Answer
$y=\sqrt{2x+3}$ is continuous on $[-3/2,\infty)$.
Work Step by Step
$$y=\sqrt{2x+3}$$
- Domain: $y$ is defined where $2x+3\ge0$, which means $x\ge-3/2$
So our domain here is $[-3/2,\infty)$
We know that $\lim_{x\to c}(2x+3)=2c+3$ on $[-3/2,\infty)$
So $y=2x+3$ is continuous on $[-3/2,\infty)$
Applying Theorem 8, if $f$ is continuous at $x=c$, then $\sqrt[n]f$ is also continuous at $x=c$ ($f$ must be defined on an open interval containing $c$, and $n\gt0$ and $n\in Z$)
Therefore, $y=\sqrt{2x+3}$ is continuous on $[-3/2,\infty)$.
