## Calculus: Early Transcendentals 8th Edition

$M(x)$ is continuous on $(-\infty,-1]\cup(0,\infty)$.
$$M(x)=\sqrt{1+\frac{1}{x}}$$ Let $f(x)=1+\frac{1}{x}$ and $g(x)=\sqrt x$, we have$$M(x)=g(f(x))$$ We see that $f(x)$ and $g(x)$ are both continuous. Therefore, by Theorem 9, $M(x)$ is continuous on its domain. *Find the continuity range of $M(x)$ $M(x)$ is defined when $$1+\frac{1}{x}\ge0$$$$\frac{x+1}{x}\ge0\hspace{.5cm}(x\ne0)$$$$\left\{\begin{array} {c l}x+1\ge0\\x>0\end{array}\right.\hspace{0.5cm}or\hspace{0.5cm}\left\{\begin{array} {c l}x+1\le0\\x<0\end{array}\right.$$$$\left\{\begin{array} {c l}x\ge-1\\x>0\end{array}\right.\hspace{0.5cm}or\hspace{0.5cm}\left\{\begin{array} {c l}x\le-1\\x<0\end{array}\right.$$$$x\gt0\hspace{.5cm}or\hspace{.5cm}x\le-1$$ Therefore, the domain of $g(f(x))$ is $(-\infty,-1]\cup(0,\infty)$. In conclusion, $M(x)$ is continuous on $(-\infty,-1]\cup(0,\infty)$.