新高考數(shù)學一輪復(fù)習講義 第15講 導數(shù)與函數(shù)的單調(diào)性（含解析）

第15講導數(shù)與函數(shù)的單調(diào)性（精講）題型目錄一覽①導數(shù)與原函數(shù)圖像之間的聯(lián)系②不含參數(shù)的函數(shù)單調(diào)性③含參數(shù)的函數(shù)單調(diào)性一次函數(shù)型二次函數(shù)型Ⅰ（可因式分解）二次函數(shù)型Ⅱ（不可因式分解）指數(shù)函數(shù)型④函數(shù)單調(diào)性中的參數(shù)值（范圍）問題★【文末附錄-導數(shù)與函數(shù)的單調(diào)性思維導圖】一、知識點梳理一、知識點梳理一、單調(diào)性基礎(chǔ)問題1．函數(shù)的單調(diào)性函數(shù)單調(diào)性的判定方法：設(shè)函數(shù)SKIPIF1<0在某個區(qū)間內(nèi)可導，如果SKIPIF1<0，則SKIPIF1<0為增函數(shù)；如果SKIPIF1<0，則SKIPIF1<0為減函數(shù)．2．已知函數(shù)的單調(diào)性問題=1\*GB3①若SKIPIF1<0在某個區(qū)間上單調(diào)遞增，則在該區(qū)間上有SKIPIF1<0恒成立（但不恒等于0）；反之，要滿足SKIPIF1<0，才能得出SKIPIF1<0在某個區(qū)間上單調(diào)遞增；=2\*GB3②若SKIPIF1<0在某個區(qū)間上單調(diào)遞減，則在該區(qū)間上有SKIPIF1<0恒成立（但不恒等于0）；反之，要滿足SKIPIF1<0，才能得出SKIPIF1<0在某個區(qū)間上單調(diào)遞減．二、討論單調(diào)區(qū)間問題類型一：不含參數(shù)單調(diào)性討論（1）求導化簡定義域(化簡應(yīng)先通分，盡可能因式分解；定義域需要注意是否是連續(xù)的區(qū)間)；（2）變號保留定號去(變號部分：導函數(shù)中未知正負，需要單獨討論的部分．定號部分：已知恒正或恒負，無需單獨討論的部分)；（3）求根做圖得結(jié)論(如能直接求出導函數(shù)等于0的根，并能做出導函數(shù)與x軸位置關(guān)系圖，則導函數(shù)正負區(qū)間段已知，可直接得出結(jié)論)；（4）未得結(jié)論斷正負(若不能通過第三步直接得出結(jié)論，則先觀察導函數(shù)整體的正負)；（5）正負未知看零點(若導函數(shù)正負難判斷，則觀察導函數(shù)零點)；（6）一階復(fù)雜求二階(找到零點后仍難確定正負區(qū)間段，或一階導函數(shù)無法觀察出零點，則求二階導)；求二階導往往需要構(gòu)造新函數(shù)，令一階導函數(shù)或一階導函數(shù)中變號部分為新函數(shù)，對新函數(shù)再求導．（7）借助二階定區(qū)間(通過二階導正負判斷一階導函數(shù)的單調(diào)性，進而判斷一階導函數(shù)正負區(qū)間段)；類型二：含參數(shù)單調(diào)性討論（1）求導化簡定義域(化簡應(yīng)先通分，然后能因式分解要進行因式分解，定義域需要注意是否是一個連續(xù)的區(qū)間)；（2）變號保留定號去(變號部分：導函數(shù)中未知正負，需要單獨討論的部分．定號部分：已知恒正或恒負，無需單獨討論的部分)；（3）恒正恒負先討論(變號部分因為參數(shù)的取值恒正恒負)；然后再求有效根；（4）根的分布來定參(此處需要從兩方面考慮：根是否在定義域內(nèi)和多根之間的大小關(guān)系)；（5）導數(shù)圖像定區(qū)間；【常用結(jié)論】①使SKIPIF1<0的離散點不影響函數(shù)的單調(diào)性，即當SKIPIF1<0在某個區(qū)間內(nèi)離散點處為零，在其余點處均為正（或負）時，SKIPIF1<0在這個區(qū)間上仍舊是單調(diào)遞增（或遞減）的.例如，在SKIPIF1<0上，SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0；當SKIPIF1<0時，SKIPIF1<0，而顯然SKIPIF1<0在SKIPIF1<0上是單調(diào)遞增函數(shù).②若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0（SKIPIF1<0不恒為0），反之不成立.因為SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，當SKIPIF1<0時，函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增.當SKIPIF1<0時，SKIPIF1<0在這個區(qū)間為常值函數(shù)；同理，若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0（SKIPIF1<0不恒為0），反之不成立.這說明在一個區(qū)間上函數(shù)的導數(shù)大于零，是這個函數(shù)在該區(qū)間上單調(diào)遞增的充分不必要條件.于是有如下結(jié)論：SKIPIF1<0SKIPIF1<0單調(diào)遞增；SKIPIF1<0單調(diào)遞增SKIPIF1<0；SKIPIF1<0SKIPIF1<0單調(diào)遞減；SKIPIF1<0單調(diào)遞減SKIPIF1<0.