1. Section 8.2
Functions

2. Objectives Define relation, domain, and range Identify functions
Use function notation
Find the domain of a function
Graph linear functions
Write equations of linear functions
Evaluate Polynomial Functions

3. Objective 1: Define Relation, Domain, and Range
We can display the data in the table as a set of ordered pairs, where the first component represents the session of Congress and the second component represents the number of women representatives serving during that session:
Sets of ordered pairs like this are called relations. The set of all first components is called the domain of the relation, and the set of all second components is called the range of the relation. A relation may consist of a finite (countable) number of ordered pairs or an infinite (unlimited) number of ordered pairs.

4. EXAMPLE 1
Find the domain and range of the relation:
{(3, 2), (5, –7), (–8, 2), (9, 0)}
Strategy
We will identify the first components and the second components of the ordered pairs.
Why
The set of all first components is the domain of the relation, and the set of all second components is the range.

5. EXAMPLE 1
Find the domain and range of the relation:
{(3, 2), (5, –7), (–8, 2), (9, 0)}
Solution
The first components of the ordered pairs are highlighted in red, and the second components are highlighted in blue:
The elements of the domain and range are usually listed in increasing order, and if a value is repeated, it is only listed once.

6. Objective 2: Identify Functions
The relation above is defined by a set of ordered pairs. Relations can also be defined using an arrow or mapping diagram. The same data from the above table is shown below as an arrow diagram.
Notice that to each session of Congress, there corresponds exactly one number of women representatives. That is, to each member of the domain there corresponds exactly one member of the range. Relations that have this characteristic are called functions.

7. Objective 2: Identify Functions
A function is a set of ordered pairs (a relation) in which to each first component there corresponds exactly one second component. The set of first components is called the domain of the function, and the set of second components is called the range of the function.
Given a relation in x and y, if to each value of x in the domain there corresponds exactly one value of y in the range, then y is said to be a function of x.
In the previous definition, since y depends on x, we call x the independent variable and y the dependent variable. The set of all possible values that can be used for the independent variable is the domain of the function, and the set of all values of the dependent variable is the range of the function.

8. Objective 2: Identify Functions
A function can also be defined by an equation. For example, sets up a rule in which to each value of x there corresponds exactly one value of y. To find the y-value (called an output) that corresponds to the x-value 4 (called an input), we substitute 4 for x and evaluate the right side of the equation.
For the function defined by , a y-value of 5 corresponds to an
x-value of 4.

9. EXAMPLE 2
Determine whether the relation defines y to be a function of x.
Strategy
We will determine whether there is more than one value of y that corresponds to a single value of x.
Why
If to any x-value there corresponds more than one y-value, then y is not a function of x.

10. EXAMPLE 2
Determine whether the relation defines y to be a function of x.
Solution
a. The arrow diagram defines a function because to each value of x there corresponds exactly one value of y.

11. EXAMPLE 2
Determine whether the relation defines y to be a function of x.
Solution
b. The table does not define a function, because to the x-value 8 there corresponds to more than one y-value.
• In the first row, to the x-value 8, there corresponds the y-value 2.
• In the third row, to the same x-value 8, there corresponds a different y-value, 3.
When the correspondence in the table is written as a set of ordered pairs, it is apparent that the relation does not define a function:

12. EXAMPLE 2
Determine whether the relation defines y to be a function of x.
Solution
c. Since to each value of x, there corresponds exactly one value of y, the set of ordered pairs defines y to be a function of x.
• (−2,3) To the x-value –2, there corresponds exactly one y-value, 3.
• (−1,3) To the x-value –1, there corresponds exactly one y-value, 3.
• (0,3) To the x-value 0, there corresponds exactly one y-value, 3.
• (1,3) To the x-value 1, there corresponds exactly one y-value, 3.
In this case, the same y-value, 3, corresponds to each x-value.
The results from parts (b) and (c) illustrate an important fact: Two different ordered pairs of a function can have the same y-value, but they cannot have the same x-value.

