新高考數(shù)學(xué)一輪復(fù)習(xí)第3章 第11講 拓展四 導(dǎo)數(shù)中的隱零點問題 精講+精練（教師版）

第11講拓展四：導(dǎo)數(shù)中的隱零點問題(精講+精練）目錄第一部分：知識點精準(zhǔn)記憶第二部分：典型例題剖析第三部分：第11講拓展四：導(dǎo)數(shù)中的隱零點問題（精練）第一部分：知識點精準(zhǔn)記憶第一部分：知識點精準(zhǔn)記憶1、不含參函數(shù)的隱零點問題已知不含參函數(shù)SKIPIF1<0，導(dǎo)函數(shù)方程SKIPIF1<0的根存在，卻無法求出，設(shè)方程SKIPIF1<0的根為SKIPIF1<0，則有：①關(guān)系式SKIPIF1<0成立；②注意確定SKIPIF1<0的合適范圍.2、含參函數(shù)的隱零點問題已知含參函數(shù)SKIPIF1<0，其中SKIPIF1<0為參數(shù)，導(dǎo)函數(shù)方程SKIPIF1<0的根存在，卻無法求出，設(shè)方程SKIPIF1<0的根為SKIPIF1<0，則有①有關(guān)系式SKIPIF1<0成立，該關(guān)系式給出了SKIPIF1<0的關(guān)系；②注意確定SKIPIF1<0的合適范圍，往往和SKIPIF1<0的范圍有關(guān).3、函數(shù)零點的存在性（1）函數(shù)零點存在性定理：設(shè)函數(shù)SKIPIF1<0在閉區(qū)間SKIPIF1<0上連續(xù)，且SKIPIF1<0，那么在開區(qū)間SKIPIF1<0內(nèi)至少有函數(shù)SKIPIF1<0的一個零點，即至少有一點SKIPIF1<0，使得SKIPIF1<0.①若SKIPIF1<0，則SKIPIF1<0的零點不一定只有一個，可以有多個②若SKIPIF1<0，那么SKIPIF1<0在SKIPIF1<0不一定有零點③若SKIPIF1<0在SKIPIF1<0有零點，則SKIPIF1<0不一定必須異號(3)若SKIPIF1<0在SKIPIF1<0上是單調(diào)函數(shù)且連續(xù)，則SKIPIF1<0在SKIPIF1<0的零點唯一.第二部分：典型例題剖析第二部分：典型例題剖析1．（2022·全國·模擬預(yù)測（文））已知函數(shù)SKIPIF1<0．(1)若曲線SKIPIF1<0在SKIPIF1<0處的切線經(jīng)過點SKIPIF1<0，求實數(shù)a的值；(2)若對任意SKIPIF1<0，都有SKIPIF1<0（e為自然對數(shù)的底），求證：SKIPIF1<0．【答案】(1)SKIPIF1<0(2)證明見解析【解析】(1)SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，

所以曲線SKIPIF1<0在點SKIPIF1<0處的切線方程為SKIPIF1<0，因為切線經(jīng)過點SKIPIF1<0，所以SKIPIF1<0解得SKIPIF1<0．(2)設(shè)SKIPIF1<0，則SKIPIF1<0，

設(shè)SKIPIF1<0，則SKIPIF1<0，因為SKIPIF1<0在SKIPIF1<0上遞增，所以當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0所以SKIPIF1<0在SKIPIF1<0上遞減，在SKIPIF1<0上遞增，所以SKIPIF1<0，

令SKIPIF1<0，則SKIPIF1<0所以SKIPIF1<0在SKIPIF1<0遞減，因為SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0．【點睛】關(guān)鍵點點睛：此題考查導(dǎo)數(shù)的綜合應(yīng)用，考查導(dǎo)數(shù)的幾何意義，考查利用導(dǎo)數(shù)證明不等式，解題的關(guān)鍵是構(gòu)造函數(shù)SKIPIF1<0，利用導(dǎo)數(shù)求得SKIPIF1<0，再利用函數(shù)SKIPIF1<0的單調(diào)性結(jié)合SKIPIF1<0可證得結(jié)論，考查數(shù)學(xué)轉(zhuǎn)化思想，屬于較難題2．（2022·甘肅·一模（文））已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)判斷函數(shù)SKIPIF1<0的單調(diào)性；(2)當(dāng)SKIPIF1<0時，關(guān)于x的不等式SKIPIF1<0恒成立，求實數(shù)b的取值范圍．【答案】(1)答案見解析；(2)SKIPIF1<0.【解析】(1)SKIPIF1<0的定義域為SKIPIF1<0，求導(dǎo)得：SKIPIF1<0，若SKIPIF1<0時，則SKIPIF1<0，此時SKIPIF1<0在SKIPIF1<0單調(diào)遞增；若SKIPIF1<0時，則當(dāng)SKIPIF1<0時SKIPIF1<0，SKIPIF1<0在SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時，SKIPIF1<0，f(x)在SKIPIF1<0單調(diào)遞增.(2)當(dāng)SKIPIF1<0時，SKIPIF1<0，由題意SKIPIF1<0在SKIPIF1<0上恒成立，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上遞增，又SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上有唯一零點SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0即SKIPIF1<0，SKIPIF1<0單調(diào)遞減；SKIPIF1<0時，SKIPIF1<0即SKIPIF1<0，SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0為SKIPIF1<0在定義域內(nèi)的最小值.