Prove that $|f'(x)| \le \frac{A}2 \forall x \in [0,1] $ 
Let $f$ be twice differentiable in $[0,1]$
$f(0) = f(1) = 0$,
$|f''(x)|\le A$.
Prove that $|f'(x)| \le  \frac{A}2, \forall x \in [0,1] $.

Well this is what I came up with,
$f'(c_1) = \dfrac{f(x) -f(0)}{x-0} = \dfrac{f(x)}{x}$
$f'(c_2) = \dfrac{f(x) -f(1)}{x-1} = \dfrac{f(x)}{x-1}$
for $0\lt c_1,c_2 \lt 1$
from rolle's theorem we know that there is : $0\lt c_3 \lt 1$
such that
$f'(c_3) =0$
 A: Proof by contradiction. Suppose there exists a point $x^*$ such that $ f'(x^*) > \frac{A}{2}$. (Note: WLOG, the value is positive.)
From FTC, we get that $ \int _0^1 f'(x) \, dx  = f(1) - f(0) = 0$.
Apply the Mean Value Theorem to $f$ on $[0,1]$, then there exists some $c \in [0,1]$ such that $f'(c) = 0 $.
Apply the Mean Value Theorem to $f'$ for the interval between $x^*$ and $c$, since $|f''(x)| \leq A$, we get that $|x^* - c | > \frac{1}{2}$. In particular, if we have 2 values of $c$ such that $ f' (c_1) = f'(c_2) = 0 $ and $ c_1 \leq \frac{1}{2} \leq c_2$, then this leads to a contradiction. 
WLOG, $c \leq \frac{1}{2} \leq x^*$. This must hold for all values of $c$ such that $ f'(c) = 0 $. Let $c^*$ be the largest possible value of $c$ such that $f'(c) = 0 $. (Note:$f'(c^*) = 0$ by continuity of $f'$.) This means that for $ x > c^*$, $f'(x) > 0 $.
Since $|f''(x) | < A,$ this tells us that $f'(x)$ cannot decrease too quickly, especially around $ x^*$. Observe that $ \int_{c^*} ^ {1} f'(x) \, dx \geq \int_{x^*- \frac{1}{2}}^{x^*} f'(x) \, dx > \frac{A}{8}.$ (Draw a triangle from the point $\left( x^*, f'(x^*)\right)$.
Similarly, $f'(x)$ cannot increase too quickly, especially around $c^*$. Observe that $\left| \int_0^{c^*} f'(x) \, dx \right| \leq \int_0^{c^*} | f'(x)| \, dx < \frac{A}{8}$. (Draw a triangle from the point $\left(c^*, f'(c^*) = 0 \right)$.
This contradicts the initial fact that $ \int_0^1 f'(x) \, dx = 0 $, since the (potentially) negative portion cannot cancel out the (guaranteed) positive portion.
A: I manage an answer using both time a taylor expansion center at a point $x \in \left [0, 1 \right ]$. I get 
$$
f(h) = f(x)+f'(x)(h-x) + \frac{1}{2} f''(\xi(h))(h-x)^2
$$
now for h = 0 and h = 1, I get 
$$0=f(x)-xf'(x)+\frac{1}{2}x^2f''(\xi(0))$$
and
$$0=f(x)+f'(x)-xf'(x)+\frac{1}{2}(1-x)^2f''(\xi(1)).$$
Subtracting one to the other to get $f(x)$ out and a little manipulation yield:
$$|f'(x)| = \frac{1}{2}|x^2f''(\xi(0))-(1-x)^2f''(\xi(1))|\leq\frac{1}{2}(|x^2f''(\xi(0))|+|(1-x)^2f''(\xi(1))|)\leq \frac{A}{2}(|x^2|+|1-x|^2) $$
since $x \in \left [0, 1 \right ], (|x^2|+|1-x|^2) \leq1 $ and we get the result.
