EASTERN MINDORO COLLEGE, INC.
Bagong Bayan II, Bongabong, Oriental Mindoro
Email Address: emc_1945@yahoo.com; Tel. No.: (043) 283-5479
LEARNING MODULE IN GENERAL MATHEMATICS
S.Y. 2021 – 2022
GMATH11-M1-2021
MODULE
1
FUNCTIONS
First Quarter – Week 1
Allotted Time: 4 hours
Content Standard
The learner demonstrates understanding of key
concepts of functions.
Performance Standard
The learner shall be able to accurately construct
mathematical models to represents real-life situations
using function.
Most Essential Learning Competencies:

Represents real-life situations using functions, including piece-wise functions.

Evaluates a function

Performs addition, subtraction, multiplication, division, and composition of functions.

Solves problems involving functions.
Objectives:

Evaluate functions.

Perform computations involving addition, subtraction, multiplication, division and composition of
functions.

Solve problems involving functions.

Represent real-life situations using functions, including piece-wise functions.
General Instructions




Read every detail in this module with comprehension.
Answer the activities diligently and intelligently.
Be honest at all times in answering the activities and assessments in this module.
Use the given answer sheet for all activities in the last page of this module.
 INTRODUCTION
Functions relate an input to an output. Function is
defined as a relationship between two sets that
associates each element of the first set to one and only
one element in the second set. A one-to-one and
many-to-one relations are considered functions, but
not one-to-many. Thus, all functions are relations, but
not all relations are functions. Functions are used to
model real-life situations. Some of the commonly
known functions are linear, quadratic, trigonometric,
exponential, and logarithmic functions. Functions can
also be in pieces. These are functions called
piecewise function. Absolute value function is an
example of a piecewise function.
 Motivational Activity
Draw a machine or any that has input and output process. Explain how input and output works. Do this
activity in a short bond paper. Guided rubrics are shown below.
Above
Excellent
Average
Need
Unsatisfactory
Average
Grade
Improvement
incomplete
(19 – 20
(14 – 15
(16 – 18
pts.)
pts.)
(11 – 13 pts.)
(0 – 10 pts.)
pts.)
20% - Deadline
Project was turned in on
time.
20% - Explanation
Precisely explained based
on its function.
20% - Value
Effective rendering and use
of shading techniques
Page 1 of 10
MR. CHRISTIAN G. GALAY | Mathematics Teacher
ENRICHING MINDS OF CHAMPIONS
EASTERN MINDORO COLLEGE, INC.
Bagong Bayan II, Bongabong, Oriental Mindoro
Email Address: emc_1945@yahoo.com; Tel. No.: (043) 283-5479
LEARNING MODULE IN GENERAL MATHEMATICS
S.Y. 2021 – 2022
GMATH11-M1-2021
based on light and shadow.
(Value scale)
20% Proportion/Observational
Objects are drawn
accurately according to
their viewpoint.
20% - Creativity
The final output is
presentable as a finish
artwork.
Attached your output on the evaluation sheet.
 DISCUSSION
Introduction to Function
A relation is a set of ordered pairs. The domain of a relation is the set of first coordinates. The range is the set
of second coordinates.
A function is a relation in which each element of the domain corresponds to exactly once element of the range.
A function can be represented in different ways.
1. Functions as ordered pairs, table of values, or mapping diagram
If we are given a set of ordered pairs, table of values, or a mapping diagram, we can easily determine whether
the relation is a function or not by simple looking if each element is used only once in the given set. The
following characteristics will help us identify if a given set of ordered pairs, table of values, or mapping diagram
define a function:
 Each element in domain 𝑥 must be matched with exactly one element in range 𝑦.
 Some elements in 𝑦 may not be matched with any element in 𝑥.
 Two or more elements in 𝑥 may be matched with the same element in 𝑦.
For illustration:
Which of the following represents a function?
a) 𝑓 = {(0,1), (1, −2), (2,0), (3,2)}
b) 𝑔 = {(2,2), (1, −2), (3,0), (0,2), (1,1)}
c)
d)
x
-5
-1
1
0
6
x
-2
-1
0
1
2
y
66
18
0
1
55
y
1/4
1/2
1
2
4
e)
f)
g)
h)
Answer: All are relations, but among the 8, the functions are (a), (c),
(d), (e), and (g). The rest are not functions because there is at least
one element in 𝑥 for which there is more than one corresponding y –
value.
