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 ≠ 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) = \(\frac{-1\pm\sqrt{y+6}}{3}\)