There are many ways of expressing a relationship between two quantities. For example, if you are paid $10 per hour, then the relation between the number of hours, x, that you work and the amount of money, y, that you earn may be expressed by the equation $y=10x.$ Replacing x with 40 yields $y=400$ and indicates that 40 hours of work corresponds to a $400 paycheck. A special relationship such as $y=10x$ in which to each element x in one set there corresponds a unique element y in another set is called a function. We say that your pay is a function of the number of hours you work. Because the value of y depends on the value of x, y is called the dependent variable and x is called the independent variable. Any symbol may be used for the dependent and independent variables.

As an example of an assignment that is not a function, consider the correspondence between U.S. and European airports. A U.S. airport is connected to a European airport if there is a direct flight between them. This is not a function because, for instance, the New York airport is connected by a direct flight to London, Paris, Barcelona, and other airports.

A rule of assignment may be described by a correspondence diagram, in which an arrow points from each domain element to the corresponding element or elements in the range.

To decide whether a correspondence diagram defines a function, we must check whether any domain element is paired with more than one range element. If, for some element of X, there are two or more corresponding elements of Y, then the relation is not a function. See the correspondence diagrams in Figure 2.36.

We usually use single letters such as f, F, g, G, h, and H as the name of a function. Some special function names use more than one letter, like sin (the sine function) and log (the logarithm function). If we choose f as the name of a function, then for each x in the domain of f, there corresponds a unique y in its range. The number y is denoted by f(x), read as “f of x” or as “f at x.” We call f(x) the value of f at the number x and say that f assigns the value f(x) to x.

If the function shown in Figure 2.36(a) is named f, then the correspondence diagram may be described by:

Another way to illustrate the concept of a function is by visualizing a “machine,” as shown in Figure 2.37. If x is in the domain of a function, then the machine accepts x as an “input” and produces the value f(x) as the “output.” A hand-held bar code scanner that reads the bar codes and outputs the price of items is an example of a function.

Functions can be prescribed in many ways: directly by describing the assignment to each element of the domain, by formula, or by algorithm.

Functions Defined by Tables and Graphs

Tables and graphs can be used to describe relations and functions. We can display in a tabular form the list of ordered pairs (x, y) of a relation, where x is in the domain and the corresponding y is in the range. For example, the relations r and s of Example 1 described in tabular form are shown in Tables 2.6 and 2.7, respectively.

x	$y=r(x)$
$-1$	2
1	3
5	2
$-1$	$-3$
Relation r

x	$y=s(x)$
$-2$	1
$-1$	$-3$
0	2
2	1
Relation s

Alternatively, we can display geometrically the list of ordered pairs (x, y) in a relation by plotting each pair (x, y) as a point in the coordinate plane. The geometric display is the graph of the relation. Graphs of relations r and s are given in Figure 2.40.

The graph of the points in the table is called a scatterplot.

Functions Defined by Equations

When the relation defined by an equation in two variables is a function, we can often solve the equation for the dependent variable in terms of the independent variable. For example, the equation $y-x^2=0$ can be solved for y in terms of x:

$$y=x^2.$$

Now we can replace the dependent variable, in this case, y, with functional notation f(x) and express the function defined by the equation $y-x^2=0$ as

$$f(x)=x^2(\text{read }“f \text{of} x \text{equals} x^2”).$$

Here f(x) plays the role of y and is the value of the function f at x. For example, if $x=3$ is an element (input value) in the domain of f, the corresponding element (output value) in the range is found by replacing x with 3 in the equation

$$\begin{array}{lrcll}& f(x)& =& x^2,& \\ \text{so that}& f(3)& =& 3^2=9.& \text{Replace }x\text{ with }3.\end{array}$$

We say that the value of the function f at 3 is 9 or that 9 is the evaluated function value f(3). See Figure 2.41. In other words, the number 9 in the range corresponds to the number 3 in the domain, and the ordered pair (3, 9) is an ordered pair of the function f.

If a function g is defined by an equation such as $y=x^2-6x+8,$ the notations

$$y=x^2-6x+8\text{and }\text{g}(x)=x^2-6x+8$$

define the same function.

Side Note

The definition of a function is independent of the letters used to denote the function and the variables. For example,

$$\begin{array}{l}s(x)=x^2, g(y)=y^2,\\ \text{and} h(t)=t^2\end{array}$$

represent the same function.

Example 2 Determining Whether an Equation Defines a Function

In each equation, determine whether y is a function of x.

1. $6x^2-3y=12$

2. $y^2-x^2=4$

Solution

Solve each equation for y in terms of x. If more than one value of y corresponds to the same value of x, then y is not a function of x.

