2025版 數(shù)學(xué)《高中全程復(fù)習(xí)方略》（提升版）人教A版第四章 第四節(jié) 第2課時(shí)　導(dǎo)數(shù)的函數(shù)零點(diǎn)問題含答案

1版數(shù)學(xué)《高中全程復(fù)習(xí)方略》（提升版）人教A版第四章第四節(jié)第2課時(shí)導(dǎo)數(shù)的函數(shù)零點(diǎn)問題第2課時(shí)導(dǎo)數(shù)的函數(shù)零點(diǎn)問題【命題分析】函數(shù)零點(diǎn)問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?？疾榛境醯群瘮?shù)、三次函數(shù)與復(fù)合函數(shù)的零點(diǎn)問題,以及函數(shù)零點(diǎn)與其他知識(shí)的交匯問題,一般作為解答題的壓軸題出現(xiàn).【核心考點(diǎn)·分類突破】題型一利用導(dǎo)數(shù)探究函數(shù)的零點(diǎn)個(gè)數(shù)[例1]設(shè)函數(shù)f(x)=lnx+mx,m∈R,討論函數(shù)g(x)=f'(x)-x3【解析】由題意知g(x)=f'(x)-x3=1x-mx2-x3(x>0),令g(x)=0,得m=-13x3設(shè)φ(x)=-13x3+x(x則φ'(x)=-x2+1=-(x-1)(x+1).當(dāng)x∈(0,1)時(shí),φ'(x)>0,φ(x)在(0,1)上單調(diào)遞增;當(dāng)x∈(1,+∞)時(shí),φ'(x)<0,φ(x)在(1,+∞)上單調(diào)遞減.所以x=1是φ(x)的唯一極值點(diǎn),且是極大值點(diǎn),所以x=1也是φ(x)的最大值點(diǎn),所以φ(x)的最大值為φ(1)=23結(jié)合y=φ(x)的圖象(如圖)可知,①當(dāng)m>23時(shí),函數(shù)g(x②當(dāng)m=23時(shí),函數(shù)g(x③當(dāng)0<m<23時(shí),函數(shù)g(x④當(dāng)m≤0時(shí),函數(shù)g(x)有且只有一個(gè)零點(diǎn).綜上所述,當(dāng)m>23時(shí),函數(shù)g(x當(dāng)m=23或m≤0時(shí),函數(shù)g(x當(dāng)0<m<23時(shí),函數(shù)g(x)有兩個(gè)零點(diǎn)【解題技法】利用導(dǎo)數(shù)確定函數(shù)零點(diǎn)或方程的根的個(gè)數(shù)的方法(1)構(gòu)造函數(shù):構(gòu)建函數(shù)g(x)(要求g'(x)易求,g'(x)=0可解),轉(zhuǎn)化為確定g(x)的零點(diǎn)個(gè)數(shù)問題求解,利用導(dǎo)數(shù)研究該函數(shù)的單調(diào)性、極值,并確定定義區(qū)間端點(diǎn)值的符號(hào)(或變化趨勢(shì))等,畫出g(x)的圖象草圖,數(shù)形結(jié)合求解函數(shù)零點(diǎn)的個(gè)數(shù).(2)應(yīng)用定理:利用零點(diǎn)存在定理,先用該定理判斷函數(shù)在某區(qū)間上有零點(diǎn),然后利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性、極值(最值)及區(qū)間端點(diǎn)值的符號(hào),進(jìn)而判斷函數(shù)在該區(qū)間上零點(diǎn)的個(gè)數(shù).【對(duì)點(diǎn)訓(xùn)練】(2023·鄭州質(zhì)檢)已知函數(shù)f(x)=ex-ax+2a,a∈R.(1)討論函數(shù)f(x)的單調(diào)性;(2)求函數(shù)f(x)的零點(diǎn)個(gè)數(shù).【解析】(1)f(x)=ex-ax+2a,定義域?yàn)镽,且f'(x)=ex-a,當(dāng)a≤0時(shí),f'(x)>0,則f(x)在R上單調(diào)遞增;當(dāng)a>0時(shí),令f'(x)=0,則x=lna,當(dāng)x<lna時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)x>lna時(shí),f'(x)>0,f(x)單調(diào)遞增.綜上所述,當(dāng)a≤0時(shí),f(x)在R上單調(diào)遞增;當(dāng)a>0時(shí),f(x)在(-∞,lna)上單調(diào)遞減,在(lna,+∞)上單調(diào)遞增.