1.3 Introduction to Functions

. Introduction to Functions. Introduction to Functions One of the core concepts in Collee Alebra is the function. There are man was to describe a function and we bein b definin a function as a special kind of relation. Definition.6. A relation in which each -coordinate is matched with onl one -coordinate is said to describe as a function of. Eample... Which of the followin relations describe as a function of?. R {(, ), (, ), (, ), (, )}. R {(, ), (, ), (, ), (, )} Solution. A quick scan of the points in R reveals that the -coordinate is matched with two different -coordinates: namel and. Hence in R, is not a function of. On the other hand, ever -coordinate in R occurs onl once which means each -coordinate has onl one correspondin -coordinate. So, R does represent as a function of. Note that in the previous eample, the relation R contained two different points with the same -coordinates, namel (, ) and (, ). Remember, in order to sa is a function of, we just need to ensure the same -coordinate isn t used in more than one point. To see what the function concept means eometricall, we raph R and R in the plane. The raph of R The raph of R The fact that the -coordinate is matched with two different -coordinates in R presents itself raphicall as the points (, ) and (, ) lin on the same vertical line,. If we turn our attention to the raph of R, we see that no two points of the relation lie on the same vertical line. We can eneralize this idea as follows Theorem.. The Vertical Line Test: A set of points in the plane represents as a function of if and onl if no two points lie on the same vertical line. We will have occasion later in the tet to concern ourselves with the concept of bein a function of. In this case, R represents as a function of ; R does not.

Relations and Functions It is worth takin some time to meditate on the Vertical Line Test; it will check to see how well ou understand the concept of function as well as the concept of raph. Eample... Use the Vertical Line Test to determine which of the followin relations describes as a function of. The raph of R The raph of S Solution. Lookin at the raph of R, we can easil imaine a vertical line crossin the raph more than once. Hence, R does not represent as a function of. However, in the raph of S, ever vertical line crosses the raph at most once, so S does represent as a function of. In the previous test, we sa that the raph of the relation R fails the Vertical Line Test, whereas the raph of S passes the Vertical Line Test. Note that in the raph of R there are infinitel man vertical lines which cross the raph more than once. However, to fail the Vertical Line Test, all ou need is one vertical line that fits the bill, as the net eample illustrates. Eample... Use the Vertical Line Test to determine which of the followin relations describes as a function of. The raph of S The raph of S

. Introduction to Functions 5 Solution. Both S and S are sliht modifications to the relation S in the previous eample whose raph we determined passed the Vertical Line Test. In both S and S, it is the addition of the point (, ) which threatens to cause trouble. In S, there is a point on the curve with -coordinate just below (, ), which means that both (, ) and this point on the curve lie on the vertical line. (See the picture below and the left.) Hence, the raph of S fails the Vertical Line Test, so is not a function of here. However, in S notice that the point with -coordinate on the curve has been omitted, leavin an open circle there. Hence, the vertical line crosses the raph of S onl at the point (, ). Indeed, an vertical line will cross the raph at most once, so we have that the raph of S passes the Vertical Line Test. Thus it describes as a function of. S and the line The raph of G for E... Suppose a relation F describes as a function of. The sets of - and -coordinates are iven special names which we define below. Definition.7. Suppose F is a relation which describes as a function of. The set of the -coordinates of the points in F is called the domain of F. The set of the -coordinates of the points in F is called the rane of F. We demonstrate findin the domain and rane of functions iven to us either raphicall or via the roster method in the followin eample. Eample... Find the domain and rane of the function F {(, ), (0, ), (, ), (5, )} and of the function G whose raph is iven above on the riht. Solution. The domain of F is the set of the -coordinates of the points in F, namel {, 0,, 5} and the rane of F is the set of the -coordinates, namel {, }. To determine the domain and rane of G, we need to determine which and values occur as coordinates of points on the iven raph. To find the domain, it ma be helpful to imaine collapsin the curve to the -ais and determinin the portion of the -ais that ets covered. This is called projectin the curve to the -ais. Before we start projectin, we need to pa attention to two

6 Relations and Functions subtle notations on the raph: the arrowhead on the lower left corner of the raph indicates that the raph continues to curve downwards to the left forever more; and the open circle at (, ) indicates that the point (, ) isn t on the raph, but all points on the curve leadin up to that point are. project down project up The raph of G The raph of G We see from the fiure that if we project the raph of G to the -ais, we et all real numbers less than. Usin interval notation, we write the domain of G as (, ). To determine the rane of G, we project the curve to the -ais as follows: project riht project left The raph of G The raph of G Note that even thouh there is an open circle at (, ), we still include the value of in our rane, since the point (, ) is on the raph of G. We see that the rane of G is all real numbers less than or equal to, or, in interval notation, (, ].

. Introduction to Functions 7 All functions are relations, but not all relations are functions. Thus the equations which described the relations in Section. ma or ma not describe as a function of. The alebraic representation of functions is possibl the most important wa to view them so we need a process for determinin whether or not an equation of a relation represents a function. (We dela the discussion of findin the domain of a function iven alebraicall until Section..) Eample..5. Determine which equations represent as a function of.. +. +. Solution. For each of these equations, we solve for and determine whether each choice of will determine onl one correspondin value of.. + etract square roots ± If we substitute 0 into our equation for, we et ± 0 ±, so that (0, ) and (0, ) are on the raph of this equation. Hence, this equation does not represent as a function of.. + For ever choice of, the equation returns onl one value of. Hence, this equation describes as a function of.. + ( + ) factor + For each choice of, there is onl one value for, so this equation describes as a function of. We could tr to use our raphin calculator to verif our responses to the previous eample, but we immediatel run into trouble. The calculator s Y menu requires that the equation be of the form some epression of. If we wanted to verif that the first equation in Eample..5

8 Relations and Functions does not represent as a function of, we would need to enter two separate epressions into the calculator: one for the positive square root and one for the neative square root we found when solvin the equation for. As predicted, the resultin raph shown below clearl fails the Vertical Line Test, so the equation does not represent as a function of. Thus in order to use the calculator to show that + does not represent as a function of we needed to know analticall that was not a function of so that we could use the calculator properl. There are more advanced raphin utilities out there which can do implicit function plots, but ou need to know even more Alebra to make them work properl. Do ou et the point we re trin to make here? We believe it is in our best interest to learn the analtic wa of doin thins so that ou are alwas smarter than our calculator.

. Introduction to Functions 9.. Eercises In Eercises -, determine whether or not the relation represents as a function of. Find the domain and rane of those relations which are functions.. {(, 9), (, ), (, ), (0, 0), (, ), (, ), (, 9)}. {(, 0), (, 6), (, ), (, ), ( 5, 6), (, 9), (6, )}. {(, 0), ( 7, 6), (5, 5), (6, ), (, 9), (, 0)}. {(, ), (, ), (9, 6), (6, 8), (5, 0), (6, ),...} 5. {(, ) is an odd inteer, and is an even inteer} 6. {(, ) is an irrational number} 7. {(, 0), (, ), (, ), (8, ), (6, ), (, 5),... } 8. {..., (, 9), (, ), (, ), (0, 0), (, ), (, ), (, 9),... } 9. {(, ) < < } 0. {(, ) < }. { (, ) is a real number}. { (, ) is a real number} In Eercises -, determine whether or not the relation represents as a function of. Find the domain and rane of those relations which are functions...

