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Question: For $f \in \mathbb{R}[x]$ is it true that there is a local maximum of $g(x) = \frac{f(x)^2}{(x^2+1)^{d+1}}$ in each connected component of $f \neq 0$ where $d = \deg(f)$?

Comments: I haven't (yet) found a counter-example for some $f$ and am looking for some ideas for proving whether this is true. It does not appear this can simply be proved by doing calculus and considering $g'(x) = 0 \wedge g(x) \neq 0 \implies g''(x) < 0$.

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    $\begingroup$ Yes, $g$ is bounded on the whole real line and eventually decreasing as $x\to\pm \infty$, hence $g$ has a local minimum on the components with large $\|x\|$. Now note that any local maximum point of of $g$ must lie in the interior of $\{x:f\neq\0\}$ since $g\ge 0$ everywhere. Conversely, every connected component of $\{x:f\neq\0\}$ has a local maximum, since $f$ cannot be identically 0 on the entire component. $\endgroup$
    – J.C. Ottem
    Commented Apr 27, 2011 at 20:40
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    $\begingroup$ This might be better suited for math.stackexchange.com by the way.. $\endgroup$
    – J.C. Ottem
    Commented Apr 27, 2011 at 20:41
  • $\begingroup$ I agree with some of what you said, however there are choices of $f$ where each connected component of $f \neq 0$ contain one and only one critical point of $g$ (all of which are local maxima). For example, $f = 4-10x^2$. The following figure has three connected components where $f$ and $g$ are the red and blue curves, respectively: i.imgur.com/xRhLV.png $\endgroup$
    – James Rohal
    Commented Apr 27, 2011 at 21:40
  • $\begingroup$ How does that contradict what I wrote? $\endgroup$
    – J.C. Ottem
    Commented Apr 27, 2011 at 22:29

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