1. $$\begin{array}{rcll}6x^2-3y& =& 12& \text{Original equation}\\ 6x^2-3y+3y-12& =& 12+3y-12& \text{Add }3y-12\text{ to both sides.}\\ 6x^2-12& =& 3y& \text{Simplify.}\\ 2x^2-4& =& y& \text{Divide by }3.\\ y& =& 2x^2-4& \text{Interchange sides}\end{array}$$

   The last equation shows that only one value of y corresponds to each value of x. For example, if $x=0,$ then $y=0-4=-4.$ So y is a function of x.

2. $$\begin{array}{rcll}{y}^2-x^2& =& 4& \text{Original equation}\\ y^2-x^2+x^2& =& 4+x^2& \text{Add }{x}^2\text{ to both sides.}\\ y^2& =& x^2+4& \text{Simplify.}\\ y& =& \pm \sqrt{x^2+4}& \text{Square root property}\end{array}$$

   The last equation shows that two values of y correspond to each value of x. For example, if $x=0,$ then $y=\pm \sqrt{0^2+4}=\pm \sqrt{4}=\pm 2.$ Both $y=2$ and $y=-2$ correspond to $x=0.$ Therefore, y is not a function of x. However, the equation $y^2-x^2=4$ defines two functions:

   $f_1(x)=\sqrt{x^2+4};$ domain: $(-\infty ,\infty );$ range: $[2,\infty ),$ and

   $f_2(x)=-\sqrt{x^2+4};$ domain: $(-\infty ,\infty );$ range $(-\infty ,-2].$ See Figure 2.42.

   

Practice Problem 2

1. In each equation, determine whether y is a function of x.

   1. $2x^2-y^2=1$

   2. $x-2y=5$

Example 3 Evaluating a Function

Let g be the function defined by the equation

$$y=x^2-6x+8.$$

Find each function value.

1. g(3)

2. $g(-2)$

3. $g\left({\displaystyle \frac{1}{2}}\right)$

4. $g(a+2)$

5. $g(x+h)$

Solution

Since the function is named g, we replace y with g(x) and write

$$g(x)=x^2-6x+8\text{Replace }y\text{ with }g(x).$$

In this notation, the independent variable x is a placeholder. We can write $g(x)=x^2-6x+8$ as

$$g()={()}^2-6()+8.$$

1. $$\begin{array}{rcll}g(x)& =& {(x)}^2-6(x)+8& \text{The given equation}\\ g(3)& =& {(3)}^2-6(3)+8& \text{Replace }x\text{ with }3\text{ at each occurrence of }x.\\ & =& 9-18+8=-1& \text{Simplify.}\end{array}$$

   The statement $g(3)=-1$ means that the value of the function g at 3 is $-1.$

   Just as in part a, we can evaluate g at any value x in its domain:

2. $\begin{array}{ll}g(x)=x^2-6x+8& \text{Original equation}\\ g(-2)={(-2)}^2-6(-2)+8=24& \text{Replace }x\text{ with }-2\text{ and simplify.}\end{array}$

3. $\begin{array}{ll}g\left({\displaystyle \frac{1}{2}}\right)={\left({\displaystyle \frac{1}{2}}\right)}^2-6\left({\displaystyle \frac{1}{2}}\right)+8={\displaystyle \frac{21}{4}}& \text{Replace }x\text{ with }{\displaystyle \frac{1}{2}}\text{ and simplify.}\end{array}$

4. $\begin{array}{llll}g(a+2)& =& {(a+2)}^2-6(a+2)+8& \text{Replace }x\text{ with }(a+2)\text{ in }g(x).\\ & =& a^2+4a+4-6a-12+8& \text{Recall: }{(x+y)}^2=x^2+2xy+y^2.\\ & =& a^2-2a& \text{Simplify.}\end{array}$

5. $\begin{array}{rcll}g(x+h)& =& {(x+h)}^2-6(x+h)+8& \text{Replace }x\text{ with }(x+h)\text{ in }g(x).\\ & =& x^2+2xh+h^2-6x-6h+8& \text{Simplify.}\end{array}$

Practice Problem 3

1. Let g be the function defined by the equation $y=-2x^2+5x.$ Find each function value.

   1. g(0)

   2. $g(-1)$

   3. $g(x+h)$

Example 4 Finding the Area of a Rectangle

In Figure 2.43, find the area of the rectangle PLMN.

Solution

$$\begin{array}{llll}\text{Area of the rectangle} PLMN& =& (\text{length})(\text{height})& \\ & =& (3-1)(f(1))& P=(1, f(1))\\ & =& (2)(2)& f(1)=1^3-3(1)+4=2\\ & =& 4 \text{square units}& \end{array}$$

Practice Problem 4

1. In Figure 2.43 , find the area of the rectangle TLMS.

Sometimes a function is given by an equation without a predefined domain. In such cases, we use the following agreement.

When we use our agreement to find the domain of a function, first we usually find the values of the variable that do not result in real number outputs. Then we exclude those numbers from the domain. Remember that

1. division by zero is undefined.

2. the square root (or any even root) of a negative number is not a real number.