# Consider the func

We have f : R+ [–9, ∞) and f(x) = 5x2 + 6x – 9.

Recall that a function is invertible only when it is both one-one and onto.

First, we will prove that f is one-one.

Let x1, x2ϵ R+ (domain) such that f(x1) = f(x2)

5x12 + 6x1 – 9 = 5x22 + 6x2 – 9

5x12 + 6x1 = 5x22 + 6x2

5x12 – 5x22 + 6x1 – 6x2 = 0

5(x12 – x22) + 6(x1 – x2) = 0

5(x1 – x2)(x1 + x2) + 6(x1 – x2) = 0

(x1 – x2)[5(x1 + x2) + 6] = 0

x1 – x2 = 0 (as x1, x2ϵ R+)

x1 = x2

So, we have f(x1) = f(x2) x1 = x2.

Thus, function f is one-one.

Now, we will prove that f is onto.

Let y ϵ [–9, ∞) (co-domain) such that f(x) = y

5x2 + 6x – 9 = y

Adding to both sides, we get

Clearly, for every y ϵ [–9, ∞), there exists x ϵ R+ (domain) such that f(x) = y and hence, function f is onto.

Thus, the function f has an inverse.

We have f(x) = y x = f-1(y)

But, we found f(x) = y

Hence,

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