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第03講導(dǎo)數(shù)與函數(shù)的極值、最值(精講+精練)目錄第一部分:知識(shí)點(diǎn)精準(zhǔn)記憶第二部分:課前自我評(píng)估測(cè)試第三部分:典型例題剖析高頻考點(diǎn)一:函數(shù)圖象與極值(點(diǎn))的關(guān)系高頻考點(diǎn)二:求已知函數(shù)的極值(點(diǎn))高頻考點(diǎn)三:根據(jù)函數(shù)的極值(點(diǎn))求參數(shù)高頻考點(diǎn)四:求函數(shù)的最值(不含參)高頻考點(diǎn)五:求函數(shù)的最值(含參)高頻考點(diǎn)六:根據(jù)函數(shù)的最值求參數(shù)高頻考點(diǎn)七:函數(shù)的單調(diào)性、極值、最值的綜合應(yīng)用第四部分:高考真題感悟第五部分:第03講導(dǎo)數(shù)與函數(shù)的極值、最值(精練)第一部分:知識(shí)點(diǎn)精準(zhǔn)記憶第一部分:知識(shí)點(diǎn)精準(zhǔn)記憶1、函數(shù)的極值一般地,對(duì)于函數(shù)SKIPIF1<0,(1)若在點(diǎn)SKIPIF1<0處有SKIPIF1<0,且在點(diǎn)SKIPIF1<0附近的左側(cè)有SKIPIF1<0,右側(cè)有SKIPIF1<0,則稱SKIPIF1<0為SKIPIF1<0的極小值點(diǎn),SKIPIF1<0叫做函數(shù)SKIPIF1<0的極小值.(2)若在點(diǎn)SKIPIF1<0處有SKIPIF1<0,且在點(diǎn)SKIPIF1<0附近的左側(cè)有SKIPIF1<0,右側(cè)有SKIPIF1<0,則稱SKIPIF1<0為SKIPIF1<0的極大值點(diǎn),SKIPIF1<0叫做函數(shù)SKIPIF1<0的極大值.(3)極小值點(diǎn)與極大值點(diǎn)通稱極值點(diǎn),極小值與極大值通稱極值.注:極大(?。┲迭c(diǎn),不是一個(gè)點(diǎn),是一個(gè)數(shù).2、函數(shù)的最大(?。┲狄话愕?,如果在區(qū)間SKIPIF1<0上函數(shù)SKIPIF1<0的圖象是一條連續(xù)不斷的曲線,那么它必有最大值與最小值.設(shè)函數(shù)SKIPIF1<0在SKIPIF1<0上連續(xù),在SKIPIF1<0內(nèi)可導(dǎo),求SKIPIF1<0在SKIPIF1<0上的最大值與最小值的步驟為:(1)求SKIPIF1<0在SKIPIF1<0內(nèi)的極值;(2)將函數(shù)SKIPIF1<0的各極值與端點(diǎn)處的函數(shù)值SKIPIF1<0,SKIPIF1<0比較,其中最大的一個(gè)是最大值,最小的一個(gè)是最小值.3、函數(shù)的最值與極值的關(guān)系(1)極值是對(duì)某一點(diǎn)附近(即局部)而言,最值是對(duì)函數(shù)的定義區(qū)間SKIPIF1<0的整體而言;(2)在函數(shù)的定義區(qū)間SKIPIF1<0內(nèi),極大(?。┲悼赡苡卸鄠€(gè)(或者沒(méi)有),但最大(?。┲抵挥幸粋€(gè)(或者沒(méi)有);(3)函數(shù)SKIPIF1<0的極值點(diǎn)不能是區(qū)間的端點(diǎn),而最值點(diǎn)可以是區(qū)間的端點(diǎn);(4)對(duì)于可導(dǎo)函數(shù),函數(shù)的最大(小)值必在極大(小)值點(diǎn)或區(qū)間端點(diǎn)處取得.第二部分:課前自我評(píng)估測(cè)試第二部分:課前自我評(píng)估測(cè)試一、判斷題1.(2021·全國(guó)·高二課前預(yù)習(xí))函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上連續(xù),則SKIPIF1<0在區(qū)間SKIPIF1<0上一定有最值,但不一定有極值.()【答案】正確2.(2021·全國(guó)·高二課前預(yù)習(xí))函數(shù)的最大值不一定是函數(shù)的極大值.()【答案】正確3.(2021·全國(guó)·高二課前預(yù)習(xí))函數(shù)的極大值一定大于極小值.()【答案】錯(cuò)誤4.(2021·全國(guó)·高二課前預(yù)習(xí))有極值的函數(shù)一定有最值,有最值的函數(shù)不一定有極值.()【答案】錯(cuò)誤二、單選題1.(2022·廣東·高州市長(zhǎng)坡中學(xué)高二階段練習(xí))函數(shù)SKIPIF1<0在閉區(qū)間SKIPIF1<0上的最大值、最小值分別是(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】CSKIPIF1<0,令SKIPIF1<0得:SKIPIF1<0或SKIPIF1<0,令SKIPIF1<0得:SKIPIF1<0,故SKIPIF1<0在SKIPIF1<0處取得極大值,在SKIPIF1<0處取得極小值,且SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以函數(shù)SKIPIF1<0在閉區(qū)間SKIPIF1<0上的最大值、最小值分別是3,-17.故選:C2.(2022·黑龍江·牡丹江市第三高級(jí)中學(xué)高二期末)函數(shù)y=SKIPIF1<0的最大值為(

