高考數(shù)學(xué)人教A版理科第一輪復(fù)習(xí)課件33導(dǎo)數(shù)的綜合應(yīng)用

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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?

-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)3-5-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(duì)參數(shù)分類,分類后相當(dāng)于增加了一個(gè)已知條件,在增加了條件的情況下,對(duì)參數(shù)的各個(gè)范圍逐個(gè)驗(yàn)證是否符合題意,符合題意的范圍即為所求范圍.-7-考點(diǎn)1考點(diǎn)2考點(diǎn)3對(duì)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.-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)3-10-考點(diǎn)1考點(diǎn)2考點(diǎn)3綜上,當(dāng)m≤0時(shí),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].-11-考點(diǎn)1考點(diǎn)2考點(diǎn)3例2(2017貴州貴陽(yáng)一模)設(shè)f(x)=xex,g(x)=x2+x.(1)令F(x)=f(x)+g(x),求F(x)的最小值;(2)若任意x1,x2∈[-1,+∞),且x1>x2,有m[f(x1)-f(x2)]>g(x1)-g(x2)恒成立,求實(shí)數(shù)m的取值范圍.思考利用導(dǎo)數(shù)解決不等式恒成立問(wèn)題的基本思路是什么?解:(1)∵F(x)=f(x)+g(x)=xex+x2+x,∴F'(x)=(x+1)(ex+1),令F'(x)>0,解得x>-1,令F'(x)<0,解得x<-1,∴F(x)在(-∞,-1)內(nèi)遞減,在(-1,+∞)內(nèi)遞增.-12-考點(diǎn)1考點(diǎn)2考點(diǎn)3(2)∵任意x1,x2∈[-1,+∞),且x1>x2,有m[f(x1)-f(x2)]>g(x1)-g(x2)恒成立,∴mf(x1)-g(x1)>mf(x2)-g(x2)恒成立.解題心得利用導(dǎo)數(shù)解決不等式恒成立問(wèn)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(wèn)題轉(zhuǎn)化為函數(shù)的最值問(wèn)題.-13-考點(diǎn)1考點(diǎn)2考點(diǎn)3(2)若在區(qū)間(1,+∞)上,函數(shù)f(x)的圖象恒在直線y=2ax的下方,求a的取值范圍.-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)3-16-考點(diǎn)1考點(diǎn)2考點(diǎn)3例3(2017全國(guó)Ⅰ,理21)已知函數(shù)f(x)=ae2x+(a-2)ex-x.(1)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.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)的定義域?yàn)?-∞,+∞),f'(x)=2ae2x+(a-2)ex-1=(aex-1)(2ex+1).(ⅰ)若a≤0,則f'(x)<0,所以f(x)在(-∞,+∞)內(nèi)單調(diào)遞減.(ⅱ)若a>0,則由f'(x)=0得x=-ln

a.當(dāng)x∈(-∞,-ln

a)時(shí),f'(x)<0;當(dāng)x∈(-ln

a,+∞)時(shí),f'(x)>0,所以f(x)在(-∞,-ln

a)內(nèi)單調(diào)遞減,在(-ln

a,+∞)內(nèi)單調(diào)遞增.-17-考點(diǎn)1考點(diǎn)2考點(diǎn)3(2)(ⅰ)若a≤0,由(1)知,f(x)至多有一個(gè)零點(diǎn).(ⅱ)若a>0,由(1)知,當(dāng)x=-ln

a時(shí),f(x)取得最小值,又f(-2)=ae-4+(a-2)e-2+2>-2e-2+2>0,故f(x)在(-∞,-ln

a)內(nèi)有一個(gè)零點(diǎn).-18-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(wèn)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(gè)熟悉函數(shù)的圖象交點(diǎn)問(wèn)題