第四章第三節(jié)　導(dǎo)數(shù)與函數(shù)的極值最值

第三節(jié)導(dǎo)數(shù)與函數(shù)的極值、最值【課程標(biāo)準(zhǔn)】1.借助函數(shù)圖象,了解函數(shù)在某點(diǎn)取得極值的必要條件和充分條件.2.會(huì)用導(dǎo)數(shù)求函數(shù)的極大值、極小值.3.會(huì)求閉區(qū)間上函數(shù)的最大值、最小值.【考情分析】考點(diǎn)考法:高考命題以考查函數(shù)的極值、最值的概念,求函數(shù)的極值、最值為重點(diǎn)內(nèi)容,對(duì)參數(shù)分類(lèi)討論,是每年的必考內(nèi)容,三種題型都可能出現(xiàn),題目難度較大.核心素養(yǎng):數(shù)學(xué)抽象、邏輯推理、數(shù)學(xué)運(yùn)算【必備知識(shí)·逐點(diǎn)夯實(shí)】【知識(shí)梳理·歸納】1.函數(shù)的極值與導(dǎo)數(shù)條件f'(x0)=0在點(diǎn)x=x0附近的左側(cè)f'(x)>0,右側(cè)f'(x)<0在點(diǎn)x=x0附近的左側(cè)f'(x)<0,右側(cè)f'(x)>0圖象極值f(x0)為極大值f(x0)為極小值極值點(diǎn)x0為極大值點(diǎn)x0為極小值點(diǎn)【微點(diǎn)撥】①函數(shù)的極大值和極小值都可能有多個(gè),極大值和極小值的大小關(guān)系不確定.②對(duì)于可導(dǎo)函數(shù)f(x),“f'(x0)=0”是“函數(shù)f(x)在x=x0處有極值”的必要不充分條件.2.函數(shù)的最值與導(dǎo)數(shù)(1)函數(shù)f(x)在[a,b]上有最值的條件一般地,如果在區(qū)間[a,b]上函數(shù)y=f(x)的圖象是一條連續(xù)不斷的曲線,那么它必有最大值和最小值.(2)求函數(shù)y=f(x)在區(qū)間[a,b]上的最大值與最小值的步驟①求函數(shù)y=f(x)在區(qū)間(a,b)內(nèi)的極值;②將函數(shù)y=f(x)的各極值與端點(diǎn)處的函數(shù)值f(a),f(b)比較,其中最大的一個(gè)是最大值,最小的一個(gè)是最小值.【微點(diǎn)撥】函數(shù)的最值是對(duì)定義域而言的整體概念,而極值是局部概念,在指定區(qū)間上極值可能不止一個(gè),也可能一個(gè)也沒(méi)有,而最值最多有一個(gè),并且有最值的未必有極值;有極值的未必有最值.【基礎(chǔ)小題·自測(cè)】類(lèi)型辨析改編易錯(cuò)高考題號(hào)13421.(多維辨析)(多選題)下列結(jié)論錯(cuò)誤的是()A.對(duì)于可導(dǎo)函數(shù)f(x),若f'(x0)=0,則x0為極值點(diǎn)B.函數(shù)的極大值不一定是最大值,最小值也不一定是極小值C.函數(shù)f(x)在區(qū)間(a,b)上不存在最值D.函數(shù)f(x)在區(qū)間[a,b]上一定存在最值【解析】選AC.A反例f(x)=x3,f'(x)=3x2,f'(0)=0,但x=0不是f(x)=x3的極值點(diǎn).×C反例f(x)=x2在區(qū)間(1,2)上的最小值為0.×2.(2022·全國(guó)甲卷)當(dāng)x=1時(shí),函數(shù)f(x)=alnx+bx取得最大值2,則f'(2)=(A.1 B.12 C.12 【解析】選B.因?yàn)楹瘮?shù)fx的定義域?yàn)?0,+∞),所以依題可知,f1=2,f'1=0,而f'x=axbx2,所以b=2,ab=0,即a=2,b=2,所以f'x=2x+2x2,因此函數(shù)fx在0,1上單調(diào)遞增,在1,+∞上單調(diào)遞減,當(dāng)x=1時(shí)取最大值,3.(選擇性必修二·P98T6·變形式)已知f(x)=x312x+1,x∈[13,1],則f(x)的最大值為13427,最小值為【解析】f'(x)=3x212=3(x2)(x+2),因?yàn)閤∈[13,1],所以f'(x故f(x)在[13,1]上單調(diào)遞減所以f(x)的最大值為f(13)=13427,最小值為f(1)=4.(忽視極值的存在條件)若函數(shù)f(x)=x3+ax2+bx+a2,在x=1處取得極值10,則a=4,b=11.

