人教A版高中數(shù)學(xué)（選擇性必修二）同步講義第12講 5.3.1函數(shù)的單調(diào)性（教師版）

第03講5.3.1函數(shù)的單調(diào)性課程標(biāo)準(zhǔn)學(xué)習(xí)目標(biāo)①理解導(dǎo)數(shù)與函數(shù)的單調(diào)性的關(guān)系。②掌握利用導(dǎo)數(shù)判斷函數(shù)單調(diào)性的方法。③能利用導(dǎo)數(shù)求不超過三次多項式函數(shù)的單調(diào)區(qū)間。④會利用導(dǎo)數(shù)證明一些簡單的不等式問題。⑤掌握利用導(dǎo)數(shù)研究含參數(shù)的單調(diào)性的基本方法。通過本節(jié)課要求能利用函數(shù)的導(dǎo)數(shù)判斷函數(shù)的單調(diào)性，會求簡單函數(shù)的單調(diào)區(qū)間，能證明簡單的不等式，會利用導(dǎo)數(shù)解決單調(diào)性與含參數(shù)相關(guān)的問題.知識點01：函數(shù)的單調(diào)性與導(dǎo)數(shù)的關(guān)系（導(dǎo)函數(shù)看正負(fù)，原函數(shù)看增減）函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)可導(dǎo)，(1)若SKIPIF1<0，則SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)是單調(diào)遞增函數(shù)；(2)若SKIPIF1<0，則SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)是單調(diào)遞減函數(shù)；(3)若恒有SKIPIF1<0，則SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)是常數(shù)函數(shù)．注意：討論函數(shù)的單調(diào)性或求函數(shù)的單調(diào)區(qū)間的實質(zhì)是解不等式，求解時，要堅持“定義域優(yōu)先”原則條件恒有結(jié)論函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上可導(dǎo)SKIPIF1<0SKIPIF1<0在SKIPIF1<0內(nèi)單調(diào)遞增SKIPIF1<0SKIPIF1<0在SKIPIF1<0內(nèi)單調(diào)遞減SKIPIF1<0SKIPIF1<0在SKIPIF1<0內(nèi)是常數(shù)函數(shù)【即學(xué)即練1】（2023下·新疆巴音郭楞·高二校考期末）如圖所示是函數(shù)SKIPIF1<0的導(dǎo)函數(shù)SKIPIF1<0的圖象，則下列判斷中正確的是（

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A．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是減函數(shù)B．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是減函數(shù)C．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是減函數(shù)D．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是增函數(shù)【答案】A【詳解】對于選項A：當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故A正確；對于選項B：當(dāng)SKIPIF1<0時，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0；則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，故B錯誤；對于選項C：當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故C錯誤；對于選項D：當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故D錯誤；故選：A.知識點02：求已知函數(shù)（不含參）的單調(diào)區(qū)間①求SKIPIF1<0的定義域②求SKIPIF1<0③令SKIPIF1<0，解不等式，求單調(diào)增區(qū)間④令SKIPIF1<0，解不等式，求單調(diào)減區(qū)間注：求單調(diào)區(qū)間時，令SKIPIF1<0（或SKIPIF1<0）不跟等號.【即學(xué)即練2】（2023下·四川資陽·高二統(tǒng)考期末）函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0【答案】A【詳解】因為SKIPIF1<0，所以函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，所以SKIPIF1<0，由SKIPIF1<0有：SKIPIF1<0，所以函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，故B，C，D錯誤.故選：A.知識點03：由函數(shù)SKIPIF1<0的單調(diào)性求參數(shù)的取值范圍的方法1、已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)①已知SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增SKIPIF1<0SKIPIF1<0，SKIPIF1<0恒成立.②已知SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減SKIPIF1<0SKIPIF1<0，SKIPIF1<0恒成立.注：已知單調(diào)性，等價條件中的不等式含等號.2、已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在單調(diào)區(qū)間①已知SKIPIF1<0在區(qū)間SKIPIF1<0上存在單調(diào)增區(qū)間SKIPIF1<0SKIPIF1<0使得SKIPIF1<0有解②已知SKIPIF1<0在區(qū)間SKIPIF1<0上存在單調(diào)減區(qū)間SKIPIF1<0SKIPIF1<0使得SKIPIF1<0有解3、已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)SKIPIF1<0SKIPIF1<0，使得SKIPIF1<0有變號零點【即學(xué)即練3】（2023上·新疆·高三校聯(lián)考期中）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0的取值范圍為.