50 Relations and Functions 5. 6. 5 7. 8. 5 6 7 8 9 9. 0. 5 5. 9 8 7 6 5 5. 5 5 5 5

. Introduction to Functions 5.. 5 5 5 5 5 5 6 5 5. 6. 7. 8. 9. 0.

5 Relations and Functions.. In Eercises - 7, determine whether or not the equation represents as a function of... 5. 6. 7. 9 8. 6 9. + 0. +. +... + 5. + 6. 7. 8. Eplain wh the population P of Sasquatch in a iven area is a function of time t. What would be the rane of this function? 9. Eplain wh the relation between our classmates and their email addresses ma not be a function. What about phone numbers and Social Securit Numbers? The process iven in Eample..5 for determinin whether an equation of a relation represents as a function of breaks down if we cannot solve the equation for in terms of. However, that does not prevent us from provin that an equation fails to represent as a function of. What we reall need is two points with the same -coordinate and different -coordinates which both satisf the equation so that the raph of the relation would fail the Vertical Line Test.. Discuss with our classmates how ou miht find such points for the relations iven in Eercises 50-5. 50. + 0 5. + 5. + 5. ( + ) +

. Introduction to Functions 5.. Answers. Function domain {,,, 0,,,} rane {0,,, 9}. Function domain { 7,,,, 5, 6} rane {0,, 5, 6, 9}. Not a function. Function domain {,, 9, 6, 5, 6,...} { is a perfect square} rane {,, 6, 8, 0,,...} { is a positive even inteer} 5. Not a function 6. Function domain { is irrational} rane {} 7. Function domain { n for some whole number n} rane { is an whole number} 8. Function domain { is an inteer} rane { n for some inteer n} 9. Not a function 0. Function domain [, ), rane {}. Function domain (, ) rane [0, ). Function domain {,,,, 0, } rane {, 0,,,, } 5. Function domain (, ) rane [, ) 7. Function domain [, ) rane [0, ). Not a function. Not a function 6. Not a function 8. Function domain (, ) rane (0, ] 9. Not a function 0. Function domain [ 5, ) (, ) rane (, ) [0, )

5 Relations and Functions. Function domain [, ) rane [, ). Function domain [ 5, ) rane [, ) 5. Function domain (, ) rane (, ] 7. Function domain [, ) rane (, ] 9. Function domain (, 0] (, ) rane (, ] {}. Not a function. Function domain [0, ) (, 6] rane (, ] [0, ] 6. Function domain (, ) rane (, ] 8. Function domain (, ) rane (, ) 0. Function domain [, ] rane [, ]. Not a function. Function domain (, ) rane {}. Function. Function 5. Function 6. Not a function 7. Function 8. Not a function 9. Not a function 0. Function. Not a function. Function. Not a function. Function 5. Function 6. Function 7. Not a function

. Function Notation 55. Function Notation In Definition.6, we described a function as a special kind of relation one in which each - coordinate is matched with onl one -coordinate. In this section, we focus more on the process b which the is matched with the. If we think of the domain of a function as a set of inputs and the rane as a set of outputs, we can think of a function f as a process b which each input is matched with onl one output. Since the output is completel determined b the input and the process f, we smbolize the output with function notation: f(), read f of. In other words, f() is the output which results b applin the process f to the input. In this case, the parentheses here do not indicate multiplication, as the do elsewhere in Alebra. This can cause confusion if the contet is not clear, so ou must read carefull. This relationship is tpicall visualized usin a diaram similar to the one below. f Domain (Inputs) f() Rane (Outputs) The value of is completel dependent on the choice of. For this reason, is often called the independent variable, or arument of f, whereas is often called the dependent variable. As we shall see, the process of a function f is usuall described usin an alebraic formula. For eample, suppose a function f takes a real number and performs the followin two steps, in sequence. multipl b. add If we choose 5 as our input, in step we multipl b to et (5)() 5. In step, we add to our result from step which ields 5 + 9. Usin function notation, we would write f(5) 9 to indicate that the result of applin the process f to the input 5 ives the output 9. In eneral, if we use for the input, applin step produces. Followin with step produces + as our final output. Hence for an input, we et the output f() +. Notice that to check our formula for the case 5, we replace the occurrence of in the formula for f() with 5 to et f(5) (5) + 5 + 9, as required.

56 Relations and Functions Eample... Suppose a function is described b applin the followin steps, in sequence. add. multipl b Determine (5) and find an epression for (). Solution. Startin with 5, step ives 5 + 9. Continuin with step, we et ()(9) 7. To find a formula for (), we start with our input. Step produces +. We now wish to multipl this entire quantit b, so we use a parentheses: ( + ) +. Hence, () +. We can check our formula b replacin with 5 to et (5) (5) + 5 + 7. Most of the functions we will encounter in Collee Alebra will be described usin formulas like the ones we developed for f() and () above. Evaluatin formulas usin this function notation is a ke skill for success in this and man other Math courses. Eample... Let f() + +. Find and simplif the followin. (a) f( ), f(0), f() (b) f(), f() (c) f( + ), f() +, f() + f(). Solve f(). Solution.. (a) To find f( ), we replace ever occurrence of in the epression f() with f( ) ( ) + ( ) + () + ( ) + 0 Similarl, f(0) (0) + (0) +, and f() () + () + + 6 + 6. (b) To find f(), we replace ever occurrence of with the quantit f() () + () + ( ) + (6) + + 6 + The epression f() means we multipl the epression f() b f() ( + + ) + 6 + 8

. Function Notation 57 (c) To find f( + ), we replace ever occurrence of with the quantit + f( + ) ( + ) + ( + ) + ( + + ) + ( + 6) + + + 6 + + 6 To find f() +, we add to the epression for f() f() + ( + + ) + + + 6 From our work above, we see f() 6 so that f() + f() ( + + ) + 6 + + 0. Since f() + +, the equation f() is equivalent to + +. Solvin we et + 0, or ( + ) 0. We et 0 or, and we can verif these answers b checkin that f(0) and f(). A few notes about Eample.. are in order. First note the difference between the answers for f() and f(). For f(), we are multiplin the input b ; for f(), we are multiplin the output b. As we see, we et entirel different results. Alon these lines, note that f( + ), f()+ and f()+f() are three different epressions as well. Even thouh function notation uses parentheses, as does multiplication, there is no eneral distributive propert of function notation. Finall, note the practice of usin parentheses when substitutin one alebraic epression into another; we hihl recommend this practice as it will reduce careless errors. Suppose now we wish to find r() for r(). Substitution ives 9 r() () () 9 6 0, which is undefined. (Wh is this, aain?) The number is not an allowable input to the function r; in other words, is not in the domain of r. Which other real numbers are forbidden in this formula? We think back to arithmetic. The reason r() is undefined is because substitution results in a division b 0. To determine which other numbers result in such a transression, we set the denominator equal to 0 and solve 9 0 9 9 etract square roots ±

58 Relations and Functions As lon as we substitute numbers other than and, the epression r() is a real number. Hence, we write our domain in interval notation as (, ) (, ) (, ). When a formula for a function is iven, we assume that the function is valid for all real numbers which make arithmetic sense when substituted into the formula. This set of numbers is often called the implied domain of the function. At this stae, there are onl two mathematical sins we need to avoid: division b 0 and etractin even roots of neative numbers. The followin eample illustrates these concepts. Eample... Find the domain of the followin functions.. (). h() 5. f(). F () + 5. r(t) 6 t + 6. I() Solution.. The potential disaster for is if the radicand is neative. To avoid this, we set 0. From this, we et or. What this shows is that as lon as, the epression 0, and the formula () returns a real number. Our domain is (, ].. The formula for h() is hauntinl close to that of () with one ke difference whereas the epression for () includes an even indeed root (namel a square root), the formula for h() involves an odd indeed root (the fifth root). Since odd roots of real numbers (even neative real numbers) are real numbers, there is no restriction on the inputs to h. Hence, the domain is (, ).. In the epression for f, there are two denominators. We need to make sure neither of them is 0. To that end, we set each denominator equal to 0 and solve. For the small denominator, we et 0 or. For the lare denominator See the Eercises for Section.. or, implicit domain The word implied is, well, implied. The radicand is the epression inside the radical.

