Existence and Uniqueness Theorem explanation I am trying to understand the role that $\frac{b}{M}$ plays in the statment $h \leq \min \left \{a, \frac{b}{M} \right \}$ from the following theorem 
We cannot have $h > a$ since then $I_h$ would extend beyond the right/left sides of $R$. However, for $\frac{b}{M}$, what does this describe? What is the purpose of division by $M$?
 A: Let me give you some intuition why $\frac{b}{M}$ appears in the statement of the theorem. Like you noticed, we cannot have $h > a$ because then the equation is not defined. However, there is another "hidden" condition that appears in the statement of the problem that puts another a priori restriction on $h$ (which might or might not be relevant in a real case).
We want to find an interval $[x_0 - h, x_0 + h]$ on which the solution $y(x)$ is defined, satisfies $y(x_0) = y_0$ and $y'(x) = F(x,y(x))$ for all $x \in [x_0 - h, x_0 + h]$. In particular, such a solution must satisfy $y(x) \in [y_0 - b, y_0 + b]$ for otherwise, the right hand side of $y'(x) = F(x,y(x))$ stops making sense! 
Now, the only thing we know about $F(x,y)$ is the bound $|F(x,y)| \leq M$ so let us assume for a second that $F(x,y) \equiv M$ (but we think of $F$ as defined only on the rectangle $[x_0 - a, x_0 + b] \times [y_0 - b, y_0 + b]$). What is the solution of this problem? We want $y'(x) = F(x,y) 
\equiv M$, $y(x_0) = y_0$ so the unique solution is $y(x) = y_0 + M(x - x_0)$ but we also want that $y(x)$ will stay in $[y_0 - b, y_0 + b]$ when $x \in [x_0 - h, x_0 + h]$ which means that $y_0 - b \leq y_0 + M(x - x_0) \leq y_0 + b$ or $-\frac{b}{M} \leq x - x_0 \leq \frac{b}{M}$. That is, the maximal interval on which the solution of our original problem exists is obtained when $h = \min \left \{ a, \frac{b}{M} \right \}$. 
The solution of the initial problem $y'(x) \equiv M, y(x_0) = y_0$ (without any further restrictions) is given by $y(x) = y_0 + M(x - x_0)$ and exists for all $x \in \mathbb{R}$ but the maximal solution of the problem "
$y(x_0) = y_0, y'(x) \equiv M, x \in [x_0 - a, x_0 + a]$ and $|y(x) - y_0| \leq b_0$ about which the theorem speaks is given by
$$y \colon \left[ x_0 - \min \left \{ a, \frac{b}{M} \right \}, x_0  + \min \left \{ a, \frac{b}{M} \right \}\right] \rightarrow \mathbb{R}, \,\,\, y(x) = y_0 + M(x - x_0) $$
and cannot be extended further!
