For a given real polynomial $$p(x)=a_0 + a_1x + a_2x^2+\dots +a_nx^n,\quad\text{with $a_0=0$, $x\in \mathbb{R}$},$$ is there any proof that all local extrema will always lie between two roots of $p(x)=0$?

I was messing around with some visualizations and tried to proof it using derivatives but could not solve it. Internet search wasn't helpful either. Thanks for your help!

  • 1
    $\begingroup$ It's not true, as the answer below gives an excellent counterexample. However, it is true that between any two roots there is at least one local extremum, which is almost the same thing, but not quite. $\endgroup$ – Arthur Jan 18 '18 at 9:36
  • $\begingroup$ Thank you. The fact that I was asking for the case without constant term $a_0=0$ was somehow removed from my original question. Does that leave other counterexamples? $\endgroup$ – Sarem Seitz Jan 18 '18 at 9:49

That is not true. Take for example $$p(x)=6\int_0^x (t-1)(t-2)dt=2x^3-9x^2+12x$$ It has local extrema at $1$, and at $2$ but $p$ has just one root, that is $0$.

However, by Gauss-Lucas theorem, for any non-constant polynomial $p$, all zeros of a $p'$ (the derivative is zero at local extrema) belong to the convex hull in the complex plane of the set of complex zeros of $p$.

  • $\begingroup$ Thanks, but I was specifically asking for the case where there is no constant term. (that was somehow edited out from my question) $\endgroup$ – Sarem Seitz Jan 18 '18 at 9:46
  • $\begingroup$ @SaremSeitz: why this remark ? This is addressed in the answer. $\endgroup$ – Yves Daoust Jan 18 '18 at 10:00
  • $\begingroup$ Wow, that second paragraph was a surprising turn! $\endgroup$ – Bananach Jan 18 '18 at 10:01
  • $\begingroup$ @Robert Z: Thanks, that was enlightening. $\endgroup$ – Sarem Seitz Jan 18 '18 at 10:11

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