Question: Suppose some real function is differentiable $n$ times on $ℝ$. Assume that there are $n+ 1$ distinct point $\{x_1, x_2, ..., x_n, x_{n+1}\}$ such that $ x_1 < x_2 < . . . < x_n < x_{n+1}$ and $f(x_i) = 0 $for all $i = 1, 2, . . . , n, n + 1.$ Prove that there exists at least one point $y$ such that $f^{(n)} (y) = 0.$
First thing I know that I probably need to use the rolles theorem due to the derivative $=0$ and since we have $f^{(n)} (y) = 0.$, I think I have to apply this theorem many times.
Definition of Rolle's theorem:
Rolle's theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval $(a, b)$ such that $f(a) = f(b)$, then $f′(c) = 0$ for some $c$ within the open interval $(a, b)$
I don't know how $f(x_i) = 0 $ comes into play? And also I don't know how to "end" the proof if I apply the Rolles theorem. How do you start? Assume that a point in between (a,b) exist and then use the extreme value theorem? I am very confused. Many thanks for the help.
Hint: Use Rolle's Theorem to conclude that there are $n$ points such that $f' = 0$. These $n$ points each lie within the intervals $(x_1,x_2), (x_2,x_3), \dots (x_n, x_{n+1}).$ Then use the same idea to conclude that there are $n-1$ points such that $f'' = 0$. Keep this process going until you have the existence of one point $y$ such that $f^{(n)}(y) = 0$.