Unclear Answer Book on Calculus by Michael Spivak (3rd edition) Question 11-26. The question goes as follows

Suppose that $f'(x)\geq M>0$ $\forall x\in [0,1]$. Show that there is an interval of length $\frac{1}{4}$ on which $|f|\geq M/4$.

and the answer book states

Note that $f$ is increasing. If $f(1/2)\geq 0$, then $f(3/4)\geq M/4$, so certainly $f\geq M/4$ on the interval $[3/4,1]$. On the other hand, if $f(1/2)\leq 0$, then $f(1/4)\leq -M/4$, so $f\leq -M/4$ on the interval $[0,1/4]$.


How does $f(1/2)\geq 0\Rightarrow f(3/4)\geq M/4$ given that $f$ is increasing on $[0,1]$? (the same question goes for $f(1/2)\leq 0$).

Essentially, you can "integrate" the expression $f'(x)\geq M$ to deduce for $x>a$ $$f(x) \geq M \cdot (x-a) + f(a) \space \space [*].$$  To prove this, use the mean value theorem: assuming the usual conditions are met if $x>a$ then $\exists c $ with $a \leq c \leq x$ such that $$\frac {f(x)-f(a)}{x-a} = f'(c)$$ which rearranges to $$f(x)=f'(c) \cdot (x-a) + f(a).$$  Given $f'(c) \geq M$ and $x-a > 0$, the result $[*]$ follows.
Applying $[*]$ to the question, we have: 
using $a = 1/2$, for $x \geq 3/4$, $x-a > 1/4$  so $$f(x) \geq M \cdot \frac 14 + f(1/2)$$ or using $x=1/2$, for $a \leq 1/4$, $x-a>1/4$ so $$f(1/2) \geq M \cdot \frac 14 + f(a)$$  From these, we can deduce respectively that 
if $f(1/2) \geq 0$ then $f(x) \geq \frac M4$ for $x \geq 3/4$
if $f(1/2) < 0$ then $f(a) < - \frac M4$ for $a \leq 1/4$.
