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第01講函數(shù)的概念1、函數(shù)的概念(1)一般地,給定非空數(shù)集SKIPIF1<0,SKIPIF1<0,按照某個對應(yīng)法則SKIPIF1<0,使得SKIPIF1<0中任意元素SKIPIF1<0,都有SKIPIF1<0中唯一確定的SKIPIF1<0與之對應(yīng),那么從集合SKIPIF1<0到集合SKIPIF1<0的這個對應(yīng),叫做從集合SKIPIF1<0到集合SKIPIF1<0的一個函數(shù).記作:SKIPIF1<0,SKIPIF1<0.集合SKIPIF1<0叫做函數(shù)的定義域,記為SKIPIF1<0,集合SKIPIF1<0,SKIPIF1<0叫做值域,記為SKIPIF1<0.(2)函數(shù)的實質(zhì)是從一個非空集合到另一個非空集合的映射.2、函數(shù)的三要素(1)函數(shù)的三要素:定義域、對應(yīng)關(guān)系、值域.(2)如果兩個函數(shù)的定義域相同,并且對應(yīng)關(guān)系完全一致,則這兩個函數(shù)為同一個函數(shù).3、函數(shù)的表示法表示函數(shù)的常用方法有解析法、圖象法和列表法.4、分段函數(shù)若函數(shù)在其定義域的不同子集上,因?qū)?yīng)關(guān)系不同而分別用幾個不同的式子來表示,這種函數(shù)稱為分段函數(shù).【解題方法總結(jié)】1、基本的函數(shù)定義域限制求解函數(shù)的定義域應(yīng)注意:(1)分式的分母不為零;(2)偶次方根的被開方數(shù)大于或等于零:(3)對數(shù)的真數(shù)大于零,底數(shù)大于零且不等于1;(4)零次冪或負(fù)指數(shù)次冪的底數(shù)不為零;(5)三角函數(shù)中的正切SKIPIF1<0的定義域是SKIPIF1<0且SKIPIF1<0;(6)已知SKIPIF1<0的定義域求解SKIPIF1<0的定義域,或已知SKIPIF1<0的定義域求SKIPIF1<0的定義域,遵循兩點:①定義域是指自變量的取值范圍;=2\*GB3②在同一對應(yīng)法則∫下,括號內(nèi)式子的范圍相同;(7)對于實際問題中函數(shù)的定義域,還需根據(jù)實際意義再限制,從而得到實際問題函數(shù)的定義域.2、基本初等函數(shù)的值域(1)SKIPIF1<0的值域是SKIPIF1<0.(2)SKIPIF1<0的值域是:當(dāng)SKIPIF1<0時,值域為SKIPIF1<0;當(dāng)SKIPIF1<0時,值域為SKIPIF1<0.(3)SKIPIF1<0的值域是SKIPIF1<0.(4)SKIPIF1<0且SKIPIF1<0的值域是SKIPIF1<0.(5)SKIPIF1<0且SKIPIF1<0的值域是SKIPIF1<0.題型一:函數(shù)的概念例1.(2023·山東濰坊·統(tǒng)考一模)存在函數(shù)SKIPIF1<0滿足:對任意SKIPIF1<0都有(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】對于A,當(dāng)SKIPIF1<0時,SKIPIF1<0;當(dāng)SKIPIF1<0時,SKIPIF1<0,不符合函數(shù)定義,A錯誤;對于B,令SKIPIF1<0,則SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0,不符合函數(shù)定義,B錯誤;對于C,令SKIPIF1<0,則SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0,不符合函數(shù)定義,C錯誤;對于D,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,則存在SKIPIF1<0時,SKIPIF1<0,符合函數(shù)定義,即存在函數(shù)SKIPIF1<0滿足:對任意SKIPIF1<0都有SKIPIF1<0,D正確,故選:D例2.