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Let $p(x)$ be a polynomial with real coefficients such that $a\le c \le b$ where $a,b$ are two consecutive roots of $p(x)$. Show that there exists at least one c for which $$p'(c)+100p(c)=0$$

Okay, so this is what I tried, By Lagrange's Mean Value Theorem we can find a point $c$ in the interval $[a,b]$ such that $p'(c)=0$ but at that time, $100p(c)$ is not $0$. I tried some other ways but I don't think those will work. Any help is appreciated.

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  • $\begingroup$ Which contest is it? $\endgroup$
    – user10354138
    Jun 24, 2019 at 15:45
  • $\begingroup$ local school math contest $\endgroup$
    – Anurag Saha
    Jun 24, 2019 at 15:46
  • $\begingroup$ Please give proper citation reference. $\endgroup$
    – user10354138
    Jun 24, 2019 at 15:47

3 Answers 3

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Let $f(x)=e^{100x}p(x)$. Since $f(a)=f(b)=0$, Lagrange's Mean Value Theorem implies that there exists $c$, $a\le c\le b$, such that $f'(c)=0$, then \begin{align*} e^{100c}p'(c)+100e^{100c}p(c)=0 \end{align*} Since $e^{100c}\neq 0$ it follows $$p'(c)+100p(c)=0$$

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    $\begingroup$ Very nice! +1 for an elegant solution. $\endgroup$
    – Adrian Keister
    Jun 24, 2019 at 15:59
  • $\begingroup$ elegant indeed!! $\endgroup$
    – Anurag Saha
    Jun 24, 2019 at 16:13
  • $\begingroup$ Elegant but unnatural. $\endgroup$
    – nonuser
    Jun 24, 2019 at 16:15
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    $\begingroup$ @Angelo Mario Gallegos, how did you come up with this? Were you thinking in terms of Integration Factor $e^{\int P dx}$ that is used to solve differential equations? $\endgroup$
    – Anurag Saha
    Jun 24, 2019 at 16:25
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    $\begingroup$ @Aqua I dissagree. I think that the solution is natural if one follows the following heuristical line: $$p'(c)+100p(c)=0 \Leftrightarrow \frac{p'(c)}{p(c)} =-100\Leftrightarrow \left( \ln p \right)' (c)=-100 \Leftrightarrow \left( \ln p(x) +100x \right)' (c)=0 \Leftrightarrow \left( \ln \left( p(x)e^{100x}\right) \right)' (c)=0$$ Thus it is natural to apply Role to the above function. $\endgroup$
    – N. S.
    Jun 24, 2019 at 17:03
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You can factorize $p$: $$ p(x)=(x-a)^{n}(x-b)^mq(x);\quad n,m\geq1 $$ Then: $$ p'(x)=(n(x-a)^{n-1}(x-b)^m+m(x-a)^n(x-b)^{m-1})q(x)+(x-a)^{n}(x-b)^mq'(x) $$ Caculate the limits: $$ \lim_{x\rightarrow b^-}\frac{p'(x)}{p(x)}=\lim_{x\rightarrow b^-}\left(\frac{q'(x)}{q(x)}+\frac{n}{x-a}+\frac{m}{x-b}\right)=-\infty\\ \lim_{x\rightarrow a^+}\frac{p'(x)}{p(x)}=\lim_{x\rightarrow a^+}\left(\frac{q'(x)}{q(x)}+\frac{n}{x-a}+\frac{m}{x-b}\right)=+\infty $$ Therefore, there exists a $c\in(a,b)$ such that: $$ \frac{p'(c)}{p(c)}=-100\\ p'(c)+100p(c)=0 $$

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    $\begingroup$ This is much more elementary and natural! +1 $\endgroup$
    – nonuser
    Jun 24, 2019 at 16:03
  • $\begingroup$ Thanks, though I consider the mean value theorem elementary calculus because it's not hard to prove and is very useful. $\endgroup$
    – J_P
    Jun 24, 2019 at 16:04
  • $\begingroup$ Wow this was actually really slick $\endgroup$
    – Brenton
    Jun 24, 2019 at 16:10
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    $\begingroup$ If $p'(b)=0$ then $c=b$ is one solution but it's not the only one and not the one my proof deals with. The limit is always $-\infty$ even if $p'(b)=0$. Though I see another issue, my proof assumes that $p'(x)/p(x)>-100$ somewhere which I will have to fix. $\endgroup$
    – J_P
    Jun 24, 2019 at 18:37
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    $\begingroup$ There, I added a second limit which is $+\infty$ so that by continuity we have $p'/p=-100$ somewhere. $\endgroup$
    – J_P
    Jun 24, 2019 at 18:39
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$p(x)$ is a smooth function, thus $p'(x) + 100p(x)$ is a smooth function as well.

Case 1: $0 \ne \text{sign}(p'(a)) \neq \text{sign}(p'(b)) \neq 0$. Thus two points exist where the function is positive and negative: $p'(a) + 100p(a) < 0$ and $p'(b) + 100p(b) > 0$ (or vice versa). As the function is smooth, it is continuous everywhere, hence it has to assume $0$ between $a$ and $b$.

Case 2: $\text{sign}(p'(a)) = 0$ or $\text{sign}(p'(b)) = 0$. In this case $p'(a) + 100p(a) = 0$ or $p'(b) + 100p(b) = 0$ as $a$ and $b$ are roots.

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  • $\begingroup$ Hmm, how do we conclude that if the signs of $p'$ are different, there must exist two points where the signs of $p'+100p$ are different? $\endgroup$
    – J_P
    Jun 24, 2019 at 16:08
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    $\begingroup$ because $a$ and $b$ are roots, i.e. $100p(a) = 0$. thus sign of $p'(a) + 100p(a)$ only depends on the sign of $p'(a)$. Simlarly for $b$. Additionally if the first derivative at one root is non-zero, the sign at any surrounding root has to have either a different sign, or has to be zero, it cannot have the same sign. $\endgroup$
    – Attila Kinali
    Jun 24, 2019 at 16:17
  • $\begingroup$ Right, of course... Nice solution, +1 $\endgroup$
    – J_P
    Jun 24, 2019 at 16:20

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