即SKIPIF1<0令SKIPIF1<0，則方程SKIPIF1<0等價于SKIPIF1<0，又易知SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，即SKIPIF1<0所以，SKIPIF1<0的最小值SKIPIF1<0所以SKIPIF1<0，即實數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0【點睛】利用導(dǎo)數(shù)研究不等式恒成立或存在型問題，首先要構(gòu)造函數(shù)，利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性，求出最值，進而得出相應(yīng)的含參不等式，從而求出參數(shù)的取值范圍；也可分離變量，構(gòu)造函數(shù)，直接把問題轉(zhuǎn)化為函數(shù)的最值問題.3．（2022·重慶巴蜀中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0（SKIPIF1<0為自然對數(shù)的底數(shù)）.(1)求SKIPIF1<0的極值；(2)（i）證明∶SKIPIF1<0與SKIPIF1<0有相同的零點；（ii）若SKIPIF1<0恒成立，求整數(shù)a的最大值.【答案】(1)SKIPIF1<0的極小值為SKIPIF1<0，無極大值；(2)（i）證明見解析，（ii）SKIPIF1<0【解析】(1)由題意可知，SKIPIF1<0，令SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)的遞增；當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)的遞減；當(dāng)SKIPIF1<0時，SKIPIF1<0取得極小值為SKIPIF1<0，無極大值；(2)（i）由SKIPIF1<0知SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；由SKIPIF1<0知SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；又SKIPIF1<0，故必存在唯一SKIPIF1<0使得SKIPIF1<0，即有SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0與SKIPIF1<0有相同的唯一零點SKIPIF1<0；（ii）由SKIPIF1<0，得SKIPIF1<0恒成立，SKIPIF1<0在SKIPIF1<0恒成立，令SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，由（i）知SKIPIF1<0單調(diào)遞增且SKIPIF1<0存在唯一零點SKIPIF1<0；則當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞增；故SKIPIF1<0；由（i）知SKIPIF1<0；又SKIPIF1<0，故進一步確定SKIPIF1<0；故SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，又SKIPIF1<0；所以整數(shù)a的最大值為SKIPIF1<0.【點睛】求解不等式問題的關(guān)鍵：適當(dāng)變形，靈活轉(zhuǎn)化，結(jié)合題設(shè)條件，有時需要對不等式進行“除法”變形，從而分離參數(shù)，有時需要進行移項變形，可使不等式兩邊具有相同的結(jié)構(gòu)特點；構(gòu)造函數(shù)，利用導(dǎo)數(shù)求解，若分離參數(shù)，則直接構(gòu)造函數(shù)，并借助導(dǎo)數(shù)加以求解，若轉(zhuǎn)化為不等式兩邊具有相同的結(jié)構(gòu)特點，則可根據(jù)該結(jié)構(gòu)特點構(gòu)造函數(shù)，并借助導(dǎo)數(shù)加以求解.4．（2022·四川南充·二模（理））已知SKIPIF1<0.(1)求SKIPIF1<0在SKIPIF1<0的切線方程；(2)求證：SKIPIF1<0僅有一個極值；(3)若存在SKIPIF1<0，使SKIPIF1<0對任意SKIPIF1<0恒成立，求實數(shù)SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0(2)證明見解析(3)SKIPIF1<0【解析】(1)當(dāng)SKIPIF1<0時，SKIPIF1<0得SKIPIF1<0，又SKIPIF1<0.所以SKIPIF1<0在SKIPIF1<0的切線方程為：SKIPIF1<0.即SKIPIF1<0；(2)SKIPIF1<0令SKIPIF1<0，由于SKIPIF1<0，得SKIPIF1<0.所以SKIPIF1<0在SKIPIF1<0單調(diào)遞減.SKIPIF1<0所以存在唯一SKIPIF1<0，使得SKIPIF1<0.所以SKIPIF1<0于SKIPIF1<0單調(diào)遞增，SKIPIF1<0單調(diào)遞減.SKIPIF1<0，無極小值.所以SKIPIF1<0僅有一個極值.(3)任意SKIPIF1<0，則SKIPIF1<0.由（2）知SKIPIF1<0.又SKIPIF1<0，則SKIPIF1<0.