. Function Notation 59 ()( ) 0 ( ) ( ) clear denominators So we et two real numbers which make denominators 0, namel and. Our domain is all real numbers ecept and : (, ) (, ) (, ).. In findin the domain of F, we notice that we have two potentiall hazardous issues: not onl do we have a denominator, we have a fourth (even-indeed) root. Our strate is to determine the restrictions imposed b each part and select the real numbers which satisf both conditions. To satisf the fourth root, we require + 0. From this we et or. Net, we round up the values of which could cause trouble in the denominator b settin the denominator equal to 0. We et 0, or ±. Hence, in order for a real number to be in the domain of F, but ±. In interval notation, this set is [, ) (, ). 5. Don t be put off b the t here. It is an independent variable representin a real number, just like does, and is subject to the same restrictions. As in the previous problem, we have double daner here: we have a square root and a denominator. To satisf the square root, we need a non-neative radicand so we set t + 0 to et t. Settin the denominator equal to zero ives 6 t + 0, or t + 6. Squarin both sides ives t + 6, or t. Since we squared both sides in the course of solvin this equation, we need to check our answer. 5 Sure enouh, when t, 6 t + 6 6 0, so t will cause problems in the denominator. At last we can find the domain of r: we need t, but t. Our final answer is [, ) (, ). 6. It s temptin to simplif I(), and, since there are no loner an denominators, claim that there are no loner an restrictions. However, in simplifin I(), we are assumin 0, since 0 0 is undefined.6 Proceedin as before, we find the domain of I to be all real numbers ecept 0: (, 0) (0, ). It is worth reiteratin the importance of findin the domain of a function before simplifin, as evidenced b the function I in the previous eample. Even thouh the formula I() simplifies to 5 Do ou remember wh? Consider squarin both sides to solve t +. 6 More precisel, the fraction 0 is an indeterminant form. Calculus is required tame such beasts. 0

60 Relations and Functions, it would be inaccurate to write I() without addin the stipulation that 0. It would be analoous to not reportin taable income or some other sin of omission... Modelin with Functions The importance of Mathematics to our societ lies in its value to approimate, or model real-world phenomenon. Whether it be used to predict the hih temperature on a iven da, determine the hours of daliht on a iven da, or predict population trends of various and sundr real and mthical beasts, 7 Mathematics is second onl to literac in the importance humanit s development. 8 It is important to keep in mind that antime Mathematics is used to approimate realit, there are alwas limitations to the model. For eample, suppose rapes are on sale at the local market for $.50 per pound. Then one pound of rapes costs $.50, two pounds of rapes cost $.00, and so forth. Suppose we want to develop a formula which relates the cost of buin rapes to the amount of rapes bein purchased. Since these two quantities var from situation to situation, we assin them variables. Let c denote the cost of the rapes and let denote the amount of rapes purchased. To find the cost c of the rapes, we multipl the amount of rapes b the price $.50 dollars per pound to et c.5 In order for the units to be correct in the formula, must be measured in pounds of rapes in which case the computed value of c is measured in dollars. Since we re interested in findin the cost c iven an amount, we think of as the independent variable and c as the dependent variable. Usin the lanuae of function notation, we write c().5 where is the amount of rapes purchased (in pounds) and c() is the cost (in dollars). For eample, c(5) represents the cost, in dollars, to purchase 5 pounds of rapes. In this case, c(5).5(5) 7.5, so it would cost $7.50. If, on the other hand, we wanted to find the amount of rapes we can purchase for $5, we would need to set c() 5 and solve for. In this case, c().5, so solvin c() 5 is equivalent to solvin.5 5 Doin so ives 5.5.. This means we can purchase eactl. pounds of rapes for $5. Of course, ou would be hard-pressed to bu eactl. pounds of rapes, 9 and this leads us to our net topic of discussion, the applied domain 0 of a function. Even thouh, mathematicall, c().5 has no domain restrictions (there are no denominators and no even-indeed radicals), there are certain values of that don t make an phsical sense. For eample, corresponds to purchasin pounds of rapes. Also, unless the local market mentioned is the State of California (or some other eporter of rapes), it also doesn t make much sense for 500,000,000, either. So the realit of the situation limits what can be, and 7 See Sections.5,., and 6.5, respectivel. 8 In Carl s humble opinion, of course... 9 You could et close... within a certain specified marin of error, perhaps. 0 or, eplicit domain Mabe this means returnin a pound of rapes?

. Function Notation 6 these limits determine the applied domain of. Tpicall, an applied domain is stated eplicitl. In this case, it would be common to see somethin like c().5, 0 00, meanin the number of pounds of rapes purchased is limited from 0 up to 00. The upper bound here, 00 ma represent the inventor of the market, or some other limit as set b local polic or law. Even with this restriction, our model has its limitations. As we saw above, it is virtuall impossible to bu eactl. pounds of rapes so that our cost is eactl $5. In this case, bein sensible shoppers, we would most likel round down and purchase pounds of rapes or however close the market scale can read to. without bein over. It is time for a more sophisticated eample. Eample... The heiht h in feet of a model rocket above the round t seconds after lift-off is iven b { 5t h(t) + 00t, if 0 t 0 0, if t > 0. Find and interpret h(0) and h(60).. Solve h(t) 75 and interpret our answers. Solution.. We first note that the independent variable here is t, chosen because it represents time. Secondl, the function is broken up into two rules: one formula for values of t between 0 and 0 inclusive, and another for values of t reater than 0. Since t 0 satisfies the inequalit 0 t 0, we use the first formula listed, h(t) 5t + 00t, to find h(0). We et h(0) 5(0) + 00(0) 500. Since t represents the number of seconds since lift-off and h(t) is the heiht above the round in feet, the equation h(0) 500 means that 0 seconds after lift-off, the model rocket is 500 feet above the round. To find h(60), we note that t 60 satisfies t > 0, so we use the rule h(t) 0. This function returns a value of 0 reardless of what value is substituted in for t, so h(60) 0. This means that 60 seconds after lift-off, the rocket is 0 feet above the round; in other words, a minute after lift-off, the rocket has alread returned to Earth.. Since the function h is defined in pieces, we need to solve h(t) 75 in pieces. For 0 t 0, h(t) 5t + 00t, so for these values of t, we solve 5t + 00t 75. Rearranin terms, we et 5t 00t + 75 0, and factorin ives 5(t 5)(t 5) 0. Our answers are t 5 and t 5, and since both of these values of t lie between 0 and 0, we keep both solutions. For t > 0, h(t) 0, and in this case, there are no solutions to 0 75. In terms of the model rocket, solvin h(t) 75 corresponds to findin when, if ever, the rocket reaches 75 feet above the round. Our two answers, t 5 and t 5 correspond to the rocket reachin this altitude twice once 5 seconds after launch, and aain 5 seconds after launch. What oes up...

