Let a function f:(0,∞)→(0,∞) be defined by f(x) = |1 - 1/x|. Then f is:

Let the function f:(0,∞)→(0,∞) be defined by f(x) = |1 - 1/x|.
We analyze the injectivity and surjectivity of the function.

Injectivity:
A function is injective if and only if f(x₁) = f(x₂) implies x₁ = x₂ for all x₁, x₂ in the domain.
Let's consider two values x₁ and x₂ such that f(x₁) = f(x₂).
|1 - 1/x₁| = |1 - 1/x₂|
This implies that either (1 - 1/x₁) = (1 - 1/x₂) or (1 - 1/x₁) = -(1 - 1/x₂).
The first case gives 1/x₁ = 1/x₂, which implies x₁ = x₂.
The second case gives 1 - 1/x₁ = -1 + 1/x₂, which implies 2 = 1/x₁ + 1/x₂.
If we choose x₁ = 1 and x₂ = 1, then f(1) = 0, which is not in the codomain (0, ∞). If we let x₁ = 2, then 2 = 1/2 + 1/x₂, which gives 1/x₂ = 3/2, so x₂ = 2/3. In this case, f(2) = |1 - 1/2| = 1/2 and f(2/3) = |1 - 3/2| = 1/2. Thus f(2) = f(2/3) but 2 ≠ 2/3, so the function is not injective.

Surjectivity:
A function is surjective if for every y in the codomain, there exists at least one x in the domain such that f(x) = y.
Let y ∈ (0, ∞). We want to find x such that |1 - 1/x| = y.
If 1 - 1/x = y, then 1/x = 1 - y, so x = 1/(1 - y). This is defined for y ∈ (-∞, 1) ∩ (1,∞)
If 1 - 1/x = -y, then 1/x = 1 + y, so x = 1/(1 + y). This is defined for y ∈ (-∞, ∞) since y∈(0,∞)
Since we are given that the codomain is (0, ∞), then y ∈ (0, ∞).
For any y ∈ (0, ∞), we can find an x such that f(x) = y.
If y < 1, then x = 1/(1 - y) is in (0, ∞), and if y > 1, then x = 1/(y - 1) is in (0, ∞).
Therefore, the function is surjective.

Conclusion:
The function f(x) = |1 - 1/x| is not injective but is surjective.