2025年新高考數(shù)學(xué)一輪復(fù)習(xí)講義：第三章一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用（學(xué)生版）

§3.1導(dǎo)數(shù)的概念及其意義、導(dǎo)數(shù)的運(yùn)算課標(biāo)要求1.了解導(dǎo)數(shù)的概念、掌握基本初等函數(shù)的導(dǎo)數(shù).2.通過(guò)函數(shù)圖象，理解導(dǎo)數(shù)的幾何意義.3.能夠用導(dǎo)數(shù)公式和導(dǎo)數(shù)的運(yùn)算法則求簡(jiǎn)單函數(shù)的導(dǎo)數(shù)，能求簡(jiǎn)單的復(fù)合函數(shù)的導(dǎo)數(shù)．知識(shí)梳理1．導(dǎo)數(shù)的概念(1)函數(shù)y＝f(x)在x＝x0處的導(dǎo)數(shù)記作________或____________．f′(x0)＝eq\o(lim,\s\do4(Δx→0))eq\f(Δy,Δx)＝__________________________________.(2)函數(shù)y＝f(x)的導(dǎo)函數(shù)(簡(jiǎn)稱導(dǎo)數(shù))f′(x)＝y(tǒng)′＝eq\o(lim,\s\do4(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________，相應(yīng)的切線方程為________________________．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)＝__________f(x)＝xα(α∈R，且α≠0)f′(x)＝__________f(x)＝sinxf′(x)＝__________f(x)＝cosxf′(x)＝__________f(x)＝ax(a>0，且a≠1)f′(x)＝__________f(x)＝exf′(x)＝__________f(x)＝logax(a>0，且a≠1)f′(x)＝__________f(x)＝lnxf′(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)]′＝________________________；eq\b\lc\[\rc\](\a\vs4\al\co1(\f(fx,gx)))′＝eq\f(f′xgx－fxg′x,[gx]2)(g(x)≠0)；[cf(x)]′＝____________.5．復(fù)合函數(shù)的定義及其導(dǎo)數(shù)復(fù)合函數(shù)y＝f(g(x))的導(dǎo)數(shù)與函數(shù)y＝f(u)，u＝g(x)的導(dǎo)數(shù)間的關(guān)系為y′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．常用結(jié)論1．在點(diǎn)處的切線與過(guò)點(diǎn)的切線的區(qū)別(1)在點(diǎn)處的切線，該點(diǎn)一定是切點(diǎn)，切線有且僅有一條．(2)過(guò)點(diǎn)的切線，該點(diǎn)不一定是切點(diǎn)，切線至少有一條．2.eq\b\lc\[\rc\](\a\vs4\al\co1(\f(1,fx)))′＝eq\f(－f′x,[fx]2)(f(x)≠0)．自主診斷1．判斷下列結(jié)論是否正確．(請(qǐng)?jiān)诶ㄌ?hào)中打“√”或“×”)(1)f′(x0)是函數(shù)y＝f(x)在x＝x0附近的平均變化率．()(2)與曲線只有一個(gè)公共點(diǎn)的直線一定是曲線的切線．()(3)f′(x0)＝[f(x0)]′.()(4)(e－x)′＝－e－x.()2．若函數(shù)f(x)＝3x＋sin2x，則()A．f′(x)＝3xln3＋2cos2xB．f′(x)＝3x＋2cos2xC．f′(x)＝eq\f(3x,ln3)＋cos2xD．f′(x)＝eq\f(3x,ln3)－2cos2x3．(選擇性必修第二冊(cè)P70T7改編)曲線y＝eq\f(1,2)x2－2在點(diǎn)eq\b\lc\(\rc\)(\a\vs4\al\co1(1，－\f(3,2)))處的切線的傾斜角是________．4．(選擇性必修第二冊(cè)P82T11改編)設(shè)曲線y＝e2ax在點(diǎn)(0,1)處的切線與直線2x－y＋1＝0垂直，則a的值為________.題型一導(dǎo)數(shù)的運(yùn)算例1(1)(多選)下列求導(dǎo)正確的是()A．[(3x＋5)3]′＝9(3x＋5)2B．(x3lnx)′＝3x2lnx＋x2C.eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(2sinx,x2)))′＝eq\f(2xcosx＋4sinx,x3)D．(ln2x)′＝eq\f(1,2x)(2)(2023·河南聯(lián)考)已知函數(shù)f(x)滿足f(x)＝2f′(1)lnx＋eq\f(x,e)(f′(x)為f(x)的導(dǎo)函數(shù))，則f(e)等于()A．e－1 B.eq\f(2,e)＋1C．1 D．－eq\f(2,e)＋1跟蹤訓(xùn)練1(多選)下列命題正確的是()A．若f(x)＝xsinx－cosx，則f′(x)＝sinx－xcosx＋sinxB．設(shè)函數(shù)f(x)＝xlnx，若f′(x0)＝2，則x0＝eC．已知函數(shù)f(x)＝3x2ex，則f′(1)＝12eD．設(shè)函數(shù)f(x)的導(dǎo)函數(shù)為f′(x)，且f(x)＝x2＋3xf′(2)＋lnx，則f′(2)＝－eq\f(9,4)題型二導(dǎo)數(shù)的幾何意義命題點(diǎn)1求切線方程例2(1)(2023·全國(guó)甲卷)曲線y＝eq\f(ex,x＋1)在點(diǎn)eq\b\lc\(\rc\)(\a\vs4\al\co1(1，\f(e,2)))處的切線方程為()A．y＝eq\f(e,4)x B．y＝eq\f(e,2)xC．y＝eq\f(e,4)x＋eq\f(e,4) D．y＝eq\f(e,2)x＋eq\f(3e,4)(2)(2022·新高考全國(guó)Ⅱ)曲線y＝ln|x|過(guò)坐標(biāo)原點(diǎn)的兩條切線的方程為________________，__________________.