Introduction to functions mcTYintrofns Afunctionisarulewhichoperatesononenumbertogiveanoth ernumber

Presentation on theme: "Introduction to functions mcTYintrofns Afunctionisarulewhichoperatesononenumbertogiveanoth ernumber"— Presentation transcript:

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Introduction to functions mc-TY-introfns-2009-1 Afunctionisarulewhichoperatesononenumbertogiveanoth ernumber.However,notevery ruledescribesavalidfunction. Thisunitexplainshowtose ewhetheragivenruledescribesa validfunction,andintroducessomeofthemathematicalter msassociatedwithfunctions. Inordertomasterthetechniquesexplainedhereitisvitalt hatyouundertakeplentyofpractice exercisessothattheybecomesecondnature. Afterreadingthistext,and/orviewingthevideotutorialo nthistopic,youshouldbeableto: recognisewhenaruledescribesavalidfunction, beabletoplotthegraphofapartofafunction,

findasuitabledomainforafunction,andfindthecorrespondi ngrange. Contents 1. Whatisafunction? 2 2. Plottingthegraphofafunction 3 3. Whenisafunctionvalid? 4 4. Somefurtherexamples 6 www.mathcentre.ac.uk 1 math centre2009
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1. What is a function? Hereisadefinitionofafunction. Afunctionisarulewhichmapsanumbertoanotheruniquenumb er. Inotherwords,ifwestartoffwithaninput,andweapplythefu nction,wegetanoutput. Forexample,wemighthaveafunctionthatadded3toanynumbe r.Soifweapplythisfunction tothenumber2,wegetthenumber5. Ifweapplythisfunctiont othenumber8,wegetthe

number11.Ifweapplythisfunctiontothenumber ,wegetthenumber + 3 Wecanshowthismathematicallybywriting ) = + 3 Thenumber thatweusefortheinputofthefunctioniscalledthe argument ofthefunction. Soifwechooseanargumentof2,weget (2) = 2 + 3 = 5 Ifwechooseanargumentof8,weget (8) = 8 + 3 = 11 Ifwechooseanargumentof ,weget 6) = 6 + 3 = Ifwechooseanargumentof ,weget ) = + 3 Ifwechooseanargumentof ,weget ) = + 3 Atfirstsight,itseemsthatwecanpickanynumberwechoosefo rtheargument.However,that isnotthecase,asweshallseelater. Butbecausewedohaveso mechoiceinthenumberwe canpick,wecalltheargumentthe

independentvariable .Theoutputofthefunction,e.g. (5) ,etc.dependsupontheargument,andsothisiscalledthe dependentvariable Key Point Afunctionisarulethatmapsanumbertoanotheruniquenumbe r. Theinputtothefunctioniscalledthe independentvariable ,andisalsocalledthe argument of thefunction.Theoutputofthefunctioniscalledthe dependentvariable www.mathcentre.ac.uk 2 math centre2009
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2. Plotting the graph of a function Ifwehaveafunctiongivenbyaformula,wecantrytoplotitsg raph. Suppose,forexample, thatwehaveafunction definedby ) = 3 Theargumentofthefunction(theindependentvariable)is

,andtheoutput(thedependent variable)is .Sowecancalculatetheoutputofthefunctionfordifferenta rguments: (0) = 3 4 = (1) = 3 4 = (2) = 3 4 = 8 1) = 3 1) 4 = 2) = 3 2) 4 = 8 Wecanputthisinformationintoatabletohelpusplotthegra phofthefunction. 2 1 2 Wecanusethegraphofthefunctiontofindtheoutputcorrespo ndingtoagivenargument.For instance,ifwehaveanargumentof2,westartonthehorizont alaxisatthepointwhere = 2 andwefollowthelineupuntilwereachthegraph. Thenwefoll owthelineacrosssothatwe canreadoffthevalueof ontheverticalaxis.Inthiscase,thevalueof is8.Ofcourse wealreadyknowthis,because = 2

isoneofthevaluesinourtable. 2 1 2 www.mathcentre.ac.uk 3 math centre2009
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Butwecanalsousethegraphforvaluesof whicharenotinourtable.Ifwehaveanargument of1.5,wefollowthelineuptothegraph,andthenacrosstoth everticalaxis. Theresultisa numberbetween2and3. 2 1 2 Ifwewanttocalculatethisnumberexactly,wecansubstitut e1.5intotheformula: (1 5) = 2 75 3. When is a function valid? Ourdefinitionofafunctionsaysthatitisarulemappinganum bertoanotheruniquenumber. Sowecannothaveafunctionwhichgivestwodifferentoutputs forthesameargument. One

