To Show: that f is invertible

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) = \(\frac{(3x+1)}{2}\).So f(x₁) = f(x₂) → \(\frac{(3x_1+1)}{2}\)= \(\frac{(3x_2+1)}{2}\) → x₁=x₂

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

Given co-domain of f(x) is R.

Let y = f(x) = \(\frac{(3x+1)}{2}\), So x = \(\frac{2y-1}{3}\)[Range of f(x) = Domain of y]

So Domain of y is R = Range of f(x)

Hence, Range of f(x) = co-domain of f(x) = R

So, f(x) is onto function

As it is bijective function. So it is invertible

Invers of f(x) is f^-1(y) = \(\frac{2y-1}{3}\)