6 Relations and Functions The tpe of function in the previous eample is called a piecewise-defined function, or piecewise function for short. Man real-world phenomena (e.. postal rates, income ta formulas ) are modeled b such functions. B the wa, if we wanted to avoid usin a piecewise function in Eample.., we could have used h(t) 5t + 00t on the eplicit domain 0 t 0 because after 0 seconds, the rocket is on the round and stops movin. In man cases, thouh, piecewise functions are our onl choice, so it s best to understand them well. Mathematical modelin is not a one-section topic. It s not even a one-course topic as is evidenced b underraduate and raduate courses in mathematical modelin bein offered at man universities. Thus our oal in this section cannot possibl be to tell ou the whole stor. What we can do is et ou started. As we stud new classes of functions, we will see what phenomena the can be used to model. In that respect, mathematical modelin cannot be a topic in a book, but rather, must be a theme of the book. For now, we have ou eplore some ver basic models in the Eercises because ou need to crawl to walk to run. As we learn more about functions, we ll help ou build our own models and et ou on our wa to applin Mathematics to our world. See the United States Postal Service website http://www.usps.com/prices/first-class-mail-prices.htm See the Internal Revenue Service s website http://www.irs.ov/pub/irs-pdf/i00tt.pdf

. Function Notation 6.. Eercises In Eercises - 0, find an epression for f() and state its domain.. f is a function that takes a real number and performs the followin three steps in the order iven: () multipl ; () add ; () divide b.. f is a function that takes a real number and performs the followin three steps in the order iven: () add ; () multipl b ; () divide b.. f is a function that takes a real number and performs the followin three steps in the order iven: () divide b ; () add ; () multipl b.. f is a function that takes a real number and performs the followin three steps in the order iven: () multipl ; () add ; () take the square root. 5. f is a function that takes a real number and performs the followin three steps in the order iven: () add ; () multipl ; () take the square root. 6. f is a function that takes a real number and performs the followin three steps in the order iven: () add ; () take the square root; () multipl b. 7. f is a function that takes a real number and performs the followin three steps in the order iven: () take the square root; () subtract ; () make the quantit the denominator of a fraction with numerator. 8. f is a function that takes a real number and performs the followin three steps in the order iven: () subtract ; () take the square root; () make the quantit the denominator of a fraction with numerator. 9. f is a function that takes a real number and performs the followin three steps in the order iven: () take the square root; () make the quantit the denominator of a fraction with numerator ; () subtract. 0. f is a function that takes a real number and performs the followin three steps in the order iven: () make the quantit the denominator of a fraction with numerator ; () take the square root; () subtract. In Eercises - 8, use the iven function f to find and simplif the followin: f() f( ) f f() f() f( ) f( ) f() f ( )

6 Relations and Functions. f() +. f(). f(). f() + 5. f() 6. f() 7. f() 6 8. f() 0 In Eercises 9-6, use the iven function f to find and simplif the followin: f() f( ) f(a) f(a) f(a + ) f(a) + f() f a f(a) f(a + h) 9. f() 5 0. f() 5. f(). f() +. f() +. f() 7 5. f() 6. f() In Eercises 7 -, use the iven function f to find f(0) and solve f() 0 7. f() 8. f() 5 9. f() 6 0. f(). f() +. f(). f() 5. Let f() + 5, 9, < + 5, >. f() Compute the followin function values. (a) f( ) (b) f( ) (c) f() (d) f(.00) (e) f(.00) (f) f()

. Function Notation 65 if 6. Let f() if < if > Compute the followin function values. (a) f() (b) f( ) (c) f() (d) f(0) (e) f( ) (f) f( 0.999) In Eercises 7-6, find the (implied) domain of the function. 7. f() + 56 9 8. f() + 9. f() +. f() +. f() + 6 0. f() +. f(). f() 5. f() 6. f() + 5 7. f() 9 + 8. f() 9. f() 6 50. f() 5. f() 6 5. f() 7 + 6 6 6 6 5. f() 6 6 55. s(t) t t 8 57. b(θ) θ θ 8 6 5. f() + 6 r 56. Q(r) r 8 58. A() 7 + 9 59. α() 8 60. (v) v 6. T (t) t 8 5 t 6. u(w) w 8 5 w

66 Relations and Functions 6. The area A enclosed b a square, in square inches, is a function of the lenth of one of its sides, when measured in inches. This relation is epressed b the formula A() for > 0. Find A() and solve A() 6. Interpret our answers to each. Wh is restricted to > 0? 6. The area A enclosed b a circle, in square meters, is a function of its radius r, when measured in meters. This relation is epressed b the formula A(r) πr for r > 0. Find A() and solve A(r) 6π. Interpret our answers to each. Wh is r restricted to r > 0? 65. The volume V enclosed b a cube, in cubic centimeters, is a function of the lenth of one of its sides, when measured in centimeters. This relation is epressed b the formula V () for > 0. Find V (5) and solve V () 7. Interpret our answers to each. Wh is restricted to > 0? 66. The volume V enclosed b a sphere, in cubic feet, is a function of the radius of the sphere r, when measured in feet. This relation is epressed b the formula V (r) π r for r > 0. Find V () and solve V (r) π. Interpret our answers to each. Wh is r restricted to r > 0? 67. The heiht of an object dropped from the roof of an eiht stor buildin is modeled b: h(t) 6t + 6, 0 t. Here, h is the heiht of the object off the round, in feet, t seconds after the object is dropped. Find h(0) and solve h(t) 0. Interpret our answers to each. Wh is t restricted to 0 t? 68. The temperature T in derees Fahrenheit t hours after 6 AM is iven b T (t) t + 8t + for 0 t. Find and interpret T (0), T (6) and T (). 69. The function C() 0 + 7 models the cost, in hundreds of dollars, to produce thousand pens. Find and interpret C(0), C() and C(5). 70. Usin data from the Bureau of Transportation Statistics, the averae fuel econom F in miles per allon for passener cars in the US can be modeled b F (t) 0.0076t + 0.5t + 6, 0 t 8, where t is the number of ears since 980. Use our calculator to find F (0), F () and F (8). Round our answers to two decimal places and interpret our answers to each. 7. The population of Sasquatch in Portae Count can be modeled b the function P (t) 50t t+5, where t represents the number of ears since 80. Find and interpret P (0) and P (05). Discuss with our classmates what the applied domain and rane of P should be. 7. For n copies of the book Me and m Sasquatch, a print on-demand compan chares C(n) dollars, where C(n) is determined b the formula 5n if n 5 C(n).50n if 5 < n 50 n if n > 50 (a) Find and interpret C(0).

