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Consider a function defined in $\mathbb{R}$ such that $f^{(3)}$ is contiuous 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 $f^ {(3)}(c) \geq 24.$

I tried to apply the Taylor polynomial at zero and $1$ . But I am getting anywhere. Someone could help me?

Thanks in advance

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  • $\begingroup$ How can you write the Taylor expansion when you possibly don't have all the derivatives? $\endgroup$ – YiFan Jul 18 at 23:51
  • $\begingroup$ There was a typo in the question . the correct is $f^{(3)}(c) \geq 24$ . Please forgive me $\endgroup$ – math student Jul 19 at 0:34
  • $\begingroup$ I’ve edited back to the very first version to match with the answers. The version before editing already has an answer here:math.stackexchange.com/questions/3297300/… $\endgroup$ – Feng Shao Jul 19 at 3:20
  • $\begingroup$ People spent time to find counter examples for your first-version question, which is a good question after all. You can ask your desired question in another post but should leave the mistakes in the first version here. $\endgroup$ – Feng Shao Jul 19 at 3:23
  • $\begingroup$ If the problem is $f^{(3)}(c) \geq 24$ instead of $f(c)\ge{24}$ then the question becomes much less easy but more interesting. $\endgroup$ – Piquito Jul 19 at 16:50
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I'm afraid you're wrong. It is not difficult to calculate a polynomial that satisfies your conditions. Take $$f(x)=6x^5-15x^4+10x^3$$ so you have $$f'(x)=30x^4-60x^3+30x^2\\f''(x)=120x^3-180x^2+60x$$ You can verify that $f(0) = f^{'}(0) = f^{''}(0)=f^{'}(1) =f^{''}(1)=0$ and that $f(1)=1$ All your condition are satisfied for all $c$ in $[0,1]$ you have $f(c)\lt24$ (the maximum is almost equal to $1$).

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  • $\begingroup$ Obviously that the third derivative is continuous. I have omitted this by distraction but I have though about it when constructing the polynomial. $\endgroup$ – Piquito Jul 19 at 0:06
  • $\begingroup$ There was a typo in the question . the correct is $f^{(3)}(c) \geq 24$ . Please forgive me $\endgroup$ – math student Jul 19 at 0:35
  • $\begingroup$ Nothing to forgive. Those typos they usually happen. Regards. $\endgroup$ – Piquito Jul 19 at 11:18
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Unfortunately wrong. $\exists f\in C^{3}[0,1]\ \forall c \in [0,1]\colon f(c) < 24.$

Here's such an $f$ as a counterexample:

Consider the polynomial $f(x) = 6x^5 -15x^4 + 10x^3 \in C^\infty(\mathbb R)$ $$ \begin{aligned} f(x) &= (6x^2 -15x + 10)x^3 &f(0)&=0 &f(1)=1\\ f'(x) &= 30(x-1)^2x^2 &f'(0)&=0 &f'(1)=0\\ f''(x) &= 120(x^2-1,5+1)x &f''(0)&=0 &f''(1)=0 \end{aligned} $$

If we assume $\exists c\in[0,1]\colon f(c)\ge24$, then there must $\exists\xi\in(0,1)$ such that $f(\xi)$ is an extremum because $f(0) < f(1) < 24$ and $f$ is continuous. Since $f$ is continuous and differentiable: $f'(\xi) = 0$ must hold. But we can simply check $$f'(x)=0 \implies x\in\{0,1\} \implies x\notin(0,1) \implies \nexists\xi\in(0,1)\colon\ f'(\xi) = 0 \implies \nexists c\in[0,1]\colon f(c)\ge24.$$

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  • $\begingroup$ Nice! Small nitpick: your statement (in the logic symbols) is technically wrong, because it doesn't specify that $f$ satisfies the conditions of the question. $\endgroup$ – YiFan Jul 18 at 23:54
  • $\begingroup$ Sorry, but I can't spot an error. Can you please explain more precisely what you think is wrong or just edit the answer? $\endgroup$ – quiliup Jul 19 at 0:22
  • $\begingroup$ There was a typo in the question . the correct is $f^{(3)}(c) \geq 24$ . Please forgive me $\endgroup$ – math student Jul 19 at 0:35

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