Operations on Functions
Transcript of Operations on Functions
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GENERAL MATHEMATICS | UNIT 2
Operations on Functions
Table of Contents
Introduction ......................................................................................................................................... 3
Test Your Prerequisite Skills ............................................................................................................. 4
Objectives ............................................................................................................................................ 5
Lesson 1: Addition and Subtraction of Functions
- Warm Up! ................................................................................................................................. 5
- Learn about It! ........................................................................................................................ 6
- Letβs Practice! .......................................................................................................................... 7
- Check Your Understanding! ................................................................................................ 12
Lesson 2: Multiplication of Functions
- Warm Up! ............................................................................................................................... 13
- Learn about It! ...................................................................................................................... 13
- Letβs Practice! ........................................................................................................................ 14
- Check Your Understanding! ................................................................................................ 17
Lesson 3: Division of Functions
- Warm Up! ............................................................................................................................... 18
- Learn about It! ...................................................................................................................... 19
- Letβs Practice! ........................................................................................................................ 20
- Check Your Understanding! ................................................................................................ 23
Lesson 4: Composition of Functions
- Warm Up! ............................................................................................................................... 25
- Learn about It! ...................................................................................................................... 25
- Letβs Practice! ........................................................................................................................ 26
- Check Your Understanding! ................................................................................................ 29
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Challenge Yourself! .......................................................................................................................... 30
Performance Task ............................................................................................................................ 31
Wrap-up ............................................................................................................................................. 33
Key to Letβs Practice! ......................................................................................................................... 34
References ........................................................................................................................................ 35
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GRADE 11|GENERAL MATHEMATICS
UNIT 2
Operations on Functions
Mathematics is everywhere from computers to aircrafts, from
the kitchen to the most sophisticated laboratories. Math is
evident in all fields of sciences, and its application to daily life
is abundant. Hence, it is imperative to have a good
understanding of one of the most important concepts in
mathematics, functions.
We define a function as a relation between given sets
of elements. These relationships are evident either
numerically, algebraically, or through numerous
models of real-life situations. Everyday objects like
washing machine, oven toaster, computer, and
smartphone in your pockets revolve around the idea of relations and functions. The
underlying concepts in these machines may be technological, but ultimately they are
mathematical in nature.
In journey through Mathematics, you have encountered linear,
quadratic, and even polynomial functions. These functions, like real
numbers, can also be added, subtracted, multiplied, or divided with
each other.
This unit will help you learn the four different operations on
functions, along with a special operation known as composition.
Click Home icon to go back to
Table of Contents
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Before you get started, answer the following items on a separate sheet of paper. This will
help you assess your prior knowledge and practice some skills that you will need in
studying the lessons in this unit. Show your complete solution.
1. Perform the indicated operations.
a. (3 β ππ₯ β 3π₯2) β (βππ₯ β 2π₯2 β 4)
b. (π₯ + π¦)(π₯ β π¦)
c. (π₯ + 3)(π₯ β 2)
d. β156π₯2π¦4π§3
26π₯2π¦2π§2
e. (7π₯ + 12 + π₯2) Γ· (π₯ + 3)
2. Factor the following expressions completely.
a. 36π₯2π¦2 β 30π₯π¦3
b. π 2 β 81
c. π‘3 β 27
d. 64 + 34π + π2
e. π₯2 + 5π₯ + 6
3. Evaluate the following functions based on the given value of π₯.
a. π(π₯) = π₯2 β 4π₯ + 5; π₯ = β1
b. π(π₯) = π₯3 β π₯2 β 5π₯ + 12; π₯ = 3
c. π(π₯) = π₯4 β 4π₯2 β π₯ + 5; π₯ = β2
d. β(π₯) = 4(βπ₯4 + π₯ β 3π₯2); π₯ = 2
e. β(π₯) =π₯2+3π₯β2
π₯+5; π₯ = 5
β’ Adding, subtracting, multiplying, and dividing polynomials
β’ Evaluating functions
β’ Factoring polynomials
β’ Solving word problems involving functions and polynomials
Test Your Prerequisite Skills
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4. A farmer is given a rectangular piece of land that measures (π₯ + 3) by (2π₯2 + 1)
meters. What is the current area of the land owned by the farmer? If the
dimensions were doubled, what would be the new area?
At the end of this unit, you should be able to
β’ perform addition, subtraction, multiplication, division, and composition of
functions; and
β’ solve problems involving functions.
Letβs Roll!
Materials Needed: colored papers, pair of dice, pen, cartolina, marker
Instructions:
1. This activity can be played by pair or by group.
2. Your teacher will prepare 6 different sets of algebraic
expressions to be written on colored papers.
3. Each expression will be assigned a number from one
to six.
4. You will be given time to roll the pair of dice 5 times.
5. Record the results of your rolls for each pair and write the corresponding
expression.
Lesson 1: Addition and Subtraction of Functions
Objectives
Warm Up!
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6. For each pair of expressions, roll a dice again. If the resulting number is odd,
add the expressions. However, if the result is even, subtract the expressions.
7. Perform the indicated operation based on your roll and show your solution
by writing it on your paper or cartolina.
8. Your teacher will call volunteers to explain their work in front of the class.
Letβs say that the expressions you have encountered in the Warm Up! activity denote a
function. As a recall, we define a function as a special type of relation wherein each input
only has one output.
Functions can be used to model situations in real life. Since they are algebraic in nature,
the basic operations can be applied as what you have seen in Warm Up!.
We can denote the first expression from your first roll as π(π₯) while the second as π(π₯).
When given two functions π(π₯) and π(π₯), their sum is denoted by (π + π)(π₯). This is
defined by (π + π)(π₯) = π(π₯) + π(π₯).
By the definition of the sum of two functions, we just add the two given expressions for
π(π₯) and π(π₯).
