第17講　導(dǎo)數(shù)與函數(shù)的極值、最值（解析版）

第17講導(dǎo)數(shù)與函數(shù)的極值、最值基礎(chǔ)知識(shí)1.函數(shù)的極值與導(dǎo)數(shù)條件f(x)在x0處可導(dǎo),f'(x0)=0一般地,設(shè)函數(shù)y=f(x)的定義域?yàn)镈,設(shè)x0∈D,如果對(duì)于x0附近的任意不同于x0的x,都有f(x)<f(x0)f(x)>f(x0)極值f(x)在x0處取值

f(x)在x0處取值

極值點(diǎn)為極大值點(diǎn)

為極小值點(diǎn)

2.函數(shù)的最值(1)在閉區(qū)間[a,b]上連續(xù)的函數(shù)f(x)在[a,b]上必有最大值與最小值.(2)若函數(shù)f(x)在[a,b]上單調(diào)遞增,則為函數(shù)的最小值,為函數(shù)的最大值;若函數(shù)f(x)在[a,b]上單調(diào)遞減,則為函數(shù)的最大值,為函數(shù)的最小值.

3.實(shí)際應(yīng)用題理解題意、建立函數(shù)模型,使用導(dǎo)數(shù)方法求解函數(shù)模型,根據(jù)求解結(jié)果回答實(shí)際問(wèn)題.1.極大極小x0x02.(2)f(a)f(b)f(a)f(b)常用結(jié)論導(dǎo)數(shù)研究不等式的關(guān)鍵是函數(shù)的單調(diào)性和最值,各類(lèi)不等式與函數(shù)最值的關(guān)系如下:不等式類(lèi)型與最值的關(guān)系?x∈D,f(x)>M?x∈D,f(x)min>M?x∈D,f(x)<M?x∈D,f(x)max<M?x0∈D,f(x0)>M?x∈D,f(x)max>M?x0∈D,f(x0)<M?x∈D,f(x)min<M?x∈D,f(x)>g(x)?x∈D,[f(x)-g(x)]min>0?x∈D,f(x)<g(x)?x∈D,[f(x)-g(x)]max<0?x1∈D1,?x2∈D2,f(x1)>g(x2)?x1∈D1,?x2∈D2,f(x1)min>g(x2)max?x1∈D1,?x2∈D2,f(x1)>g(x2)?x1∈D1,?x2∈D2,f(x1)min>g(x2)min?x1∈D1,?x2∈D2,f(x1)>g(x2)?x1∈D1,?x2∈D2,f(x1)max>g(x2)max?x1∈D1,?x2∈D2,f(x1)>g(x2)?x1∈D1,?x2∈D2,f(x1)max>g(x2)min(注:上述的大于、小于分別改為不小于、不大于,相應(yīng)的與最值關(guān)系對(duì)應(yīng)的不等號(hào)也改變)分類(lèi)訓(xùn)練探究點(diǎn)一利用導(dǎo)數(shù)解決函數(shù)的極值問(wèn)題微點(diǎn)1由圖象判斷函數(shù)極值例1設(shè)三次函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),函數(shù)y=x·f'(x)的部分圖象如圖3-17-1所示,則下列說(shuō)法正確的是 ()圖3-17-1A.f(x)的極大值為f(3),極小值為f(-3)B.f(x)的極大值為f(-3),極小值為f(3)C.f(x)的極大值為f(3),極小值為f(-3)D.f(x)的極大值為f(-3),極小值為f(3)例1[思路點(diǎn)撥]由y=x·f'(x)的圖象可以得出y=f'(x)在各區(qū)間上的正負(fù)情況,然后可得f(x)在各區(qū)間上的單調(diào)性,進(jìn)而可得極值.C[解析]由圖象可知,當(dāng)x=-3和x=3時(shí),x·f'(x)=0,則f'(-3)=f'(3)=0;當(dāng)x<-3時(shí),x·f'(x)>0,則f'(x)<0;當(dāng)-3<x<0時(shí),x·f'(x)<0,則f'(x)>0;當(dāng)0<x<3時(shí),x·f'(x)>0,則f'(x)>0;當(dāng)x>3時(shí),x·f'(x)<0,則f'(x)<0.所以f(x)在(-∞,-3)上單調(diào)遞減;在(-3,0),(0,3)上單調(diào)遞增;在(3,+∞)上單調(diào)遞減.所以f(x)的極小值為f(-3),極大值為f(3).故選C.[總結(jié)反思]可導(dǎo)函數(shù)在極值點(diǎn)處的導(dǎo)數(shù)一定為零,是否為極值點(diǎn)以及是極大值點(diǎn)還是極小值點(diǎn)要看在極值點(diǎn)左、右兩側(cè)導(dǎo)數(shù)的符號(hào).微點(diǎn)2已知函數(shù)求極值例2函數(shù)f(x)=lnx-x的極大值是.

