Today I will be presenting the piecewise function continuity, how can we prove it and the
applications of the continuity.
Before we get into the actual continuity let’s remember what is a piecewise function.
For example, a piecewise function can represent the income tax. The income tax is based on a
graduated tax calculation. In a graduated tax system, the first let say $10,000 you earn may not
be taxed the next $30,000 at a 10% rate, the next $30,000 at a higher, rate and so on.
Another example can be leasing a car for a period of years with a certain kilometers included
every year, and for every Km driven over the included amount of kilometres you get charged for
every extra km.
Another one can be your cellphone internet usage, with limits set, and everything that goes over
the limits it is charged extra.
And the list can go on.
A piecewise function is a function where more than one formula is used to define the output.
Each formula has its own domain, and the domain of the function is the union of all of these
smaller domains.
𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝟏, 𝒊𝒇 𝒙 𝒊𝒔 𝒊𝒏 𝒕𝒉𝒆 𝒅𝒐𝒎𝒂𝒊𝒏 𝟏
𝒇(𝒙) = |𝒙| = {𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝟐, 𝒊𝒇 𝒙 𝒊𝒔 𝒊𝒏 𝒕𝒉𝒆 𝒅𝒐𝒎𝒂𝒊𝒏 𝟏
𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝟑, 𝒊𝒇 𝒙 𝒊𝒔 𝒊𝒏 𝒕𝒉𝒆 𝒅𝒐𝒎𝒂𝒊𝒏 𝟏
A common example is the absolute value function
−𝒙,
𝒇(𝒙) = |𝒙| = { 𝟎,
𝒙,
𝒇𝒐𝒓 𝒙 < 𝟎
𝒇𝒐𝒓 𝒙 = 𝟎
𝒇𝒐𝒓 𝒙 > 𝟎
A real life piecewise function can model the cost of the shipping package, based of the weight of
the package:
𝟒,
𝑪(𝒙) = { 𝟎. 𝟓(𝒙 − 𝟑) + 𝟒,
𝟎. 𝟓(𝒙 − 𝟏𝟎) + 𝟔,
𝟎< 𝒙≤𝟑
𝟑 < 𝒙 ≤ 𝟏𝟎
𝟏𝟎 < 𝒙 ≤ 𝟏𝟓
Or the most recent example that you can think of is the piecewise function that you used to
model your rollercoaster CPT project.
Now that we’ve seen some real life examples of piecewise functions, let’s get into what is
continuity of these functions and why is important.
The most general form to describe the continuity of a function is that a function is continuous on
an interval if we can draw the graph of the function from start to finish without ever once picking
up our pencil. If we lift the pencil to draw another section of the function’s graph that we have a
discontinuity at that point (lifting your hand because your hand gets tired, doesn’t count)
To better understand what is actually happening with the function at that point we will have to
use “limits”
Limits in math are defined as the values that a function approaches the output for the given
input values.
Limits are unique real numbers. Let us consider a real-valued function f(x) and the real number
“a”, the limit is normally defined as
𝐥𝐢𝐦 𝒇(𝒙) = 𝒃
𝒙→𝒂
It is read as “the limit of f of x, as x approaches ”a” equals “b”, b –real number.
For a piecewise function is important to know what is happening with the function at a certain
value of x, is the function continuous or discontinuous at that point.
To investigate the behavior of the function at a certain point we will have to investigate the
function on both sides of the point, and we will have to see what is happening with the value of
the function when x approaches from the left and from the right side of the point, and for this we
will use the limit from the left side and from the right side.
Ex: lim− 𝑓(𝑥) – limit of 𝑓(𝑥) when 𝑥 approaches 2 from the left side
𝑥→2
lim 𝑓(𝑥) – limit of 𝑓(𝑥) when 𝑥 approaches 2 from the left side
𝑥→2+
To prove that a function is continuous, the function has to pass the 3 step continuity test
1. The function has to be defined at some point a
𝒇(𝒂) =b
2. The limit of the function f(x) approaching point a exists
Is 𝐥𝐢𝐦 𝒇(𝒙) 𝐞𝐱𝐢𝐬𝐭𝐬?
