If the graph of $f(x)$ has $n+1 $ co-linear points, then prove that there exists $c\in(a,b)$ such that $f^{(n)}(c)=0 $ Let $n>1$ be an integer, and let $f:[a,b] \to \mathbb{R}$ be a continuous function which is $n-times$ differentiable on $(a,b)$. If the graph of $f(x)$ has $n+1 $ co-linear points, then prove that there exists $c\in(a,b)$ such that $f^{(n)}(c)=0 $
My Attempt:
If $f(x)$ has $n+1$ co linear points that means $f(x)$ is a polynomial with  $n-1$ degree, i.e, $n-1$ th derivative is constant hence $n$th derivative is $0$ at every point. How to show this using MVT?
 A: If the graph of $f$ has $n+1$ co-linear points, then it intersect a line
$$
y=rx+s
$$
at $n+1$ point. Equivalently,
$$
g(x)=f(x)-(ax+b)
$$
has $n+1$ distinct roots in $[a,b]$, say $a\le x_{0,1}<x_{0,2}<\cdots<x_{0,n+1}\le b$. Rolle's Theorem provides that $g'(x)=f'(x)-a$ has has $n$ distinct roots in $(x_{0,1},x_{0,n+1})\subset (a,b)$:
$$
a\le x_{0,1}<x_{1,1}<x_{0,2}<x_{1,2}<\cdots<x_{1,n}<x_{0,n+1}\le b.
$$
Then Rolle's Theorem again provides that $g''(x)=f''(x)$ has has $n-1$ distinct roots in $(x_{1,1},x_{1,n})\subset (a,b)$:
$$
a\le x_{1,1}<x_{2,1}<x_{1,2}<x_{2,2}<\cdots<x_{2,n-1}<x_{1,n}\le b.
$$
After $n-2$ more such applications of Rolle's Theorem we finally obtain that
$g^{(n)}=f^{(n)}$ has a root $x_{n,1}\subset (x_{n-1,1},x_{n-1,2})\subset(a,b)$.
A: Mean value theorem is the key here. If there are $n+1$ points on the graph of $f$ all of which lie on the same straight line then let the $x$ coordinates of these points be denoted by $$x_0,x_1,x_2,\dots,x_n$$ Each pair of consecutive points (this means $n$ pairs in total) leads to a chord on the graph with the same slope. By mean value theorem this gives us $n$ distinct points where the slope of tangent is also the same. Thus the derivative $f'$ takes same values at $n$ distinct points.
Applying Rolle's theorem the second derivative $f''$ vanishes at $n-1$ distinct points. And continuing in this fashion we can see that $f^{(n)} $ vanishes at some point $c\in(a, b) $.
