20xx高考數(shù)學(xué)文人教a版一輪復(fù)習(xí)學(xué)案：32-第2課時(shí)-利用導(dǎo)數(shù)研究函數(shù)的極值、最大(小)值-在線瀏覽
2025-04-03 03:28本頁(yè)面
【正文】 解題心得洛必達(dá)法則:如果當(dāng)x→x0(x0也可以是177。(x)≠0.(2)limx→x0f(x)=limx→x0g(x)=0.(3)limx→x0f39。(x)=a,則有l(wèi)imx→x0f(x)g(x)=limx→x0f39。(x)=a.定理2:若函數(shù)f(x)和g(x)滿足條件:(1)f(x)和g(x)在x0的某個(gè)去心鄰域內(nèi)可導(dǎo),且g39。(x)g39。(x)g39。(2)當(dāng)x0時(shí),若k為整數(shù),且x+1(kx)[f(x)+x+1],求k的最大值.考點(diǎn)已知函數(shù)的極值求參數(shù)的取值范圍【例4】設(shè)函數(shù)f(x)=exx2k2x+ln x(k為常數(shù),e= 28…是自然對(duì)數(shù)的底數(shù)).(1)當(dāng)k≤0時(shí),求函數(shù)f(x)的單調(diào)區(qū)間。(x)=0是f(x)有極值的必要不充分條件,例如函數(shù)f(x)=x3,f39。(0)=0,但x=0不是函數(shù)f(x)=x3的極值點(diǎn).對(duì)點(diǎn)訓(xùn)練4(2020江西名校大聯(lián)考,理21)已知函數(shù)f(x)=lnx+ax+x(a∈R).若函數(shù)f(x)在區(qū)間(1,+∞)上有極值,求實(shí)數(shù)a的取值范圍.考點(diǎn)利用導(dǎo)數(shù)求實(shí)際問(wèn)題中的最值【例5】(2020江蘇,17)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所示:谷底O在水平線MN上,橋AB與MN平行,OO39。在AB上).經(jīng)測(cè)量,左側(cè)曲線AO上任一點(diǎn)D到MN的距離h1(單位:米)與D到OO39。右側(cè)曲線BO上任一點(diǎn)F到MN的距離h2(單位:米)與F到OO39。的距離為40米.(1)求橋AB的長(zhǎng)度。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(單位:萬(wàn)元),橋墩CD每米造價(jià)32k(單位:萬(wàn)元)(k0),問(wèn)O39。第2課時(shí) 利用導(dǎo)數(shù)研究函數(shù)的極值、最大(小)值必備知識(shí)(x)0 f39。(x)0 f39。(x)=4x212x+8=4(x2)(x1)=0得x=1或x=2,當(dāng)x1時(shí),f39。當(dāng)1x2時(shí),f39。當(dāng)x2時(shí),f39。(x)=ex+xex=(1+x)39。(x)0,當(dāng)x1時(shí),f39。(x)=1x1=1xx,當(dāng)x∈(0,1)時(shí),f39。當(dāng)x∈(1,e]時(shí),f39。(x)=(xc)2+2x(xc),f39。學(xué)案突破例1解f(x)的定義域?yàn)?1,+∞),f39。(x)0在(1,+∞)上恒成立,則f(x)在(1,+∞)上單調(diào)遞增,即當(dāng)a=0時(shí),函數(shù)無(wú)極值點(diǎn).當(dāng)a≠0時(shí),①由Δ=a(9a8)≤0,得0a≤89,此時(shí)g(x)≥0,則f39。(x)0,f(x)單調(diào)遞增,當(dāng)x∈(x1,x2)時(shí),g(x)0,則f39。(x)0,f(x)單調(diào)遞增,因此,當(dāng)a89,函數(shù)有兩個(gè)極值點(diǎn)。(x)0,f(x)單調(diào)遞增,當(dāng)x∈(x2,+∞)時(shí),g(x)0,則f39。當(dāng)0≤a≤89時(shí),函數(shù)無(wú)極值點(diǎn)。(x)=2x+bx+1=2x2+2x+bx+1.令g(x)=2x2+2x+b,則Δ=12時(shí),Δ0,所以g(x)在(1,+∞)上恒大于0,所以f39。(x)≥0在(1,+∞)上恒成立,所以函數(shù)f(x)在(1,+∞)上單調(diào)遞增,從而函數(shù)f(x)在(1,+∞)上無(wú)極值點(diǎn).③當(dāng)Δ0,即b12時(shí),g(x)=0有兩個(gè)不相等的根x1=112b2,x2=1+12b2,其中x1x2.為確定兩個(gè)根是否都在定義域(1,+∞)上需要對(duì)參數(shù)b分類討論.當(dāng)b0時(shí),x1=112b21,x2=1+12b21,由f39。(x)0,可得1xx2,所以f(x)在(1,x2)上單調(diào)遞減,在(x2,+∞)上單調(diào)遞增,所以當(dāng)b0時(shí),f(x)在(1,+∞)上有唯一極小值點(diǎn)x2=1+12b2.當(dāng)0b12時(shí),x1=112b21,x2=1+12b21,由f39。(x)0,可得x1xx2,所以f(x)在(1,x1)上單調(diào)遞增,在(x1,x2)上單調(diào)遞減,在(x2,+∞)上單調(diào)遞增,所以當(dāng)0b12時(shí),f(x)在(1,+∞)上有一個(gè)極大值點(diǎn)x1=112b2和一個(gè)極小值點(diǎn)x2=1+12b2.綜上所述,當(dāng)b0時(shí),f(x)在(1,+∞)上有唯一的極小值點(diǎn)x2=1+12b2。當(dāng)b≥12時(shí),f(x)無(wú)極值點(diǎn).例2解函數(shù)f(x)的定義域?yàn)?0,+∞),f39。(x)0,函數(shù)f(x)在[1,2]上單調(diào)遞增,所以[f(x)]min=f(1)=0.②當(dāng)k0時(shí),由f39。(x)0,可得x1k,所以f(x)在0,1k上單調(diào)遞增,在1k,+∞上單調(diào)遞減.于是f(x)在[1,2]上的最小
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