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Let $f$ be a function that is $C^3$ on an open interval containing $[0,1]$ - that is, the third derivative $f'''$ exists and is continuous on an open interval containing $[0,1]$. Assume that $f(0) = f'(0) = f''(0) = 0$ and that $f'(1) = f''(1) = 0$. If $f(1) = 1$, prove that there is some $c\in\left(0,1\right)$ such that $f'''(c) \geq 24$.

Not sure how to approach this problem, been stuck on it for a while. Any help would be nice.

I tried using the Mean Value Theorem, and got that at some point $c_1\in\left[0,1\right], f'(c_1) = \frac{f(1)-f(0)}{1-0}=1$ and there is some point $c_2 \in [c_1,1]$ such that $f''(c_2) = \frac{f'(1)-f'(c_1)}{1-c_1}=\frac{-1}{1-c_1}<-1$ and there is some point $c_3\in[c_2,1]$ such that $f'''(c_3) = \frac{f''(1)-f''(c_2)}{1-c_2}>1$. I'm not sure how to extend this to $24$ though, or even if this method will work.

Is there a way to use Taylor Polynomials perhaps?

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  • $\begingroup$ What have you tried? $\endgroup$ – saulspatz Nov 24 '18 at 20:45
  • $\begingroup$ I've tried using the mean value theorem but couldn't come up with anything. $\endgroup$ – You Zhou Nov 24 '18 at 20:55
  • $\begingroup$ I don't know how to do it off the top of my head, but I suspect that Taylor polynomials come into it. I advise you to edit the question to explain what you have tried. The question may well be closed if you don't give any more context than you have. Many people will vote to close without reading the comments, so please edit the question body. $\endgroup$ – saulspatz Nov 24 '18 at 21:04
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    $\begingroup$ The function $f(x)=6x^2-8x^3+3x^4$ satisfies the conditions. Note that $f'''(x)=-48+72x$, so there is no $c \in (0,1)$ where $f$ has large enough third derivative, but $f'''(1)=24$. That is to say, I think there's a typo somewhere and $c$ should be allowed to be an endpoint. $\endgroup$ – Micah Nov 24 '18 at 22:20
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    $\begingroup$ @Micah If $g(x)=6 x^2 - 8 x^3 + 3 x^4$ then $g''(x)=12 - 48x + 36x^2$ and $g''(0)=12\neq0,$ so this does not satisfy all the conditions. $\endgroup$ – David K Nov 25 '18 at 20:12
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I think the idea is that if $f'''(x) < 24$ when $0 < x < 1,$ then the conditions $f'(0)=f''(0)=0$ imply that $f'(x) < 12x^2$ when $0 < x \leq \frac12,$ the conditions $f'(1)=f''(1)=0$ imply that $f'(x) < 12(x-1)^2$ when $\frac12 \leq x < 1,$ and together these imply that $f(1) - f(0) < 1.$

Note that in order for $f'(x)$ to get close to these limits when $x \approx\frac12,$ we need $f''(x)$ to change quickly from something near $12$ to something near $-12,$ which requires $f'''(x)$ to be much less than $-24.$ If the conclusion were of the form $\lvert f'''(x)\rvert \geq L$ then I think we could set $L = 32,$ though this seems a bit harder to prove.

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  • $\begingroup$ How on Earth this answer did not receive any upvote yet? +1 $\endgroup$ – Oldboy Nov 26 '18 at 21:48
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Here's a proof.

We will prove it by contradiction. Let us assume that $f(x)$ satisfies the conditions, and that $f'''(x) < 24$ for $0 < x < 1$.

First, we show that we can assume that the function is symmetric around the point $(\frac{1}{2},\frac{1}{2})$.

We first note that if a function $f$ satisfies the conditions (including $f'''(x) < 24$), then $g(x) = 1 - f(1-x)$ also does. Why is this true? Because when you reflect horizontally — taking $f(x)$ to $f(1-x)$ — you negate all the odd derivatives. And when you reflect vertically — taking $f(1-x)$ to $1-f(1-x)$ — you negate all the derivatives. So $g'''(x) = f'''(1-x) < 24$ as well.

Now, consider the function $h(x) = \frac{f(x) + g(x)}{2}$. This is symmetric around the point $(\frac{1}{2},\frac{1}{2})$, and also satisfies the conditions. So $h(x)$ is also a counterexample, and $h(\frac{1}{2})=\frac{1}{2}$.

Consider the function $p(x) = 4x^3,\ 0 \leq x \leq \frac{1}{2}$.

This has $p(\frac{1}{2})=\frac{1}{2}$, the third derivative $p'''$ is 24 for $0 \leq x \leq \frac{1}{2}$, and $p(0)=p'(0) = p''(0)$. So if $h'''(x) < 24$ for $0 \leq x \leq \frac{1}{2}$, then $h(x) < p(x)$ for $0 < x \leq \frac{1}{2}$. This contradicts $h(\frac{1}{2})=\frac{1}{2}$.

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