Determining Whether a Relation Represents a Function

A relation is a set of ordered pairs. The set consisting of the first components of each ordered pair is called the domain and the set consisting of the second components of each ordered pair is called the range. Consider the following set of ordered pairs. The first numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first.

The domain is [latex]\left\{1,\,2,\,3,\,4,\,5\right\}.[/latex]
The range is [latex]\left\{2,\,4,\,6,\,8,\,10\right\}.[/latex]

Note that each value in the domain is also known as an input value, or independent variable, and is often labeled with the lowercase letter[latex]\,x.\,[/latex]Each value in the range is also known as an output value, or dependent variable, and is often labeled lowercase letter[latex]\,y.[/latex]

A function[latex]\,f\,[/latex]is a relation that assigns a single element in the range to each element in the domain. In other words, no x-values are repeated. For our example that relates the first five natural numbers to numbers double their values, this relation is a function because each element in the domain, [latex]\left\{1,\,2,\,3,\,4,\,5\right\},[/latex] is paired with exactly one element in the range, [latex]\left\{2,\,4,\,6,\,8,\,10\right\}.[/latex]

Now let’s consider the set of ordered pairs that relates the terms “even” and “odd” to the first five natural numbers. It would appear as

Notice that each element in the domain, [latex]\left\{\text{even,}\,\text{odd}\right\}[/latex]
is not paired with exactly one element in the range, [latex]\left\{1,\,2,\,3,\,4,\,5\right\}.[/latex]
For example, the term “odd” corresponds to three values from the domain, [latex]\left\{1,\,3,\,5\right\}[/latex]
and the term “even” corresponds to two values from the range, [latex]\left\{2,\,4\right\}.[/latex]
This violates the definition of a function, so this relation is not a function.

(Figure) compares relations that are functions and not functions.

Figure 1. (a) This relationship is a function because each input is associated with a single output. Note that input[latex]\,q\,[/latex]and[latex]\,r\,[/latex]both give output[latex]\,n.\,[/latex](b) This relationship is also a function. In this case, each input is associated with a single output. (c) This relationship is not a function because input[latex]\,\,q\,[/latex]is associated with two different outputs.

Determining If Menu Price Lists Are Functions

The coffee shop menu, shown in (Figure) consists of items and their prices.

1. Is price a function of the item?
2. Is the item a function of the price?

Figure 2.

Show Solution

1. Let’s begin by considering the input as the items on the menu. The output values are then the prices.

   Each item on the menu has only one price, so the price is a function of the item.

2. Two items on the menu have the same price. If we consider the prices to be the input values and the items to be the output, then the same input value could have more than one output associated with it. See (Figure).
   

   Figure 3.

   Therefore, the item is a not a function of price.

Finding Input and Output Values of a Function

When we know an input value and want to determine the corresponding output value for a function, we evaluate the function. Evaluating will always produce one result because each input value of a function corresponds to exactly one output value.

When we know an output value and want to determine the input values that would produce that output value, we set the output equal to the function’s formula and solve for the input. Solving can produce more than one solution because different input values can produce the same output value.

Evaluation of Functions in Algebraic Forms

When we have a function in formula form, it is usually a simple matter to evaluate the function. For example, the function[latex]\,f\left(x\right)=5-3{x}^{2}\,[/latex]can be evaluated by squaring the input value, multiplying by 3, and then subtracting the product from 5.

How To

Given the formula for a function, evaluate.

1. Replace the input variable in the formula with the value provided.
2. Calculate the result.

Evaluating Functions at Specific Values

Evaluate[latex]\,f\left(x\right)={x}^{2}+3x-4\,[/latex]at

1. [latex]2[/latex]
2. [latex]a[/latex]
3. [latex]a+h[/latex]
4. [latex]\frac{f\left(a+h\right)-f\left(a\right)}{h}[/latex]

Show Solution

Replace the[latex]\,x\,[/latex]
in the function with each specified value.

