Given second and first derivatives at 2 points, prove that some point in between them has third derivative greater than or equal to 24. 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?
 A: 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.
A: 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 $h(\frac{1}{2})=\frac{1}{2}$.
