Basic Calculus

Module 4: Continuity of a Function

1
 Lesson

4 Continuity of a Function

Flowing water, long winding roads and waiting for forever are just examples of
uninterrupted actions or continuity. There are times wherein some actions were
meant to be disjointed or disconnected. Somehow, the concept of continuity is also
present in Calculus and it will be presented to you in details as you go on with this
module.

2
 What’s In

Answer the following questions.

1. Complete the table of values and illustrate the graph of the function
𝑓(𝑥) = 𝑥 − 2 .
Graph: Y

x-values y-values
0
1
2
3
X

2. Complete the Table of Values and illustrate the graph of the function
𝑥2−1
𝑓(𝑥) = .
𝑥−1 Graph:

Y
x-values y-values

-1

2
X
3

3
 What’s New

Observe the functions below with their respective graphs.

2 𝑥2 − 9 2
(𝑏) 𝑓(𝑥) = (𝑐) 𝑓(𝑥) =
(𝑎) 𝑓(𝑥) = 𝑥 + 5𝑥 + 6 𝑥−3 𝑥

On a sheet of paper, copy the graphs above. As you sketch the graphs, try not to lift
your pens. Is it possible with the three graphs? Based on what you did, what can you
say about the behavior of the first graph? How about the last two graphs?

The behavior of the graphs above has something to do with our lesson. I think you
already have an idea of what our topic is all about. So, let’s go and explore the lesson
together.

4
 What is It

There are different ways on how we can represent a function. One of those is its
graph. If the graph of a function has no gap or holes, then we can say that the
function is continuous. Otherwise, it’s discontinuous.

In addition, a function is said to be continuous at a point x = a if all of the following

conditions are satisfied (Comandante, 2008):

1. If 𝑓(𝑥) is defined, that is, exists at x=a.

A number exists if it is defined. Division by zero yields to undefined
values. The same with square root of negative numbers. Thus, these
cases must be checked.
2. If lim 𝑓(𝑥) exists.
𝑥→𝑎

Based on our previous lesson on limits, the limit of a function f(x) exists
if the left-hand limit is equal to the right-hand limit. That is,
lim 𝑓(𝑥) = lim 𝑓(𝑥)
𝑥→𝑎− +
𝑥→𝑎
3. If lim 𝑓(𝑥) = 𝑓(𝑐)
𝑥→𝑎

Upon checking whether a function satisfies the first two conditions,

check whether the obtained values from these conditions are equal to
each other.

If any of the above-mentioned conditions failed or violated, then the function is

discontinuous. The following are the different types of discontinuity.

a. Removable Discontinuity

A function is said to have a removable discontinuity at x = a, if

a. lim 𝑓(𝑥) exists b. lim 𝑓(𝑥) ≠ 𝑓(𝑎)

𝑥→𝑎 𝑥→𝑎

Either because 𝑓(𝑎) is undefined or the value of 𝑓(𝑎) differs from the value of
the limit (Comandante, 2008).

5
Example:

𝑥2−4
𝑓(𝑥) =
𝑥+2

b. Jump Discontinuity

A function is said to have a jump discontinuity at 𝑥 = 𝑎 if lim 𝑓(𝑥) ≠ lim 𝑓(𝑥)

−
𝑥→𝑎 + 𝑥→𝑎

for a any positive integer. (Comandante, 2008). This type of discontinuity can
usually be found on piecewise functions.

Example:

3 𝑖𝑓 𝑥 ≤ 0
{
𝑥 − 1 𝑖𝑓 𝑥 > 0

c. Infinite Discontinuity

This type of discontinuity exists if a function has one or more infinite limits.
Many rational functions exhibit this type of behavior (Milefoot.com
Mathematics, n.d.).

Example:

1
𝑓(𝑥) =
𝑥

6
For deeper understanding of the lesson, let us have the following illustrative
examples:

Example1

Identify whether the function 𝑓(𝑥) = 𝑥2 + 5𝑥 + 6 is continuous at 𝑥 = −1. Justify your

answer.

Solution:

Check whether the function will satisfy the three conditions for continuity at
the given point 𝑥 = −1.

1. 𝑓(𝑥) is defined, that is, exists at x=a.

