高考數(shù)學(xué)一輪復(fù)習(xí)第三章導(dǎo)數(shù)及其應(yīng)用33導(dǎo)數(shù)的綜合應(yīng)用文新人教A版課件

3.3

導(dǎo)數(shù)的綜合應(yīng)用3.3導(dǎo)數(shù)的綜合應(yīng)用-2-考點(diǎn)1考點(diǎn)2考點(diǎn)3

例1設(shè)f(x)=xlnx-ax2+(2a-1)x,a∈R.(1)令g(x)=f'(x),求g(x)的單調(diào)區(qū)間;(2)已知f(x)在x=1處取得極大值.求實(shí)數(shù)a的取值范圍.思考如何求與函數(shù)極值有關(guān)的參數(shù)范圍?-2-考點(diǎn)1考點(diǎn)2考點(diǎn)3例1設(shè)f(x)=xlnx--3-考點(diǎn)1考點(diǎn)2考點(diǎn)3-3-考點(diǎn)1考點(diǎn)2考點(diǎn)3-4-考點(diǎn)1考點(diǎn)2考點(diǎn)3-4-考點(diǎn)1考點(diǎn)2考點(diǎn)3-5-考點(diǎn)1考點(diǎn)2考點(diǎn)3-5-考點(diǎn)1考點(diǎn)2考點(diǎn)3-6-考點(diǎn)1考點(diǎn)2考點(diǎn)3解題心得依據(jù)題意,對參數(shù)分類,分類后相當(dāng)于增加了一個已知條件,在增加了條件的情況下,對參數(shù)的各個范圍逐個驗(yàn)證是否符合題意,符合題意的范圍即為所求范圍.-6-考點(diǎn)1考點(diǎn)2考點(diǎn)3解題心得依據(jù)題意,對參數(shù)分類,分類后-7-考點(diǎn)1考點(diǎn)2考點(diǎn)3對點(diǎn)訓(xùn)練1設(shè)函數(shù)f(x)=x2-2x+mlnx+1,其中m為常數(shù).(2)若函數(shù)f(x)有唯一極值點(diǎn),求實(shí)數(shù)m的取值范圍.-7-考點(diǎn)1考點(diǎn)2考點(diǎn)3對點(diǎn)訓(xùn)練1設(shè)函數(shù)f(x)=x2-2x-8-考點(diǎn)1考點(diǎn)2考點(diǎn)3-8-考點(diǎn)1考點(diǎn)2考點(diǎn)3-9-考點(diǎn)1考點(diǎn)2考點(diǎn)3-9-考點(diǎn)1考點(diǎn)2考點(diǎn)3-10-考點(diǎn)1考點(diǎn)2考點(diǎn)3綜上,當(dāng)m≤0時,函數(shù)f(x)有唯一極值點(diǎn),即f(x)有唯一極值點(diǎn),故實(shí)數(shù)m的取值范圍為(-∞,0].-10-考點(diǎn)1考點(diǎn)2考點(diǎn)3綜上,當(dāng)m≤0時,函數(shù)f(x)有唯-11-考點(diǎn)1考點(diǎn)2考點(diǎn)3例2已知函數(shù)f(x)=-x3+x2,g(x)=alnx(a≠0,a∈R).(1)求f(x)的極值;(2)若對任意x∈[1,+∞),使得f(x)+g(x)≥-x3+(a+2)x恒成立,求實(shí)數(shù)a的取值范圍;思考利用導(dǎo)數(shù)解決不等式恒成立問題的基本思路是什么?-11-考點(diǎn)1考點(diǎn)2考點(diǎn)3例2已知函數(shù)f(x)=-x3+x2-12-考點(diǎn)1考點(diǎn)2考點(diǎn)3-12-考點(diǎn)1考點(diǎn)2考點(diǎn)3-13-考點(diǎn)1考點(diǎn)2考點(diǎn)3-13-考點(diǎn)1考點(diǎn)2考點(diǎn)3-14-考點(diǎn)1考點(diǎn)2考點(diǎn)3-14-考點(diǎn)1考點(diǎn)2考點(diǎn)3-15-考點(diǎn)1考點(diǎn)2考點(diǎn)3解題心得利用導(dǎo)數(shù)解決不等式恒成立問題,首先要構(gòu)造函數(shù),利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性,然后求出最值,進(jìn)而得出相應(yīng)的含參不等式,最后求出參數(shù)的取值范圍;也可分離變量,構(gòu)造函數(shù),直接把問題轉(zhuǎn)化為函數(shù)的最值問題.