# Show $|f'(x)| \leq (A/2)$ if $|f''(x)|\leq A$

Problem: Given that $f$ is differentiable at $[0,1]$ and $f(0)=f(1)=0$. If $\forall x\in (0,1)$ $|f''(x)|\leq A$ show that $\forall x \in [0,1]$ $|f'(x)| \leq (A/2)$.

My attempt was to to develop a Taylor series for $f(x)$ and $f'(x)$ around point $c \in (0,1)$ where $f'(c)=0$.

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A hint: For a given $c\in[0,1]$ compute $f(0)$ and $f(1)$ by means of a Taylor expansion at $c$ (with Lagrange remainder term) and draw conclusions.
Actually I found a proof without using Taylor expansion. MVT theorem is enough to prove the above statement. I just assume $c$ (where $f'(c) = 0$) is $\leq 1/2$ and go from there. It also works in the case $\geq 1/2$ from symmetry. – Ma.H Jan 8 '11 at 20:00
How did you use the assumption $f(0)=f(1)=0\thinspace$? – Christian Blatter Jan 9 '11 at 12:31