高三數(shù)學(xué)一輪總復(fù)習(xí)第2課時(shí)利用導(dǎo)數(shù)證明不等式課件

第2課時(shí)利用導(dǎo)數(shù)證明不等式考點(diǎn)一作差構(gòu)造函數(shù)法證明不等式例1(12分)(2023·新高考Ⅰ,19)已知函數(shù)f(x)=a(ex+a)-x.(1)討論f(x)的單調(diào)性;突破口:結(jié)合ex>0,對a分類討論.(2)證明:當(dāng)a>0時(shí),f(x)>2ln

a+.關(guān)鍵點(diǎn):構(gòu)造函數(shù),確定最值,證明不等式.審題指導(dǎo):(1)求導(dǎo)數(shù)后,結(jié)合ex>0從而對參數(shù)a分類討論,確定f'(x)的符號(hào),得到函數(shù)的單調(diào)性.(2)思路1:結(jié)合(1)中函數(shù)的單調(diào)性得到函數(shù)的最小值為1+a2+ln

a,從而將欲證不等式化為a2--ln

a>0,然后構(gòu)造函數(shù)通過最值證得結(jié)果.思路2:首先證明ex≥x+1成立,然后借助該不等式將f(x)進(jìn)行放縮,轉(zhuǎn)化為證明不等式a2--ln

a>0,再構(gòu)造函數(shù)通過最值證得結(jié)果.不要漏掉這種情形

當(dāng)a>0時(shí),令f'(x)=aex-1=0,解得x=-ln

a,當(dāng)x<-ln

a時(shí),f'(x)<0,f(x)在(-∞,-ln

a)上單調(diào)遞減;當(dāng)x>-ln

a時(shí),f'(x)>0,f(x)在(-ln

a,+∞)上單調(diào)遞增.分類討論后要將結(jié)果進(jìn)行綜述

借助(1)中單調(diào)性得到函數(shù)的最小值

將欲證不等式轉(zhuǎn)化構(gòu)造函數(shù)

確定函數(shù)最小值證得結(jié)果

首先證明常用放縮不等式ex≥x+1對f(x)放縮

構(gòu)造函數(shù)將欲證不等式進(jìn)行轉(zhuǎn)化

構(gòu)造函數(shù)

確定函數(shù)最小值證得結(jié)果

[對點(diǎn)訓(xùn)練1](12分)(2024·北京密云模擬)已知函數(shù)f(x)=xln(x+1).(1)求曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程;考點(diǎn)二拆分構(gòu)造兩個(gè)函數(shù)法證明不等式當(dāng)x∈(0,e)時(shí),h'(x)>0,h(x)單調(diào)遞增,當(dāng)x∈(e,+∞)時(shí),h'(x)<0,h(x)單調(diào)遞減,[對點(diǎn)訓(xùn)練2](2024·河南開封模擬)設(shè)函數(shù)f(x)=(x2-2x)ex,g(x)=e2ln

x-aex.(1)若函數(shù)g(x)在(e,+∞)上存在最大值,求實(shí)數(shù)a的取值范圍;(2)當(dāng)a=2時(shí),求證:f(x)>g(x).(2)證明

當(dāng)a=2時(shí),g(x)=e2ln

x-2ex,且函數(shù)g(x)的定義域?yàn)?0,+∞),要證明f(x)>g(x),即證明當(dāng)x>0時(shí),(x2-2x)ex>e2ln

x-2ex,只需要證明當(dāng)x>0時(shí),設(shè)φ(x)=(x-2)ex+2e,則φ'(x)=(x-1)ex,令φ'(x)=0,得x=1,當(dāng)0<x<1時(shí),φ'(x)<0,當(dāng)x>1時(shí),φ'(x)>0,所以φ(x)在(0,1)內(nèi)單調(diào)遞減,在(1,+∞)上單調(diào)遞增,故h'(x)=0,得x=e,當(dāng)0<x<e時(shí),h'(x)>0,當(dāng)x>e時(shí),h'(x)<0,所以h(x)在(0,e)內(nèi)單調(diào)遞增,在(e,+∞)上單調(diào)遞減,故h(x)≤h(e)=e,且當(dāng)x=e時(shí),等號(hào)成立.綜上可得,當(dāng)x>0時(shí),φ(x)≥h(x),且等號(hào)不同時(shí)成立,所以當(dāng)x>0時(shí),φ(x)>h(x),故當(dāng)a=2時(shí),f(x)>g(x)得證.考點(diǎn)三放縮法證明不等式例3(2024·福建廈門模擬)已知函數(shù)f(x)=aex+2x-1(其中常數(shù)e=2.71828…是自然對數(shù)的底數(shù)).(1)討論函數(shù)f(x)的單調(diào)性;(2)證明:對任意的a≥1,當(dāng)x>0時(shí),f(x)≥(x+ae)x.[對點(diǎn)訓(xùn)練3](2024·福建泉州模擬)已知函數(shù)f(x)=ex-axsin

x-x-1(a∈R).(1)當(dāng)a=0時(shí),討論函數(shù)f(x)的單調(diào)性;(2)當(dāng)a=時(shí),證明:對任意的x∈(0,+∞),f(x)>0.(1)解

當(dāng)a=0時(shí),f(x)=ex-x-1,f'(x)=ex-1.令f'(x)>0,解得x>0,f(x)在(0,+∞)上單調(diào)遞增;令f'(x)<0,解得x<0,f(x)在(-∞,0)上單調(diào)遞減.考點(diǎn)四特殊化處理證明數(shù)列不等式例4(2022·新高考Ⅱ,22)已知函數(shù)f(x)=xeax-ex.(1)當(dāng)a=1時(shí),討論f(x)的單調(diào)性;(2)當(dāng)x>0時(shí),f(x)<-1,求實(shí)數(shù)a的取值范圍;(1)解

當(dāng)a=1時(shí),f(x)=xex-ex,x∈R,則f'(x)=xex,當(dāng)x>0時(shí),f'(x)>0,當(dāng)x<0時(shí),f'(x)<0,∴函數(shù)f(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增.(2)解

函數(shù)f(x)的定義域?yàn)镽,f'(x)=(1+ax)eax-ex.對于x∈(0,+∞),當(dāng)a≥1時(shí),f'(x)=(1+ax)eax-ex>eax-ex≥ex-ex=0,∴f'(x)>0,f(x)在(0,+∞)上單調(diào)遞增.∵f(0)=-1,∴f(x)>-1,不滿足題意.當(dāng)a≤0時(shí),f'(x)≤eax-ex≤1-ex<0且等號(hào)不恒成立,∴f(x)在(0,+∞)上單調(diào)遞減.∵f(0)=-1,∴f(x)<-1,滿足題意.令h(x)=1+ax-e(1-a)x,則h'(x)=a+(a-1)e(1-a)x.∵h(yuǎn)'(x)為減函數(shù),又h'(0)=2a-1>0,x→+∞,h'(x)<0,∴?x0∈(0,+∞),使h'(x0)=0,∴當(dāng)x∈(0,x0)時(shí),h'(x)>0