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1. GENERAL MATHEMATICS
ANGELO S. REYES
SAMUEL CHRISTIAN COLLEGE

2. UNDERSTANDING
FUNCTIONS

3. UNDERSTANDING FUNCTIONS
•REVISITING FUNCTIONS
•EVALUATING FUNCTIONS
•PERFORMING OPERATIONS ON FUNCTIONS
•COMPOSITION OF FUNCTIONS
•SOLVING PROBLEMS INVOLVING FUNCTIONS

7. Why functions are important in
representing real-life situations?
Essential Questions:

8. REVISITING
FUNCTIONS

9. AT THE END OF THE LESSON
•I can represent real-life situations using functions,
including piecewise functions.
•I can differentiate functions from mere relations
•I can be representing rational functions through its (a)
table of values (b) graph (c) equations
•I can find the domain and range of rational functions;
and
•I can describe and graph different types of functions.

10. DEFINITION:
RELATIONS ∥ DOMAIN ∥
RANGE ∥ FUNCTION

11. WHAT IS RELATION, DOMAIN, RANGE?
•A 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 is a set of ordered pairs.
•The 𝑑𝑜𝑚𝑎𝑖𝑛 of a relation is the set of the first
coordinates.
•The 𝑟𝑎𝑛𝑔𝑒 is the set of the second coordinates.
• 𝑅𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠 are often defined by equations with no
domain stated.
•If the 𝑑𝑜𝑚𝑎𝑖𝑛 is not stated, we agree that the domain
consists of all real numbers that, when used to
substitute the independent variable, produce real
numbers for the dependent variable.

12. EXAMPLE:

13. EXAMPLE:

14. WHAT IS FUNCTION?
•A 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 is a relation that assigns to each member of
the domain exactly one member of the range.
•It is a 𝑠𝑒𝑡 𝑜𝑓 𝑜𝑟𝑑𝑒𝑟𝑒𝑑 𝑝𝑎𝑖𝑟𝑠 (𝑥, 𝑦) in which no two
distinct ordered pairs have the same first member.
•The set of all permissible values of 𝑥 is called the
𝑑𝑜𝑚𝑎𝑖𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.
•And the set of all resulting values of 𝑦 is called the
𝑟𝑎𝑛𝑔𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛.

15. WHAT IS FUNCTION?
•We can also define a function as a rule that
assigns to each element x in set A exactly one
element, called 𝒇(𝒙), in set B.
•We usually represent a function by the equation
𝒚 = 𝒇(𝒙) where 𝒚 is the dependent variable and
𝒙 is the independent variable.
•Take note: that for every input (domain), there
corresponds a unique output (range).

17. WAYS ON HOW TO
DETERMINE IF A
RELATION IS A FUNCTION
OR NOT

24. FUNCTIONS DEFINED BY
GRAPH
VERTICAL LINE TEST

25. VERTICAL LINE TEST
•The graph of a function can be intersected by a
vertical line in at most one point, since in a
function; there is a unique value of to each value
of in the domain of the function.
•But if the vertical line intersects the graph at more
than one point, the graph is not that of a function
(it is only a relation).
•We usually represent a function by the equation

26. Graph the relation
𝒇 𝒙 = 𝒙𝟐
+ 𝟐𝒙 − 𝟑
Using Vertical Line Test
𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒐𝒓 𝑵𝒐𝒕?
Answer:
𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏

27. Graph the relation
𝒙𝟐
+ 𝒚𝟐
= 𝟏𝟔
Using Vertical Line Test
𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒐𝒓 𝑵𝒐𝒕?
Answer:
𝑵𝒐𝒕

28. Using Vertical Line Test
𝑭𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒐𝒓 𝑵𝒐𝒕?
Answer:
𝑵𝒐𝒕

29. TYPES OF FUNCTION

30. CONSTANT
FUNCTION
LINEAR
FUNCTION
QUADRATIC
FUNCTION

31. CUBIC
FUNCTION
ABSOLUTE
FUNCTION
SQUARE ROOT
FUNCTION

32. EXPONENTIAL
FUNCTION
RATIONAL
FUNCTION
LOGARITHMIC
FUNCTION

33. INVERSE
FUNCTION
FLOORING
FUNCTION
CEILING
FUNCTION

34. SIGNUM
FUNCTION
OSCILLATING
FUNCTION

35. PIECEWISE FUNCTION

36. PIECEWISE FUNCTION
•Is a function defined by two or more formulas on
different parts of its domain.

37. Example
A merchant sells rice at 45 pesos a kilo. If, however,
a buyer buys more than 150 kilos of rice. The
merchant gives a 10% discount.
•If x represents the number of kilos of rice
purchased, express the amount to be paid by f(x).
•How much should be paid for 100 kilos of rice?
•How much should be paid for 200 kilos of rice?

