Let A = R − {3} and B = R − {1}. Consider the function f: A → B defined by f(x) = (x − 3)/(x − 1). Show that f is one-one and onto and hence find f⁻¹

Let x₁, x₂ ∈ A. Now, f(x₁) = f(x₂) ⇒ (x₁ − 3)/(x₁ − 1) = (x₂ − 3)/(x₂ − 1)
(x₁ − 3)(x₂ − 1) = (x₁ − 1)(x₂ − 3)
x₁x₂ − x₁ − 3x₂ + 3 = x₁x₂ − 3x₁ − x₂ + 3
−x₁ − 3x₂ = −3x₁ − x₂
2x₁ = 2x₂ ⇒ x₁ = x₂
Hence f is a one-one function.
For Onto: Let y = (x − 3)/(x − 1)
y(x − 1) = x − 3
xy − y = x − 3
xy − x = y − 3
x(y − 1) = y − 3
x = (y − 3)/(y − 1) — (1)
From above it is obviously that ∀y except 1, i.e., ∀y ∈ B = R − {1} ∃x ∈ A. Hence f is an onto function.
Thus f is a one-one function.
If f⁻¹ is the inverse of f then
f⁻¹(y) = (y − 3)/(y − 1) [from (1)]