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 the square root (the radicand) must be non - negative, i.e., \(u\geq0\).
In the function \(f(x)=\sqrt{\frac{1}{2}x - 10}+3\), the radicand is \(\frac{1}{2}x-10\).

Step2: Determine the inequality

To find the domain of \(f(x)\), we need to ensure that the radicand \(\frac{1}{2}x - 10\) is non - negative. So we set up the inequality \(\frac{1}{2}x-10\geq0\).

Answer:

\(\frac{1}{2}x - 10\geq0\) (corresponding to the option \(\boldsymbol{\frac{1}{2}x-10\geq0}\))