Is $\sqrt{-x^2-\frac{1}{x}}$ a rational function?

Question

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.

Answer 1

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.

Answer 2

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 $x^k$.