人教版導(dǎo)與練總復(fù)習(xí)數(shù)學(xué)一輪教學(xué)課件：第二章第6節(jié)　對(duì)數(shù)與對(duì)數(shù)函數(shù)

1、第6節(jié)對(duì)數(shù)與對(duì)數(shù)函數(shù)課程標(biāo)準(zhǔn)要求1.理解對(duì)數(shù)的概念和運(yùn)算性質(zhì),知道用換底公式能將一般對(duì)數(shù)轉(zhuǎn)化成自然對(duì)數(shù)或常用對(duì)數(shù).2.通過(guò)具體實(shí)例,了解對(duì)數(shù)函數(shù)的概念.能用描點(diǎn)法或借助計(jì)算工具畫出具體對(duì)數(shù)函數(shù)的圖象,探索并了解對(duì)數(shù)函數(shù)的單調(diào)性與特殊點(diǎn).3.知道對(duì)數(shù)函數(shù)y=logax與指數(shù)函數(shù)y=ax互為反函數(shù)(a0,且a1).必備知識(shí)課前回顧 回歸教材 夯實(shí)四基關(guān)鍵能力課堂突破 類分考點(diǎn) 落實(shí)四翼必備知識(shí)課前回顧 回歸教材 夯實(shí)四基知識(shí)梳理1.對(duì)數(shù)對(duì)數(shù)x=logaNNlogaM+logaNlogaM-logaNnlogaM2.對(duì)數(shù)函數(shù)的圖象與性質(zhì)a10a1時(shí), ;當(dāng)0 x1時(shí), ;當(dāng)0 x0y0y0增減3.指

2、數(shù)函數(shù)與對(duì)數(shù)函數(shù)的關(guān)系一般地,指數(shù)函數(shù)y=ax(a0,且a1)與對(duì)數(shù)函數(shù)y=logax(a0,且a1)互為反函數(shù),它們的定義域與值域正好 ,圖象關(guān)于直線 對(duì)稱.互換y=x重要結(jié)論1.換底公式及其推論(3)logablogbclogcd=logad.2.對(duì)數(shù)函數(shù)的圖象與底數(shù)大小的比較如圖,作直線y=1,則該直線與四個(gè)函數(shù)圖象交點(diǎn)的橫坐標(biāo)為相應(yīng)的底數(shù),故0cd1a1時(shí)與0a1或0a1這兩種不同的情況.角度三 對(duì)數(shù)函數(shù)圖象的應(yīng)用解題策略求解與對(duì)數(shù)型方程、不等式有關(guān)的恒成立問題,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.針對(duì)訓(xùn)練 1.當(dāng)0a1時(shí),在同一平面直角坐標(biāo)系中,函數(shù)y=a-x與y=log

3、ax的圖象是()2.函數(shù)y=|log2x|的圖象是()考點(diǎn)三對(duì)數(shù)函數(shù)的性質(zhì)及其應(yīng)用角度一 利用對(duì)數(shù)函數(shù)的單調(diào)性比較大小例2-1 (1)已知a=log36,b=log510,c=log714,則a,b,c的大小關(guān)系為()A.abc B.cbaC.cab D.bclog52log72,故cba.故選B.解題策略比較對(duì)數(shù)式大小的類型及相應(yīng)的方法(1)若底數(shù)為同一常數(shù),則可由對(duì)數(shù)函數(shù)的單調(diào)性直接進(jìn)行判斷;若底數(shù)為同一字母,則需對(duì)底數(shù)進(jìn)行分類討論.(2)若底數(shù)不同,真數(shù)相同,則可以先用換底公式化為同底后,再進(jìn)行比較,或利用圖象數(shù)形結(jié)合求解.(3)若底數(shù)與真數(shù)都不同,則常借助1,0等中間量進(jìn)行比較.(4)

4、若不能夠使用以上三種方法比較大小,則需要將已知的對(duì)數(shù)式變形或利用對(duì)數(shù)的運(yùn)算性質(zhì)確定對(duì)數(shù)值的取值范圍,或利用作差(或作商)比較法以及利用結(jié)論logn+1(n+2)1的解集為 .解題策略簡(jiǎn)單對(duì)數(shù)不等式問題的求解策略(1)解決簡(jiǎn)單的對(duì)數(shù)不等式,應(yīng)先利用對(duì)數(shù)的運(yùn)算性質(zhì)化為同底數(shù)的對(duì)數(shù)值,再利用對(duì)數(shù)函數(shù)的單調(diào)性轉(zhuǎn)化為一般不等式求解.求解時(shí)不要忘記對(duì)數(shù)函數(shù)的定義域.(2)對(duì)數(shù)函數(shù)的單調(diào)性和底數(shù)a的值有關(guān),在研究對(duì)數(shù)函數(shù)的單調(diào)性時(shí),要按0a1進(jìn)行分類討論.角度三 對(duì)數(shù)型復(fù)合函數(shù)的定義域與值域例2-3 若函數(shù)f(x)=lg(ax2-2x+a)的值域?yàn)镽,則實(shí)數(shù)a的取值范圍為()A.(-1,0)B.(0,1)

5、C.0,1 D.(1,+)解題策略對(duì)數(shù)型復(fù)合函數(shù)的定義域與值域的求解策略:函數(shù)y=logaf(x)的值域是R,這說(shuō)明函數(shù)y=f(x)可以取遍所有大于0的數(shù),而函數(shù)y=logaf(x)的定義域是R,則說(shuō)明f(x)0在R上恒成立.針對(duì)訓(xùn)練 1.已知x=lg 2,y=ln 3,z=log23,則()A.xzy B.zyxC.xyz D.zxy當(dāng)x2時(shí),log3(x-1)2,即log3(x-1)log39,所以0 x-19,解得1x10,所以2x10.綜上所述,g(x)2的解集為(-,1)2,10).答案:(-,1)2,10)3.已知函數(shù)f(x)=log2(x2-2ax+3).若函數(shù)f(x)的定義域?yàn)?/p>

6、R,則實(shí)數(shù)a的取值范圍是;若函數(shù)f(x)的值域?yàn)镽,則實(shí)數(shù)a的取值范圍是.備選例題例1例2 已知函數(shù)f(x)=ln x+ln(2-x),則()A.f(x)在(0,2)上單調(diào)遞增B.f(x)在(0,2)上單調(diào)遞減C.y=f(x)的圖象關(guān)于直線x=1對(duì)稱D.y=f(x)的圖象關(guān)于點(diǎn)(1,0)對(duì)稱解析:由于f(x)=lnx(2-x)=ln(-x2+2x)=ln-(x-1)2+1.令t=-(x-1)2+1,結(jié)合函數(shù)的定義域(0,2)可知,函數(shù)t在區(qū)間(0,1)上單調(diào)遞增,在區(qū)間(1,2)上單調(diào)遞減,由y=ln t在區(qū)間(0,+)上單調(diào)遞增可知,函數(shù)f(x)=ln x+ln(2-x)在區(qū)間(0,1)上單調(diào)遞增,在區(qū)間(1,2)上單調(diào)遞減,因此A,B錯(cuò)誤;由題意知,f(2-x)=ln(2-x)+ln x=f(x),所以f(x)的圖象關(guān)于直線x=1對(duì)稱,C正確,D錯(cuò)誤.故