2. Functions as graph
Vertical Line Test
A graph represents a function if and only if any vertical line intersect the graph at most once. It is a visual way
to determine if a curve is a graph of a function or not.
Page 2 of 10
MR. CHRISTIAN G. GALAY | Mathematics Teacher
ENRICHING MINDS OF CHAMPIONS
EASTERN MINDORO COLLEGE, INC.
Bagong Bayan II, Bongabong, Oriental Mindoro
Email Address: emc_1945@yahoo.com; Tel. No.: (043) 283-5479
LEARNING MODULE IN GENERAL MATHEMATICS
S.Y. 2021 – 2022
GMATH11-M1-2021
For illustration:
Which of the following graphs represent a function?
Answer: Graphs (a), (b), (d), and (f) are functions while (e), (g), (c), and (h) are not because they do not satisfy
the vertical line test.
3. Function as equations
Not every set of ordered pairs defines a function. Similarly, not all equations with the variables 𝑥 and 𝑦 define
a function. If an equation is solved for 𝑦 and more than one value of 𝑦 can be obtained for a particular 𝑥 value,
then the equation does not define 𝑦 as a function of 𝑥.
For illustration
Which of the following equations represent a function?
a) 4𝑥 = 𝑦 + 3
b) 𝑦 = 𝑥 2 + 6𝑥 + 9
c) 𝑥2 + 𝑦2 = 4
d) 𝑦 = log2 𝑥
𝑥+5
e) 𝑦 = 𝑥−2
𝑥 − 2, 𝑥 < 0
f) 𝑦 = {
𝑥 + 5, 𝑥 ≥ 0
Answer: Among the given equations, (c) is not a function because if we substitute a real number to the 𝑥 variable,
we can find more than one y – value (ex. if 𝑥 = 0, then 𝑦 can be +2 or -2). Hence, it violates the definition of
function. Equation (c) is an equation of a circle.
If a relation is a function, then y can be replaced with f(x) to denote that the value of y depends on the value of x.
The functions in the examples above, replacing f with other letters to distinguish the functions from each other:
a) 𝑓(𝑥) = 4𝑥 − 3
b) 𝑔(𝑥) = 𝑥 2 + 6𝑥 + 9
d) ℎ(𝑥) = log2 𝑥
𝑥+5
e) 𝑟(𝑥) = 𝑥−2
𝑥 − 2, 𝑥 < 0
f) 𝑝(𝑥) = {
𝑥 + 5, 𝑥 ≥ 0
There are different types of functions, equation (a) is a linear function, (b) is a quadratic function, (d) is a logarithmic
function, (e) is a rational function, and (f) is an example of a piecewise function – defined function.
We also perform operations on functions, such as addition, subtraction, multiplication, division, and composition
of functions.
 Lesson 1 | Linear & Piece-wise Functions
Linear Function
A function 𝑓 is a linear function if 𝑓(𝑥) = 𝑚𝑥 + 𝑏, where 𝑚 and 𝑏 are real numbers, and 𝑚 and 𝑓(𝑥) are not
both equal to zero.
Page 3 of 10
MR. CHRISTIAN G. GALAY | Mathematics Teacher
Commented [Mr. Galay1]: We will use the slope
formula.
ENRICHING MINDS OF CHAMPIONS
EASTERN MINDORO COLLEGE, INC.
Bagong Bayan II, Bongabong, Oriental Mindoro
Email Address: emc_1945@yahoo.com; Tel. No.: (043) 283-5479
LEARNING MODULE IN GENERAL MATHEMATICS
S.Y. 2021 – 2022
GMATH11-M1-2021
Example No. 1
To sell more T-shirts, the class needs to charge a lower price as indicated in the following table:
Target No.
Price per
of
T-shirt
Shirt Sales
500
₱540
900
₱460
1 300
₱380
1 700
₱300
2 100
₱220
2 500
₱140
The price for which you can sell x printed T-shirts is called the price function p(x). p(x) represents each data
point in the table.
Solution: To write a linear equation with two variables when given two points using the slope-intercept method,
we have:
𝑦 −𝑦
- find the slope m of the line using the slope formula 𝑚 = 2 1 and
𝑥2 −𝑥1
-
write the linear equation with two variables by substituting the values of m and (x1, y1) to the formula
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )−the point-slope form of linear equation.
