Example: Determining If Menu Price Lists Are Functions

The coffee shop menu consists of items and their prices.

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

Answer:

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

Example: Determining If Class Grade Rules Are Functions

In a particular math class, the overall percent grade corresponds to a grade point average. Is grade point average a function of the percent grade? Is the percent grade a function of the grade point average? The table below shows a possible rule for assigning grade points.

Percent Grade	0–56	57–61	62–66	67–71	72–77	78–86	87–91	92–100
Grade Point Average	0.0	1.0	1.5	2.0	2.5	3.0	3.5	4.0

Answer: For any percent grade earned, there is an associated grade point average, so the grade point average is a function of the percent grade. In other words, if we input the percent grade, the output is a specific grade point average. In the grading system given, there is a range of percent grades that correspond to the same grade point average. For example, students who receive a grade point average of 3.0 could have a variety of percent grades ranging from 78 all the way to 86. Thus, percent grade is not a function of grade point average.

Example: Identifying Tables that Represent Functions

Which table, a), b), or c), represents a function (if any)?

Table A
Input	Output
2	1
5	3
8	6

Table B
Input	Output
–3	5
0	1
4	5

Table C
Input	Output
1	0
5	2
5	4

Answer: a) and b) define functions. In both, each input value corresponds to exactly one output value. c) does not define a function because the input value of 5 corresponds to two different output values. When a table represents a function, corresponding input and output values can also be specified using function notation. The function represented by a) can be represented by writing

[latex]f\left(2\right)=1,f\left(5\right)=3,\text{and }f\left(8\right)=6[/latex]

Similarly, the statements [latex]g\left(-3\right)=5,g\left(0\right)=1,\text{and }g\left(4\right)=5[/latex] represent the function in b). c) cannot be expressed in a similar way because it does not represent a function.

Example: Evaluating Functions

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

Answer:

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{cases}\text{ }h\left(p\right)={p}^{2}+2p\hfill \\ \text{ }h\left(4\right)={\left(4\right)}^{2}+2\left(4\right)\hfill \\ \text{ }=16+8\hfill \\ \text{ }=24\hfill \end{cases}[/latex]

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

Example: 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]

Answer: Replace the [latex]x[/latex] in the function with each specified value.

1. Because the input value is a number, 2, we can use algebra to simplify.
   [latex]\begin{cases}f\left(2\right)={2}^{2}+3\left(2\right)-4\hfill \\ =4+6 - 4\hfill \\ =6\hfill \end{cases}[/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{cases}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{cases}[/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{cases}\begin{cases}\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 \end{cases}\hfill \\ \begin{cases}\text{ }=\frac{2ah+{h}^{2}+3h}{h}\hfill & \hfill \\ \text{ }=\frac{h\left(2a+h+3\right)}{h}\hfill & \begin{cases}{cc}\begin{cases}{cc}& \end{cases}& \end{cases}\text{Factor out }h.\hfill \\ \text{ }=2a+h+3\hfill & \begin{cases}{cc}\begin{cases}{cc}& \end{cases}& \end{cases}\text{Simplify}.\hfill \end{cases}\hfill \end{cases}[/latex]

Example: Solving Functions

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

Answer:

[latex]\begin{cases}\text{ }h\left(p\right)=3\hfill & \hfill & \hfill & \hfill \\ \text{ }{p}^{2}+2p=3\hfill & \hfill & \hfill & \text{Substitute the original function }h\left(p\right)={p}^{2}+2p.\hfill \\ \text{ }{p}^{2}+2p - 3=0\hfill & \hfill & \hfill & \text{Subtract 3 from each side}.\hfill \\ \text{ }\left(p+3\text{)(}p - 1\right)=0\hfill & \hfill & \hfill & \text{Factor}.\hfill \end{cases}[/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{cases}\left(p+3\right)=0,\hfill & p=-3\hfill \\ \left(p - 1\right)=0,\hfill & p=1\hfill \end{cases}[/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 5. The graph verifies that [latex]h\left(1\right)=h\left(-3\right)=3[/latex] and [latex]h\left(4\right)=24[/latex].

Example: Evaluating and Solving a Tabular Function

Using the table below,

1. Evaluate [latex]g\left(3\right)[/latex].
2. Solve [latex]g\left(n\right)=6[/latex].

n	1	2	3	4	5
g(n)	8	6	7	6	8

Answer:

• Evaluating [latex]g\left(3\right)[/latex] means determining the output value of the function [latex]g[/latex] for the input value of [latex]n=3[/latex]. The table output value corresponding to [latex]n=3[/latex] is 7, so [latex]g\left(3\right)=7[/latex].
• Solving [latex]g\left(n\right)=6[/latex] means identifying the input values, [latex]n[/latex], that produce an output value of 6. The table below shows two solutions: [latex]n=2[/latex] and [latex]n=4[/latex].

n	1	2	3	4	5
g(n)	8	6	7	6	8

When we input 2 into the function [latex]g[/latex], our output is 6. When we input 4 into the function [latex]g[/latex], our output is also 6.

Example: Applying the Horizontal Line Test

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

Answer: The function in (a) is not one-to-one. The horizontal line shown below intersects the graph of the function at two points (and we can even find horizontal lines that intersect it at three points.) The function in (b) is one-to-one. Any horizontal line will intersect a diagonal line at most once.