若存在SKIPIF1<0，使SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0，轉(zhuǎn)化為SKIPIF1<0.SKIPIF1<0SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0或SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0；所以SKIPIF1<0于SKIPIF1<0單調(diào)遞增，SKIPIF1<0單調(diào)遞減，SKIPIF1<0單調(diào)遞增當(dāng)SKIPIF1<0時，SKIPIF1<0由于SKIPIF1<0，則SKIPIF1<0，又SKIPIF1<0.SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0綜上：當(dāng)SKIPIF1<0時，SKIPIF1<0得SKIPIF1<0.故SKIPIF1<0.【點睛】不等式恒成立問題、存在性問題的求解，有相同點：即分離常數(shù)法.也有不同點，如SKIPIF1<0恒成立問題，轉(zhuǎn)化為SKIPIF1<0；SKIPIF1<0能成立問題，則轉(zhuǎn)化為SKIPIF1<0.5．（2022·河南洛陽·模擬預(yù)測（理））已知函數(shù)SKIPIF1<0．(1)討論函數(shù)SKIPIF1<0的單調(diào)性；(2)當(dāng)SKIPIF1<0時，求證：SKIPIF1<0．【答案】(1)當(dāng)SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上為單調(diào)遞增；當(dāng)SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上為單調(diào)遞減，在SKIPIF1<0上為單調(diào)遞增.(2)證明見見解析.【解析】(1)由已知條件得函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，SKIPIF1<0，因為SKIPIF1<0①當(dāng)SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上恒成立，故SKIPIF1<0在SKIPIF1<0上為單調(diào)遞增.②當(dāng)SKIPIF1<0時，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0故SKIPIF1<0在SKIPIF1<0上為單調(diào)遞減，在SKIPIF1<0上為單調(diào)遞增；綜上所述：當(dāng)SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上為單調(diào)遞增當(dāng)SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上為單調(diào)遞減，在SKIPIF1<0上為單調(diào)遞增(2)當(dāng)SKIPIF1<0時，SKIPIF1<0要證原式成立，需證SKIPIF1<0成立，即需證SKIPIF1<0成立，令SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，SKIPIF1<0，SKIPIF1<0，由零點存在性定理可知，存在SKIPIF1<0使SKIPIF1<0，則在SKIPIF1<0上SKIPIF1<0，在SKIPIF1<0上SKIPIF1<0，即在SKIPIF1<0上SKIPIF1<0，在SKIPIF1<0上SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0單調(diào)遞增，在SKIPIF1<0處取得最小值，由SKIPIF1<0可得SKIPIF1<0，即SKIPIF1<0，兩邊同取對數(shù)SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0的最小值為SKIPIF1<0，即SKIPIF1<0成立，故當(dāng)SKIPIF1<0時，SKIPIF1<0成立.【點睛】關(guān)鍵點睛：本題考查利用導(dǎo)數(shù)討論函數(shù)的單調(diào)性，利用導(dǎo)數(shù)證明不等式.解答本題的關(guān)鍵是構(gòu)造函數(shù)SKIPIF1<0，分析其單調(diào)性，得出其最小值，從而得出函數(shù)在在SKIPIF1<0處取得最小值，而SKIPIF1<0滿足SKIPIF1<0，兩邊同取對數(shù)得SKIPIF1<0，從而得出最小值為0，從而得證.屬于難題.6．（2022·全國·模擬預(yù)測（理））已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時，若SKIPIF1<0滿足SKIPIF1<0，討論函數(shù)SKIPIF1<0的單調(diào)性；(2)當(dāng)SKIPIF1<0時，若SKIPIF1<0恒成立，試比較a和1.5625的大小．參考數(shù)據(jù)：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．【答案】(1)SKIPIF1<0時SKIPIF1<0單調(diào)遞減；SKIPIF1<0時，SKIPIF1<0單調(diào)遞增(2)SKIPIF1<0【解析】(1)SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0與SKIPIF1<0均單調(diào)遞增，從而SKIPIF1<0是SKIPIF1<0上的增函數(shù)，又SKIPIF1<0滿足SKIPIF1<0，所以SKIPIF1<0是SKIPIF1<0在SKIPIF1<0上的唯一零點，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞增．