命題點(diǎn)2求參數(shù)的值(范圍)例3(1)(2024·瀘州模擬)若直線y＝kx＋1為曲線y＝lnx的一條切線，則實(shí)數(shù)k的值是()A．eB．e2C.eq\f(1,e)D.eq\f(1,e2)(2)(2022·新高考全國(guó)Ⅰ)若曲線y＝(x＋a)ex有兩條過(guò)坐標(biāo)原點(diǎn)的切線，則a的取值范圍是____________________．思維升華(1)處理與切線有關(guān)的問(wèn)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(guò)點(diǎn)P的切線”．跟蹤訓(xùn)練2(1)(2023·深圳質(zhì)檢)已知f(x)為偶函數(shù)，當(dāng)x<0時(shí)，f(x)＝x3－x，則曲線y＝f(x)在點(diǎn)(1,0)處的切線方程是()A．2x－y－2＝0 B．4x－y－4＝0C．2x＋y－2＝0 D．4x＋y－4＝0(2)若函數(shù)f(x)＝x－eq\f(1,x)＋alnx存在與x軸平行的切線，則實(shí)數(shù)a的取值范圍是________．題型三兩曲線的公切線例4(1)(2024·青島模擬)已知定義在區(qū)間(0，＋∞)上的函數(shù)f(x)＝－2x2＋m，g(x)＝－3lnx－x，若以上兩函數(shù)的圖象有公共點(diǎn)，且在公共點(diǎn)處切線相同，則m的值為()A．2B．5C．1D．0(2)若兩曲線y＝lnx－1與y＝ax2存在公切線，則正實(shí)數(shù)a的取值范圍是()A．(0,2e] B.eq\b\lc\[\rc\)(\a\vs4\al\co1(\f(1,2)e－3，＋∞))C.eq\b\lc\(\rc\](\a\vs4\al\co1(0，\f(1,2)e－3)) D．[2e，＋∞)__________________________跟蹤訓(xùn)練3(1)(2023·青島模擬)若曲線C1：f(x)＝x2＋a和曲線C2：g(x)＝4lnx－2x存在有公共切點(diǎn)的公切線，則a＝________.(2)已知f(x)＝ex－1，g(x)＝lnx＋1，則f(x)與g(x)的公切線有()A．0條B．1條C．2條D．3條

§3.2導(dǎo)數(shù)與函數(shù)的單調(diào)性課標(biāo)要求1.結(jié)合實(shí)例，借助幾何直觀了解函數(shù)的單調(diào)性與導(dǎo)數(shù)的關(guān)系.2.能利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性，會(huì)求函數(shù)的單調(diào)區(qū)間(其中多項(xiàng)式函數(shù)一般不超過(guò)三次).3.會(huì)利用函數(shù)的單調(diào)性判斷大小，求參數(shù)的取值范圍等簡(jiǎn)單應(yīng)用．知識(shí)梳理1．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)在區(qū)間(a，b)上____________________f′(x)<0f(x)在區(qū)間(a，b)上____________________f′(x)＝0f(x)在區(qū)間(a，b)上是____________________2.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)f(x)的______________；第2步，求出導(dǎo)數(shù)f′(x)的______________；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．常用結(jié)論1．若函數(shù)f(x)在(a，b)上單調(diào)遞增，則當(dāng)x∈(a，b)時(shí)，f′(x)≥0恒成立；若函數(shù)f(x)在(a，b)上單調(diào)遞減，則當(dāng)x∈(a，b)時(shí)，f′(x)≤0恒成立．2．若函數(shù)f(x)在(a，b)上存在單調(diào)遞增區(qū)間，則當(dāng)x∈(a，b)時(shí)，f′(x)>0有解；若函數(shù)f(x)在(a，b)上存在單調(diào)遞減區(qū)間，則當(dāng)x∈(a，b)時(shí)，f′(x)<0有解．自主診斷1．判斷下列結(jié)論是否正確．(請(qǐng)?jiān)诶ㄌ?hào)中打“√”或“×”)(1)如果函數(shù)f(x)在某個(gè)區(qū)間內(nèi)恒有f′(x)＝0，則f(x)在此區(qū)間內(nèi)沒有單調(diào)性．()(2)在(a，b)內(nèi)f′(x)≤0且f′(x)＝0的根有有限個(gè)，則f(x)在(a，b)內(nèi)單調(diào)遞減．()(3)若函數(shù)f(x)在定義域上都有f′(x)>0，則f(x)在定義域上一定單調(diào)遞增．()(4)函數(shù)f(x)＝x－sinx在R上是增函數(shù)．()2.(選擇性必修第二冊(cè)P86例2改編)(多選)如圖是函數(shù)y＝f(x)的導(dǎo)函數(shù)y＝f′(x)的圖象，則下列判斷正確的是()A．在區(qū)間(－2,1)上f(x)單調(diào)遞增B．在區(qū)間(2,3)上f(x)單調(diào)遞減C．在區(qū)間(4,5)上f(x)單調(diào)遞增D．在區(qū)間(3,5)上f(x)單調(diào)遞減3．(選擇性必修第二冊(cè)P97習(xí)題5.3T2(4)改編)已知f(x)＝x3＋x2－x的單調(diào)遞增區(qū)間為________________________．4．已知f(x)＝2x2－ax＋lnx在區(qū)間(1，＋∞)上單調(diào)遞增，則實(shí)數(shù)a的取值范圍是________________________________________________________________________．題型一不含參函數(shù)的單調(diào)性例1(1)函數(shù)f(x)＝xlnx－3x＋2的單調(diào)遞減區(qū)間為________________．(2)若函數(shù)f(x)＝eq\f(lnx＋1,ex)，則函數(shù)f(x)的單調(diào)遞增區(qū)間為________________．