# Let $f(x)$ be continuously differentiable twice on $[0,1]$ and $f(0)=f(1)=0$, and $|f''(x)| ≤A$ on $(0,1)$, prove $|f'(x)| ≤\frac{A}{2}$ on $(0,1)$.

Let $$f(x)$$ be continuously differentiable twice on $$[0,1]$$ and $$f(0)=f(1)=0$$, and $$|f''(x)| ≤A$$ on $$(0,1)$$, prove $$|f'(x)| ≤\frac{A}{2}$$ on $$[0,1]$$.

I know thanks to Rolle's theorem that there exists a $$c$$ such that $$f'(c)=0$$, I'll use the mean value theorem on the derivative: $$|\frac{f'(c)-f'(0)}{c}| ≤A$$ and $$|\frac{f'(1)-f'(c)}{1-c}| ≤A$$

I'm not sure how this helps me.

You have to use the Lagrange mean value theorem more often than you have currently done. The point is that because $$f'' \leq A$$ uniformly, applying the Lagrange mean value theorem can give you a lot more information on the variation than merely applying it at the two extreme points.
For example, suppose that $$a \neq b$$ are arbitrary points in $$[0,1]$$. By the mean value theorem applied to $$f'$$ in the interval $$[a,b]$$ (or $$[b,a]$$), we get that $$f'(b)-f'(a)= |b-a|f'(c)$$ (or $$f'(a)-f'(b) = |b-a|f'(c)$$) for some $$c \in (a,b)$$ (or $$(b,a)$$), but because $$|f''(c)| \leq A$$ we have respectively $$-|b-a|A \leq f'(b) - f'(a) \leq |b-a|A$$ (or $$-|b-a|A \leq f'(a) - f'(b) \leq |b-a|A$$ )Combining these, we get for all $$a \neq b$$ that $$f'(a) - f'(b) \leq |b-a|A \\ f'(a) - f'(b) \geq -|b-a|A$$
In particular, let us now FIX an $$a \in (0,1)$$. Integrating the variable $$b$$ in the first inequality from $$0$$ to $$1$$ gives $$\int_0^1 (f'(a)-f'(b))db \leq \int_{0}^1 |b-a|A db$$ the LHS is, by FTC $$f'(a) - f(1) + f(0) = f'(a)$$ while the RHS is $$A\int_0^1 |b-a|db = A \int_0^a (a-b)db + \int_a^1 (b-a)db = A\left(a^2-a+\frac 12\right) \leq \frac{A}{2}$$ because $$a^2 -a\leq 0$$ if $$a \in [0,1]$$. Thus, we are led to $$f'(a) \leq \frac{A}{2}$$.
By similarly integrating the bound $$f'(a) - f'(b) \geq -|b-a|A$$ with respect to $$b$$ from $$0$$ to $$1$$, we get $$f'(a) \geq -\frac{A}{2}$$, leading to the answer, because $$a$$ was arbitrary in $$[0,1]$$, although fixed for the argument above.