. Piecewise Functions YOU WILL NEED graph paper graphing calculator GOAL Understand, interpret, and graph situations that are described b piecewise functions. LEARN ABOUT the Math A cit parking lot uses the following rules to calculate parking fees: A flat rate of $5. for an amount of time up to and including the first hour A flat rate of $.5 for an amount of time over h and up to and including h A flat rate of $3 plus $3 per hour for each hour after h? How can ou describe the function for parking fees in terms of the number of hours parked? EXAMPLE Representing the problem using a graphical model Use a graphical model to represent the function for parking fees. Solution Time (h) Parking Fee ($).5 5..5 5.. 5..5.5.5.5..5.5.5 3... 9. Create a table of values.. Piecewise Functions
. Plot the points in the table of values. Use a solid dot to include a value in an interval. Use an open dot to eclude a value from an interval. Cost ($) 3 Time (h) 5 The domain of this piecewise function is $. The function is linear over the domain, but it is discontinuous at 5,, and. There is a solid dot at (, ) and an open dot at (, 5) because the parking fee at h is $.. There is a closed dot at (, 5) and an open dot at (,.5) because the parking fee at h is $5.. There is a closed dot at (,.5) and an open dot at (, 3) because the parking fee at h is $.5. The last part of the graph continues in a straight line since the rate of change is constant after h. piecewise function a function defined b using two or more rules on two or more intervals; as a result, the graph is made up of two or more pieces of similar or different functions Each part of a piecewise function can be described using a specific equation for the interval of the domain. EXAMPLE Representing the problem using an algebraic model Use an algebraic model to represent the function for parking fees. Solution 5 5 5 3 5.5 5 3 3 if 5 if, # if, # if. Write the relation for each rule. Chapter 7
f () 5 μ, if 5 5, if, #.5, if, # 3 3, if. Combine the relations into a piecewise function. The domain of the function is $. The function is discontinuous at 5,, and because there is a break in the function at each of these points. Reflecting A. How do ou sketch the graph of a piecewise function? B. How do ou create the algebraic representation of a piecewise function? C. How do ou determine from a graph or from the algebraic representation of a piecewise function if there are an discontinuities? APPLY the Math EXAMPLE 3 Representing a piecewise function using a graph Graph the following piecewise function.,if, f () 5 e 3, if $ Solution Create a table of values. f () 5 f () 5 3 f() f() 7 3 9 5 3 5 From the equations given, the graph consists of part of a parabola that opens up and a line that rises from left to right. Both tables include 5 since this is where the description of the function changes.. Piecewise Functions
. = f() Plot the points, and draw the graph. A solid dot is placed at (, 7) since 5 is included with f() 5 3. An open dot is placed at (, ) since 5 is ecluded from f() 5. f () is discontinuous at 5. EXAMPLE = f () Representing a piecewise function using an algebraic model Determine the algebraic representation of the following piecewise function. Solution f () 5,if #, if. The graph is made up of two pieces. One piece is part of the reciprocal function defined b 5 when #. The other piece is a horizontal line defined b 5 when.. The solid dot indicates that point Q, R belongs with the reciprocal function. Chapter 9
EXAMPLE 5 Reasoning about the continuit of a piecewise function Is this function continuous at the points where it is pieced together? Eplain., if # g() 5, if,, 3, if $ 3 Solution The function is continuous at the points where it is pieced together if the functions being joined have the same -values at these points. Calculate the values of the function at 5 using the relevant equations: 5 5 5 5 5 () 5 Calculate the values of the function at 5 3 using the relevant equations: 5 5 (3) 5 7 5 5 3 55 The function is discontinuous, since there is a break in the graph at 5 3. The graph is made up of three pieces. One piece is part of an increasing line defined b 5 when #. The second piece is an increasing line defined b 5 when,, 3. The third piece is part of a parabola that opens down, defined b 5 when $ 3. The two -values are the same, so the two linear pieces join each other at 5. The two -values are different, so the second linear piece does not join with the parabola at 5 3. Tech Support For help using a graphing calculator to graph a piecewise function, see Technical Appendi, T-. Verif b graphing. 5. Piecewise Functions
. In Summar Ke Ideas Some functions are represented b two or more pieces. These functions are called piecewise functions. Each piece of a piecewise function is defined for a specific interval in the domain of the function. Need to Know To graph a piecewise function, graph each piece of the function over the given interval. A piecewise function can be either continuous or not. If all the pieces of the function join together at the endpoints of the given intervals, then the function is continuous. Otherwise, it is discontinuous at these values of the domain. CHECK Your Understanding. Graph each piecewise function., if,,if # a) f () 5 e d) f () 5 e 3,if $, if., if, f () 5 e e), if $, if # c) f () 5 e f) f () 5,if,,if.,if $. State whether each function in question is continuous or not. If not, state where it is discontinuous. 3. Write the algebraic representation of each piecewise function, using function notation. a) = f () f () 5 e!,if,,if $ = f (). State the domain of each piecewise function in question 3, and comment on the continuit of the function. Chapter 5
