答:f(x)的定义域为[0,1]E(x)=f(x+m)+f(x-m),m>0则有:0<=x+m<=10<=x-m<=1所以:-m<=x<=1-mm<=x<=1+m所以:m<=x<=1-m当m>1-m即m>1/2时,E(x)定义域为空集当m<=1-m即0