-15-考點(diǎn)1考點(diǎn)2考點(diǎn)3解題心得利用導(dǎo)數(shù)解決不等式恒成立問-16-考點(diǎn)1考點(diǎn)2考點(diǎn)3對點(diǎn)訓(xùn)練2(2017遼寧大連一模)已知函數(shù)f(x)=ax-lnx.(1)過原點(diǎn)O作函數(shù)f(x)圖象的切線,求切點(diǎn)的橫坐標(biāo);(2)對?x∈[1,+∞),不等式f(x)≥a(2x-x2)恒成立,求實(shí)數(shù)a的取值范圍.解:(1)設(shè)切點(diǎn)為M(x0,f(x0)),切線方程為y-f(x0)=k(x-x0),又切線過原點(diǎn)O,∴-ax0+lnx0=-ax0+1,由lnx0=1,解得x0=e,∴切點(diǎn)的橫坐標(biāo)為e.-16-考點(diǎn)1考點(diǎn)2考點(diǎn)3對點(diǎn)訓(xùn)練2(2017遼寧大連一模)-17-考點(diǎn)1考點(diǎn)2考點(diǎn)3(2)∵不等式ax-lnx≥a(2x-x2)恒成立,∴等價(jià)于a(x2-x)≥lnx對?x∈[1,+∞)恒成立.設(shè)y1=a(x2-x),y2=lnx,由于x∈[1,+∞),且當(dāng)a≤0時,y1≤y2,故a>0.設(shè)g(x)=ax2-ax-lnx,當(dāng)0<a<1時,g(3)=6a-ln3≥0不恒成立,當(dāng)a≥1,x=1時,g(x)≥0恒成立;-17-考點(diǎn)1考點(diǎn)2考點(diǎn)3(2)∵不等式ax-lnx≥a(-18-考點(diǎn)1考點(diǎn)2考點(diǎn)3例3已知函數(shù)f(x)=(x-2)ex+a(x-1)2.(1)討論f(x)的單調(diào)性;(2)若f(x)有兩個零點(diǎn),求a的取值范圍.思考如何利用導(dǎo)數(shù)求與函數(shù)零點(diǎn)有關(guān)的參數(shù)范圍?解(1)f'(x)=(x-1)ex+2a(x-1)=(x-1)(ex+2a).(ⅰ)設(shè)a≥0,則當(dāng)x∈(-∞,1)時,f'(x)<0;當(dāng)x∈(1,+∞)時,f'(x)>0.所以f(x)在(-∞,1)內(nèi)單調(diào)遞減,在(1,+∞)內(nèi)單調(diào)遞增.-18-考點(diǎn)1考點(diǎn)2考點(diǎn)3例3已知函數(shù)f(x)=(x-2)e-19-考點(diǎn)1考點(diǎn)2考點(diǎn)3-19-考點(diǎn)1考點(diǎn)2考點(diǎn)3-20-考點(diǎn)1考點(diǎn)2考點(diǎn)3-20-考點(diǎn)1考點(diǎn)2考點(diǎn)3-21-考點(diǎn)1考點(diǎn)2考點(diǎn)3(ⅱ)設(shè)a=0,則f(x)=(x-2)ex,所以f(x)只有一個零點(diǎn).又當(dāng)x≤1時f(x)<0,故f(x)不存在兩個零點(diǎn);單調(diào)遞增.又當(dāng)x≤1時f(x)<0,故f(x)不存在兩個零點(diǎn).綜上,a的取值范圍為(0,+∞).