)A.e-1 B.e C.e2 D.10【答案】A令SKIPIF1<0當(dāng)SKIPIF1<0時(shí),SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0所以函數(shù)得極大值為SKIPIF1<0,因?yàn)樵诙x域內(nèi)只有一個(gè)極值,所以SKIPIF1<0故選:A.3.(2022·河北邢臺(tái)·高二階段練習(xí))已知函數(shù)SKIPIF1<0的導(dǎo)函數(shù)的圖象如圖所示,則SKIPIF1<0極值點(diǎn)的個(gè)數(shù)為(

)A.4 B.5 C.6 D.7【答案】A對(duì)于處處可導(dǎo)的函數(shù),函數(shù)的極值點(diǎn)要滿足兩個(gè)條件,一個(gè)是該點(diǎn)的導(dǎo)數(shù)為0,另一個(gè)是該點(diǎn)左、右的導(dǎo)數(shù)值異號(hào),由圖象可知,導(dǎo)函數(shù)與SKIPIF1<0軸有5個(gè)交點(diǎn),因?yàn)樵?附近的左側(cè)SKIPIF1<0,右側(cè)SKIPIF1<0,所以0不是SKIPIF1<0極值點(diǎn).其余四個(gè)點(diǎn)的左、右的導(dǎo)數(shù)值異號(hào),所以是極值點(diǎn),故SKIPIF1<0極值點(diǎn)的個(gè)數(shù)是4.故選:A.4.(2022·天津市濱海新區(qū)塘沽第一中學(xué)高二階段練習(xí))若函數(shù)SKIPIF1<0在SKIPIF1<0處取得極值,則SKIPIF1<0(

)A.1 B.2 C.3 D.4【答案】A解:因?yàn)楹瘮?shù)SKIPIF1<0在SKIPIF1<0處取得極值,SKIPIF1<0,所以SKIPIF1<0,解得SKIPIF1<0,檢驗(yàn)當(dāng)SKIPIF1<0時(shí),函數(shù)SKIPIF1<0在SKIPIF1<0處取得極大值,所以SKIPIF1<0.故選:A.第三部分:典型例題剖析第三部分:典型例題剖析高頻考點(diǎn)一:函數(shù)圖象與極值(點(diǎn))的關(guān)系1.(2022·黑龍江·牡丹江市第三高級(jí)中學(xué)高二開學(xué)考試)已知函數(shù)SKIPIF1<0的導(dǎo)函數(shù)SKIPIF1<0的圖像如圖所示,則下列結(jié)論正確的是(

)A.當(dāng)SKIPIF1<0時(shí),函數(shù)SKIPIF1<0取得極小值B.函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是單調(diào)遞增的C.當(dāng)SKIPIF1<0時(shí),函數(shù)SKIPIF1<0取得極大值D.函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是單調(diào)遞增的【答案】A由圖像可知,SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0單調(diào)遞減,故B錯(cuò)誤;SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0單調(diào)遞增,所以當(dāng)SKIPIF1<0時(shí),函數(shù)SKIPIF1<0取得極小值,故A正確;當(dāng)SKIPIF1<0時(shí),函數(shù)SKIPIF1<0取得極大值,不是SKIPIF1<0的極值,故C錯(cuò)誤;導(dǎo)函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在SKIPIF1<0使得SKIPIF1<0,所以函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是先減后增,故D錯(cuò)誤;故選:A.2.(2022·全國(guó)·高三專題練習(xí))設(shè)函數(shù)SKIPIF1<0的導(dǎo)函數(shù)為SKIPIF1<0,函數(shù)SKIPIF1<0的圖像如圖所示,則(

