# 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 $$|f|\geq M/4$$.

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 $$f(1/2)\leq 0$$).

• Do you have access to mean value theorem at this point of the book? In that case, the inequality should be a straightforward application of mvt. Commented Mar 14, 2020 at 4:05
• Hint: As $f$ is differentiable and continuous on $[0,1]$, apply the mean value theorem. First, observe that $f(\frac{3}{4})-f(\frac{1}{4})=f'(c)(\frac{3}{4}-\frac{1}{4})\geq \frac{M}{2}$ and that $f$ is increasing on $[0,1]$. Consider two cases for and $f(\frac{1}{4})\leq-\frac{M}{4}$ and $f(\frac{1}{4})>-\frac{M}{4}$. Commented Mar 14, 2020 at 4:15
• Thank you, this answers my question. Commented Mar 14, 2020 at 4:26

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 $$a \leq 1/4$$.