高三分段函數(shù)復(fù)習(xí)材料

1、1分段函數(shù)分段函數(shù)一、一、分段函數(shù)定義分段函數(shù)定義分段函數(shù)是指自變量在兩個或兩個以上不同的范圍內(nèi)有不同的對應(yīng)法則的函數(shù)它是一個函數(shù)它的定義域是各段函數(shù)定義域的并集其值域也是各段函數(shù)值域的并集。二、二、分段函數(shù)的考查點分段函數(shù)的考查點1、求分段函數(shù)的定義域和值域求分段函數(shù)的定義域和值域例1求函數(shù)的定義域、1222[10]()(02)3[2)xxfxxxx??????????????值域.【解析】作圖利用“數(shù)形結(jié)合”易知的定義()fx域為值

2、域為.[1)???(13]?練1、設(shè)函數(shù)，，則的值域)(2)(2Rxxxg????????????)()()(4)()(xgxxxgxgxxxgxf)(xf是（）ABCD)1(]049[????)0[??)49[???)2(]049[????2求分段函數(shù)的函數(shù)值求分段函數(shù)的函數(shù)值例2已知函數(shù)求.2|1|2(||1)()1(||1)1xxfxxx???????????12[()]ff【解析】因為所以.311222()|1|2f?????3

3、12223214[()]()1()13fff??????練2、定義在R上的函數(shù)滿足=?????????0)2()1(0)4(log2xxfxfxx，)(xf)(xf則的值為（）)3(fAB.C.1D.2123、給出函數(shù)，則（）?????????)4()1()4()21()(xxfxxfx?)3(log2f11o3221yx13個單位再沿軸向下平移1個單位得解析式所以y2(2)1124yxx??????綜上可得故選A.12()2([02]

4、)fxxx???222(10)()2(02)xxxfxx???????????5作分段函數(shù)的圖像作分段函數(shù)的圖像例5函數(shù)的圖像大致是（）|ln||1|xyex???A11oyxByx11OCyxO11DyxO116判斷分段函數(shù)的奇偶性判斷分段函數(shù)的奇偶性例6判斷函數(shù)的奇偶性.22(1)(0)()(1)(0)xxxfxxxx???????????【解析】當(dāng)時0x?0x??22()()(1)(1)()fxxxxxfx?????????當(dāng)時，

5、當(dāng)0x?(0)(0)0ff???0x?0x??因此對于任意都有22()()(1)(1)()fxxxxxfx?????????xR?()()fxfx??所以為偶函數(shù).()fx7判斷分段函數(shù)的單調(diào)性（考查具有隱含性，例如：與分段函數(shù)不等式有關(guān)的問題）判斷分段函數(shù)的單調(diào)性（考查具有隱含性，例如：與分段函數(shù)不等式有關(guān)的問題）例7判斷函數(shù)的單調(diào)性.32(0)()(0)xxxfxxx??????????【解析】顯然連續(xù).當(dāng)時恒成立所以是單調(diào)遞()f