In the original operation of real number, we define the new operation "⊕" as follows: when a ≥ B, a ⊕ B = a; when a < B, a ⊕ B = B & # 178 Let f (x) = (1 ⊕ x) - (2 ⊕ x) x, X ∈ [- 2,2] 1, find the analytic expression 2 of F (x), and find the range of F (x)

In the original operation of real number, we define the new operation "⊕" as follows: when a ≥ B, a ⊕ B = a; when a < B, a ⊕ B = B & # 178 Let f (x) = (1 ⊕ x) - (2 ⊕ x) x, X ∈ [- 2,2] 1, find the analytic expression 2 of F (x), and find the range of F (x)

1. When - 2 ≤ x ≤ 1: F (x) = 1-2x
When 1 ＜ x ≤ 2: F (x) = x & # 178; - 2x
So the analytic expression of function f (x) is f (x) = 1-2x, X ∈ [- 2,1]
f﹙x﹚=x²-2x,x∈﹙1,2]
2. When - 2 ≤ x ≤ 1, f (x) = [- 1,5]
When 1 < x ≤ 2, f (x) = (- 1,0]
So the range of function f (x) is [- 1,5]