Question

Consider f: R₊ → [−5, ∞) given by f(x) = 9x² + 6x − 5. Show that f is invertible with `f^(-1)(y) = ((sqrt(y +6) - 1)/3)`

Answer 1

f: R₊ → [−5, ∞) is given as f(x) = 9x² + 6x − 5.

Let y be an arbitrary element of [−5, ∞).

Let y = 9x² + 6x − 5.

`=> y = (3x + 1)^2 - 1  - 5 = (3x + 1)^2 - 6`

`=> (3x + 1)^2 = y + 6`

`=> 3x + 1 = sqrt(y+6)`          [as `y >= 5 => y + 6 > 0`]

`=> x = (sqrt(y + 6) - 1)/3`

∴f is onto, thereby range f = [−5, ∞).

Let us define  g: [−5, ∞) → R₊ as g(y) = `(sqrt(y+6) - 1)/3`

We now have:

`(gof) (x) = g(f(x)) = g(9x^2 + 6x - 5)`

`= g((3x + 1)^2 - 6)`

`= sqrt((3x + 1)^2 - 6 + 6 - 1)/3`

`= (3x + 1 - 1)/3 = x`

And `(fog)(y) = f(g(y)) = f((sqrt(y + 6)-1)/3)`

`= [3((sqrt(y + 6) - 1)/3)+1]^2 - 6`

`= (sqrt(y + 6))^2 - 6 = y + 6 - 6 = y`

`:. gof = I_R and fog = I_((-5, oo)`

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

`f^(-1)  (y) = g(y)= (sqrt(y + 6) - 1)/3`