Question

if ( f(x) = sqrt{\frac{1}{2}x - 10} + 3 ), which inequality can be used to find the domain of ( f(x) )?

( \bigcirc sqrt{\frac{1}{2}x} geq 0 )

( \bigcirc \frac{1}{2}x geq 0 )

( \bigcirc \frac{1}{2}x - 10 geq 0 )

( \bigcirc sqrt{\frac{1}{2}x - 10} + 3 geq 0 )

Explanation:

Step1: Recall Domain of Square Root

For a square root function $\sqrt{u}$, the expression inside (radicand) $u$ must be non - negative, i.e., $u\geq0$.

Step2: Identify Radicand in $f(x)$

In the function $f(x)=\sqrt{\frac{1}{2}x - 10}+3$, the radicand is $\frac{1}{2}x-10$.

Step3: Set Up Inequality for Domain

To find the domain, we set the radicand greater than or equal to zero. So the inequality is $\frac{1}{2}x - 10\geq0$.

Answer:

$\frac{1}{2}x - 10\geq0$ (corresponding to the option with this inequality)