二、題型分類精講二、題型分類精講題型一導數(shù)與原函數(shù)圖像之間的聯(lián)系策略方法原函數(shù)的單調(diào)性與導函數(shù)的函數(shù)值的符號的關(guān)系，原函數(shù)SKIPIF1<0單調(diào)遞增SKIPIF1<0導函數(shù)SKIPIF1<0（導函數(shù)等于0，只在離散點成立，其余點滿足SKIPIF1<0）；原函數(shù)單調(diào)遞減SKIPIF1<0導函數(shù)SKIPIF1<0（導函數(shù)等于0，只在離散點成立，其余點滿足SKIPIF1<0）.【典例1】已知函數(shù)SKIPIF1<0的圖象如圖所示（其中SKIPIF1<0是函數(shù)SKIPIF1<0的導函數(shù)），則SKIPIF1<0的圖象大致是（

）A． B．C． D．【答案】C【分析】由SKIPIF1<0的圖象得到SKIPIF1<0的取值情況，即可得到SKIPIF1<0的單調(diào)性，即可判斷.【詳解】由SKIPIF1<0的圖象可知當SKIPIF1<0時SKIPIF1<0，則SKIPIF1<0，當SKIPIF1<0時SKIPIF1<0，則SKIPIF1<0，當SKIPIF1<0時SKIPIF1<0，則SKIPIF1<0，當SKIPIF1<0時SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，故符合題意的只有C.故選：C.【題型訓練】一、單選題1．（2023·浙江紹興·統(tǒng)考模擬預(yù)測）如圖是函數(shù)SKIPIF1<0的導函數(shù)SKIPIF1<0的圖象，若SKIPIF1<0，則SKIPIF1<0的圖象大致為（

）A． B．C． D．【答案】D【分析】根據(jù)導函數(shù)的圖象在區(qū)間SKIPIF1<0內(nèi)的函數(shù)的范圍，判斷出函數(shù)SKIPIF1<0區(qū)間SKIPIF1<0上各點處切線的斜率的范圍，根據(jù)導函數(shù)的圖象得導函數(shù)函數(shù)值的符號，得函數(shù)SKIPIF1<0的單調(diào)性，再結(jié)合四個選項可得答案.【詳解】由SKIPIF1<0的圖象可知，當SKIPIF1<0時，SKIPIF1<0，則在區(qū)間SKIPIF1<0上，函數(shù)SKIPIF1<0上各點處切線的斜率在區(qū)間SKIPIF1<0內(nèi)，對于A，在區(qū)間SKIPIF1<0上，函數(shù)SKIPIF1<0上各點處切線的斜率均小于0，故A不正確；對于B，在區(qū)間SKIPIF1<0上，函數(shù)SKIPIF1<0上存在點，在該點處切線的斜率大于1，故B不正確；對于C，在區(qū)間SKIPIF1<0上，函數(shù)SKIPIF1<0上存在點，在該點處切線的斜率大于1，故C不正確；對于D，由SKIPIF1<0的圖象可知，當SKIPIF1<0時，SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，所以函數(shù)SKIPIF1<0上各點處切線的斜率在區(qū)間SKIPIF1<0內(nèi)，在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，而函數(shù)SKIPIF1<0的圖象均符合這些性質(zhì)，故D正確.故選：D2．（2023·全國·高三專題練習）設(shè)SKIPIF1<0是函數(shù)SKIPIF1<0的導函數(shù)，SKIPIF1<0的圖象如圖所示，則SKIPIF1<0的圖象最有可能的是（

）A． B．C． D．【答案】C【分析】根據(jù)導函數(shù)的圖象得出函數(shù)的單調(diào)區(qū)間，根據(jù)函數(shù)SKIPIF1<0的單調(diào)性即可判斷.【詳解】由導函數(shù)的圖象可得當SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增；當SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞減；當SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增.只有C選項的圖象符合.故選：C.3．（2023·陜西西安·校聯(lián)考一模）已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0的大致圖像如圖所示，SKIPIF1<0是SKIPIF1<0的導函數(shù)，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】分SKIPIF1<0、SKIPIF1<0兩種情況求解即可.【詳解】若SKIPIF1<0，則SKIPIF1<0單調(diào)遞減，圖像可知，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0單調(diào)遞增，由圖像可知SKIPIF1<0，故不等式SKIPIF1<0的解集為SKIPIF1<0．故選：C二、多選題4．