Extrema of a Function

Subsection 5.5.1 Relative Extrema ... A relative maximum point on a function is a point (x,y) ( x , y ) on the graph of the function whose y y -coordinate is ... Skiptomaincontent\(\renewcommand{\mybuildpath}{./} \renewcommand{\labelitemi}{\scriptsize$\blacktriangleright$} \def\ds{\displaystyle} \def\R{\mathbb{R}} \def\arraystretch{2.5} \renewcommand{\Heq}{\overset{H}{=}} \renewcommand{\vect}{\textbf} \renewcommand{\longvect}{\overrightarrow} \newcommand{\diff}[2]{\dfrac{d#1}{d#2}} \newcommand{\diffp}[2]{\dfrac{\partial#1}{\partial#2}} \newcommand{\lt}{} \newcommand{\amp}{&} \) FrontMatter Copyright Acknowledgements OpenEducationalResources(OER)Support:CorrectionsandSuggestions Dedication Introduction ProblemSolvingStrategies 1Review Algebra AnalyticGeometry Trigonometry AdditionalExercises 2Functions AllAboutFunctions Symmetry,TransformationsandCompositions ExponentialFunctions InverseFunctions LogarithmicFunctions TrigonometricFunctions EconomicModels AdditionalExercises 3LimitsandContinuity TheLimit PreciseDefinitionofaLimit ComputingLimits:Graphically ComputingLimits:Algebraically LimitsatInfinity,InfiniteLimitsandAsymptotes TheSqueezeTheorem ContinuityandIVT 4Derivatives TheRateofChangeofaFunction TheDerivativeFunction DerivativeRules TheChainRule DerivativeRulesforTrigonometricFunctions DerivativesofExponential&LogarithmicFunctions ImplicitandLogarithmicDifferentiation DerivativesofInverseFunctions AdditionalExercises 5ApplicationsofDerivatives ElasticityofDemand RelatedRates LinearandHigherOrderApproximations IndeterminateForm&L'Hôpital'sRule ExtremaofaFunction TheMeanValueTheorem CurveSketching OptimizationProblems 6ThreeDimensions TheCoordinateSystem Vectors TheDotProduct TheCrossProduct LinesandPlanes OtherCoordinateSystems 7Multi-VariableCalculus FunctionsofSeveralVariables LimitsandContinuity PartialDifferentiation TheChainRule DirectionalDerivatives MaximaandMinima LagrangeMultipliers BackMatter Index AuthoredinPreTeXt Section5.5ExtremaofaFunction ¶Incalculus,thereismuchemphasisplacedonanalyzingthebehaviourofafunction\(f\)onaninterval\(I\text{.}\)Does\(f\)haveamaximumvalueon\(I\text{?}\)Doesithaveaminimumvalue?Howdoestheinterval\(I\)impactourdiscussionofextrema? Subsection5.5.1RelativeExtrema ¶Arelativemaximumpointonafunctionisapoint\((x,y)\)onthegraphofthefunctionwhose\(y\)-coordinateislargerthanallother\(y\)-coordinatesonthegraphatpoints“closeto”\((x,y)\text{.}\)Moreprecisely,\((x,f(x))\)isarelativemaximumifthereisaninterval\((a,b)\)with\(a\ltx\ltb\)and\(f(x)\gef(z)\)forevery\(z\)in\((a,b)\text{.}\)Similarly,\((x,y)\)isarelativeminimumpointifithaslocallythesmallest\(y\)-coordinate.Againbeingmoreprecise:\((x,f(x))\)isarelativeminimumifthereisaninterval\((a,b)\)with\(a\ltx\ltb\)and\(f(x)\lef(z)\)forevery\(z\)in\((a,b)\text{.}\)Arelativeextremumiseitherarelativeminimumorarelativemaximum. Note: Thepluralofextremumisextremaandsimilarlyformaximumandminimum. Becausearelativeextremumis“extreme”locallybylookingatpoints“closeto”it,itisalsoreferredtoasalocalextremum. Definition5.46.RelativeMaximaandMinima. Areal-valuedfunction\(f\)hasarelativemaximumat\(x_0\)if\(f(x_0)\geqf(x)\)forall\(x\)insomeopenintervalcontaining\(x_0\text{.}\) Areal-valuedfunction\(f\)hasarelativeminimumat\(x_0\)if\(f(x_0)\leqf(x)\)forall\(x\)insomeopenintervalcontaining\(x_0\text{.}\) Relativemaximumandminimumpointsarequitedistinctiveonthegraphofafunction,andarethereforeusefulinunderstandingtheshapeofthegraph.Inmanyappliedproblemswewanttofindthelargestorsmallestvaluethatafunctionachieves(forexample,wemightwanttofindtheminimumcostatwhichsometaskcanbeperformed)andsoidentifyingmaximumandminimumpointswillbeusefulforappliedproblemsaswell.SomeexamplesofrelativemaximumandminimumpointsareshowninFigure 5.14. Figure5.14.Somerelativemaximumpoints(\(A\))andminimumpoints(\(B\)).If\((x,f(x))\)isapointwhere\(f(x)\)reachesarelativemaximumorminimum,andifthederivativeof\(f\)existsat\(x\text{,}\)thenthegraphhasatangentlineandthetangentlinemustbehorizontal.Thisisimportantenoughtostateasatheorem. Theproofissimpleenoughandweincludeithere,butyoumayacceptFermat'sTheorembasedonitsstrongintuitiveappealandcomebacktoitsproofatalatertime. Theorem5.47.Fermat'sTheorem. If\(f(x)\)hasarelativeextremumat\(x=a\)and\(f\)isdifferentiableat\(a\text{,}\)then\(f'(a)=0\text{,}\)providedthat\(f'(a)\)exists.Proof.Weshallgivetheproofforthecasewhere\(f\left(x\right)\)hasarelativemaximumat\(x=a\text{.