To determine the domain of the function [tex]\( y = \sqrt[3]{x-1} \)[/tex], we need to understand the conditions under which the function is defined.
The function involves a cube root, [tex]\( \sqrt[3]{x-1} \)[/tex]. One important property of the cube root is that it is defined for all real numbers, unlike the square root which is only defined for non-negative numbers.
In this function, we are taking the cube root of [tex]\( x-1 \)[/tex], so we need to consider when [tex]\( x-1 \)[/tex] is a real number. Since the cube root is defined for every real number, it means that [tex]\( x-1 \)[/tex] can be any real number.
To further clarify, any real number subtracted by 1 will itself also be a real number. This means that there are no restrictions on [tex]\( x \)[/tex]; it can be any real number.
Thus, the domain of the function [tex]\( y = \sqrt[3]{x-1} \)[/tex] is the set of all real numbers, which is mathematically denoted as:
[tex]\[ (-\infty, \infty) \][/tex]
This means that for every real number [tex]\( x \)[/tex], the expression [tex]\( \sqrt[3]{x-1} \)[/tex] is well-defined and will yield a real number result. Hence, the domain of [tex]\( y = \sqrt[3]{x-1} \)[/tex] includes all real numbers.
