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

Explanation:

Step1: Recall the property of cube - root function

For the cube - root function (y = \sqrt[3]{u}), the domain of (y=\sqrt[3]{u}) is all real numbers because for any real number (u), the cube - root (\sqrt[3]{u}) is well - defined. In the given function (y=\sqrt[3]{x - 1}), let (u=x - 1).

Step2: Determine the domain of (x)

Since (u=x - 1) can be any real number (because of the property of the cube - root function), and there is no restriction on (x) from the equation (y=\sqrt[3]{x - 1}) (unlike a square - root function (y=\sqrt{u}) where (u\geq0)). The function (y=\sqrt[3]{x - 1}) is defined for all real values of (x). Mathematically, we can write the domain in interval notation as ((-\infty,\infty)), which is equivalent to (-\infty<x<\infty) in inequality notation.

Answer:

(-\infty<x<\infty) (the first option)