20xx屆二輪復(fù)習(xí)------函數(shù)的單調(diào)性極值點(diǎn)極值最值--學(xué)案(全國(guó)通用)(編輯修改稿)
2025-04-03 03:23 本頁(yè)面
【文章內(nèi)容簡(jiǎn)介】 39。(x)=1x10得0x1,由G39。(x)=1x10得x1,則G(x)在區(qū)間(0,1)上單調(diào)遞增,在區(qū)間(1,+∞)上單調(diào)遞減,即F39。(x)在區(qū)間(0,1)上單調(diào)遞增,在區(qū)間(1,+∞)上單調(diào)遞減,∴F39。(x)≤F39。(1)=0,∴F(x)在定義域(0,+∞)上單調(diào)遞減.【例2】解f(x)的定義域是(0,+∞).f39。(x)=1x+2a(1a)x2(1a)=2a(1a)x22(1a)x+1x.令g(x)=2a(1a)x22(1a)x+1,為確定函數(shù)g(x)的函數(shù)類型對(duì)a進(jìn)行分類討論.(1)當(dāng)a=1時(shí),g(x)是常數(shù)函數(shù),此時(shí)g(x)=10,f39。(x)=1x0,于是f(x)在(0,+∞)上單調(diào)遞增.(2)當(dāng)a≠1時(shí),g(x)是二次函數(shù),首先討論f39。(x)=0是否有實(shí)根,方程g(x)=0對(duì)應(yīng)的Δ=4(a1)(3a1).①當(dāng)Δ0,即13a1時(shí),g(x)=0無(wú)實(shí)根,g(x)的圖象在x軸上方,即f39。(x)0,f(x)在(0,+∞)上單調(diào)遞增.②當(dāng)Δ=0,即a=13時(shí),g(x)=0有兩個(gè)相等的實(shí)根x1=x2=32,于是f39。(x)≥0,所以f(x)在(0,+∞)上單調(diào)遞增.③當(dāng)Δ0,即0a13或a1時(shí),g(x)=0有兩個(gè)不相等的實(shí)根分別為x1=12a(a1)(3a1)2a(1a),x2=12a+(a1)(3a1)2a(1a).因?yàn)閤1+x2=1a,x1x2=12a(1a),所以當(dāng)0a13時(shí),有x1+x20且x1x20,所以x10,x20.由x1與x2的表達(dá)式知x1x2,由f39。(x)0,可得0xx1或xx2,所以f(x)在(0,x1)和(x2,+∞)上單調(diào)遞增。由f39。(x)0,可得x1xx2,所以f(x)在(x1,x2)上單調(diào)遞減.當(dāng)a1時(shí),有x1+x20且x1x20,此時(shí)x20x1,由f39。(x)0,可得0xx1,所以f(x)在(0,x1)上單調(diào)遞增。由f39。(x)0可得xx1,所以f(x)在(x1,+∞)上單調(diào)遞減.綜上所述,當(dāng)0a13時(shí),f(x)在(0,x1)和(x2,+∞)上單調(diào)遞增,在(x1,x2)上單調(diào)遞減。當(dāng)13≤a≤1時(shí),f(x)在(0,+∞)上單調(diào)遞增。當(dāng)a1時(shí),f(x)在(0,x1)上單調(diào)遞增,在(x1,+∞)上單調(diào)遞減.其中x1=12a(a1)(3a1)2a(1a),x2=12a+(a1)(3a1)2a(1a).對(duì)點(diǎn)訓(xùn)練2解設(shè)h(x)=f(x)2xc,則h(x)=2lnx2x+1c,其定義域?yàn)?0,+∞),h39。(x)=2x2.(1)當(dāng)0x1時(shí),h39。(x)0。當(dāng)x1時(shí),h39。(x)(x)在區(qū)間(0,1)單調(diào)遞增,在區(qū)間(1,+∞)=1時(shí),h(x)取得最大值,最大值為h(1)=1c.故當(dāng)且僅當(dāng)1c≤0,即c≥1時(shí),f(x)≤2x+c.所以c的取值范圍為[1,+∞).(2)g(x)=f(x)f(a)xa=2(lnxlna)xa,x∈(0,a)∪(a,+∞).g39。(x)=2xax+lnalnx(xa)2=21ax+lnax(xa)2.取c=1得h(x)=2lnx2x+2,h(1)=0,則由(1)知,當(dāng)x≠1時(shí),h(x)0,即1x+lnx∈(0,a)∪(a,+∞)時(shí),1ax+lnax0,從而g39。(x)(x)在區(qū)間(0,a),(a,+∞)單調(diào)遞減.【例3】解定義域?yàn)?1,+∞),∴f39。(x)=1x+1+a(2x1)=1x+1(2ax2+ax+1a),∵1x+10,令g(x)=2ax2+ax+1a(x1),當(dāng)a=0時(shí),g(x)=1,則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)
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