𝑓(𝑥) = 𝑥2 + 5𝑥 + 6, 𝑥 = −1 (Given)

𝑓(−1) = (−1)2 + 5(−1) + 6 (Substituted the x-value to the function)

𝑓(−1) = 1 − 5 + 6 (Simplified form)
𝑓(−1) = 2

Since f(x) exists, then the first condition is satisfied.

2. lim 𝑓(𝑥) exists.

𝑥→𝑎

𝑓(𝑥) = 𝑥2 + 5𝑥 + 6 (Given)

lim 𝑓(𝑥) = lim (𝑥2) + lim (5𝑥) + lim 6 (Evaluated the limit)
𝑥→𝑎 𝑥→−1 𝑥→−1 𝑥→−1

lim 𝑓(𝑥) = (−1)2 + 5(−1) + 6 (Simplified form)

𝑥→𝑎

lim 𝑓(𝑥) = 2
𝑥→𝑎

Since lim 𝑓(𝑥) exists, then the second condition is satisfied.

𝑥→𝑎

3. lim 𝑓(𝑥) = 𝑓(𝑐)

𝑥→𝑎

Using the obtained values above, let us check if the third condition will
be satisfied.
lim 𝑓(𝑥) = 𝑓(𝑐) (Third Condition)
𝑥→𝑎

2=2 (Obtained values from the solutions above)

Since the two values are equal, then the third condition is satisfied.
Since all of the three conditions were satisfied, then we can say that the function
𝑓(𝑥) = 𝑥2 + 5𝑥 + 6 is continuous at 𝑥 = −1.

7
To illustrate that the function 𝑓(𝑥) = 𝑥2 + 5𝑥 + 6 is continuous at 𝑥 = −1, simply
graph the function which includes the x-value −1.

Continuous at x = -1

𝑓(𝑥) = 𝑥2 + 5𝑥 + 6

Example 2
𝑥2−9
At 𝑥 = 3, is the function 𝑓(𝑥) = continuous at that particular point? Show your
𝑥−3
solution.

Solution:
Check whether the function will satisfy the three conditions for continuity at
the given point 𝑥 = 3.

1. 𝑓(𝑥) is defined, that is, exists at x=a.

𝑥2−9
𝑓(𝑥) = ,𝑥=3 (Given)
𝑥−3
32−9
𝑓(3) = (Substituted the x-value to the function)
3−3
9−9
𝑓(3) = (Simplified form)
3−3
0
𝑓(3) =
0

Since f(x) is indeterminate, then the first condition NOT satisfied.

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 2. lim 𝑓(𝑥) exists.
𝑥→𝑎
𝑥2−9
𝑓(𝑥) = (Given)
𝑥−3

𝑥2−9
lim 𝑓(𝑥) = lim (Evaluated the limit)
𝑥→𝑎 𝑥→3 𝑥−3

𝑥2−9
lim = lim
(𝑥+3)(𝑥−3)
(Factored out the numerator)
𝑥→𝑎 𝑥−3 𝑥→3 𝑥−3

= lim 𝑥 + 3 (Simplified form)

𝑥→3

= 3+3 (Evaluated the limit)

𝑥2−9
lim =6
𝑥→𝑎 𝑥−3

Since lim 𝑓(𝑥) exists, then the second condition is satisfied.

𝑥→𝑎

3. lim 𝑓(𝑥) = 𝑓(𝑐)

𝑥→𝑎

Using the obtained values above, let us check if the third condition will
be satisfied.
lim 𝑓(𝑥) = 𝑓(𝑐) (Third Condition)
𝑥→𝑎

indeterminate ≠ 6 (Obtained values from the solutions above)

Since two of the conditions are NOT satisfied, then we can say that the
𝑥2−9
function 𝑓(𝑥) = is not continuous at x = 3.
𝑥−3

𝑥2−9
To illustrate that the function 𝑓(𝑥) = is not continuous at 𝑥 = 3, simply graph
𝑥−3

the function which includes the x-value 3.

Graph of
𝑥 2−9
𝑓(𝑥) =
𝑥−3
There’s a hole
on the graph,
that’s why it’s
not continuous
at x = 3

*Also, we can see that this is a Removable Discontinuity.

