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Consider a function defined in $\mathbb{R}$ such that $f^{(3)}$ is continuous in $[0,1].$ Suppose that $f(0)=f′(0)=f′′(0)=f′(1)=f′′(1)=0$ and $f(1)=1$. Prove that there exists $c \in [0,1]$ such that $f^{(3)}(c)\geq 24$.

The question was originally posted here Inequality related to Taylor polynomial

But there was some mistakes in the statement. Thus maybe it is better to write as an new question

Thanks in advance

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  • $\begingroup$ Err.... how is that not the exact same statement? You edited there, and then posted it back here? $\endgroup$ – Clement C. Jul 19 at 1:08
  • $\begingroup$ Possible duplicate of Inequality related to Taylor polynomial $\endgroup$ – Clement C. Jul 19 at 1:10
  • $\begingroup$ (What I mean is -- either you don't edit the statement there (as it invalidated the current answers) and post a new question with the corrected statement; or you edit it there (as you just did) and don't post it again. Doing what you have done (edit there but still repost) is bad on all aspects. $\endgroup$ – Clement C. Jul 19 at 1:25
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    $\begingroup$ @ClementC. I agree with you. I have edited the other post to match the answers there, waiting for the OP to review my edit. $\endgroup$ – Feng Shao Jul 19 at 1:34
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    $\begingroup$ @FengShao I suggest trying the edit again but this time in the comment link to this question and summarize the reasons given above. I regret that I did not have this information when I reviewed the edit, because I assumed the OP was trying to correct the question without also reposting it, which also would have been a reasonable course of action at the time. (But I think the door has since been shut on that.) $\endgroup$ – David K Jul 19 at 3:09
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Use Taylor’s formula to obtain $$f(\frac12)=f(0)+\frac12f’(0)+\frac{f’’(0)}{2!}(\frac12)^2+\frac{f’’’(\xi_1)}{3!}(\frac12)^3,$$ $$f(\frac12)=f(1)-\frac12f’(1)+\frac{f’’(1)}{2!}(\frac12)^2-\frac{f’’’(\xi_2)}{3!}(\frac12)^3,$$ where $\xi_1\in[0,1/2]$ and $\xi_2\in[1/2,1]$. Doing the subtraction we get $$f’’’(\xi_1)+f’’’(\xi_2)=48,$$ so one of them will be no less than $24$.

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