I. Functions
Functions and It’s Concept
Relation, Dependent Variable and
Independent Variable
Relation is a set of ordered pairs. It is defined as a
pairing of the elements of the first set to the elements in the
second set that follow a certain rule of correspondence.
Relation, Dependent Variable and
Independent Variable
Distance Traveled (x)
0m
Taxi Fare (y)
₱40
300 m
₱43.50
600 m
₱47
900 m
₱50.50
1500 m
₱57.50
Which value depends on the other value?
Relation, Dependent Variable and
Independent Variable
Distance Traveled (x)
0m
Taxi Fare (y)
₱40
300 m
₱43.50
600 m
₱47
900 m
₱50.50
1500 m
₱57.50
Dependent Variable (y) is the set of all values in a relation called the range;
while
Independent Variable (x) is the set of all values in a relation called the
domain.
Functions
A function is a relation wherein each element of the first set is paired
with exactly one element of the second set.
Examples:
• The temperature in Celsius that corresponds to a temperature in Fahrenheit
• The height of an object thrown straight upward and the time in seconds since
it was thrown
• The total cost of buying vaccine for an entire locality and the cost per dose of
the vaccine
Functions as a Machine
• If a function is defined by a rule or by an
equation using independent and dependent
variables x and y, respectively, then we say
that
“y is a function of x.”
• In symbols, we denote this function
notation as
𝑦=𝑓 𝑥 .
• Here, we take note that 𝑓 𝑥 is equivalent
to y, thus denoting the dependent variable.
Functions
Amount of Gasoline
Total Price
0L
₱0
1L
₱42.50
2L
₱85
3L
₱127.50
4L
₱170
𝒇 𝒙 = 𝟒𝟐. 𝟓𝟎𝒙
Functions
Amount of Gasoline
Total Price
0L
₱0
₱42.50 0 = ₱0
1L
₱42.50
₱42.50 1 = ₱42.50
2L
₱85
₱42.50 2 = ₱85
3L
₱127.50
₱42.50 3 = ₱127.50
4L
₱170
₱42.50 4 = ₱170
𝒇 𝒙 = 𝟒𝟐. 𝟓𝟎𝒙
Vertical Line Test
Not a function
A function
Vertical Line Test
• The vertical line test is used to determine if a curve is a graph of
a function or not.
• A function can have only one output, y, for each unique input, x.
• If all vertical lines intersect a curve at most once, the curve
represents a function.
• However, if a vertical line is drawn over a graph of a relation
and the line intersects the graph in more than one point, then the
relation is not a function.
Example 1.1. Identifying Independent and
Dependent Variable
a.
The calendar month and the number of days in each month in
a non-leap year
b.
A real number and its cube
c.
Year
Estimated Population in Town A
2023
2025
2028
2032
97 600
102 800
110 600
123 600
Example 1.1. Identifying Independent and
Dependent Variable
d.
The graph shows the relation between the number of orders of
a homemade dessert and its price per pack (in ₱)
e.
y = 2x -5
Example 1.2. Determining the domain and range
a.
The calendar month and the number of days in each month in
a non-leap year
b.
A real number and its cube
c.
Year
Estimated Population in Town A
2023
2025
2028
2032
97 600
102 800
110 600
123 600
Example 1.2. Determining the domain and range
d.
e.
y = 2x -5
Example 1.3. Determining whether a relation is
a function or not
Decide whether each given relation defines a function or not.
a.
b.
Time (in years)
3
5
7
9
Height (in cm)
94
108
122
133
Example 1.3. Determining whether a relation is
a function or not
Decide whether each given relation defines a function or not.
c.
{(5,5), (-1,-7), (-3,-3), (5,-5), (-7,-1)}
d.
f x = 𝑥2 + 1
e.
𝑦2 = 𝑥
Example 1.4. Evaluating functions
*To evaluate a function means to find the value of the function at a
given value of x or any variable.
Let us evaluate the following functions for the indicated value of x.
a.
𝑓 𝑥 = 2𝑥 − 6 ; 𝑥 = 5
b. ℎ 𝑥 = 𝑥 2 + 4𝑥 ; 𝑥 = −3
c.
𝑔 𝑎 =
𝑎2 + 𝑎
𝑎−1
;
𝑎 = −2
Evaluate the following functions, where f and q are as
defined in example 1:
1. 𝑓(3𝑥 − 1)
2. 𝑞(2𝑥 + 3)
Formative Assessment Part 1
I.
Determine whether each given relation defines a function or not, then
identify the domain and range of each given.(3 points each)
1)
2)
{(5,5), (-1,6), (3,-2), (-5,3), (5,8)}
x
y
-3
5
8
4
3
7
2
5
1
3)
II.
Evaluate the following functions:(3 points each)
1)
f x = 4x 3 + 3x 2 − 5x − 12;
2)
f x = x + 2 2x − 1 ;
x = −3
x=5
I. Functions
Operations on Functions
Operations of Functions
Function can also be combined to create new functions using
different operations. We use the following definitions of sum,
difference, product, quotient, and composition of functions.
Let 𝑓(𝑥) and 𝑔(𝑥) be functions.
• Sum: 𝑓 + 𝑔 𝑥 = 𝑓 𝑥 + 𝑔(𝑥)
• Difference: 𝑓 − 𝑔 𝑥 = 𝑓 𝑥 − 𝑔(𝑥)
• Product: 𝑓 ∙ 𝑔 𝑥 = 𝑓 𝑥 ∙ 𝑔(𝑥)
• Quotient:
𝑓
𝑔
𝑥 =
𝑓(𝑥)
; 𝑔(𝑥)
𝑔(𝑥)
≠0
• Composition: (𝑓 ◦ 𝑔)(𝑥) = 𝑓(𝑔(𝑥))
Example 1.5. Performing Operations on Functions
Perform the indicated operations on the following functions.
