First Part of the problem: Let $n_1 < . . . < n_k$ be non-negative integers, let $a_1, . . . , a_k$ be positive real numbers and let $f(x) = a_1x^ {n_1} + · · · + a_kx^{n_k}$ . Suppose $0 < s < t$ and $f(s) = f(t) = 0$. Show that the polynomial
$g(x) = a_1 + a_2x^{n_2−n_1} + · · · + a_kx^{n_k−n_1} = \frac{1}{x^{n_1}}f(x)$
satisfies $g'(x) = 0$ for some $x$ between $s$ and $t$. (Hint: Use Rolle’s Theorem).
My proof: We know that polynomials are continuous and differentiable, $\therefore$ we use Rolle's Theorem, since $f(s) = f(t)$, $\exists c \in (s,t)$ with $f'(c) = 0$.
Now we take the derivative of $g(x)$ using the product rule:
$g'(x) = (\frac{1}{x^n})'f(x) + (\frac{1}{x^n})f'(x)=$
$-\frac{n}{x^{-(1+n_1)}} + (\frac{1}{x^n})f'(x)$
From Rolle's theorem we know there is an x $\in (s,t)$ such that $f'(x) = 0$ and if $n_1=0$ then $g'(x) = 0$ for that same $x$.
Second Part of the problem: Use induction on $k$ and Question 7 to show that the polynomial
$f(x) = a_1x^{n_1} + · · · + a_kx^{n_k}$
has at most $k$ positive solutions to $f(x) = 0$.
My attempt: Quite honestly I'm not sure how to start. I know I can use the base case $k = 1$ (and there is only one solution to that: $x = 0$), so I can assume for $k$ terms there are $k$ solutions. But I don't know where to go from there. Do I just keep taking derivatives and prove $k-1$? How can I use induction to prove $k+1$?
Since $g$ is everywhere continuous, it suffices to note $g(s)=g(t)=0$.