（2023·全國·高三專題練習）已知函數(shù)SKIPIF1<0的導函數(shù)SKIPIF1<0的圖象如圖所示，則下列說法正確的是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)有SKIPIF1<0個極值點D．SKIPIF1<0的圖象在點SKIPIF1<0處的切線的斜率大于SKIPIF1<0【答案】ACD【分析】根據(jù)導函數(shù)的正負可得SKIPIF1<0單調(diào)性，由單調(diào)性可判斷AB正誤；由極值點定義可知C正確；由SKIPIF1<0可知D正確.【詳解】由圖象可知：當SKIPIF1<0時，SKIPIF1<0；當SKIPIF1<0時，SKIPIF1<0；SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增；在SKIPIF1<0上單調(diào)遞減；對于A，SKIPIF1<0，SKIPIF1<0，A正確；對于B，SKIPIF1<0，SKIPIF1<0，B錯誤；對于C，由極值點定義可知：SKIPIF1<0為SKIPIF1<0的極大值點；SKIPIF1<0為SKIPIF1<0的極小值點，即SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)有SKIPIF1<0個極值點，C正確；對于D，當SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0在點SKIPIF1<0處的切線的斜率大于SKIPIF1<0，D正確.故選：ACD.三、填空題5．（2023·全國·高三專題練習）函數(shù)SKIPIF1<0的圖象如圖所示，記SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，則SKIPIF1<0、SKIPIF1<0、SKIPIF1<0最大的是________.【答案】SKIPIF1<0【分析】根據(jù)導數(shù)的幾何意義結(jié)合SKIPIF1<0的圖象分析判斷即可【詳解】根據(jù)導數(shù)的幾何意義，SKIPIF1<0、SKIPIF1<0、SKIPIF1<0分別為SKIPIF1<0處的切線斜率，又SKIPIF1<0與SKIPIF1<0處的切線單調(diào)遞增，SKIPIF1<0處的切線單調(diào)遞減，且SKIPIF1<0處的切線比SKIPIF1<0處的切線更陡峭，∴SKIPIF1<0，故最大為SKIPIF1<0.故答案為：SKIPIF1<06．（2023春·上海·高三統(tǒng)考開學考試）已知定義在SKIPIF1<0上的奇函數(shù)SKIPIF1<0的導函數(shù)是SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0的圖象如圖所示，則關(guān)于x的不等式SKIPIF1<0的解集為______.【答案】SKIPIF1<0【分析】先判斷出SKIPIF1<0的單調(diào)性，然后求得SKIPIF1<0的解集.【詳解】依題意SKIPIF1<0是奇函數(shù)，圖象關(guān)于原點對稱，由圖象可知，SKIPIF1<0在區(qū)間SKIPIF1<0遞減，SKIPIF1<0；SKIPIF1<0在區(qū)間SKIPIF1<0遞增，SKIPIF1<0.所以SKIPIF1<0的解集SKIPIF1<0.故答案為：SKIPIF1<0題型二不含參數(shù)的函數(shù)單調(diào)性策略方法求函數(shù)單調(diào)區(qū)間的步驟(1)確定函數(shù)f(x)的定義域．(2)求f′(x)．(3)在定義域內(nèi)解不等式f′(x)＞0，得單調(diào)遞增區(qū)間．(4)在定義域內(nèi)解不等式f′(x)＜0，得單調(diào)遞減區(qū)間．【典例1】函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】求出導函數(shù)SKIPIF1<0，在定義域內(nèi)解不等式SKIPIF1<0可得單調(diào)遞增區(qū)間．【詳解】由已知得SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，∴函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0．故選：B．【題型訓練】一、單選題1．（2023·全國·高三專題練習）函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0和SKIPIF1<0【答案】C【分析】根據(jù)給定的函數(shù)，利用導數(shù)求出單調(diào)減區(qū)間作答.【詳解】函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，求導得SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，所以函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間是SKIPIF1<0.故選：C2．（2023·全國·高三專題練習）已知SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，現(xiàn)給出如下結(jié)論：①SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0；④SKIPIF1<0．其中正確結(jié)論個數(shù)為（