. Function Notation 67 (b) How much does it cost to order 50 copies of the book? What about 5 copies? (c) Your answer to 7b should et ou thinkin. Suppose a bookstore estimates it will sell 50 copies of the book. How man books can, in fact, be ordered for the same price as those 50 copies? (Round our answer to a whole number of books.) 7. An on-line comic book retailer chares shippin costs accordin to the followin formula {.5n +.5 if n S(n) 0 if n 5 where n is the number of comic books purchased and S(n) is the shippin cost in dollars. (a) What is the cost to ship 0 comic books? (b) What is the sinificance of the formula S(n) 0 for n 5? 7. The cost C (in dollars) to talk m minutes a month on a mobile phone plan is modeled b { 5 if 0 m 000 C(m) 5 + 0.(m 000) if m > 000 (a) How much does it cost to talk 750 minutes per month with this plan? (b) How much does it cost to talk 0 hours a month with this plan? (c) Eplain the terms of the plan verball. 75. In Section.. we defined the set of inteers as Z {...,,,, 0,,,,...}. 5 The reatest inteer of, denoted b, is defined to be the larest inteer k with k. (a) Find 0.785, 7,.00, and π + 6 (b) Discuss with our classmates how ma be described as a piecewise defined function. HINT: There are infinitel man pieces! (c) Is a + b a + b alwas true? What if a or b is an inteer? Test some values, make a conjecture, and eplain our result. 76. We have throuh our eamples tried to convince ou that, in eneral, f(a + b) f(a) + f(b). It has been our eperience that students refuse to believe us so we ll tr aain with a different approach. With the help of our classmates, find a function f for which the followin properties are alwas true. (a) f(0) f( + ) f( ) + f() 5 The use of the letter Z for the inteers is ostensibl because the German word zahlen means to count.

68 Relations and Functions (b) f(5) f( + ) f() + f() (c) f( 6) f(0 6) f(0) f(6) (d) f(a + b) f(a) + f(b) reardless of what two numbers we ive ou for a and b. How man functions did ou find that failed to satisf the conditions above? Did f() work? What about f() or f() + 7 or f()? Did ou find an attribute common to those functions that did succeed? You should have, because there is onl one etremel special famil of functions that actuall works here. Thus we return to our previous statement, in eneral, f(a + b) f(a) + f(b).

. Function Notation 69.. Answers. f() + Domain: (, ). f() (+) Domain: (, ) +. f() ( + ) + 6 Domain: (, ). f() + Domain: [, ) 5. f() ( + ) + 6 Domain: [, ) 6. f() + Domain: [, ) 7. f() Domain: [0, 69) (69, ) 9. f() Domain: (0, ) 8. f() Domain: (, ) 0. f() Domain: (0, ). For f() + f() 7 f( ) f ( ) f() 8 + f() 8 + f( ) + f( ) 7 f() f ( ) +. For f() f() 9 f( ) 7 f ( ) f() 6 f() 6 f( ) + f( ) 9 f() f ( )

70 Relations and Functions. For f() f() 7 f( ) f f() 6 f() 8 f( ) f( ) +8 f() f ( ). For f() + f() f( ) 6 f f() 6 + f() + 8 f( ) + + f( ) + 0 f() f ( ) + 5. For f() f() f( ) f ( ) f() f() f( ) + f( ) 5 f() f ( ) 6. For f() f() 7 f( ) f 6 7 f() f() 8 f( ) f( ) ( ) +8 6 7. For f() 6 f() ( f ) 6 f() 6 f( ) 6 f ( ) 6 f() 6 f() f( ) 6 f( ) 6 f() f ( ) 6

. Function Notation 7 8. For f() 0 f() 0 f( ) 0 f ( ) 0 f() 0 f() 0 f( ) 0 f( ) 0 f() f ( ) 0 9. For f() 5 f() f( ) 9 f(a) a 5 f(a) a 0 f(a + ) a f(a) + f() a 6 f a a 5 5a a f(a) a 5 f(a + h) a + h 5 0. For f() 5 f() f( ) 9 f(a) 5 a f(a) 0 a f(a + ) a f(a) + f() 6 a f a 5 a 5a a f(a) 5 a f(a + h) 5 a h. For f() f() 7 f( ) 7 f(a) 8a f(a) a f(a + ) a + 8a + 7 f(a) + f() a + 6 f a 8 a 8 a a f(a) a f(a + h) a + ah + h

7 Relations and Functions. For f() + f() 6 f( ) f(a) a + 6a f(a) 6a + 6a f(a+) a +5a+6 f(a)+f() a +a+ f a + 6 a a +6a a a f(a) a +a f(a + h) a + 6ah + h + a + h. For f() + f() 5 f( ) is not real f(a) a + f(a) a + f(a + ) a + 5 f(a)+f() a + + 5 f ( a ) a + a+ a f(a) a+ f(a+h) a + h +. For f() 7 f() 7 f( ) 7 f(a) 7 f(a) f(a + ) 7 f(a) + f() f a 7 f(a) 7 f(a + h) 7 5. For f() f() f( ) f(a) a f(a) a f(a + ) a+ f(a) + f() a + a+ f a a f(a) a f(a + h) a+h

. Function Notation 7 6. For f() f() f( ) f(a) a f(a) a f(a + ) a+ f(a) + f() a + a+ f a a f(a) f(a + h) a a+h 7. For f(), f(0) and f() 0 when 8. For f() 5, f(0) and f() 0 when 5 9. For f() 6, f(0) 6 and f() 0 when ± 0. For f(), f(0) and f() 0 when or. For f() +, f(0) and f() 0 when. For f(), f(0) and f() 0 when. For f(), f(0) and f() is never equal to 0. For f(), f(0) 0 and f() 0 when 0 or 5. (a) f( ) (b) f( ) (c) f() 0 (d) f(.00).999 (e) f(.00).999 (f) f() 5 6. (a) f() (b) f( ) 9 (c) f() 0 (d) f(0) (e) f( ) (f) f( 0.999) 0.07 7. (, ) 8. (, ) 9. (, ) (, ) 0. (, ) (, ) (, ). (, ). (, ) (, ) (, ). (, 6) ( 6, 6) (6, ). (, ) (, ) 5. (, ] 6. [ 5, )

7 Relations and Functions 7. [, ) 8. (, 7] 9. [, ) 50. (, ) 5. (, ) 5. [, ) (, ) 5. [, 6) (6, ) 5. (, ) 55. (, 8) (8, ) 56. [0, 8) (8, ) 57. (8, ) 58. [7, 9] 59. (, 8) (8, ) 60. (, ) (, 0) ( 0, ) (, ) 6. [0, 5) (5, ) 6. [0, 5) (5, ) 6. A() 9, so the area enclosed b a square with a side of lenth inches is 9 square inches. The solutions to A() 6 are ±6. Since is restricted to > 0, we onl keep 6. This means for the area enclosed b the square to be 6 square inches, the lenth of the side needs to be 6 inches. Since represents a lenth, > 0. 6. A() π, so the area enclosed b a circle with radius meters is π square meters. The solutions to A(r) 6π are r ±. Since r is restricted to r > 0, we onl keep r. This means for the area enclosed b the circle to be 6π square meters, the radius needs to be meters. Since r represents a radius (lenth), r > 0. 65. V (5) 5, so the volume enclosed b a cube with a side of lenth 5 centimeters is 5 cubic centimeters. The solution to V () 7 is. This means for the volume enclosed b the cube to be 7 cubic centimeters, the lenth of the side needs to centimeters. Since represents a lenth, > 0. 66. V () 6π, so the volume enclosed b a sphere with radius feet is 6π cubic feet. The solution to V (r) π is r. This means for the volume enclosed b the sphere to be π cubic feet, the radius needs to feet. Since r represents a radius (lenth), r > 0. 67. h(0) 6, so at the moment the object is dropped off the buildin, the object is 6 feet off of the round. The solutions to h(t) 0 are t ±. Since we restrict 0 t, we onl keep t. This means seconds after the object is dropped off the buildin, it is 0 feet off the round. Said differentl, the object hits the round after seconds. The restriction 0 t restricts the time to be between the moment the object is released and the moment it hits the round. 68. T (0), so at 6 AM (0 hours after 6 AM), it is Fahrenheit. T (6), so at noon (6 hours after 6 AM), the temperature is Fahrenheit. T () 7, so at 6 PM ( hours after 6 AM), it is 7 Fahrenheit.

