導(dǎo)數(shù)中的求參數(shù)取值范圍問題

1、令g(x)x(a2)xa,1(a2)a0,1(a2)a0(2)假設(shè)函數(shù)f(x)在R上單調(diào)遞減,那么即x2(a2)xxxx2,e0,x(a令g(x)x(a2)xa,v圖象開口向上不可能對(duì)xR都成立假設(shè)函數(shù)f(x)在R上單調(diào)遞減,那么f(x)0對(duì)xR都成立,即x2(a2)xae-x0對(duì)xR都成立,xc2,八,一、Ie0,x(a2)xa0對(duì)xR都成立.22,(a2)24aa240故函數(shù)f(x)不可能在R上單調(diào)遞增.綜上可知,函數(shù)f(x)不可能是R上的單調(diào)函數(shù)例2：函數(shù)fxalnxax3aR假設(shè)函數(shù)yf(x)的圖像在點(diǎn)(2,f(2)處的切幫你歸納總結(jié)(五)：導(dǎo)數(shù)中的求參數(shù)取值范圍問題一、常見基此題型：

2、(1)函數(shù)單調(diào)性,求參數(shù)的取值范圍,如函數(shù)f(x)增區(qū)間,那么在此區(qū)間上導(dǎo)函數(shù)f(x)0,如函數(shù)f(x)減區(qū)間,那么在此區(qū)間上導(dǎo)函數(shù)f(x)0.(2)不等式恒成立,求參數(shù)的取值范圍問題,可轉(zhuǎn)化為求函數(shù)的最值問題.2x例1.aR,函數(shù)f(x)(xax)e.(xR,e為自然對(duì)數(shù)的底數(shù))(1)假設(shè)函數(shù)f(x)在(1,1)內(nèi)單調(diào)遞減,求a的取值范圍；(2)函數(shù)f(x)是否為R上的單調(diào)函數(shù),假設(shè)是,求出a的取值范圍；假設(shè)不是,請(qǐng)說明理由.解：(1)ff(x)(x2ax)e-xf(x)(2xa)e-x(x2ax)(e-x)=x2(a2)xae-x.要使f(x)在-1,1上單調(diào)遞減,那么f(x)0對(duì)x(1,

3、1)都成立,2x(a2)xa0對(duì)x(1,1)都成立.g(1)0,那么g(1)0.3a一.2f(x)0對(duì)xR都成立ae-x0對(duì)xR都成立.2)xa0對(duì)xR都成立2線的傾余角為45；,對(duì)于任意t1,2,函數(shù)gx32/,、m_、X3x2f/(x)在區(qū)間(t,3)上總不是單調(diào)函數(shù),求m的取值范圍；解：由f/(2)a1,a22f(x)2lnx2x3g(x)x3(m2)x22x,g/(x)3x2(m4)x22人/2令g(x)0得,(m4)240故g(x)0兩個(gè)根一正一負(fù),即有且只有一個(gè)正根7函數(shù)gxx3x2f/(x)m在區(qū)間(t,3)上總不是單調(diào)函數(shù)2g/(x)0在(t,3)上有且只有實(shí)數(shù)根g/(0)20

4、,g/(t)0,g/(3)037兀2m一,(m4)t23t故m4-3t,3t一2而y33t在t1,2單調(diào)減,例3.函數(shù)f(x)lnx1x1.44x(I)求函數(shù)f(x)的單調(diào)區(qū)間；(n)設(shè)g(x)x22bx4,假設(shè)對(duì)任意m9,綜合得一m93Xi(0,2),x21,2,不等式f(xi)g(xz)恒成立,求實(shí)數(shù)b的取值范圍.1斛：(I)f(x)lnx-x4f(x)1的定義域是(0,4x34xx23224x4x由x0及f(x)0得1x3；由x0及f(x)0得0故函數(shù)f(x)的單調(diào)遞增區(qū)間是(1,3);單調(diào)遞減區(qū)間是(0,1),(3,)(II)假設(shè)對(duì)任意x1(0,2),x21,2,不等式f(x1)g(x

