Question

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

Answer 1

The given function is f (x) = |x| - |x + 1|.
The two functions, g and h, are defined as
g(x) = |x| and h(x) = |x + 1|
Then, f = g - h
The continuity of g and h is examined first.
g(x) = |x| can be written as
g(x) = $\{\begin{array}{c}-x,ifx<0\\ x,if\ge 0\end{array}$
Clearly, g is defined for all real numbers.
Let c be a real number.
Case I
If c < 0, then g(c) = -c and $\underset{x\to c}{lim}$ g(x) = $\underset{x\to c}{lim}$ (-x) = -c
$\therefore$ $\underset{x\to c}{lim}$ g(x) = g(c)
Therefore, g is a continuous at all points x, such that x < 0
Case II
If c > , then g(c) = c and $\underset{x\to c}{lim}$ g(x) = $\underset{x\to c}{lim}$ x = c
$\therefore$ $\underset{x\to c}{lim}$ g(x) = g(c)
Therefore, g is continuous at all points x, such that x > 0
Case III
If c = 0, then g(c) = g(0) = 0
$\underset{x\to 0}{lim}$ g(x) = $\underset{x\to 0}{lim}$ (-x) = 0
$\underset{x\to 0}{lim}$ g(x) = $\underset{x\to 0}{lim}$ (x) = 0
$\therefore$ $\underset{x\to 0}{lim}$ g(x) = $\underset{x\to 0}{lim}$ (x) = g(0)
Therefore, g is continuous at x = 0
From the above three observation, it can be concluded that g is continuous at all points.
h(x) = |x + 1| can be written as
h(x) $\{\begin{array}{c}-(x+1),ifc<-1\\ x+1,ifx\ge -1\end{array}$
Clearly, h is defined for every real number.
Let c be a real number.
Case I :
If c < -1, then h(c) = -(c + 1) and $\underset{x\to c}{lim}$ h(x) = $\underset{x\to c}{lim}$ [-(x + 1)] = -(c + 1)
$\therefore$ $\underset{x\to c}{lim}$ h(x) = h(c)
Therefore, h is continuous at all points x, such that x < -1
Case II:
If c > -1, then h(c) = c + 1 and $\underset{x\to c}{lim}$ h(x) = $\underset{x\to c}{lim}$ (x + 1) = c + 1
$\therefore$ $\underset{x\to c}{lim}$ h(x) = h(c)
Therefore, h is continuous at all points x such that x > -1.
Case III
If c = -1, then h(c) = h(-1) = -1 + 1 = 0
$\underset{x\to -1}{lim}$ h(x) = $\underset{x\to -1}{lim}$ [-(x + 1)] = -(-1 + 1) = 0
$\underset{x\to -1}{lim}$ h(x) = $\underset{x\to -1}{lim}$ (x + 1) = (-1 + 1) = 0
$\therefore$ $\underset{x\to -1}{lim}$ h(x) = $\underset{x\to -1}{lim}$ h(x) = h(-1)
Therefore, h is continuous at x = 1
From the above three observations, it can be concluded that h is continuous at all points of the real line.
g and h are continuous functions. Therefore, f = g h is also a continuous function.
Therefore, f has no point of discontinuity.