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# Col Alg for Math,Science,Engin MTH 111C

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Date Created: 10/19/15

Haberman MTH 1110 Section IV Power Polynomial and Rational Functions Module 5 Graphing Rational Functions In the previous module we studied the longrun behavior of rational functions Now we will study the shortrun behavior so that we can sketch the complete graphs of rational functions Like polynomial functions the shortterm behavior of rational functions includes roots and y intercepts In addition we must also pay attention to the values that are excluded from the domains of rational functions To get started let s consider some simple rational functions In the next two examples we will sketch graphs of fx i and gx The reason to look at these examples is they x x are very similar both have horizontal asymptote y 0 and domain xxeRandx at 0 The only difference between these functions is that the factor of x in the denominator is squared in g but not in f 80 these examples will help us determine how the degrees of the factors in the denominator affect the shape of the graphs of rational functions L EXAMPLE Sketch a graph of the rational function fx x SOLUTION Let s first establish the longrun behavior Clearly as x gt ioo fx gets closer and closer to zero so y 0 is the horizontal asymptote When x gt oo fx is positive so the graph is above the horizontal asymptote y 0 but when x gt oo fx is negative so the graph is below the horizontal asymptote y 0 This information leads to the following graph of the longrun behavior of fx For the shortrun behavior we need to find the roots and the yintercepts but this function has NO roots and NO yintercept There is no yintercept since 0 is not in the domain of f and there are no roots since i at 0 for all x So the only other thing to x look for are values that must be excluded from the domain of f Obviously 0 must be excluded from the domain of f Let s investigate what happens to the graph of f as x gets closer and closer to 0 In Table 1 we ll see what happens as x gets close to zero from the right side of zero ie from above 0 and in Table 2 we ll see what happens as x gets close to zero from the left side of zero ie from below 0 Table 1 Table 2 x fx x fx 10 710 100 7100 W 1000 w 71000 W 10000 W 710000 M 100000 m 7100000 We can see in Table 1 that as x gets close to 0 from the right side of 0 fx gets bigger and bigger so f increases without bound From Table 2 we can see that as x gets close to 0 from the left side of 0 fx gets smaller and smaller so f decreases without bound We can use this information to sketch the complete graph of f see Figure 2 L x Figure 2 The graph of fx The line x 0 Le the yaxis is an important feature in the graph of f the graph gets closer and closer to this line but never crosses it This vertical line is an asymptote Since it is a vertical line it is called a vertical asymptote DEFINITION A vertical asymptote is a vertical line that the graph of a function never crosses As the graph of a function approaches this vertical line the outputs either increase without bound or decrease without bound EXAMPLE Sketch a graph of the rational function gx x SOLUTION Just as with fx in the example above gx gets closer and closer to 0 as x gt ioo so y 0 is the horizontal asymptote But unlike f as x gt ioo gx is always positive so the graph is always above the line y 0 The longrun behavior of g is graphed in Figure 3 1 Just like fx the function gx 2 has no roots and no y intercept And since g x isn t defined at 0 it probably has vertical asymptote x 0 just like f Let s study at a couple of appropriate tables to make sure Table3 Table4 T x gm 100 100 10000 10000 W 1000000 m 1000000 m 100000000 W 100000000 Tahtes 3 arm 4 suggestthat W dues th taet have a vettteat asymptute at x 0 x Eutunhke x t g increases thhuutbuund eth betth the tett a d hght stue ufU We eah use thts thtethhatteth tet sketch a graph ett g th thute 4 hetetw Figure 4 The graph ett Eefure we muve Elm tet s ubsENE what we cart team truth these tast tvvu examptes Bath 1 1 x 7 and x have vertteat asymptettes at x 0 X2 thts ts because x ts a femur th the dEanmatDr th mhetteth f x s a ttheattaetettahu the behavtur atthevertteat asymptute x 0 ts dt eteht eth the tett a d hght stues ett 0t VWHE th mhetteth g x ts a quadrattcfa ur t e t tt ts squared arm the behavtur ts the same eth bum stues ett 0 tt turns nut that thts ts atways the ease tt the femur ts squared then the b avmr ts the same eth bum stues ett the asymptute but tt the taetett ts ttheah the behavtur ts dWErEnt eth bum stues ett the asymptute Yuu ve ptehahty aheauy rEahZEd that tt the taetett has any EVEN petweh the behavmrs tht be has an MP hPha mr vth be uttteteht eth betth stues ett the asymptute EXAMPLE Sketch the graph uf the tuhetteth m 2x2 8 x a 1 SOLUT ON th the prevmus meuute we determtned that the Dngrrun behavtur tut ht see thute 5 hetetw 2x 78 x712 hunzunta asymptute y ho Tu sketch the sndmdn behavmr uf h we need m nnd the mats yrmtercept and the venma asymptutes Tu nnd the mats we eed m nnd 3 x suen that x 0 Of nurse the NW way that hx 0 s mne numerator uf h s 0 d we need m selwe 2x2 8 0 2x2780 2x2740 2x72x20 Thus 2 and 72 are rum Tu ndthe yrmtercept vva need m Eva uate hm 2 h020 2 0 1 8 5d the ynntereept s 0 e 8 Tu nnd the venma asymptutes we need m determme wnat Va uES make the denummatur 0 se we need m sedve x 71 8 Easy m see thatthe un y sendndn s 1 Therefure x1 m m m 8 x mm d n nnnm new us thts vertteat asymptute ts a quadrant femur t e r t ts squared we knew that the graph has the same behavmr uh ppth stg s at the vertteat asymptute When we ptut aH at the thtprrhattph we ve dEtErWWEd thus tar see thure B we eah see that m urder tn avmd the BMW pusstpthty ts that the graph decreases wthput ppuhg uh ppth stges at the vertteat asymptute 2x 78 ptmpW m Ed by the mats yrmtErEE 3 g pt vertteat ahg huthhtat asymptutes we eah sketch the eprhptete graph at h see thure 7 Figure7 The graph pt atuhg wth asyrhptutes v and x 1 Be sure to convince yourself that based on what we discovered about the longterm 2 2 8 behavior and the roots and vertical asymptote the graph of kx MUST look x like the one pictured in Figure 7 You should also graph this function yourself on your graphing calculator for practice Notice that we were forced to cross the horizontal asymptote at x m 25 in order to have the appropriate longterm behavior as well as the root x 2 amp Key Point It is perfectly reasonable for the graph of a rational function to cross its horizontal asymptote when the xvalues aren t very large EXAMPLE Sketch the graph of the function kx 2 x 25 x SOLUTION In the previous module we determined that the longrun behavior for k see Figure 8 Figure 839The longrun behavior x of kx x2 25 To sketch the shortrun behavior of k we need to find the roots yintercept and the vertical asymptotes To find the roots we need to find all x such that kx 0 Of course the only way that kx 0 is if the numerator of k is 0 which obviously occurs when x 0 so 0 is the only root To find the yintercept we need to evaluate k0 Since 0 is a root we know that kO 0 so the yintercept is 0 0 So the xintercept and the yintercept are the same point To find the vertical asymptotes we need to determine what values make the denominator 0 so we need to solve x2 25 0 7250 xe5x5o F5 prx h retpre the We x 5 and x 75 are the vertteat asymptutes at k Let s ptpt 3 at the graphteat behavmrvve have determmed thusfar uf k x amn Wth 3 7 25 a tts the vemeat asymptutes x75 ahg X5 and the pumt 0 0 g facturs x e 5 and x 5 have an hwstpte EXpDnEnt pt 1 we knuvvthat the graph has the gtttereht behavmr uh ppth stges at the vertteat asymptute 5 t the graph mErEaSES x re r ia d the same tstrue abuut x e5 Usmg Whatvve ve ptptteg h thure B WE eah kmg sure that 3 at the behavmrvve ve dEtErmmEd ts saus ed hhtsh the graph t k by rha ahg avmdmg any Euntradmtmns