2.2函數(shù)的單調(diào)性與最值課標(biāo)要求精細(xì)考點(diǎn)素養(yǎng)達(dá)成1.理解函數(shù)的單調(diào)性、最大值、最小值及其幾何意義2.會(huì)運(yùn)用基本初等函數(shù)的圖象分析函數(shù)的性質(zhì)函數(shù)的單調(diào)性主要通過(guò)函數(shù)單調(diào)性(區(qū)間)的確定,考查數(shù)學(xué)抽象與數(shù)學(xué)運(yùn)算核心素養(yǎng)函數(shù)單調(diào)性的應(yīng)用通過(guò)應(yīng)用單調(diào)性求最值、比較大小、解不等式等,提升邏輯推理和數(shù)學(xué)運(yùn)算素養(yǎng)函數(shù)的值域(最值)通過(guò)求函數(shù)的值域(最值)問(wèn)題,培養(yǎng)數(shù)學(xué)運(yùn)算素養(yǎng)和深化分類討論思想1.(概念辨析)下列結(jié)論正確的個(gè)數(shù)為().①若定義在R上的函數(shù)f(x),有f(1)<f(3),則函數(shù)f(x)在R上為增函數(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為單調(diào)遞減函數(shù);④所有的單調(diào)函數(shù)都有最大值或最小值A(chǔ).0 B.1 C.2 D.32.(對(duì)接教材)如果函數(shù)f(x)=x2+2x3,x∈[0,2],那么函數(shù)f(x)的值域?yàn)?).A.[4,+∞) B.[4,5]C.[3,5] D.[0,5]3.(對(duì)接教材)已知函數(shù)f(x)=4x2kx8在[5,20]上具有單調(diào)性,則實(shí)數(shù)k的取值范圍為.

4.(易錯(cuò)自糾)已知函數(shù)f(x)=(a-2)x,x≥2,12x-1,x<2滿足對(duì)任意的實(shí)數(shù)x1≠x2,5.(真題演練)(2023·新高考Ⅰ卷)設(shè)函數(shù)f(x)=2x(xa)在區(qū)間(0,1)上單調(diào)遞減,則a的取值范圍是().A.(∞,2] B.[2,0)C.(0,2] D.[2,+∞)確定函數(shù)的單調(diào)性(區(qū)間)典例1(1)函數(shù)y=x2+2|x|+1的單調(diào)遞增區(qū)間為,單調(diào)遞減區(qū)間為.

(2)設(shè)函數(shù)f(x)=1,x>0,0,x=0,-1,x<0,g(x)=x2f(x1),訓(xùn)練1(1)函數(shù)f(x)=|x2|x的單調(diào)遞減區(qū)間是.

(2)函數(shù)f(x)=ln(x22x8)的單調(diào)遞增區(qū)間是().A.(∞,2) B.(∞,1)C.(1,+∞) D.(4,+∞)訓(xùn)練2試討論函數(shù)f(x)=axx-1(a≠0)在(1,1)典例2已知f(x)是定義在(0,+∞)上的單調(diào)遞增函數(shù),且滿足f(xy)=f(x)+f(y),f(3)=1,則當(dāng)f(x)+f(x8)≤2時(shí),實(shí)數(shù)x的取值范圍是().(8,+∞) B.(8,9]C.[8,9] D.(0,8)確定函數(shù)單調(diào)性的方法(1)定義法和導(dǎo)數(shù)法;(2)復(fù)合函數(shù)法,復(fù)合函數(shù)單調(diào)性的規(guī)律是“同增異減”;(3)利用單調(diào)函數(shù)的運(yùn)算性質(zhì).訓(xùn)練3已知定義在R上的函數(shù)f(x)滿足:①f(x+y)=f(x)+f(y)+1;②當(dāng)x>0時(shí),f(x)>1.求f(0)的值.(2)證明:f(x)在R上是增函數(shù).函數(shù)單調(diào)性的應(yīng)用典例3(1)已知函數(shù)f(x)的圖象向左平移1個(gè)單位長(zhǎng)度后關(guān)于y軸對(duì)稱,當(dāng)x2>x1>1時(shí),[f(x2)f(x1)]·(x2x1)<0恒成立,設(shè)a=f-12,b=f(2),c=f(3),則a,b,c的大小關(guān)系為(A.c>a>b B.c>b>aC.a>c>b D.b>a>c(2)已知函數(shù)f(x)為R上的增函數(shù),若f(a2a)>f(a+3),則實(shí)數(shù)a的取值范圍是.

