The function $g(x) = \sqrt{1 - x^2}$ involves a square root, which means we need to ensure that the expression inside the square root is non-negative. This constraint is essential because the square root of a negative number is not defined in the real number system.

To find the domain, let's consider the condition:

$1 - x^2 \geq 0$

Step 1: Solve the inequality

1. Rearrange the inequality to: $x^2 \leq 1$

2. Take the square root of both sides: $-1 \leq x \leq 1$

Step 2: Determine the domain

The domain of $g(x)$ is all values of $x$ that satisfy $-1 \leq x \leq 1$. This ensures that $1 - x^2$ remains non-negative, allowing the square root to be defined.

Conclusion

The domain of $g(x) = \sqrt{1 - x^2}$ is:

$[-1, 1]$

Would you like further details or have any questions?

Here are five related questions for further exploration:

1. What is the range of $g(x) = \sqrt{1 - x^2}$?
2. How would the domain change if the function were $h(x) = \sqrt{x^2 - 1}$?
3. How can we find the domain of functions involving higher roots, like $f(x) = \sqrt[3]{1 - x^2}$?
4. What would the graph of $g(x) = \sqrt{1 - x^2}$ look like on a coordinate plane?
5. How do we find the domain of a function involving both square roots and fractions?

Tip: For functions with square roots, always ensure the expression under the square root is non-negative to determine the domain.