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Let $f \colon \mathbb{R} \to \mathbb{R}$ be an infinitely differentiable function and suppose that for some $ n ≥ 1$ - $f(1)=f(0)=f'(0)=f''(0)=f'''(0)=\dots=f^{(n)}(0)=0$,

I have to prove that there exists $x \in (0,1)$ such that $f^{(n+1)}(x)=0$.

Here is what I did:

Since $f(1)=f(0)$ Mean-value theorem says that there exists at least one $x \in (0,1)$, say $x_1$, such that $f^{(1)}(x_1)=0$

With that $x_1$ if we again apply MVT , on $f^{(1)}(x_1)=f'(0)=0$ we can say that there exists atleast one $x \in (0,1) \,\,\text{say } x_2$ such that $f^{(2)}(x_2)=0$

Repeating the above process, we will get a $x_n \in (0,1)$ such that $f^{(n+1)}(x_n)=0$.

Here is the catch. In the question mentioned above, it says that $f$ is only differentiable, and I assumed it to be continuous in order to apply MVT.

I cannot think of any other way to prove this.

DRPR
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    Your proof is good. Differentiable functions are continuous. They said that $f$ was differentiable, with no point specified. That language is understood as differentiable at every point. –  Apr 27 '18 at 21:05
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    You should specify that $x_2\in (0,x_1)$, not $(0,1)$. –  Apr 27 '18 at 21:07
  • I forgot the fact that differentiable functions were contionuous. I got confused because I had recently asked this question on MSE https://math.stackexchange.com/questions/2702443/example-of-continuous-non-differentiable-function – DRPR Apr 27 '18 at 21:28

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