【解析】f'(x)=3x2+2ax+b,依題意得f(1解得a=4,經(jīng)驗(yàn)證,當(dāng)a=3,b=3時(shí),f'(x)=3x26x+3=3(x1)2≥0,故f(x)在R上單調(diào)遞增,所以a=-3,當(dāng)a=4,b=11時(shí),符合題意.【核心考點(diǎn)·分類(lèi)突破】考點(diǎn)一利用導(dǎo)數(shù)求函數(shù)的極值問(wèn)題【考情提示】函數(shù)極值是導(dǎo)數(shù)在研究函數(shù)中的一個(gè)重要應(yīng)用,在高考中也是重點(diǎn)考查的內(nèi)容,主要考查導(dǎo)數(shù)與函數(shù)單調(diào)性、極值或方程、不等式的綜合應(yīng)用,既有選擇題、填空題,也有解答題.角度1根據(jù)導(dǎo)函數(shù)圖象判斷極值[例1](多選題)(2023·石家莊模擬)函數(shù)y=f(x)的導(dǎo)函數(shù)y=f'(x)的圖象如圖所示,則()A.3是函數(shù)y=f(x)的極值點(diǎn)B.1是函數(shù)y=f(x)的極小值點(diǎn)C.y=f(x)在區(qū)間(3,1)上單調(diào)遞增D.2是函數(shù)y=f(x)的極大值點(diǎn)【解析】選AC.根據(jù)導(dǎo)函數(shù)的圖象可知,當(dāng)x∈(∞,3)時(shí),f'(x)<0,當(dāng)x∈(3,1)時(shí),f'(x)≥0,所以函數(shù)y=f(x)在(∞,3)上單調(diào)遞減,在(3,1)上單調(diào)遞增,可知3是函數(shù)y=f(x)的極值點(diǎn),所以A正確.因?yàn)楹瘮?shù)y=f(x)在(3,1)上單調(diào)遞增,可知1不是函數(shù)y=f(x)的極小值點(diǎn),2也不是函數(shù)y=f(x)的極大值點(diǎn),所以B錯(cuò)誤,C正確,D錯(cuò)誤.角度2已知函數(shù)解析式求極值或極值點(diǎn)[例2](1)(2023·西安模擬)已知f(x)=3xex,則f(x)A.在(∞,+∞)上單調(diào)遞增B.在(∞,1)上單調(diào)遞減C.有極大值3e,D.有極小值3e,【解析】選C.因?yàn)閒(x)=3x所以f'(x)=3·ex當(dāng)x>1時(shí),f'(x)<0,f(x)在區(qū)間(1,+∞)上單調(diào)遞減,故A錯(cuò)誤;當(dāng)x<1時(shí),f'(x)>0,f(x)在區(qū)間(∞,1)上單調(diào)遞增,故B錯(cuò)誤;當(dāng)x=1時(shí),f(x)=3xex取得極大值3e,無(wú)極小值,故C(2)函數(shù)f(x)=1x+ln|x|的極值點(diǎn)為1【解析】當(dāng)x>0時(shí),f(x)=1x+lnx,所以f'(x)=1x2+1x=x-1x2,所以當(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,所以f(x)在x=1處取得極小值,故f(x)的極值點(diǎn)為1;當(dāng)x<0時(shí),f(x)=1x+ln(x),所以f'(x)=1x2+1x=x-1x2<0,所以f(x)單調(diào)遞減,(3)已知函數(shù)f(x)=lnxax(a∈R).①當(dāng)a=12時(shí),求f(x)的極值②討論函數(shù)f(x)在定義域內(nèi)極值點(diǎn)的個(gè)數(shù).【解析】①當(dāng)a=12時(shí),f(x)=lnx12x,函數(shù)的定義域?yàn)?0,+∞)且f'(x)=1令f'(x)=0,得x=2,于是當(dāng)x變化時(shí),f'(x),f(x)的變化情況如表.x(0,2)2(2,+∞)f'(x)+0f(x)單調(diào)遞增ln21單調(diào)遞減故f(x)在定義域上的極大值為f(2)=ln21,無(wú)極小值.