prove that between any two consecutive roots of $\phi(x)=0$, there lies a root of $\phi'(x)+\lambda \phi(x)=0$

Question

Let $\phi(x)$ is a polynomial function, with real coefficients. Let $\alpha$ and $\beta$ be any two consecutive roots of $\phi(x)=0$, then prove that there lies a root of $\phi'(x)+\lambda \phi(x)=0$ in the interval $(\alpha, \beta)$. (here $\lambda$ is a fixed constant).

My approach:

We need to prove that there exists $x\in(\alpha,\beta), $ such that $\phi'(x)=-\lambda \phi(x)$.

Since $\phi(\alpha)=0$ and $\phi(\beta)=0$ and $\phi(x)$ is a polynomial, hence $\phi(x)$ attains a maxima or a minima in $(\alpha, \beta)$; hence, $\phi'(c)=0$ for some $c\in(\alpha, \beta)$.

I do not know how to proceed further. I have been able to verify the above statement for specific cases, but haven't been able to generalize from there.

Also, I don't think the above statement should be true. for example what if $\beta$ was a repeated root of $\phi$ and $\phi'$. something like this

unfortunately, I haven't been able to generate a specific polynomial that behaves in this way. so I'm stuck

edit: based on @Hagen von Eitzen's answer, can we use

1. $f(x)=\phi(x)\cdot x^{\lambda}$ to prove $x\cdot\phi'(x)+\lambda \phi(x)=0$
2. $f(x)=\phi(x)\cdot e^{tan^{-1}x}$ to prove $(1+x^2)\cdot\phi'(x)+\phi(x)=0$
3. $f(x)=\phi(x)\cdot e^{x^2/2}$ to prove $\phi'(x)+x\cdot\phi(x)=0$

Answer 1

Hint: Consider $$f(x)=\phi(x)e^{\lambda x}$$ and apply Rolle's theorem.

Answer 2

First, note that since $\phi$ is a polynomial, we can write $\phi(x) = (x-\alpha)^n(x-\beta)^m\tilde\phi(x)$ and consider only $\tilde\phi(x)$. Look at cases, as you mentioned. Notice that $\tilde\phi'(\alpha)$ and $\tilde\phi'(\beta)$ cannot both have the same nonzero sign (why?). Let $g(x) = \tilde\phi'(x) + \lambda\tilde\phi(x)$. Then $g(\alpha) = \tilde\phi'(\alpha)$ and $g(\beta) = \tilde\phi'(\beta)$. But these have opposite signs, so we can apply the Intermediate Value Theorem on g. Since $g$ has a $0$ in $(\alpha,\beta)$, so will $g(x)(x-\alpha)^n(x-\beta)^m = \phi'(x) + \lambda\phi(x)$.