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-閱讀頁(yè)
2025-04-03 03:28本頁(yè)面
【正文】 值為f(1)=0或f(2)=ln2k.(ⅰ)當(dāng)0ln2k,即0kln2時(shí),[f(x)]min=f(1)=0.(ⅱ)當(dāng)0≥ln2k,即k≥ln2時(shí),[f(x)]min=f(2)=ln2k.綜上所述,當(dāng)kln2時(shí),[f(x)]min=f(1)=0。(x)=2x,設(shè)切點(diǎn)為(x0,12x02),則2x0=2,即x0=1,所以切點(diǎn)為(1,11),由點(diǎn)斜式可得切線方程為y11=2(x1),即2x+y13=0.(2)顯然t≠0,因?yàn)閥=f(x)在點(diǎn)(t,12t2)處的切線方程為y(12t2)=2t(xt),令x=0,得y=t2+12,令y=0,得x=t2+122t,所以S(t)=12(t2+12)(t)=143t2+24144t2=3(t4+8t248)4t2=3(t24)(t2+12)4t2=3(t2)(t+2)(t2+12)4t2,由S39。(t)0,得0t2,所以S(t)在(0,2)上單調(diào)遞減,在(2,+∞)上單調(diào)遞增,所以當(dāng)t=2時(shí),S(t)取得極小值,也是最小值為S(2)=16+244+1448=32.例3解令t=|x|∈[0,1),則ln1+|x|1|x|≥a|x|對(duì)x∈(1,1)恒成立等價(jià)于ln1+t1t≥at對(duì)t∈[0,1)恒成立.(方法1 分離參數(shù)法)當(dāng)t=0時(shí),不等式恒成立,當(dāng)t0時(shí),有a≤ln1+t1tt對(duì)t∈(0,1)恒成立.令G(t)=ln1+t1tt,則G39。(t)=2+2t2(1t2)221t2=4t2(1t2)20,所以H(t)在(0,1)上單調(diào)遞增,于是H(t)H(0)=0,即G39。(t)=21t2a=at2+2a1t2.①當(dāng)2a≥0,即a≤2時(shí),F39。(t)0可得0≤ta2a,所以函數(shù)F(t)在0,a2a上單調(diào)遞減,所以F(t)≤F(0)=0,a的最大值為2.對(duì)點(diǎn)訓(xùn)練3解(1)由f39。(ln2)=1,即2mln2+n=ln2,2m=1,解得m=1,n=2.(2)由(1)知f(x)=exx2,由x+1(kx)[f(x)+x+1],得x+1(kx)(ex1),故當(dāng)x0時(shí),等價(jià)于kx+1ex1+x(x0),①令g(x)=x+1ex1+x,則g39。(x)=ex10.∴函數(shù)h(x)=exx2在(0,+∞)上單調(diào)遞增.而h(1)0,h(2)0,所以h(x)在(0,+∞)上存在唯一的零點(diǎn),故g39。(x)0,g(x)單調(diào)遞減。(x)0,g(x)單調(diào)遞增.所以g(x)在(0,+∞)上的最小值為g(α),又由g39。(x)=xex2exx3k2x2+1x=(x2)(exkx)x3.當(dāng)k≤0時(shí),exkx0.所以當(dāng)0x2時(shí),f39。當(dāng)x2時(shí),f39。(x)=0在(0,2)上有兩個(gè)不同的實(shí)數(shù)根.(方法1)f39。(x)=exk.當(dāng)k≤1時(shí),h39。(x)0可得xlnk,由h39。(x)=0在(0,2)上有兩個(gè)不同的實(shí)數(shù)根等價(jià)于exkx=0在(0,2)上有兩個(gè)不同的實(shí)數(shù)根,等價(jià)于k=exx在(0,2)上有兩個(gè)不同的實(shí)數(shù)根,等價(jià)于y=k與g(x)=exx在(0,2)上有兩個(gè)不同的交點(diǎn).g39。(x)0。(x)0.g(1)=e,g(2)=e22,當(dāng)x→0+時(shí),g(x)→+∞.畫出g(x)在(0,2)上的圖象(圖略),可知要使y=k與g(x)在(0,2)上有兩個(gè)不同的交點(diǎn),k的取值范圍為e,e22.對(duì)點(diǎn)訓(xùn)練4解f(x)=lnx+ax+x(x1),f39。(x)=2x1x=2x21x,當(dāng)x∈(1,+∞)時(shí),F39。(x)0,f(x)在(1,+∞)上單調(diào)遞增,無(wú)極值。(x)=2x1x1=2x2x1x,當(dāng)x2時(shí),G39。B=40時(shí),BB1=1800403+640=160,則AA1=160.由140O39。A=80.所以AB=O39。B=80+40=120(米).(2)以O(shè)為原點(diǎn),MN為x軸,OO39。C=80x.設(shè)D(x80,y1),則y1=140(80x)2,所以CD=160y1=160140(80x)2=140x2+4x.記橋墩CD和EF的總造價(jià)為f(x),則f(x)=k160+1800x36x+32k140x2+4x=k1800x3380x2+160(0x40).f39。(x)=0,得x=20.x(0,20)20(20,40)f39。(2)當(dāng)O39。22x=24(x360x2+900x),0x30,則V39。(x)=0,解得x1=10,或x2=30(舍去).當(dāng)x∈(0,10)時(shí),V39。當(dāng)x∈(10,30)時(shí),V3
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