第三章一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用第4節(jié)導(dǎo)數(shù)與函數(shù)的極值、最值學(xué)習(xí)導(dǎo)航站核心知識(shí)庫(kù)：重難考點(diǎn)總結(jié)，梳理必背知識(shí)、歸納重點(diǎn)考點(diǎn)1函數(shù)的極值★★★☆☆考點(diǎn)2函數(shù)的最大(小)值★★★☆☆【知識(shí)拓展】三次函數(shù)的對(duì)稱(chēng)性★★★☆☆（星級(jí)越高，重要程度越高）限時(shí)【變式訓(xùn)練】挑戰(zhàn)場(chǎng)：感知真題，檢驗(yàn)成果，考點(diǎn)追溯【知識(shí)梳理】1.函數(shù)的極值★★★☆☆(1)函數(shù)的極小值函數(shù)y=f(x)在點(diǎn)x=a的函數(shù)值f(a)比它在點(diǎn)x=a附近其他點(diǎn)的函數(shù)值都小,f'(a)=0;而且在點(diǎn)x=a附近的左側(cè)f'(x)<0,右側(cè)f'(x)>0.則a叫做函數(shù)y=f(x)的極小值點(diǎn),f(a)叫做函數(shù)y=f(x)的極小值.(2)函數(shù)的極大值函數(shù)y=f(x)在點(diǎn)x=b的函數(shù)值f(b)比它在點(diǎn)x=b附近其他點(diǎn)的函數(shù)值都大,f'(b)=0;而且在點(diǎn)x=b附近的左側(cè)f'(x)>0,右側(cè)f'(x)<0.則b叫做函數(shù)y=f(x)的極大值點(diǎn),f(b)叫做函數(shù)y=f(x)的極大值.(3)極小值點(diǎn)、極大值點(diǎn)統(tǒng)稱(chēng)為極值點(diǎn),極小值和極大值統(tǒng)稱(chēng)為極值.2.函數(shù)的最大(小)值★★★☆☆(1)函數(shù)f(x)在區(qū)間[a,b]上有最值的條件: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y=f(x)在區(qū)間[a,b]上的最大(小)值的步驟:①求函數(shù)y=f(x)在區(qū)間(a,b)上的極值;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.【名師點(diǎn)撥】1.求最值時(shí),應(yīng)注意極值點(diǎn)和所給區(qū)間的關(guān)系,關(guān)系不確定時(shí),需要分類(lèi)討論,不可想當(dāng)然認(rèn)為極值就是最值.2.函數(shù)最值是“整體”概念,而函數(shù)極值是“局部”概念,極大值與極小值之間沒(méi)有必然的大小關(guān)系.【教材回歸】概念思考辨析+教材經(jīng)典改編1.思考辨析(在括號(hào)內(nèi)打“√”或“×”)(1)對(duì)于可導(dǎo)函數(shù)f(x),若f'(x0)=0,則x0為極值點(diǎn).()(2)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.()(3)函數(shù)f(x)在區(qū)間(a,b)上不存在最值.()(4)連續(xù)函數(shù)f(x)在區(qū)間[a,b]上一定存在最值.()2.(人教B選修三P100T1改編)如圖是f(x)的導(dǎo)函數(shù)f'(x)的圖象,則f(x)的極小值點(diǎn)的個(gè)數(shù)為()A.1 B.2C.3 D.43.(湘教選修二P49T7改編)已知f(x)=x3-12x+1,x∈?13,1,則f(x)的最大值為,最小值為4.(人教A選修二P104T9改編)函數(shù)f(x)=x(x-c)2有極值,則實(shí)數(shù)c的取值范圍是.

