Can we say that $x=0$ is a minimum of the function $f(x)=|x|$? The derivative of a function $f(x)$ is said to exist at $x=x_0$ if both the left-derivative $Lf^\prime(x)$ and the right-derivative $Rf^\prime(x)$ exists and they are equal i.e. $$\lim\limits_{h\to 0^-}\frac{f(x_0+h)-f(x_0)}{h}=\lim\limits_{h\to 0^+}\frac{f(x_0+h)-f(x_0)}{h}.$$ However, if either of them is nonexistant or not equal to each other, the function does not have a unique derivative at $x=x_0$. The derivative is said to not exists at $x=x_0$. 
For example, $f(x)=|x|$ does not have a unique derivative at $x=0$ because the left-derivative and right-derivative are unequal. In this case, can we say that $x=0$ is a minimum of the function $f(x)=|x|$? Looks like the criterion of determining maximum or minimum is in trouble. Sorry for my sloppy mathematics language.
 A: Sometimes calculus classes skip over the important step of defining what a minimum is:

A minimum of a function $f$ is some $x$ in the domain such that $f(x) \leq f(y)$ for all $y$ in the domain.

One can also define a local minimum as:

A local minimum of a function $f$ is some $x$ in the domain such that, for some $\varepsilon > 0$ we have $f(x)\leq f(y)$ whenever $|x-y|<\varepsilon$.

Notice that there are no derivatives involved here - you could take something nasty like the Weierstrass function, which isn't differentiable anywhere, and talk about the minimum value it obtains.
It's a theorem that if a function $f$ is differentiable (from both sides) at $x$, then $x$ can only be a local minimum (or maximum) if $f'(x)=0$ - and the proof is easy: if $f'(x) < 0$, you can see that $f(x+\alpha)$ has to be less than $f(x)$ for some small $\alpha$ due to the definition of the derivative. Likewise, if $f'(x) > 0$, one has that $f(x-\alpha)$ has to be less than $f(x)$. Either way, $f(x)$ cannot have been a local minimum. Sometimes you'll see this stated as saying that local minima/maxima may only occur at zeros of a functions derivative, points where the function is not differentiable, and endpoints  of the domain.
Basically, all we can conclude from this theorem is that for $f(x)=|x|$ the only possible location for a minimum is at $x=0$. We would have to argue by other methods why this really is a minimum - but that's easy for this example because $f(0)=0$ and $0\leq f(y)$ for every $y$ since $|y|$ is always non-negative by definition.
A: The derivative is just a tool that's sometimes useful when looking for a minimum. In this case it's irrelevant. You know $|x| \ge 0$ for all $x$ and that the value is $0$ only when $x=0$ so that't the minimum value.
A: A function does not need to be differentiable to have a minimum.
For example, the Weierstrass function https://en.wikipedia.org/wiki/Weierstrass_function
is differentiable nowhere but has many local minima.
In general, $x^*$ is a local minimiser of $f$ is there is some interval $I$ with $x^*$
in its interior such that $f(x) \ge f(x^*)$ for all $x \in I$.
If in addition the function is differentiable at $x=0$ then we must have $f'(x^*) = 0$.
The function $x \mapsto |x|$ is convex and there is a generalised concept of
derivative for convex functions called the subdifferential. In a similar manner to
differentiable functions, if $x^*$ is a minimiser of a convex function $f$ then we
have $0 \in \partial f(x^*)$.
In the case of the absolute value function, we have 
$\partial f(x) = \begin{cases} \{-1\},& x<0 \\
[-1,1],& x = 0 \\
\{1\},& x>0
\end{cases}$
A: The places in a function to check for minimas are:


*

*Where the derivative is 0

*Where the function is discontinuous

*Where the derivative does not exist

*At the edges of the valid domain (note that on a continuous function, the "edge" is infinity)


Here is a function where the global minima occurs for case #3.
A: The global minimum is the lowest value achieved by the function in its whole domain. For $|x|$, it is indisputably $0$, because $|x|\ge0$ and $|0|=0$.
For a differentiable function in an interval, the global minimum is achieved either at an endpoint of the interval or at a stationary point.
For a piecewise differentiable function, you need to consider all subintervals (here, $(-\infty,0]\cup[0,\infty)$).

In a case like
$$\frac{x^2}{|x|},$$
all values are strictly positive and zero is never reached, but we can go as close as you want, hence we have an infimum.
