2. © William James Calhoun, 2001 5-5: Functions OBJECTIVES: You will be able to determine whether a given relation is a function and find the value of a function for a given element of the domain. The last sections have been about relations. Specifically, the relationship between a given (or chosen) x-variable and a resulting y-variable. This was a relationship between an independent and dependent variable. Now, a FUNCTION is a special relation in which each value of the domain only has one value for a range. The formal definition follows along with an in-depth description.

3. © William James Calhoun, 2001 5-5: Functions A function is a relation in which each element of the domain is paired with exactly one element of the range. 5.5.1 DEFINITION OF A FUNCTION Functions can be recognized by the following: (1) They are a relation between an independent and dependent variable. (2) They never have fewer domains than ranges. (3) They never have repeats in their domains. (4) Their mappings never have multiple arrows leading out from the domain. (5) Their graphs pass the vertical line test (explained later.) The first type of problems you will face will ask you determine if a relation is a function.

4. © William James Calhoun, 2001 5-5: Functions EXAMPLE 1: Determine whether each relation is a function. Explain your answer. A. {(2, 3), (3, 0), (5, 2), (-1, -2), (4, 1)} B. C. 3 4 -3 -8 0 -6 10 XY Each element of the domain only has one element of the range. There are no repeats in the domain. No repeats in the domain means this is a function. If you listed this relation out in ordered pairs, you would get: {(3, 0), (4, -6), (-3, 10), (-8, 10)} Notice there are no repeats in the domain. Also, there are no multiple arrows leaving from a single element of the domain. No repeats in the domain means this is a function. In this relation, 5 in the x pairs up with both 2 and 3 in the y. That means one domain pairs with two different ranges. There is a repeat in the domain. A repeat in the domain means this is not a function.

5. © William James Calhoun, 2001 Now we need to discuss the VERTICAL LINE TEST. If a relation passes this test, it is a function. If it fails, it is not a function. (1) To use the vertical line test, the relation must be graphed. (2) Then, draw a bunch of vertical lines through the coordinate plane. (A) If a single vertical line passes through two points of the graph, the relation fails the test. (B) If you cannot draw a single vertical line through any two points of the graph, the relation passes the test. Also, just so you know: All linear equations are functions except the vertical line functions which always look like: x = #. 5-5: Functions

6. © William James Calhoun, 2001 5-5: Functions EXAMPLE 2: Determine whether x - 4y = 12 is a function. METHOD 1 Graph it. METHOD 2 Believe me. Is it a linear equation? Yes. Is the linear equation a vertical line? No. Then believe what I said on the last slide: All non-vertical linear equations are functions. So, this is a function. Here is the graph: Try to draw a vertical line through any two points. Since you cannot draw a vertical line through any two points, it passes the test. This relation is a function.

7. © William James Calhoun, 2001 Some relations are already graphed and all you have to do is use the VLT to determine if they are functions. EXAMPLE 3: Use the vertical line test to determine if each relation is a function. A. B. C. Fails. Not a function. Passes. Function. Fails. Not a function. 5-5: Functions

8. © William James Calhoun, 2001 5-5: Functions Equations that are functions can be written in functional notation. When working with functional notation, it helps to remember that the f(x) is really just another way to write “y”. y is the dependent variable…it depends on x…therefore y is a function of x…so, y = f(x). The equation:In functional notation: y = 3x + 7f(x) = 3x + 7 Now, if we want to find the value of a function for a given domain, that is the same thing as finding out the y-value for a given x-value. So, in the above, if we want to find out what “f of 4” is - f(4) - plug in 4 everywhere you see an x. f(x) = 3x + 7 f(4) = 3(4) + 7f(4) = 12 + 7f(4) = 19 Now for the second type of problem.

9. © William James Calhoun, 2001 5-5: Functions EXAMPLE 4: If f(x) = 2x - 9, find each value. A. f(6) B. f(-2) C. f(k + 1) Plug in 6 everywhere there is an x. Plug in -2 everywhere there is an x. Plug in (k + 1) everywhere there is an x. f(x) = 2x - 9 f(6) = 2(6) - 9 = 12 - 9 = 3 f(x) = 2x - 9 f(-2) = 2(-2) - 9 = -4 - 9 = -13 f(x) = 2x - 9 f(k + 1) = 2(k + 1) - 9 = 2k + 2 - 9 = 2k - 7 You can use functional notation with non-linear functions as in the next example. The process is the same. There will just be some exponents or variables in denominators.

10. © William James Calhoun, 2001 5-5: Functions EXAMPLE 5: If h(z) = z 2 - 4z + 9, find each value. A. h(-3) B. h(5c) C. 5[h(c)] h(-3) = (-3) 2 - 4(-3) + 9 = 9 + 12 + 9 = 30 h(5c) = (5c) 2 - 4(5c) + 9 = 25c 2 - 20c + 9 No like terms to combine, so that is the answer. First find h(c), then multiply by the whole thing by 5. h(c) = (c) 2 - 4(c) + 9 = c 2 - 4c + 9 No like terms to combine, so multiply by 5. 5[h(c)] = 5c 2 - 20c + 45

11. © William James Calhoun, 2001 5-5: Functions Some things to remember about the functional notation problems: (1) If there is a letter in the parenthesis, your answer will have letters in it. (2) If there is only a number in the parenthesis, you end up with a number only. (3) Instead of ending up with a “y” or other letter on the left-hand-side, you will end up with something with parenthesis included. Do NOT try to divide your answer by the value in the parenthesis!

12. © William James Calhoun, 2001 5-5: Functions HOMEWORK Page 292 #19 - 49 odd