Q. 75.0( 1 Vote )

# Consider f : R^{+}→ [4, ∞) given by f(x) = x^{2} + 4. Show that f is invertible with f^{-1} of f given by, where R^{+} is the set of all non-negative real numbers.

Answer :

We have f : R^{+}→ [4, ∞) and f(x) = x^{2} + 4.

Recall that a function is invertible only when it is both one-one and onto.

First, we will prove that f is one-one.

Let x_{1}, x_{2}ϵ R^{+} (domain) such that f(x_{1}) = f(x_{2})

⇒ x_{1}^{2} + 4 = x_{2}^{2} + 4

⇒ x_{1}^{2} = x_{2}^{2}

∴ x_{1} = x_{2} (x_{1}≠–x_{2}as x_{1}, x_{2}ϵ R^{+})

So, we have f(x_{1}) = f(x_{2}) ⇒ x_{1} = x_{2}.

Thus, function f is one-one.

Now, we will prove that f is onto.

Let y ϵ [4, ∞) (co-domain) such that f(x) = y

⇒ x^{2} + 4 = y

⇒ x^{2} = y – 4

Clearly, for every y ϵ [4, ∞), there exists x ϵ R^{+} (domain) such that f(x) = y and hence, function f is onto.

Thus, the function f has an inverse.

We have f(x) = y ⇒ x = f^{-1}(y)

But, we found f(x) = y ⇒

Hence,

Thus, f(x) is invertible and

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