We now need to move into the second topic of this
chapter. Before we do that however we
need a quick definition taken care of.
Definition of Relation
A relation is a
set of ordered pairs.

This seems like an odd definition but we’ll need it for the
definition of a function (which is the main topic of this section). However, before we actually give the
definition of a function let’s see if we can get a handle on just what a
relation is.
Think back to Example
1 in the Graphing section of this chapter.
In that example we constructed a set of ordered pairs we used to sketch
the graph of . Here are the ordered pairs that we used.
Any of the following are then relations because they consist
of a set of ordered pairs.
There are of course many more relations that we could form
from the list of ordered pairs above, but we just wanted to list a few possible
relations to give some examples. Note as
well that we could also get other ordered pairs from the equation and add those
into any of the relations above if we wanted to.
Now, at this point you are probably asking just why we care
about relations and that is a good question.
Some relations are very special and are used at almost all levels of
mathematics. The following definition
tells us just which relations are these special relations.
Definition of a
Function
A function is a
relation for which each value from the set the first components of the
ordered pairs is associated with exactly one value from the set of second
components of the ordered pair.

Okay, that is a mouth full.
Let’s see if we can figure out just what it means. Let’s take a look at the following example
that will hopefully help us figure all this out.
Example 1 The
following relation is a function.
Solution
From these ordered pairs we have the following sets of
first components (i.e. the first
number from each ordered pair) and second components (i.e. the second number from each ordered pair).
For the set of second components notice that the “3”
occurred in two ordered pairs but we only listed it once.
To see why this relation is a function simply pick any
value from the set of first components.
Now, go back up to the relation and find every ordered pair in which
this number is the first component and list all the second components from
those ordered pairs. The list of
second components will consist of exactly one value.
For example let’s choose 2 from the set of first
components. From the relation we see
that there is exactly one ordered pair with 2 as a first component, . Therefore the list of second components (i.e. the list of values from the set
of second components) associated with 2 is exactly one number, 3.
Note that we don’t care that 3 is the second component of
a second ordered par in the relation.
That is perfectly acceptable.
We just don’t want there to be any more than one ordered pair with 2
as a first component.
We looked at a single value from the set of first
components for our quick example here but the result will be the same for all
the other choices. Regardless of the
choice of first components there will be exactly one second component
associated with it.
Therefore this relation is a function.

In order to really get a feel for what the definition of a
function is telling us we should probably also check out an example of a
relation that is not a function.
Example 2 The
following relation is not a function.
Solution
Don’t worry about where this relation came from. It is just one that we made up for this example.
Here is the list of first and second components
From the set of first components let’s choose 6. Now, if we go up to the relation we see
that there are two ordered pairs with 6 as a first component : and . The list of second components associated
with 6 is then : 10, 4.
The list of second components associated with 6 has two
values and so this relation is not a function.
Note that the fact that if we’d chosen 7 or 0 from the
set of first components there is only one number in the list of second
components associated with each. This
doesn’t matter. The fact that we found
even a single value in the set of first components with more than one second
component associated with it is enough to say that this relation is not a
function.
As a final comment about this example let’s note that if
we removed the first and/or the fourth ordered pair from the relation we
would have a function!

So, hopefully you have at least a feeling for what the
definition of a function is telling us.
Now that we’ve forced you to go through the actual
definition of a function let’s give another “working” definition of a function
that will be much more useful to what we are doing here.
The actual definition works on a relation. However, as we saw with the four relations we
gave prior to the definition of a function and the relation we used in Example
1 we often get the relations from some equation.
It is important to note that not all relations come from
equations! The relation from the second
example for instance was just a set of ordered pairs we wrote down for the
example and didn’t come from any equation.
This can also be true with relations that are functions. They do not have to come from equations.
However, having said that, the functions that we are going
to be using in this course do all come from equations. Therefore, let’s write down a definition of a
function that acknowledges this fact.
Before we give the “working” definition of a function we
need to point out that this is NOT the actual definition of a function, that is
given above. This is simply a good
“working definition” of a function that ties things to the kinds of functions that
we will be working with in this course.
“Working Definition”
of Function
A function is
an equation for which any x that
can be plugged into the equation will yield exactly one y out of the equation.

