Relations, Mappings & Functions.

Brilliant Maths - Ngozi Orevaoghene PhD
7 Aug 202130:55
EducationalLearning
32 Likes 10 Comments

TLDRThis video from BrilliantMaths.com introduces key concepts in mathematics, focusing on mappings, relations, and functions. It explains the difference between one-to-one, one-to-many, many-to-one, and many-to-many relations, and how these relate to mappings. The lesson further delves into functions as a type of mapping where each domain element corresponds to exactly one range element, covering inverse functions and composite functions. The video emphasizes the importance of understanding these concepts to excel in mathematics, providing examples and encouraging viewers to practice more using Brilliant Math Module 8.

Takeaways
  • πŸ“š A relation is a connection between two sets that associates elements of one set to elements of another.
  • πŸ—ΊοΈ The domain of a relation is the set of elements from which the relation originates, while the codomain is the set of possible outputs.
  • πŸ” One-to-one relations ensure that each element in the domain has a unique image in the codomain.
  • 🌐 One-to-many relations occur when one element in the domain can have multiple images in the codomain.
  • πŸ”’ Many-to-one relations mean multiple elements in the domain can map to a single element in the codomain.
  • πŸ” Many-to-many relations involve elements in the domain having multiple images in the codomain, with no single element in the domain mapping to more than one element in the codomain.
  • 🌟 A mapping is a relation where each member of the domain maps onto exactly one member of the codomain.
  • πŸ”„ Functions are mappings where each element in the domain has one and only one image in the range, allowing for inverse functions.
  • πŸ”„ To find the inverse of a function, switch the roles of x and y (replace f(x) with y and solve for x), then replace y with the inverse function notation f^(-1)(x).
  • πŸ”§ Composite functions involve the combination of two or more functions, where one function is applied to the result of another (e.g., g(f(x))).
  • πŸ“ˆ Understanding relations, mappings, and functions is crucial for excelling in mathematics, and resources like Brilliant Math Module 8 can aid in this learning process.
Q & A
  • What is the main topic of the lesson?

    -The main topic of the lesson is mappings, relations, and functions in mathematics.

  • What is a relation in the context of this lesson?

    -A relation is a connection between two sets that associates the elements of one set to the elements of another set.

  • What are the two sets involved in a relation?

    -The two sets involved in a relation are the domain (set X) and the codomain (set Y).

  • What is the image of a domain in a relation?

    -The image of a domain in a relation is the set of elements in the codomain that are associated with the elements in the domain.

  • What are the different types of relations discussed in the lesson?

    -The different types of relations discussed are one-to-one, one-to-many, many-to-one, and many-to-many.

  • What is a mapping?

    -A mapping is a relation in which each member in the domain maps onto only one member in the codomain.

  • What is a function in the context of mathematics?

    -A function is a mapping in which each element in the domain has one and only one image in the range.

  • How do you find the rule of a mapping?

    -To find the rule of a mapping, you observe the pattern or operation that transforms elements from the domain to the codomain.

  • How do you find the inverse of a function?

    -To find the inverse of a function, you switch the roles of x and y (make x the subject, then replace x with f^(-1)(x) and f(x) with y).

  • What are composite functions?

    -Composite functions are functions that involve more than one function, where one function is applied to the result of another function.

  • How do you evaluate a composite function at a specific value?

    -To evaluate a composite function at a specific value, you substitute the input value into the function, perform the necessary operations, and then substitute the result back into the other function as required.

Outlines
00:00
πŸ“š Introduction to Mappings, Relations, and Functions

This paragraph introduces the concepts of mappings, relations, and functions from the Brilliant Maths Module 8. It explains that a relation is a connection between two sets, associating elements of one set with elements of another. The video provides examples of different types of relations, including one-to-one, one-to-many, many-to-one, and many-to-many, using real-life scenarios to illustrate these concepts. It sets the stage for a deeper exploration of mathematical mappings and their properties.

