- #36
archaic
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This can't be as it has a positive zero.fresh_42 said:##x^n−x^2=:P(x)##
I have shown in my second solution post that ##P(x)## is positive after its greatest zero, which is negative, given the premises, so ##|P(x)|=P(x)## for ##x\geq0##.
##f(x)=c/x^n## where ##c\in(1,\,\infty)##, any number does the job, and ##g(x)=1/f(x)##.
Why any ##c## in that interval does the job? Because (for ##n\geq2##):
$$\lim_{x\to\infty}\left(g(x)-P(x)\right)=\lim_{x\to\infty}\left(\frac{x^n}{c}-x^n-a_{n-1}x^{n-1}+...+a_0\right)=-\infty<0$$
This tells me that, at some point, ##P(x)>g(x)\Leftrightarrow \frac{1}{P(x)}<f(x)##.
EDIT: I have initially written the result of that limit as ##(1-c)/c##, even in the previous post (I corrected it), sorry!
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