導(dǎo)數(shù)雙變量專題

1、1導(dǎo)數(shù)-雙變量問(wèn)題1. 構(gòu)造函數(shù)利用單調(diào)性證實(shí)2. 任意性與存在性問(wèn)題3. 整體換元一雙變單4. 極值點(diǎn)偏移5. 賦值法構(gòu)造函數(shù)利用單調(diào)性證實(shí)形式如：| f (xj - f (x2) |亠 m | 捲-x2 |方法：將相同變量 移到一邊,構(gòu)造函數(shù)_ -*1 '1.函數(shù)f (xH(x2 3)(x 9)對(duì)任意Xi,x 1-1,0 1,不等式I f (xj - f (X2)|乞m恒成立, 試求m的取值范圍.2.函數(shù)f (x) =(a +1)ln x +ax2 +1 .設(shè)ac ,如果對(duì)P心x?亡(0母),有J / <=/( II f(x1)- f(X2)|_4|x1-X2|,求實(shí)數(shù) a

2、的取值范圍.3.函數(shù)f(x)=al n(x1)-x2區(qū)間(0,1)內(nèi)任取兩個(gè)實(shí)數(shù)P2,且Pq時(shí),假設(shè)不等式f ( p 1) - f (q 1) 1恒成立,求實(shí)數(shù)a的取值范圍.p - qII4.函數(shù)f(x)x2-2alnx (a-2)x,aR .是否存在實(shí)數(shù)a,對(duì)任意的2x1,x20：,且X2=X1,有f (x2) - f (xJ . a,恒成立,假設(shè)存在求出a的取值范圍,假設(shè)不x2 _捲存在,說(shuō)明理由.練習(xí)1 :函數(shù)f(x)=alnx x2,假設(shè)a 0 ,且對(duì)任意的X1,x 1,e,都有I f (xj -f儀2)卜：|丄-丄|,求實(shí)數(shù)a的取值范圍.練習(xí)2.設(shè)函數(shù)f (x) = Inx m,m

3、R.假設(shè)對(duì)任意b a 0rfb)':1恒成立, xb a求m的取值范圍.5.函數(shù) f (x) rqx2 - ax a -1 In x,a 1(1 )討論函數(shù)的單調(diào)性(2 )證實(shí)：假設(shè)a : 5,那么對(duì)任意的為兀三0：,且x2 = x1,有f(X2) - f (xjd恒成立> _ IX? - x6.設(shè)函數(shù) f x = emx - x2 -mx(1) 證實(shí)：f x在：,0單調(diào)遞減,在0：?jiǎn)握{(diào)遞增;(2)假設(shè)對(duì)于任意X1,X2T,1】,都有| f(xj 一 f (X2)|_e-1,求m的取值范圍W jijk''lj任意與存在性問(wèn)題2函數(shù) f(x)= x+王,g(x)=x

4、 + l nx,其中 a>0 -x(1) 假設(shè)函數(shù)y二f X在1,e上的圖像恒在y二gx的上方,求實(shí)數(shù)a的取值范圍.(2) 假設(shè)對(duì)任意的x1,x2 1, el ( e為自然對(duì)數(shù)的底數(shù))都有f捲> g x2成立,求實(shí)數(shù)a的取值范圍.2g(x)二-x 2x af (x) =lx3+x2 _3x + 12.函數(shù)3(1)討論方程f(x)=k ( k為常數(shù))的實(shí)根的個(gè)數(shù).(2)假設(shè)對(duì)任意x 10,21,恒有f(x)-a成立,求a的取值范圍.(3)假設(shè)對(duì)任意X0,2丨,恒有f(x)-g X成立,求a的取值范圍.(4)假設(shè)對(duì)任意0,2 存在x20,2】,恒有f(X1)_g X2成立,求a的取值范