(2023·重慶·二模)任給SKIPIF1<0,對應(yīng)關(guān)系SKIPIF1<0使方程SKIPIF1<0的解SKIPIF1<0與SKIPIF1<0對應(yīng),則SKIPIF1<0是函數(shù)的一個充分條件是(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】根據(jù)函數(shù)的定義,對任意SKIPIF1<0,按SKIPIF1<0,在SKIPIF1<0的范圍中必有唯一的值與之對應(yīng),SKIPIF1<0,則SKIPIF1<0,則SKIPIF1<0的范圍要包含SKIPIF1<0,故選:A.例3.(2023·全國·高三專題練習(xí))如圖,可以表示函數(shù)SKIPIF1<0的圖象的是(

)A. B.C. D.【答案】D【解析】根據(jù)函數(shù)的定義,對于一個SKIPIF1<0,只能有唯一的SKIPIF1<0與之對應(yīng),只有D滿足要求故選:D變式1.(2023·全國·高三專題練習(xí))函數(shù)y=f(x)的圖象與直線SKIPIF1<0的交點個數(shù)(

)A.至少1個 B.至多1個 C.僅有1個 D.有0個、1個或多個【答案】B【解析】若1不在函數(shù)f(x)的定義域內(nèi),y=f(x)的圖象與直線SKIPIF1<0沒有交點,若1在函數(shù)f(x)的定義域內(nèi),y=f(x)的圖象與直線SKIPIF1<0有1個交點,故選:B.【解題方法總結(jié)】利用函數(shù)概念判斷題型二:同一函數(shù)的判斷例4.(2023·高三課時練習(xí))下列各組函數(shù)中,表示同一個函數(shù)的是(

).A.SKIPIF1<0,SKIPIF1<0B.SKIPIF1<0,SKIPIF1<0C.SKIPIF1<0,SKIPIF1<0D.SKIPIF1<0,SKIPIF1<0【答案】C【解析】對于A:SKIPIF1<0的定義域為SKIPIF1<0,SKIPIF1<0的定義域為SKIPIF1<0.因為定義域不同,所以SKIPIF1<0和SKIPIF1<0不是同一個函數(shù).故A錯誤;對于B:SKIPIF1<0的定義域為SKIPIF1<0,SKIPIF1<0的定義域為SKIPIF1<0.因為定義域不同,所以SKIPIF1<0和SKIPIF1<0不是同一個函數(shù).故B錯誤;對于C:SKIPIF1<0的定義域為SKIPIF1<0,SKIPIF1<0的定義域為SKIPIF1<0,所以定義域相同.又對應(yīng)關(guān)系也相同,所以為同一個函數(shù).故C正確;對于D:SKIPIF1<0的定義域為SKIPIF1<0,SKIPIF1<0的定義域為SKIPIF1<0.因為定義域不同,所以SKIPIF1<0和SKIPIF1<0不是同一個函數(shù).故D錯誤;故選:C例5.(2023·全國·高三專題練習(xí))下列四組函數(shù)中,表示同一個函數(shù)的一組是(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】對于SKIPIF1<0,SKIPIF1<0和SKIPIF1<0的定義域都是SKIPIF1<0,對應(yīng)關(guān)系也相同,是同一個函數(shù),故選項SKIPIF1<0正確;對于SKIPIF1<0,函數(shù)SKIPIF1<0的定義域為SKIPIF1<0,函數(shù)SKIPIF1<0的定義域為SKIPIF1<0,定義域不同,不是同一個函數(shù),故選項SKIPIF1<0錯誤;對于SKIPIF1<0,函數(shù)SKIPIF1<0的定義域為SKIPIF1<0,函數(shù)SKIPIF1<0的定義域為SKIPIF1<0,定義域不同,不是同一個函數(shù),故選項SKIPIF1<0錯誤;對于SKIPIF1<0,函數(shù)SKIPIF1<0的定義域為SKIPIF1<0,函數(shù)SKIPIF1<0的定義域為SKIPIF1<0,定義域不同,不是同一個函數(shù),故選項SKIPIF1<0錯誤,故選:SKIPIF1<0.例6.(2023·全國·高三專題練習(xí))下列各組函數(shù)中,表示同一函數(shù)的是(