}\)Theprooffortherelativeminimumcaseissimilar. Since\(f\left(x\right)\)hasarelativemaximumat\(x=a\text{,}\)thereisanopeninterval\(\left(c,d\right)\)with\(c\lta\ltd\)and\(f\left(x\right)\leqf\left(a\right)\)forevery\(x\)in\(\left(c,d\right)\text{.}\)So,\(f\left(x\right)-f\left(a\right)\leq0\)forallsuch\(x\text{.}\)Letusnowlookatthesignofthedifferencequotient\(\dfrac{f\left(x\right)-f\left(a\right)}{x-a}\text{.}\)Weconsidertwocasesaccordingas\(x>a\)or\(x\lta\text{.}\) If\(x>a\text{,}\)then\(x-a>0\)andso,\(\dfrac{f\left(x\right)-f\left(a\right)}{x-a}\leq0\text{.}\)Takinglimitas\(x\)approach\(a\)fromtheright,weget \begin{equation*} \lim_{x\rightarrowa^{+}}\frac{f\left(x\right)-f\left(a\right)}{x-a}\leq0\text{.} \end{equation*} Ontheotherhand,if\(x\lta\text{,}\)then\(x-a\lt0\)andso,\(\dfrac{f\left(x\right)-f\left(a\right)}{x-a}\geq0\text{.}\)Takinglimitas\(x\)approach\(a\)fromtheleft,weget \begin{equation*} \lim_{x\rightarrowa^{-}}\frac{f\left(x\right)-f\left(a\right)}{x-a}\geq0\text{.} \end{equation*} Since\(f\)isdifferentiableat\(a\text{,}\) \begin{equation*} f^{\prime}\left(a\right)=\lim\limits_{x\rightarrowa^{+}}\dfrac{f\left(x\right)-f\left(a\right)}{x-a}=\lim\limits_{x\rightarrowa^{-}}\dfrac{f\left(x\right)-f\left(a\right)}{x-a}\text{.} \end{equation*} Therefore,wehaveboth\(f^{\prime}\left(a\right)\leq0\)and\(f^{\prime}\left(a\right)\geq0\text{.}\)So,\(f^{\prime}\left(a\right)=0\text{.}\) Thus,theonlypointsatwhichafunctioncanhavearelativemaximumorminimumarepointsatwhichthederivativeiszero,asintheleft-handgraphinFigure 5.14,orthederivativeisundefined,asintheright-handgraph.Thisleadsustodefinethesespecialpoints. Definition5.48.CriticalPoint. Anyvalueof\(x\)inthedomainof\(f\)forwhich\(f'(x)\)iszeroorundefinediscalledacriticalpointof\(f\text{.}\) The\(x\)-values\(a\text{,}\)\(b\)and\(c\)aboveareplacesforwhich\(f'(x)\)iszero,andthe\(x\)-values\(d\)and\(e\)aboveareplacesforwhich\(f'(x)\)isundefined.Note:Whenlookingforrelativemaximumandminimumpoints,youarelikelytomaketwosortsofmistakes. Youmayforgetthatamaximumorminimumcanoccurwherethederivativedoesnotexist.Youshouldthereforecheckwhetherthederivativeexistseverywhere. Youmightalsoassumethatanyplacethatthederivativeiszeroisarelativemaximumorminimumpoint,butthisisnottrue.Aportionofthegraphof\(\dsf(x)=x^3\)isshowninFigure 5.15.Thederivativeof\(f\)is\(f'(x)=3x^2\text{,}\)and\(f'(0)=0\text{,}\)butthereisneitheramaximumnorminimumat\((0,0)\text{.}\)Inotherwords,theconverseofFermat'sTheorem—if\(f'(a)=0\)atsomepoint\(x=a\text{,}\)then\(f\)musthavearelativeextremumatthatpoint—isnottrue. Figure5.15.Norelativeextremaeventhoughthederivativeiszeroat\(x=0\text{.}\)Sincethederivativeiszeroorundefinedatbothrelativemaximumandrelativeminimumpoints,weneedawaytodeterminewhich,ifeither,actuallyoccurs.Themostelementaryapproach,butonethatisoftentediousordifficult,istotestdirectlywhetherthe\(y\)-coordinates“near”thepotentialmaximumorminimumareaboveorbelowthe\(y\)-coordinateatthepointofinterest.Ofcourse,therearetoomanypoints“near”thepointtotest,butalittlethoughtshowsweneedonlytesttwoprovidedweknowthat\(f\)iscontinuous(recallthatthismeansthatthegraphof\(f\)hasnojumpsorgaps). Suppose,forexample,thatwehaveidentifiedthreepointsatwhich\(f'\)iszeroornonexistent:\(\ds(x_1,y_1)\text{,}\)\(\ds(x_2,y_2)\text{,}\)\(\ds(x_3,y_3)\text{,}\)and\(\dsx_1\ltx_2\ltx_3\)(seeFigure 5.16).Supposethatwecomputethevalueof\(f(a)\)for\(\dsx_1\lta\ltx_2\text{,}\)andthat\(\dsf(a)\ltf(x_2)\text{.}\)Whatcanwesayaboutthegraphbetween\(a\)and\(\dsx_2\text{?}\)Couldtherebeapoint\(\ds(b,f(b))\text{,}\)\(\dsa\ltb\ltx_2\)with\(\dsf(b)>f(x_2)\text{?}\)No:iftherewere,thegraphwouldgoupfrom\((a,f(a))\)to\((b,f(b))\)thendownto\(\ds(x_2,f(x_2))\)andsomewhereinbetweenwouldhavearelativemaximumpoint.(Thisisnotobvious;itisaresultoftheExtremeValueTheoremstatedinthenextsection.)Butatthatrelativemaximumpointthederivativeof\(f\)wouldbezeroornonexistent,yetwealreadyknowthatthederivativeiszeroornonexistentonlyat\(\dsx_1\text{,}\)\(\dsx_2\text{,}\)and\(\dsx_3\text{.