. Function Notation 75 69. C(0) 7, so to make 0 pens, it costs 6 $700. C(), so to make 000 pens, it costs $00. C(5), so to make 5000 pens, it costs $000. 70. F (0) 6.00, so in 980 (0 ears after 980), the averae fuel econom of passener cars in the US was 6.00 miles per allon. F () 0.8, so in 99 ( ears after 980), the averae fuel econom of passener cars in the US was 0.8 miles per allon. F (8).6, so in 008 (8 ears after 980), the averae fuel econom of passener cars in the US was.6 miles per allon. 7. P (0) 0 which means in 80 (0 ears after 80), there are no Sasquatch in Portae Count. P (05) 075 9.77, so in 008 (05 ears after 80), there were between 9 and 0 Sasquatch in Portae Count. 7. (a) C(0) 00. It costs $00 for 0 copies of the book. (b) C(50) 675, so it costs $675 for 50 copies of the book. C(5) 6, so it costs $6 for 5 copies of the book. (c) 56 books. 7. (a) S(0) 7.5, so it costs $7.50 to ship 0 comic books. (b) There is free shippin on orders of 5 or more comic books. 7. (a) C(750) 5, so it costs $5 to talk 750 minutes per month with this plan. (b) Since 0 hours 00 minutes, we substitute m 00 and et C(00) 5. It costs $5 to talk 0 hours per month with this plan. (c) It costs $5 for up to 000 minutes and 0 cents per minute for each minute over 000 minutes. 75. (a) 0.785 0, 7 7,.00, and π + 6 9 6 This is called the fied or start-up cost. We ll revisit this concept on pae 8.

76 Relations and Functions.5 Function Arithmetic In the previous section we used the newl defined function notation to make sense of epressions such as f() + and f() for a iven function f. It would seem natural, then, that functions should have their own arithmetic which is consistent with the arithmetic of real numbers. The followin definitions allow us to add, subtract, multipl and divide functions usin the arithmetic we alread know for real numbers. Function Arithmetic Suppose f and are functions and is in both the domain of f and the domain of. a The sum of f and, denoted f +, is the function defined b the formula (f + )() f() + () The difference of f and, denoted f, is the function defined b the formula (f )() f() () The product of f and, denoted f, is the function defined b the formula (f)() f()() The quotient of f and, denoted f, is the function defined b the formula provided () 0. f () f() (), a Thus is an element of the intersection of the two domains. In other words, to add two functions, we add their outputs; to subtract two functions, we subtract their outputs, and so on. Note that while the formula (f + )() f() + () looks suspiciousl like some kind of distributive propert, it is nothin of the sort; the addition on the left hand side of the equation is function addition, and we are usin this equation to define the output of the new function f + as the sum of the real number outputs from f and. Eample.5.. Let f() 6 and ().. Find (f + )( ). Find (f)(). Find the domain of f then find and simplif a formula for ( f)().

.5 Function Arithmetic 77. Find the domain of f then find and simplif a formula for Solution. ( f ) ().. To find (f + )( ) we first find f( ) 8 and ( ). B definition, we have that (f + )( ) f( ) + ( ) 8 +.. To find (f)(), we first need f() and (). Since f() 0 and () 5, our formula ields (f)() f()() (0) ( 5 ) 50.. One method to find the domain of f is to find the domain of and of f separatel, then find the intersection of these two sets. Owin to the denominator in the epression (), we et that the domain of is (, 0) (0, ). Since f() 6 is valid for all real numbers, we have no further restrictions. Thus the domain of f matches the domain of, namel, (, 0) (0, ). A second method is to analze the formula for ( f)() before simplifin and look for the usual domain issues. In this case, ( f)() () f() so we find, as before, the domain is (, 0) (0, ). ( ) ( 6 ), Movin alon, we need to simplif a formula for ( f)(). In this case, we et common denominators and attempt to reduce the resultin fraction. Doin so, we et ( f)() () f() ( ) ( 6 ) 6 + 6 + 6 6 + et common denominators. As in the previous eample, we have two was to approach findin the domain of f. First, we can find the( domain ) of and f separatel, and find the intersection of these two sets. In addition, since f () () f(), we are introducin a new denominator, namel f(), so we need to uard aainst this bein 0 as well. Our previous work tells us that the domain of is (, 0) (0, ) and the domain of f is (, ). Settin f() 0 ives 6 0

78 Relations and Functions or 0,. As a result, the domain of f is all real numbers ecept 0 and, or (, 0) ( 0, ) (, ). Alternativel, we ma proceed as above and analze the epression f () () f() before simplifin. In this case, () () f f() 6 We see immediatel from the little denominator that 0. To keep the bi denominator awa from 0, we solve 6 0 and et 0 or. Hence, as before, we find the domain of f to be (, 0) ( 0, ) (, ). Net, we find and simplif a formula for f (). () () f f() 6 6 ( ) (6 ) (6 ) ( ) ( ) ( ) simplif compound fractions factor cancel Please note the importance of findin the domain of a function before simplifin its epression. In number in Eample.5. above, had we waited to find the domain of f until after simplifin, we d just have the formula to o b, and we would (incorrectl!) state the domain as (, 0) (0, ), since the other troublesome number,, was canceled awa. We ll see what this means eometricall in Chapter.

.5 Function Arithmetic 79 Net, we turn our attention to the difference quotient of a function. Definition.8. Given a function f, the difference quotient of f is the epression f( + h) f() h We will revisit this concept in Section., but for now, we use it as a wa to practice function notation and function arithmetic. For reasons which will become clear in Calculus, simplifin a difference quotient means rewritin it in a form where the h in the definition of the difference quotient cancels from the denominator. Once that happens, we consider our work to be done. Eample.5.. Find and simplif the difference quotients for the followin functions. f(). () +. r() Solution.. To find f( + h), we replace ever occurrence of in the formula f() with the quantit ( + h) to et f( + h) ( + h) ( + h) + h + h h. So the difference quotient is f( + h) f() h ( + h + h h ) ( ) + h + h h + + h h + h h h h ( + h ) h h ( + h ) h h factor cancel + h.

80 Relations and Functions. To find ( + h), we replace ever occurrence of in the formula () quantit ( + h) to et + with the which ields ( + h) ( + h) + + h +, ( + h) () h + h + + h + h + + h ( + ) ( + h + ) h( + h + )( + ) 6 + 6 6h h( + h + )( + ) 6h h( + h + )( + ) 6 h h( + h + )( + ) 6 ( + h + )( + ). ( + h + )( + ) ( + h + )( + ) Since we have manaed to cancel the oriinal h from the denominator, we are done.. For r(), we et r( + h) + h so the difference quotient is r( + h) r() h + h h In order to cancel the h from the denominator, we rationalize the numerator b multiplin b its conjuate. Rationalizin the numerator!? How s that for a twist!