(2)令f(x)=0,得ex=a(x-2),當(dāng)a=0時(shí),ex=a(x-2)無解,所以f(x)無零點(diǎn),當(dāng)a≠0時(shí),1a=x令φ(x)=x-2ex,x∈R,所以φ'(當(dāng)x∈(-∞,3)時(shí),φ'(x)>0;當(dāng)x∈(3,+∞)時(shí),φ'(x)<0,所以φ(x)在(-∞,3)上單調(diào)遞增,在(3,+∞)上單調(diào)遞減,且φ(x)max=φ(3)=1e又x→+∞時(shí),φ(x)→0,x→-∞時(shí),φ(x)→-∞,所以φ(x)的大致圖象如圖所示.當(dāng)1a>1e3,即0<a<e3時(shí),f當(dāng)1a=1e3,即a=e3時(shí),f當(dāng)0<1a<1e3,即a>e3時(shí),f當(dāng)1a<0,即a<0時(shí),f(x)有一個(gè)零點(diǎn)綜上所述,當(dāng)a∈(0,e3)時(shí),f(x)無零點(diǎn);當(dāng)a∈(-∞,0)∪{e3}時(shí),f(x)有一個(gè)零點(diǎn);當(dāng)a∈(e3,+∞)時(shí),f(x)有兩個(gè)零點(diǎn).【加練備選】已知函數(shù)f(x)=xex+ex.(1)求函數(shù)f(x)的單調(diào)區(qū)間和極值;(2)討論函數(shù)g(x)=f(x)-a(a∈R)的零點(diǎn)的個(gè)數(shù).【解析】(1)函數(shù)f(x)的定義域?yàn)镽,且f'(x)=(x+2)ex,令f'(x)=0得x=-2,則f'(x),f(x)的變化情況如表所示:x(-∞,-2)-2(-2,+∞)f'(x)-0+f(x)單調(diào)遞減-1單調(diào)遞增所以f(x)的單調(diào)遞減區(qū)間是(-∞,-2),單調(diào)遞增區(qū)間是(-2,+∞).當(dāng)x=-2時(shí),f(x)有極小值,為f(-2)=-1e2(2)令f(x)=0,得x=-1,當(dāng)x<-1時(shí),f(x)<0;當(dāng)x>-1時(shí),f(x)>0,且f(x)的圖象經(jīng)過點(diǎn)(-2,-1e2當(dāng)x→-∞時(shí),f(x)→0;當(dāng)x→+∞時(shí),f(x)→+∞,根據(jù)以上信息,畫出f(x)大致圖象如圖所示.函數(shù)g(x)=f(x)-a(a∈R)的零點(diǎn)的個(gè)數(shù)為y=f(x)的圖象與直線y=a的交點(diǎn)個(gè)數(shù),所以關(guān)于函數(shù)g(x)=f(x)-a(a∈R)的零點(diǎn)個(gè)數(shù)有如下結(jié)論:當(dāng)a<-1e2時(shí),零點(diǎn)的個(gè)數(shù)為0;當(dāng)a=-1e當(dāng)-1e2<a題型二利用函數(shù)零點(diǎn)問題求參數(shù)范圍[例2]已知函數(shù)f(x)=ex-a(x+2).(1)當(dāng)a=1時(shí),討論f(x)的單調(diào)性;(2)若f(x)有兩個(gè)零點(diǎn),求a的取值范圍.【解析】(1)當(dāng)a=1時(shí),f(x)=ex-x-2,則f'(x)=ex-1.當(dāng)x<0時(shí),f'(x)<0;當(dāng)x>0時(shí),f'(x)>0.所以f(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增;(2)f'(x)=ex-a.當(dāng)a≤0時(shí),f'(x)>0,所以f(x)在(-∞,+∞)上單調(diào)遞增,故f(x)至多存在1個(gè)零點(diǎn),不合題意;當(dāng)a>0時(shí),由f'(x)=0可得x=lna.當(dāng)x∈(-∞,lna)時(shí),f'(x)<0;當(dāng)x∈(lna,+∞)時(shí),f'(x)>0.所以f(x)在(-∞,lna)上單調(diào)遞減,在(lna,+∞)上單調(diào)遞增,故當(dāng)x=lna時(shí),f(x)取得最小值,最小值為f(lna)=-a(1+lna).(i)若0<a≤1e,則f(lna)≥0,f(x(ii)若a>1e,則f(lna)<0因?yàn)閒(-2)=e-2>0,所以f(x)在(-∞,lna)上存在唯一零點(diǎn).