)A.SKIPIF1<0的極大值為SKIPIF1<0,極小值為SKIPIF1<0B.SKIPIF1<0的極大值為SKIPIF1<0,極小值為SKIPIF1<0C.SKIPIF1<0的極大值為SKIPIF1<0,極小值為SKIPIF1<0D.SKIPIF1<0的極大值為SKIPIF1<0,極小值為SKIPIF1<0【答案】D當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0單調(diào)遞減;同理可得,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0單調(diào)遞增;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0單調(diào)遞減.∴SKIPIF1<0的極大值是SKIPIF1<0,SKIPIF1<0的極小值是SKIPIF1<0.故選:D.3.(2022·寧夏·銀川二中高二期末(文))已知函數(shù)SKIPIF1<0的導(dǎo)函數(shù)SKIPIF1<0的圖象如圖所示,則下列結(jié)論正確的是(

).A.函數(shù)SKIPIF1<0在SKIPIF1<0上是增函數(shù)B.SKIPIF1<0C.SKIPIF1<0D.SKIPIF1<0是函數(shù)SKIPIF1<0的極小值點(diǎn)【答案】B解:根據(jù)函數(shù)SKIPIF1<0的導(dǎo)函數(shù)SKIPIF1<0的圖象,可得SKIPIF1<0或SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0或SKIPIF1<0時(shí),SKIPIF1<0,所以函數(shù)SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上遞減,在SKIPIF1<0和SKIPIF1<0上遞增,故A錯(cuò)誤;SKIPIF1<0,故B正確;SKIPIF1<0,故C錯(cuò)誤;SKIPIF1<0是函數(shù)SKIPIF1<0的極大值點(diǎn),故D錯(cuò)誤.故選:B.4.(2022·全國(guó)·高二)如圖是函數(shù)SKIPIF1<0的大致圖象,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】C由圖示可知:SKIPIF1<0經(jīng)過(guò)(0,0)、(1,0)、(2,0),所以有:SKIPIF1<0,即SKIPIF1<0,解得:SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0.由圖示可知SKIPIF1<0是SKIPIF1<0的極值點(diǎn),所以SKIPIF1<0是SKIPIF1<0的兩根.所以SKIPIF1<0.故選:C.高頻考點(diǎn)二:求已知函數(shù)的極值(點(diǎn))1.(2022·山東師范大學(xué)附中高二階段練習(xí))函數(shù)SKIPIF1<0,SKIPIF1<0有(

)A.極大值25,極小值SKIPIF1<0 B.極大值25,極小值SKIPIF1<0C.極大值25,無(wú)極小值 D.極小值SKIPIF1<0,無(wú)極大值【答案】D由SKIPIF1<0,得SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0(舍去),當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0在SKIPIF1<0上遞減,在SKIPIF1<0上遞增,所以當(dāng)SKIPIF1<0時(shí),SKIPIF1<0取得極小值,無(wú)極大值,極小值為SKIPIF1<0,故選:D2.(2022·江蘇·海門中學(xué)高二期末)已知函數(shù)SKIPIF1<0在SKIPIF1<0處取得極值,則SKIPIF1<0的極大值為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B解:因?yàn)镾KIPIF1<0,所以SKIPIF1<0,依題意可得SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0定義域?yàn)镾KIPIF1<0,且SKIPIF1<0,令SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0,令SKIPIF1<0解得SKIPIF1<0,即SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增,在SKIPIF1<0上單調(diào)遞減,即在SKIPIF1<0處取得極大值,在SKIPIF1<0處取得極小值,所以SKIPIF1<0;故選:B3.(2022·全國(guó)·高二)已知函數(shù)SKIPIF1<0,則SKIPIF1<0在定義域上(

