Download A , B

SETS
Sets are denoted by
Capital letters
Sets use “curly” brackets
A = {1, 3, 2, 5}
n(A) = | A | = 4
The number of elements
in Set A is 4
3 A
7 A
3 is an element of A
7 is not an element of A
A set is a distinct collection of objects. The objects are
called elements.
Order does not matter. If a set
{1, 2, 3, 4} = {2, 3, 1, 4} contains the same elements as
another set, the sets are equal.
{1,2,3,2,3,2,3,5,5}
5}
{1, 3,
{1,
2}
In ascending
order
We
never repeat
elements in a set.
This symbol means "is a subset of"
AB
This is read "A is a subset of B".
A = {1, 2, 3}
B = {1, 2, 3, 4, 5}
If a set doesn't contain any elements it is called the
empty set or the null set. It is denoted by  or { }.
NOT {} 
It is agreed that the empty set is a subset of all other sets
so:

A
A where A is any set.
List all of the subsets of {1, 2, 3}.

{1}
{2}
{3}
{1, 2}
Notice the empty
set is NOT in set
brackets.
{1, 3}
{2, 3}
{1, 2, 3}
Number of
Elements in Set
Possible Subsets
Total Number of
Possible Subsets
1. {A}
{A} 
2
2. {A , B}
{A , B} {A} {B} 
4
3. {A , B , C}
{A , B , C} {A , B} {A , C}
{B , C} {A} {B} {C} 
8
4. {A , B , C, D}
{A , B , C , D} {A , B , C} {A ,
B , D} {A , C , D}
{B , C , D} {A , B} {A , C}
{A , D} {A , B} …… {D} 
?
16
2n
The number of possible subsets of a set of size n is ?
A = {1, 2, 3, 4, 5} B = {1, 3, 5, 7, 9}
AB
= {1, 2, 3, 4, 5, 7, 9}
Remember we do
not list elements
more than once.
This is the union symbol. It means the set that consists of all
elements of set A and all elements of set B.
AB
= {1, 3, 5}
This is the intersect symbol. It means the set
containing all elements that are in both A and B.
These sets can be visualized with circles in what is called a
Venn Diagram.
A
A
B
B
AB
Everything that is in
A or B.
A
B
AB
Everything that is in
A AND B.
Often will have a set that contains all elements that we
wish to consider. This is called the universal set. All other
sets are subsets of this set.
AB=
Universal Set
AA
A
BA
There are no
elements in
both A and B.
When this is
the case they
are called
disjoint sets.
This means the complement of A, and
means the set of all elements in the
universal set that are not in A.
100 people were surveyed. 52 people in a survey owned a
cat. 36 people owned a dog. 24 did not own a dog or cat.
Draw a Venn diagram.
52 + 36 = 88 so
there must be
88 - 76 = 12
people that own
both a dog and
a cat.
universal set is 100 people surveyed
C
40
12
D
24
24
Set C is the cat owners and Set D is the dog
owners. The sets are NOT disjoint. Some
people could own both a dog and a cat.
Counting Formula:
n(A  B) = n(A) + n(B) - n(A  B)
Since 24
did not own
a dog or
cat, there
must be 76
that do.
n(C  D) =
76
This n means the
number of elements
in the set
Competition Problems
{2, 4, 6, 8}  { 21, 22 , 23, 24}
Answer: {2,4,8}
How many total number of possible
subsets are there in:
{2, 4, 6, 8}  { 21, 22 , 23, 24}
Answer:
{2,4,8}
2n ,n=3
23 =
8
If M={even whole numbers less than 10},
N={non-negative integers},
P={primes less than10},
What is (N  M )  P ?
A. {2,3,4,5,6,7,8,9}
B. {0,2,3,4,5,6,7,8}
C. {0,2,3,4,5,6,7,8,9}
D. {2,3,4,5,6,7,8}
M={0,2,4,6,8},
N={0, positive integers..},
P={2,3,5,7},
What is (N  M )  P ?
N  M = {0,2,4,6,8}
{0,2,4,6,8}  {2,3,5,7}
Answer: B {0,2,3,4,5,6,7,8}
M={0,2,4,6,8},
N={0, positive integers..},
P={2,3,5,7},
What is (N  M )  P ?
N  M = {0,2,4,6,8}
{0,2,4,6,8}  {2,3,5,7}
Answer: B {0,2,3,4,5,6,7,8}
Bonus: How many possible subsets?
M={0,2,4,6,8},
N={0, positive integers..},
P={2,3,5,7},
What is (N  M )  P ?
N  M = {0,2,4,6,8}
{0,2,4,6,8}  {2,3,5,7}
Answer: B {0,2,3,4,5,6,7,8}
Bonus: How many possible subsets?
28 = 256
Relations
And
Functions
Objective
 To determine whether a relation is a function and
identify the domain and range of a function.
 Be able to define domain, range, relation, and
function.
 Generate ordered pairs for the equation:
y=x + 3 for x = -2,-1,0,1,and 2
Can you graph this???
Do you know quadrants?
This can be represented as a graph, set of points, table,
or mapping
A relation is a set of ordered pairs.
The domain is the set of all x values in the relation
domain = {-1,0,2,4,9}
These are the x values written in a set from smallest to largest
{(2,3), (-1,5), (4,-2), (9,9), (0,-6)}
These are the y values written in a set from smallest to largest
range = {-6,-2,3,5,9}
The range is the set of all y values in the relation
This is a
relation
A relation assigns the x’s with y’s
1
2
3
4
2
4
6
5
8
10
Domain (set of all x’s)
Range (set of all y’s)