)A.SKIPIF1<0,SKIPIF1<0B.SKIPIF1<0C.SKIPIF1<0,SKIPIF1<0D.SKIPIF1<0,SKIPIF1<0,0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,0,SKIPIF1<0【答案】D【解析】對于A:SKIPIF1<0的定義域是SKIPIF1<0,SKIPIF1<0的定義域是SKIPIF1<0,兩個函數(shù)的定義域不相同,不是同一函數(shù),對于B:SKIPIF1<0,SKIPIF1<0,SKIPIF1<0的定義域是SKIPIF1<0,兩個函數(shù)的定義域不相同,不是同一函數(shù),對于C:SKIPIF1<0的定義域為SKIPIF1<0,SKIPIF1<0的定義域是SKIPIF1<0,兩個函數(shù)的定義域不相同,不是同一函數(shù),對于D:SKIPIF1<0對應(yīng)點的坐標(biāo)為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0對應(yīng)點的坐標(biāo)為SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,兩個函數(shù)對應(yīng)坐標(biāo)相同,是同一函數(shù),故選:D.【解題方法總結(jié)】當(dāng)且僅當(dāng)給定兩個函數(shù)的定義域和對應(yīng)法則完全相同時,才表示同一函數(shù),否則表示不同的函數(shù).題型三:給出函數(shù)解析式求解定義域例7.(2023·北京·高三專題練習(xí))函數(shù)SKIPIF1<0的定義域為________.【答案】SKIPIF1<0【解析】令SKIPIF1<0,可得SKIPIF1<0,解得SKIPIF1<0.故函數(shù)SKIPIF1<0的定義域為SKIPIF1<0.故答案為:SKIPIF1<0.例8.(2023·全國·高三專題練習(xí))若SKIPIF1<0,則SKIPIF1<0_________.【答案】SKIPIF1<0或SKIPIF1<0【解析】由SKIPIF1<0有意義可得SKIPIF1<0,所以SKIPIF1<0或SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,SKIPIF1<0,故答案為:SKIPIF1<0或SKIPIF1<0.例9.(2023·高三課時練習(xí))函數(shù)SKIPIF1<0的定義域為______.【答案】SKIPIF1<0【解析】要使函數(shù)有意義,則SKIPIF1<0,解得SKIPIF1<0.所以函數(shù)的定義域為SKIPIF1<0.故答案為:SKIPIF1<0.變式2.(2023·全國·高三專題練習(xí))已知正數(shù)a,b滿足SKIPIF1<0,則函數(shù)SKIPIF1<0的定義域為___________.【答案】SKIPIF1<0【解析】由SKIPIF1<0可得SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,代入SKIPIF1<0即SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0(舍),則SKIPIF1<0所以SKIPIF1<0SKIPIF1<0解得SKIPIF1<0所以函數(shù)定義域為SKIPIF1<0故答案為:SKIPIF1<0變式3.(2023·全國·高三專題練習(xí))已知等腰三角形的周長為SKIPIF1<0,底邊長SKIPIF1<0是腰長SKIPIF1<0的函數(shù),則函數(shù)的定義域為()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【解析】由題設(shè)有SKIPIF1<0,由SKIPIF1<0得SKIPIF1<0,故選A.【解題方法總結(jié)】對求函數(shù)定義域問題的思路是:(1)先列出使式子SKIPIF1<0有意義的不等式或不等式組;(2)解不等式組;(3)將解集寫成集合或區(qū)間的形式.題型四:抽象函數(shù)定義域例10.