5、2"l1成立,問題等價(jià)于f(x)ming(x)max,由(I)可知,在(0,2)上,x1是函數(shù)極小值點(diǎn),這個(gè)極小值是唯一的極值點(diǎn),故也是最小值點(diǎn),所以f(x)min2g(x)x2bx4,x1.2當(dāng)b1時(shí),g(x)maxg(1)2b5；當(dāng)1b2時(shí),g(x)maxg(b)b24；當(dāng)b2時(shí),g(x)maxg(2)4b8;b1問題等價(jià)于1122bb21 2b24b8解得b,142rr14.一即b,所以實(shí)數(shù)b的取值范圍是22一一.2例4.設(shè)函數(shù)f(x)xmlnx,h(x)xxa,(1)當(dāng)a=0時(shí),f(x)Rh(x)在(1,+°°)上恒成立,求實(shí)數(shù)m的取值范圍;(2)當(dāng)m2時(shí)

6、,假設(shè)函數(shù)k(x)=f(x)-h(x)在1,3上恰有兩個(gè)不同零點(diǎn),求實(shí)數(shù)a的取值范圍.解：(1)由a=0,f(x)>h(x),可得一mnx>x,xC(1,十0°)即m<-x-.lnx、x一記6(x)=j,那么f(x)>h(x)在(1,+°°)上恒成立等價(jià)于()(x)min.lnx,lnx1求得()(x)=ln2x當(dāng)xC(1,e),6'(x)v0；當(dāng)xC(e,+8)時(shí),6'(x)>0.故6(x)在x=e處取得極小值,也是最小值,即6(x)min=6(e)=e,故m<e.(2)函數(shù)k(x)=f(x)h(x)在1,3上

7、恰有兩個(gè)不同的零點(diǎn)等價(jià)于方程x-21nx=a,在1,3上恰有兩個(gè)相異實(shí)根.令g(x)=x2ln,貝Ug'(x)v12.x當(dāng)xC1,2)時(shí),g'(x)<0;當(dāng)xC(2,3時(shí),g'(x)>0.g(x)在(1,2)上是單調(diào)遞減函數(shù),在(2,3上是單調(diào)遞增函數(shù).故g(x)min=g(2)=22ln2.又g(1)=1,g(3)=32ln3,-g(1)>g(3),只需g(2)vawg(3).故a的取值范圍是(2ln2,32ln3.二、針對(duì)性練習(xí)1.函數(shù)f(x)x2alnx.假設(shè)函數(shù)g(x)f(x)2x在1,4上是減函數(shù),求實(shí)數(shù)a的取值范圍._,22_a2解：由g(

8、x)xalnx一,得g(x)2x.xxx一一,92又函數(shù)g(x)xalnx一為1,4上的單倜減函數(shù).x那么g(x)0在1,4上恒成立,所以不等式2xa20在1,4上恒成立.xxrr22.即a2x在1,4上恒成立.x2_2設(shè)(x)2x,顯然(x)在1,4上為減函數(shù),x所以(x)的最小值為(4)63.2a的取值范圍是a63.22.函數(shù)f(x)ex1x4.(1)右存在x1,ln-,使ae1x0成立,求a的取值范圍;3(2)當(dāng)X0時(shí),f(x)tx2恒成立,求t的取值范圍.解：(1)aex2,不符合題意.綜上可得x,即af(x).f'(x)ex10,x0.x0時(shí),f(x)0,x0時(shí),f(x)0.

9、f(x)在(,0)上減,在(0,)上增.x0又41,1n3時(shí),f(x)的最大值在區(qū)間端點(diǎn)處取到4一31n14一3,4-3Inf,1一e111eD/kf4-3In1-31一e4-3In14-31一e4-3InfD/kf441f(1)fln,f(x)1,ln13在3上最大值為e1故a的取值范圍是e,(3)由得x0時(shí),exx1tx20恒成立,x2'x設(shè)g(x)ex1tx.g(x)e12tx.x由(2)知e1x,當(dāng)且僅當(dāng)x0時(shí)等號(hào)成立,'故g(x)x2tx(12t)x,從而當(dāng)12t0,t1即2時(shí),g(x)0(x0),g(x)為增函數(shù),又g(0)0,2t1于是當(dāng)x0時(shí),g(x)0,即f(x)tx,2時(shí)符合題意1xxt由e1x(x0)可得e1x(x0),從而當(dāng)2時(shí),g(x)ex12t(ex1)ex(ex1)(ex2t),故當(dāng)x(0,ln2t)時(shí),g(x)0,g(x)為減函數(shù),又g(0)0,1t的取值范圍為,213.函數(shù)f(x)1n(1%),設(shè)h(x)xf(x)xxax3在(0,2)上有極值,求a的取值范圍.解：由h(x)xf(x)xax3可得,hN11-M3"'*3"+1)n(x)-1-3gx9假設(shè)a20,對(duì)任總工2)出(工)<0,b(x)在(0,2