As we mentioned in Section GraphsofFunctions, in this chapter, we are studying functions in a more abstract and general setting. In this section, we begin our study of what can be considered as the algebra of functions by defining function arithmetic.

Given two real numbers, we have four primary arithmetic operations available to us: addition, subtraction, multiplication, and division (provided we don’t divide by \(0\).) Since the functions we study in this text have ranges which are sets of real numbers, it makes sense we can extend these arithmetic notions to functions.

For example, to add two functions means we add their outputs; to subtract two functions, we subtract their outputs, and so on and so forth. More formally, given two functions \(f\) and \(g\), we define a new function \(f+g\) whose rule is determined by adding the outputs of \(f\) and \(g\). That is \((f+g)(x) = f(x) + g(x)\). While this looks suspiciously like some kind of distributive property, it is nothing of the sort. The ‘\(+\)’ sign in the expression ‘\(f+g\)’ is part of the name of the function we are defining, whereas the plus sign ‘\(+\)’ sign in the expression \(f(x) + g(x)\) represents real number addition: we are adding the output from \(f\), \(f(x)\) with the output from \(g\), \(g(x)\) to determine the output from the sum function, \((f+g)(x)\).

Of course, in order to define \((f+g)(x)\) by the formula \((f+g)(x) = f(x) + g(x)\), both \(f(x)\) and \(g(x)\) need to be defined in the first place; that is, \(x\) must be in the domain of \(f\) and the domain of \(g\). You’ll recall this means \(x\) must be in the intersection of the domains of \(f\) and \(g\). We define the following.

We put these definitions to work for us in the next example.

A few remarks are in order. First, in number funcarithfindvaluesex parts threefunctionsfirstex through threefunctionslastex, we first encountered combinations of three functions despite Definition functionarithmeticdefn only addressing combinations of two functions at a time. It turns out that function arithmetic inherits many of the same properties of real number arithmetic. For example, we showed above that \(((s+g)+h)(3) = (s+(g+h))(3)\). In general, given any three functions \(f\), \(g\), and \(h\), \((f+g)+h = f +(g+h)\) that is, function addition is assocative. To see this, choose an element \(x\) common to the domains of \(f\), \(g\), and \(h\). Then

\[ \begin{array}{rclr} ((f+g)+h)(x) & = & (f+g)(x)+h(x) & \text {definiton of $((f+g)+h)(x)$} \\ & = & (f(x)+g(x))+h(x) & \text {definition of $(f+g)(x)$} \\ & = & f(x) + (g(x)+h(x)) & \text {associative property of real number addition} \\ & = & f(x) + (g+h)(x) & \text {definition of $(g+h)(x)$} \\ & = & (f+(g+h))(x) & \text {definition of $(f+(g+h))(x)$} \\ \end{array} \]

The key step to the argument is that \((f(x)+g(x))+h(x) = f(x) + (g(x)+h(x))\) which is true courtesy of the associative property of real number addition. And just like with real number addition, because function addition is associative, we may write \(f+g+h\) instead of \((f+g)+h\) or \(f+(g+h)\) even though, when it comes down to computations, we can only add two things together at a time.

For completeness, we summarize the properties of function arithmetic in the theorem below. The proofs of the properties all follow along the same lines as the proof of the associative property and are left to the reader. We investigate some additional properties in the exercises.

In the next example, we decompose given functions into sums, differences, products and/or quotients of other functions. Note that there are infinitely many different ways to do this, including some trivial ones. For example, suppose we were instructed to decompose \(f(x) = x+2\) into a sum or difference of functions. We could write \(f = g+h\) where \(g(x) = x\) and \(h(x) = 2\) or we could choose \(g(x) = 2x+3\) and \(h(x) = -x-1\). More simply, we could write \(f = g+h\) where \(g(x) = x+2\) and \(h(x) = 0\). We’ll call this last decomposition a ‘trivial’ decomposition. Likewise, if we ask for a decomposition of \(f(x) = 2x\) as a product, a nontrivial solution would be \(f = gh\) where \(g(x) = 2\) and \(h(x) = x\) whereas a trivial solution would be \(g(x) = 2x\) and \(h(x) = 1\). In general, non-trivial solutions to decomposition problems avoid using the additive identity, \(0\), for sums and differences and the multiplicative identity, \(1\), for products and quotients.

