TITLE: The relation between a polynomial's multiplicity and that of its derivative.
QUESTION [5 upvotes]: Say a polynomial $p(x)$ has $n$ real roots with a multiplicity of $k$. 
It can be shown that $p'(x)$ has a multiplicity of $k-1$. 
$$ p(x) = (x-a)^k\cdot h(x) $$
$$ p'(x) = k\cdot (x-a)^{k-1}\cdot h(x) + (x-a)^k\cdot h'(x)$$
$$ p'(x) = (x-a)^{k-1}(k\cdot h(x) + (x-a)\cdot h'(x))$$
Is it true in reverse? If I show that the derivative of a function $f(x)$ has a multiplicity of $k-1$, does this mean that $f(x)$ having multiplicity of $k$ is true?

REPLY [5 votes]: If $f$ has a zero there at all, then yes. Otherwise no. Consider for example $f(x)=x^k+1$ at $x=0$: the derivative has a zero of order $k-1$ but the function itself has no zero at all.