05:02
πŸ” Understanding Mappings and Function Rules

This section delves into the specifics of mappings, which are relations where each domain element maps to exactly one element in the codomain. It differentiates between one-to-one and many-to-one mappings, providing examples for each. The paragraph also teaches how to identify the rule of a mapping, using a numerical pattern to demonstrate how the mapping operates. This knowledge is crucial for comprehending the structure and behavior of functions.

10:05
πŸ“ˆ Functions and Their Applications

This part of the script focuses on functions as a special type of mapping where each domain element has a unique corresponding element in the range. It clarifies that only one-to-one and many-to-one relations qualify as functions. The video introduces the concept of a function's expression, such as f(x) = 3x - 2, and explains how to evaluate a function for specific values of x. It also presents a method for solving equations when a function is given a specific value, illustrating this with a linear equation example.

15:06
πŸ”„ Finding Inverse Functions

This paragraph introduces the concept of inverse functions, which are essentially the reverse of a given function. It outlines the steps to find the inverse of a function, emphasizing the process of making x the subject and then swapping the variables. Two examples are provided to demonstrate the process, showing how to solve for the inverse function and evaluate it at a specific point. This section is essential for understanding how functions can be reversed and how to work with them in various mathematical contexts.

20:09
πŸ€” Evaluating Composite Functions

This section explores composite functions, which involve applying one function to the result of another. The video explains the concept with an example, showing how to find the composite functions (g ∘ f)(x) and (f ∘ g)(x) by replacing the x in one function with the other function. It also demonstrates how to evaluate a composite function for a specific input by working through a more complex example involving two functions. This part is crucial for understanding the interaction between multiple functions and their combined effects.

25:09
πŸ“ Solving Composite Functions with Specific Inputs

This paragraph continues the discussion on composite functions, focusing on evaluating them for specific values of x. It provides a step-by-step solution for finding the value of a composite function g(h(x)) for x = -3, using a given pair of functions. The process involves finding the function g(h(x)), simplifying the expression, and then substituting the specific value of x to obtain the result. This detailed example helps to solidify the understanding of how to work with composite functions and their applications.

30:11
πŸŽ“ Conclusion and Encouragement for Maths Learning

In the concluding paragraph, the video wraps up the lessons on relations, mappings, functions, inverse functions, and composite functions from Brilliant Maths Module 8. It encourages viewers to practice and understand these mathematical concepts, emphasizing that maths can be fun and that anyone can excel in it. The video ends with a call to action to subscribe to Brilliant Maths and follow their social media for further learning and success in maths.

Mindmap
gh(x) evaluation
g(f(x)) and f(g(x))
f(x) = 3x - 1/2
f(x) = 2x - 3
Examples
Definition
Examples
Finding the Inverse
Definition
f(2) and f(-3)
Many-to-One Mapping
One-to-One Mapping
Many-to-Many
Many-to-One
One-to-Many
One-to-One
Composite Functions
Inverse Functions
Finding Function Values
Function Notation
Definition
Examples
Definition
Types of Relations
Domain and Codomain
Definition
Social Media
Brilliant Maths Module 8
Composite Functions
Inverse Functions
Function Mastery
Identifying Mappings
Understanding of Relations
Functions
Mappings
Relations
Instructor
Context
Additional Resources
Learning Outcomes
Key Concepts
Introduction
Mathematical Concepts: Relations, Mappings, and Functions
Alert
Keywords
πŸ’‘Mappings
Mappings refer to the process of associating elements from one set to another according to specific rules. In the context of the video, mappings are a foundational concept for understanding relations and functions. For instance, the video describes a mapping where people in a set (Kemi, Dan, Amma, Tom, and Ngazi) are associated with their countries of origin (Nigeria, France, Ghana, and Togo), illustrating how mappings connect two sets based on a defined relationship.
πŸ’‘Relations
Relations are connections between two sets that associate elements from one set to elements of another set. The video emphasizes that a relation can be of various types, such as one-to-one, one-to-many, many-to-one, or many-to-many, depending on how elements in the domain correspond to elements in the codomain. Relations are fundamental to understanding mappings and functions, as they define the rules by which elements are associated.
πŸ’‘Functions
Functions are a specific type of mapping where each element in the domain is associated with exactly one element in the range. This means that functions are a subset of relations that follow the one-to-one and many-to-one patterns. Functions are often represented by expressions, such as f(x) = 3x - 2, where 'f' denotes the function, 'x' is the input variable, and the expression defines the output for each input.
πŸ’‘Inverse Functions
Inverse functions are the 'reverse' of a given function, effectively 'undoing' the original function's operation. If a function takes an input and produces an output, its inverse will take that output and return the original input. The video explains that to find the inverse of a function, one must switch the roles of x and y (or f(x)) and solve for the new y (or x), which represents the inverse function.
πŸ’‘Composite Functions
Composite functions occur when one function is applied to the result of another function. It is the process of nesting functions, where the output of one function becomes the input for another. The video clarifies that to evaluate a composite function like (g ∘ f)(x), one must first apply function f to x and then apply function g to the result.
πŸ’‘Domain
The domain of a relation or function is the set of all possible input values that the relation or function can accept. It defines the range of values that can be input into the function without leading to undefined results. In the context of the video, the domain is crucial for understanding which elements can be mapped or associated with elements in the codomain.
πŸ’‘Codomain
The codomain of a relation or function is the set of all possible output values that the relation or function can produce. It is the range of results one can expect when all elements in the domain are mapped according to the relation or function's rules. The codomain is essential for understanding the complete set of possible outcomes from a given domain.
πŸ’‘Range
The range of a function is the set of all output values that result from applying the function to every element in its domain. It is the counterpart to the domain and represents the complete set of values that the function can produce. The video emphasizes that the range is a subset of the codomain, which is the set of all possible outputs without considering the domain's restrictions.
πŸ’‘One-to-One Relation
A one-to-one relation is a type of relation where each element in the domain corresponds to a unique element in the codomain. This means that no two elements in the domain have the same associated element in the codomain, ensuring a one-to-one correspondence. One-to-one relations are important because they can be inverted to find the inverse function, which is a key concept in many mathematical applications.
πŸ’‘Many-to-One Relation
A many-to-one relation is a type of relation where multiple elements in the domain can correspond to a single element in the codomain. This means that one output value can be associated with several inputs, which is common in situations where there is a summarization or aggregation of data. Many-to-one relations are useful in creating mappings but do not have inverse functions because the single output does not uniquely identify the input.
πŸ’‘Rule of a Mapping
The rule of a mapping is the formula or pattern that defines how elements from the domain are associated with elements in the codomain. It is the mathematical expression that dictates the transformation from input to output. Understanding the rule of a mapping is crucial for predicting the outcome of any given input within the domain.
Highlights

Introduction to mappings, relations, and functions from Brilliant Maths Module 8.

A relation is a connection between two sets that associates elements of one set to elements of another.

The domain of a relation is the set of elements which are associated, and the codomain is the set of possible associations.

Different types of relations include one-to-one, one-to-many, many-to-one, and many-to-many.

A mapping is a relation where each member of the domain maps onto only one member of the codomain.

Examples of mappings include a times four rule and the number of days in different months.

A function is a mapping where each element in the domain has one and only one image in the range.

Inverse functions are the reverse of a function, and can be found by swapping x and f(x) and solving for the new f(x).

Composite functions involve applying one function to the result of another, also known as function of a function.

To find the value of a composite function, replace the x in the inner function with the result of the outer function.

Examples and exercises are provided to help understand and apply the concepts of mappings, relations, functions, and their inverses.

The lesson emphasizes the importance of understanding the relationships and properties of mathematical mappings, relations, and functions.

The concept of the image of a domain and codomain is introduced, which refers to the actual elements associated in a relation.

A detailed explanation of how to find the rule of a mapping is provided, using a step-by-step approach.

The lesson explains how to evaluate a function at a specific value by substituting the value into the function's expression.

The process of finding the inverse of a function is outlined, including the necessary algebraic steps.

The lesson concludes with an encouragement to practice and apply the learned concepts to excel in mathematics.

Transcripts
00:01

[Music]

00:18

hi there

00:19

this is brilliantmath.com where every

00:21

student has an opportunity to learn

00:23

and excel in maths today's lesson is

00:26

taken from brilliant maths module 8

00:29

and we're going to learn mappings

00:31

relations

00:32

and functions get a copy of the book and

00:34

let's go to class

00:36

in this lesson we want to look at

00:38

relations

00:39

mappings and functions a relation

00:43

is a connection between two sets

00:46

it associates the elements of one set

00:49

to the elements of another set

00:52

so for example we have cami

00:56

dan amma tom and ngozi

00:59

they're all friends in a particular set

01:03

let's just set x and we have

01:06

the countries from which they come from

01:08

ghana nigeria togo

01:11

and france the set x

01:14

is the domain of the relation and the

01:17

set y

01:18

is the codomain of the relation

01:22

so from the diagram we have on the board

01:25

kemi is from nigeria

01:29

dan is from france

01:32

amma is from ghana tom

01:36

is from france and ungazi

01:39

is from nigeria from the domain

01:43

there is nobody in the code domain from

01:46

togo

01:48

so the image of this domain

01:53

would be the countries that actually

01:54

have people from them that's

01:56

ghana nigeria and france

02:00

with the exception of togo so the image

02:03

of this domain

02:04

will be ghana nigeria and france

02:08

there are different types of relations

02:12

we have one-to-one relation where

02:15

every element in the domain

02:19

has an image in the code domain

02:22

when you have that that is called a one

02:25

to one relation

02:27

for example we have the set x with one

02:30

two three

02:32

and the code domain three six

02:35

nine the relation is

02:38

times three so one

02:41

times three is three two times three

02:45

is six and three times three is nine

02:48

so in this type of relation each

02:52

member of the domain has

02:55

a unique image in the

02:58

code domain next we have

03:02

one-to-many relation here

03:06

the domain is three and four and the

03:08

code domain is

03:09

9 18 8 and 6. and the relation between

03:13

the two sets

03:14

is is a factor of so

03:17

three is a factor of nine

03:21

so three has an image in the codomain

03:25

three is also a factor of eighteen so

03:28

that's why it's called

03:30

one to many

03:33

one member one element in the domain

03:36

to many in the code domain

03:40

one element has more than one image so

03:43

three has two images

03:45

four is a factor of eight and four is a

03:47

factor of sixteen

03:49

one too many so the next type of

03:53

relation is the many to one

03:55

relation and here our relation is

03:58

a multiple off so we have 12 15 and 36

04:04

it's a multiple of so 12 is a multiple

04:07

of 12.

04:09

15 is a multiple of 5. 36

04:13

is a multiple of 12 and a multiple of

04:15

nine so

04:17

many to one signifying that 12 and 36

04:21

that is more than one member of the

04:24

domain

04:25

has one image in the co-domain

04:29

so many to one 12 and 36 have

04:33

one image so that's an example of a many

04:36

to one

04:37

relation and then the fourth type is

04:39

many to many

04:41

and this relationship here is greater

04:44

than

04:44

so we have 6 and 7 in the domain and we

04:47

have 2 3

04:48

and 5 in the codomain 6

04:52

is greater than 2 6 is greater than

04:54

three

04:55

and six is greater than five so

04:58

six one element or one member of the

05:02

domain

05:02

has three images in the code domain

05:08

seven is greater than two seven is

05:11

greater than three

05:13

and seven is greater than five so

05:16

we have many to many relation

05:19

this is very interesting just take your

05:22

time

05:22

to understand what the relationship is

05:27

so that you can identify the members

05:30

of the co-domain we're looking at

05:33

relations

05:34

mappings and functions we have already

05:37

defined

05:38

relations a mapping is a relation in

05:41

which

05:42

each member in the domain maps

05:45

onto only one member

05:48

in the code domain the first set

05:52

is the domain and the second set is the

05:54

codomain so what that means that

05:56

is that each member each element

06:00

in the domain has only one image

06:04

in the core domain so when you have that

06:07

then it is a mapping so that implies

06:10

that a one-to-one

06:12

relation and a many-to-one relation

06:15

are mappings we should note that a

06:19

one to many and in many to many

06:21

relations

06:22

are not mappings we defined all of this

06:27

in the first segment so examples of

06:30

mappings

06:31

this is a one-to-one mapping one maps to

06:34

four

06:35

two maps to eight and three maps to

06:37

twelve the relation is times four

06:40

that's the rule of the mapping so one

06:42

times four is four

06:43

two times four is eight and three times

06:46

four is twelve

06:48

another mapping here is

06:51

month of the year

06:54

and the number of days so january

06:58

june and july the relation is has

07:00

january

07:01

has 31 days

07:06

june has 30 days and july has

07:09

31 days so this is a many to one

07:13

mapping more than one member of the

07:17

domain

07:19

mapped to one member of the core

07:22

domain so these are two examples of

07:26

mappings next we want to learn how to

07:29

find

07:30

the rule of a mapping example

07:33

what is the rule of this mapping x maps

07:36

to y

07:37

one maps to one two maps to three

07:41

three maps to five four maps to seven

07:44

and five maps to nine

07:48

study the mapping what do we notice

07:52

the domain is one two three four five

07:56

and the code domain is one three five

07:59

seven

08:00

nine so what is the rule what happens to

08:03

the number in the domain

08:04

to produce the number in the core

08:08

domain by inspection

08:12

we can see that each number is mapped

08:15

to twice itself minus one

08:19

so two times one is two

08:22

minus one is one two times two is four

08:26

four minus one is three two times three

08:28

is six

08:29

six minus one is five so obviously

08:33

that is the rule for this mapping

08:36

for any element x the rule of the

08:39

mapping is

08:41

two x minus one x maps to

08:44

two x minus one so to check we've

08:47

already done that when x is one

08:50

two times one is two two minus one is

08:52

one and when x is four

08:54

two times four is eight and eight minus

08:56

one is seven so the rule for this

08:59

mapping

08:59

is two x minus one

09:03

for more examples get a copy of

09:05

brilliant math

09:07

module 8 we're still looking at

09:09

relations

09:10

mappings and functions a function is a

09:13

mapping

09:14

in which each element in the domain

09:17

has one and only one

09:20

image in the range

09:25

note that only one to one

09:28

and many to one relations

09:31

are functions that means that

09:35

one to many and many too many relations

09:38

are not functions

09:42

an expression in the form 3x minus 2

09:45

in which the variable is x is called a

09:49

function of x

09:50

so that function is in terms of x the

09:54

numerical value of the expression

09:57

depends

09:58

on the value of x so this is written as

10:01

f of x is equal to three x minus two or

10:05

f x maps to three x

10:08

minus two so for example if f of x is

10:11

equal to four

10:12

x minus two find the value of

10:16

f two and f minus three what does that

10:19

mean

10:20

it means find the value of the function

10:22

when x

10:23

is equal to two and find the value of

10:26

the function

10:27

when x is equal to minus three

10:30

so let's quickly do that so the solution

10:33

the function is f of x is equal to 4x

10:38

minus 2 so when x is equal to 2

10:41

we simply replace x with

10:44

2 so 4x becomes 4

10:48

times 2. so f2 is 4

10:51

times 2 minus 2

10:55

so that is 8 minus 2 which is equal to 6

10:58

so that the numerical value of the

11:00

function

11:01

when x is equal to 2 is equal to 6.

11:05

so next to find the numerical value of

11:07

the function

11:08

when x is equal to -3

11:11

f minus 3 we simply replace the value of

11:15

x with minus three

11:16

so it's four

11:20

times minus three

11:24

minus two and 4 times minus 3 is minus

11:29

12

11:30

minus 12 minus 2 is equal to

11:34

minus 14.

11:38

so let's look at other examples so let's

11:41

look at this example

11:43

if f of x is equal to 6 x plus 16

11:47

and f of x is equal to 10 solve for x

11:51

that is very easy what have we been

11:54

given

11:56

we have that f of x

12:01

is equal to 6 x plus 16

12:05

and also f of x is equal to 10 that

12:08

implies that 6x plus 16 is equal to 10.

12:12

so 6x plus 16

12:15

is equal to 10. we have a linear

12:18

equation in

12:19

x so we simply solve that equation

12:25

6x is equal to 10 minus

12:28

16 and 10 minus 16 is minus 6.

12:32

so 6x is equal to -6 so

12:35

x is equal to minus 6

12:38

over 6 which is equal to minus

12:42

1. see how easy it is you can get more

12:45

practice

12:46

from brilliant maths module 8. so here

12:49

we want to look at

12:50

inverse functions what is the inverse of

12:54

a function

12:55

the inverse of a function is reverse

12:59

it's as simple as that the reverse of a

13:01

function

13:02

is the inverse so the inverse of a

13:04

function f of s is written

13:06

as f to the power -1

13:10

x just called f

13:13

inverse of x

13:17

example find the inverse of the

13:18

following functions the first function

13:21

is f of x is equal to 2x minus 3

13:23

and the second function is f of x is

13:25

equal to 3x

13:26

minus 1 over 2. the steps to finding

13:30

the inverse of a function as follows

13:34

first make x the subject

13:39

then after you have made x the subject

13:41

then you

13:42

change your x to f prime of x

13:45

or f inverse of x and then change your f

13:49

of x

13:50

to x since the reverse

13:53

is the inverse is very simple

13:56

so f of x is equal to 2x minus 3.

14:00

so solution f of x is 2x minus 3 you

14:03

write out the function again

14:05

make x the subject so the target is to

14:08

make x the subject

14:10

so 2x is equal to we take minus 3 to the

14:13

other side of the equation so two x is

14:16

equal to f of x

14:18

plus three so x

14:21

is equal to f of x plus three over two

14:25

so since the reverse is the inverse

14:29

we now change our x to

14:33

the inverse f prime of x and then

14:36

change the f of x to x so f

14:40

prime of x is equal to x

14:43

plus three over two so the inverse of

14:48

two x minus three is x

14:51

plus three over two the second example

14:55

f of x is equal to three x minus

14:59

two so solution

15:06

make x the subject so we have

15:09

two f of x is equal to

15:13

three x minus one

15:17

so three x is equal to

15:22

two f of x

15:25

plus 1 because the minus 1 crosses over

15:28

to the other side of the equation so

15:31

that our

15:32

x is equal to 2

15:35

f of x plus 1

15:38

over 3. so the final step is to swap

15:42

your x which is now our inverse

15:48

is equal to swap the f of x and make it

15:51

x 2x plus 1

15:55

over 3. you can try many more

15:59

exercises and examples from brilliant

16:01

maths

16:02

module 8. still looking at inverse

16:05

functions

16:07

if g of x is equal to three x minus two

16:10

over five evaluate g inverse

16:13

of five so find numerical value of the

16:17

inverse

16:17

when x is equal to five so the first

16:20

task is to find the

16:22

inverse of the function now remember the

16:24

steps

16:26

first make x the subject of the function

16:30

and then swap your x change your x

16:34

to the inverse and change the function

16:37

to x so first step

16:40

make x the subject so cross multiply we

16:44

have

16:44

5 g of x

16:47

is equal to 3x minus two

16:52

we want to make x the subject so three x

16:56

is equal to five g of x

17:01

plus two so that

17:05

x is equal to 5

17:08

g of x plus 2

17:12

over 3. so that's the inverse now so we

17:16

swap

17:18

g inverse of x is equal to

17:21

five x plus two over three

17:26

we are done with that so the next task

17:29

is to find g inverse of 5.

17:36

so next we are going to find the value

17:38

of the

17:39

inverse when x is equal to 5.

17:43

so g inverse of five

17:47

will be equal to five times

17:50

five because the value of x is now five

17:55

five times five plus two

17:59

over 3.

18:02

so g inverse of 5 5 times 5 is 25

18:07

25 plus 2 is 27

18:11

27 over 3 which is equal to 9

18:15

so g inverse

18:18

of 5 is equal to 9.

18:22

this is actually very easy take your

18:25

time

18:25

to solve some more problems so here

18:28

we're looking at

18:29

composite functions composite functions

18:33

deal with more than one function

18:35

it is also called function of a function

18:38

so for example given that f of x is

18:41

equal to three x

18:43

and g of x is equal to x minus one

18:46

find g f of x

18:50

and find f g of x

18:53

how does this work assembly means

18:59

g f of x

19:04

becomes

19:08

g replace that with

19:12

what you have as f of x

19:16

3x and then next

19:20

g3 x that means that the value of x now

19:23

is 3x

19:24

so in the function g of x

19:28

anywhere you find x replace it with 3x

19:32

and that would be the answer so g3s will

19:38

be equal to

19:40

x minus 1 becomes 3x

19:43

minus 1. so g

19:47

f of x is equal to 3x

19:51

minus 1. so let's now see

19:55

f g of x

20:02

f g of x so in the same manner

20:09

we replace g of x with x minus one

20:13

so that f can now act on it

20:16

x minus one and in our

20:20

function f of x

20:24

wherever we see x we replace it with

20:28

x minus 1. so

20:31

f of x minus 1 is

20:35

3x and that becomes

20:39

3 into x minus 1. if we simplify that

20:44

so we have that f g

20:47

of x is equal to three x

20:50

minus three we can see that

20:55

g f of x is different from

20:59

f g of x so just take your time

21:03

and work more examples let's look at

21:06

another example

21:08

let's look at one more example working

21:11

with composite

21:12

functions remember that composite

21:15

functions deal with more than one

21:17

function

21:19

evaluate gh of minus 3

21:22

if g of x is equal to 6 into

21:25

minus x plus 2 and h of x is equal to 3

21:30

plus 2x

21:31

over 5. please pay attention

21:34

this is a little complicated

21:38

so g of x is equal to 6 into minus

21:41

x plus 2 and h of x is three plus two x

21:45

over five so we want to find g

21:48

h of minus three so first we'll find g h

21:51

of x

21:53

so g h of x what's our h of x

21:56

we replace this h of x

21:59

with three plus two x over five

22:03

so we have g

22:06

into three plus

22:09

two x over 5.

22:12

so in the function g now wherever i

22:16

find x we'll replace it with 3 plus 2x

22:21

over 5. so that's equal to

22:27

six into

22:31

minus three plus

22:35

two x over five

22:39

then plus 2 because this is now our x

22:45

plus 2. so next

22:49

i'd like to open up the first bracket

22:55

so 6 times 3

22:59

that will be minus 18

23:04

minus 12x

23:08

all over five plus

23:11

six times two plus twelve

23:16

so next i can simply

23:20

make both of them a denominator of 5

23:23

so that that becomes

23:28

minus 18

23:32

minus 12x plus

23:37

60 over 5

23:40

by simply multiplying 5 by 12 to give me

23:44

60.

23:45

if i divide this 60 here by 5 to take it

23:48

out of the denominator it will give

23:50

me 12. so collect like terms now

23:55

that would be 60 minus

23:59

18 minus

24:02

12x all over 5.

24:07

so my final answer 60 minus 18

24:10

will be 42 42

24:14

minus 12x over

24:18

five this is equal to g h

24:21

of x

24:26

but the value of x should be minus three

24:29

so substitute minus 3 into this

24:33

expression so minus 3 times minus 12

24:37

will be positive 36

24:41

so that we have 42

24:46

plus 36

24:50

over 5 and what does that become

24:54

78

24:59

78 over 5. and 78 over 5 is equal to

25:03

what

25:05

5 into 7 will be 1 remainder 2

25:09

into 28 will be 5

25:13

25 remember three

25:16

three fifths which is the same as

25:19

fifteen

25:20

three fifths or fifteen

25:24

point six so that is the value of

25:28

g h of minus three and that's the final

25:32

answer for more exercises and more

25:35

examples

25:36

get a copy of brilliant maths module

25:38

eight

25:40

from here we have

25:44

minus 18

25:47

minus 12x

25:51

over five but i'd like to put all of

25:54

them under one denominator

25:57

so that lcm is 5. so what do i do

26:02

if i'm putting all of that over 5

26:05

then i'll multiply the 12 by 5

26:09

to give me 60. so plus

26:13

60. so now i have minus 18

26:17

minus 12x plus 60

26:20

over 5. collect like terms now

26:24

so i have 60

26:27

minus 18 minus 12x

26:32

over 5. 60 minus 18

26:36

is 42 so i have 42

26:40

minus 12x over 5. so this is equal to

26:45

42 minus 12x

26:49

over 5. so this is my

26:52

g h of x

26:55

so that is g h

26:58

of x what were we supposed to find

27:01

it says evaluate g h of minus three

27:06

so meaning when x is minus 3.

27:09

so next we're going to substitute the

27:12

value of

27:13

x so x is supposed to be minus 3.

27:16

so what do we have we have that equal to

27:21

so we have g h of minus 3

27:25

is equal to 42 minus

27:30

12 times 3 over

27:34

five

27:38

twelve times minus three please

27:41

so 42 minus

27:45

twelve times minus three over 5

27:49

so we have

27:52

42 minus 12 times minus 3 will give us

27:57

plus 36 so 42

28:01

plus 36

28:04

over 5. so our g

28:07

h of -3 will be

28:14

g h minus 3

28:17

is equal to 42 plus 36

28:20

will be 78 78

28:24

over 5. and 78 over 5

28:29

if we want to convert that to a mixed

28:32

number

28:33

five into seven is one

28:37

remainder two five into twenty eight

28:42

is five 28

28:46

minus 25 is 3

28:50

so that's 15 3 over 5

28:53

or 15 3 5

28:56

is the same as point six

29:01

or fifteen point six we have

29:04

answered the question in full because we

29:07

are told to evaluate

29:10

g h of minus 3 if g x

29:13

if g of x is equal to 6 into minus x

29:16

plus 2

29:17

and h of x is equal to 3 plus

29:20

2 x over 5. so we evaluated

29:24

g h of x and we found g h of x to be

29:28

equal to

29:29

42 minus 12 x over five

29:32

having found that we then input the

29:36

value of

29:37

x which is minus three into our

29:40

composite

29:41

function so when we put that value the

29:44

value of x which is minus 3

29:46

we get gh of minus 3 to be equal to 78

29:49

over 5

29:51

or 15 three fifths as a mixed fraction

29:55

or 15.6 as a decimal

29:59

number for more examples and more

30:02

exercises

30:03

get a copy of brilliant maths module 8.

30:08

hello i hope you enjoyed the lessons

30:11

on relations mappings and functions

30:14

how to identify the rule of a function

30:18

and when is a mapping a function and

30:20

when is it not a function

30:22

we also learnt how to find the inverse

30:24

of a function

30:25

and we learnt composite functions

30:27

finding the function

30:29

of a function all of these we are taking

30:31

from brilliant math

30:33

module 8. subscribe to

30:34

brilliantmaths.com follow us on

30:36

all our social media handles and you

30:38

will be on your way to success

30:41

in maths my name is ngozi aravoggini

30:44

and always remember that maths is fun

30:47

and you can excel in it

30:54

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