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,\(\left( {2, - 3} \right)\). 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 pair 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.

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

\[{1^{{\mbox{st}}}}{\mbox{ components : }}\left\{ {6, - 7,0} \right\}\hspace{0.25in}\hspace{0.25in}{2^{{\mbox{nd}}}}{\mbox{ components : }}\left\{ {10,3,4, - 4} \right\}\]

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 : \(\left( {6,10} \right)\) and \(\left( {6, - 4} \right)\). 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!

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.

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 values 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.

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 \(x\).

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 \(y\).

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 y² 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.

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 \(f\left( 3 \right)\) we will use the function \(f\left( x \right)\) and for \(g\left( 3 \right)\) we will use \(g\left( x \right)\). In other words, we just need to make sure that the variables match up.

Here are the evaluations for this part.

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.

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.

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.

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 \(f\left( {t + 1} \right)\) first. This one works exactly the same as the previous part did. All the \(x\)’s on the left will get replaced with \(t + 1\). 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 \(f\left( {x + 1} \right)\). With the exception of the \(x\) this is identical to \(f\left( {t + 1} \right)\) 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.

Again, don’t get excited about the \(x\)’s in the parenthesis here. Just evaluate it as if it were a number.

One more evaluation and this time we’ll use the other function.

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.

In this case -6 satisfies the top inequality and so we’ll use the top equation for this evaluation.

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.

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,

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.

For the final evaluation in this example the number satisfies the bottom inequality and so we’ll use the bottom equation for the evaluation.

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.

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 \(x = - 5\) or \(x = 2\). 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,

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,

In this case we’ve got a fraction, but notice that the denominator will never be zero for any real number since x² 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

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 \(x \ge \frac{1}{2}\) 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 \(x = - 4\) 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, \(x = 4\) does satisfy the inequality. This means that it is okay to plug \(x = 4\) 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,