38. Step 1: Express the situation, using function
If the number of kilos of purchased rice exceeds 150. That’s
𝑥 > 150. Then the number of kilos in excess of 150 kilos is
given a 10% discount
If the number of kilos of purchased rice does not exceed
150. That is 0 ≤ 𝑥 ≤ 150 then 𝑓 𝑥 = 45𝑥

39. Step 1: Express the situation, using function
𝑓 𝑥 =
45𝑥 𝑖𝑓 0 ≤ 𝑥 ≤ 150
40.50 + 675𝑥 𝑖𝑓 𝑥 > 150
So, the function that describes the situation is:
Amount of the first 150
kilos of rice 45 150 = 6 750
Amount in excess of 150
kilos
𝟒𝟓 𝟎. 𝟗𝟎 𝒙 − 𝟏𝟓𝟎 = 𝟒𝟎. 𝟓𝟎(𝒙 − 𝟏𝟓𝟎)
Total Amount 𝟔 𝟕𝟓𝟎 + 𝟒𝟎. 𝟓𝟎 𝒙 − 𝟏𝟓𝟎 = 𝟒𝟎. 𝟓𝟎𝒙 + 𝟔𝟕𝟓

40. Step 2: Find the amount to be paid for 100
kilos of rice.
𝑓 𝑥 = 45𝑥
𝑓 100 = 45 100 = 4 500
Find f(100) by substituting x = 100 in the first piece or
chunk of the function because 100 < 150.
So, the amount to be paid for 100 kilos of rice is Php 4,500

41. Step 2: Find the amount to be paid for 200
kilos of rice.
𝑓 𝑥 = 40.50𝑥 + 675
𝑓 200 = 40.50 200 + 675
𝑓 200 = 8 775
Find f(200) by substituting x=200 in the second piece or
chunk of the function because 200>150.
So, the amount to be paid for 200 kilos of rice is Php 8 775.

42. GRAPHING
PIECEWISE FUNCTION

43. Graph: 𝒇 𝒙 =
𝑥 − 2 , 𝑥 ≤ −2
3𝑥2 − 5 , 𝑥 > −2
The function is described by
two intervals. Take note that
when x=-2, f(x)=x-2 is
defined. That is why the
point (-2,-4) is shaded.

44. Graph: 𝒇 𝒙 =
𝑥 − 2 , 𝑥 ≤ −2
3𝑥2 − 5 , 𝑥 > −2
In 𝑓 𝑥 = 3𝑥2
− 5 when x=-2
the point (-2,7) is not shaded
because it is not defined.
Clearly, by vertical line test,
the piecewise is indeed a
function.

45. EVALUATING
FUNCTION

46. Example: If 𝑓 𝑥 = 4𝑥2 − 5𝑥 + 3 evaluate the
following: 𝑓 5 𝑎𝑛𝑑 𝑓 0
𝑓 𝑥 = 4𝑥2
− 5𝑥 + 3
𝑓 5 = 4 5 2
− 5 5 + 3
𝑓 5 = 100 − 25 + 3
𝑓 5 = 78
𝑓 𝑥 = 4𝑥2
− 5𝑥 + 3
𝑓 0 = 4 0 2
− 5 0 + 3
𝑓 0 = 0 − 0 + 3
𝑓 0 = 3

47. Example: If 𝑓 𝑥 = 4𝑥2 − 5𝑥 + 3 evaluate the
following: 𝑓
1
4
𝑎𝑛𝑑 𝑓
2
2
𝑓 𝑥 = 4𝑥2
− 5𝑥 + 3
𝑓
1
4
= 4
1
4
2
− 5
1
4
+ 3
𝑓
1
4
=
1
4
−
5
4
+ 3
𝑓
1
4
= 2
𝑓 𝑥 = 4𝑥2
− 5𝑥 + 3
𝑓
2
2
= 4
2
2
2
− 5
2
2
+ 3
𝑓
2
2
= 4
1
2
−
5 2
4
+ 3
𝑓
2
2
= −
5 2
4
+ 5

