How can I get the domain of a fraction under a square root? I can find the domain of a fraction function under a square root through drawing the function on graph. Yet, I seem to not be able to find it algebrically.
To be precise, I want to find the domain of this function:
$$f(x) = \sqrt{1+1/x}$$
Thank you.
 A: I usually start by assuming the domain is all real numbers, then removing the values of $x$ that don't make sense.  What remains is the domain.  In your case, if $x=0$ then you are dividing by zero.  That's bad, so remove $x=0$ from the domain.  What else can go wrong?  Taking the square root of a negative number will cause problems (unless you're working in the complex numbers).  So figure out when is $1 + \frac{1}{x} < 0$.  All those values of $x$ are not in the domain.
A: Hint: 
Basically, the conditions are
$$x\ne 0\quad\text{and}\quad 1+\frac 1x=\frac{x+1}x\ge 0. $$
Now the sign of the fraction is the sign of the product $x(x+1)$, and you have a theorem about the sign of a quadratic polynomial…
A: When working with real numbers, you can't take a square root of a negative number. So we need whatever is under the square root to be $\geq 0$. Start there and then manipulate.
In your example:
(First note that since a $\frac{1}{x}$ appears, $x \neq 0$)
$1 + \frac{1}{x} \geq 0$
$\frac{1}{x} \geq -1$
Now if $x$ is positive this always holds. If $x$ is negative, then we get:
$1 \leq -x$
$x \leq -1$
So the domain is $(-\infty, -1] \cup (0, \infty)$
