函数f(x)的定义域为[-1/2,1/2]
则函数f(2x)-1/2<=2x<=1/2
-1/4<=x<=1/4
则函数f(x-1/2) -1/2<=x-1/2<=1/2
0<=x<=1
综上求函数y=f(2x)+f(x-1/2)的定义域[0,1/4]