There it is. That is
the definition of functions that we’re going to use and will probably be easier
to decipher just what it means.
Before we examine this a little more note that we used the
phrase “x that can be plugged into”
in the definition. This tends to imply
that not all x’s can be plugged into
an equation and this is in fact correct.
We will come back and discuss this in more detail towards the end of
this section, however at this point just remember that we can’t divide by zero
and if we want real numbers out of the equation we can’t take the square root
of a negative number. So, with these two
examples it is clear that we will not always be able to plug in every x into any equation.
Further, when dealing with functions we are always going to
assume that both x and y will be real numbers. In other words, we are going to forget that
we know anything about complex numbers for a little bit while we deal with this
section.
Okay, with that out of the way let’s get back to the
definition of a function and let’s look at some examples of equations that are
functions and equations that aren’t functions.
Example 3 Determine
which of the following equations are functions and which are not functions.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
Solution
The “working” definition of function is saying is that if
we take all possible values of x
and plug them into the equation and solve for y we will get exactly one value for each value of x.
At this stage of the game it can be pretty difficult to actually show
that an equation is a function so we’ll mostly talk our way through it. On the other hand it’s often quite easy to
show that an equation isn’t a function.
(a)
So, we need to show that no matter what x we plug into the equation and solve
for y we will only get a single
value of y. Note as well that the value of y will probably be different for each
value of x, although it doesn’t
have to be.
Let’s start this off by plugging in some values of x and see what happens.
So, for each of these value of x we got a single value of y
out of the equation. Now, this isn’t
sufficient to claim that this is a function.
In order to officially prove that this is a function we need to show
that this will work no matter which value of x we plug into the equation.
Of course we can’t plug all possible value of x into the equation. That just isn’t physically possible. However, let’s go back and look at the ones
that we did plug in. For each x, upon plugging in, we first
multiplied the x by 5 and then
added 1 onto it. Now, if we multiply a
number by 5 we will get a single value from the multiplication. Likewise, we will only get a single value
if we add 1 onto a number. Therefore,
it seems plausible that based on the operations involved with plugging x into the equation that we will only
get a single value of y out of the
equation.
So, this equation is a function.
[Return to Problems]
(b)
Again, let’s plug in a couple of values of x and solve for y to see what happens.
Now, let’s think a little bit about what we were doing
with the evaluations. First we squared
the value of x that we plugged
in. When we square a number there will
only be one possible value. We then
add 1 onto this, but again, this will yield a single value.
So, it seems like this equation is also a function.
Note that it is okay to get the same y value for different x’s. For example,
We just can’t get more than one y out of the equation after we plug in the x.
[Return to Problems]
(c)
As we’ve done with the previous two equations let’s plug
in a couple of value of x, solve
for y and see what we get.
Now, remember that we’re solving for y and so that means that in the first and last case above we will
actually get two different y values
out of the x and so this equation
is NOT a function.
Note that we can have values of x that will yield a single y
as we’ve seen above, but that doesn’t matter.
If even one value of x
yields more than one value of y
upon solving the equation will not be a function.
What this really means is that we didn’t need to go any
farther than the first evaluation, since that gave multiple values of y.
[Return to Problems]
(d)
With this case we’ll use the lesson learned in the
previous part and see if we can find a value of x that will give more than one value of y upon solving. Because
we’ve got a y^{2} in the
problem this shouldn’t be too hard to do since solving will eventually mean
using the square root property which
will give more than one value of y.
So, this equation is not a function. Recall, that from the previous section this
is the equation of a circle. Circles
are never functions.
[Return to Problems]