跟蹤訓(xùn)練1已知函數(shù)f(x)＝xsinx＋cosx，x∈[0,2π]，則f(x)的單調(diào)遞減區(qū)間為()A.eq\b\lc\(\rc\)(\a\vs4\al\co1(0，\f(π,2))) B.eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)，\f(3π,2)))C．(π，2π) D.eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)，2π))題型二含參數(shù)的函數(shù)的單調(diào)性例2已知函數(shù)g(x)＝(x－a－1)ex－(x－a)2，討論函數(shù)g(x)的單調(diào)性．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________思維升華(1)研究含參數(shù)的函數(shù)的單調(diào)性，要依據(jù)參數(shù)對(duì)不等式解集的影響進(jìn)行分類討論．(2)劃分函數(shù)的單調(diào)區(qū)間時(shí)，要在函數(shù)定義域內(nèi)討論，還要確定導(dǎo)數(shù)為零的點(diǎn)和函數(shù)的間斷點(diǎn)．跟蹤訓(xùn)練2(2023·北京模擬)已知函數(shù)f(x)＝eq\f(2x－a,x＋12).(1)當(dāng)a＝0時(shí)，求曲線y＝f(x)在點(diǎn)(0，f(0))處的切線方程；(2)求函數(shù)f(x)的單調(diào)區(qū)間．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________題型三函數(shù)單調(diào)性的應(yīng)用命題點(diǎn)1比較大小或解不等式例3(1)(多選)(2024·深圳模擬)若0<x1<x2<1，則()A．>lneq\f(x2＋1,x1＋1)B．<lneq\f(x2＋1,x1＋1)C．D．常見組合函數(shù)的圖象在導(dǎo)數(shù)的應(yīng)用中常用到以下函數(shù)，記住以下的函數(shù)圖象對(duì)解題有事半功倍的效果．典例(多選)如果函數(shù)f(x)對(duì)定義域內(nèi)的任意兩實(shí)數(shù)x1，x2(x1≠x2)都有eq\f(x1fx1－x2fx2,x1－x2)>0，則稱函數(shù)y＝f(x)為“F函數(shù)”．下列函數(shù)不是“F函數(shù)”的是()A．f(x)＝ex B．f(x)＝x2C．f(x)＝lnx D．f(x)＝sinx(2)(2023·成都模擬)已知函數(shù)f(x)＝ex－e－x－2x＋1，則不等式f(2x－3)＋f(x)>2的解集為____________________．命題點(diǎn)2根據(jù)函數(shù)的單調(diào)性求參數(shù)例4已知函數(shù)f(x)＝lnx－eq\f(1,2)ax2－2x(a≠0)．(1)若f(x)在[1,4]上單調(diào)遞減，求實(shí)數(shù)a的取值范圍；________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2)若f(x)在[1,4]上存在單調(diào)遞減區(qū)間，求實(shí)數(shù)a的取值范圍．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________跟蹤訓(xùn)練3(1)(2024·鄭州模擬)函數(shù)f(x)的圖象如圖所示，設(shè)f(x)的導(dǎo)函數(shù)為f′(x)，則f(x)·f′(x)>0的解集為()A．(1,6) B．(1,4)C．(－∞，1)∪(6，＋∞) D．(1,4)∪(6，＋∞)(2)已知函數(shù)f(x)＝(1－x)lnx＋ax在(1，＋∞)上不單調(diào)，則a的取值范圍是()A．(0，＋∞) B．(1，＋∞)C．[0，＋∞) D．[1，＋∞)

§3.3導(dǎo)數(shù)與函數(shù)的極值、最值課標(biāo)要求1.借助函數(shù)圖象，了解函數(shù)在某點(diǎn)取得極值的必要和充分條件.2.會(huì)用導(dǎo)數(shù)求函數(shù)的極大值、極小值.3.掌握利用導(dǎo)數(shù)研究函數(shù)最值的方法.4.會(huì)用導(dǎo)數(shù)研究生活中的最優(yōu)化問(wèn)題．知識(shí)梳理1．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處的函數(shù)值f(a)比它在點(diǎn)x＝a附近其他點(diǎn)處的函數(shù)值都小，f′(a)＝0；而且在點(diǎn)x＝a附近的左側(cè)____________，右側(cè)____________，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處的函數(shù)值f(b)比它在點(diǎn)x＝b附近其他點(diǎn)處的函數(shù)值都大，f′(b)＝0；而且在點(diǎn)x＝b附近的左側(cè)____________，右側(cè)____________，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________，極小值和極大值統(tǒng)稱為________．2．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____________的曲線，那么它必有最大值和最小值．(2)求函數(shù)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)內(nèi)的_______________________________________________；②將函數(shù)y＝f(x)的各極值與________________________________比較，其中最大的一個(gè)是最大值，最小的一個(gè)是最小值．常用結(jié)論對(duì)于可導(dǎo)函數(shù)f(x)，“f′(x0)＝0”是“函數(shù)f(x)在x＝x0處有極值”的必要不充分條件．