新高考數(shù)學(xué)二輪復(fù)習(xí)強化練習(xí)專題04 導(dǎo)數(shù)的基本應(yīng)用（講）（解析版）

第一篇熱點、難點突破篇專題04導(dǎo)數(shù)的基本應(yīng)用（講）真題體驗感悟高考1．（2022·全國·高考真題（理））當(dāng)SKIPIF1<0時，函數(shù)SKIPIF1<0取得最大值SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．1【答案】B【分析】根據(jù)題意可知SKIPIF1<0，SKIPIF1<0即可解得SKIPIF1<0，再根據(jù)SKIPIF1<0即可解出．【詳解】因為函數(shù)SKIPIF1<0定義域為SKIPIF1<0，所以依題可知，SKIPIF1<0，SKIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，因此函數(shù)SKIPIF1<0在SKIPIF1<0上遞增，在SKIPIF1<0上遞減，SKIPIF1<0時取最大值，滿足題意，即有SKIPIF1<0．故選：B.2．（2022·全國·高考真題（文））函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0的最小值、最大值分別為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】利用導(dǎo)數(shù)求得SKIPIF1<0的單調(diào)區(qū)間，從而判斷出SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值和最大值.【詳解】SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0和SKIPIF1<0上SKIPIF1<0，即SKIPIF1<0單調(diào)遞增；在區(qū)間SKIPIF1<0上SKIPIF1<0，即SKIPIF1<0單調(diào)遞減，又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0上的最小值為SKIPIF1<0，最大值為SKIPIF1<0.故選：D3．（2022·全國·高考真題（理））已知SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】由SKIPIF1<0結(jié)合三角函數(shù)的性質(zhì)可得SKIPIF1<0;構(gòu)造函數(shù)SKIPIF1<0,利用導(dǎo)數(shù)可得SKIPIF1<0,即可得解.【詳解】[方法一]：構(gòu)造函數(shù)因為當(dāng)SKIPIF1<0故SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0；設(shè)SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增，故SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故選A[方法二]：不等式放縮因為當(dāng)SKIPIF1<0，取SKIPIF1<0得：SKIPIF1<0，故SKIPIF1<0SKIPIF1<0，其中SKIPIF1<0，且SKIPIF1<0當(dāng)SKIPIF1<0時，SKIPIF1<0，及SKIPIF1<0此時SKIPIF1<0，SKIPIF1<0故SKIPIF1<0SKIPIF1<0，故SKIPIF1<0所以SKIPIF1<0，所以SKIPIF1<0，故選A[方法三]：泰勒展開設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，計算得SKIPIF1<0，故選A.[方法四]：構(gòu)造函數(shù)因為SKIPIF1<0，因為當(dāng)SKIPIF1<0，所以SKIPIF1<0,即SKIPIF1<0，所以SKIPIF1<0；設(shè)SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0單調(diào)遞增，則SKIPIF1<0,所以SKIPIF1<0，所以SKIPIF1<0,所以SKIPIF1<0，故選：A．[方法五]：【最優(yōu)解】不等式放縮因為SKIPIF1<0，因為當(dāng)SKIPIF1<0，所以SKIPIF1<0,即SKIPIF1<0，所以SKIPIF1<0；因為當(dāng)SKIPIF1<0，取SKIPIF1<0得SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0．故選：A．【整體點評】方法4：利用函數(shù)的單調(diào)性比較大小，是常見思路，難點在于構(gòu)造合適的函數(shù)，屬于通性通法；方法5：利用二倍角公式以及不等式SKIPIF1<0放縮，即可得出大小關(guān)系，屬于最優(yōu)解．4．（2021·全國·高考真題）若過點SKIPIF1<0可以作曲線SKIPIF1<0的兩條切線，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】解法一：根據(jù)導(dǎo)數(shù)幾何意義求得切線方程，再構(gòu)造函數(shù)，利用導(dǎo)數(shù)研究函數(shù)圖象，結(jié)合圖形確定結(jié)果；解法二：畫出曲線SKIPIF1<0的圖象，根據(jù)直觀即可判定點SKIPIF1<0在曲線下方和SKIPIF1<0軸上方時才可以作出兩條切線.