Question

已知函数f(x)的定义域为[-1/2,3/2],求函数g(x)=f(3x)+f(x/3)的定义域

Answer 1

f(x)的定义域为[-1/2,3/2] 则有：

-3/2≤3x≤9/2
可得：f(3x)的定义域为[-3/2,9/2]
-1/6≤x/3≤1/2
可得：f(x/3)的定义域为[-1/6,1/2]
综上可得：
g(x)=f(3x)+f(x/3)的定义域为[-1/6,1/2]

Answer 2

设3x=t1,x/3=t2
因为f(t1),f(t2)的定义域都是[-1/2,3/2]
所以-1/2<=t1<=3/2,-1/2<=t2<=3/2
-1/2<=3x<=3/2,-1/2<=x/3<=3/2
-1/6<=x<=1/2,-3/2<=x<=9/2
取交集，得到定义域是[-1/6,1/2]

Answer 3

-1/2<=3x<=3/2且-1/2<=x/3<=3/2
解得：[-1/6,1/2]