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>> Good morning, or afternoon,
or evening,
whenever you're watching
this show.
My name is Professor Shannon
Gracie, and today we'll be
covering Introduction
to Functions.
We are using the Blitzer
Introductory
and Intermediate Algebra text
for college students.
And here we go.
We will be learning how
to find the domain and range
of a relation,
and determining whether a
relation is a function,
as well as evaluating a
function today.
So let's warm
up by evaluating this
equation: Y equals the
opposite of X squared,
minus 22 times X,
plus 5 at X equals negative 3.
So I know you guys can do
this one.
Pause the movie.
Give it a shot.
[ Pause ]
Let's see how you did.
At X equals negative 3,
we have Y equals the opposite
of negative 3 squared,
minus 22, times negative 3,
plus 5.
So all we done is,
we've been given this value
of X equal to negative 3,
so any time we saw an X
in a V equation,
we placed a negative 3.
And it's a little tricky here
with all the negative signs
in the problem.
So let's see how you did.
This will yield a result
of Y equals the opposite of 9,
because when you square
that negative 3,
that negative goes away,
but the negative
on the outside does not.
Minus to minus is plus,
so we get plus 66, plus 5,
so at the end of the day,
we get an end result of 62.
[ Pause ]
How did we do on that guy?
Awesome. All right.
So here we go.
We move right
in to learning the definition
of a relation.
Relation, very simple.
Relation is any set
of ordered pairs.
The set of all first
components
of the ordered pairs is called
the domain.
So get used to this word.
You'll see it for the rest
of your math life.
The set of all second
components is called the range
of the relation.
So just a very real world
example for you here.
Find the domain and range
of the relation, right?
So here, right there,
these are all domain values.
And here we have, actually,
let's modify
that a little bit.
These are the elements
of the domain.
The type of vehicle.
And then over here,
a number of wheels.
That's the elements.
Those are the elements
of the range.
So putting
that in a more organized
fashion, we can say domain,
it's a set,
so we put set brackets,
and we've got a car,
a motorcycle, and a boat.
[ Pause ]
The range is the number
of wheels.
So we've got 4, 2, and 0,
and we're done.
[ Pause ]
Whoops.
[ Pause ]
See, that's not so bad.
But you do need
to learn this vocabulary
because it is going to be used
in definitions later
down the road.
Namely, definition
of a function.
Definition
of a function is a
correspondence
from a first set called the?
You got it, domain,
to a second set called the?
Range. Awesome.
Such that each element
in the domain corresponds
to exactly 1 element
in the range.
Example, to illustrate
with a picture and a chart,
if a relation is a function,
so remember the trick is each
of the domain values has
to map to exactly 1
range value.
The range values can be
repeated, but you can't have 2
different domain values map
to the same range value,
and have it be a function.
All right?
So let's take a look
at what we have.
The domain, if we just,
kind of, made a little
list here.
D for domain.
We've got negative 6,
negative 1, 0, 1 and 2,
so I'm just listing the first
element in each order pair,
and then the range is just 1,
and we won't repeat that 1.
Now look what's
happening here.
They each map each
of these domain values map
to this repeated range value.
So here this actually is a
function because each
of these guys
on the domain maps
to exactly one range value.
It's okay that that range
value is repeated.
So it fits the definition.
Another way to look at this is
to make a picture of it.
So if you look over here,
let's label our graph.
You always want
to label your graph.
I'm going to have it have a
regular scale, so each one
of these tick marks has a
value of 1.
And if we graph it
to those ordered pairs,
that's negative 6-1,
negative 1-1, 0-1,
1-1, and 2-1.
Notice that there are,
if you look up and down, okay?
Up and down,
you won't have 2 input values,
2 X values that go
to different Y values.
So that's another way to see,
kind of a picture
of what it would look like.
Now let's put our result in.
So we have ascertained
that this is a function,
right?
So the relation is a function.
The domain, I'll just put D
for domain, is negative 6,
negative 1, 0, 1, and 2.
And the range is the set
of just 1 element 1.
Okay? Okay.