【詳解】在曲線SKIPIF1<0上任取一點SKIPIF1<0，對函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0，所以，曲線SKIPIF1<0在點SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0，由題意可知，點SKIPIF1<0在直線SKIPIF1<0上，可得SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0.當(dāng)SKIPIF1<0時，SKIPIF1<0，此時函數(shù)SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時，SKIPIF1<0，此時函數(shù)SKIPIF1<0單調(diào)遞減，所以，SKIPIF1<0，由題意可知，直線SKIPIF1<0與曲線SKIPIF1<0的圖象有兩個交點，則SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖象如下圖所示：由圖可知，當(dāng)SKIPIF1<0時，直線SKIPIF1<0與曲線SKIPIF1<0的圖象有兩個交點.故選：D.解法二：畫出函數(shù)曲線SKIPIF1<0的圖象如圖所示，根據(jù)直觀即可判定點SKIPIF1<0在曲線下方和SKIPIF1<0軸上方時才可以作出兩條切線.由此可知SKIPIF1<0.故選：D.【點睛】解法一是嚴格的證明求解方法，其中的極限處理在中學(xué)知識范圍內(nèi)需要用到指數(shù)函數(shù)的增長特性進行估計，解法二是根據(jù)基于對指數(shù)函數(shù)的圖象的清晰的理解與認識的基礎(chǔ)上，直觀解決問題的有效方法.5．（2022·全國·高考真題（文））已知函數(shù)SKIPIF1<0，曲線SKIPIF1<0在點SKIPIF1<0處的切線也是曲線SKIPIF1<0的切線．(1)若SKIPIF1<0，求a；(2)求a的取值范圍．【答案】(1)3(2)SKIPIF1<0【分析】（1）先由SKIPIF1<0上的切點求出切線方程，設(shè)出SKIPIF1<0上的切點坐標，由斜率求出切點坐標，再由函數(shù)值求出SKIPIF1<0即可；（2）設(shè)出SKIPIF1<0上的切點坐標，分別由SKIPIF1<0和SKIPIF1<0及切點表示出切線方程，由切線重合表示出SKIPIF1<0，構(gòu)造函數(shù)，求導(dǎo)求出函數(shù)值域，即可求得SKIPIF1<0的取值范圍.（1）由題意知，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0在點SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0，設(shè)該切線與SKIPIF1<0切于點SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0；（2）SKIPIF1<0，則SKIPIF1<0在點SKIPIF1<0處的切線方程為SKIPIF1<0，整理得SKIPIF1<0，設(shè)該切線與SKIPIF1<0切于點SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，則切線方程為SKIPIF1<0，整理得SKIPIF1<0，則SKIPIF1<0，整理得SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，則SKIPIF1<0變化時，SKIPIF1<0的變化情況如下表：SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<00SKIPIF1<01SKIPIF1<0SKIPIF1<0SKIPIF1<00SKIPIF1<00SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0則SKIPIF1<0的值域為SKIPIF1<0，故SKIPIF1<0的取值范圍為SKIPIF1<0.總結(jié)規(guī)律預(yù)測考向（一）規(guī)律與預(yù)測1.高考對本部分的要求一般有三個層次：第一層次主要考查求導(dǎo)公式，求導(dǎo)法則與導(dǎo)數(shù)的幾何意義；第二層次是導(dǎo)數(shù)的簡單應(yīng)用，包括求函數(shù)的單調(diào)區(qū)間、極值、最值等；第三層次是綜合考查，如研究函數(shù)零點、證明不等式、恒成立問題、求參數(shù)等，包括解決應(yīng)用問題，將導(dǎo)數(shù)內(nèi)容和傳統(tǒng)內(nèi)容中有關(guān)不等式、數(shù)列及函數(shù)單調(diào)性有機結(jié)合，設(shè)計綜合題.2.導(dǎo)數(shù)的計算和幾何意義是高考命題的熱點，多以選擇題、填空題形式考查，難度較小.3.應(yīng)用導(dǎo)數(shù)研究函數(shù)的單調(diào)性、極值、最值多在選擇題、填空題靠后的位置考查，難度中等偏上，屬綜合性問題．4.涉及導(dǎo)數(shù)的幾何意義、單調(diào)性、極（最）值的求參數(shù)取值范圍問題，是?？碱}型.（二）本專題考向展示考點突破典例分析考向一導(dǎo)數(shù)的幾何意義【核心知識】1.基本初等函數(shù)的導(dǎo)數(shù)公式原函數(shù)導(dǎo)函數(shù)f(x)＝c(c為常數(shù))f′(x)＝0f(x)＝xn(n∈Q*)f′(x)＝nxn－1f(x)＝sinxf′(x)＝cosxf(x)＝cosxf′(x)＝－sinxf(x)＝axf′(x)＝axlnaf(x)＝exf′(x)＝exf(x)＝logaxf′(x)＝eq\f(1,xlna)f(x)＝lnxf′(x)＝eq\f(1,x)2．