②由①知,函數(shù)f(x)的定義域?yàn)?0,+∞),f'(x)=1xa=1當(dāng)a≤0時(shí),f'(x)>0在(0,+∞)上恒成立,則函數(shù)在(0,+∞)上單調(diào)遞增,此時(shí)函數(shù)在定義域上無(wú)極值點(diǎn);當(dāng)a>0時(shí),若x∈(0,1a),則f'(x若x∈(1a,+∞),則f'(x故函數(shù)在x=1a處有極大值綜上可知,當(dāng)a≤0時(shí),函數(shù)f(x)無(wú)極值點(diǎn),當(dāng)a>0時(shí),函數(shù)y=f(x)有一個(gè)極大值點(diǎn),且為1a角度3已知極值(點(diǎn))求參數(shù)(規(guī)范答題)[例3](1)已知函數(shù)f(x)=x(xc)2在x=2處有極小值,則c的值為()A.2 B.4 C.6 D.2或6【解析】選A.由題意,f'(x)=(xc)2+2x(xc)=(xc)·(3xc),則f'(2)=(2c)(6c)=0,所以c=2或c=6.若c=2,則f'(x)=(x2)(3x2),當(dāng)x∈(∞,23)時(shí),f'(x)>0,f(x)單調(diào)遞增當(dāng)x∈(23,2)時(shí),f'(x)<0,f(x)單調(diào)遞減當(dāng)x∈(2,+∞)時(shí),f'(x)>0,f(x)單調(diào)遞增,函數(shù)f(x)在x=2處有極小值,滿足題意;若c=6,則f'(x)=(x6)(3x6),當(dāng)x∈(∞,2)時(shí),f'(x)>0,f(x)單調(diào)遞增;當(dāng)x∈(2,6)時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)x∈(6,+∞)時(shí),f'(x)>0,f(x)單調(diào)遞增,函數(shù)f(x)在x=2處有極大值,不符合題意.綜上,c=2.(2)(2023·南京模擬)已知函數(shù)f(x)=x(lnxax)在(0,+∞)上有兩個(gè)極值,則實(shí)數(shù)a的取值范圍為(0,12)【解析】f'(x)=lnx+12ax,由題意知lnx+12ax=0在(0,+∞)上有兩個(gè)不相等的實(shí)根,則2a=lnx設(shè)g(x)=lnx+1x,則g'(x當(dāng)0<x<1時(shí),g'(x)>0,g(x)單調(diào)遞增;當(dāng)x>1時(shí),g'(x)<0,g(x)單調(diào)遞減,所以g(x)的極大值為g(1)=1,又當(dāng)x>1時(shí),g(x)>0,當(dāng)x→+∞時(shí),g(x)→0,當(dāng)x→0時(shí),g(x)→∞,所以0<2a<1,即0<a<12(3)(12分)(2023·新高考Ⅱ卷)①證明:當(dāng)0<x<1時(shí),xx2<sinx<x;②已知函數(shù)f(x)=cosaxln(1x2),若x=0是f(x)的極大值點(diǎn),求a的取值范圍.審題導(dǎo)思破題點(diǎn)·柳暗花明①思路:通過(guò)構(gòu)造函數(shù)并借助導(dǎo)數(shù)得到單調(diào)性,進(jìn)而證明不等式②思路:通過(guò)第①問(wèn)鋪設(shè)好的不等式,利用導(dǎo)數(shù)討論函數(shù)的單調(diào)性,進(jìn)而得到極值規(guī)范答題微敲點(diǎn)·水到渠成【解析】①設(shè)h(x)=sinxx,則h'(x)=cosx1, …………[1分]當(dāng)0<x<1時(shí),h'(x)<0,所以當(dāng)0<x<1時(shí),h(x)單調(diào)遞減,h(x)<h(0)=0,即sinx<x(0<x<1). …………[2分]

