新高考數(shù)學二輪復習考點講義第九講導數(shù)與函數(shù)的單調(diào)性（原卷版）

第九講：導數(shù)與函數(shù)的單調(diào)性【考點梳理】1、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.2、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,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,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（SKIPIF1<0為變號零點）3、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（導函數(shù)中有分母通分）第三步：確定導函數(shù)有效部分，記為SKIPIF1<0對于SKIPIF1<0進行求導得到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函數(shù)有效部分，只有該部分決定SKIPIF1<0的正負.第四步：確定導函數(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：通過分析導函數(shù)有效部分，討論SKIPIF1<0的單調(diào)性【典型題型講解】考點一：求函數(shù)的單調(diào)區(qū)間（不含參）【典例例題】例1.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例2.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【方法技巧與總結(jié)】函數(shù)單調(diào)區(qū)間的求法：解不等式法，列表格法【變式訓練】1.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<02.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<03.已知函數(shù)f(x)滿足SKIPIF1<0，則f(x)的單調(diào)遞減區(qū)間為（

）A．(-,0) B．(1,+∞) C．(-,1) D．(0,+∞)4．函數(shù)SKIPIF1<0的單調(diào)增區(qū)間是(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間為__________．【典型題型講解】考點二：已知含量參函數(shù)在區(qū)間上單調(diào)或不單調(diào)或存在單調(diào)區(qū)間，求參數(shù)范圍【典例例題】例1.如果函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0例2.若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)存在單調(diào)遞增區(qū)間，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0例3.函數(shù)SKIPIF1<0在SKIPIF1<0上不單調(diào)的一個充分不必要條件是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【方法技巧與總結(jié)】（1）已知函數(shù)在區(qū)間上單調(diào)遞增或單調(diào)遞減，轉(zhuǎn)化為導函數(shù)恒大于等于或恒小于等于零求解，先分析導函數(shù)的形式及圖像特點，如一次函數(shù)最值落在端點，開口向上的拋物線最大值落在端點，開口向下的拋物線最小值落在端點等.（2）已知區(qū)間上函數(shù)不單調(diào)，轉(zhuǎn)化為導數(shù)在區(qū)間內(nèi)存在變號零點，通常用分離變量法求解參變量范圍.（3）已知函數(shù)在區(qū)間上存在單調(diào)遞增或遞減區(qū)間，轉(zhuǎn)化為導函數(shù)在區(qū)間上大于零或小于零有解.【變式訓練】1．若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)單調(diào)遞減，則實數(shù)SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02.已知函數(shù)SKIPIF1<0在SKIPIF1<0上為單調(diào)遞增函數(shù)，則實數(shù)m的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．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<03.已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在單調(diào)遞增區(qū)間，則實數(shù)SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04.已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)，則實數(shù)SKIPIF1<0的取值范圍為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)，則實數(shù)SKIPIF1<0的取值范圍是（）A．(-∞，-3] B．(-3，1)C．[1，+∞) D．(-∞，-3]∪[1，+∞)考點三：含參問題討論單調(diào)性【典例例題】例1.已知函數(shù)SKIPIF1<0，其中SKIPIF1<0.求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；例題2．設(shè)函數(shù)SKIPIF1<0，求SKIPIF1<0的單調(diào)區(qū)間．例3.已知函數(shù)SKIPIF1<0.討論SKIPIF1<0的單調(diào)性；例4.已知函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0的導函數(shù)為SKIPIF1<0．討論函數(shù)SKIPIF1<0的單調(diào)性；【方法技巧與總結(jié)】1．關(guān)于含參函數(shù)單調(diào)性的討論問題，要根據(jù)導函數(shù)的情況來作出選擇，通過對新函數(shù)零點個數(shù)的討論，從而得到原函數(shù)對應(yīng)導數(shù)的正負，最終判斷原函數(shù)的增減．（注意定義域的間斷情況）．2．需要求二階導的題目，往往通過二階導的正負來判斷一階導函數(shù)的單調(diào)性，結(jié)合一階導函數(shù)端點處的函數(shù)值或零點可判斷一階導函數(shù)正負區(qū)間段．3．利用草稿圖像輔助說明．【變式訓練】1.已知函數(shù)SKIPIF1<0.(1)當SKIPIF1<0時，求函數(shù)SKIPIF1<0在SKIPIF1<0處的切線方程；(2)求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；2．（2022·廣東深圳·高三期末）已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0．(1)求SKIPIF1<0的單調(diào)遞增區(qū)間；(2)對于SKIPIF1<0，若不等式SKIPIF1<0恒成立，求a的取值范圍．3.已知函數(shù)SKIPIF1<0．(1)當SKIPIF1<0時，求曲線SKIPIF1<0在SKIPIF1<0處的切線方程；(2)求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；4．已知函數(shù)SKIPIF1<0討論f（x）的單調(diào)性；5.已知函數(shù)SKIPIF1<0，記SKIPIF1<0的導函數(shù)為SKIPIF1<0討論SKIPIF1<0的單調(diào)性；6.（2022·廣東深圳·一模）已知函數(shù)SKIPIF1<0（SKIPIF1<0）．(1)求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(2)若函數(shù)SKIPIF1<0有兩個零點SKIPIF1<0，SKIPIF1<0．（i）求實數(shù)a的取值范圍；（ii）求證：SKIPIF1<0．【鞏固練習】一、單選題1．已知函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<02．已知函數(shù)SKIPIF1<0，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<03．“函數(shù)SKIPIF1<0在SKIPIF1<0上是增函數(shù)”是“SKIPIF1<0”的（

）A．充分不必要條件 B．必要不充分條件C．充分必要條件 D．既不充分也不必要條件4．已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在單調(diào)減區(qū)間，則實數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0二、多選題5．已知SKIPIF1<0，下列說法正確的是（

）A．SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0 B．SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0C．SKIPIF1<0的極大值為SKIPIF1<0 D．方程SKIPIF1<0有兩個不同的解6．已知函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，其導函數(shù)為SKIPIF1<0，對于任意SKIPIF1<0，都有SKIPIF1<0，則使不等式SKIPIF1<0成立的SKIPIF1<0