導(dǎo)數(shù)的運算法則（1）[f(x)±g(x)]′＝f′(x)±g′(x)；（2）[f(x)·g(x)]′＝f′(x)g(x)＋f(x)g′(x)；（3）(g(x)≠0)．3.函數(shù)f(x)在x0處的導(dǎo)數(shù)是曲線f(x)在點P(x0，f(x0))處的切線的斜率，曲線f(x)在點P處的切線的斜率k＝f′(x0)，相應(yīng)的切線方程為y－f(x0)＝f′(x0)·(x－x0)．警示:求曲線的切線方程時，要注意是在點P處的切線還是過點P的切線，前者點P為切點，后者點P不一定為切點.【典例分析】典例1.（2022·貴州遵義·高三期中（理））若直線SKIPIF1<0與曲線SKIPIF1<0相切，則切點的坐標為_____________．【答案】SKIPIF1<0【分析】設(shè)切點為SKIPIF1<0，求出函數(shù)的導(dǎo)函數(shù)，即可得到方程組，解得即可.【詳解】解：設(shè)切點為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，∴切點坐標為SKIPIF1<0.故答案為：SKIPIF1<0典例2.（2021·全國·高考真題）已知函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0的圖象在點SKIPIF1<0和點SKIPIF1<0的兩條切線互相垂直，且分別交y軸于M，N兩點，則SKIPIF1<0取值范圍是_______．【答案】SKIPIF1<0【分析】結(jié)合導(dǎo)數(shù)的幾何意義可得SKIPIF1<0，結(jié)合直線方程及兩點間距離公式可得SKIPIF1<0，SKIPIF1<0，化簡即可得解.【詳解】由題意，SKIPIF1<0，則SKIPIF1<0，所以點SKIPIF1<0和點SKIPIF1<0，SKIPIF1<0,所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，同理SKIPIF1<0,所以SKIPIF1<0.故答案為：SKIPIF1<0【點睛】關(guān)鍵點點睛：解決本題的關(guān)鍵是利用導(dǎo)數(shù)的幾何意義轉(zhuǎn)化條件SKIPIF1<0，消去一個變量后，運算即可得解.典例3.（2019·江蘇·高考真題）在平面直角坐標系SKIPIF1<0中，P是曲線SKIPIF1<0上的一個動點，則點P到直線x+y=0的距離的最小值是_____.【答案】4.【分析】將原問題轉(zhuǎn)化為切點與直線之間的距離，然后利用導(dǎo)函數(shù)確定切點坐標可得最小距離【詳解】當(dāng)直線SKIPIF1<0平移到與曲線SKIPIF1<0相切位置時，切點Q即為點P到直線SKIPIF1<0的距離最小.由SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0，即切點SKIPIF1<0，則切點Q到直線SKIPIF1<0的距離為SKIPIF1<0，故答案為SKIPIF1<0．【點睛】本題考查曲線上任意一點到已知直線的最小距離，滲透了直觀想象和數(shù)學(xué)運算素養(yǎng).采取導(dǎo)數(shù)法和公式法，利用數(shù)形結(jié)合和轉(zhuǎn)化與化歸思想解題.【總結(jié)提升】1.求曲線y＝f(x)的切線方程的三種類型及方法(1)已知切點P(x0，y0)，求y＝f(x)過點P的切線方程：求出切線的斜率f′(x0)，由點斜式寫出方程．(2)已知切線的斜率為k，求y＝f(x)的切線方程：設(shè)切點P(x0，y0)，通過方程k＝f′(x0)解得x0，再由點斜式寫出方程．(3)已知切線上一點(非切點)，求y＝f(x)的切線方程：設(shè)切點P(x0，y0)，利用導(dǎo)數(shù)求得切線斜率f′(x0)，然后由斜率公式求得切線斜率，列方程(組)解得x0，再由點斜式或兩點式寫出方程．2．一些距離類最值，可以轉(zhuǎn)化為求一條直線上的點到一條曲線上的點的最小值，此時與已知直線平行的曲線的切線到已知直線的距離即為最小值.考向二利用導(dǎo)數(shù)的幾何意義求參數(shù)【核心知識】主要涉及公切線問題、兩直線位置關(guān)系問題、切點坐標、切線的斜率（切線方程）問題以及與切線相關(guān)的距離問題.【典例分析】典例4.（2022·河南·一模（理））已知曲線SKIPIF1<0在點SKIPIF1<0處的切線與直線SKIPIF1<0垂直，則實數(shù)SKIPIF1<0的值為（

）．A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】根據(jù)導(dǎo)數(shù)幾何意義和垂直關(guān)系可得SKIPIF1<0，解方程即可.【詳解】令SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0在點SKIPIF1<0處的切線與SKIPIF1<0垂直，SKIPIF1<0，解得：SKIPIF1<0.故選：A.典例5.（2021·全國·高考真題）若過點SKIPIF1<0可以作曲線SKIPIF1<0的兩條切線，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】解法一：根據(jù)導(dǎo)數(shù)幾何意義求得切線方程，再構(gòu)造函數(shù)，利用導(dǎo)數(shù)研究函數(shù)圖象，結(jié)合圖形確定結(jié)果；解法二：畫出曲線SKIPIF1<0的圖象，根據(jù)直觀即可判定點SKIPIF1<0在曲線下方和SKIPIF1<0軸上方時才可以作出兩條切線.