【答案】SKIPIF1<0【詳解】因為SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，所以當(dāng)SKIPIF1<0時，SKIPIF1<0恒成立，即SKIPIF1<0在SKIPIF1<0恒成立，又SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0.【即學(xué)即練4】（2023上·貴州貴陽·高三清華中學(xué)?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0存在單調(diào)遞減區(qū)間，則實數(shù)SKIPIF1<0的取值范圍是.【答案】SKIPIF1<0【詳解】函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，求導(dǎo)得SKIPIF1<0，依題意，不等式SKIPIF1<0在SKIPIF1<0上有解，等價于SKIPIF1<0在SKIPIF1<0上有解，而SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時取等號，則SKIPIF1<0，所以實數(shù)a的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<0．知識點04：含參問題討論單調(diào)性第一步：求SKIPIF1<0的定義域第二步：求SKIPIF1<0（導(dǎo)函數(shù)中有分母通分）第三步：確定導(dǎo)函數(shù)有效部分，記為SKIPIF1<0對于SKIPIF1<0進(jìn)行求導(dǎo)得到SKIPIF1<0，對SKIPIF1<0初步處理（如通分），提出SKIPIF1<0的恒正部分，將該部分省略，留下的部分則為SKIPIF1<0的有效部分（如：SKIPIF1<0，則記SKIPIF1<0為SKIPIF1<0的有效部分）.接下來就只需考慮導(dǎo)函數(shù)有效部分，只有該部分決定SKIPIF1<0的正負(fù).第四步：確定導(dǎo)函數(shù)有效部分SKIPIF1<0的類型：①SKIPIF1<0為一次型（或可化為一次型）②SKIPIF1<0為二次型（或可化為二次型）第五步：通過分析導(dǎo)函數(shù)有效部分，討論SKIPIF1<0的單調(diào)性題型01求函數(shù)的單調(diào)區(qū)間【典例1】（2022下·湖北·高二統(tǒng)考期末）函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】解：因為SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，故選：B【典例2】（2023下·河北滄州·高二?？茧A段練習(xí)）函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】SKIPIF1<0，定義域為SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減.故選：D.【變式1】（多選）（2023下·吉林長春·高二長春外國語學(xué)校校考期中）函數(shù)SKIPIF1<0的一個單調(diào)遞增區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】ABD【詳解】由題意SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，因此SKIPIF1<0的增區(qū)間是SKIPIF1<0，因此ABD正確，C錯誤．故選：ABD．題型02函數(shù)與導(dǎo)函數(shù)圖象間的關(guān)系【典例1】（2023·高二課時練習(xí)）已知函數(shù)SKIPIF1<0的導(dǎo)函數(shù)SKIPIF1<0的圖象如圖，則下列結(jié)論正確的是（

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A．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增B．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減C．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增D．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增【答案】C【詳解】由導(dǎo)數(shù)的圖象可知，當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，故C正確；當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，故A、B、D錯誤；故選：C．【典例2】（2022下·廣東深圳·高二統(tǒng)考期末）設(shè)SKIPIF1<0是函數(shù)SKIPIF1<0的導(dǎo)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【詳解】由導(dǎo)函數(shù)的圖象可得當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞減；當(dāng)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.【變式1】（2023下·四川成都·高二四川省成都市新都一中校聯(lián)考期中）已知函數(shù)SKIPIF1<0的導(dǎo)函數(shù)SKIPIF1<0的圖像如圖所示，則函數(shù)SKIPIF1<0（

）A．在SKIPIF1<0上單調(diào)遞增 B．在SKIPIF1<0上單調(diào)遞減C．在SKIPIF1<0上單調(diào)遞增 D．在SKIPIF1<0上單調(diào)遞減【答案】D【詳解】由圖可知：當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞增；故選：D.【變式2】（2022·湖南·校聯(lián)考二模）設(shè)函數(shù)SKIPIF1<0在定義域內(nèi)可導(dǎo)，SKIPIF1<0的圖象如圖所示，則其導(dǎo)函數(shù)SKIPIF1<0的圖象可能是（

