<s>[INST] Prove that there is $x \in [0,1]$ such that $|f''(x)| > 4$ 
Assume $f$ has a continuous second derivative with $f(0) = f'(0) = f'(1) = 0$ and $f(1) = 1$. Prove that there is $x \in [0,1]$ such that $|f''(x)| > 4$. 

We must also have that $f'$ is continuous by differentiability. Therefore, by Rolles theorem there exists a $c$ in $[0,1]$ such that $f''(c) = 0$. I am not sure how to use the fact that $f(1) = 1$ to show that $|f''(x)| > 4$.
 [/INST] $$f(\frac{1}{2}) = f(0) + f'(0)(\frac{1}{2}) + \frac{f''(\theta_1)}{2}(\frac{1}{2})^2 = f''(\theta_1)\frac{1}{8}$$
$$f(\frac{1}{2}) = f(1) - f'(1)(\frac{1}{2}) + \frac{f''(\theta_2)}{2}(\frac{1}{2})^2 = 1 + f''(\theta_2)\frac{1}{8}$$
Here, $0\le \theta_1 \le 1/2$, and $1/2 \le \theta_2 \le 1$
Combining both, we have
$$ 1 = f''(\theta_1)\frac{1}{8} - f''(\theta_2)\frac{1}{8}$$
Take absolute value,
$$ 1 \le max \{|f''(\theta_1)|, |f''(\theta_2)|\} \frac{1}{4}$$
So c is one of $\theta_1, \theta_2$.
Let 
$$f(x) = \begin{cases} 2x^2\, \text{ for } 0 \le x \le 1/2\\ -2(x-1)^2 +1\, \text{ for } 1/2 < x \le 1 \end{cases}$$
$$f''(x) = 4$$, so the equality holds for this f.
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