例2[思路點(diǎn)撥]確定函數(shù)f(x)的定義域,求出f'(x),進(jìn)而得出單調(diào)性,即可得到極大值.-1[解析]∵f(x)=lnx-x,∴定義域?yàn)?0,+∞),f'(x)=1x-1.令f'(x)=0,解得x=1當(dāng)0<x<1時(shí),f'(x)>0,當(dāng)x>1時(shí),f'(x)<0,∴f(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,故f(x)在x=1處取得極大值,極大值為f(1)=ln1-1=-1.[總結(jié)反思]求函數(shù)極值的一般步驟:①先求函數(shù)f(x)的定義域,再求函數(shù)f(x)的導(dǎo)函數(shù);②求f'(x)=0的根;③判斷在f'(x)=0的根的左、右兩側(cè)f'(x)的符號(hào),確定極值點(diǎn);④求出具體極值.微點(diǎn)3已知極值求參數(shù)例3若函數(shù)f(x)=ax22+(1-2a)x-2lnx在區(qū)間(12A.(-∞,-1e) C.(-2,-1) D.(-∞,-2)例3[思路點(diǎn)撥]求出f'(x),根據(jù)f(x)在(12,1)上有極小值可得f'(x)的正負(fù)情況,從而可求得aC[解析]f'(x)=ax+1-2a-2x=ax2+(1-2a)x-2x,由題意知f'(x)在區(qū)間(12,1)上有零點(diǎn),且在該零點(diǎn)的左側(cè)附近,有f'(x)<0,在右側(cè)附近,有f'(x)>0,則h(x)=ax2+(1-2a)x-2在區(qū)間(12,1)上有零點(diǎn),且在該零點(diǎn)的左側(cè)附近,有h(x)<0,在右側(cè)附近,有h(x)>0.當(dāng)a>0時(shí),h(x)的圖象為開(kāi)口向上的拋物線(xiàn),且h(0)=-2,故h(12)<0,h(1[總結(jié)反思]根據(jù)極值求參數(shù)的值(或取值范圍)就是根據(jù)極值點(diǎn)處的導(dǎo)數(shù)等于零、極值點(diǎn)處的函數(shù)值即極值列出關(guān)于參數(shù)的方程組(或不等式組),通過(guò)解方程組(或不等式組)求得參數(shù)的值(或取值范圍).?應(yīng)用演練1.【微點(diǎn)1】若f(x)是R上的可導(dǎo)函數(shù),x=-1為函數(shù)y=f(x)ex的一個(gè)極值點(diǎn),則下列圖象一定不可能為函數(shù)f(x)圖象的是 ()圖3-17-21.D[解析][f(x)ex]'=[f'(x)+f(x)]ex,令g(x)=f'(x)+f(x),則x=-1為函數(shù)y=f(x)ex的一個(gè)極值點(diǎn)等價(jià)于g(-1)=0,且g(x)在x=-1的左右兩側(cè)取值可異號(hào).對(duì)于選項(xiàng)A,f'(-1)=0,f(-1)=0,g(-1)=0,且g(x)在x=-1的左右兩側(cè)取值可異號(hào),符合條件.對(duì)于選項(xiàng)B,f'(-1)=0,f(-1)=0,g(-1)=0,且g(x)在x=-1的左右兩側(cè)取值可異號(hào),符合條件.對(duì)于選項(xiàng)C,f'(-1)>0,f(-1)<0,可能有g(shù)(-1)=0,g(x)在x=-1的左右兩側(cè)取值可異號(hào),故可能符合條件.對(duì)于選項(xiàng)D,f'(-1)>0,f(-1)>0,因此g(-1)≠0,不滿(mǎn)足條件.故選D.2.【微點(diǎn)2】函數(shù)f(x)=x2-6x+2ex的極值點(diǎn)所在的區(qū)間為 ()A.(0,1) B.(-1,0) C.(1,2) D.(-2,-1)2.A[解析]∵f(x)=x2-6x+2ex,∴f'(x)=2x-6+2ex,且函數(shù)f'(x)單調(diào)遞增.又f'(0)=-6+2e0=-4<0,f'(1)=-4+2e>0,∴函數(shù)f'(x)在區(qū)間(0,1)內(nèi)存在唯一的零點(diǎn),即函數(shù)f(x)的極值點(diǎn)在區(qū)間(0,1)內(nèi).