(2)SKIPIF1<0，當(dāng)SKIPIF1<0時，原不等式可轉(zhuǎn)化為SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，∴SKIPIF1<0，又SKIPIF1<0，由于SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上存在唯一零點SKIPIF1<0，故當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞增．故SKIPIF1<0，又SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，由于SKIPIF1<0，即SKIPIF1<0．【點睛】利用導(dǎo)數(shù)研究函數(shù)的單調(diào)區(qū)間、極值、最值，當(dāng)求導(dǎo)一次無法解決問題時，可考慮利用二次求導(dǎo)來進行求解.7．（2022·湖北·石首市第一中學(xué)高二階段練習(xí)）已知函數(shù)SKIPIF1<0的圖象在點SKIPIF1<0處的切線方程為SKIPIF1<0．(1)判斷函數(shù)SKIPIF1<0的單調(diào)性．(2)證明：當(dāng)SKIPIF1<0時，SKIPIF1<0．【答案】(1)答案見解析(2)證明見解析【解析】(1)解：因為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0．函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0；令SKIPIF1<0，得SKIPIF1<0．所以函數(shù)SKIPIF1<0的增區(qū)間為SKIPIF1<0，減區(qū)間為SKIPIF1<0.(2)證明：要證SKIPIF1<0，即證SKIPIF1<0，只需證SKIPIF1<0．令SKIPIF1<0，其中SKIPIF1<0，則SKIPIF1<0．令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．因為SKIPIF1<0，SKIPIF1<0，所以存在SKIPIF1<0，使SKIPIF1<0，可得SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時，SKIPIF1<0，即SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．所以SKIPIF1<0．所以SKIPIF1<0，所以SKIPIF1<0．【點睛】方法點睛：利用導(dǎo)數(shù)證明不等式問題，方法如下：（1）直接構(gòu)造函數(shù)法：證明不等式SKIPIF1<0（或SKIPIF1<0）轉(zhuǎn)化為證明SKIPIF1<0（或SKIPIF1<0），進而構(gòu)造輔助函數(shù)SKIPIF1<0；（2）適當(dāng)放縮構(gòu)造法：一是根據(jù)已知條件適當(dāng)放縮；二是利用常見放縮結(jié)論；（3）構(gòu)造“形似”函數(shù)，稍作變形再構(gòu)造，對原不等式同解變形，根據(jù)相似結(jié)構(gòu)構(gòu)造輔助函數(shù).8．（2022·河南·平頂山市教育局教育教學(xué)研究室高二開學(xué)考試（文））已知函數(shù)SKIPIF1<0．(1)討論SKIPIF1<0的單調(diào)性．(2)當(dāng)SKIPIF1<0時，證明：SKIPIF1<0對SKIPIF1<0恒成立．【答案】(1)SKIPIF1<0單調(diào)區(qū)間、單調(diào)性見解析；(2)證明見解析.【解析】(1)因為SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0時，由SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時取“=”，則SKIPIF1<0在R上單調(diào)遞減，當(dāng)SKIPIF1<0時，由SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增.(2)當(dāng)SKIPIF1<0，SKIPIF1<0時，SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，顯然SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，SKIPIF1<0，即存在SKIPIF1<0，使得SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，于是得SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，即SKIPIF1<0，而SKIPIF1<0，即SKIPIF1<0，因此，SKIPIF1<0，而SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0對SKIPIF1<0恒成立.【點睛】思路點睛：涉及雙變量的不等式證明問題，將所證不等式等價轉(zhuǎn)化，構(gòu)造新函數(shù)，再借助導(dǎo)數(shù)探討函數(shù)的單調(diào)性、極(最)值問題處理.9．