高等數(shù)學(xué)-復(fù)旦大學(xué)出版社-課后習(xí)題答案

1、1.解:(1)相等.因為兩函數(shù)的定義域相同都是實數(shù)集R由2xx?知兩函數(shù)的對應(yīng)法則也相同所以兩函數(shù)相等.(2)相等.因為兩函數(shù)的定義域相同都是實數(shù)集R由已知函數(shù)關(guān)系式顯然可得兩函數(shù)的對應(yīng)法則也相同所以兩函數(shù)相等.(3)不相等.因為函數(shù)()fx的定義域是1xxx??R而函數(shù)()gx的定義域是實數(shù)集R兩函數(shù)的定義域不同所以兩函數(shù)不相等.2.解:(1)要使函數(shù)有意義必須400xx??????即40xx?????所以函數(shù)的定義域是(0)(04]

2、???.(2)要使函數(shù)有意義必須30lg(1)010xxx???????????即301xxx?????????所以函數(shù)的定義域是[30)∪(01).(3)要使函數(shù)有意義必須210x??即1x??所以函數(shù)的定義域是(1)(11)(1)????????.(4)要使函數(shù)有意義必須12sin1x???即11sin22x???即ππ2π2π66kxk?????或5π7π2π2π66kxk????(k為整數(shù)).也即ππππ66kxk?????(k

3、為整數(shù)).所以函數(shù)的定義域是ππ[ππ]66kk???k為整數(shù).3.解:由已知顯然有函數(shù)的定義域為(∞∞)又當(dāng)0x?時1x可以是不為零的任意實數(shù)此時1sinx可以取遍[11]上所有的值所以函數(shù)的值域為[11].4.解:10(0)110f????1()1()1()1xxfxxx?????????1111().111xxfxxx??????5.解:1110101(1).(1)101213xxfxxxxx???????????????????

4、???6.解:()ln(())22gxxxfgx??又由1212121222221212()(1)11(1)(1)xxxxxxyyxxxx??????????知當(dāng)12xx?且121xx?時12yy?而當(dāng)12xx?且121xx?時12yy?.故函數(shù)21xyx??在定義域內(nèi)不單調(diào).(2)函數(shù)的定義域為(0∞)100Mx?????且12e0MxMx????使2lnxM?.取012maxxxx?則有0012lnln2xxxxMM?????所以函

5、數(shù)lnyxx??在定義域內(nèi)是無界的.又當(dāng)120xx??時有12120lnln0xxxx????故1211221212(ln)(ln)()(lnln)0yyxxxxxxxx??????????.即當(dāng)120xx??時恒有12yy?所以函數(shù)lnyxx??在(0)??內(nèi)單調(diào)遞增.11.解:(1)124(1)yx??是由1241yuux???復(fù)合而成.(2)2sin(12)yx??是由2sin12yuuvvx????復(fù)合而成.(3)512(110

6、)xy???是由152110wyuuvvwx??????復(fù)合而成.(4)11arcsin2yx??是由11arcsin2yuuvvwwx??????復(fù)合而成.12.證:(1)設(shè)()()()Fxfxfx???則()x??????有()()()()FxfxfxFx?????故()()fxfx??為偶函數(shù).(2)設(shè)()()()Gxfxfx???則()x??????有()()()[()()]()GxfxfxfxfxGx???????????故(

7、)()fxfx??為奇函數(shù).13.解:設(shè)年銷售批數(shù)為x則準(zhǔn)備費為103x又每批有產(chǎn)品610x件庫存數(shù)為6102x件庫存費為6100.052x?元.設(shè)總費用為則63100.05102yxx???.14.解:當(dāng)x能被20整除即[]2020xx?時郵資0.802025xxy???當(dāng)x不能被20整除時即[]2020xx?時由題意知郵資0.80120xy?????????.綜上所述有020002520200.8002000.1202020xxxx