𝒙→𝒂
For this limit to exist the limit approaching from the left has to be equal to the limit
approaching from the right.
𝐥𝐢𝐦 𝒇(𝒙) = 𝐥𝐢𝐦+ 𝒇(𝒙)
𝒙→𝒂−
𝒙→𝒂
Only if this condition is met, the limit lim 𝑓(𝑥) exists.
𝑥→𝑎
3. The limit as x approaches a from either side is equal to f(a)
𝐥𝐢𝐦 𝒇(𝒙) = 𝒇(𝒂)
𝒙→𝒂
Let see some examples
Example 1
Let the piecewise function be:
𝒇(𝒙) = {
𝒙𝟐 ,
𝟒,
𝒙<𝟐
𝒙≥𝟐
Step 1 – Is the function defined for 𝑥 =2
When 𝑥 =2, we will use the 2nd function
𝒇(𝟐) = 𝟒 -function is defined
Step 2 – Does the limit of the function approaching 𝑥 =2 exists
𝐥𝐢𝐦 𝒇(𝟐) = 𝐥𝐢𝐦+ 𝒇(𝟐)
𝒙→𝟐−
𝒙→𝟐
𝐥𝐢𝐦 𝒇(𝟐) = 𝟐𝟐 = 𝟒 – using first function
𝒙→𝟐−
𝐥𝐢𝐦 𝒇(𝟐) = 𝟒 – using second function
𝒙→𝟐+
Second condition met – the limit of the function exist because the limit of the function
approaching 𝑥 =2, from the left side and right side are equal.
Step 3 - The limit as x approaches 2 from either side is equal to f(2)
𝐥𝐢𝐦 𝒇(𝟐) = 𝐥𝐢𝐦+ 𝒇(𝟐) = 𝒇(𝟐) = 𝟒
𝒙→𝟐−
𝒙→𝟐
In conclusion the function is continuous at 𝑥 =2.
The graph of the function will be:
Example 2:
Let the piecewise function be:
𝒙,
𝒇(𝒙) = {𝒙𝟐 − 𝟐,
𝟐𝒙 + 𝟓,
𝒙<𝟐
𝟐≤𝒙<𝟑
𝒙≥𝟑
To check continuity at 𝒙 =2:
Step 1 – Is the function defined for 𝑥 =2
When 𝑥 =2, we will use the second function
𝒇(𝟐) =𝟐𝟐 − 𝟐 = 𝟐 -function is defined
Step 2 – Does the limit of the function approaching 𝑥 =2 exist
Is 𝐥𝐢𝐦− 𝒇(𝟐) = 𝐥𝐢𝐦+ 𝒇(𝟐) ?
𝒙→𝟐
𝒙→𝟐
𝐥𝐢𝐦 𝒇(𝟐) = 𝟐 – using first function
𝒙→𝟐−
𝐥𝐢𝐦 𝒇(𝟐) = 𝟐𝟐 − 𝟐 = 𝟐 – using second function
𝒙→𝟐+
Second condition met – the limit of the function exist because the limit of the function
approaching 𝑥 =2, from the left side and right side are equal.
o check continuity at 𝒙 =3:
Step 1 – Is the function defined for 𝑥 =3
When 𝑥 =3, we will use the 3rd function
𝒇(𝟑) = 𝟐(𝟑)+ 𝟓 = 𝟏𝟏 -function is defined
Step 2 – Does the limit of the function approaching 𝑥 =2 exists
Is 𝐥𝐢𝐦− 𝒇(𝟑) = 𝐥𝐢𝐦+ 𝒇(𝟑) ?
