Domain of $\sqrt{1+\frac1x}$ I solved this as follows:
$1+\frac1x \ge 0$
$\frac1x \ge -1$ (subtract $1$ from both sides)
$1 \ge -x$ (multiply $x$ to both sides, cancel out $x$ from bottom of left side)
$-1 \le x$ (multiply both sides by $-1$, change sign)
$x \ge -1$
Another way to solve is:
$1+\frac1x \ge 0$
$\frac{(x+1)}x \ge 0$
$x+1 \ge 0$
$x \ge -1$
The domain should be $x\ge -1$, not including $0$.  However, I see from WolframAlpha and searching online that the domain is $x\le -1$ or $x>0$.  Why is my solution wrong?

Here's a better way at looking at this:
$$\begin{align*}
1+\frac{1}{x} &\ge 0 \\
x+1 &\ge 0 \tag{multiply both sides by $x$ when $x >0$} \\
&\textrm{or} \\
x+1 &\le 0 \tag{multiply both sides by $x$ when $x < 0$}
\end{align*}$$
In the first case, we restrict $x>0$, so every permissible value of $x$ satisfies the inequality.
In the second case, we get $x \le -1$, as expected.