(2023·全國·高三專題練習(xí))已知函數(shù)SKIPIF1<0的定義域為SKIPIF1<0,則函數(shù)SKIPIF1<0的定義域為_____【答案】SKIPIF1<0【解析】令SKIPIF1<0,由SKIPIF1<0得:SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,所以,函數(shù)SKIPIF1<0的定義域為SKIPIF1<0.故答案為:SKIPIF1<0例11.(2023·高三課時練習(xí))已知函數(shù)SKIPIF1<0的定義域為SKIPIF1<0,則函數(shù)SKIPIF1<0的定義域為______.【答案】SKIPIF1<0【解析】因為函數(shù)SKIPIF1<0的定義域為SKIPIF1<0,所以在函數(shù)SKIPIF1<0中,SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0,故函數(shù)SKIPIF1<0的定義域為SKIPIF1<0.故答案為:SKIPIF1<0.例12.(2023·全國·高三專題練習(xí))已知函數(shù)SKIPIF1<0定義域為SKIPIF1<0,則函數(shù)SKIPIF1<0的定義域為_______.【答案】SKIPIF1<0【解析】因SKIPIF1<0的定義域為SKIPIF1<0,則當(dāng)SKIPIF1<0時,SKIPIF1<0,即SKIPIF1<0的定義域為SKIPIF1<0,于是SKIPIF1<0中有SKIPIF1<0,解得SKIPIF1<0,所以函數(shù)SKIPIF1<0的定義域為SKIPIF1<0.故答案為:SKIPIF1<0變式4.(2023·全國·高三專題練習(xí))已知函數(shù)SKIPIF1<0的定義域為SKIPIF1<0,則函數(shù)SKIPIF1<0的定義域為______【答案】SKIPIF1<0【解析】由函數(shù)SKIPIF1<0的定義域是SKIPIF1<0,得到SKIPIF1<0,故SKIPIF1<0即SKIPIF1<0.解得:SKIPIF1<0;所以原函數(shù)的定義域是:SKIPIF1<0.故答案為:SKIPIF1<0.變式5.(2023·全國·高三專題練習(xí))已知函數(shù)SKIPIF1<0的定義域為SKIPIF1<0,則函數(shù)SKIPIF1<0的定義域為__________.【答案】SKIPIF1<0【解析】由SKIPIF1<0解得SKIPIF1<0,所以函數(shù)SKIPIF1<0的定義域為SKIPIF1<0.故答案為:SKIPIF1<0【解題方法總結(jié)】1、抽象函數(shù)的定義域求法:此類型題目最關(guān)鍵的就是法則下的定義域不變,若SKIPIF1<0的定義域為SKIPIF1<0,求SKIPIF1<0中SKIPIF1<0的解SKIPIF1<0的范圍,即為SKIPIF1<0的定義域,口訣:定義域指的是SKIPIF1<0的范圍,括號范圍相同.已知SKIPIF1<0的定義域,求四則運算型函數(shù)的定義域2、若函數(shù)是由一些基本函數(shù)通過四則運算結(jié)合而成的,其定義域為各基本函數(shù)定義域的交集,即先求出各個函數(shù)的定義域,再求交集.題型五:函數(shù)定義域的應(yīng)用例13.(2023·全國·高三專題練習(xí))若函數(shù)SKIPIF1<0的定義域為R,則實數(shù)a的取值范圍是__________.【答案】SKIPIF1<0【解析】SKIPIF1<0的定義域是R,則SKIPIF1<0恒成立,SKIPIF1<0時,SKIPIF1<0恒成立,SKIPIF1<0時,則SKIPIF1<0,解得SKIPIF1<0,綜上,SKIPIF1<0.故答案為:SKIPIF1<0.例14.(2023·全國·高三專題練習(xí))已知SKIPIF1<0的定義域為SKIPIF1<0,那么a的取值范圍為_________.【答案】SKIPIF1<0【解析】依題可知,SKIPIF1<0的解集為SKIPIF1<0,所以SKIPIF1<0,解得SKIPIF1<0.故答案為:SKIPIF1<0.例15.(2023·全國·高三專題練習(xí))函數(shù)SKIPIF1<0的定義域為SKIPIF1<0,則實數(shù)a的取值范圍是___________.