No, the domain of the function $f(x) = \sqrt[3]{x}$ is not restricted to $[0, \infty)$. This statement is false.

The cube root function $\sqrt[3]{x}$ is defined for all real numbers because any real number, whether positive, negative, or zero, has a real cube root. Therefore, the correct domain of $f(x) = \sqrt[3]{x}$ is:

$\text{Domain} = (-\infty, \infty)$

Unlike square roots, cube roots can handle negative inputs because the cube of a negative number is also negative. For example, $\sqrt[3]{-8} = -2$.

Would you like more details or have any other questions?

Here are 5 related questions:

1. What is the domain of $f(x) = \sqrt[4]{x}$?
2. What is the range of $f(x) = \sqrt[3]{x}$?
3. How do cube root and square root functions differ in terms of their domains?
4. Can the cube root function be expressed as a power function?
5. What is the inverse function of $f(x) = \sqrt[3]{x}$?

Tip: When determining the domain of a function, check whether there are any restrictions on the input values, such as square roots or logarithms, which only allow non-negative values.