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

Answer:

A. $-\infty < y<\infty$

Explanation:

Step1: Recall cube - root function property

The cube - root function $y = \sqrt[3]{u}$ has a domain and range of all real numbers.

Step2: Analyze the given function

In the function $y=\sqrt[3]{x + 8}$, let $u=x + 8$. Since $u$ can take on any real - number value (because $x\in(-\infty,\infty)$ and $u=x + 8$ is a linear function), and the cube - root function $y=\sqrt[3]{u}$ can output any real number for any real input $u$. So the range of $y=\sqrt[3]{x + 8}$ is $(-\infty,\infty)$, which can be written as $-\infty < y<\infty$.