【詳解】在曲線SKIPIF1<0上任取一點SKIPIF1<0，對函數(shù)SKIPIF1<0求導(dǎo)得SKIPIF1<0，所以，曲線SKIPIF1<0在點SKIPIF1<0處的切線方程為SKIPIF1<0，即SKIPIF1<0，由題意可知，點SKIPIF1<0在直線SKIPIF1<0上，可得SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0.當(dāng)SKIPIF1<0時，SKIPIF1<0，此時函數(shù)SKIPIF1<0單調(diào)遞增，當(dāng)SKIPIF1<0時，SKIPIF1<0，此時函數(shù)SKIPIF1<0單調(diào)遞減，所以，SKIPIF1<0，由題意可知，直線SKIPIF1<0與曲線SKIPIF1<0的圖象有兩個交點，則SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖象如下圖所示：由圖可知，當(dāng)SKIPIF1<0時，直線SKIPIF1<0與曲線SKIPIF1<0的圖象有兩個交點.故選：D.解法二：畫出函數(shù)曲線SKIPIF1<0的圖象如圖所示，根據(jù)直觀即可判定點SKIPIF1<0在曲線下方和SKIPIF1<0軸上方時才可以作出兩條切線.由此可知SKIPIF1<0.故選：D.典例6.（2022·全國·高考真題）若曲線SKIPIF1<0有兩條過坐標原點的切線，則a的取值范圍是___________．【答案】SKIPIF1<0【分析】設(shè)出切點橫坐標SKIPIF1<0，利用導(dǎo)數(shù)的幾何意義求得切線方程，根據(jù)切線經(jīng)過原點得到關(guān)于SKIPIF1<0的方程，根據(jù)此方程應(yīng)有兩個不同的實數(shù)根，求得SKIPIF1<0的取值范圍.【詳解】∵SKIPIF1<0，∴SKIPIF1<0，設(shè)切點為SKIPIF1<0,則SKIPIF1<0,切線斜率SKIPIF1<0,切線方程為：SKIPIF1<0,∵切線過原點，∴SKIPIF1<0,整理得：SKIPIF1<0,∵切線有兩條，∴SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0,∴SKIPIF1<0的取值范圍是SKIPIF1<0,故答案為：SKIPIF1<0【總結(jié)提升】利用導(dǎo)數(shù)的幾何意義求參數(shù)的基本方法利用切點的坐標、切線的斜率、切線方程等得到關(guān)于參數(shù)的方程(組)或者參數(shù)滿足的不等式(組)，進而求出參數(shù)的值或取值范圍．考向三利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性【核心知識】導(dǎo)數(shù)與單調(diào)性的關(guān)系1．f′(x)>0是f(x)為增函數(shù)的充分不必要條件，如函數(shù)f(x)＝x3在(－∞，＋∞)上單調(diào)遞增，但f′(x)≥0；2．f′(x)≥0是f(x)為增函數(shù)的必要不充分條件，當(dāng)函數(shù)在某個區(qū)間內(nèi)恒有f′(x)＝0時，則f(x)為常數(shù)，函數(shù)不具有單調(diào)性．【典例分析】典例7.（2022·全國·高考真題）設(shè)SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】構(gòu)造函數(shù)SKIPIF1<0，導(dǎo)數(shù)判斷其單調(diào)性，由此確定SKIPIF1<0的大小.【詳解】方法一：構(gòu)造法設(shè)SKIPIF1<0，因為SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時SKIPIF1<0，所以函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0,令SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，又SKIPIF1<0，所以當(dāng)SKIPIF1<0時，SKIPIF1<0，所以當(dāng)SKIPIF1<0時，SKIPIF1<0，函數(shù)SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0故選：C.方法二：比較法解：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，①SKIPIF1<0，令SKIPIF1<0則SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，可得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0；②SKIPIF1<0，令SKIPIF1<0則SKIPIF1<0，令SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，可得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，可得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0故SKIPIF1<0【注：此類問題已連年考查】典例8.（2022·廣西北?！ひ荒＃ㄎ模┖瘮?shù)SKIPIF1<0的增區(qū)間為____________.【答案】SKIPIF1<0##SKIPIF1<0【分析】解不等式SKIPIF1<0即得解.【詳解】由題得SKIPIF1<0，可得SKIPIF1<0.故函數(shù)的增區(qū)間為SKIPIF1<0.故答案為：SKIPIF1<0典例9.（2021·全國·高考真題（文））設(shè)函數(shù)SKIPIF1<0，其中SKIPIF1<0.（1）討論SKIPIF1<0的單調(diào)性；（2）若SKIPIF1<0的圖象與SKIPIF1<0軸沒有公共點，求a的取值范圍.