easywaytocheckthisisfromthegraphofthefunction,byusi ngaruler.Iftherulerisaligned vertically,thenitonlyevercrossesthegraphonce;nomore andnoless. Thismeansthatthe graphrepresentsavalidfunction. Whathappensifwetrytodefineafunctionwithmorethanoneou tputforthesameargument? Let’stryanexample.Supposewetrytodefineafunctionbysay ingthat ) = x. Inthesamewayasbefore,wecanproduceatableofresultstoh elpusplotthegraphofthe function: (0) = 0 (1) = (2) = to1d.p. (3) = to1d.p. (4) = (Ifwetrytouseanynegativearguments,weendupintroubleb ecausewearetryingtofind

thesquarerootofanegativenumber.)Plottingtheresultsf romthetable,wegetthefollowing graph. www.mathcentre.ac.uk 4 math centre2009
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2 3 4 ruler Usingtheruler,itisquiteclearthattherearetwovaluesfo rallofthepositivearguments. So asitstands,thisisnotavalidfunction. Onewayaroundthisproblemistodefine totakeonlythepositivevalues,orzero: thisis sometimescalledthe positivesquareroot of . However,thereisstilltheissuethatwecannot chooseanegativeargument. Soweshouldalsochoosetorestr ictthechoiceofargumentto positivevalues,orzero. 2 3 4 When considering these kinds of restrictions, itis

importa nt tousetheright mathematical language.Wesaythatthesetofpossibleinputsiscalledthe domain ofthefunction,andtheset ofcorrespondingoutputsiscalledthe range .Intheexampleabove,wehavedefinedthefunction asfollows: ) = x x , f sothatthedomainofthefunctionisthesetofnumbers ,andtherangeisthecorresponding setofnumbers Key Point The domain ofafunctionisthesetofpossibleinputs. The range ofafunctionisthesetof correspondingoutputs. www.mathcentre.ac.uk 5 math centre2009
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4. Some further examples Example Considerthefunction ) = 2 + 5

Tomakesurethatthefunctionisvalid,weneedtocheckwheth erwegetexactlyoneoutput foreachinput,andwhetherthereneedstobeanyrestriction onthedomain.Asbefore,wecan calculatetheoutputofthisfunctionatsomespecificvalues tohelpuswithplottingourgraph: (0) = 2 0 + 5 = 5 (1) = 2 1 + 5 = 2 3 + 5 = 4 (2) = 2 2 + 5 = 8 6 + 5 = 7 (3) = 2 3 + 5 = 18 9 + 5 = 14 1) = 2 1) 1) + 5 = 2 + 3 + 5 = 10 Nowwecanputthisintoatable,andplotthegraph. 10 14 14 10 -1 www.mathcentre.ac.uk 6 math centre2009
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Averticalruleralwayscrossesthegraphonce,andsothedom ainneedsnorestrictions,and thefunctionisvalid.

Wecanalsoseefromthegraphthatthem inimumoutputoccurswhen = 0 75 ,andthatiswhen ) = 3 875 .Sotherangeofthefunctionis 875 Example Whataboutthefunction ) = Asusual,thefirststepistochecksomevalues. (1) = 1 , f (2) = , f (3) = , f (4) = 1) = 1) , f 2) = 2) 3) = 3) , f 4) = 4) Whenwetrytocalculate (0) wehaveaproblem,becausewecannotdividebyzero. Sowe havetorestrictthedomaintoexclude = 0 2 1 2 3 4 Becauseofthisproblemwhen = 0 ,wehavetorestrictthedomaintomakethefunctionvalid. Youcanalsoseefromthegraphthatthereisnovalueof where ) = 0 ,sozeroisalso

excludedfromtherange.Thefunctionisthereforedefinedby ) = 1 /x x = 0 , f = 0 Youmightwanttoknowwhatexactlyisgoingonatthispointwh en = 0 . Onewaytofind outistolookatwhatishappeningveryclosetozero. Solet’s trysomepositivevaluesforthe argumentgettingcloserandclosertozero,inordertoseewh athappens: (1) = 1 , f = 2 , f 10 10 = 10 000 000 = 1 000 , f 000 000 000 000 = 1 000 000 Soyoucanseethatasweapproachzerofromtheright,theoutp utapproachesinfinity. www.mathcentre.ac.uk 7 math centre2009
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Nowlet’strysomenegativevaluesfortheargument,getting

closerandclosertozerofromthe left-handside,inordertoseewhathappens: 1) = 1) , f 2) , f 10 10) 10 000 000) 000 , f 000 000 000 000) 000 000 Youcanseethatasweapproachzerofromtheleft,theoutputa pproachesnegativeinfinity.So inthiscase,approachingzerofromtheleftisverydifferent fromapproachingitfromtheright. Example Forourfinalexample,takethefunction ) = 2) Asusual,wemustcalculatesomevalues: 2) = 2) 16 1) = 2) (0) = (0 2) (1) = (1 2) = 1 (2) = (2 2) dividingbyzero (3) = (3 2) = 1 (4) = (4 2) (5) = (5 2) (6) = (6 2) 16 Nowwecanconstructatable,andplotthegraphofthefunctio n. 16

