Show that the function f in defined as is one-one and onto. Hence find f^-1.

Given:

For one – one

Let f(x₁) = f(x₂)

⇒ (4x₁ + 3)(6x₂ – 4) = (4x₂ + 3)(6x₁ – 4)

⇒ 24x₁x₂ – 16x₁ + 18x₂ – 12 = 24x₁x₂ – 16x₂ + 18x₁ – 12

⇒ -16x₁ + 16x₂ + 18x₂ – 18x₁ = 0

⇒ -34x₁ + 34x₂ = 0

⇒ -34(x₁ – x₂) = 0

⇒ x₁ – x₂ = 0

⇒ x₁ = x₂

∴ f(x) is one – one

For onto

Since, is a real number, therefore, for every y in the co – domain of f there exists a number x in R - such that

∴, f(x) is onto

Hence, f^-1 exists.

Now, let

⇒ y(6x – 4) = 4x + 3

⇒ 6xy – 4y = 4x + 3

⇒ 6xy – 4x = 4y + 3

⇒ x(6y – 4) = 4y + 3

[Interchanging the variables x and y]

Put y = f^-1(x)