故選A.3.【微點(diǎn)2】函數(shù)f(x)=x6-3x2有 ()A.一個(gè)極大值和一個(gè)極小值B.兩個(gè)極大值和一個(gè)極小值C.一個(gè)極大值和兩個(gè)極小值D.兩個(gè)極大值和兩個(gè)極小值3.C[解析]由題意知函數(shù)f(x)=x6-3x2,則f'(x)=6x5-6x=6x(x2+1)(x-1)(x+1),令f'(x)>0,即6x(x2+1)(x-1)(x+1)>0,解得-1<x<0或x>1,令f'(x)<0,即6x(x2+1)(x-1)(x+1)<0,解得x<-1或0<x<1,所以函數(shù)f(x)在區(qū)間(-1,0),(1,+∞)上單調(diào)遞增,在區(qū)間(-∞,-1),(0,1)上單調(diào)遞減,所以當(dāng)x=±1時(shí),函數(shù)f(x)取得極小值,當(dāng)x=0時(shí),函數(shù)f(x)取得極大值,即函數(shù)f(x)有一個(gè)極大值和兩個(gè)極小值.故選C.4.【微點(diǎn)3】若函數(shù)f(x)=x3-3bx+3在(-1,2)內(nèi)有極值,則實(shí)數(shù)b的取值范圍是 ()A.(0,4) B.[0,4) C.[1,4) D.(1,4)4.A[解析]f'(x)=3x2-3b,令f'(x)=0,得x2=b.∵f(x)在(-1,2)內(nèi)有極值,∴0≤b<4.若b=0,則f(x)=x3+3在R上單調(diào)遞增,沒(méi)有極值,故0<b<4.故選A.5.【微點(diǎn)3】已知函數(shù)f(x)=xlnx-12(m+1)x2-x有兩個(gè)極值點(diǎn),則實(shí)數(shù)m的取值范圍為A.(-1e,0) B.(-1,1C.(-∞,1e-1) 5.B[解析]因?yàn)楹瘮?shù)f(x)=xlnx-12(m+1)x2-x,所以f'(x)=lnx-(m+1)x,因?yàn)楹瘮?shù)f(x)=xlnx-12(m+1)x2-x有兩個(gè)極值點(diǎn),所以f'(x)=lnx-(m+1)x有兩個(gè)變號(hào)零點(diǎn),即m+1=lnxx有兩個(gè)不同的根,令g(x)=lnxx,則該問(wèn)題等價(jià)于g(x)與y=m+1的圖象有兩個(gè)不同的交點(diǎn).因?yàn)間'(x)=1-lnxx2,所以當(dāng)0<x<e時(shí),g'(x)>0,當(dāng)x>e時(shí),g'(x)<0,所以g(x)max=g(e)=1e,當(dāng)x→0時(shí),g(x)→-∞,當(dāng)x→+∞時(shí),g(x)→0,則0<m+1<1探究點(diǎn)二利用導(dǎo)數(shù)解決函數(shù)的最值問(wèn)題例4已知函數(shù)f(x)=ax+lnx,其中常數(shù)a<0.(1)當(dāng)a=-1時(shí),求f(x)的最大值;(2)若f(x)在區(qū)間(0,e]上的最大值為-3,求a的值.例4[思路點(diǎn)撥](1)利用導(dǎo)數(shù)求出函數(shù)f(x)在a=-1時(shí)的極大值,此極大值即為函數(shù)的最大值;(2)求出函數(shù)f(x)的導(dǎo)數(shù)f'(x)=ax+1x,對(duì)a分-1e≤解:(1)易知函數(shù)f(x)的定義域?yàn)?0,+∞),當(dāng)a=-1時(shí),f(x)=lnx-x,f'(x)=1x-1=1令f'(x)=0,得x=1.當(dāng)0<x<1時(shí),f'(x)>0;當(dāng)x>1時(shí),f'(x)<0.所以函數(shù)f(x)在區(qū)間(0,1)上單調(diào)遞增,在區(qū)間(1,+∞)上單調(diào)遞減.此時(shí),函數(shù)f(x)在x=1處取得極大值,亦為最大值.因此,f(x)max=f(1)=-1.(2)因?yàn)閒(x)=ax+lnx,所以f'(x)=a+1x=ax令f'(x)=0,得x=-1a>0.