PRACTISING 5. Graph the following piecewise functions. Determine whether each K function is continuous or not, and state the domain and range of the function., if, a) f () 5 e c) 3, if $ f () 5 e, if,, if $, if, f () 5 e,if # d) f () 5, if # # 3,if. 5, if. 3. Graham s long-distance telephone plan includes the first 5 min per A month in the $5. monthl charge. For each minute after 5 min, Graham is charged $.. Write a function that describes Graham s total long-distance charge in terms of the number of long distance minutes he uses in a month. 7. Man income ta sstems are calculated using a tiered method. Under a certain ta law, the first $ of earnings are subject to a 35% ta; earnings greater than $ and up to $5 are subject to a 5% ta. An earnings greater than $5 are taed at 55%. Write a piecewise function that models this situation.. Find the value of k that makes the following function continuous. T Graph the function. f () 5 e k,if,, if $ 9. The fish population, in thousands, in a lake at an time,, in ears is modelled b the following function:,if # # f () 5 e, if. This function describes a sudden change in the population at time 5, due to a chemical spill. a) Sketch the graph of the piecewise function. Describe the continuit of the function. c) How man fish were killed b the chemical spill? d) At what time did the population recover to the level it was before the chemical spill? e) Describe other events relating to fish populations in a lake that might result in piecewise functions. 5. Piecewise Functions
. Create a flow chart that describes how to plot a piecewise function with two pieces. In our flow chart, include how to determine where the function is continuous... An absolute value function can be written as a piecewise function that C involves two linear functions. Write the function f () 5 3 as a piecewise function, and graph our piecewise function to check it.. The demand for a new CD is described b D(p) 5 p,if, p # 5, if p. 5 where D is the demand for the CD at price p, in dollars. Determine where the demand function is discontinuous and continuous. Etending 3. Consider a function, f (), that takes an element of its domain and rounds it down to the nearest. Thus, f (5.) 5, while f (.7) 5 and f (3) 5 3. Draw the graph, and write the piecewise function. You ma limit the domain to P[, 5). Wh do ou think graphs like this one are often referred to as step functions?. Eplain wh there is no value of k that will make the following function continuous. 5, if, f () 5 k, if # # 3,if. 3 5. The greatest integer function is a step function that is written as f () 5 3, where f () is the greatest integer less than or equal to. In other words, the greatest integer function rounds an number down to the nearest integer. For eample, the greatest integer less than or equal to the number [5.3] is 5, while the greatest integer less than or equal to the number 35.3 is. Sketch the graph of f () 5 3.. a) Create our own piecewise function using three different transformed parent functions. Graph the function ou created in part a). c) Is the function ou created continuous or not? Eplain. d) If the function ou created is not continuous, change the interval or adjust the transformations used as required to change it to a continuous function. Chapter 53
.7 Eploring Operations with Functions GOAL Eplore the properties of the sum, difference, and product of two functions. A popular coffee house sells iced cappuccino for $ and hot cappuccino for $3. The manager would like to predict the relationship between the outside temperature and the total dail revenue from each tpe of cappuccino sold. The manager discovers that ever C increase in temperature leads to an increase in the sales of cold drinks b three cups per da and to a decrease in the sales of hot drinks b five cups per da. The function f () 5 3 can be used to model the number of iced cappuccinos sold. Number of iced cappuccinos sold 5 5 f() = 3+ 3 5 Outside temperature (C) The function g() 55 can be used to model the number of hot cappuccinos sold. Number of hot cappuccinos sold g() = 5 + 3 Outside temperature (C) In both functions, represents the dail average outside temperature. In the first function, f () represents the dail average number of iced cappuccinos sold. In the second function, g() represents the dail average number of hot cappuccinos sold.? How does the outside temperature affect the dail revenue from cappuccinos sold? A. Make a table of values for each function, with the temperature in intervals of 5, from to. B. What does h() 5 f () g() represent? 5.7 Eploring Operations with Functions