易知,當(dāng)x>2時(shí),ex-x-2>0,所以當(dāng)x>4且x>2ln(2a)時(shí),f(x)=ex2·ex2-a(x+2)>eln(2a)·x2+2-a故f(x)在(lna,+∞)上存在唯一零點(diǎn),從而f(x)在(-∞,+∞)上有兩個(gè)零點(diǎn).綜上,a的取值范圍是1e【解題技法】由函數(shù)零點(diǎn)求參數(shù)范圍的策略(1)涉及函數(shù)的零點(diǎn)(方程的根)問題,主要利用導(dǎo)數(shù)確定函數(shù)的單調(diào)區(qū)間和極值點(diǎn),根據(jù)函數(shù)零點(diǎn)的個(gè)數(shù)尋找函數(shù)在給定區(qū)間的極值以及區(qū)間端點(diǎn)的函數(shù)值與0的關(guān)系,從而求得參數(shù)的取值范圍.(2)解決此類問題的關(guān)鍵是將函數(shù)零點(diǎn)、方程的根、曲線交點(diǎn)相互轉(zhuǎn)化,突出導(dǎo)數(shù)的工具作用,體現(xiàn)轉(zhuǎn)化與化歸的思想方法.(3)含參數(shù)的函數(shù)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),可將參數(shù)分離出來后,用x表示參數(shù)的函數(shù),作出該函數(shù)圖象,根據(jù)圖象特征求參數(shù)的范圍.【對(duì)點(diǎn)訓(xùn)練】(一題多法)(2020·全國Ⅰ卷)已知函數(shù)f(x)=ex-a(x+2).(1)當(dāng)a=1時(shí),討論f(x)的單調(diào)性;(2)若f(x)有兩個(gè)零點(diǎn),求a的取值范圍.【解析】(1)當(dāng)a=1時(shí),f(x)=ex-(x+2),f'(x)=ex-1,令f'(x)<0,解得x<0,令f'(x)>0,解得x>0,所以f(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增.(2)方法一:當(dāng)a≤0時(shí),f'(x)=ex-a>0恒成立,f(x)在(-∞,+∞)上單調(diào)遞增,不符合題意;當(dāng)a>0時(shí),令f'(x)=0,解得x=lna,當(dāng)x∈(-∞,lna)時(shí),f'(x)<0,f(x)單調(diào)遞減,當(dāng)x∈(lna,+∞)時(shí),f'(x)>0,f(x)單調(diào)遞增,所以f(x)的極小值也是最小值為f(lna)=a-a(lna+2)=-a(1+lna).又當(dāng)x→-∞時(shí),f(x)→+∞,當(dāng)x→+∞時(shí),f(x)→+∞;所以要使f(x)有兩個(gè)零點(diǎn),只要f(lna)<0即可,則1+lna>0,可得a>1e.綜上,若f(x)有兩個(gè)零點(diǎn),則a的取值范圍是(1e方法二:若f(x)有兩個(gè)零點(diǎn),即ex-a(x+2)=0有兩個(gè)解,顯然x=-2不成立,即a=exx+2(x≠-2)有兩個(gè)解,令h(x)=e則有h'(x)=ex(x令h'(x)>0,解得x>-1,令h'(x)<0,解得x<-2或-2<x<-1,所以函數(shù)h(x)在(-∞,-2)和(-2,-1)上單調(diào)遞減,在(-1,+∞)上單調(diào)遞增,且當(dāng)x<-2時(shí),h(x)<0,而當(dāng)x→(-2)+(從右側(cè)趨近于-2)時(shí),h(x)→+∞,當(dāng)x→+∞時(shí),h(x)→+∞,所以當(dāng)a=exx+2(x≠-2)有兩個(gè)解時(shí),有a>h所以滿足條件的a的取值范圍是(1e,+∞)【加練備選】已知函數(shù)f(x)=xlnx,g(x)=(-x2+ax-3)ex(a∈R).(1)當(dāng)a=4時(shí),求曲線y=g(x)在x=0處的切線方程;(2)如果關(guān)于x的方程g(x)=2exf(x)在區(qū)間[1e,e]上有兩個(gè)不等實(shí)根,求實(shí)數(shù)a的取值范圍【解析】(1)當(dāng)a=4時(shí),g(x)=(-x2+4x-3)ex,g(0)=-3,g'(x)=(-x2+2x+1)ex,g'(0)=1,所以所求的切線方程為y+3=x-0,即y=x-3.