see thure m x Figure10 The graph at m 725 a g wtth t the vemeat asymptutes x 75 and x 5 Be sure to convince yourself that based on what we discovered about the longrun x 5 MUST look behavior and the roots and vertical asymptote the graph of kx 2 x like the one pictured in Figure 10 You should also graph this function yourself on your graphing calculator for practice x3 16x EXAMPLE Sketch a graph of the function mx 2 64 x SOLUTION In the previous module we determined that the longrun behavior for m see Figure 11 Figure of x3 16x x2 mx along with its oblique asymptote y x To sketch the shortrun behavior of m we need to find the roots yintercept and the vertical asymptotes To find the roots we need to find all x such that mx 0 So we need to solve x3 16x 0 x3 16x 0 2 xx2 160 xx 4x40 x0 or x4 or x 4 Thus 0 4 and 7 4 are roots To find the yintercept we need to evaluate m0 Since 0 is a root we know that m0 0 so the yintercept is 0 0 10 Tu nd the vemca t asymptutes we need In determme mhatvames make the denummatur needtu sutve x2 7 64 0 7640 x78x 0 x T erefme the hhes x 8 and x 78 are the vemeat asymptutes uf m Let s met 3 uf the graphteat behavmrvve have determmed thusfar x3716 Figure12 The tuhgrmh behavturuf mx 2 6 x 7 4 atuhg thh ts ubhque asymptute y s A ts vemeat asyrh tutes x 78 and x 8 and the pmhts o 0 04 t and 0 74 12 m a t s 3H uf the behavmr We ve determmed S satts ed and avmdmg any nntradmtmns see gure 13 x 716x Fi ure13 The Va h uf mx t 9 a p X2764 atuhg thh tts ubhque asymptute ts VENEa asymptutes x 78 and x 8 Be sure to convince yourself that based on what we discovered about the longrun 3 16 behavior and the roots and vertical asymptote the graph of mx MUST look x like the one pictured in Figure 13 You should also graph this function yourself on your graphing calculator for practice EXAMPLE Find an algebraic rule for the function g graphed in Figure 14 Figure 14 The graph of y gx SOLUTION Since the graph has vertical asymptotes we know that g is a rational function Since the vertical asymptotes are x 4 and x 4 we know that the factors x 4 and x 4 must appear in the denominator Since g has roots 73 and 5 its rule must have factors x 3 and x 5 in the numerator This tells us that the rule for g looks like k x 3 x 5 gm 2 x x 4x 4 Now we need to find k Let s use the horizontal asymptote y 4 to find k Since y 4 is the horizontal asymptote we know that for extreme x values ie as x gt ioo gx m 4 So kx3x 5NE gx x4x 4Nx2 k 439 4x 3x 5 Therefore gx x 4x 4 39 i Try this one yourself and check your answer Find an algebraic rule for the function h graphed in Figure 15 SOLUTION Since the graph has vertical asymptotes we know that h is a rational function Like the graph of h above since the vertical asymptotes are x 74 and x 4 we know that the factors x 4 and x 7 4 must appear in the denominator But in this case since the behavior is the same on both sides of the asymptote x 4 we know that the factor x 7 4 must be squared or raised to any even power Since I has roots 72 and 8 its rule must have factors x 2 and x 7 8 in the numerator Thistells usthat the rule for h looks like Mx kx 2x 7 82 x 4x 7 4 Now we need to find k In this case we cannot use the horizontal asymptote y 0 to find k We need to use a graph feature that is NOT on the x axis You should try to nd k using the horizontal asymptote y 0 to convince yourself that it isn t possible Instead let s use the y intercept 0 72 Since 0 72 is the y intercept we know that h072 So k28 h o 72 0 404Z 716k7 W 5 772 5 k8 8x 2x 7 8 Therefore hx x 4Xx 7 4f Habermaanling MTHlllc Section 1 Sets and Functions Module 2 Introduction to Functions A function is a special type of binary relation 80 before we discuss what a function is we need to define binary relation DEFINITION A binary relation is a rule that establishes a relationship between two sets Note that this is an informal definition The two sets involved in a binary relation play different roles these roles are determined by the rule of the relation see the first example The two roles that the sets play are the first setquot or the set of inputs and the second setquot or the set of outputsquot The set of inputs is called the domain the set of outputs is called the range 4 EXAMPLE a If the rule of the relation is Associate each person with hisher social security number then the domain is the set of people and the range is the set of social security numbers b If the rule of the relation is Associate each social security number with the person who has been assigned to that numberquot then the domain is the set of social security numbers and the range is the set of people EXAMPLE Consider a binary relation defined on the following two sets Set of inputs domain Peter Kandace Elijah Logan Set of outputs range Cat Dog Mouse The rule that relates these sets is that each person in the set of inputs is related to an animal from the set of outputs that he or she has as a petquot The arrow diagram below defines this relation Inputs Outputs dz Hg 3 Peter R l quot 39 l Cat x V quotquot x f Kandace Elijah quot I quot 4 Mouse 3 quotk Logan j 2 39 fquot 1 The arrow diagram tells us for example that Logan is related to Dogquot since there is an arrow from Logan to Dog and this means that Logan has a dog We also see that Logan is related to Mousequot so Logan also has a mouse Another way to convey the information contained in a relation is to use ordered pairs To represent the fact that Logan is related to Dogquot we use the ordered pair Logan Dog The first term in each ordered pair is an element of the set of inputs and the second term in each ordered pair is an element of the set of outputs The entire relation represented by the arrow diagram above can be represented by the set of ordered pairs below Peter Dog Kandace Cat Kandace Dog Kandace Mouse Elijah Cat Logan Dog Logan Mouse corresponds to exactly one element in the range quotDEFINITION A function is a binary relation in which each element in the domain EXAMPLE The arrow diagram below represents a binary relation on the sets Steve Nancy Kyra Atticus and Boy Girl The quotrulequot that defines this relation is Associate each person with hisher genderquot This relation is a function since each person has exactly one gender On the arrow diagram we can tell that this is a function since there is exactly one arrow coming out of each element in the domain Notice that in the example above more than one arrow comes from some elements in the domain so the diagram does not represent a function Steve A a Domain Rang dd h 7 P V I z 39a f ll BOY y Nancy Kyra f j A l Glrl Atticus f 2 2 H 1 We can represent this function by the following set of ordered pairs Steve Boy Nancy Girl Kyra Girl Atticus Boy EXAMPLE Determine pnetner eaen pt tne quuvvirig reiatipns describes a mnetipn vntn dumain 1 2 3 4 and range 5 67 8 a 1 5 2 s 3 7 4 8 b 1 7 1 8 2r 5 3r 6 c 1 5 2 5 3 5 4 5 SOLUT ONS a 1 5 2 s 3 7 4 8 represents amnetipn because eaen eiernentintne dumain eprresppngs tp examy one eiernent in tne range b 1 7 182 5 3 6 due nut represent a functiun because tne number 1 eprresppngs tp two drtterent eiernents pttne range c 1525 35 4 5 is a mnetipn because eaen eiernent in tne dumain currespunds tn exaLM one eiernent in tne range Nate tnat it is may tn repeat eiernents in tne range EXAMPLE Tne tapie beiqu defines the functiun Where the inputs eurne frum A7 4737271 01 2 345 and the uutputs eurne frum 3 7573 01 234 9 13 Tnus tne set A istne domain at minie tne set 3 is tne range at f mien We ean represent vntn tne nutatiun AgtB it sprnepne were tp ask yuu Where dues sand 727 yuu Wpuig hupequy say y send 72 tn 4 since tnere is a epiurnn in tne tapie snpvnng tnat tne eiernent 72 in tne dumain is reiateg tn 4 m tne range Function notation uffers anutnerway tn eurnrnunieate tnis infurrnatiun e24 rneans y sendermAt Use tne tapie abuve tp Evamate tne quuvvirig a Evamate 43 b Evamate 1 c Evamate 9 u CLICK HERE FORANSWER Sometimes