Using the point P1(500, 540) and P2(900, 460), we have:
460 − 540
80
1
𝑚=
=−
𝑜𝑟 −
900 − 500
400
5
1
Now, substitute one of the ordered pairs and the slope, − 5 to the point-slope form.
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
1
𝑦 − 540 = − (𝑥 − 500)
5
1
𝑦 − 540 = − 𝑥 + 100
5
1
𝑦 = − 𝑥 + 100 + 540
5
1
𝑦 = − 𝑥 + 640
5
𝑦 = 640 − 0.2𝑥
Thus, the price function is 𝑝(𝑥) = 640 − 0.2𝑥.
 Note: To check, verify that the function 𝑝(𝑥) = 640 − 0.2𝑥 fits each data in the table.
A piecewise function or compound function is a function defined by multiple sub-functions, where each subfunction applies to a certain interval of the main function’s domain.
Some situations can only be described by more than one formula, depending on the value of the independent
variable.
For illustration
The following are examples of piecewise functions.
2,
𝑖𝑓 𝑥 > −4
−𝑥,
a. 𝑓(𝑥) = {
b. 𝑓(𝑥) = {
−5,
𝑖𝑓 𝑥 < −4
𝑥,
𝑖𝑓 𝑥 ≤ 2
𝑖𝑓 𝑥 > 2
Look at the following examples given by the problem.
Example No. 2
A user is charged ₱300 monthly for a particular mobile plan, which includes 100 free text messages. Message in
excess of 100 are charged ₱1 each. Represent the amount a consumer pays each month as a function of the
number of messages m sent in a month.
Solution: Let 𝐴(𝑚) represent the amount paid by the consumer each month. It can be expressed by the
Solution: Let
A(m) represent the amount paid by the consumer each month. It can be expressed by the piecewise function.
piecewise
function.
No. of
messages sent
in a month
300,
𝑖𝑓 0 < 𝑚 ≤ 100
𝐴(𝑚) = {
300 + 𝑚, 𝑖𝑓 𝑚 > 100
Amount charged if a user consumed 0
or less than 100 text messages.
Page 4 of 10
MR. CHRISTIAN G. GALAY | Mathematics Teacher
ENRICHING MINDS OF CHAMPIONS
EASTERN MINDORO COLLEGE, INC.
Bagong Bayan II, Bongabong, Oriental Mindoro
Email Address: emc_1945@yahoo.com; Tel. No.: (043) 283-5479
LEARNING MODULE IN GENERAL MATHEMATICS
S.Y. 2021 – 2022
GMATH11-M1-2021
Example No. 3
You are a crew at ACTS 12:7 Convenience Store that pays an hourly wage of ₱45.00 and 1.5 times the hourly
wage for the extra hours if you work for more than 40 hours a week.
a. Write a piecewise function that gives the weekly pay P in terms of the number of hours h you work.
b. Graph the piecewise function.
Solution:
a. For up to 40 hours, your pay is given by 45 h.
b. The graph of the function is shown
below.
For over 40 hours, your pay is given by:
45(40) + 1.5(45)(ℎ − 40)
= 1 800 + 67.5(ℎ − 40)
= 1 800 + 67.5ℎ − 2 700
= 67.5ℎ − 900
Thus, the piecewise function is:
𝑃(ℎ) = {
45ℎ,
𝑖𝑓 0 ≤ 𝑚 ≤ 40
67.5ℎ − 900,
𝑖𝑓 ℎ > 40
 Lesson 2 | Operations of Function
Recall rules and steps in each mathematical operations.
Addition and Subtraction
 Find the least common denominator (LCD) of both fractions.
 Rewrite the fractions as equivalent fractions with the same LCD.
 The LCD is the denominator of the resulting fraction.
 The sum or difference of the numerators is the numerator of the resulting fraction.
Example No. 4
Multiplication
 Rewrite the enumerator and denominator in terms of its prime factors.
 Common factors in the enumerator and denominator can be simplified as (this is often called cancelling).
 Multiply the enumerators together to get the new numerator.
 Multiply the denominator together to get the new denominator.
Example No. 5
10
15
1. Find the product of and . Use cancellation of factors when convenient.
21
8
Solution:
Express the numerators and
denominators of the two fractions into
their prime factors.