自主診斷1．判斷下列結(jié)論是否正確．(請(qǐng)?jiān)诶ㄌ?hào)中打“√”或“×”)(1)函數(shù)的極值可能不止一個(gè)，也可能沒有．()(2)函數(shù)的極小值一定小于函數(shù)的極大值．()(3)函數(shù)的極小值一定是函數(shù)的最小值．()(4)函數(shù)的極大值一定不是函數(shù)的最小值．()2.(選擇性必修第二冊(cè)P98T4改編)如圖是f(x)的導(dǎo)函數(shù)f′(x)的圖象，則f(x)的極小值點(diǎn)的個(gè)數(shù)為()A．1B．2C．3D．43．若函數(shù)f(x)＝x3－ax2＋2x－1有兩個(gè)極值點(diǎn)，則實(shí)數(shù)a的取值范圍是___________________．4．(選擇性必修第二冊(cè)P93例6改編)函數(shù)f(x)＝eq\f(1,3)x3－4x＋4在區(qū)間[0,3]上的最大值是________，最小值是________．題型一利用導(dǎo)數(shù)求解函數(shù)的極值問(wèn)題命題點(diǎn)1根據(jù)函數(shù)圖象判斷極值例1(多選)(2023·連云港模擬)如圖是函數(shù)y＝f(x)的導(dǎo)函數(shù)f′(x)的圖象，下列說(shuō)法正確的是()A．f(1)為函數(shù)f(x)的極大值B．當(dāng)x＝－1時(shí)，f(x)取得極小值C．f(x)在(－1,2)上單調(diào)遞增，在(2,4)上單調(diào)遞減D．當(dāng)x＝3時(shí)，f(x)取得極小值命題點(diǎn)2求已知函數(shù)的極值例2設(shè)函數(shù)f(x)＝(x2＋ax＋a)ex，討論f(x)的單調(diào)性并判斷f(x)有無(wú)極值，若有極值，求出f(x)的極值．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________命題點(diǎn)3已知極值(點(diǎn))求參數(shù)例3(1)(2024·成都模擬)若函數(shù)f(x)＝x(x＋a)2在x＝1處有極大值，則實(shí)數(shù)a的值為()A．1 B．－1或－3C．－1 D．－3(2)(2023·威海模擬)若函數(shù)f(x)＝ex－ax2－2ax有兩個(gè)極值點(diǎn)，則實(shí)數(shù)a的取值范圍為()A.eq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(1,2)，0)) B.eq\b\lc\(\rc\)(\a\vs4\al\co1(－∞，－\f(1,2)))C.eq\b\lc\(\rc\)(\a\vs4\al\co1(0，\f(1,2))) D.eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,2)，＋∞))跟蹤訓(xùn)練1(1)已知函數(shù)f(x)＝x3＋ax2＋bx－a2－7a在x＝1處取得極大值10，則a＋b的值為()A．－1或3 B．1或－3C．3 D．－1(2)(2023·商丘模擬)已知函數(shù)f(x)＝x2－aln(2x＋1)在定義域內(nèi)不存在極值點(diǎn)，則實(shí)數(shù)a的取值范圍是__________________．題型二利用導(dǎo)數(shù)求函數(shù)的最值問(wèn)題命題點(diǎn)1不含參函數(shù)的最值例4(2022·全國(guó)乙卷)函數(shù)f(x)＝cosx＋(x＋1)sinx＋1在區(qū)間[0,2π]的最小值、最大值分別為()A．－eq\f(π,2)，eq\f(π,2) B．－eq\f(3π,2)，eq\f(π,2)C．－eq\f(π,2)，eq\f(π,2)＋2 D．－eq\f(3π,2)，eq\f(π,2)＋2命題點(diǎn)2含參函數(shù)的最值例5已知函數(shù)f(x)＝ax2－(a＋2)x＋lnx．當(dāng)a>0時(shí)，f(x)在區(qū)間[1，e]上的最小值為－2，求實(shí)數(shù)a的取值范圍．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________思維升華求含有參數(shù)的函數(shù)的最值，需先求函數(shù)的定義域、導(dǎo)函數(shù)，通過(guò)對(duì)參數(shù)分類討論，判斷函數(shù)的單調(diào)性，從而得到函數(shù)f(x)的最值．跟蹤訓(xùn)練2(1)(2021·新高考全國(guó)Ⅰ)函數(shù)f(x)＝|2x－1|－2lnx的最小值為________．(2)(2024·上饒模擬)已知函數(shù)f(x)＝lnx＋ax2＋1.當(dāng)0<x≤e2時(shí)，g(x)＝f(x)－ax2－3＋eq\f(a,x)有最小值2，求a的值．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

§3.4函數(shù)中的構(gòu)造問(wèn)題重點(diǎn)解讀函數(shù)中的構(gòu)造問(wèn)題是高考考查的一個(gè)熱點(diǎn)內(nèi)容，經(jīng)常以客觀題出現(xiàn)，同構(gòu)法構(gòu)造函數(shù)也在解答題中出現(xiàn)，通過(guò)已知等式或不等式的結(jié)構(gòu)特征，構(gòu)造新函數(shù)，解決比較大小、解不等式、恒成立等問(wèn)題．題型一利用f(x)與x構(gòu)造函數(shù)例1(2023·信陽(yáng)統(tǒng)考)已知f(x)是定義在R上的偶函數(shù)，當(dāng)x>0時(shí)，xf′(x)－f(x)<0，且f(－2)＝0，則不等式eq\f(fx,x)>0的解集是()A．(－2,0)∪(0,2)B．(－∞，－2)∪(2，＋∞)C．(－2,0)∪(2，＋∞)D．(－∞，－2)∪(0,2)思維升華(1)出現(xiàn)nf(x)＋xf′(x)形式，構(gòu)造函數(shù)F(x)＝xnf(x)．(2)出現(xiàn)xf′(x)－nf(x)形式，構(gòu)造函數(shù)F(x)＝eq\f(fx,xn).跟蹤訓(xùn)練1(多選)(2023·郴州統(tǒng)考)已知函數(shù)f(x)是定義在R上的奇函數(shù)，當(dāng)x>0時(shí)，xf′(x)＋2f(x)>0恒成立，則()A．