【答案】SKIPIF1<0【解析】因為函數(shù)SKIPIF1<0的定義域為R,所以SKIPIF1<0的解為R,即函數(shù)SKIPIF1<0的圖象與x軸沒有交點,(1)當(dāng)SKIPIF1<0時,函數(shù)SKIPIF1<0與x軸沒有交點,故SKIPIF1<0成立;(2)當(dāng)SKIPIF1<0時,要使函數(shù)SKIPIF1<0的圖象與x軸沒有交點,則SKIPIF1<0,解得SKIPIF1<0.綜上:實數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為:SKIPIF1<0變式6.(2023·全國·高三專題練習(xí))若函數(shù)SKIPIF1<0的定義域是R,則實數(shù)SKIPIF1<0的取值范圍是__________.【答案】SKIPIF1<0【解析】由函數(shù)SKIPIF1<0的定義域為R,得SKIPIF1<0恒成立,化簡得SKIPIF1<0恒成立,所以由SKIPIF1<0解得:SKIPIF1<0.故答案為:SKIPIF1<0.【解題方法總結(jié)】對函數(shù)定義域的應(yīng)用,是逆向思維問題,常常轉(zhuǎn)化為恒成立問題求解,必要時對參數(shù)進(jìn)行分類討論.題型六:函數(shù)解析式的求法例16.(2023·全國·高三專題練習(xí))求下列函數(shù)的解析式:(1)已知SKIPIF1<0,求SKIPIF1<0的解析式;(2)已知SKIPIF1<0,求SKIPIF1<0的解析式;(3)已知SKIPIF1<0是一次函數(shù)且SKIPIF1<0,求SKIPIF1<0的解析式;(4)已知SKIPIF1<0滿足SKIPIF1<0,求SKIPIF1<0的解析式.【解析】(1)設(shè)SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0∵SKIPIF1<0∴SKIPIF1<0,SKIPIF1<0即SKIPIF1<0,SKIPIF1<0(2)∵SKIPIF1<0由勾型函數(shù)SKIPIF1<0的性質(zhì)可得,其值域為SKIPIF1<0所以SKIPIF1<0(3)由f(x)是一次函數(shù),可設(shè)f(x)=ax+b(a≠0),∴3[a(x+1)+b]-2[a(x-1)+b]=2x+17,即ax+(5a+b)=2x+17,∴SKIPIF1<0解得SKIPIF1<0∴f(x)的解析式是f(x)=2x+7.(4)∵2f(x)+f(-x)=3x,①∴將x用SKIPIF1<0替換,得SKIPIF1<0,②由①②解得f(x)=3x.例17.(2023·全國·高三專題練習(xí))根據(jù)下列條件,求SKIPIF1<0的解析式(1)已知SKIPIF1<0滿足SKIPIF1<0(2)已知SKIPIF1<0是一次函數(shù),且滿足SKIPIF1<0;(3)已知SKIPIF1<0滿足SKIPIF1<0【解析】(1)令SKIPIF1<0,則SKIPIF1<0,故SKIPIF1<0,所以SKIPIF1<0;(2)設(shè)SKIPIF1<0,因為SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0;(3)因為SKIPIF1<0①,所以SKIPIF1<0②,SKIPIF1<0②SKIPIF1<0①得SKIPIF1<0,所以SKIPIF1<0.例18.(2023·全國·高三專題練習(xí))根據(jù)下列條件,求函數(shù)SKIPIF1<0的解析式.(1)已知SKIPIF1<0,則SKIPIF1<0的解析式為__________.(2)已知SKIPIF1<0滿足SKIPIF1<0,求SKIPIF1<0的解析式.(3)已知SKIPIF1<0,對任意的實數(shù)x,y都有SKIPIF1<0,求SKIPIF1<0的解析式.【解析】(1)方法一(換元法):令SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0.所以SKIPIF1<0,所以函數(shù)SKIPIF1<0的解析式為SKIPIF1<0.方法二(配湊法):SKIPIF1<0.因為SKIPIF1<0,所以函數(shù)SKIPIF1<0的解析式為SKIPIF1<0.(2)將SKIPIF1<0代入SKIPIF1<0,得SKIPIF1<0,因此SKIPIF1<0,解得SKIPIF1<0.