①當(dāng)0<-1a<e,即a<-1若0<x<-1a,則f'(x)>0若-1a<x≤e,則f'(x)<0此時(shí),函數(shù)f(x)在x=-1a處取得極大值,亦為最大值,即f(x)max=-1+ln(-1a)解得a=-e2,符合題意.②當(dāng)-1a≥e,即-1e≤a<0時(shí),對(duì)任意的x∈(0,e],f'(x)≥此時(shí),函數(shù)f(x)在(0,e]上單調(diào)遞增,則f(x)max=f(e)=ae+1=-3,得a=-4e綜上所述,a=-e2.[總結(jié)反思](1)函數(shù)在閉區(qū)間上的最值在端點(diǎn)處或區(qū)間內(nèi)的極值點(diǎn)處取得,上述值中最大的即為最大值、最小的即為最小值.如果函數(shù)在一個(gè)區(qū)間上(不論區(qū)間的類(lèi)型)有唯一的極值點(diǎn),則該點(diǎn)也是最值點(diǎn).(2)注意把不等式恒成立問(wèn)題轉(zhuǎn)化為函數(shù)的最值問(wèn)題.變式題(1)已知函數(shù)f(x)=x2-2lnx+a的最小值為3,則a=.

(2)若函數(shù)f(x)=alnx-x2-2,x>0,A.[0,2e2] B.[0,2e3] C.(0,2e2] D.(0,2e3]變式題(1)2(2)B[解析](1)∵函數(shù)f(x)=x2-2lnx+a,∴x>0,f'(x)=2x-2x=2x2-2x.令f'(x)=0得x=1或x=-1(舍去),∴當(dāng)x∈(0,1)時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)x∈(1,+∞)時(shí),f'(x)>0,f(x)∴f(x)min=f(1)=1+a.∵f(x)的最小值為3,∴1+a=3,解得a=2.(2)f(-1)=-2+a,由題意可得alnx-x2-2≤-2+a在x>0時(shí)恒成立,即a(1-lnx)≥-x2在x>0時(shí)恒成立.當(dāng)x=e時(shí),0>-e2顯然成立.當(dāng)0<x<e時(shí),有1-lnx>0,可得a≥x2lnx設(shè)g(x)=x2lnx則g'(x)=2x(lnx-1)-x(lnx-1)2=x(2lnx-3)(lnx-1)2,當(dāng)0<x<e時(shí),2lnx<2<3,則g'(x)<0,g(x)在(0,e)上單調(diào)遞減,且g(x)<0,可得a≥0.當(dāng)x>e時(shí),有1-lnx<0,可得a≤x2lnx-1,設(shè)h(x)=x2lnx-1(x>e),則h'(x)=x(2lnx-3)(lnx-1)2,可得a≤2e3.綜上可得0≤a≤2e3.故選B.例5已知函數(shù)f(x)=ax+ex-xlna(a>0,a≠1),對(duì)任意x1,x2∈[0,1],不等式|f(x2)-f(x1)|≤a-2恒成立,則a的取值范圍為 ()A.12,e2 B.[ee,+∞)C.12,+∞ D.[e2,ee]例5[思路點(diǎn)撥]對(duì)任意x1,x2∈[0,1],不等式|f(x1)-f(x2)|≤a-2恒成立即為f(x)max-f(x)min≤a-2,因此需求得f(x)在[0,1]上的最值,由導(dǎo)數(shù)可得f(x)在[0,1]上單調(diào)遞增,得出a-2≥a+e-lna-2,解出a即可.B[解析]因?yàn)閒(x)=ax+ex-xlna,所以f'(x)=axlna+ex-lna=(ax-1)lna+ex.當(dāng)a>1時(shí),對(duì)任意的x∈[0,1],ax-1≥0,lna>0,恒有f'(x)>0;當(dāng)0<a<1時(shí),對(duì)任意的x∈[0,1],ax-1≤0,lna<0,恒有f'(x)>0.所以f(x)在[0,1]上單調(diào)遞增.對(duì)任意的x1,x2∈[0,1],不等式|f(x2)-f(x1)|≤a-2恒成立,則f(x)max-f(x)min≤a-2,f(x)max=f(1)=a+e-lna,f(x)min=f(0)=1+1=2,所以a-2≥a+e-lna-2,即lna≥e,得a≥ee.故選B.變式題(1)已知函數(shù)f(x)=aex-x+2a2-3的值域?yàn)镸,集合I=(0,+∞),若I?M,則實(shí)數(shù)a的取值范圍是.

