How to prove nondifferentiability? Prove that the the function $f(x) = |x-1|+|x-2|$ is not differentiable at $x = 1$ and $x = 2$.
This is actually an objective-type question where there are $3$ other options  $(\{0\},\{1\},\{3\},\{1,2\})$, now the problem is graph is not really much of a choice, hence I guess I have to break the functions into ranges and then solve; any ideas?
 A: Show that the limits
$$\lim_{h\to 0}\frac{f(1+h)-f(1)}{h}$$
and
$$\lim_{h\to 0}\frac{f(2+h)-f(2)}{h}$$
do not exist.
In each case, the easiest thing will be to consider the one sided limits, as $h\to 0^+$ and as $h\to 0^{-}$; if you can show that the one-sided limits are different from each other or at least one does not exist (including the case that they equal $\infty$ or $-\infty$), (each of the two limits separately, of course), then this will prove the function is not differentiable, because $f$ is differentiable at $a$ if and only if the limit $\lim\limits_{h\to 0}(f(a+h)-f(a))/h$ exists. 
After the edit: There is a very big difference between "How can I choose the right answer from a given list of possibilities quickly?" and "How do I prove that the function is not differentiable at $x=1$ and $x=2$?" Please try to ask the question you mean to ask and not something different, lest you get an answer for the question you ask instead of for the question you mean.
The key here is that you should know by now that $y=|t|$ is differentiable everywhere except at $t=0$ (the standard example of a function that is continuous at a point but not differentiable there). That means that a horizontal translate of the function by $a$ (that is, the function $y=|t-a|$, which is the function $y=|t|$ "translated" $a$ units to the right) will necessarily be differentiable everywhere except at $t=a$.
So, $y=|x-1|$ is differentiable everywhere except at $x=1$, and $y=|x-2|$ is differentiable everywhere except at $x=2$. The sum of two functions is differentiable everywhere that the two functions are differentiable, and there is a chance that if neither function is differentiable at a point $a$, then their sum will be differentiable (for example, take $y=|x|$ and $y=-|x|$; neither is differentiable at $x=0$, but $|x|-|x|$ is). However, if exactly one of $f(x)$ and $g(x)$ are differentiable at $x=a$, then $f(x)+g(x)$ is not differentiable at $x=a$: say $f$ is differentiable at $x=a$, and so is $f+g$. Then $g= (f+g)-f$ is also differentiable at $x=a$, being the sum of two functions that are differentiable at $x=a$.
So, here, you have that $y=|x-1|+|x-2|$ is certainly differentiable at every $x$ that is not equal to either $1$ or $2$, because both $|x-1|$ and $|x-2|$ are differentiable there; and it is certainly not differentiable at either $x=1$ or $x=2$, because one and only one of $|x-1|$ and $|x-2|$ is differentiable at each of those points. So that tells you quickly that the right answer has to be $\{1,2\}$. 
In general, you should know at a glance that $y=|x-a|$ is differentiable everywhere except at $x=a$. 
A: Hint: Look at the function $f(x)=|x|$. What happens to the graph when $x=0$? If we try finding the derivative, we have
$f^\prime(x)=\lim_{\epsilon \to 0^+} \frac{|x+\epsilon|-|x|}{\epsilon}$
This approaches 1, but if we try to compute $\lim_{\epsilon \to 0^-} \frac{|x+\epsilon|-|x|}{\epsilon}$ we get -1. However, for the derivative to exist, the limit must be unique.