(2)由g(x)=2exf(x),可得2xlnx=-x2+ax-3,a=x+2lnx+3x設(shè)h(x)=x+2lnx+3x(x所以h'(x)=1+2x-3x2=(x+3)(x-1)x2,所以x[1e1(1,e]h'(x)-0+h(x)單調(diào)遞減極小值(最小值)單調(diào)遞增又h(1e)=1e+3e-2,h(1)=4,h(e)=且h(e)-h(1e)=4-2e+2e<0,所以實(shí)數(shù)a的取值范圍為(4,e+2+3題型三與函數(shù)零點(diǎn)有關(guān)的證明[例3](2022·新高考Ⅰ卷改編)已知函數(shù)f(x)=ex-x,g(x)=x-lnx.(1)判斷直線y=b與曲線y=f(x)和y=g(x)的交點(diǎn)分別有幾個(gè);(2)證明:曲線y=f(x)和y=g(x)有且只有一個(gè)公共點(diǎn);(3)證明:存在直線y=b,其與兩條曲線y=f(x)和y=g(x)共有三個(gè)不同的交點(diǎn),并且從左到右的三個(gè)交點(diǎn)的橫坐標(biāo)成等差數(shù)列.【解析】(1)設(shè)S(x)=ex-x-b,S'(x)=ex-1,當(dāng)x<0時(shí),S'(x)<0,當(dāng)x>0時(shí),S'(x)>0,故S(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增,所以S(x)min=S(0)=1-b.當(dāng)b<1時(shí),S(x)min=1-b>0,S(x)無零點(diǎn);當(dāng)b=1時(shí),S(x)min=1-b=0,S(x)有1個(gè)零點(diǎn);當(dāng)b>1時(shí),S(x)min=1-b<0,而S(-b)=e-b>0,S(b)=eb-2設(shè)u(b)=eb-2b,其中b>1,則u'(b)=eb-2>0,故u(b)在(1,+∞)上單調(diào)遞增,故u(b)>u(1)=e-2>0,故S(b)>0,故S(x)=ex-x-b有兩個(gè)不同的零點(diǎn).設(shè)T(x)=x-lnx-b,T'(x)=x-當(dāng)0<x<1時(shí),T'(x)<0,當(dāng)x>1時(shí),T'(x)>0,故T(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,所以T(x)min=T(1)=1-b.當(dāng)b<1時(shí),T(x)min=1-b>0,T(x)無零點(diǎn);當(dāng)b=1時(shí),T(x)min=1-b=0,T(x)有1個(gè)零點(diǎn);當(dāng)b>1時(shí),T(x)min=1-b<0,而T(e-b)=e-b>0,T(eb)=e所以T(x)=x-lnx-b有兩個(gè)不同的零點(diǎn).綜上可知,當(dāng)b<1時(shí),直線y=b與曲線y=f(x)和y=g(x)的交點(diǎn)個(gè)數(shù)都是0;當(dāng)b=1時(shí),直線y=b與曲線y=f(x)和y=g(x)的交點(diǎn)個(gè)數(shù)都是1;當(dāng)b>1時(shí),直線y=b與曲線y=f(x)和y=g(x)的交點(diǎn)個(gè)數(shù)都是2.(2)由f(x)=g(x)得ex-x=x-lnx,即ex+lnx-2x=0,設(shè)h(x)=ex+lnx-2x,其中x>0,故h'(x)=ex+1x-2.設(shè)s(x)=ex-x-1,x>0,則s'(x)=ex故s(x)在(0,+∞)上單調(diào)遞增,故s(x)>s(0)=0,即ex>x+1,所以h'(x)>x+1x-1≥2-1>0,所以h(x而h(1)=e-2>0,h(1e3)=e1e3故h(x)在(0,+∞)上有且只有一個(gè)零點(diǎn)x0,且1e3<x0<1,當(dāng)0<x<x0時(shí),h(x)<0,即ex-x<x-lnx,即f(x)<g(x),當(dāng)x>x0時(shí),h(x)>0,即ex-x>x-lnx,即f(x)>g(x),所以曲線y=f(x)和y=g(x(3)由(2)知,若存在直線y=b與曲線y=f(x),y=g(x)有三個(gè)不同的交點(diǎn),則b=f(x0)=g(x0)>1,此時(shí)ex-x=b有兩個(gè)不同的解x1,x0(x1<0<x0),x-lnx=b有兩個(gè)不同的解x0,x2(0<x0<1<x2),故ex1-x1=b,ex0-x0=b,x2-lnx2-b=0,x0-lnx所以x2-b=lnx2,即ex2-b=x2,即ex2-b故x2-b為方程ex-x=b的解,同理x0-b也為方程ex-x=b的解,所以{x1,x0}={x0-b,x2-b},而b>1,故x0=x2-b,x1=【解題技法】1.