C. Simplif h() 5 (3 ) (5 )..7 D. Make a table of values for the function in part C, with the temperature in intervals of 5, from to. How do the values compare with the values in each table ou made in part A? How do the domains of f (), g(), and h() compare? E. What does h() 5 f () g() represent? F. Simplif h() 5 (3 ) (5 ). G. Make a table of values for the function in part F, with the temperature in intervals of 5, from to. How do the values compare with the values in each table ou made in part A? How do the domains of f (), g(), and h() compare? H. What does R() 5 f () 3g() represent? I. Simplif R() 5 (3 ) 3(5 ). J. Make a table of values for the function in part I, with the temperature in intervals of 5, from to. How do the values compare with the values in each table ou made in part A? How do the domains of f (), g(), and R() compare? K. How does temperature affect the dail revenue from cappuccinos sold? Reflecting L. Eplain how the sum function, h(), would be different if a) both f () and g() were increasing functions both f () and g() were decreasing functions M. What does the function k() 5 g() f () represent? Is its graph identical to the graph of h() 5 f () g()? Eplain. N. Determine the function h() 5 f () 3 g(). Does this function have an meaning in the contet of the dail revenue from cappuccinos sold? Eplain how the table of values for this function is related to the tables of values ou made in part A. O. If ou are given the graphs of two functions, eplain how ou could create a graph that represents a) the sum of the two functions the difference between the two functions c) the product of the two functions Chapter 55
In Summar Ke Idea If two functions have domains that overlap, the can be added, subtracted, or multiplied to create a new function on that shared domain. Need to Know Two functions can be added, subtracted, or multiplied graphicall b adding, subtracting, or multipling the values of the dependent variable for identical values of the independent variable. Two functions can be added, subtracted, or multiplied algebraicall b adding, subtracting, or multipling the epressions for the dependent variable and then simplifing. The properties of each original function have an impact on the properties of the new function. FURTHER Your Understanding. Let f 5 5(, ), (, ), (, 3), (3, 5), (, ) and g 5 5(, ), (, ), (, ), (, ), (, ), (, ). Determine: a) f g f g c) g f d) fg. Use the graphs of f and g to sketch the graphs of f g. a) f f g g 3. Use the graphs of f and g to sketch the graphs of f g. a) f g f g 5.7 Eploring Operations with Functions
.7. Use the graphs of f and g to sketch the graphs of fg. a) f g g f 5. Determine the equation of each new function, and then sketch its graph. a) h() 5 f () g(), where f () 5 and g() 5 p() 5 m() n(), where m() 5 and n() 57 c) r() 5 s() t(), where s() 5 and t() 5 d) a() 5 b() 3 c(), where b() 5 and c() 5. a) Using the graphs ou sketched in question 5, compare and contrast the relationship between the properties of the original functions and the properties of the new function. Which properties of the original functions determined the properties of the new function? 7. Let f () 5 3 and g() 5 5, PR. a) Sketch each graph on the same set of aes. Make a table of values for 3 # # 3, and determine the corresponding values of h() 5 f () 3 g(). c) Use the table to sketch h() on the same aes. Describe the shape of the graph. d) Determine the algebraic model for h(). What is its degree? e) What is the domain of h()? How does this domain compare with the domains of f () and g()?. Let f () 5 and g() 5, PR. a) Sketch each graph on the same set of aes. Make a table of values for 3 # # 3, and determine the corresponding values of h() 5 f () 3 g(). c) Use the table to sketch h() on the same aes. Describe the shape of the graph. d) Determine the algebraic model for h(). What is its degree? e) What is the domain of h()? Chapter 57
Chapter Review Stud Aid See Lesson., Eamples, 3, and. Tr Chapter Review Questions 7,, and 9. FREQUENTLY ASKED Questions Q: In what order are transformations performed on a function? Function Perform all stretches/compressions and reflections. Perform all translations. A: All stretches/compressions (vertical and horizontal) and reflections can be applied at the same time b multipling the - and -coordinates on the parent function b the appropriate factors. Both vertical and horizontal translations can then be applied b adding or subtracting the relevant numbers to the - and -coordinates of the points. Stud Aid See Lesson.5, Eamples,, and 3. Tr Chapter Review Questions to 3. Q: How do ou find the inverse relation of a function? A: You can find the inverse relation of a function numericall, graphicall, or algebraicall. To find the inverse relation of a function numericall, using a table of values, switch the values for the independent and dependent variables. f() f (, ) (, ) To find the inverse relation graphicall, reflect the graph of the function in the line 5. This is accomplished b switching the - and -coordinates in each ordered pair. To find the algebraic representation of the inverse relation, interchange the positions of the - and -variables in the function and solve for. 5 Chapter Review
