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What is the domain of the function [tex]$y=\sqrt[3]{x-1}$[/tex]?

A. [tex]-\infty\ \textless \ x\ \textless \ \infty[/tex]
B. [tex]-1\ \textless \ x\ \textless \ \infty[/tex]
C. [tex]0 \leq x\ \textless \ \infty[/tex]
D. [tex]1 \leq x\ \textless \ \infty[/tex]


Sagot :

To determine the domain of the function [tex]\( y = \sqrt[3]{x-1} \)[/tex], we need to consider the nature of the cube root function.

The cube root function, denoted as [tex]\( \sqrt[3]{x} \)[/tex] or [tex]\( x^{1/3} \)[/tex], is defined for all real numbers. This means that you can take the cube root of any real number, positive or negative, and still obtain a real number as a result.

Given the function [tex]\( y = \sqrt[3]{x-1} \)[/tex]:

1. The expression inside the cube root is [tex]\( x - 1 \)[/tex].
2. Since the cube root function can accept any real number, [tex]\( x - 1 \)[/tex] can be any real number.
3. As a result, we do not need to place any specific restrictions on [tex]\( x \)[/tex]. We simply need [tex]\( x - 1 \)[/tex] to be a real number, which [tex]\( x \)[/tex] can always satisfy.

Thus, the domain of [tex]\( y = \sqrt[3]{x-1} \)[/tex] includes all real numbers [tex]\( x \)[/tex]. Therefore, the domain can be expressed as:
[tex]\[ -\infty < x < \infty \][/tex]

Thus, the correct option is:
[tex]\[ -\infty < x < \infty \][/tex]