證明與零點(diǎn)有關(guān)的不等式,函數(shù)的零點(diǎn)本身就是一個(gè)條件,即零點(diǎn)對(duì)應(yīng)的函數(shù)值為0;2.證明的思路一般對(duì)條件等價(jià)轉(zhuǎn)化,構(gòu)造合適的新函數(shù),利用導(dǎo)數(shù)知識(shí)探討該函數(shù)的性質(zhì)(如單調(diào)性、極值情況等),再結(jié)合函數(shù)圖象來解決.【對(duì)點(diǎn)訓(xùn)練】已知函數(shù)f(x)=lnx-x+2sinx,f'(x)為f(x)的導(dǎo)函數(shù).(1)求證:f'(x)在(0,π)上存在唯一零點(diǎn);(2)求證:f(x)有且僅有兩個(gè)不同的零點(diǎn).【證明】(1)設(shè)g(x)=f'(x)=1x-1+2cosx當(dāng)x∈(0,π)時(shí),g'(x)=-2sinx-1x2<0,所以g(又因?yàn)間(π3)=3π-1+1>0,g(π2所以g(x)在(0,π)上有唯一的零點(diǎn).(2)設(shè)f'(x)在(0,π)上的唯一零點(diǎn)為α,由(1)知π3<α<π①當(dāng)x∈(0,α)時(shí),f'(x)>0,f(x)單調(diào)遞增;當(dāng)x∈(α,π)時(shí),f'(x)<0,f(x)單調(diào)遞減;所以f(x)在(0,π)上存在唯一的極大值點(diǎn)α,所以f(α)>f(π2)=lnπ2-π2又因?yàn)閒(1e2)=-2-1e2+2sin所以f(x)在(0,α)上恰有一個(gè)零點(diǎn).又因?yàn)閒(π)=lnπ-π<2-π<0,所以f(x)在(α,π)上也恰有一個(gè)零點(diǎn).②當(dāng)x∈[π,2π)時(shí),sinx≤0,f(x)≤lnx-x,設(shè)h(x)=lnx-x,h'(x)=1x所以h(x)在[π,2π)上單調(diào)遞減,所以h(x)≤h(π)<0,所以當(dāng)x∈[π,2π)時(shí),f(x)≤h(x)≤h(π)<0恒成立,所以f(x)在[π,2π)上沒有零點(diǎn).③當(dāng)x∈[2π,+∞)時(shí),f(x)≤lnx-x+2.設(shè)φ(x)=lnx-x+2,φ'(x)=1x所以φ(x)在[2π,+∞)上單調(diào)遞減,所以φ(x)≤φ(2π)<0,所以當(dāng)x∈[2π,+∞)時(shí),f(x)≤φ(x)≤φ(2π)<0恒成立,所以f(x)在[2π,+∞)上沒有零點(diǎn).綜上,f(x)有且僅有兩個(gè)不同的零點(diǎn).第3課時(shí)導(dǎo)數(shù)的不等式問題【命題分析】導(dǎo)數(shù)中的不等式證明經(jīng)常被考查,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,題目難度較大,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,采用不同的解題方法,可以達(dá)到事半功倍的效果.【核心考點(diǎn)·分類突破】題型一作差法構(gòu)造函數(shù),證明不等式[例1]設(shè)f(x)=2xlnx+1.(1)求f(x)的最小值;(2)證明:f(x)≤x2-x+1x+2ln【解析】(1)f'(x)=2(lnx+1),所以當(dāng)x∈(0,1e)時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)x∈(1e,+∞)時(shí),f'(x)>0,f(x)單調(diào)遞增,所以當(dāng)x=1e時(shí),f(x)取得最小值f(1(2)x2-x+1x+2lnx-f(x)=x(x-1)-x-1x-2(x-1)lnx=(x-1)(x-令g(x)=x-1x-2lnx,則g'(x)=1+1x2-2x=(x-1)2x2≥0,所以g(x)在(0,+∞)上單調(diào)遞增,又g(1)=0,所以當(dāng)0<x<1時(shí),g(x)<0,當(dāng)x>1時(shí),g即f(x)≤x2-x+1x+2ln【解題技法】作差法構(gòu)造函數(shù),證明不等式的策略(1)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);(2)有時(shí)對(duì)復(fù)雜的式子要進(jìn)行變形,借助所構(gòu)造函數(shù)的單調(diào)性和最值求解,利用導(dǎo)數(shù)研究其單調(diào)性和最值.