48. Example: If 𝑓 𝑥 = 4𝑥2 − 5𝑥 + 3 evaluate the
following: 𝑓(𝑥 + 1)
𝑓 𝑥 = 4𝑥2
− 5𝑥 + 3
𝑓 𝑥 + 1 = 4 𝑥 + 1 2
− 5 𝑥 + 1 + 3
𝑓 𝑥 + 1 = 4 𝑥2
+ 2𝑥 + 1 − 5𝑥 − 5 + 3
𝑓 𝑥 + 1 = 4𝑥2
+ 8𝑥 + 4 − 5 − 5𝑥 − 2
𝑓 𝑥 + 1 = 4𝑥2
+ 3𝑥 − 3

49. Example: If 𝑓 𝑥 = 3𝑥 + 2 evaluate the
following:
𝑓 𝑥+ℎ −𝑓(𝑥)
ℎ
𝑓 𝑥 = 3𝑥 + 2
𝑓 𝑥 + ℎ = 3 𝑥 + ℎ + 2
𝑓 𝑥 + ℎ = 3𝑥 + 3ℎ + 2
𝑓 𝑥 + ℎ − 𝑓 𝑥 = 3𝑥 + 3ℎ + 2 − 3𝑥 + 2
𝑓 𝑥 + ℎ − 𝑓 𝑥 = 3𝑥 + 3ℎ + 2 − 3𝑥 − 2
𝑓 𝑥 + ℎ − 𝑓 𝑥 = 3𝑥 − 3𝑥 + 3ℎ + 2 − 2
𝑓 𝑥 + ℎ − 𝑓 𝑥 = 3ℎ
𝑓 𝑥 + ℎ − 𝑓 𝑥
ℎ
=
3ℎ
ℎ
= 3

50. EVALUATING
PIECEWISE FUNCTION

51. Example: Given the piecewise below
Find the following: ℎ −10 , ℎ
1
2
, ℎ(6)

52. Find: ℎ(−10)
Step 1: Determine the piece that should be used.
Since -10 belongs to the interval x < -5. therefore,
we shall use the third piece of function
Step 2: Find h(-10) using the 3rd piece of the function
ℎ 𝑥 = 2 𝑥 + 4 2
+ 10
ℎ −10 = 2 −10 + 4 2
+ 10
ℎ −10 = 2 36 + 10
ℎ −10 = 82

53. Find: ℎ
1
2
Step 1: Determine the piece that should be used.
Since 1/2 belongs to the interval −5 ≤ 𝑥 < 6.
therefore, we shall use the second piece of
function
Step 2: Find h(1/2) using the 2nd piece of the function
ℎ 𝑥 =
3
2𝑥
ℎ
1
2
=
3
2
1
2
ℎ
1
2
= 3

54. Find: ℎ(6)
Step 1: Determine the piece that should be used.
Since x = 6 belongs to the interval 𝑥 ≥ 6. therefore,
we shall use the first piece of function
Step 2: Find h(6) using the 1st piece of the function
ℎ 𝑥 = 𝑥3
+ 2𝑥
ℎ 6 = 6 3
+ 2(6)
ℎ 6 = 216 + 12
ℎ 6 = 228

55. PERFORMING
OPERATIONS ON FUNCTION

56. SUM, DIFFERENCE, PRODUCT, AND
QUOTIENT OF FUNCTIONS
Given two functions 𝑓(𝑥) and 𝑔(𝑥), we define the
following.
•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

57. Example: If 𝑓 𝑥 = 6𝑥2 − 12𝑥 − 19 and
𝑔 𝑥 = 𝑥3 − 𝑥2 + 8𝑥 − 1 find the
following:
𝑓 + 𝑔 𝑥 𝑎𝑛𝑑 (𝑓 − 𝑔)(𝑥)

58. Find: (𝑓 + 𝑔)(𝑥)
Step 1: Apply the definition for adding functions.
𝑓 + 𝑔 𝑥 = 𝑓 𝑥 + 𝑔 𝑥
𝑓 + 𝑔 𝑥 = 6𝑥2
− 12𝑥 − 19 + (𝑥3
− 𝑥2
+ 8𝑥 − 1)
Step 2: Group similar terms and make sure that the terms
are arranged in descending order:
𝑓 + 𝑔 𝑥 = 6𝑥2
− 12𝑥 − 19 + 𝑥3
− 𝑥2
+ 8𝑥 − 1
𝑓 + 𝑔 𝑥 = 𝑥3
+ 6𝑥2
− 𝑥2
+ −12𝑥 + 8𝑥 + (−19 − 1)