Given the functions π(π₯) = 5π₯ + 2 and π(π₯) = 8 + 6π₯ β 2π₯2, let us find (π + π)(π₯).
(π + π)(π₯) = π(π₯) + π(π₯)
(π + π)(π₯) = (5π₯ + 2) + (8 + 6π₯ β 2π₯2) Substitute the given functions.
(π + π)(π₯) = (2 + 8) + (5π₯ + 6π₯) β 2π₯2 Group similar terms together.
(π + π)(π₯) = 10 + 11π₯ β 2π₯2 Combine similar terms.
Learn about It!
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Hence, (π + π)(π₯) = 10 + 11π₯ β 2π₯2 or (π + π)(π₯) = β2π₯2 + 11π₯ + 10.
On the other hand, if we are given two functions π(π₯) and π(π₯), then we can find their
difference, denoted by (π β π)(π₯) as (π β π)(π₯) = π(π₯) β π(π₯).
Since a function is being subtracted from another function, every term of the subtrahend
is being subtracted from the minuend. Recall that when we subtract algebraic
expressions, we copy the minuend, then change the operation from subtraction to
addition, and we change the signs of all the terms in the subtrahend.
Given the functions π(π₯) = 5π₯ + 2 and π(π₯) = 8 + 6π₯ β 2π₯2, let us find (π β π)(π₯).
(π β π)(π₯) = π(π₯) β π(π₯)
(π β π)(π₯) = (5π₯ + 2) β (8 + 6π₯ β 2π₯2) Substitute the given functions.
(π β π)(π₯) = (5π₯ + 2) + (β8 β 6π₯ + 2π₯2) Change the operation from
subtraction to addition. Change the
signs of all the terms in the
subtrahend.
(π β π)(π₯) = (2 β 8) + (5π₯ β 6π₯) + 2π₯2 Group similar terms together.
(π β π)(π₯) = β6 β π₯ + 2π₯2 Combine similar terms.
Hence, (π β π)(π₯) = β6 β π₯ + 2π₯2 or (π β π)(π₯) = 2π₯2 β π₯ β 6.
Let us study some other examples below to better understand addition and subtraction of
functions.
Example 1: Given the functions π(π₯) = 8 + 6π₯ β 2π₯2 and β(π₯) = 4(βπ₯4 + π₯ β 3π₯2), find
(π + β)(π₯).
Letβs Practice!
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Solution: We find (π + β)(π₯) by adding the two given expressions for π(π₯) and β(π₯).
(π + β)(π₯) = π(π₯) + β(π₯)
(π + β)(π₯) = (8 + 6π₯ β 2π₯2) + 4(βπ₯4 + π₯ β 3π₯2) Substitute the given functions.
(π + β)(π₯) = (8 + 6π₯ β 2π₯2) + (β4π₯4 + 4π₯ β 12π₯2) Distribute 4 to all of the terms
inside the second parentheses.
(π + β)(π₯) = (β4π₯4) + (β2π₯2 β 12π₯2) Group similar terms together.
+(4π₯ + 6π₯) + (8)
(π + β)(π₯) = β4π₯4 β 14π₯2 + 10π₯ + 8 Combine similar terms.
Try It Yourself!
Given the functions π(π₯) = π₯3 β π₯4 + 6π₯ β 3 and β(π₯) = 2(βπ₯4 β 2π₯ + 4π₯2), find (π +
β)(π₯).
Example 2: What is (π β π)(4) if π(π₯) = 5π₯ + 2 and π(π₯) = 15π₯ + 6?
Solution:
Step 1: Subtract the two given expressions for π(π₯) and π(π₯).
(π β π)(π₯) = π(π₯) β π(π₯)
(π β π)(π₯) = (5π₯ + 2) β (15π₯ + 6) Substitute the given functions.
(π β π)(π₯) = (5π₯ + 2) + (β15π₯ β 6) Change the operation from
subtraction to addition. Change the
signs of all the terms in the
subtrahend.
(π β π)(π₯) = (2 β 6) + (5π₯ β 15π₯) Group similar terms together.
(π β π)(π₯) = β4 β 10π₯ Combine similar terms.
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Step 2: Substitute π₯ = 4 to the resulting function.
(π β π)(4) = β4 β 10(4)
(π β π)(4) = β48
Try It Yourself!
What is (π β π)(3) if π(π₯) = 5π₯2 β 3 and π(π₯) = π₯2 β 15π₯ + 6
Example 3: Find (π + π)(π₯), given π(π₯) =π₯+7
π₯+1 and π(π₯) =
π₯β2
π₯+3.
Solution:
Step 1: Let us first look for the LCD of the two functions. The LCD is (π₯ + 1)(π₯ + 3) or
π₯2 + 4π₯ + 3.
Step 2: To find (π + π)(π₯), we solve for (π + π)(π₯) = π(π₯) + π(π₯).
(π + π)(π₯) = π(π₯) + π(π₯)
(π + π)(π₯) =π₯+7
π₯+1+
π₯β2
π₯+3 Substitute the given functions.
(π + π)(π₯) =(π₯+7)(π₯+3)+(π₯β2)(π₯+1)
π₯2+4π₯+3 Add the functions using the LCD.
(π + π)(π₯) =(π₯2+10π₯+21)+(π₯2βπ₯β2)
π₯2+4π₯+3 Simplify the numerator.
(π + π)(π₯) =(π₯2+π₯2)+(10π₯βπ₯)+(21β2)
π₯2+4π₯+3 Combine similar terms in the
numerator.
(π + π)(π₯) =2π₯2+9π₯+19
π₯2+4π₯+3
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Try It Yourself!
Find (π β π)(π₯), given π(π₯) =π₯+7
π₯+1 and π(π₯) =
π₯β2
π₯+3.
