Suppose that $f'(x) \geq 10$ for all $x$ in $[0,5]$. Find $a>0$, as large as possible, such that there is guaranteed to be an interval $I$ of length $a$ contained in $[0,5]$ such that $|f|\geq 2$ on $I$. Justify your answer using the mean value theorem.
My professor gave us the solution, however, I am looking for more intuition on how it works. Let $b<d$ be in the interval $[0,5]$, then there is $c$ between $b$ and $d$ such that $f'(c) = \frac{f(b)-f(d)}{b-d}$
Use $f'\geq 10$, so $\frac{f(b)-f(d)}{b-d}\geq 10$
Now choose $f(b) = -2$, $f(d)=2$, then $\frac{-4}{b-d}\geq10$
length = $d-b\leq\frac{2}{5}$, use $I$ to be longer piece of $(0,b)$ and $(d,5)$. He also asked us to explain why it still works even if a $b,d$ such that $f(b)=-2$,$f(d)=2$.
I am confused on the choice of $f(b)$ and $f(d)$. Why did we choose those precise values? Is it because it is the smallest allowed value that $f$ can be on $I$? I am just looking for more intuition on how this problem works in general. Also I am guessing it still works if $f(b)\neq-2$ and $f(d)\neq2$ as we can just can take a bigger interval for $I$?