How do i work this function?

i'm supposed to find the domain, f(x) = 1/ square root of x-3How do i work this function?
If x = 3, x - 3 becomes zero and 1/0 is not defined. Therefore x = 3 is not included in the domain of the function.


again if x is less than 3, x - 3 becomes negative and sqrt(x-3) does not have real values as the sqrt of a negative quantity is imaginary and not real. Therefore x cannot have values less than three.





Now x cannot be equal to 3 and cannot have values less than three.


therefore x should be greater than 3


The domain therefore, is [3 to +infinity)


Note that the solution set is open on the left. It means the set does not include 3.How do i work this function?
I'm guessing that this is a 'pre-calculus' or lower division course and the domain of f is a subset of the reals.





Find all of the points in the real numbers which, when plugged in for x, yield a defined answer.





HINT #1: If the denominator of a fraction is 0, the value of the fraction is undefined.





HINT #2: Negative numbers have no square roots in the real numbers.





Have fun.





Doug
The domain is just all the x values that you could plug into x and still be okay, for instance you could not plug in any x values that are less than or equal to three because you would either get a zero in the denominator (x=3) which is a no no or a negative square root which produce imaginary numbers so the domain must be x%26gt;3.
The domain is all values that keep the denominator positive. x%26gt;3 is the domain. x=3 is not part of the domain because the denominator can't be zero. x%26lt;3 is not part of the domain because you can't take the square root of a negative number.
If you plug in a value and f(x) is not undefined, the value is on the domain. The only time this equation is undefined is when x = 3. So the domain is x=/=3.