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A． B．C． D．【答案】D【詳解】由SKIPIF1<0的圖象可知，SKIPIF1<0在SKIPIF1<0上為單調(diào)遞減函數(shù)，故SKIPIF1<0時，SKIPIF1<0，故排除A，C；當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0的圖象是先遞增，再遞減，最后再遞增，所以SKIPIF1<0的值是先正，再負(fù)，最后是正，因此排除B，故選：D.題型03已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)，求參數(shù)【典例1】（2023上·廣西·高三南寧三中校聯(lián)考階段練習(xí)）若函數(shù)SKIPIF1<0是SKIPIF1<0上的減函數(shù)，則實數(shù)SKIPIF1<0的最大值為．【答案】SKIPIF1<0/SKIPIF1<0【詳解】由函數(shù)SKIPIF1<0是SKIPIF1<0上的減函數(shù)，則SKIPIF1<0在SKIPIF1<0上恒成立，即SKIPIF1<0在SKIPIF1<0上恒成立，設(shè)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞減當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，可得SKIPIF1<0，所以SKIPIF1<0，即實數(shù)SKIPIF1<0的最大值為SKIPIF1<0．故答案為：SKIPIF1<0.【典例2】（2023·海南·校聯(lián)考模擬預(yù)測）設(shè)SKIPIF1<0且SKIPIF1<0，若函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則a的取值范圍是．【答案】SKIPIF1<0【詳解】由題意知：SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，要使SKIPIF1<0，則SKIPIF1<0，整理得SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0．故答案為：SKIPIF1<0【典例3】（2023上·遼寧大連·高三大連市金州高級中學(xué)?？计谥校┤艉瘮?shù)SKIPIF1<0在SKIPIF1<0具有單調(diào)性，則a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】由SKIPIF1<0，當(dāng)函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞增時，SKIPIF1<0恒成立，得SKIPIF1<0，設(shè)SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0，因此有SKIPIF1<0，當(dāng)函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞減時，SKIPIF1<0恒成立，得SKIPIF1<0，設(shè)SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時，SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0，顯然無論SKIPIF1<0取何實數(shù)，不等式SKIPIF1<0不能恒成立，綜上所述，a的取值范圍是SKIPIF1<0，故選：C【變式1】（2023上·江蘇蘇州·高三常熟中學(xué)?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則實數(shù)a的取值范圍是．【答案】SKIPIF1<0【詳解】由SKIPIF1<0得SKIPIF1<0，由于函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0在SKIPIF1<0上恒成立，因此在SKIPIF1<0對任意的SKIPIF1<0恒成立，所以SKIPIF1<0，故答案為：SKIPIF1<0【變式2】（2023·海南省直轄縣級單位·?？寄M預(yù)測）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則實數(shù)SKIPIF1<0的取值范圍為．【答案】SKIPIF1<0【詳解】由題設(shè)SKIPIF1<0在SKIPIF1<0上恒成立．設(shè)SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上恒成立，又SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時等號成立，所以SKIPIF1<0，即實數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0．故答案為：SKIPIF1<0題型04已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在單調(diào)區(qū)間，求參數(shù)【典例1】（2023上·浙江寧波·高二鎮(zhèn)海中學(xué)?？计谥校┤艉瘮?shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有單調(diào)遞增區(qū)間，則實數(shù)SKIPIF1<0的取值范圍是．【答案】SKIPIF1<0【詳解】SKIPIF1<0，由題意SKIPIF1<0在SKIPIF1<0上有解，即SKIPIF1<0在SKIPIF1<0上有解，根據(jù)對勾函數(shù)的性質(zhì)可知，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以在SKIPIF1<0時取最大值，故SKIPIF1<0，故實數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<0【典例2】（2023下·江西撫州·高二江西省臨川第二中學(xué)?？茧A段練習(xí)）函數(shù)SKIPIF1<0在SKIPIF1<0上存在單調(diào)遞增區(qū)間，則SKIPIF1<0的取值范圍是.【答案】SKIPIF1<0【詳解】函數(shù)SKIPIF1<0，∴SKIPIF1<0，∵函數(shù)SKIPIF1<0在SKIPIF1<0上存在單調(diào)遞增區(qū)間，SKIPIF1<0，即SKIPIF1<0有解，令SKIPIF1<0,SKIPIF1<0,∴當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0即可．故答案為:SKIPIF1<0【變式1】（2023下·廣西·高二校聯(lián)考期中）若函數(shù)SKIPIF1<0在SKIPIF1<0存在單調(diào)遞減區(qū)間，則a的取值范圍為．【答案】SKIPIF1<0【詳解】SKIPIF1<0，等價于SKIPIF1<0在SKIPIF1<0有解，即SKIPIF1<0在SKIPIF1<0有解，即SKIPIF1<0在SKIPIF1<0有解，所以SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上是增函數(shù)，∴SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0.