【考向核心題型】考點(diǎn)一利用導(dǎo)數(shù)求函數(shù)的極值角度1根據(jù)函數(shù)圖象判斷極值【典例】1(多選)如圖是函數(shù)y=f(x)的導(dǎo)函數(shù)f'(x)的圖象,下列說(shuō)法正確的是()A.f(1)為函數(shù)f(x)的極大值B.當(dāng)x=-1時(shí),f(x)取得極小值C.f(x)在(-1,2)上單調(diào)遞增,在(2,4)上單調(diào)遞減D.當(dāng)x=3時(shí),f(x)取得極小值角度2求已知函數(shù)的極值【典例】2已知函數(shù)f(x)=lnx+2ax2+2(a+1)x(a≠0),討論函數(shù)f(x)的極值.角度3由函數(shù)的極值求參數(shù)【典例】3(2024·新高考Ⅱ卷節(jié)選)已知函數(shù)f(x)=ex-ax-a3,若f(x)有極小值,且極小值小于0,求a的取值范圍.【變式訓(xùn)練】1(1)(2025·咸陽(yáng)模擬)已知函數(shù)f(x)=2cos2x2+a2x2,若x=0是f(x)的唯一極小值點(diǎn),A.[1,+∞) B.(0,1)C.[-1,+∞) D.(-∞,1](2)已知函數(shù)f(x)=lnx-ax(a∈R),討論函數(shù)f(x)在定義域內(nèi)極值點(diǎn)的個(gè)數(shù).考點(diǎn)二利用導(dǎo)數(shù)求函數(shù)的最值角度1求已知函數(shù)的最值【典例】4(2022·全國(guó)乙卷)函數(shù)f(x)=cosx+(x+1)sinx+1在區(qū)間[0,2π]上的最小值、最大值分別為()A.-π2,π2 B.-3πC.-π2,π2+2 D.-3π2,角度2由函數(shù)的最值求參數(shù)【典例】5(2025·福建名校聯(lián)考)已知函數(shù)f(x)=x3-3x2+3在區(qū)間(a,a+6)上存在最小值,則實(shí)數(shù)a的取值范圍為()A.[-1,2) B.?C.?2,32 D.[-1【變式訓(xùn)練】2(1)(2025·南京、鹽城模擬)用min{x,y}表示x,y中較小的數(shù).已知函數(shù)f(x)=xex,則min{f(x),f(x+lnA.2e2 B.C.ln22 D.ln(2)(2025·河北名校聯(lián)考)已知函數(shù)f(x)=ax-lnx的最小值為0,則a=.

【知識(shí)拓展】三次函數(shù)的對(duì)稱(chēng)性★★★☆☆結(jié)論1:三次函數(shù)f(x)=ax3+bx2+cx+d(a≠0)的圖象關(guān)于點(diǎn)?b3結(jié)論2:已知三次函數(shù)f(x)=ax3+bx2+cx+d(a≠0)中心對(duì)稱(chēng)點(diǎn)的橫坐標(biāo)為x0,兩個(gè)極值點(diǎn)分別為x1,x2,則f(x1)?f(x2)x1?x2=23結(jié)論3:若y=f(x)圖象關(guān)于點(diǎn)(m,n)對(duì)稱(chēng),則y=f'(x)圖象關(guān)于軸x=m對(duì)稱(chēng),點(diǎn)對(duì)稱(chēng)函數(shù)的導(dǎo)數(shù)是軸對(duì)稱(chēng)函數(shù),軸對(duì)稱(chēng)函數(shù)的導(dǎo)數(shù)是點(diǎn)對(duì)稱(chēng)函數(shù),奇函數(shù)的導(dǎo)數(shù)是偶函數(shù),偶函數(shù)的導(dǎo)數(shù)是奇函數(shù),周期函數(shù)的導(dǎo)數(shù)還是周期函數(shù).【典例】已知任意三次函數(shù)的圖象必存在唯一的對(duì)稱(chēng)中心,若函數(shù)f(x)=x3+ax2+bx+c,且M(x0,f(x0))為曲線y=f(x)的對(duì)稱(chēng)中心,則必有g(shù)'(x0)=0(其中函數(shù)g(x)=f'(x)).若實(shí)數(shù)m,n滿足m3+6m2+13A.-4 B.-3 C.-2 D.-1【變式訓(xùn)練】(多選)(2024·新高考Ⅱ卷)設(shè)函數(shù)f(x)=2x3-3ax2+1,則()A.當(dāng)a>1時(shí),f(x)有三個(gè)零點(diǎn)B.