）A．1個 B．2個 C．3個 D．4個【答案】D【分析】利用導數(shù)判定三次函數(shù)的圖象與性質(zhì)，結(jié)合圖象即可一一判定結(jié)論.【詳解】求導函數(shù)可得SKIPIF1<0SKIPIF1<0當SKIPIF1<0時，SKIPIF1<0；當SKIPIF1<0，或SKIPIF1<0時，SKIPIF1<0所以SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0和SKIPIF1<0單調(diào)遞減區(qū)間為SKIPIF1<0所以SKIPIF1<0極大值SKIPIF1<0SKIPIF1<0，SKIPIF1<0極小值SKIPIF1<0SKIPIF1<0要使SKIPIF1<0有三個解SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，那么結(jié)合函數(shù)SKIPIF1<0草圖可知：SKIPIF1<0及函數(shù)有個零點SKIPIF1<0在SKIPIF1<0之間，所以SKIPIF1<0SKIPIF1<0，且SKIPIF1<0SKIPIF1<0所以SKIPIF1<0SKIPIF1<0，SKIPIF1<0，故①正確；SKIPIF1<0SKIPIF1<0，SKIPIF1<0，即②③正確；SKIPIF1<0，SKIPIF1<0SKIPIF1<0SKIPIF1<0，SKIPIF1<0（i），SKIPIF1<0（ii），把（ii）代入（i）式的平方化簡得：SKIPIF1<0；即④正確；故選：D．3．（2023·全國·高三專題練習）已知函數(shù)SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】首先根據(jù)題干條件SKIPIF1<0，得SKIPIF1<0，化簡整理得SKIPIF1<0，然后構(gòu)造函數(shù)SKIPIF1<0，借助導數(shù)求解SKIPIF1<0的最小值，即可求出SKIPIF1<0的最小值.【詳解】由SKIPIF1<0，得SKIPIF1<0，化簡整理得：SKIPIF1<0；令SKIPIF1<0（SKIPIF1<0），SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0.當SKIPIF1<0時，SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當SKIPIF1<0時，SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；即SKIPIF1<0，故SKIPIF1<0故選：D二、多選題4．（2023·全國·模擬預(yù)測）已知函數(shù)SKIPIF1<0，則（

）A．SKIPIF1<0的單調(diào)遞減區(qū)間是SKIPIF1<0 B．SKIPIF1<0有4個零點C．SKIPIF1<0的圖象關(guān)于點SKIPIF1<0對稱 D．曲線SKIPIF1<0與SKIPIF1<0軸不相切【答案】CD【分析】對A直接求導，令導函數(shù)小于0，解出即可，對B，通過求出極大值和極小值，結(jié)合其單調(diào)性即可判斷，對C選項利用函數(shù)奇偶性和函數(shù)平移的原則即可判斷，對D，利用函數(shù)極大值、極小值的符號即可判斷.【詳解】A選項：易知SKIPIF1<0的定義域為SKIPIF1<0，SKIPIF1<0，令SKIPIF1<00，解得SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0和SKIPIF1<0，A錯誤；B選項：令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0，和SKIPIF1<0上單調(diào)遞增，所以當SKIPIF1<0時，SKIPIF1<0取得極大值，因為SKIPIF1<0，且SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0在SKIPIF1<0上沒有零點，當SKIPIF1<0時，SKIPIF1<0取得極小值，因為SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上至多有兩個零點，B錯誤；C選項：設(shè)SKIPIF1<0，函數(shù)定義域為SKIPIF1<0，關(guān)于原點對稱，且SKIPIF1<0，則SKIPIF1<0為奇函數(shù)，所以SKIPIF1<0的圖象關(guān)于原點對稱，將SKIPIF1<0的圖象向下平移2個單位長度得到SKIPIF1<0的圖象，所以SKIPIF1<0的圖象關(guān)于點SKIPIF1<0對稱，C正確；D選項：因為SKIPIF1<0的極小值SKIPIF1<0，極大值SKIPIF1<0，所以曲線SKIPIF1<0與SKIPIF1<0軸不相切，D正確.故選：CD.三、填空題5．（2023·云南·校聯(lián)考二模）函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為____________．