Real-World Problems
Example 4: The HR Department informed Karl that his starting
salary as a car sales agent will be
π(π₯) = 10π₯2 β 5π₯ + 12 pesos. He will receive
π(π₯) = 2π₯2 + 3π₯ β 5 pesos as incentive if he is able
to meet the company quota for six months. What
will be his new salary after six months, assuming
that he meets the quota?
Solution: To find his new salary after six months, add his starting salary and incentive.
(π + π)(π₯) = π(π₯) + π(π₯)
(π + π)(π₯) = (10π₯2 β 5π₯ + 12) + (2π₯2 + 3π₯ β 5) Substitute the given functions.
(π + π)(π₯) = (10π₯2 + 2π₯2) + (β5π₯ + 3π₯) + (12 β 5) Group similar terms together.
(π + π)(π₯) = 12π₯2 β 2π₯ + 7 Combine similar terms.
Hence, Karlβs monthly salary will be (12π₯2 β 2π₯ + 7) pesos if he will be able to
meet the quota after six months.
Example 5: Christian is making a cabinet. He
bought a slab of wood that is
π(π₯) = 9π₯2 β 3π₯ + 2 feet long. What is
the length of the wood after
π(π₯) = 2π₯2 + 5π₯ β 4 feet have been
cut off?
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Solution: Let π(π₯) = 9π₯2 β 3π₯ + 2 be the original length of the wood and
π(π₯) = 2π₯2 + 5π₯ β 4 be the part that was cut off. To find the length of the
remaining wood, subtract the measure of the part of the wood that was cut
off from the original length.
(π β π)(π₯) = π(π₯) β π(π₯)
(π β π)(π₯) = (9π₯2 β 3π₯ + 2) β (2π₯2 + 5π₯ β 4) Substitute the given functions.
(π β π)(π₯) = (9π₯2 β 3π₯ + 2) + (β2π₯2 β 5π₯ + 4) Change the operation from
subtraction to addition. Change
the signs of all the terms in the
subtrahend.
(π β π)(π₯) = (9π₯2 β 2π₯2) + (β3π₯ β 5π₯) + (2 + 4) Group similar terms together.
(π β π)(π₯) = 7π₯2 β 8π₯ + 6 Combine similar terms.
Therefore, the length of the remaining wood is (7π₯2 β 8π₯ + 6) feet.
Try It Yourself!
Johnny is making a balance sheet for his monthly allowance. He
already had two entries on his debit and three entries on his
credit. If his debit entries are π₯5 and 4π₯4, while the entries for his
credit are π₯3 + 3, βπ₯2, and π₯3, express his total allowance (π β
π)(5) in terms of his total debit and total credit defined by π(π₯)
and π(π₯), respectively. (Assume that all values are in pesos.)
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1. Consider the following functions:
π(π₯) = π₯2 + 7π₯ β 8
π(π₯) =π₯+7
π₯+2
π(π₯) =2π₯β3
π₯+2
π(π₯) = π₯ + 3
π(π₯) = 2π₯ β 7
π(π₯) = π₯2 + 5π₯ + 4
π(π₯) =π₯β2
π₯β3
β(π₯) =π₯β1
π₯+1
π(π₯) = π₯3 β 3π₯2 β 2π₯ + 8
Find:
a. (π + π)(π₯)
b. (π + π)(β3)
c. (π + π)(π₯)
d. (π β π)(π₯)
e. (π + π)(2)
f. (π β π)(β1)
g. (π β π)()
h. (π + β)(π₯)
i. (π β π)(1)
2. Consider the following functions:
π(π₯) = π₯2 β 2π₯ + 1
π(π₯) =2π₯+1
π₯β1
π(π₯) = π₯3 β π₯2 + 3π₯ + 1
π (π₯) = π₯2 + 3π₯ + 5
π‘(π₯) = 2π₯ + 1
π’(π₯) = 2π₯ β 1
π£(π₯) = π₯3 β 2π₯ β 1
a. Express π1(π₯) = π₯2 + 2 as a sum or difference of any two given functions.
b. Express π2(π₯) = π₯3 as a sum or difference of any two given functions.
c. Express π3(π₯) = βπ₯2 + 4π₯ as a sum or difference of any two given functions.
d. Express π4(π₯) = β5π₯ β 4 as a sum or difference of any two given functions.
e. Express π5(π₯) = π₯3 β π₯2 β 5π₯ β 6 as a sum or difference of any two given
functions.
3. Joanna is making a cake. The flour mixture is π(π₯) =π₯+3
π₯β5 grams while the other
ingredients are π(π₯) = π₯2 β 25 grams. What is the total weight π(π₯) of the mixture if
the flour is added to the other ingredients?
Check Your Understanding!
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Look, Pair, Share!
Materials Needed: colored papers, paper and pen
Instructions:
1. This activity can be played by the whole class.
2. Your teacher has hidden different colored papers under your armchair
containing algebraic expressions.
3. When your teacher gives the signal, collect the colored paper assigned to you and
look for someone with the same colored paper. Once you are with your partner,
multiply the two expressions written on the paper.
4. The first five pairs to give the correct answer win.
As what we have discussed previously, functions can be added or subtracted from each
other. Furthermore, aside from these two operations, they can also be multiplied.
If we denote the expression that you got from the activity in Warm Up! with π(π₯) and the
expression that your classmate got with πΆ(π₯), then we will have two functions. To find
their product, we multiply the two expressions.
Lesson 2: Multiplication of Functions
Warm Up!
Learn about It!
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When given two functions, say π(π₯) and πΆ(π₯), or π(π₯) and π(π₯), their product, denoted by
(π β π)(π₯), is the function defined by (π β π)(π₯) = π(π₯) β π(π₯). By the definition of the product
of two functions, we just multiply the two given expressions for π(π₯) and π(π₯).
