習(xí)題課指數(shù)型函數(shù)、對(duì)數(shù)型函數(shù)的性質(zhì)的綜合問(wèn)1：你能熟練說(shuō)出指數(shù)、對(duì)數(shù)函數(shù)的圖象和性質(zhì)嗎？問(wèn)2：你能求指數(shù)型、對(duì)數(shù)型函數(shù)的單調(diào)性和值域嗎？問(wèn)3：你會(huì)判斷指數(shù)型、對(duì)數(shù)型函數(shù)的單調(diào)性嗎？一指數(shù)型函數(shù)的單調(diào)性問(wèn)題

例

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延伸探究

2.若本例不變,求f(x)的值域.

(1)求指數(shù)型函數(shù)的單調(diào)區(qū)間,首先求出函數(shù)的定義域,然后把函數(shù)分解成y=f(u),u=φ(x),通過(guò)考察f(u)和φ(x)的單調(diào)性,利用同增異減原則,求出y=f(φ(x))的單調(diào)性.(2)關(guān)于指數(shù)型函數(shù)y=af(x)(a>0,且a≠1)的單調(diào)性由兩點(diǎn)決定,一是底數(shù)a>1還是0<a<1;二是f(x)的單調(diào)性,它由兩個(gè)函數(shù)y=au,u=f(x)復(fù)合而成.

反思感悟

跟蹤訓(xùn)練

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√二對(duì)數(shù)型函數(shù)的單調(diào)性問(wèn)題

例

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(2)求函數(shù)y=(log0.4x)2-2log0.4x+2的單調(diào)區(qū)間.令t=log0.4x,則它在(0,+∞)上單調(diào)遞減.y=t2-2t+2=(t-1)2+1在(1,+∞)上單調(diào)遞增,在(-∞,1)上單調(diào)遞減.由t=log0.4x>1得0<x<0.4;由t=log0.4x<1得x>0.4,故所求函數(shù)的單調(diào)遞增區(qū)間為(0.4,+∞),單調(diào)遞減區(qū)間為(0,0.4).

反思感悟函數(shù)單調(diào)性的判定方法與策略(1)定義法:一般步驟:設(shè)元→作差→變形→判斷符號(hào)→得出結(jié)論.(2)圖象法:如果函數(shù)f(x)是以圖象形式給出或函數(shù)f(x)的圖象易作出,結(jié)合圖象可求得函數(shù)的單調(diào)區(qū)間.(3)y=f(g(x))型函數(shù):先將函數(shù)y=f(g(x))分解為y=f(t)和t=g(x),再討論這兩個(gè)函數(shù)的單調(diào)性,最后根據(jù)復(fù)合函數(shù)“同增異減”的規(guī)則進(jìn)行判定.

跟蹤訓(xùn)練

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三函數(shù)的綜合應(yīng)用

例

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若把本例改為y=4x-2x+1-3.求函數(shù)的值域和單調(diào)區(qū)間.延伸探究函數(shù)y=4x-2x+1-3的定義域?yàn)镽,設(shè)t=2x,則t>0.因?yàn)閥=4x-2x+1-3=(2x)2-2×2x-3=t2-2t-3=(t-1)2-4≥-4,所以函數(shù)y=4x-2x+1-3的值域?yàn)閇-4,+∞).因?yàn)閥=t2-2t-3在(-∞,1]上單調(diào)遞減,此時(shí)由t≤1得x≤0.又指數(shù)函數(shù)t=2x在(-∞,0]上單調(diào)遞增,所以函數(shù)y=4x-2x+1-3的單調(diào)遞減區(qū)間為(-∞,0].同理,因?yàn)閥=t2-2t-3在[1,+∞)上單調(diào)遞增,此時(shí)由t≥1得x≥0.又指數(shù)函數(shù)t=2x在[0,+∞)上單調(diào)遞增,所以函數(shù)y=4x-2x+1-3的單調(diào)遞增區(qū)間為[0,+∞).

反思感悟?qū)τ谛稳鐈=logaf(x)(a>0，且a≠1)的復(fù)合函數(shù)，其值域的求解步驟如下：(1)分解成y=logau,u=f(x)兩個(gè)函數(shù)；(2)求f(x)的定義域；(3)求u的取值范圍；(4)利用y=logau的單調(diào)性求解.

求下列函數(shù)的值域:(1)f(x)=log2(3x+1);跟蹤訓(xùn)練

3f(x)的定義域?yàn)镽.∵3x>0,∴3x+1>1.∵y=log2x在(0,+∞)上單調(diào)遞增,∴l(xiāng)og2(3x+1)>log21=0,∴f(x)的值域?yàn)?0,+∞).

1.知識(shí)清單:(1)指數(shù)型函數(shù)的單調(diào)性.(2)對(duì)數(shù)型函數(shù)的單調(diào)性.(3)函數(shù)的綜合應(yīng)用.2.方法歸納:換元法.3.常見(jiàn)誤區(qū):求對(duì)數(shù)型函數(shù)的單調(diào)性易忽視定義域.隨堂演練四1.函數(shù)f(x)=loga|x-1|在(0,1)上單調(diào)遞減,那么f(x)在(1,+∞)上A.單調(diào)遞增且無(wú)最大值 B.單調(diào)遞減且無(wú)最小值C.單調(diào)遞增且有最大值 D.單調(diào)遞減且有最小值設(shè)u=|x-1|,∵u在(0,1)上單調(diào)遞減,且函數(shù)f(x)=loga|x-1|在(0,1)上單調(diào)遞減,則a>1,∵u在(1,+∞)上單調(diào)遞增,∴f(x)在(1,+∞)上單調(diào)遞增且無(wú)最大值.√1234