# Show that $p'(c)+100p(c)=0$

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.

• Which contest is it? Jun 24, 2019 at 15:45
• local school math contest Jun 24, 2019 at 15:46
• Please give proper citation reference. Jun 24, 2019 at 15:47

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 $$p'(c)+100p(c)=0$$

• Very nice! +1 for an elegant solution. Jun 24, 2019 at 15:59
• elegant indeed!! Jun 24, 2019 at 16:13
• Elegant but unnatural. Jun 24, 2019 at 16:15
• @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? Jun 24, 2019 at 16:25
• @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. Jun 24, 2019 at 17:03

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: $$\frac{p'(c)}{p(c)}=-100\\ p'(c)+100p(c)=0$$

• This is much more elementary and natural! +1 Jun 24, 2019 at 16:03
• Thanks, though I consider the mean value theorem elementary calculus because it's not hard to prove and is very useful.
– J_P
Jun 24, 2019 at 16:04
• Wow this was actually really slick Jun 24, 2019 at 16:10
• 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.
– J_P
Jun 24, 2019 at 18:37
• There, I added a second limit which is $+\infty$ so that by continuity we have $p'/p=-100$ somewhere.
– J_P
Jun 24, 2019 at 18:39

$$p(x)$$ 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.

• 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?
– J_P
Jun 24, 2019 at 16:08
• 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. Jun 24, 2019 at 16:17
• Right, of course... Nice solution, +1
– J_P
Jun 24, 2019 at 16:20