新高考數(shù)學(xué)一輪復(fù)習(xí)知識點(diǎn)總結(jié)與題型精練專題10 利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性、極值和最值(原卷版)

專題10利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性、極值和最值【考綱要求】了解函數(shù)的單調(diào)性與導(dǎo)數(shù)的關(guān)系；2、能利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性，會求函數(shù)的單調(diào)區(qū)間(其中多項式函數(shù)一般不超過三次).3、了解函數(shù)在某點(diǎn)取得極值的必要條件和充分條件；會用導(dǎo)數(shù)求函數(shù)的極大值、極小值；4、會求閉區(qū)間上函數(shù)的最大值、最小值一、利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性【思維導(dǎo)圖】【考點(diǎn)總結(jié)】1．函數(shù)的導(dǎo)數(shù)與單調(diào)性的關(guān)系函數(shù)y＝f(x)在某個區(qū)間內(nèi)可導(dǎo)，則(1)若f′(x)＞0，則f(x)在這個區(qū)間內(nèi)單調(diào)遞增。(2)若f′(x)＜0，則f(x)在這個區(qū)間內(nèi)單調(diào)遞減。(3)若f′(x)＝0，則f(x)在這個區(qū)間內(nèi)是常數(shù)函數(shù)。二、利用導(dǎo)數(shù)研究函數(shù)的極值與最值【思維導(dǎo)圖】【考點(diǎn)總結(jié)】1、函數(shù)的極小值若函數(shù)y＝f(x)在點(diǎn)x＝a處的函數(shù)值f(a)比它在點(diǎn)x＝a附近其他點(diǎn)的函數(shù)值都小，且f′(a)＝0，而且在點(diǎn)x＝a附近的左側(cè)f′(x)＜0，右側(cè)f′(x)＞0，則x＝a叫做函數(shù)的極小值點(diǎn)，f(a)叫做函數(shù)的極小值。2、函數(shù)的極大值若函數(shù)y＝f(x)在點(diǎn)x＝b處的函數(shù)值f(b)比它在點(diǎn)x＝b附近其他點(diǎn)的函數(shù)值都大，且f′(b)＝0，而且在點(diǎn)x＝b附近的左側(cè)f′(x)＞0，右側(cè)f′(x)＜0，則x＝b叫做函數(shù)的極大值點(diǎn)，f(b)叫做函數(shù)的極大值，極大值和極小值統(tǒng)稱為極值。3、函數(shù)的最值與導(dǎo)數(shù)(1)函數(shù)f(x)在[a，b]上有最值的條件：一般地，如果在區(qū)間[a，b]上，函數(shù)y＝f(x)的圖象是一條連續(xù)不斷的曲線，那么它必有最大值和最小值。(2)求函數(shù)y＝f(x)在[a，b]上的最大值與最小值的步驟為：①求函數(shù)y＝f(x)在(a，b)內(nèi)的極值；②將函數(shù)y＝f(x)的各極值與端點(diǎn)處的函數(shù)值f(a)，f(b)比較，其中最大的一個是最大值，最小的一個是最小值。三、利用導(dǎo)數(shù)研究零點(diǎn)與不等式恒成立問題【思維導(dǎo)圖】【題型匯編】題型一：利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性題型二：利用導(dǎo)數(shù)研究函數(shù)的極值題型三：利用導(dǎo)數(shù)研究函數(shù)的最值題型四：導(dǎo)數(shù)中的極值點(diǎn)偏移問題【題型講解】題型一：利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性一、單選題1．（2022·全國·高考真題）設(shè)SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2022·江西贛州·二模（理））已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2022·上海松江·二模）下列函數(shù)中，與函數(shù)SKIPIF1<0的奇偶性和單調(diào)性都一致的函數(shù)是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．（2022·四川省瀘縣第一中學(xué)二模（文））下列函數(shù)中為奇函數(shù)且在SKIPIF1<0單調(diào)遞增的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<05．（2022·北京·首都師范大學(xué)附屬中學(xué)三模）下列函數(shù)中，既是偶函數(shù)又在SKIPIF1<0上單調(diào)遞減的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<05．（2022·江西師大附中三模（理））下列函數(shù)中既是奇函數(shù)又是增函數(shù)的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．（2022·云南曲靖·二模（文））設(shè)SKIPIF1<0是函數(shù)SKIPIF1<0的導(dǎo)函數(shù)，SKIPIF1<0是函數(shù)SKIPIF1<0的導(dǎo)函數(shù)，若對任意SKIPIF1<0恒成立，則下列選項正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0二、多選題1．（2022·湖南·長沙市明德中學(xué)二模）已知SKIPIF1<0，若SKIPIF1<0（SKIPIF1<0為自然對數(shù)的底數(shù)），則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0三、解答題1．（2022·北京·高考真題）已知函數(shù)SKIPIF1<0．(1)求曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程；(2)設(shè)SKIPIF1<0，討論函數(shù)SKIPIF1<0在SKIPIF1<0上的單調(diào)性；2．（2022·浙江·高考真題）設(shè)函數(shù)SKIPIF1<0．(1)求SKIPIF1<0的單調(diào)區(qū)間；題型二：利用導(dǎo)數(shù)研究函數(shù)的極值一、單選題1．（2022·陜西商洛·一模（文））已知函數(shù)SKIPIF1<0，則SKIPIF1<0的極大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2022·甘肅平?jīng)觥ざ＃ㄎ模┮阎瘮?shù)SKIPIF1<0的導(dǎo)函數(shù)的圖象如圖所示，則SKIPIF1<0的極值點(diǎn)的個數(shù)為（

