Answer
$A(t)$ is continous on $[-1,0]$.
Work Step by Step
$$A(t)=\arcsin(1+2t)$$
1) Find the domain of $A(t)$
$\arcsin(1+2t)$ is defined only when $$-1\leq(1+2t)\leq1$$$$-2\leq2t\leq0$$$$-1\leq t\leq0$$
In other words, the domain of $A(t)$ is $t\in[-1,0]$.
2) According to Theorem 7, $A(t)$ is continous on its domain.
Therefore, $A(t)$ is continous on $[-1,0]$.