}\)Theupshotisthatonecomputationtellsusthat\(\ds(x_2,f(x_2))\)hasthelargest\(y\)-coordinateofanypointonthegraphnear\(\dsx_2\)andtotheleftof\(\dsx_2\text{.}\)Wecanperformthesametestontheright.Ifwefindthatonbothsidesof\(\dsx_2\)thevaluesaresmaller,thentheremustbearelativemaximumat\(\ds(x_2,f(x_2))\text{;}\)ifwefindthatonbothsidesof\(\dsx_2\)thevaluesarelarger,thentheremustbearelativeminimumat\(\ds(x_2,f(x_2))\text{;}\)ifwefindoneofeach,thenthereisneitherarelativemaximumorminimumat\(\dsx_2\text{.}\) Figure5.16.Testingforamaximumorminimum.Itisnotalwayseasytocomputethevalueofafunctionataparticularpoint.Thetaskismadeeasierbytheavailabilityofcalculatorsandcomputers,buttheyhavetheirowndrawbacks—theydonotalwaysallowustodistinguishbetweenvaluesthatareveryclosetogether.Nevertheless,becausethismethodisconceptuallysimpleandsometimeseasytoperform,youshouldalwaysconsiderit. Example5.49.TestingforRelativeExtremainCubicFunction. Findallrelativemaximumandminimumpointsforthefunction\(\dsf(x)=x^3-x\text{.}\) SolutionThederivativeis\(\dsf'(x)=3x^2-1\text{.}\)Thisisdefinedeverywhereandiszeroat\(\dsx=\pm\sqrt{3}/3\text{.}\)Lookingfirstat\(\dsx=\sqrt{3}/3\text{,}\)weseethat\(\dsf(\sqrt{3}/3)=-2\sqrt{3}/9\text{.}\)Nowwetesttwopointsoneithersideof\(\dsx=\sqrt{3}/3\text{,}\)choosingonepointintheinterval\((-\sqrt{3}/3,\sqrt{3}/3)\)andonepointintheinterval\((\sqrt{3}/3,\infty)\text{.}\)Since\(\dsf(0)=0>-2\sqrt{3}/9\)and\(\dsf(1)=0>-2\sqrt{3}/9\text{,}\)theremustbearelativeminimumat\(\dsx=\sqrt{3}/3\text{.}\)For\(\dsx=-\sqrt{3}/3\text{,}\)weseethat\(\dsf(-\sqrt{3}/3)=2\sqrt{3}/9\text{.}\)Thistimewecanuse\(x=0\)and\(x=-1\text{,}\)andwefindthat\(\dsf(-1)=f(0)=0\lt2\sqrt{3}/9\text{,}\)sotheremustbearelativemaximumat\(\dsx=-\sqrt{3}/3\text{.}\) Ofcoursethisexampleismadeverysimplebyourchoiceofpointstotest,namely\(x=-1\text{,}\)\(0\text{,}\)\(1\text{.}\)Wecouldhaveusedothervalues,say\(-5/4\text{,}\)\(1/3\text{,}\)and\(3/4\text{,}\)butthiswouldhavemadethecalculationsconsiderablymoretedious,andweshouldalwayschooseverysimplepointstotestifwecan. Example5.50.TestingforRelativeExtremainTrigonometricFunction. Findallrelativemaximumandminimumpointsfor\(f(x)=\sinx+\cosx\text{.}\) Solution Thederivativeis\(f'(x)=\cosx-\sinx\text{.}\)Thisisalwaysdefinedandiszerowhenever\(\cosx=\sinx\text{.}\)Recallingthatthe\(\cosx\)and\(\sinx\)arethe\(x\)-and\(y\)-coordinatesofpointsonaunitcircle,weseethat\(\cosx=\sinx\)when\(x\)is\(\pi/4\text{,}\)\(\pi/4\pm\pi\text{,}\)\(\pi/4\pm2\pi\text{,}\)\(\pi/4\pm3\pi\text{,}\)etc.Sincebothsineandcosinehaveaperiodof\(2\pi\text{,}\)weneedonlydeterminethestatusof\(x=\pi/4\)and\(x=5\pi/4\text{.}\)Wecanuse\(0\)and\(\pi/2\)totestthecriticalvalue\(x=\pi/4\text{.}\)Wefindthat\(\dsf(\pi/4)=\sqrt{2}\text{,}\)\(\dsf(0)=1\lt\sqrt{2}\)and\(\dsf(\pi/2)=1\text{,}\)sothereisarelativemaximumwhen\(x=\pi/4\)andalsowhen\(x=\pi/4\pm2\pi\text{,}\)\(\pi/4\pm4\pi\text{,}\)etc.Wecansummarizethismoreneatlybysayingthattherearerelativemaximaat\(\pi/4\pm2k\pi\)foreveryinteger\(k\text{.}\) Weuse\(\pi\)and\(\frac{3\pi}{2}\)totestthecriticalvalue\(x=5\pi/4\text{.}\)Therelevantvaluesare\(\dsf(5\pi/4)=-\sqrt2\text{,}\)\(\dsf(\pi)=-1>-\sqrt2\text{,}\)\(\dsf(3\pi/2)=1>-\sqrt2\text{,}\)sothereisarelativeminimumat\(x=5\pi/4\text{,}\)\(5\pi/4\pm2\pi\text{,}\)\(5\pi/4\pm4\pi\text{,}\)etc.Moresuccinctly,therearerelativeminimaat\(5\pi/4\pm2k\pi\)foreveryinteger\(k\text{.}\) Example5.51.TestingforRelativeExtremainPowerFunction. Findallrelativemaximumandminimumpointsfor\(g\left(x\right)=x^{2/3}\text{.}\) SolutionThederivativeis\(g^{\prime}\left(x\right)=\frac{2}{3}x^{-1/3}\text{.}\)Thisisundefinedwhen\(x=0\)andisnotequaltozeroforany\(x\)inthedomainof\(g^{\prime}\text{.}\)Nowwetesttwopointsoneithersideof\(x=0\text{.}\)Weuse\(x=-1\)and\(x=1\text{.}\)Since\(g\left(0\right)=0\text{,}\)\(g\left(-1\right)=1>0\)and\(g\left(1\right)=1>0\text{,}\)theremustbearelativeminimumat\(x=0\text{.}\) ExercisesforSection 5.5.1. Exercise5.5.1.Findallrelativemaximumandminimumpoints\((x,y)\)bythemethodofthissection. \(\dsy=x^2-x\) Answerminat\(x=1/2\) Solution Let\(y=f(x)\text{.}\)Then\(f(x)=x^2-x\)isapolynomial,andsoitsdomainisallrealnumbers.Differentiating,wefind: \begin{equation*} f'(x)=2x-1\text{.