新高考數(shù)學(xué)一輪復(fù)習(xí)講義 第07講 函數(shù)的基本性質(zhì)Ⅰ-單調(diào)性與最值（原卷版）

第07講函數(shù)的基本性質(zhì)Ⅰ-單調(diào)性與最值（精講）題型目錄一覽①函數(shù)單調(diào)性的判斷與證明②求函數(shù)的單調(diào)區(qū)間③復(fù)合函數(shù)的單調(diào)性④函數(shù)單調(diào)性的應(yīng)用⑤函數(shù)的最值（值域）★【文末附錄-函數(shù)的單調(diào)性與最值思維導(dǎo)圖】一、知識(shí)點(diǎn)梳理一、知識(shí)點(diǎn)梳理1.函數(shù)的單調(diào)性

（1）增函數(shù)：若對(duì)于定義域內(nèi)的某個(gè)區(qū)間上的任意兩個(gè)自變量、，當(dāng)時(shí)，都有，那么就說(shuō)函數(shù)在區(qū)間上是增函數(shù)；（2）減函數(shù)：若對(duì)于定義域內(nèi)的某個(gè)區(qū)間上的任意兩個(gè)自變量、，當(dāng)時(shí)，都有，那么就說(shuō)函數(shù)在區(qū)間上是減函數(shù).（3）【特別提醒】①單調(diào)區(qū)間只能用區(qū)間表示，不能用不等式或集合表示．②有多個(gè)單調(diào)區(qū)間應(yīng)分別寫，不能用符號(hào)“∪”連接，也不能用“或”連接，只能用“逗號(hào)”或“和”連接．2.函數(shù)的最值（1）最大值：一般地，設(shè)函數(shù)的定義域?yàn)椋绻嬖趯?shí)數(shù)滿足：=1\*GB3①對(duì)于任意的，都有；=2\*GB3②存在，使得.那么，我們稱是函數(shù)的最大值.（2）最小值：一般地，設(shè)函數(shù)的定義域?yàn)椋绻嬖趯?shí)數(shù)滿足：=1\*GB3①對(duì)于任意的，都有；=2\*GB3②存在，使得.那么，我們稱是函數(shù)的最小值.（3）函數(shù)最值存在的兩個(gè)結(jié)論①閉區(qū)間上的連續(xù)函數(shù)一定存在最大值和最小值．②開(kāi)區(qū)間上的“單峰”函數(shù)一定存在最大(小)值．【常用結(jié)論】1.?x1，x2∈D(x1≠x2)，SKIPIF1<0?f(x)在D上是增函數(shù)；SKIPIF1<0?f(x)在D上是減函數(shù)．2.對(duì)勾函數(shù)y＝SKIPIF1<0(a＞0)的增區(qū)間為(－∞，－SKIPIF1<0]和[SKIPIF1<0，＋∞)，減區(qū)間為[－SKIPIF1<0，0)和(0，SKIPIF1<0]．3.當(dāng)f(x)，g(x)都是增(減)函數(shù)時(shí)，f(x)＋g(x)是增(減)函數(shù)．4.若k＞0，則kf(x)與f(x)單調(diào)性相同；若k＜0，則kf(x)與f(x)的單調(diào)性相反．5.函數(shù)y＝f(x)在公共定義域內(nèi)與y＝SKIPIF1<0的單調(diào)性相反．6.復(fù)合函數(shù)y＝f[g(x)]的單調(diào)性與函數(shù)y＝f(u)和u＝g(x)的單調(diào)性關(guān)系是“同增異減”．二、題型分類精講二、題型分類精講題型一函數(shù)單調(diào)性的判斷與證明策略方法1.定義法證明函數(shù)單調(diào)性的步驟2.判斷函數(shù)單調(diào)性的四種方法(1)圖象法；(2)性質(zhì)法；(3)導(dǎo)數(shù)法；(4)定義法．3.證明函數(shù)單調(diào)性的兩種方法(1)定義法；(2)導(dǎo)數(shù)法．【典例1】設(shè)函數(shù)SKIPIF1<0，指出SKIPIF1<0在SKIPIF1<0上的單調(diào)性，并證明你的結(jié)論．【題型訓(xùn)練】一、單選題1．設(shè)函數(shù)SKIPIF1<0滿足：對(duì)任意的SKIPIF1<0都有SKIPIF1<0，則SKIPIF1<0與SKIPIF1<0大小關(guān)系是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．設(shè)函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，已知SKIPIF1<0為SKIPIF1<0上的減函數(shù)，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0是SKIPIF1<0的（

