find out all the points of discontinuity of f defined by f(x) = |x-1| + |x-2|
Answers
Step-by-step explanation:
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) = {
−x,ifx<0
x,if≥0
Clearly, g is defined for all real numbers.
Let c be a real number.
Case I
If c < 0, then g(c) = -c and
x→c
lim
g(x) =
x→c
lim
(-x) = -c
∴
x→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
x→c
lim
g(x) =
x→c
lim
x = c
∴
x→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
x→0
lim
g(x) =
x→0
lim
(-x) = 0
x→0
lim
g(x) =
x→0
lim
(x) = 0
∴
x→0
lim
g(x) =
x→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) {
−(x+1),ifc<−1
x+1,ifx≥−1
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
x→c
lim
h(x) =
x→c
lim
[-(x + 1)] = -(c + 1)
∴
x→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
x→c
lim
h(x) =
x→c
lim
(x + 1) = c + 1
∴
x→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
x→−1
lim
h(x) =
x→−1
lim
[-(x + 1)] = -(-1 + 1) = 0
x→−1
lim
h(x) =
x→−1
lim
(x + 1) = (-1 + 1) = 0
∴
x→−1
lim
h(x) =
x→−1
lim
h(x) = h(-1)
Therefore, h is continuous at x = 1