I am currently working on Spivak. I am in need of help or a hint on this question, because every which way I try this question whether it be with Cauchy MVT or just plain MVT, I get to some promising point and end up coming back out with nothing. I've been trying at this problem since last Friday.
The problem:
Prove that if $f$ is a twice differentiable function with $f(0)=0$ and $f(1)=1$ and $f'(0) = f'(1) = 0$, then $|f''(x)|\geq4$ for some $x \in (0, 1).$ In more picturesque terms: A particle which travels a unit distance in a unit time, and starts and ends with velocity $0$, has at some time an acceleration $>4$. Hint: Prove that either $f''(x)\geq4$ for some $x\in(0,1)$, or else $f''(x)\leq4$ for some $x\in(0,1)$.
I really don't know how else to approach this. I've tried arbitrary intervals say $(0,x)$, intervals of $(0,1/2)$ and $(1/2,1)$ via Cauchy MVT and 'regular' MVT, I've tried his "hint", I've tried breaking it into cases where $f'(1/2)<1$, $f'(1/2)>1$ and $f'(1/2)=1$ with no real luck.
I'm sure there's something I'm not seeing that I should be seeing, but I figure after a week I should probably ask around, say, not at school. Any help is much appreciated.
If $\bigl(\forall x\in[0,1]\bigr):\bigl\lvert f''(x)\bigr\rvert<4$, then there a $x_0\in\left[0,\frac12\right]$ such that\begin{align}f\left(\frac12\right)&=f(0)+\frac12f'(0)+\left(\frac12\right)^2\frac{f''(x_0)}2\\&=\left(\frac12\right)^2\frac{f''(x_0)}2\\&=\frac{f''(x_0)}8\\&<\frac48\\&=\frac12\end{align}and there is a $x_1\in\left[\frac12,1\right]$ such that\begin{align}f\left(\frac12\right)&=f(1)+\left(\frac12-1\right)f'(1)+\left(\frac12-1\right)^2\frac{f''(x_1)}2\\&= 1+\frac{f''(x_1)}8\\&>1-\frac48\\&=\frac12.\end{align}