13. Objective 3: Use Function Notation
The notation y = ƒ(x) indicates that y is a function of x.
In the following example, the function is read as “ƒ of x is equal to 2x minus 5”
Function notation provides a compact way of representing the output value that corresponds to some input value. If ƒ(x) = 2x – 5, the value that corresponds to x = 6 is represented by ƒ(6).
Thus, ƒ(6) = 7. We read this result as “ƒ of 6 equals 7.” The output 7 is called a function value.

14. EXAMPLE 4 Strategy Strategy Why
Let ƒ(x) = 4x + 3 and g(t) = t2 – 2t. Find:
a. ƒ(3) b. ƒ(–1) c. ƒ(r + 1) d. g(–2.4)
Strategy
Strategy
We will substitute 3, –1, and r + 1 for each x in ƒ(x) = 4x + 3 and evaluate the right side. We will substitute –2.4 for each t in g(t) = t2 – 2t and evaluate the right side.
Why
Whatever appears within the parentheses in ƒ( ) is to be substituted for each x in ƒ(x) = 4x + 3. Whatever expression appears within the parentheses in g( ) is to be substituted for each t in g(t) = t2 – 2t.

15. EXAMPLE 4 Solution a. To find ƒ(3), we replace each x with 3:
Let ƒ(x) = 4x + 3 and g(t) = t2 – 2t. Find:
a. ƒ(3) b. ƒ(–1) c. ƒ(r + 1) d. g(–2.4)
Solution
a. To find ƒ(3), we replace each x with 3:
b. To find ƒ(–1), we replace each x with –1:

16. EXAMPLE 4 Solution c. To find ƒ(r + 1), we replace each x with r + 1:
Let ƒ(x) = 4x + 3 and g(t) = t2 – 2t. Find:
a. ƒ(3) b. ƒ(–1) c. ƒ(r + 1) d. g(–2.4)
Solution
c. To find ƒ(r + 1), we replace each x with r + 1:
d. To find g(–2.4), we replace each t with –2.4:

17. Objective 4: Find the Domain of a Function
We can think of a function as a machine that takes some input x and turns it into some output ƒ(x), as shown in figure (a). The machine shown in figure (b) turns the input –6 into the output –11. The set of numbers that we put into the machine is the domain of the function, and the set of numbers that comes out is the range.

18. EXAMPLE 6
Find the domain of each function:
Strategy
We will ask, “What values of x are permissible replacements for x in 3x + 1 and ?”
Why
These values of x form the domain of the function.

19. EXAMPLE 6
Find the domain of each function:
Solution
a. We will be able to evaluate 3x + 1 for any value of x that is a real number. Thus, the domain of the function is the set of real numbers, which can be represented by the symbol .
b. Since division by 0 is undefined, we will not be
able to evaluate for any number x that
makes the denominator equal to 0. To find such x-values, we set the denominator equal to 0 and solve for x.

20. EXAMPLE 6
Find the domain of each function:
Solution
We have found that 2 must be excluded from the domain of the function because it makes the denominator 0. However, all other real numbers are permissible replacements for x. Thus, the domain of the function is the set of all real numbers except 2.

21. Objective 5: Graph Linear Functions
A linear function is a function that can be defined by an equation of the form ƒ(x) = mx + b, where m and b are real numbers. The graph of a linear function is a straight line with slope m and y-intercept (0, b).
The most basic linear function is ƒ(x) = x. It is called the identity function because it assigns each real number to itself. The graph of the identity function is a line with slope 1 and y-intercept (0,0), as shown in figure (a).
A linear function defined by ƒ(x) = b is called a constant function, because for any input x, the output is the constant b. The graph of a constant function is a horizontal line. The graph of ƒ(x) = 2 is shown in figure (b).

22. Objective 6: Write Equations of Linear Functions
The slope–intercept and point–slope forms for the equation of a line that we studied in Sections 3.5 and 3.6 can be adapted to write equations of linear functions.

