新高考數(shù)學(xué)一輪復(fù)習(xí)精講精練9.3 利用導(dǎo)數(shù)求極值最值（基礎(chǔ)版）（解析版）

9.3利用導(dǎo)數(shù)求極值最值（精講）（基礎(chǔ)版）思維導(dǎo)圖思維導(dǎo)圖考點(diǎn)呈現(xiàn)考點(diǎn)呈現(xiàn)例題剖析例題剖析考點(diǎn)一極值【例1-1】（2022·崇左模擬）函數(shù)SKIPIF1<0的極小值是．【答案】2【解析】由題意可得SKIPIF1<0．由SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0；由SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0．故答案為：2【例1-2】（2022·遼陽(yáng)二模）設(shè)函數(shù)SKIPIF1<0，則下列不是函數(shù)SKIPIF1<0極大值點(diǎn)的是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由題可得SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，SKIPIF1<0，則當(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，SKIPIF1<0，SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0，SKIPIF1<0，SKIPIF1<0上單調(diào)遞減，故不是函數(shù)SKIPIF1<0極大值點(diǎn)的是SKIPIF1<0.故答案為：D.【例1-3】（2022·安康模擬）若函數(shù)SKIPIF1<0有兩個(gè)極值點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由SKIPIF1<0，得SKIPIF1<0.因?yàn)楹瘮?shù)SKIPIF1<0有兩個(gè)極值點(diǎn)，所以SKIPIF1<0有兩個(gè)不同的解，即SKIPIF1<0有兩個(gè)不同的解轉(zhuǎn)化為SKIPIF1<0與SKIPIF1<0的圖象有兩個(gè)交點(diǎn)；設(shè)SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，即SKIPIF1<0，解得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)遞增，在SKIPIF1<0上單調(diào)遞減.分別作出函數(shù)SKIPIF1<0與SKIPIF1<0的圖象，如圖所示由圖可知，0SKIPIF1<0，解得SKIPIF1<0.所以實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.故答案為：D.【一隅三反】1（2022高三上·襄陽(yáng)期末）已知函數(shù)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0所有極值點(diǎn)的和為（）A．SKIPIF1<0 B．13π C．17π D．SKIPIF1<0【答案】D【解析】SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0兩側(cè)異號(hào)，所以SKIPIF1<0是函數(shù)SKIPIF1<0的極值點(diǎn)，又SKIPIF1<0，所以極值點(diǎn)SKIPIF1<0，所以SKIPIF1<0所有極值點(diǎn)的和為SKIPIF1<0，故答案為：D.2．（2022·昆明模擬）若SKIPIF1<0是函數(shù)SKIPIF1<0的極值點(diǎn)，則SKIPIF1<0的極大值為（）A．-1 B．SKIPIF1<0 C．SKIPIF1<0 D．1【答案】C【解析】因?yàn)镾KIPIF1<0，故可得SKIPIF1<0，因?yàn)镾KIPIF1<0是函數(shù)SKIPIF1<0的極值點(diǎn)，故可得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，此時(shí)SKIPIF1<0令SKIPIF1<0，解得SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0或SKIPIF1<0；由SKIPIF1<0可得SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞增，在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增，故SKIPIF1<0的極大值點(diǎn)為SKIPIF1<0，則SKIPIF1<0的極大值為SKIPIF1<0。故答案為：C.3（2022·河西模擬）若函數(shù)SKIPIF1<0在SKIPIF1<0處取得極值，則SKIPIF1<0．【答案】1【解析】SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0在SKIPIF1<0處取得極值，所以，SKIPIF1<0，解得SKIPIF1<0，此時(shí)，SKIPIF1<0，故當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0和SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；所以，函數(shù)SKIPIF1<0在SKIPIF1<0處取得極小值，滿足題意，所以，SKIPIF1<0所以SKIPIF1<0故答案為：1考點(diǎn)二最值【例2】（2021·浙江）已知函數(shù)SKIPIF1<0，則SKIPIF1<0的最大值是_____，最小值是______．【答案】SKIPIF1<0；SKIPIF1<0.