10
21
×
15
8
=
=
2. Find the product of
Solution:
𝑥2 − 4𝑥 − 5
𝑥2
− 3𝑥 + 2
×
𝑥 2 −4𝑥−5
𝑥 2 −3𝑥+2
𝑥2 − 5𝑥 + 6
𝑥2 − 3𝑥 − 10
=
=
Page 5 of 10
and
𝑥 2 −5𝑥+6
𝑥 2 −3𝑥−10
2∙5
3∙5
×
3∙7 2∙2∙2
2
/ ∙5∙3
/ ∙5
3
/ ∙7∙2
/ ∙2∙2
=
25
28
Cancel out common factors and multiply
the enumerator and the denominator to
reduce the final answer to lowest terms.
.
(𝑥 + 1)(𝑥 − 5) (𝑥 − 2)(𝑥 − 3)
×
(𝑥 − 2)(𝑥 − 1) (𝑥 − 5)(𝑥 + 2)
Express the numerators and denominators of the
two rational expressions into their prime factors.
(𝑥 + 1)(x-5)
/ (x-2)
/ (𝑥 − 3)
Cancel out common factors and multiply the
enumerator and the denominator to reduce the
final answer to lowest terms.
ENRICHING MINDS OF CHAMPIONS
(x-2)
/ (𝑥 − 1)(x-5)
/ (𝑥 + 2)
(𝑥 + 1)(𝑥
− 3)
MR. CHRISTIAN G. GALAY | Mathematics
Teacher
=
(𝑥 − 1)(𝑥 + 2)
𝑥 2 − 2𝑥 − 3
= 2
𝑥 +𝑥−2
1. Find the product of
10
21
15
and . Use cancellation of factors when convenient.
8
EASTERN MINDORO COLLEGE, INC.
Solution:
Bagong Bayan II, Bongabong, Oriental Mindoro
Express the numerators and
Email Address:
emc_1945@yahoo.com;
10 15
2∙5
3 ∙ 5 Tel. No.: (043) 283-5479
denominators of the two fractions into
×
=
×
21 8
3∙7 2∙2∙2
their prime factors.
LEARNING MODULE
MATHEMATICS
GMATH11-M1-2021
Cancel out common factors and multiply
2
/ ∙IN
5 ∙GENERAL
3
/ ∙5
25
= S.Y. 2021 – 2022
=
the enumerator and the denominator to
3
/ ∙7∙2
/ ∙ 2 ∙ 2 28
reduce the final answer to lowest terms.
2. Find the product of
Solution:
𝑥2 − 4𝑥 − 5
𝑥2
− 3𝑥 + 2
×
𝑥 2 −4𝑥−5
𝑥 2 −3𝑥+2
𝑥2 − 5𝑥 + 6
𝑥2 − 3𝑥 − 10
=
=
and
𝑥 2 −5𝑥+6
𝑥 2 −3𝑥−10
.
(𝑥 + 1)(𝑥 − 5) (𝑥 − 2)(𝑥 − 3)
×
(𝑥 − 2)(𝑥 − 1) (𝑥 − 5)(𝑥 + 2)
Express the numerators and denominators of the
two rational expressions into their prime factors.
(𝑥 + 1)(x-5)
/ (x-2)
/ (𝑥 − 3)
Cancel out common factors and multiply the
enumerator and the denominator to reduce the
final answer to lowest terms.
(x-2)
/ (𝑥 − 1)(x-5)
/ (𝑥 + 2)
(𝑥 + 1)(𝑥 − 3)
=
(𝑥 − 1)(𝑥 + 2)
𝑥 2 − 2𝑥 − 3
= 2
𝑥 +𝑥−2
Division
 To divide two fractions or rational expressions, multiply the dividend with the reciprocal of the divisor.
 Sum, Difference, Product, and Quotient of Functions
Let f and g be functions.
(a) Their sum, denoted by f + g, is the function defined by (𝑓 + 𝑔)(𝑥) = 𝑓(𝑥) + 𝑔(𝑥).
(b) Their difference, denoted by f – g, is the function defined by (𝑓 − 𝑔)(𝑥) = 𝑓(𝑥) − 𝑔(𝑥).
(c) Their product, denoted by f · g, is the function defined by (𝑓 ∙ 𝑔)(𝑥) = 𝑓(𝑥) ∙ 𝑔(𝑥).
