Question

what is the domain of the function $y = \sqrt3{x - 1}$?
\\(\circ\\) \\(-\infty < x < \infty\\)
\\(\circ\\) \\(-1 < x < \infty\\)
\\(\circ\\) \\(0 \leq x < \infty\\)
\\(\circ\\) \\(1 \leq x < \infty\\)

Explanation:

Step1: Recall cube root domain

The cube root function $\sqrt[3]{u}$ is defined for all real numbers $u$ (since we can take the cube root of negative numbers, zero, and positive numbers).

Step2: Analyze the inner function

In the function $y = \sqrt[3]{x - 1}$, the expression inside the cube root is $u=x - 1$. Since $x-1$ is a linear function (defined for all real $x$), and the cube root is defined for all real $u$, the domain of $y=\sqrt[3]{x - 1}$ is all real numbers. So the domain is $-\infty < x < \infty$.

Answer:

A. $-\infty < x < \infty$