**We know that**

f(x_{1}) = f(x_{2})

It can be written as

½ (3x_{1} + 1) = ½ (3x_{2} + 1)

So we get

3x_{1} + 1 = 3x_{2} + 1

**On further calculation**

3x_{1} = 3x_{2} where x_{1} = x_{2}

f is one-one.

Take y = ½ (3x + 1)

**It can be written as**

2y = 3x + 1

We get

2y – 1 = 3x

So x = (2y – 1)/3

If y ∈ R there exists x = (2y – 1)/3 ∈ R

We know that

f(x) = f([2y – 1]/ 3) = ½ (3([2y – 1]/ 3) + 1) = y

**f is onto**

**Here, f is one-one and invertible.**

Take y = f(x)

So y = (3x + 1)/2

It can be written as

x = (2y – 1)/ 3

**Hence, we define f **^{-1}: R → R: f ^{-1}(y) = (2y – 1)/ 3 for all y ∈ R