Appendix A.6 Functions

Transcription

1 A. Functions 539 RELATIONS: DOMAIN AND RANGE Appendi A. Functions A relation is a set of ordered pairs. A relation can be a simple set of just a few ordered pairs, such as {(0, ), (1, 3), (, )}, or it can be infinite, such as the set of all points ton a line or a curve. The domain of a relation is the set of all (possible) -values, and the range is the set of all -values. (, ) (0, ) (3, ) (-1, ) (-5, -) - (3, -) - - (-1, -) (3, 1) {0, 3, } [-5, 3] (-, )* Range: {,, } Range: [-, ] Range: [1, ) The domain and range are sets of numbers and we can represent each in one of several was. In this appendi, ou might come across an of these solution sets: Words Interval Notation Set Builder Notation Smbolicall all real numbers (-, ) { is a real number } is between -5 and 3, inclusive is between -5 and 3, eclusive is greater than or equal to -1 [-5, 3] { -5 3 } -5 3 (-5, 3) { -5 < < 3 } -5 < < 3 [-1, ) { -1 } -1 is less than (-, ) { < } < is not 7 (-, 7) ( 7, ) { 7 } { 7 } A. Functions 539 Robert H. Prior, 01 Coping is prohibited.

2 50 A. Functions Some domains, called restricted domains, are truncated for one reason or another. For eample, the equation p p = 0.1t + 10 is a linear equation that represents the price, p dollars, of a cell phone plan, based on the number of tet messages sent, t, (up to 100). (0, 10) (100, 0) In this case, t cannot be negative, and values of t that are more than 100 do not add to the price (and are not considered here). So,... t [0, 100] and Range: [10, 0] You Tr It 1 For each relation, identif both the domain and the range. a) (, ) b) c) (, ) (-, 1) (8, 1) (1, -) (, -) Range: Range: Range: d) (3, 5) e) f) (-, 7) (-5, 0) (, 1) Range: Range: Range: A. Functions 50 Robert H. Prior, 01 Coping is prohibited.

3 A. Functions 51 FUNCTION DEFINITION A function is a relation such that, for ever there is onl one. A more formal definition of function is: for ever element, i, of the domain of f, there is eactl one element, i, in the range of f to which i is assigned. The vertical line test is a wa to visuall identif whether the graph of a relation is a function. If an vertical line can cross the graph in more than one place, then the graph is not a function; otherwise, it is a function. The basic idea behind the vertical line test is that we are visuall checking each and ever -value in the domain, making sure that it corresponds to onl one -value (0, 0) This relation is a function. This relation is not a function. This relation is not a function. ONE-TO-ONE FUNCTIONS A function is one-to-one if, for ever there is onl one that corresponds to it. A function that is not one-to-one is called man-to-one. For eample, a slanted line is one-to-one, but a parabola is not. Below are three tables showing the correspondence between domain and range values of a line, a parabola, and a sine wave. Line: = Parabola: = Sine wave: = sin() ! 0! ! 5! 1 13! This function is one-to-one. This function is man-to-one. This function is man-to-one. A. Functions 51 Robert H. Prior, 01 Coping is prohibited.

4 5 A. Functions The horizontal line test is a wa to visuall identif whether the graph of a function is one-to-one or man-to-one. If an horizontal line can cross the graph in more than one place, then the function is man-to-one; otherwise, it is one-to-one. = = = sin (0, 0) -! -! 1! -1! 3!! - This function is one-to-one. This function is man-to-one. This function is man-to-one. If a function has an turn-around points (such as the verte of a parabola), then the function is manto-one. The bottom line is, a function is one-to-one if it is either... all increasing: or all decreasing: This graph continuall rises from its verte. This graph continuall decreases (from left to right). A. Functions 5 Robert H. Prior, 01 Coping is prohibited.