)A.有極小值SKIPIF1<0 B.有極大值SKIPIF1<0 C.有最大值 D.無(wú)最小值【答案】A解:因?yàn)镾KIPIF1<0定義域?yàn)镾KIPIF1<0,所以SKIPIF1<0,令SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,即SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,令SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,即SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,所以SKIPIF1<0時(shí)函數(shù)取得極小值即最小值,所以SKIPIF1<0,故有極小值SKIPIF1<0,最小值SKIPIF1<0,無(wú)極大值與最大值;故選:A4.(2022·全國(guó)·高二)函數(shù)SKIPIF1<0的極大值與極小值之和為(

)A.SKIPIF1<0 B.3 C.SKIPIF1<0 D.SKIPIF1<0【答案】D根據(jù)題意SKIPIF1<0,今SKIPIF1<0,∴SKIPIF1<0或1,當(dāng)SKIPIF1<0或SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0極小值SKIPIF1<0,SKIPIF1<0極大值SKIPIF1<0,所以極大值與極小值之和為SKIPIF1<0.故選:D.5.(2022·全國(guó)·高二課時(shí)練習(xí))若SKIPIF1<0是函數(shù)SKIPIF1<0的極值點(diǎn),則函數(shù)SKIPIF1<0(

)A.有極小值1 B.有極大值1 C.有極小值-1 D.有極大值-1【答案】A因?yàn)閤=1是函數(shù)SKIPIF1<0的極值點(diǎn),所以SKIPIF1<0,SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0時(shí),SKIPIF1<0,函數(shù)SKIPIF1<0單調(diào)遞增,SKIPIF1<0時(shí),SKIPIF1<0,函數(shù)SKIPIF1<0單調(diào)遞減,所以函數(shù)SKIPIF1<0有極小值SKIPIF1<0,故選:A.高頻考點(diǎn)三:根據(jù)函數(shù)的極值(點(diǎn))求參數(shù)1.(2022·河南新鄉(xiāng)·二模(文))已知SKIPIF1<0,函數(shù)SKIPIF1<0的極小值為SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.1 C.SKIPIF1<0 D.SKIPIF1<0【答案】CSKIPIF1<0,則SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞減,在SKIPIF1<0上單調(diào)遞增,所以SKIPIF1<0,則SKIPIF1<0,則SKIPIF1<0.故選:C2.(2022·全國(guó)·高三專題練習(xí))已知SKIPIF1<0在SKIPIF1<0處取得極值,則SKIPIF1<0的最小值為___________.【答案】3SKIPIF1<0,因?yàn)镾KIPIF1<0在SKIPIF1<0處取得極值,所以SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0.所以SKIPIF1<0SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào).把SKIPIF1<0,SKIPIF1<0代入SKIPIF1<0檢驗(yàn)得,SKIPIF1<0是SKIPIF1<0的極值點(diǎn),故SKIPIF1<0的最小值為3.故答案為:3.3.(2022·全國(guó)·高三專題練習(xí))若函數(shù)SKIPIF1<0不存在極值點(diǎn),則SKIPIF1<0的取值范圍是______.【答案】SKIPIF1<0∵SKIPIF1<0,∴SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0恒成立,SKIPIF1<0在SKIPIF1<0上為增函數(shù),滿足條件;若SKIPIF1<0,則SKIPIF1<0時(shí),即SKIPIF1<0時(shí),SKIPIF1<0恒成立,SKIPIF1<0在SKIPIF1<0上為增函數(shù),滿足條件;綜上可得SKIPIF1<0,即SKIPIF1<0.故答案為:SKIPIF1<0.4.(2022·江西南昌·高二期末(文))已知函數(shù)SKIPIF1<0在SKIPIF1<0處有極值2,則SKIPIF1<0______.