新高考數(shù)學(xué)二輪復(fù)習(xí)講義專題10 導(dǎo)數(shù)在函數(shù)中的應(yīng)用 （原卷版）

專題10講：導(dǎo)數(shù)在函數(shù)中的應(yīng)用【考點(diǎn)解密】1．導(dǎo)數(shù)的概念(1)如果當(dāng)Δx→0時(shí)，平均變化率eq\f(Δy,Δx)無限趨近于一個(gè)確定的值，即eq\f(Δy,Δx)有極根，則稱y＝f(x)在x＝x0處可導(dǎo)，并把這個(gè)確定的值叫做y＝f(x)在x＝x0處的導(dǎo)數(shù)(也稱瞬時(shí)變化率)，記作f′(x0)或SKIPIF1<0，即f′(x0)＝eq\o(lim,\s\do6(Δx→0))eq\f(Δy,Δx)＝eq\o(lim,\s\do6(Δx→0))eq\f(fx0＋Δx－fx0,Δx).(2)當(dāng)x＝x0時(shí)，f′(x0)是一個(gè)唯一確定的數(shù)，當(dāng)x變化時(shí)，y＝f′(x)就是x的函數(shù)，我們稱它為y＝f(x)的導(dǎo)函數(shù)(簡(jiǎn)稱導(dǎo)數(shù))，記為f′(x)(或y′)，即f′(x)＝y(tǒng)′＝eq\o(lim,\s\do6(Δx→0))eq\f(fx＋Δx－fx,Δx).2．導(dǎo)數(shù)的幾何意義函數(shù)y＝f(x)在x＝x0處的導(dǎo)數(shù)的幾何意義就是曲線y＝f(x)在點(diǎn)P(x0，f(x0))處的切線的斜率，相應(yīng)的切線方程為y－f(x0)＝f′(x0)(x－x0)．3．基本初等函數(shù)的導(dǎo)數(shù)公式基本初等函數(shù)導(dǎo)函數(shù)f(x)＝c(c為常數(shù))f′(x)＝0f(x)＝xα(α∈Q，α≠0)f′(x)＝αxα－1f(x)＝sinxf′(x)＝cosxf(x)＝cosxf′(x)＝－sinxf(x)＝ax(a>0且a≠1)f′(x)＝axlnaf(x)＝exf′(x)＝exf(x)＝logax(a>0且a≠1)f′(x)＝eq\f(1,xlna)f(x)＝lnxf′(x)＝eq\f(1,x)4．導(dǎo)數(shù)的運(yùn)算法則若f′(x)，g′(x)存在，則有[f(x)±g(x)]′＝f′(x)±g′(x)；[f(x)g(x)]′＝f′(x)g(x)＋f(x)g′(x)；eq\b\lc\[\rc\](\a\vs4\al\co1(\f(fx,gx)))′＝eq\f(f′xgx－fxg′x,[gx]2)(g(x)≠0)；[cf(x)]′＝cf′(x)．5．復(fù)合函數(shù)的定義及其導(dǎo)數(shù)(1)一般地，對(duì)于兩個(gè)函數(shù)y＝f(u)和u＝g(x)，如果通過中間變量u，y可以表示成x的函數(shù)，那么稱這個(gè)函數(shù)為函數(shù)y＝f(u)與u＝g(x)的復(fù)合函數(shù)，記作y＝f(g(x))．(2)復(fù)合函數(shù)y＝f(g(x))的導(dǎo)數(shù)和函數(shù)y＝f(u)，u＝g(x)的導(dǎo)數(shù)間的關(guān)系為y′x＝y(tǒng)′u·u′x，即y對(duì)x的導(dǎo)數(shù)等于y對(duì)u的導(dǎo)數(shù)與u對(duì)x的導(dǎo)數(shù)的乘積．6．函數(shù)的單調(diào)性與導(dǎo)數(shù)的關(guān)系條件恒有結(jié)論函數(shù)y＝f(x)在區(qū)間(a，b)上可導(dǎo)f′(x)>0f(x)在(a，b)上單調(diào)遞增f′(x)<0f(x)在(a，b)上單調(diào)遞減f′(x)＝0f(x)在(a，b)上是常數(shù)函數(shù)7．利用導(dǎo)數(shù)判斷函數(shù)單調(diào)性的步驟第1步，確定函數(shù)的定義域；第2步，求出導(dǎo)數(shù)f′(x)的零點(diǎn)；第3步，用f′(x)的零點(diǎn)將f(x)的定義域劃分為若干個(gè)區(qū)間，列表給出f′(x)在各區(qū)間上的正負(fù)，由此得出函數(shù)y＝f(x)在定義域內(nèi)的單調(diào)性．8．函數(shù)的極值(1)函數(shù)的極小值：函數(shù)y＝f(x)在點(diǎn)x＝a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，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a叫做函數(shù)y＝f(x)的極小值點(diǎn)，f(a)叫做函數(shù)y＝f(x)的極小值．(2)函數(shù)的極大值：函數(shù)y＝f(x)在點(diǎn)x＝b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，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b叫做函數(shù)y＝f(x)的極大值點(diǎn)，f(b)叫做函數(shù)y＝f(x)的極大值．(3)極小值點(diǎn)、極大值點(diǎn)統(tǒng)稱為極值點(diǎn)，極小值和極大值統(tǒng)稱為極值．9．