【答案】SKIPIF1<0/SKIPIF1<0【分析】通過二次求導，證明當SKIPIF1<0時，SKIPIF1<0，即得解.【詳解】由題得函數(shù)定義域為SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，又SKIPIF1<0，所以當SKIPIF1<0時，SKIPIF1<0，故SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0（或SKIPIF1<0）．故答案為：SKIPIF1<06．（2023春·江西·高三校聯(lián)考階段練習）函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為__________.【答案】SKIPIF1<0【分析】求導數(shù)SKIPIF1<0，令SKIPIF1<0，解不等式即可得函數(shù)的單調(diào)遞增區(qū)間.【詳解】函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，故函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0.故答案為：SKIPIF1<0.7．（2023秋·山東東營·高三東營市第一中學?？计谀┖瘮?shù)SKIPIF1<0的單調(diào)遞增區(qū)間為___________．【答案】SKIPIF1<0，SKIPIF1<0【分析】對函數(shù)求導，判斷導函數(shù)的正負，導函數(shù)分子無法判斷正負，再對分子求導，利用導函數(shù)的單調(diào)性來判斷導函數(shù)的正負，進而得出原函數(shù)的單調(diào)區(qū)間.【詳解】因為函數(shù)SKIPIF1<0，則SKIPIF1<0．設(shè)SKIPIF1<0，則SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以當SKIPIF1<0時，SKIPIF1<0，則當SKIPIF1<0時，SKIPIF1<0．所以SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0，故答案為：SKIPIF1<0，SKIPIF1<0.8．（2023·福建·統(tǒng)考模擬預(yù)測）函數(shù)SKIPIF1<0的單調(diào)增區(qū)間是_______．【答案】SKIPIF1<0(或SKIPIF1<0也對)【分析】SKIPIF1<0，由復(fù)合函數(shù)單調(diào)性知：SKIPIF1<0的增區(qū)間即為所求.【詳解】SKIPIF1<0，由復(fù)合函數(shù)單調(diào)性知：SKIPIF1<0的增區(qū)間即為所求，SKIPIF1<0．故答案為：SKIPIF1<0(或SKIPIF1<0也對）9．（2023春·安徽亳州·高三?？茧A段練習）函數(shù)SKIPIF1<0有兩個零點，則SKIPIF1<0的取值范圍是__.【答案】SKIPIF1<0【分析】函數(shù)SKIPIF1<0有兩個零點，即方程SKIPIF1<0有兩個根，構(gòu)造函數(shù)SKIPIF1<0，利用導數(shù)求出函數(shù)的單調(diào)區(qū)間，從而可畫出函數(shù)SKIPIF1<0的大致圖像，根據(jù)圖象即可得解.【詳解】SKIPIF1<0函數(shù)SKIPIF1<0有兩個零點，SKIPIF1<0方程SKIPIF1<0有兩個根，即方程SKIPIF1<0有兩個根，設(shè)SKIPIF1<0，則函數(shù)SKIPIF1<0與SKIPIF1<0的圖像有兩個交點，SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，SKIPIF1<0函數(shù)SKIPIF1<0在SKIPIF1<0時，取得最大值SKIPIF1<0，又SKIPIF1<0當SKIPIF1<0時，SKIPIF1<0；當SKIPIF1<0時，SKIPIF1<0且SKIPIF1<0，SKIPIF1<0函數(shù)SKIPIF1<0的大致圖像，如圖所示，由圖像可知，SKIPIF1<0，SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<0.題型三含參數(shù)的函數(shù)單調(diào)性策略方法解決含參數(shù)的函數(shù)的單調(diào)性問題應(yīng)注意兩點(1)研究含參數(shù)的函數(shù)的單調(diào)性，要依據(jù)參數(shù)對不等式解集的影響進行分類討論．(2)劃分函數(shù)的單調(diào)區(qū)間時，要在函數(shù)定義域內(nèi)討論，還要確定導數(shù)為0的點和函數(shù)的間斷點．【典例1】已知函數(shù)SKIPIF1<0（其中a為參數(shù)）.求函數(shù)SKIPIF1<0的單調(diào)區(qū)間.