【答案】6解:SKIPIF1<0,因?yàn)楹瘮?shù)SKIPIF1<0在SKIPIF1<0處有極值2,所以SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,則SKIPIF1<0,故當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以函數(shù)SKIPIF1<0在SKIPIF1<0處有極大值,所以SKIPIF1<0,所以SKIPIF1<0.故答案為:6.5.(2022·全國(guó)·高二課時(shí)練習(xí))函數(shù)SKIPIF1<0在x=1處有極值為10,則b的值為__.【答案】SKIPIF1<0SKIPIF1<0,SKIPIF1<0,依題意可知SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0在區(qū)間SKIPIF1<0遞減;在區(qū)間SKIPIF1<0遞增,所以SKIPIF1<0是SKIPIF1<0的極小值,符合題意.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0在SKIPIF1<0上遞增,沒(méi)有極值.所以SKIPIF1<0.故答案為:SKIPIF1<06.(2022·四川省綿陽(yáng)南山中學(xué)高二階段練習(xí)(文))若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有兩個(gè)極值點(diǎn),則實(shí)數(shù)a的取值范圍是______.【答案】SKIPIF1<0由題意,函數(shù)SKIPIF1<0,可得SKIPIF1<0,因?yàn)楹瘮?shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有兩個(gè)極值點(diǎn),即SKIPIF1<0在SKIPIF1<0上有兩個(gè)不等的實(shí)數(shù)根,即SKIPIF1<0在SKIPIF1<0上有兩個(gè)不等的實(shí)數(shù)根,即函數(shù)SKIPIF1<0和SKIPIF1<0的圖象有兩個(gè)交點(diǎn),又由SKIPIF1<0,可得SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0單調(diào)遞增;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0單調(diào)遞減,所以SKIPIF1<0,且當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0,解得SKIPIF1<0,即實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為:SKIPIF1<0.7.(2022·寧夏·平羅中學(xué)高二期末(文))若函數(shù)SKIPIF1<0,函數(shù)SKIPIF1<0有極值SKIPIF1<0.(1)求函數(shù)SKIPIF1<0的解析式;(2)求函數(shù)SKIPIF1<0的單調(diào)區(qū)間.【答案】(1)SKIPIF1<0(2)單調(diào)增區(qū)間為SKIPIF1<0,SKIPIF1<0;單調(diào)減區(qū)間SKIPIF1<0.(1)解:因?yàn)镾KIPIF1<0,所以SKIPIF1<0,由題意知SKIPIF1<0,解得SKIPIF1<0,SKIPIF1<0,SKIPIF1<0所求的解析式為SKIPIF1<0;(2)解:由(1)可得SKIPIF1<0,令SKIPIF1<0,解得SKIPIF1<0、SKIPIF1<0,令SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0,函數(shù)的單調(diào)增區(qū)間為SKIPIF1<0,SKIPIF1<0;令SKIPIF1<0,解得SKIPIF1<0,所以函數(shù)的單調(diào)減區(qū)間SKIPIF1<0.高頻考點(diǎn)四:求函數(shù)的最值(不含參)1.(2022·四川·攀枝花七中高二階段練習(xí)(理))已知SKIPIF1<0是SKIPIF1<0的極值點(diǎn),則SKIPIF1<0在SKIPIF1<0上的最大值是(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A由題意,SKIPIF1<0SKIPIF1<0且SKIPIF1<0SKIPIF1<0則SKIPIF1<0,SKIPIF1<0當(dāng)SKIPIF1<0時(shí),SKIPIF1<0單調(diào)遞減;當(dāng)

