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

Apr 18, 2015

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 $\frac{12}{x}$ is defined, and that the square root is defined as well.

The first request is quite easy, since $\frac{12}{x}$ is a non-zero-denominator fraction if and only if $x \setminus \ne 0$

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

Now, if $x$ is positive, $\frac{12}{x}$ is positive as well, and so $\frac{12}{x} + 9$ will be positive because it's the sum of two positive numbers.

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

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

$D = \left\{x \setminus \in \setminus m a t h \boldsymbol{R} : \setminus x \setminus \le - \frac{4}{3}\right\} \setminus \cup \left\{x \setminus \in \setminus m a t h \boldsymbol{R} : \setminus x > 0\right\} = \left(- \infty , - \frac{4}{3}\right] \setminus \cup \left(0 , \setminus \infty\right)$