A *combination of functions* is the result of performing arithmetic operations on the outputs of two or more functions.

Examples include:

- \(f(x) + g(x)\)
The ouputs of two functions are added together.

- \(\frac{f(x)}{g(x)}\)
The ouputs of two functions are divided.

- \(f(x)^{g(x)}\)
The output of one function is used as an exponent on the output of the other function.

Note that this is different from the composition of functions from chapter 7, where the output of one function was used as the input of another function.

Typical combinations in this course are adding, subtracting, multiplying or dividing functions, though any operation on two or more numbers would correspond to a function combination.

#####
Example8.2.2

Let \(w\) and \(h\) be two different functions, and suppose \(w(4) = 9\) and \(h(4) = 2\text{.}\) We can create a new function \(S(x) = w(x) + h(x)\text{,}\) which is just the sum of the functions \(w(x)\) and \(h(x)\text{.}\)

This new function \(S\) can be evaluated, provided that we can evaluate each of the functions \(w\) and \(h\text{.}\)

For instance,
\begin{align*}
S(4) \amp= w(4) + h(4)\\
\amp= 9 + 2\\
\amp= 11
\end{align*}

#####
Exercise8.2.3

An example of function combinations involves *Profit*, *Revenue* and *Cost*.

#####
Example8.2.4

It costs money to manufacture a product, and you can make money by selling it. The *profit* is the difference between how much money you collect from selling the product (called the *revenue*), and how much it costs you to manufacture it.

Let \(C(u)\) be the cost, in thousands of dollars, of manufacturing \(u\) tons of a product. Let \(R(u)\) be the revenue, in thousands of dollars, collected through selling \(u\) tons of the product. Finally, \(P(u)\) is the profit, in thousands of dollars, from making and selling \(u\) tons of the product.

Using function notation, we would write:
\begin{equation*}
P(u) = R(u) - C(u)
\end{equation*}

#####
Exercise8.2.5

#####
Example8.2.6

As a business owner, you know the value in giving your customers a good deal, especially when they buy a lot from you.

This month at your pet supply store, you decide to run the following special on your most popular cat food:

The retail price for one container of the food is $\(24\text{,}\) but if a customer orders two of them, you will reduce the unit price by $\(0{.}25\text{,}\) so they would each cost only $\(23{.}75\text{.}\)

If a customer orders three containers, the unit cost would be reduced another $\(0{.}25\text{,}\) so they would each cost only $\(23{.}50\text{.}\)

Thus, as the number of containers increases, the cost of each one will decrease.

Write a formula for the unit cost, \(C(x)\text{,}\) of each container when the customer orders \(x\) containers. Note that this is a *linear* function of \(x\text{.}\)

If a customer ordered \(6\) containers, we could evaluate \(C(6)\) to find how much they would be charged for each container. How much would they pay in total for all six containers?

If a customer ordered \(x\) containers, how much would they pay in total for all the containers? Let \(T(x)\) represent the total they would pay for \(x\) containers.

Solution 1The Cost function is linear, beginning at \(24\) when \(x = 1\text{,}\) and with a slope of \(-0.25\text{.}\) So, the function can be written as:
\begin{align*}
C(x) \amp= 24 - 0.25(x-1)\\
\amp= 24.25 - 0.25x
\end{align*}

Solution 2If a customer ordered \(6\) containers, the unit cost would be:
\begin{align*}
C(6) \amp= 24.25 - 0.25(6)\\
\amp= 22.75
\end{align*}

You would charge $\(22{.}75\) for each container, so the customer would pay \(6(22.75) = \) $\(136{.}50\) in total.

Solution 3In general, if a customer ordered \(x\) containers, then they would have to pay the unit cost (given by the function \(C(x)\)) multiplied by the number of containers they ordered (which is just \(x\)).

Therefore, the customer's total cost for ordering \(x\) containers is a new function:
\begin{align*}
T(x) \amp= x \cdot C(x)\\
\amp= x(24.25 - 0.25x)\\
\amp= 24.25x - 0.25x^2
\end{align*}

#
Subsection8.2.1Graphing Combinations

Because there are an infinite number of ways to combine two functions, there is no straightforward way to describe all possible graphs we may see as a result. However, one thing *is* true for any function combination we may create: *We are just doing arithmetic on the outputs of the original functions.*

This means that for a combination such as
\begin{equation*}
W(x) = \left(g(x)+h(x)\right)^3 + 4g(x)
\end{equation*}
in order to evaluate a particular value such as \(W(12)\text{,}\) the number \(12\) must at least be in the domains of both functions \(f\) and \(g\text{.}\)

If \(f(12) = 2\) and \(g(12) = 3\text{,}\) then \(W(x)\) would be:
\begin{equation*}
(3+2)^3 + 4(3) = 137
\end{equation*}

In the next exercise, you will see four different combinations of the functions \(f(x) = x-1\) and \(g(x) = 2-x\text{.}\)

#####
Exercise8.2.7

#
Subsection8.2.2Combinations vs. Composition

It may have occurred to you that function combinations look very much like composite functions. For example, the expression
\begin{equation*}
\left(f(x)\right)^2 + 3f(x)
\end{equation*}
is a combination of just the function \(f(x)\text{.}\) However, it may also be thought of as the *composition* \(g(f(x))\text{,}\) where \(g(x) = x^2 + 3x\text{.}\)

In general, this is true. A composite function is really just a “combination of itself”.

Now you would be justified to ask: If combinations and compositions are so similar, why do we have two different names for them? The answer is that, while similar, they represent different things we may do with function outputs, especially in a real context. Here is an example:

#####
Example8.2.8

As the temperature drops, so does the speed of sound. The function
\begin{equation*}
v = f(T) = 740.4 + 1.34T
\end{equation*}
models the speed of sound (in miles-per-hour) when the temperature is \(T\) degrees Celsius.

On a particular day between \(6{:}00\) P.M. and midnight, suppose the temperature outside is modeled by the function
\begin{equation*}
T = g(h) = 24 - 3h
\end{equation*}
where \(h\) is the number of hours after \(6{:}00\) P.M.

Now, we certainly could create both combinations and compositions using the functions \(f\) and \(g\text{,}\) but what would actually makes sense?

Since we know that \(g\) changes *hours* into *degrees Celsius*, and \(f\) changes *degrees Celsius* into *miles-per-hour*, it seems natural to use the composition
\begin{equation*}
f(g(h)) = 740.4+1.34(24-3h)
\end{equation*}
to turn *hours* into *miles-per-hour*.

It would not be sensible to make a combination of the functions \(f\) and \(g\text{,}\) because they do not use the same input variable. We would have to use the same input value (with the same units) into both functions, but this cannot be.

However, outside of any context, the functions \(f(x) = 740.4 + 1.34x\) and \(g(x) = 24 - 3x\) could be combined in any variety of ways. These combinations, however, would have no applicable meaning.

#
Subsection8.2.3Conclusion

Now you have seen that function combinations are just different ways of doing arithmetic with the outputs of different functions, provided the same input may be used for each function. Any arithmetic expression you could write with numbers could be done with function outputs.

The domain of a function combination is the set of numbers that are in both of the domains of the original functions, except for those inputs which cause the combination to be undefined (division by zero, negative number under a square root, etc.).

Even beginning with simple functions, different combinations may produce very complicated graphs. In general, though, the graph of a combination is found by combining the outputs (y-values) of the functions involved.

Composition is really just some combination of a single function.

Formulas for Combinations

##### 6

##### 7

##### 8

##### 9

##### 10

##### 11

##### 12

##### 13

##### 14

##### 15

##### 16