f(1)<4f(2) B．f(－1)<4f(－2)C．16f(4)<9f(3) D．4f(－2)>9f(－3)題型二利用f(x)與ex構(gòu)造函數(shù)例2(2024·吉安模擬)已知定義在R上的函數(shù)f(x)滿足f(x)<f′(x)－2，則()A．f(2023)－ef(2022)<2(e－1)B．f(2023)－ef(2022)>2(e－1)C．f(2023)－ef(2022)>2(e＋1)D．f(2023)－ef(2022)<2(e＋1)跟蹤訓(xùn)練2(2023·南昌模擬)已知定義在R上的函數(shù)f(x)滿足f(x)＋f′(x)>0，且有f(3)＝3，則f(x)>3e3－x的解集為________．題型三利用f(x)與sinx，cosx構(gòu)造函數(shù)例3設(shè)f(x)是定義在(－π，0)∪(0，π)上的奇函數(shù)，其導(dǎo)函數(shù)為f′(x)，且當(dāng)x∈(0，π)時(shí)，f′(x)sinx－f(x)cosx<0，則關(guān)于x的不等式f(x)<2f

eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,6)))sinx的解集為_____________________________．跟蹤訓(xùn)練3已知定義在R上的奇函數(shù)f(x)，其導(dǎo)函數(shù)為f′(x)，且當(dāng)x∈(0，＋∞)時(shí)，f′(x)sinx＋f(x)cosx<0，若a＝eq\f(\r(2),2)f

eq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(π,6)))，b＝－f

eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,4)))，則a與b的大小關(guān)系為________．(用“<”連接)

§3.5利用導(dǎo)數(shù)研究恒(能)成立問(wèn)題課標(biāo)要求恒(能)成立問(wèn)題是高考的?？伎键c(diǎn)，其中不等式的恒(能)成立問(wèn)題經(jīng)常與導(dǎo)數(shù)及其幾何意義、函數(shù)、方程等相交匯，綜合考查分析問(wèn)題、解決問(wèn)題的能力，一般作為壓軸題出現(xiàn)，試題難度略大．題型一分離參數(shù)求參數(shù)范圍例1已知函數(shù)f(x)＝ex－ax－1.(1)當(dāng)a＝1時(shí)，求f(x)的單調(diào)區(qū)間與極值；________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2)若f(x)≤x2在(0，＋∞)上有解，求實(shí)數(shù)a的取值范圍．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________跟蹤訓(xùn)練1已知函數(shù)f(x)＝ax－ex(a∈R)，g(x)＝eq\f(lnx,x).(1)當(dāng)a＝1時(shí)，求函數(shù)f(x)的極值；(2)若存在x∈(0，＋∞)，使不等式f(x)≤g(x)－ex成立，求實(shí)數(shù)a的取值范圍．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________題型二等價(jià)轉(zhuǎn)化求參數(shù)范圍例2(2023·柳州模擬)已知函數(shù)f(x)＝ax－lnx.(1)討論函數(shù)f(x)的單調(diào)性；(2)若x＝1為函數(shù)f(x)的極值點(diǎn)，當(dāng)x∈[e，＋∞)時(shí)，不等式x[f(x)－x＋1]≤m(e－x)恒成立，求實(shí)數(shù)m的取值范圍．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________跟蹤訓(xùn)練2(2024·咸陽(yáng)模擬)已知函數(shù)f(x)＝lnx＋x＋eq\f(2,ax)(a≠0)．(1)當(dāng)a＝1時(shí)，求f(x)的極值；(2)若對(duì)?x∈(e－1，e)，f(x)<x＋2，求實(shí)數(shù)a的取值范圍．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________題型三雙變量的恒(能)成立問(wèn)題例3(2023·濟(jì)南模擬)已知函數(shù)f(x)＝eq\f(2elnx,x)－1(其中e為自然對(duì)數(shù)的底數(shù))，函數(shù)g(x)＝x3＋ax2＋1.(1)求曲線y＝f(x)在點(diǎn)(1，f(1))處的切線方程；(2)若對(duì)?x1，x2∈[1，e]，不等式f(x1)≤g(x2)恒成立，求實(shí)數(shù)a的取值范圍．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________思維升華“雙變量”的恒(能)成立問(wèn)題一定要正確理解其實(shí)質(zhì)，深刻挖掘內(nèi)含條件，進(jìn)行等價(jià)變換，常見的等價(jià)變換有對(duì)于某一區(qū)間I(1)?x1，x2∈I，f(x1)>g(x2)?f(x)min>g(x)max.(2)?x1∈I1，?x2∈I2，f(x1)>g(x2)?f(x)min>g(x)min.(3)?x1∈I1，?x2∈I2，f(x1)>g(x2)?f(x)max>g(x)max.跟蹤訓(xùn)練3已知函數(shù)f(x)＝eq\f(ax2－x－1,ex)(x∈R)，a為正實(shí)數(shù)．(1)求函數(shù)f(x)的單調(diào)區(qū)間；________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2)若?x1，x2∈[0,4]，不等式|f(x1)－f(x2)|<1恒成立，求實(shí)數(shù)a的取值范圍．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