1 The Arithmetic of Change

Recall the average rate of change of a function over the interval \([a,b]\) is the slope of the line connecting the two points \((a, f(a))\) and \((b, f(b))\) and is given by

\[ \frac {\Delta [f(x)]}{\Delta x} = \frac {f(b)-f(a)}{b-a}.\]

For the purposes of this section, consider a function \(f\) defined over an interval containing \(x\) and \(x+\Delta x\) where \(\Delta x \neq 0\). The average rate of change of \(f\) over the interval \([x,x+\Delta x]\) is thus given by the formula:

\[ \frac {\Delta [f(x)]}{\Delta x} = \frac {f(x+\Delta x)-f(x)}{\Delta x}, \quad \Delta x \neq 0.\]

Our aim in this section is to develop formulas which relate the rate of change of arithmetic combinations of functions to the rates of change of the component functions. Our first step is to study the difference operator ‘\(\Delta \)’ and how it works with the standard arithmetic operations.

In general, if \(u\) is some quantity which assumes two values in a particular order, say \(u_{1}\) (the ‘first’ or ‘initial’ value) and \(u_{2}\) (the ‘second’ or ‘final’ value), then \(\Delta u = u_{2} - u_{1}\). For example, if \(u\) represents the temperature of an object before (\(u_{1}\)) heat is applied and after (\(u_{2}\)) heat is applied, \(\Delta u = u_{2} - u_{1}\) represents the increase in temperature of the object.

In the context of functions and rates of change, \(u\) is the function \(f\) defined on the interval \([x, x + \Delta x]\) with \(u_{1} = f(x)\) and \(u_{2} = f(x+\Delta x)\). Here, \(\Delta u = u_{2} - u_{1} = f(x+\Delta x) - f(x) = \Delta [f(x)]\).

Suppose we have two quantities, \(u\) and \(v\) with \(\Delta u = u_{2} - u_{1}\) and \(\Delta v = v_{2} - v_{1}\). What do we mean by \(\Delta [u+v]\)? The initial value of the sum \(u+v\) would be the sum of the initial values \(u_{1} + v_{1}\). Likewise, the final value of the sum would be the sum of the final values \(u_{2} + v_{2}\). Hence:

\[ \begin{array}{rcl} \Delta [u+v] & = & (u_{2} + v_{2}) - (u_{1} + v_{1}) \\ & = & u_{2} + v_{2} - u_{1} - v_{1} \\ & = & (u_{2} - u_{1}) + (v_{2} - v_{1}) \\ & = & \Delta u + \Delta v \\ \end{array} \]

A similar calculation gives \(\Delta [u - v] = \Delta u - \Delta v\).

Let’s turn our attention to products. We have \(\Delta [uv] = u_{2} \, v_{2} - u_{1} \, v_{1}\). We’d like to express \(\Delta [uv]\) in terms of \(\Delta u\) and \(\Delta v\) and there seems to be no way obvious way to do that. We take to some geometric reasoning for inspiration. Let’s assume the all the quantities we’re working with are positive.

We imagine the product \(u_{1} \, v_{1}\) as being the area of a rectangle with width \(u_{1}\) and length \(v_{1}\). Likewise, the product \(u_{2} \, v_{2}\) is the area of a (larger) rectangle with width \(u_{2}\) and length \(v_{2}\).

From \(\Delta u = u_{2} - u_{1}\), we get \(u_{2} = u_{1} + \Delta u\) and, likewise, \(v_{2} = v_{1} + \Delta v\). Doing so allows us to decompose the larger rectangle into four smaller rectangles.

Using this schematic, we see the area \(u_{2} \, v_{2}\) is the sum of the areas of four smaller rectangles:

\[ u_{2} \, v_{2} = u_{1} \, v_{1} + v_{1} \, \Delta u + u_{1} \, \Delta v + \Delta u \, \Delta v \]

Hence, \(\Delta [uv] = u_{2} \, v_{2} - u_{1} \, v_{1} = v_{1} \, \Delta u + u_{1} \, \Delta v + \Delta u \, \Delta v\).

