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.

Question

Philoid · Accepted Answer

We have f : R⁺→ [4, ∞) and f(x) = x² + 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 x₁, x₂ϵ R⁺ (domain) such that f(x₁) = f(x₂)

⇒ x₁² + 4 = x₂² + 4

⇒ x₁² = x₂²

∴ x₁ = x₂ (x₁≠–x₂as x₁, x₂ϵ R⁺)

So, we have f(x₁) = f(x₂) ⇒ x₁ = x₂.

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

⇒ x² + 4 = y

⇒ x² = 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

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

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