(3)令SKIPIF1<0,得SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0.變式7.(2023·全國·高三專題練習(xí))已知SKIPIF1<0,求SKIPIF1<0的解析式.【解析】由SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0.變式8.(2023·廣東深圳·高三深圳外國語學(xué)校??茧A段練習(xí))寫出一個滿足:SKIPIF1<0的函數(shù)解析式為______.【答案】SKIPIF1<0【解析】SKIPIF1<0中,令SKIPIF1<0,解得SKIPIF1<0,令SKIPIF1<0得SKIPIF1<0,故SKIPIF1<0,不妨設(shè)SKIPIF1<0,滿足要求.故答案為:SKIPIF1<0變式9.(2023·全國·高三專題練習(xí))已知定義在SKIPIF1<0上的單調(diào)函數(shù)SKIPIF1<0,若對任意SKIPIF1<0都有SKIPIF1<0,則方程SKIPIF1<0的解集為_______.【答案】SKIPIF1<0.【解析】∵定義在SKIPIF1<0上的單調(diào)函數(shù)SKIPIF1<0,對任意SKIPIF1<0都有SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0,在上式中令SKIPIF1<0,則SKIPIF1<0,解得SKIPIF1<0,故SKIPIF1<0,由SKIPIF1<0得,SKIPIF1<0即SKIPIF1<0,在同一坐標(biāo)系中作出函數(shù)SKIPIF1<0和SKIPIF1<0的圖像,可知這兩個圖像有2個交點,即SKIPIF1<0和SKIPIF1<0,則方程SKIPIF1<0的解集為SKIPIF1<0.故答案為:SKIPIF1<0.【解題方法總結(jié)】求函數(shù)解析式的常用方法如下:(1)當(dāng)已知函數(shù)的類型時,可用待定系數(shù)法求解.(2)當(dāng)已知表達(dá)式為SKIPIF1<0時,可考慮配湊法或換元法,若易將含SKIPIF1<0的式子配成SKIPIF1<0,用配湊法.若易換元后求出SKIPIF1<0,用換元法.(3)若求抽象函數(shù)的解析式,通常采用方程組法.(4)求分段函數(shù)的解析式時,要注意符合變量的要求.(5)當(dāng)出現(xiàn)大基團(tuán)換元轉(zhuǎn)換繁瑣時,可考慮配湊法求解.(6)若已知成對出現(xiàn)SKIPIF1<0,SKIPIF1<0或SKIPIF1<0,SKIPIF1<0,類型的抽象函數(shù)表達(dá)式,則常用解方程組法構(gòu)造另一個方程,消元的方法求出SKIPIF1<0.題型七:函數(shù)值域的求解例19.(2023·全國·高三專題練習(xí))求下列函數(shù)的值域(1)SKIPIF1<0;(2)SKIPIF1<0;(3)SKIPIF1<0;(4)SKIPIF1<0;(5)SKIPIF1<0;(6)SKIPIF1<0;(7)SKIPIF1<0;(8)SKIPIF1<0(9)SKIPIF1<0;(10)SKIPIF1<0.【解析】(1)分式函數(shù)SKIPIF1<0,定義域為SKIPIF1<0,故SKIPIF1<0,所有SKIPIF1<0,故值域為SKIPIF1<0;(2)函數(shù)SKIPIF1<0中,分母SKIPIF1<0,則SKIPIF1<0,故值域為SKIPIF1<0;(3)函數(shù)SKIPIF1<0中,令SKIPIF1<0得SKIPIF1<0,易見函數(shù)SKIPIF1<0和SKIPIF1<0都是減函數(shù),故函數(shù)SKIPIF1<0在SKIPIF1<0時是遞減的,故SKIPIF1<0時SKIPIF1<0,故值域為SKIPIF1<0;(4)SKIPIF1<0,故值域為SKIPIF1<0且SKIPIF1<0;(5)SKIPIF1<0,SKIPIF1<0而SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0,故值域為SKIPIF1<0;(6)函數(shù)SKIPIF1<0,定義域為SKIPIF1<0,令SