To Show: that f is one-one and onto

To Find: Inverse of f

[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function f : A → B is said to be a one-one function or injective mapping if different elements of A have different images in B. Thus for x₁, x₂ ∈ A & f(x₁), f(x₂) ∈ B, f(x₁) = f(x₂) ↔ x₁= x₂ or x₁ ≠ x₂ ↔ f(x₁) ≠ f(x₂)

onto function: If range = co-domain then f(x) is onto functions.

So, We need to prove that the given function is one-one and onto.

Let x₁, x₂ ∈ Q and f(x) = \(\frac{(4x+3)}{(6x-4)}\).So f(x₁) = f(x₂) → \(\frac{(4x_1+3)}{(6x_1-4)}=\frac{(4x_2+3)}{(6x_2-4)}\) → = on solving we get x₁=x₂

So f(x1) = f(x2) ↔ x1= x2, f(x) is one-one

Given co-domain of f(x) is R except 3x-2=0.

Let y = f(x) = \(\frac{(4x+3)}{(6x-4)}\) So x = \(\frac{4x+3}{6x-4}\) [Range of f(x) = Domain of y]

So Domain of y is R (except 3x-2=0) = Range of f(x)

Hence, Range of f(x) = co-domain of f(x) = R except 3x-2=0

So, f(x) is onto function

As it is bijective function. So it is invertible

Invers of f(x) is f^-1(y) = \(\frac{4x+3}{6x-4}\)