Hopefully these examples have given you a better feel for
what a function actually is.
We now need to move onto something called function notation. Function notation will be used heavily
throughout most of the remaining chapters in this course and so it is important
to understand it.
Let’s start off with the following quadratic equation.
We can use a process similar to what we used in the previous
set of examples to convince ourselves that this is a function. Since this is a function we will denote it as
follows,
So, we replaced the y
with the notation . This is read as “f of x”. Note that there is
nothing special about the f we used
here. We could just have easily used
any of the following,
The letter we use does not matter. What is important is the “ ” part.
The letter in the parenthesis must match the variable used on the right
side of the equal sign.
It is very important to note that is really nothing more than a really fancy way
of writing y. If you keep that in mind you may find that
dealing with function notation becomes a little easier.
Also, this is NOT
a multiplication of f by x! This is one of the more common mistakes people
make when they first deal with functions.
This is just a notation used to denote functions.
Next we need to talk about evaluating functions.
Evaluating a function is really nothing more than asking what its value
is for specific values of x. Another way of looking at it is that we are
asking what the y value for a given x is.
Evaluation is really quite simple. Let’s take the function we were looking at
above
and ask what its value is for . In terms of function notation we will “ask”
this using the notation . So, when there is something other than the
variable inside the parenthesis we are really asking what the value of the
function is for that particular quantity.
Now, when we say the value of the function we are really
asking what the value of the equation is for that particular value of x.
Here is .
Notice that evaluating a function is done in exactly the
same way in which we evaluate equations.
All we do is plug in for x
whatever is on the inside of the parenthesis on the left. Here’s another evaluation for this function.
So, again, whatever is on the inside of the parenthesis on
the left is plugged in for x in the
equation on the right. Let’s take a look
at some more examples.
Example 4 Given
and evaluate each of the following.
(a) and [Solution]
(b) and [Solution]
(c) [Solution]
(d) [Solution]
(e) and [Solution]
(f) [Solution]
(g) [Solution]
Solution
(a) and
Okay we’ve got two function evaluations to do here and
we’ve also got two functions so we’re going to need to decide which function
to use for the evaluations. The key
here is to notice the letter that is in front of the parenthesis. For we will use the function and for we will use . In other words, we just need to make sure
that the variables match up.
Here are the evaluations for this part.
[Return to Problems]
(b) and
This one is pretty much the same as the previous part with
one exception that we’ll touch on when we reach that point. Here are the evaluations.
Make sure that you deal with the negative signs properly
here. Now the second one.
We’ve now reached the difference. Recall that when we first started talking
about the definition of functions we stated that we were only going to deal
with real numbers. In other words, we
only plug in real numbers and we only want real numbers back out as answers. So, since we would get a complex number out
of this we can’t plug 10 into this function.
[Return to Problems]
(c)
Not much to this one.
Again, don’t forget that this isn’t multiplication! For some reason students like to think of this
one as multiplication and get an answer of zero. Be careful.
[Return to Problems]
(d)
The rest of these evaluations are now going to be a little
different. As this one shows we don’t
need to just have numbers in the parenthesis.
However, evaluation works in exactly the same way. We plug into the x’s on the right side of the equal sign whatever is in the
parenthesis. In this case that means
that we plug in t for all the x’s.
Here is this evaluation.
Note that in this case this is pretty much the same thing
as our original function, except this time we’re using t as a variable.
[Return to Problems]
(e) and
Now, let’s get a
little more complicated, or at least they appear to be more complicated. Things aren’t as bad as they may appear
however. We’ll evaluate first.
This one works exactly the same as the previous part did. All the x’s
on the left will get replaced with . We will have some simplification to do as
well after the substitution.
Be careful with parenthesis in these kinds of
evaluations. It is easy to mess up
with them.
Now, let’s take a look at . With the exception of the x this is identical to and so it works exactly the same way.
Do not get excited about the fact that we reused x’s in the evaluation here. In many places where we will be doing this
in later sections there will be x’s
here and so you will need to get used to seeing that.
[Return to Problems]
(f)
Again, don’t get excited about the x’s in the parenthesis here.
Just evaluate it as if it were a number.
[Return to Problems]
(g)
One more evaluation and this time we’ll use the other
function.
[Return to Problems]

Function evaluation is something that we’ll be doing a lot
of in later sections and chapters so make sure that you can do it. You will find several later sections very
difficult to understand and/or do the work in if you do not have a good grasp
on how function evaluation works.
While we are on the subject of function evaluation we should
now talk about piecewise functions. We’ve actually already seen an example of a
piecewise function even if we didn’t call it a function (or a piecewise
function) at the time. Recall the
mathematical definition of absolute value.
This is a function and if we use function notation we can
write it as follows,
This is also an example of a piecewise function. A piecewise function is nothing more than a
function that is broken into pieces and which piece you use depends upon value
of x.
So, in the absolute value example we will use the top piece if x is positive or zero and we will use
the bottom piece if x is negative.
Let’s take a look at evaluating a more complicated piecewise
function.
Example 5 Given,
evaluate each of the following.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
(e) [Solution]
Solution
Before starting the evaluations here let’s notice that
we’re using different letters for the function and variable than the ones
that we’ve used to this point. That
won’t change how the evaluation works.
Do not get so locked into seeing f
for the function and x for the
variable that you can’t do any problem that doesn’t have those letters.
Now, to do each of these evaluations the first thing that
we need to do is determine which inequality the number satisfies, and it will
only satisfy a single inequality. When
we determine which inequality the number satisfies we use the equation
associated with that inequality.
So, let’s do some evaluations.
(a)
In this case 6 satisfies the top inequality and so we’ll
use the top equation for this evaluation.
[Return to Problems]
(b)
Now we’ll need to be a little careful with this one since
4 shows up in two of the inequalities.
However, it only satisfies the top inequality and so we will once
again use the top function for the evaluation.
[Return to Problems]
(c)
In this case the number, 1, satisfies the middle
inequality and so we’ll use the middle equation for the evaluation. This evaluation often causes problems for
students despite the fact that it’s actually one of the easiest evaluations
we’ll ever do. We know that we
evaluate functions/equations by plugging in the number for the variable. In this case there are no variables. That isn’t a problem. Since there aren’t any variables it just
means that we don’t actually plug in anything and we get the following,
[Return to Problems]
(d)
Again, like with the second part we need to be a little
careful with this one. In this case
the number satisfies the middle inequality since that is the one with the
equal sign in it. Then like the
previous part we just get,
Don’t get excited about the fact that the previous two
evaluations were the same value. This
will happen on occasion.
[Return to Problems]
(e)
For the final evaluation in this example the number
satisfies the bottom inequality and so we’ll use the bottom equation for the
evaluation.
[Return to Problems]

