The function `f(x) = 1/(x - 6)` and `g(x) = sqrt(x - 1)`

`f(g(x)) = f(sqrt(x - 1)) = 1/(sqrt(x - 1) - 6)`

The domain of `f(g(x))` is the set of real numbers that x can take on for which `f(g(x))` is real and defined.

As the square root...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

The function `f(x) = 1/(x - 6)` and `g(x) = sqrt(x - 1)`

`f(g(x)) = f(sqrt(x - 1)) = 1/(sqrt(x - 1) - 6)`

The domain of `f(g(x))` is the set of real numbers that x can take on for which `f(g(x))` is real and defined.

As the square root of a negative number is not real `x - 1 >= 0 => x >= 1`

Also, `sqrt(x - 1) - 6 != 0`

=> `sqrt(x - 1) != 6`

=> `x - 1 != 36`

=> `x != 37`

**The domain is **`[1, oo) - {37}`