we don t know what the input is In such a case we can call the input x or any other variable where x represents a generic element of the domain then fx represents the element in the range that x is associated with under the function f In this example the domain of f is the set A and the range is the set B Thus xeA and fx 6 B With this new notation we can rename the rows in ourtable x I 5 4 3 2 1012345 fltxgt4 3 s41309 32 13 Solve the following for x d Solve fx1 e Solve fx 3 f Solve fx5 CLICK HERE A CLICK HERE CLICK HERE FORANSWER f FOR ANSWER 39 i FOR ANSWER EQUATIONS AS FUNCTIONS Functions are often defined by equations For example the equation y x 3 defines a function where the inputs are represented by x and the outputs are represented by y It is often helpful to write equations like y x 3 in function notation yx3 3 fxx3 which is read quotfof x equals x plus 3quot Similarly the equation u 3r 5 can be written in function notation as gt 3r 5 This is read g of t equals 3t minus 5quot Cautlon The notation fx does not mean multiplication of f by x Function notation involving fx is more compact than notation involving quoty quot For example With function notation If fx x 3 then f5 8 Without function notation If y x3then the value of y is 8 when x is 5 The independent variable refers to the variable representing possible values in the domain and the dependent variable refers to the variable representing possible values in the range Thus in our usual ordered pair notationx y x is the independent variable and y is the dependent variable It is customary to plot the independent variable on the horizontal axis and the dependent variable on the vertical axis Q EXAMPLE Fwd the IndIcatEd functIun venue a 7 If x 3x7 5 x4 b gEZ If gx c hm If m 71 5 a k0 If km wf 13m35 7 62m 6 SOLUTIONS 3 f0375 175 c h717715 15 6 d 1c005 13036 7 620 6 6 1 CLICK HERE EONA FUNCTION NOTATION EXAMPLE CLICK HERE EON ANOTNEN EuNcTION NOTATION EXAMPLE EXAMPLE Suppuse x a f2a b a 5 c fab SOLUT ONS a f2a 7 2a2 7 2a 7 4112 7 2a b fa 5 1752 701775 abab7a7b a22ab527a7b c aba27ab EXAMPLE Suppuse gx 3x Fwd a gx 2 b gx 2 c g2x a g2x 7 2 SOLUT ONS a gx 2 7 30 2 x b gx23x2 c g2x 32x 6x d g2x 2 32x 2 6x 6 EXAMPLE The function f is graphed in Figure 1 below Note that there is an arrow on the lefthand side of the graph since the function doesn t end when the graph ends If the graph ends it will be denoted with a filledin circle as there is on the righthand side Determine the following a f0 339 the domain of f 5quot any xvalues for which fx 2 D any xvalues for which fx 1 e the range of f Figure 1 Graph of y fx SOLUTIONS a f1 3 which can be determined because the point 1 3 is on the graph b The domain of f is 00 3 since the graph suggests that there is a point above or below every xvalue less than or equal to 3 every xvalue less than or equal to 3 is quotusedquot by the function so it39s in the domain c fx 2 when x 22 because f 22 m 2 Notice that we use to denote the fact that we are approximating the answer d fx 1whenx lorx3 e The range of f is 3 00 since the graph suggests that there is a point to the left or right of every yvalue greater than or equal to 3 every yvalue greater than or equal to 3 is quotusedquot by the function so it39s in the range EXAMPLE The function p is graphed in Figure 2 below Find the following a p1 IT the domain of p c all xvalues for which px 2 d the range of p SOLUTIONS a p1 1 because the point 1 1 is on the graph Note that there is a hole at the point 1 3 so the point 1 3 is NOT on the graph b The domain of p is 5 3 in interval notation or xlxe R and 5 S x S 3 in set builder notation c The xvalues for which px 2 are x 4 and x m 15 d The range of p is 3 1 U 0 4 in interval notation 9 Recall that the definition of a function requires that each input has only one corresponding output Since the inputs are xvalues and the outputs are y values a function can have only one y value for each xvalue If we translate this information to the context of the graph of a function we quickly realize that since each xvalue is represented by a vertical line the graph of a function can have at most one point on each vertical line If a vertical line x a passes through a graph more than one time then the xvalue a has more than one corresponding output so the graph cannot represent a function This leads us to the Vertical Line Test The VerticalLine Test If it is possible for a vertical line to cross a graph more than once then the graph does not represent a function EXAMPLE In which of the following graphs is y a function of x SOLUTION The graph in Figure 4 is the only one that passes the vertical line test so this is the only graph in which y a function of x 10 EXAMPLE The mhetmh y gx ts de ned m thure B the mhetteh y fx ts de ned m Tame 1 and hx x 1 Fwd each uf the quuvvthg a 1 e h2x3 b gm r x fx7 c M73 9 x tt gx 0 d g4 h x M 5 SOLUT ONS a 1 71 usmgthetabte b g714 usmgthe graph Nuts that gm 3 SWEE 713 ts hut a pumt eh the graph The suhd etretes represent pmnts eh the graph whereas the DpEn mrde at 71 3 represents a hate m the graph c M73 73 1 1 10 d g4 fen mt catcmate g4 usmgthe graph then mama s71 homthetabte P n 9 h2x 3 2x 32 1 the input is the entire expression 2x 3 2x 32x 3 1 4x26x6x91 4x2 12x 10 Since f2 7 if fx 7 then x 2 Look for an output value of 7 in the table To solve this equation we need to find an output value of 0 on the graph of y gx So we are looking forthe x intercepts on the graph of y gx Since g 3 0 and g3 0 if gx 0 then x 3 or x 3 Another way of writing this using set notation is xe 3 3 or we could say The solution set for the equation gx 0 is 3 3 hx5 2 x215 x24 3x2 or x 2 3 HabermanKling MTH lllc Section II Exponential and Logarithmic Functions Module 2 Exponential Models EXAMPLE Suppose that the function mt 500945 models the population of a certain species of monkeys in South America I years after 1998 Describe this monkey population SOLUTION Based on our study in Module 1 we know that an exponential function of the form fx a1 rquot has initial value a and growth rate r per unit of x Thus we can determine immediately that the monkey population modeled by mt 500945 has initial value 50 and since 0945 1 0055 the growth rate is 55 per year since the growth rate is negative the population is decreasing CONCLUSION There were 50 monkeys in South America in 1998 and the population is decreasing at the rate of 55 per year EXAMPLE Suppose that the population of a certain species of snakes in South America was 136 in 1998 If this snake population is increasing at the rate of 27 per year find a function that models this snake population SOLUTION Based on our study in Module 1 we know that an exponential function of the form fx a1 rquot has initial value a and growth rate r per unit of x Thus we can determine immediately that this snake population is modeled by the exponential function st 1361027t where t is years after 1998 EXAMPLE Suppuse that the pppu1a0pn pr 3 enam s 2112 pr rudent 1n Suuth Arnenea Was 69961n 19 8 Suppuse funherthat m 2002 the pppu1a0pn pr 1715 same spee1es pr rudent 1n Suuth Arnenea was 8077 1r 171s rudent pppu1a0pn 1s mereasmg exppnenuauu mm a rummn that mpue1sm1s rudent pppu1a0pn 1 years after 1998 SOLUT ON Smee we are 1pm thatthe rudentpupu atmn1s1ncreas1ng exppnenuauu we knuvvthat an exppnenua1 rummn wH made the pppu1a0pn 5 we 8an that the rummn we are 1mm fur has rprrn r0 a b Smee z 0 represents 1998 we knuvvthat the 1mm vame 151172 rudent pppu1a0pn m 1998 5 we ean 1mmed1ate y see that a 6996 r and that r0 6996 0 7p mm b we ean use the fact that m 2002 the pppu1a0pn pr 1715 same spee1es pr rudent 1n Suuth Arnenea Was 8077 Smee 2002 currespunds thh z 4 WE SEE that r4 8077 2 6996 A 8077 j 50 8177 Thus 1 r0 6996 6996 m underthe Expunent and the umer une1sn u EXAMPLE Fwd the rum fur an EXpunEnUa funmun passmg mmth the We pumts 1 6 and 3 24 SOLUT ON pans mm Equatmnsmvu vmg 1 6 2 f0 3 24 2 3 24 fxa AX f1a A 3 a A3 24 Let s use the rst