Piecewise functions do not arise all that often in an
Algebra class however, the do arise in several places in later classes and so
it is important for you to understand them if you are going to be moving on to
more math classes.
As a final topic we need to come back and touch on the fact
that we can’t always plug every x
into every function. We talked briefly
about this when we gave the definition of the function and we saw an example of
this when we were evaluating functions.
We now need to look at this in a little more detail.
First we need to get a couple of definitions out of the
way.
Domain and Range
The domain of
an equation is the set of all x’s
that we can plug into the equation and get back a real number for y.
The range of an equation is
the set of all y’s that we can ever
get out of the equation.

Note that we did mean to use equation in the definitions
above instead of functions. These are
really definitions for equations.
However, since functions are also equations we can use the definitions
for functions as well.
Determining the range of an equation/function can be pretty
difficult to do for many functions and so we aren’t going to really get into
that. We are much more interested here
in determining the domains of functions.
From the definition the domain is the set of all x’s that we can plug into a function and get back a real
number. At this point, that means that
we need to avoid division by zero and taking square roots of negative numbers.
Let’s do a couple of quick examples of finding domains.
Example 6 Determine
the domain of each of the following functions.
(a) [Solution]
(b) [Solution]
(c) [Solution]
(d) [Solution]
Solution
The domains for these functions are all the values of x for which we don’t have division by
zero or the square root of a negative number.
If we remember these two ideas finding the domains will be pretty
easy.
(a)
So, in this case there are no square roots so we don’t
need to worry about the square root of a negative number. There is however a possibility that we’ll
have a division by zero error. To determine
if we will we’ll need to set the denominator equal to zero and solve.
So, we will get division by zero if we plug in or . That means that we’ll need to avoid those
two numbers. However, all the other
values of x will work since they
don’t give division by zero. The
domain is then,
[Return to Problems]
(b)
In this case we won’t have division by zero problems since
we don’t have any fractions. We do
have a square root in the problem and so we’ll need to worry about taking the
square root of a negative numbers.
This one is going to work a little differently from the
previous part. In that part we
determined the value(s) of x to
avoid. In this case it will be just as
easy to directly get the domain. To
avoid square roots of negative numbers all that we need to do is require that
This is a fairly simple linear inequality that we should
be able to solve at this point.
The domain of this function is then,
[Return to Problems]
(c)
In this case we’ve got a fraction, but notice that the
denominator will never be zero for any real number since x^{2} is guaranteed to be positive or zero and adding 4
onto this will mean that the denominator is always at least 4. In other words, the denominator won’t ever
be zero. So, all we need to do then is
worry about the square root in the numerator.
To do this we’ll require,
Now, we can actually plug in any value of x into the denominator, however, since
we’ve got the square root in the numerator we’ll have to make sure that all x’s satisfy the inequality above to
avoid problems. Therefore, the domain
of this function is
[Return to Problems]
(d)
In this final part we’ve got both a square root and
division by zero to worry about. Let’s
take care of the square root first since this will probably put the largest
restriction on the values of x. So, to keep the square root happy (i.e. no square root of negative
numbers) we’ll need to require that,
So, at the least we’ll need to require that in order to avoid problems with the square
root.
Now, let’s see if we have any division by zero
problems. Again, to do this simply set
the denominator equal to zero and solve.
Now, notice that doesn’t satisfy the inequality we need for
the square root and so that value of x
has already been excluded by the square root.
On the other hand does satisfy the inequality. This means that it is okay to plug into the square root, however, since it
would give division by zero we will need to avoid it.
The domain for this function is then,
[Return to Problems]
