高等數(shù)學復(fù)旦大學出版社習題答案一

1、1習題一習題一1.下列函數(shù)是否相等為什么2222(1)()()(2)sin(31)sin(31)1(3)()()1.1fxxgxyxutxxfxgxxx???????????解:(1)相等.因為兩函數(shù)的定義域相同都是實數(shù)集R由知兩函數(shù)的對應(yīng)法則也相同所以2xx?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)的定義域是而函數(shù)的定義域是實數(shù)集R兩

2、函數(shù)()fx1xxx??R()gx的定義域不同所以兩函數(shù)不相等.2.求下列函數(shù)的定義域211(1)4arctan(2)3lg(1)(3)(4)arccos(2sin).1yxyxxxxyyxx??????????解:(1)要使函數(shù)有意義必須即400xx??????40xx?????所以函數(shù)的定義域是.(0)(04]???(2)要使函數(shù)有意義必須即30lg(1)010xxx???????????301xxx?????????所以函數(shù)的定義

3、域是[30)∪(01).(3)要使函數(shù)有意義必須即210x??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3故函數(shù)的反函數(shù)是這與是同一個函3()21fxx??31()2xyx???R31()2xgx??數(shù)所以和互為反函數(shù).3()21fxx??31()2xgx??8.求下列函數(shù)的反函數(shù)及其定義域:2531(1)(2)ln(2)11(3)3(4)1cos[0π].xxyyxxyyxx???????

4、????解:(1)由解得11xyx???11yxy???所以函數(shù)的反函數(shù)為.11xyx???1(1)1xyxx?????(2)由得ln(2)1yx???1e2yx???所以函數(shù)的反函數(shù)為.ln(2)1yx???1e2()xyx????R(3)由解得253xy??31(log5)2xy??所以函數(shù)的反函數(shù)為.253xy??31(log5)(0)2yxx???(4)由得又故.31cosyx??3cos1xy??[0π]x?3arccos1x

5、y??又由得1cos1x???301cos2x???即故可得反函數(shù)的定義域為[02]所以函數(shù)的反函02y??31cos[0π]yxx???數(shù)為.3arccos1(02)yxx????9.判斷下列函數(shù)在定義域內(nèi)的有界性及單調(diào)性:2(1)(2)ln1xyyxxx????解:(1)函數(shù)的定義域為(∞∞)當時有當時有0x?201xx??0x?21122xxxx???故有.即函數(shù)有上界.()x??????12y?21xyx??又因為函數(shù)為奇函數(shù)所