第四章　第四節(jié)　第3課時(shí)　導(dǎo)數(shù)的不等式問(wèn)題

PAGE溫馨提示：此套題為Word版，請(qǐng)按住Ctrl,滑動(dòng)鼠標(biāo)滾軸，調(diào)節(jié)合適的觀看比例，答案解析附后。板塊。第3課時(shí)導(dǎo)數(shù)的不等式問(wèn)題【命題分析】導(dǎo)數(shù)中的不等式證明經(jīng)常被考查,常與函數(shù)的性質(zhì)、函數(shù)的零點(diǎn)與極值、數(shù)列等相結(jié)合,題目難度較大,解題方法多種多樣,如構(gòu)造函數(shù)法、放縮法等,針對(duì)不同的題目,采用不同的解題方法,可以達(dá)到事半功倍的效果.【核心考點(diǎn)·分類突破】考點(diǎn)一作差法構(gòu)造函數(shù),證明不等式[例1]設(shè)f(x)=2xlnx+1.(1)求f(x)的最小值;【解析】(1)f'(x)=2(lnx+1),所以當(dāng)x∈(0,1e)時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)x∈(1e,+∞)時(shí),f'(x)>0,f(x)單調(diào)遞增,所以當(dāng)x=1e時(shí),f(x)取得最小值f(1(2)證明:f(x)≤x2-x+1x+2ln【解析】(2)x2-x+1x+2lnx-f(x)=x(x-1)-x-1x-2(x-1)lnx=(x-1)(x-令g(x)=x-1x-2lnx,則g'(x)=1+1x2-2x=(x-1)2x2≥0,所以g(x)在(0,+∞)上單調(diào)遞增,又g(1)=0,所以當(dāng)0<x<1時(shí),g(x)<0,當(dāng)x>1時(shí),g即f(x)≤x2-x+1x+2ln解題技法作差法構(gòu)造函數(shù),證明不等式的策略(1)待證不等式的兩邊含有同一個(gè)變量時(shí),一般地,可以直接構(gòu)造“左減右”的函數(shù);(2)有時(shí)對(duì)復(fù)雜的式子要進(jìn)行變形,借助所構(gòu)造函數(shù)的單調(diào)性和最值求解,利用導(dǎo)數(shù)研究其單調(diào)性和最值.對(duì)點(diǎn)訓(xùn)練已知函數(shù)f(x)=x+xlnx,求證:f(x)>3(x-1).【證明】令g(x)=f(x)-3(x-1),即g(x)=xlnx-2x+3(x>0).g'(x)=lnx-1,由g'(x)=0,得x=e.由g'(x)>0,得x>e;由g'(x)<0,得0<x<e.所以g(x)在(0,e)上單調(diào)遞減,在(e,+∞)上單調(diào)遞增,所以g(x)在(0,+∞)上的最小值為g(e)=3-e>0.于是在(0,+∞)上,都有g(shù)(x)≥g(e)>0,所以f(x)>3(x-1).考點(diǎn)二分拆函數(shù)法證明不等式[例2]已知函數(shù)f(x)=elnx-ax(x>0).(1)討論函數(shù)f(x)的單調(diào)性;【解析】(1)f'(x)=ex-a(x①若a≤0,則f'(x)>0,f(x)在(0,+∞)上單調(diào)遞增;②若a>0,則當(dāng)0<x<ea時(shí),f'(x當(dāng)x>ea時(shí),f'(x)<0,所以f(x)在(0,ea)上單調(diào)遞增,在(e綜上,當(dāng)a≤0時(shí),f(x)在(0,+∞)上單調(diào)遞增;當(dāng)a>0時(shí),f(x)在(0,ea)上單調(diào)遞增,在(ea(2)當(dāng)a=e時(shí),證明:xf(x)-ex+2ex≤0.【解析】(2)方法一:因?yàn)閤>0,所以只需證f(x)≤ex當(dāng)a=e時(shí),由(1)知,f(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,所以f(x)max=f(1)=-e.記g(x)=exx-2e(則g'(x)=(x所以當(dāng)0<x<1時(shí),g'(x)<0,當(dāng)x>1時(shí),g'(x)>0,故g(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,所以g(x)min=g(1)=-e.