Let us study the example given below.
Given the functions π(π₯) = π₯2 and π(π₯) = 5π₯ + 2, find (π β π)(π₯).
(π β π)(π₯) = π(π₯) β π(π₯)
(π β π)(π₯) = (π₯2)(5π₯ + 2) Substitute the given functions.
(π β π)(π₯) = (π₯2)(5π₯) + (π₯2)(2) Multiply the expressions.
(π β π)(π₯) = 5π₯3 + 2π₯2 Simplify.
Let us study other examples below to better understand multiplication of functions.
Example 1: What is (π β π)(π₯) if π(π₯) = 5π₯ + 2 and π(π₯) = 8 + 6π₯ β 2π₯2?
Solution: Multiply the two given expressions for π(π₯) and π(π₯).
(π β π)(π₯) = π(π₯) β π(π₯)
(π β π)(π₯) = (5π₯ + 2)(8 + 6π₯ β 2π₯2) Substitute the given functions.
(π β π)(π₯) = (5π₯)(8) + (5π₯)(6π₯) + (5π₯)(β2π₯2) Multiply the expressions.
+(2)(8) + (2)(6π₯) + (2)(β2π₯2)
(π β π)(π₯) = 40π₯ + 30π₯2 β 10π₯3 Simplify.
+16 + 12π₯ β 4π₯2
(π β π)(π₯) = (β10π₯3) + (30π₯2 β 4π₯2) Group similar terms together.
+(40π₯ + 12π₯) + (16)
(π β π)(π₯) = β10π₯3 + 26π₯2 + 52π₯ + 16 Combine similar terms.
Letβs Practice!
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Try It Yourself!
What is (π β π)(π₯) if π(π₯) = 3π₯ β 2 and π(π₯) = 3π₯2 + 2π₯ β 1?
Example 2: What is (π β π)(2) if π(π₯) = 6 β 2π₯ and π(π₯) = 4 β 3π₯?
Solution:
Step 1: Multiply the two given expressions for π(π₯) and π(π₯).
(π β π)(π₯) = π(π₯) β π(π₯)
(π β π)(π₯) = (6 β 2π₯ )(4 β 3π₯) Substitute the given functions.
(π β π)(π₯) = (6)(4) + (6)(β3π₯) Multiply the expressions.
+(β2π₯)(4) + (β2π₯)(β3π₯)
(π β π)(π₯) = 24 β 18π₯ β 8π₯ + 6π₯2 Simplify.
(π β π)(π₯) = 6π₯2 + (β18π₯ β 8π₯) + 24 Group similar terms together.
(π β π)(π₯) = 6π₯2 β 26π₯ + 24 Combine similar terms.
Step 2: Substitute π₯ = 2 to the resulting function.
(π β π)(2) = 6(2)2 β 26(2) + 24 Evaluate.
(π β π)(2) = 24 β 52 + 24 Simplify. (π β π)(2) = β4
Try It Yourself!
What is (π β β)(2) if π(π₯) = 20 β π₯2 and β(π₯) = 5 β π₯?
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Example 3: Find (π β π)(π₯), given π(π₯) =π₯+7
π₯+1 and π(π₯) =
π₯β2
π₯+3.
Solution: Multiply the two given expressions for π(π₯) and π(π₯).
(π β π)(π₯) = π(π₯) β π(π₯)
(π β π)(π₯) = (π₯+7
π₯+1) (
π₯β2
π₯+3) Substitute the given functions.
(π β π)(π₯) =(π₯+7)(π₯β2)
(π₯+1)(π₯+3) Multiply the numerator and the
denominator.
(π β π)(π₯) =(π₯)(π₯)+(π₯)(β2)+(7)(π₯)+(7)(β2)
(π₯)(π₯)+(π₯)(3)+(1)(π₯)+(1)(3 ) Multiply the expressions on the
numerator and denominator.
(π β π)(π₯) =π₯2β2π₯+7π₯β14
π₯2+3π₯+π₯+3 Simplify.
(π β π)(π₯) =π₯2+5π₯β14
π₯2+4π₯+3 Combine similar terms.
Try It Yourself!
Find (π β π)(π₯), given π(π₯) =π₯β5
π₯β3 and π(π₯) =
π₯β1
π₯+3.
Real-World Problems
Example 4: A farmer has a rectangular plot of
land. The length of the plot of land
can be expressed as a function
π(π₯) = π₯ + 2 meters and its width
can be expressed as π€(π₯) = 3π₯ + 21 meters. Find the area of the plot.
Solution: Use the formula for the area of a rectangle, which is π΄ = π β π€.
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(π β π€)(π₯) = π(π₯) β π€(π₯)
(π β π€)(π₯) = (π₯ + 2)(3π₯ + 21)
(π β π€)(π₯) = (π₯)(3π₯) + (π₯)(21) + (2)(3π₯) + (2)(21)
(π β π€)(π₯) = π₯2 + 21π₯ + 6π₯ + 42
(π β π€)(π₯) = π₯2 + 27π₯ + 42
Thus, the area of the farmerβs plot of land is (π₯2 + 27π₯ + 42 ) square meters.
Try It Yourself!
If the plot of land in Example 4 will be converted into a
warehouse, what would be the total volume π(π₯) of the
structure if the height of the walls from the floor to the flat
roof is (π₯ + 5) meters high.
1. Consider the following functions:
π(π₯) = π₯ + 1
π(π₯) = π₯2 β 1
π(π₯) = 2π₯ + 3
π(π₯) = 3π₯ β 5
π(π₯) = 3π₯2 + 4
π(π₯) = 5π₯ β 12
π(π₯) =π₯2β1
3π₯+5
β(π₯) =π₯2 β 2π₯ + 1
π₯ + 1
π(π₯) =π₯2 β 1
3π₯2 + π₯ β 5
Find:
a. (π β π)(π₯)
b. (π β π)(β2)
c. (π β π)(π₯)
d. (π β π)(π₯)
e. (π β π)(3)
f. (π β β)(β3)
g. (π β π)(π₯)
h. (π β π)(π₯)
i. (π β β)(2)
j. (β β π)(β2)
Check Your Understanding!
