Question

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

Explanation:

Step1: Recall the domain of cube root function

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

Step2: Analyze the inner function

In the function \( y=\sqrt[3]{x - 1}\), the expression inside the cube root is \( u=x - 1 \). Since the cube root is defined for all real \( u \), we need to find all real \( x \) such that \( x-1 \) is a real number. But \( x-1 \) is a real number for all real \( x \), because subtracting 1 from any real number \( x \) gives a real number.

Step3: Determine the domain

So, \( x \) can be any real number. In interval notation, this is \( -\infty

Answer:

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