§3.6利用導(dǎo)數(shù)證明不等式課標(biāo)要求導(dǎo)數(shù)中的不等式證明是高考的常考題型，常與函數(shù)的性質(zhì)、函數(shù)的零點(diǎn)與極值、數(shù)列等相結(jié)合，雖然題目難度較大，但是解題方法多種多樣，如構(gòu)造函數(shù)法、放縮法等，針對(duì)不同的題目，靈活采用不同的解題方法，可以達(dá)到事半功倍的效果．題型一將不等式轉(zhuǎn)化為函數(shù)的最值問(wèn)題例1(12分)(2023·新高考全國(guó)Ⅰ)已知函數(shù)f(x)＝a(ex＋a)－x.(1)討論f(x)的單調(diào)性；[切入點(diǎn)：求導(dǎo)，討論a的正負(fù)](2)證明：當(dāng)a>0時(shí)，f(x)>2lna＋eq\f(3,2).[方法一關(guān)鍵點(diǎn)：作差法比較f(x)min與2lna＋eq\f(3,2)的大小][方法二關(guān)鍵點(diǎn)：利用不等式ex≥x＋1把函數(shù)f(x)中的指數(shù)換成一次函數(shù)][思路分析](1)求f′(x)→分a>0，a≤0判斷f′(x)的符號(hào)→f(x)的單調(diào)性(2)方法一：求f(x)min→構(gòu)造函數(shù)g(a)＝f(x)min－eq\b\lc\(\rc\)(\a\vs4\al\co1(2lna＋\f(3,2)))→求g(a)最小值方法二：證明不等式ex≥x＋1→aex＝ex＋lna≥x＋lna＋1→f(x)≥a2＋lna＋1→構(gòu)造函數(shù)g(a)＝a2＋lna＋1－eq\b\lc\(\rc\)(\a\vs4\al\co1(2lna＋\f(3,2)))→求g(a)最小值思維升華待證不等式的兩邊含有同一個(gè)變量時(shí)，一般地，可以直接構(gòu)造“左減右”的函數(shù)，有時(shí)對(duì)復(fù)雜的式子要進(jìn)行變形，利用導(dǎo)數(shù)研究其單調(diào)性和最值，借助所構(gòu)造函數(shù)的單調(diào)性和最值即可得證．(1)解因?yàn)閒(x)＝a(ex＋a)－x，定義域?yàn)镽，所以f′(x)＝aex－1，(1分)當(dāng)a≤0時(shí)，由于ex>0，則aex≤0，故f′x＝aex－1<0恒成立，①處判斷f′(x)的符號(hào)所以f(x)是減函數(shù)；(2分)當(dāng)a>0時(shí)，令f′(x)＝aex－1＝0，解得x＝－lna，eq\x(\a\al\vs4\co1(當(dāng)x<－lna時(shí)，f′x<0，,則fx在－∞，－lna上單調(diào)遞減；,當(dāng)x>－lna時(shí)，f′x>0，,則fx在－lna，＋∞上單調(diào)遞增.))(4分)②處判斷f′(x)的符號(hào)綜上，當(dāng)a≤0時(shí)，f(x)是減函數(shù)；當(dāng)a>0時(shí)，f(x)在(－∞，－lna)上單調(diào)遞減，在(－lna，＋∞)上單調(diào)遞增．(5分)(2)證明方法一由(1)得，當(dāng)a>0時(shí)，eq\x(\a\al\vs4\co1(fxmin＝f－lna＝ae－lna＋a＋lna,＝1＋a2＋lna，))(7分)③處利用單調(diào)性求f(x)min要證f(x)>2lna＋eq\f(3,2)，即證1＋a2＋lna>2lna＋eq\f(3,2)，即證a2－eq\f(1,2)－lna>0恒成立，(8分)eq\x(\a\al\vs4\co1(令g(a)＝a2－\f(1,2)－lna(a>0)，))(9分)④處構(gòu)造函數(shù)ga＝fxmin－eq\b\lc\(\rc\)(\a\vs4\al\co1(2lna＋\f(3,2)))則g′(a)＝2a－eq\f(1,a)＝eq\f(2a2－1,a)，令g′(a)<0，則0<a<eq\f(\r(2),2)；令g′(a)>0，則a>eq\f(\r(2),2)，所以g(a)在eq\b\lc\(\rc\)(\a\vs4\al\co1(0，\f(\r(2),2)))上單調(diào)遞減，在eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(\r(2),2)，＋∞))上單調(diào)遞增，(11分)eq\x(\a\al\vs4\co1(所以g(a)min＝g\b\lc\(\rc\)(\a\vs4\al\co1(\f(\r(2),2)))＝\b\lc\(\rc\)(\a\vs4\al\co1(\f(\r(2),2)))2－\f(1,2)－ln

\f(\r(2),2)＝ln

\r(2)>0，))⑤處求gamin并判斷其符號(hào)則g(a)>0恒成立，所以當(dāng)a>0時(shí)，f(x)>2lna＋eq\f(3,2)恒成立，證畢．(12分)方法二eq\x(令hx＝ex－x－1，)⑥處構(gòu)造函數(shù)證明ex≥x＋1則h′(x)＝ex－1，由于y＝ex是增函數(shù)，所以h′(x)＝ex－1是增函數(shù)，又h′(0)＝e0－1＝0，所以當(dāng)x<0時(shí)，h′(x)<0；當(dāng)x>0時(shí)，h′(x)>0，所以h(x)在(－∞，0)上單調(diào)遞減，在(0，＋∞)上單調(diào)遞增，故h(x)≥h(0)＝0，則ex≥x＋1，當(dāng)且僅當(dāng)x＝0時(shí)，等號(hào)成立，(6分)eq\x(\a\al\vs4\co1(因?yàn)閒x＝aex＋a－x＝aex＋a2－x,＝ex＋lna＋a2－x≥x＋lna＋1＋a2－x，))⑦處通過(guò)不等式ex≥x＋1放縮函數(shù)fx當(dāng)且僅當(dāng)x＋lna＝0，即x＝－lna時(shí)，等號(hào)成立，所以要證f(x)>2lna＋eq\f(3,2)，即證x＋lna＋1＋a2－x>2lna＋eq\f(3,2)，即證a2－eq\f(1,2)－lna>0，(8分)eq\x(\a\al\vs4\co1(令g(a)＝a2－\f(1,2)－lna(a>0)，))(9分)⑧處構(gòu)造函數(shù)ga則g′(a)＝2a－eq\f(1,a)＝eq\f(2a2－1,a)，令g′(a)<0，則0<a<eq\f(\r(2),2)；令g′(a)>0，則a>eq\f(\r(2),2)，所以g(a)在eq\b\lc\(\rc\)(\a\vs4\al\co1(0，\f(\r(2),2)))上單調(diào)遞減，在eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(\r(2),2)，＋∞))上單調(diào)遞增，(11分)所以g(a)min＝geq\b\lc\(\rc\)(\a\vs4\al\co1(\f(\r(2),2)))＝eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(\r(2),2)))2－eq\f(1,2)－lneq\f(\r(2),2)＝lneq\r(2)>0，⑨處求gamin并判斷其符號(hào)則g(a)>0恒成立，所以當(dāng)a>0時(shí)，f(x)>2lna＋eq\f(3,2)恒成立，證畢．