【對(duì)點(diǎn)訓(xùn)練】(2023·濰坊模擬)已知函數(shù)f(x)=ex-ax-a,a∈R.(1)討論f(x)的單調(diào)性;(2)當(dāng)a=1時(shí),令g(x)=2f(x)x2.證明:當(dāng)x【解析】(1)函數(shù)f(x)=ex-ax-a的定義域?yàn)镽,求導(dǎo)得f'(x)=ex-a,當(dāng)a≤0時(shí),f'(x)>0恒成立,即f(x)在(-∞,+∞)上單調(diào)遞增;當(dāng)a>0時(shí),令f'(x)=ex-a>0,解得x>lna,令f'(x)<0,解得x<lna,即f(x)在(-∞,lna)上單調(diào)遞減,在(lna,+∞)上單調(diào)遞增,所以當(dāng)a≤0時(shí),f(x)在(-∞,+∞)上單調(diào)遞增,當(dāng)a>0時(shí),f(x)在(-∞,lna)上單調(diào)遞減,在(lna,+∞)上單調(diào)遞增.(2)當(dāng)a=1時(shí),g(x)=2(當(dāng)x>0時(shí),2(ex-x-1)x2>1令F(x)=12x2+x+1ex-1,則F(x)在(0,+∞)上單調(diào)遞減,F(x)<F(0)=1-1=0,因此12所以當(dāng)x>0時(shí),g(x)>1,即原不等式得證.題型二轉(zhuǎn)化為兩個(gè)函數(shù)的最值證明不等式[例2](2023·武漢模擬)已知函數(shù)f(x)=alnx+x.(1)討論f(x)的單調(diào)性;(2)當(dāng)a=1時(shí),證明:xf(x)<ex.【解析】(1)f(x)的定義域?yàn)?0,+∞),f'(x)=ax+1=x當(dāng)a≥0時(shí),f'(x)>0,所以f(x)在(0,+∞)上單調(diào)遞增.當(dāng)a<0時(shí),若x∈(-a,+∞),則f'(x)>0,若x∈(0,-a),則f'(x)<0,所以f(x)在(-a,+∞)上單調(diào)遞增,在(0,-a)上單調(diào)遞減.綜上所述,當(dāng)a≥0時(shí),f(x)在(0,+∞)上單調(diào)遞增;當(dāng)a<0時(shí),f(x)在(-a,+∞)上單調(diào)遞增,在(0,-a)上單調(diào)遞減.(2)當(dāng)a=1時(shí),要證xf(x)<ex,即證x2+xlnx<ex,即證1+lnxx<令函數(shù)g(x)=1+lnxx,則g'(x)=令g'(x)>0,得x∈(0,e);令g'(x)<0,得x∈(e,+∞).所以g(x)在(0,e)上單調(diào)遞增,在(e,+∞)上單調(diào)遞減,所以g(x)max=g(e)=1+1e令函數(shù)h(x)=exx2,則h'(x當(dāng)x∈(0,2)時(shí),h'(x)<0;當(dāng)x∈(2,+∞)時(shí),h'(x)>0.所以h(x)在(0,2)上單調(diào)遞減,在(2,+∞)上單調(diào)遞增,所以h(x)min=h(2)=e2因?yàn)閑24-(1+1e)>0,所以h(x)min>g(x即1+lnxx<exx2,從而xf(【解題技法】最值法證明不等式的策略(1)構(gòu)造“形似”函數(shù):對(duì)原不等式同解變形,如移項(xiàng)、通分、取對(duì)數(shù),把不等式左、右兩邊轉(zhuǎn)化為相同結(jié)構(gòu)的式子,根據(jù)“相同結(jié)構(gòu)”構(gòu)造輔助函數(shù);(2)最值法:欲證f(x)<g(x),有時(shí)可以證明f(x)max<g(x)min.【對(duì)點(diǎn)訓(xùn)練】(2023·蘇州模擬)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f(x)-exx【解析】(1)函數(shù)的定義域?yàn)?0,+∞),因?yàn)閒'(x)=ex-a=e-ax所以當(dāng)a≤0時(shí),f'(x)>0在(0,+∞)上恒成立,故函數(shù)f(x)在區(qū)間(0,+∞)上單調(diào)遞增;當(dāng)a>0時(shí),由f'(x)>0,得0<x<ea,由f'(x)<0,得x>ea,即函數(shù)f(x)在區(qū)間(0,ea)上單調(diào)遞增,在(綜上,當(dāng)a≤0時(shí),f(x)在區(qū)間(0,+∞)上單調(diào)遞增;當(dāng)a>0時(shí),f(x)在區(qū)間(0,ea)上單調(diào)遞增,在(ea(2)要證明f(x)-exx+2e≤0,只需證明f(x)≤由(1)知,當(dāng)a=e時(shí),函數(shù)f(x)在區(qū)間(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,所以f(x)max=f(1)=-e.