9
Example 3
2
Determine if the function𝑓(𝑥) = continuous at 𝑥 = 0
𝑥

Solution:
2
Step 1: Evaluate the function 𝑓(𝑥) = at 𝑥 = 0
𝑥

2 2
𝑓(0) = = or undefined
(0) 0

2
Step 2: Find lim ( )
𝑥→0 𝑥

2 2
lim ( ) = −∞ lim ( ) = +∞
𝑥→0− 𝑥 𝑥→0+ 𝑥

𝟐
𝐥𝐢𝐦 ( ) 𝑫𝑵𝑬, 𝑜𝑟 𝑫𝒐𝒆𝒔 𝑵𝒐𝒕 𝑬𝒙𝒊𝒔𝒕
𝒙→𝟎 𝒙

Step 3: Observe that 𝑓(0) ≠ lim 𝑓( 𝑥)

𝑥→0

Since all three conditions were not satisfied, then we can say that the function
2
𝑓(𝑥) = is not continuous at 𝑥 = 0.
𝑥
2
To illustrate that the function 𝑓(𝑥) = is not continuous at 𝑥 = 0, simply graph the
𝑥
function which includes the x-value 0

The graph is This part of the

not graph is going
continuous positive infinity
at x = 0 due
to the
vertical
This part of the
asymptote
(in dash line) graph is going
negative infinity

*Also, we can see that this is an Infinite Discontinuity.

10
  A function is said to be continuous at a closed interval [a, b] if its right
endpoint, open interval and left endpoint has no breakage, holes or
discontinuity. (see figure below)

Y-axis Left endpoint “b”

Open
Right endpoint “a“ interval
(a, b)
Graph of the
function
𝑓(𝑥)

X-axis
a b
[a, b] Closed interval

 The following are the conditions needed to be satisfied to be able to know

whether the function is continuous or not on a closed interval.
1. The function 𝑓(𝑥) needs to be continuous at the open interval (a, b).
2. The function 𝑓(𝑥) needs to be continuous at the left endpoint [a].
a. If 𝑓(𝑎) is defined.
b. If lim 𝑓(𝑥) exists.
+
𝑥→𝑎
c. If 𝑓(𝑎) = lim 𝑓(𝑥)
+
𝑥→𝑎
3. The function 𝑓(𝑥) needs to be continuous at the right endpoint [b].
a. If 𝑓(𝑏) is defined.
b. If lim 𝑓(𝑥) exists.
−
𝑥→𝑏
c. If 𝑓(𝑏) = lim 𝑓(𝑥)
−
𝑥→𝑏

Example 4

Is the function 𝑓(𝑥) = 𝑥2 + 5𝑥 + 6 is continuous on the closed interval [−4, −1] ? Prove
your answer by showing complete solution.
Solution:

Step 1: The function 𝑓(𝑥) needs to be verified continuous at the open interval (−4, −1)
by creating its table of values or simply knowing its restrictions with regards to its
domain.

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Table of values
x-values 𝑓(𝑥) = 𝑥2 + 5𝑥 + 6 y-values
−3.9 = (−3.9)2 + 5(−3.9) + 6 1.71
-3 = (−3)2 + 5(−3) + 6 0
-2.5 = (−2.5)2 + 5(−2.5) + 6 -0.25
-2 = (−2)2 + 5(−2) + 6 0
-1.0001 = (−1.0001)2 + 5(−1.0001) + 6 1.99970001

Based on the table, all domains (x-values) in between (-4, -1) has its own defined y-
value. Therefore, we can say that it is continuous on the open interval (-4, -1).
Another justification that this function is continuous on the open interval (-4, -1) is
that, all polynomial functions are continuous at every point or any interval {𝑥|𝑥 𝜖 ℝ }
Step 2: The function 𝑓(𝑥) needs to be continuous at the left endpoint [-4].

a. Evaluate the function 𝑓(𝑥) = 𝑥2 + 5𝑥 + 6 at 𝑥 = −4

𝑓(−4) = (−4)2 + 5(−4) + 6 = 𝟐

b. Find lim ( 𝑥2 + 5𝑥 + 6)
+
𝑥→−1

lim (𝑥2) + lim +(5𝑥) + lim + 6 = (−4)2 + 5(−4) + 6 = 𝟐

𝑥→−4+ 𝑥→−4 𝑥→−4

c. Observe that 𝑓(−4) = lim 𝑓(𝑥)

