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.

Wronsonia8g
2022-07-07
Answered

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.

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A problem in Dummit and Foote states:

Let k be a field and let k(x) be the field of rational functions in x with coefficients from k. Let$t\u03f5k\left(x\right)$ be the rational function $\frac{P\left(x\right)}{Q\left(x\right)}$ with relatively ' polynomials $P\left(x\right),Q\left(x\right)\u03f5k\left[x\right]$ , with $Q\left(x\right)\ne 0$ . Then k(x) is an extension of k(t) and to compute its degree it is necessary to compute the minimal polynomial with coefficients in k(t) satisfied by x.

By k(t), do they mean k adjoin t, i.e. the set of polynomials$k}_{0}+{k}_{1}\frac{P\left(x\right)}{Q\left(x\right)}+\dots +{k}_{n}{\left(\frac{P\left(x\right)}{Q\left(x\right)}\right)}^{n$ ? Or do they mean the set of rational functions in t, e.g. $\frac{\frac{P\left(x\right)}{Q\left(x\right)}+1}{2{\left(\frac{P\left(x\right)}{Q\left(x\right)}\right)}^{2}+3}$

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Let k be a field and let k(x) be the field of rational functions in x with coefficients from k. Let

By k(t), do they mean k adjoin t, i.e. the set of polynomials

Thanks for answer!

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Let