Question

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

Explanation:

Step1: Recall cube root domain rule

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

Step2: Apply to the given function

For the function \( y = \sqrt[3]{x - 1} \), the expression inside the cube root is \( u=x - 1 \). Since \( x-1 \) is a real number for any real number \( x \) (because subtracting 1 from any real \( x \) gives a real number), there are no restrictions on \( x \). So the domain is all real numbers, which is \( -\infty < x < \infty \).

Answer:

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