函數(shù)的最大(小)值(1)函數(shù)f(x)在區(qū)間[a，b]上有最值的條件：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y＝f(x)在區(qū)間[a，b]上的最大(小)值的步驟：①求函數(shù)y＝f(x)在區(qū)間(a，b)上的極值；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(gè)是最大值，最小的一個(gè)是最小值．【方法技巧】1．(1)處理與切線有關(guān)的參數(shù)問題，關(guān)鍵是根據(jù)曲線、切線、切點(diǎn)的三個(gè)關(guān)系列出參數(shù)的方程：①切點(diǎn)處的導(dǎo)數(shù)是切線的斜率；②切點(diǎn)在切線上；③切點(diǎn)在曲線上．(2)注意區(qū)分“在點(diǎn)P處的切線”與“過點(diǎn)P處的切線”：在“點(diǎn)P處的切線”，說明點(diǎn)P為切點(diǎn)，點(diǎn)P既在曲線上，又在切線上；“過點(diǎn)P處的切線”，說明點(diǎn)P不一定是切點(diǎn)，點(diǎn)P一定在切線上，不一定在曲線上．2．根據(jù)函數(shù)單調(diào)性求參數(shù)的一般思路(1)利用集合間的包含關(guān)系處理：y＝f(x)在(a，b)上單調(diào)，則區(qū)間(a，b)是相應(yīng)單調(diào)區(qū)間的子集．(2)f(x)為增(減)函數(shù)的充要條件是對(duì)任意的x∈(a，b)都有f′(x)≥0(f′(x)≤0)，且在(a，b)內(nèi)的任一非空子區(qū)間上，f′(x)不恒為零，應(yīng)注意此時(shí)式子中的等號(hào)不能省略，否則會(huì)漏解．(3)函數(shù)在某個(gè)區(qū)間上存在單調(diào)區(qū)間可轉(zhuǎn)化為不等式有解問題．3．函數(shù)極值的兩類熱點(diǎn)問題(1)求函數(shù)f(x)極值的一般解題步驟①確定函數(shù)的定義域．②求導(dǎo)數(shù)f′(x)．③解方程f′(x)＝0，求出函數(shù)定義域內(nèi)的所有根．④列表檢驗(yàn)f′(x)在f′(x)＝0的根x0左右兩側(cè)值的符號(hào)．(2)根據(jù)函數(shù)極值情況求參數(shù)的兩個(gè)要領(lǐng)①列式：根據(jù)極值點(diǎn)處導(dǎo)數(shù)為0和極值這兩個(gè)條件列方程組，利用待定系數(shù)法求解．②驗(yàn)證：求解后驗(yàn)證根的合理性．【核心題型】題型一：由函數(shù)的單調(diào)區(qū)間求參數(shù)1．（2022·黑龍江佳木斯·佳木斯一中校考三模）已知函數(shù)SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0在SKIPIF1<0單調(diào)遞增，a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2020·遼寧大連·大連二十四中?？寄M預(yù)測(cè)）已知SKIPIF1<0，若對(duì)于SKIPIF1<0且SKIPIF1<0都SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2019·四川達(dá)州·統(tǒng)考一模）若SKIPIF1<0是SKIPIF1<0上的減函數(shù)，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題型二：由函數(shù)在區(qū)間上單調(diào)性求參數(shù)4．（2022·寧夏吳忠·吳忠中學(xué)?？既＃┤艉瘮?shù)SKIPIF1<0，在定義域內(nèi)任取兩個(gè)不相等的實(shí)數(shù)SKIPIF1<0，不等式SKIPIF1<0恒成立，則實(shí)數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2022·安徽·南陵中學(xué)校聯(lián)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，若當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則實(shí)數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．（2022·黑龍江齊齊哈爾·統(tǒng)考二模）設(shè)函數(shù)SKIPIF1<0，若SKIPIF1<0為SKIPIF1<0上的單調(diào)函數(shù)，則實(shí)數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0題型三：含參數(shù)的分類討論問題7．（2023·全國(guó)·高三專題練習(xí)）若SKIPIF1<0是函數(shù)SKIPIF1<0的極大值點(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<08．（2023秋·四川宜賓·高三四川省宜賓市第四中學(xué)校?？计谀┮阎瘮?shù)SKIPIF1<0，若SKIPIF1<0有四個(gè)不同的零點(diǎn)，則a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<09．