How do you find the domain of this function `y = sqrt( (12/x) + 9)`?

The domain of a function is the set of values for which the function is evaluable. To guarantee that, we must make sure that:

1. No fraction has a zero denominator
2. No square root has a (strictly) negative argument
3. No logarithm has a negative (or zero) argument.

So, in your case, you must make sure that the fraction `12/x` is defined, and that the square root is defined as well.

The first request is quite easy, since `12/x` is a non-zero-denominator fraction if and only if `x\ne 0`

As for the whole root, we must make sure that `12/x+9` is greater to zero, or at most exactly zero. In fact, if we would choose a value of `x` such that `12/x+9<0`, we would be calculating the square root of a negative number, which is impossible to do using real numbers.

Now, if `x` is positive, `12/x` is positive as well, and so `12/x+9` will be positive because it's the sum of two positive numbers.

If `x` is negative, we must solve `12/x+9 \ge 0`. Subtracting `9` from both sides, we get `12/x\ge -9`. Since `x\ne 0`, we can multiply both terms by `x`, but since `x` is negative we must invert the inequality, obtaining `12\le -9x`; and again dividing by `-9` both terms, we have `12/{-9}\ge x`, which means `x \le -4/3`.

So, every positive number is ok, and amongst the negative ones we can accept only those which are lesser or equal to `-4/3`. This means that the domain of the function is

`D={ x \in \mathbb{R}:\ x\le -4/3}\cup { x \in \mathbb{R}:\ x> 0} = (-infty, -4/3] \cup (0,\infty)`