Proving that if $|f''(x)| \le A$ then $|f'(x)| \le A/2$ 
Suppose that $f(x)$ is differentiable on $[0,1]$ and $f(0) = f(1) = 0$. It is also
  known that $|f''(x)| \le A$ for every $x \in (0,1)$. Prove that
  $|f'(x)| \le A/2$ for every $x \in [0,1]$.

I'll explain what I did so far. First using Rolle's theorem, there is some point $c \in [0,1]$ so $f'(c) = 0$.
EDIT: My first preliminary solution was wrong so I tried something else.
EDIT2: Another revision :\
I define a Taylor series of a second order around the point $1$:
$$ f(x) = f(1) + f'(1)(x-1) + \frac12 f''(d_1)(x-1)^2 $$
$$ f(0) = f(1) + f'(1)(-1) + \frac12 f''(d_1)(-1)^2 $$
$$ |f'(1)| = \frac12 |f''(d_1)| <= \frac12 A $$
Now I develop a Taylor series of a first order for $f'(x)$ around $1$:
$$ f'(x) = f'(1) + f''(d_2)(x-1) $$
$$ |f'(x)| = |f'(1)| + x*|f''(d_2)|-|f''(d_2)| \leq \frac{A}{2} + A - A = \frac{A}{2} $$
It looks correct to me, what do you guys think?
Note: I cannot use integrals, because we have not covered them yet.

If $f''$ exists on $[0,1]$ and $|f''(x)| \le A$, then $f'$ is Lipschitz continuous, which implies that $f'$ is absolutely continuous. So $f'(a)-f'(b)=\int_{a}^{b}f''(t)\,dt$ for any $0 \le a, b \le 1$. And, of course, the same is true of $f$ because $f$ is continuously differentiable.
Because $f(0)=0$, one has the following for $0 \le x \le 1$:
$$
     f(x) = \int_{0}^{x}f'(t)\,dt = tf'(t)|_{0}^{x}-\int_{0}^{x}tf''(t)\,dt=xf'(x)-\int_{0}^{x}tf''(t)\,dt.
$$
Similarly, because $f(1)=0$, one has the following for $0 \le x \le 1$:
$$
     f(x)=\int_{1}^{x}f'(t)\,dt = (t-1)f'(t)|_{1}^{x}-\int_{1}^{x}(t-1)f''(t)\,dt=(x-1)f'(x)-\int_{1}^{x}(t-1)f''(t)\,dt.
$$
Subtracting the second equation from the first gives
$$
    0=f'(x)-\int_{0}^{x}tf''(t)\,dt+\int_{1}^{x}(t-1)f''(t)\,dt.
$$
Therefore,
$$
            |f'(x)| \le A\left(\int_{0}^{x}t\,dt + \int_{x}^{1}(1-t)\,dt\right)=\frac{A}{2}(x^{2}+(1-x)^{2}).
$$
The maximum of the expression on the right occurs at $x=0$ and $x=1$, with a value of $A/2$. Therefore, $|f'(x)| \le A/2$.