} \end{equation*} Hence,\(f'(x)=0\)when\(x=0.5\text{,}\)andtherearenopointsinthedomainof\(f\)forwhich\(f'(x)\)isundefined.Therefore,theonlycriticalpointisat\(x=0.5\text{.}\)Weusethetestpoints\(x=0\)and\(x=1\text{:}\) \begin{equation*} \begin{array}{c|c}x\ampf(x)\\\hline0\amp0\\0.5\amp-0.75\\1\amp0\end{array} \end{equation*} Andso\(f\)hasarelativeminimumatthepoint\((0.5,-0.75)\text{.}\) \(\dsy=2+3x-x^3\) Answerminat\(x=-1\text{,}\)maxat\(x=1\) Solution Thefunction\(2+3x-x^3\)isapolynomial,andsohasdomain\((-\infty,\infty)\text{.}\)Nowdifferentiate: \begin{equation*} y'=3-3x^2\text{.} \end{equation*} Then\(y'=0\)hastwosolutions:\(x=\pm1\text{.}\)Therearenopointsforwhich\(y'\)isundefined,so\(y\)hastwocriticalpointsat\(x=1\)and\(x=-1\text{.}\)Since \begin{equation*} \begin{array}{c|c}x\ampy\\\hline-2\amp4\\-1\amp0\\0\amp2\\1\amp4\\2\amp0\end{array} \end{equation*} wemusthavethat\(x=-1\)isarelativeminimum,and\(x=1\)isarelativemaximum. \(\dsy=x^3-9x^2+24x\) Answermaxat\(x=2\text{,}\)minat\(x=4\) Solution Thefunction\(x^3-9x^2+24x\)isapolynomial,andsoitsdomainis\((-\infty,\infty)\text{.}\)Wenowcalculate\(y'\text{:}\) \begin{equation*} y'=3x^2-18x+24\text{.} \end{equation*} Therefore, \begin{equation*} \begin{split}y'\amp=0\\3x^2-18x+24\amp=0\\3(x^2-6x+8)\amp=0\\3(x-2)(x-4)\amp=0\\x\amp=2,4.\end{split} \end{equation*} Therearenopointsinthedomainof\(y\)suchthat\(y'\)isundefined,andso\(x=2\)and\(x=4\)aretheonlycriticalpointsof\(y\text{.}\)Since \begin{equation*} \begin{array}{c|c}x\ampy\\\hline1\amp16\\2\amp20\\3\amp18\\4\amp16\\5\amp20\end{array} \end{equation*} weseethat\(y\)hasarelativemaximumat\(x=2\)andarelativeminimumat\(x=4\text{.}\) \(\dsy=x^4-2x^2+3\) Answerminat\(x=\pm1\text{,}\)maxat\(x=0\) Solution Let\(y=f(x)\text{.}\)Thedomainof\(f\)isallrealnumbers.Calculate\(f'(x)\text{:}\) \begin{equation*} f'(x)=\diff{}{x}\left(x^4-2x^2+3\right)=4x^3-4x=4x(x^2-1)\text{.} \end{equation*} Therefore,\(f'(x)=0\)when\(x=-1,0,1\)andisdefinedovertheentiredomainof\(f\text{.}\)Noticethat\(f(x)\)isanevenfunction(thatis,\(f(x)=f(-x)\)).Therefore,weneedonlyclassifytherelativemaximaandminimafor\(x\leq0\text{.}\)Nowselectappropriatetestpointsandevaluate\(f\text{:}\) \begin{equation*} \begin{array}{c|c}x\ampf(x)\\\hline-2\amp11\\-1\amp2\\-0.5\amp2.5625\\0\amp3\end{array} \end{equation*} Hence,\(f(x)\)hasrelativeminimaat\(x=\pm1\text{,}\)andarelativemaximumat\(x=0\text{.}\) \(\dsy=3x^4-4x^3\) Answerminat\(x=1\) Solution Let\(y=f(x)\text{.}\)Thedomainof\(f\)is\((-\infty,\infty)\text{.}\)Nowdifferentiate: \begin{equation*} f'(x)=12x^3-12x^2=12x^2(x-1)\text{.} \end{equation*} Hence,\(f'(x)=0\)at\(x=0\)and\(x=1\text{.}\)Since\(f'(x)\)isdefinedforallrealnumbers,theonlycriticalpointsof\(f\)areat\(x=0\)and\(x=1\text{.}\)Wenowevaluate\(f(x)=3x^4-4x^3\)atthefollowingtestpoints: \begin{equation*} \begin{array}{c|c}x\ampf(x)\\\hline-1\amp7\\0\amp0\\1\amp-1\\2\amp16\end{array} \end{equation*} Therefore,\(f(x)\)hasarelativeminimumat\(x=1\text{,}\)andneitherarelativemaximumnorarelativeminimumat\(x=0\text{.}\) \(\dsy=(x^2-1)/x\) Answernone Solution Let\(y=f(x)\text{.}\)Thedomainof\(f\)is\(\{x\in\mathbb{R}|x\neq0\}\text{.}\)Calculate\(f'(x)\text{:}\) \begin{equation*} f'(x)=\diff{}{x}\frac{x^2-1}{x}=\frac{1+x^2}{x^2}\text{.} \end{equation*} Weseethat\(f'(x)\)isundefinedat\(x=0\text{,}\)whichisnotinthedomainof\(f\text{.}\)So\(x=0\)cannotbeacriticalpointof\(f\text{.}\)Further, \begin{equation*} f'(x)=1+x^2=0 \end{equation*} hasnorealsolutions.Weconcludethat\(f\)hasnorelativemaximaorminima. \(\dsy=3x^2-(1/x^2)\) Answernone Solution Let\(y=f(x)\text{.}\)Noticethatthedomainof\(f\)isallrealnumberssuchthat\(x\neq0\text{.}\)Tofindtherelativeextrema,wefirstcalculate\(f'(x)\text{:}\) \begin{equation*} f'(x)=6x+\frac{2}{x^3}\text{.} \end{equation*} \(f'(x)\)isundefinedat\(x=0\text{,}\)butthiscannotbearelativeextremumsinceitisnotinthedomainof\(f\text{.}\)Wenexttrysolving\(f'(x)=0\text{:}\) \begin{equation*} \begin{split}0\amp=f'(x)=6x+\frac{2}{x^3}\\0\amp=6x^4+2\\x^4\amp=-\frac{1}{3},\end{split} \end{equation*} whichgivesnosolutions.Weconcludethat\(f\)hasnorelativemaximaorminima. \(y=\cos(2x)-x\) Answerminat\(x=7\pi/12+k\pi\text{,}\)maxat\(x=-\pi/12+k\pi\text{,}\)forinteger\(k\) Solution Let\(y=f(x)\text{.