Differentiability of a piecewise defined discontinuous function 
Let
$ f(x) = \begin{cases}
x-4 & \text{if } x \lt 1; \\
x+1 & \text{if } x > 1; \\
0 & \text{if } x = 1.
\end{cases}$
Why isn’t this function differentiable at 1? Why isn’t its derivative=1 at x=1?

Both left and right sides have their derivatives equal to $1$ as $x$ approaches $1$. I can see that the left hand and right hand derivatives do not agree at $x=1$. However, could someone explain that to me intuitively, that is, without using the definition of the derivative?
 A: If one is to begin from the geometric concept of the derivative, then the solution is clear, there is no tangent line. The issue is only in trying to understand it algebraically , which in that case you have to do it through the definition of limits.
In the algebraic standpoint, we have that the derivative exists at a point $a$ if the following mentioned limits exist and equalities hold:
$$  f'(a)=\lim_{h \to 0} \frac{f(a) - f(a-h)}{h} = \lim_{h \to 0} \frac{f(a+h) - f(a) }{h}$$
The beauty here is that the algebraic definition is made in such a way that it lines up with the basic conceptual intuition of the derivative we have from geometry. It fails if the tangent line doesn't actually exist!
A: You could also think that differentiability implies continuity. This is an assertion of the form $A \Rightarrow B$, where $A$ and $B$ are propositions ("being differentiable" and "being continuous", in this case). It is not very difficult to see that this kind of statements are logically equivalent to the ones of the form $\neg B \Rightarrow \neg A$, where $\neg B$ means "not being continuous" and $\neg A$ means "not being differentiable". That would explain why in $x=1$ the function $f(x)$ is not differentiable, if you don't want to use strictly any definition.
A: Also, in addition to more literal calculus-level ways of looking at this, it is worthwhile (I think) to ask what the distributional derivative (in an intuitive sense!?!?) would be. This is relevant even in elementary solution of differential equations, and very relevant in Heaviside's signal processing, Dirac's quantum physics, Green-Maxwell classical physics, etc.
So, first, yes, the function is not classically differentiable at $x=1$, as in the other answers. But, the jump in the values of the function at $x=1$ does have an interpretation in terms of the derivative... namely, the derivative includes a scalar multiple of a Dirac delta (generalized) function at $x=1$, in addition to being just $=1$ to the left and to the right of $x=1$.