59. Find: (𝑓 + 𝑔)(𝑥)
Step 3: Combined similar terms
𝑓 + 𝑔 𝑥 = 6𝑥2
− 12𝑥 − 19 + 𝑥3
− 𝑥2
+ 8𝑥 − 1
𝑓 + 𝑔 𝑥 = 𝑥3
+ 6𝑥2
− 𝑥2
+ −12𝑥 + 8𝑥 + −19 − 1
𝑓 + 𝑔 𝑥 = 𝒙𝟑
+ 𝟓𝒙𝟐
− 𝟒𝒙 − 𝟐𝟎

60. Find: (𝑓 − 𝑔)(𝑥)
Step 1: Apply the definition for subtracting functions.
𝑓 − 𝑔 𝑥 = 𝑓 𝑥 − 𝑔 𝑥
𝑓 − 𝑔 𝑥 = 6𝑥2
− 12𝑥 − 19 − (𝑥3
− 𝑥2
+ 8𝑥 − 1)
Step 2: Group similar terms and make sure that the terms
are arranged in descending order:
𝑓 − 𝑔 𝑥 = 6𝑥2
− 12𝑥 − 19 − 𝑥3
− 𝑥2
+ 8𝑥 − 1
𝑓 − 𝑔 𝑥 = 6𝑥2
− 12𝑥 − 19 − 𝑥3
+ 𝑥2
− 8𝑥 + 1
𝑓 − 𝑔 𝑥 = −𝑥3
+ 6𝑥2
+ 𝑥2
+ −12𝑥 − 8𝑥 + (−19 + 1)

61. Find: (𝑓 + 𝑔)(𝑥)
Step 3: Combined similar terms
𝑓 − 𝑔 𝑥 = 6𝑥2
− 12𝑥 − 19 − 𝑥3
− 𝑥2
+ 8𝑥 − 1
𝑓 − 𝑔 𝑥 = 6𝑥2
− 12𝑥 − 19 − 𝑥3
+ 𝑥2
− 8𝑥 + 1
𝑓 − 𝑔 𝑥 = −𝑥3
+ 6𝑥2
+ 𝑥2
+ −12𝑥 − 8𝑥 + −19 + 1
𝑓 − 𝑔 𝑥 = −𝒙𝟑
+ 𝟕𝒙𝟐
− 𝟐𝟎𝒙 − 𝟏𝟖

62. Example: If 𝑓 𝑥 =
3
𝑥2−1
and 𝑔 𝑥 =
𝑥
𝑥−1
find 𝑓 + 𝑔 𝑥 .

63. Find: (𝑓 + 𝑔)(𝑥)
Step 1: Apply the definition for adding functions.
𝑓 + 𝑔 𝑥 = 𝑓 𝑥 + 𝑔 𝑥
𝑓 + 𝑔 𝑥 =
3
𝑥2 − 1
+
𝑥
𝑥 − 1
Step 2: Find the LCD of 𝑥2
− 1 𝑎𝑛𝑑 𝑥 − 1, the 𝐿𝐶𝐷 𝑖𝑠
𝑥 + 1 𝑥 − 1
𝑓 + 𝑔 𝑥 =
3
𝑥2 − 1
+
𝑥
𝑥 − 1
𝑓 + 𝑔 𝑥 =
3 + 𝑥(𝑥 + 1)
(𝑥 + 1)(𝑥 − 1)

64. Find: (𝑓 + 𝑔)(𝑥)
Step 3: Simplify the expression:
𝑓 + 𝑔 𝑥 =
3
𝑥2 − 1
+
𝑥
𝑥 − 1
𝑓 + 𝑔 𝑥 =
3 + 𝑥(𝑥 + 1)
(𝑥 + 1)(𝑥 − 1)
𝑓 + 𝑔 𝑥 =
𝒙𝟐
+ 𝒙 + 𝟑
(𝒙 + 𝟏)(𝒙 − 𝟏)