題型05已知函數(shù)SKIPIF1<0在的單調(diào)區(qū)間為（是）SKIPIF1<0，求參數(shù)【典例1】（2023下·高二課時練習(xí)）已知函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間是SKIPIF1<0，則SKIPIF1<0.【答案】SKIPIF1<0【詳解】SKIPIF1<0，因為函數(shù)SKIPIF1<0單調(diào)遞減區(qū)間是SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，所以函數(shù)SKIPIF1<0單調(diào)遞減區(qū)間是SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0.題型06已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)，求參數(shù)【典例1】（2023下·湖北·高二校聯(lián)考階段練習(xí)）若函數(shù)SKIPIF1<0在其定義域的一個子區(qū)間SKIPIF1<0內(nèi)不是單調(diào)函數(shù)，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因為函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0（舍去），因為SKIPIF1<0在定義域的一個子區(qū)間SKIPIF1<0內(nèi)不是單調(diào)函數(shù)，所以SKIPIF1<0，得SKIPIF1<0，綜上，SKIPIF1<0，故選：A【典例2】（2022上·河南·高三校聯(lián)考階段練習(xí)）已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不是單調(diào)函數(shù)，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因為SKIPIF1<0在區(qū)間SKIPIF1<0上不是單調(diào)函數(shù)，所以SKIPIF1<0在區(qū)間SKIPIF1<0上有解，即SKIPIF1<0在區(qū)間SKIPIF1<0上有解．令SKIPIF1<0，則SKIPIF1<0．當(dāng)SKIPIF1<0時，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0．故SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．又因為SKIPIF1<0，且當(dāng)SKIPIF1<0時，SKIPIF1<0所以SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，解得SKIPIF1<0.故選：A【變式1】（2022·全國·高二專題練習(xí)）已知函數(shù)SKIPIF1<0．若SKIPIF1<0在SKIPIF1<0內(nèi)不單調(diào)，則實數(shù)a的取值范圍是．【答案】SKIPIF1<0【詳解】由SKIPIF1<0，得SKIPIF1<0，當(dāng)SKIPIF1<0在SKIPIF1<0內(nèi)為減函數(shù)時，則SKIPIF1<0在SKIPIF1<0內(nèi)恒成立，所以SKIPIF1<0在SKIPIF1<0內(nèi)恒成立，當(dāng)SKIPIF1<0在SKIPIF1<0內(nèi)為增函數(shù)時，則SKIPIF1<0在SKIPIF1<0內(nèi)恒成立，所以SKIPIF1<0在SKIPIF1<0內(nèi)恒成立，令SKIPIF1<0，因為SKIPIF1<0在SKIPIF1<0內(nèi)單調(diào)遞增，在SKIPIF1<0內(nèi)單調(diào)遞減，所以SKIPIF1<0在SKIPIF1<0內(nèi)的值域為SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0內(nèi)單調(diào)時，a的取值范圍是SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上不單調(diào)時，實數(shù)a的取值范圍是SKIPIF1<0．故答案為：SKIPIF1<0．【變式2】（2022下·福建漳州·高二福建省漳州第一中學(xué)?？茧A段練習(xí)）若函數(shù)SKIPIF1<0在區(qū)間(1，4)上不單調(diào)，則實數(shù)a的取值范圍是.【答案】(4，5)【詳解】解：SKIPIF1<0函數(shù)SKIPIF1<0，SKIPIF1<0，若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)，則SKIPIF1<0在SKIPIF1<0上存在變號零點，由SKIPIF1<0得SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0遞減，在SKIPIF1<0遞增，而SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0.題型07含參問題討論單調(diào)性(導(dǎo)函數(shù)有效部分是一次型)【典例1】（2023上·陜西咸陽·高三統(tǒng)考期中）已知函數(shù)SKIPIF1<0.(1)若SKIPIF1<0，求函數(shù)SKIPIF1<0的極值；(2)求函數(shù)SKIPIF1<0的單調(diào)區(qū)間.【答案】(1)函數(shù)SKIPIF1<0的極大值為SKIPIF1<0，無極小值(2)答案見解析【詳解】（1）當(dāng)SKIPIF1<0時，SKIPIF1<0，其定義域為SKIPIF1<0，SKIPIF1<0.令SKIPIF1<0，則SKIPIF1<0.SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，SKIPIF1<0函數(shù)SKIPIF1<0的極大值為SKIPIF1<0，無極小值.（2）SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時，由SKIPIF1<0，得SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0,若SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0，綜上，當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0.【典例2】（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0．