23. EXAMPLE 7
a. Write an equation for the linear function whose graph has slope and y-intercept (0, 2).
c. Write an equation for the linear function whose graph passes through (–16, –12) and is perpendicular to the graph of g (x) = – 8x – 1.
b. Write an equation for the linear function whose graph passes through (6, 4) with slope 5.
Strategy
We will use either the slope–intercept or point–slope form to write each equation.
Why
Writing equations of linear functions is similar to writing equations of linear equations in two variables, except that we replace y with ƒ(x).

24. EXAMPLE 7
a. Write an equation for the linear function whose graph has slope and y-intercept (0, 2).
c. Write an equation for the linear function whose graph passes through (–16, –12) and is perpendicular to the graph of g (x) = – 8x – 1.
b. Write an equation for the linear function whose graph passes through (6, 4) with slope 5.
Solution
a. If the slope is and the y-intercept is (0, 2), then m = and b = 2.

25. EXAMPLE 7
a. Write an equation for the linear function whose graph has slope and y-intercept (0, 2).
c. Write an equation for the linear function whose graph passes through (–16, –12) and is perpendicular to the graph of g (x) = – 8x – 1.
b. Write an equation for the linear function whose graph passes through (6, 4) with slope 5.
Solution
b. If the slope is 5 and the graph passes through (6, 4), then m = 5 and (x1, y1) = (6, 4).

26. EXAMPLE 7
a. Write an equation for the linear function whose graph has slope and y-intercept (0, 2).
c. Write an equation for the linear function whose graph passes through (– 16, – 12) and is perpendicular to the graph of g (x).= – 8x – 1.
b. Write an equation for the linear function whose graph passes through (6, 4) with slope 5.
Solution
c. The slope of the line represented by g (x) = – 8x – 1 is the coefficient of x: – 8. Since the desired function is to have a graph that is perpendicular to the graph of g (x) = – 8x – 1, its slope must be the negative reciprocal of – 8, which is .

27. EXAMPLE 7
a. Write an equation for the linear function whose graph has slope and y-intercept (0, 2).
c. Write an equation for the linear function whose graph passes through (– 16, – 12) and is perpendicular to the graph of g (x).= – 8x – 1.
b. Write an equation for the linear function whose graph passes through (6, 4) with slope 5.
Solution

28. Objective 7: Evaluate Polynomial Functions
A polynomial function is a function whose equation is defined by a polynomial in one variable.
We have seen that linear functions are defined by equations of the form ƒ(x) = mx + b. Some examples of linear functions are
Another example of a polynomial function is ƒ(x) = x2 + 6x – 8. This is a second degree polynomial function, called a quadratic function.
An example of a third-degree polynomial function is ƒ(x) = x3 – 3x2 – 9x + 2. Third-degree polynomial functions, also called cubic functions.
To evaluate a polynomial function at a specific value, we replace the variable in the defining equation with that value, called the input. Then we simplify the resulting expression to find the output.

29. EXAMPLE 8 Strategy Why We will find ƒ(3).
Packaging. To make boxes, a manufacturer cuts equal sized squares from each corner of 10 in.  12 in. pieces of cardboard and then folds up the sides. The polynomial function ƒ(x) = 4x3 – 44x x gives the volume (in cubic inches) of the resulting box when a square with sides x inches long is cut from each corner. Find the volume of a box if 3-inch squares are cut out.
Strategy
We will find ƒ(3).
Why
The notation ƒ(3) represents the volume of the box when 3-inch squares are cut out of the corners of the piece of
cardboard.

30. EXAMPLE 8
Packaging. To make boxes, a manufacturer cuts equal sized squares from each corner of 10 in.  12 in. pieces of cardboard and then folds up the sides. The polynomial function ƒ(x) = 4x3 – 44x x gives the volume (in cubic inches) of the resulting box when a square with sides x inches long is cut from each corner. Find the volume of a box if 3-inch squares are cut out.
Solution
If 3-inch squares are cut out of the corners, the resulting box has a volume of 72 in.3.