To Show: that f is invertible

To Find: Inverse of f

[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 x₂f(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) (9² + 6 – 5) = (9² + 6 – 5) on solving we get x₁=x₂

So f(x₁) = f(x₂) x₁= x_2, f(x) is one-one

Given co-domain of f(x) is [-5, ∞]

Let y = f(x) = (9x² + 6x – 5), So x = [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) =.