𝒙→𝟑
𝒙→𝟑
𝐥𝐢𝐦 𝒇(𝟑) = 𝟑𝟐 − 𝟑 = 𝟔 – using second function
𝒙→𝟑−
𝐥𝐢𝐦 𝒇(𝟑) = 𝟐(𝟑) + 𝟓 = 𝟏𝟏 – using second function
𝒙→𝟑+
Second condition is not met – the limit of the function does not exist because the limit of the
function approaching 𝑥=3, from the left side is not equal with the limit on the right side.
𝐥𝐢𝐦 𝒇(𝟑) ≠ 𝐥𝐢𝐦+ 𝒇(𝟑)
𝒙→𝟑−
𝒙→𝟑
Step 3 – is not necessary because one of the previous steps did not met the continuity test.
In conclusion the function will have a discontinuity at 𝑥=3, called jump discontinuity
The graph of the function will be:
Example 3:
Let the piecewise function be:
𝒙𝟐 ,
𝒇(𝒙) = { 𝟏
,
𝒙−𝟐
𝒙≤𝟐
𝒙>𝟐
To check continuity at 𝒙 =2:
Step 1 – Is the function defined for 𝑥 =2
When 𝑥 =2, we will use the first function
𝒇(𝟐) =𝟐𝟐 = 𝟒 -function is defined
Step 2 – Does the limit of the function approaching 𝑥 =2 exists
Is 𝐥𝐢𝐦− 𝒇(𝟐) = 𝐥𝐢𝐦+ 𝒇(𝟐) ?
𝒙→𝟐
𝒙→𝟐
𝐥𝐢𝐦 𝒇(𝟐) = 𝟐𝟐 = 𝟒 – using first function
𝒙→𝟐−
𝟏
𝟏
𝐥𝐢𝐦 𝒇(𝟐) = 𝟐−𝟐 = 𝟎 = 𝒖𝒏𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒐𝒓 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝒊𝒏𝒇𝒊𝒏𝒊𝒕𝒚 – using second function
𝒙→𝟐+
Second condition is not met – the limit of the function does not exist because the limit of the
function approaching 𝑥=2, from the left side is not equal with the limit on the right side.
𝐥𝐢𝐦 𝒇(𝟐) ≠ 𝐥𝐢𝐦+ 𝒇(𝟐)
𝒙→𝟐−
𝒙→𝟐
Step 3 – is not necessary because one of the previous steps did not met the continuity test.
In conclusion the function will have a discontinuity at 𝑥=2, called infinite discontinuity.
The graph of the function will be:
Example 4:
Let the piecewise function be:
𝒙,
𝒇(𝒙) = {𝒙𝟐 − 𝟐,
𝟐𝒙 + 𝟓,
𝒙<𝟐
𝟐≤𝒙<𝟑
𝒙≥𝟑
To check continuity at 𝒙 =2:
Step 1 – Is the function defined for 𝑥 =2
When 𝑥 =2, we will use the second function
𝒇(𝟐) =𝟐𝟐 − 𝟐 = 𝟐 -function is defined
Step 2 – Does the limit of the function approaching 𝑥 =2 exist
Is 𝐥𝐢𝐦− 𝒇(𝟐) = 𝐥𝐢𝐦+ 𝒇(𝟐) ?
𝒙→𝟐
𝒙→𝟐
𝐥𝐢𝐦 𝒇(𝟐) = 𝟐 – using first function
𝒙→𝟐−
𝐥𝐢𝐦 𝒇(𝟐) = 𝟐𝟐 − 𝟐 = 𝟐 – using second function
𝒙→𝟐+
Second condition met – the limit of the function exist because the limit of the function
approaching 𝑥 =2, from the left side and right side are equal.
o check continuity at 𝒙 =3:
Step 1 – Is the function defined for 𝑥 =3
When 𝑥 =3, we will use the 3rd function
𝒇(𝟑) = 𝟐(𝟑)+ 𝟓 = 𝟏𝟏 -function is defined
Step 2 – Does the limit of the function approaching 𝑥 =2 exists
Is 𝐥𝐢𝐦− 𝒇(𝟑) = 𝐥𝐢𝐦+ 𝒇(𝟑) ?
