Consider f : R+ → [–9, ∞] given by f(x) = 5x^2 + 6x – 9. Prove that f is invertible with f^{-1}(y) = (√(54+5y) - 3)/5

Question

Consider f : R+ → [–9, ∞] given by f(x) = 5x² + 6x – 9. Prove that f is invertible with \(f^{-1}(y) = \) \((\frac{\sqrt{54+5y}\,-\,3}{5})\).

Answer 1

To prove f is invertible, it is sufficient to prove f is one - one onto.

Here,

f(x) = 5x² + 6x - 9

One - one : 

Let x₁,x₂ ∈ R,. then,

f(x₁) = f(x₂)

⇒ 5x₁² + 6x₁ - 9 = 5x₂² + 6x₂ - 9

⇒ 5x₁² + 6x₁  - 5x₂² - 6x₂ = 0

⇒ 5(x₁² - x₂²) + 6(x₁ - x₂) = 0 

⇒ 5(x₁ - x₂)(x₁ + x₂) +6(x₁ - x₂) = 0

⇒ (x₁ - x₂) (5x₁ + 5x₂ + 6) = 0

⇒ x₁ - x₂ = 0 [ ∵ 5x¹ + 5x² + 6 ≠ 0]

⇒ x₁ = x₂

i.e., f is one - one function.

Onto : 

Let f(x) = y

∴ y = 5x² + 6x -9 

⇒ 5x² + 6x - (9 + y) = 0 

\(⇒x = {-6 \pm \sqrt{36\,+\,4\,\times\,5(9\,+\,y)} \over 10}\)

\(⇒x = {-6 \pm \sqrt{216\,+\,20y} \over 10}\) 

\(⇒x = {\pm \sqrt{54\,+\,5y\,-\,3} \over 5}\) 

\(⇒x = \frac{\sqrt{54+5y}\,-\,3}{5}\)  [ ∵ x ∈ R₊]

Obviously, 

∀ y ∈ [-9, ∞] the value of x ∈ R₊

⇒ f is onto function.

Hence, f is one - one onto function, i.e., invertible.

Also, f is invertible with

\(f^{-1}(y) = \frac{\sqrt{54+5y}\,-\,3}{5}\)