Find all points of discontinuity of the function $f$ defined by $f(x) = |x| - |x + 1|$.

(NONE) The given function is $f(x) = |x| - |x + 1|$.
Define two functions $g$ and $h$ as follows:
$g(x) = |x|$ and $h(x) = |x + 1|$.
Therefore,$f = g - h$.
First,let us check the continuity of $g$ and $h$.
The function $g(x) = |x|$ can be written as:
$g(x) = \begin{cases} -x, & \text{if } x < 0 \\ x, & \text{if } x \ge 0 \end{cases}$
It is clear that $g$ is defined for all real numbers.
Let $c$ be a real number.
Case $I$: If $c < 0$,then $g(c) = -c$ and $\lim_{x \to c} g(x) = \lim_{x \to c} (-x) = -c$. Thus,$g$ is continuous for $x < 0$.
Case $II$: If $c > 0$,then $g(c) = c$ and $\lim_{x \to c} g(x) = \lim_{x \to c} (x) = c$. Thus,$g$ is continuous for $x > 0$.
Case $III$: If $c = 0$,then $g(0) = 0$. The left-hand limit is $\lim_{x \to 0^-} (-x) = 0$ and the right-hand limit is $\lim_{x \to 0^+} (x) = 0$. Thus,$g$ is continuous at $x = 0$.
Thus,$g$ is continuous at every point.
Similarly,$h(x) = |x + 1|$ can be written as:
$h(x) = \begin{cases} -(x + 1), & \text{if } x < -1 \\ x + 1, & \text{if } x \ge -1 \end{cases}$
It can be proved in the same way that $h$ is continuous for every real number.
The difference of two continuous functions is also a continuous function. Therefore,$f = g - h$ is also continuous at every point.
Thus,$f$ has no points of discontinuity.