綜上,當(dāng)x>0時(shí),f(x)≤g(x),即f(x)≤exx-2e,即xf(x)-ex+2ex方法二:由題意知,即證exlnx-ex2-ex+2ex≤0,從而等價(jià)于lnx-x+2≤ex設(shè)函數(shù)G(x)=lnx-x+2,則G'(x)=1x-1所以當(dāng)x∈(0,1)時(shí),G'(x)>0,當(dāng)x∈(1,+∞)時(shí),G'(x)<0,故G(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,從而G(x)在(0,+∞)上的最大值為g(1)=1.設(shè)函數(shù)h(x)=exex,則h'(x所以當(dāng)x∈(0,1)時(shí),h'(x)<0,當(dāng)x∈(1,+∞)時(shí),h'(x)>0,故h(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,從而h(x)在(0,+∞)上的最小值為h(1)=1.綜上,當(dāng)x>0時(shí),g(x)≤h(x),即xf(x)-ex+2ex≤0.解題技法1.當(dāng)直接求導(dǎo)后導(dǎo)數(shù)式比較復(fù)雜或無(wú)從下手時(shí),可將待證式進(jìn)行變形,構(gòu)造兩個(gè)函數(shù),從而找到可以傳遞的中間量.在證明過(guò)程中,等價(jià)轉(zhuǎn)化是關(guān)鍵,此處g(x)min≥f(x)max恒成立,從而f(x)≤g(x)恒成立.2.等價(jià)變形的目的是求導(dǎo)后簡(jiǎn)單地找到極值點(diǎn),一般地,ex與lnx要分離,常構(gòu)造xn與lnx,xn與ex的積、商形式.對(duì)點(diǎn)訓(xùn)練證明:對(duì)一切x∈(0,+∞),都有l(wèi)nx>1ex-2【證明】問(wèn)題等價(jià)于證明xlnx>xex-2e(x∈(0,+∞)).設(shè)f(x)=xlnx,f'(x)=1+lnx,易知x=1e為則f(x)=xlnx(x∈(0,+∞))的最小值是-1e,當(dāng)且僅當(dāng)x=1e設(shè)m(x)=xex-2e(x∈(0,+∞)),則m'(x)=1-xex,由m'(x由m'(x)>0得0<x<1時(shí),m(x)單調(diào)遞增,易知m(x)max=m(1)=-1e,當(dāng)且僅當(dāng)x=1時(shí)取到.從而對(duì)一切x∈(0,+∞),xlnx≥-1e≥xex-2e,兩個(gè)等號(hào)不同時(shí)取到,所以對(duì)一切x∈(0,+∞)都有l(wèi)nx>考點(diǎn)三放縮構(gòu)造函數(shù)證明不等式[例3]f(x)=ex.(1)求曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程;【解析】(1)由f(x)=ex,得f(0)=1,f'(x)=ex,則f'(0)=1,即曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程為y-1=x-0,所以所求切線方程為x-y+1=0.(2)當(dāng)x>-2時(shí),求證:f(x)>ln(x+2).【解析】(2)設(shè)g(x)=f(x)-(x+1)=ex-x-1(x>-2),則g'(x)=ex-1,當(dāng)-2<x<0時(shí),g'(x)<0;當(dāng)x>0時(shí),g'(x)>0,即g(x)在(-2,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增,于是當(dāng)x=0時(shí),g(x)min=g(0)=0,因此f(x)≥x+1(當(dāng)x=0時(shí)取等號(hào)).令h(x)=x+1-ln(x+2)(x>-2),則h'(x)=1-1x+2=則當(dāng)-2<x<-1時(shí),h'(x)<0,當(dāng)x>-1時(shí),h'(x)>0,即h(x)在(-2,-1)上單調(diào)遞減,在(-1,+∞)上單調(diào)遞增,于是當(dāng)x=-1時(shí),h(x)min=h(-1)=0,因此x+1≥ln(x+2)(當(dāng)x=-1時(shí)取等號(hào)),所以當(dāng)x>-2時(shí),f(x)>ln(x+2).