1. Because the input value is a number, 2, we can use simple algebra to simplify.
   [latex]\begin{array}{ccc}\hfill f\left(2\right)& =& {2}^{2}+3\left(2\right)-4\\ & =& 4+6-4\hfill \\ & =& 6\hfill \end{array}[/latex]
2. In this case, the input value is a letter so we cannot simplify the answer any further.
   [latex]f\left(a\right)={a}^{2}+3a-4[/latex]
3. With an input value of[latex]\,a+h,\,[/latex]we must use the distributive property.
   [latex]\begin{array}{ccc}\hfill f\left(a+h\right)& =& {\left(a+h\right)}^{2}+3\left(a+h\right)-4\hfill \\ & =& {a}^{2}+2ah+{h}^{2}+3a+3h-4\hfill \end{array}[/latex]
4. In this case, we apply the input values to the function more than once, and then perform algebraic operations on the result. We already found that
   [latex]f\left(a+h\right)={a}^{2}+2ah+{h}^{2}+3a+3h-4[/latex]

   and we know that

   [latex]f\left(a\right)={a}^{2}+3a-4[/latex]

   Now we combine the results and simplify.

   [latex]\begin{array}{ccc}\hfill \frac{f\left(a+h\right)-f\left(a\right)}{h}& =& \frac{\left({a}^{2}+2ah+{h}^{2}+3a+3h-4\right)-\left({a}^{2}+3a-4\right)}{h}\hfill \\ & =& \frac{2ah+{h}^{2}+3h}{h}\hfill \\ & =& \frac{h\left(2a+h+3\right)}{h}\phantom{\rule{2em}{0ex}}\text{Factor out }h.\hfill \\ & =& 2a+h+3\phantom{\rule{3.3em}{0ex}}\text{Simplify}.\hfill \end{array}[/latex]

Evaluating Functions

Given the function[latex]\,h\left(p\right)={p}^{2}+2p,\,[/latex]evaluate[latex]\,h\left(4\right).\,[/latex]

Show Solution

To evaluate[latex]\,h\left(4\right),\,[/latex]we substitute the value 4 for the input variable[latex]\,p\,[/latex]in the given function.

[latex]\begin{array}{ccc}\hfill h\left(p\right)& =& {p}^{2}+2p\hfill \\ \hfill h\left(4\right)& =& {\left(4\right)}^{2}+2\left(4\right)\hfill \\ & =& 16+8\hfill \\ & =& 24\hfill \end{array}[/latex]

Therefore, for an input of 4, we have an output of 24.

Try It

Given the function[latex]\,g\left(m\right)=\sqrt{m-4},\,[/latex]evaluate[latex]\,g\left(5\right).[/latex]

Show Solution

[latex]\,g\left(5\right)=1\,[/latex]

Solving Functions

Given the function[latex]\,h\left(p\right)={p}^{2}+2p,\,[/latex]solve for[latex]\,h\left(p\right)=3.[/latex]

Show Solution

[latex]\begin{array}{cccc}\hfill h\left(p\right)& =& 3\hfill & \\ \hfill {p}^{2}+2p& =& 3\hfill & \phantom{\rule{2em}{0ex}}\text{Substitute the original function }h\left(p\right)={p}^{2}+2p.\hfill \\ \hfill {p}^{2}+2p-3& =& 0\hfill & \phantom{\rule{2em}{0ex}}\text{Subtract 3 from each side}\text{.}\hfill \\ \hfill \left(p+3\text{)(}p-1\right)& =& 0\hfill & \phantom{\rule{2em}{0ex}}\text{Factor}.\hfill \end{array}[/latex]

If[latex]\,\left(p+3\right)\left(p-1\right)=0,\,[/latex]either[latex]\,\left(p+3\right)=0\,[/latex]or[latex]\,\left(p-1\right)=0\,[/latex](or both of them equal 0). We will set each factor equal to 0 and solve for[latex]\,p\,[/latex]in each case.