(2)若關(guān)于x的不等式ax-2a>2x-lnx-4有且只有兩個(gè)整數(shù)解,則實(shí)數(shù)a的取值范圍是 ()A.(2-ln3,2-ln2] B.(-∞,2-ln2)C.(-∞,2-ln3] D.(-∞,2-ln3)變式題(1)(-∞,1](2)C[解析](1)由題意可得f(x)的最小值小于或等于0.f'(x)=aex-1,若a≤0,則f'(x)<0,f(x)在R上單調(diào)遞減,當(dāng)x→-∞時(shí),f(x)→+∞,當(dāng)x→+∞時(shí),f(x)→-∞,故f(x)的值域?yàn)镽,滿(mǎn)足題意;若a>0,則易得函數(shù)f(x)在(-∞,-lna)上單調(diào)遞減,在(-lna,+∞)上單調(diào)遞增,所以當(dāng)x=-lna時(shí),函數(shù)f(x)取得極小值f(-lna)=lna-2+2a2,令g(a)=lna-2+2a2,則g'(a)=4a+1a>0恒成立,故g(a)在(0,+∞)上單調(diào)遞增,且g(1)=0,要使得g(a)≤0,則a≤1,故0<a≤1.綜上可得,a的取值范圍是(-∞,1].(2)由題意可知,ax-2a>2x-lnx-4有且只有兩個(gè)整數(shù)解.設(shè)g(x)=2x-lnx-4,h(x)=ax-2a,由g'(x)=2-1x=2x-1x,可知g(x)=2x-lnx-4在（0,12）上為減函數(shù),在（1當(dāng)a≤0時(shí),原不等式有且只有兩個(gè)整數(shù)解;當(dāng)a>0時(shí),若原不等式有且只有兩個(gè)整數(shù)解,則滿(mǎn)足a>0,h(1)解得0<a≤2-ln3.綜上可得a≤2-ln3,故選C.探究點(diǎn)三利用導(dǎo)數(shù)解決實(shí)際問(wèn)題例6某地準(zhǔn)備在山谷中建一座橋梁,橋址位置的豎直截面圖如圖3-17-3所示.谷底O在水平線(xiàn)MN上,橋AB與MN平行,OO'為鉛垂線(xiàn)(O'在AB上).經(jīng)測(cè)量,左側(cè)曲線(xiàn)AO上任一點(diǎn)D到MN的距離h1(米)與D到OO'的距離a(米)之間滿(mǎn)足關(guān)系式h1=140a2;右側(cè)曲線(xiàn)BO上任一點(diǎn)F到MN的距離h2(米)與F到OO'的距離b(米)之間滿(mǎn)足關(guān)系式h2=-1800b(1)求橋AB的長(zhǎng)度.(2)計(jì)劃在谷底兩側(cè)建造平行于OO'的橋墩CD和EF,且CE為80米,其中C,E在AB上(不包括端點(diǎn)).橋墩EF每米造價(jià)k(萬(wàn)元),橋墩CD每米造價(jià)32圖3-17-3例6解:(1)如圖,設(shè)AA1,BB1,CD1,EF1都與MN垂直,A1,B1,D1,F1是相應(yīng)垂足.由條件知,當(dāng)O'B=40時(shí),BB1=-1800×403+6×40=160,則AA1=160由140O'A2=160,得O'A=80所以AB=O'A+O'B=80+40=120(米).(2)以O(shè)為原點(diǎn),OO'為y軸建立平面直角坐標(biāo)系xOy(如圖所示).設(shè)F(x,y2),x∈(0,40),則y2=-1800x3+6xEF=160-y2=160+1800x3-6x.因?yàn)镃E=80,所以O(shè)'C=80-x.設(shè)D(x-80,y1),則y1=140(80-x)2所以CD=160-y1=160-140(80-x)2=-140記橋墩CD和EF的總造價(jià)為f(x),則f(x)=k（160+1800x3-6x）+32k（-140x2+4x）=k（1800x3-380xf'(x)=k（3800x2-340x）=3令f'(x)=0,得x=20.當(dāng)x變化時(shí),f'(x),f(x)的變化情況如下:x(0,20)20(20,40)f'(x)-0+f(x)↘極小值↗所以當(dāng)x=20時(shí),f(x)取得最小值.答:(1)橋AB的長(zhǎng)度為120米;(2)當(dāng)O'E為20米時(shí),橋墩CD和EF的總造價(jià)最低.