SKIPIF1<0或SKIPIF1<0時(shí),SKIPIF1<0單調(diào)遞增,SKIPIF1<0在SKIPIF1<0上,SKIPIF1<0單調(diào)遞增;

SKIPIF1<0單調(diào)遞減,又因?yàn)镾KIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0SKIPIF1<0在SKIPIF1<0上最大值是SKIPIF1<0,故選:A.2.(多選)(2022·山東省東明縣第一中學(xué)高二階段練習(xí))函數(shù)SKIPIF1<0在SKIPIF1<0上的最值情況為(

)A.最大值為12 B.最大值為5C.最小值為SKIPIF1<0 D.最小值為SKIPIF1<0【答案】AC由題意得:SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0或SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0>0.,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,故SKIPIF1<0是函數(shù)的極大值點(diǎn),則函數(shù)的極大值也即在SKIPIF1<0上的最大值為SKIPIF1<0,故A正確,B錯(cuò)誤;而當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,故函數(shù)在SKIPIF1<0上的最小值為SKIPIF1<0,故C正確,D錯(cuò)誤,故選:AC3.(2022·福建·啟悟中學(xué)高二階段練習(xí))已知函數(shù)SKIPIF1<0(1)求SKIPIF1<0在SKIPIF1<0處的切線方程;(2)求SKIPIF1<0在SKIPIF1<0上的最值.【答案】(1)SKIPIF1<0(2)最大值為2,最小值為-25(1)SKIPIF1<0,SKIPIF1<0,又SKIPIF1<0,SKIPIF1<0在SKIPIF1<0處的切線方程為SKIPIF1<0,即SKIPIF1<0(2)SKIPIF1<0,SKIPIF1<0令SKIPIF1<0,得SKIPIF1<0,令SKIPIF1<0,得SKIPIF1<0,故SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,在SKIPIF1<0上單調(diào)遞減.又SKIPIF1<0,SKIPIF1<0SKIPIF1<0,故SKIPIF1<0在SKIPIF1<0上的最大值為2,最小值為-25.4.(2022·廣東·深圳市南山區(qū)華僑城中學(xué)高二階段練習(xí))已知關(guān)于x的函數(shù)SKIPIF1<0,且函數(shù)f(x)在SKIPIF1<0處有極值-SKIPIF1<0.(1)求實(shí)數(shù)b,c的值;(2)求函數(shù)f(x)在[-1,2]上的最大值和最小值.【答案】(1)SKIPIF1<0,SKIPIF1<0(2)最大值為SKIPIF1<0,最小值為SKIPIF1<0(1)因?yàn)镾KIPIF1<0,所以SKIPIF1<0.因?yàn)楹瘮?shù)f(x)在SKIPIF1<0處有極值-SKIPIF1<0.所以SKIPIF1<0,解得SKIPIF1<0,或SKIPIF1<0.(i)當(dāng)SKIPIF1<0,SKIPIF1<0時(shí),SKIPIF1<0,所以f(x)在R上單調(diào)遞減,不存在極值.(ii)當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,f(x)單調(diào)遞增;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,f(x)單調(diào)遞減.所以f(x)在SKIPIF1<0處存在極大值,符合題意.綜上所述,SKIPIF1<0,SKIPIF1<0(2)由(1)知.SKIPIF1<0,則SKIPIF1<0,令SKIPIF1<0,得SKIPIF1<0,SKIPIF1<0.當(dāng)x變化時(shí),SKIPIF1<0,f(x)在[-1,2]的變化情況如下表:x-1(-1,1)1(1,2)2SKIPIF1<0+0-f(x)SKIPIF1<0單調(diào)遞增SKIPIF1<0單調(diào)遞減SKIPIF1<0所以f(x)在[-1,2]上的最大值為SKIPIF1<0,最小值為SKIPIF1<0.5.(2022·廣東·高州市長(zhǎng)坡中學(xué)高二階段練習(xí))已知函數(shù)SKIPIF1<0.(SKIPIF1<0為常數(shù))(1)當(dāng)SKIPIF1<0時(shí),求函數(shù)SKIPIF1<0的最值;【答案】(1)函數(shù)SKIPIF1<0的最小值為1,無(wú)最大值.(1)當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,定義域?yàn)镾KIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,故SKIPIF1<0在SKIPIF1<0處取得極小值,也是最小值,SKIPIF1<0,綜上:函數(shù)SKIPIF1<0的最小值為1,無(wú)最大值.6.(2022·遼寧·朝陽(yáng)市第二高級(jí)中學(xué)高二階段練習(xí))已知SKIPIF1<0.(1)若SKIPIF1<0在SKIPIF1<0處取得極值,求SKIPIF1<0的最小值;【答案】(1)SKIPIF1<0(1)∵SKIPIF1<0,∴SKIPIF1<0,∵SKIPIF1<0在SKIPIF1<0處取得極值,SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0.∴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.又∵當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0的最小值為SKIPIF1<0.7.(2022·江蘇·常熟中學(xué)高二階段練習(xí))已知函數(shù)SKIPIF1<0.(1)若SKIPIF1<0,求SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值;【答案】(1)SKIPIF1<0(1)當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0在區(qū)間SKIPIF1<0遞減;在區(qū)間SKIPIF1<0遞增.SKIPIF1<0,所以SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值為SKIPIF1<0.高頻考點(diǎn)五:求函數(shù)的最值(含參)1.(2022·廣西·高二期末(文))已知函數(shù)SKIPIF1<0.(1)若SKIPIF1<0,討論函數(shù)SKIPIF1<0的單調(diào)性;(2)當(dāng)SKIPIF1<0時(shí),求SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值和最大值.