Answer :

It is given that f: R+ → [–5, ∞) given by f(x) = 9x^{2} + 6x – 5.

Let y be any element of [-5, ∞)

Now, let y = 9x^{2} + 6x – 5

⇒ y = (3x+1)^{2} -1-5 = (3x+1)2 – 6

⇒ 3x + 1 =

⇒ f is onto and it’s range is f = [-5, ∞)

Now, Let us define g: [-5, ∞) → R_{+} as g(y) =

Now, we have:

(gof)(x) = g(f(x)) = g(9x^{2} + 6x – 5)

= g((3x+1)^{2} – 6)

And, (fog)(y) = f(g(y)) =

Thus, gof = I_{R} anf fog = I(-5, ∞)

Therefore, f is invertible and the inverse of f is given by

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