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2. Consider the following functions:
π(π₯) = 2π₯ + 4
π(π₯) = π₯2 + 2π₯ β 3
π(π₯) = π₯ β 1
π (π₯) = π₯2 β 3
π‘(π₯) = π₯ + 2
π’(π₯) = 2π₯ β 1
π£(π₯) = π₯2 + 2π₯
a. Express π1(π₯) = π₯2 + π₯ β 2 as a product of any two given functions.
b. Express π2(π₯) = 2π₯2 + 2π₯ β 4 as a product of any two given functions.
c. Express π3(π₯) = 2π₯3 + 8π₯2 + 8π₯ as a product of any two given functions.
d. Express π4(π₯) = 2π₯3 + 3π₯2 β 3π₯ as a product of any two given functions.
e. Express π5(π₯) = π₯4 + 2π₯3 β 6π₯2 β 6π₯ + 9 as a product of any two given functions.
3. The area of the base of a cylindrical container is defined by π(π₯) = π₯2 β π₯ + 12. Find
the volume of water that can be stored inside the container if the height is defined
by β(π₯) = π₯ β 1.
First Three!
Materials Needed: paper and pen
Instructions:
1. This activity may be done individually or by pair.
2. Your teacher will present a problem on the board. There will be 2 questions per
category.
Lesson 3: Division of Functions
Warm Up!
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3. For the easy category, you will be asked to answer the problem mentally and
only the final answer will be considered. For the average category, you will be
asked to show your solution using synthetic division. For the difficult category,
you will be asked to solve using long division.
4. The first five persons or pairs to give the correct answer gain a point or
incentive.
5. Individuals or pairs who have correctly answered the first question for each
category may be exempted to give chance to others.
Sample Questions:
Easy: (π₯2 β 25) Γ· (π₯ + 5)
Average: (π₯3 + 3π₯2 β 13π₯ β 15) Γ· (π₯ + 5)
Difficult: (2π₯4 + 8π₯3 β 20π₯2 β 56π₯ β 30) Γ· (2π₯ + 2)
Algebraic expressions can be used to represent rules for relations and consequently,
functions. In this lesson, we will discuss how to find the quotient of two given functions.
When given two functions π(π₯) and π(π₯), their quotient, denoted by (π
π) (π₯), is the function
defined by (π
π) (π₯) =
π(π₯)
π(π₯), where π(π₯) β 0. By the definition of the quotient of two
functions, we just divide the two given expressions for π(π₯) and π(π₯).
Let us study the example given below.
Given π(π₯) = 5π₯3 + 12π₯2 + 4π₯ and π(π₯) = π₯2, find (π
π) (π₯).
Learn about It!
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(π
π) (π₯) =
π(π₯)
π(π₯)
(π
π) (π₯) =
5π₯3+12π₯2+4π₯
π₯2 Substitute the given functions.
(π
π) (π₯) =
π₯(5π₯2+12π₯+4)
π₯2 Factor both the dividend and the divisor
(π
π) (π₯) =
5π₯2+12π₯+4
π₯ Cancel the common factor/s.
Keep in mind that aside from factoring, there are also other ways to simplify and find
quotients of functions like long division and synthetic division.
Let us study some other examples below to better understand division of functions.
Example 1: What is (π
π) (π₯), if π(π₯) = π₯2 β 1 and π(π₯) = π₯ + 1?
Solution: Divide the two given expressions for π(π₯) and π(π₯).
(π
π) (π₯) =
π(π₯)
π(π₯)
(π
π) (π₯) =
π₯2β1
π₯+1 Substitute the given functions.
(π
π) (π₯) =
(π₯+1)(π₯β1)
π₯+1 Factor the dividend.
(π
π) (π₯) = π₯ β 1 Divide/ Simplify the common factor.
Try It Yourself!
What is (π
π) (π₯) =
π(π₯)
π(π₯), if π(π₯) = π₯2 + 6π₯ + 5 and π(π₯) = π₯ + 1?
Letβs Practice!
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Example 2: Find (π
π) (π₯) if π(π₯) = 2π₯3 β 7π₯2 + 2π₯ + 3 and π(π₯) = π₯ β 3.
Solution: Divide the two given expressions for π(π₯) and π(π₯).
Step 1: Substitute the given functions.
(π
π) (π₯) =
π(π₯)
π(π₯)
(π
π) (π₯) =
2π₯3 β 7π₯2 + 2π₯ + 3
π₯ β 3
Step 2: We can try factoring the numerator. However, it may be time-consuming so
let us try dividing the functions directly. We can use long division or synthetic
division to accomplish this.
3 2 β7 2 3
6 β3 β3
2 β1 β1 0
Since we have a remainder of zero, we can conclude that the quotient of the
two given functions is 2π₯2 β π₯ β 1.
Try It Yourself!
Find (π
π) (π₯) if π(π₯) = 2π₯4 β 5π₯3 β 5π₯2 + 5π₯ + 3 and π(π₯) = π₯ β 3.
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Example 3: Find (π
π) (5) if π(π₯) = π₯3 β 1 and π(π₯) = π₯2 β 1.
Solution:
Step 1: Divide the two given expressions for π(π₯) and π(π₯).
(π
π) (π₯) =
π(π₯)
π(π₯)
(π
π) (π₯) =
π₯3β1
π₯2β1 Substitute the given functions.
(π
π) (π₯) =
(π₯β1)(π₯2+π₯+1)
(π₯β1)(π₯+1) Factor the dividend and divisor.