【答案】(1)SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增,在SKIPIF1<0上單調(diào)遞減.(2)答案見解析.(1)函數(shù)定義域?yàn)镾KIPIF1<0,SKIPIF1<0,SKIPIF1<0時(shí),SKIPIF1<0或SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,SKIPIF1<0時(shí),SKIPIF1<0或SKIPIF1<0,SKIPIF1<0時(shí),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.(2)因?yàn)镾KIPIF1<0,由(1)知,SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,在SKIPIF1<0上單調(diào)遞增,所以SKIPIF1<0的最小值為SKIPIF1<0,又因?yàn)镾KIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,此時(shí)最小值為SKIPIF1<0,最大值為SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,此時(shí)最小值為SKIPIF1<0,最大值為SKIPIF1<0.2.(2022·北京市朝陽(yáng)區(qū)人大附中朝陽(yáng)分校模擬預(yù)測(cè))設(shè)函數(shù)SKIPIF1<0.(1)求曲線SKIPIF1<0在SKIPIF1<0處的切線方程;(2)若函數(shù)SKIPIF1<0有最大值并記為SKIPIF1<0,求SKIPIF1<0的最小值;【答案】(1)SKIPIF1<0(2)SKIPIF1<0取得最小值SKIPIF1<0(1)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以函數(shù)SKIPIF1<0在SKIPIF1<0處的切線方程是SKIPIF1<0;(2)SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞減,函數(shù)沒(méi)有最大值,故舍去;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,得SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,函數(shù)單調(diào)遞增,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,函數(shù)單調(diào)遞減,所以當(dāng)SKIPIF1<0時(shí),函數(shù)取得最大值SKIPIF1<0,SKIPIF1<0,得SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,函數(shù)SKIPIF1<0單調(diào)遞減,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,函數(shù)SKIPIF1<0單調(diào)遞增,所以當(dāng)SKIPIF1<0時(shí),函數(shù)SKIPIF1<0取得最小值,SKIPIF1<0.3.(2022·河南·模擬預(yù)測(cè)(理))已知函數(shù)SKIPIF1<0.(1)當(dāng)SKIPIF1<0時(shí),判斷函數(shù)SKIPIF1<0的單調(diào)性;(2)證明函數(shù)SKIPIF1<0存在最小值SKIPIF1<0,并求出函數(shù)SKIPIF1<0的最大值.【答案】(1)在SKIPIF1<0上單調(diào)遞減,在SKIPIF1<0上單調(diào)遞增(2)證明見解析,SKIPIF1<0(1)由題意知,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.所以函數(shù)SKIPIF1<0單調(diào)遞增.又SKIPIF1<0,所以當(dāng)SKIPIF1<0時(shí)SKIPIF1<0,函數(shù)SKIPIF1<0單調(diào)遞減;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,函數(shù)SKIPIF1<0單調(diào)遞增.所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,在SKIPIF1<0上單調(diào)遞增.(2)由題意知,SKIPIF1<0,SKIPIF1<0.所以函數(shù)SKIPIF1<0單調(diào)遞增.令SKIPIF1<0,則SKIPIF1<0.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,函數(shù)SKIPIF1<0單調(diào)遞增;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,函數(shù)SKIPIF1<0單調(diào)遞減.所以SKIPIF1<0,即SKIPIF1<0.所以SKIPIF1<0,即SKIPIF1<0.另一方面,SKIPIF1<0,所以存在SKIPIF1<0,使得SKIPIF1<0,①即當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0單調(diào)遞減,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0單調(diào)遞增.所以函數(shù)SKIPIF1<0存在最小值SKIPIF1<0.由①式,得SKIPIF1<0.所以SKIPIF1<0(當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0時(shí),等號(hào)成立).所以SKIPIF1<0,即為所求.4.(2022·山東·菏澤一中高二階段練習(xí))已知函數(shù)SKIPIF1<0,SKIPIF1<0.(1)求函數(shù)SKIPIF1<0的單調(diào)區(qū)間;(2)求SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值.【答案】(1)答案見解析(2)答案見解析(1)根據(jù)題意,函數(shù)SKIPIF1<0,其導(dǎo)數(shù)SKIPIF1<0.①當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則SKIPIF1<0在SKIPIF1<0上為增函數(shù);②當(dāng)SKIPIF1<0時(shí),令SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0,則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;③當(dāng)SKIPIF1<0時(shí),令SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0,則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.