【答案】（1）SKIPIF1<0的減區(qū)間為SKIPIF1<0，增區(qū)間為SKIPIF1<0；（2）SKIPIF1<0.【分析】（1）求出函數(shù)的導(dǎo)數(shù)，討論其符號后可得函數(shù)的單調(diào)性.（2）根據(jù)SKIPIF1<0及（1）的單調(diào)性性可得SKIPIF1<0，從而可求a的取值范圍.【詳解】（1）函數(shù)的定義域為SKIPIF1<0，又SKIPIF1<0，因為SKIPIF1<0，故SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0；所以SKIPIF1<0的減區(qū)間為SKIPIF1<0，增區(qū)間為SKIPIF1<0.（2）因為SKIPIF1<0且SKIPIF1<0的圖與SKIPIF1<0軸沒有公共點，所以SKIPIF1<0的圖象在SKIPIF1<0軸的上方，由（1）中函數(shù)的單調(diào)性可得SKIPIF1<0，故SKIPIF1<0即SKIPIF1<0.【點睛】方法點睛：不等式的恒成立問題，往往可轉(zhuǎn)化為函數(shù)的最值的符號來討論，也可以參變分離后轉(zhuǎn)化不含參數(shù)的函數(shù)的最值問題，轉(zhuǎn)化中注意等價轉(zhuǎn)化.【規(guī)律方法】1.求函數(shù)的單調(diào)區(qū)間，只需在函數(shù)的定義域內(nèi)解(證)不等式SKIPIF1<0(或SKIPIF1<0).2.利用導(dǎo)數(shù)比較大小或解不等式，常常要構(gòu)造新函數(shù)，把比較大小或求解不等式的問題轉(zhuǎn)化為利用導(dǎo)數(shù)研究函數(shù)單調(diào)性的問題，再由單調(diào)性比較大小或解不等式．常見構(gòu)造的輔助函數(shù)有：g(x)＝xf(x)，g(x)＝SKIPIF1<0，g(x)＝exf(x)，g(x)＝SKIPIF1<0，g(x)＝f(x)lnx，g(x)＝SKIPIF1<0等．3.溫馨提醒：（1）在利用導(dǎo)數(shù)討論函數(shù)的單調(diào)區(qū)間時，首先要確定函數(shù)的定義域．（2）單調(diào)區(qū)間的劃分要注意對導(dǎo)數(shù)等于零的點的確認．（3）所求函數(shù)的單調(diào)區(qū)間不止一個，這些區(qū)間之間不能用并集“∪”及“或”連接，只能用“，”“和”字隔開．考向四由函數(shù)的單調(diào)性求參數(shù)取值范圍【核心知識】(1)可導(dǎo)函數(shù)在區(qū)間(a，b)上單調(diào)，實際上就是在該區(qū)間上f′(x)≥0(或f′(x)≤0)恒成立，得到關(guān)于參數(shù)的不等式，從而轉(zhuǎn)化為求函數(shù)的最值問題，求參數(shù)的取值范圍；(2)可導(dǎo)函數(shù)在區(qū)間(a，b)上存在單調(diào)區(qū)間，實際上就是f′(x)>0(或f′(x)<0)在該區(qū)間上存在解集，即f′(x)max>0(或f′(x)min<0)在該區(qū)間上有解，從而轉(zhuǎn)化為不等式問題，求參數(shù)的取值范圍.【典例分析】典例10.（2022·上海市進才中學(xué)高三期中）已知SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，則實數(shù)SKIPIF1<0的取值范圍是__________.【答案】SKIPIF1<0.【分析】求導(dǎo)后得到SKIPIF1<0在SKIPIF1<0上恒成立，參變分離后得到SKIPIF1<0在SKIPIF1<0上恒成立，利用導(dǎo)函數(shù)求出SKIPIF1<0，從而求出實數(shù)SKIPIF1<0的取值范圍.【詳解】SKIPIF1<0，SKIPIF1<0，故只需SKIPIF1<0在SKIPIF1<0上恒成立，則SKIPIF1<0在SKIPIF1<0上恒成立，其中SKIPIF1<0在SKIPIF1<0上恒成立，故SKIPIF1<0，所以SKIPIF1<0，故答案為：SKIPIF1<0.典例11.（2022·陜西·蒲城縣蒲城中學(xué)高三階段練習(xí)（理））已知函數(shù)SKIPIF1<0在SKIPIF1<0上不單調(diào)，則SKIPIF1<0的取值范圍是__________.【答案】SKIPIF1<0【分析】結(jié)合函數(shù)的導(dǎo)數(shù)討論單調(diào)性，確定函數(shù)在SKIPIF1<0上既有增區(qū)間又有減區(qū)間即可求解.【詳解】由題可知，SKIPIF1<0，令SKIPIF1<0，因為SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0恒成立，即SKIPIF1<0恒成立，則函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，不滿足題意;若SKIPIF1<0，則SKIPIF1<0，因為SKIPIF1<0，SKIPIF1<0,所以SKIPIF1<0，所以由零點的唯一性定理可知，SKIPIF1<0在SKIPIF1<0必定存在唯一的零點記為SKIPIF1<0，所以當(dāng)SKIPIF1<0時SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0時SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0時單調(diào)遞增，SKIPIF1<0時單調(diào)遞減，滿足題意;綜上得SKIPIF1<0，故答案為:SKIPIF1<0.典例12.（2019·北京·高考真題（理））設(shè)函數(shù)f（x）=ex+ae?x（a為常數(shù)）．若f（x）為奇函數(shù)，則a=________；若f（x）是R上的增函數(shù)，則a的取值范圍是___________．【答案】

