Given second and first derivatives at 2 points, prove that some point in between them has third derivative greater than or equal to 24.

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.

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 $h(\frac{1}{2})=\frac{1}{2}$.