dividingbyzero 16 www.mathcentre.ac.uk 8 math centre2009
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1 2 3 4 5 6 Theverticallineat = 2 iscalledanasymptote;itisalinewhichisapproachedbythe graphbut neverreached.Althoughitisclearthatwewillhavetoomit = 2 fromthedomain,thistimethe situationisslightlydifferent.Youcanseethatateithersi deof = 2 isapproachinginfinity. Youshouldalsonoticethatwecannothaveanynegativevalue sfor ,oreven ) = 0 .So wehavetodefineourfunctionas ) = 2) = 2 , f Exercises 1.Considerthefunction ) = 2 + 5 (a)Writedowntheargumentofthisfunction.

(b)Writedownthedependentvariableintermsoftheargumen t. (c)Useatableofvaluestohelpyouplotthegraphofthefunct ion. (d)Fromyourgraph,estimate (1 5) (e)Useyourfunctiontocalculate (1 5) exactly. (f)Writedownthedomainandrangeofthefunction. (g)Re-writethefunctionwithargument 2.Considerthefunction ) = 3) (a)Plotthegraphofthefunction. (b)Writedownthedomainandrangeofthefunction. (c)Re-writethefunctionwithargument (d)Useyourgraphtoestimate (1) (e)Usethefunctiontocalculate (1) exactly. (f)Writedownanotherfunctionwhere = 4 hastobeomittedfromthedomain. 3.Considerthefunction ) = 3 x.

(a)Whatassumptionsmustbemadeaboutthefunctiontoensur evalidity? (b)Plotthegraphofthefunction. www.mathcentre.ac.uk 9 math centre2009
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(c)Writedownthedomainandrangeofthefunction. (d)Calculate (3) (e)Whathappensifyoutrytocalculate 2) 4.Considerthefunction ) = (a)Plotthegraphofthefunction. (b)Writedownthedomainandrangeofthefunction. (c)Whathappenstotheoutputofthefunctionastheargument approacheszero? (d)Isapproachingzerofromtheleftdifferenttoapproachin gzerofromtheright?Ifyes,why? (e)Calculate (2) 10) and 5.Inthefollowinglist,youshouldwritedownthedomainand

rangeforeachfunction,andthen pairupfunctionsthatsharethesamedomainandrange. ) = 2 sin ) = + 9 ) = 2 ) = 4 ) = 2 + 3 ) = 3 ) = 2 cos 2 ) = 4 16 + 16 Answers 1. (a)Theargumentis (b)Thedependentvariableis + 5 intermsof (c)Anexampleofatableofvaluesmightbe 15 4 2 4 www.mathcentre.ac.uk 10 math centre2009
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(d)Drawalineupfrom = 1 ,andreadoffthevalueonthe -axis. (e) (1 5) = 9 (f)Readingofffromthegraph, 125 ,andalso < x < (g)Thefunctionis ) = 2 + 5 usinganargument www.mathcentre.ac.uk 11 math centre2009
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2. (a) 2 4 6 8 (b)Thedomainis = 3 ,therangeis

(c)Thefunctionis ) = 3) usinganargument (d)Drawalineupfrom = 1 ,andreadoffthevalueonthe -axis. (e) (1) = (1 3) (f)Onesuchfunctionis ) = 4) ,buttherearemanyothers. 3. (a)Wemustassumethatwetakethepositivesquareroot,andt hatwetake (b) 5 10 15 10 (c)Thedomainis ,therangeis (d) (3) = 3 3 = 5 (to1d.p.). (e)Ifyoutrytocalculate 2) ,youwillbeattemptingtofindthesquarerootofanegative number.Thishasnorealsolutions. www.mathcentre.ac.uk 12 math centre2009
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4. (a) 2 4 6 8 (b)Thedomainis = 0 ,therangeis = 0 (c)Theoutputapproaches

(d)Approachingfromtheleftisdifferenttoapproachingfro mtheright.Approachingfromthe left,theoutputdecreasesrapidlytonegativeinfinity. App roachingfromtheright,theoutput increasesrapidlytopositiveinfinity. (e) (2) = 10) = 10 ) = 1 /z 5. domain range ) = 2 sin < x < ) = + 9 < x < ) = 2 < x < ) = 4 < x < ) = 2 + 3 < x < ) = 3 < x < ) = 2 cos 2 < x < ) = 4 16 + 16 < x < Sothepairsare ) = 2 sin( and ) = 2 cos 2 ) = + 9 and ) = 4 16 + 16 ) = 2 and ) = 3 ) = 4 and ) = 2 + 3 www.mathcentre.ac.uk 13 math centre2009