Show there's a $c\in [0,1]$ such that $|f'(c)| \ge 4\int_0^1 f$. Let $f:[0,1]\to\mathbb{R}$, differentiable such that $f(0)=f(1)=0$. For any $x$: $f(x)\ge 0$. Show there's a $c$ such that:
$$\left|f'(c)\right|\ge 4\int_0^1 f$$
Now, my tutor claimed that this demand is equivalent to: "if $\forall x: |f'(x)| < M$ then $\int_0^1 f \lt \frac{M}{4}$" and I don't understand why.
I will be glad for a clarification.
Thanks.
 A: Let $\mathcal{F}$ be the set of all non-negative differentiable functions on $[0,1]$ such that $f(0) = f(1) = 0$. The transposition of the statement given by your tutor is
$$
\forall f \in \mathcal{F} \int_0^1 f(x) dx \ge \frac{M}{4} \, \Rightarrow \,   \exists c \in [0,1] \, \ni \, |f'(c)| \ge M \, .
$$
Suppose this is true. Set $M = 4 \int_0^1 f$. It follows that there is a $c$ such that $|f'(c)| \ge M = 4 \int_0^1f$. 
On the other hand, suppose you know that for all $f \in \mathcal{F}$
$$
\exists c \in [0,1] \ni |f'(c)| \ge 4 \int_0^1 f \, .
$$
Pick any $f \in \mathcal{F}$. Assume that $\int_0^1 f \ge \frac{M}{4}$. Then for some $c$, you know that $|f'(c)| \ge 4 \int_0^1 f \ge 4 \frac{M}{4} =M$. Done.
A: In general, $x \ge y$ iff for all $a >x$ implies $a >y$.
Suppose $|f'(x)| < M$ for all $x$, then
$|f(x) -f(y)| < M |x-y|$ for all $x \neq y$.
Hence, taking $y=0,1$ (and $x \notin \{0,1\}$) we have $f(x) < Mx$ and $f(x) < M (1-x)$. 
Then $\int f < \int_0^{1 \over 2} Mx dx + \int_{1 \over 2}^1 M(1-x) dx  = {M \over 4}$.
