We know that
f(x1) = f(x2)
It can be written as
2x1 + 3 = 2x2 + 3
On further calculation
2x1 = 2x2
So we get
x1 = x2
Hence, f is one-one.
Consider y = 2x + 3
It can be written as
y – 3 = 2x
So we get
x = (y – 3)/ 2
If y ∈ R, there exists x = (y – 3)/ 2 ∈ R
f (x) = f ([y-3]/2) = 2([y – 3]/ 2) +3 = y
f is onto
Here, f is one-one onto and invertible.
Take y = f(x)
It can be written as
y = 2x + 3
So we get
x = (y-3)/ 2
So f -1 (y) = (y – 3)/ 2
Hence, we define f -1: R → R: f -1(y) = (y – 3)/ 2 for all y ∈ R