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,對稱軸方程為SKIPIF1<0,所以SKIPIF1<0時,函數(shù)SKIPIF1<0,故值域為SKIPIF1<0;(7)由題意得SKIPIF1<0,解得SKIPIF1<0,則SKIPIF1<0,故SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,由y的非負(fù)性知,SKIPIF1<0,故函數(shù)的值域為SKIPIF1<0;(8)函數(shù)SKIPIF1<0,定義域為SKIPIF1<0,SKIPIF1<0,故SKIPIF1<0,即值域為SKIPIF1<0;(9)函數(shù)SKIPIF1<0,定義域為SKIPIF1<0,故SKIPIF1<0,所有SKIPIF1<0,故值域為SKIPIF1<0;(10)函數(shù)SKIPIF1<0,令SKIPIF1<0,則由SKIPIF1<0知,SKIPIF1<0,SKIPIF1<0,根據(jù)對勾函數(shù)SKIPIF1<0在SKIPIF1<0遞減,在SKIPIF1<0遞增,可知SKIPIF1<0時,SKIPIF1<0,故值域為SKIPIF1<0.例20.(2023·全國·高三專題練習(xí))若函數(shù)SKIPIF1<0的值域是SKIPIF1<0,則函數(shù)SKIPIF1<0的值域為__.【答案】SKIPIF1<0【解析】因為函數(shù)SKIPIF1<0的值域是SKIPIF1<0,所以函數(shù)SKIPIF1<0的值域為SKIPIF1<0,則SKIPIF1<0的值域為SKIPIF1<0,所以函數(shù)SKIPIF1<0的值域為SKIPIF1<0.故答案為:SKIPIF1<0.例21.(2023·全國·高三專題練習(xí))函數(shù)SKIPIF1<0的值域為_____【答案】SKIPIF1<0【解析】SKIPIF1<0表示點SKIPIF1<0與點SKIPIF1<0連線的斜率,SKIPIF1<0的軌跡為圓SKIPIF1<0,SKIPIF1<0表示圓SKIPIF1<0上的點與點SKIPIF1<0連線的斜率,由圖象可知:過SKIPIF1<0作圓SKIPIF1<0的切線,斜率必然存在,則設(shè)過SKIPIF1<0的圓SKIPIF1<0的切線方程為SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0圓心SKIPIF1<0到切線的距離SKIPIF1<0,解得:SKIPIF1<0,結(jié)合圖象可知:圓SKIPIF1<0上的點與點SKIPIF1<0連線的斜率的取值范圍為SKIPIF1<0,即SKIPIF1<0的值域為SKIPIF1<0.故答案為:SKIPIF1<0.變式10.(2023·重慶渝中·高三重慶巴蜀中學(xué)校考階段練習(xí))函數(shù)SKIPIF1<0的最大值為______.【答案】SKIPIF1<0/SKIPIF1<0【解析】因為SKIPIF1<0,令SKIPIF1<0,則SKIPIF1<0,令SKIPIF1<0,SKIPIF1<0,因為函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,所以SKIPIF1<0,即SKIPIF1<0,則SKIPIF1<0,即函數(shù)SKIPIF1<0的最大值為SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時取等號.故答案為:SKIPIF1<0變式11.(2023·全國·高三專題練習(xí))函數(shù)SKIPIF1<0的值域為______.【答案】SKIPIF1<0【解析】由SKIPIF1<0有意義可得SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0的定義域為SKIPIF1<0,SKIPIF1<0SKIPIF1<0,設(shè)SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0.故答案為:SKIPIF1<0.【解題方法總結(jié)】函數(shù)值域的求法主要有以下幾種(1)觀察法:根據(jù)最基本函數(shù)值域(如SKIPIF1<0≥0,SKIPIF1<0及函數(shù)的圖像、性質(zhì)、簡單的計算、推理,憑觀察能直接得到些簡單的復(fù)合函數(shù)的值域.