Consider f : R+ → {4, ∞) given by f (x) = x2 + 4. Show that f is i

1.	Consider f : R+ → {4, ∞) given by f (x) = x2 + 4. Show that f is invertible with f –1 = \(\sqrt{y-4}\) where R+ is the set of all non-negative real numbers.
Answer» For the function f to be invertible, f has to be one-one onto. One-One Let x₁, x₂ ∈ R⁺ such that f (x₁) = f (x₂) ⇒ x₁² + 4 = x₂² + 4 ⇒ x₁² = x₂² ⇒ (x₁ – x₂) (x₁ + x₂) = 0 ⇒ x₁ – x₂= 0 ⇒ x₁ = x₂ (∵ x₁ + x₂ ≠ 0 as x₁, x₂ ∈ R⁺) ⇒ f is one-one Onto Let y= f (x) = x² + 4 ⇒ x² = y – 4 ⇒ x = ±\(\sqrt{y-4}\) ⇒ x = f^–1(y) = \(\sqrt{y-4}\) ⇒ f^–1(x) = \(\sqrt{x-4}\) (∵ x ∈ R + so we ignore–ve value) For every element y ∈ [4, ∞], there exists a pre image f^–1(x) ∈ R⁺. So f is onto. Hence f being one-one onto is invertible.

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

Answer»

For the function f to be invertible, f has to be one-one onto.

One-One

Let x₁, x₂ ∈ R⁺ such that f (x₁) = f (x₂)

⇒ x₁² + 4 = x₂² + 4

⇒ x₁² = x₂²

⇒ (x₁ – x₂) (x₁ + x₂) = 0

⇒ x₁ – x₂= 0

⇒ x₁ = x₂ (∵ x₁ + x₂ ≠ 0 as x₁, x₂ ∈ R⁺)

⇒ f is one-one

Onto

Let y= f (x) = x² + 4 ⇒ x² = y – 4

⇒ x = ±\(\sqrt{y-4}\)

⇒ x = f^–1(y) = \(\sqrt{y-4}\)

⇒ f^–1(x) = \(\sqrt{x-4}\) (∵ x ∈ R + so we ignore–ve value)

For every element y ∈ [4, ∞], there exists a pre image f^–1(x) ∈ R⁺.

So f is onto.

Hence f being one-one onto is invertible.