【解析】SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0令SKIPIF1<0，得SKIPIF1<0；令SKIPIF1<0，得SKIPIF1<0.SKIPIF1<0SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，SKIPIF1<0SKIPIF1<0的最大值是2；最小值是SKIPIF1<0.故答案為：SKIPIF1<0；SKIPIF1<0.【一隅三反】1．（2021·全國(guó)專(zhuān)題練習(xí)）函數(shù)SKIPIF1<0的最大值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】函數(shù)的定義域?yàn)镾KIPIF1<0，則令SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則函數(shù)單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則函數(shù)單調(diào)遞減，則當(dāng)SKIPIF1<0時(shí)，函數(shù)有最大值，為SKIPIF1<0，故選:D.2（2021·江蘇）已知函數(shù)SKIPIF1<0，則SKIPIF1<0的最小值是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由題得SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減.所以SKIPIF1<0取得最小值時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；所以SKIPIF1<0的最小值是SKIPIF1<0.故選：C3．（2021·甘肅蘭州市）函數(shù)SKIPIF1<0的最大值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由SKIPIF1<0，得SKIPIF1<0，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上遞減，在SKIPIF1<0上遞增，因?yàn)镾KIPIF1<0，所以函數(shù)SKIPIF1<0的最大值為SKIPIF1<0，故選：B考點(diǎn)三已知極值最值求參數(shù)【例3-1】（2022·新疆三模）若函數(shù)SKIPIF1<0在SKIPIF1<0處有極值10，則SKIPIF1<0（）A．6 B．-15 C．-6或15 D．6或-15【答案】B【解析】SKIPIF1<0SKIPIF1<0，SKIPIF1<0又SKIPIF1<0時(shí)SKIPIF1<0有極值10SKIPIF1<0SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0此時(shí)SKIPIF1<0在SKIPIF1<0處無(wú)極值，不符合題意經(jīng)檢驗(yàn)，SKIPIF1<0時(shí)滿足題意SKIPIF1<0故答案為：B【例3-2】（2022·涼山模擬）函數(shù)SKIPIF1<0，若SKIPIF1<0在SKIPIF1<0上有最小值，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】A【解析】由題意，函數(shù)SKIPIF1<0，可得SKIPIF1<0，若SKIPIF1<0時(shí)，當(dāng)SKIPIF1<0時(shí)，可得SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，此時(shí)函數(shù)SKIPIF1<0在SKIPIF1<0沒(méi)有最小值，不符合題意；當(dāng)SKIPIF1<0時(shí)，令SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0與SKIPIF1<0的交點(diǎn)，畫(huà)出函數(shù)SKIPIF1<0與SKIPIF1<0的圖象，如圖所示，結(jié)合圖象，可得存在SKIPIF1<0，使得SKIPIF1<0，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，此時(shí)函數(shù)SKIPIF1<0在SKIPIF1<0上有最小值，符合題意，綜上可得，實(shí)數(shù)a的取值范圍是SKIPIF1<0.故答案為：A.【例3-3】（2022高三上·開(kāi)封開(kāi)學(xué)考）已知函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上值的集合是SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上值的集合為SKIPIF1<0，因函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，于是得SKIPIF1<0，則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：D【一隅三反】1（2021高三上·江西月考）設(shè)函數(shù)f(x)=x3?3x，x≤a，?2x，x>a，若SKIPIF1<0無(wú)最大值，則實(shí)數(shù)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因?yàn)閒(x)=作出函數(shù)SKIPIF1<0與直線SKIPIF1<0的圖象，它們的交點(diǎn)是SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0，則令SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0是SKIPIF1<0的極大值點(diǎn)，SKIPIF1<0是SKIPIF1<0的極小值點(diǎn)，由圖象可知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有最大值SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，有SKIPIF1<0，此時(shí)SKIPIF1<0無(wú)最大值，故實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.