Wronsonia8g

2022-07-07

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.

Tanner Hamilton

Beginner2022-07-08Added 12 answers

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

$f(x)=\sqrt{-{x}^{2}-\frac{1}{x}}$

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 $x$, $-{x}^{2}-\frac{1}{x}$ will never be a perfect square, even for your range of values. Especially for the fact where x=0 because $\sqrt{-(0{)}^{2}-\frac{1}{0}}$ cannot be a real root (because the $\frac{1}{0}$ part is indeterminate). I credit the commenter of this post for the edit.

$f(x)=\sqrt{-{x}^{2}-\frac{1}{x}}$

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 $x$, $-{x}^{2}-\frac{1}{x}$ will never be a perfect square, even for your range of values. Especially for the fact where x=0 because $\sqrt{-(0{)}^{2}-\frac{1}{0}}$ cannot be a real root (because the $\frac{1}{0}$ part is indeterminate). I credit the commenter of this post for the edit.

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