（2022·海南·嘉積中學(xué)高三階段練習(xí)）已知函數(shù)SKIPIF1<0(SKIPIF1<0為自然對數(shù)的底數(shù)，SKIPIF1<0)．(1)求SKIPIF1<0的單調(diào)區(qū)間和極值；(2)設(shè)SKIPIF1<0，若對任意的SKIPIF1<0，都有SKIPIF1<0恒成立，求SKIPIF1<0的取值范圍.【答案】(1)答案見解析；(2)SKIPIF1<0﹒【解析】(1)SKIPIF1<0，x＞0，當(dāng)SKIPIF1<0時，－a≥0，SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，無極值；當(dāng)SKIPIF1<0時，令SKIPIF1<0，得SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0；∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0單調(diào)遞增，極小值為SKIPIF1<0，無極大值.綜上，當(dāng)SKIPIF1<0時，SKIPIF1<0的增區(qū)間為SKIPIF1<0，無減區(qū)間，無極值；當(dāng)SKIPIF1<0時，SKIPIF1<0的減區(qū)間為SKIPIF1<0，增區(qū)間為SKIPIF1<0，極小值為SKIPIF1<0，無極大值.(2)∵對任意的SKIPIF1<0，不等式SKIPIF1<0恒成立，即SKIPIF1<0在SKIPIF1<0上恒成立，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上為增函數(shù)，又∵SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，使得SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，可得SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞減；當(dāng)SKIPIF1<0時，SKIPIF1<0，可得SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，∴SKIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，又由SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，∴SKIPIF1<0，可得SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，綜上所述，滿足條件SKIPIF1<0的取值范圍是SKIPIF1<0.【點睛】本題關(guān)鍵是參變分離不等式SKIPIF1<0，將問題轉(zhuǎn)化為求SKIPIF1<0在SKIPIF1<0時的最小值，轉(zhuǎn)化為通過導(dǎo)數(shù)SKIPIF1<0研究F(x)的單調(diào)性和最小值.在求解過程中，需要對導(dǎo)數(shù)二次求導(dǎo)，從而判斷導(dǎo)數(shù)的零點，該零點為隱零點，故需采用隱零點的討論方法求解.在處理方程SKIPIF1<0時，還需要采用同構(gòu)思想構(gòu)造函數(shù)SKIPIF1<0，達到簡化的目的.10．（2022·全國·哈師大附中模擬預(yù)測（理））已知函數(shù)SKIPIF1<0（SKIPIF1<0，e為自然對數(shù)的底數(shù)）．(1)若SKIPIF1<0在x=0處的切線與直線y=ax垂直，求a的值；(2)討論函數(shù)SKIPIF1<0的單調(diào)性；(3)當(dāng)SKIPIF1<0時，求證：SKIPIF1<0．【答案】(1)SKIPIF1<0(2)答案見解析(3)證明見解析【解析】(1)SKIPIF1<0，則SKIPIF1<0，由已知SKIPIF1<0，解得SKIPIF1<0(2)SKIPIF1<0（?。┊?dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減；（ⅱ）當(dāng)SKIPIF1<0時，令SKIPIF1<0，得SKIPIF1<0，①SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，SKIPIF1<0，則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；②SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；③SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，SKIPIF1<0，則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．綜上，SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減；SKIPIF1<0時，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；SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增；SKIPIF1<0時，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．(3)方法一：SKIPIF1<0等價于SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0令SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增