【答案】答案見解析【分析】求出原函數(shù)的導函數(shù)，然后對SKIPIF1<0分類求得函數(shù)的單調(diào)區(qū)間；【詳解】SKIPIF1<0，SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)遞增，當SKIPIF1<0時，令SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，SKIPIF1<0時，SKIPIF1<0單調(diào)遞增；綜上：SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上遞增，無減區(qū)間，當SKIPIF1<0時，SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，單調(diào)遞增區(qū)間為SKIPIF1<0；【典例2】已知函數(shù)SKIPIF1<0，SKIPIF1<0.討論函數(shù)SKIPIF1<0的單調(diào)性.【答案】答案見解析【分析】對SKIPIF1<0求導，然后分SKIPIF1<0和SKIPIF1<0兩種情況討論即可；【詳解】函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，所以SKIPIF1<0.當SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當SKIPIF1<0時，令SKIPIF1<0得SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減：在SKIPIF1<0上單調(diào)遞增.綜上，當SKIPIF1<0時，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增.【典例3】設(shè)函數(shù)SKIPIF1<0，求函數(shù)SKIPIF1<0的單調(diào)區(qū)間.【答案】當SKIPIF1<0時，SKIPIF1<0的單調(diào)減區(qū)間為SKIPIF1<0；當SKIPIF1<0時，SKIPIF1<0的單調(diào)減區(qū)間為SKIPIF1<0和SKIPIF1<0，單調(diào)增區(qū)間為SKIPIF1<0.【分析】對SKIPIF1<0求導，分SKIPIF1<0和SKIPIF1<0和SKIPIF1<0三類討論導數(shù)的正負，即可得出SKIPIF1<0的單調(diào)區(qū)間.【詳解】由題意得SKIPIF1<0的定義域為SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，①當SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，②當SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，③當SKIPIF1<0時，令SKIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增.綜上，當SKIPIF1<0時，SKIPIF1<0的單調(diào)減區(qū)間為SKIPIF1<0；當SKIPIF1<0時，SKIPIF1<0的單調(diào)減區(qū)間為SKIPIF1<0和SKIPIF1<0，單調(diào)增區(qū)間為SKIPIF1<0.【典例4】已知函數(shù)SKIPIF1<0（SKIPIF1<0為自然對數(shù)的底數(shù)，SKIPIF1<0）．求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；【答案】答案見解析【分析】先求得SKIPIF1<0，結(jié)合SKIPIF1<0和SKIPIF1<0討論SKIPIF1<0的正負，進而求解.【詳解】函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，則SKIPIF1<0.①當SKIPIF1<0時，對任意的SKIPIF1<0，SKIPIF1<0，此時函數(shù)SKIPIF1<0的減區(qū)間為SKIPIF1<0，無增區(qū)間；②當SKIPIF1<0時，由SKIPIF1<0可得SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，此時函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，遞減區(qū)間為SKIPIF1<0；綜上所述，當SKIPIF1<0時，函數(shù)SKIPIF1<0的減區(qū)間為SKIPIF1<0，無增區(qū)間；當SKIPIF1<0時，函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，遞減區(qū)間為SKIPIF1<0.【題型訓練】一、單選題1．（2023·全國·高三專題練習）已知函數(shù)SKIPIF1<0，若不等式SKIPIF1<0在區(qū)間SKIPIF1<0上恒成立，則實數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】SKIPIF1<0即為SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，求出函數(shù)SKIPIF1<0的導函數(shù)，分解SKIPIF1<0和SKIPIF1<0討論函數(shù)SKIPIF1<0的單調(diào)性，求出函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值，即可得解.