故答案為：A.2．（2022金臺(tái)月考）已知函數(shù)SKIPIF1<0有兩個(gè)極值點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由題意SKIPIF1<0有兩個(gè)不等實(shí)根，即SKIPIF1<0有兩個(gè)不等實(shí)根，設(shè)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0遞減，SKIPIF1<0時(shí)，SKIPIF1<0為極大值也是最大值，SKIPIF1<0時(shí)，SKIPIF1<0，且SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，直線SKIPIF1<0與SKIPIF1<0的圖象有兩個(gè)交點(diǎn)，即SKIPIF1<0有兩個(gè)不等實(shí)根．故答案為：B3（2022濰坊期中）若函數(shù)SKIPIF1<0在SKIPIF1<0上無(wú)極值，則實(shí)數(shù)SKIPIF1<0的取值范圍（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由SKIPIF1<0可得SKIPIF1<0，SKIPIF1<0恒成立，SKIPIF1<0為開(kāi)口向上的拋物線，若函數(shù)SKIPIF1<0在SKIPIF1<0上無(wú)極值，則SKIPIF1<0恒成立，所以SKIPIF1<0，解得：SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0，故答案為：D.4．（2021·全國(guó)高三專(zhuān)題練習(xí)）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在最小值，則SKIPIF1<0的取值范圍是A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】SKIPIF1<0，SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0；令SKIPIF1<0，解得SKIPIF1<0.故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，函數(shù)SKIPIF1<0在SKIPIF1<0處取得極小值，由于函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)取到最小值，則SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，可得SKIPIF1<0,即SKIPIF1<0，解得SKIPIF1<0.因此，實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：C.9.3利用導(dǎo)數(shù)求極值最值（精練）（基礎(chǔ)版）題組一題組一極值1．（2022太原期中）若SKIPIF1<0是函數(shù)SKIPIF1<0的極值點(diǎn)，則函數(shù)（）A．有最小值SKIPIF1<0，無(wú)最大值B．有最大值SKIPIF1<0，無(wú)最小值C．有最小值SKIPIF1<0，最大值SKIPIF1<0D．無(wú)最大值，無(wú)最小值【答案】A【解析】由題設(shè)SKIPIF1<0且SKIPIF1<0，∴SKIPIF1<0，可得SKIPIF1<0.∴SKIPIF1<0且SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，SKIPIF1<0遞減；當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，SKIPIF1<0遞增；∴SKIPIF1<0有極小值SKIPIF1<0，無(wú)極大值.綜上，有最小值SKIPIF1<0，無(wú)最大值。故答案為：A2．（2022湖北期中）已知函數(shù)SKIPIF1<0(SKIPIF1<0且SKIPIF1<0，SKIPIF1<0)的一個(gè)極值點(diǎn)為2，則SKIPIF1<0的最小值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．7【答案】B【解析】對(duì)SKIPIF1<0求導(dǎo)得：SKIPIF1<0，因函數(shù)SKIPIF1<0的一個(gè)極值點(diǎn)為2，則SKIPIF1<0，此時(shí)，SKIPIF1<0，SKIPIF1<0，因SKIPIF1<0，即SKIPIF1<0，因此，在2左右兩側(cè)鄰近的區(qū)域SKIPIF1<0值一正一負(fù)，2是函數(shù)SKIPIF1<0的一個(gè)極值點(diǎn)，則有SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，于是得SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取“=”，所以SKIPIF1<0的最小值為SKIPIF1<0.故答案為：B3．