∵SKIPIF1<0，∴存在SKIPIF1<0，使得SKIPIF1<0，即SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增∴SKIPIF1<0∴SKIPIF1<0，故SKIPIF1<0方法二：當(dāng)SKIPIF1<0時，SKIPIF1<0SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0∴SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，SKIPIF1<0上單調(diào)遞增．∴SKIPIF1<0，即SKIPIF1<0∴SKIPIF1<0，【關(guān)鍵點點睛】解決本題的關(guān)鍵：一是導(dǎo)數(shù)幾何意義的運用，二是通過導(dǎo)函數(shù)等于零，比較方程的根對問題分類討論，三是隱零點的運用及放縮法的運用.11．（2022·河北衡水·高三階段練習(xí)）已知函數(shù)SKIPIF1<0.(1)若SKIPIF1<0，求SKIPIF1<0在SKIPIF1<0上的最大值與最小值之差；(2)若SKIPIF1<0，證明：SKIPIF1<0【答案】(1)SKIPIF1<0(2)證明見解析【解析】(1)由題意得SKIPIF1<0，則SKIPIF1<0，由于SKIPIF1<0都是遞增函數(shù)，故SKIPIF1<0是遞減函數(shù)，則SKIPIF1<0，故SKIPIF1<0為遞減函數(shù)，則SKIPIF1<0，故SKIPIF1<0；(2)證明：由SKIPIF1<0，SKIPIF1<0，可得SKIPIF1<0，設(shè)SKIPIF1<0，令SKIPIF1<0，故SKIPIF1<0單調(diào)遞增，又SKIPIF1<0，故存在SKIPIF1<0，使得SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0單調(diào)遞增，故SKIPIF1<0，由于SKIPIF1<0，則SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0.12．（2022·江西宜春·模擬預(yù)測（理））已知函數(shù)SKIPIF1<0．(1)討論函數(shù)SKIPIF1<0的單調(diào)性；(2)若SKIPIF1<0有兩個不等實根SKIPIF1<0，證明：SKIPIF1<0．【答案】(1)單調(diào)遞減區(qū)間為SKIPIF1<0，單調(diào)遞增區(qū)間為SKIPIF1<0(2)證明見解析【解析】(1)解：因為SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0時，SKIPIF1<0單調(diào)遞減；SKIPIF1<0時，SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0單調(diào)遞減區(qū)間為SKIPIF1<0，單調(diào)遞增區(qū)間為SKIPIF1<0；(2)由題意可知SKIPIF1<0，不妨設(shè)SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，于是原式SKIPIF1<0等價于SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0，下面先證明，當(dāng)SKIPIF1<0時，SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，又∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，在SKIPIF1<0時恒成立，∴欲證SKIPIF1<0，只需證明SKIPIF1<0即可．∴SKIPIF1<0，變形得，SKIPIF1<0，當(dāng)SKIPIF1<0時，顯然成立，所以SKIPIF1<0得證．【點睛】關(guān)鍵點點睛：本題第二問關(guān)鍵是由SKIPIF1<0兩邊取對數(shù)，得到SKIPIF1<0，再結(jié)合SKIPIF1<0，不妨設(shè)SKIPIF1<0，令SKIPIF1<0，轉(zhuǎn)化為SKIPIF1<0而得證.13．（2022·江蘇·蘇州市蘇州高新區(qū)第一中學(xué)高二期中）已知函數(shù)SKIPIF1<0的圖象在點SKIPIF1<0(SKIPIF1<0為自然對數(shù)的底數(shù))處的切線斜率為SKIPIF1<0．(1)求實數(shù)SKIPIF1<0的值;(2)若SKIPIF1<0，且存在SKIPIF1<0使SKIPIF1<0成立，求SKIPIF1<0的最小值．【答案】(1)1(2)4【解析】(1)由題意知：SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0；(2)由（1）知：SKIPIF1<0，存在SKIPIF1<0使SKIPIF1<0成立等價于SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單增，又SKIPIF1<0，故存在SKIPIF1<0使SKIPIF1<0，即SKIPIF1<0，故當(dāng)SKIPIF1<0時，SKIPIF1<0單減，故當(dāng)SKIPIF1<0時，SKIPIF1<0單增，故SKIPIF1<0，故SKIPIF1<0，又SKIPIF1<0且SKIPIF1<0，故SKIPIF1<0的最小值為4.14．（2022·安徽省桐城中學(xué)高三階段練習(xí)（理））已知函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0處取得最大值.