Given the function:

f(x) = sqrt(x-2) / sqrt(3-x) we need to find the domain.

Since the function if a quotient, then the denominator can not be zero.

Also, the square root must be a values greater than zero.

Then the domain is all x values such that the numerator is equal or greater than zero and the denominator is greater than zero.

Let us find the domain of the denominator:

==> sqrt(3-x) > 0

==> 3-x > 0

==> -x > -3

==> x < 3

==> x belongs to the interval ( -inf, 3)..........(1)

Now we will determine the domain for the numerator.

==> sqrt(x-2) >= 0

==> x-2 >= 0

==> x >= 2

==> x belongs to the interval [2, inf)............(1)

Then we conclude that the domain of f(x) is (1) and (2).

**==> x belongs to the interval (-inf, 3) and [2, inf) = [2,3)**

**==> x = [ 2, 3 )**

We need to find the domain of the function f(x) = sqrt(x-2)/sqrt(3-x).

The domain of a function is all values of x for which the function has a real value.

For f(x) to be real the numerator and the denominator have to be real. And the denominator cannot be equal to 0 as that would make the value of f(x) undefined.

This means x - 2 cannot be less than 0 and 3-x has to be greater than 0.

So we have x - 2 >=0 and 3-x > 0

=> x >= 2 and 3 > x

=> x >= 2 and x < 3

**Therefore the domain of the function is all values of x that lie in [ 2, 3).**