# Determine the interval (s) on which the vector-value function is conti

Determine the interval (s) on which the vector-value function is continious. (Enter your answer using interval notation.)
$$\displaystyle{r}{\left({t}\right)}=\sqrt{{{t}}}{i}+\sqrt{{{t}-{1}}}{k}$$

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Step 1
$$\displaystyle{r}{\left({t}\right)}={\left(\sqrt{{{t}}}\right)}{i}+{\left(\sqrt{{{t}-{1}}}\right)}{k}$$
we know for a function to be continuous in Interval then it should be defined at every point in that interval.
we know that inside a root function there can't be a negative number so
t should greater than or equal to 0 and t - 3 should greater than or equal to 0
by taking the intersection of both t should greater than or equal to 3
But at 0 and 3 limit does not exist as the left-hand side limit is not equal to the right-hand side limit.
Step 2
Therefore, the interval is(1, co) on which the vector-valued function is continuous.