Question

The inverse of `f(x) = sqrt(3x^2 - 4x + 5)` is

Options

• `(- oo, sqrt(11/3))`

• `(- oo, sqrt(11/5))`

• `(sqrt(11/3), oo)`

• `(sqrt(11/5), oo)`

Answer 1

`(sqrt(11/3), oo)`

Explanation:

`f(x)` is defined if `3x^2 - 4x + 5 ≥ 0`

⇒ `3[x^2 - 4/3x + 5/3] ≥ 0`

⇒ `3[(x - 2/3)^2 + 11/9] ≥ 0`

Which is true for all real x

∴ Domain of `f(x) = (- oo, oo)`

⇒ Let `y = sqrt(3x^2 - 4x + 5)`

⇒ `y^2 = 3x^2 - 4x + 5`

i.e. `3x^2 - 4x + (5 - y^2)` = 0

For x to be real, `16 - 12 (5 - y^2)  ≥ 0`

⇒ `y ≥ sqrt(11/3)`

∴ Range of `y = (sqrt(11/3, oo))`