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20xx專題五:函數(shù)與導(dǎo)數(shù)(含近年高考試題)(編輯修改稿)

2025-05-13 08:53 本頁(yè)面
 

【文章內(nèi)容簡(jiǎn)介】 的切線平行于x軸,得f′(1)=0,即1-=0,解得a=e.(2)f′(x)=1-,①當(dāng)a≤0時(shí),f′(x)0,f(x)為(-∞,+∞)上的增函數(shù),所以函數(shù)f(x)無極值.②當(dāng)a0時(shí),令f′(x)=0,得ex=a,即x=ln a.x∈(-∞,ln a),f′(x)0;x∈(ln a,+∞),f′(x)0,所以f(x)在(-∞,ln a)上單調(diào)遞減,在(ln a,+∞)上單調(diào)遞增,故f(x)在x=ln a處取得極小值,且極小值為f(ln a)=ln a,無極大值.綜上,當(dāng)a ≤0時(shí),函數(shù)f(x)無極值;當(dāng)a0時(shí),f(x)在x=ln a處取得極小值ln a,無極大值.[針對(duì)訓(xùn)練]設(shè)f(x)=2x3+ax2+bx+1的導(dǎo)數(shù)為f′(x),若函數(shù)y=f′(x)的圖像關(guān)于直線x=-對(duì)稱,且f′(1)=0.(1)求實(shí)數(shù)a,b的值;(2)求函數(shù)f(x)的極值.解:(1)因?yàn)閒(x)=2x3+ax2+bx+1,故f′(x)=6x2+2ax+b,從而f′(x)=62+b-,即y=f′(x)關(guān)于直線x=-對(duì)稱.從而由題設(shè)條件知-=-,即a=3.又由于f′(1)=0,即6+2a+b=0,得b=-12.(2)由(1)知f(x)=2x3+3x2-12x+1,所以f′(x)=6x2+6x-12=6(x-1)(x+2),令f′(x)=0,即6(x-1)(x+2)=0,解得x=-2或x=1,當(dāng)x∈(-∞,-2)時(shí),f′(x)0,即f(x)在(-∞,-2)上單調(diào)遞增;當(dāng)x∈(-2,1)時(shí),f′(x)0,即f(x)在(-2,1)上單調(diào)遞減;當(dāng)x∈(1,+∞)時(shí),f′(x)0,即f(x)在(1,+∞)上單調(diào)遞增.從而函數(shù)f(x)在x=-2處取得極大值f(-2)=21,在x=1處取得極小值f(1)=-6.考點(diǎn)五 運(yùn)用導(dǎo)數(shù)解決函數(shù)的最值問題 [典例] 已知函數(shù)f(x)=ln x-ax(a∈R).(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)當(dāng)a0時(shí),求函數(shù)f(x)在[1,2]上的最小值.[解] (1)f′(x)=-a(x0),①當(dāng)a≤0時(shí),f′(x)=-a0,即函數(shù)f(x)的單調(diào)增區(qū)間為(0,+∞).②當(dāng)a0時(shí),令f′(x)=-a=0,可得x=,當(dāng)0x時(shí),f′(x)=0;當(dāng)x時(shí),f′(x)=0,故函數(shù)f(x)的單調(diào)遞增區(qū)間為,單調(diào)遞減區(qū)間為.(2)①當(dāng)≤1,即a≥1時(shí),函數(shù)f(x)在區(qū)間[1,2]上是減函數(shù),∴f(x)的最小值是f(2)=ln 2-2a.②當(dāng)≥2,即0a≤時(shí),函數(shù)f(x)在區(qū)間[1,2]上是增函數(shù),∴f(x)的最小值是f(1)=-a.③當(dāng)12,即a1時(shí),函數(shù)f(x)在上是增函數(shù),在上是減函數(shù).又f(2)-f(1)=ln 2-a,∴當(dāng)aln 2時(shí),最小值是f(1)=-a;當(dāng)ln 2≤a1時(shí),最小值為f(2)=ln 2-2a.綜上可知,當(dāng)0aln 2時(shí),函數(shù)f(x)的最小值是-a;當(dāng)a≥ln 2時(shí),函數(shù)f(x)的最小值是ln 2-2a.[針對(duì)訓(xùn)練]設(shè)函數(shù)f(x)=aln x-bx2(x0),若函數(shù)f(x)在x=1處與直線y=-相切, (1)求實(shí)數(shù)a,b的值;(2)求函數(shù)f(x)在上的最大值.解:(1)f′(x)=-2bx,∵函數(shù)f(x)在x=1處與直線y=-相切,∴解得(2)f(x)=ln x-x2,f′(x)=-x=,∵當(dāng)≤x≤e時(shí),令f′(x)0得≤x1;令f′(x)0,得1x≤e,∴f(x)在上單調(diào)遞增,在[1,e]上單調(diào)遞減,∴f(x)max=f(1)=-.考點(diǎn)六:用導(dǎo)數(shù)解決函數(shù)極值、最值問題[典例] (2013北京豐臺(tái)高三期末)已知函數(shù)f(x)=(a0)的導(dǎo)函數(shù)y=f′(x)的兩個(gè)零點(diǎn)為-3和0.(1)求f(x)的單調(diào)區(qū)間;(2)若f(x)的極小值為-e3,求f(x)在區(qū)間[-5,+∞)上的最大值.[解] (1)f′(x)==,令g(x)=-ax2+(2a-b)x+b-c,因?yàn)閑x0,所以y=f′(x)的零點(diǎn)就是g(x)=-ax2+(2a
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