令g(x)=exx-2e(x>0),則g'(x)=所以當(dāng)x∈(0,1)時(shí),g'(x)<0,函數(shù)g(x)單調(diào)遞減;當(dāng)x∈(1,+∞)時(shí),g'(x)>0,函數(shù)g(x)單調(diào)遞增,所以g(x)min=g(1)=-e,所以當(dāng)x>0,a=e時(shí),f(x)-exx題型三放縮構(gòu)造函數(shù)證明不等式[例3]f(x)=ex.(1)求曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程;(2)當(dāng)x>-2時(shí),求證:f(x)>ln(x+2).【解析】(1)由f(x)=ex,得f(0)=1,f'(x)=ex,則f'(0)=1,即曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程為y-1=x-0,所以所求切線方程為x-y+1=0.(2)設(shè)g(x)=f(x)-(x+1)=ex-x-1(x>-2),則g'(x)=ex-1,當(dāng)-2<x<0時(shí),g'(x)<0;當(dāng)x>0時(shí),g'(x)>0,即g(x)在(-2,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增,于是當(dāng)x=0時(shí),g(x)min=g(0)=0,因此f(x)≥x+1(當(dāng)x=0時(shí)取等號(hào)).令h(x)=x+1-ln(x+2)(x>-2),則h'(x)=1-1x+2=x+1x+2,則當(dāng)-2<x<-1時(shí),h'(x)<0,當(dāng)x即h(x)在(-2,-1)上單調(diào)遞減,在(-1,+∞)上單調(diào)遞增,于是當(dāng)x=-1時(shí),h(x)min=h(-1)=0,因此x+1≥ln(x+2)(當(dāng)x=-1時(shí)取等號(hào)),所以當(dāng)x>-2時(shí),f(x)>ln(x+2).【解題技法】放縮法證明不等式的策略導(dǎo)數(shù)方法證明不等式的問題中,最常見的是ex和lnx與其他代數(shù)式結(jié)合的問題,對(duì)于這類問題,可以考慮先對(duì)ex和lnx進(jìn)行放縮,使問題簡(jiǎn)化,簡(jiǎn)化后再構(gòu)造函數(shù)進(jìn)行證明.常見的放縮公式如下:(1)ex≥1+x,當(dāng)且僅當(dāng)x=0時(shí)取等號(hào);(2)lnx≤x-1,當(dāng)且僅當(dāng)x=1時(shí)取等號(hào).【對(duì)點(diǎn)訓(xùn)練】f(x)=aex-1-ln(1)若a=1,求f(x)在(1,f(1))處的切線方程;(2)證明:當(dāng)a≥1時(shí),f(x)≥0.【解析】(1)當(dāng)a=1時(shí),f(x)=ex-1-lnxf'(x)=ex-1-1x,又f(1)=0,所以切點(diǎn)為(1,0).所以切線方程為y-0=0(x-1),即y=0.(2)因?yàn)閍≥1,所以aex-1所以f(x)≥ex-1-ln方法一:令φ(x)=ex-1-lnx所以φ'(x)=ex-1令h(x)=ex-1-1x,所以h'(x)=所以φ'(x)在(0,+∞)上單調(diào)遞增,又φ'(1)=0,所以當(dāng)x∈(0,1)時(shí),φ'(x)<0;當(dāng)x∈(1,+∞)時(shí),φ'(x)>0,所以φ(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,所以φ(x)min=φ(1)=0,所以φ(x)≥0,所以f(x)≥φ(x)≥0,即f(x)≥0.方法二:令g(x)=ex-x-1,所以g'(x)=ex-1.當(dāng)x∈(-∞,0)時(shí),g'(x)<0;當(dāng)x∈(0,+∞)時(shí),g'(x)>0,所以g(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增,所以g(x)min=g(0)=0,故ex≥x+1,當(dāng)且僅當(dāng)x=0時(shí)取“=”.