𝑓
𝑓
𝑓(𝑥)
(d) Their quotient, denoted by ⁄𝑔, is the function defined by ( ⁄𝑔) (𝑥) =
⁄𝑔(𝑥),
excluding the values of x where 𝑔(𝑥) = 0.
Use these functions to develop the examples below.
 𝑓(𝑥) = 𝑥 + 3
 𝑝(𝑥) = 2𝑥 − 7
 𝑣(𝑥) = 𝑥 2 + 5𝑥 + 4
Example No. 6
(a) (𝑣 + 𝑔)(𝑥)
(b) (𝑓 ∙ 𝑝)(𝑥)
•
•
𝑔(𝑥) = 𝑥 2 + 2𝑥 − 8
𝑥+7
ℎ(𝑥) = 2−𝑥
•
𝑡(𝑥) = 𝑥+3
𝑥−2
(c) (𝑓 + ℎ)(𝑥)
(d) (𝑝 − 𝑓)(𝑥)
`
(e) (𝑣⁄𝑔)(𝑥)
Solution:
(a) (𝑣 + 𝑔)(𝑥) = (𝑥2 + 5𝑥 + 4) + (𝑥2 + 2𝑥 − 8) = 𝑥2 + 5𝑥 + 4 + 𝑥2 + 2𝑥 − 8 = 𝟐𝒙𝟐 + 𝟕𝒙 − 𝟒
(b) (𝑓 ∙ 𝑝)(𝑥) = (𝑥 + 3) ∙ (2𝑥 − 7) = 𝟐𝒙𝟐 − 𝒙 − 𝟐𝟏
(c) (𝑓 + ℎ)(𝑥) = (𝑥 + 3) + (
𝑥+7
2−𝑥
) = (𝑥 + 3) ∙
2−𝑥
2−𝑥
+
𝑥+7
2−𝑥
=
(𝑥+3)(2−𝑥) +(𝑥+7)
(2−𝑥)
−𝑥 2 − 𝑥 + 6 + 𝑥 + 7 −𝑥 2 + 13 −1 𝒙𝟐 − 𝟏𝟑
=
∙
=
2−𝑥
−𝑥 + 2 −1
𝒙−𝟐
(d) (𝑝 − 𝑓)(𝑥) = (2𝑥 − 7) − (𝑥 + 3) = 2𝑥 − 7 − 𝑥 − 3 = 𝒙 − 𝟏𝟎
𝟐
𝒙 +𝟓𝒙+𝟒
(e) (𝑣⁄𝑔)(𝑥) = 𝟐
𝒙 +𝟐𝒙−𝟖
=
Use the following functions:
 𝑓(𝑥) = 2𝑥 + 1
• 𝑞(𝑥) = 𝑥 2 − 2𝑥 + 2
Example No. 7
(a) Express the function 𝑓1 (𝑥) = 𝑥2 + 3 as a sum or difference of the functions above.
• 𝑟(𝑥) =
2𝑥+1
𝑥−1
Solution: The solution can involve some trial and error. Add q(x) and f(x) and check if the sum is 𝑥2 + 3.
𝑞(𝑥) + 𝑓(𝑥) = (𝑥 2 − 2𝑥 + 2) + (2𝑥 + 1)
= 𝑥 2 − 2𝑥 + 2 + 2𝑥 + 1
= 𝑥2 + 3
= 𝑓1 (𝑥)
Thus, 𝑓1 (𝑥) = 𝑞(𝑥) + 𝑓(𝑥).
 Practice
𝒇𝟏 (𝒙) = 𝟐𝒙 + 𝟏
1. Express the function 𝑓2 (𝑥) = 𝑥 2 − 4𝑥 + 1 as a sum or difference of the function above.
2. Express the function 𝑔1 (𝑥) = 2𝑥 3 − 3𝑥 2 + 2𝑥 + 2 as a product or a quotient of the
function above.
Page 6 of 10
MR. CHRISTIAN G. GALAY | Mathematics Teacher
Commented [Mr. Galay2]: As a sum: 𝑥 2 − 2𝑥 + 2
As a difference: −𝑥 2 + 6𝑥
As a product: 2𝑥 4 − 4𝑥 3 + 𝑥 2 + 6𝑥 + 2
As a quotient: 𝑥 2 − 2𝑥 + 2
ENRICHING MINDS OF CHAMPIONS
EASTERN MINDORO COLLEGE, INC.