Equatmn and suwe fur a s As2 a a b A aA324 2A324 2 2224 Notemat ahhough both 2 and 72 sche A2 4 omy 2 y vesem A sweethe base 0 an exponenha hmchon 5 Ways posmve Smce A 2 we knuvvthat the uesweu functer 5 fx a AX We can use enher une uf the gwen urdered pawsm nd a 16 2 1a2 6 2 a 3 Thus 7 3 2 Checktms yuurse i Try this one yourself and check your answer If you know that f is an exponential function and you know that f73 and f2 20 nd a rule for f SOLUTION Since we know that the desired function is exponential we know that it has form fx abquot We can now use this definition of f to create equations base on the given information f73 ab 3 f2 20 abz We can use the first equation to solve for a 727 72 31 3 a b 7 8 gt a b b 8 I Now we can substitute this expression for a into the second equation and solve for 12 ab2 20 b3 42 20 5 5 7 5 b 7 20 8 39 b5 84 5532 5 122 U Since b 2 we know that the desired function is fx aZ We now use the fact that f2 20 to find a f220a2Z 2045 gt 515 So it looks like fx 52quot 23 Check f 3 m and f2 522 2 5i 5 39 4 3 2 20 2 Both function values check Therefore fx 52quot Try this one yourself and check your answer Find the equation of the exponential function gx that passes through 2 5 and 5 63 SOLUTION Since both of these points should satisfy the equation gx ab we can use them to obtain a system of equations 25 2 g25 2 ab25 1 563 2 g563 2 abs 63 2 Solving the second equation 2 for a we obtain a ii and we can use this equation to substitute for a in the first equation 1 abs 63 2 135 63 We can now solve this equation for b b32135 63 5b3 63 b3 g 5 13 b E 5 b m 2326967 So 2326967 can be substituted for the constant b in the equation gx ab gx m a2326967 We can now find the value of the constant b by using the fact that the ordered pair 2 5 satisfies the function gx m a2326967 5 m a23269672 5 e 541477561 5 m a 5414775 a m 0923399 So an approximation for the exponential function that passes through 2 5 and 5 63 is gx m 09233992326867 Click here some practice problems for this module HabermanKling MTHlllc Section II Exponential and Logarithmic Functions Module 3 Comparing Linear and Exponential Functions EXAMPLE 1 The table below gives values for the functions f and g Determine which is a linear function and which is an exponential function x 75 o 5 1o 15 f x 7 22 37 52 67 gm 4 2 64 2048 65536 Before deciding let s study tables representing a functions that we know are either linear or exponential EXAMPLE 1a Let kx 4x 3 a linear function x 72 71 o 1 2 3 8 kx 5 1 3 7 11 15 35 Notice that each time we make the same change in the xvalue the corresponding y values have a constant difference ie when you subtract the numbers the result is always the same For example if we increase xvalues by 1 the difference of corresponding yvalues is always 4 Change x from 71 to 0 k0 k 1 3 1 Change x from 0 to 1 k1 k07 3 4 Change x from lto 2 k2 k111 7 4 Note that we see this same behavior where corresponding yvalues have a constant difference with any constant change in xvalues Let s see what happens if we change our xvalues by 5 Change x from 72 to 3 k3 k 2 15 5 20 Change x from 3 to 8 k8 k3 35 15 20 So if we change our xvalues by 5 the difference of corresponding y values is 20 In general if a function is linear and you make a constant change in xvalues the difference of corresponding yvalues is constant EXAMPLE 1b Let jx 23 an exponential function 2 3 8 x 72 71 0 1 M 2 6 18 54 13122 Notice that each time we make the same change in the xvalue the corresponding y values have a constant ratio ie when you divide the yvalues the result is always the same For example if we increase xvalues by 1 the ratio of corresponding yvalues is 3 Change x from 71 to 0 f0 L J391 3 Change x from 0 to 1 10 3 10 2 3 Change x from 1to 2 12 2 g 11 6 Note that we see this same behavior where corresponding yvalues have a constant ratio with any constant change in xvalues Let s see what happens if we change our x values by 5 Change x from 72 to 3 f3 J392 243 Change x from 3 to 8 j8 13122 13 54 243 So if we change our xvalues by 5 the ratio of corresponding yvalues is 243 In general if a function is exponential and you make a constant change in xvalues the ratio of corresponding yvalues is constant Returning to the EXAMPLE 1 above we are now ready to determine which function is linear and which is exponential x 75 0 5 10 15 f x 7 22 37 52 67 gx X6 2 64 2048 65536 First let s study y fx When we change our xvalues by 5 the difference of the corresponding yvalues is 15 Change x from 75 to 0 f0f5227 15 Change x from 0 to 5 f5 f0 37 22 15 Change x from 5 to 10 f10 f5 52 37 15 Change x from 10to 15 f15 f10 67 52 15 Since the difference of the corresponding y values is constant f is a linearfunction Now we ll study y gx When we change our xvalues by 5 the ratio of corresponding y values is 32 Change x from 75 to 0 g L g5 X6 32 Change x from 0 to 5 go 2 g g0 2 32 Change x from 5 to 10 g10 2048 g5 64 32 Change x from 10to 15 g15 2 65536 g10 2048 32 Since the ratio of the corresponding y values is constant g is an exponential function Notice that it isn t correct to say Exponential functions increase faster than linear functionsquot In fact there is an interval during which the linear function increases faster than the exponential function EXAMPLEZ Fwd a gebramru esfurbuth and g mm EXAMPLE 1 Fm Et s nd the rmefur Smce shnear we knuvvthat fxmxb Smce 0 22 saus es f we knuvvthat fx mx 22 We can use the pawl 0 22 and 5 37 tu ndthe s upe m The x 3x 22 Nuvvvve H nd the rmefer g Smee g s EXpunEnUaLWE knuvvthat gxa AX Smee 0 2 saus es g We knuvvthat gx 2 H saus es g me nd A We can use any emer urdered pawmat 564 g564255 32 5 The gx 2 2 Haberman MTH 1110 Section IV Power Polynomial and Rational Functions Module 4 Introduction to Rational Functions Including the LongRun Behavior of their Graphs DEFINITION A rational function is a ratio of polynomial functions Thus if p and q are polynomial functions then rx px is a rational function Note q x that qx at 0 EXAMPLE The four functions given below are all rational functions x87x5x 20 x S a ax x U x2 8x7 6x7 15x6 36x5 x2xl 5 C qpx3 6xm x x2xl Since rational functions involve division we need to be sure not to divide by zero In order to avoid division by zero we need to exclude from the domain of a rational function any numbers that make the denominator zero x S 7 EXAMPLE What is the domain of the function ax 2 x 8x 7 SOLUTION In order to find the domain we need to determine which numbers make the denominator of a zero 80 we need to solve x2 8x 7 0 2 x 8x 7 0 3 x 7x 10 3 x 7 or x 1 Since both 7 and 1 make the denominator zero we must exclude them from the domain of 0 Thus the domain of a is xlxe R and x at 7 and x 1 ie all real numbers except 7 and 1 In the next module we will study what happens to the graph of a rational function near the values excluded from its domain Although 7 is not in the domain of the function x 5 x2 8x 7 extremely close to 7 but the input cannot equal 7 In the next module we ll study what happens to the graph of a as x get closer and closer to 7 Before we study this we will investigate the longrun behavior of rational functions ax values like 701 and 699 are valid inputs So the inputs can get THE LONGRUN BEHAVIOR OF RATIONAL FUNCTIONS The longrun behavior of rational functions refers to what happens to graph of the function as the x values get really big or really small which we can denote by the symbols x gt 00 or x gt oo Since a rational function is a ratio of polynomial functions we can use what we learned about polynomial functions here Recall that the longrun behavior of a polynomial function is determined by its leading term Thus the longrun behavior of a rational function can be found by comparing the leading terms of the polynomials in the numerator and denominator 2 2 8 EXAMPLE Determine the longrun behavior of the function hx x SOLUTION The longrun behavior of a function concerns what happens as the inputs get extreme Le x gt ioo As we learned when we studied polynomials both the numerator and denominator are dominated by their leading terms when x gt ioo Thus as x gt ice 2 2 hog 295 m Ziz 2 x 1 x This tells us that when our inputs are really big or really small the outputs will be about 2 which means that the graph of h gets closer and closer to the line y 2 as x gt ioo so in the longrun the graph of h will look like the graph of y 2 In order to decide if h approaches y 2 from above 2 or below 2 let s evaluate h for extreme inputs and observe if the outputs are greater than 2 or less than 2 21000002 8 x100000 2 h100000 2 100000 1 e 200004 Soas x gt oo hx gt 2 so as x gets really big hx approaches 2 from above 2 2 1000002 8 e 19999 100000 12 x 100000 3 h 100000 Soas x gt oo hx lt 2 so as x gets really small hx approaches 2 from below 2 V th this information we can sketch the longrun behavior of h see Figure 1 Figure1 The graph of y 2 and the longrun behavior of 2 105 x 1 Note that in the next module we will study the short term behavior of rational functions and finish sketching the graph of h The line y 2 plays a crucial role in the longterm behavior of the rational function h Thus this line is called a horizontal asymptote Hopefully you recall the following definition fromthe Section III Module 1 Introduction to Exponential Functions DEFINITION A horizontal asymptote is a horizontal line that the graph of a function gets arbitrarily close to as the input values get very large or very small EXAMPLE Determine the longrun behavior of the function kx What is the x2 25 horizontal asymptote of k SOLUTION To find the longrun behavior of this rational function we need recognize that as x gt ice the leading terms of the numerator and denominator determine the nature of the outputs 1 kx x ml x2 25 x2 x So as x gtioo the graph of k looks like y But we know that as x gtioo gets closer and closer to zero So we see that as x gt ioo kx m 0 which means that y 0 is the horizontal asymptote To decide if k approaches y 0 from above 0 or below 0 let s evaluate k for extreme inputs and observe if the outputs are greater than 0 or less than 0 x 100000 2 k100000 w z 000001 100000 25 Soas x gt oo kx gt 0 so as x gets really big kx approaches 0 from above 0 x 100000 2 k 100000 w z 000001 1000002 25 Soas x gt oo kxlt 0 so as x gets really small kx approaches 0 from below 0 V th this information we can sketch the longrun behavior of k see Figure 2 Figure 2 The long run behavior of y kx We will finish the graph of k in the next module gt 3 EXAMPLE Determine the longrun behavior of the function mx x SOLUTION To find the longrun behavior of this rational function we need recognize that as x gt ice the leading terms of the numerator and denominator determine the nature of the outputs So as x gt ioo m looks like y x This isn t a horizontal line but it is an asymptote nonetheless If an asymptote is a nonhorizontal line it is often called oblique Thus y x is an oblique asymptote for the function m In order to sketch the longrun behavior of m we need to decide if the graph of m approaches its oblique asymptote y x from above or below 1000003 16 x100000 2 m100000 2 100000 64 m 100000000486 gt 100000 Soas x gtoo mxgtx so as x gets really big mx approachesy x from abovey x 1000003 16 x 100000 2 m 100000 2 100000 64 m 100000000486 lt 100000 Soas x gt oo mxlt x so as x gets really small mx approachesy x from belowy x V th this information we can sketch the longrun behavior of m see Figure 3 Figure y x term behavior of y mx We will finish the graph of m in the next module HabermanKling MTH 1110 Section III Exponential and Logarithmic Functions Module 1 Introduction to Exponential Functions Exponential functions are functions in which the variable appears in the exponent For example fx 801035quot is an exponential function since the independent variable x appears in the exponent One way to characterize exponential functions is to say that they represent quantities that change at a constantpercentage rate Q EXAMPLE When Rodney first got his job in 1993 he earned 21000 per year After every year Rodney receives a 10 raise 39 After one year Rodney gets a 10 raise his salary becomes 21000 21000010 210001 010 21000110 T T T olif raise olif satiny new satiny times 10 satiny So to find his new salary we must multiply his original salary by 110 39 After one year of receiving 21000110 dollars ie after his second year Rodney gets another raise of 10 His salary becomes 21000110 21000110010 210001101 010 210001102 T T T olif raise olif satiny new satiny times 10 satiny So to find his new salary we must multiply his original salary by 1102 39 After his third year Rodney gets another raise of 10 His salary becomes 210001102 210001102010 2100011021 010 210001103 T T olif raise olif satiny new satiny times 10 satiny So to find his new salary we must multiply his original salary by 1103 We can now write a formula for Rodney s salary st in dollars after he has worked at the job t years st 21000110t This is obviously an exponential function since the variable is in the exponent Thus we can see why exponential functions represent quantities that change at a constant percent rate Note that this function works when t 0 because s0 210001100 210001 21000 and 21000 is Rodney s initial salary Below is another example that shows us that exponential functions represent quantities that change at a constant percentage rate EXAMPLE Suppose that the population of the Expo Nation this year is 150000 If the population decreases at a rate of 8 each year find a function p that represents the population of the Expo Nation t years from now population this year 150000 population after 1 year 150000 150000008 1500001 008 150000092 population after 2 years 150000092 150000092008 1500000921 008 150000092092 1500000922 population after 3 years 1500000922 1500000922008 1500000922 1 008 1500000922092 1500000923 Observing the pattern above we can deduce that the population of the Expo Nation after t years is given by the function pt 150000092 Again note that this function works when t 0 because p0 1500000920 1500001 150000 and the initial population of Expo Nation was 150000 In order to generalize about exponential functions we need to analyze the structure of both of the exponential functions st 21000110 and pt150000092 that we found in the examples above 0 In both functions the initial valuequot Rodney s initial salary of 21000 and Expo Nation s initial population of 150000 plays the same role st 21000010t and pt150000092t 0 Also in both functions the number under the exponent the base of the exponential function is 1 r where r is the decimal representation of the percent rate of change per units of t st 21000110 2 110 1 010 2 Rodney s raise 10 per year pt 15000009239 2 092 1 008 2 Population loss 8 per year We can use the information above to obtain a definition of an exponential function DEFINITION An exponential function has the form fx abx where a is the initial value Le a f0 and b is the growth factor and b 1 r where r is the decimal representation of the percent rate of change per unit of x NOTE If r gt 0 then b gt 1 and the resulting function exhibits exponential growth If 1 lt r lt 0 then b lt 1 and the resulting function exhibits exponential decay ALSO NOTE b is always positive I 1 r and we know that r 2 1 since the rate of change cannot be less than 400 Le we cannot lose more than 100 per unit of time GRAPHS OF EXPONENTIAL FUNCTIONS We already know what happens to the graphs of functions when we multiply their rules by positive and negative constants Thus all we need to determine is the shape of a generic exponential function and we will then be able to determine the shape of any exponential function There are basically two classes of exponential functions 1 fxab with bgt1 2 fxab with0ltblt1 The next two examples will help us determine the shape of the graphs of these two classes of