Answer :
To determine the domain of the function [tex]\(y = \sqrt[3]{x-1}\)[/tex], we need to analyze where the expression under the cube root is defined and produces real numbers.
The function given is [tex]\(y = \sqrt[3]{x-1}\)[/tex]. Let's break it down step by step:
1. Identify the Core Operation: The function involves a cube root, specifically [tex]\(\sqrt[3]{x-1}\)[/tex].
2. Characteristics of Cube Root Function: The cube root function, [tex]\(\sqrt[3]{u}\)[/tex], is defined for all real numbers [tex]\(u\)[/tex]. This means that there are no restrictions on [tex]\(u\)[/tex] because the cube root of any real number [tex]\(u\)[/tex] (positive, negative, or zero) is also a real number.
3. Translate to the Given Function: For the function [tex]\(y = \sqrt[3]{x-1}\)[/tex], we substitute [tex]\(u = x-1\)[/tex]. Given that the cube root function does not impose any restrictions, [tex]\(u = x-1\)[/tex] can be any real number.
4. Solving for [tex]\(x\)[/tex]: Since [tex]\(x-1\)[/tex] can be any real number, solving for [tex]\(x\)[/tex] yields:
[tex]\[ x-1 \in \mathbb{R} \][/tex]
where [tex]\(\mathbb{R}\)[/tex] denotes the set of all real numbers. Adding 1 to both sides, we get:
[tex]\[ x \in \mathbb{R} \][/tex]
Thus, the domain of the function [tex]\(y = \sqrt[3]{x-1}\)[/tex] is all real numbers. In interval notation, this is expressed as:
[tex]\[ (-\infty, \infty) \][/tex]
Therefore, the correct answer is:
[tex]\[ -\infty < x < \infty \][/tex]
The function given is [tex]\(y = \sqrt[3]{x-1}\)[/tex]. Let's break it down step by step:
1. Identify the Core Operation: The function involves a cube root, specifically [tex]\(\sqrt[3]{x-1}\)[/tex].
2. Characteristics of Cube Root Function: The cube root function, [tex]\(\sqrt[3]{u}\)[/tex], is defined for all real numbers [tex]\(u\)[/tex]. This means that there are no restrictions on [tex]\(u\)[/tex] because the cube root of any real number [tex]\(u\)[/tex] (positive, negative, or zero) is also a real number.
3. Translate to the Given Function: For the function [tex]\(y = \sqrt[3]{x-1}\)[/tex], we substitute [tex]\(u = x-1\)[/tex]. Given that the cube root function does not impose any restrictions, [tex]\(u = x-1\)[/tex] can be any real number.
4. Solving for [tex]\(x\)[/tex]: Since [tex]\(x-1\)[/tex] can be any real number, solving for [tex]\(x\)[/tex] yields:
[tex]\[ x-1 \in \mathbb{R} \][/tex]
where [tex]\(\mathbb{R}\)[/tex] denotes the set of all real numbers. Adding 1 to both sides, we get:
[tex]\[ x \in \mathbb{R} \][/tex]
Thus, the domain of the function [tex]\(y = \sqrt[3]{x-1}\)[/tex] is all real numbers. In interval notation, this is expressed as:
[tex]\[ (-\infty, \infty) \][/tex]
Therefore, the correct answer is:
[tex]\[ -\infty < x < \infty \][/tex]