Bagong Bayan II, Bongabong, Oriental Mindoro
Email Address: emc_1945@yahoo.com; Tel. No.: (043) 283-5479
LEARNING MODULE IN GENERAL MATHEMATICS
S.Y. 2021 – 2022
GMATH11-M1-2021
 Lesson 3 | Composition of Function and Evaluating Function
Composition of Functions
Let 𝑓 and 𝑔 be functions of the variable 𝑥. The composite function denoted by (𝑓 ∘ 𝑔) is defined by
(𝑓 ∘ 𝑔)(𝑥) = 𝑓(𝑔(𝑥))
The process of obtaining a composite function is called function composition.
Example No. 8
Given 𝑓(𝑥) = 𝑥 + 4 and 𝑔(𝑥) = 𝑥 2 − 2𝑥 − 3, find (𝑓 ∘ 𝑔)(𝑥).
Solution
(𝑓 ∘ 𝑔)(𝑥) = 𝑓(𝑔(𝑥))
= 𝑔(𝑥) + 4  substitute
= (𝑥 2 − 2𝑥 − 3) + 4  replace value of 𝑔(𝑥)
= 𝑥 2 − 2𝑥 − 3 + 4  distribute
(𝑓 ∘ 𝑔)(𝑥) = 𝑥 2 − 2𝑥 + 1  simplify
Is (𝑓 ∘ 𝑔)(𝑥) = (𝑔 ∘ 𝑓)(𝑥)?
Example No. 9
Given 𝑓(𝑥) = 𝑥 + 4 and 𝑔(𝑥) = 𝑥 2 − 2𝑥 − 3, find (𝑔 ∘ 𝑓)(𝑥).
Solution
(𝑔 ∘ 𝑓)(𝑥) = 𝑔(𝑓(𝑥))
2
= (𝑓(𝑥)) − 2(𝑓(𝑥)) − 3
= (𝑥 + 4)2 − 2(𝑥 + 4) − 3
= (𝑥 2 + 8𝑥 + 16) − 2𝑥 − 8 − 3
= 𝑥 2 + 8𝑥 − 2𝑥 + 16 − 8 − 3
(𝑔 ∘ 𝑓)(𝑥) = 𝑥 2 + 6𝑥 + 5
From our examples, we can say that (𝑓 ∘ 𝑔)(𝑥) ≠ (𝑔 ∘ 𝑓)(𝑥).
The process of plugging in an argument (given number or expression) for the function’s variable and simplifying
the resulting equation is called evaluating functions.
Evaluating Functions
Evaluating
functionmeans
means
replacing
the variable
the function,
this
with from
a value
from the function’s
Evaluating aa function
replacing
the variable
in theinfunction,
in this in
case
x, case
with ax,value
the function’s
domain
andcomputing
computing
result.
To denote
are evaluating
f at aa for
some
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domain
of f, we
domain and
forfor
thethe
result.
To denote
that wethat
are we
evaluating
f at a for some
in the
domain
f, we
write f(a).
write f(a).
The function
function notation
tells
youyou
thatthat
y is ya function
of x. Ifofthere
a ruleisrelating
x, such to
as x,
y =such
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tells
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can also
then
youwrite:
can also write:
𝑓(𝑥) = 3𝑥 + 1
The name of the function is f. Other
letter may be used to name functions,
especially g and h.
f(x) is read as “f of x”, and this
represents the value of the function
at x.
Example No. 10
2𝑥−1
Given ℎ(𝑥) = 𝑥+1 and 𝑚(𝑥) = 2𝑥 − 2, find (ℎ ∘ 𝑚)(𝑥).
Solution
(ℎ ∘ 𝑚)(𝑥) = ℎ(𝑚(𝑥))
2(𝑚(𝑥)) − 1
=
(𝑚(𝑥)) + 1
2(2𝑥 − 2) − 1
=
(2𝑥 − 2) + 1
4𝑥 − 4 − 1
=
2𝑥 − 2 + 1
4𝑥 − 5
(ℎ ∘ 𝑚)(𝑥) =
2𝑥 − 1
Commented [Mr. Galay3]: −
 Practice
1. Perform the indicated operation.
3𝑥+2
𝑔(𝑥) = 1−2𝑥 and ℎ(𝑥) = 4𝑥 + 3. Find (ℎ ∘ 𝑔)(−5)
Page 7 of 10
MR. CHRISTIAN G. GALAY | Mathematics Teacher
ENRICHING MINDS OF CHAMPIONS
19
11
EASTERN MINDORO COLLEGE, INC.
