Answer :

It is given that f : R_{+}→ [4, ∞) given by f (x) = x^{2} + 4.

Now, Let f(x) = f(y)

⇒ x^{2} + 4 = y^{2} + 4

⇒ x^{2} = y^{2}

⇒ x = y

⇒ f is one-one function.

Now, for y ϵ [4, ∞), let y = x^{2} + 4.

⇒ x^{2} = y -4 ≥ 0

⇒ for any y ϵ R, there exists x = ϵ R such that

= y -4 + 4 = y.

⇒ f is onto function.

Therefore, f is one–one and onto function, so f-1 exists.

Now, let us define g: [4, ∞) → R+ by,

g(y) =

Now, gof(x) = g(f(x)) = g(x^{2} + 4) =

And, fog(y) = f(g(y)) = =

Therefore, gof = gof = I_{R}.

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

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

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