5 A. Functions 53 The Algebra of Functions RESTRICTIONS ON THE DOMAIN Some functions have natural restrictions. In particular, (i) a denominator can never be 0 (The numerator is unaffected b this restriction.) (ii) the radicand of a square root can never be negative. We do not consider the option of imaginar numbers because the - and -aes are real number aes. Eample 1: Identif the domain of each function. a) = + 1 b) = + 3 c) = + 1 Procedure: Identif whether the function has a natural domain restriction or otherwise. Answer: a) The radicand cannot be negative: b) The denominator cannot be zero: c) For polnomial functions, the domain is all real numbers (unless it has a given domain restriction) Note: For some functions, the range is not intuitive and is often found onl after the function has been graphed. You Tr It For each function, identif the domain. (You are not asked to find the range.) a) = 9 3 b) = c) = 5 A. Functions 53 Robert H. Prior, 01 Coping is prohibited.

6 5 A. Functions FUNCTIONAL VALUES Definitions: 1. The smbol f() means the function of and is called the function notation.. In function notation, the value within the parentheses is called the argument of the function. 3. The variable in an argument can be replaced b a number, called a replacement value. A replacement value must be an element (member) of the domain. Caution: The parentheses used in the function notation, f(), do not mean multipl. When a replacement value is used, each and ever occurrence of the variable gets replaced b this value. For eample, if f() = 5 +, then a) f(3) = (3) 5(3) + b) f(-) = (-) 5(-) + = (9) 15 + = (1) = = = 7 = 5 You Tr It 3 Given f() = , find the following: a) f () b) f ( 1 3 ) c) f (-) You Tr It Given g() = 1, find the following: a) g(0) b) g(5) c) g(-) You Tr It 5 Given h() = - + 5, find the following: a) h() b) h( 5 ) d) h(-3) A. Functions 5 Robert H. Prior, 01 Coping is prohibited.

7 A. Functions 55 We can also replace the argument with other variable arguments. For eample, if Eample : Given g() = 3, find the following: a) g(a) b) g(a) c) g(a + 1) Procedure: For each, replace with the requested argument. Answer: a) Replace each with a: c) Replace each with the quantit (a + 1): g(a) = a 3a g(a + 1) = (a + 1) 3(a + 1) = a + a + 1 3a 3 b) Replace each with a; simplif. g(a) = (a) 3(a) g(a + 1) = a a g(a) = a a You Tr It Given h() = 1, find the following: a) h(a ) b) h(a ) COMPOSITE FUNCTIONS We can also replace an argument in one function, f(), with another function, g(). This is called the composition of two functions. The result is another function. f() composed with g() is written g() composed with f() is written f g () = f [g()] g f () = g [f()] It is even possible to consider a function composed with itself: f f () = f [f()] A. Functions 55 Robert H. Prior, 01 Coping is prohibited.

8 5 A. Functions Eample 3: Given f() = 3 1 and g() = +, find the following: a) f g () b) g f () c) f f () Procedure: For each, the replacement value is a full function. Answer: a) f g () f [g()] = f( + ) = 3( + ) 1 = = b) g f () g [f()] = g(3 1) = (3 1) + = = c) f f () f [g()] = f(3 1) = 3(3 1) 1 = = 9 You Tr It 7 Given f() = 3 and g() =, find the following: a) f g () b) g f () You Tr It 8 Given f() = 3 1 and g() = 3 + 3, find the following: a) f g () b) g f () A. Functions 5 Robert H. Prior, 01 Coping is prohibited.

9 A. Functions 57 EVEN AND ODD FUNCTIONS Some functions, not all, have the propert of being either odd or even based on these criteria: Function Tpe Algebraic Propert Graph Smmetr Even f(-) = f() about the -ais Odd f(-) = - f() about the origin Eamples of the graphs of even and odd functions: Even Even Odd Odd f() = g() = + 3 h() = 3 k() = 3 (-3, 9) (3, 9) (, 8) (3, 15) (-, 3) (, 3) (-1, 1) (1, 1) (1, 1) (-, 0) (-1, 0) (1, 0) (-1, -1) (, 0) (-, -8) (-3, -15) At right is a function, p(), that is neither even nor odd; it has smmetr about the point (1, 0) but not about the origin or the -ais You might have noticed the eamples of even functions have onl even powers. Certainl f() = has an even power. So, too, does g(), which can be written g() = Similarl, odd functions have onl odd powers, such as h() = 3 and k() = 3 1. Because p() = has a mi of odd and even powers, it is neither odd nor even. p() = (0, ) (3, ) (, -) (-, -18) In-Class Eample : What tpe of function is it that has smmetr about the -ais? (Hint: Draw a diagram.) A. Functions 57 Robert H. Prior, 01 Coping is prohibited.