(2)由(1)可得,當(dāng)SKIPIF1<0或SKIPIF1<0,SKIPIF1<0.①當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí),SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,此時(shí)SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值為SKIPIF1<0;②當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí),SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,在SKIPIF1<0內(nèi)單調(diào)遞增,此時(shí)SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值為SKIPIF1<0;③當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí),SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,此時(shí)SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值為SKIPIF1<0.綜上可得:當(dāng)SKIPIF1<0時(shí),SKIPIF1<0的最小值為SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0的最小值頭SKIPIF1<0;當(dāng)SKIPIF1<0時(shí),SKIPIF1<0的最小值為SKIPIF1<0.5.(2022·河南·模擬預(yù)測(cè)(文))已知函數(shù)SKIPIF1<0.(1)若SKIPIF1<0僅有一個(gè)零點(diǎn),求a的取值范圍;(2)若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值與最小值之差為SKIPIF1<0,求SKIPIF1<0的最小值.【答案】(1)SKIPIF1<0(2)SKIPIF1<0(1)SKIPIF1<0①SKIPIF1<0時(shí),SKIPIF1<0恒成立,SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,易知其有1個(gè)零點(diǎn),滿足題意②SKIPIF1<0時(shí),SKIPIF1<0時(shí)SKIPIF1<0,SKIPIF1<0時(shí)SKIPIF1<0故SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增,在SKIPIF1<0上單調(diào)遞減SKIPIF1<0,SKIPIF1<0由題意SKIPIF1<0僅有1個(gè)零點(diǎn),故SKIPIF1<0,解得SKIPIF1<0綜上,SKIPIF1<0的取值范圍是SKIPIF1<0(2)由(1)可知①SKIPIF1<0時(shí),SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增,SKIPIF1<0②SKIPIF1<0即SKIPIF1<0時(shí),SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減,SKIPIF1<0③SKIPIF1<0即SKIPIF1<0時(shí),SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減,在SKIPIF1<0上單調(diào)遞增SKIPIF1<0,SKIPIF1<0,SKIPIF1<0故SKIPIF1<0綜上,SKIPIF1<0可得SKIPIF1<0高頻考點(diǎn)六:根據(jù)函數(shù)的最值求參數(shù)1.(2022·河南·模擬預(yù)測(cè)(文))已知函數(shù)SKIPIF1<0無(wú)最大值,則實(shí)數(shù)a的取值范圍是(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D令SKIPIF1<0,則SKIPIF1<0,令SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0;令SKIPIF1<0,解得SKIPIF1<0,∴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)遞增,g(-1)=2,g(1)=-2,據(jù)此,作出SKIPIF1<0和y=-2x的圖像,由圖可知,當(dāng)x=a<-1時(shí),函數(shù)f(x)無(wú)最大值.故選:D.2.(2022·陜西安康·高二期末(文))已知SKIPIF1<0,函數(shù)SKIPIF1<0的最小值為SKIPIF1<0,則SKIPIF1<0(

)A.1或2 B.2 C.1或3 D.2或3【答案】A由SKIPIF1<0(SKIPIF1<0),得SKIPIF1<0(SKIPIF1<0,SKIPIF1<0),當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,在SKIPIF1<0上單調(diào)遞增,故SKIPIF1<0,得SKIPIF1<0,解得SKIPIF1<0或2.故選:A3.(2022·河南開封·高二階段練習(xí)(理))已知函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上有最小值,則實(shí)數(shù)a的取值范圍是______.【答案】SKIPIF1<0由題知,SKIPIF1<0,SKIPIF1<0,因?yàn)镾KIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增,若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0有最小值,則SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,所以實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為:SKIPIF1<0.4.(2022·河北·武安市第三中學(xué)高二階段練習(xí))已知函數(shù)SKIPIF1<0.(1)若SKIPIF1<0,求SKIPIF1<0的極值;(2)若SKIPIF1<0在SKIPIF1<0上的最大值為SKIPIF1<0,求實(shí)數(shù)SKIPIF1<0的值.【答案】(1)有極大值e,無(wú)極小值(2)SKIPIF1<0(1)若SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0時(shí),SKIPIF1<0;SKIPIF1<0時(shí),SKIPIF1<0.所以SKIPI

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