# A function is differentiable $n$ times. Assume there are $n+1$ distinct points. Prove that $\exists$ one point $y$ such that $f^{(n)}(y)=0$

So I'm stuck on this question, I have an idea on the question but I missed the lecture which it pertained to. So I'm unsure of the theory behind it so, it'd be appreciated if someone could help me out! :)

Question

Suppose the function $f : \mathbb{R}\rightarrow\mathbb{R}$ is $n$ times differentiable on $\mathbb{R}$. Assume there are $n+1$ distinct points {$x_1, x_2,...,x_n,x_{n+1}$} 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 $f^{(n)}(y)=0$.

Note

Unfortunately I don't really have an attempt as I've been sitting on it for 2 hours now unsure where to even start, because as I said I missed the lecture, however I have come to the conclusion that it could possibly involve doing Rolle's theorem multiple times but I don't really know how to actually apply it, etc. Anyway, ANY help would GREATLY be appreciated! :)

• I feel like this is a duplicate, but I can't find something it would be a dupe of. May 1 '17 at 23:46
• See also this. Actually I think the answers here are better. But this is a standard application of Rolle's theorem. No upvotes from me :-( May 2 '17 at 4:05

Yes, you are going right way, applying Rolle's Theorem to the interval $(x_1,x_{n+1})$ we get that $f^{(1)}(x)$ must have at least $n$ roots in the interval $(x_1,x_{n+1})$

Now again apply Rolle's Theorem for the function $f^{(1)}(x)$ from which you get that $f^{(2)}(x)$ must have at least $n-1$ roots in the interval $(x_1,x_{n+1})$ .

Doing so you'll get that $f^{(i)}(x)$ must have at least $n+1-i$ roots in the interval $(x_1,x_{n+1})$ .

Now for $i=n$ you'll have at least $n+1-n$ root, I.e. at least $1$ root in the given interval .

For applying Rolle's Theorem you need :

For interval $[a,b]$ if $f$ is continuous and differentiable, and $f(a) =f(b)$ then $f'(x)$ must be zero for at least on $x$ in the interval $[a,b]$

• How does applying Rolles Theorem show that there at least n roots in the interval?
– user431606
May 3 '17 at 14:46
• @PixelRain Apply Rolle's Theorem n times for the interval $(x_i,x_{i+1})$ You'll get the result that $f'(x)$ must be zero at least one time in each of these interval.where $i \in \{1,2,\ldots ,n\}$ May 4 '17 at 1:29

Rolle's theorem states that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ with $f(a)=f(b)$, then there is a point $c$ in $(a,b)$ such that $f^{\prime}(c)=0$. In particular, this is the case if $f(a)=f(b)=0$.

In your problem, you can apply Rolle's theorem on each interval $[x_i,x_{i+1}]$ for $i=1,2,\dots,n$. This will yield points $y_1<\dots<y_n$ for which $f^{\prime}(y_i)=0$.

Then simply repeat the process by applying Rolle's theorem to $f^{\prime}$, and so on.

By Rolle's Theorem, there exist $c_i$ such that $x_i < c_i < x_{i+1}$ and $f'(c_i) = 0$. Now apply Rolle's Theorem in this way to the $c_i$'s. Keep doing this.