We know that
f(x1) = f(x2)
It can be written as
3x1 – 4 = 3x2 – 4
So we get
3x1 = 3x2
Where x1 = x2
f is one-one.
Take y = 3x – 4
It can be written as
y + 4 = 3x
So we get
x = (y + 4)/ 3
If y ∈ R there exists x = (y + 4)/ 3 ∈ R
We know that f (x) = f([y + 4]/ 3) = 3 ([y + 4]/ 3) – 4 = y
f is onto
Here, f is one-one onto and invertible.
Take y = f(x)
So we get
y = 4x – 3
It can be written as
x = (y + 4)/ 3
So f -1 (y) = (y + 4)/ 3
Hence, we define f -1: R → R: f -1(y) = (y + 4)/ 3 for all y ∈ R