同理可證lnx≤x-1,當(dāng)且僅當(dāng)x=1時(shí)取“=”.由ex≥x+1?ex-1≥x由x-1≥lnx?x≥lnx+1(當(dāng)且僅當(dāng)x=1時(shí)取“=”),所以ex-1≥x≥lnx+1,即e即ex-1-lnx即f(x)≥0.【加練備選】(2023·南充模擬)已知函數(shù)f(x)=ax-sinx.(1)若函數(shù)f(x)為增函數(shù),求實(shí)數(shù)a的取值范圍;(2)求證:當(dāng)x>0時(shí),ex>2sinx.【解析】(1)f(x)=ax-sinx,所以f'(x)=a-cosx,由函數(shù)f(x)為增函數(shù),則f'(x)=a-cosx≥0恒成立,即a≥cosx在R上恒成立,因?yàn)閥=cosx∈[-1,1],所以a≥1,即實(shí)數(shù)a的取值范圍是[1,+∞).(2)由(1)知,當(dāng)a=1時(shí),f(x)=x-sinx為增函數(shù),當(dāng)x>0時(shí),f(x)>f(0)=0?x>sinx,要證當(dāng)x>0時(shí),ex>2sinx,只需證當(dāng)x>0時(shí),ex>2x,即證ex-2x>0在(0,+∞)上恒成立.設(shè)g(x)=ex-2x(x>0),則g'(x)=ex-2,令g'(x)=0解得x=ln2,所以g(x)在(0,ln2)上單調(diào)遞減,在(ln2,+∞)上單調(diào)遞增,所以g(x)min=g(ln2)=eln2-2ln2=2(1-ln2)>0,所以g(x)≥g(ln2)>0,所以ex>2x成立,故當(dāng)x>0時(shí),ex>2sinx.【重難突破】泰勒公式在比較大小中的應(yīng)用比較大小的選擇題是近年高考的常見題型,一般情況下我們會(huì)構(gòu)造函數(shù)模型代入數(shù)值進(jìn)行比較和運(yùn)算,但是對(duì)學(xué)生來說函數(shù)模型的選擇是非常有難度的,因此在選擇題中我們可以選擇利用泰勒公式計(jì)算近似值的辦法進(jìn)行比較大小.【教材探源】在人教A必修一教材中三角函數(shù)一章第256頁“拓廣探索”中第26題.英國數(shù)學(xué)家泰勒給出如下公式:sinx=x-x33!+x55cosx=1-x22!+x44其中n!=1×2×3×4×…×n.這些公式被編入計(jì)算工具,計(jì)算工具計(jì)算足夠多的項(xiàng)就可以確保顯示值的精確性.比如,用前三項(xiàng)計(jì)算cos0.3,就得到cos0.3≈1-0.322!+【教材拓展】下面給出高中階段常用的泰勒公式:(1)ex=1+x+12!x2+…+1n!xn+…,(2)sinx=x-x33!+…+(-1)k-1x2k-1(3)cosx=1-x22!+…+(-1)k1(2k)!x2k(4)ln(1+x)=x-12x2+…+(-1)n-11nxn+…,-1<(5)(1+x)a=1+ax+a(a-1)2!x2+…+a(類型一利用ex、ln(1+x)的泰勒展開式比較[例1](2022·新高考Ⅰ卷)設(shè)a=0.1e0.1,b=19,c=-ln0.9,則(A.a<b<c B.c<b<aC.c<a<b D.a<c<b【解析】選C.b=19≈0.由公式ex=1+x+x22可得e0.1≈1+0.1+0.12則a=0.1e0.1≈0.1105,c=-ln0.9=ln109=ln(1+1由公式ln(1+x)=x-x22+x得c=ln(1+19)≈19-(1所以c<a<b.類型二利用sinx,cosx的泰勒展開式比較[例2](2022·全國甲卷)已知a=3132,b=cos14,c=4sin14A.c>b>a B.b>a>cC.a>b>c D.a>c>b【解析】選A.由公式sinx=x-x33!+…得c=4sin14≈4(14-1由公式cosx=1-x22!+得b=cos14≈1-1422+(1排除BCD.【一題多解】選A.構(gòu)造函數(shù)h(x)=1-12x2-cosx,x∈[0,π2],則g(x)=h'(x)=-x+sing'(x)=-1+cosx≤0,所以g(x)≤g(0)=0,因此h(x)在[0,π2所以h(14)=a-b<h(0)=0,即a<bcb=4sin14cos14=tan14所以cb>1,即b<c,綜上c>b>類型三利用ln(1+x),(1+x)a的泰勒展開式比較[例3](2021·全國乙卷)設(shè)a=2ln1.01,b=ln1.02,