This relation can be written {(1,6), (2,2), (3,4), (4,8), (5,10)}
AAfunction
function fffrom
fromset
setAAto
toset
setBBisisaarule
ruleof
ofcorrespondence
correspondence
that
thatassigns
assigns to
toeach
eachelement
element xxin
in the
theset
setAAexactly
exactly one
element
element yyin
inthe
theset
setB.
B.
1
2
3
4
5
2
4
6
8
10
Set A is the domain
What
This is a Whew!
function
did that
---it meets
oursay?
conditions
Set B is the range
Must use all the x’s
The x value can only be assigned to one y
Let’s look at another relation and decide if it is a function.
The second condition says each x can have only one y, but it CAN
be the same y as another x gets assigned to.
1
2
3
4
5
2
4
6
8
10
Set A is the domain
This is a function
---it meets our
conditions
Set B is the range
Must use all the x’s
The x value can only be assigned to one y
A good example that you can “relate” to is students in our
maths class this semester are set A. The grade they earn out
of the class is set B. Each student must be assigned a grade
and can only be assigned ONE grade, but more than one
student can get the same grade (we hope so---we want lots
of A’s). The example show on the previous screen had each
student getting the same grade. That’s okay.
1
2
3
4
5
2
4
6
8
10
2 was assigned both 4 and 10
Is the relation shown above a function?
NO
Why not???
Check this relation out to determine if it is a function.
It is not---3 didn’t get assigned to anything
Comparing to our example, a student in maths must receive a grade
1
2
3
4
5
Set A is the domain
2
4
6
8
10
Set B is the range
This is not a
Must use all the x’s
function---it
doesn’t assign
each x with a y The x value can only be assigned to one y
Check this relation out to determine if it is a function.
This is fine—each student gets only one grade. More than one can
get an A and I don’t have to give any D’s (so all y’s don’t need to be
used).
1
2
3
4
5
2
4
6
8
10
Set A is the domain
This is a function
Set B is the range
Must use all the x’s
The x value can only be assigned to one y
If a vertical line passes through a graph more than
once, the graph is not the graph of a function.
Hint:
Pass a pencil across
the graph held
vertically to represent
a vertical line.
The pencil crosses the
graph more than once.
This is not a function
because there are two
y-values for the same
x-value.
Review: Objective
 To determine whether a relation is a function and
identify the domain and range of a function.
 Be able to define domain, range, relation, and
function.
Be able to evaluate a function for given input
values
Think back!!!
I asked:
 Generate ordered pairs for the equation:
y=x + 3 for x = -2,-1,0,1,and 2
How do I write this as a function???
Warm-up lets think
How can we use a function table / rule to calculate how much
Money you will earn for working specific amounts of time?
At your job you earn 12.00 an hour (h).
Time
worked
f(h)
1
2
3
6
Amount
earned
Write an equation ….and now we will write a function!!!
We commonly call functions by letters. Because function
starts with f, it is a commonly used letter to refer to
functions.
f  x   2 x  3x  6
2
This means
the right
hand side is
a function
called f
This means
the right hand
side has the
variable x in it
The left side DOES NOT
MEAN f times x like
brackets usually do, it
simply tells us what is on
the right hand side.
The left hand side of this equation is the function notation.
It tells us two things. We called the function f and the
variable in the function is x.
Remember---this tells you what
is on the right hand side---it is
not something you work. It says
that the right hand side is the
function f and it has x in it.
f  x   2 x  3x  6
2
f 2   22   32   6
2
f 2  24  32  6  8  6  6  8
So we have a function called f that has the variable x in it.
Using function notation we could then ask the following:
This means to find the function f and instead of
having an x in it, put a 2 in it. So let’s take the
Find f (2).
function above and make brackets everywhere
the x was and in its place, put in a 2.
Don’t forget order of operations---powers, then
multiplication, finally addition & subtraction
Find f (-2).
f  x   2 x  3x  6
2
f  2  2 2  3 2  6
2
f  2  24  3 2  6  8  6  6  20
This means to find the function f and instead of having an x
in it, put a -2 in it. So let’s take the function above and make
brackets everywhere the x was and in its place, put in a -2.
Don’t forget order of operations---powers, then
multiplication, finally addition & subtraction
f  x   2 x  3x  6
2
Find f (k).
f k   2k   3k   6
2
 
