Let $f:[0,1]\to \mathbb R$ be a function such that its third derivative $f'''(x)$ is continuous on $[0,1]$. Suppose that $f(0)=f'(0)=f''(0)=f'(1)=f''(1)=0$ and $f(1)=1$, show that there exists a point $c\in (0,1)$ such that $f'''(c)\ge 24$.
When I apply Taylor's theorem to $f$, I can only determine that there is a $c\in (0,1)$ such that $f'''(c)=6$. I attempted another approach using $f({1\over 2})$ and Taylor's theorem: $f({1\over 2})=f(0)+{f'''(c_1)\over 6}{1\over 2}$ for some $c_1\in(0,{1\over 2})$and $f({1\over 2})=f(1)-{f'''(c_2)\over 6}(-{1\over 2})$. By some calculation, if $f({1\over 2})\le -1$ or $f({1\over 2})\ge 2$, then one of $f'''(c_1)$ and $f'''(c_2)$ is greater than or equal to 24? But what if $-1<f({1\over 2})<2$?