It is given that f : R → R given by f (x) = 4x + 3

Let f(x) = f(y)

⇒ 4x +3 = 4y +3

⇒ 4x = 4y

⇒ x = y

⇒ f is one- one function.

Now, for y ϵ R, Let y = 4x +3

⇒ for any y ϵ R, there exists x = ϵ R

such that, f(x) =

⇒ F is onto function.

Since, f is one –one and onto

⇒ f^-1 exists.

Let us define g: R → R by g(x) =

Now, (gof)(x) = g(f(x)) = g(4x + 3) =

(fog)(y) = f(g(xy)) =

Therefore, gof = fog = I_R

Therefore, f is invertible and the inverse of f is given by

f^-1 (y) = g(y) = .