To find the domain of the function $y = \frac{\sqrt{x^2 - 25}}{3x}$, we need to analyze both the square root and the denominator.

Step 1: Analyzing the square root

The expression inside the square root $x^2 - 25$ must be non-negative for the function to be defined (since the square root of a negative number is not real). Therefore, we solve the inequality: $x^2 - 25 \geq 0$ $x^2 \geq 25$ $x \leq -5 \quad \text{or} \quad x \geq 5$ Thus, $x \in (-\infty, -5] \cup [5, \infty)$ for the square root to be defined.

Step 2: Analyzing the denominator

The denominator is $3x$, which means $x \neq 0$ to avoid division by zero.

Step 3: Combining the conditions

The domain of the function is the intersection of the two conditions:

• $x \in (-\infty, -5] \cup [5, \infty)$ from the square root.
• $x \neq 0$ from the denominator.

Thus, the domain is: $x \in (-\infty, -5] \cup [5, \infty)$

Set-builder notation:

$\{x \,|\, x \leq -5 \text{ or } x \geq 5\}$

Would you like more details or have any questions?

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Here are 5 related questions for practice:

1. What is the domain of $y = \frac{\sqrt{x^2 - 36}}{4x}$?
2. How do you solve inequalities involving square roots and quadratic expressions?
3. What is the domain of $y = \frac{1}{\sqrt{x^2 - 25}}$?
4. How would the domain change if $y = \frac{\sqrt{25 - x^2}}{x}$?
5. What is the range of the function $y = \frac{\sqrt{x^2 - 25}}{3x}$?

Tip: Always check for both square roots and denominators to find the domain of a function.