(1)求a的取值范圍；(2)當(dāng)SKIPIF1<0時，求證：SKIPIF1<0.【答案】(1)SKIPIF1<0(2)證明見解析【解析】(1)顯然SKIPIF1<0，由已知SKIPIF1<0得SKIPIF1<0.故SKIPIF1<0.若SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0；當(dāng)正數(shù)SKIPIF1<0時，SKIPIF1<0.SKIPIF1<0有最小值，不符合題意.若SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0.SKIPIF1<0有最大值SKIPIF1<0，故a的取值范圍為SKIPIF1<0.(2)由（1）知SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0.當(dāng)SKIPIF1<0時，因為SKIPIF1<0，只需證SKIPIF1<0，即證SKIPIF1<0令SKIPIF1<0，SKIPIF1<0設(shè)SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上為增函數(shù).所以SKIPIF1<0，所以存在SKIPIF1<0，使得SKIPIF1<0，此時SKIPIF1<0.當(dāng)SKIPIF1<0時，SKIPIF1<0，即SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，即SKIPIF1<0.故SKIPIF1<0.又因為SKIPIF1<0在SKIPIF1<0為減函數(shù)，且SKIPIF1<0，所以SKIPIF1<0故當(dāng)SKIPIF1<0時，SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0.綜上，當(dāng)SKIPIF1<0時，SKIPIF1<0.解法二：由（1）知SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0.當(dāng)SKIPIF1<0時，因為SKIPIF1<0，只需證SKIPIF1<0，即證SKIPIF1<0.令SKIPIF1<0在SKIPIF1<0上單遞增，所以SKIPIF1<0；令SKIPIF1<0，由SKIPIF1<0得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.當(dāng)SKIPIF1<0時，SKIPIF1<0，故SKIPIF1<0所以SKIPIF1<0綜上，當(dāng)SKIPIF1<0時，SKIPIF1<0.【點睛】不等式證明問題是近年高考命題的熱點，利用導(dǎo)數(shù)證明不等式的方法主要有兩個：（1）不等式兩邊作差構(gòu)造函數(shù)，利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性，求出函數(shù)最值即可；（2）觀察不等式的特點，結(jié)合已解答問題把要證的不等式變形，并運用已證結(jié)論先行放縮，再化簡或者進一步利用導(dǎo)數(shù)證明.第三部分：第11講拓展四：導(dǎo)數(shù)中的隱零點問題（精練）第三部分：第11講拓展四：導(dǎo)數(shù)中的隱零點問題（精練）1．（2022·天津市寧河區(qū)蘆臺第一中學(xué)模擬預(yù)測）已知函數(shù)SKIPIF1<0，SKIPIF1<0.(1)若SKIPIF1<0，求函數(shù)SKIPIF1<0的極值；(2)若關(guān)于SKIPIF1<0的不等式SKIPIF1<0恒成立，求整數(shù)SKIPIF1<0的最小值；(3)若SKIPIF1<0，正實數(shù)SKIPIF1<0滿足SKIPIF1<0，證明：SKIPIF1<0.【答案】(1)極大值為SKIPIF1<0，無極小值；(2)SKIPIF1<0；(3)證明見解析.【解析】(1)∵SKIPIF1<0，∴SKIPIF1<0，此時SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，∴SKIPIF1<0的單調(diào)增區(qū)間為SKIPIF1<0，單調(diào)減區(qū)間為SKIPIF1<0，∴SKIPIF1<0有極大值為SKIPIF1<0，無極小值；(2)由SKIPIF1<0恒成立，得SKIPIF1<0在SKIPIF1<0上恒成立，問題等價于SKIPIF1<0在SKIPIF1<0上恒成立.令SKIPIF1<0，只要SKIPIF1<0.∵SKIPIF1<0.令SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0在SKIPIF1<0上單調(diào)遞減.∵SKIPIF1<0，SKIPIF1<0，∴在(0，＋)上存在唯一的SKIPIF1<0SKIPIF1<0，使得SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0.∴當(dāng)SKIPIF1<0時，SKIPIF1<0，g(x)單調(diào)遞增，當(dāng)SKIPIF1<0時，SKIPIF1<0，g(x)單調(diào)遞減，∴SKIPIF1<0SKIPIF1<0，即SKIPIF1<0，∵SKIPIF1<0，∴整數(shù)SKIPIF1<0的最小值為SKIPIF1<0；(3)由題可知SKIPIF1<0，SKIPIF1<0.當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0.∵SKIPIF1<0