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

Question

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.

Answer 1

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

Answer 2

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

Answer 3

$p(x)$ 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.