Bagong Bayan II, Bongabong, Oriental Mindoro
Email Address: emc_1945@yahoo.com; Tel. No.: (043) 283-5479
GMATH11-M1-2021
LEARNING MODULE IN GENERAL MATHEMATICS
S.Y. 2021 – 2022
 Lesson 4 | Functions as Representations of Real-life Situations
Functions may be used as models in depicting real-life situations. The goal of functions is not necessarily to
produce actual results or accurately predict outcomes; the merely approximate the behavior of certain variables
to make us better understand the actual circumstances.
Problem No. 1
A car rental company charges ₱3,000 a day and ₱15 per kilometer for renting a car. Find a mathematical model
that will represent the relationship of the rental fee in a day and the distance traveled by a car.
Solution
We have the following given:
 There is a fix rental feel a day which is ₱3,000.
 An additional ₱15 per kilometer.
Let 𝑅(𝑥) be the rental fee and 𝑥 represent the distance traveled by the car in a day.
Thus, the function will be;
𝑅(𝑥) = 3,000 + 15𝑥
Problem No. 2
A user is charged ₱1,800 monthly for a particular mobile plan, which includes 200 free text messages.
Messages in excess of 200 are charge ₱1.00 each.
a) Represent the monthly cost for text messaging using function 𝑝(𝑚), where 𝑚 is the number of
messages sent in a month.
b) If a user sent 250 texts messages in a certain month, what is the total cost does he/she needs to pay at
the end of the month?
c) How many text messages should a user send if she/he intend to pay ₱2,100 at the end of the month?
Solution
a) Let 𝑝(𝑚) represent the amount paid by the user each month. The equation can be expressed as
piecewise function.
𝑝(𝑚) = {
1,800
, 0 ≤ 𝑚 ≤ 200
1,800 + 1(𝑚 − 200), 0 > 𝑚 ≥ 200
b) Given: the number of text messages is 250 (𝑚 = 250)
Required: total cost to pay at the end of the month
Since 𝑚 > 200, thus, we will use the second sub-function above
𝑝(250) = 1800 + 1.00(250 − 200)
𝑝(250) = 1,800 + 1.00(50)
𝑝(250) = 1850
Therefore, if a user sent 250 texts messages in a certain month, the total cost that she/he needs to pay
at the end of the month is ₱1,850.
c) Given: the total cost to pay at the end of the month is ₱2,100 (𝑝(𝑚) = 2,100)
Required: total number of text messages sent in a month
2100 = 1800 + 1.00(𝑚 − 200)
2100 = 1800 + 𝑚 − 200
2100 = 1600 + 𝑚
2100 − 1600 = 𝑚
500 = 𝑚
Thus, if a user intends to pay ₱2,100 at the end of the month, he/she can spend up to 500 messages.
Problem No. 3
A breakfast cereal company manufactures boxes to package their product. To comply with standards, the box
must have a length 3 times its width and a height 5 times its width.
a) Find a function that models the volume of the box.
b) Find the volume of the box if the width is 1.5 in.
c) For what width is the 90 𝑖𝑛3?
d) For what width is the volume greater than 60 𝑖𝑛3?
Solution
Given:
 Rectangular box
 The length is 3 times the width
 The height is 5 times the width
The volume of a rectangular box is
𝑣𝑜𝑙𝑢𝑚𝑒 = 𝑙𝑒𝑛𝑔ℎ𝑡ℎ ∗ 𝑤𝑑𝑡ℎ ∗ ℎ𝑒𝑖𝑔ℎ𝑡
Let w be the width of the box. Then,
𝑙𝑒𝑛𝑔𝑡ℎ = 3𝑤
Page 8 of 10
MR. CHRISTIAN G. GALAY | Mathematics Teacher
ENRICHING MINDS OF CHAMPIONS
EASTERN MINDORO COLLEGE, INC.