-21-考點(diǎn)1考點(diǎn)2考點(diǎn)3(ⅱ)設(shè)a=0,則f(x)=(x--22-考點(diǎn)1考點(diǎn)2考點(diǎn)3解題心得與函數(shù)零點(diǎn)有關(guān)的參數(shù)范圍問題,往往利用導(dǎo)數(shù)研究函數(shù)的單調(diào)區(qū)間和極值點(diǎn),并結(jié)合特殊點(diǎn),從而判斷函數(shù)的大致圖象,討論其圖象與x軸的位置關(guān)系(或者轉(zhuǎn)化為兩個熟悉函數(shù)的圖象交點(diǎn)問題),進(jìn)而確定參數(shù)的取值范圍.-22-考點(diǎn)1考點(diǎn)2考點(diǎn)3解題心得與函數(shù)零點(diǎn)有關(guān)的參數(shù)范圍問-23-考點(diǎn)1考點(diǎn)2考點(diǎn)3對點(diǎn)訓(xùn)練3(2017天津六校聯(lián)考)設(shè)函數(shù)f(x)=lnx-ax2-bx.(1)當(dāng)a=b=時,求函數(shù)f(x)的單調(diào)區(qū)間;(2)當(dāng)a=0,b=-1時,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.解:(1)依題意,知f(x)的定義域?yàn)?0,+∞),令f'(x)=0,解得x=1或x=-2(舍去),當(dāng)0<x<1時,f'(x)>0;當(dāng)x>1時,f'(x)<0,所以f(x)的單調(diào)增區(qū)間為(0,1),減區(qū)間為(1,+∞).-23-考點(diǎn)1考點(diǎn)2考點(diǎn)3對點(diǎn)訓(xùn)練3(2017天津六校聯(lián)考)-24-考點(diǎn)1考點(diǎn)2考點(diǎn)3(2)當(dāng)a=0,b=-1時,f(x)=lnx+x,由f(x)=mx,得lnx+x=mx,又x>0,由g'(x)<0,得x>e,所以g(x)在區(qū)間[1,e]上是增函數(shù),在區(qū)間[e,e2]上是減函數(shù).-24-考點(diǎn)1考點(diǎn)2考點(diǎn)3(2)當(dāng)a=0,b=-1時,f(x-25-考點(diǎn)1考點(diǎn)2考點(diǎn)31.利用導(dǎo)數(shù)證明不等式,就是利用不等式與函數(shù)之間的聯(lián)系,先結(jié)合不等式的結(jié)構(gòu)特征,直接或等價(jià)變形后構(gòu)造相應(yīng)的函數(shù),再通過導(dǎo)數(shù)運(yùn)算判斷出函數(shù)的單調(diào)性,利用單調(diào)性證明,或利用導(dǎo)數(shù)運(yùn)算來求出函數(shù)的最值,利用最值證明.2.求解不等式恒成立問題時,可以考慮將參數(shù)分離出來,將參數(shù)范圍問題轉(zhuǎn)化為研究新函數(shù)的值域問題.3.研究函數(shù)圖象的交點(diǎn)、方程的根、函數(shù)的零點(diǎn),一般是通過數(shù)形結(jié)合的思想找到解題思路,使用的知識是函數(shù)的性質(zhì),如單調(diào)性、極值等.-25-考點(diǎn)1考點(diǎn)2考點(diǎn)31.利用導(dǎo)數(shù)證明不等式,就是利用不3.3

導(dǎo)數(shù)的綜合應(yīng)用3.3導(dǎo)數(shù)的綜合應(yīng)用-27-考點(diǎn)1考點(diǎn)2考點(diǎn)3