exponential functions are EXAMPLE Sketch a graph of hx 2quot Note that this is an exponential function of the form hx abx where a 1 and b 2 SOLUTION In order to graph h we will create a table of values that we can use to form ordered pairs Then we will plot the ordered pairs and connect our dots in an appropriate manner Table 1 2quot hx x hx 1A 12 i 12 1 1 DEFINITION A horizontal asymptote is a horizontal line that the graph of a function gets arbitrarily close to as the input values get very large or very small The function hx 2quot graphed in the previous example has a horizontal asymptote at y 0 the x axis EXAMPLE Sketch a graph er W e Nate that this is an expeneniiai functiuri er thefurm w e a AX Were a 1 and A SOLUT ON in urder in grapn g we WiH create a iabie er vaiues that we ean use in rerrn urdered pairs Nate that the horizontal asyrmioierer y go isine KraXiS s e Wu exarnpies abeve we ean eeneiuee that the grapn er an expenenuai X is increasing r A gt1 and decreasing r 0 ltA lt1 we fur inis increasingdecreasing J Ea en en h funmiun ufthe rerrn fxa A TechnicaHy we need in aise siaie that a is pesi behaviur the next exarnpie rnigni eianiy Whythis is se EXAMPL Use yuur undersianmng er grapn transfurmatiuns in predict huvvthe grapns er m 4 2 and m 77 2 are hx X On a graphing eaieuiaierer Either graphing unity sketch grapns er h m and n in e nrirrn uur predictiuns FinaHy draw a eeneiusien abuut the reie er 11 en the grapn er an expenenuai functiun ufthe farmx eernpare win the 8 H SOLUTION We aren t going to graph these functions here since you can do that yourself on your graphing calculator but we will discuss how we can use graph transformations to predict how the graph will look To compare mx 42quot with hx 2quot we need to write m in terms of h mx 42quot 4hx Since mx is hx multiplied by 4 on the outside we know that to graph y mx we need to stretch the graph of y hx vertically by a factor of 4 So if we perform this transformation to the yintercept of y hx which is 0 1 by a factor of 4 we see that the yintercept of y mx is 0 4 To compare nx 72quot with hx 2quot we need to write nx in terms of hx nx 72x 7hx Since nx is hx multiplied by 77 on the outside we know that to graph y nx we need to reflect the graph of y hx about the xaxis and stretch the graph of y hx vertically by a factor of 7 So if we perform these transformations to the yintercept of y hx which is 0 1 we see that the yintercept of y nx is 0 7 to see a video in which these functions are graphed on a Tl89 Be sure to turn up the volume on your computer Notice that the number that plays the role of a in the rules for mx and nx is the y coordinate of the yintercept for both functions mx 42quot 3 a 4 3 yintercept 0 4 nx 72quot 3 a 7 3 yintercept 0 7 Of course we already should have expected this since we know that a represents the initial value The yintercept of an exponential function of the form fx abx is 0 a HabermanKling MTH lllc Section 1 Sets an 1 Functions mal um NH 4 I mnuvH W W Module 5 Function Composition In The Algebra of Functions Section 1 Module 4 we discussed adding subtracting multiplying and dividing functions In this module we will study another way to combine functions function composition EXAMPLE When Peter was younger and people asked about his age he never had to think he had my age memorized But now he s older and his age has taken enough different values that he sometimes lose track and need to do some calculations to find his age In this example we ll discuss the function that Peter uses to calculate his age when he can t remember the age function Let s call the age function a Since his goal is to determine his age we need to input Peter into the function a Thus it makes sense to define the domain of a to be the set of all living people So what does the function a need to do to determine a person s age First a needs to find the person s birthdate and then it needs to calculate how long ago the person s birthdate occurred Since a needs to do these two things we say that a is the composition of two functions the birthdate function b and the how long ago this date occurred function h So a is the function that computes a person s age I is the function that finds a person s birthdate h is the function that calculates how long ago measured in complete years a date occurred The diagram below represents how function a works J 7 V 4 quot x x I i i V it I 1 e i I b i I 7 i r i o 39 quot I Iquot 3quot Years I Peter I u 122m I I I I I 3951 xv 74 I I Vrguw ill V 5 Emma J J I J 391 i 39i3977 x w I Ly 44 4 I Living People Birthidates How long ago We can express this function symbolically as follows aPeter hbPeter h12271971 35 so Peter is 35 years old and if x represents a generic person then the age of person x can be calculated as follows ax hbx As mentioned above a is the composition of two functions b and h We have special notation for the composition of two functions ax hobx hbx aKEY POINT The composition of functions is denoted by the symbol The composition of functions f and g is the function fog defined as follows fogx fgx The notation fog can be translated as f composed with gquot or the composition of f and gquot BE CAREFUL fog does not mean the same thing as fg which is the product of f and g fogXx fgx While f39gx fx39gx EXAMPLE Table 1 shows the temperature C in degrees Celsius as a function of the temperature in degrees Fahrenheit F Table 2 shows the temperature in degrees Kelvin K as a function of the temperature in degrees Celsius C The Kelvin scale is the temperature scale devised by Lord Kelvin in 1848 Table 1 Celsius temperature vs Fahrenheit temperature F 731713 32 68149212 CF735725 0 20 65 100 Table 2 Kelvin temperature vs Celsius temperature C 735725 0 20 65 100 KC I23815 24815 27315 29315 33815 37315 Suppose we want a table that shows direct conversions from temperatures in degrees Fahrenheit to temperatures in degrees Kelvin Table 3 shows the temperature in degrees Kelvin K as a function of the temperature in degrees Fahrenheit F Table 3 is easy to obtain using Table 1 and Table 2 because the outputs of Table 1 are the same as the inputs of Table 2 Table 3 Kelvin temperature VS Fahrenheit temperature F e31e13 32 68149 212 l KCF I23815 24515 27315 29315 33815 37315 Since the output of Table 1 is used as the input of Table 2 we write the new function in Table 3 as KCF The new function is formed by composing the other two functions The mathematical expression for this composition is K0CF Therefore KoCF KCF EXAMPLE Given CF F 32 and KC C 27315 find KoCF the function which converts temperature in degrees Fahrenheit directly to temperatures in degrees Kelvin SOLUTION K o CF K CF KF 32 Replace CF with F 32 F 32 2735 Replace the input variable C in the formula KF C 2735 with F 32 EXAMPLE Use Tamezzm Eva uate fog3 and goz EXp am Wny fog4 s unde ned Remember 7 thh nu a gebrau me ur graph the venues m the evn J tame are the WW va uesvve kn SOLUT ONS S MegG 4 snaeA1 g 3 gm smce 3 2 0 was g2 0 pew glt4gt EM 9 e undehned becausetheve e no mpmvame ow n Tab eA Theveiove fog4 e undehned EXAMPLE Usethe graph m ngure 1 mm the va uesfur xterm and muk2 and y kg sthe hne SOLUT ONS WW We km Wehndthat ma 71 on nguve1sowevepace m3 W 71 2 the hneavmnchon m ngme1 shws mm k7172 WW2 mklt2gt m4 Wehndmat k2 4 on nguve1sowevepace k2 WM 4 2 the pavebohc mnchon m ngme1 shws usthat quot14 2 Q EXAMPLE w mg 3x 7 5 and m 2x2 1 nd and swmphiythe quuvwng a monx b nomx c m WW SOLUT ONS 1 m 0 ma nomx nmx n3x 5 23x 521 29x2 30x 25 1 18x2 60x501 18x2 60x 51 c momx mmx m3x 5 33x 5 5 9x 15 5 9x 20 aKEY POINT As the example above suggests fogx and gofx are typically different Although it is possible that they are equal in general fogx gofx In fact in a key pointquot below we notice that some functions cannot even be composed in both ways Try this one yourself If fx x2 5x 4 and gx 2x 3 find and simplify fogx Click Here to Check Your Answer ypd ask tne deater tet appty ttrst7 