91Ó°ÊÓ

Analyze the expression under the square root: \( \frac{x}{x^{2}-9} \). In a square root, the value inside, referred to as the radicand, cannot be negative because the square root of a negative number does not fall within the real number system. Therefore, we get that \( \frac{x}{x^{2}-9} \geq 0 \). Separate the cases to consider \( x > 0 \) and \( x < 0 \). Since a multiplication of a number with the same sign results in a positive nnumber, it means that when x is positive, the denominator \( x^{2} - 9 \) must also be positive. Therefore \( x^{2} - 9 > 0 \) which simplifies to \( x < -3 \) or \( x > 3 \). Similarly, when x is negative, the denominator \( x^{2} - 9 \) must be negative for the result to be positive. Therefore \( x^{2} - 9 < 0 \) which simplifies to \( -3 < x < 3 \). Combining these two we get \( x < -3 \) or \( -3 < x < 3 \) or \( x > 3 \). The next step is to look at the restrictions from the denominator.

Next, we need to check when the denominator \( x^{2}-9 \) equals zero because the function is undefined at these values. Solving \( x^{2}-9 = 0 \) gives \( x = -3 \) and \( x = 3 \). So, these points have to be excluded from the domain.

Now we combine the results from the radicand and the denominator. We exclude \( x = -3 \) and \( x = 3 \) from our solution in Step 1. This leaves us with the final domain of \( x < -3 \) or \( -3 < x < 3 \) or \( x > 3 \). In interval notation, the domain is \( (-\infty, -3) \cup (-3, 3) \cup (3, \infty) \).

Enter the function \( \sqrt{\frac{x}{x^{2}-9}} \) into a graphing utility and observe the values of x at which the function exists. This should match with the domain obtained earlier. If it does, then your domain is correct.