Q: Is an inverse of a function alwas a function? A: No; if an element in the domain of the original function corresponds to more than one number in the range, then the inverse relation is not a function. Q: What is a piecewise function? A: A piecewise function is a function that has two or more function rules for different parts of its domain.,if, For eample, the function defined b f () 5 e, if $ consists of two pieces. The first equation defines half of a parabola that opens down when,. The second equation defines a decreasing line with a -intercept of when $. The graph confirms this. Stud Aid Chapter Review See Lesson.5, Eamples,, and 3. Tr Chapter Review Questions to 3. Stud Aid See Lesson., Eamples,, 3, and. Tr Chapter Review Questions to 7. Q: If ou are given the graphs or equations of two functions, how can ou create a new function? A: You can create a new function b adding, subtracting, or multipling the two given functions. This can be done graphicall b adding, subtracting, or multipling the -coordinates in each pair of ordered pairs that have identical -coordinates. This can be done algebraicall b adding, subtracting, or multipling the epressions for the dependent variable and then simplifing. Stud Aid See Lesson.7. Tr Chapter Review Questions to. Chapter 59
PRACTICE Questions Lesson.. Determine whether each relation is a function, and state its domain and range. a) c) d). A cell phone compan charges a monthl fee of $3, plus $. per minute of call time. a) Write the monthl cost function, C(t), where t is the amount of time in minutes of call time during a month. Find the domain and range of C. Lesson...5 5 3. Graph f () 5 3, and state the domain and range.. Describe this interval using absolute value notation. 3 3 3 5.5. Lesson.3 5. For each pair of functions, give a characteristic that the two functions have in common and a characteristic that distinguishes them. a) f () 5 and g() 5 sin f () 5 and g() 5 c) f () 5 and g() 5 d) f () 5 and g() 5. Identif the intervals of increase/decrease, the smmetr, and the domain and range of each function. a) c) Lesson. f () 5 3 f () 5 f () 5 7. For each of the following equations, state the parent function and the transformations that were applied. Graph the transformed function. a) 5 5.5"3( 7) c) 5 sin (3), # # 3 d) 5 3. The graph of 5 is horizontall stretched b a factor of, reflected in the -ais, and shifted 3 units down. Find the equation that results from the transformation, and graph it. 9. (, ) is a point on the graph of 5 f (). Find the corresponding point on the graph of each of the following functions. a) 5f () 5 f (( 9)) 7 c) 5 f ( ) d) 5.3f (5( 3)) e) 5 f ( ) f) 5f (( )) Chapter Review
Lesson.5. For each point on a function, state the corresponding point on the inverse relation. a) (, ) d) f (5) 5 7 (, 9) e) g() 53 c) (, 7) f) h() 5. Given the domain and range of a function, state the domain and range of the inverse relation. a) D 5 5PR, R 5 5 PR,, D 5 5PR $ 7, R 5 5 PR,. Graph each function and its inverse relation on the same set of aes. Determine whether the inverse relation is a function. a) f () 5 g() 5 3. Find the inverse of each function. a) f () 5 g() 5 3 Lesson.. Graph the following function. Determine whether it is discontinuous and, if so, where. State the domain and the range of the function., if, f () 5 e, if $ 5. Write the algebraic representation for the following piecewise function, using function notation. 3 5. If f () 5 e, if, 3,if $ is f () continuous at 5? Eplain. 7. A telephone compan charges $3 a month and gives the customer free call minutes. After the min, the compan charges $.3 a minute. a) Write the function using function notation. Find the cost for talking 35 min in a month. c) Find the cost for talking min in a month. Lesson.7. Given f 5 5(, ), (, 3), (, 7), (5, ) and g 5 5(, ), (, ), (, 3), (, ), (, 9), determine the following. a) f () g() f () g() c) 3 f ()3g() 9. Given f () 5, # # 3 and g() 5, 3 # # 5, graph the following. a) f d) f g g e) fg c) f g. f () 5 and g() 5. Match the answer with the operation. Answer: Operation: a) 3 3 A f () g() B f () g() c) 3 C g() f () d) D f () 3 g(). f () 5 3 and g() 5, a) Complete the table. 3 f() g() (f g)() Chapter Review Use the table to graph f () and g() on the same aes. c) Graph (f g)() on the same aes as part. d) State the equation of (f g)(). e) Verif the equation of (f g)() using two of the ordered pairs in the table. Chapter
Chapter Self-Test 5 3 5 3. Consider the graph of the relation shown. a) Is the relation a function? Eplain. State the domain and range.. Given the following information about a function: D 5 5PR R 5 5 PR $ decreasing on the interval (`, ) increasing on the interval (, `) a) What is a possible parent function? Draw a possible graph of the function. c) Describe the transformation that was performed. 3. Show algebraicall that the function f () 5 3 is an even function.. Both f () 5 and g() 5 have a domain of all real numbers. List as man characteristics as ou can to distinguish the two functions...5..5.5. 5. Describe the transformations that must be applied to 5 to obtain the function f () 5( 3) 5, then graph the function.. Given the graph shown, describe the transformations that were performed to get this function. Write the algebraic representation, using function notation. 7. (3, 5) is a point on the graph of 5 f (). Find the corresponding point on the graph of each of the following relations. a) 5 3f ( ) 5 f (). Find the inverse of f () 5( ). 9. A certain ta polic states that the first $5 of income is taed at 5% and an income above $5 is taed at %. a) Calculate the ta on $5. Write a function that models the ta polic.. a) Sketch the graph of where f () 5 e, if, f ()! 3, if $ Is f () continuous over its entire domain? Eplain. c) State the intervals of increase and decrease. d) State the domain and range of this function. Chapter Self-Test
