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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?

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    $\begingroup$ 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
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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 $p$.

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