Question

unit 7 review: cube & cube root
what is the domain of
the function
a(x) = \sqrt6{x + 2} - 3?
options: ( x in mathbb{r} ), ( x geq -2 ), ( x geq 2 ), ( x geq -3 ), ( x geq 3 )

Explanation:

Step1: Recall domain of even - root functions

For an even - root function of the form $\sqrt[n]{u}$, where $n$ is even (in this case, $n = 6$ which is even), the expression inside the root (the radicand) must be non - negative. That is, $u\geq0$.
In the function $A(x)=\sqrt[6]{x + 2}-3$, the radicand is $x + 2$. So we set up the inequality:
$x+2\geq0$

Step2: Solve the inequality

To solve the inequality $x + 2\geq0$, we subtract 2 from both sides of the inequality.
$x+2-2\geq0 - 2$
$x\geq - 2$

Answer:

$x\geq - 2$ (corresponding to the option "x ≥ -2")