(π
π) (π₯) =
π₯2+π₯+11
π₯+1 Divide/Simplify the common factor.
Step 2: Substitute π₯ = 5 to the resulting function.
(π
π) (5) =
(5)2
+5+1
5+1 Evaluate.
(π
π) (5) =
31
6 Simplify.
Try It Yourself!
Find (π
π) (5) if π(π₯) = π₯3 β 1 and π(π₯) = π₯2 β 2π₯ + 1.
Real-World Problems
Example 4: A train traveling at top speed reached its destination in π‘(π₯) = π₯ β 2 hours. If
the distance traveled by the train is π(π₯) = π₯2 + 3π₯ β 10 kilometers, find the
top speed π£(π₯) of the train.
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Solution: Distance divided by time is equal to speed, denoted by π£(π₯) = (π
π‘) (π₯) =
π(π₯)
π‘(π₯).
Thus, we can obtain the required speed by dividing the given functions that
represent distance and time.
From the problem, we have π(π₯) = π₯2 + 3π₯ β 10 and π‘(π₯) = π₯ β 2.
2 1 3 β10
2 10
1 5 0
Therefore, the speed of the train is (π₯ + 5 ) kilometers per hour.
Try It Yourself!
An electronically controlled printing machine can print a number
of words defined by π€(π₯) = π₯4 β π₯3 + 2π₯2 β π₯ β 1. If a sheet of
paper fed in the printer will only accommodate π(π₯) = π₯ β 1
words, how many sheets of paper can the machine print on? Use
the function π (π₯) to define the number of sheets.
1. Consider the following functions:
π(π₯) = π₯ + 1
π(π₯) = π₯ β 2
π(π₯) = π₯ + 3
π(π₯) = π₯2 + 4π₯ + 1
π(π₯) = π₯2 β 125
π(π₯) = 5π₯ β 10
π(π₯) =π₯2 β 1
π₯ β 5
β(π₯) =π₯2 + 2π₯ + 1
π₯ β 2
π(π₯) =π₯2 β 1
π₯2 + π₯ β 6
π(π₯) = π₯3 + 2π₯2 β 6π₯ β 2
Check Your Understanding!
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Find:
a. (π
π) (0)
b. (π
π) (β2)
c. (π
π) (1)
d. (π
π) (β3)
e. (π
π) (3)
f. (π
π) (β3)
g. (β
π) (β2)
h. (π
π) (1)
i. (π
π) (2)
j. (π
π) (β1)
2. Consider the following functions:
π(π₯) = 2π₯ + 1
π(π₯) = 2π₯ β 2
β(π₯) = 2π₯2 β 5π₯ β 3
π(π₯) = 2π₯3 β 7π₯2 + 2π₯ + 3
π(π₯) = π₯2 β 2π₯ + 1
π(π₯) =2π₯ + 1
π₯ β 1
a. Express π1(π₯) = π₯ β 1 as a quotient of any two given functions.
b. Express π2(π₯) = 1
π₯β1 as a quotient of any two given functions.
c. Express π3(π₯) =π₯β1
2 as a quotient of any two given functions.
d. Express π4(π₯) = π₯ β 3 a quotient of any two given functions.
e. Express π5(π₯) = 2π₯2β5π₯β3
2 as a quotient of any two given functions.
3. A train travels from Tutuban to Calamba station. The distance between Tutuban
and Calamba is defined by the function π(π₯) = π₯4 + π₯3 + 3π₯2 + 4π₯ β π₯ kilometers. If
the train travels from Tutuban to Calamba for π‘(π₯) = π₯ hours without stopping at
any station, how fast is the train traveling?
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Input-Process-Output
Materials Needed: paper and pen, marker and cartolina, drawing and coloring
materials
Instructions:
1. This activity may be done individually or by group.
2. The teacher will let you roam around the school to observe or imagine your day-
to-day activities. Recall all activities that involve processes wherein there is an
input material and an output product.
3. You will be given time to make an illustration regarding the said process
focusing on the flow: INPUT-PROCESS-OUTPUT.
4. A volunteer will be called to show and briefly explain their work.
The activity in Warm Up! is a great way on how to envision functions in general. It
functions the same way as other machines that you may encounter in real life. As what we
have elaborated in the previous activity, some machines require an input in their process
that will result in a specific output. If you put dirty clothes in a washing machine and let it
tumble and wash, then you will have washed clothes.
Lesson 4: Composition of Functions
Warm Up!
Learn about It!
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The same idea is also present when we evaluate functions in terms of the value of π₯.
However, how will we proceed with our solution if the input to our function is another
function?
The process of combining functions, where the output of one function is used as the input
of the other function, is known as function composition. Given two functions π(π₯) and
π(π₯), their composite is denoted by (π β π)(π₯) = π(π(π₯)). It is read as βf composed with g of
xβ or βf of g of xβ.
To have a better understanding of composition of functions, letβs study the examples
below.
Given two functions π(π₯) = 5π₯ + 2 and π(π₯) = π₯2, find (π β π)(π₯).
(π β π)(π₯) = π(π(π₯))
(π β π)(π₯) = π(π₯2) Substitute the inner function.
(π β π)(π₯) = 5(π₯2) + 2 Evaluate the main function in terms of the inner function.
(π β π)(π₯) = 5π₯2 + 2 Simplify.
Notice that if there are more terms in the inner function, the main function gains more
terms.
Example 1: What is (π β π)(π₯) if π(π₯) = 5π₯3 + 12π₯2 + 4π₯ and π(π₯) = π₯2.
Solution:
(π β π)(π₯) = π(π(π₯))
(π β π)(π₯) = π(π₯2) Substitute the given input function.
Letβs Practice!
