Consider f : R+ → [–9, ∞] given by f(x) = 5x2 + 6x – 9. Prove

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

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»

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

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