Case 1: all roots are distinct

In this case, assume $f$ has total of $n$ real roots. I can take any consecutive pair of roots $a,b$, $a < b$ and say that since $f(a) = f(b) = 0$, using Rolle's theorem, $\exists c \in \mathbb{R}$ such that $f'(c) = 0, a < c < b$. Then I end up with $n-1$ real roots in $f'$. How do I show that $f'$ has total of $n-1$ roots?

Case 2: repeated roots

Then for some order $k$ at $x_0$, $f(x) = (x-x_0)^kq(x)$ where $q(x_0)\neq 0$. Then $f'(x) = k(x-x_0)^{k-1}q(x) + (x-x_0)^k q'(x)$. How many root are there in $f'$ and how do I show that they are all real?

  • 1
    $\begingroup$ What is the assumption on$f$?. It is not true that roots of the derivative of any polynomial are real. Ex. $f(x)=3x+x^{3}$. $\endgroup$ – Kavi Rama Murthy Apr 30 '18 at 9:35
  • $\begingroup$ that $f$ is a polynomial with real roots $\endgroup$ – MoneyBall Apr 30 '18 at 9:36

There is a theorem of Guass and Lucas which says that for any polynomial $p$ the roots of $p'$ are all convex combinations of the roots of $p$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.