-1;

SKIPIF1<0.【分析】首先由奇函數(shù)的定義得到關(guān)于SKIPIF1<0的恒等式，據(jù)此可得SKIPIF1<0的值，然后利用導(dǎo)函數(shù)的解析式可得a的取值范圍.【詳解】若函數(shù)SKIPIF1<0為奇函數(shù),則SKIPIF1<0，SKIPIF1<0對任意的SKIPIF1<0恒成立.若函數(shù)SKIPIF1<0是SKIPIF1<0上的增函數(shù),則SKIPIF1<0恒成立,SKIPIF1<0.即實數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0【總結(jié)提升】1.利用單調(diào)性確定參數(shù)的范圍.解答過程中,需利用轉(zhuǎn)化與化歸思想,轉(zhuǎn)化成恒成立問題.應(yīng)用條件SKIPIF1<0(或SKIPIF1<0)，SKIPIF1<0恒成立，解出參數(shù)的取值范圍(一般可用不等式恒成立的理論求解)，應(yīng)注意參數(shù)的取值范圍是SKIPIF1<0不恒等于0的參數(shù)的取值范圍.2.可導(dǎo)函數(shù)在區(qū)間(a，b)上存在單調(diào)區(qū)間，實際上就是f′(x)＞0(或f′(x)＜0)在該區(qū)間上存在解集，從而轉(zhuǎn)化為不等式問題，求出參數(shù)的取值范圍．再驗證參數(shù)取“＝”時f(x)是否滿足題意．3.若已知f(x)在區(qū)間I上的單調(diào)性，區(qū)間I上含有參數(shù)時，可先求出f(x)的單調(diào)區(qū)間，令I(lǐng)是其單調(diào)區(qū)間的子集，從而求出參數(shù)的取值范圍．4.函數(shù)SKIPIF1<0在SKIPIF1<0上不單調(diào)，則轉(zhuǎn)化為SKIPIF1<0在SKIPIF1<0上有解.5.特別提醒：（1）弄清參數(shù)對f′(x)符號的影響，分類討論要不重不漏．（2）已知函數(shù)單調(diào)性求參數(shù)范圍，要注意導(dǎo)數(shù)等于零的情況．考向五利用導(dǎo)數(shù)研究函數(shù)的極值、最值【核心知識】1.導(dǎo)數(shù)與極值、最值(1)函數(shù)f(x)在x0處的導(dǎo)數(shù)f′(x0)＝0且f′(x)在x0附近“左正右負”?f(x)在x0處取極大值；函數(shù)f(x)在x0處的導(dǎo)數(shù)f′(x0)＝0且f′(x)在x0附近“左負右正”?f(x)在x0處取極小值．(2)函數(shù)f(x)在一閉區(qū)間上的最大值是此函數(shù)在該區(qū)間上的極值與該區(qū)間端點處函數(shù)值中的“最大者”；函數(shù)f(x)在一閉區(qū)間上的最小值是此函數(shù)在該區(qū)間上的極值與該區(qū)間端點處函數(shù)值中的“最小者”．2．求函數(shù)f(x)在[a，b]上的最大值和最小值的步驟(1)求函數(shù)在(a，b)內(nèi)的極值．(2)求函數(shù)在區(qū)間端點處的函數(shù)值f(a)，f(b)．(3)將函數(shù)f(x)的各極值與f(a)，f(b)比較，其中最大的一個為最大值，最小的一個為最小值．【典例分析】典例13.【多選題】（2022·全國·高考真題）已知函數(shù)SKIPIF1<0的圖像關(guān)于點SKIPIF1<0中心對稱，則（

）A．SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞減B．SKIPIF1<0在區(qū)間SKIPIF1<0有兩個極值點C．直線SKIPIF1<0是曲線SKIPIF1<0的對稱軸D．直線SKIPIF1<0是曲線SKIPIF1<0的切線【答案】AD【分析】根據(jù)三角函數(shù)的性質(zhì)逐個判斷各選項，即可解出．【詳解】由題意得：SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0時，SKIPIF1<0，故SKIPIF1<0．對A，當(dāng)SKIPIF1<0時，SKIPIF1<0，由正弦函數(shù)SKIPIF1<0圖象知SKIPIF1<0在SKIPIF1<0上是單調(diào)遞減；對B，當(dāng)SKIPIF1<0時，SKIPIF1<0，由正弦函數(shù)SKIPIF1<0圖象知SKIPIF1<0只有1個極值點，由SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0為函數(shù)的唯一極值點；對C，當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，直線SKIPIF1<0不是對稱軸；對D，由SKIPIF1<0得：SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0,從而得：SKIPIF1<0或SKIPIF1<0,所以函數(shù)SKIPIF1<0在點SKIPIF1<0處的切線斜率為SKIPIF1<0，切線方程為：SKIPIF1<0即SKIPIF1<0．故選：AD．典例14.（2019·全國·高考真題（文））已知函數(shù)SKIPIF1<0.（1）討論SKIPIF1<0的單調(diào)性；（2）當(dāng)SKIPIF1<0時，記SKIPIF1<0在區(qū)間SKIPIF1<0的最大值為SKIPIF1<0，最小值為SKIPIF1<0，求SKIPIF1<0的取值范圍.【答案】(1)見詳解；(2)SKIPIF1<0.