So why don't we do this.
Go ahead and pause the movie,
and make a little chart.
Identify domain and range.
Make the picture of the graph,
and then you try to figure
out if this guy is a function
or not.
So pause the movie
and do your thing.
[ Pause ]
All right.
Let's see how you did.
So, again, if we make a list
of domain and range, we have,
let's go from least
to greatest.
We've got negative 2 is the
smallest domain value.
It's repeated.
And 3 and 4,
and then the range we have
negative 5, 0 is repeated,
so we'll just put it once,
and 3.
So negative 2 corresponds
to 0, and, look,
negative 2 corresponds
to negative 5.
So do you see this violates
the definition for a function.
You still have a relation,
but you have a domain value
that is going
to two different range values.
Therefore,
this is not a function.
But let's continue our mapping
just for fun.
3 maps to 3 and 4 maps to 0.
So it's because of
that negative 2 mapping
to both negative 5 and 0
that we have a relation,
which is not a function.
So let's make our picture.
[ Pause ]
So we had 3-3, negative 2-0,
4-0, and negative 2
negative 5.
Now notice
in the picture, right?
The other one,
if you drew vertical lines,
you would never hit two
ordered pairs.
You'd never hit the
graph twice.
With this one,
if I were to draw
in vertical lines,
do you see I would hit
right here.
I would hit 2 points
if I did a vertical line here.
All right?
So we'll be learning
that shortly.
There's something called the
vertical line test
that we'll learn.
So the relation is not
a function.
[ Pause ]
And we put our results
in set notation.
[ Pause ]
All right.
Ready to go?
Now here's some information.
I mean function
for a little hard sometimes
for people to grasp.
So I thought we would just
talk a bit about them.
So functions are often given
in terms of equations.
All right?
So we were working with a list
of order pairs,
but we often are given
that in terms of equations,
rather than as sets
of ordered pairs.
[ Pause ]
So if you consider this
equation below,
which describes the position
of an object in feet dropped
from a height of 500 feet,
after X second.
This is an equation,
so it has infinitely many
ordered pairs.
Right? That's
when we start connecting all
the dots.
All right?
So the variable X is a
function of the variable Y.
For each value of X,
there is one, and only one,
value of Y. The variable X is
called the independent
variable because it can be
assigned any value
from the domain.
The variable Y is called the
dependent variable
because its value depends
on X.
[ Pause ]
When an equation represents a
function, the function is
often named by a letter,
such as f, g, h, or F,
G or H. Any letter can be used
to name a function.
The domain is the set
of the function's inputs.
What you can put
in to the function.
And the range is the set
of the function's output.
Just like a calculator.
What comes
out after you put a number in,
and perform an operation.
So if we were
to name our function F,
the input is represented by X,
and the output is represented
by F at X.
So the way you read this is
read either F of X
or F at X. I will often say F
at X so that you don't think
that these are
being multiplied.
They are not being multiplied.
It is a evaluating the
function F
at X equals the sum value.
So this equation here,
Y equals negative 16 X
squared, plus 500,
it's an equation notation.
Okay? If we put it
in function notation,
since the Y is isolated,
we could just write it as F
at X equals negative 16 X
squared, plus 500.
So the nice thing
about this notation is
that it's got sort
of a place holder,
so you see exactly what you're
plugging in for X. All right?
So now here I have a wee bit
of a typo here.
Let's evaluate our function
after 1 second.
After 1 second.
So we would be evaluating F
at 1.
That will equal
to negative 16,
times 1 squared, plus 500.
So all we did was wherever
there was an X, we put in a 1.
So F at 1 will be,
well negative 16 times 1 is
negative 16,
so we will get 484.
Well what does that mean?
After 1 second,
the object is 484 feet
above the ground.
So this, you'll see this
if you take further math
and it's really a special case
of a position function.
Position function.
Where they've already inputted
certain things.
So that was our result,
and that is what it means.
[ Pause ]
All right.
So here is a little bit
of practice
evaluating functions.
So we've been given a little
bit of a complicated function,
and we're evaluating this
function at several
different values.
