Calculus I - Critical Points - Pauls Online Math Notes

We say that x=c x = c is a critical point of the function f(x) f ( x ) if f(c) f ( c ) exists and if either of the following are true. 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Section4-2:CriticalPoints Criticalpointswillshowupthroughoutamajorityofthischaptersowefirstneedtodefinethemandworkafewexamplesbeforegettingintothesectionsthatactuallyusethem. Definition Wesaythat\(x=c\)isacriticalpointofthefunction\(f\left(x\right)\)if\(f\left(c\right)\)existsandifeitherofthefollowingaretrue. \[f'\left(c\right)=0\hspace{0.5in}{\mbox{OR}}\hspace{0.5in}f'\left(c\right)\,\,\,{\mbox{doesn'texist}}\] Notethatwerequirethat\(f\left(c\right)\)existsinorderfor\(x=c\)toactuallybeacriticalpoint.Thisisanimportant,andoftenoverlooked,point.Whatthisisreallysayingisthatallcriticalpointsmustbeinthedomainofthefunction.Ifapointisnotinthedomainofthefunctionthenitisnotacriticalpoint. Noteaswellthat,atthispoint,weonlyworkwithrealnumbersandsoanycomplexnumbersthatmightariseinfindingcriticalpoints(andtheywillariseonoccasion)willbeignored.Thereareportionsofcalculusthatworkalittledifferentlywhenworkingwithcomplexnumbersandsoinafirstcalculusclasssuchasthisweignorecomplexnumbersandonlyworkwithrealnumbers.Calculuswithcomplexnumbersisbeyondthescopeofthiscourseandisusuallytaughtinhigherlevelmathematicscourses. Themainpointofthissectionistoworksomeexamplesfindingcriticalpoints.So,let’sworksomeexamples. Example1Determineallthecriticalpointsforthefunction. \[f\left(x\right)=6{x^5}+33{x^4}-30{x^3}+100\] ShowSolution Wefirstneedthederivativeofthefunctioninordertofindthecriticalpointsandsolet’sgetthatandnoticethatwe’llfactoritasmuchaspossibletomakeourlifeeasierwhenwegotofindthecriticalpoints. \[\begin{align*}f'\left(x\right)&=30{x^4}+132{x^3}-90{x^2}\\&=6{x^2}\left({5{x^2}+22x-15}\right)\\&=6{x^2}\left({5x-3}\right)\left({x+5}\right)\end{align*}\] Now,ourderivativeisapolynomialandsowillexisteverywhere.Therefore,theonlycriticalpointswillbethosevaluesof\(x\)whichmakethederivativezero.So,wemustsolve. \[6{x^2}\left({5x-3}\right)\left({x+5}\right)=0\] Becausethisisthefactoredformofthederivativeit’sprettyeasytoidentifythethreecriticalpoints.Theyare, \[x=-5,\,\,\,\,\,\,x=0,\,\,\,\,\,\,x=\frac{3}{5}\] Polynomialsareusuallyfairlysimplefunctionstofindcriticalpointsforprovidedthedegreedoesn’tgetsolargethatwehavetroublefindingtherootsofthederivative. Mostofthemore“interesting”functionsforfindingcriticalpointsaren’tpolynomialshowever.Solet’stakealookatsomefunctionsthatrequirealittlemoreeffortonourpart. Example2Determineallthecriticalpointsforthefunction. \[g\left(t\right)=\sqrt[3]{{{t^2}}}\left({2t-1}\right)\] ShowSolution