[總結(jié)反思](1)利用導(dǎo)數(shù)研究生活中的優(yōu)化問(wèn)題的關(guān)鍵:理清數(shù)量關(guān)系、選取合適的自變量建立函數(shù)模型.(2)注意:函數(shù)的定義域由實(shí)際問(wèn)題確定,最后要把求解的數(shù)量結(jié)果“翻譯”為實(shí)際問(wèn)題的答案.變式題如圖3-17-4所示,正方形ABCD的邊長(zhǎng)為2,切去陰影部分后圍成一個(gè)正四棱錐,則當(dāng)正四棱錐的體積最大時(shí),該正四棱錐外接球的表面積為 ()圖3-17-4A.22B.52π25C.169π25D.338π變式題D[解析]由題意,正方形ABCD的邊長(zhǎng)為2,可得對(duì)角線(xiàn)長(zhǎng)的一半為2,折成正四棱錐后,設(shè)正四棱錐的底面邊長(zhǎng)為a,高為h,可得h2=2-2a(0<a<2),正四棱錐的體積V=13a2·h=1設(shè)y=2a4-2a5,則y'=8a3-52a4,令y'=0,可得a=852.由y'>0,得0<a<852,可知y=2a4-2a5在（0,852）故當(dāng)a=852時(shí),正四棱錐的體積取得最大值,此時(shí)h=105設(shè)正四棱錐外接球的半徑為R,得（105-R）2+（45）2=R2,解得R2=所以正四棱錐外接球的表面積S=4πR2=338125π.故選D.同步作業(yè)1.函數(shù)f(x)的導(dǎo)函數(shù)f'(x)的圖象如圖K17-1所示,則 ()圖K17-1A.12為f(x)B.-2為f(x)的極大值點(diǎn)C.2為f(x)的極大值點(diǎn)D.45為f(x)1.A[解析]當(dāng)x<-2時(shí),f'(x)<0,當(dāng)-2<x<12時(shí),f'(x)>0,當(dāng)12<x<2時(shí),f'(x)<0,當(dāng)x>2時(shí),f'(x)>0,所以-2為f(x)的極小值點(diǎn),12為f(x)2.對(duì)于函數(shù)f(x)=lnxx,下列說(shuō)法正確的是A.有極小值-1e C.有最小值1e D.有最大值2.D[解析]由題意得f'(x)=1-lnxx2,令f'(x)>0得0<x<e,令f'(x)<0得x>e,故f(x)在(0,e)上單調(diào)遞增,在(e,+∞)上單調(diào)遞減,所以當(dāng)x=e時(shí),f(x)3.函數(shù)f(x)=x+2cosx在[0,π]上的極小值點(diǎn)為 ()A.0 B.π6 C.5π6 3.C[解析]f'(x)=1-2sinx,令f'(x)=0,得x=π6或x=5π6,所以f(x)=x+2cosx在區(qū)間[0,π6)上是增函數(shù),在區(qū)間[π6,5π6)上是減函數(shù),在[5π64.函數(shù)f(x)=kx-lnx的極值點(diǎn)為x=2,則k的值為 ()A.2 B.1C.12 D.-4.C[解析]因?yàn)閒(x)=kx-lnx,所以f'(x)=k-1x,又f(x)=kx-lnx的極值點(diǎn)為x=2,所以f'(2)=0,即k=12.經(jīng)驗(yàn)證,可知k=5.若函數(shù)f(x)=x(x-c)2在x=2處有極大值,則常數(shù)c為 ()A.2 B.6 C.2或6 D.-2或-65.B[解析]∵函數(shù)f(x)=x(x-c)2=x3-2cx2+c2x,∴f'(x)=3x2-4cx+c2,由題意知,f'(2)=12-8c+c2=0,∴c=6或c=2,又函數(shù)f(x)=x(x-c)2在x=2處有極大值,故導(dǎo)數(shù)值在x=2處左側(cè)為正數(shù),右側(cè)為負(fù)數(shù).當(dāng)c=2時(shí),f'(x)=3x2-8x+4=3(x-23)(x-2),不滿(mǎn)足導(dǎo)數(shù)值在x=2處左側(cè)為正數(shù),右側(cè)為負(fù)數(shù).當(dāng)c=6時(shí),f'(x)=3x2-24x+36=3(x2-8x+12)=3(x-2)(x-6),滿(mǎn)足導(dǎo)數(shù)值在x=2處左側(cè)為正數(shù),右側(cè)為負(fù)數(shù).故c=66.函數(shù)f(x)=ex-2x的最小值為.

