## Calculus 8th Edition

$t\in (-1,3)$
1) $t+1\gt0$, since $ln(x)$ is undefined when $x\leq0$ 2) Therefore, $t\gt-1$ 3) $9-t^2\gt0$, since $sqrt(x)$ is undefined when $x\leq0$ 4) Therefore, $t^2\lt9$ 5) $t\lt\pm3$ 6) $-3\lt t\lt3$ 7) Since $t$ must be greater than $-1$ (see step 2) and less than $3$, we conclude that $t\in(1,3)$