f k   2 k  3k   6  2k  3k  6
2
2
This means to find the function f and instead of having an x
in it, put a k in it. So let’s take the function above and make
brackets everywhere the x was and in its place, put in a k.
Don’t forget order of operations---powers, then
multiplication, finally addition & subtraction
Lets try…..
f(x) = 5x +1
find f(5) and f(-2)
Find Range and list as
Ordered Pairs…
f(x) = ½ x – 4
The domain is -4, -2,0,2,4
Find Range and list as
Ordered Pairs…
f(x) = 3x + 3
The domain is -3,-1,0,1,3
Find Range and list as
Ordered Pairs…
f(x) = |x-2|
The domain is 0,1,2,3,4
More…
 g(x) = 3x + 5 if g(4), g(-4)
 c(r) = 4r + 7 if c(10)
Why are all these functions & why do we call it function
notation?
f  x   2 x  3x  6
2
Find f (2k).
f 2k   22k   32k   6
2
 
f 2k   2 4k  32k   6  8k  6k  6
2
2
This means to find the function f and instead of having an x in
it, put a 2k in it. So let’s take the function above and make
brackets everywhere the x was and in its place, put in a 2k.
Don’t forget order of operations---powers, then
multiplication, finally addition & subtraction
Let's try a new function
Find g(1)+ g(-4).
g x   x  2 x
2
g 1  1  21  1
2
g  4   4  2 4  16  8  24
2
So g 1  g  4  1  24  23
The last thing we need to learn about functions for
this section is something about their domain. Recall
domain meant "Set A" which is the set of values you
plug in for x.
For the functions we will be dealing with, there
are two "illegals":
1. You can't divide by zero (denominator (bottom)
of a fraction can't be zero)
2. You can't take the square root (or even root) of
a negative number
When you are asked to find the domain of a function,
you can use any value for x as long as the value
won't create an "illegal" situation.
Find the domain for the following functions:
Since no matter what value you
choose for x, you won't be dividing
f x  2x 1 by zero or square rooting a negative
number, you can use anything you
Note: There is
want so we say the answer is:
nothing wrong with
the top = 0 just means All real numbers x.

the fraction = 0
x3
g x  
x2
illegal if this
is zero
If you choose x = 2, the denominator
will be 2 – 2 = 0 which is illegal
because you can't divide by zero.
The answer then is:
All real numbers x such that x ≠ 2.
means does not equal
Let's find the domain of another one:
h x   x  4
Can't be negative so must be ≥ 0
x4 0
solve
this
x4
We have to be careful what x's we use so that the second
"illegal" of square rooting a negative doesn't happen. This
means the "stuff" under the square root must be greater
than or equal to zero (maths way of saying "not negative").
So the answer is:
All real numbers x such that x ≥ 4
Average Rate
of Change
Average Rate of Change
• Functions are often used to model changing
quantities.
• In this section, we learn how to:
– Find the rate at which the values of a function
change as the input variable changes.
Average Rate of Change
We are all familiar with the concept
of speed.
If you drive a distance of 120 miles in 2 hours, then
your average speed, or rate of travel,
is:
120 mi
 60 mi/h
2h
Average Rate of Change
Now, suppose you take a car trip and record
the distance that you travel every few
minutes.
– The distance s you have traveled is
a function of the time t:
s(t) = total distance traveled at time t
Average Rate of Change
We graph the function s as shown.
The graph shows that you have traveled
a total of:
50 miles after 1 hour
75 miles after 2 hours
140 miles after 3 hours
and so on.
Average Rate of Change
To find your average speed between any
two points on the trip, we divide the distance
traveled by the time elapsed.
– Let’s calculate your average speed between
1:00 P.M. and 4:00 P.M.
– The time elapsed is 4 – 1 = 3 hours.
Average Rate of Change
• To find the distance you traveled, we subtract
the distance at 1:00 P.M. from the distance
at 4:00 P.M.,
that is,
200 – 50 = 150 mi
Average Rate of Change
Thus, your average speed is:
distance traveled
average speed 
time elapsed
150 mi

