A function is a special kind of relation . We will begin this lesson by discussing relations.

1. A function is a special kind of relation. We will begin this lesson by discussing relations. A relation is a set of ordered pairs. This set may be finite or infinite. 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: Example:The graph of an equation in two variables is also a relation, since each point of the graph represents an ordered pair. y = x+1 y x = y2 y These graphs are relations. x x Note that the equations for these graphs specify exactly howx and y are related. Also note that since the equations generatethe graphs, it’s reasonable to refer to the equations themselves 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. Next Slide

2. Your Turn Problem #1 Determine the domain and range of: B = { (-2,2), (-2,1), (-1,0), (0,-1) }. Now we are ready to consider a special kind of relation called a function. A function is a relation in which no two ordered pairs have the same x-coordinate. 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. Solution: D = {0, 1, 2}, R = {2, 3, 4} Next Slide Example 1. Determine the domain and range of A = { (0,2), (1,3), (2,4) }. Answer: D = {-2, -1, 0 }, R = {2, 1, 0, -1} 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?

3. Example. Consider the two relations we saw in graph form (below). Are they functions? y = x+1 y y x = y2 Solution: x x The Vertical Line Test To determine if a graph represents a function we consider (or imagine) all vertical lines that intersect the graph. (Recall that vertical lines are those lines parallel to the y-axis.) If any vertical line can touch the graph at more than one point, then the graph does not represent a function. If no vertical lines are able to touch the graph at more than one point, then the graph does represent a function. The relation y = x + 1 is a function since for any value of x there will be exactly one value for y. For example, if x = 3, then y must be equal to 4 (since y = x + 1). It’s simply not possible for there to be another ordered pair in this relation that has an x-coordinate of 3 with 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 in this set (as are (4,2) and (4,-2)). Two ordered pairs with the same x-coordinate is in direct violation of our definition for a function. Instead of having to find these particular ordered pairs, there’s an easier way to see if a graph represents a function. We use the vertical line test.

4. Example 2. For x = y2 (to the right), note that a vertical line can touch the graph in more than one point. These two points have the same x-coordinate and different y-coordinates. Therefore, the relation x = y2 is not a function. Your Turn Problem #2 Use the vertical line test to determine whether the following graphs represent functions. y y (a) (b) x x Solution: y y (a) (b) Graph (a) passes the vertical line test. It does represent a function. Graph (b) fails the vertical line test. It is not a function. x x x = y2 (x1,y1) x (x1,y2)

5. Function Vocabulary and Special Notation Consider once again the function y = x + 1. Notice that if you replaced x with a number, y is quickly determined. For this reason, we can say that y depends on x. Yet another way of saying this 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 the function. For example, for y = x + 1, when x = 5, the value of this function is 6! Altogether, f(x) represents the y-coordinate in an ordered pair for any given value of x. Next Slide 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 write f(x) = x + 1. Notice that y has been replaced by f(x), which is read, “f of x”. (This is not f times x!)“f” is simply the name of the function. For now, we will use mostly f, g, and h to name our functions, although any letter or symbol could be used. The letter x in f(x) is simply the letter we have chosen to represent a value in the domain of function f.

6. Example 3a. If g(x) = 2x – 1, find the value of g when x = 4. Solution: 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”. If f(x) = x2 – 3, find f(– 5). Example 3b. 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 Your Turn Problem #3 Answers:

7. 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 functions containing an infinite number of ordered pairs. These functions will be described in equation form using function notation. Finding Domains To find the domain of a function in an equation form, we have to notice what values can legally replace x. For our purposes in this course, there are only two kinds of illegal replacements. • 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, sixth root, etc.). Next Slide • It is illegal to replace x with any value that would cause division by zero.

8. There’s no way to make an illegal substitution for x. Therefore the domain of f is all (real) #’s. x + 4 = 0, x = – 4. Example 4a. Find the domain of f(x)= – 3x2 + 2x – 5. Since x = – 4 causes division by zero, the number – 4 must be excluded from the domain. The domain of g(x) is all real numbers except – 4. Next Slide Since there is no division present, division by zero can’t happen. Since there are no radicals present, there can’t be any even-indexed radicals with negative radicands. Since the function has division by a variable expression, you must determine if there are any values of x that will cause division by zero and then exclude these values from the domain. Set the denominator equal to zero and solve for x.

9. Solution: 2x = 1 or x = – 6 Since there is no division present, we don’t have to worry about division by zero. But, we do have to ensure that the radicand, 2x+5, is nonnegative! Solution: 2x – 1 = 0 or x + 6 = 0 As before, we find the values of x that cause division by zero. We set the denominator equal to zero and solve for x. This value (or these values) will be excluded from the domain. (2x – 1)(x + 6) = 0 Set the radicand to be greater than or equal to zero, then solve for x.

10. Your Turn Problem #4 Answers: The End. B.R. 1-01-07