}\)Wefirstnoticethatthedomainof\(f(x)\)isallrealnumbers.Differentiating,wehave \begin{equation*} f'(x)=-2\sin(2x)-1\text{.} \end{equation*} Since\(f'(x)\)isdefinedforallrealnumbers,allcriticalpointsof\(f\)willbesolutionsto\(f'(x)=0\text{:}\) \begin{equation*} -2\sin(2x)-1=0\implies\sin(2x)=\frac{1}{2}\text{.} \end{equation*} Sointheinterval\([-\pi/2,\pi/2]\text{,}\)thesolutiontothisequationis \begin{equation*} x=\frac{1}{2}\sin^{-1}\left(\frac{1}{2}\right)=\frac{1}{2}\left(\frac{\pi}{6}\right)\text{.} \end{equation*} Therefore,since\(\sin(2x)\)hasaperiodof\(\pi\text{,}\)weget \begin{equation*} f'(x)=0\impliesx=\frac{\pi}{12}+k\pi,\k\in\mathbb{Z}\text{.} \end{equation*} Thisgivesaninfinitenumberofcriticalpoints. \(\dsf(x)=\begin{cases}x-1\ampx\lt2\\x^2\ampx\geq2\end{cases}\) Answernone Solution Thegraphof\(f\)hasacornerat\(x=2\text{,}\)andso\(f'(2)\)isundefined.For\(x\neq2\text{,}\)wefind \begin{equation*} f'(x)=\begin{cases}1\amp\text{for}x\lt2\\2x\amp\text{for}x>2\end{cases} \end{equation*} So\(f'(x)=0\)hasnosolutions:\(f'(x)\)isclearlynonzeroforall\(x\lt2\text{,}\)and\(f'(x)=2x=0\)hasnosolutionsfor\(x>2\text{.}\)Therefore\(x=2\)isouronlycriticalpoint.Wenowevaluate\(f(x)\)atappropriatetestpointstodetermineif\(x=2\)correspondstoarelativeextremumornot.Since \begin{equation*} f(1)=(1)-1=0,\\f(2)=(2)^2=4,\\\text{and}\\f(3)=(3)^2=9\text{,} \end{equation*} weconcludethat\(f\)hasnorelativeextrema. \(\dsf(x)=\begin{cases}x-3\ampx\lt3\\x^3\amp3\leqx\leq5\\1/x\ampx>5\end{cases}\) Answerrelativemaxat\(x=5\) Solution Thedomainof\(f(x)\)isallrealnumbers.Wecompute: \begin{equation*} f'(x)=\begin{cases}1\amp\text{for}x\lt3\\3x^2\amp\text{for}3\ltx\lt5\\-\frac{1}{x^2}\amp\text{for}x>5\end{cases} \end{equation*} with\(f'(x)\)undefinedat\(x=3\)and\(x=5\text{,}\)only.Nowset\(f'(x)=0\text{:}\)For\(x\lt3\text{,}\)\(f(x)=x-3=0\impliesx=3\text{,}\)sonosolutions(since\(x\neq3\)).For\(3\ltx\lt5\text{,}\)\(x^3=0\impliesx=0\text{,}\)andsoagainnosolutions.Therearefurthermorenosolutionsfor\(x>5\text{,}\)andsotheonlycriticalpointsareat\(x=3\)and\(x=5\text{.}\)Check: \begin{equation*} \begin{array}{c|c}x\ampf(x)\\\hline2\amp2-3=-1\\3\amp3^3=27\\4\amp4^3=64\\5\amp5^3=125\\6\amp1/6\end{array} \end{equation*} Therefore,\(f(x)\)hasarelativemaximumat\(x=5\text{.}\) \(\dsf(x)=x^2-98x+4\) Answerrelativeminat\(x=49\) Solution \(f(x)\)isapolynomial,andsohasdomain\((-\infty,\infty)\text{.}\)Differentiating,weget \begin{equation*} f'(x)=2x-98\text{.} \end{equation*} Since\(f'(x)=0\)at\(x=49\)and\(f'(x)\)isdefinedforallrealnumbers,\(x=49\)istheonlycriticalpointof\(f\text{.}\)Nowusethetestpoints: \begin{equation*} \begin{array}{c|c}x\ampf(x)\\\hline0\amp4\\49\amp-2397\\100\amp204\end{array} \end{equation*} Therefore,\(f(x)\)hasarelativeminimumat\(x=49\text{.}\) \(\dsf(x)=\begin{cases}-2\ampx=0\\1/x^2\ampx\neq0\end{cases}\) Answerrelativeminat\(x=0\) Solution Wecompute: \begin{equation*} f'(x)=\begin{cases}\text{undefined}\ampx=0\\-\frac{2}{x^3}\ampx\neq0\end{cases} \end{equation*} Therefore,\(x=0\)istheonlycriticalpointof\(f\text{.}\)Since \begin{equation*} \begin{array}{c|c}x\ampf(x)\\\hline-1\amp1/(-1)^2=1\\0\amp-2\\1\amp1/1^2=1\end{array} \end{equation*} weseethat\(f(x)\)hasarelativeminimumat\(x=0\text{.}\) Exercise5.5.2.Foranyrealnumber\(x\)thereisauniqueinteger\(n\)suchthat\(n\leqx\ltn+1\text{,}\)andthegreatestintegerfunctionisdefinedas\(\ds\lfloorx\rfloor=n\text{.}\)Wherearethecriticalvaluesofthegreatestintegerfunction?Whicharerelativemaximaandwhicharerelativeminima? Answernone Solution Noticethatthedomainofthegreatestinteger(orfloor)function,\(f(x)=\left\lfloor{x}\right\rfloor\)is\(\mathbb{R}\text{,}\)whiletherangeis\(\mathbb{Z}\text{.}\) Wevisualize\(\left\lfloor{x}\right\rfloor\)below: Hence,\(f(x)=\left\lfloor{x}\right\rfloor\)hasnorelativeextrema. Exercise5.5.3.Explainwhythefunction\(f(x)=1/x\)hasnorelativemaximaorminima. Solution Thefunction\(f(x)=\dfrac{1}{x}\)hasdomain\(\{x\in\mathbb{R}|x\neq0\}\text{.}\)Wecalculate\(f'(x)\text{:}\) \begin{equation*} f'(x)=\diff{}{x}\frac{1}{x}=-\frac{1}{x^2}\text{.} \end{equation*} Therefore,\(f'(x)\)isundefinedonlyat\(x=0\text{,}\)whichisnotinthedomainof\(f\text{,}\)andisneverequaltozero.Thus,\(f(x)\)hasnorelativemaximaorminima. Exercise5.5.4.Howmanycriticalpointscanaquadraticpolynomialfunctionhave? Answerone Solution Anypolymomial\(P(x)\)hasdomain\((-\infty,\infty)\text{.