Answer
$$\left\{ {t:\left| t \right| \leqslant 2} \right\}{\text{ }}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \sqrt {t + 2} {\bf{i}} + \sqrt {2 - t} {\bf{j}} \cr
& {\bf{r}}\left( t \right) = \left\langle {\sqrt {t + 2} ,\sqrt {2 - t} } \right\rangle \cr
& {\text{the vector value is of the form }}{\bf{r}}\left( t \right) = \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle \cr
& {\text{The domain of }}{\bf{r}}{\text{ is the largest set of values of }}t{\text{ on which all of }} \cr
& f\left( t \right){\text{, }}g\left( t \right){\text{ }}{\text{, and }}h\left( t \right){\text{ are defined}}{\text{. then }} \cr
& f\left( t \right) = \sqrt {t + 2} {\text{ and }}g\left( t \right) = \sqrt {2 - t} \cr
& {\text{for }}f\left( t \right) = \sqrt {t + 2} {\text{ the radicand cannot be negative}}{\text{, then}} \cr
& t + 2 \geqslant 0,\,\,\,\,\,\,t \geqslant - 2,\,\,\,\,\left[ { - 2,\infty } \right) \cr
& {\text{for }}g\left( t \right) = \sqrt {2 - t} {\text{ the radicand cannot be negative}}{\text{, then}} \cr
& 2 - t \geqslant 0,\,\,\,\,\,\,t \leqslant 2,\,\,\,\,\left( { - \infty ,2} \right] \cr
& {\text{The domain of }}{\bf{r}}\left( t \right){\text{ is the intersection of the domains of }}f\left( t \right){\text{ and }}h\left( t \right) \cr
& \left( { - \infty ,2} \right] \cap \left[ { - 2,\infty } \right) \cr
& \left[ { - 2,2} \right] \cr
& {\text{The domain is}}:{\text{ }}\left\{ {t: - 2 \leqslant t \leqslant 2} \right\} = \left\{ {t:\left| t \right| \leqslant 2} \right\}{\text{ }} \cr} $$