𝒙→𝟑
𝒙→𝟑
𝐥𝐢𝐦 𝒇(𝟑) = 𝟑𝟐 − 𝟑 = 𝟔 – using second function
𝒙→𝟑−
𝐥𝐢𝐦 𝒇(𝟑) = 𝟐(𝟑) + 𝟓 = 𝟏𝟏 – using second function
𝒙→𝟑+
Second condition is not met – the limit of the function does not exist because the limit of the
function approaching 𝑥=3, from the left side is not equal with the limit on the right side.
𝐥𝐢𝐦 𝒇(𝟑) ≠ 𝐥𝐢𝐦+ 𝒇(𝟑)
𝒙→𝟑−
𝒙→𝟑
Step 3 – is not necessary because one of the previous steps did not met the continuity test.
In conclusion the function will have a discontinuity at 𝑥=3, called jump discontinuity
The graph of the function will be:
I will finish my presentation with a more interesting situation where the function is:
𝒙 + 𝟒,
𝒇(𝒙) = {𝒂𝒙𝟐 + 𝒃𝒙 + 𝟐,
𝟔𝒙 + 𝒂 − 𝒃,
𝒙<𝟏
𝟏≤𝒙<𝟑
𝟑≤𝒙
We have to find the constants a and b for the function to be continuous at both 𝑥=1 and 𝑥=3.
Let the piecewise function be:
To check continuity at 𝒙 =1 and 𝒙 =3 :
Step 1 – Is the function defined for 𝑥 =1
Cannot check the condition at this stage
Step 2 – For the function to be continuous the limit of the function approaching 𝑥 =1 exists, left and
right limits are equal
𝐥𝐢𝐦 𝒇(𝟏) = 𝐥𝐢𝐦+ 𝒇(𝟏)
𝒙→𝟏−
𝒙→𝟏
𝐥𝐢𝐦 𝒇(𝟏) = 𝟏 + 𝟒 = 𝟓 – using first function
𝒙→𝟏−
𝐥𝐢𝐦 𝒇(𝟏) = 𝒂(𝟏)𝟐 + 𝒃(𝟏) + 𝟐 – using second function
𝒙→𝟏+
Then: 𝒂 + 𝒃 + 𝟐 = 𝟓
𝒂 + 𝒃 + 𝟐 = 𝟓 (1)
– For the function to be continuous the limit of the function approaching 𝑥 =3 exists, left and
right limits are equal
𝐥𝐢𝐦 𝒇(𝟑) = 𝐥𝐢𝐦+ 𝒇(𝟑)
𝒙→𝟑−
𝒙→𝟑
𝐥𝐢𝐦 𝒇(𝟑) = 𝒂(𝟑)𝟐 + 𝒃(𝟑) + 𝟐 – using second function
𝒙→𝟑−
𝐥𝐢𝐦 𝒇(𝟑) = 𝟗𝒂 + 𝟑𝒃 + 𝟐 (2)
𝒙→𝟑−
𝐥𝐢𝐦 𝒇(𝟑) = 𝟔(𝟑) + 𝒂 − 𝒃 – using third function
𝒙→𝟑+
𝐥𝐢𝐦 𝒇(𝟑) = 𝒂 − 𝒃 + 𝟏𝟖 (3)
𝒙→𝟑+
Equation (2) equal equation (3):
𝟗𝒂 + 𝟑𝒃 + 𝟐 = 𝒂 − 𝒃 + 𝟏𝟖
𝟖𝒂 + 𝟒𝒃 − 𝟏𝟔 = 𝟎 – divide by 4
𝟐𝒂 + 𝒃 − 𝟒 = 𝟎 (4)
Multiply equation (1) with -2 and add it to equation (4)
-𝒃 + 𝟐 = 𝟎
𝒃=𝟐
Then replacing 𝒃 in (1):
𝒂+𝟐−𝟑=𝟎
𝒂=𝟏
Check the limits:
𝐥𝐢𝐦 𝒇(𝟏) = 𝟏 + 𝟐 + 𝟐 = 𝟓
𝒙→𝟏−
𝐥𝐢𝐦 𝒇(𝟏) = 𝟏 + 𝟐 + 𝟐 = 𝟓
𝒙→𝟏+
Second condition met – the limit of the function exist because the limit of the function approaching
𝑥 = 1, from the left side and right side are equal.