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(π β π)(π₯) = 5(π₯2 )3 + 12(π₯2 )2 + 4(π₯2) Evaluate in terms of the input
function.
(π β π)(π₯) = 5π₯6 + 12π₯4 + 4π₯2 Simplify.
Try It Yourself!
What is (π β π)(π₯) if π(π₯) = 2π₯3 β 4π₯2 + 12π₯ + 3 and π(π₯) = 4π₯?
Example 2: What is (π β π)(π₯) if π(π₯) = 5π₯3 + 12π₯2 + 4π₯ and π(π₯) = π₯2 + 6.
Solution:
(π β π)(π₯) = π(π(π₯))
(π β π)(π₯) = π(5π₯3 + 12π₯2 + 4π₯) Substitute.
(π β π)(π₯) = (5π₯3 + 12π₯2 + 4π₯)2 + 6 Evaluate and simplify.
(π β π)(π₯) = 25π₯6 + 120π₯5 + 184π₯4 Simplify.
+96π₯3 + 16π₯2 + 6
Try It Yourself!
What is (π β π)(π₯) if π(π₯) = π₯3 β π₯2 + 4 and π(π₯) = π₯ + 1?
Example 3: Find (π β π)(2) given π(π₯) = 2π₯3 β 4π₯2 β 5 and π(π₯) = π₯2.
Solution:
(π β π)(π₯) = π(π(π₯))
(π β π)(π₯) = π(π₯2) Substitute the given input function.
(π β π)(π₯) = 2(π₯2 )3 β 4(π₯2)2 β 5 Evaluate in terms of the input
function.
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(π β π)(π₯) = 2π₯6 β 4π₯4 β 5 Simplify.
(π β π)(2) = 2(2)6 β 4(2)4 β 5 Evaluate in terms of the given value of
π₯.
(π β π)(2) = 128 β 64 β 5 Simplify.
(π β π)(2) = 59
Try It Yourself!
What is (π β π)(4) if π(π₯) = π₯2 + 3π₯ β 12 and π(π₯) = π₯ + 2.
Real-World Problems
Example 4: Angelo works 45 hours a week at a printing press. He
receives a weekly salary of β±1500 plus 2%
commission for every 1000 printouts. Assuming that
Angelo makes enough printouts this week to earn his
commission, what is the composition that represents
his commission, given the functions π(π₯) = 0.02π₯ and
π (π₯) = π₯ β 1000?
Solution: The composition (π β π )(π₯) means to take the number of printouts π₯,
subtract the 1000 printouts, and then multiply by 2%.
(π β π )(π₯) = π(π (π₯))
(π β π )(π₯) = π(π₯ β 1000) Substitute the given input function.
(π β π )(π₯) = 0.02(π₯ β 1000) Evaluate in terms of the input function.
(π β π )(π₯) = 0.02π₯ β 20 Simplify.
Thus, the composite function for his commission is (π β π )(π₯) = 0.02π₯ β 20.
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Try It Yourself!
The radius of a spherical balloon inflates at a rate of π(π‘) = 3π‘
centimeters per π‘ seconds. What will be the total increase in
volume of the balloon after π‘ seconds? How large is the balloon
from one side to the other side if it can be totally inflated in 5
seconds? (Hint: π =4
3ππ3)
1. Consider the following functions:
π(π₯) = 2π₯ β 1
π(π₯) = 3π₯
π(π₯) = π₯2 β 1
π(π₯) = π₯2 β 2π₯ + 1
π(π₯) =π₯ + 1
2π₯
π(π₯) =π₯2 β π₯ + 5
π₯ β 2
Find:
a. π(π(π₯))
b. π(π(π₯))
c. π(π(π₯))
d. π(π(π₯))
e. π(π(π₯))
f. π(π(π₯))
g. π(π(π₯))
h. π(π(π₯))
i. π(π(π₯))
j. π(π(π₯))
2. Consider the following functions:
π(π₯) = 9 β π₯ π(π₯) = π₯2 β π₯ β(π₯) = π₯ + 2
a. Evaluate (π β β)(β3).
b. Evaluate (β β π)(2).
c. Evaluate (π β π)(1).
d. Evaluate (π β π)(β2).
e. Evaluate (π β π)(3).
f. Evaluate (π β β)(5).
Check Your Understanding!
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STUDY GUIDE
3. Every person needs to pay taxes due to him or her. Suppose a freelance editor is
accountable to pay 10% tax each month. Jill, a freelance editor, earns π₯ pesos daily
for 20 days each month. How much tax will she pay? Express your answer as a
composition of functions.
1. In physics, waves may combine constructively or destructively, as shown in the
following diagrams.
Write two functions π(π₯) and π(π₯) (do not choose linear functions). Then, calculate
their sum (π + π)(π₯) and difference (π β π)(π₯). Graph all of these functions. How do
these graphs compare with the diagrams of constructive and destructive
interference?
2. The following table shows the average price per kilogram of fish, chicken, and beef
for the years 2013 to 2015, as well as the total amount of each type consumed by a
family (in kilograms).
Year Price of
Fish
Quantity
of Fish
Price of
Chicken
Quantity
of Chicken
Price of
Beef
Quantity
of Beef
2013 β±80 40 β±250 22 β±270 18
2014 β±85 55 β±280 25 β±300 25
2015 β±90 50 β±300 28 β±320 27
Challenge Yourself!
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How much money was spent on all the food described in the table? Create a
model to describe the expenditure for each year. Compare the differences
between the years on the table given.
3. βAll functions are relations but not all relations are functionsβ. Explain briefly the
context of the given statement, and give 5 examples using the concepts you have
learned in this unit to support your answer. Show your complete solution.