(3)已知函數(shù)f(x)=(3a-1)x+4a,x<1,logax,x≥1滿足對(duì)任意的實(shí)數(shù)x1≠x2A.(0,1) B.0,13C.17,利用單調(diào)性比較大小,應(yīng)根據(jù)函數(shù)的性質(zhì)(如圖象的對(duì)稱性等)將自變量轉(zhuǎn)化到函數(shù)的同一個(gè)單調(diào)區(qū)間上.注意:(1)若函數(shù)在區(qū)間[a,b]上是單調(diào)的,則該函數(shù)在此區(qū)間的任意子集上也是單調(diào)的;(2)分段函數(shù)的單調(diào)性,除注意各段的單調(diào)性外,還要注意銜接點(diǎn)的取值.訓(xùn)練4已知函數(shù)f(x)=x2+12a-2,x≤1,ax-a,x>1,若f(x)求函數(shù)的最值(值域)典例4求下列函數(shù)的值域:(配方法)y=x22x+3,x∈[0,3);(2)(分離常數(shù)法)y=2x(換元法)y=2xx-1;(4)(單調(diào)性法)y=x+1求函數(shù)最值的四種常用方法(1)單調(diào)性法:先確定函數(shù)的單調(diào)性,再由單調(diào)性求最值.(2)圖象法:先作出函數(shù)的圖象,再觀察其最高點(diǎn)、最低點(diǎn),求出最值.(3)換元法:對(duì)比較復(fù)雜的函數(shù)可通過(guò)換元轉(zhuǎn)化為熟悉的函數(shù),再用相應(yīng)的方法求最值.(4)分離常數(shù)法:形如求y=cx+dax+b(ac訓(xùn)練5(1)函數(shù)f(x)=13xlog2(x+2)在區(qū)間[1,1]上的最大值為(2)函數(shù)y=1+x1?2x的值域?yàn)?)A.-∞,32 B.-∞,32C.(3)函數(shù)f(x)=3x3x構(gòu)造簡(jiǎn)單函數(shù)利用單調(diào)性解決問(wèn)題典例已知函數(shù)f(x)是定義在R上的奇函數(shù),若對(duì)任意的x1,x2∈[0,+∞),且x1≠x2,都有x1f(x1)-x2f(x2)x1-x2<0成立,則不等式mf(m)13,1 B.(∞,1)C.(1,+∞) D.-∞,13∪(單調(diào)性是函數(shù)的重要性質(zhì)之一,具有廣泛的應(yīng)用,通過(guò)構(gòu)造輔助單調(diào)函數(shù),將原問(wèn)題轉(zhuǎn)化為研究輔助單調(diào)函數(shù)的問(wèn)題,并利用函數(shù)的單調(diào)性解決問(wèn)題,這是一種創(chuàng)造性的思維過(guò)程,具有較強(qiáng)的靈活性和技巧性.訓(xùn)練已知f(x)為R上的奇函數(shù),f(2)=2,若?x1,x2∈(0,+∞),當(dāng)x1>x2時(shí),都有(x1x2)·f(x1)x2-f(x2)x1<0,則不等式(x+A.(3,1)B.(3,1)∪(1,1)C.(∞,1)∪(1,1)D.(∞,3)∪(1,+∞)一、單選題1.已知函數(shù)f(x)在(0,+∞)上單調(diào)遞減,則不等式f(2x1)<f(1x)的解集為().A.0,23 B.23,1 C.2.(2024·江蘇南通期初)已知函數(shù)f(x)=12x2-ax+3在區(qū)間[0,1]上是減函數(shù),則aA.(∞,2] B.(∞,0] C.[2,+∞) D.[0,+∞)3.函數(shù)f(x)=|x2|(x4)的減區(qū)間是().A.[2,4] B.[2,3]C.[2,+∞) D.[3,+∞)4.已知函數(shù)f(x)=-x2+2ax-a,x≤1,2x,x>1A.(∞,1] B.[1,3]C.[3,+∞) D.(∞,1]∪[3,+∞)二、多選題5.已知函數(shù)f(x)=ax2(a>0且a≠1)為減函數(shù),則函數(shù)y=loga|x+2|在下列區(qū)間中單調(diào)遞增的是().A.(2,+∞) B.(4,2)C.(4,2) D.(5,3)6.下列函數(shù)中滿足“對(duì)任意x1,x2∈(0,+∞),且x1≠x2,都有f(x1)-f(xA.f(x)=3x+1 B.f(x)=2x C.f(x)=x2+4x+3 D.f(x)=x1x三、填空題7.函數(shù)y=2x+x-1的最小值為8.已知函數(shù)f(x)=x2-x,0≤x≤2,2x-1,x>2,則函數(shù)四、解答題9.已知函數(shù)f(x)=xx2+4,x∈(2,(1)用定義證明函數(shù)f(x)在(2,2)上為增函數(shù);(2)若f(a+2)>f(2a1),求實(shí)數(shù)a的取值范圍.10.已知函數(shù)f(x)=2xax的定義域?yàn)?0,1](a為實(shí)數(shù))(1)當(dāng)a=1時(shí),求函數(shù)y=f(x)的值域;(2)求函數(shù)y=f(x)在區(qū)間(0,1]上的最大值及最小值,并求出當(dāng)函數(shù)f(x)取得最值時(shí)x的值.11.定義:min{a,b}表示a,b兩數(shù)中較小的數(shù),例如min{2,4}=2.已知f(x)=min{x2,x2},g(x)=2x+x+m(m∈R),若對(duì)任意的x1∈[2,0],存在x2∈[1,2],都有f(x1)≤g(x2)成立,則實(shí)數(shù)m的取值范圍為().A.[4,+∞) B.[6,+∞)C.[7,+∞) D.[10,+∞)12.(2023·山東濟(jì)南期末)對(duì)