已知函数f(x)的定义域为[-1/2,3/2],求……………………

解：1、因为f(x)的定义域为[-1/2,3/2]
g(x)=f[2/(x+1)]
则-1/2<=2/(x+1)<=3/2
解得：x<=-5或者x>=1/3
所以g(x)的定义域为：(负无穷，-5]并[1/3,正无穷)

2、-1/2<=3x<=3/2且-1/2<=3/x<=3/2
前者解得：-1/6<=x<=1/2,后者解得：x<=-6或者x>=2
两者无交集，所以g(x)的定义域为空集。

f(x)的定义域为[-1/2,3/2],

f[2/(x+1)]的定义域
-1/2<=2/(x+1)<=3/2
-1/2<=2/(x+1)
若x+1>0,x>-1
-(x+1)<=4
x>=-5
所以x>=-1
若x+1<0,x<-1
-(x+1)>=4
x<=-5
所以x<-5
2/(x+1)<=3/2
若x+1>0,x>-1
4<=3(x+1)
x>=1/3
所以x>=1/3
若x+1<0,x<-1
4>=3(x+1)
x<=1/3
所以x<-1
取交集
所以定义域(-∞,-5]

g(x)=f(3x)+f(3/x)
-1/2<=3x<=3/2
-1/6<=x<=1/2
-1/2<=3/x<=3/2
若x<0,则3/x<=3/2成立
-1/2<=3/x,-x>=6,x<=-6
所以x<=-6
若x>0,则3/x>=-1/2成立
3/x>=3/2
3x<=6
0<x<=2
所以x<=-6,0<