f（x+2) 1= 1=< x+2<=4 -1=< x<=2 f(x+2)定义域[-1,2]
1=< x+2<=4 -1=< x<=2
f(x+2)定义域[-1,2]
f(2x-1） 3=<2x-1<=5
2=<2x<=4 1= f(2x-1)定义域[1,2]
f(2x-1)定义域[1,2]
0= f(x）定义域[1,4]
f(x）定义域[1,4]
2= f(x^2)定义域[-3,-2]∪[2,3]
f(x^2)定义域[-3,-2]∪[2,3]
帮我p张图吧
x+2的定义域为1到4闭区间,再求解x
