最全函數(shù)零點(diǎn)問(wèn)題處理

1、1最全函數(shù)零點(diǎn)問(wèn)題處理最全函數(shù)零點(diǎn)問(wèn)題處理函數(shù)中的賦值問(wèn)題函數(shù)中的賦值問(wèn)題...........................................................................................1第一講第一講賦值的意義賦值的意義...........................................................................

2、..........1第二講第二講賦值的依據(jù)和方法賦值的依據(jù)和方法...........................................................................4第三講第三講賦值的若干經(jīng)典問(wèn)題賦值的若干經(jīng)典問(wèn)題...................................................................10導(dǎo)數(shù)大題的常用找點(diǎn)技巧和常見(jiàn)模型導(dǎo)數(shù)

3、大題的常用找點(diǎn)技巧和常見(jiàn)模型.........................................................15常用的放縮公式常用的放縮公式（考試時(shí)需給出證明過(guò)程）（考試時(shí)需給出證明過(guò)程）.............................................17第一組：對(duì)數(shù)放縮第一組：對(duì)數(shù)放縮..............................................

4、...........................................17第二組：指數(shù)放縮第二組：指數(shù)放縮.........................................................................................17第三組：指對(duì)放縮第三組：指對(duì)放縮....................................................

5、.....................................17第四組：三角函數(shù)放縮第四組：三角函數(shù)放縮.................................................................................18幾個(gè)經(jīng)典函數(shù)模型幾個(gè)經(jīng)典函數(shù)模型..............................................................

6、...........................18導(dǎo)數(shù)零點(diǎn)不可求的四種破解策略導(dǎo)數(shù)零點(diǎn)不可求的四種破解策略.................................................................22法一法一:利用零點(diǎn)存在性定理利用零點(diǎn)存在性定理.........................................................................

7、..22法二法二:利用函數(shù)與方程思想利用函數(shù)與方程思想...........................................................................23法三法三:構(gòu)造新的函數(shù)構(gòu)造新的函數(shù).......................................................................................24法四法四:利用極限思

8、想利用極限思想.......................................................................................25導(dǎo)數(shù)壓軸題之隱零點(diǎn)問(wèn)題導(dǎo)數(shù)壓軸題之隱零點(diǎn)問(wèn)題.............................................................................26直擊函數(shù)壓軸題中零點(diǎn)問(wèn)題直擊函數(shù)壓軸題中零點(diǎn)問(wèn)

9、題.........................................................................413Oxy1AAe()lnfxxax??1ae??O()fxxy2真題探究真題探究題目題目2（2013江蘇20）設(shè)函數(shù)，其中為實(shí)數(shù)()ln()exfxxaxgxax????a（1）略；（2）若在上是單調(diào)增函數(shù)，求的零點(diǎn)個(gè)數(shù)，并證明你的結(jié)論()gx(1)???()fx（2）解：由在上單調(diào)增，

10、得(過(guò)程略)()gx(1)???1ea≤時(shí)，，10a?≤1()0()fxafxx????A而，且圖像不間斷，11(e)(1e)10(e)1e0aafafa?????????()fx依據(jù)零點(diǎn)定理，有且只有一個(gè)零點(diǎn)()fx【分析時(shí)，由(極大值點(diǎn))，】0a?1()0fxxa????max1()ln1fxa??時(shí)令12ea??1()lnefxxx??11()0eefxxx?????且，e()00e()0xfxxfx???????所以是的極大值點(diǎn)

11、，也是最大值點(diǎn)，ex?()fx所以，當(dāng)且僅當(dāng)()(e)0fxf?≤()0xefx??故有唯一零點(diǎn)()fxex?時(shí)，令列表：130ea???11()0fxaxxa?????x1(0)a1a1()a??()fx??0?()fxAmax()fxA所以max11()()ln10fxfaa????①在上，且單調(diào)，所以有且只有一個(gè)零點(diǎn)；1(0)a(1)0fa???()fx()fx②在上，顯然，注意到的結(jié)論，1()a??211aa?2?1(ln)ex