【詳解】解：由已知可得SKIPIF1<0即為SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，當SKIPIF1<0時，顯然SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上也成立，所以SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0恒成立；當SKIPIF1<0時，當SKIPIF1<0時，SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，于是，存在SKIPIF1<0，使得SKIPIF1<0，不滿足SKIPIF1<0，舍去此情況，綜上所述，SKIPIF1<0.故選：A.2．（2023秋·四川宜賓·高三四川省宜賓市第四中學校?？计谀┮阎瘮?shù)SKIPIF1<0，若SKIPIF1<0有四個不同的零點，則a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】討論SKIPIF1<0、SKIPIF1<0，應(yīng)用導數(shù)研究單調(diào)性，要使SKIPIF1<0有四個不同的解，即當兩個區(qū)間均存在兩個零點時，求a的范圍即可.【詳解】由題意知：SKIPIF1<0有四個不同的零點，∴SKIPIF1<0，則SKIPIF1<0有四個不同的解，當SKIPIF1<0時，SKIPIF1<0，其零點情況如下：1）當SKIPIF1<0或SKIPIF1<0時，有SKIPIF1<0；2）當SKIPIF1<0或SKIPIF1<0時，SKIPIF1<0或SKIPIF1<0；當SKIPIF1<0時，SKIPIF1<0，則有如下情況：1）當SKIPIF1<0時SKIPIF1<0，即SKIPIF1<0單調(diào)遞增，不可能出現(xiàn)兩個零點，不合題意；2）當SKIPIF1<0時，在SKIPIF1<0上SKIPIF1<0，SKIPIF1<0單調(diào)遞增，在SKIPIF1<0上SKIPIF1<0，SKIPIF1<0單調(diào)遞減，而SKIPIF1<0有SKIPIF1<0，SKIPIF1<0有SKIPIF1<0，所以只需SKIPIF1<0，得SKIPIF1<0時，SKIPIF1<0必有兩個零點.∴綜上，有SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0、SKIPIF1<0上各有兩個零點，即共有四個不同的零點.故選：A.【點睛】關(guān)鍵點點睛：應(yīng)用分類討論，利用導數(shù)研究函數(shù)的單調(diào)性，求在滿足零點個數(shù)的情況下參數(shù)范圍.二、填空題3．（2023·全國·高三專題練習）已知SKIPIF1<0，若對任意SKIPIF1<0，都有SKIPIF1<0，則實數(shù)SKIPIF1<0的取值范圍是______.【答案】SKIPIF1<0【分析】SKIPIF1<0，則易得出SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，與SKIPIF1<0矛盾，故SKIPIF1<0，求出函數(shù)的導函數(shù)，對SKIPIF1<0進行討論，判斷函數(shù)的單調(diào)性，從而可得出答案.【詳解】解：SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，則當SKIPIF1<0時，SKIPIF1<0，這與SKIPIF1<0矛盾，故SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，則當SKIPIF1<0時，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上遞減，所以SKIPIF1<0符合題意；若SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0上遞增，故當當SKIPIF1<0時，SKIPIF1<0，這與SKIPIF1<0矛盾，綜上所述SKIPIF1<0.故答案為：SKIPIF1<0.三、解答題4．（2023·全國·高三專題練習）已知函數(shù)SKIPIF1<0．討論SKIPIF1<0的單調(diào)性；【答案】答案見解析【分析】對函數(shù)求導，注意定義域，討論SKIPIF1<0、SKIPIF1<0研究導數(shù)的符號，進而確定區(qū)間單調(diào)性.【詳解】由題設(shè)SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0恒成立，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增當SKIPIF1<0時，令SKIPIF1<0得：SKIPIF1<0，令SKIPIF1<0得：SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減綜上，當SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減；當SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；5．