Question

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

Explanation:

Step1: Recall cube root function range

The cube root function \( y = \sqrt[3]{u} \) has a range of all real numbers, i.e., \( -\infty < y < \infty \), because for any real number \( u \), we can find a real cube root, and as \( u \) takes all real values (from \( -\infty \) to \( \infty \)), \( \sqrt[3]{u} \) also takes all real values.

Step2: Analyze the given function

In the function \( y = \sqrt[3]{x + 8} \), let \( u = x + 8 \). The domain of \( u = x + 8 \) is all real numbers (\( -\infty < x < \infty \)), so \( u \) can take any real value (since adding 8 to any real \( x \) still gives any real number). Then, since \( y = \sqrt[3]{u} \) and \( u \) is all real numbers, the range of \( y \) is the same as the range of the cube root function, which is all real numbers.

Answer:

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