Suppose $f$ is a twice-differentiable function with $f(0) = 0$, $f\left(\frac12\right) = \frac12$ and $f'(0) = 0$. Prove that $|f''(x)| \ge 4$ for some $x \in \left[0,\frac12\right]$.
I've been stuck on this question for a while now without any idea on how to get started. Is it possible for someone to help me?
Thanks!
Use Taylor's theorem to get $$f\left(\frac{1}{2}\right)=f(0)+\frac{1}{2}\cdot f'(0)+\frac{1}{8}\cdot f''(c)$$ for some $c\in(0,1/2)$ and your answer follows.
Note that the result can not be obtained by mean value theorem directly, but rather one should observe that proofs of mean value theorem and Taylor's theorem depend on Rolle's theorem.
[Paramanand's answer is definitely the way to go. I just want to share another elementary approach in the $f''$-is-continuous case.]
If $f$ is twice continuously differentiable then by the (second) Fundamental theorem of Calculus we can write $$ f(x)=0+\int_0^xf'(y)\, dy=\int_0^x\left(0+\int_0^yf''(z)\, dz\right)dy. $$ Plugging in $x=1/2$ and assuming that $|f''(z)|<4$ for $z\in[0,1/2]$ we arrive at the contradiction $$ \frac{1}{2}=\int_0^{1/2}\int_0^yf''(z)\, dz\, dy<\int_0^{1/2}\int_0^y4\, dz\, dy=\int_0^{1/2}4y\, dy=\frac{1}{2}. $$