Show that f disappears at n points Let $a, b \in \mathbb{R}$ with $a<b$ and $f:[a,b] \rightarrow \mathbb{R}$,  and that there's some point $x \in [a,b]$ with $f(x)$ nonzero.  If there exists $n \in \mathbb{N}$ such that for all $k\leq n, \int_{a}^{b} t^kf(t)dt = 0$.  I need to show that there are $n+1$ distinct points where f vanishes and changes sign.  I've proved the case for $n=0$.  For $n=1$, I was trying the following:
From the n=0 case, we know that $f$ must vanish at at least one point $c \in [a, b]$.  Assume that $c$ is the only such point where $f$ vanishes.  Then $\int_{a}^{c} f(t)dt = -\int_{c}^{b} f(t)dt$.
Since $t\rightarrow t$ is strictly increasing, this implies that $|\int_{a}^{c} tf(t)dt| \neq |\int_{c}^{b} tf(t)dt|$, and therefore there must be some other point $d$ where $f$ vanishes such that 
$\int_{a}^{d} tf(t)dt = -\int_{d}^{b} tf(t)dt$
Does this method work?  Could I just repeat that process to generalize for all $n \in \mathbb{N}$?
 A: From the problem statement, we believe that $f$ should be continuous for the question to make sense.  Furthermore, $f$ has to be non-zero (otherwise $f$ vanishes everywhere but changes signs nowhere).  
Let $t\in (a,b)$ and suppose that $g:[a,b]\to\Bbb R$ is a non-zero continuous function on $[a,b]$ s.t. $g(t)=0$.  Set $$t_-=\sup\Big(\{a\}\cup\big\{s\in[a,t):g(s)\ne0\big\}\Big)$$ and $$t_+=\inf\Big(\{s\in(t,b]:g(s)\ne 0\big\}\cup\{b\}\Big).$$  We say that $g$ changes signs at $t$ if there exists $\epsilon>0$ such that $$a\le t_--\epsilon<t_-\le t\le t_+<t_++\epsilon\le b$$ and $$g(s_-)g(s_+)< 0$$ for all $s_-\in (t_--\epsilon,t_-)$ and $s_+\in(t_+,t_++\epsilon)$.
Suppose on the contrary that there are exactly $m\leq n$ distinct values $x_1,x_2,\ldots,x_m$ in $[a,b]$ s.t. for $i=1,2,\ldots,m$, $f(x_i)=0$ and $f(x)$ changes signs at $x=x_i$.  Then take $$p(x)=(x-x_1)(x-x_2)\cdots (x-x_m)$$ and define
$$F(x)=p(x)f(x).$$
Note that either $F(x)\ge 0$ for all $x\in[a,b]$, or $F(x)\le 0$ for all $x\in [a,b]$, and $F$ is a non-zero continuous function.  Wlog, suppose that $F(x)\ge 0$ for all $x\in[a,b]$.  Therefore $$\sum_{k=0}^mp_k\int_a^b x^k f(x)dx=\int_a^b F(x)dx>0,$$
if $p(x)=\sum_{k=0}^m p_kx^k$. But this is a contradiction as $\int_a^b x^k f(x)dx=0$ for all $k=0,1,2,\ldots,m$ (recalling that $m\le n$).