）A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分也不必要條件SKIPIF1<0二、填空題3．若SKIPIF1<0，則函數(shù)在SKIPIF1<0上的值域是______________．4．對(duì)于函數(shù)SKIPIF1<0定義域內(nèi)的任意SKIPIF1<0且SKIPIF1<0，給出下列結(jié)論：（1）SKIPIF1<0（2）SKIPIF1<0（3）SKIPIF1<0（4）SKIPIF1<0其中正確結(jié)論為：__．三、解答題5．根據(jù)定義證明函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增.6．已知函數(shù)SKIPIF1<0.(1)求SKIPIF1<0的解析式;(2)判斷并證明函數(shù)SKIPIF1<0在SKIPIF1<0上的單調(diào)性.7．設(shè)SKIPIF1<0對(duì)任意的SKIPIF1<0有SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.(1)求證SKIPIF1<0是SKIPIF1<0上的減函數(shù)；(2)若SKIPIF1<0，求SKIPIF1<0在SKIPIF1<0上的最大值與最小值.題型二求函數(shù)的單調(diào)區(qū)間策略方法求復(fù)合函數(shù)單調(diào)區(qū)間的一般步驟(1)求函數(shù)的定義域(定義域先行)．(2)求簡(jiǎn)單函數(shù)的單調(diào)區(qū)間．(3)求復(fù)合函數(shù)的單調(diào)區(qū)間，其依據(jù)是“同增異減”．【典例1】已知函數(shù)SKIPIF1<0(1)畫出函數(shù)圖象(2)結(jié)合圖象寫出函數(shù)的單調(diào)增區(qū)間和的單調(diào)減區(qū)間.【題型訓(xùn)練】一、單選題1．函數(shù)SKIPIF1<0，SKIPIF1<0的單調(diào)減區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．函數(shù)SKIPIF1<0的單調(diào)增區(qū)間是（

）A．SKIPIF1<0和SKIPIF1<0 B．SKIPIF1<0和SKIPIF1<0C．SKIPIF1<0和SKIPIF1<0 D．SKIPIF1<0和SKIPIF1<03．如果函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是減函數(shù)，且函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上是增函數(shù)，那么稱函數(shù)SKIPIF1<0是區(qū)間SKIPIF1<0上的“可變函數(shù)”，區(qū)間SKIPIF1<0叫做“可變區(qū)間”.若函數(shù)SKIPIF1<0是區(qū)間SKIPIF1<0上的“可變函數(shù)”，則“可變區(qū)間”SKIPIF1<0為（

）A．SKIPIF1<0和SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0二、填空題4．函數(shù)SKIPIF1<0的單調(diào)減區(qū)間為_(kāi)__________.5．函數(shù)SKIPIF1<0的單調(diào)增區(qū)間是___________.三、解答題6．已知二次函數(shù)SKIPIF1<0的最小值為1，且滿足SKIPIF1<0，SKIPIF1<0，點(diǎn)SKIPIF1<0在冪函數(shù)SKIPIF1<0的圖像上．(1)求SKIPIF1<0和SKIPIF1<0的解析式；(2)定義函數(shù)SKIPIF1<0試畫出函數(shù)SKIPIF1<0的圖象，并求函數(shù)SKIPIF1<0的定義域、值域和單調(diào)區(qū)間．7．已知函數(shù)SKIPIF1<0．（其中SKIPIF1<0）(1)求函數(shù)SKIPIF1<0的單調(diào)增區(qū)間；(2)若對(duì)任意SKIPIF1<0，使得SKIPIF1<0恒成立，求實(shí)數(shù)a的取值范圍．8．已知函數(shù)SKIPIF1<0（a為正常數(shù)），且函數(shù)SKIPIF1<0與SKIPIF1<0的圖象在y軸上的截距相等．(1)求a的值；(2)求函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間；題型三復(fù)合函數(shù)的單調(diào)性策略方法集合運(yùn)算三步驟【典例1】函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【題型訓(xùn)練】一、單選題1．函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．已知SKIPIF1<0在SKIPIF1<0上是減函數(shù)，則實(shí)數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．已知函數(shù)SKIPIF1<0則SKIPIF1<0的大致圖像是（