（2021高三上·三門(mén)峽期中）“SKIPIF1<0”是“函數(shù)SKIPIF1<0在SKIPIF1<0上有極值”的（）A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分也不必要條件【答案】B【解析】SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，所以，函數(shù)SKIPIF1<0在SKIPIF1<0處取得極小值，若函數(shù)SKIPIF1<0在SKIPIF1<0上有極值，則SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0SKIPIF1<0，但是由SKIPIF1<0推不出SKIPIF1<0，因此SKIPIF1<0是函數(shù)SKIPIF1<0在SKIPIF1<0上有極值的必要不充分條件．故答案為：B．40．（2022·鎮(zhèn)江）已知等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，公差SKIPIF1<0，SKIPIF1<0和SKIPIF1<0是函數(shù)SKIPIF1<0的極值點(diǎn)，則SKIPIF1<0（）A．－38 B．38 C．－17 D．17【答案】A【解析】由題意，函數(shù)SKIPIF1<0，其中SKIPIF1<0，可得SKIPIF1<0令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，又SKIPIF1<0和SKIPIF1<0是函數(shù)SKIPIF1<0的極值點(diǎn)，且公差SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0.故答案為：A.題組二最值題組二最值1．（2022·淮北模擬）函數(shù)SKIPIF1<0的最大值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．3【答案】B【解析】因?yàn)镾KIPIF1<0所以SKIPIF1<0令SKIPIF1<0則SKIPIF1<0則SKIPIF1<0令SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；SKIPIF1<0時(shí)SKIPIF1<0所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0取得最大值，此時(shí)SKIPIF1<0所以SKIPIF1<0故答案為：B2．（2022高三上·安徽開(kāi)學(xué)考）函數(shù)SKIPIF1<0的值域是．【答案】[2，+∞)【解析】SKIPIF1<0，令SKIPIF1<0，易得當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，且SKIPIF1<0為增函數(shù).記SKIPIF1<0，則SKIPIF1<0，易知當(dāng)SKIPIF1<0時(shí)．SKIPIF1<0為減函數(shù)；當(dāng)SKIPIF1<0時(shí)．SKIPIF1<0為增函數(shù)．∴SKIPIF1<0，∴SKIPIF1<0的值域?yàn)閇2，+∞)．故答案為：[2，+∞)3．（2021·全國(guó)高考真題）函數(shù)SKIPIF1<0的最小值為_(kāi)_____.【答案】1【解析】由題設(shè)知：SKIPIF1<0定義域?yàn)镾KIPIF1<0，∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，有SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，有SKIPIF1<0，此時(shí)SKIPIF1<0單調(diào)遞增；又SKIPIF1<0在各分段的界點(diǎn)處連續(xù)，∴綜上有：SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增；∴SKIPIF1<0故答案為：1.4．（2021·江西高三二模）已知函數(shù)SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上的最大值是__________．【答案】SKIPIF1<0【解析】由題意可知，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0．故答案為：SKIPIF1<05（2021·湖南）函數(shù)SKIPIF1<0的最小值為_(kāi)________.【答案】SKIPIF1<0【解析】SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，SKIPIF1<0的最小值為SKIPIF1<06．（2022·西藏）設(shè)函數(shù)SKIPIF1<0，直線SKIPIF1<0是曲線SKIPIF1<0的切線，則SKIPIF1<0的最大值是【答案】SKIPIF1<0【解析】由題得SKIPIF1<0.設(shè)切點(diǎn)SKIPIF1<0，則SKIPIF1<0;則切線方程為SKIPIF1<0即SKIPIF1<0又因?yàn)镾KIPIF1<0是曲線SKIPIF1<0的切線所以SKIPIF1<0則SKIPIF1<0.令SKIPIF1<0.則SKIPIF1<0.則有SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上遞減;SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上遞增﹐所以SKIPIF1<0時(shí)，SKIPIF1<0取最大值SKIPIF1<0即SKIPIF1<0的最大值為SKIPIF1<0.7．（2021·全國(guó)高三專(zhuān)題練習(xí)（理））已知函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則實(shí)數(shù)SKIPIF1<0的最小值是【答案】SKIPIF1<0【解析】由SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，得SKIPIF1<0SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0是單調(diào)遞減函數(shù)，SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0的最小值為SKIPIF1<0.8．