解題技法放縮法證明不等式的策略導(dǎo)數(shù)方法證明不等式的問(wèn)題中,最常見(jiàn)的是ex和lnx與其他代數(shù)式結(jié)合的問(wèn)題,對(duì)于這類問(wèn)題,可以考慮先對(duì)ex和lnx進(jìn)行放縮,使問(wèn)題簡(jiǎn)化,簡(jiǎn)化后再構(gòu)造函數(shù)進(jìn)行證明.常見(jiàn)的放縮公式如下:(1)ex≥1+x,當(dāng)且僅當(dāng)x=0時(shí)取等號(hào);(2)lnx≤x-1,當(dāng)且僅當(dāng)x=1時(shí)取等號(hào).對(duì)點(diǎn)訓(xùn)練已知f(x)=aex-1-ln(1)若a=1,求f(x)在(1,f(1))處的切線方程;【解析】(1)當(dāng)a=1時(shí),f(x)=ex-1-lnxf'(x)=ex-1-1x,又f(1)=0,所以切點(diǎn)為(1,0).所以切線方程為y-0=0(x-1),即y=0.(2)證明:當(dāng)a≥1時(shí),f(x)≥0.【解析】(2)因?yàn)閍≥1,所以aex-1所以f(x)≥ex-1-ln方法一:令φ(x)=ex-1-lnx所以φ'(x)=ex-1令h(x)=ex-1-1x,所以h'(x)=所以φ'(x)在(0,+∞)上單調(diào)遞增,又φ'(1)=0,所以當(dāng)x∈(0,1)時(shí),φ'(x)<0;當(dāng)x∈(1,+∞)時(shí),φ'(x)>0,所以φ(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,所以φ(x)min=φ(1)=0,所以φ(x)≥0,所以f(x)≥φ(x)≥0,即f(x)≥0.方法二:令g(x)=ex-x-1,所以g'(x)=ex-1.當(dāng)x∈(-∞,0)時(shí),g'(x)<0;當(dāng)x∈(0,+∞)時(shí),g'(x)>0,所以g(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增,所以g(x)min=g(0)=0,故ex≥x+1,當(dāng)且僅當(dāng)x=0時(shí)取“=”.同理可證lnx≤x-1,當(dāng)且僅當(dāng)x=1時(shí)取“=”.由ex≥x+1?ex-1≥x由x-1≥lnx?x≥lnx+1(當(dāng)且僅當(dāng)x=1時(shí)取“=”),所以ex-1≥x≥lnx+1,即e即ex-1-lnx即f(x)≥0.【加練備選】(2023·南充模擬)已知函數(shù)f(x)=ax-sinx.(1)若函數(shù)f(x)為增函數(shù),求實(shí)數(shù)a的取值范圍;【解析】(1)f(x)=ax-sinx,所以f'(x)=a-cosx,由函數(shù)f(x)為增函數(shù),則f'(x)=a-cosx≥0恒成立,即a≥cosx在R上恒成立,因?yàn)閥=cosx∈[-1,1],所以a≥1,即實(shí)數(shù)a的取值范圍是[1,+∞).(2)求證:當(dāng)x>0時(shí),ex>2sinx.【解析】(2)由(1)知,當(dāng)a=1時(shí),f(x)=x-sinx為增函數(shù),當(dāng)x>0時(shí),f(x)>f(0)=0?x>sinx,要證當(dāng)x>0時(shí),ex>2sinx,只需證當(dāng)x>0時(shí),ex>2x,即證ex-2x>0在(0,+∞)上恒成立.設(shè)g(x)=ex-2x(x>0),則g'(x)=ex-2,令g'(x)=0解得x=ln2,所以g(x)在(0,ln2)上單調(diào)遞減,在(ln2,+∞)上單調(diào)遞增,所以g(x)min=g(ln2)=eln2-2ln2=2(1-ln2)>0,所以g(x)≥g(ln2)>0,所以ex>2x成立,故當(dāng)x>0時(shí),ex>2sinx.