𝑥→−4+

Step 3: The function 𝑓(𝑥)needs to be continuous at the right endpoint [-1].

a. Evaluate the function 𝑓(𝑥) = 𝑥2 + 5𝑥 + 6 at 𝑥 = −1

𝑓(−1) = (−1)2 + 5(−1) + 6 = 𝟐

b. Find lim ( 𝑥2 + 5𝑥 + 6)
𝑥→2−

lim (𝑥2) + lim −(5𝑥) + lim − 6 = (−1)2 + 5(−1) + 6 = 𝟐

𝑥→−1 𝑥→−1 𝑥→−1

c. Observe that 𝑓(−1) = lim 𝑓(𝑥)

− 𝑥→−1

Conclusion: Since all of the three conditions were satisfied, then we can say that the
function 𝑓(𝑥) = 𝑥2 + 5𝑥 + 6 is continuous on the closed interval [−4, −1].

To illustrate that the function 𝑓(𝑥) = 𝑥2 + 5𝑥 + 6 is continuous on the closed

interval [−4, −1 ], simply graph the function which includes the x-values from −4
to −1.

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 Graph of 𝑓(𝑥) = 𝑥2 + 5𝑥 + 6

Continuous on the closed interval [-4, -1]

Example 2

Question: Check if the function 𝑓(𝑥) = √𝑥 + 3 is continuous on the closed

interval [−4, 1]? Present your solution.

Solution:
Step 1: The function 𝑓(𝑥) needs to be verified continuous at the open interval(−4, 1)
by creating its table of values or simply knowing its restrictions with regards to its
domain.
Table of values

x-values 𝑓(𝑥) = √𝑥 + 3 y-values

−3.9 = √(−3.9) + 3 (*imaginary
number)
0 √(0) + 3 1.732050808
0.9999 = √(0.999) + 3 1.999749984

Based on the table, all domains (x-values) in between (−4 to 1) has its own defined
y-value except −3.9. Therefore, we can say that it is not continuous on the open
interval (−4, 1).
Another justification that this function is not continuous on the open interval (−4,1)
is that, all radical functions are continuous only within the domain, if beyond
restrictions, then it won’t exist.

The domain of the function 𝑓(𝑥) = √𝑥 + 3 is 𝑥 ≥ −3, since the open interval being
asked is (−4, 1) therefore it is not continuous on (−4, −3).

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Step 2: The function 𝑓(𝑥) needs to be continuous at the left endpoint [-4].

a. Evaluate the function 𝑓(𝑥) = √𝑥 + 3 at 𝑥 = −4

𝑓(−4) = √(−4) + 3 = √(−𝟏) = 𝒊𝒎𝒂𝒈𝒊𝒏𝒂𝒓𝒚 𝒏𝒖𝒎𝒃𝒆𝒓

b. Find lim ( √𝑥 + 3)
𝑥→−4+

√( lim + 𝑥 + 3) = √(−4) + 3 = √(−𝟏) = 𝒊𝒎𝒂𝒈𝒊𝒏𝒂𝒓𝒚 𝒏𝒖𝒎𝒃𝒆𝒓

c. Observe that 𝑓(−4) 𝑎𝑛𝑑 lim 𝑓(𝑥) does not exist.

𝑥→−4+

Step 3: The function 𝑓(𝑥) needs to be continuous at the right endpoint [1].

a. Evaluate the function 𝑓(𝑥) = √𝑥 + 3 at 𝑥 = 1

𝑓(1) = √(1) + 3 = √(4) = 𝟐

b. Find lim ( √𝑥 + 3)
𝑥→1−

√( lim− 𝑥 + 3) = √(1) + 3 = √(4) = 𝟐

𝑥→1

c. Observe that 𝑓(1) = lim 𝑓(𝑥)

𝑥→1−

Conclusion: Only one of the three conditions was satisfied, then we can say that the
function 𝑓(𝑥) = √𝑥 + 3 is not continuous on the closed interval [−4, 1].

To illustrate that the function 𝑓(𝑥) = √𝑥 + 3 is not continuous on the closed

interval [−4, 1 ], simply graph the function which includes the x-values from -4 to 1.

