# Relation between roots of function and roots of its derivative

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?

• Consider $f(x)=x^3-x$. How many roots does $f'$ have? – 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.
• @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. – Paul Frost Dec 20 '18 at 13:25