To determine the domain of the function [tex]\( y = \sqrt[3]{x-1} \)[/tex], we need to establish the set of [tex]\( x \)[/tex]-values for which the function is defined. Let’s carefully analyze the function:
1. The function under consideration is [tex]\( y = \sqrt[3]{x-1} \)[/tex]. This is a cube root function.
2. Cube roots are unique among roots because they are defined for all real numbers. This means that we can take the cube root of any real number, whether it's positive, negative, or zero.
3. The expression inside the cube root is [tex]\( x - 1 \)[/tex]. We need to consider if there are any restrictions on [tex]\( x \)[/tex].
- For cube roots, unlike square roots, there is no restriction on the expression inside. The cube root of any real number is always a real number. Thus, [tex]\( x - 1 \)[/tex] can be any real number.
4. Since there are no restrictions on [tex]\( x \)[/tex], it follows that any real number can be plugged into the function [tex]\( y = \sqrt[3]{x-1} \)[/tex].
Thus, the domain of the function [tex]\( y = \sqrt[3]{x-1} \)[/tex] is all real numbers, which can be written in interval notation as [tex]\( (-\infty, \infty) \)[/tex].
Therefore, the correct answer is:
[tex]\[ -\infty < x < \infty \][/tex]