𝐥𝐢𝐦 𝒇(𝟑) = 𝟗 + 𝟑(𝟐) + 𝟐 = 𝟏𝟕
𝒙→𝟑−
𝐥𝐢𝐦 𝒇(𝟑) = 𝟏 − 𝟐 + 𝟏𝟖 = 𝟏𝟕
𝒙→𝟑+
Second condition met – the limit of the function exist because the limit of the function approaching
𝑥 = 3, from the left side and right side are equal.
Step 3 - The limit as x approaches 1 from either side is equal to 𝑓(1) and the limit as x
approaches 3 from either side is equal to 𝑓(3).
𝒇(𝟏) = 𝟏 + 𝟐 + 𝟐 = 𝟓
𝒇(𝟑) = 𝟔(𝟑) + 𝟏 − 𝟐 = 𝟏𝟕
First condition met, the function is defined.
𝐥𝐢𝐦 𝒇(𝟏) = 𝐥𝐢𝐦+ 𝒇(𝟏) = 𝒇(𝟏) = 𝟓
𝒙→𝟏−
𝒙→𝟏
And
𝐥𝐢𝐦 𝒇(𝟑) = 𝐥𝐢𝐦+ 𝒇(𝟑) = 𝒇(𝟑) = 𝟏𝟕
𝒙→𝟑−
𝒙→𝟑
In conclusion the function is continuous at 𝑥 =1 and 𝑥 =3.
The function will be:
𝒙 + 𝟒,
𝒇(𝒙) = { 𝒙𝟐 + 𝟐𝒙 + 𝟐,
𝟔𝒙 − 𝟏,
The graph of the function will be:
𝒙<𝟏
𝟏≤𝒙<𝟑
𝟑≤𝒙
Example 2:
Let the piecewise function be:
−𝟑,
𝒇(𝒙) = { |𝒙 − 𝟏| ,
𝟐𝒙 + 𝟓,
𝒙 < −𝟏
−𝟏≤𝒙<𝟒
𝒙>𝟒
To check continuity at 𝒙 =-1:
Step 1 – Is the function defined for 𝑥 =-1
When 𝑥 =-1, we will use the second function
𝒇(−𝟏) = |−𝟏 − 𝟏|= 𝟐 -function is defined
Step 2 – Does the limit of the function approaching 𝑥 =-1 exist
Is 𝐥𝐢𝐦− 𝒇(−𝟏) = 𝐥𝐢𝐦+ 𝒇(−𝟏) ?
𝒙→−𝟏
𝒙→−𝟏
𝐥𝐢𝐦 𝒇(−𝟏) = −𝟑 – using first function
𝒙→−𝟏−
𝐥𝐢𝐦 𝒇(−𝟏) = |−𝟏 − 𝟏| = 𝟐 – using second function
𝒙→−𝟏+
Second condition not met – the limit of the function does not exist because the limit of the
function approaching 𝑥 =-1, from the left side and right side are not equal.
Step 3 – is not necessary because one of the previous steps did not pass the continuity test.
In conclusion the function will have a discontinuity at 𝑥=-1, called jump discontinuity
To check continuity at 𝒙 =4:
Step 1 – Is the function defined for 𝑥 =4
𝒇(𝟒) 𝒏𝒐𝒕 𝒅𝒆𝒇𝒊𝒏𝒆𝒅
First condition is not met.
Steps 2 and 3 – are not necessary because step one did not pass the continuity test.
In conclusion the function will have a discontinuity at 𝑥=4, called point or hole discontinuity
The graph of the function will be: