Relations, Mappings & Functions.
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
- π A relation is a connection between two sets that associates elements of one set to elements of another.
- πΊοΈ The domain of a relation is the set of elements from which the relation originates, while the codomain is the set of possible outputs.
- π One-to-one relations ensure that each element in the domain has a unique image in the codomain.
- π One-to-many relations occur when one element in the domain can have multiple images in the codomain.
- π’ Many-to-one relations mean multiple elements in the domain can map to a single element in the codomain.
- π 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.
- π A mapping is a relation where each member of the domain maps onto exactly one member of the codomain.
- π Functions are mappings where each element in the domain has one and only one image in the range, allowing for inverse functions.
- π 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).
- π§ 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))).
- π 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
π 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.
π 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.
π 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.
π 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.
π€ 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 β f)(x) and (f β 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.
π 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.
π 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
Keywords
π‘Mappings
π‘Relations
π‘Functions
π‘Inverse Functions
π‘Composite Functions
π‘Domain
π‘Codomain
π‘Range
π‘One-to-One Relation
π‘Many-to-One Relation
π‘Rule of a Mapping
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
[Music]
hi there
this is brilliantmath.com where every
student has an opportunity to learn
and excel in maths today's lesson is
taken from brilliant maths module 8
and we're going to learn mappings
relations
and functions get a copy of the book and
let's go to class
in this lesson we want to look at
relations
mappings and functions a relation
is a connection between two sets
it associates the elements of one set
to the elements of another set
so for example we have cami
dan amma tom and ngozi
they're all friends in a particular set
let's just set x and we have
the countries from which they come from
ghana nigeria togo
and france the set x
is the domain of the relation and the
set y
is the codomain of the relation
so from the diagram we have on the board
kemi is from nigeria
dan is from france
amma is from ghana tom
is from france and ungazi
is from nigeria from the domain
there is nobody in the code domain from
togo
so the image of this domain
would be the countries that actually
have people from them that's
ghana nigeria and france
with the exception of togo so the image
of this domain
will be ghana nigeria and france
there are different types of relations
we have one-to-one relation where
every element in the domain
has an image in the code domain
when you have that that is called a one
to one relation
for example we have the set x with one
two three
and the code domain three six
nine the relation is
times three so one
times three is three two times three
is six and three times three is nine
so in this type of relation each
member of the domain has
a unique image in the
code domain next we have
one-to-many relation here
the domain is three and four and the
code domain is
9 18 8 and 6. and the relation between
the two sets
is is a factor of so
three is a factor of nine
so three has an image in the codomain
three is also a factor of eighteen so
that's why it's called
one to many
one member one element in the domain
to many in the code domain
one element has more than one image so
three has two images
four is a factor of eight and four is a
factor of sixteen
one too many so the next type of
relation is the many to one
relation and here our relation is
a multiple off so we have 12 15 and 36
it's a multiple of so 12 is a multiple
of 12.
15 is a multiple of 5. 36
is a multiple of 12 and a multiple of
nine so
many to one signifying that 12 and 36
that is more than one member of the
domain
has one image in the co-domain
so many to one 12 and 36 have
one image so that's an example of a many
to one
relation and then the fourth type is
many to many
and this relationship here is greater
than
so we have 6 and 7 in the domain and we
have 2 3
and 5 in the codomain 6
is greater than 2 6 is greater than
three
and six is greater than five so
six one element or one member of the
domain
has three images in the code domain
seven is greater than two seven is
greater than three
and seven is greater than five so
we have many to many relation
this is very interesting just take your
time
to understand what the relationship is
so that you can identify the members
of the co-domain we're looking at
relations
mappings and functions we have already
defined
relations a mapping is a relation in
which
each member in the domain maps
onto only one member
in the code domain the first set
is the domain and the second set is the
codomain so what that means that
is that each member each element
in the domain has only one image
in the core domain so when you have that
then it is a mapping so that implies
that a one-to-one
relation and a many-to-one relation
are mappings we should note that a
one to many and in many to many
relations
are not mappings we defined all of this
in the first segment so examples of
mappings
this is a one-to-one mapping one maps to
four
two maps to eight and three maps to
twelve the relation is times four
that's the rule of the mapping so one
times four is four
two times four is eight and three times
four is twelve
another mapping here is
month of the year
and the number of days so january
june and july the relation is has
january
has 31 days
june has 30 days and july has
31 days so this is a many to one
mapping more than one member of the
domain
mapped to one member of the core
domain so these are two examples of
mappings next we want to learn how to
find
the rule of a mapping example
what is the rule of this mapping x maps
to y
one maps to one two maps to three
three maps to five four maps to seven
and five maps to nine
study the mapping what do we notice
the domain is one two three four five
and the code domain is one three five
seven
nine so what is the rule what happens to
the number in the domain
to produce the number in the core
domain by inspection
we can see that each number is mapped
to twice itself minus one
so two times one is two
minus one is one two times two is four
four minus one is three two times three
is six
six minus one is five so obviously
that is the rule for this mapping
for any element x the rule of the
mapping is
two x minus one x maps to
two x minus one so to check we've
already done that when x is one
two times one is two two minus one is
one and when x is four
two times four is eight and eight minus
one is seven so the rule for this
mapping
is two x minus one
for more examples get a copy of
brilliant math
module 8 we're still looking at
relations
mappings and functions a function is a
mapping
in which each element in the domain
has one and only one
image in the range
note that only one to one
and many to one relations
are functions that means that
one to many and many too many relations
are not functions
an expression in the form 3x minus 2
in which the variable is x is called a
function of x
so that function is in terms of x the
numerical value of the expression
depends
on the value of x so this is written as
f of x is equal to three x minus two or
f x maps to three x
minus two so for example if f of x is
equal to four
x minus two find the value of
f two and f minus three what does that
mean
it means find the value of the function
when x
is equal to two and find the value of
the function
when x is equal to minus three
so let's quickly do that so the solution
the function is f of x is equal to 4x
minus 2 so when x is equal to 2
we simply replace x with
2 so 4x becomes 4
times 2. so f2 is 4
times 2 minus 2
so that is 8 minus 2 which is equal to 6
so that the numerical value of the
function
when x is equal to 2 is equal to 6.
so next to find the numerical value of
the function
when x is equal to -3
f minus 3 we simply replace the value of
x with minus three
so it's four
times minus three
minus two and 4 times minus 3 is minus
12
minus 12 minus 2 is equal to
minus 14.
so let's look at other examples so let's
look at this example
if f of x is equal to 6 x plus 16
and f of x is equal to 10 solve for x
that is very easy what have we been
given
we have that f of x
is equal to 6 x plus 16
and also f of x is equal to 10 that
implies that 6x plus 16 is equal to 10.
so 6x plus 16
is equal to 10. we have a linear
equation in
x so we simply solve that equation
6x is equal to 10 minus
16 and 10 minus 16 is minus 6.
so 6x is equal to -6 so
x is equal to minus 6
over 6 which is equal to minus
1. see how easy it is you can get more
practice
from brilliant maths module 8. so here
we want to look at
inverse functions what is the inverse of
a function
the inverse of a function is reverse
it's as simple as that the reverse of a
function
is the inverse so the inverse of a
function f of s is written
as f to the power -1
x just called f
inverse of x
example find the inverse of the
following functions the first function
is f of x is equal to 2x minus 3
and the second function is f of x is
equal to 3x
minus 1 over 2. the steps to finding
the inverse of a function as follows
first make x the subject
then after you have made x the subject
then you
change your x to f prime of x
or f inverse of x and then change your f
of x
to x since the reverse
is the inverse is very simple
so f of x is equal to 2x minus 3.
so solution f of x is 2x minus 3 you
write out the function again
make x the subject so the target is to
make x the subject
so 2x is equal to we take minus 3 to the
other side of the equation so two x is
equal to f of x
plus three so x
is equal to f of x plus three over two
so since the reverse is the inverse
we now change our x to
the inverse f prime of x and then
change the f of x to x so f
prime of x is equal to x
plus three over two so the inverse of
two x minus three is x
plus three over two the second example
f of x is equal to three x minus
two so solution
make x the subject so we have
two f of x is equal to
three x minus one
so three x is equal to
two f of x
plus 1 because the minus 1 crosses over
to the other side of the equation so
that our
x is equal to 2
f of x plus 1
over 3. so the final step is to swap
your x which is now our inverse
is equal to swap the f of x and make it
x 2x plus 1
over 3. you can try many more
exercises and examples from brilliant
maths
module 8. still looking at inverse
functions
if g of x is equal to three x minus two
over five evaluate g inverse
of five so find numerical value of the
inverse
when x is equal to five so the first
task is to find the
inverse of the function now remember the
steps
first make x the subject of the function
and then swap your x change your x
to the inverse and change the function
to x so first step
make x the subject so cross multiply we
have
5 g of x
is equal to 3x minus two
we want to make x the subject so three x
is equal to five g of x
plus two so that
x is equal to 5
g of x plus 2
over 3. so that's the inverse now so we
swap
g inverse of x is equal to
five x plus two over three
we are done with that so the next task
is to find g inverse of 5.
so next we are going to find the value
of the
inverse when x is equal to 5.
so g inverse of five
will be equal to five times
five because the value of x is now five
five times five plus two
over 3.
so g inverse of 5 5 times 5 is 25
25 plus 2 is 27
27 over 3 which is equal to 9
so g inverse
of 5 is equal to 9.
this is actually very easy take your
time
to solve some more problems so here
we're looking at
composite functions composite functions
deal with more than one function
it is also called function of a function
so for example given that f of x is
equal to three x
and g of x is equal to x minus one
find g f of x
and find f g of x
how does this work assembly means
g f of x
becomes
g replace that with
what you have as f of x
3x and then next
g3 x that means that the value of x now
is 3x
so in the function g of x
anywhere you find x replace it with 3x
and that would be the answer so g3s will
be equal to
x minus 1 becomes 3x
minus 1. so g
f of x is equal to 3x
minus 1. so let's now see
f g of x
f g of x so in the same manner
we replace g of x with x minus one
so that f can now act on it
x minus one and in our
function f of x
wherever we see x we replace it with
x minus 1. so
f of x minus 1 is
3x and that becomes
3 into x minus 1. if we simplify that
so we have that f g
of x is equal to three x
minus three we can see that
g f of x is different from
f g of x so just take your time
and work more examples let's look at
another example
let's look at one more example working
with composite
functions remember that composite
functions deal with more than one
function
evaluate gh of minus 3
if g of x is equal to 6 into
minus x plus 2 and h of x is equal to 3
plus 2x
over 5. please pay attention
this is a little complicated
so g of x is equal to 6 into minus
x plus 2 and h of x is three plus two x
over five so we want to find g
h of minus three so first we'll find g h
of x
so g h of x what's our h of x
we replace this h of x
with three plus two x over five
so we have g
into three plus
two x over 5.
so in the function g now wherever i
find x we'll replace it with 3 plus 2x
over 5. so that's equal to
six into
minus three plus
two x over five
then plus 2 because this is now our x
plus 2. so next
i'd like to open up the first bracket
so 6 times 3
that will be minus 18
minus 12x
all over five plus
six times two plus twelve
so next i can simply
make both of them a denominator of 5
so that that becomes
minus 18
minus 12x plus
60 over 5
by simply multiplying 5 by 12 to give me
60.
if i divide this 60 here by 5 to take it
out of the denominator it will give
me 12. so collect like terms now
that would be 60 minus
18 minus
12x all over 5.
so my final answer 60 minus 18
will be 42 42
minus 12x over
five this is equal to g h
of x
but the value of x should be minus three
so substitute minus 3 into this
expression so minus 3 times minus 12
will be positive 36
so that we have 42
plus 36
over 5 and what does that become
78
78 over 5. and 78 over 5 is equal to
what
5 into 7 will be 1 remainder 2
into 28 will be 5
25 remember three
three fifths which is the same as
fifteen
three fifths or fifteen
point six so that is the value of
g h of minus three and that's the final
answer for more exercises and more
examples
get a copy of brilliant maths module
eight
from here we have
minus 18
minus 12x
over five but i'd like to put all of
them under one denominator
so that lcm is 5. so what do i do
if i'm putting all of that over 5
then i'll multiply the 12 by 5
to give me 60. so plus
60. so now i have minus 18
minus 12x plus 60
over 5. collect like terms now
so i have 60
minus 18 minus 12x
over 5. 60 minus 18
is 42 so i have 42
minus 12x over 5. so this is equal to
42 minus 12x
over 5. so this is my
g h of x
so that is g h
of x what were we supposed to find
it says evaluate g h of minus three
so meaning when x is minus 3.
so next we're going to substitute the
value of
x so x is supposed to be minus 3.
so what do we have we have that equal to
so we have g h of minus 3
is equal to 42 minus
12 times 3 over
five
twelve times minus three please
so 42 minus
twelve times minus three over 5
so we have
42 minus 12 times minus 3 will give us
plus 36 so 42
plus 36
over 5. so our g
h of -3 will be
g h minus 3
is equal to 42 plus 36
will be 78 78
over 5. and 78 over 5
if we want to convert that to a mixed
number
five into seven is one
remainder two five into twenty eight
is five 28
minus 25 is 3
so that's 15 3 over 5
or 15 3 5
is the same as point six
or fifteen point six we have
answered the question in full because we
are told to evaluate
g h of minus 3 if g x
if g of x is equal to 6 into minus x
plus 2
and h of x is equal to 3 plus
2 x over 5. so we evaluated
g h of x and we found g h of x to be
equal to
42 minus 12 x over five
having found that we then input the
value of
x which is minus three into our
composite
function so when we put that value the
value of x which is minus 3
we get gh of minus 3 to be equal to 78
over 5
or 15 three fifths as a mixed fraction
or 15.6 as a decimal
number for more examples and more
exercises
get a copy of brilliant maths module 8.
hello i hope you enjoyed the lessons
on relations mappings and functions
how to identify the rule of a function
and when is a mapping a function and
when is it not a function
we also learnt how to find the inverse
of a function
and we learnt composite functions
finding the function
of a function all of these we are taking
from brilliant math
module 8. subscribe to
brilliantmaths.com follow us on
all our social media handles and you
will be on your way to success
in maths my name is ngozi aravoggini
and always remember that maths is fun
and you can excel in it
[Music]
Browse More Related Video
Algebra Basics: What Are Functions? - Math Antics
Finding Inverse Functions (Precalculus - College Algebra 51)
Functions - Vertical Line Test, Ordered Pairs, Tables, Domain and Range
Introduction to Functions (Precalculus - College Algebra 2)
One to One Functions (Precalculus - College Algebra 50)
06 - What is a Function in Math? (Learn Function Definition, Domain & Range in Algebra)
5.0 / 5 (0 votes)
Thanks for rating: