## Calculus: Early Transcendentals 8th Edition

$A(t)$ is continous on $[-1,0]$.
$$A(t)=\arcsin(1+2t)$$ 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 $[-1,0]$.