}\)Sincethederivativeofapolynomialisonceagainapolynomial,\(P'(x)\)isdefinedoverallrealnumbers.Sincethegeneralformofaquadraticpolynomialis: \begin{equation*} P_2(x)=ax^2+bx+c\text{,} \end{equation*} forsomerealconstants\(a\text{,}\)\(b\text{,}\)\(c\text{,}\)suchthat\(a\neq0\text{.}\)Thegeneralformofthederivativeisthen: \begin{equation*} P'_2(x)=2ax+b\text{.} \end{equation*} Thismeansthat \begin{equation*} P'_2(x)=2ax+b=0\impliesx=-\frac{b}{2a}\text{.} \end{equation*} Hence,anyquadraticpolynomialcanhasexactlyonecriticalpoint. Exercise5.5.5.Showthatacubicpolynomialcanhaveatmosttwocriticalpoints.Giveexamplestoshowthatacubicpolynomialcanhavezero,one,ortwocriticalpoints. Solution Anypolymomial\(P(x)\)hasdomain\((-\infty,\infty)\text{.}\)Sincethederivativeofapolynomialisonceagainapolynomial,\(P'(x)\)isdefinedoverallrealnumbers.Sincethegeneralformofacubicpolynomialis: \begin{equation*} P_3(x)=ax^3+bx^2+cx+d\text{,} \end{equation*} with\(a\neq0\text{.}\)Thenthegeneralformofthederivativeis \begin{equation*} P'_3(x)=3ax^2+2bx+c\text{.} \end{equation*} Hence, \begin{equation*} P'_3(x)=3ax^2+2bx+c=0 \end{equation*} hassolutionswhicharegivenbythequadraticformula: \begin{equation*} x=\frac{-2b\pm\sqrt{4b^2-12ac}}{6a}\text{.} \end{equation*} Hence,\(P_3(x)\)canhavezero,oneortwocriticalpoints.Forexample: \begin{equation*} y=x^3-2x+1 \end{equation*} hastwocriticalpoints, \begin{equation*} y=x^3 \end{equation*} hasonecriticalpointand \begin{equation*} y=x^3+x^2+x+1 \end{equation*} hasnocriticalpoints. Exercise5.5.6.Explorethefamilyoffunctions\(\dsf(x)=x^3+cx+1\)where\(c\)isaconstant.Howmanyandwhattypesofrelativeextremesarethere?Youranswershoulddependonthevalueof\(c\text{,}\)thatis,differentvaluesof\(c\)willgivedifferentanswers. Solution First,notethatthedomainof\(f(x)=x^3+cx+1\)is\((-\infty,\infty)\text{,}\)andthat\(f'(x)=3x^2+c\)isdefinedeverywhere(\(c\)issomeconstant).Wenowtrytofindsolutionsto\(f'(x)=0\text{.}\) \begin{equation*} \begin{split}3x^2+c\amp=0\\3x^2\amp=-c\\x^2\amp=-\frac{c}{3}\end{split} \end{equation*} Weneedtoconsiderthreecasesforthevalueoftheconstant\(c\text{:}\) CASE1:\(c>0\) Then\(x^2=-\frac{c}{3}\lt0\)hasnosolutions,and\(f\)hasnorelativeextrema. CASE2:\(c=0\) Thenouronlycriticalpointis\(x=0\text{.}\)Toclassifythispoint,weconsiderthefollowingvaluesof\(f(x)=x^3+1\text{,}\) \begin{equation*} f(-1)=0,\\f(0)=1,\\f(1)=2\text{,} \end{equation*} andconcludethatthisisneitherarelativemaximumnorminimum. CASE3:\(c\lt0\) Then\(x^2-\frac{c}{3}>0\)hastwouniquesolutions, \begin{equation*} x=\pm\sqrt{\frac{-c}{3}}=\pm\sqrt{\frac{|c|}{3}}\cdot \end{equation*} Byonceagaincomparingpointsoneithersideofeachcriticalpoint,wefindthat\(-\sqrt{\frac{-c}{3}}\)isarelativemaximum,and\(\sqrt{\frac{-c}{3}}\)isarelativeminimum(seegraphbelowforexamples). Exercise5.5.7.Wegeneralizetheprecedingtwoquestions.Let\(n\)beapositiveintegerandlet\(f\)beapolynomialofdegree\(n\text{.}\)Howmanycriticalpointscan\(f\)have?HintRecalltheFundamentalTheoremofAlgebra,whichsaysthatapolynomialofdegree\(n\)hasatmost\(n\)roots. Solution Anypolymomial\(P(x)\)hasdomain\((-\infty,\infty)\text{.}\)Sincethederivativefapolynomialisonceagainapolynomial,\(P'(x)\)isdefinedoverallrealnumbers.Now,if\(f(x)\)isapolynomialofdegree\(n\text{,}\)ithastheform \begin{equation*} f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0\text{.} \end{equation*} Hence, \begin{equation*} f'(x)=na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+...+2a_2x+a_1\text{.} \end{equation*} Thisisapolynomialofdegree\((n-1)\text{.}\)Sobythefundamentaltheoremofalgebra,\(f'(x)\)hasatmost\(n-1\)realroots.Hence,\(f(x)\)hasatmost\(n-1\)criticalpoints. Subsection5.5.2AbsoluteExtrema ¶Unlikearelativeextremum,whichisonly“extreme”relativetopoints“closeto”it,anabsoluteextremumis“extreme”comparedtoallotherpointsintheintervalunderconsideration.SomeexamplesofabsolutemaximumandminimumpointsareshowninFigure 5.17.Thisleadsustothefollowingdefinitions. Definition5.52.AbsoluteMaximaandMinima. Areal-valuedfunction\(f\)hasanabsolutemaximumat\(x_0\)if\(f(x_0)\geqf(x)\)forall\(x\)inthedomainof\(f\text{.}\) Areal-valuedfunction\(f\)hasanabsoluteminimumat\(x_0\)if\(f(x_0)\leqf(x)\)forall\(x\)inthedomainof\(f\text{.