第二章函數(shù)第2節(jié)單調(diào)性與最大(小)值學(xué)習(xí)導(dǎo)航站核心知識庫：重難考點總結(jié)，梳理必背知識、歸納重點考點1函數(shù)的單調(diào)性★★☆☆☆考點2函數(shù)的最值★★★☆☆（星級越高，重要程度越高）限時【變式訓(xùn)練】挑戰(zhàn)場：感知真題，檢驗成果，考點追溯【知識梳理】考點1函數(shù)的單調(diào)性(1)單調(diào)函數(shù)的定義增函數(shù)減函數(shù)定義一般地,設(shè)函數(shù)f(x)的定義域為I,區(qū)間D?I,如果?x1,x2∈D當(dāng)x1<x2時,都有f(x1)<f(x2),那么就稱函數(shù)f(x)在區(qū)間D上單調(diào)遞增,特別地,當(dāng)函數(shù)f(x)在它的定義域上單調(diào)遞增時,我們就稱它是增函數(shù)當(dāng)x1<x2時,都有f(x1)>f(x2),那么就稱函數(shù)f(x)在區(qū)間D上單調(diào)遞減,特別地,當(dāng)函數(shù)f(x)在它的定義域上單調(diào)遞減時,我們就稱它是減函數(shù)圖象描述自左向右看圖象是上升的自左向右看圖象是下降的(2)單調(diào)區(qū)間的定義如果函數(shù)y=f(x)在區(qū)間D上單調(diào)遞增或單調(diào)遞減,那么就說函數(shù)y=f(x)在這一區(qū)間具有(嚴(yán)格的)單調(diào)性,區(qū)間D叫做y=f(x)的單調(diào)區(qū)間.考點2函數(shù)的最值前提設(shè)函數(shù)y=f(x)的定義域為I,如果存在實數(shù)M滿足條件(1)?x∈I,都有f(x)≤M;(2)?x0∈I,使得f(x0)=M(1)?x∈I,都有f(x)≥M;(2)?x0∈I,使得f(x0)=M結(jié)論M為最大值M為最小值【名師點撥】1.有關(guān)單調(diào)性的常用結(jié)論在公共定義域內(nèi),增函數(shù)+增函數(shù)=增函數(shù);減函數(shù)+減函數(shù)=減函數(shù);增函數(shù)-減函數(shù)=增函數(shù);減函數(shù)-增函數(shù)=減函數(shù).2.函數(shù)y=f(x)(f(x)≠0)在公共定義域內(nèi)與y=-f(x),y=1f(【教材回歸】概念思考辨析+教材經(jīng)典改編1.思考辨析(在括號內(nèi)打“√”或“×”)(1)對于函數(shù)y=f(x),若f(1)<f(3),則f(x)為增函數(shù).()(2)函數(shù)y=f(x)在[1,+∞)上是增函數(shù),則函數(shù)的單調(diào)遞增區(qū)間是[1,+∞).()(3)函數(shù)y=1x的單調(diào)遞減區(qū)間是(-∞,0)∪(0,+∞(4)對于函數(shù)f(x),x∈D,若對任意x1,x2∈D,且x1≠x2有(x1-x2)[f(x1)-f(x2)]>0,則函數(shù)f(x)在區(qū)間D上是增函數(shù).()2.(人教A必修一P86T7改編)函數(shù)f(x)=x2?2x的單調(diào)遞增區(qū)間是3.(人教B必修一P140T2(1)改編)函數(shù)f(x)=4x2(x∈[-2,-1]),則f(x)的最小值為,最大值為4.(蘇教必修一P122T4改編)函數(shù)y=f(x)是定義在[-2,2]上的減函數(shù),且f(a+1)<f(2a),則實數(shù)a的取值范圍是.

【考向核心題型】考點一函數(shù)的單調(diào)性(區(qū)間)【典例】1.函數(shù)y=x2+2x?24【典例】2.試討論函數(shù)f(x)=axx?1(a≠0)在(-1【變式訓(xùn)練】1.下列函數(shù)在R上為增函數(shù)的是()A.y=x2 B.y=xC.y=-x D.y=1【變式訓(xùn)練】2.(2025·西安模擬)已知函數(shù)f(x)=|x2-5x+6|,則函數(shù)f(x)的單調(diào)遞增區(qū)間是()A.?∞,5C.2,52和(3,+∞) D.(-∞,考點二求函數(shù)的最值(值域)【典例】2.(多選)下列函數(shù)中,值域正確的是()A.當(dāng)x∈[0,3)時,函數(shù)y=x2-2x+3的值域為[2,6)B.函數(shù)y=2x+1C.函數(shù)y=2x-x?1的值域為D.函數(shù)y=x+1+x?1的值域為[2【變式訓(xùn)練】3.(2025·北京懷柔模擬)已知函數(shù)f(x)=4x22x2+1,則對任意實數(shù)xA.(0,2) B.(0,2] C.[0,2) D.[0,2]【變式訓(xùn)練】4.對于任意實數(shù)a,b,定義min{a,b}=a設(shè)函數(shù)f(x)=-x+3,g(x)=log2x,則函數(shù)h(x)=min{f(x),g(x)}的最大值是.

考點三函數(shù)單調(diào)性的應(yīng)用角度1比較大小【典例】3.已知f(x)=2x-1x?1,a=f(2),b=f(3),c=f(5)A.a>b>c B.a>c>bC.c>a>b D.c>b>a角度2解函數(shù)不等式【典例】4.已知函數(shù)f(x)=lnx+2x,若f(x2-4)<2,則實數(shù)x的取值范圍是.

