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I am doing an exercise and came to a point where I'd need to know this:

As a consequence of Rolle's theorem, the derivative of a function has a zero whenever our function has more than one zero. But can we say that if all the roots of a polynomial, $p(n)$, are real, all the roots of p'(n) are real?

3 Answers 3

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Yes. Suppose we have a polynomial $p(x)$ of degree $d$, with all roots real (meaning it has $d$ real roots counting multiplicity). If $p(x)$ has a root of multiplicity $m$ at some point $x_0$, then p'(x) has a root of multiplicity $m-1$ at $x_0$, as can be seen by applying product rule: \frac{d}{dx}(x-r)^mq(x)=(x-r)^{m-1}(mq(x)+(x-r)q'(x)) Also, between any pair of distinct roots of $p(x)$ there must be a root of p'(x) by Rolle's theorem. Thus the number of real roots of p'(x), counting multiplicity, is the number of distinct roots of $p(x)$ minus 1, plus $m-1$ roots for each repeated root of multiplicity $m$, giving a total of $d-1$ (since the number of distinct roots of $p(x)$ plus $m-1$ roots for each repeated root of multiplicity $m$ is the total number of roots of $p(x)$, which is $d$). This is the total number of roots of p'(x), so all its roots must be real.

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If all zeros of $p$ lie in the interval $[a,b]$, then all zeros of p' lie in $[a,b]$ by the Gauss–Lucas theorem. More generally, each zero of p' lies in the convex hull of the set of zeros of $p$.

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    (Of course $p$ should be nonconstant.)2011-12-29
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HINT $\rm\ \qquad \ f\ =\ (x-r_1)^{n_1}\:\cdots\:(x-r_k)^{n_k},\qquad\qquad \rm r_1 < r_2 <\cdots < r_k \in \mathbb R$

\rm\qquad\quad \Rightarrow\: \qquad f\:\:' =\: (x-r_1)^{n_1-1}\:\cdots\ (x-r_k)^{n_k-1}\: g\quad for\ some\ g\in \mathbb R[x]

\rm\qquad\quad \Rightarrow\ \ \ \#roots\ f\:\:' \ge\: (n_1\!\!-1)+\cdots+(n_k\!\!-1)\ +\ k\!-\!1\ \ \ by\ f\:\:'\ has\ root\ in\ (r_i,r_{i+1})\ by\ Rolle

\rm\phantom{\rm\ \qquad\quad \Rightarrow\ \ \#roots\ f'} \ge\: deg\ f - 1

\rm But\ also\:\:\ \ \#roots\ f\:\:' \le deg\ f - 1\ =\ deg\ f\:\:'\ since $\rm\:\mathbb R\:$ is a domain.