Tofindthederivativeit’sprobablyeasiesttodoalittlesimplificationbeforeweactuallydifferentiate.Let’smultiplytherootthroughtheparenthesisandsimplifyasmuchaspossible.Thiswillallowustoavoidusingtheproductrulewhentakingthederivative. \[g\left(t\right)={t^{\frac{2}{3}}}\left({2t-1}\right)=2{t^{\frac{5}{3}}}-{t^{\frac{2}{3}}}\] Nowdifferentiate. \[g'\left(t\right)=\frac{{10}}{3}{t^{\frac{2}{3}}}-\frac{2}{3}{t^{-\frac{1}{3}}}=\frac{{10{t^{\frac{2}{3}}}}}{3}-\frac{2}{{3{t^{\frac{1}{3}}}}}\] Wewillneedtobecarefulwiththisproblem.Whenfacedwithanegativeexponentitisoftenbesttoeliminatetheminussignintheexponentaswedidabove.Thisisn’treallyrequiredbutitcanmakeourlifeeasieronoccasionifwedothat. Noticeaswellthateliminatingthenegativeexponentinthesecondtermallowsustocorrectlyidentifywhy\(t=0\)isacriticalpointforthisfunction.Oncewemovethesecondtermtothedenominatorwecanclearlyseethatthederivativedoesn’texistat\(t=0\)andsothiswillbeacriticalpoint.Ifyoudon’tgetridofthenegativeexponentinthesecondtermmanypeoplewillincorrectlystatethat\(t=0\)isacriticalpointbecausethederivativeiszeroat\(t=0\).Whilethismayseemlikeasillypoint,afterallineachcase\(t=0\)isidentifiedasacriticalpoint,it issometimesimportanttoknowwhyapointisacriticalpoint.Infact,inacoupleofsectionswe’llseeafactthatonlyworksforcriticalpointsinwhichthederivativeiszero. So,we’vefoundonecriticalpoint(wherethederivativedoesn’texist),butwenowneedtodeterminewherethederivativeiszero(provideditisofcourse…).Tohelpwiththisit’susuallybesttocombinethetwotermsintoasinglerationalexpression.So,gettingacommondenominatorandcombininggivesus, \[g'\left(t\right)=\frac{{10t-2}}{{3{t^{\frac{1}{3}}}}}\] Noticethatwestillhave\(t=0\)asacriticalpoint.Doingthiskindofcombiningshouldneverlosecriticalpoints,it’sonlybeingdonetohelpusfindthem.Aswecanseeit’snowbecomemucheasiertoquicklydeterminewherethederivativewillbezero.Recallthatarationalexpressionwillonlybezeroifitsnumeratoriszero(andprovidedthedenominatorisn’talsozeroatthatpointofcourse). So,inthiscasewecanseethatthenumeratorwillbezeroif\(t=\frac{1}{5}\)andsotherearetwocriticalpointsforthisfunction. \[t=0\hspace{0.5in}\,{\mbox{and}}\hspace{0.5in}t=\frac{1}{5}\] Example3Determineallthecriticalpointsforthefunction. \[R\left(w\right)=\frac{{{w^2}+1}}{{{w^2}-w-6}}\] ShowSolution We’llleaveittoyoutoverifythatusingthequotientrule,alongwithsomesimplification,wegetthatthederivativeis, \[R'\left(w\right)=\frac{{-{w^2}-14w+1}}{{{{\left({{w^2}-w-6}\right)}^2}}}=-\frac{{{w^2}+14w-1}}{{{{\left({{w^2}-w-6}\right)}^2}}}\] Noticethatwefactoreda“-1”outofthenumeratortohelpalittlewithfindingthecriticalpoints.Thisnegativeoutinfrontwillnotaffectthederivativewhetherornotthederivativeiszeroornotexistbutwillmakeourworkalittleeasier. Now,wehavetwoissuestodealwith.Firstthederivativewillnotexistifthereisdivisionbyzerointhedenominator.Soweneedtosolve, \[{w^2}-w-6=\left({w-3}\right)\left({w+2}\right)=0\] Wedidn’tbothersquaringthissinceifthisiszero,thenzerosquaredisstillzeroandifitisn’tzerothensquaringitwon’tmakeitzero. So,wecanseefromthisthatthederivativewillnotexistat\(w=3\)and\(w=-2\).However,theseareNOTcriticalpointssincethefunctionwillalsonotexistatthesepoints.Recallthatinorderforapointtobeacriticalpointthefunctionmustactuallyexistatthatpoint. Atthispointweneedtobecareful.Thenumeratordoesn’tfactor,butthatdoesn’tmeanthattherearen’tanycriticalpointswherethederivativeiszero.Wecanusethequadraticformulaonthenumeratortodetermineifthefractionasawholeiseverzero. \[w=\frac{{-14\pm\sqrt{{{\left({14}\right)}^2}-4\left(1\right)\left({-1}\right)}}}{{2\left(1\right)}}=\frac{{-14\pm\sqrt{200}}}{2}=\frac{{-14\pm10\sqrt2}}{2}=-7\pm5\sqrt2\] So,wegettwocriticalpoints.Also,thesearenot“nice”integersorfractions.Thiswillhappenonoccasion.Don’tgettoolockedintoanswersalwaysbeing“nice”.Oftentheyaren’t. Noteaswellthatweonlyuserealnumbersforcriticalpoints.So,ifuponsolvingthequadraticinthenumerator,wehadgottencomplexnumberthesewouldnothavebeenconsideredcriticalpoints. Summarizing,wehavetwocriticalpoints.Theyare, \[w=-7+5\sqrt2,\,\,\,\,w=-7-5\sqrt2\] Again,rememberthatwhilethederivativedoesn’texistat\(w=3\)and\(w=-2\)neitherdoesthefunctionandsothesetwopointsarenotcriticalpointsforthisfunction. Inthepreviousexamplewehadtousethequadraticformulatodeterminesomepotentialcriticalpoints.Weknowthatsometimeswewillgetcomplexnumbersoutofthequadraticformula.Justrememberthat,asmentionedatthestartofthissection,whenthathappenswewillignorethecomplexnumbersthatarise. Sofaralltheexampleshavenothadanytrigfunctions,exponentialfunctions,etc.inthem.Weshouldn’texpectthattoalwaysbethecase.So,let’stakealookatsomeexamplesthatdon’tjustinvolvepowersof\(x\). Example4Determineallthecriticalpointsforthefunction. \[y=6x-4\cos\left({3x}\right)\] ShowSolution Firstgetthederivativeanddon’tforgettousethechainruleonthesecondterm. \[y'=6+12\sin\left({3x}\right)\] Now,thiswillexisteverywhereandsotherewon’tbeanycriticalpointsforwhichthederivativedoesn’texist.Theonlycriticalpointswillcomefrompointsthatmakethederivativezero.Wewillneedtosolve, \[\begin{align*}6+12\sin\left({3x}\right)&=0\\\sin\left({3x}\right)&=-\frac{1}{2}\end{align*}\] Solvingthisequationgivesthefollowing. \[\begin{align*}3x&=3.6652+2\pin,\hspace{0.25in}n=0,\pm1,\pm2,\ldots\\3x&=5.7596+2\pin,\hspace{0.25in}n=0,\pm1,\pm2,\ldots\end{align*}\] Don’tforgetthe\(2\pin\)onthese!Therewillbeproblemsdowntheroadinwhichwewillmisssolutionswithoutthis!Alsomakesurethatitgetsputonatthisstage!Nowdivideby3togetallthecriticalpointsforthisfunction. \[\begin{align*}x&=1.2217+\frac{{2\pin}}{3},\hspace{0.5in}n=0,\pm1,\pm2,\ldots\\x&=1.9199+\frac{{2\pin}}{3},\hspace{0.5in}n=0,\pm1,\pm2,\ldots\end{align*}\] Noticethatinthepreviousexamplewegotaninfinitenumberofcriticalpoints.Thatwillhappenonoccasionsodon’tworryaboutitwhenithappens. Example5Determineallthecriticalpointsforthefunction. \[h\left(t\right)=10t{{\bf{e}}^{3-{t^2}}}\] ShowSolution