65. Example: If 𝑓 𝑥 =
𝑥−3
𝑥+4
and 𝑔 𝑥 =
𝑥+1
𝑥+4
find
𝑓 − 𝑔 𝑥 .

66. Find: (𝑓 − 𝑔)(𝑥)
Step 1: Apply the definition for subtracting functions.
𝑓 − 𝑔 𝑥 = 𝑓 𝑥 − 𝑔 𝑥
𝑓 − 𝑔 𝑥 =
𝑥 − 3
𝑥 + 4
−
𝑥 + 1
𝑥 + 4
Step 2: Simplify
𝑓 − 𝑔 𝑥 =
𝑥 − 3
𝑥 + 4
−
𝑥 + 1
𝑥 + 4
𝑓 − 𝑔 𝑥 =
𝑥 − 3 − (𝑥 + 1)
𝑥 + 4
=
𝑥 − 3 − 𝑥 − 1
𝑥 + 4
=
−𝟒
𝒙 + 𝟒

67. Example:
If 𝑓 𝑥 = 𝑥3 − 27, 𝑔 𝑥 = 𝑥2 + 3𝑥 + 9
and ℎ 𝑥 = 4𝑥 − 1. .
Find the following: (𝑓 ∗ ℎ)(𝑥),
𝑓
𝑔
(𝑥),
𝑓
ℎ
(𝑥)

68. Find: (𝑓 ∗ ℎ)(𝑥)
Step 1: Apply the definition for multiplying functions.
𝑓 ∗ ℎ 𝑥 = 𝑓 𝑥 ∗ ℎ 𝑥
𝑓 ∗ ℎ 𝑥 = 𝑥3
− 27 (4𝑥 − 1)
Step 2: Find the product of polynomials using the
distributive law.
𝑓 ∗ ℎ 𝑥 = 𝑥3
− 27 4𝑥 − 1
𝑓 ∗ ℎ 𝑥 = 𝟒𝒙𝟒
− 𝒙𝟑
− 𝟏𝟎𝟖𝒙 + 𝟐𝟕

69. Find:
𝑓
𝑔
(𝑥)
Step 1:
Apply the definition
for dividing functions.
𝑓
𝑔
𝑥 =
𝑓 𝑥
𝑔 𝑥
Step 2: Factor the numerator
𝑓
𝑔
𝑥 =
𝑥3
− 27
𝑥2 + 3𝑥 + 9
𝑓
𝑔
𝑥 =
𝑥3
− 27
𝑥2 + 3𝑥 + 9
𝑓
𝑔
𝑥 =
(𝑥 − 3)(𝑥2
+ 3𝑥 + 9)
𝑥2 + 3𝑥 + 9
Step 3: Cancel the common
factors and simplify
𝑓
𝑔
𝑥 =
(𝑥 − 3)(𝑥2
+ 3𝑥 + 9)
𝑥2 + 3𝑥 + 9
𝑓
𝑔
𝑥 = 𝒙 − 𝟑

70. Find:
𝑓
ℎ
(𝑥)
Step 1:
Apply the definition
for dividing functions.
𝑓
ℎ
𝑥 =
𝑓 𝑥
𝑔 𝑥
Step 2: Since there are no
common factors in the
numerator and the
denominator, this is
already the simplest
form
𝑓
ℎ
𝑥 =
4𝑥 − 1
𝑥3 − 27
𝑓
ℎ
𝑥 =
𝟒𝒙 − 𝟏
𝒙𝟑 − 𝟐𝟕

71. Example:
If 𝑓 𝑥 =
𝑥2−3𝑥−28
𝑥2−8𝑥+16
, 𝑔 𝑥 =
𝑥2−5𝑥+4
𝑥2−49
.
Find the following: 𝑓 ∗ 𝑔 (𝑥) and
𝑓
𝑔
(𝑥)

72. Find: (𝑓 ∗ 𝑔)(𝑥)
Step 1: Apply the definition for multiplying functions.
𝑓 ∗ 𝑔 𝑥 = 𝑓 𝑥 ∗ 𝑔 𝑥
𝑓 ∗ 𝑔 𝑥 =
𝑥2
− 3𝑥 − 28
𝑥2 − 8𝑥 + 16
∗
𝑥2
− 5𝑥 + 4
𝑥2 − 49
Step 2: Factor the numerator and denominator of both
rational function.
𝑓 ∗ 𝑔 𝑥 =
𝑥2
− 3𝑥 − 28
𝑥2 − 8𝑥 + 16
∗
𝑥2
− 5𝑥 + 4
𝑥2 − 49
𝑓 ∗ 𝑔 𝑥 =
(𝑥 − 7)(𝑥 + 4)
(𝑥 − 4)(𝑥 − 4)
∗
(𝑥 − 1)(𝑥 − 4)
(𝑥 − 7)(𝑥 + 7)