3h
 50 mi/h
Average Rate of Change
The average speed we have calculated can be
expressed using function notation:
s  4   s 1
average speed 
4 1
200  50

3
 50 mi/h
Average Rate of Change
Note that the average speed is
different over different time
intervals.
Average Rate of Change
For example, between 2:00 P.M. and
3:00 P.M., we find that:
average speed
s 3   s  2

32
140  75

1
 65 mi/h
Average Rate of Change—Significance
Finding average rates of change is important in
many contexts.
For instance, we may be interested in knowing:
How quickly the air temperature is dropping
as a storm approaches.
How fast revenues are increasing from
the sale of a new product.
Average Rate of Change—Significance
So, we need to know how to determine
the average rate of change of the functions
that model these quantities.
In fact, the concept of average rate of change can be
defined for any function.
Average Rate of Change—Definition
The average rate of change of the function
y = f(x) between x = a and x = b is:
change in y
average rate of change 
change in x
f  b   f a 

ba
Average Rate of Change—Definition
The average rate of change is the slope
of the secant line between x = a and x = b
on the graph of f.
This is the line
that passes
through (a, f(a))
and (b, f(b)).
Average Rate of Change
The graphs show that, if a function is:
Increasing on an interval, then the average rate
of change between any two points is positive.
Decreasing on an interval, then the average rate
of change between any two points is negative.
Linear Functions Have
Constant Rate of Change
Linear Functions
For a linear function f(x) = mx + b, the average
rate of change between any two points is the
same constant m.
The slope of a line is the average rate of change
of y with respect to x.
Linear Functions Have Constant Rate of
Change
Let f(x) = 3x – 5.
Find the average rate of change of f
between the following points.
x = 0 and x = 1
x = 3 and x = 7
x = a and x = a + h
What conclusion can you draw from your answers?
Linear Functions
Average rate of change
f 1  f  0 

1 0
3 1 5  3  0  5

1
 2    5 

1
3
Linear Functions
Average rate of change
f 7  f 3

73
3  7  5  3  3  5

4
16  4

4
3
Linear Functions
Average rate of change
f a  h   f a 

a  h   a
3  a  h   5  3a  5 


h
3a  3h  5  3a  5

h
3h

3
h
Linear Functions Have Constant Rate of
Change
It appears that the average rate of change
is always 3 for this function.
In fact, the rate of change between any two arbitrary
points x = a and x = a + h is 3.
Competition Problems
Find the average rate of change for the
function f(x)=x²+9x from
x1 =3 to x2 =7
a) 16
b)19
c)18
d)-19
Find the average rate of change for the function
f(x)=x²+9x from
x1 =3 to x2 =7
f(7)=7²+9·7 = 112
f(3)=3²+9·3 = 36
f(7) - f(3)
7–3
112 - 36
7-3
Answer: b)19
Solve for f(g(x²)),if
f(x)=2x+5 and g(x)=3x²-1
f(x)=2x+5 and g(x)=3x²-1
Solve for f(g(x²))
g(x²) = 3(x²)²-1
= 3x4 – 1
f(3x4 – 1)=2(3x4 – 1) +5
=6x4 – 2 + 5
4
6x +
3
G(t)= t3 – t2 -2t
H(t)= t - 1
3
Find G(H(3))
find G(H(3))
H(3)= 3 – 1 =2/3
3
G(2/3)= (2/3)3 – (2/3)2 -2(2/3)
= 8/27 – 4/9 – 4/3
= 8/27 – 12/27 – 36/27
-40/27
If f(x) = (x + 1) and g(x) = (x2 - 2)2 ,
Find g( f(3) )
Find g( f(3) )
f(3) = (3 + 1) = 4
g(4) = (42 - 2)2 = (14)2 = 196
196
If a set has 128 subsets, how many
elements does it have?
n
2
= 128
n
2
7
2
=
n=7