To prove this formula holds in general, we can substitute \(u_{2} = u_{1} + \Delta u\) and \(v_{2} = v_{1} + \Delta v\) into \(\Delta [uv] = u_{2} \, v_{2} - u_{1} \, v_{1}\) and simplify. We leave the details to the reader.

Next, we turn our attention to quotients. We begin with: \(\Delta \left [ \frac {u}{v} \right ] = \frac {u_{2}}{v_{2}} - \frac {u_{1}}{v_{1}}\).

Instead of appealing to geometric reasoning here, we take a cue from the previous discussion and substitute \(u_{2} = u_{1} + \Delta u\) and \(v_{2} = v_{1} + \Delta v\) and set about getting a common denominator:

\[ \begin{array}{rcl} \Delta \left [ \frac {u}{v} \right ]& = & \frac {u_{2}}{v_{2}} - \frac {u_{1}}{v_{1}} \\ & = & \frac { u_{1} + \Delta u}{ v_{1} + \Delta v} - \frac {u_{1}}{v_{1}} \\ & = & \frac {(u_{1} + \Delta u) \, v_{1}}{(v_{1} + \Delta v) \, v_{1}} - \frac {u_{1} \, (v_{1} + \Delta v)}{v_{1} \, (v_{1} + \Delta v) }\\ & = & \frac {u_{1} \, v_{1} + v_{1} \, \Delta u- u_{1} \, v_{1} - u_{1} \, \Delta v}{v_{1} \, (v_{1} + \Delta v)}\\ & = & \frac { v_{1} \, \Delta u - u_{1} \, \Delta v}{v_{1} \, (v_{1} + \Delta v_{1})}\\ \end{array} \]

We summarize these results in the following theorem.

You’ll note we’ve called out a special case for the Product Rule, the Constant Multiple Rule. If one of the factors is a constant, \(c\), then \(\Delta c = 0\) (since constants don’t change.)

In the following example, we use the Quotient Rule for Change to help approximate the propagated error when using measured quantities (with associated uncertainties) in calculations.

In order to establish formulas for the average rate of change for functions, we substitute \(f(x)\) for \(u_{1}\) and \(g(x)\) for \(v_{1}\) and divide each of the expressions in Theorem 2 by \(\Delta x\). For example, if we to find an expression for the average rate of change of \(fg\) in terms of \(f\), \(g\), and their respective average rates of change:

\[ \frac {\Delta [(fg)(x)]}{\Delta x} = \frac { \Delta [f(x)]g(x) + f(x)\Delta [g(x)] + \Delta [f(x)]\Delta [g(x)]}{\Delta x} = \frac {\Delta [f(x)]}{\Delta x} \, g(x) + f(x) \, \frac {\Delta [g(x)]}{\Delta x} + \frac {\Delta [f(x)]\Delta [g(x)]}{\Delta x}.\]

Note that with the last term, we may associate the ‘\(\Delta x\)’ with either of the factors in the numerator:

\[ \frac {\Delta [f(x)]\Delta [g(x)]}{\Delta x} = \frac {\Delta [f(x)]}{\Delta x} \, \Delta [g(x)] = \Delta [f(x)] \, \frac {\Delta [g(x)]}{\Delta x}.\]

Either way, we’ve managed to express the average rate of change of the function \(fg\) in terms of the changes and rates of change of \(f\) and \(g\).

In the result below, we abbreviate the average rate of change as ‘ARoC’ for convenience.

Our final example revisits the scenario in Exercise ?? in Section ??.

Note that in Example 4, since \(\Delta x = 1\), the average rate of change for the cost moving from producing \(70\) to \(71\) systems, \( \frac {C(71) - C(70)}{71-70}\) is the same numerical value as the additional cost incurred by producing the \(71\)st system, \(C(71) - C(70)\). The same goes for the revenue and profit calculations. This is the concept of marginal analysis is studied at length in Economics and Business Calculus classes. For us, it’s time for some Exercises.