例1設(shè)f(x)=xlnx-ax2+(2a-1)x,a∈R.(1)令g(x)=f'(x),求g(x)的單調(diào)區(qū)間;(2)已知f(x)在x=1處取得極大值.求實(shí)數(shù)a的取值范圍.思考如何求與函數(shù)極值有關(guān)的參數(shù)范圍?-2-考點(diǎn)1考點(diǎn)2考點(diǎn)3例1設(shè)f(x)=xlnx--28-考點(diǎn)1考點(diǎn)2考點(diǎn)3-3-考點(diǎn)1考點(diǎn)2考點(diǎn)3-29-考點(diǎn)1考點(diǎn)2考點(diǎn)3-4-考點(diǎn)1考點(diǎn)2考點(diǎn)3-30-考點(diǎn)1考點(diǎn)2考點(diǎn)3-5-考點(diǎn)1考點(diǎn)2考點(diǎn)3-31-考點(diǎn)1考點(diǎn)2考點(diǎn)3解題心得依據(jù)題意,對參數(shù)分類,分類后相當(dāng)于增加了一個已知條件,在增加了條件的情況下,對參數(shù)的各個范圍逐個驗(yàn)證是否符合題意,符合題意的范圍即為所求范圍.-6-考點(diǎn)1考點(diǎn)2考點(diǎn)3解題心得依據(jù)題意,對參數(shù)分類,分類后-32-考點(diǎn)1考點(diǎn)2考點(diǎn)3對點(diǎn)訓(xùn)練1設(shè)函數(shù)f(x)=x2-2x+mlnx+1,其中m為常數(shù).(2)若函數(shù)f(x)有唯一極值點(diǎn),求實(shí)數(shù)m的取值范圍.-7-考點(diǎn)1考點(diǎn)2考點(diǎn)3對點(diǎn)訓(xùn)練1設(shè)函數(shù)f(x)=x2-2x-33-考點(diǎn)1考點(diǎn)2考點(diǎn)3-8-考點(diǎn)1考點(diǎn)2考點(diǎn)3-34-考點(diǎn)1考點(diǎn)2考點(diǎn)3-9-考點(diǎn)1考點(diǎn)2考點(diǎn)3-35-考點(diǎn)1考點(diǎn)2考點(diǎn)3綜上,當(dāng)m≤0時,函數(shù)f(x)有唯一極值點(diǎn),即f(x)有唯一極值點(diǎn),故實(shí)數(shù)m的取值范圍為(-∞,0].-10-考點(diǎn)1考點(diǎn)2考點(diǎn)3綜上,當(dāng)m≤0時,函數(shù)f(x)有唯-36-考點(diǎn)1考點(diǎn)2考點(diǎn)3例2已知函數(shù)f(x)=-x3+x2,g(x)=alnx(a≠0,a∈R).(1)求f(x)的極值;(2)若對任意x∈[1,+∞),使得f(x)+g(x)≥-x3+(a+2)x恒成立,求實(shí)數(shù)a的取值范圍;思考利用導(dǎo)數(shù)解決不等式恒成立問題的基本思路是什么?-11-考點(diǎn)1考點(diǎn)2考點(diǎn)3例2已知函數(shù)f(x)=-x3+x2-37-考點(diǎn)1考點(diǎn)2考點(diǎn)3-12-考點(diǎn)1考點(diǎn)2考點(diǎn)3-38-考點(diǎn)1考點(diǎn)2考點(diǎn)3-13-考點(diǎn)1考點(diǎn)2考點(diǎn)3-39-考點(diǎn)1考點(diǎn)2考點(diǎn)3-14-考點(diǎn)1考點(diǎn)2考點(diǎn)3-40-考點(diǎn)1考點(diǎn)2考點(diǎn)3解題心得利用導(dǎo)數(shù)解決不等式恒成立問題,首先要構(gòu)造函數(shù),利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性,然后求出最值,進(jìn)而得出相應(yīng)的含參不等式,最后求出參數(shù)的取值范圍;也可分離變量,構(gòu)造函數(shù),直接把問題轉(zhuǎn)化為函數(shù)的最值問題.-15-考點(diǎn)1考點(diǎn)2考點(diǎn)3解題心得利用導(dǎo)數(shù)解決不等式恒成立問-41-考點(diǎn)1考點(diǎn)2考點(diǎn)3對點(diǎn)訓(xùn)練2(2017遼寧大連一模)已知函數(shù)f(x)=ax-lnx.(1)過原點(diǎn)O作函數(shù)f(x)圖象的切線,求切點(diǎn)的橫坐標(biāo);(2)對?x∈[1,+∞),不等式f(x)≥a(2x-x2)恒成立,求實(shí)數(shù)a的取值范圍.