Let $f: \mathbb{R} \rightarrow \mathbb{R}$. We say that $x \in \mathbb{R}$ is a real root of $f$ if $f(x)=0$. Show that if $f$ is differentiable and has $k$ distinct real roots, then $f^{\prime}$ has at least $k-1$ real roots. [Rolle's theorem may be used, provided you state it clearly.]

Let $p(x)=\sum_{i=1}^{n} a_{i} x^{d_{i}}$ be a polynomial in $x$, where all $a_{i} \neq 0$ and $d_{i+1}>d_{i}$. (In other words, the $a_{i}$ are the nonzero coefficients of the polynomial, arranged in order of increasing power of $x$.) The number of sign changes in the coefficients of $p$ is the number of $i$ for which $a_{i} a_{i+1}<0$. For example, the polynomial $x^{5}-x^{3}-x^{2}+1$ has 2 sign changes. Show by induction on $n$ that the number of positive real roots of $p$ is less than or equal to the number of sign changes in its coefficients.