= Solution
The <Rolle theorem> states that if a real function is continuous on $[u,v]$, differentiable on $(u,v)$, and has equal endpoint values, then its derivative vanishes somewhere in $(u,v)$. Order the $k$ distinct real roots as $r_1<\cdots<r_k$. The theorem gives a root of $f'$ in each interval $(r_j,r_{j+1})$. These intervals are disjoint, so the obtained roots are distinct:
$$
\boxed{f'\text{ has at least }k-1\text{ distinct real roots}.}
$$
We prove the requested bound for <positive roots of a polynomial> by <mathematical induction> on the number of nonzero terms. In fact, the proof also works with roots counted according to their <multiplicity>. Write $V(p)$ for the number of sign changes and $N_+(p)$ for the positive-root count.
A one-term polynomial $a_1x^{d_1}$ has no positive roots and no sign changes. For the induction step, factor out the smallest power of $x$:
$$
q(x)=x^{-d_1}p(x)
=a_1+\sum_{j=2}^na_jx^{e_j},
\qquad e_j=d_j-d_1>0.
$$
For $x>0$ this factor is nonzero, so $N_+(q)=N_+(p)$, including multiplicities, and $V(q)=V(p)$. Differentiation removes exactly the constant term:
$$
q'(x)=\sum_{j=2}^ne_ja_jx^{e_j-1}.
$$
The exponents remain strictly increasing, and the positive multipliers $e_j$ do not change coefficient signs. The induction hypothesis therefore gives
$$
N_+(q')\le V(q'),\qquad
V(q')=V(a_2,\ldots,a_n).
$$
First suppose $a_1a_2<0$. Removing $a_1$ removes exactly one sign change, so $V(q)=V(q')+1$. If $q$ has positive roots $r_1<\cdots<r_k$ of multiplicities $m_1,\ldots,m_k$, its derivative has multiplicity $m_j-1$ at $r_j$ and, by the <Rolle theorem>, an additional root between each pair. Thus the <Rolle root count with multiplicities> gives
$$
N_+(q')\ge\sum_{j=1}^k(m_j-1)+(k-1)=N_+(q)-1
$$
when there is a positive root. Consequently
$$
N_+(q)\le N_+(q')+1\le V(q')+1=V(q).
$$
If there is no positive root, the desired bound is automatic.
Now suppose $a_1a_2>0$, so $V(q)=V(q')$. Again the zero-root case is immediate. Otherwise, multiplying $q$ by $-1$ if necessary lets us assume $a_1>0$ and $a_2>0$ without changing roots or sign changes. For sufficiently small $x>0$,
$$
q(x)>0,\qquad q'(x)>0,
$$
since $q(0)=a_1$ and the lowest-power term of $q'$ has positive coefficient. Choose such a point $u$ before the first positive root $r_1$. The <mean value theorem> on $[u,r_1]$ gives a point $v\in(u,r_1)$ with
$$
q'(v)=\frac{q(r_1)-q(u)}{r_1-u}<0.
$$
But $q'(u)>0$. The <intermediate value theorem> applied to the polynomial $q'$ therefore gives an extra root in $(u,v)$, before $r_1$. It is distinct from the roots at $r_j$ and those between consecutive roots. Hence
$$
N_+(q')\ge N_+(q),
\qquad
N_+(q)\le N_+(q')\le V(q')=V(q).
$$
Both sign cases complete the induction:
$$
\boxed{N_+(p)\le V(p).}
$$
\b[An initial sign change allows one extra positive root; without that sign change, the derivative must already turn before the first positive root.] This is the upper-bound part of the <Descartes' rule of signs>. The multiplicity claim used above follows by differentiating $q(x)=(x-r)^mh(x)$ with $h(r)\ne0$: the derivative has a factor $(x-r)^{m-1}$ and its remaining factor is nonzero at $r$.
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