5、圍整體換元一一雙變單1.函數(shù) f(x)=ax2 7n x.(I)求f(x)的單調(diào)區(qū)間；(H)當(dāng)a=0時(shí),設(shè)斜率為k的直線與函數(shù)y二f(x)相交于兩點(diǎn)A(x1, y1)> B(x2,y2)(X2 - X1),求證: 捲：-:x2.k歡送共閱練習(xí)1.函數(shù)f (X)=x2 2X, g(x) =iogax(a . 0,且a = 1),其中a為常數(shù),如果2h(x f(x) g(x)在其定義域上是增函數(shù),且h(x)存在零點(diǎn)(h(x)為h(x)的導(dǎo)函數(shù)).(I )求a的值；(II )設(shè) A(m, g(m), B(n,g(n)(m : n)是函數(shù) y=g(x)的圖象上兩點(diǎn),g(x0) = g(n) -

6、g(m) n m(g (x)為g(x)的導(dǎo)函數(shù)),證實(shí):m ： x° ： n.練習(xí)2.函數(shù)f(x) =ln x-ax+ 1,g(x) = _ x2 , a R ;2(1) a ：2 , h(x)二 f (x) g(x),求 h(x)的單調(diào)區(qū)間；I)|(2) a =1,假設(shè) Ocxjvxzcl , f "(t) =_(xjvtvx?),求證：t c xi 十x2X2 Xi21A'練習(xí) 3.函數(shù)f(x戶ex,xR,設(shè)acb,比擬f(a)十f(b)與f(b)- f (a )的大小,并說(shuō) 2b a明理由.2.函數(shù)f x =ln x a -x有且只有一個(gè)零點(diǎn),其中 a>

7、0.(I )求a的值；(II )設(shè)hx = fx x,對(duì)任意為公2三(-1*c紋=x2 ,證實(shí)：不等式12 >JXjX2 +論 +x2 +1 恒成立.hX )-叭)1 2123.f(x)=2lnx-x2+ax在(0,畑)內(nèi)有兩個(gè)零點(diǎn)x(,x2,求證:f'( _ )<0.2練習(xí).函數(shù)f (x) = Inx mx(m R),假設(shè)函數(shù)f (x)有兩個(gè)不同的零點(diǎn)X1, X2,求證:2X1X2> e .歡送共閱4.函數(shù) f (x) = x2 In ax a 0,2(1)假設(shè)f' x "x對(duì)任意的x 0恒成立,求a的取值范圍(2)當(dāng) a =1 時(shí),設(shè)函數(shù) g(x

8、)二F,假設(shè) X1,X21,1,X1X2： 1,求證：x1xX)x24.對(duì)稱軸問(wèn)題X1 X2的證實(shí)歡送共閱1.函數(shù)f x二xe-x.(1)求函數(shù)f x的單調(diào)區(qū)間和極值； 函數(shù) 八gx的圖象與函數(shù)y=f x的圖象關(guān)于直線x=1對(duì)稱.證實(shí)：當(dāng)x 1時(shí),f (x)Ag(x )；(3)如果 X2 = %,且 f Xi = f x2,證實(shí)：xi x2 22.函數(shù) f x 二 ax x2xln a a 0,a = 1(1)求函數(shù)f x的單調(diào)區(qū)間；(2) a 1,證實(shí):當(dāng) x 三 10,亠 j 時(shí),f x j,f ； -x(3)假設(shè)對(duì)任意x2 =治,且當(dāng)f x1 v-f x2時(shí),有x, x2 ： 0,求a的取值范圍.練習(xí).函數(shù)f(x)=xlnx .(1)求函數(shù)f x的單調(diào)區(qū)間和極值；,*i "iJ(2)如果 X2X1,且 f(X1)=f(X2 ),證實(shí)：X1+X2'e賦值法1. 函數(shù)f x二rx -xr 1 -r x 0,其中r為有理數(shù),且0 ： r : 1(1)求f (x )的最小值；(2) 試用(1)的結(jié)果證實(shí)：假設(shè)印_ 0,a2 一 0,bi,b2為正有理數(shù),假設(shè)b 6=1,那么a1bla2)a1b1 - a2b(3)將(2)中的命題推廣到一般形式,并用數(shù)學(xué)歸納法證實(shí).2. 函數(shù) f x =1 nx,g x =ln x,1-；7|nx,"0