To find the domain of the function \( p(x) = \sqrt{12 - \frac{x^2}{4}} \), we need to ensure that the expression inside the square root is non-negative, as the square root of a negative number is not defined in the set of real numbers.

The expression inside the square root is:
\[
12 - \frac{x^2}{4}
\]

We need to solve the inequality:
\[
12 - \frac{x^2}{4} \geq 0
\]

1. First, isolate \( \frac{x^2}{4} \):
   \[
   12 \geq \frac{x^2}{4}
   \]

2. Multiply both sides by 4 to eliminate the fraction:
   \[
   48 \geq x^2
   \]

3. This inequality can be rewritten as:
   \[
   x^2 \leq 48
   \]

4. Taking the square root of both sides (and considering both positive and negative roots):
   \[
   -\sqrt{48} \leq x \leq \sqrt{48}
   \]

Since \( \sqrt{48} = 4\sqrt{3} \), the domain of \( p(x) \) is:
\[
x \in [-4\sqrt{3}, 4\sqrt{3}]
\]

So, the domain of \( p(x) \) is \( [-4\sqrt{3}, 4\sqrt{3}] \).

Would you like more details or have any questions?

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**Relative Questions:**
1. How do you simplify the expression \( \sqrt{48} \)?
2. Can you explain the properties of square roots used in solving this inequality?
3. What would the domain be if the square root function was not present?
4. How would the domain change if the expression inside the square root was \( 12 + \frac{x^2}{4} \)?
5. How would the domain change if \( p(x) = \sqrt{\frac{x^2}{4} - 12} \)?
6. What is the range of \( p(x) \) for this domain?
7. How would the domain change if the square root was squared, i.e., \( p(x) = \left(\sqrt{12 - \frac{x^2}{4}}\right)^2 \)?
8. What happens to the domain if you subtract a constant, say 3, from \( p(x) \)?

**Tip:** Always ensure that expressions under a square root are non-negative when determining the domain of a function.

Learn how to find the domain of the function p(x) = √(12 - x^2/4) using inequalities and square root properties. Understand why the expression inside the square root must be non-negative and how to solve for the domain interval. Suitable for advanced high school students.

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