This part of the interval

violates the domain of the
func tion 𝑓(𝑥) = √𝑥 + 3

Not continuous on the closed interval

14
 What’s More

Answer the following questions. Show all possible solutions on a separate sheet of
paper.

1. Determine whether the function 𝑓(𝑥) = 𝑥2 + 6𝑥 + 9 is continuous at 𝑥 = −2.

Sketch its graph illustrating its continuity or discontinuity.

Solution:

Step 1: Evaluate the function 𝑓(𝑥) = 𝑥2 + 6𝑥 + 9 at 𝑥 = −2

Step 2: Find lim ( 𝑥2 + 6𝑥 + 9) = _
𝑥→−2

Step 3: What did you observe between the values of 𝑓(−2) and lim 𝑓( 𝑥) ?
𝑥→−2

Conclusion:

Illustrate the graph:

15
 3
2. Is the function 𝑓(𝑥) = continuous on the closed interval [−1, 1] ? Draw its
𝑥
graph illustrating its continuity or discontinuity.

Solution:
Step 1: The function 𝑓(𝑥) needs to be verified continuous at the open interval (−1, 1)
by creating its table of values or simply knowing its restrictions with regard to its
domain.
A. Table of values

X-values 3 y-values
𝑓(𝑥) =
𝑥
−0.999
0
0.9999

B. Domain restriction of the function:

Step 2: The function 𝑓(𝑥) needs to be continuous at the left endpoint [-1].
3
a. Evaluate the function 𝑓(𝑥) = at 𝑥 = −1
𝑥
3
b. Find lim + ( )
𝑥→−1 𝑥
c. What did you observe between the values of 𝑓(−1) and lim 𝑓( 𝑥) ?
𝑥→−1+

Step 3: The function 𝑓(𝑥)needs to be continuous at the right endpoint [1].

3
a. Evaluate the function 𝑓(𝑥) = at 𝑥 = 1
𝑥
3
b. Find lim− ( ) =
𝑥→1 𝑥
c. What did you observe between the values of 𝑓(1) and lim 𝑓( 𝑥) ?
𝑥→1−

Conclusion:

16
Illustrate the graph:
Y

𝑥2−3𝑥−4
3. Check whether the function 𝑓(𝑥) = is continuous at 𝑥 = 4 . Sketch its
𝑥−4
graph illustrating its continuity or discontinuity.

4. Prove whether the function 𝑓(𝑥) = √𝑥 − 2 is continuous on the closed

interval [2, 4] or not? Sketch the graph illustrating its continuity or
discontinuity.

What I Have Learned

Express what you have learned in this lesson by answering the questions below.
Write your answers on a separate sheet of paper.

1. How to determine whether the function is continuous or not at a certain point?

2. How to determine whether the function is continuous or not on a closed interval?
3. How to illustrate the continuity or discontinuity of a function?

17
 What I Can Do

Read, understand and answer the given questions below. Write your answers on a
separate sheet of paper.

 Last summer, a PENELCO electrician conducts a record of the electrical

voltage supply readings within 5 hours at a certain locality. (See table below)

Voltage readings in volts (v) Time (t)

220 v 1st hour
223 v 2nd hour
0v 3rd hour
(No voltage due to power interruption)
221 v 4th hour
220 v 5th hour

1. Represent a graph showing the voltage readings as y-values and the time as
x-values.

2. Is the graph continuous or not? If your answer is not continuous, at what

point or interval does the discontinuity happen?

Assessment

Identify whether the following functions are continuous or not. Write your solutions
on a separate sheet of paper.

18
Justify your answer whether the following functions are continuous or not at a
certain point or interval. Write your answers on a separate sheet of paper.
𝑥2+𝑥−2
11. 𝑓(𝑥) = at 𝑥 = −2
𝑥+2
𝑥
12. 𝑓(𝑥) = at 𝑥 = 1
𝑥−1

13. 𝑓(𝑥) = 𝑥2 + 7𝑥 + 6 on [−2, 3]

𝑥2−2𝑥−8
14. 𝑓(𝑥) = on [0, 5]
𝑥−4

15. 𝑓(𝑥) = √1 − 𝑥2 on [−1, 1]

19