.5 Function Arithmetic 8 r( + h) r() h + h h ( ) ( ) + h + h + ( ) Multipl b the conjuate. h + h + ( + h ) ( ) h ( + h + ) ( + h) h ( + h + ) h h ( + h + ) h h ( + h + ) + h + Difference of Squares. Since we have removed the oriinal h from the denominator, we are done. As mentioned before, we will revisit difference quotients in Section. where we will eplain them eometricall. For now, we want to move on to some classic applications of function arithmetic from Economics and for that, we need to think like an entrepreneur. Suppose ou are a manufacturer makin a certain product. Let be the production level, that is, the number of items produced in a iven time period. It is customar to let C() denote the function which calculates the total cost of producin the items. The quantit C(0), which represents the cost of producin no items, is called the fied cost, and represents the amount of mone required to bein production. Associated with the total cost C() is cost per item, or averae cost, denoted C() and read C-bar of. To compute C(), we take the total cost C() and divide b the number of items produced to et C() C() On the retail end, we have the price p chared per item. To simplif the dialo and computations in this tet, we assume that the number of items sold equals the number of items produced. From a Not reall, but entrepreneur is the buzzword of the da and we re trin to be trend. Poorl desined resin Sasquatch statues, for eample. Feel free to choose our own entrepreneurial fantas.

8 Relations and Functions retail perspective, it seems natural to think of the number of items sold,, as a function of the price chared, p. After all, the retailer can easil adjust the price to sell more product. In the lanuae of functions, would be the dependent variable and p would be the independent variable or, usin function notation, we have a function (p). While we will adopt this convention later in the tet, 5 we will hold with tradition at this point and consider the price p as a function of the number of items sold,. That is, we reard as the independent variable and p as the dependent variable and speak of the price-demand function, p(). Hence, p() returns the price chared per item when items are produced and sold. Our net function to consider is the revenue function, R(). The function R() computes the amount of mone collected as a result of sellin items. Since p() is the price chared per item, we have R() p(). Finall, the profit function, P () calculates how much mone is earned after the costs are paid. That is, P () (R C)() R() C(). We summarize all of these functions below. Summar of Common Economic Functions Suppose represents the quantit of items produced and sold. The price-demand function p() calculates the price per item. The revenue function R() calculates the total mone collected b sellin items at a price p(), R() p(). The cost function C() calculates the cost to produce items. The value C(0) is called the fied cost or start-up cost. The averae cost function C() C() Here, we necessaril assume > 0. calculates the cost per item when makin items. The profit function P () calculates the mone earned after costs are paid when items are produced and sold, P () (R C)() R() C(). It is hih time for an eample. Eample.5.. Let represent the number of dopi media plaers ( dopis 6 ) produced and sold in a tpical week. Suppose the cost, in dollars, to produce dopis is iven b C() 00 + 000, for 0, and the price, in dollars per dopi, is iven b p() 50 5 for 0 0.. Find and interpret C(0).. Find and interpret C(0).. Find and interpret p(0) and p(0).. Solve p() 0 and interpret the result. 5. Find and simplif epressions for the revenue function R() and the profit function P (). 6. Find and interpret R(0) and P (0). 7. Solve P () 0 and interpret the result. 5 See Eample 5.. in Section 5.. 6 Pronounced dopes...

.5 Function Arithmetic 8 Solution.. We substitute 0 into the formula for C() and et C(0) 00(0) + 000 000. This means to produce 0 dopis, it costs $000. In other words, the fied (or start-up) costs are $000. The reader is encouraed to contemplate what sorts of epenses these miht be.. Since C() C() C(0), C(0) 0 000 0 00. This means when 0 dopis are produced, the cost to manufacture them amounts to $00 per dopi.. Pluin 0 into the epression for p() ives p(0) 50 5(0) 50. This means no dopis are sold if the price is $50 per dopi. On the other hand, p(0) 50 5(0) 50 which means to sell 0 dopis in a tpical week, the price should be set at $50 per dopi.. Settin p() 0 ives 50 5 0. Solvin ives 0. This means in order to sell 0 dopis in a tpical week, the price needs to be set to $0. What s more, this means that even if dopis were iven awa for free, the retailer would onl be able to move 0 of them. 7 5. To find the revenue, we compute R() p() (50 5) 50 5. Since the formula for p() is valid onl for 0 0, our formula R() is also restricted to 0 0. For the profit, P () (R C)() R() C(). Usin the iven formula for C() and the derived formula for R(), we et P () ( 50 5 ) (00+000) 5 +50 000. As before, the validit of this formula is for 0 0 onl. 6. We find R(0) 0 which means if no dopis are sold, we have no revenue, which makes sense. Turnin to profit, P (0) 000 since P () R() C() and P (0) R(0) C(0) 000. This means that if no dopis are sold, more mone ($000 to be eact!) was put into producin the dopis than was recouped in sales. In number, we found the fied costs to be $000, so it makes sense that if we sell no dopis, we are out those start-up costs. 7. Settin P () 0 ives 5 + 50 000 0. Factorin ives 5( 0)( 0) 0 so 0 or 0. What do these values mean in the contet of the problem? Since P () R() C(), solvin P () 0 is the same as solvin R() C(). This means that the solutions to P () 0 are the production (and sales) fiures for which the sales revenue eactl balances the total production costs. These are the so-called break even points. The solution 0 means 0 dopis should be produced (and sold) durin the week to recoup the cost of production. For 0., thins are a bit more complicated. Even thouh. satisfies 0 0, and hence is in the domain of P, it doesn t make sense in the contet of this problem to produce a fractional part of a dopi. 8 Evaluatin P () 5 and P () 0, we see that producin and sellin dopis per week makes a (sliht) profit, whereas producin just one more puts us back into the red. While breakin even is nice, we ultimatel would like to find what production level (and price) will result in the larest profit, and we ll do just that... in Section.. 7 Imaine that! Givin somethin awa for free and hardl anone takin advantae of it... 8 We ve seen this sort of thin before in Section...

8 Relations and Functions.5. Eercises In Eercises - 0, use the pair of functions f and to find the followin values if the eist. (f + )() (f )( ) ( f)() (f) f (0) f ( ). f() + and (). f() and () +. f() and (). f() and () 5. f() + and () 6. f() and () + 7. f() and () + 8. f() and () 9. f() and () 0. f() + and () + In Eercises - 0, use the pair of functions f and to find the domain of the indicated function then find and simplif an epression for it. (f + )() (f )() (f)() f (). f() + and (). f() and (). f() and (). f() and () 7 5. f() and () + 6 6. f() + + 6 and () 9 7. f() and () 8. f() and () 9. f() and () + 0. f() 5 and () f() 5 In Eercises - 5, find and simplif the difference quotient f( + h) f() h. f() 5. f() + 5. f() 6. f() 5. f() + 6. f() for the iven function.

.5 Function Arithmetic 85 7. f() 8. f() + 9. f() m + b where m 0 0. f() a + b + c where a 0. f(). f(). f(). f() + 5 5. f() 6. f() + 7. f() 9 8. f() + 9. f() 9 0. f() +. f() + 5. f(). f() a + b, where a 0.. f() 5. f(). HINT: (a b) ( a + ab + b ) a b In Eercises 6-50, C() denotes the cost to produce items and p() denotes the price-demand function in the iven economic scenario. In each Eercise, do the followin: Find and interpret C(0). Find and interpret p(5) Find and simplif P (). Find and interpret C(0). Find and simplif R(). Solve P () 0 and interpret. 6. The cost, in dollars, to produce I d rather be a Sasquatch T-Shirts is C() + 6, 0 and the price-demand function, in dollars per shirt, is p() 0, 0 5. 7. The cost, in dollars, to produce bottles of 00% All-Natural Certified Free-Trade Oranic Sasquatch Tonic is C() 0 + 00, 0 and the price-demand function, in dollars per bottle, is p() 5, 0 5. 8. The cost, in cents, to produce cups of Mountain Thunder Lemonade at Junior s Lemonade Stand is C() 8 + 0, 0 and the price-demand function, in cents per cup, is p() 90, 0 0. 9. The dail cost, in dollars, to produce Sasquatch Berr Pies C() + 6, 0 and the price-demand function, in dollars per pie, is p() 0.5, 0.