）A．B．C． D．二、填空題4．函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間是____________.5．已知SKIPIF1<0在SKIPIF1<0上是嚴(yán)格減函數(shù)，則實(shí)數(shù)a的取值范圍是______.6．已知函數(shù)SKIPIF1<0且SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減，則實(shí)數(shù)a的取值范圍是___________.三、解答題7．已知函數(shù)SKIPIF1<0為奇函數(shù)．(1)求常數(shù)SKIPIF1<0的值；(2)判斷函數(shù)SKIPIF1<0在SKIPIF1<0上的單調(diào)性．8．已知函數(shù)SKIPIF1<0(1)若SKIPIF1<0，求SKIPIF1<0的定義域.(2)若函數(shù)在區(qū)間SKIPIF1<0上是減函數(shù)，求實(shí)數(shù)a的取值范圍.題型四函數(shù)單調(diào)性的應(yīng)用策略方法1．比較函數(shù)值大小的解題思路比較函數(shù)值的大小時(shí)，若自變量的值不在同一個(gè)單調(diào)區(qū)間內(nèi)，要利用其函數(shù)性質(zhì)，轉(zhuǎn)化到同一個(gè)單調(diào)區(qū)間內(nèi)進(jìn)行比較，對(duì)于選擇題、填空題能數(shù)形結(jié)合的盡量用圖象法求解．2．求解含“f”的函數(shù)不等式的解題思路先利用函數(shù)的相關(guān)性質(zhì)將不等式轉(zhuǎn)化為f(g(x))＞f(h(x))的形式，再根據(jù)函數(shù)的單調(diào)性去掉“f”，得到一般的不等式g(x)＞h(x)(或g(x)＜h(x))．此時(shí)要特別注意函數(shù)的定義域．3．利用單調(diào)性求參數(shù)的范圍(或值)的策略(1)視參數(shù)為已知數(shù)，依據(jù)函數(shù)的圖象或單調(diào)性定義，確定函數(shù)的單調(diào)區(qū)間，與已知單調(diào)區(qū)間比較求參數(shù)．(2)解決分段函數(shù)的單調(diào)性問(wèn)題，要注意上、下段端點(diǎn)函數(shù)值的大小關(guān)系．【典例1】已知函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)，則實(shí)數(shù)k的取值范圍為（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【典例2】已知函數(shù)SKIPIF1<0，若SKIPIF1<0，則不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【題型訓(xùn)練】一、單選題1．“SKIPIF1<0”是“SKIPIF1<0在SKIPIF1<0上單調(diào)遞增”的（

）A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分也不必要條件2．若函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則實(shí)數(shù)SKIPIF1<0的范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．已知函數(shù)SKIPIF1<0在SKIPIF1<0上是減函數(shù)，則實(shí)數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．已知函數(shù)SKIPIF1<0滿足對(duì)任意SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)都有SKIPIF1<0成立，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．已知函數(shù)SKIPIF1<0是定義域?yàn)镾KIPIF1<0的減函數(shù)，若SKIPIF1<0，則實(shí)數(shù)m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．已知函數(shù)f(x)的圖象關(guān)于y軸對(duì)稱，且f(x)在(－∞,0]上單調(diào)遞減，則滿足SKIPIF1<0的實(shí)數(shù)x的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<07．已知偶函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且對(duì)于任意SKIPIF1<0均有SKIPIF1<0成立，若SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<08．已知函數(shù)SKIPIF1<0，若SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<09．已知SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù)，且在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、填空題10．已知函數(shù)SKIPIF1<0與SKIPIF1<0在區(qū)間SKIPIF1<0上都是減函數(shù)，那么SKIPIF1<0__________.11．已知函數(shù)SKIPIF1<0,在SKIPIF1<0為單調(diào)函數(shù)，則實(shí)數(shù)a的取值范圍為_(kāi)_____．12．已知函數(shù)SKIPIF1<0，則不等式SKIPIF1<0的解集是_________．13．奇函數(shù)f(x)是定義域?yàn)?－1，1)上的減函數(shù)，且f(2a－1)＋f(a－1)>0，則a的取值范圍是________．三、解答題14．已知函數(shù)SKIPIF1<0，其中SKIPIF1<0為常數(shù).(1)該函數(shù)在SKIPIF1<0嚴(yán)格單調(diào)，求SKIPIF1<0的取值范圍；(2)若對(duì)任意的SKIPIF1<0恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍；15．設(shè)SKIPIF1<0，其中SKIPIF1<0．(1)若函數(shù)SKIPIF1<0是奇函數(shù)，求SKIPIF1<0的值；(2)若函數(shù)SKIPIF1<0在SKIPIF1<0上是嚴(yán)格減函數(shù)，求SKIPIF1<0的取值范圍．16．已知函數(shù)SKIPIF1<0，且SKIPIF1<0為奇函數(shù)．(1)判斷函數(shù)SKIPIF1<0的單調(diào)性并證明；(2)解不等式：SKIPIF1<0．題型五函數(shù)的最值（值域）策略方法求函數(shù)最值的五種常用方法【典例1】已知二次函數(shù)SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0.(1)求函數(shù)SKIPIF1<0的解析式；(2)求函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的值域．【典例2】函數(shù)SKIPIF1<0在SKIPIF1<0時(shí)有最大值為SKIPIF1<0，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【典例3】已知SKIPIF1<0為正的常數(shù)，若不等式SKIPIF1<0對(duì)一切非負(fù)實(shí)數(shù)SKIPIF1<0恒成立，則SKIPIF1<0的最大值為_(kāi)_______.【題型訓(xùn)練】一、單選題1．函數(shù)SKIPIF1<0的最小值為（

）A．2 B．SKIPIF1<0 C．3 D．以上都不對(duì)2．已知函數(shù)SKIPIF1<0，若SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值為28，則實(shí)數(shù)k的值可以是（