73. Find: (𝑓 ∗ 𝑔)(𝑥)
Step 3: Cancel common factors and simplify.
𝑓 ∗ 𝑔 𝑥 =
𝑥2
− 3𝑥 − 28
𝑥2 − 8𝑥 + 16
∗
𝑥2
− 5𝑥 + 4
𝑥2 − 49
𝑓 ∗ 𝑔 𝑥 =
(𝑥 − 7)(𝑥 + 4)
(𝑥 − 4)(𝑥 − 4)
∗
(𝑥 − 1)(𝑥 − 4)
(𝑥 − 7)(𝑥 + 7)
𝑓 ∗ 𝑔 𝑥 =
(𝒙 + 𝟒)(𝒙 − 𝟏)
(𝒙 − 𝟒)(𝒙 + 𝟕)

74. Find:
𝑓
𝑔
(𝑥)
Step 1: Apply the definition for dividing functions.
𝑓
𝑔
𝑥 =
𝑓 𝑥
𝑔 𝑥
→
Step 2: Apply the rule in dividing rational expressions.
𝑓
𝑔
𝑥 =
𝑥2
− 3𝑥 − 28
𝑥2 − 8𝑥 + 16
∗
𝑥2
− 49
𝑥2 − 5𝑥 + 4
𝑓
𝑔
𝑥 =
𝑥2
− 3𝑥 − 28
𝑥2 − 8𝑥 + 16
÷
𝑥2
− 5𝑥 + 4
𝑥2 − 49
𝑓
𝑔
𝑥 =
𝑥2
− 3𝑥 − 28
𝑥2 − 8𝑥 + 16
÷
𝑥2
− 5𝑥 + 4
𝑥2 − 49

75. Find: (𝑓 ∗ 𝑔)(𝑥)
Step 3: Factor the numerator and denominator of both
rational functions.
𝑓
𝑔
𝑥 =
(𝑥 − 7)(𝑥 + 4)
(𝑥 − 4)(𝑥 − 4)
∗
(𝑥 − 7)(𝑥 + 7)
(𝑥 − 4)(𝑥 − 1)
𝑓
𝑔
𝑥 =
𝑥2
− 3𝑥 − 28
𝑥2 − 8𝑥 + 16
∗
𝑥2
− 49
𝑥2 − 5𝑥 + 4

76. Find: (𝑓 ∗ 𝑔)(𝑥)
Step 4: There are no common factors to cancel. So we just
simplify
𝑓
𝑔
𝑥 =
(𝑥 − 7)(𝑥 + 4)
(𝑥 − 4)(𝑥 − 4)
∗
(𝑥 − 7)(𝑥 + 7)
(𝑥 − 4)(𝑥 − 1)
𝑓
𝑔
𝑥 =
𝒙 − 𝟕 𝟐
(𝒙 + 𝟕)(𝒙 + 𝟒)
𝒙 − 𝟒 𝟑(𝒙 − 𝟏)

77. COMPOSITION
OF FUNCTION

78. Definition of Composition of Function
and the domain of (𝑓 ∘ 𝑔)(𝑥) is the set of all numbers x in
the domain of 𝑔(𝑥), such that 𝑔(𝑥) is the domain of 𝑓(𝑥).
Given two function 𝑓(𝑥) and 𝑔(𝑥), the composition of
function is denoted by (𝑓 ∘ 𝑔)(𝑥) is defined by:
𝒇 ∘ 𝒈 𝒙 = 𝒇(𝒈(𝒙))
Similarly, we define (𝑔 ∘ 𝑓)(𝑥) as follows:
𝒈 ∘ 𝒇 𝒙 = 𝒈(𝒇(𝒙))