10 58 A. Functions Here, again, are the algebraic properties of even and odd functions: Eample 5: A function, f(), is even if f(-) = f() for all domain values. A function, f(), is odd if f(-) = - f() for all domain values. Determine whether the function is even, odd, or neither. a) f() = 3 b) g() = + 8 c) h() = ( 3 + ) Procedure: Find f(-) and simplif. If the resulting function is the same as f(), then the function is even; if it is the opposite of f(), then it is odd; it might also be neither of these options. Answer: a) f() = 3 Replace each in the function with - and simplif. f(-) = (-) 3 (-) Evaluate each term. = Because the lead term is negative, factor out -1. = -1( 3 ) In the parentheses is the original f(), so... f(-) = - f() f() is an odd function. b) g() = + 8 Replace each in the function with - and simplif. g(-) = (-) + (-) 8 = 8 Evaluate each term. The lead term is the same as it is for g(), so g is not odd; however, the middle terms of g() and g(-) are different, so... g(-) - g() g() is neither even nor odd. c) h() = ( 3 + ) h(-) = [(-) 3 + (-)] Replace each in the function with - and simplif. Evaluate each term inside the brackets. = [- 3 ] Inside the brackets, the lead term is negative, and we can factor out -1 from those two terms. = [-1( 3 + ) ] Use the distributive rule of eponents, (ab) n = a n b n. = (-1) ( 3 + ) Simplif: (-1) = +1. = ( 3 + ) This is the same as h(), so... h(-) = h() h() is an even function. A. Functions 58 Robert H. Prior, 01 Coping is prohibited.

11 A. Functions 59 In part c), h() = ( 3 + ) is a bit confusing. It appears as though h() might be an odd function because of the odd power on each. However, because the quantit is being squared, it puts a twist on the even/odd notion of eponents, so mabe it s neither. In part c), we learn that h() is actuall an even function. Here is another technique that could be used: In-Class Eample : Show that h() = ( 3 + ) is an even function b first squaring out the quantit. You Tr It Answers YTI 1: a) - 8 b) c) Range: - Range: - Range: d) e) -5 c) - Range: 5 Range: 0 Range: 1 7 YTI : a) 1 b) c) 5 YTI 3: a) 9 b) c) 3 YTI : a) -1 b) -7 c) 0 YTI 5: a) -11 b) - 5 c) - YTI : a) h(a ) = a a 1 b) h(a ) = a a a 3 YTI 7: a) f g () = 3 3 b) g f () = YTI 8: a) f g () = b) g f () = A. Functions 59 Robert H. Prior, 01 Coping is prohibited.

12 550 A. Functions Focus Eercises Given the graph of f(), determine its domain and range. 1.. (3, 5) (-3, ) 3.. (-5, 0) (, -1) Identif the domain of the function. Keep in mind an possible restriction the domain ma have. 5. f() =. h() = g() = 3 8. f() = 8 9. k() = 10. f() = + 1 A. Functions 550 Robert H. Prior, 01 Coping is prohibited.

13 A. Functions 551 Find the requested functional value for f() = + 5 and g() = f() 1. f(10) 13. f (- 1 ) 1. f(-) 15. g(0) 1. g(5) 17. g(-1) 18. g(-3) Given f() = 3 and g() = +, find the following. 19. f(3w) 0. g(c) 1. g( ). f( ) A. Functions 551 Robert H. Prior, 01 Coping is prohibited.

14 55 A. Functions 3. f g (). g f () For each, use the algebraic properties to determine whether the function is even, odd, or neither. 5. f() = - 3. g() = h() = 3 8. f() = g() = 30. h() = ( ) A. Functions 55 Robert H. Prior, 01 Coping is prohibited.