（2023·全國·高三專題練習）已知函數(shù)SKIPIF1<0．討論函數(shù)SKIPIF1<0的單調(diào)性；【答案】答案見解析【分析】求出導函數(shù)SKIPIF1<0，分類討論，由SKIPIF1<0得增區(qū)間，由SKIPIF1<0得減區(qū)間．【詳解】∵SKIPIF1<0，SKIPIF1<0（1）當SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，（2）當SKIPIF1<0時，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0上單調(diào)遞減，綜上，當SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減6．（2023·全國·高三專題練習）已知函數(shù)SKIPIF1<0．討論SKIPIF1<0在SKIPIF1<0上的單調(diào)性；【答案】答案見解析【分析】討論SKIPIF1<0，SKIPIF1<0，SKIPIF1<0三種情況，結(jié)合導數(shù)得出SKIPIF1<0在SKIPIF1<0上的單調(diào)性；【詳解】由題意得SKIPIF1<0．因為SKIPIF1<0，所以SKIPIF1<0．當SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減．當SKIPIF1<0時，令SKIPIF1<0，則SKIPIF1<0．①若SKIPIF1<0，則SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；②若SKIPIF1<0，則SKIPIF1<0，當SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．綜上，當SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增7．（2023·全國·高三專題練習）設(shè)函數(shù)SKIPIF1<0.當SKIPIF1<0時，討論函數(shù)SKIPIF1<0的單調(diào)性；【答案】見解析【分析】由題求導得SKIPIF1<0，分SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，三種情況討論其單調(diào)性即可.【詳解】由題知，函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，所以求導得SKIPIF1<0，若SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0或SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0，和SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，若SKIPIF1<0，恒有SKIPIF1<0，當且僅當SKIPIF1<0時取等號，因此函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，若SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0或SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，所以當SKIPIF1<0時，函數(shù)SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減；當SKIPIF1<0時，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當SKIPIF1<0時，函數(shù)SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減.8．（2023·全國·高三專題練習）已知函數(shù)SKIPIF1<0.討論函數(shù)SKIPIF1<0的單調(diào)性；【答案】答案見解析【分析】先求出函數(shù)的定義域，再求導，分SKIPIF1<0和SKIPIF1<0兩種情況討論，即可得解.【詳解】由SKIPIF1<0，可知定義域SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，①當SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0成立，即SKIPIF1<0成立，所以SKIPIF1<0的單調(diào)增區(qū)間為SKIPIF1<0；②當SKIPIF1<0時，令SKIPIF1<0，得SKIPIF1<0，記SKIPIF1<0，SKIPIF1<0，當SKIPIF1<0變化時，SKIPIF1<0，SKIPIF1<0的變化情況如下表SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0+0-0+SKIPIF1<0↗極大值↘極小值↗所以SKIPIF1<0的增區(qū)間為SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，減區(qū)間為SKIPIF1<0，綜上，當SKIPIF