Let $f(x) = \sqrt{x-2}$. Use interval notation to indicate where $f(x)$ is continuous. I don't know how to solve it.
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$\begingroup$$[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 $\infty$.
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