费玉清污段子打麻将:已知f(x+1)定义域为[-2,3],则f(2*x^2-2)定义域是

来源:百度文库 编辑:高校问答 时间:2024/05/10 05:37:56

f(x+1)定义域为[-2,3],即
-2<=x<=3
<=>-1<=x+1<=4
so:
f(x)定义域为[-1,4],

令-1<=2*x^2-2<=4,
得到:
-根号3〈=X〈=-(根号2)/2 或
(根号2)/2〈=X〈= 根号3

2<=x<3

3<=x+1<4

3<=2*x^2-2<4

5<=2*x^2<6

5/2<=x^2<3

根号5/2<=x<根号3
-根号3<x<=-根号5/2

f(x+1)定义域为[-2,3],即
-2<=x<=3
<=>-1<=x+1<=4
so:
f(x)定义域为[-1,4],

令-1<=2*x^2-2<=4,
得到:
-根号3〈=X〈=-(根号2)/2 或
(根号2)/2〈=X〈= 根号3

对就是这样的。