【分析】(1)先求SKIPIF1<0的導(dǎo)數(shù)，再根據(jù)SKIPIF1<0的范圍分情況討論函數(shù)單調(diào)性；(2)討論SKIPIF1<0的范圍，利用函數(shù)單調(diào)性進行最大值和最小值的判斷，最終求得SKIPIF1<0的取值范圍.【詳解】(1)對SKIPIF1<0求導(dǎo)得SKIPIF1<0.所以有當(dāng)SKIPIF1<0時，SKIPIF1<0區(qū)間上單調(diào)遞增，SKIPIF1<0區(qū)間上單調(diào)遞減，SKIPIF1<0區(qū)間上單調(diào)遞增；當(dāng)SKIPIF1<0時，SKIPIF1<0區(qū)間上單調(diào)遞增；當(dāng)SKIPIF1<0時，SKIPIF1<0區(qū)間上單調(diào)遞增，SKIPIF1<0區(qū)間上單調(diào)遞減，SKIPIF1<0區(qū)間上單調(diào)遞增.(2)若SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞減，在區(qū)間SKIPIF1<0單調(diào)遞增，所以區(qū)間SKIPIF1<0上最小值為SKIPIF1<0.而SKIPIF1<0，故所以區(qū)間SKIPIF1<0上最大值為SKIPIF1<0.所以SKIPIF1<0，設(shè)函數(shù)SKIPIF1<0，求導(dǎo)SKIPIF1<0當(dāng)SKIPIF1<0時SKIPIF1<0從而SKIPIF1<0單調(diào)遞減.而SKIPIF1<0，所以SKIPIF1<0.即SKIPIF1<0的取值范圍是SKIPIF1<0.若SKIPIF1<0，SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞減，在區(qū)間SKIPIF1<0單調(diào)遞增，所以區(qū)間SKIPIF1<0上最小值為SKIPIF1<0而SKIPIF1<0，故所以區(qū)間SKIPIF1<0上最大值為SKIPIF1<0.所以SKIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0.即SKIPIF1<0的取值范圍是SKIPIF1<0.綜上得SKIPIF1<0的取值范圍是SKIPIF1<0.【特別提醒】1.不能忽略函數(shù)SKIPIF1<0的定義域.2.SKIPIF1<0是可導(dǎo)函數(shù)SKIPIF1<0在SKIPIF1<0處取得極值的必要不充分條件.3.函數(shù)的極小值不一定比極大值小.4.若函數(shù)在區(qū)間SKIPIF1<0上有唯一的極值點，則這個極值點也是最值點，此結(jié)論在導(dǎo)數(shù)的實際應(yīng)用中經(jīng)常用到.考向六函數(shù)的極（最）值相關(guān)參數(shù)問題【核心知識】由函數(shù)極值（個數(shù)）求參數(shù)的值或范圍．討論極值點有無(個數(shù))問題，轉(zhuǎn)化為討論f′(x)＝0根的有無(個數(shù))．然后由已知條件列出方程或不等式求出參數(shù)的值或范圍，特別注意：極值點處的導(dǎo)數(shù)為0，而導(dǎo)數(shù)為0的點不一定是極值點，要檢驗極值點兩側(cè)導(dǎo)數(shù)是否異號．【典例分析】典例15.（2021·全國·高考真題（理））設(shè)SKIPIF1<0，若SKIPIF1<0為函數(shù)SKIPIF1<0的極大值點，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】先考慮函數(shù)的零點情況，注意零點左右附近函數(shù)值是否變號，結(jié)合極大值點的性質(zhì)，對進行分類討論，畫出圖象，即可得到SKIPIF1<0所滿足的關(guān)系，由此確定正確選項.【詳解】若SKIPIF1<0，則SKIPIF1<0為單調(diào)函數(shù)，無極值點，不符合題意，故SKIPIF1<0.SKIPIF1<0有SKIPIF1<0和SKIPIF1<0兩個不同零點，且在SKIPIF1<0左右附近是不變號，在SKIPIF1<0左右附近是變號的.依題意，為函數(shù)的極大值點，SKIPIF1<0在SKIPIF1<0左右附近都是小于零的.當(dāng)SKIPIF1<0時，由SKIPIF1<0，SKIPIF1<0，畫出SKIPIF1<0的圖象如下圖所示：由圖可知SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0.當(dāng)SKIPIF1<0時，由SKIPIF1<0時，SKIPIF1<0，畫出SKIPIF1<0的圖象如下圖所示：由圖可知SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0.綜上所述，SKIPIF1<0成立.故選：D典例16.（2022·廣東實驗中學(xué)高三階段練習(xí)）設(shè)SKIPIF1<0，若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0有極值點，則SKIPIF1<0取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】先對函數(shù)求導(dǎo)，根據(jù)函數(shù)在區(qū)間有極值點，轉(zhuǎn)化為導(dǎo)函數(shù)有零點，再由零點存在定理列出不等式求解即可.【詳解】SKIPIF1<0，SKIPIF1<0為單調(diào)函數(shù)，所以函數(shù)在區(qū)間SKIPIF1<0有極值點，即SKIPIF1<0，代入解得SKIPIF1<0，解得SKIPIF1<0取值范圍為SKIPIF1<0，故選：B．典例17.