f : R → R given by f(x) = 4x + 3.
f(x1) = 4x1 + 3, f(x2) = 4x2 + 3
If f(x1) = f(x2),
then 4x1 + 3 = 4x2 + 3
or 4x1 = 4x2
⇒ x1 = x2.
⇒ f is one-one.
Aslo, let y = 4x + 3
or 4x = y – 3
∴ x = \(\frac { y-3 }{ 4 }\).
For each value of y e R and belonging to codomain of y, there is a pre-image in its domain.
∴f is onto.
i. e., f is one-one and onto.
∴ f is invertible.
So, f-1(y) = g(y) = \(\frac { y-3 }{ 4 }\)
