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

[tex]
\begin{array}{l}
-\infty-10 \leq x<\infty \\
1 \leq x<\infty
\end{array}
\]



Answer :

To determine the domain of the function [tex]\( y = \sqrt[3]{x-1} \)[/tex], we must identify the set of all permissible input values [tex]\( x \)[/tex] for which the function is defined.

The given function involves the cube root of [tex]\( x-1 \)[/tex]. The cube root, denoted by [tex]\( \sqrt[3]{\cdot} \)[/tex], is defined for all real numbers. Unlike the square root, the cube root can take both positive and negative numbers, as well as zero. This is a key characteristic of cubic roots: they have a real number output for any real number input.

Thus, we need to ensure that [tex]\( x-1 \)[/tex] can take any real value. Since there are no restrictions that preclude any particular value of [tex]\( x \)[/tex], the expression [tex]\( x-1 \)[/tex] can indeed be any real number.

By setting [tex]\( x - 1 \)[/tex] to range over all real numbers, we observe that:

[tex]\[ x \in (-\infty, \infty) \][/tex]

Therefore, we conclude that the domain of [tex]\( y = \sqrt[3]{x-1} \)[/tex] is:

[tex]\[ -\infty < x < \infty \][/tex]

Thus, the correct choice from the given options is:

[tex]\[ -\infty < x < \infty \][/tex]