Chapter Task Modelling with Functions In 95, a team of chemists led b Dr. W. F. Libb developed a method for determining the age of an natural specimen, up to approimatel ears of age. Dr. Libb s method is based on the fact that all living materials contain traces of carbon-. His method involves measuring the percent of carbon- that remains when a specimen is found. The percent of carbon- that remains in a specimen after various numbers of ears is shown in the table below. Years Carbon- Remaining (%) 5 73 5. 5. 7 9.5 9.5 5 3.5 3 3.55? 573 How can ou use the function P(t) 5 (.5) to model this situation and determine the age of a natural specimen? t A. What percent of carbon is remaining for t 5? What does this mean in the contet of Dr. Libb s method? B. Draw a graph of the function P(t) 5 (.5), using the given table of values. C. What is a reasonable domain for P(t)? What is a reasonable range? D. Determine the approimate age of a specimen, given that P(t) 5 7. E. Draw the graph of the inverse function. F. What information does the inverse function provide? G. What are the domain and range of the inverse function? t Task Checklist Did ou show all our steps? Did ou draw and label our graphs accuratel? Did ou determine the age of the specimen that had 7% carbon- remaining? Did ou eplain our thinking clearl? Chapter 3
Answers Chapter Getting Started, p.. a) c) 5 d) a 5a. a) ( )( ) (5 )( 3) c) ( )( ) d) (a ( ) 3. a) horizontal translation 3 units to the right, vertical translation units up; horizontal translation unit to the right, vertical translation units up; c) horizontal stretch b a factor of, vertical stretch b a factor of, reflection across the -ais; d) horizontal compression b a factor of vertical stretch b a factor of,, reflection across the -ais; 7 9 9 7. a) D 5 5PR # #, f) R 5 5PR # # D 5 5PR, R 5 5 PR $9 c) D 5 5PR, R 5 5 PR d) D 5 5PR, R 5 5 PR 3 # $ 3 e) D 5 5PR, R 5 5 PR. 5. a) This is not a function; it does not pass the vertical line test. This is a function; for each -value, there is eactl one corresponding -value. c) This is not a function; for each -value greater than, there are two corresponding -values. d) This is a function; for each -value, there is eactl one corresponding -value. e) This is a function; for each -value, there is eactl one corresponding -value.. a) about.7 7. If a relation is represented b a set of ordered pairs, a table, or an arrow diagram, one can determine if the relation is a function b checking that each value of the independent variable is paired with no more than one value of the dependent variable. If a relation is represented using a graph or scatter plot, the vertical line test can be used to determine if the relation is a function. A relation ma also be represented b a description/rule or b using function notation or an equation. In these cases, one can use reasoning to determine if there is more than one value of the dependent variable paired with an value of the independent variable. Lesson., pp. 3. a) D 5 5PR; R 5 5PR # #; This is a function because it passes the vertical line test. D 5 5PR # # 7; R 5 5 PR 3 # # ; This is a function because it passes the vertical line test. c) D 5 5,, 3, ; R 5 55,, 7, 9, ; This is not a function because is sent to more than one element in the range. d) D 5 5PR; R 5 5PR; This is a function because ever element in the domain produces eactl one element in the range. e) D 5 5, 3,, ; R 5 5,,, 3; This is a function because ever element of the domain is sent to eactl one element in the range. D 5 5PR; R 5 5PR # ; This is a function because ever element in the domain produces eactl one element in the range.. a) D 5 5PR; R 5 5 PR #3; This is a function because ever element in the domain produces eactl one element in the range. D 5 5PR 3; R 5 5 PR ; This is a function because ever element in the domain produces eactl one element in the range. c) D 5 5PR; R 5 5 PR. ; This is a function because ever element in the domain produces eactl one element in the range. d) D 5 5PR; R 5 5 PR # # ; This is a function because ever element in the domain produces eactl one element in the range. e) D 5 5PR 3 # # 3; R 5 5PR 3 # # 3; This is not a function because (, 3) and (, 3) are both in the relation. f) D 5 5PR; R 5 5PR # # ; This is a function because ever element in the domain produces eactl one element in the range. 3. a) function; D 5 5, 3, 5, 7; R 5 5,, function; D 5 5,,, 5; R 5 5, 3, c) function; D 5 5,,, 3; R 5 5, d) not a function; D 5 5,, ; R 5 5, 3, 5, 7 e) not a function; D 5 5,, ; R 5 5,,, 3 f) function; D 5 5,, 3, ; R 5 5,, 3,. a) function; D 5 5PR; R 5 5 PR $. not a function; D 5 5PR $ ; R 5 5 PR c) function; D 5 5PR; R 5 5 PR $.5 d) not a function; D 5 5PR $ ; R 5 5 PR e) function; D 5 5PR ; R 5 5 PR f) function; D 5 5PR; R 5 5 PR 5. a) 5 3 c) 5 3( ) 5 5 d) 5 5 Answers
. a). a) The length is twice the width. f () 5 ; f (7) 5 ; f () 5 5 f (l ) 5 3 l Yes, f (5) 5 f (3) 3 f (5) c) Yes, f () 5 f (3) 3 f () c) f(b) d) Yes, there are others that will work. f (a) 3 f ( 5 f (a 3 whenever a and b have no common factors other than. 3. Answers ma var. For eample: B independent numerical variable model d) length 5 m; width 5 m 7. a) 5 5 5 Time (s) D 5 5,,,,,,,,,,,, c) R 5 5, 5, d) It is a function because it passes the vertical line test. e) Height (m) Time (s) 5 5 5 Height (m) f) It is not a function because (5, ) and (5, ) are both in the relation.. a) 5(, ), (3, ), (5, ) 5(, ), (3, ), (5, ) c) 5(, ), (, 3), (5, ) 9. If a vertical line passes through a function and hits two points, those two points have identical -coordinates and different -coordinates. This means that one -coordinate is sent to two different elements in the range, violating the definition of function.. a) Yes, because the distance from (, 3) to (, ) is 5. No, because the distance from (, 5) to (, ) is not 5. c) No, because (, 3) and (, 3) are both in the relation.. a) g() 5 3 g(3) g() 5 7 5 5 g(3 ) 5 g() 5 So, g(3) g() g(3 ).. domain range dependent variable The first is not a function because it fails the vertical line test: D 5 5PR 5 # # 5; R 5 5PR 5 # # 5. The second is a function because it passes the vertical line test: D 5 5PR 5 # # 5; R 5 5PR # # 5. 5. is a function of if the graph passes the horizontal line test. This occurs when an horizontal line hits the graph at most once. Lesson., p. FUNCTION function notation. 5,, 5,, 5. a) c) e) 35 d) f) 3. a). 3 c) $ # d) 5. a) mapping model algebraic model graphical model vertical line test c) The absolute value of a number is alwas greater than or equal to. There are no solutions to this inequalit. d) 5. a) # 3 c) $. d),. a) The graphs are the same. Answers ma var. For eample, 5( ), so the are negatives of each other and have the same absolute value. 7. a) c) d). When the number ou are adding or subtracting is inside the absolute value signs, it moves the function to the left (when adding) or to the right (when subtracting) of the origin. When the number ou are Answers Answers 3
adding or subtracting is outside the absolute value signs, it moves the function down (when subtracting) or up (when adding) from the origin. The graph of the function will be the absolute value function moved to the left 3 units and up units from the origin. 9. This is the graph of g() 5 horizontall compressed b a factor of and translated unit to the left.. This is the graph of g() 5 horizontall compressed b a factor of reflected over the -ais, translated, units to the right, and translated 3 units up. Lesson.3, pp. 3 5. Answers ma var. For eample, domain because most of the parent functions have all real numbers as a domain.. Answers ma var. For eample, the end behaviour because the onl two that match are and. 3. Given the horizontal asmptote, the function must be derived from. But the asmptote is at 5, so it must have been translated up two. Therefore, the function is f () 5.. a) Both functions are odd, but their domains are different. Both functions have a domain of all real numbers, but sin () has more zeros. c) Both functions have a domain of all real numbers, but different end behaviour. d) Both functions have a domain of all real numbers, but different end behaviour. 5. a) even d) odd odd e) neither even nor odd c) odd f) neither even nor odd. a), because it is a measure of distance from a number sin (), because the heights are periodic. c), because population tends to increase eponentiall d), because there is $ on the first da, $ on the second, $3 on the third, etc. 7. a) f () 5! c) f () 5 f () 5 sin d) f () 5. a) f () 5 3 9. g() 5 sin 3 c) h() 5 3 3 5 5 5 3. a) f () 5 ( ) There is not onl one function. f () 5 3 works as well. ( ) c) There is more than one function that satisfies the propert. f () 5 and f () 5 both work.. is a smooth curve, while has a sharp, pointed corner at (, ).. See net page. 3. It is important to name parent functions in order to classif a wide range of functions according to similar behaviour and characteristics. D 5 5PR, R 5 5 f ()PR; interval of increase 5 (`, `), no interval of decrease, no discontinuities, - and -intercept at (, ), odd, S `, S `, and S `, S `. It is ver similar to f () 5. It does not, however, have a constant slope. 5. No, cos is a horizontal translation of sin.. The graph can have,, or zeros. zeros: zero: 3 zeros: Mid-Chapter Review, p.. a) function; D 5 5, 3, 5, 7, R 5 5, 3, function; D 5 5PR, R 5 5PR c) not a function; D 5 5PR 5 # # 5, R 5 5PR 5 # # 5 d) not a function; D 5 5,,, R 5 5, 3,, 7. a) Yes. Ever element in the domain gets sent to eactl one element in the range. D 5 5,,, 3,, 5,, 7,, 9, c) R 5 5,, 5, 3, 35,, 5, 5 Answers
. Parent Function f() 5 g() 5 h() 5 k() 5 m() 5! p() 5 r() 5 sin Sketch. r() = sin.5 7 9 9 7.5. Domain 5PR 5PR 5PR 5PR 5PR $ 5PR 5PR Range 5f() PR 5f()PR f() $ 5f()PR f() 5f()PR f() $ 5f()PR f() $ 5f()PR f(). 5f()PR # f() # Intervals of Increase (`, `) (, `) None (, `) (, `) (`, `) 39(k ), 9(k 3) KPZ Intervals of Decrease None (`, ) (`, ) (, `) (`, ) None None 39(k 3), 9(k ) KPZ Location of None None 5 None None 5 None Discontinuities and 5 Asmptotes Zeros (, ) (, ) None (, ) (, ) None k KPZ -Intercepts (, ) (, ) None (, ) (, ) (, ) (, ) Smmetr Odd Even Odd Even Neither Neither Odd End Behaviours S `, S ` S `, S ` S `, S S `, S ` S `, S ` S `, S ` Oscillating S `, S ` S `, S ` S `, S S `, S ` S `, S 3 Answers Answers 5
