To determine the domain of the function [tex]\( y = \sqrt[3]{x-1} \)[/tex], let's analyze the behavior of the cube root function. The cube root function, denoted as [tex]\( \sqrt[3]{x} \)[/tex], is defined for all real numbers [tex]\( x \)[/tex]. Therefore, we need to determine the value of [tex]\( x \)[/tex] for which the expression [tex]\( x - 1 \)[/tex] is defined, since this is the argument of the cube root.
The function inside the cube root, [tex]\( x - 1 \)[/tex], is a linear expression and is defined for all real numbers.
Because the cube root function [tex]\( \sqrt[3]{x} \)[/tex] is defined for all real numbers [tex]\( x \)[/tex], the same applies to [tex]\( \sqrt[3]{x - 1} \)[/tex].
Thus, there is no restriction on the values that [tex]\( x \)[/tex] can take. This means that the domain of the function [tex]\( y = \sqrt[3]{x - 1} \)[/tex] is all real numbers.
In interval notation, the domain is given by:
[tex]\[ (-\infty, \infty) \][/tex]
Therefore, the correct answer is:
[tex]\[ -\infty < x < \infty \][/tex]