Let 𝑓 𝑥 = 𝑥 − 6 and 𝑔 𝑥 = 4𝑥 2 − 2𝑥 − 1. Find the following:
a.
b.
c.
d.
e.
f.
𝑓+𝑔 𝑥
𝑔−𝑓 𝑥
𝑓∙𝑔 𝑥
𝑔
𝑓
𝑥
(𝑓 ◦ 𝑔)(𝑥)
(𝑔 ◦ 𝑓)(𝑥)
Example 1.5. Performing Operations on Functions
Perform the indicated operations on the following functions.
• 𝑓 𝑥 =𝑥+3
• 𝑔 𝑥 = 𝑥 2 + 2𝑥 − 8
• 𝑣 𝑥 = 𝑥 2 + 5𝑥 − 4
• 𝑝 𝑥 = 2𝑥 − 7
•ℎ 𝑥 =
𝑥+7
2−𝑥
Find the following:
1.
2.
3.
𝑣+𝑔 𝑥
𝑓∙𝑝 𝑥
𝑓◦ℎ 𝑥
4.
5.
𝑝−𝑓 𝑥
𝑣/𝑔 𝑥
Formative Assessment Part 2
III. Solve the following:
Let 𝑓 𝑥 = 𝑥 2 + 2𝑥 + 1
𝑔 𝑥 = 3𝑥 + 2
ℎ 𝑥 = 12𝑥 2 − 𝑥 − 6
1. (𝑓 + 𝑔)(𝑥)
2. (ℎ − 𝑔)(𝑥)
3. (𝑓 ∙ 𝑔)(𝑥)
ℎ
𝑔
4. ( )(𝑥)
I. Functions
Piecewise Functions
I. Functions
Solving Situations Involving Functions
Solve the following problems:
1. An entrepreneur projected that producing his
new product will have a variable cost of ₱395
per unit and the fixed cost will be at about
₱130,000. Let x be the number of units to be
produced.
a) Express the total cost C as a function of x.
b) If the entrepreneur plans to sell each item for
₱ 645, at what value of x will he breakeven?
Solution:
1- a. The total cost C is equal to the total variable
cost plus fixed costs. If x is the number units
produced, then
C = Total variable cost + fixed cost
C = 395x + 130 000
or C(x) = 395x + 130 000
Solution:
1-b. To breakeven means to find the value of x for
which there is no profit and no loss. That is to find the
value when revenue equals cost. The revenue R is equal
to the selling price per unit times the number of units to
be sold.
R = 645x
To breakeven, R(x) = C(x)
645x = 395x + 130 000
250x = 130 000
x = 520
Solve the following problems:
2. A rectangular land area will be fenced
and converted to a garden. One side of the
lot that is facing the house will not be
fenced. If there are 100 meters of fencing
materials available, what dimensions of the
fenced lot will enclose the maximum area?
What is the maximum area?
Solution:
2. Let x be the width of the lot. Because there are
100 meters of fencing materials, then the length
is represented by 100 – 2x.
We express the area A as the product of the
length and width.
A = (100 – 2x)(x)
= 100x – 2x 2
A = – 2x 2 + 100x
This is a quadratic function.
Solution:
To determine the maximum value, we need to
find the coordinates of the vertex. We can use the form
𝑏
− ,𝑓
2𝑎
𝑏
−
2𝑎
to find the vertex.
𝑏
−
2𝑎
=
100
−
2(−2)
= 25
𝑓
𝑏
−
2𝑎
= 𝑓(25)
= −2 25
= 1 250
2
+ 100 25
Solution:
The vertex is at (25, 1 250). The vertex is
the maximum point. So, the maximum area is the
maximum value 1250 𝑚2 , which is obtained
when the width x is 25 meters. If x = 25, then
100 – 2x = 50. The dimensions that will result in
a maximum area are 25 meters by 50 meters.
Solve the following problems:
3. An online shop advertises a promo for customers who
will buy their product. For customers who will order
fewer than 10 pieces, they will pay ₱120 as the unit
price plus the delivery fee. For customers who will
order 10-30 pieces, they will pay only ₱110 as the unit
price plus the delivery fee. For customers who will
order more than 30 pieces, the unit price will be only
₱100 with free delivery fee. Write the function that
illustrates the situation. If the shop received 9, 28, and
32 units of orders in the first hour of the promo, what is
their total revenue for the first hour?
Solution:
3. The function that illustrates the situation is a piecewise
function f(x). Let x be the number of orders and y the total price
of the orders, where x is a natural number.
For the first sub-function, we use 𝑦 = 120𝑥 + 50 for x < 10,
because each order is priced at ₱120 and the delivery fee is ₱50.
For the second sub-function, we use 𝑦 = 110𝑥 + 50 for 10≤ x ≤
30, because each order is priced at ₱110 and the delivery fee is
₱50.
For the third sub-function, we use 𝑦 = 100𝑥 for x > 30, because
each order is priced at ₱100 and the delivery is free.
Solution:
𝑓 𝑥 =
120𝑥 + 50,
𝑥 < 10
110𝑥 + 50, 10 ≤ 𝑥 ≤ 30
100𝑥,
𝑥 > 30
For 9 orders, the total price is 120 9 + 50 = ₱1 130
For 28 orders, the total price is 110 28 + 50 = ₱3 130
For 32 orders, the total price is 100 32 = ₱3 200
The total revenue of the orders is ₱7 460