Question

unit 7 review: cube & cube root
determine the domain of ( f(x) = 2 + sqrt{x - 1} ), as shown in the graph above.
options: ( x geq 0 ), ( x geq 1 ), ( x geq 2 ), ( x geq 3 )

Explanation:

Step1: Recall domain of square root function

For a square root function $\sqrt{u}$, the expression inside the square root ($u$) must be non - negative (i.e., $u\geq0$) to have real - valued outputs. In the function $f(x)=2+\sqrt{x - 1}$, the expression inside the square root is $x - 1$.

Step2: Solve the inequality for the domain

We set up the inequality $x - 1\geq0$. To solve for $x$, we add 1 to both sides of the inequality. So, $x-1 + 1\geq0 + 1$, which simplifies to $x\geq1$. We can also verify this from the graph: the graph starts at $x = 1$ (the point on the graph has an $x$ - coordinate of 1) and extends to the right, which means the domain includes all real numbers $x$ such that $x\geq1$.

Answer:

$x\geq1$ (corresponding to the option with text "x ≥ 1")