We know that
f(x1) = f(x2)
It can be written as
½ (3x1 + 1) = ½ (3x2 + 1)
So we get
3x1 + 1 = 3x2 + 1
On further calculation
3x1 = 3x2 where x1 = x2
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