• Theory and Examples

Introduction to Functions

In this chapter, we provide a rudimentary introduction to functions. Our focus will be on the basic meaning of functions, neglecting exact mathematical definitions. Moreover, we will concentrate on linear functions and discuss the handling of this type of functions. This serves to show how functions can be used to describe the relation between economic variables.

1. The notion of functions

A function describes the relation between two or more variables in a compact way. In the following, we will concentrate on the relation between just two variables. These are called x and y. The function describing the relation between x and y is called f.

Mathematically, we can think of function f as an input-output mechanism between two variables. Function f transforms input x into output y. The following figure illustrates this:

\[input\to function\to output\]


\[x\to f\to y\]

This means that input x is inserted in function f, and f then transforms this x into an output y. In short, this relation can be written as follows:


This shorthand description says: y is the output, if x is inserted in f. Thus, the function f depends on x. This dependency is highlighted by putting x into brackets and writing it right after f.

The following exercise shows us that the output y can (usually) be calculated from a specific value x, provided the (arbitrary) function is well-defined.

Exercise 1:

Let the function be defined by \[f(x)=1+\frac{1}{x}\]

a) Determine the value of that function for x=2!

Solution: input x=2 is inserted in function f. \[f(2)=1+\frac{1}{2}=1.5\] Thus, function f transformes x=2 into y=1.5.

b) For which x is the value of the function 3?

Solution: \[f(x)=3\Rightarrow 1+\frac{1}{x}=3\Rightarrow \frac{1}{x}=3-1=2\Rightarrow 1=2\cdot x\Rightarrow x=0.5\]

Thus, the value of the function is 3 for x=0.5.


2. Economic application I

In economic theory, functions map the relation between economic factors and allow us to analyze this relation. For instance, the relation between the number of hours worked and the corresponding wage of an employee can be illustrated as follows. Input x is the number of hours worked and output y is the wage. The "wage function" f assigns the wage (output) to a specific number of hours worked (input).


According to the above example, the input-output mechanism can be interpreted as follows:

\[\begin{align} & x\to f\to y \\ & 0\to f\to 0 \\ & 1\to f\to 20 \\ & 2\to f\to 40 \\ & 3\to f\to 60 \\ & 5\to f\to 100 \\ \end{align}\]

For instance, if the number of hours worked is 5, inserting 5 into the (wage) function yields an output of 100 sFr. According to the table above, wage per hour is 20 sFr. Hence, the following wage function can be derived:

\[y=20\cdot x\]

for x=0, x=1, x=2, x=3 and x=5. This means that f(x) corresponds to 20x.

According to the table, the employee has the possibility to work either 0 hours or 1 hour or 2 hours or 3 hours or 5 hours. Therefore, we call the possible inputs x the domain of the function f, which is abbreviated by \({D_f} =\){0, 1, 2, 3, 5}. The letter D in \({D_f}\) stands for the domain, and f written as a subscript, showing that the domain refers to function f. If the function is called g then the domain is written as \({D_g} =\) {0, 1, 2, 3, 5}. According to the table the values for y are either \(y=0\), \(y=20\), \(y=40\), \(y=60\), or \(y=100\). The set of values of f which is abbreviated by \({W_f} =\){0, 20, 40, 60, 100}. If we enlarge the domain of f to accept all real values for x between 0 and 24 (the maximal working hours per day), then we are able to solve the following exercise:

Exercise 2:

The wage function f(x) for the number of hours worked is given by: \[f(x)=20\cdot x\]

a) Determine the wage for 2.5 working hours.

Solution: We insert 2.5 for x into the wage function \[y=f(x)=20\cdot x.\] Thus, \[y=f(2.5)=20\cdot 2.5=50\]

The wage is 50 sFr. for 2.5 working hours.

b) Inversely, one could ask how many hours an employee has to work to earn 80 sFr.

Solution: First, we rephrase the question mathematically. For which x is \(y=180\)? Thus, we are looking for an x such that \(y=180\)d. Hence, \[y=f(x)=20\cdot x=80\]

Solving for x, we get x=4. This means that the employee has to work 4 hours to get a wage of 80 sFr.


3. Graphical illustration of functions

We can also illustrate our wage function graphically. For this purpose, the individual input/output combinations from the table above are transferred into the coordinate system and exhibited as points P(x,y). This way, we obtain the following five points:


Point \({P_3}\), for instance, can be illustrated in the coordinate system by moving two units to the right on the x-axis, and moving up 40 units on the y-axis.


