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Afunction is a special kind ofrelation. We will begin this lesson by discussing relations.

Arelation is a set of ordered pairs. This set may be finite or infinite.

Example: Both finite sets, A and B, are relations.A= { (0,2), (1,3), (2,4) }

B= { (-2,5), (-2,6), (-1,7), (0,8) }

Example:The graph of an equation in two variables is also a relation, since each point of thegraph represents an ordered pair.

x

y

Thesegraphsarerelations.

y

x

x = y2

Note that the equations for these graphs specify exactly howx and y are related. Alsonote that since the equations generatethe graphs, its reasonable to refer to the equationsthemselves as relations. Most of the relations we study will be described with an equation.

The domain of a relation is the set of all x-coordinates in the relation.

The range of a relation is the set of all y-coordinates in the relation.

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Example 1. Determine the domain and range of A = { (0,2), (1,3), (2,4) }.

Solution: D = {0, 1, 2}, R = {2, 3, 4}

Determine the domain and range of: B = { (-2,2), (-2,1), (-1,0), (0,-1) }.

Your Turn Problem #1

Answer: D = {-2, -1, 0 }, R = {2, 1, 0, -1}

Now we are ready to consider a special kind of relation called a function.

Afunction is a relation in which no two ordered pairs have the same x-coordinate.

Example:Consider once again the relations A = { (0,2), (1,3), (2,4) } and

B = { (-2,2), (-2,1), (-1,0), (0,-1) }. Is A a function? Is B a function?

Solution:

A = { (0,2), (1,3), (2,4) } is a function since all x-coordinates are different!

B = { (-2,2), (-2,1), (-1,0), (0,-1) } is not a function since the two ordered pairs (-2,2) and(-2,1) have the same x- coordinate.

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Example. Consider the two relations we saw in graph form (below). Are they functions?

y y

x x

x = y

2

Solution:

The relation y = x + 1 is a function since for any value of x there will be exactly one valuefor y. For example, if x = 3, then y must be equal to 4 (since y = x + 1). Its simply notpossible for there to be another ordered pair in this relation that has an x-coordinate of 3with a y-coordinate of something other than 4.

The relation x = y2

is not a function since (1,1) and (1,-1) are among the ordered pairs inthis set (as are (4,2) and (4,-2)). Two ordered pairs with the same x-coordinate is in directviolation of our definition for a function.

Instead of having to find these particular ordered pairs, theres an easier way to see if a graphrepresents a function. We use the vertical line test.

The Vertical Line TestTo determine if a graph represents a function we consider (or imagine) all vertical lines thatintersect the graph. (Recall that vertical lines are those lines parallel to the y-axis.) If anyvertical line can touch the graph at more than one point, then the graph does not represent afunction. If no vertical lines are able to touch the graph at more than one point, then thegraph does represent a function.

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(a) (b)y y

x x

Your Turn Problem #2Use the vertical line test to determine whether the following graphs represent functions.

Solution:(a) (b)

y y

x x

Graph (b) fails the vertical line test.It is not a function.

Graph (a) passes the vertical line test. It doesrepresent a function.

x

x = y2(x1,y1)

(x1,y2)

Example 2. For x = y2 (to the right), note that a vertical line cantouch the graph in more than one point. These two points have thesame x-coordinate and different y-coordinates. Therefore, the relation

x = y2

is not a function.

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Function Vocabulary and Special Notation

Consider once again the function y = x + 1. Notice that if you replaced x with a number, y isquickly determined. For this reason, we can say that y depends on x. Yet another way of sayingthis is y is a function of x.

From this perspective we can view x as input to our function, and y then, as the value of thefunction. For example, for y = x + 1, when x = 5, the value of this function is 6!

To make it easier to focus on this perspective (a good thing!), a special notation has been

adopted. Instead of writing y = x +1, we writef(x) = x + 1. Notice that y has been replaced by f(x), which is read, f of x. (This is not f timesx!)f is simply the name of the function. For now, we will use mostly f, g, and h to name ourfunctions, although any letter or symbol could be used. The letter x in f(x) is simply the letter wehave chosen to represent a value in the domain of function f.

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Altogether, f(x) represents the y-coordinate in an ordered pair for any given value of x.

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Example 3a. If g(x) = 2x 1, find the value of g when x = 4.

Solution:

Example 3b. If f(x) = x2 3, find f( 5).

Solution: f(5) is the short way of writing the value of f when x = 5.

f( 5) = ( 5)2 3 = 22, so f( 5) = 22

This question requires a simple substitution, 4 for x.

g(4) = 2(4) 1 = 8 1 =7

So, when x = 4, the value of g is 7. We write this as: which is read as g of 4 is 7.g(4) = 7

2a. If f x x 2x 3, find f 5 .

b. If g x 7 3x , find g 3 .

1

c. If h x , find h 3 .2x 6

Your Turn Problem #3

a. f 5 18 b. g 3 4 c. h 3 is undefined.

Answers:

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Domains and Ranges Revisited

Earlier in this lesson, we saw that the domain of the function A = {(0,2), (1,3), (2,4)}was D = {0, 1, 2}, the set of all of x-coordinates. The range of A is R = {2, 3, 4}, the set of

all of y-coordinates. This process was fairly simple because of the finite nature of function A.

We are now going to look at the process for finding domains and ranges of functionscontaining an infinite number of ordered pairs. These functions will be described in equationform using function notation.

Finding Domains

To find the domain of a function in an equation form, we have to notice what values can legallyreplace x. For our purposes in this course, there are only two kinds of illegal replacements.

1. It is illegal to replace x with any value that would cause division by zero.

2. It is illegal to replace x with any value that will cause a square root of a negative

number (or a negative radicand for any even-indexed radical, i.e., fourth root, sixthroot, etc.).

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10Find the domain of g(Example 4b. )

4 x .

x

Since the function has division by a variable expression, you must determine if there are anyvalues of x that will cause division by zero and then exclude these values from the domain. Setthe denominator equal to zero and solve for x.

Since x = 4 causes division by zero, the number 4 must be excluded from the domain. Thedomain of g(x) is all real numbers except 4.

Interval Notation: , 4 4,

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x + 4 = 0,

x = 4.

Example 4a. Find the domain of f(x)= 3x2 + 2x 5.

Since there is no division present, division by zero cant happen. Since there are no radicalspresent, there cant be any even-indexed radicals with negative radicands.

Theres no way to make an illegal substitution for x. Therefore the domain of f is all (real) #s.

D All real numbers or, using interval notation, D : ( , )

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23x 4

Find the domain of h xExample2x 11x 6

4c.

Solution: As before, we find the values of x that cause division by zero. We set the denominatorequal to zero and solve for x. This value (or these values) will be excluded from the

domain. (2x 1)(x + 6) = 0 2x 1 = 0 or x + 6 = 0

2x = 1 or x = 6

1x or x = 6

2

1Domain: All real numbers except and 6

2

1 1or, using interval notation: , 6 6, , .

2 2

Solution: Since there is no division present, we dont have to worry about division by zero.But, we do have to ensure that the radicand, 2x+5, is nonnegative!

Set the radicand to be greater than or equal to zero, then solve for x.

2x 5 0,

2x 5, 5

x2

Find the domain ofEx f(ample 4 x) 2d 5. x .

5Those values of x that are greater than or equal to cause

2the radicand to be non-negative. Those numbers are keepers!

5 5Domain= x | x or, using interval notation: , .

2 2

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a) The domain of h(x) is all real numbers. Using Interval Notation: , .

b) Domain is all real numbers except -2 and 8. Interval Notation: , 2 2,8 8, .

Answers:

c) D x | x 2 . Interval Notation: 2, .

The End.B.R.1-01-07

26x 3b) Find the domain of f x .

x 6x 16

c) Find the domain of h x 9x 18.

Your Turn Problem #4

3 2a) Find the domain of h(x) = 7x 16x 8x 19.