(12分)跟蹤訓(xùn)練1(2023·咸陽(yáng)模擬)已知函數(shù)f(x)＝eq\f(sinx,ex)(x∈R)．(1)求f(x)的圖象在點(diǎn)(0，f(0))處的切線方程；(2)求證：當(dāng)x∈[0，π]時(shí)，f(x)≤x.________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________題型二將不等式轉(zhuǎn)化為兩個(gè)函數(shù)的最值進(jìn)行比較例2已知函數(shù)f(x)＝elnx－ax(a∈R)．(1)討論f(x)的單調(diào)性；________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2)當(dāng)a＝e時(shí)，證明：xf(x)－ex＋2ex≤0.________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________思維升華若直接求導(dǎo)比較復(fù)雜或無(wú)從下手時(shí)，可將待證式進(jìn)行變形，構(gòu)造兩個(gè)函數(shù)，從而找到可以傳遞的中間量，達(dá)到證明的目標(biāo)．本例中同時(shí)含lnx與ex，不能直接構(gòu)造函數(shù)，把指數(shù)與對(duì)數(shù)分離兩邊，分別計(jì)算它們的最值，借助最值進(jìn)行證明．跟蹤訓(xùn)練2(2023·合肥模擬)已知函數(shù)f(x)＝ex＋x2－x－1.(1)求f(x)的最小值；(2)證明：ex＋xlnx＋x2－2x>0.________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________題型三雙變量不等式的證明例3已知函數(shù)f(x)＝(a＋1)lnx＋ax2＋1.(1)討論函數(shù)f(x)的單調(diào)性；(2)設(shè)a≤－2，證明：對(duì)任意x1，x2∈(0，＋∞)，|f(x1)－f(x2)|≥4|x1－x2|.________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________跟蹤訓(xùn)練3已知函數(shù)f(x)＝ax＋1(x>0)，g(x)＝lnx－eq\f(a－1,x)＋2a.(1)若a＝eq\f(1,2)，比較函數(shù)f(x)與g(x)的大??；(2)若m>n>0，求證：eq\f(m－n,lnm－lnn)>eq\r(mn).________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

§3.7利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)課標(biāo)要求函數(shù)零點(diǎn)問(wèn)題在高考中占有很重要的地位，主要涉及判斷函數(shù)零點(diǎn)的個(gè)數(shù)或范圍．高考?？疾槿魏瘮?shù)與復(fù)合函數(shù)的零點(diǎn)問(wèn)題，以及函數(shù)零點(diǎn)與其他知識(shí)的交匯問(wèn)題，一般作為解答題的壓軸題出現(xiàn)．題型一利用函數(shù)性質(zhì)研究函數(shù)的零點(diǎn)例1(2023·遼寧實(shí)驗(yàn)中學(xué)模擬)已知函數(shù)f(x)＝excosx.(1)求f(x)在區(qū)間eq\b\lc\(\rc\)(\a\vs4\al\co1(0，\f(π,2)))內(nèi)的極大值；(2)令函數(shù)h(x)＝eq\f(axfx,ex)－1，當(dāng)a>eq\f(4\r(2),π)時(shí)，證明：h(x)在區(qū)間eq\b\lc\(\rc\)(\a\vs4\al\co1(0，\f(π,2)))內(nèi)有且僅有兩個(gè)零點(diǎn)．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________跟蹤訓(xùn)練1(2023·蕪湖模擬)已知函數(shù)f(x)＝ax＋(a－1)lnx＋eq\f(1,x)－2，a∈R.(1)討論f(x)的單調(diào)性；(2)若f(x)只有一個(gè)零點(diǎn)，求a的取值范圍．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________題型二數(shù)形結(jié)合法研究函數(shù)的零點(diǎn)例2(2023·安慶模擬)已知函數(shù)f(x)＝alnx＋bx2e1－x，a，b∈R.e＝2.71828….(1)若曲線y＝f(x)在點(diǎn)(2，f(2))處的切線方程是y＝x＋ln2，求a和b的值；(2)若a＝e，討論導(dǎo)函數(shù)f′(x)的零點(diǎn)個(gè)數(shù)．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________思維升華含參數(shù)的函數(shù)的零點(diǎn)個(gè)數(shù)，可轉(zhuǎn)化為方程解的個(gè)數(shù)，若能分離參數(shù)，則可將參數(shù)分離出來(lái)后，用x表示參數(shù)的函數(shù)，作出該函數(shù)的圖象，根據(jù)圖象特征求參數(shù)的范圍或判斷零點(diǎn)個(gè)數(shù)．跟蹤訓(xùn)練2(2024·廈門模擬)設(shè)函數(shù)f(x)＝lnx－eq\f(1,2)ax2－bx(a，b∈R)．