Let $f:[0,1] \to \mathbb{R}$ be a function whose third derivative $f'''(x)$ is continuous on $[0,1]$. Suppose that $$ f(0)=f'(0)=f''(0)=f'(1)=f''(1)=0 \text{ and } f(1)=1$$ Then there exist $c\in(0,1)$ such that $f'''(c)\geq 24$
I am trying to prove the above statement via Taylor's theorem, and see if I can make use of the Darboux's theorem, but I don't know how to come up with a 24 in the last sentence of the statement. Any kind of hints are much appreciated, thank you.
I was just playing around with Taylor's theorem, and I came up with something that seems right to me. Using Taylor's theorem, we can say that there exists $c_1\in (0,\frac{1}{2})$ such that $$f\left(\frac{1}{2}\right) = f(0) + f'(0)\left(\frac{1}{2}-0\right)+ \frac{f''(0)}{2!}\left(\frac{1}{2}-0\right)^2 + \frac{f'''(c_1)}{3!}\left(\frac{1}{2} - 0\right)^3$$ which is just $$f\left(\frac{1}{2}\right) = \frac{f'''(c_1)}{48}$$ Another application of Taylor's theorem guarantees the existence of $c_2 \in(\frac{1}{2},1)$ such that $$f\left(\frac{1}{2}\right) = f(1) + f'(1)\left(\frac{1}{2}-1\right) + \frac{f''(1)}{2!}\left(\frac{1}{2}-1\right)^2 + \frac{f'''(c_2)}{3!}\left(\frac{1}{2}-1\right)^3$$ which is just $$f\left(\frac{1}{2}\right) = 1 - \frac{f'''(c_2)}{48}$$ So we have $$f'''(c_1) + f'''(c_2) = 48 \text{ where }c_1\in \left(0,\frac{1}{2}\right) \text{ and }c_2\in \left(\frac{1}{2},1\right)$$
Now, we claim that one of $f'''(c_1)$ and $f'''(c_2)$ must exceed or equal $24$. We give a proof by contradiction.
Suppose, if possible, that $f'''(c_1) < 24$ and $f'''(c_2)<24$. This would mean that $f'''(c_1)+f'''(c_2) < 48$ which is a contradiction. Hence, we have at least one of $f'''(c_1)\geq 24$ and $f'''(c_2)\geq 24$.
This shows the existence of $c\in (0,1)$ such that $f'''(c)\geq 24$.
