Is $\sqrt{-x^2-\frac{1}{x}}$ a rational function? I have to construct a rational function with the range being $[-1,0)$, which is pretty much just $-1$. I came up with the solution $\sqrt{-x^2-\frac{1}{x}}$. It works for the range, but I'm not sure if it is a rational function.
 A: The answer is simply no. A rational function cannot have a square root in their numerator (the denominator of yours is 1). Since your function $$f(x) = \sqrt{-x^2 - \frac 1x}$$ has a radical, the function isn't rational (because square roots are not polynomials, so functions with roots are not rational).
Edit:
The term inside the radical isn't a perfect square anyways, since for any value of $x$, $-x^2 - \frac 1x$ will never be a perfect square, even for your range of values. Especially for the fact where $x = 0$ because $\sqrt{-(0)^2 - \frac 10}$ cannot be a real root (because the $\frac 10$ part is indeterminate). I credit the commenter of this post for the edit.
A: Your function
$f(x) = \sqrt{-x^2 - \frac 1x}
$
is not rational.
For one thing,
it is imaginary for
$x \gt 0$
and for
$x < -1$.
One way to prove 
that $f$ is not rational
is to note that
a rational function
must behave like
$x^k$ as $x \to \infty$
and as $x \to 0$,
where $k$ is an integer
($0$, positive, or negative).
If$-1 < x < 0$,
$f(x)
=\frac1{\sqrt{-x}}\sqrt{x^3+1}
\approx\frac1{\sqrt{-x}}(1+\frac{x^3}{2})
$
as
$x \to 0^-$.
However,
$f(x)
\approx x^{-1/2}
$
as
$x \to 0^-$,
which is not of the form
$x^k$.
