第18講 第3課時(shí)　放縮法證明不等式

第18講導(dǎo)數(shù)與不等式第3課時(shí)放縮法證明不等式【備選理由】例1、例2均考查通過放縮法證明一元不等式,鞏固常用的指對(duì)數(shù)的放縮結(jié)論,進(jìn)一步熟悉放縮技巧,強(qiáng)化放縮解題意識(shí);例3雖然是考查不等式恒成立求參數(shù)問題,但是在其中穿插了比較多的三角放縮的技巧的應(yīng)用,注意三角放縮技巧在不等式恒成立或者不等式證明中的運(yùn)用;例4考查通過放縮法證明多元不等式,整體代換消元為放縮證明提供了前提.例1[配例1使用]已知函數(shù)f(x)=aex-1-lnx-1.(1)若a=1,求曲線y=f(x)在(1,f(1))處的切線方程;(2)證明:當(dāng)a≥1時(shí),f(x)≥0.解:(1)當(dāng)a=1時(shí),f(x)=ex-1-lnx-1(x>0),f'(x)=ex-1-1x,則曲線y=f(x)在(1,f(1))處的切線斜率k=f'(1)=又f(1)=0,∴切點(diǎn)坐標(biāo)為(1,0),∴切線方程為y-0=0×(x-1),即y=0.(2)證明:∵a≥1,∴aex-1≥ex-1,∴f(x)≥ex-1-lnx-1.令g(x)=ex-x-1,則g'(x)=ex-1.當(dāng)x∈(-∞,0)時(shí),g'(x)<0,當(dāng)x∈(0,+∞)時(shí),g'(x)>0,∴g(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增,∴g(x)min=g(0)=0,故ex≥x+1,當(dāng)且僅當(dāng)x=0時(shí)取等號(hào).同理可證lnx≤x-1,當(dāng)且僅當(dāng)x=1時(shí)取等號(hào).由ex≥x+1得ex-1≥x(當(dāng)且僅當(dāng)x=1時(shí)取等號(hào)),由x-1≥lnx得x≥lnx+1(當(dāng)且僅當(dāng)x=1時(shí)取等號(hào)),∴ex-1≥x≥lnx+1,即ex-1≥lnx+1,∴ex-1-lnx-1≥0(當(dāng)且僅當(dāng)x=1時(shí)取等號(hào)),即f(x)≥0.例2[配例1使用][2023·廣州一模]已知函數(shù)f(x)=ex(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)解關(guān)于x的不等式f(x)>12解:(1)由函數(shù)f(x)=ex-11+lnx,得f(x)f'(x)=ex-11+lnx-令g(x)=1+lnx-1x,則g'(x)=1x+1∴函數(shù)g(x)在定義域上單調(diào)遞增,∴當(dāng)0<x<1e或1e<x<1時(shí),f'(x)<0,函數(shù)f(x)當(dāng)x>1時(shí),f'(x)>0,函數(shù)f(x)單調(diào)遞增,∴f(x)的單調(diào)遞減區(qū)間為0,1e,1e,1(2)不等式f(x)>12x+1x,當(dāng)0<x<1e時(shí),f(x)<0,不合題意,舍去.當(dāng)x=1時(shí),不等式的左邊=右邊,不合題意,舍去∴x>1e且x≠1①當(dāng)1e<x<1時(shí),ex-1>x,要證ex-11+lnx>12x+1x,只需證令F(x)=2x21+x2-(1+則F'(x)=-(x2-1)2x(1+x2)2<0,∴函數(shù)F(x)在②當(dāng)x>1時(shí),原不等式等價(jià)于2ex-令h(x)=2ex-11+x2(x>1),u(則h'(x)=2ex-1(x-1)2(1+x2∴h(x)>h(1)=1.∵當(dāng)x>1時(shí),lnx<x-1,∴u(x)<1,∴不等式2ex-1綜上可得,不等式f(x)>12x+1x的解集為1例3[配例3使用][2023·長(zhǎng)沙長(zhǎng)郡中學(xué)一模]已知函數(shù)f(x)=(cosx-1)e-x,g(x)=ax2+(1-ex)x(a∈R).