源于教材sinx<x(0<x<1)可參考人教A版《選擇性必修第二冊(cè)》第86頁(yè)例1(2)及97頁(yè)練習(xí)第1題.設(shè)g(x)=sinxx+x2,則g'(x)=cosx1+2x,g″(x)=sinx+2>0, …………[3分]因而當(dāng)0<x<1時(shí),g'(x)>g'(0)=0,g(x)>g(0)=0,則有sinx>xx2(0<x<1).………[4分]綜上,當(dāng)0<x<1時(shí),xx2<sinx<x.…………[5分]②f(x)=cosaxln(1x2),由1x2>0得函數(shù)的定義域是(1,1).由f(x)=f(x),得函數(shù)f(x)為偶函數(shù),則f'(x)=asinax+2x1-x2f″(x)=a2cosax+2+2x2f″(0)=2a2.若2a2=0,此時(shí)x=0是f(x)的極小值點(diǎn),不符合題意,則2a2≠0,即a≠±2若f(x)在x=0處取得極大值,那么該函數(shù)在x=0處是上凸的,因而f″(0)=2a2<0,敲黑板實(shí)際上,這里蘊(yùn)含著“高觀點(diǎn)”函數(shù)f(x)在x0處具有二階導(dǎo)數(shù),且f'(x0)=0,f″(x0)≠0,則f(x)在x0處取得極大值的充分條件為f″(x0)<0.則a<2或a>2. …………[6當(dāng)a>2時(shí),取0<x<1a<1,則0<從而f'(x)=asinax+2x1-x2<a(axa2x2)+指點(diǎn)迷津利用①的結(jié)論進(jìn)行放縮,轉(zhuǎn)化成不含三角函數(shù)的形式,有助于判斷零點(diǎn).易知s(x)=a2+a3x+21-x2而s(0)=2a2<0,s1a=2因而關(guān)于x的方程f'(x)=0在0,1a上有唯一解當(dāng)x∈(0,x0)時(shí),f'(x)<0,f(x)單調(diào)遞減,當(dāng)x∈x0,1a時(shí),f'(x)>0,f(由f(x)是偶函數(shù)且連續(xù),得當(dāng)x∈(x0,0)時(shí)f(x)單調(diào)遞增,所以f(x)在x=0處取極大值.…………[9分]拓展思維也可利用f'(x)為奇函數(shù)且連續(xù),得x∈(x0,0)時(shí)f'(x)>0,進(jìn)而判斷極值點(diǎn).當(dāng)a<2時(shí),取1a<x<0,則0<從而f'(x)=asinax+2x1-x2<a(ax)+易知u(x)=a2+21-x2而u(0)=a2+2<0,由a2+21-x2=0,得x=a2-2易錯(cuò)警示注意此時(shí)的前提條件a<0,所以x=a2-2a>0要舍去因而f(x)在(a2-2a,0)(則取區(qū)間1a,0由f(x)為偶函數(shù),得f(x)在(0,a2-2a)上單調(diào)遞減,結(jié)合函數(shù)f(x)連續(xù)得f(x)在x=0處取極大值.綜上所述,a<2或a>2所以a的取值范圍為-∞,-2∪2,+∞.【解題技法】1.由圖象判斷函數(shù)y=f(x)的極值要抓住的兩點(diǎn)(1)由y=f'(x)的圖象與x軸的交點(diǎn),可得函數(shù)y=f(x)的可能極值點(diǎn).(2)由導(dǎo)函數(shù)y=f'(x)的圖象可以看出y=f'(x)的值的正負(fù),從而可得函數(shù)y=f(x)的單調(diào)性.兩者結(jié)合可得極值點(diǎn).2.運(yùn)用導(dǎo)數(shù)求函數(shù)f(x)極值的一般步驟(1)確定函數(shù)f(x)的定義域;(2)求導(dǎo)數(shù)f'(x);(3)解方程f'(x)=0,求出函數(shù)定義域內(nèi)的所有根;(4)列表檢驗(yàn)f'(x)在f'(x)=0的根x0左右兩側(cè)導(dǎo)數(shù)值的符號(hào);(5)求出極值.3.