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Chapter 3.2

Chapter 3.2. Introduction to Functions. Objectives. I will define relation, domain, and range. I will identify functions. I will use the vertical line test for functions. I will find the domain and range of a function. I will use function notation. Relation, Domain, and Range.

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Chapter 3.2

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  1. Chapter 3.2 Introduction to Functions

  2. Objectives • I will define relation, domain, and range. • I will identify functions. • I will use the vertical line test for functions. • I will find the domain and range of a function. • I will use function notation.

  3. Relation, Domain, and Range • Relation – a set of ordered pairs. • Domain of the relation is the set of all first components of the ordered pairs. • Range of the relation is the set of all second components of the ordered pairs.

  4. Example 1 • Determine the domain and range of each relation. • {(2,3), (2,4), (0,-1), (3, -1)} Domain – set of all first components {2, 0, 3} Range – Set of all second components {3,4, -1}

  5. Function • A function is a relation in which each first component in the ordered pairs corresponds to exactly one second component. • A function is a special type of relation, so all functions are relations, but not all relations are functions.

  6. Example 1, Continued • Find the domain and range of each relation. • {(2,3), (2,4), (0,-1), (3, -1)} • Input: Output: Cities Population (in thousands) Erie 109 Miami 359 Escondido 117 Waco 104 Gary 182

  7. Function • A function is a relation in which each first component in the ordered pairs corresponds to exactly one second component. • A function is a special type of relation, so all functions are relations, but not all relations are functions.

  8. Give it a try!  • Determine the domain and range of each relation. • A. {(1,6), (2,8), (0,3), (0,-2)} • Domain – {1,2,0} • Range – {6,8,3, -2}

  9. Give it a try!  • Determine the domain and range of each relation. Input: Output: States # of representatives Arkansas 4 Texas 30 Oklahoma 6 South 6 Domain: {Arkansas, Texas, Oklahoma, South Carolina} Range: {4, 30, 6}

  10. Example 2: Functions • Which of the following relations are also functions? • {(-2,5), (2,7), (-3,5), (9,9)} Yes! Each x – value is assigned to only one y-value, so this set of ordered pairs is a function.

  11. Example 2: Functions • Which of the following relations are also functions? • {(-3,2), (0,3), (1,5), (0, -2), (6, 6)} No! The x-value 0 is assigned to two y-values. This relation is not a function.

  12. Example 2: Functions • Which of the following relations are also functions? Input: Correspondence: Output: People Each person’s the set in a certain age of non city negative integers This relation is a function because although two different People may have the same age, each person has only one age!

  13. Give it a try!  • Determine whether each relation is also a function. • {(-3,7), (1,7), (2,2)} • Input: Correspondence: Output: People County Counties of in a certain that that state state a person lives in

  14. Example 3: • Is the relation y = 2x + 1 also a function? • The relation y= 2x+1 is a function if each x-value corresponds to just one y-value. Fore each x-value substituted in the equation y=2x+1, the multiplication and addition performed on each gives a single result, so only one y-value will be associate with each x-value. Thus, y = 2x +1 is a function.

  15. Example 4: • Is the relation x= y2 also a function? • Substitute y = 3 and y = -3 into the equation. • You will get, x = 9 for both values of y. Thus, x= y2 is not a function.

  16. Give it a try!  • Determine whether the following relations are functions. • A. x = y2 + 1 • B. y = 3x + 2

  17. Vertical Line Test • Vertical Line Test – If no vertical line can be drawn so that it intersects the graph more than once, the graph is the graph of a function.

  18. Example 5: - Which of the following graphs are graphs of functions? A. B. C. D. E. F.

  19. Concept Check • Determine which equations represent functions. Explain your answer. • a. y = 14 • b. x = - 5 • c. x + y = 6

  20. Example 6 • Find the domain and range of each relation. Determine whether the relation is also a function. • Look at graphs on page 146.

  21. Naming Functions • Many times letters such as f, g, and h are used to name functions. To denote that y is a function of x, we can write: y = f(x) This means that y is a function of x or that y depends on x. For this reason, y is called the dependent variable and x is called the independent variable. The notation f(x) is read “f of x” and is called function notation.

  22. Naming Functions Continued • Use function notation with the function y = 4x+3. • f(x) = 4x + 3 • The notation f(1) means to replace x with 1 and find the resulting y or function value. f(x) = 4x + 3 f(1) = 4x + 3 f(1) = 4(1) + 3 f(x) = 7

  23. Find f(2), f(0), and f(-1) f(x) = 4x +3 f(x) = 4x+3 f(x) = 4x + 3 f(2) = 4(2)+3 f(0) = 4(0) + 3 f(-1) = 4(-1) + 3 = 8 + 3 = 0 + 3 = -4 + 3 = 11 = 3 = -1 Ordered pairs: (2, 11), (0,3), (-1,1)

  24. Helpful Hint • Note that f(x) is a special symbol in mathematics used to denote a function. The symbol f(x) is read “f of x” It does not mean f times x.

  25. Example 7 • If f(x) = 7 x2 – 3x + 1 and g(x) = 3x – 2, find the following. • a. f(1) b. g(1) c. f(-2) d. g(0) • Step 1: Substitute f (1) = 7(1)2 – 3(1) + 1 Step 2: Simplify f (1) = 7 – 3 + 1 f (1) = 5

  26. Example 7 Continued g(x) = 3x – 2 g(0) = 3(0) -2 g(0) = -2 • g(x) = 3x – 2 g(1) = 3(1) -2 g(1) = 1 f (-2) = 7(-2)2 – 3(-2) + 1 f (-2) = 28 + 6 + 1 f(-2) = 35

  27. Give it a try!  • If g(x) = 4x + 5 and f(x) = 3x2 – x + 2, find: a. g(0) b. g(-5) c. f(2) d. f(-1)

  28. Concept Check • Suppose y = f(x) and we are told that f(3) = 9. Which is not true? • A. When x = 3, y = 9 • B. A possible function is f(x) = x2 • C. A point on the graph of the function is (3, 9) • D. A possible function is f(x) = 2x + 4

  29. Example 8 – Refer to pg. 149

  30. Give it a try!  • Use the graphs on page 149 to find : • A. f(2) • B. g(3) • C. Find all x-values such that f(x) = 2 • D. Find all x-values such that g(x) = 3

  31. Example 10 • Use the function f(x) = 1.558x – 3092 to predict the amount of money that will be spent by Pharmaceutical Manufacturers Association on research and development in 2010. • To predict we will use f(x) = 1.558x – 3092 and find f(2010)

  32. Example 10 - Continued • f(x) = 1.558x – 3092 • f(2010) = 1.558 (2010) – 3092 • = 39.58 • We predict that in the year 2010, $39.58 billion dollars will be spend on research and development.

  33. Give it a try!  • Use f(x) = 1.558 x – 3092 to approximate the money spend in 2009.

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