solution:

Think about f : R+ → [–5, ∞) given by f (x) = 9×2 + 6x – 5 Consider f : R+ → [4, ∞) given by f(x) = x2 + 4

Let x, y ∈ R → [-5, ∞) then, at that point f(x) = 9×2 + 6x – 5 and f(y) = 9y2 + 6y – 5

on the off chance that f(x) = f(y) 9×2 + 6x – 5 = 9y2 + 6y – 5 9(x2 – y2 ) + 6 (x – y) = 0

9{(x-y)(x+ y)} + 6 (x – y) = 0

(x – y) (9)(x+ y) + 6) = 0

either x – y = 0 or 9(x+ y) + 6 = 0

Say x – y = 0, then, at that point x = y. So f is one-one. Presently, y = f(x) = 9×2 + 6x – 5

Addressing this quadratic condition, we have

x = −6±6√????+6

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???????? ???? = √????+6−1

3

√????+6−1

√????+6−1 2 √????+6−1

Along these lines, f(x) = f(

3

) = 9(

) + 6(

3 3

) – 5

= y + 7 – 2√???? + 6 + 2 √???? + 6 – 2 – 5 = y

f(x) = y, thusly, f is onto.

f(x) is invertible and f-1(x) = √????+6−1 .

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