I'm reading a section my Calculus book that is about the relation between the roots of a polynomial function and the roots of its derivative. So:

Notice that if $x_1$ and $x_2$ are roots of $f$, so that $f(x_1) = f(x_2) = 0$, then by Rolle's Theorem there is a number $x$ between $x_1$ and $x_2$ such that $f'(x) = 0$.

Ok that makes sense. Then:

This means that if $f$ has $k$ different roots $x_1 < x_2 < ... < x_k$, then $f'$ has at least least $k-1$ roots: one between $x_1$ and $x_2$, one between $x_2$ and $x_3$, etc.

That also makes sense, but what confuses me is "at least least $k-1$ roots". Why "at least"? Didn't we just show that there are exactly $k-1$ roots for the derivative, or so to say, if we have a polynomial of degree $n$, then its derivative has $n-1$ roots?

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    $\begingroup$ Consider $f(x)=x^3-x$. How many roots does $f'$ have? $\endgroup$ – Yadati Kiran Dec 20 '18 at 0:10

If $p(x)=x^2+1$, then $p(x)$ has zero roots. However, $p'(x)=2x$, so $p'(x)$ has one root $x=0$.

That argument from your Calculus textbook proves that between any two roots of the original polynomial $p(x)$ there is at least one root of $p'(x)$ between them, but there may be other roots.

  • $\begingroup$ That makes perfect sense, thank you. Is there an upper bound on how many zeroes the derivative might have? Or is that not possible to define. $\endgroup$ – Max Dec 20 '18 at 4:43
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    $\begingroup$ There is no upper bound. You can find polynomial functions with no roots such that their derivatives have an arbitrarily large number of roots. $\endgroup$ – José Carlos Santos Dec 20 '18 at 7:44
  • $\begingroup$ @Max For $n$ odd let $p(x) = \prod_{i=1}^n (x-x_i)$ with arbitrary $x_i \in \mathbb{R}$. There exists $q(x)$ such that $q'(x) = p(x)$. Since $q(x)$ is a polynomial of even degree, it has a lower bound $b > -\infty$. Then $r(x) = q(x) - b +1$ does not have zeros, but its derivative has $n$ zeros. $\endgroup$ – Paul Frost Dec 20 '18 at 13:25

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