$\begingroup$
$\endgroup$
4
Let $f(x) = \sqrt{x-2}$. Use interval notation to indicate where $f(x)$ is continuous. I don't know how to solve it.
1 Answer
$\begingroup$
$\endgroup$
$[2,\infty)$ The function has restrictions on its domain. $x-2$ has to be greater than or equal to $0$. Solving this inequality, $x \geq 2$. Therefore, continuous from $2$ to $\infty$.