已知函數(shù)極值點(diǎn)或極值求參數(shù)的兩個(gè)要領(lǐng)(1)列式:根據(jù)極值點(diǎn)處導(dǎo)數(shù)為0和極值這兩個(gè)條件列方程組,利用待定系數(shù)法求解.(2)驗(yàn)證:因?yàn)閷?dǎo)數(shù)值等于零不是此點(diǎn)為極值點(diǎn)的充要條件,所以利用待定系數(shù)法求解后必須驗(yàn)證根的合理性.【對(duì)點(diǎn)訓(xùn)練】1.設(shè)函數(shù)f(x)在R上可導(dǎo),其導(dǎo)函數(shù)為f'(x),且函數(shù)y=(x1)f'(x)的圖象如圖所示,則下列結(jié)論正確的是()A.函數(shù)f(x)有極大值f(3)和f(3)B.函數(shù)f(x)有極小值f(3)和f(3)C.函數(shù)f(x)有極小值f(3)和極大值f(3)D.函數(shù)f(x)有極小值f(3)和極大值f(3)【解析】選D.由題圖知,當(dāng)x∈(∞,3)時(shí),y>0,x1<0?f'(x)<0,f(x)單調(diào)遞減;當(dāng)x∈(3,1)時(shí),y<0,x1<0?f'(x)>0,f(x)單調(diào)遞增;當(dāng)x∈(1,3)時(shí),y>0,x1>0?f'(x)>0,f(x)單調(diào)遞增;當(dāng)x∈(3,+∞)時(shí),y<0,x1>0?f'(x)<0,f(x)單調(diào)遞減.所以函數(shù)有極小值f(3)和極大值f(3).2.(2023·長(zhǎng)沙模擬)若1是函數(shù)f(x)=(x2+ax1)ex1的極值點(diǎn),則f(x)的極大值為5e3.

【解析】因?yàn)閒(x)=(x2+ax1)ex可得f'(x)=ex-1[x2+(a+2)x因?yàn)?是函數(shù)f(x)的極值點(diǎn),故可得f'(1)=0,即2a+2=0,解得a=1.此時(shí)f'(x)=ex1(x2+x2)=ex1(x+2)(x1).由f'(x)>0可得x<2或x>1;由f'(x)<0可得2<x<1,故f(x)的極大值點(diǎn)為2.則f(x)的極大值為f(2)=(4+21)e3=5e3.【加練備選】1.(2023·威海模擬)若函數(shù)f(x)=exax22ax有兩個(gè)極值點(diǎn),則實(shí)數(shù)a的取值范圍為()A.(12,0) B.(∞,1C.(0,12) D.(1【解析】選D.由f(x)=exax22ax,得f'(x)=ex2ax2a.因?yàn)楹瘮?shù)f(x)=exax22ax有兩個(gè)極值點(diǎn),所以f'(x)=ex2ax2a有兩個(gè)變號(hào)零點(diǎn),令f'(x)=0,得12a=設(shè)g(x)=x+1ex,y=12a;則g'令g'(x)=0,即xex=0,解得當(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)遞增,在(0,+∞)上單調(diào)遞減.分別作出函數(shù)g(x)=x+1ex與y=12由圖可知,0<12a<1,解得a>所以實(shí)數(shù)a的取值范圍為(12,+∞)2.(2023·岳陽(yáng)模擬)已知函數(shù)f(x)=cosxax2,其中a∈R.(1)當(dāng)a=2π時(shí),求曲線y=f(x)在x=π2(2)若函數(shù)f(x)在(π,π)上恰有兩個(gè)極小值點(diǎn)x1,x2,求a的取值范圍.【解析】(1)當(dāng)a=2π時(shí),f(x)=cosx+2πx則f'(x)=sinx+4πx所以f(π2)=π2,f'(所以曲線y=f(x)在x=π2處的切線方程為yπ2=xπ2,即(2)因?yàn)閒(x)=cos(x)a(x)2=cosxax2=f(x),所以f(x)是(π,π)上的偶函數(shù).