分段函數(shù)專題非常全面

分段函數(shù)是函數(shù)中比較復(fù)雜的一種函數(shù)，其要點(diǎn)在于自變量取不同范圍的值時(shí)所使用的解析式不同，所以在解決分段函數(shù)的問(wèn)題時(shí)要時(shí)刻盯著自變量的范圍是否在發(fā)生變化。即“分段函數(shù)——分段看”一、基礎(chǔ)知識(shí)：1、分段函數(shù)的定義域與值域——各段的并集2、分段函數(shù)單調(diào)性的判斷：先判斷每段的單調(diào)性，如果單調(diào)性相同，則需判斷函數(shù)是連續(xù)的還是斷開的，如果函數(shù)連續(xù)，則單調(diào)區(qū)間可以合在一起，如果函數(shù)不連續(xù)，則要根據(jù)函數(shù)在兩段分界點(diǎn)出的函數(shù)值（和臨界值）的大小確定能否將單調(diào)區(qū)間并在一起。3、分段函數(shù)對(duì)稱性的判斷：如果能夠?qū)⒚慷蔚膱D像作出，則優(yōu)先采用圖像法，通過(guò)觀察圖像判斷分段函數(shù)奇偶性。如果不便作出，則只能通過(guò)代數(shù)方法比較f■XQf?X■的關(guān)系，要注意x,IX的范圍以代入到正確的解析式。4、分段函數(shù)分析要注意的幾個(gè)問(wèn)題（1）分段函數(shù)在圖像上分為兩類，連續(xù)型與斷開型，判斷的方法為將邊界值代入每一段函數(shù)（其中一段是函數(shù)值，另外一段是臨界值），若兩個(gè)值相等，那么分段函數(shù)是連續(xù)的。否X■X■1,X■3則是斷開的。例如：f.—―■X2■4,X■3將X■3代入兩段解析式，計(jì)算結(jié)果相同，那x■1,x■3么此分段函數(shù)圖像即為一條連續(xù)的曲線，其性質(zhì)便于分析。再比如f■中■X2■1,X■3兩段解析式結(jié)果不同，進(jìn)而分段函數(shù)的圖像是斷開的兩段。（2）每一個(gè)含絕對(duì)值的函數(shù)，都可以通過(guò)絕對(duì)值內(nèi)部的符號(hào)討論，將其轉(zhuǎn)化為分段函數(shù)。.1■3,x■1例如：f■x■x-1■3，可轉(zhuǎn)化為：f■xT■X■3,X■15、遇到分段函數(shù)要時(shí)刻盯住變量的范圍，并根據(jù)變量的范圍選擇合適的解析式代入，若變量的范圍并不完全在某一段中，要注意進(jìn)行分類討論例如：6、如果分段函數(shù)每一段的解析式便于作圖，則在解題時(shí)建議將分段函數(shù)的圖像作出，以便必要時(shí)進(jìn)行數(shù)形結(jié)合。DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDx?2口口/??1-3xH4B1DDDDD了■■、典型例題R-1x■1R-1x■1例1：已知函數(shù)/(%)■■

.2?\axx■1，若/則實(shí)數(shù)〃■DDDDDDDDDDDDD00412(214(2DDDDDDDDDDDDD00412(214(2■a?2答案：a?2mmK：osMx,x10mmK：osMx,x10例：設(shè)函數(shù)/則/I勺值為DDDD/?電□□□□□□//??/??1?11XBODDDDDDDDDDDDDDDDDD浴"■隹＞■保i"■什9答案：A2Bcos——3DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD浴"■隹＞■保i"■什9答案：A2Bcos——3DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDx??1，1口□□□□□□□1DDDDDDDDDDDDDDDDDDDD■ddddddddddddddddddddddd例：函數(shù)/?DDDDDDDDDDD■ddddddddddddddddddddddd例：函數(shù)/?■■3xB4,xB2■2，則不等式/?.1的解集是r?xB2DDDDDDD/■■1口口口口口口口口口口口口DDDDDDD/■■1口口口口口口口口口口口口□:□:x■■或x■2或X■5?！跻詘■■15■X■2;當(dāng)33x■或X■5?！跻詘■■15■X■2;當(dāng)33x■2口口f■一■2^—■1■2■x■1叫x■1X■3□所以2■x■3DDDDDD答案：/H1例4：已知函數(shù)f(x)■■x■x■0x■0,則不等式x*.Mx叫'1的解集是一DDDDDDDDDDDDDDf?■1，為具體的表達(dá)式,觀察已知分段函數(shù),■H1f(x)B|.mx|0,x00x|0f?1m1里DDDDDODDDDDDDDDDDDDDDDDDDODDDDDDDDDD□□對(duì)x■1DDDDDDDDDD(1)當(dāng)x■1■0■x■?DDf■■■■|H-1?1■喊,□等x■xH■11k1■ix2■1■x■■(2叫x■1■0■x■■DDf?■■xH1H1■xDDDDDDDx■xH-1，1■x2■2x■1■0■■■<2■x■■■<2■x.事,■■<21答案：x■<21已知函數(shù)f已知函數(shù)f?■則不等式f■x■8*f理■3x，解集為DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDx■8,x2■3xDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDf?■□□□□現(xiàn)f?■Idddddddddddddddf?■□□□發(fā)□f?■Bddd0D0Dx■8,x2■3xODDODDf■■81fcf星■3x■x■8■x2■3xDODD