求函數(shù)SKIPIF1<0的單調(diào)區(qū)間．【答案】增區(qū)間為SKIPIF1<0，減區(qū)間為SKIPIF1<0．【詳解】SKIPIF1<0的定義域為SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0．令SKIPIF1<0，得SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，∴SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0．【變式1】（2023·全國·高三專題練習(xí)）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【詳解】由函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，且SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，若SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減.若SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增.若SKIPIF1<0，此時函數(shù)SKIPIF1<0為常數(shù)函數(shù)，無單調(diào)性.【變式2】（2023·全國·高三專題練習(xí)）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【詳解】由題意，得函數(shù)SKIPIF1<0的定義域為R，則SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0對任意SKIPIF1<0恒成立，∴函數(shù)SKIPIF1<0在R上單調(diào)遞減；當(dāng)SKIPIF1<0時，令SKIPIF1<0，得SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，∴函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．綜上，當(dāng)SKIPIF1<0時，SKIPIF1<0在R上單調(diào)遞減；當(dāng)SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增．題型08含參問題討論單調(diào)性(導(dǎo)函數(shù)有效部分是二次型且可因式分解)【典例1】（2023上·江蘇揚(yáng)州·高三儀征市第二中學(xué)校考期中）已知函數(shù)SKIPIF1<0，其中SKIPIF1<0．(1)若SKIPIF1<0是函數(shù)SKIPIF1<0的極值點，求a的值；(2)若SKIPIF1<0，討論函數(shù)SKIPIF1<0的單調(diào)性．【答案】(1)SKIPIF1<0(2)答案見解析【詳解】（1）SKIPIF1<0SKIPIF1<0，因為SKIPIF1<0是函數(shù)SKIPIF1<0的極值點，所以SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0.即函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，即SKIPIF1<0是函數(shù)SKIPIF1<0的極值點.故SKIPIF1<0．（2）SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時，令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0，即SKIPIF1<0時，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0或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，在SKIPIF1<0上單調(diào)遞減．當(dāng)SKIPIF1<0時，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0或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，在SKIPIF1<0上單調(diào)遞減．當(dāng)SKIPIF1<0，即SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減.綜上，當(dāng)SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上遞減，在SKIPIF1<0上遞增，在SKIPIF1<0上遞減；當(dāng)SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減．【典例2】（2024·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，其中SKIPIF1<0.求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；【答案】答案見解析【詳解】SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，則函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0時，SKIPIF1<0或SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0時，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，當(dāng)SKIPIF1<0時，SKIPIF1<0或SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0時，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.綜上所述，當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，無單調(diào)遞減區(qū)間；當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為在SKIPIF1<0，SKIPIF1<0，單調(diào)遞減區(qū)間為SKIPIF1<0.【典例3】（2023上·甘肅慶陽·高三?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0SKIPIF1<0(1)若SKIPIF1<0，求曲線SKIPIF1<0在點SKIPIF1<0處的切線方程；(2)當(dāng)SKIPIF1<0時，求函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間.