3. a) D 5 5PR, R 5 5 f ()PR; This if f () 5 sin translated down ; 5. a) f () 5, translated left function continuous 3 D 5 5PR 3 # # 3, R 5 5PR 3 # # 3; not a function c) D 5 5PR # 5, 3 3 R 5 5PR $ ; function d) D 5 5PR, R 5 5PR $; function c) This is f () 5 translated down ; 3. 3,, 3,, 5 continuous 5. a) f () 5, vertical stretch b 3 3 3 3 9. c) f () 5 sin, horizontal compression 3 of translation up 3, 3 3 3 3 3 c) 3 d). a) f () 5 f () 5 c) f () 5 " 7. a) even c) neither odd nor even even d) neither odd nor even. a) This is f () 5 translated right and up 3; discontinuous Lesson., pp. 35 37. a) translation unit down horizontal compression b a factor of, translation unit right c) reflection over the -ais, translation units up, translation 3 units right d) reflection over the -ais, vertical stretch b a factor of, horizontal compression b a factor of e) reflection over the -ais, translation 3 units down, reflection over the -ais, translation units left f) vertical compression b a factor of, translation units up, horizontal stretch b a factor of, translation 5 units right. a) a 5, k 5 d 5, c 5 3, a 5 3, k 5 d 5, c 5, 3. (, 3), (, 3), (, ), (, ), (, ), (, ). a) (, ), (, ), (, ), (, ) (5, 3), (7, 7), (, 5), (, ) c) (, 5), (, 9), (, 7), (, ) d) (, ), (3, ), (3, ), (5, 3) e) (, 5), (, ), (, 3), (, 7) f) (, ), (, ), (, ), (, 5) d) f () 5 translation up 3, e) f () 5, horizontal stretch b f) f () 5 ", horizontal compression b translation right, 3 7 5 3 3 3 3 3 3 Answers
. a) c) d) e) f) 7. a). D 5 5PR,. R 5 5 f ()PR f () $ h() D 5 5PR, R 5 5 f ()PR f () $ g() D 5 5PR, R 5 5 f ()PR # f () # D 5 5PR, R 5 5 f ()PR f () 3 D 5 5PR, R 5 5 f ()PR f (). D 5 5PR $, f() R 5 5 f ()PR f () $ 3. a) a vertical stretch b a factor of a horizontal compression b a factor of c) () 5 5. Answers ma var. For eample: horizontal stretch or compression, based on value of k The domain remains unchanged at D 5 5PR. The range must now be less than : R 5 5 f ()PR f (),. It changes from increasing on (`, `) to decreasing on (`, `). The end behaviour becomes as S `, S, and as S `, S `. c) g() 5( 3() ) 53" 5 9. a) (3, ) d) (.75, ) (.5, ) e) (, ) c) (, 9) f) (, 7). a) D 5 5PR $, R 5 5 g()pr g() $ D 5 5PR $, R 5 5h()PR h() $ c) D 5 5PR #, R 5 5k()PR k() $ d) D 5 5PR $ 5, R 5 5 j()pr j() $3. 5 5( 3) is the same as 5 5 5, not 5 5 3. vertical stretch or compression, based on value of a reflection in -ais if a, ; reflection in -ais if k, horizontal translation based on value of d vertical translation based on value of c 5. (, 5). a) horizontal compression b a factor of 3, translation units to the left because the are equivalent epressions: 3( ) 5 3 c) Lesson.5, pp. 3 5. a) (5, ) c) (, ) e) (, 3) (, 5) d) (, ) f) (7, ). a) D 5 5PR, R 5 5PR D 5 5PR, R 5 5PR $ c) D 5 5PR,, R 5 5PR $5 d) D 5 5PR 5,,, R 5 5PR, 3. A and D match; B and F match; C and E match. a) (, 9) (9, ) c) D 5 5PR, R 5 5PR d) D 5 5PR, R 5 5PR e) Yes; it passes the vertical line test. 5. a) (, ) (, ) c) D 5 5PR, R 5 5PR $ d) D 5 5PR $ R 5 5 PR e) No; (, ) and (, ) are both on the inverse relation.. a) Not a function Not a function c) Function p d) Not a function p p p p p p p 7. a) C 5 5 (F 3); this allows ou to 9 convert from Fahrenheit to Celsius. C 5 F. a) r 5 $ p; A this can be used to determine the radius of a circle when its area is known. A 5 5p cm, r 5 5 cm 9.. k 5 a) 3 c) e) 5 d) f) Answers Answers 7
. No; several students could have the same grade point average.. a) f () 5 ( ) 3 h () 5 c) g () 5 " 3 d) m () 5 5 3. a) 5 ( 3) c) 5 Å d) (., 3.55), (.,.), (3.55,.), (3., 3.) e) $ 3 because a negative square root is undefined. f) g() 5 5, but g (5) 5 or ; the inverse is not a function if this is the domain of g.. For 5", D 5 5PR $ and R 5 5PR #. For 5, D 5 5PR and R 5 5PR $. The student would be correct if the domain of 5 is restricted to D 5 5PR #. 5. Yes; the inverse of 5 " is 5 so long as the domain of this second function is restricted to D 5 5PR $.. John is correct. Algebraic: 5 3 5 3 ; ; ( ) 5 3 ; 5 " 3 ( ). Numeric: Let 5. 5 3 5 Graphical: 3 5 " 3 () 5 " 3 5. 5 5 ; 5 " 3 ( ) 5 " 3 ( ) The graphs are reflections over the line 5. 7. f () 5 k works for all kpr. 5 k Switch variables and solve for : 5 k 5 k So the function is its own inverse.. If a horizontal line hits the function in two locations, that means there are two points with equal -values and different -values. When the function is reflected over the line 5 to find the inverse relation, those two points become points with equal -values and different -values, thus violating the definition of a function. Lesson., pp. 5 53. a) c) d) e) f). a) Discontinuous at 5 Discontinuous at 5 c) Discontinuous at 5 d) Continuous e) Discontinuous at 5 f) Discontinuous at 5 and 5 3. a) 3 f() 5 e, if #, if., if, f () 5 e!, if $. a) D 5 5PR; the function is discontinuous at 5. D 5 5PR; the function is continuous. 5. a) The function is discontinuous at 5. D 5 5PR R 5 5, 3 The function is continuous. D 5 5PR R 5 5f ()PR f () $ Answers