當(dāng)a<0時(shí),x=0是f(x)的極大值點(diǎn)C.存在a,b,使得x=b為曲線y=f(x)的對(duì)稱(chēng)軸D.存在a,使得點(diǎn)(1,f(1))為曲線y=f(x)的對(duì)稱(chēng)中心【限時(shí)訓(xùn)練】（限時(shí)：60分鐘）一、單選題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=(1-x)f'(x)的圖象如圖所示,則下列結(jié)論中一定成立的是()A.函數(shù)f(x)有極大值f(2)和極小值f(1)B.函數(shù)f(x)有極大值f(-2)和極小值f(1)C.函數(shù)f(x)有極大值f(2)和極小值f(-2)D.函數(shù)f(x)有極大值f(-2)和極小值f(2)2.函數(shù)f(x)=13x3+x2-3x-A.1 B.1,?C.-3 D.(-3,8)3.函數(shù)f(x)=xcosx-sinx在區(qū)間[-π,0]上的最大值為()A.1 B.π C.32 D.4.已知函數(shù)f(x)=13x3+(a-1)x2+x+1沒(méi)有極值,則實(shí)數(shù)aA.[0,1] B.(-∞,0]∪[1,+∞)C.[0,2] D.(-∞,0]∪[2,+∞)5.(2025·重慶診斷)若f(x)=ex-2msinxcosx,x∈0,12存在極值,則實(shí)數(shù)A.12,eC.1,e126.(2025·河南五市聯(lián)考)已知函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),且f(x)=-37f'(3)lnx-f(1)x2-4x,則f(A.32或12 C.-12或32 7.(2025·佛山調(diào)研)若函數(shù)f(x)=alnx+4x+bx2(a≠A.a<0 B.b<0C.ab>-1 D.a+b>08.(2025·榆林質(zhì)檢)已知函數(shù)f(x)=x2+4x+5,x≤0,lnx,x>0,若存在實(shí)數(shù)x1,x2,x3且x1<x2<x3,使得f(x1)=f(x2)=f(x3),則x1f(x1)+x2fA.3e3-12 B.3e3-20C.5e5-12 D.5e5-20二、多選題9.(2025·運(yùn)城調(diào)研)若x=-2是函數(shù)f(x)=(x2+ax-1)ex-1的極值點(diǎn),則下面結(jié)論正確的為()A.a=-1B.f(x)的單調(diào)遞增區(qū)間為(-2,1)C.f(x)的極小值為1D.f(x)的極大值為5e-310.(2025·武漢調(diào)研)已知函數(shù)f(x)=-x3+3x2-2,則下列說(shuō)法正確的是()A.函數(shù)f(x)在(-∞,0),(2,+∞)上單調(diào)遞增B.x=2是函數(shù)f(x)的極大值點(diǎn)C.函數(shù)f(x)有3個(gè)零點(diǎn)D.若函數(shù)f(x)在區(qū)間(3a-1,a+3)上存在最小值,則實(shí)數(shù)a的取值范圍為(-3,0]11.(2025·濟(jì)南聯(lián)考)已知函數(shù)f(x)=ex(sinx+cosx)在區(qū)間(-2π,0)內(nèi)有兩個(gè)極值點(diǎn)x1,x2,且x1<x2,則()A.|x1+x2|=πB.f(x)在區(qū)間(x1,x2)上單調(diào)遞減C.f(x1)+f(x2)>0D.|f(x1)-f(x2)|<1三、填空題12.已知函數(shù)f(x)=x(lnx-ax)在(0,+∞)上有兩個(gè)極值,則實(shí)數(shù)a的取值范圍為.

13.甲、乙兩地相距240km,汽車(chē)從甲地以速度v(km/h)勻速行駛到乙地.已知汽車(chē)每小時(shí)的運(yùn)輸成本由固定成本和可變成本組成,固定成本為160元,可變成本為v36400元.為使全程運(yùn)輸成本最小,汽車(chē)應(yīng)以14.(2024·廣州模擬)設(shè)a,b為實(shí)數(shù),函數(shù)f(x)=x3+ax2+bx+a2在x=1處取得極值10,則f(-1)=.

四、解答題15.(2025·江西名校聯(lián)考)已知函數(shù)f