因?yàn)楹瘮?shù)f(x)在(π,π)上恰有兩個(gè)極小值點(diǎn),所以函數(shù)f(x)在(0,π)上恰有一個(gè)極小值點(diǎn).不妨設(shè)x1>0,f'(x)=sinx2ax,令g(x)=sinx2ax,則g'(x)=cosx2a.(ⅰ)當(dāng)a≥0時(shí),f'(x)<0,則f(x)在(0,π)上單調(diào)遞減,無(wú)極小值;(ⅱ)當(dāng)a≤12時(shí),g'(x則f'(x)在(0,π)上單調(diào)遞增,所以f'(x)>f'(0)=0.則f'(x)>0,此時(shí)f(x)在(0,π)上單調(diào)遞增,無(wú)極小值;(ⅲ)當(dāng)12<a<0時(shí),存在x0∈使g'(x0)=cosx02a=0,當(dāng)x∈(0,x0)時(shí),g'(x)<0,所以f'(x)在(0,x0)上單調(diào)遞減,當(dāng)x∈(x0,π)時(shí),g'(x)>0,所以f'(x)在(x0,π)上單調(diào)遞增.因?yàn)閒'(0)=0,f'(π)=2aπ>0,由零點(diǎn)存在定理知存在x1∈(x0,π),使得f'(x1)=sinx12ax1=0,當(dāng)x∈(0,x1)時(shí),f'(x)<0,所以f(x)在(0,x1)上單調(diào)遞減,當(dāng)x∈(x1,π)時(shí),f'(x)>0,所以f(x)在(x1,π)上單調(diào)遞增,所以函數(shù)f(x)在(0,π)上恰有一個(gè)極小值點(diǎn)x1.所以函數(shù)f(x)在(π,π)上恰有兩個(gè)極小值點(diǎn).綜上所述,a的取值范圍為(12,0)考點(diǎn)二利用導(dǎo)數(shù)求函數(shù)最值問(wèn)題[例4](1)(2022·全國(guó)乙卷)函數(shù)f(x)=cosx+(x+1)sinx+1在區(qū)間0,2π的最小值、最大值分別為(A.π2,π2 B.3πC.π2,π2+2 D.3π2【解析】選D.f'(x)=sinx+sinx+(x+1)cosx=(x+1)cosx,所以在區(qū)間(0,π2)和(3π2,2π)上f'(即f(x)單調(diào)遞增;在區(qū)間(π2,3π2)上f'(x)<0,即f(x)又f(0)=f(2π)=2,f(π2)=π2+2,f(3π2)=(3π所以f(x)在區(qū)間0,2π上的最小值為3π2,最大值為(2)設(shè)函數(shù)f(x)=x-2a,x≤0,lnx,x>0,若f(x1)=f(x2)(x1<x2),且A.12 B.C.e2 D.【解析】選B.令f(x1)=f(x2)=t,由圖象可得t∈(∞,2a],因?yàn)閤1<x2,所以x12a=t,lnx2=t,即x1=t+2a,x2=et,則2x2x1=2ett2a,令g(t)=2ett2a,則g(t)在(∞,2a]的最小值為ln2,因?yàn)間'(t)=2et1,令g'(t)=0,解得t=ln2,所以g(t)在(∞,ln2)上單調(diào)遞減,在(ln2,+∞)上單調(diào)遞增,因?yàn)閠∈(∞,2a],所以當(dāng)a≥ln22時(shí),g'(t)≤0,g(t)單調(diào)遞減,所以g(t)min=g(2a)=2e-2a=ln2,解得a=當(dāng)a<ln22時(shí),g(t)在(∞,ln2)上單調(diào)遞減在(ln2,2a)上單調(diào)遞增,則g(t)min=g(ln2)=1+ln22a=ln2,解得a=12,不符合題意.綜上,a=ln【解題技法】求函數(shù)f(x)在[a,b]上的最值的方法(1)若函數(shù)f(x)在區(qū)間[a,b]上單調(diào)遞增或單調(diào)遞減,則f(a)與f(b)一個(gè)為最大值,一個(gè)為最小值.