Why is the derivative at a jump undefined even if the slope remains the same? I've searched online and found almost nothing. What in the mathematical definition of a derivative makes it so that the derivative of the following is undefined at 0.
\begin{equation*}
f(x) =\begin{cases}
3x  & \text{if } x < 0 \\
3x+2  & \text{if } x \ge 0
\end{cases}
\end{equation*}
 A: Here's my intuitive argument for why your book says what it does:
I personally consider the derivative as the best linear approximation of a function at some given point. For this function, at $x=0$ there isn't any good sort of linear approximation so we cannot have such a derivative.
As for a more proofy argument consider the limit
$$
f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}
$$
which I'm sure you recognize as the derivative. Now note that we must have this limit exist no matter how we approach it, i.e. if $h \in \mathbb{R}^-$ or $h \in \mathbb{R}^+$ or some combination of those two statements. Let's first consider when $h$ is negative and try to evaluate what $f'(0)$ is:
$$
f'(0) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{3h - 2}{h} = + \infty 
$$
so you can see that this does not exist.
A: A function is continuous at a point if it is differentiable at it. But here you have a piecewise function, that is discontinuous at $0.$
A: Recall the following definition of a derivative:
$$f'(x) = \lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}\text{.}$$
This limit exists if and only if the two-sided limits exist and 
$$\lim\limits_{h \to 0^{-}}\dfrac{f(x+h)-f(x)}{h} = \lim\limits_{h \to 0^{+}}\dfrac{f(x+h)-f(x)}{h}\text{.}$$
Edit: My attempt of computing the limits:
$$\begin{align}
&\lim\limits_{h \to 0^{-}}\dfrac{3(0+h)- 3(0) - 2}{h} = \lim\limits_{h \to 0^{-}}\dfrac{3h-2}{h} = +\infty\\
&\lim\limits_{h \to 0^{+}}\dfrac{3(0+h)+2-3(0)-2}{h} = \lim\limits_{h \to 0^{+}}3 = 3\text{.}
\end{align}$$
A: One thing that you should always note is that if $f(x)$ is differentiable at $x=a$ then $f(x)$ is continuous at $x=a$. Thus this means if it is not continous at $x=0$ it can't be differentiable at that point either. 