4. Suppose π(π₯) = π₯ is a function that defines the total number of shoes sold at a
department store, π(π₯) = 1200 β 4π₯ is the price of shoes bought, for 0 β€ π₯ β€ 20, and
π(π₯) = 0.2π₯ is the discount given during a sale.
a. What function will represent the total revenue earned?
b. What function will represent the discount for items sold at a discounted price?
c. What function will represent the total revenue earned if there is a sale?
5. Let π₯ represent the price of a regular laptop computer.
a. Give a function π that will represent the price of the laptop when a 1000 pesos
price reduction is applied.
b. Give a function π that will represent the price of the laptop when a 10%
discount will be given.
c. Solve for π(π(π₯)) and π(ππ₯)). What do these two composite functions
represent? Which of the two is more beneficial to the customer? Explain briefly.
As a Senior High School student, you will be encountering three different types of
subjects, the CORE, SPECIALIZED, and APPLIED subjects. One of the core subjects you will
study alongside General Mathematics is Disaster Readiness and Risk Reduction.
As a collaborative approach, your teacher in General Mathematics and DRRR asked your
class to prepare a research proposal integrating the concepts about environmental
preservation and operations and composition of functions. The preliminary paper to be
passed should contain an introduction and mathematical justifications of real-life
Performance Task
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STUDY GUIDE
models of functions based on and related to the stated environmental preservation
topics written on the introduction.
You are required to pass a 2-page introduction, with 1.5-line spacing, font size of 12, and a
font style of Arial Narrow. A mathematical justification part will follow the introduction
containing, but not limited to, ten real-life word problems or situations modeling the
concepts of operations on functions and composition of functions together with their
respective solutions. The solutions for the word problems must also be included in this
part. An initial title may also be included but not required. You may be asked to write
your research proposal individually or by groups of four.
Your research proposal will be evaluated by the two teachers based on the following
criteria in the given performance task rubric as follows.
The top 5 individuals or groups with the highest points will be asked to present their work
in a mini class research symposium to highlight the relevance of environmental
preservation.
Performance Task Rubric
Criteria
Below
Expectation
(0β49%)
Needs
Improvement
(50β74%)
Successful
Performance
(75β99%)
Exemplary
Performance
(99+%)
Mathematical
Justification
Justification is
ambiguous.
Only few
concepts
learned in the
unit are
applied.
Justification are
not so clear.
Some ideas are
not connected
to each other.
Not all concepts
learned in the
unit are applied.
Justification is
clear and
informatively
delivered.
Appropriate
concepts
learned in the
unit are
applied.
Justification is
logically clear,
informative, and
professionally
delivered. The
concepts learned
in the unit are
applied.
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Accuracy
The
information
given are
erroneous and
do not show
wise use of
the concepts
from the
integrated
subjects.
The information
given are
erroneous and
show some use
of the concepts
from the
integrated
subjects.
The
information
given are
accurate and
shows use of
the concepts
from the
integrated
subjects.
The information
given are
accurate and
show a wise use
of the concepts
from the
integrated
subjects.
Impact
The
introduction is
poor.
The
introduction is
somewhat
informative.
The
introduction is
informative
and flawless,
with minimal
errors.
The introduction
is very
informative and
flawlessly done.
It is also easily
understandable.
Efficiency
The
introduction
has more than
10
grammatical,
capitalization,
spelling, or
any errors.
The
introduction has
6 to 10
grammatical,
capitalization,
spelling, or any
errors.
The
introduction
has 1 to 5
grammatical,
capitalization,
spelling, or any
errors.
The introduction
has no
grammatical,
capitalization,
spelling, or any
errors.
Operation Formula
Addition
(π + π)(π₯) = π(π₯) + π(π₯)
Subtraction
(π β π)(π₯) = π(π₯) β π(π₯)
Wrap-up
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Multiplication
(π β π)(π₯) = π(π₯) β π(π₯)
Division
(π
π) (π₯) =
π(π₯)
π(π₯)
Composition
(π β π)(π₯) = π(π(π₯))
Lesson 1
1. (π + β)(π₯) = β3π₯4 + π₯3 + 8π₯2 + 2π₯ β 3
2. (π β π)(π₯) = 72
3. (π β π)(π₯) =11π₯+23
π₯2+4π₯+3
4. (π β π)(5) = 5397. Therefore, he has β±5,397 left for his allowance.
Lesson 2
1. (π β π)(π₯) = 9π₯3 β 7π₯ + 2
2. (π β β)(2) = 48
3. (π β π)(π₯) =π₯2β6π₯+5
π₯2β9
4. The total volume of the structure is defined by π(π₯) = π₯3 + 32π₯2 + 177π₯ + 210
Lesson 3
1. (π
π) (π₯) = π₯ + 5
2. (π
π) (π₯) = 2π₯3+π₯2 β 2π₯ β 1
3. (π
π) (5) =
31
4
4. A total of (π₯3 + 2π₯ + 1) sheets are needed.
Key to Letβs Practice!
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Lesson 4
1. (π β π)(π₯) = 128π₯3 β 64π₯2 + 48π₯ + 3
2. (π β π)(π₯) = π₯3 + 2π₯2 + π₯ + 4
3. (π β π)(4) = 42
4. The volume of the balloon will be π(π₯) = 4500π or π(π₯) β 14137.16694 cubic
centimeters.
Pearson Education. βFunction Operations and Composition.β Accessed April 16, 2018.
https://www.pvamu.edu/universitycollege/wp-content/uploads/sites/71/CACFC2.pdf
Math is Fun. βFunctions.β Accessed April 16, 2018.
http://www.mathisfun.com/definitions/function.html
Math Motivation. βThe Algebra of Functions.β Accessed January 7, 2017.
http://www.mathmotivation.com/ lectures/Algebra-Of-Functions.pdf
Versoza, Debbie Marie. B., et al. General Mathematics. Quezon City: Lexicon Press, Inc.,
2016.
References