# How do you find the domain of sqrt((x/(x-2)))?

Oct 13, 2015

$x \in \left(- \infty , 0\right] \cup \left(2 , + \infty\right)$

#### Explanation:

The first thing to look out for is any value of $x$ that will maek the denominator of the fraction equal to zero. That happens when

$x - 2 = 0 \implies x = 2$

so this value of $x$ will be excluded from the domain of the function.

The second thing to look out for is the fact that you're dealing with a square root, which, for real numbers, can only be taken of positive numbers.

This means that the fraction $\frac{x}{x - 2}$ must be greater than or equal to zero.

$\frac{x}{x - 2} \ge 0$

This condition is satisfied for

$x \le 0 \implies \left\{\begin{matrix}x \le 0 \\ x - 2 < 0\end{matrix}\right. \implies \frac{x}{x - 2} \ge 0$

and

$x > 2 \implies \left\{\begin{matrix}x > 2 \\ x - 2 > 0\end{matrix}\right. \implies \frac{x}{x - 2} \ge 0$

Therefore, the domain of the function will include any value of $x$ that is smaller than or equal to $0$ or greter than $2$. In interval notation, this is equivalent to

$x \in \left(- \infty , 0\right] \cup \left(2 , + \infty\right)$

graph{sqrt(x/(x-2)) [-10, 10, -5, 5]}