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Let R+ be the set of all positive real numbers. show that the function f : R+→ [-5, ∞]: f(x) = (9x2 + 6x – 5) is invertible. Find f-1.

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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) = \(\frac{-1\pm\sqrt{y+6}}{3}\)

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