If you connect the five points and then extend that line, you get the graph of the entire wage function \[y=20\cdot x.\]


Graphically, exercise 2 can be solved as follows. In exercise a), we had to determine the wage for \(x=2.5\). This means that we choose \(x=2.5\) on the x-axis, and move up vertically to the graph of f(x). Then, we can read off the corresponding wage by moving horizontally to the left till reaching the y-axis. Thus, the wage is 50 sFr. for \(x=2.5\).

In exercise b), we have to determine the number of hours worked to get a wage of 80 sFr. For this purpose we move to \(y=80\) on the y-axis, then to the right till meeting the wage function. Then, we read off the x-value from the x-axis by moving downward vertically. This way we see that the employee has to work for 4 hours in order to obtain the wage of 80 sFr.


4. Economic application II

Suppose that a firm A produces x cars [in quantity units]. Obviously, the costs C [in sFr.] of the firm will be related to x. Therefore, we write C(x), i.e., costs as a function of x. We guess that costs are increasing for a larger number of cars. For instance, the positive relation between the two variables x und C could be described as follows: \[C(x)=3000\cdot x\]

Implicitly, we have assumed that the variable cost per unit is a constant of 3000 sFr.


This cost function implies that an additional car costs 3000 sFr. For instance, the production of two cars costs \[C(2)=3000\cdot 2=6000\]

With a budget of 12000 sFr., we can produce a total of four cars, since \[\begin{align} & K(x)=3000\cdot x=12000 \\ & \Rightarrow x=4 \\ \end{align}\]

One could argue that the costs of zero cars should not be zero since there are possibly some fixed costs. These fixed costs represent costs like the rent of a factory building and/or the machines. The fixed costs have to be paid even if the firm produces no cars. Therefore, the cost function could be extended as follows assuming fixed costs of 10000 sFr.: \[C(x)=10000+3000\cdot x\]

In comparison to the previous graphic the cost function shifts up due to the fixed costs for all x.


This new cost function shows that, evem with no production at all, costs are 10000 sFr.:

\[C(0)=10000+3000\cdot 0=10000\]

Now, suppose that another firm B has a different cost situation due to different technology and knowledge. The fixed costs are 9000 sFr. and therefore, at x=0, B's costs are lower by 1000 sFr. than the costs of firm A. At the same time, we assume that the variable costs per unit are 4000 sFr. for firm B. Therefore, firm B has the following cost function:

\[C(x)=9000+4000\cdot x\]

It is important to note that we are still analyzing the relation between the variables x and C. The cost function of firm B has the same structure as the cost function of firm A. Both firms have linear cost functions. Nonetheless, the functions look differently. The difference between these cost functions motivates to write down a general linear cost function:

\[C(x)=b+a\cdot x\]

In this general linear cost function, the fixed costs are represented by parameter b, and the variable costs per unit are represented by parameter a. For economical reasons we assume that a>0 and b≥0 hold. This means that the variable costs per unit are strictly positive and the fixed costs are nonnegative. Note that a und b are parameters. Depending on the firm, these parameter can take different values. Writing generally a and b, we do not have to specify the exact value. The parameter a determines the slope of the cost function, i.e., a larger value of a means a steeper graph. The parameter b shows the costs for x=0.


Without any knowledge of the firms' cost functions, we are able to conclude from the figure that firm A has larger fixed costs than firm B, since the costs for firm A are larger at x=0 than for firm B. On the other hand, we are able to conclude that firm A has lower variable costs per unit than firm B due to the smaller slope of firm A's cost function.

Even if we do not know the values of parameters a and b, we are able to calculate with this general cost function:

Exercise 3:

Suppose the cost function for a car producer is \[C(x)=b+a\cdot x.\]

a) Determine the costs at x=5?

Solution: \[C(5)=b+a\cdot 5\]

Therefore, the costs are b+5a (whatever the values are for parameter a and b!) (Remark: The costs, i.e., one specific number in sFr., could be explicitly calculated for a specific firm. All we need for that is the values for parameter a and b.)

b) How many cars can be produced with budget of 20000 sFr.?


\[C(x)=b+a\cdot x=20000\]


The number of cars are therefore (20000-b)/a (Whatever the values are for parameter a and b!) (Remark: The number of cars, i.e., one specific number x, could be explicitly determined for a specific firm. All we need for that is the explicit values for parameter a and b.)