vvnten eetrnpetstttetn represents yuur enettee7 Jus ufy yuur answer by Wrmng EXprEss1unsfurogP and goP SOLUT ONS a 1t P an ttt represented by P 7 CUP 0 85F 5d P 85F ts a mnettetn wnten represents tne pnee etttne Dmputer tt etntytne 15 dtseetdnt ts apptted 1t P ptt an t P7 500 5d g P P7 500 ts atdnettetn g wnten representstne pnee ttetntytne 500 repate ts apptted b Tu tnterpret gP t we need tet wer frum tne tnstde nut P represents tne etrtgtnat tee tnen g penetrrns tne 500 repate tuttetwed by wnten penetrrns tne 15 dtseetdnt Tu tnterpret goP t we agatn need tet Walk tretrn tne tnstde nut P represents tne etngtnat pneet tnen penetrrnstne 15 dtscuunt quuWEd by g wnten penetrrnstne 500 rebate f0gP fgP 085gP put gP into f by replacing the input variable off with gP 085P 500 replace gP with P 500 since gP P 500 085P 425 apply the distributive property g 0 fP gfP fP 500 put fP into g by replacing the input variable of g with fP 085P 500 replace fP with 085P sincefP 085P Since fogP 085P 425 and gofP 085P 500 it appears that gfP the 15 discount followed by the 500 rebate would be the better deal a KEY POINT In most applied problems functions cannot be composed both ways as demonstrated in the following example EXAMPLE Suppose that the function n Pt represents the population n of the Portland metropolitan area I years after 1990 and l Cy represents the carbon dioxide C02 concentration I in the atmosphere of a city of population y Which composition function C0Pt or PoCy makes sense Explain your reasoning SOLUTION COPt is the only composition that makes sense since COPI CPt and the input of C must be a population and the output of P is a population PoCy PCy doesn t make sense because the input of P must be a time in years since 1990 but the output of C is not a time EXAMPLE f fxlx3 find two new functions u and w so that fxuowx SOLUTION Essentially this example asks us to decomposequot the function fx Jx 3 into two new functions u and w Since we need fx u owx uwx we need to think of fx lx3 as consisting of a twostep process where w represents the first step of the process and u represents the second step in the process There are always many different correct choices for u and w but in this case it is most natural to consider that the two steps involved in the function fx lx 3 are 15 Add 3 to the input 2quot Extract the square root of the result of the 1St step Thus we can define the functions u and w as follows wx x 3 ux J Let s check if this choice of u and w works u o wx uwx ux 3 quotx 3 fx Since uowx fx our choice of u and w is correct 4 EXAMPLE lf gx x3 3 find two new functions u and w so that gx u owx SOLUTION ln orderto decomposequot the function gx x3 3 into two functions u and w we need to think of gx x3 3 a twostep process where w represents the first step of the process and u represents the second step in the process In this case it is most natural to consider that the two steps involved in the function gx x3 3 are 15 Cube the input 2quot Add 3 to the result of the 1st step Thuswe can define the functions u and w as follows wx x3 ux x 3 Let s check if this choice of u and w works u owx uwx a x33 Since u owx gx our choice of u and w is correct EXAMPLE lf hx2x 510 find two new functions u and w so that hx u owx SOLUTION 10 In order to decomposequot the function hx 2x 5 into two functions u and w we need to think of hx 2x 510 a twostep process where w represents the first step of the process and u represents the second step in the process In this case there are a few equally natural ways to breakdown the function into two steps We ll show two different ways here Solution A We can take the two steps involved in the function hx 2x 510 to be 1 Multiply the input by 2 and then subtract 5 from the result 2quot Raise the result of the 1St step to the power 10 Thuswe can define the functions u and w as follows wx 2x 5 ux x10 Let s check if this choice of u and w works u o wx uwx u2x 5 2x 510 hx Since u owx hx our choice of u and w is correct Solution B We can take the two steps involved in the function hx 2x 510 to be 1 Multiply the input by 2 2quot Subtract 5 from the result of the 1St step and then raise the result to the power 10 Thuswe can define the functions u and w as follows wx 2x ux x 510 Let s check if this choice of u and w works u o wx uwx u2x 2x 510 hx Since u owx hx our choice of u and w is correct In the example above we decomposed the function hx 2x 510 into two functions u and w but you may have noticed that the function reay consists of a threestep process Thus the most natural decomposition consists of three functions Let s find a threefunction decomposition of the function hx 2x 510 EXAMPLE f hx2x 510 find three new functions u v and w so that hx uovowx SOLUTION First let s notice that hx u 0 Va wx vwltxgtgt so in order to decomposequot the function hx 2x 510 into three functions u v and w we need to think of hx 2x 510 a threestep process where w represents the first step v represents the second step and u represents the third step In this case it is most natural to considerthat the three steps involved in the function hx 2x 510 are 1 Multiply the input by 2 2rd Subtract 5 from the result of the 1St step 3quot Raise the result of the 2quotd step to the power 10 Thuswe can define the functions u v and w as follows wx 2x vx x 5 ux x10 Let s check if this choice of u v and w works u ovo wx u vwx u 2x 5 2x 510 hx Since u oVowx hx our choice of u v and w is correct Haberman King MTH lllc Section III Graph Transformations Module 2 Reflections and Symmetry REFLECTIONS We have studied how adding or subtracting constants to the inside or outside of a function affects its graph We will now begin to study what happens when we multiply the inside or outside of a function by constants Before we consider what happens when we multiply by any constant we shall first focus on the consequences of multiplying by 1 Q EXAMPLE Let the graph below define the function y mx On the same coordinate plane graph y hx if hx mx Let s start with a few key points on the graph of y mx and see where they end up on the graph of y hx Let s use the points 4 1 2 1 1 2 2 0 3 3 and 4 0 Let s start by evaluating h 4 h 4 m 4 co 1 Now we ll make a table of values showing where these key points end up on the graph of y hx Let s mm the pawl WE m1 fuund FmaHy cunnectthe um Wm a pwecevwse hnear funmun smce y mx was a pwecevv se hnearfunctmn GENERALIZATIDN The graph uf y 7 m s a re ectmn uf the graph uf y m abuutthe xrast EXAMPLE Let the graph below define the function y mx On the same coordinate plane graph y kx if kx m x Let s start with a few key points on the graph of y mx and see where they end up on the graph of y kx Let s use the following points 4 1 3 1 2 1 LO 1 2 2 0 3 3 and 4 0 Let s start by evaluating k 4 k 4 m 4 M4 0 Now we ll make a table of values showing where these key points end up on the graph of y kx kx 0 4 Let s plot the points we just found and graph y kx by connecting the points with a piecewise linear function GENERALIZATIDN The graph pr y ex rs a re ectmn pr the graph pr y m appuuhe yraxrs SYMMETRY WE rust smgreg m aharyze these symmemes symmetry about the yeaxi EXAMPLE The graph pr 0 x2 rs gwen perpw Huvv dues rr eprhpare vvrth the graph pr y um h shumd pe erearrhar rrwe re emhe graph pr m x2 appunhe yraxrs We Dbtam the exam sarhe grapr Thrs eah be represented argeprareauy wrrh rhe SiatEmEnt xu i x We say that graphs like ux x2 have symmetry about the yaxis and call functions with this type of symmetry even functions It helps me to remember that ux x2 is an example of an even function since the power on x ie 2 is an even number A function f is even if its graph is symmetric about the yaxis which means that if the graph of f isn t changed if it is reflected about the yaxis An algebraic test to determine if a function is even is given below A function f is even if for all x in its domain f x fx Notice that the test is an algebraic representation