This is asking for approvation rather than asking for explainations.
Let $f$ be a polinomial. A real number $\alpha$ is a zero of $f$ with multiplicity $m$ (meaning that there are $m$ zeros for $f$.) if $f(x)=(x-\alpha)^mg(x)$ where $g(x)\neq0$
Let $r$ be the number of roots of $f$ in $[a,b]$. The proof that the k-enesim derivative of $f$ has at least $r-k$ roots in $[a,b]$ is this:
$f(x) = (x-\alpha)^mg(x)$
k-enesim derivative of $f$ is $=m-k+1(x-\alpha)^{m-k}[mg(x)+(x-\alpha)g(x)]$
Now let $mg(x)+(x-\alpha)g(x)$ be a function h(x):
k-enesim derivative of $f$ turns into $m-k+1(x-\alpha)^{m-k}h(x)$
Which is the same general form of the original $f(x)$, but now with multiplicity $m-k$.
That means also that if the k-enesim derivative of $f$ has exactly $r$ roots in $[a,b]$, the original function $f(x)$ has at least $r + k$ zeros in $[a,b]$.
Is my answer good? I bet not.
