2019屆高考數(shù)學(xué)專題五導(dǎo)數(shù)的應(yīng)用精準(zhǔn)培優(yōu)專練理201811081155

1、1培優(yōu)點(diǎn)五培優(yōu)點(diǎn)五導(dǎo)數(shù)的應(yīng)用導(dǎo)數(shù)的應(yīng)用1利用導(dǎo)數(shù)判斷單調(diào)性例1：求函數(shù)????32333exfxxxx?????的單調(diào)區(qū)間【答案】見(jiàn)解析【解析】第一步：先確定定義域，??fx定義域?yàn)镽，第二步：求導(dǎo)：????????2323363e333e9exxxfxxxxxxxx?????????????????33exxxx?????，第三步：令??0fx??，即????33e0xxxx?????，第四步：處理恒正恒負(fù)的因式，可得????330x

2、xx???，第五步：求解????303x?????，列出表格2函數(shù)的極值例2：求函數(shù)()exfxx??的極值【答案】??fx的極大值為??11ef?，無(wú)極小值【解析】????ee1exxxfxxx???????令??0fx?解得：1x?，??fx?的單調(diào)區(qū)間為：3對(duì)點(diǎn)增分集訓(xùn)對(duì)點(diǎn)增分集訓(xùn)一、單選題1函數(shù)??lnfxxx??的單調(diào)遞減區(qū)間為（）A??01B??0??C??1??D????01?????【答案】A【解析】函數(shù)lnyxx??的

3、導(dǎo)數(shù)為11yx???，令110yx???，得1x?，∴結(jié)合函數(shù)的定義域，得當(dāng)??01x?時(shí)，函數(shù)為單調(diào)減函數(shù)因此，函數(shù)lnyxx??的單調(diào)遞減區(qū)間是??01故選A2若1x?是函數(shù)??lnfxaxx??的極值點(diǎn)，則（）A??fx有極大值1?B??fx有極小值1?C??fx有極大值0D??fx有極小值0【答案】A【解析】因?yàn)?x?是函數(shù)??lnfxaxx??的極值點(diǎn)，所以??10f??，101a???，1a???，??1101fxxx???