当x属于A时
1 ≤
x
≤
b
∴把x代入方程中
0
≤
x-1
≤
b-1
0
≤
1/2(x-1)
≤
1/2(b-1)
1 ≤
1/2(x-1)+1≤
1/2(b-1)
+
1
即1 ≤
f(x)≤
1/2(b-1)
+
1
∵f(x)属于A
∴ 1 ≤
f(x) ≤
b
∴1/2(b-1)+
1
≤b
∴b≥1
当x属于A时
1 ≤
x
≤
b
∴把x代入方程中
0
≤
x-1
≤
b-1
0
≤
1/2(x-1)
≤
1/2(b-1)
1 ≤
1/2(x-1)+1≤
1/2(b-1)
+
1
即1 ≤
f(x)≤
1/2(b-1)
+
1
∵f(x)属于A
∴ 1 ≤
f(x) ≤
b
∴1/2(b-1)+
1
≤b
∴b的范围为:b≥1