(2)配方法:對于形如SKIPIF1<0的值域問題可充分利用二次函數(shù)可配方的特點,結(jié)合二次函數(shù)的定義城求出函數(shù)的值域.(3)圖像法:根據(jù)所給數(shù)學(xué)式子的特征,構(gòu)造合適的幾何模型.(4)基本不等式法:注意使用基本不等式的條件,即一正、二定、三相等.(5)換元法:分為三角換元法與代數(shù)換元法,對于形SKIPIF1<0的值城,可通過換元將原函數(shù)轉(zhuǎn)化為二次型函數(shù).(6)分離常數(shù)法:對某些齊次分式型的函數(shù)進(jìn)行常數(shù)化處理,使函數(shù)解析式簡化內(nèi)便于分析.(7)判別式法:把函數(shù)解析式化為關(guān)于x的―元二次方程,利用一元二次方程的判別式求值域,一般地,形如SKIPIF1<0,SKIPIF1<0或SKIPIF1<0的函數(shù)值域問題可運用判別式法(注意x的取值范圍必須為實數(shù)集R).(8)單調(diào)性法:先確定函數(shù)在定義域(或它的子集)內(nèi)的單調(diào)性,再求出值域.對于形如SKIPIF1<0或SKIPIF1<0的函數(shù),當(dāng)ac>0時可利用單調(diào)性法.(9)有界性法:充分利用三角函數(shù)或一些代數(shù)表達(dá)式的有界性,求出值域.因為常出現(xiàn)反解出y的表達(dá)式的過程,故又常稱此為反解有界性法.(10)導(dǎo)數(shù)法:先利用導(dǎo)數(shù)求出函數(shù)的極大值和極小值,再確定最大(?。┲担瑥亩蟪龊瘮?shù)的值域.題型八:分段函數(shù)的應(yīng)用例22.(2023·四川成都·成都七中統(tǒng)考模擬預(yù)測)已知函數(shù)SKIPIF1<0,則SKIPIF1<0(

)A.-6 B.0 C.4 D.6【答案】A【解析】由分段函數(shù)知:當(dāng)SKIPIF1<0時,周期SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0.故選:A例23.(2023·河南·統(tǒng)考模擬預(yù)測)已知函數(shù)SKIPIF1<0且SKIPIF1<0,則SKIPIF1<0(

)A.-16 B.16 C.26 D.27【答案】C【解析】當(dāng)SKIPIF1<0時,SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,所以SKIPIF1<0,故選:C例24.(2023·全國·高三專題練習(xí))已知SKIPIF1<0,滿足SKIPIF1<0,則SKIPIF1<0的取值范圍是(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0【答案】D【解析】當(dāng)SKIPIF1<0時,SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,所以,SKIPIF1<0的取值范圍是SKIPIF1<0故選:D變式12.(多選題)(2023·全國·高三專題練習(xí))已知函數(shù)SKIPIF1<0,則使SKIPIF1<0的SKIPIF1<0可以是(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】BCD【解析】①當(dāng)SKIPIF1<0時,由SKIPIF1<0,可得SKIPIF1<0,若SKIPIF1<0時,則SKIPIF1<0,此時SKIPIF1<0無解,若SKIPIF1<0時,由SKIPIF1<0,解得SKIPIF1<0;②當(dāng)SKIPIF1<0時,由SKIPIF1<0,可得SKIPIF1<0或SKIPIF1<0.若SKIPIF1<0時,則SKIPIF1<0,由SKIPIF1<0可得SKIPIF1<0,方程SKIPIF1<0無解,若SKIPIF1<0時,由SKIPIF1<0可得

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