6.2-2ln2[解析]f'(x)=ex-2,令f'(x)=0,得x=ln2,所以f(x)在(-∞,ln2)上單調(diào)遞減,在(ln2,+∞)上單調(diào)遞增,所以f(x)min=f(ln2)=2-2ln2.7.從一張圓形鐵板上剪下一個(gè)扇形,將其制成一個(gè)無(wú)底圓錐容器,當(dāng)容器的體積最大時(shí),該扇形的圓心角是.

7.263π[解析]設(shè)圓錐底面半徑為r,高為h,扇形的半徑為R,則r2+h2=R圓錐的體積V=f(h)=13πr2h=13π(R2-h2)h=13π(R2由f'(h)=13π(R2-3h2)=0,得h2=13R2,此時(shí)圓錐體積最大,r=63R.設(shè)圓心角為α,則α=2πrR8.已知函數(shù)f(x)的導(dǎo)函數(shù)f'(x)=x4(x-1)3(x-2)2(x-3),則下列結(jié)論正確的是 ()A.f(x)在x=0處有極大值B.f(x)在x=2處有極小值C.f(x)在[1,3]上單調(diào)遞減D.f(x)有3個(gè)零點(diǎn)8.C[解析]令f'(x)>0,解得x>3或x<1,令f'(x)<0,解得1<x<3,故f(x)在(-∞,1)上單調(diào)遞增,在(1,3)上單調(diào)遞減,在(3,+∞)上單調(diào)遞增,故f(x)的極大值是f(1),極小值是f(3),故A,B錯(cuò)誤,C正確.又f(1),f(3)的符號(hào)無(wú)法確定,故無(wú)法確定f(x)的零點(diǎn)個(gè)數(shù),故D錯(cuò)誤.故選C.9.若x=-2是函數(shù)f(x)=(x2+ax-1)ex-1A.-1 B.-2e-3C.5e-3 D.19.A[解析]由題可得f'(x)=(2x+a)ex-1+(x2+ax-1)ex-1=[x2+(a+2)x+a-1]ex-1,因?yàn)閒'(-2)=0,所以a=-1,f(x)=(x2-x-1)ex-1,故f'(x)=(x2+x-2)ex-1,令f'(x)>0,解得x<-2或x>1,令f'(x)<0,解得-2<x<1,所以f(x)在(-∞,-2),(1,+∞)上單調(diào)遞增,在(-2,1)上單調(diào)遞減,所以f(x)的極小值為f(1)=(1-1-1)e1-1=-1,故選A.10.函數(shù)f(x)=xex-2lnx-2x的最小值為 ()A.-2ln2 B.ln2C.2-2ln2 D.2+ln210.C[解析]f(x)=xex-2lnx-2x,設(shè)t=lnx+x,則t∈R,且f(x)化為g(t)=et-2t,則g'(t)=et-2,所以g(t)在(-∞,ln2)上單調(diào)遞減,在(ln2,+∞)上單調(diào)遞增,所以g(t)≥g(ln2)=2-2ln2,所以g(t)的最小值為2-2ln2,即f(x)的最小值為2-2ln2.故選C.11.設(shè)函數(shù)f(x)=ln(x+k)+2,g(x)=ex2+1.若實(shí)數(shù)x1,x2滿(mǎn)足f(x1)=g(x2),且2x1-x2有極小值-2,則實(shí)數(shù)kA.3 B.2 C.1 D.-111.B[解析]依題意,令f(x1)=g(x2)=t,則ln(x1+k)+2=t,ex22+1=t,且t>1,則x1=et-2-k,x2=2ln(t-1),所以2x1-x2=2et-2-2k-2ln(t-1).構(gòu)造函數(shù)h(t)=2et-2-2k-2ln(t-1),t>1,則h'(t)=2et-2-2t-1,設(shè)m(t)=h'(t),則m'(t)=2et-2+12.