of the statement a function is even if reflection about the yaxis does not change the graphquot symmetry about the origin EXAMPLE The graph of vx x3 is given below How does the graph of y vx compare with the graph of y v x Notice that if you anchor the graph of vx x3 at the origin 0 0 and rotate it 1800 in either direction the graph ends up in the same place it started We say that graphs with this sort of symmetry have symmetry about the origin We can also study this symmetry by considering reflections Let s reflect the graph of vx x3 about both the x and yaxes ie let s graph both y v x reflection about the yaxis and y vx reflection about the xaxis tetteotton ot y 7 yo about yraxts tetteotton uf y yo about Kraxts These graphs show us that when we re em Va 9 about the yraxts we obtatn the taetwth the tottowtng atgebtate statement vrx em Functtuns wtth symmetty about the ongtn ate aHEd odd functions tt hetbs me t temembetthat m 9 ts an exambte of an odd tunotton stnee the puvver on x t e t 3 t an odd number o s Atunotton y x ts odd tt tts graph is symmetric aboutthe origin hhteh means that tt you rutate tts gra h 180 about the BMW you obtatn the ongtnat gtabh Equtva en y a tunotton ts ooo tt tetteotton about the yraxts gtyes you the same gtabh Kraxts betow FuraH x tn tts dumam fez 7x NEIUEE that the 551 ts an atgebrat representattun Elf the statement a funmtun ts Edd f tetteetton about the yraxts gtyes you the same graph as tetteotton aboutthe Kraxts EXAMPLE Perform the abbtobnate atgebtate test to oetetmtne tt the tottowtng tunottons ate eyent ooot ur netthet a m b m x2 2x e 3 c x SOLUTIONS It is important to notice that the algebraic tests for both even and odd symmetry start with making the input the opposite sign so when we test for symmetry we need to start with this step simplify and observe the result We only need to perform one test and observe the result rather than performing a test for both even and odd a 2x xz 3 2x x2 3 2 96 x2 3 a x ace Since a x ax we can conclude that a is an odd function b b x x2 2 x 3 x2 2x 3 Since this is neither the original function nor the opposite of the original function ie b x 73 bx and b x 73 bx we see that b is neither even nor odd c 5 c x 3 x x x x3 x x3 x x3 x cx Since c x Cx we can conclude that c is an even function HabermanKling MTH lllc Section Sets and Functions Module 1 Sets and Numbers DEFINITION A set is a collection of objects specified in a manner that enables one to determine if a given object is or is not in the set In other words a set is a welldefined collection of objects EXAMPLE Which of the following represent a set a The students registered for MTH 95 at PCC this quarter b The good students registered for MTH 95 at PCC this quarter SOLUTIONS a This represents a set since it is well definedquot We all know what it means to be registered for a class b This does NOT represent a set since it is not well defined There are many different understandings of what it means to be a good student get an A or pass the class or attend class or avoid falling asleep in class EXAMPLE Which of the following represent a set a All of the really big numbers b All the whole numbers between 3 and 10 SOLUTION a It should be obvious why this does NOT represent a set What does it mean to be a big numberquot b This represents a set We can represent sets like b in roster notation see box at top of next page quotAll the whole numbers between 3 and 10quot 4 5 6 7 8 9 ll Roster Notation involves listing the elements in a set within curly brackets quot DEFINITION An object in a set is called an element of the set symbol quot6quot EXAMPLE 5 is an element of the set 4 5 6 7 8 9 We can express this symbolically Se4 5 6 7 8 9 DEFINITION Two sets are considered equal if they have the same elements We used this definition earlier when we wrote quotAll the whole numbers between 3 and 10quot 4 5 6 7 8 9 DEFINITION A set S is a subset of a set T denoted S g T if all elements of S are also elements of T If S and T are sets and S T then S g T Sometimes it is useful to consider a subset S of a set T that is not equal to T In such a case we write S c T and say that S is a proper subset of T EXAMPLE 4 7 8 is a subset of the set 4 5 6 7 8 9 We can express this fact symbolically by 4 7 8 g 4 5 6 7 8 9 Since these two sets are not equal 47 8 is a proper subset of 4 5 6 7 8 9 so we can write 4 7 8 c4 5 6 7 89 DEFINITION The empty set denoted 0 is the set with no elements 0 There are NO elements in 0 The empty set is a subset of all sets Note that 0 at 0 DEFINITION The union of two sets A and B denoted AUB is the set containing all of the elements in either A or B or both A and B EXAMPLE Considerthe sets 47 8 0 2 4 6 8 and 1 3 5 7 Then aZU lin 14iZQ bZUmZampQQampz4QZQ a mzaagu iinmLzaaiaza DEFINITION The intersection of two sets A and B denoted AnB is the set containing all of the elements in both A and B Q EXAMPLE Considerthe sets 47 8 0 2 4 6 8 and 1 3 5 7 Then aZnmzampQQ bzanaginw c 0 2 4 6 8nl 3 5 7 0 These sets have no elements in common so their intersection is the empty set 4 EXAMPLE All of the whole numbers positive and negative form a set This set is called the integers and is represented by the symbol Z We can express the set of integers in roster notation Z 3 2 10123 Note that Z is used to represent the integers because the German word for quotnumberquot is quotzahlenquot Now that we have the integers we can represent sets like All of the whole numbers between 3 and 10quot using setbuilder notation SETBUILDER NOTATION quotAllthe whole numbers between3and10quot xler and 3 lt x lt10 tThis vertical line means quotsuch thatquot Armed with setbuilder notation we can define important sets of numbers DEFINITIONS The set of natural numbers N 1 2 3 4 5 The set of integers Z 3 2 1 0 1 2 The set of rational numbers Q f n q e Z and q at 0 This set is sometimes described as the set of fractions The set of real numbers R All the numbers on the number line abeR and 139 The set of complex numbers C a bi Note that NC Z c Q c R c C Le the set of natural numbers is a subset of the set of integers which is a subset of the set of rational numbers which is a subset of the real numbers which is a subset of the set of complex numbers Throughout this course we will assume that the numberset in question is the real numbers R unless we are specifically asked to consider an alternative set Since we use the real numbers so often we have special notation for subsets of the real numbers interval notation Interval notation involves square or round brackets Use the examples below to understand how interval notation works 2 CLICK HERE FOR AN INTRODUCTION TO INTERVAL NOTATION 394 EXAMPLE a x Ix e R and 2 s x s 3 2 3 We use square brackets here since the endpoints are included Setbuilder Notation Interval Notation b xlxe R and 2 lt x lt 3 2 3 We use round brackets here since the endpoints are NOT included Setbuilder Notation Interval Notation c x x 6 R and 2 lt x S 3 2 3 We use a round bracket on the left since 2 is NOT included Set builder Notation Interval Notation d x x 6 R and 2 S x lt 3 2 3 We use a round bracket on the right T T since 3 is NOT included Set builder Notation Interval Notation EXAMPLE When the interval has no upper or lower bound the symbol 00 or 00 is used a xlxe R and x s 4 00 4 We ALWAYS use a round bracket with 00 T T since it is NOT a number in the set Setbuilder Notation Interval Notation b xlxe R and x Z 4 4 00 We ALWAYS use a round bracket with 00 T T since it is NOT a number in the set Setbuilder Notation Interval Notation CLICK HERE FOR SOME INTERVAL NOTATION EXAMPLES EXAMPLE a 4 oOU 8 3 Simplify the following expressions b 4 oOU oo 2 c 4 oom oo 2 d 4 00 n 10 5 SOLUTION a 4 oOU 8 3 8 00 b 4 ooU oo 2 00 00 R c 4 oom oo 2 4 2 6 d 4 oon 10 5 0 L CLICK HERE FOR A SUMMARY OF INTERSECTIONS UNIONS SETBUILDER NOTATION AND INTERVAL NOTATION WITH NUMBER LINES

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