Question on functions and derivatives I can't seem to get this subject very well. 
Let $f(x)$ be twice differentiable on $[0,1]$, and that there is a constant $A$ so that $|f''(x)|\le A$. Show that if $f(0)=f(1)=0$, then $|f'(x)|\le {A\over2}$ for all $x\in[0,1]$.
Thanks in advance for any help. Would prefer hints please for my learning. 
Thanks!
 A: Here's a hint towards a proof by contradiction. Suppose that there exists $w\in[0,1]$ such that $f'(w) > A/2$. Then the function $f'(x)$ cannot take values very much less than $A/2$ when $x$ is near $w$ - because the derivative of $f'$, namely $f''$, cannot be too large. Can you use the Mean Value Theorem to derive a lower bound on $f'(x)$ that depends on the distance $|x-w|$?
Then notice that we're supposed to have $0 = f(1)-f(0) = \int_0^1 f'(x) \, dx$. A sufficiently large lower bound for $f'(x)$ would contradict this....
A: for $x \in (0,1]$ : 
use taylors thrm:
$$f(0) = f(x-x ) =f(x) -xf^\prime(x) + \frac{x^2}{2}f^{\prime\prime}(x - h_1x).....(1)$$ 
put $x=1$ in $(1) \implies |f^\prime (1)| \leq \frac{A}{2}$ 
for $x \in[0,1)$ use taylors thrm.
$$f(1) =f(x+(1-x)) = f(x) + (1-x)f^\prime(x) +\frac{(1-x)^2}{2}f^{\prime\prime}(x  +h_2(1-x)) .....(2)$$
put $x=0$ in $(2)\implies |f^\prime (0)| \leq \frac{A}{2}$
now $$(2) - (1) \implies f^\prime(x) = \frac{1}{2}(x^2f^{\prime\prime}(x - h_1x) - f^{\prime\prime}(x  +h_2(1-x))(1-x)^2)$$ 
$$|f^\prime(x)| \leq \frac{A}{2}(2x^2 -2x+1) < \frac{A}{2} \forall x \in(0,1)$$
A: Hint  :By Rolle's theorem,  $\exists c\in[0,1] $ with $f'(c)=0$
Using the Mean value theorem on $f'$ show that this, and  the fact $|f''(x)|\leq A $ implies $|f'(x)|\leq\frac{A}{2}$
A: This is not a proof but it is a step in the correct direction.
In order to have the maximum $|f'(x)|$ at some x value you should keep $|f''(x)|$ at the maximum value for the entire range. Therefore $|f''(x)| = A$ and for sake of simplicity we will say $f''(x) = -A$.
Therefore:$f'(x) = C_1-Ax$ for some value $C_1$.
This means:$$f(x) = \int(C_1-Ax)dx = C_2+C_1x-Ax^2/2$$
$$f(0) = C_2+0=0; C_2=0$$
$$f(1) = C_2+C_1-A/2=0; C_1=A/2$$
$$f(x) = Ax/2-Ax^2/2$$
$$f'(x) = A/2-Ax$$
And you should be able to see that the $|f'(x)|$ has maxima at 0 and 1 and both have the value of A/2. This is not a true proof because I have not shown that you cannot arrive at a greater $|f'(x)|$ by using a more complex function of $f''(x)$.
