Domain of Real-Valued Functions

The only restrictions on the domain that we have run across, come from taking an even root (square root in most cases) and from dividing by zero.

Note, that it is okay to have zero in a numerator since zero divided by anything is zero. For example if you have the function f(x)=(x+3)/(x+4) the domain would be (-∞,-4)U(-4,∞). -4 makes the denominator zero so we want to skip over it. Notice that we did not skip -3, since -3 only makes the numerator zero and it is okay to have zero on top.

For a real valued function, you cannot take the square root of a negative number, so the argument of the square root must be greater than or equal to zero. For example: if I give you the real valued function f(x)=sqrt(x+1), the domain would be found by setting the argument greater than or equal to zero and solving for x: x+1≥0 leads to x≥-1. Written in interval notation this gives us: [-1,∞)

Now what happens if you have a square root in the denominator: f(x)=1/sqrt(x+1). Because it is a sqrt, you would set the x+1≥0, but since it is in the denominator we don't want the zero, so instead of x+1≥0, we have x+1>0 which leads to a solution of (-1,inf).