79. Example:
If 𝑓 𝑥 = 𝑥2 − 6 and 𝑔 𝑥 = 3𝑥 + 2.
Find the following:
(𝑓 ∘ 𝑔)(𝑥), (𝑔 ∘ 𝑓)(𝑥), (𝑔 ∘ 𝑔)(𝑥), (𝑓 ∘ 𝑓)(𝑥)

80. Find:(𝑓 ∘ 𝑔)(𝑥)
Step 1: Apply the definition of composition of functions
𝑓 ∘ 𝑔 𝑥 = 𝑓 𝑔 𝑥 → (𝑓 ∘ 𝑔)(𝑥) = 𝑓(3𝑥 + 2)
Step 2: Apply the definition of 𝑓
𝑓 ∘ 𝑔 𝑥 = 𝑓 3𝑥 + 2 → 𝑓 ∘ 𝑔 𝑥 = 3𝑥 + 2 2
− 6
Step 3: Simplify the results.
𝑓 ∘ 𝑔 𝑥 = 3𝑥 + 2 2
− 6
𝑓 ∘ 𝑔 𝑥 = 9𝑥2
+ 12𝑥 + 4 − 6
𝑓 ∘ 𝑔 𝑥 = 𝟗𝒙𝟐
+ 𝟏𝟐𝒙 − 𝟐

81. Find:(𝑔 ∘ 𝑓)(𝑥)
Step 1: Apply the definition of composition of functions
𝑔 ∘ 𝑓 𝑥 = 𝑔 𝑓 𝑥 → (𝑔 ∘ 𝑓)(𝑥) = 𝑔(𝑥2
− 6)
Step 2: Apply the definition of 𝑓
𝑔 ∘ 𝑓 𝑥 = 𝑔 𝑥2
− 6 → 𝑔 ∘ 𝑓 𝑥 = 3 𝑥2
− 6 + 2
Step 3: Simplify the results.
𝑔 ∘ 𝑓 𝑥 = 3𝑥2
− 18 + 2
𝑔 ∘ 𝑓 𝑥 = 𝟑𝒙𝟐
− 𝟏𝟔
𝑔 ∘ 𝑓 𝑥 = 3 𝑥2
− 6 + 2

82. Find:(𝑔 ∘ 𝑔)(𝑥)
Step 1: Apply the definition of composition of functions
𝑔 ∘ 𝑔 𝑥 = 𝑔 𝑔 𝑥 → (𝑔 ∘ 𝑔)(𝑥) = 𝑔(3𝑥 + 2)
Step 2: Apply the definition of 𝑓
𝑔 ∘ 𝑔 𝑥 = 𝑔 3𝑥 + 2 → 𝑔 ∘ 𝑔 𝑥 = 3 3𝑥 + 2 + 2
Step 3: Simplify the results.
𝑔 ∘ 𝑔 𝑥 = 9𝑥 + 6 + 2
𝑔 ∘ 𝑔 𝑥 = 𝟑𝒙𝟐
+ 𝟖
𝑔 ∘ 𝑔 𝑥 = 3 3𝑥 + 2 + 2

83. Find:(𝑓 ∘ 𝑓)(𝑥)
Step 1: Apply the definition of composition of functions
𝑓 ∘ 𝑓 𝑥 = 𝑓 𝑓 𝑥 → (𝑓 ∘ 𝑓)(𝑥) = 𝑓(𝑥2
− 6)
Step 2: Apply the definition of 𝑓
𝑓 ∘ 𝑓 𝑥 = 𝑓 𝑥2
− 6 → 𝑓 ∘ 𝑓 𝑥 = 𝑥2
− 6 2
− 6
Step 3: Simplify the results.
𝑔 ∘ 𝑓 𝑥 = 𝑥4
− 12𝑥2
+ 36 − 6
𝑔 ∘ 𝑓 𝑥 = 𝒙𝟒
− 𝟏𝟐𝒙𝟐
+ 𝟑𝟎
𝑓 ∘ 𝑓 𝑥 = 𝑥2
− 6 2
− 6

84. Example:
Given the following functions 𝑓 𝑥 = 𝑥3,
𝑔 𝑥 =
𝑥+2
3𝑥−1
,and ℎ 𝑥 = 6𝑥 − 7
Find the following:
𝑓 ∘ 𝑓 𝑥 𝑎𝑛𝑑 (𝑓 ∘ 𝑓)(2) , 𝑔 ∘ ℎ 𝑥 𝑎𝑛𝑑 (𝑔 ∘ ℎ)(−1)

85. Find:(𝑓 ∘ 𝑓)(𝑥)
Step 1: Apply the definition of composition of functions
𝑓 ∘ 𝑓 𝑥 = 𝑓 𝑓 𝑥 → (𝑓 ∘ 𝑓)(𝑥) = 𝑓(𝑥3
)
Step 2: Apply the definition of 𝑓
𝑓 ∘ 𝑓 𝑥 = 𝑓 𝑥3
→ 𝑓 ∘ 𝑓 𝑥 = 𝑥3 3
Step 3: Simplify the results.
𝑔 ∘ 𝑓 𝑥 = 𝒙𝟗
𝑓 ∘ 𝑓 𝑥 = 𝑥3 3
𝑔 ∘ 𝑓 2 = (𝟐)𝟗
= 𝟓𝟏𝟐