[latex]\begin{array}{cccccc}\hfill \left(p+3\right)& =& 0,\hfill & \hfill \phantom{\rule{0.5em}{0ex}}p& =& -3\hfill \\ \hfill \left(p-1\right)& =& 0,\hfill & \hfill \phantom{\rule{0.5em}{0ex}}p& =& 1\hfill \end{array}[/latex]

This gives us two solutions. The output[latex]\,h\left(p\right)=3\,[/latex]when the input is either[latex]\,p=1\,[/latex]or[latex]\,p=-3.\,[/latex]We can also verify by graphing as in (Figure). The graph verifies that[latex]\,h\left(1\right)=h\left(-3\right)=3\,[/latex]and[latex]\,h\left(4\right)=24.[/latex]

Figure 4.

Try It

Given the function[latex]\,g\left(m\right)=\sqrt{m-4},\,[/latex]solve[latex]\,g\left(m\right)=2.[/latex]

Show Solution

[latex]\,m=8\,[/latex]

Finding Function Values from a Graph

Evaluating a function using a graph also requires finding the corresponding output value for a given input value, only in this case, we find the output value by looking at the graph. Solving a function equation using a graph requires finding all instances of the given output value on the graph and observing the corresponding input value(s).

Reading Function Values from a Graph

Given the graph in (Figure),

1. Evaluate[latex]\,f\left(2\right).[/latex]
2. Solve[latex]\,f\left(x\right)=4.[/latex]

Figure 5.

Show Solution

1. To evaluate[latex]\,f\left(2\right),\,[/latex]locate the point on the curve where[latex]\,x=2,\,[/latex]then read the y-coordinate of that point. The point has coordinates[latex]\,\left(2,1\right),\,[/latex]so[latex]\,f\left(2\right)=1.\,[/latex]See (Figure).
   

   Figure 6.

2. To solve[latex]\,f\left(x\right)=4,\,[/latex]we find the output value[latex]\,4\,[/latex]on the vertical axis. Moving horizontally along the line[latex]\,y=4,\,[/latex]we locate two points of the curve with output value[latex]\,4:[/latex] [latex]\left(-1,4\right)\,[/latex]and[latex]\,\left(3,4\right).\,[/latex]These points represent the two solutions to[latex]\,f\left(x\right)=4:[/latex] [latex]-1\,[/latex] or [latex]\,3.\,[/latex]This means[latex]\,f\left(-1\right)=4\,[/latex]and[latex]\,f\left(3\right)=4,\,[/latex]or when the input is[latex]\,-1\,[/latex]or[latex]\text{3,}\,[/latex]the output is[latex]\,\text{4}\text{.}\,[/latex]See (Figure).
   

   Figure 7.

Try It

Using (Figure), solve[latex]\,f\left(x\right)=1.[/latex]

Show Solution

[latex]x=0\,[/latex]or[latex]\,x=2\,[/latex]

Determining Whether a Function is One-to-One

Some functions have a given output value that corresponds to two or more input values. For example, in the stock chart shown in (Figure) at the beginning of this chapter, the stock price was $1000 on five different dates, meaning that there were five different input values that all resulted in the same output value of $1000.

However, some functions have only one input value for each output value, as well as having only one output for each input. We call these functions one-to-one functions. As an example, consider a school that uses only letter grades and decimal equivalents, as listed in (Figure).

This grading system represents a one-to-one function because each letter input yields one particular grade-point average output and each grade-point average corresponds to one input letter.

To visualize this concept, let’s look again at the two simple functions sketched in (Figure)(a) and (Figure)(b). The function in part (a) shows a relationship that is not a one-to-one function because inputs[latex]\,q\,[/latex]and[latex]\,r\,[/latex]both give output[latex]\,n.\,[/latex]The function in part (b) shows a relationship that is a one-to-one function because each input is associated with a single output.

Using the Vertical Line Test

As we have seen in some examples above, we can represent a function using a graph. Graphs display a great many input-output pairs in a small space. The visual information they provide often makes relationships easier to understand. By convention, graphs are typically constructed with the input values along the horizontal axis and the output values along the vertical axis.

