(shí),常利用數(shù)形結(jié)合思想. (2)一些對(duì)數(shù)型方程、不等式問題常轉(zhuǎn)化為相應(yīng)的函數(shù)圖象問題,利用數(shù)形結(jié)合法求解.,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,對(duì)點(diǎn)訓(xùn)練2(1)(2017福建泉州一模,文7)函數(shù)f(x)=ln(x+1)+ln(x-1) +cos x的圖象大致是( ),A,D,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,解析: (1)函數(shù)f(x)=ln(x+1)+ln(x-1)+cos x,則函數(shù)的定義域?yàn)閤1,故排除C,D; -1cos x1,當(dāng)x+時(shí),f(x)+,故選A.,設(shè)曲線y=x2-2x在x=0處的切線l的斜率為k, 由y'=2x-2,可知k=y'|x=0=-2. 要使|f(x)|ax,則直線y=ax的傾斜角要大于等于直線l的傾斜角,小于等于,即a的取值范圍是-2,0.,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,對(duì)數(shù)函數(shù)的性質(zhì)及其應(yīng)用(多考向) 考向1 比較含對(duì)數(shù)的函數(shù)值的大小 例3(2017天津,文6)已知奇函數(shù)f(x)在R上是增函數(shù), 若a=-f ,b=f(log24.1),c=f(20.8),則a,b,c的大小關(guān)系為( ) A.abc B.bac C.cba D.cab,C,log25log24.1log24=2,20.8log24.120.8. 又f(x)在R上是增函數(shù), f(log25)f(log24.1)f(20.8),即abc.故選C.,思考如何比較兩個(gè)含對(duì)數(shù)的函數(shù)值的大小?,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考向2 解含對(duì)數(shù)的函數(shù)不等式,C,C,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,思考如何解簡(jiǎn)單對(duì)數(shù)不等式?,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考向3 對(duì)數(shù)型函數(shù)的綜合問題 例5已知f(x)=loga(ax-1)(a0,且a1). (1)求f(x)的定義域; (2)討論函數(shù)f(x)的單調(diào)性.,解 (1)由ax-10,得ax1. 當(dāng)a1時(shí),x0;當(dāng)01時(shí),f(x)的定義域?yàn)?0,+); 當(dāng)01時(shí),設(shè)0x1x2,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,思考在判斷對(duì)數(shù)型復(fù)合函數(shù)的單調(diào)性時(shí)需要注意哪些條件? 解題心得1.比較含對(duì)數(shù)的函數(shù)值的大小,首先應(yīng)確定對(duì)應(yīng)函數(shù)的單調(diào)性,然后比較含對(duì)數(shù)的自變量的大小,同底數(shù)的可借助函數(shù)的單調(diào)性;底數(shù)不同、真數(shù)相同的可以借助函數(shù)的圖象;底數(shù)、真數(shù)均不同的可借助中間值(0或1). 2.解簡(jiǎn)單對(duì)數(shù)不等式,先統(tǒng)一底數(shù),再利用函數(shù)的單調(diào)性,要注意對(duì)底數(shù)a的分類討論. 3.在判斷對(duì)數(shù)型復(fù)合函數(shù)的單調(diào)性時(shí),一定要明確底數(shù)a對(duì)增減性的影響,以及真數(shù)必須為正的限制條件.,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,對(duì)點(diǎn)訓(xùn)練3(1)已知定義在R上的函數(shù)f(x)=2|x-m|-1(m為實(shí)數(shù))為偶函數(shù).記a=f(log0.53),b=f(log25),c=f(2m),則a,b,c的大小關(guān)系為( ) A.a0,且a1. 求f(x)的定義域; 判斷f(x)的奇偶性,并予以證明; 當(dāng)a1時(shí),求使f(x)0的x的取值范圍.,C,A,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,1.多個(gè)對(duì)數(shù)函數(shù)圖象比較底數(shù)大小的問題,可通過圖象與直線y=1交點(diǎn)的橫坐標(biāo)進(jìn)行判定. 2.研究對(duì)數(shù)型函數(shù)的圖象時(shí),一般從最基本的對(duì)數(shù)函數(shù)的圖象入手,通過平移、伸縮、對(duì)稱變換得到.特別地,要注意底數(shù)a1和0a1的兩種不同情況.有些復(fù)雜的問題,借助于函數(shù)圖象來解決,就變得簡(jiǎn)單了,這是數(shù)形結(jié)合思想的重要體現(xiàn). 3.利用對(duì)數(shù)函數(shù)單調(diào)性可解決比較大小、解不等式、求最值等問題,其基本方法是“同底法”,即把不同底的對(duì)數(shù)式化為同底的對(duì)數(shù)式，然后根據(jù)單調(diào)性來解決.,考點(diǎn)一,考點(diǎn)二,考點(diǎn)三,