【答案】(1)SKIPIF1<0(2)答案見解析【詳解】（1）當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0，所以切線的斜率SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0．（2）因為SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，又SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0恒成立，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．當(dāng)SKIPIF1<0時，SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0；綜上所述，當(dāng)SKIPIF1<0時，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，SKIPIF1<0.【變式1】（2023上·河南南陽·高三?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時，求曲線SKIPIF1<0在點SKIPIF1<0處的切線方程；(2)當(dāng)SKIPIF1<0時，討論函數(shù)SKIPIF1<0的單調(diào)性；【答案】(1)SKIPIF1<0(2)在區(qū)間SKIPIF1<0上單調(diào)遞增，在區(qū)間SKIPIF1<0上單調(diào)遞減【詳解】（1）當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0，所以，當(dāng)SKIPIF1<0時，SKIPIF1<0，又SKIPIF1<0，所以，由導(dǎo)數(shù)的幾何意義知曲線SKIPIF1<0在點SKIPIF1<0處的切線方程為SKIPIF1<0．（2）因為SKIPIF1<0，易知，SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0,當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，在區(qū)間SKIPIF1<0上單調(diào)遞減．【變式2】（2023上·北京順義·高三楊鎮(zhèn)第一中學(xué)校考階段練習(xí)）已知函數(shù)SKIPIF1<0.(1)當(dāng)SKIPIF1<0時，求函數(shù)SKIPIF1<0的最小值；(2)當(dāng)SKIPIF1<0時，討論SKIPIF1<0的單調(diào)性.【答案】(1)SKIPIF1<0(2)答案見解析【詳解】（1）當(dāng)SKIPIF1<0時：SKIPIF1<0，令SKIPIF1<0解得SKIPIF1<0，又因為當(dāng)SKIPIF1<0，SKIPIF1<0，此時函數(shù)SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0，SKIPIF1<0，此時函數(shù)SKIPIF1<0單調(diào)遞增.所以SKIPIF1<0的最小值為SKIPIF1<0SKIPIF1<0.（2）SKIPIF1<0，當(dāng)SKIPIF1<0時，由SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0.①若SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；②若SKIPIF1<0，則SKIPIF1<0.故當(dāng)SKIPIF1<0時，SKIPIF1<0或SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0.所以SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減.③若SKIPIF1<0，則SKIPIF1<0.故當(dāng)SKIPIF1<0時，SKIPIF1<0或SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0.所以SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減.綜上所述，當(dāng)SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；當(dāng)SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減.當(dāng)SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減.【變式2】（2023·全國·高三專題練習(xí)）已知SKIPIF1<0，求SKIPIF1<0的單調(diào)遞減區(qū)間.【答案】答案見解析【詳解】易得SKIPIF1<0的定義域為SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0或SKIPIF1<0.當(dāng)SKIPIF1<0時，因為SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0，所以SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0.當(dāng)SKIPIF1<0時，①若SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0,當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0；②若SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0恒成立，SKIPIF1<0沒有單調(diào)遞減區(qū)間；③若SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時SKIPIF1<0，所以SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0.綜上所述，當(dāng)SKIPIF1<0時，SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0無單調(diào)遞減區(qū)間；當(dāng)SKIPIF1<0時，SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0.題型09含參問題討論單調(diào)性(導(dǎo)函數(shù)有效部分是二次型且不可因式分解)【典例1】（2023·全國·高三專題練習(xí)）設(shè)函數(shù)SKIPIF1<0，其中SKIPIF1<0為常數(shù)，討論函數(shù)SKIPIF