重難突破泰勒公式在比較大小的應(yīng)用比較大小的選擇題是近年高考的常見(jiàn)題型,一般情況下我們會(huì)構(gòu)造函數(shù)模型代入數(shù)值進(jìn)行比較和運(yùn)算,但是對(duì)學(xué)生來(lái)說(shuō)函數(shù)模型的選擇是非常有難度的,因此在選擇題中我們可以選擇利用泰勒公式計(jì)算近似值的辦法進(jìn)行比較大小.【教材探源】在人教A必修一教材中三角函數(shù)一章第256頁(yè)“拓廣探索”中第26題.英國(guó)數(shù)學(xué)家泰勒給出如下公式:sinx=x-x33!+xcosx=1-x22!+x其中n!=1×2×3×4×…×n.這些公式被編入計(jì)算工具,計(jì)算工具計(jì)算足夠多的項(xiàng)就可以確保顯示值的精確性.比如,用前三項(xiàng)計(jì)算cos0.3,就得到cos0.3≈1-0.322!+【教材拓展】下面給出高中階段常用的泰勒公式:(1)ex=1+x+12!x2+…+1n!xn+…,(2)sinx=x-x33!+…+(-1)kx∈R;(3)cosx=1-x22!+…+(-1)k1(x∈R;(4)ln(1+x)=x-12x2+…+(-1)n-11nxn+…,-1<(5)(1+x)a=1+ax+a(a-1)2!x2+…+類型一利用ex、ln(1+x)的泰勒展開(kāi)式比較[例1](2022·新高考Ⅰ卷)設(shè)a=0.1e0.1,b=19,c=-ln0.9,則(A.a<b<c B.c<b<aC.c<a<b D.a<c<b【解析】選C.b=19≈0.由公式ex=1+x+x2可得e0.1≈1+0.1+0.12則a=0.1e0.1≈0.1105,c=-ln0.9=ln109=ln(1+1由公式ln(1+x)=x-x22+得c=ln(1+19)≈19-(1所以c<a<b.類型二利用sinx,cosx的泰勒展開(kāi)式比較[例2](2022·全國(guó)甲卷)已知a=3132,b=cos14,c=4sin14A.c>b>a B.b>a>cC.a>b>c D.a>c>b【解析】選A.由公式sinx=x-x3得c=4sin14≈4(14-1436由公式cosx=1-x22!得b=cos14≈1-1422+(1排除BCD.【一題多法】構(gòu)造函數(shù)h(x)=1-12x2-cosx,x∈[0,π2],則g(x)=h'(x)=-x+sinx,g'(=-1+cosx≤0,所以g(x)≤g(0)=0,因此h(x)在[0,π2]所以h(14)=a-b<h(0)=0,即a<bcb=4sin14cos14=tan141所以cb>1,即b<c,綜上c>b>類型三利用ln(1+x),(1+x)a的泰勒展開(kāi)式比較[例3](2021·全國(guó)乙卷)設(shè)a=2ln1.01,b=ln1.02,c=1.04-1,則(A.a<b<c B.b<c<aC.b<a<c D.c<a<b【解析】選B.因?yàn)閍=2ln1.01=ln1.012=ln1.0201>ln1.02,所以a>b.排除A,D.根據(jù)選項(xiàng)B,C可知,只需比較a,c的大小即可.由公式ln(1+x)=x-x2得a=2ln(1+0.01)≈2×(0.01-0.≈0.02-0.0001=0.0199,由公式(1+x)a=1+ax+a(a-得c=(1+0.04)12-1≈12×0≈0.02-0.0002=0.0198,所以a>c.排除C.解題技法通過(guò)以上示例可以看出,利用泰勒公式近似計(jì)算求解難度比較大的試題確實(shí)可以提高解題速度,運(yùn)用該法的難點(diǎn)是要利用數(shù)字特征構(gòu)造對(duì)應(yīng)的函數(shù).對(duì)點(diǎn)訓(xùn)練1.已知a=e0.02,b=1.02,c=ln2.02,則()A.c>a>b B.a>b>cC.a>c>b D.b>a>c【解析】選B.方法一(泰勒公式):設(shè)x=0.02,則a=e0.02=1+0.02+0.顯然a>b>1>c.方法二(構(gòu)造函數(shù)):令f(x)=ex-(1+x),令f'(x)=ex-1=0,得x=0,當(dāng)x∈(-∞,0)時(shí),f'(x)<0,函數(shù)f(x)在(-∞,0)上單調(diào)遞減,當(dāng)x∈(0,+∞)時(shí),f'(x)>0,函數(shù)f(x)在(0,+∞)上單調(diào)遞增,所以f(0.02)>f(0)=0,從而e0.02>1+0.02=1.