Here’sthederivativeforthisfunction. \[h'\left(t\right)=10{{\bf{e}}^{3-{t^2}}}+10t{{\bf{e}}^{3-{t^2}}}\left({-2t}\right)=10{{\bf{e}}^{3-{t^2}}}-20{t^2}{{\bf{e}}^{3-{t^2}}}\] Now,thislooksunpleasant,howeverwithalittlefactoringwecancleanthingsupalittleasfollows, \[h'\left(t\right)=10{{\bf{e}}^{3-{t^2}}}\left({1-2{t^2}}\right)\] Thisfunctionwillexisteverywhere,sonocriticalpointswillcomefromthederivativenotexisting.Determiningwherethisiszeroiseasierthanitlooks.Weknowthatexponentialsareneverzeroandsotheonlywaythederivativewillbezeroisif, \[\begin{align*}1-2{t^2}&=0\\1&=2{t^2}\\\frac{1}{2}&={t^2}\end{align*}\] Wewillhavetwocriticalpointsforthisfunction. \[t=\pm\frac{1}{{\sqrt2}}\] Example6Determineallthecriticalpointsforthefunction. \[f\left(x\right)={x^2}\ln\left({3x}\right)+6\] ShowSolution Beforegettingthederivativelet’snoticethatsincewecan’ttakethelogofanegativenumberorzerowewillonlybeabletolookat\(x>0\). Thederivativeisthen, \[\begin{align*}f'\left(x\right)&=2x\ln\left({3x}\right)+{x^2}\left({\frac{3}{{3x}}}\right)\\&=2x\ln\left({3x}\right)+x\\&=x\left({2\ln\left({3x}\right)+1}\right)\end{align*}\] Now,thisderivativewillnotexistif\(x\)isanegativenumberorif\(x=0\),butthenagainneitherwillthefunctionandsothesearenotcriticalpoints.Rememberthatthefunctionwillonlyexistif\(x>0\)andnicelyenoughthederivativewillalsoonlyexistif\(x>0\)andsotheonlythingweneedtoworryaboutiswherethederivativeiszero. Firstnotethat,despiteappearances,thederivativewillnotbezerofor\(x=0\).Asnotedabovethederivativedoesn’texistat\(x=0\)becauseofthenaturallogarithmandsothederivativecan’tbezerothere! So,thederivativewillonlybezeroif, \[\begin{align*}2\ln\left({3x}\right)+1&=0\\\ln\left({3x}\right)&=-\frac{1}{2}\end{align*}\] Recallthatwecansolvethisbyexponentiatingbothsides. \[\begin{align*}{{\bf{e}}^{\ln\left({3x}\right)}}&={{\bf{e}}^{-\frac{1}{2}}}\\3x&={{\bf{e}}^{-\frac{1}{2}}}\\x&=\frac{1}{3}{{\bf{e}}^{-\frac{1}{2}}}=\frac{1}{{3\sqrt{\bf{e}}}}\end{align*}\] Thereisasinglecriticalpointforthisfunction. Let’sworkonemoreproblemtomakeapoint. Example7Determineallthecriticalpointsforthefunction. \[f\left(x\right)=x{{\bf{e}}^{{x^2}}}\] ShowSolution NotethatthisfunctionisnotmuchdifferentfromthefunctionusedinExample5.Inthiscasethederivativeis, \[f'\left(x\right)={{\bf{e}}^{{x^2}}}+x{{\bf{e}}^{{x^2}}}\left({2x}\right)={{\bf{e}}^{{x^2}}}\left({1+2{x^2}}\right)\] Thisfunctionwillneverbezeroforanyrealvalueof\(x\).Theexponentialisneverzeroofcourseandthepolynomialwillonlybezeroif\(x\)iscomplexandrecallthatweonlywantrealvaluesof\(x\)forcriticalpoints. Therefore,thisfunctionwillnothaveanycriticalpoints. Itisimportanttonotethatnotallfunctionswillhavecriticalpoints!Inthiscoursemostofthefunctionsthatwewillbelookingatdohavecriticalpoints.Thatisonlybecausethoseproblemsmakeformoreinterestingexamples.Donotletthisfactleadyoutoalwaysexpectthatafunctionwillhavecriticalpoints.Sometimestheydon’tasthisfinalexamplehasshown.