Question

termine the domain
f(x) = -sqrt{x - 2} + 4 .

x ≤ -2

x ≤ -4

x ≥ 2

x ≥ 4

Explanation:

Step1: Recall domain of square root

For a square root function $\sqrt{u}$, the expression inside the square root ($u$) must be non - negative, i.e., $u\geq0$. In the function $f(x)=-\sqrt{x - 2}+4$, the expression inside the square root is $x - 2$.
So we set up the inequality $x-2\geq0$.

Step2: Solve the inequality

To solve $x - 2\geq0$, we add 2 to both sides of the inequality.
$x-2 + 2\geq0+2$
Which simplifies to $x\geq2$.

Answer:

$x\geq2$ (corresponding to the option with text "x ≥ 2")