The most common graphs name the input value[latex]\,x\,[/latex]and the output value[latex]\,y,\,[/latex]and we say[latex]\,y\,[/latex]is a function of[latex]\,x,\,[/latex]or[latex]\,y=f\left(x\right)\,[/latex]when the function is named[latex]\,f.\,[/latex]The graph of the function is the set of all points[latex]\,\left(x,y\right)\,[/latex]in the plane that satisfies the equation[latex]y=f\left(x\right).\,[/latex]If the function is defined for only a few input values, then the graph of the function consists of only a few points, where the x-coordinate of each point is an input value and the y-coordinate of each point is the corresponding output value. For example, the black dots on the graph in (Figure) tell us that[latex]\,f\left(0\right)=2\,[/latex]and[latex]\,f\left(6\right)=1.\,[/latex]However, the set of all points[latex]\,\left(x,y\right)\,[/latex]satisfying[latex]\,y=f\left(x\right)\,[/latex]is a curve. The curve shown includes[latex]\,\left(0,2\right)\,[/latex]and[latex]\,\left(6,1\right)\,[/latex]because the curve passes through those points.

Figure 8.

The vertical line test can be used to determine whether a graph represents a function. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because a function has only one output value for each input value. See (Figure).

Figure 9.

Applying the Vertical Line Test

Which of the graphs in (Figure) represent(s) a function[latex]\,y=f\left(x\right)?[/latex]

Figure 10.

Show Solution

If any vertical line intersects a graph more than once, the relation represented by the graph is not a function. Notice that any vertical line would pass through only one point of the two graphs shown in parts (a) and (b) of (Figure). From this we can conclude that these two graphs represent functions. The third graph does not represent a function because, at most x-values, a vertical line would intersect the graph at more than one point, as shown in (Figure).

Figure 11.

Try It

Does the graph in (Figure) represent a function?

Figure 12.

Show Solution

yes

Using the Horizontal Line Test

Once we have determined that a graph defines a function, an easy way to determine if it is a one-to-one function is to use the horizontal line test. Draw horizontal lines through the graph. If any horizontal line intersects the graph more than once, then the graph does not represent a one-to-one function.

Applying the Horizontal Line Test

Consider the functions shown in (Figure)(a) and (Figure)(b). Are either of the functions one-to-one?

Show Solution

The function in (Figure)(a) is not one-to-one. The horizontal line shown in (Figure) intersects the graph of the function at two points (and we can even find horizontal lines that intersect it at three points.)

Figure 13.

The function in (Figure)(b) is one-to-one. Any horizontal line will intersect a diagonal line at most once.

Identifying Basic Toolkit Functions

In this text, we will be exploring functions—the shapes of their graphs, their unique characteristics, their algebraic formulas, and how to solve problems with them. When learning to read, we start with the alphabet. When learning to do arithmetic, we start with numbers. When working with functions, it is similarly helpful to have a base set of building-block elements. We call these our “toolkit functions,” which form a set of basic named functions for which we know the graph, formula, and special properties. Some of these functions are programmed to individual buttons on many calculators. For these definitions we will use[latex]\,x\,[/latex]as the input variable and[latex]\,y=f\left(x\right)\,[/latex]as the output variable.

We will see these toolkit functions, combinations of toolkit functions, their graphs, and their transformations frequently throughout this book. It will be very helpful if we can recognize these toolkit functions and their features quickly by name, formula, graph, and basic table properties. The graphs and sample table values are included with each function shown in (Figure).

Toolkit Functions
Name	Function	Graph
Constant	[latex]f\left(x\right)=c,\,[/latex]where [latex]c[/latex] is a constant
Identity	[latex]f\left(x\right)=x[/latex]
Absolute value	[latex]f\left(x\right)=|x|[/latex]
Quadratic	[latex]f\left(x\right)={x}^{2}[/latex]
Cubic	[latex]f\left(x\right)={x}^{3}[/latex]
Reciprocal	[latex]f\left(x\right)=\frac{1}{x}[/latex]
Reciprocal squared	[latex]f\left(x\right)=\frac{1}{{x}^{2}}[/latex]
Square root	[latex]f\left(x\right)=\sqrt{x}[/latex]
Cube root	[latex]f\left(x\right)=\sqrt[3]{x}[/latex]