角度3求參數(shù)的取值范圍【典例】5.(2025·保定調(diào)研)已知a>0,且a≠1,函數(shù)f(x)=3a?x,x<2,logA.(1,+∞) B.1C.23,1 【變式訓(xùn)練】5.(2024·新高考Ⅰ卷)已知函數(shù)f(x)=?x2?2ax?a,A.(-∞,0] B.[-1,0]C.[-1,1] D.[0,+∞)【變式訓(xùn)練】6.已知函數(shù)f(x)=?x2?2x,x≥0,x2?2x,x<0,若f(3一、求復(fù)合函數(shù)的單調(diào)區(qū)間【典例】1.已知函數(shù)f(x)=x2-2x-3,g(x)=f(5-x2),試求g(x)的單調(diào)區(qū)間.二、由復(fù)合函數(shù)的單調(diào)性求參數(shù)【典例】2.設(shè)f(x)=x2+1,g(x)=f(f(x)),F(x)=g(x)-λf(x).問是否存在實數(shù)λ,使F(x)在區(qū)間?∞,?22上單調(diào)遞減且在區(qū)間?22,0【變式訓(xùn)練】1.已知函數(shù)f(x)在R上是增函數(shù),則下列說法正確的是()A.y=-f(x)在R上是減函數(shù)B.y=1f(xC.y=[f(x)]2在R上是增函數(shù)D.y=af(x)(a為實數(shù))在R上是增函數(shù)【變式訓(xùn)練】2.已知函數(shù)f(x)是R上的減函數(shù),若f(ax2-2x)在(1,+∞)上是增函數(shù),則實數(shù)a的取值范圍是.

【限時訓(xùn)練】（限時：60分鐘）一、單選題1.下列函數(shù)中,在區(qū)間(0,1)上單調(diào)遞增的是()A.y=-x2+1 B.y=xC.y=1x D.y=3-2.)已知函數(shù)f(x)=-x2+ax+1在(2,6)上不單調(diào),則a的取值范圍是()A.(2,6)B.(-∞,2]∪[6,+∞)C.(4,12)D.(-∞,4]∪[12,+∞)3.若函數(shù)f(x)=(m-1)x+1在R上是增函數(shù),則f(m)與f(1)的大小關(guān)系是()A.f(m)<f(1) B.f(m)>f(1)C.f(m)≤f(1) D.f(m)≥f(1)4.(2025·南京質(zhì)檢)若定義在R上的函數(shù)f(x)滿足f(x)=3f(|x|)+x2-2x,則f(x)的單調(diào)遞增區(qū)間為()A.(-∞,-10]和[0,1] B.(-∞,-5]和[0,1]C.[-10,0]和[1,+∞) D.[-5,0]和[1,+∞)5.已知函數(shù)f(x)=x2,x≥t,x,0<xA.{1} B.(0,+∞)C.(1,+∞) D.[1,+∞)6.已知函數(shù)y=x2-2x+3在區(qū)間[0,m]上有最大值3,最小值2,則實數(shù)m的取值范圍是()A.[1,+∞) B.[0,2]C.(-∞,2] D.[1,2]7.如果函數(shù)y=f(x)在區(qū)間I上是減函數(shù),且函數(shù)y=f(x)x在區(qū)間I上是增函數(shù),那么稱函數(shù)y=f(x)是區(qū)間I上的“可變函數(shù)”,區(qū)間I叫作“可變區(qū)間”.若函數(shù)f(x)=x2-4x+2是區(qū)間I上的“可變函數(shù)”,則“A.(-∞,-2]和[2,2] B.[2,2]C.(0,2] D.[1,3]8.(2024·新高考Ⅱ卷)設(shè)函數(shù)f(x)=(x+a)ln(x+b).若f(x)≥0,則a2+b2的最小值為()A.18 B.1C.12 二、多選題9.下列說法中,正確的是()A.若對任意x1,x2∈I,當(dāng)x1<x2時,f(x1)?f(x2)x1B.函數(shù)y=x2在R上是增函數(shù)C.函數(shù)y=-1xD.函數(shù)y=1x的單調(diào)遞減區(qū)間是(-∞,0)和(0,+∞10.已知函數(shù)f(x)=bx+3ax+2在區(qū)間(-2,+∞)上單調(diào)遞增,則aA.a=-1,b=2 B.a=2,b=1C.a=1,b>32 D.0<a≤1,b11.(2025·杭州調(diào)研)設(shè)函數(shù)f(x),g(x)的定義域都為R,且f(x)>0,g(x)>0,f(x)是減函數(shù),g(x)是增函數(shù),則下列說法正確的有()A.g(x)+f(x)是增函數(shù)B.f(x)-g(x)是減函數(shù)C.f(x)g(x)是增函數(shù)D.f(三、填空題12.函數(shù)y=f(x)是定義在[-2,2]上的減函數(shù),且f(a+1)<f(2a),則實數(shù)a的取值范圍是.

13.設(shè)函數(shù)f(x)=1,x>0,0,x=0,?1,x<0,g(x)=x2f(x-1),14.已知命題p:“若f(x)<f(4)對任意的x∈(0,4)都成立,則f(x)在(0,4)上單調(diào)遞增”.能說明命題p為假命題的一個函數(shù)是.

四、解答題15.給定函數(shù)f(x)=12x,g(x)=-x2+4x+1,x∈(1)在同一直角坐標(biāo)系中畫出函數(shù)f(x)和g(x)的圖象;(2)?