So here we go.
F at 0 will be negative 0
cubed, minus 0 squared,
minus 0, plus 10.
So everywhere that I saw an X,
I put in a 0 on this one.
So when I complete this
operation,
do you see this is going
to be 0 minus 0,
minus 0, plus 10.
So we get F at 0 equals 10.
[ Pause ]
So why don't you go ahead
and pause the movie and try B,
C, and D, and then start her
up and see how you did.
Good luck.
[ Pause ]
Okay. Let's see how you did.
F at 2 will be negative 2
cubed, minus 2 squared,
minus 2, plus 10.
So, again,
wherever there was an X,
this time I put in a 2.
Okay? So F
at 2 equals negative 2 cubed
will be negative 8,
2 squared is 4,
so we're subtracting 4,
and then minus 2, plus 10,
so F at 2 will have an end
result of negative 4.
[ Pause ]
Let's look at negative 2.
So now we're going
to have the opposite
of negative 2 cubed,
minus negative 2 squared,
minus negative 2, plus 10,
and just double check
that you didn't mess
up a minus.
Negative 2 on each of those.
And so we'll get F
at negative 2 is equal to,
minus a minus becomes
positive, that's all
within what's being cubed,
so we're going
to get a positive 8.
Negative 2, times negative 2,
is positive 4,
so we'll be subtracting 4,
minus a minus 2 is plus 4,
and then plus - ooh, my bad,
I was making
up my own math again.
Notice that this will be minus
a minus 2 is plus 2,
and then we'll have our plus
10 hanging out.
And so at the end of the day,
F at negative 2 will be equal
to - 4 plus 2 is 6,
plus 10 is 16.
Now it turns
out that you can
add functions.
So you do the same things
that you've been doing
in these other problems,
right?
But you're going to end
up doing it twice
and summing each
of the results.
So just take it a step
at a time.
You're going to have,
so F at 1 will be the opposite
of 1 cubed, minus 1 squared,
minus 1, plus 10
from our original,
so that there,
I'm going to put brackets
around it, is F at 1.
F at negative 1 will be the
opposite of negative 1 cubed,
minus negative 1 squared,
minus negative 1, plus 10.
So we'll get positive,
actually, negative 1
when we cube that, minus 1,
minus 1, plus 10,
so that's our F at positive 1.
F at negative 1 will end
up with a positive 1
because minus
and minus is plus,
and that's all inside what's
being cubed.
And then we'll get minus 1,
then this negative will go
away after we square it.
Minus and minus 1 is plus 1,
and then we'll get plus 10.
So, sorry,
we'll keep our notation going.
[ Pause ]
So we'll end up with 7,
plus 11, with an end result
of 18.
So just to keep track
of what all we did
on this guy, F at 1 ended
up being all of this,
and when we worked it out,
we got 7.
F at negative 1 ended
up being all of this,
and when we worked it all out,
we got 11.
And then we added them
together to get 18.
All right.
So let's box our -
put our happy little box
around the answer.
Okay. So here we go.
The next example, actually,
this will be Example 4,
will be finding the indicated
function and domain values,
and we'll be using this
information on the table.
So we don't have an equation
for H, but we have 5 sets
of ordered pairs
from evaluating the function
H. So if I have an input
of negative 2,
what is my output?
Good. It's 2
because of this ordered
pair here.
Negative 2 goes in,
2 comes out.
H of 1, what do you think?
Great. It's going to be 1.
That's from over here,
1 goes in, a 1 comes out.
Now this one's a
little tricky.
It's giving you the function
value, the Y value,
and it's asking you what X
values correspond.
So with this guy,
do you see that H of X is 1.
H of X is 1 here and here.
So it's asking you
which X values correspond?
So it's gone backwards.
So when X equals negative 1
and positive 1.
[ Pause ]
All right.
And that wraps it
up for today.
I think that one's a
relatively short lecture.
So have a fabulous day.
Make sure you do some
of the assignments from 8.1.
And I'm sure you'll be hearing
me shortly.
All right.
Bye.