c). Answers ma var. For eample: Plot the function for the left interval. Plot the function for the right interval.. Answers ma var. For eample: 3, if, a) f () 5, if # #!, if. 5 The function is continuous. D 5 5PR R 5 5f ()PR f () $ d) The function is continuous. D 5 5PR R 5 5 f ()PR # f () # 5 5, if # # 5. f () 5 e 5., if $ 5 7. f () 5.35, if # #.5, if, # 5.55, if. 5. Determine if the plots for the left and right intervals meet at the -value that serves as the common end point for the intervals; if so, the function is continuous at this point. Determine continuit for the two intervals using standard methods. 3, if $3 f () 5 3 5 e 3, if,3. discontinuous at p 5 and p 5 5; continuous at, p, 5 and p. 5 3 3 3 c) The function is not continuous. The last two pieces do not have the same value for 5. 3, if, d) f () 5, if # #!, if. Lesson.7, pp. 5 57. a) 5(, ), (, 5), (, 5), (, ) 5(, ), (, 3), (, ), (, ) c) 5(, ), (, 3), (, ), (, ) d) 5(, ), (, ), (, ), (, ). a). k 5 9. a) 7 5 3 3 3 5 The function is discontinuous at 5. c) 3 fish d) 5 ; 5 5; 5 e) Answers ma var. For eample, three possible events are environmental changes, introduction of a new predator, and increased fishing. 3., if #,, if #, f () 5 f, if #, 3 3, if 3 #,, if #, 5 5 3 3 5 It is often referred to as a step function because the graph looks like steps.. To make the first two pieces continuous, 5() 5 k, so k 5. But if k 5, the graph is discontinuous at 5 3. 5. 3. a) Answers Answers 9
c) 5. a) 3 3 3. a) 5. a) 5 d) 5 3. a) Answers ma var. For eample, properties of the original graphs such as intercepts and sign at various values of the independent variable figure prominentl in the shape of the new function. 7. a) 3 3 3 3 c) f() g() h() 5 f() 3 g() 3 7 77 3 3 3 3 3 7 77 3 3 d) h() 5 ( )( ) 5 ; degree is e) D 5 5PR 5 7 3 5 5 5 5 f() g() h() 5 f() 3 g() 3 3 5 5 5 5 3 c) 5 3 5 3 5 d) h() 5 ( 3)( 5) 5 3 3 5 5; degree is 3 e) D 5 5PR; this is the same as the domain of both f and g. Chapter Review, pp.. a) function; D 5 5PR; R 5 5 PR function; D 5 5PR; R 5 5 PR # 3 c) not a function; D 5 5PR # # ; R 5 5 PR d) function; D 5 5PR. ; R 5 5 PR. a) C(t) 5 3.t D 5 5tPR t $, R 5 5C(t)PR C(t) $ 3 3. D 5 5PR, R 5 5 f ()PR f () $ Answers
., 5. a) Both functions have a domain of all real numbers, but the ranges differ. Both functions are odd but have different domains. c) Both functions have the same domain and range, but is smooth and has a sharp corner at (, ). d) Both functions are increasing on the entire real line, but has a horizontal asmptote while does not.. a) Increasing on (`, `); odd; D 5 5PR; R 5 5 f ()PR Decreasing on (`, ); increasing on (, `); even; D 5 5PR; R 5 5 f ()PR f () $ c) Increasing on (`, `); neither even nor odd; D 5 5PR; R 5 5 f ()PR f (). 7. a) Parent: 5 ; translated left c) Parent: 5 sin ; reflected across the -ais, epanded verticall b a factor of, compressed horizontall b a factor of translated up b 3, 7 Parent: 5!; compressed verticall b a factor of.5, reflected across the -ais, compressed horizontall b a factor of and translated left 7 3, 9 9 7. d) Parent: 5 ; reflected across the -ais, compressed horizontall b a factor of and translated down b 3, 5Q 3 R 9. a) (, ) (, ) c) (, 3) d) a 7 5,.3b e) f) (, ) (9, ). a) (, ) (9, ) c) (7, ) d) (7, 5) e) (3, ) f) (, ). a) D 5 5PR,,, R 5 5PR D 5 5PR,, R 5 5PR $ 7. a) The inverse relation is not a function. 3. a) f () 5 g () 5 " 3. The function is continuous; D 5 5PR, R 5 5PR 3, if # 5. f () 5 e, if. ; the function is discontinuous at 5.. In order for f () to be continuous at 5, the two pieces must have the same value when 5. When 5, 5 and 3 5 3. The two pieces are not equal when 5, so the function is not continuous at 5. 3, if # 7. a) f () 5 e.3, if. $3.5 c) $3. a) 5(, 7), (, 5) 5(, ), (, ) c) 5(, ), (, 5) 9. a) Answers The inverse relation is a function. c) 3 Answers
d) e). a) D C c) A d) B. a) c) d) 3 e) Answers ma var. For eample, (, ) belongs to f, (, ) belongs to g and (, ) belongs to f g. Also, (, 3) belongs to f, (, 5) belongs to g and (, ) belongs to f g. Chapter Self-Test, p.. a) Yes. It passes the vertical line test. D 5 5PR; R 5 5PR $. a) f () 5 or f () 5 5 5 3 3 5 5 3 f() 9 3 g() 9 7 5 (f g)() 3 3 or 5 5 5 5 5PR. c) The graph was translated units down. 3. f () 5 3() () 5 3 5 f (). has a horizontal asmptote while does not. The range of is while the range of is 5PR $. is increasing on the whole real line and has an interval of decrease and an interval of increase. 5. reflection over the -ais, translation down 5 units, translation left 3 units. horizontal stretch b a factor of, translation unit up; f () 5 if 7. a) (, 7) (5, 3). 9. f () 5 a) $9.5, if # 5 f () 5 e., if. 5. a) f () is discontinuous at 5 because the two pieces do not have the same value when 5. When 5, 5 and " 3 5 3. c) Intervals of increase: (`, ), (, `); no intervals of decrease d) D 5 5PR, R 5 5PR,, or $ 3 Chapter Getting Started, p.. a) 3 7. a) Each successive first difference is times the previous first difference. The function is eponential. The second differences are all. The function is quadratic. 3. a) 3 c) 5, 5, d) 7, 9. a) vertical compression b a factor of vertical stretch b a factor of, horizontal translation units to the right c) vertical stretch b a factor of 3, reflection across -ais, vertical translation 7 units up d) vertical stretch b a factor of 5, horizontal translation 3 units to the right, vertical translation units down, 5. a) A 5 (.) t $59.7 c) No, since the interest is compounded each ear, each ear ou earn more interest than the previous ear.. a) 5 m; m s c) 5 m 7. Linear relations constant; same as slope of line; Lesson., pp. 7 7. a) 9 c) 3 e). 5 d) f).. a) i) 5 m> s ii) 5 ms > Nonlinear relations variable; can be positive, positive for negative, or Rates of Change lines that for slope up from left to different parts of the right; negative for same relation lines that slope down from left to right; for horizontal lines. Answers