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?
 A: You can't multiply by $x$ unless you take care to note if $x > 0$ or $x < 0$.  Separate the cases and you can get a correct analysis.  You know right off that $x = 0$ is not in the domain; you can exclude it immediately.
A: 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.
A: $$\dfrac 1x \ge-1\implies 1 \geq -x\tag{1}$$
$$\dfrac{x+1}{x} \ge 0 \implies x+1 \geq 0\tag{2}$$
In your work (both $(1), \;(2)\;$), you incorrectly assume $x$ is positive. You cannot know what the sign of $x$ is. If $x\lt 0$, then the direction of the inequality will change. In general, not knowing the sign of a variable should scream "be careful when operating on the inequality. This is especially the case when you have a rational function, as you have here.
So you have two cases to consider if we exclude $x = 0$ from the "get-go" knowing $1 + \frac 1x$ is undefined at $x = 0$, and hence, $x = 0$ is not in the domain. Those two cases are 


*

*either $\;x \lt 0;\;$ (This is the case you missed.)

*OR

*else $\;x \gt 0.\;$ (This is what you assumed. Be careful with multiplying $x \cdot 0$: you may lose some information!)

A: In your attempts there are two mistakes: you can't go from $1/x\ge-1$ to $1\ge-x$, because these two inequalities are not equivalent. Similarly, in $x+1/x\ge0$ you can't ”multiply” by $x$. Let's see why.
The inequality $1+1/x\ge0$ can be written as
$$
\frac{x+1}{x}\ge0
$$
and this holds whenever numerator and denominator are either both positive or both negative (or the numerator is zero).
So, let's study separately $x+1>0$ and $x>0$. So you can make a diagram
$$
\begin{matrix}
\text{(numerator)}   & <0 &&   >0 &&  >0 \\\hline
                      &&-1&    &0&\\\hline
\text{(denominator)} & <0 && <0 && >0
\end{matrix}
$$
Thus the fraction is nonnegative for $x\le -1$ or $x>0$.
For instance, when $x=-2$, you have $1+1/x=1-1/2=1/2>0$; on the other hand $1$ is not greater than $-(-2)$, so for $x=-2$ the inequality $1+1/x\ge0$ is satisfied, while $1\ge-x$ isn't and the two are not equivalent.
A: This is not really an answer, but a piece of advice for when you have your solution and the official solution, and these two do not match (to be more precise: you have some solutions outside of the official solution).
Pick any number from your solution which is not included in the official solution, put it in all your expressions, and see what went wrong.
In this case, the "offending" set is $\langle -1, 0 \rangle$, so let's pick $x = -1/2$. Now, go through your steps:
\begin{align*}
1 + (1/x) \ge 0 \quad &\Rightarrow \quad 1 + \frac{1}{-1/2} = 1 - 2 = -1 \not\ge 0, \\
1/x \ge −1 \quad \text{(subtract 1 from both sides)} \quad &\Rightarrow \quad \frac{1}{-1/2} = -2 \not\ge -1, \\
1 \ge −x \quad \text{(multiply $x\dots$)} \quad &\Rightarrow \quad 1 \ge -(-1/2) = \frac{1}{2}, \\
−1 \le x \quad \text{(multiply both sides by $−1,\dots$)} \quad &\Rightarrow \quad -1 \le -1/2, \\
x \ge −1 \quad &\Rightarrow \quad -1/2 \ge -1.
\end{align*}
See when it turned from "not correct" to "correct"? That's where you've made a mistake and introduced a wrong solution. And, you also have an example ($x = -1/2$) that lets you look into that specific line (multiplying by $x$ when it is negative) to see why it went wrong.
Your "another way to solve":
\begin{align*}
1 + (1/x) \ge 0 \quad &\Rightarrow \quad 1 + \frac{1}{-1/2} = 1 - 2 = -1 \not\ge 0, \\
(x+1) / x \ge 0 \quad &\Rightarrow \quad \frac{(-1/2)+1}{-1/2} = \frac{1/2}{-1/2} = -1 \not\ge 0, \\
x+1 \ge 0 \quad &\Rightarrow \quad (-1/2)+1 = 1/2 \ge 0, \\
x \ge −1 \quad &\Rightarrow \quad -1/2 \ge -1.
\end{align*}
Here, it is even more obvious. When you multiplied $(x+1) / x \ge 0$ by a negative number, you switched from (the wrong) $-1 \ge 0$ to (the right) $1/2 \ge 0$. This is also a strong indicator that you should've switched the inequality sign as well. Now, try such multiplication of any inequality by any negative number and you'll see that it always has to be done. For example, try multiplying $2 \ge 1$ by $-1$. Is $-2 \ge -1$ or $-2 \le -1$?
Of course, in this process it is possible that the $x$ you pick satisfies the inequality from the beginning. In that case, you've either picked the $x$ that is actually included in the official solution (instead of just yours), or you've made some mistake substituting it in the inequality, or the official solution is wrong and you've found an $x$ outside of it that proves it.
As for the correct solution of your inequality, you've got plenty of those in the other answers, so I won't repeat that. I hope this helps you in the future.
