1. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-1 Chapter 1 Section 8-4 An Introduction to Functions: Linear Functions, Applications, and Models

2. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-2 An Introduction to Functions: Linear Functions, Applications, and Models Relations and Functions Domain and Range Graphs of Relations Graphs of Functions Function Notation Linear Functions Modeling with Linear Functions

3. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-3 Terminology If the value of the variable y depends on the value of the variable x, the y is the dependent variable and x the independent variable. (x, y) Independent variable Dependent variable

4. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-4 Relation A relation is a set of ordered pairs. For example, the sets F = {(1, 2), (–1, 5), (4, 3)} and G = {(1, 5), (9, 0), (9, 8)} are both relations.

5. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-5 Function A function is a relation in which for each value of the first component of the ordered pairs there is exactly one value of the second component. Of the two sets, F = {(1, 2), (–1, 5), (4, 3)} and G = {(1, 5), (9, 0), (9, 8)}, only F is a function.

6. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-6 Example: Determining Independent and Dependent Variables Determine the independent and dependent variables for the following. The procedure by which someone uses a calculator that finds square roots. Solution The independent variable (input) is a nonnegative real number. The dependent variable (output) is the nonnegative square root. For example (81, 9) belongs to this function.

7. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-7 Domain and Range In a relation, the set of all values of the independent variable (x) is the domain. The set of all values of the dependent variable (y) is the range.

8. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-8 Example: Determining Domain and Range Give the domain and range of the square root function from the previous slide. Solution The domain is restricted to the nonnegative numbers: The range also is

9. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-9 Graphs of Relations The graph of a relation is the graph of its ordered pairs. The graph gives a picture of the relation, which can be used to determine its domain and range.

10. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-10 Example: Determining Domain and Range y x y x Domain {-3, -2, 0, 1} Range {-2, -1, 2, 3} Domain [-1, 1] Range [-3, 3]

11. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-11 Example: Determining Domain and Range y x y x Domain Range Domain Range

12. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-12 Agreement on Domain The domain of a relation is assumed to be all real numbers that produce real numbers when substituted for the independent variable.

13. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-13 Graphs of Functions In a function each value of x leads to only one value of y, so any vertical line drawn through the graph of a function must intersect the graph in at most one point. This is called the vertical line test for a function.

14. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-14 Vertical Line Test If a vertical line intersects the graph of a relation in more than one point, then the relation is not a function.

15. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-15 Example: Vertical Line Test y x y x Not a function – the same x-value corresponds to multiple y-values Function – each x-value corresponds to only one y-value

16. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-16 Example: Determining Whether a Relation is a Function Determine whether each equation defines a function and give the domain. Solution a) Yes, to find the domain note that The domain is

17. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-17 Example: Determining Whether a Relation is a Function Solution (continued) b) No, (16, 4) and (16, –4) both work. The domain is c) No, if x = 1, infinitely many y values satisfy it. The domain is d) Yes, the domain is since we avoid 0 in the denominator.

18. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-18 Variations of the Definition of Function 1. A function is a relation in which for each value of the first component of the ordered pairs there is exactly one value of the second component. 2. A function is a set of distinct ordered pairs in which no first component is repeated. 3. A function is a rule or correspondence that assigns exactly one range value to each domain value.

19. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-19 Function Notation When a function f is defined with a rule or an equation using x and y for the independent and dependent variables, we say “y is a function of x” to emphasize that y depends on x. We use the notation y = f (x), called function notation, to express this and read f (x) as “f of x.” For example if y = 3x + 1, we write f (x) = 3x + 1.

20. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-20 Function Notation Note that f (x) is just another name for the dependent variable y. If f (x) = 3x + 1, we find f (2) by replacing x with 2, f (2) = 3(2) + 1 = 7 Read f (2) as “f of 2” or “f at 2.”

21. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-21 Example: Using Function Notation Let f (x) = x 2 + 3x – 1. Find the following. a) f (2) b) f (0)c) f (2x) Solution a) f (2) = 2 2 + 3(2) – 1 = 9 b) f (0) = 0 + 0 – 1 = –1 c) f (2x) = (2x) 2 + 3(2x) – 1 = 4x 2 + 6x – 1

22. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-22 Linear Functions A function that can be written in the form f (x) = mx + b for real numbers m and b is a linear function.

23. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-23 Example: Graphing Linear Functions y x y x Graph each linear function. a) f (x) = –2x + 1b) f (x) = 2 Solution

24. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-24 Example: Modeling with Linear Functions A company produces DVDs of live concerts. The company pays $200 for advertising the DVDs. Each DVD costs $12 to produce and the company charges $20 per disk. a) Express the cost C as a function of x, the number of DVDs produced. b) Express the revenue R as a function of x, the number of DVDs sold. c) When will the company break-even? That is, for what value of x does revenue equal cost?

25. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-25 Example: Modeling with Linear Functions a) The fixed cost is $200 and for each DVD produced, the variable cost is $12. The cost C can be expressed as a function of x, the number of DVDs produced: C(x) = 12x + 200. b) Each DVD sells for $20, so revenue R is given by: R(x) = 20x. Solution

26. © 2008 Pearson Addison-Wesley. All rights reserved 8-4-26 Example: Modeling with Linear Functions c) The company will just break even (no profit and no loss) as long as revenue just equals cost, or C(x) = R(x). This is true whenever 12x + 200 = 20x 200 = 8x 25 = x. If 25 DVDs are produced and sold, the company will break even. Solution (continued)