（2020·全國(guó)·高三專題練習(xí)）已知不等式ex﹣x﹣1＞m[x﹣ln（x+1）]對(duì)一切正數(shù)x都成立，則實(shí)數(shù)m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．（﹣∞，1] D．（﹣∞，e]題型四：根據(jù)極值（點(diǎn)）求參數(shù)問題10．（2021秋·四川瀘州·高三四川省瀘縣第二中學(xué)?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的圖像上恰有兩對(duì)關(guān)于x軸對(duì)稱的點(diǎn)，則實(shí)數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<011．（2022·陜西咸陽·武功縣普集高級(jí)中學(xué)統(tǒng)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0存在極大值點(diǎn)和極小值點(diǎn)，則實(shí)數(shù)SKIPIF1<0可以取的一個(gè)值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<012．（2022·陜西西安·西安中學(xué)?？级＃┮阎瘮?shù)SKIPIF1<0有兩個(gè)極值點(diǎn)SKIPIF1<0，若SKIPIF1<0，則關(guān)于SKIPIF1<0的方程SKIPIF1<0的不同實(shí)根個(gè)數(shù)為（

）A．2 B．3 C．4 D．5題型五：由導(dǎo)數(shù)求函數(shù)的最值問題13．（2022·安徽·巢湖市第一中學(xué)校聯(lián)考模擬預(yù)測(cè)）已知不等式SKIPIF1<0對(duì)SKIPIF1<0恒成立，則實(shí)數(shù)a的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．114．（2022秋·湖南郴州·高三?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0若方程SKIPIF1<0恰有3個(gè)不同的實(shí)根，則實(shí)數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<015．（2021秋·河南駐馬店·高三校考階段練習(xí)）已知函數(shù)SKIPIF1<0，對(duì)任意SKIPIF1<0，不等式SKIPIF1<0恒成立，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0題型六：由函數(shù)最值求參數(shù)問題16．（2022·浙江·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，若SKIPIF1<0恒成立，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<017．（2022·遼寧丹東·統(tǒng)考一模）設(shè)SKIPIF1<0，若函數(shù)SKIPIF1<0的最小值為SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<018．（2022秋·河南洛陽·高三校聯(lián)考階段練習(xí)）設(shè)函數(shù)SKIPIF1<0，若SKIPIF1<0，且SKIPIF1<0的最小值為SKIPIF1<0，則a的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0題型七：函數(shù)的單調(diào)性極值和最值問題綜合19．（2023·全國(guó)·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)求函數(shù)SKIPIF1<0的最值；(2)若關(guān)于x的不等式SKIPIF1<0恒成立，求實(shí)數(shù)k的取值范圍．20．（2023·陜西·西安市西光中學(xué)校聯(lián)考一模）已知函數(shù)SKIPIF1<0，其中SKIPIF1<0為常數(shù)，SKIPIF1<0為自然對(duì)數(shù)的底數(shù).(1)當(dāng)SKIPIF1<0時(shí)，求SKIPIF1<0的單調(diào)區(qū)間；(2)若SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值為SKIPIF1<0，求SKIPIF1<0的值.21．（2023·廣東廣州·統(tǒng)考二模）已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0.(1)若SKIPIF1<0，討論SKIPIF1<0的單調(diào)性；(2)若SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，不等式SKIPIF1<0恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍.【高考必刷】一、單選題22．（2023·云南昆明·昆明一中?？