(1)當(dāng)a＝2，b＝1時(shí)，求函數(shù)f(x)的單調(diào)區(qū)間；(2)當(dāng)a＝0，b＝－1時(shí)，方程f(x)＝mx在區(qū)間[1，e2]上有唯一實(shí)數(shù)解，求實(shí)數(shù)m的取值范圍．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________題型三構(gòu)造函數(shù)法研究函數(shù)的零點(diǎn)例3已知函數(shù)f(x)＝ex＋x＋4ln(2－x)．(1)求函數(shù)f(x)的圖象在點(diǎn)(0，f(0))處的切線方程；(2)判斷函數(shù)f(x)的零點(diǎn)個(gè)數(shù)，并說(shuō)明理由．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________跟蹤訓(xùn)練3(2021·全國(guó)甲卷)已知a>0且a≠1，函數(shù)f(x)＝eq\f(xa,ax)(x>0)．(1)當(dāng)a＝2時(shí)，求f(x)的單調(diào)區(qū)間；(2)若曲線y＝f(x)與直線y＝1有且僅有兩個(gè)交點(diǎn)，求a的取值范圍．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

培優(yōu)點(diǎn)2指對(duì)同構(gòu)問(wèn)題把一個(gè)等式或不等式通過(guò)變形，使左右兩邊結(jié)構(gòu)形式完全相同，構(gòu)造函數(shù)，利用函數(shù)的單調(diào)性進(jìn)行處理，找到這個(gè)函數(shù)模型的方法就是同構(gòu)法．同構(gòu)法主要解決含有指數(shù)、對(duì)數(shù)混合的等式或不等式問(wèn)題．題型一同構(gòu)法的理解例1(1)若ea＋a>b＋lnb(a，b為變量)成立，則下列選項(xiàng)正確的是()A．a(chǎn)＞lnb B．a(chǎn)＜lnbC．lna＞b D．lna＜b(2)若關(guān)于a的方程aea－2＝e4和關(guān)于b的方程b(lnb－2)＝e3λ－1(a，b∈R＋)可化為同構(gòu)方程，則ab的值為()A．e8B．eC．ln6D．1思維升華利用恒等式x＝lnex和x＝elnx，通過(guò)冪轉(zhuǎn)指或冪轉(zhuǎn)對(duì)進(jìn)行等價(jià)變形，構(gòu)造函數(shù)，然后由構(gòu)造的函數(shù)的單調(diào)性進(jìn)行研究．跟蹤訓(xùn)練1已知不等式ax＋eax＞ln(bx)＋bx進(jìn)行指對(duì)同構(gòu)時(shí)，可以構(gòu)造的函數(shù)是()A．f(x)＝lnx＋x B．f(x)＝xlnxC．f(x)＝xex D．f(x)＝eq\f(x,ex)題型二同構(gòu)法的應(yīng)用命題點(diǎn)1alna與xex同構(gòu)例2設(shè)實(shí)數(shù)k>0，對(duì)于任意的x>1，不等式kekx≥lnx恒成立，則k的最小值為________．命題點(diǎn)2beb與xlnx同構(gòu)例3(2023·南京模擬)設(shè)a，b都為正數(shù)，e為自然對(duì)數(shù)的底數(shù)，若aea<blnb，則()A．a(chǎn)b>e B．b>eaC．a(chǎn)b<e D．b<ea命題點(diǎn)3eq\f(c,ec)與eq\f(lnx,x)同構(gòu)例4若關(guān)于x的不等式eq\f(x＋lna,ex)－eq\f(alnx,x)>0對(duì)?x∈(0,1)恒成立，則實(shí)數(shù)a的取值范圍為()A.eq\b\lc\(\rc\](\a\vs4\al\co1(－∞，\f(1,e))) B.eq\b\lc\[\rc\)(\a\vs4\al\co1(\f(1,e)，＋∞))C.eq\b\lc\[\rc\)(\a\vs4\al\co1(\f(1,e)，1)) D.eq\b\lc\(\rc\](\a\vs4\al\co1(0，\f(1,e)))命題點(diǎn)4d＋lnd與x＋ex同構(gòu)例5對(duì)于任意的x>0，ex≥(a－1)x＋ln(ax)恒成立，則a的最大值是________．思維升華常見的同構(gòu)函數(shù)有：①f(x)＝eq\f(lnx,x)；②f(x)＝xlnx；③f(x)＝xex；④f(x)＝eq\f(x,ex).其中①④可以借助eq\f(lnx,x)＝eq\f(lnx,elnx)＝eq\f(t,et)，②③可以借助xex＝(lnex)ex＝(lnt)t＝tlnt進(jìn)行指對(duì)互化．跟蹤訓(xùn)練2(1)(2024·武漢模擬)已知a>0，若在(1，＋∞)上存在x使得不等式ex－x≤xa－alnx成立，則a的最小值為________．(2)若對(duì)任意x∈[e，＋∞)，滿足2x3lnx－≥0恒成立，則實(shí)數(shù)m的取值范圍是________________________．1．設(shè)x>0，y>0，若ex＋lny>x＋y，則下列選項(xiàng)正確的是()A．x>y B．x>lnyC．x<y D．x<lny2．若ex－ax≥－x＋ln(ax)，則正實(shí)數(shù)a的取值范圍為()A.eq\b\lc\(\rc\)(\a\vs4\al\co1(0，\f(1,e))) B．(0，e]C.eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,e)，＋∞)) D．(e，＋∞)3．已知函數(shù)f(x)＝x2ex－a(x＋2lnx)有兩個(gè)零點(diǎn)，則a的取值范圍是()A．a(chǎn)≥1 B．a(chǎn)≤2C．a(chǎn)≤e D．a(chǎn)>e4．(多選)若不相等的正數(shù)a，b滿足aa＝bb，則()A．a(chǎn)＞1B．b＜1C．a(chǎn)＋b>eq\f(2,e)D.(n∈N*)5．若?x∈(0，＋∞)，ln2x－eq\f(aex,2)≤lna恒成立，則a的取值范圍為________________________．6．(2024·漳州質(zhì)檢)已知函數(shù)f(x)＝aex＋x＋1.(1)討論f(x)的單調(diào)性