Consider f : R₊→ [4, ∞) given by f(x) = x² + 4. Show that f is invertible with the inverse f^–1 of given by where R₊ is the set of all non-negative real numbers.

A function is said to be invertible if it is one-one onto.

To prove that f(x) is invertible we need to prove that it is one-one onto.

As f(x) = x² + 4

And domain is all non-negative real numbers.

∴ No two values of x is going to give same results as x is a non-negative.

So for every unique value of x there is only one unique value of f(x)

Hence, we say that f(x) is one – one.

f(x) = x² + 4 ≥ 4

range of f(x) = [4, ∞)

As, f : R⁺→ [4, ∞) [given mapping of f]

∴ Range = [4,∞) = co-domain

∴ f(x) is onto

As, f(x) is one-one and onto.

∴ f(x) is invertible.

Now, we need to find the inverse of the function f(x)

Let, y = x² + 4

⇒ x² = y – 4

⇒ x = ± √(y-4)

As domain of f is [0,∞) {given}

∴ x > 0

∴ x = √(y-4)

Or we can write: f^-1(y) = √(y-4)