解:(1)設(shè)切點(diǎn)為M(x0,f(x0)),切線方程為y-f(x0)=k(x-x0),又切線過原點(diǎn)O,∴-ax0+lnx0=-ax0+1,由lnx0=1,解得x0=e,∴切點(diǎn)的橫坐標(biāo)為e.-16-考點(diǎn)1考點(diǎn)2考點(diǎn)3對點(diǎn)訓(xùn)練2(2017遼寧大連一模)-42-考點(diǎn)1考點(diǎn)2考點(diǎn)3(2)∵不等式ax-lnx≥a(2x-x2)恒成立,∴等價(jià)于a(x2-x)≥lnx對?x∈[1,+∞)恒成立.設(shè)y1=a(x2-x),y2=lnx,由于x∈[1,+∞),且當(dāng)a≤0時,y1≤y2,故a>0.設(shè)g(x)=ax2-ax-lnx,當(dāng)0<a<1時,g(3)=6a-ln3≥0不恒成立,當(dāng)a≥1,x=1時,g(x)≥0恒成立;-17-考點(diǎn)1考點(diǎn)2考點(diǎn)3(2)∵不等式ax-lnx≥a(-43-考點(diǎn)1考點(diǎn)2考點(diǎn)3例3已知函數(shù)f(x)=(x-2)ex+a(x-1)2.(1)討論f(x)的單調(diào)性;(2)若f(x)有兩個零點(diǎn),求a的取值范圍.思考如何利用導(dǎo)數(shù)求與函數(shù)零點(diǎn)有關(guān)的參數(shù)范圍?解(1)f'(x)=(x-1)ex+2a(x-1)=(x-1)(ex+2a).(ⅰ)設(shè)a≥0,則當(dāng)x∈(-∞,1)時,f'(x)<0;當(dāng)x∈(1,+∞)時,f'(x)>0.所以f(x)在(-∞,1)內(nèi)單調(diào)遞減,在(1,+∞)內(nèi)單調(diào)遞增.-18-考點(diǎn)1考點(diǎn)2考點(diǎn)3例3已知函數(shù)f(x)=(x-2)e-44-考點(diǎn)1考點(diǎn)2考點(diǎn)3-19-考點(diǎn)1考點(diǎn)2考點(diǎn)3-45-考點(diǎn)1考點(diǎn)2考點(diǎn)3-20-考點(diǎn)1考點(diǎn)2考點(diǎn)3-46-考點(diǎn)1考點(diǎn)2考點(diǎn)3(ⅱ)設(shè)a=0,則f(x)=(x-2)ex,所以f(x)只有一個零點(diǎn).又當(dāng)x≤1時f(x)<0,故f(x)不存在兩個零點(diǎn);單調(diào)遞增.又當(dāng)x≤1時f(x)<0,故f(x)不存在兩個零點(diǎn).綜上,a的取值范圍為(0,+∞).-21-考點(diǎn)1考點(diǎn)2考點(diǎn)3(ⅱ)設(shè)a=0,則f(x)=(x--47-考點(diǎn)1考點(diǎn)2考點(diǎn)3解題心得與函數(shù)零點(diǎn)有關(guān)的參數(shù)范圍問題,往往利用導(dǎo)數(shù)研究函數(shù)的單調(diào)區(qū)間和極值點(diǎn),并結(jié)合特殊點(diǎn),從而判斷函數(shù)的大致圖象,討論其圖象與x軸的位置關(guān)系(或者轉(zhuǎn)化為兩個熟悉函數(shù)的圖象交點(diǎn)問題),進(jìn)而確定參數(shù)的取值范圍.-22-考點(diǎn)1考點(diǎn)2考點(diǎn)3解題心得與函數(shù)零點(diǎn)有關(guān)的參數(shù)范圍問-48-考點(diǎn)1考點(diǎn)2考點(diǎn)3對點(diǎn)訓(xùn)練3(2017天津六校聯(lián)考)設(shè)函數(shù)f(x)=lnx-ax2-bx.(1)當(dāng)a=b=時,求函數(shù)f(x)的單調(diào)區(qū)間;(2)當(dāng)a=0,b=-1時,方程