1. Chapter 3 Section 5 Copyright © 2011 Pearson Education, Inc.

2. 1 1 2 2 3 3 Introduction to Functions Distinguish between independent and dependent variables. Define and identify relations and functions. Find the domain and range. Identify functions defined by graphs and equations. Use function notation. Graph linear and constant functions. 4 43.53.5 5 5 6 6

3. Slide 3.5- 3 Copyright © 2011 Pearson Education, Inc. 1 Objective Distinguish between independent and dependent variables.

4. Slide 3.5- 4 Copyright © 2011 Pearson Education, Inc. We often describe one quantity in terms of another: The amount of your paycheck if you are paid hourly depends on the number of hours you worked. The cost at the gas station depends on the number of gallons of gas you pumped into your car. The distance traveled by a car moving at a constant speed depends on the time traveled. If the value of the variable y depends on the value of the variable x, then y is the dependent variable and x is the independent variable.

5. Slide 3.5- 5 Copyright © 2011 Pearson Education, Inc. Objective 2 Define and identify relations and functions.

6. Slide 3.5- 6 Copyright © 2011 Pearson Education, Inc. (x, y) Dependent VariableIndependent Variable Relation A relation is a set of ordered pairs. 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.

7. Slide 3.5- 7 Copyright © 2011 Pearson Education, Inc. EXAMPLE 1 Determine whether each relation defines a function. a. {(0, 3), (–1, 2), (–1, 3)} No, the same x-value is paired with a different y- value. In a function, no two ordered pairs can have the same first component and different second components. b. {(5, 4), (6, 4), (7, 4)} Yes, each different x-value is paired with a y-value. This does not violate the definition of a function.

8. Slide 3.5- 8 Copyright © 2011 Pearson Education, Inc. Relations and functions can also be expressed as a correspondence or mapping from one set to another. 123123 256256 1212 2626 Function Not a function

9. Slide 3.5- 9 Copyright © 2011 Pearson Education, Inc. Objective 3 Find the domain and range.

10. Slide 3.5- 10 Copyright © 2011 Pearson Education, Inc. 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.

11. Slide 3.5- 11 Copyright © 2011 Pearson Education, Inc. EXAMPLE 2 Give the domain and range of the relation represented by the table for cellular telephone subscribers. Does it define a function? YearSubscribers 199986,047 2000109,478 2001128,375 2002140,767 2003158,722 2004182,140 Domain: {1999, 2000, 2001, 2002, 2003, 2004} Range: {86,047, 109,478, 128,375, 140,767, 158,722, 182,140} Yes; it is a function.

12. Slide 3.5- 12 Copyright © 2011 Pearson Education, Inc. EXAMPLE 3 Give the domain and range of the relation. Range Range: ( , 4] Domain: ( ,  ) The arrowheads indicate that the line extends indefinitely left and right.

13. Slide 3.5- 13 Copyright © 2011 Pearson Education, Inc. Agreement on Domain Unless specified otherwise, the domain of a relation is assumed to be all real numbers that produce real numbers when substituted for the independent variable.

14. Slide 3.5- 14 Copyright © 2011 Pearson Education, Inc. Objective 4 Identify functions defined by graphs and equations.

15. Slide 3.5- 15 Copyright © 2011 Pearson Education, Inc. Vertical Line Test If every vertical line intersects the graph of a relation in no more than one point, then the relation is a function.

16. Slide 3.5- 16 Copyright © 2011 Pearson Education, Inc. Function Not a Function

17. Slide 3.5- 17 Copyright © 2011 Pearson Education, Inc. EXAMPLE 4 Use the vertical line test to decide whether the relation shown below is a function. Yes, the relation is a function.

18. Slide 3.5- 18 Copyright © 2011 Pearson Education, Inc. EXAMPLE 5 Decide whether each equation defines y as a function of x, and give the domain. a. y = –2x + 7 y is always found by multiplying by negative two and adding 7. Each value of x corresponds to just one value of y. b. Yes, ( ,  ) Yes,

19. Slide 3.5- 19 Copyright © 2011 Pearson Education, Inc. continued c. y 4 = x d. y  4x + 2 e. No, [0,  ) No, ( ,  ) Yes The denominator would be zero and this is undefined so it is not included in the domain.

20. Slide 3.5- 20 Copyright © 2011 Pearson Education, Inc. Variations of the Definition of a 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.

21. Slide 3.5- 21 Copyright © 2011 Pearson Education, Inc. Objective 5 Use function notation.

22. Slide 3.5- 22 Copyright © 2011 Pearson Education, Inc. 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.” y = f(x) = 9x – 5 Name of the function Name of the independent variable Value of the function Defining expression

23. Slide 3.5- 23 Copyright © 2011 Pearson Education, Inc. EXAMPLE 6 Let Find the following a. f(–3)b. f(t)

24. Slide 3.5- 24 Copyright © 2011 Pearson Education, Inc. EXAMPLE 7 Let g(x) = 5x – 1. Find and simplify g(m + 2). g(x) = 5x – 1 g(m + 2) = 5(m + 2) – 1 = 5m + 10 – 1 = 5m + 9

25. Slide 3.5- 25 Copyright © 2011 Pearson Education, Inc. EXAMPLE 8 Find f(2) for each function. a.b. f = {(2, 6), (4, 2)} xf(x)f(x) –416 –24 00 24 416 f(2) = 4 f(2) = 6

26. Slide 3.5- 26 Copyright © 2011 Pearson Education, Inc. continued c. f(x) = –x 2 d. The function graphed. f(2) = –2 2 f(2) =  4 f(2) = 3

27. Slide 3.5- 27 Copyright © 2011 Pearson Education, Inc. Finding an Expression for f(x) Step 1 Solve the equation for y. Step 2 Replace y with f(x).

28. Slide 3.5- 28 Copyright © 2011 Pearson Education, Inc. EXAMPLE 9 Rewrite the equation using function notation f(x). Then find f(1) and f(a). x 2 – 4y = 3 Step 1: Solve for y.

29. Slide 3.5- 29 Copyright © 2011 Pearson Education, Inc. continued f(1) f(a)

30. Slide 3.5- 30 Copyright © 2011 Pearson Education, Inc. Objective 6 Graph linear and constant functions.

31. Slide 3.5- 31 Copyright © 2011 Pearson Education, Inc. Linear Function A function that can be defined by f(x) = ax + b for real numbers a and b is a linear function. The value of a is the slope m of the graph of the function.

32. Slide 3.5- 32 Copyright © 2011 Pearson Education, Inc. EXAMPLE 10 Graph y-intercept Slope