86 Relations and Functions 50. The monthl cost, in hundreds of dollars, to produce custom built electric scooters is C() 0 + 000, 0 and the price-demand function, in hundreds of dollars per scooter, is p() 0, 0 70. In Eercises 5-6, let f be the function defined b and let be the function defined f {(, ), (, ), (, 0), (0, ), (, ), (, ), (, )} {(, ), (, 0), (, ), (0, 0), (, ), (, ), (, )}. Compute the indicated value if it eists. 5. (f + )( ) 5. (f )() 5. (f)( ) 5. ( + f)() 55. ( f)() 56. (f)( ) 57. f ( ) 58. f ( ) 59. f () 60. ( f ) ( ) 6. ( f ) () 6. ( f ) ( )

.5 Function Arithmetic 87.5. Answers. For f() + and () (f + )() 9 (f )( ) 7 ( f)() (f) 5 f (0) f ( ) 6 5. For f() and () + (f + )() (f )( ) ( f)() (f) ( ) 0 f (0) 0 f ( ) 5. For f() and () (f + )() 0 (f )( ) 9 ( f)() (f) 7 6 f (0) 0 f ( ). For f() and () (f + )() 5 (f )( ) 0 ( f)() 8 (f) 7 6 f (0) 0 f ( ) 6 5. For f() + and () (f + )() + 5 (f )( ) + ( f)() (f) ( ) 0 f (0) f ( ) 5 6. For f() and () + (f + )() + (f )( ) + 5 ( f)() 0 (f) 5 f (0) f ( ) 0

88 Relations and Functions 7. For f() and () + (f + )() 5 (f )( ) ( f)() 5 (f) f (0) 0 f ( ) 8. For f() and () (f + )() 7 (f )( ) 8 5 ( f)() (f) 8 f (0) 0 f ( ) 8 9. For f() and () (f + )() 7 (f )( ) 0 ( f)() 0 (f) ( ) f (0) is undefined. f ( ) 6 0. For f() + and () + (f + )() 6 5 (f )( ) ( f)() (f) ( ) f (0) f ( ) 5. For f() + and () (f + )() Domain: (, ) (f)() Domain: (, ) (f )() + Domain: (, ) f () + Domain: (, ) (, ). For f() and () (f + )() Domain: (, ) (f)() 8 + 6 Domain: (, ) (f )() 6 Domain: (, ) f () Domain: (, ) (, )

.5 Function Arithmetic 89. For f() and () (f + )() + Domain: (, ) (f)() Domain: (, ) (f )() + Domain: (, ) f () Domain: (, ) (, ). For f() and () 7 (f + )() + 6 Domain: (, ) (f)() 7 7 Domain: (, ) (f )() 8 Domain: (, ) f () 7 Domain: (, 0) (0, ) 5. For f() and () + 6 (f + )() + + Domain: (, ) (f)() + 6 Domain: (, ) (f )() 0 Domain: (, ) f () Domain: (, ) (, ) 6. For f() + + 6 and () 9 (f + )() Domain: (, ) (f)() + + 5 9 5 Domain: (, ) (f )() + + 5 Domain: (, ) f () + + Domain: (, ) (, ) (, ) 7. For f() and () (f + )() + Domain: (, 0) (0, ) (f)() Domain: (, 0) (0, ) (f )() Domain: (, 0) (0, ) f () Domain: (, 0) (0, )

90 Relations and Functions 8. For f() and () (f + )() + Domain: (, ) (, ) (f)() Domain: (, ) (, ) (f )() Domain: (, ) (, ) f () + Domain: (, ) (, ) 9. For f() and () + (f + )() + + Domain: [, ) (f)() + Domain: [, ) (f )() + Domain: [, ) f () + Domain: (, ) 0. For f() 5 and () f() 5 (f + )() 5 Domain: [5, ) (f)() 5 Domain: [5, ) (f )() 0 Domain: [5, ) f () Domain: (5, )... 0. 6 + h 5. h + 6. 8 + h 7. h + 8. + h + h 9. m 0. a + ah + b.. 5. ( + h). ( + h) ( + h). ( )( + h ) 6. ( h)( ) ( + 5)( + h + 5) ( + )( + h + )

.5 Function Arithmetic 9 7. 9... 5. 9 ( 9)( + h 9) 8. + h 9 + 9 0. h + 5 + + 5. a a + ah + b + a + b. ( + h) / + ( + h) / / + / + h + + h ( + )( + h + ) + h + + + h + + h + h ( + h) / + / 6. C(0) 6, so the fied costs are $6. C(0).6, so when 0 shirts are produced, the cost per shirt is $.60. p(5) 0, so to sell 5 shirts, set the price at $0 per shirt. R() + 0, 0 5 P () + 8 6, 0 5 P () 0 when and. These are the break even points, so sellin shirt or shirts will uarantee the revenue earned eactl recoups the cost of production. 7. C(0) 00, so the fied costs are $00. C(0) 0, so when 0 bottles of tonic are produced, the cost per bottle is $0. p(5) 0, so to sell 5 bottles of tonic, set the price at $0 per bottle. R() + 5, 0 5 P () + 5 00, 0 5 P () 0 when 5 and 0. These are the break even points, so sellin 5 bottles of tonic or 0 bottles of tonic will uarantee the revenue earned eactl recoups the cost of production. 8. C(0) 0, so the fied costs are 0 or $.0. C(0), so when 0 cups of lemonade are made, the cost per cup is. p(5) 75, so to sell 5 cups of lemonade, set the price at 75 per cup. R() + 90, 0 0 P () + 7 0, 0 0 P () 0 when and 0. These are the break even points, so sellin cups of lemonade or 0 cups of lemonade will uarantee the revenue earned eactl recoups the cost of production.

9 Relations and Functions 9. C(0) 6, so the dail fied costs are $6. C(0) 6.6, so when 0 pies are made, the cost per pie is $6.60. p(5) 9.5, so to sell 5 pies a da, set the price at $9.50 per pie. R() 0.5 +, 0 P () 0.5 + 9 6, 0 P () 0 when 6 and. These are the break even points, so sellin 6 pies or pies a da will uarantee the revenue earned eactl recoups the cost of production. 50. C(0) 000, so the monthl fied costs are 000 hundred dollars, or $00,000. C(0) 0, so when 0 scooters are made, the cost per scooter is 0 hundred dollars, or $,000. p(5) 0, so to sell 5 scooters a month, set the price at 0 hundred dollars, or $,000 per scooter. R() + 0, 0 70 P () + 0 000, 0 70 P () 0 when 0 and 50. These are the break even points, so sellin 0 scooters or 50 scooters a month will uarantee the revenue earned eactl recoups the cost of production. 5. (f + )( ) 5. (f )() 5. (f)( ) 0 5. ( + f)() 0 55. ( f)() 56. (f)( ) 8 57. f ( ) does not eist 58. f ( ) 0 59. f () 60. ( f ) ( ) does not eist 6. ( f ) () 6. ( f ) ( )