Given the equation: f(x) =1/sqrt(x-3)

We need to find the domain of f(x).

We know that the domain is all x values such that f(x) is defined.

Since f(x) is a quotient, then the denominator can not be zero.

Also, we notice that the denominator is a square root.

Then (x-3) must be positive values.

==> sqrt(x-3) > 0

==> x-3 > 0

==> x > 3

**Then the domain is x = ( 3, inf)**

f(x) = 1/sqrt(x-3). To find the domain.

The domain of the function f(x) = 1/sqrt(x-3) is the set of all values of x for which the function is defined and is real.

If x is less than 3, then f(x) can not be real. Therefore the x must be greater than or equal to 3 for which f(x) is defined.

**Therefore the domain of the function f(x) = 1/sqrt(x-3) is the set {x : x> = 3}. **

**Or the domain of the function is the interval (3, infinity**).