Show that the function f in A = R - {2/3} defined as f(x) = 4x + 36x is one-one and onto. Hence find f⁻¹.

The function f is one-one.
Reason: Let x₁, x₂ ∈ A be such that f(x₁) = f(x₂)
4x₁ + 36x₁ = 4x₂ + 36x₂
x₁x₂(4x₁ + 36x₂) = 24x₁x₂ + 18x₁ + 18x₂
⇒ 34x₂ = 34x₁ ⇒ x₂ = x₁ ⇒ f is one-one.
The function f is onto:
We shall find the range of the function f.
For the range of f, let y = f(x)
y = 4x + 36x
⇒ 6xy = 4x + 3 ⇒ (6y)x = 4y + 3 ⇒ x = (4y + 3)/6y
but x ∈ A ⇒ 6y ≠ 0 ⇒ y ≠ 2/3
Range of f is R - {2/3} = A
Range of f = Co-domain of f
f is onto function.
Thus, the function f is one-one and onto, f is invertible.
To find f⁻¹:
Let f(x) = y
y = 4x + 36x
6xy = 4x + 3
(6y)x = 4y + 3
x = (4y + 3)/6y
Now, f(x) = y and f is inversible.
f⁻¹(y) = x ⇒ f⁻¹(y) = (4y + 3)/6y
Thus, the function f⁻¹: A → A is given by f⁻¹(y) = (4y + 3)/6y
i.e., f⁻¹ = (4y + 3)/6y
So, f⁻¹ = (4y + 3)/6y