Bagong Bayan II, Bongabong, Oriental Mindoro
Email Address: emc_1945@yahoo.com; Tel. No.: (043) 283-5479
GMATH11-M1-2021
a)
b)
c)
d)
Page 9 of 10
LEARNING MODULE IN GENERAL MATHEMATICS
S.Y. 2021 – 2022
ℎ𝑒𝑖𝑔ℎ𝑡 = 5𝑤
𝑣𝑜𝑙𝑢𝑚𝑒 = 3𝑤 ∗ 𝑤 ∗ 5𝑤
The function that models the volume of the box is
𝑣(𝑤) = 15𝑤 3
Required:
Find the volume of the box of the width is 1.5 in
𝑣(𝑤) = 15𝑤 3
𝑣(1.5) = 15(1.5)3
𝑣(1.5) = 50.625 𝑖𝑛3
Answer: The volume of the box if the width is 1.5 in. is 50.625 in3.
Required:
For what width is the 90 in3?
𝑣(𝑤) = 15𝑤 3
90 𝑖𝑛3 = 15𝑤 3
6 𝑖𝑛3 = 𝑤 3
3
√6𝑖𝑛 = 𝑤
3
3
Answer: The volume is 90 in when the width is √6 𝑖𝑛.
Required:
For what width is the volume greater than 60 in3?
𝑣(𝑤) = 15𝑤 3
15𝑤 3 > 60 𝑖𝑛3
𝑤 3 > 4 𝑖𝑛3
3
𝑤 > √4 𝑖𝑛
3
Answer: The volume will be greater than √4 𝑖𝑛.
MR. CHRISTIAN G. GALAY | Mathematics Teacher
ENRICHING MINDS OF CHAMPIONS
EASTERN MINDORO COLLEGE, INC.
Bagong Bayan II, Bongabong, Oriental Mindoro
Email Address: emc_1945@yahoo.com; Tel. No.: (043) 283-5479
LEARNING MODULE IN GENERAL MATHEMATICS
S.Y. 2021 – 2022
GMATH11-M1-2021
 WORKSHEET
MDL1
Name:
Date Submitted:
Section:
I.
Perform any 3 the following operations by the given functions.
Functions: 𝑓 (𝑥 ) =
1. 𝑓 + 𝑔
3. 𝑓 × 𝑔
2. 𝑓 − 𝑔
4.
5.
𝑥−2
𝑥+2
1
; 𝑔(𝑥 ) = 𝑥
𝑓
𝑔
𝑔
𝑓
Write your solution here.
II.
1.
(𝑓 + 𝑔)(𝑥) =
2.
(𝑓 − 𝑔)(𝑥) =
3.
(𝑓 ∗ 𝑔)(𝑥) =
4.
( ) (𝑥) =
5.
( ) (𝑥) =
𝑥−2
𝑥+2
1
𝒙𝟐 −𝒙+𝟐
𝑥
𝒙𝟐+𝟐𝒙
+( )=
𝒙𝟐 −𝟑𝒙−𝟐
𝒙𝟐 +𝟐𝒙
𝒙−𝟐
𝒙𝟐 +𝟐𝒙
𝑓
𝑥 2 −2𝑥
𝑔
𝑥+2
𝑔
𝑥+2
𝑓
𝑥 2 −2𝑥
If 𝒇(𝒙) = 𝒙𝟐 + 𝟑𝒙 𝐚𝐧𝐝 𝒈(𝒙) = 𝒙 − 𝟐, find and simplify by the given composite functions:
A. 𝑓 ∘ 𝑔
B. 𝑓 ∘ 𝑓
(𝑓 ∘ 𝑔)(𝑥) = (𝑥 − 2)2 + 3(𝑥 − 2)
= 𝑥 2 − 4𝑥 + 4 + 3𝑥 − 6
= 𝒙𝟐 − 𝒙 − 𝟐
(𝑓 ∘ 𝑓)(𝑥) = (𝑥 2 + 3𝑥)2 + 3(𝑥 2 + 3𝑥)
= 𝑥 4 + 6𝑥 3 + 9𝑥 2 + 3𝑥 2 + 9𝑥
= 𝒙𝟒 + 𝟔𝒙𝟑 + 𝟏𝟐𝒙𝟐 + 𝟗𝒙
III. Answer the given problem.
6. The cost of hiring a catering service to serve food for a party is ₱150 per head for 20 persons or less, ₱130
per head for 21 to 50 persons, and ₱110 per head for 51 to 100 persons. For 100 or more persons, the cost
is at ₱100 per head. Represent the total cost as a piecewise function of the number of attendees of the party.
(Show your solution at the back of this sheet.)
Page 10 of 10
MR. CHRISTIAN G. GALAY | Mathematics Teacher
ENRICHING MINDS OF CHAMPIONS