University Calculus: Early Transcendentals (3rd Edition)

$y=\sqrt{2x+3}$ is continuous on $[-3/2,\infty)$.
$$y=\sqrt{2x+3}$$ - 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 $[-3/2,\infty)$.