（2022·全國·高考真題（理））已知SKIPIF1<0和SKIPIF1<0分別是函數(shù)SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）的極小值點和極大值點．若SKIPIF1<0，則a的取值范圍是____________．【答案】SKIPIF1<0【分析】法一：依題可知，方程SKIPIF1<0的兩個根為SKIPIF1<0，即函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個不同的交點，構(gòu)造函數(shù)SKIPIF1<0，利用指數(shù)函數(shù)的圖象和圖象變換得到SKIPIF1<0的圖象，利用導(dǎo)數(shù)的幾何意義求得過原點的切線的斜率，根據(jù)幾何意義可得出答案.【詳解】[方法一]：【最優(yōu)解】轉(zhuǎn)化法，零點的問題轉(zhuǎn)為函數(shù)圖象的交點因為SKIPIF1<0，所以方程SKIPIF1<0的兩個根為SKIPIF1<0，即方程SKIPIF1<0的兩個根為SKIPIF1<0，即函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個不同的交點，因為SKIPIF1<0分別是函數(shù)SKIPIF1<0的極小值點和極大值點，所以函數(shù)SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上遞減，在SKIPIF1<0上遞增，所以當(dāng)時SKIPIF1<0SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0圖象在SKIPIF1<0上方當(dāng)SKIPIF1<0時，SKIPIF1<0，即SKIPIF1<0圖象在SKIPIF1<0下方SKIPIF1<0，圖象顯然不符合題意，所以SKIPIF1<0．令SKIPIF1<0，則SKIPIF1<0，設(shè)過原點且與函數(shù)SKIPIF1<0的圖象相切的直線的切點為SKIPIF1<0，則切線的斜率為SKIPIF1<0，故切線方程為SKIPIF1<0，則有SKIPIF1<0，解得SKIPIF1<0，則切線的斜率為SKIPIF1<0，因為函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的圖象有兩個不同的交點，所以SKIPIF1<0，解得SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，綜上所述，SKIPIF1<0的取值范圍為SKIPIF1<0．[方法二]：【通性通法】構(gòu)造新函數(shù)，二次求導(dǎo)SKIPIF1<0=0的兩個根為SKIPIF1<0因為SKIPIF1<0分別是函數(shù)SKIPIF1<0的極小值點和極大值點，所以函數(shù)SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上遞減，在SKIPIF1<0上遞增，設(shè)函數(shù)SKIPIF1<0，則SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，此時若SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，此時若有SKIPIF1<0和SKIPIF1<0分別是函數(shù)SKIPIF1<0且SKIPIF1<0的極小值點和極大值點，則SKIPIF1<0，不符合題意；若SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，此時若SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，令SKIPIF1<0，則SKIPIF1<0，此時若有SKIPIF1<0和SKIPIF1<0分別是函數(shù)SKIPIF1<0且SKIPIF1<0的極小值點和極大值點，且SKIPIF1<0，則需滿足SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0故SKIPIF1<0，所以SKIPIF1<0.【整體點評】法一：利用函數(shù)的零點與兩函數(shù)圖象交點的關(guān)系，由數(shù)形結(jié)合解出，突出“小題小做”，是該題的最優(yōu)解；法二：通過構(gòu)造新函數(shù)，多次求導(dǎo)判斷單調(diào)性，根據(jù)極值點的大小關(guān)系得出不等式，解出即可，該法屬于通性通法．典例18.（2022·北京海淀·高三期中）已知函數(shù)SKIPIF1<0．①當(dāng)SKIPIF1<0時，SKIPIF1<0的極值點個數(shù)為__________；②若SKIPIF1<0恰有兩個極值點，則SKIPIF1<0的取值范圍是__________．【答案】

SKIPIF1<0

SKIPIF1<0【分析】①驗證分段處函數(shù)值可知SKIPIF1<0為連續(xù)函數(shù)，由單調(diào)性可確定SKIPIF1<0和SKIPIF1<0是SKIPIF1<0的極值點，由此可得極值點個數(shù)；②驗證分段處函數(shù)值可知SKIPIF1<0為連續(xù)函數(shù)，根據(jù)一次函數(shù)和二次函數(shù)單調(diào)性可確定SKIPIF1<0和SKIPIF1<0必為SKIPIF1<0的兩個極值點，得到SKIPIF1<0；根據(jù)二次函數(shù)的單調(diào)性，結(jié)合極值點定義可知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，即SKIPIF1<0；由此可得SKIPIF1