The function $f(x) = \sqrt{x}$ involves a square root, so we need to determine the values of $x$ for which the expression inside the square root is defined and real.

Analysis:

1. For $f(x) = \sqrt{x}$, the expression inside the square root, $x$, must be non-negative because the square root of a negative number is not a real number.

2. Therefore, the condition is: $x \geq 0$

Natural Domain:

The natural domain of $f(x) = \sqrt{x}$ is: $[0, \infty)$

This means $x$ can take any real number value starting from 0 and extending to infinity.

Would you like further details or have any specific questions?

Related Questions:

1. What is the domain for $f(x) = \sqrt{x - 5}$?
2. How does the domain change if we have $f(x) = \sqrt{x^2}$?
3. What is the domain and range of $f(x) = \sqrt{x + 1} + 2$?
4. How do we find the domain of a square root function with a linear expression inside, like $f(x) = \sqrt{2x - 3}$?
5. What is the natural domain of a cube root function, such as $f(x) = \sqrt[3]{x}$?

Tip:

For any function involving a square root, the expression inside must be non-negative to ensure the output is a real number.