Question

若f(x+1)的定义域是[-2,3)求f(1//x+2)的定义域

Answer 1

f(t)=f(x+1)
x属于[-2,3)
=>t属于[-1,4)
f(1/(x+2))=f(t)
=> -1<=1/(x+2)<4
=>x<=-3或x>-7/4
即定义域为(-无穷，-3]U(-7/4,+无穷)
觉得好请采纳 祝学习进步

追问

不好意思是f(（1/x）+2)

追答

-1-3-1/3<=x<0

Answer 2

-2<=x<3得-1<=x+1<4, 即函数f(t)的定义域为[-1,4), 即-1<=t<4
函数f(1/x+2)要有意义必须-1<=1/x+2<2, 即-3<=1/x<0, 即x>-1/3, 即其定义域为[-1/3,+∞)