寄M預(yù)測(cè)）函數(shù)SKIPIF1<0，則滿足不等式SKIPIF1<0的實(shí)數(shù)x的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<023．（2023·遼寧·遼寧實(shí)驗(yàn)中學(xué)?？寄M預(yù)測(cè)）已知函數(shù)f（x）為定義在R上的偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<024．（2023·甘肅蘭州·校考一模）已知函數(shù)SKIPIF1<0的極值點(diǎn)為SKIPIF1<0，函數(shù)SKIPIF1<0的最大值為SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<025．（2023·內(nèi)蒙古赤峰·統(tǒng)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0存在唯一的極值點(diǎn)，則實(shí)數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<026．（2023·全國(guó)·模擬預(yù)測(cè)）函數(shù)SKIPIF1<0恰有3個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<027．（2023·吉林·長(zhǎng)春十一高校聯(lián)考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）在區(qū)間SKIPIF1<0上總存在零點(diǎn)，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<028．（2022秋·新疆·高三校聯(lián)考階段練習(xí)）已知函數(shù)SKIPIF1<0對(duì)SKIPIF1<0均滿足SKIPIF1<0，其中SKIPIF1<0是SKIPIF1<0的導(dǎo)數(shù)，則下列不等式恒成立的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<029．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0有四個(gè)不同的零點(diǎn)，從小到大依次為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、多選題30．（2023·全國(guó)·唐山市第十一中學(xué)?？寄M預(yù)測(cè)）已知SKIPIF1<0存在兩個(gè)極小值點(diǎn),則SKIPIF1<0的取值可以是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<031．（2023·遼寧·遼寧實(shí)驗(yàn)中學(xué)校考模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0在SKIPIF1<0處有極值，且極值為8，則（

）A．SKIPIF1<0有三個(gè)零點(diǎn)B．SKIPIF1<0C．曲線SKIPIF1<0在點(diǎn)SKIPIF1<0處的切線方程為SKIPIF1<0D．函數(shù)SKIPIF1<0為奇函數(shù)32．（2023·湖北·宜昌市一中校聯(lián)考模擬預(yù)測(cè)）已知SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<033．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，則下列結(jié)論正確的是（

）A．當(dāng)m＞0時(shí)，函數(shù)SKIPIF1<0的圖象在點(diǎn)SKIPIF1<0處的切線的斜率為SKIPIF1<0B．當(dāng)m＝l時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減C．當(dāng)m＝l時(shí)，函數(shù)SKIPIF1<0的最小值為1D．若SKIPIF1<0對(duì)SKIPIF1<0恒成立，則SKIPIF1<034．（2023·全國(guó)·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，SKIPIF1<0，則下列說法正確的是（

）A．SKIPIF1<0在SKIPIF1<0上是增函數(shù)B．SKIPIF1<0，不等式SKIPIF1<0恒成立，則正實(shí)數(shù)SKIPIF1<0的最小值為SKIPIF1<0C．若SKIPIF1<0有兩個(gè)零點(diǎn)SKIPIF1<0，則SKIPIF1<0D．若SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最大值為SKIPIF1<0三、填空題35．（2023·全國(guó)·模擬預(yù)測(cè)）已知SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上的最小值為2，則實(shí)數(shù)SKIPIF1<0__________.36．（2023·遼寧·校聯(lián)考模擬預(yù)測(cè)）已知SKIPIF1<0和SKIPIF1