(2)若函數(shù)f(x)在區(qū)間[a,b]上有極值,則先求出函數(shù)在[a,b]上的極值,再與f(a),f(b)比較,最大的是最大值,最小的是最小值,可列表完成.(3)函數(shù)f(x)在區(qū)間(a,b)上有唯一一個(gè)極值點(diǎn),這個(gè)極值點(diǎn)就是最值點(diǎn),此結(jié)論在導(dǎo)數(shù)的實(shí)際應(yīng)用中經(jīng)常用到.【對(duì)點(diǎn)訓(xùn)練】1.(2023·蘇州模擬)若函數(shù)f(x)=13x3+x223在區(qū)間(a,a+5)內(nèi)存在最小值,則實(shí)數(shù)a的取值范圍是(A.[5,0) B.(5,0)C.[3,0) D.(3,0)【解析】選C.由題意,f'(x)=x2+2x=x(x+2),當(dāng)x<2或x>0時(shí),f'(x)>0;當(dāng)2<x<0時(shí),f'(x)<0.故f(x)在(∞,2),(0,+∞)上單調(diào)遞增,在(2,0)上單調(diào)遞減,所以函數(shù)f(x)的極小值為f(0)=23令13x3+x223=23得x3+3x2=0,解得x=0或作其圖象如圖,結(jié)合圖象可知-解得a∈[3,0).2.已知函數(shù)f(x)=ax+lnx,其中a為常數(shù).(1)若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.【解析】(1)f(x)的定義域?yàn)?0,+∞),當(dāng)a=1時(shí),f(x)=x+lnx,f'(x)=1+1x=1-xx,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(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,所以f(x)max=f(1)=1.所以當(dāng)a=1時(shí),函數(shù)f(x)在(2)f'(x)=a+1x,x∈(0,e]時(shí),1x∈[1e,+∞).①若a≥1e,則f'(x)≥0,從而f(x)在(0,e]上單調(diào)遞增,所以f(x)max=f(e)=ae+1≥0,不合題意.②若a<1e,令f'(x)>0得a+1x>0,結(jié)合x(chóng)∈(0,e],解得0<x<1a;令f'(x)<0得a+1x<0,結(jié)合x(chóng)∈(0,e],解得1a<x≤e,從而f(x)在(0,1a)上單調(diào)遞增,在(1a,e]上單調(diào)遞減,所以f(x)max=f(1a)=1+ln(1a).令1+ln(1a)=3,得ln(1a)=2,即a=e2.因?yàn)閑2<考點(diǎn)三生活中的優(yōu)化問(wèn)題[例5]某村莊擬修建一個(gè)無(wú)蓋的圓柱形蓄水池(不計(jì)厚度),設(shè)該蓄水池的底面半徑為r米,高為h米,體積為V立方米.假設(shè)建造成本僅與表面積有關(guān),側(cè)面的建造成本為100元/平方米,底面的建造成本為160元/平方米,該蓄水池的總建造成本為12000π元(π為圓周率).(1)將V表示成r的函數(shù)V(r),并求該函數(shù)的定義域;(2)討論函數(shù)V(r)的單調(diào)性,并確定r和h為何值時(shí),該蓄水池的體積最大.【解析】(1)因?yàn)樾钏氐膫?cè)面的總成本為100×2πrh=200πrh(元),底面的總成本為160πr2元,所以蓄水池的總成本為(200πrh+160πr2)元.由題意得200πrh+160πr2=12000π,所以h=15r(3004r2從而V(r)=πr2h=π5(300r4r3)由h>0,且r>0,可得0<r<53.故函數(shù)V(r)的定義域?yàn)?0,53).(2