Question

what is the range of the function $y = \sqrt3{x + 8}$?
options:
$2 \leq y < \infty$
$-\infty < y < \infty$
$-8 < y < \infty$
$0 \leq y < \infty$

Explanation:

Step1: Recall the cube root function property

The cube root function \( y = \sqrt[3]{u} \) has a domain of all real numbers (\( -\infty < u < \infty \)) and a range of all real numbers (\( -\infty < y < \infty \)) because we can take the cube root of any real number (positive, negative, or zero), and the result can be any real number.

Step2: Analyze the inner function

In the given function \( y=\sqrt[3]{x + 8} \), let \( u=x + 8 \). The domain of \( u=x + 8 \) is all real numbers (\( -\infty < x < \infty \)), so the range of \( u \) (the set of all possible values of \( x + 8 \)) is also all real numbers (\( -\infty < u<\infty \)).

Step3: Determine the range of the cube root function

Since the cube root function \( y = \sqrt[3]{u} \) can take any real number \( u \) as input and produce any real number \( y \) as output, and \( u=x + 8 \) can be any real number, the range of \( y=\sqrt[3]{x + 8} \) is all real numbers, which is \( -\infty < y<\infty \).

Answer:

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