To find the domain of the function $p(x) = \sqrt{4 - x^2}$, we need to ensure that the expression inside the square root, $4 - x^2$, is non-negative. This is because the square root of a negative number is undefined in the real number system.

Step-by-Step Solution

1. Set up the inequality for the expression inside the square root: $4 - x^2 \geq 0$

2. Rearrange the inequality: $x^2 \leq 4$

3. Take the square root of both sides, remembering to consider both the positive and negative roots: $-2 \leq x \leq 2$

Conclusion

The domain of $p(x) = \sqrt{4 - x^2}$ is all $x$ values between -2 and 2, inclusive. In interval notation, this is:

$[-2, 2]$

Would you like a detailed explanation of the steps or have any further questions?

Here are five related questions:

1. How would the domain change if the function were $p(x) = \sqrt{9 - x^2}$?
2. What would the domain be if the function were $p(x) = \sqrt{x^2 - 4}$?
3. How do you determine the range of a function like $p(x) = \sqrt{4 - x^2}$?
4. Why does the square root function restrict the domain to non-negative values?
5. How do we handle domains for functions with even powers inside the square root?

Tip: For square root functions, always check that the expression inside the root is greater than or equal to zero to ensure real values.