**We know that**

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

It can be written as

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

On further calculation

2x_{1} = 2x_{2}

So we get

x_{1} = x_{2}

**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