}\)Note: Noticethatthedefinitionofabsoluteextremaentailsthatanabsoluteextremum,unlikearelativeextremum,canfallonanendpointasshowninFigure 5.17. Becauseofthe“global”natureofanabsoluteextremumitisalsooftenreferredtoasaglobalextremum. Figure5.17.ClassificationofExtrema Example5.53.AbsoluteExtremaUsingaGraph. Findtheabsoluteextremaofthefollowingfunctionsusingtheirgraphs. \(f(x)=x^2\)ontheinterval\((-\infty,\infty)\text{.}\) \(f(x)=|x|\)ontheinterval\([-1,2]\text{.}\) \(f(x)=\cosx\)ontheinterval\([0,\pi]\text{.}\) Solution Thisparabolahasanabsoluteminimumat\(x=0\text{.}\)However,itdoesnothaveanabsolutemaximum. Thisgraphlookslikeacheckmark.Ithasanabsoluteminimumat\(x=0\)andanabsolutemaximumat\(x=2\text{.}\) \(f(x)=\cos(x)\)hasanabsoluteminimumat\(x=\pi\)andanabsolutemaximumat\(x=0\)ontheinterval\([0,\pi]\text{.}\) LikeFermat'sTheorem,thefollowingtheoremhasanintuitiveappeal.However,unlikeFermat'sTheorem,theproofreliesonamoreadvancedconceptcalledcompactness,whichwillonlybecoveredinacoursetypicallyentitledAnalysis.So,wewillbecontentwithunderstandingthestatementofthetheorem. Theorem5.54.ExtremeValueTheorem. Ifafunction\(f\)iscontinuousonaclosedinterval\([a,b]\text{,}\)then\(f\)hasbothanabsolutemaximumandanabsoluteminimumon\([a,b]\text{.}\)Althoughthistheoremtellsusthatanabsoluteextremumexists,itdoesnottelluswhatitisorhowtofindit. Notethatifanabsoluteextremumisinsidetheinterval(i.e.notanendpoint),thenitmustalsobearelativeextremum.Thisimmediatelytellsusthattofindtheabsoluteextremaofafunctiononaninterval,weneedonlyexaminetherelativeextremainsidetheinterval,andtheendpointsoftheinterval.Wecandeviseamethodforfindingabsoluteextremaforafunction\(f\)onaclosedinterval\([a,b]\text{.}\) GuidelineforFindingAbsoluteExtremaGivenContinuityof\(f\)andClosedInterval. Verifythefunctioniscontinuouson\([a,b]\text{.}\) Findthederivativeanddetermineallcriticalvaluesof\(f\)thatarein\((a,b)\text{.}\) EvaluatethefunctionatthecriticalvaluesfoundinStep2andtheendpoints\(x=a\)and\(x=b\)oftheinterval. Theabsolutemaximumvalueandabsoluteminimumvalueof\(f\)correspondtothelargestandsmallest\(y\)-valuesrespectivelyfoundinStep3. Whymustafunctionbecontinuousonaclosedintervalinordertousethistheorem?Considerthefollowingexample. Example5.55.AbsoluteExtremaandContinuity. Findanyabsoluteextremafor\(f(x)=1/x\)ontheinterval\([-1,1]\text{.}\) Solution Thefunction\(f\)isnotcontinuousat\(x=0\text{.}\)Since\(0\in[-1,1]\text{,}\)\(f\)isnotcontinuousontheclosedinterval: \begin{align*} \lim_{x\to0^+}f(x)\amp=+\infty\\ \lim_{x\to0^-}f(x)\amp=-\infty\,\text{,} \end{align*} soweareunabletoapplytheExtremeValueTheorem.Therefore,\(f(x)=1/x\)doesnothaveanabsolutemaximumoranabsoluteminimumon\([-1,1]\text{.}\) However,ifweconsiderthesamefunctiononanintervalwhereitiscontinuous,thetheoremwillapply.Thisisillustratedinthefollowingexample. Example5.56.AbsoluteExtremaandContinuity. Findanyabsoluteextremafor\(f(x)=1/x\)ontheinterval\([1,2]\text{.}\) SolutionThefunction\(f\)iscontinousontheinterval,sowecanapplytheExtremeValueTheorem.Webeginwithtakingthederivativetobe\(f'(x)=-1/x^2\)whichhasacriticalvalueat\(x=0\text{,}\)butsincethiscriticalvalueisnotin\([1,2]\)weignoreit.Theonlypointswhereanextremacanoccuraretheendpointsoftheinterval.Tofindthemaximumorminimumwecansimplyevaluatethefunction:\(f(1)=1\)and\(f(2)=1/2\text{,}\)sotheabsolutemaximumisat\(x=1\)andtheabsoluteminimumisat\(x=2\text{.}\) Whymustanintervalbeclosedinordertousetheabovetheorem?Recallthedifferencebetweenopenandclosedintervals.Considerafunction\(f\)ontheopeninterval\((0,1)\text{.}\)Ifwechoosesuccessivevaluesof\(x\)movingcloserandcloserto\(1\text{,}\)whathappens?Since1isnotincludedintheintervalwewillnotattainexactlythevalueof1.Supposewereachavalueof0.9999—isitpossibletogetcloserto1?Yes:Thereareinfinitelymanyrealnumbersbetween0.9999and1.Infact,anyconceivablerealnumbercloseto1willhaveinfinitelymanyrealnumbersbetweenitselfand1.Now,suppose\(f\)isdecreasingon\((0,1)\text{:}\)Asweapproach1,\(f\)willcontinuetodecrease,evenifthedifferencebetweensuccessivevaluesof\(f\)isslight.Similarlyif\(f\)isincreasingon\((0,1)\text{.}\) Considerafewmoreexamples: Example5.57.DeterminingAbsoluteExtrema. Determinetheabsoluteextremaof\(f(x)=x^3-x^2+1\)ontheinterval\([-1,2]\text{.