DDDDDDDDDIIIDDDDDDDDDDDDDDDD%進(jìn)行分類討論將不等式轉(zhuǎn)變岫口等□求□DDDDDDDDDIIIDDDDDDDDDDDDDDDD%進(jìn)行分類討論將不等式轉(zhuǎn)變岫口等□求□DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD□□對(duì)(比如例3,例4DDDDDDDDDDD5)。M_l%22X,X0例：已知函數(shù)f..■若fIfc.f■.2fJ則a的取值范圍是IX2■2X,X■0.*。?.■!,1■.%,1?.睢,2■路:DDDDDaDDDDDDDDDf4.f..2f■!]□□□□□□□,但可從■a,aDDDDDDDDDDf?%□□□□□□□□□□f?%□□,□以f4■f?■所解不等式變?yōu)閒?■f1?再由圖像可得只需a■1,口■■a■1答案：DDDDDD(1DDDDDDDf■(咻□□□□□□□□□□□□□□□□□□□□□□□□□□a,■aDDDDDDDDDDDDDDDDDDDDDDDDDD(2)由于f?%□□□□□□,DDDDDf?%□□□□□□□□□DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDf?Qf■1%■,□□□□□□□□□□□,XDDDDDx■■,■J，:喊■■,0■■f■■x2■2x,fIX"MX■■2■Hfar■x2■2xdDDf■%*f■■%■即f■%%DDtg(力f)x■(gX例7：已知函數(shù)f(X)■1■2x2,g(X)■x2■2x,若F(X)■耳:二、二/、，則F(X)的W(X),f(X)■g(X)值域是解析：F■■Idddddddddddddf?嘎?■Idddddddddddd好xDDDDDDDDDf■■g■■1■2X2■X2■2XDDDDB1■X■1D口

M2■2x,■1■x■1DDDDDDDDDDDDDDDDF??l31DDDDDDDDDDDDDDDD!■2x2,x■■-orxI13?a■2)x?例8?a■2)x?例8：已知函數(shù)f(x)?|Ilogxa，若f(x)在■?■■單調(diào)遞增，則實(shí)數(shù)a的(x■1)取值范圍是□路□若f?■■??^1口,則在R□□取x■x，□有f?■f?%任取1212DDDDx,xDDDDDDDDDDDDDDDDDDRDDDDDDDDDDDDDD12DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDx,xDDDDDDDD12DDDDDDDDDDDDDDDDDDDDDDDDDIDDDDDDDDDDDDDDDDDDD1a■1DDDDx■x□□有f??f??□□□□□□

1212x■1DDDD■DDDDa■2■1■log1■0■a■3aDDDDDDDa■■,3■答案：例：已知f??■■1.x■%,0■；2■1,例：已知f??■■1.x■%,0■；2■1,xE.則下列選項(xiàng)錯(cuò)誤的是()①是f??1■圖像②是f?L1的圖像④是f.,圖像③是f什1的圖像DDDDDDDDf??：dDDDDDDD,DDDDDDDDDDDDA.f??1?f?咻□□□□□□□DDDDB.f■1f?咻JDDDDDDDDDDDC.f■1■f?/□□□□□□DDDDDDDDJDDDDDDDDDDDD.fDDDDDDDDf??：dDDDDDDD,DDDDDDDDDDDDA.f??1?f?咻□□□□□□□DDDDB.f■1f?咻JDDDDDDDDDDDC.f■1■f?/□□□□□□DDDDDDDDJDDDDDDDDDDDD.f?咻f?■xDDDD在xDDDDDDDDDDDDDDxDDDDDDDDf?"I1DDDDDD答案:例：函數(shù)f■x?二,sin，則下列結(jié)論正確的是（函數(shù)f■x睢■,■?上為增函數(shù)函數(shù)f?B的最小正周期為函數(shù)f?1是奇函數(shù)函數(shù)f?■最小值DDDDDDDf?■RiddddddddddddddddddddddddddddDDDDDDDBCDDDDDDDDf?%□□值f?■fclEDD以DDDDADDDDDDj■2sinxDDDDDD：2DDDD答案：DDDDDDD1DDDDDDDDDDDDDDDDDDDDf?*□□□□,DDDDDDDDDDDDDDDDDDDD2DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDx■■DDD三、近年模擬題題目精選ix■aB3,x■1,.、已知函數(shù)fx^l■■若f科f?■則a■■gl—■i|x■1,

2、已知口（口）■2、已知口（口）■若□[□（口0）]■3，貝U□0■>2sin口,（0■口?■x,x■0,、（，湖州中學(xué)期中）函數(shù)若/"（a）],則實(shí).2,x■0,數(shù)。的取值范圍為（）.（*,0].[mo].B,*].[?,■?]則f.■■的解集為|^x■a,x■15、（2015,北京）設(shè)函數(shù)f■1■■..■a■IV2a嗎I1①若a■1,則f?1的最小值為②若f?,有2個(gè)零點(diǎn)，則實(shí)數(shù)a的取值范圍是6、⑵⑸福建）若函數(shù)于■■nMj.0M.,值域是■,■■則實(shí)a數(shù)a的取值范圍是TOC\o"1-5"\h\z?■_?Blog■■x1!1■1.?_7、（2015,新課標(biāo)II）設(shè)函數(shù)f..■2，則f?2.f110go12,（）律x■,x■12A.3B,6C,9D.12，x■1,x■18、（2015,山東）設(shè)函數(shù)f■,I，則滿足f4■1■2f?，勺a的取值范圍*x,xI1是、）TOC\o"1-5"\h\zA,??；B.?,