Q. 9

# Consider f: R+ <s

It is given that f: R+ [5, ) given by f(x) = 9x2 + 6x 5.

Let y be any element of [-5, ∞)

Now, let y = 9x2 + 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(9x2 + 6x – 5)

= g((3x+1)2 – 6)

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

Thus, gof = IR anf fog = I(-5, ∞)

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

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