y=f(x+1)的定义域是【-2，3】即-2<=x<=3-1<=x+1<=4f(x)括号内的范围相等-1<=2x-1<=40<=x<=5/2函数y=f(2x-1)的定义域为[0,5/2]
【-5，5】。具体解法：-2<=x+1<=3，解得-3<=x<=2，所以-5<=2x-1<=5
