[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function f : A → B is said to be a one-one function or injective mapping if different elements of A have different images in B. Thus for x₁, x₂ ∈ A & f(x₁), f(x₂) ∈ B, f(x₁) = f(x₂) ↔ x₁= x₂ or x₁ ≠ x₂ ↔ f(x₁) ≠ f(x₂)

onto function: If range = co-domain then f(x) is onto functions.

So, We need to prove that the given function is one-one and onto.

Let x1, x2 ∈ R and f(x) = (9x² + 6x – 5).So f(x₁) = f(x₂) (9x₁² + 6x₁ – 5) = (9x₂² + 6x₂ – 5) on solving we get x₁=x₂

So f(x₁) = f(x₂) x₁= x₂, f(x) is one-one

Given co-domain of f(x) is [-5, ∞]

Let y = f(x) = (9x² + 6x – 5), So x = \(\frac{-1\pm\sqrt{y+6}}{3}\)[Range of f(x) = Domain of y]

So Domain of y = Range of f(x) = [-5, ∞]

Hence, Range of f(x) = co-domain of f(x) =[-5, ∞]

So, f(x) is onto function

As it is bijective function. So it is invertible

Invers of f(x) is f^-1(y) = \(\frac{-1\pm\sqrt{y+6}}{3}\)