(1)當(dāng)x∈(0,π)時(shí),求函數(shù)f(x)的最小值;(2)當(dāng)x∈-π2,+∞時(shí),不等式xf(x)≥g(x解:(1)由題意得f'(x)=1-2sinx+π4ex,當(dāng)x∈(0,π)時(shí),令f'當(dāng)x∈0,π2時(shí),f'(x)<0,當(dāng)x∈π2,π時(shí),所以f(x)min=fπ2=-e(2)由xf(x)≥g(x)ex,得x(cosx-1)e-x≥e-x[ax2+(1-即xcosx-x≥ax2+(1-ex)x,即x(ex+cosx-ax-2)≥0.記h(x)=ex+cosx-ax-2,則xh(x)≥0恒成立,且h'(x)=ex-sinx-a.令k(x)=h'(x),則k'(x)=ex-cosx.當(dāng)a>1,x∈[0,+∞)時(shí),k'(x)≥0,k(x)=h'(x)在[0,+∞)上單調(diào)遞增,且h'(0)=1-a<0,h'(1+a)=e1+a-sin(1+a)-a≥e1+a-1-a,令y=ex-x-1,x>0,則y'=ex-1>0,故y=ex-x-1在(0,+∞)上單調(diào)遞增,又當(dāng)x=0時(shí),y=0,所以當(dāng)x>0時(shí),ex-x-1>0,所以e1+a-(1+a)>e1+a-(1+a)-1>0,即h'(1+a)>0,故存在唯一的x0∈(0,1+a),使得h'(x0)=0,當(dāng)x∈(0,x0)時(shí),h'(x)<0,h(x)單調(diào)遞減,所以h(x)<h(0)=0,此時(shí)xh(x)<0,不合題意.當(dāng)a≤1時(shí),若x>0,則h'(x)>1+x-sinx-a.令y=x-sinx,x>0,則y'=1-cosx≥0,故y=x-sinx在(0,+∞)上單調(diào)遞增,又當(dāng)x=0時(shí),y=0,所以當(dāng)x>0時(shí),x-sinx>0,所以h'(x)>1+x-sinx-a>1-a≥0,故h(x)在(0,+∞)上單調(diào)遞增,所以當(dāng)x>0時(shí),h(x)>h(0)=0恒成立,故xh(x)>0恒成立.若x∈-π2,0,則令t(x)=k'(x),故t'(x)=ex+sin又t'(0)=1,t'-π2=e-π2-1<0,所以存在唯一的x1∈-π2,0當(dāng)x∈-π2,x1時(shí),t'(x)<0,t(x)單調(diào)遞減,當(dāng)x∈(x1,0]時(shí),t'(x)>0,t又t-π2=e-π2>0,t(0)=0,故存在唯一的x2∈-π2,0,使得t(x當(dāng)x∈-π2,x2時(shí),t(x)=k'(x)>0,k(x)=h'(x)單調(diào)遞增,當(dāng)x∈(x2,0]時(shí),t(x)=k'(x)≤0,k(x)=h'又k-π2=e-π2+1-a>0,k所以當(dāng)x∈-π2,0時(shí),k(x)=h'(x)≥0,則h(x)在-π2,0上單調(diào)遞增,所以當(dāng)x∈-π2,0時(shí),h(x綜上,當(dāng)a≤1,x∈-π2,+∞時(shí),xh(x)≥0恒成立,故實(shí)數(shù)a例4[補(bǔ)充使用]已知函數(shù)f(x)=12x2+lnx+mx,m∈R(1)若f(x)存在兩個(gè)極值點(diǎn),求實(shí)數(shù)m的取值范圍;(2)若x1,x2為f(x)的兩個(gè)極值點(diǎn),證明:f(x1)+解:(1)由題可得f'(x)=x+1x+m,x>0,若f(x)存在兩個(gè)極值點(diǎn),則f'(x)=0在(0,+∞)上有兩個(gè)根所以m=-x+1x有兩個(gè)根,即直線y=m與y=-x+1x,x>0的圖象有兩個(gè)交點(diǎn),由y=-x+1當(dāng)x∈(0,1)時(shí),y'>0,y=-x+1x單調(diào)遞增,當(dāng)x∈(1,+∞)時(shí),y'<0,y=-所以當(dāng)x=1時(shí),ymax=-2,又當(dāng)x→0時(shí),y→-∞,當(dāng)x→+∞時(shí),y→-∞,所以m的取值范圍為(-∞,-2).(2)證明:由