# Find the domain of the function:  f(x) = sqrt(x-2)/sqrt(3-x) 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...

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).

Approved by eNotes Editorial Team 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 )

Approved by eNotes Editorial Team