關(guān)于函數(shù)f(x)=|x-1|-lnx,下列說(shuō)法正確的是 ()A.f(x)在(1eB.f(x)有極小值0,無(wú)極大值C.f(x)的值域?yàn)?-1,+∞)D.f(x)的圖象關(guān)于直線(xiàn)x=1對(duì)稱(chēng)12.B[解析]f(x)=|x-1|-lnx=x-1-∴當(dāng)x>1時(shí),f'(x)>0,f(x)單調(diào)遞增,當(dāng)0<x<1時(shí),f'(x)<0,f(x)單調(diào)遞減,故A錯(cuò)誤;f(x)在(1,+∞)上單調(diào)遞增,在(0,1)上單調(diào)遞減,則當(dāng)x=1時(shí),f(x)有極小值f(1)=0,無(wú)極大值,故B正確;∵f(x)≥f(1)=0,∴f(x)∈[0,+∞),故C錯(cuò)誤;f(12)=|12-1|-ln12=12+ln2,f(32)=|32-1|-ln32=113.(多選題)已知函數(shù)f(x)=xlnx-12ax2-1,當(dāng)a>0時(shí),函數(shù)f(x)的極值點(diǎn)的個(gè)數(shù)可能是A.0 B.1 C.2 D.313.AC[解析]由題意,令f'(x)=1+lnx-ax=0可得a=1+lnxx,令g(x)=1+lnxx,則g'(x)=-lnxx2,易得函數(shù)g(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,故當(dāng)x=1時(shí)函數(shù)g(x)取得最大值g(1)=1,又當(dāng)x→0時(shí),g(x)→-∞,當(dāng)x→+∞時(shí),g(x)→0,所以當(dāng)0<a<1時(shí),函數(shù)f(x)有2個(gè)極值點(diǎn),當(dāng)a≥14.函數(shù)f(x)=sin3x-2sinxcos2x(x∈[0,π2])的最大值為14.69[解析]函數(shù)f(x)=sin3x-sin2xcosx=sin2xcosx+cos2xsinx-sin2xcosx=cos2xsinx=sinx-2sin3令sinx=t,則t∈[0,1],則y=t-2t3,y'=1-6t2,令y'=0可得t=66,當(dāng)t∈[0,66)時(shí),y'>0,當(dāng)t∈(66,1]時(shí),y'<0,所以當(dāng)t=66時(shí),y取得最大值6915.若函數(shù)f(x)=2x2-lnx+3在其定義域內(nèi)的一個(gè)子區(qū)間(a-1,a+1)內(nèi)存在極值,則實(shí)數(shù)a的取值范圍是.

15.[1,32)[解析]函數(shù)f(x)的定義域是(0,+∞),f'(x)=4x-1x=4x2-1x=4(x+12)(x-12)x,當(dāng)0<x<12時(shí),f'(x)<0,f(x)單調(diào)遞減,當(dāng)x>12時(shí),f'(x)>0,f(x)單調(diào)遞增,∴f(x)只有一個(gè)極小值點(diǎn)12.由12∈(a-1,a+1),得a-16.已知函數(shù)f(x)=a(x+1)ex(a≠0).(1)求f(x)的最值;(2)若x>0時(shí),恒有f(x)≥x2-x-2,求實(shí)數(shù)a的取值范圍.16.解:(1)f'(x)=a(x+2)ex,當(dāng)a>0時(shí),f(x)在(-∞,-2)上單調(diào)遞減,在(-2,+∞)上單調(diào)遞增,所以f(x)min=f(-2)=-ae當(dāng)a<0時(shí),f(x)在(-∞,