How to decide whether the relation x=y2 defines a function?

Question

Alleryexerimivn3 · Accepted Answer

This is a function of x and y. Can be wriiten as f(x)=y2 Given solution: Generally speaking, a function is a relationship between two variables.

Sanai Navarro · Accepted Answer

Given→usistherelationship:x = y2.Whetheritdef∈esafunctionisup→us→decide.Ifnomaerwt^hevalueofthefirstvariab≤, x,thereis≺iselyo≠valueofthesec⁡ondvariab≤, y,co∩ected→it∈sidetherelationship−−thenitwillbeafunction.Ifthisbreaksdownforeveno≠valueofthefirstvariab≤,itwillfail→beafunction.Ti^s→say,ifforsomevalueofthefirstvariab≤,therearetwoormorevalues(ornovalues)ofthesec⁡ondvariab≤co∩ected→it∈sidetherelationship,thenitwill¬beafunction.Note−−∈≥≠ral,thereisnoprocedure→decideifanarbitrarilygivenrelationisfunctional[−−isafunctionor¬].Thetruthis,∈≥≠ral,therearenosuchprocedures.Ourcase,thankfl―y,turnsout→be∼p≤enough→makethedecision,≤t′ssay,usin⁡gg∞d∈st∈cts!!Wehave:x = y2.Weask,∈ourmind,foragivenvalueof  x,howmanyvaluesof  y  areco∩ected→it∈therelationship−−o≠,ormorethano≠?(displaystyletext{}{T}hat{{i}}{s}to{s}{a}{y},{f}{quadtext{or}quad}{a}{g}{i}{v}{e}{n}{v}{a}{l}{u}{e}{o}{f}text{} {x},text{}{h}{o}{w}{m}{a}{n}{y}{s}{o}{l}{u}{t}{i}{o}{n}{s}text{}  {y} )arethere→therelation: x = y2  ?−−o≠,ormorethano≠?Forexa∓≤,for  x  tak∈gthevalue 1,howmanysolutions  yarethere→therest―∈grelation:1⏟x = y2  ?−−o≠,ormorethano≠−−?Thisis,thankfl―y(!),easy→decide!!Weproceed,l∞k∈gatthesolutionsof:1 = y2.y2 = 1.y = ±1.y = −1,1.(displaystyletext{}{S}{o},{f}{quadtext{or}quad}text{}  {x}  text{}{t}{a}{k}in{g}{t}{h}{e}{v}{a}{l}{u}{e}text{} {1},text{}{t}{h}{e}{r}{e}{a}{r}{e}{t}{w}{o}{v}{a}{l}{u}{e}{s}{f}{quadtext{or}quad}text{}  {y} )co∩ected→it∈thegivenrelation: −1,1.  So,morethano≠valuefor  y, forthisvalueof  x.  Thisendsthedecisionrighthere.PS&quot;&quot;We can stop immediately now -- and conclude that the given&quot;&quot;ZSKrelationis¬afunction.Thisisourrest―:therelationx = y2is¬afunction.Iwant→makeaperhapsvaluab≤¬e,→keepperspective.(displaystyletext{}{I}{f}in{t}{h}{e}{a}{b}{o}{v}{e}{w}{quadtext{or}quad}{k},{w}{e}{h}{a}{d}pi{c}{k}{e}{d}{t}{h}{e}{v}{a}{l}{u}{e}{o}{f}text{}  {0}  text{}{f}{quadtext{or}quad}text{}  {x} )→take∈therelation,andthenl∞ked→seehowmanysolutions  y  thereare→therest―∈grelation:  0 = y2,wewod―havel∞kedatthesolutionsof:0 = y2.y2 = 0.y = 0,only.Andwewod―haveconcludedt,^for  x  tak∈gthevalue 0,thereisexactlyo≠value  y  co∩ected→it∈thegivenrelation:  0.  Exactlyo≠valuefor  y, co∩ected→thisvalueof  x.Wd^oesthistellusaboutwhetherthegivenrelationisafunction?NOTHING!!Becausethereisexactlyo≠valuefor  y  forthisvalueof  x,&amp;lt;br&amp;gt;weca∩otexcludetherelationombe∈gafunction,aswedidaboveusin⁡gthevalueof  1  for  x.Wealsoca∩otsayomthistt^herelationisafunction,either.Why?Theworkhere→lduswh^appe≠dwiththevaluesfor  y  co∩ectedwiththevalue  0  for  x  −−exactlyo≠valuefor  y.  Butit→ldus¬h∈gaboutthevaluesfor  y  co∩ectedwithanyothervaluefor  x.  Othervaluesfor  x  mighthaveexactlyo≠valuefor  y  co∩ected→it,mighthavemorethano≠valuefor  y  co∩ected→it,ormighthavenovaluesfor  y  co∩ected→it.Weca∩otknowun≤sswegobackandcheckvaluesfor  x,otherthan  0.Wo^thervaluesfor  x,shod―wecheck−−otherthan  0  ?Thetruthis,∈≥≠ral,thereisnoway→determinew^othervaluesfor  x  (ifthereareany)weshod―check.Wewereluckyweπckedthevalue  1  for  x  above−−whichallowedus→makeadecisiononthisrelation.Forcerta∈typesofrelations,thereareways→determineothervalues→check.In≥≠ral,thereisnosuchprocedureforf∈d∈gsuchluck−−justhope,andg∞d∈st∈cts!!

How to decide whether the relation x=y^2 defines a function?

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