）A．0 B．1 C．2 D．33．（2022·陜西西安·二模（理））函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，其導(dǎo)函數(shù)SKIPIF1<0的圖像如圖所示，則函數(shù)SKIPIF1<0極值點(diǎn)的個數(shù)為（

）A．2 B．3 C．4 D．54．（2022·寧夏·吳忠中學(xué)三模（理））下列函數(shù)中，既是奇函數(shù)又存在極值的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2022·內(nèi)蒙古包頭·一模（理））設(shè)SKIPIF1<0，若SKIPIF1<0為函數(shù)SKIPIF1<0的極小值點(diǎn)，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．（2022·內(nèi)蒙古呼倫貝爾·二模（文））已知SKIPIF1<0是函數(shù)SKIPIF1<0的極值點(diǎn)，則SKIPIF1<0（

）A．1 B．2 C．SKIPIF1<0 D．SKIPIF1<07．（2022·江西南昌·一模（理））已知函數(shù)SKIPIF1<0，若不等式SKIPIF1<0的解集為SKIPIF1<0，且SKIPIF1<0，則函數(shù)SKIPIF1<0的極大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．0 D．SKIPIF1<0二、多選題1．（2022·重慶八中模擬預(yù)測）設(shè)函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，SKIPIF1<0是SKIPIF1<0的極小值點(diǎn)，以下結(jié)論一定正確的是（

）A．SKIPIF1<0是SKIPIF1<0的最小值點(diǎn)B．SKIPIF1<0是SKIPIF1<0的極大值點(diǎn)C．SKIPIF1<0是SKIPIF1<0的極大值點(diǎn)D．SKIPIF1<0是SKIPIF1<0的極大值點(diǎn)2．（2022·湖南懷化·一模）下列函數(shù)中，存在極值點(diǎn)的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0三、解答題1．（2022·天津·耀華中學(xué)二模）已知函數(shù)SKIPIF1<0.(1)若SKIPIF1<0，求函數(shù)SKIPIF1<0的單調(diào)區(qū)間；(2)若SKIPIF1<0存在兩個極小值點(diǎn)SKIPIF1<0，求實數(shù)SKIPIF1<0的取值范圍.2．（2022·北京市第十二中學(xué)三模）已知函數(shù)SKIPIF1<0.(1)當(dāng)SKIPIF1<0時，求函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間；(2)設(shè)函數(shù)SKIPIF1<0，若SKIPIF1<0在SKIPIF1<0上存在極值，求a的取值范圍.題型三：利用導(dǎo)數(shù)研究函數(shù)的最值一、單選題1．（2022·河南鄭州·三模（文））SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值是（

）A．SKIPIF1<0 B．1 C．SKIPIF1<0 D．SKIPIF1<02．（2022·廣西南寧·二模（文））已知函數(shù)SKIPIF1<0，SKIPIF1<0，則函數(shù)SKIPIF1<0的最大值是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．-1 D．SKIPIF1<03．（2022·江西萍鄉(xiāng)·三模（理））已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0，對任意SKIPIF1<0，當(dāng)SKIPIF1<0時，都有SKIPIF1<0，若存在SKIPIF1<0，使不等式SKIPIF1<0成立，則實數(shù)SKIPIF1<0的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2022·內(nèi)蒙古·滿洲里市教研培訓(xùn)中心三模（文））設(shè)直線SKIPIF1<0與函數(shù)SKIPIF1<0的圖像分別交于點(diǎn)SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2022·山西晉城·三模（理））已知函數(shù)SKIPIF1<0，若對任意SKIPIF1<0，SKIPIF1<0，SKIPIF1<0恒成立，則m的最大值為（

）A．－1 B．0 C．1 D．e二、解答題1．（2022·天津·二模）已知函數(shù)SKIPIF1<0．(1)當(dāng)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；(3)當(dāng)SKIPIF1<0時，求函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值．2．（2022·北京房山·二模）已知函數(shù)SKIPIF1<0.(1)當(dāng)SKIPIF1<0時，求曲線SKIPIF1<0在SKIPIF1<0處的切線方程；(2)求函數(shù)SKIPIF1<0在SKIPIF1<0上的最小值.題型四：導(dǎo)數(shù)中的極值點(diǎn)偏移問題一、解答題1．（2022·天津河?xùn)|·二模）已知函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）．(1)SKIPIF1<0，求函數(shù)SKIPIF1<0在SKIPIF1<0處的切線方程．(2)討論函數(shù)SKIPIF1<0的單調(diào)性；(3)若函數(shù)SKIPIF1<0有兩個零點(diǎn)SKIPIF1<0