（2021·天津）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在最大值，則實(shí)數(shù)SKIPIF1<0的取值范圍為.【答案】SKIPIF1<0【解析】因?yàn)镾KIPIF1<0，且函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上存在最大值，故只需SKIPIF1<0滿足SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0．題組三題組三已知極值最值求參數(shù)1．（2022·莆田三模）已知函數(shù)SKIPIF1<0的最小值是4．則SKIPIF1<0（）A．3 B．4 C．5 D．6【答案】A【解析】由題，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0單調(diào)遞增，又SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0為SKIPIF1<0最小值點(diǎn)，即SKIPIF1<0，解得SKIPIF1<0，故答案為：A2．（2021高三上·湖北期中）若函數(shù)SKIPIF1<0（SKIPIF1<0為常數(shù)）有兩個(gè)不同的極值點(diǎn)，則實(shí)數(shù)SKIPIF1<0取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】SKIPIF1<0，函數(shù)SKIPIF1<0（SKIPIF1<0為常數(shù)）有兩個(gè)不同的極值點(diǎn)，等價(jià)于函數(shù)SKIPIF1<0與SKIPIF1<0的圖象有兩個(gè)不同的交點(diǎn)，SKIPIF1<0，因?yàn)镾KIPIF1<0為增函數(shù)，且SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為減函數(shù)，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為增函數(shù)，所以SKIPIF1<0，故SKIPIF1<0.故答案為：C3．（2022湖南）已知f(x)＝SKIPIF1<0x3＋(a－1)x2＋x＋1沒(méi)有極值，則實(shí)數(shù)a的取值范圍是（）A．[0，1] B．(－∞，0]∪[1，＋∞)C．[0，2] D．(－∞，0]∪[2，＋∞)【答案】C【解析】由SKIPIF1<0得SKIPIF1<0，根據(jù)題意得SKIPIF1<0，解得SKIPIF1<0。故答案為：C4．（2022遼寧月考）已知函數(shù)SKIPIF1<0在SKIPIF1<0上恰有兩個(gè)極值點(diǎn)，則SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0，根據(jù)題意得SKIPIF1<0在SKIPIF1<0有2個(gè)變號(hào)零點(diǎn)，當(dāng)SKIPIF1<0時(shí)，顯然不合題意，當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0等價(jià)于SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，因?yàn)镾KIPIF1<0，解得SKIPIF1<0，可得SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0單調(diào)遞增，又因?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0，要使SKIPIF1<0與SKIPIF1<0的圖像有2個(gè)不同的交點(diǎn)，需要滿足SKIPIF1<0，解得SKIPIF1<0。故答案為：D．5．（2022河南月考）已知函數(shù)SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）存在極大值和極小值，且極大值與極小值互為相反數(shù)，則（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】SKIPIF1<0設(shè)SKIPIF1<0是方程SKIPIF1<0的兩個(gè)實(shí)數(shù)根，根據(jù)題意可知SKIPIF1<0，不妨設(shè)SKIPIF1<0則SKIPIF1<0，且SKIPIF1<0，即SKIPIF1<0化簡(jiǎn)得：SKIPIF1<0將SKIPIF1<0代入化簡(jiǎn)計(jì)算得SKIPIF1<0，SKIPIF1<0，B符合題意，ACD不符合題意故答案為：B.6．（2021高三上·邢臺(tái)月考）若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0有最小值，則實(shí)數(shù)SKIPIF1<0的取值范圍為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0，①當(dāng)SKIPIF1<0時(shí)，可得函數(shù)SKIPIF1<0的增區(qū)間為SKIPIF1<0，SKIPIF1<0，減區(qū)間為SKIPIF1<0，若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0有最小值，必有SKIPIF1<0，有SKIPIF1<0，由SKIPIF1<0，有SKIPIF1<0，SKIPIF1<0，不合題意；②當(dāng)SKIPIF1<0時(shí)，此時(shí)函數(shù)SKIPIF1<0的增區(qū)間為SKIPIF1<0，SKIPIF1<0，減區(qū)間