86. Find:(𝑔 ∘ ℎ)(𝑥)
Step 1: Apply the definition of composition of functions
𝑔 ∘ ℎ 𝑥 = 𝑔 ℎ 𝑥 → (𝑔 ∘ ℎ)(𝑥) = 𝑔 6𝑥 + 7
Step 2: Apply the
definition of 𝑓
𝑔 ∘ ℎ 𝑥 = 𝑔 6𝑥 + 7
𝑔 ∘ ℎ 𝑥 =
6𝑥 − 7 + 2
3 6𝑥 − 7 − 1
Step 3: Simplify the results.
𝑔 ∘ ℎ 𝑥 =
6𝑥 − 7 + 2
3 6𝑥 − 7 − 1
𝑔 ∘ ℎ 𝑥 =
𝟔𝒙 − 𝟓
𝟏𝟖𝒙 − 𝟐𝟐
𝑔 ∘ ℎ −1 =
6(−1) − 5
18(−1) − 22
= −
𝟏𝟏
𝟒

87. SOLVING PROBLEMS
INVOLVING FUNCTION

88. The following steps will be helpful in
solving problems involving functions.
• Read and comprehend the problem carefully.
• Identify the function to be used (linear, quadratic,
cubic, etc.) in representing the given problem.
• Write the function that describes the problem.
• Use the function to solve the problem.
• Use the function to solve the problem.
• Check the answer.

89. EXAMPLE: A Science major student earned
6 units after enrolling for a semester. Then,
he took an academic leave for one year.
After his leave, he enrolled again, ever
more desirous this time to finish all the 150
units that he needs. In how many
semesters does he need to enroll if he takes
9 units of Science subjects every semester?

90. Step:1
Represent the given and unknown
quantities in the problem.
Let:
𝑦 = total number of units to be taken in x
semesters.
𝑥 = number of semesters.
Step:2
Determine the function or
equation that relates the
quantities or variables in the
problem. The problem can be
described best by a linear
function which relates the
number of semesters (x) to the
total number of units (y).
The function that describes the
relationship between x and y is:
𝑦 = 9𝑥 + 16

91. Step:3
Solve the problem. We want to know
the number of semesters that the
student needs to take in order to
finish all 150 units.
𝑦 = 150
The student needs 16 semesters to
complete the program.
150 = 9𝑥 + 6
150 − 6 = 9𝑥
144 = 9𝑥
𝑥 = 16

92. EXAMPLE: The area of a rectangle is
420 𝑐𝑚2. The length of the rectangle is
one less than thrice its width. What are
the dimensions of the given rectangle?

93. Step:1
Represent the given and unknown
quantities in the problem. Express
the width and the length of the
rectangle in terms of one variable.
Let:
𝑦 = area of the rectangle
𝑥 = width
3𝑥 − 1 = Length
Step:2
Use function to relate the quantities
or variables in the problem.
The area of the rectangle is given by:
𝐴 = 𝑙𝑤; where 𝑙 = length and
𝑤 = width
Using the representation in Step 1,
we have:
𝑦 = 𝑥 3𝑥 − 1
𝑦 = 3𝑥2 − 𝑥
The function that describes the
relationship is 𝑦 = 3𝑥2 − 𝑥

94. Step:3
Solve the problem.
The area of rectangle is 𝑦 = 420
Substitute the value in 𝑦 = 3𝑥2 − 𝑥
3𝑥2
− 𝑥 − 420 = 0 Find the length and the width by substituting
the value of x in the representation found in
Step 1. Reject the negative value of x because
the measurement of width cannot be negative.
3𝑥 + 35 𝑥 − 12 = 0
420 = 3𝑥2 − 𝑥
3𝑥 + 35 = 0
If 𝑥 = 12, then:
width = 𝑥 = 12
Length = 3𝑥 − 1 = 3 12 − 1 = 35
Thus the dimensions of the given rectangle
is 12 and 35.
3𝑥 = −35
𝒙 = −
𝟑𝟓
𝟑
𝑥 − 12 = 0
𝒙 = 𝟏𝟐
Step:4