}\) Solution First,notice\(f\)iscontinuousontheclosedinterval\([-1,2]\text{,}\)sowe'reabletouseTheorem 5.54todeterminetheabsoluteextrema.Thederivativeis\(f'(x)=3x^2-2x\text{,}\)andthecriticalvaluesare\(x=0,2/3\)whicharebothintheinterval\([-1,2]\text{.}\)Inordertofindtheabsoluteextrema,wemustconsiderallcriticalvaluesthatliewithintheinterval(thatis,in\((-1,2)\))andtheendpointsoftheinterval. \begin{align*} f(-1)\amp=(-1)^3-(-1)^2+1=-1\\ f(0)\amp=(0)^3-(0)^2+1=1\\ f(2/3)\amp=(2/3)^3-(2/3)^2+1=23/27\\ f(2)\amp=(2)^3-(2)^2+1=5 \end{align*} Theabsolutemaximumisat(2,5)andtheabsoluteminimumisat(-1,-1). Example5.58.DeterminingAbsoluteExtrema. Determinetheabsoluteextremaof\(f(x)=-9/x-x+10\)ontheinterval\([2,6]\text{.}\) Solution First,notice\(f\)iscontinuousontheclosedinterval\([2,6]\text{,}\)sowe'reabletouseTheorem 5.54todeterminetheabsoluteextrema.Thefunctionisnotcontinuousat\(x=0\text{,}\)butwecanignorethisfactsince0isnotin\([2,6]\text{.}\)Thederivativeis\(f'(x)=9/x^2-1\text{,}\)andthecriticalvaluesare\(x=\pm3\text{,}\)butonly\(x=+3\)isintheinterval.Inordertofindtheabsoluteextrema,wemustconsiderallcriticalvaluesthatliewithintheintervalandtheendpointsoftheinterval. \begin{align*} f(2)\amp=-9/(2)-(2)+10=7/2=3.5\\ f(3)\amp=-9/(3)-(3)+10=4\\ f(6)\amp=-9/(6)-(6)+10=5/2=2.5 \end{align*} Theabsolutemaximumisat(3,4)andtheabsoluteminimumisat(6,2.5). Whenwearetryingtofindtheabsoluteextremaofafunctiononanopeninterval,wecannotusetheExtremeValueTheorem.However,ifthefunctioniscontinuousontheinterval,manyofthesameideasapply.Inparticular,ifanabsoluteextremumexists,itmustalsobearelativeextremum.Inadditiontocheckingvaluesattherelativeextrema,wemustcheckthebehaviourofthefunctionasitapproachestheendsoftheinterval. Someexamplestoillustratethismethod. Example5.59.DeterminingAbsoluteExtrema. Findtheextremaof\(y=\sec(x)\)on\((-\pi/2,\pi/2)\text{.}\) Solution Notice\(\sec(x)\)iscontinuouson\((-\pi/2,\pi/2)\)andhasonerelativeminimumat0.Also \begin{equation*} \lim_{x\to(-\pi/2)^+}\sec(x)=\lim_{x\to(\pi/2)^-}\sec(x)=+\infty\,\text{,} \end{equation*} so\(\sec(x)\)hasnoabsolutemaximum,butthepoint\((0,1)\)istheabsoluteminimum. Asimilarapproachcanbeusedforinfiniteintervals. Example5.60.DeterminingAbsoluteExtrema. Findtheextremaof\(y=\ds\frac{x^2}{x^2+1}\)on\((-\infty,\infty)\text{.}\) SolutionSince\(x^2+1\neq0\)forall\(x\)in\((-\infty,\infty)\)thefunctioniscontinuousonthisinterval.Thisfunctionhasonlyonecriticalvalueat\(x=0\text{,}\)whichistherelativeminimumandalsotheabsoluteminimum.Now,\(\ds\lim_{x\to\pm\infty}\frac{x^2}{x^2+1}=1\text{,}\)sothefunctiondoesnothaveanabsolutemaximum:Itcontinuestoincreasetowards1,butdoesnotattainthisexactvalue. ExercisesforSection 5.5.2. Exercise5.5.8.Findtheabsoluteextremaforthefollowingfunctionsoverthegiveninterval. \(f(x)=-\frac{x+4}{x-4}\)on\([0,3]\) AnswerAbsolutemaximum\((3,7)\text{;}\)Absoluteminimum\((0,1)\) Solution Wefirstlookforanyrelativeextremaof\(f\)intheinterval\([0,3]\text{.}\) \begin{equation*} f'(x)=\diff{}{x}\left(-\frac{x+4}{x-4}\right)=\frac{8}{(x-4)^2}\text{.} \end{equation*} So\(f'(x)=0\)hasnosolutionsand\(f\)isdefinedforall\(x\in[0,3]\text{.}\)Theabsoluteextremamustthereforeoccurattheendpoints.Since \begin{equation*} f(0)=-\frac{4}{-4}=1\\\text{and}\\\f(3)=-\frac{7}{-1}=7\text{,} \end{equation*} weconcludethat\((0,1)\)istheabsoluteminimumand\((3,7)\)istheabsolutemaximum. \(f(x)=x^3+4x^2+4\)on\([-4,1]\) AnswerAbsolutemaximum\((0,4)\text{;}\)Absoluteminimum\((-4,4)\) Solution Wecompute: \begin{equation*} f'(x)=3x^2+8x \end{equation*} Thisisanotherpolynomial,andso\(f'(x)\)isdefinedforall\(x\in[-4,1]\text{.}\)Nowsolve: \begin{equation*} f'(x)=x(3x+8)=0\impliesx=0,-8/3\text{.} \end{equation*} Bothofthesetwopointsareinthedomainof\(f\text{,}\)andso\(f\)hastwocriticalpoints.Wenowcomparethevalueof\(f\)atbothendpointsandateachcriticalpoint: \begin{equation*} f(-4)=4,\\f(-8/3)=\frac{367}{27},\\f(0)=4,\\f(1)=9\text{.} \end{equation*} Hence,\(f(x)\)hasanabsolutemaximumat\(\left(-\frac{8}{3},\frac{367}{37}\right)\)andabsoluteminimaat\((-4,4)\)and\((0,4)\text{.}\) \(f(x)=\csc(x)\)on\([0,\pi]\) AnswerAbsoluteminimum\((\pi/2,1)\text{;}\)Noabsolutemaximum Solution Wefirstrewrite \begin{equation*} f(x)=\csc(x)=\frac{1}{\sin(x)}\text{.} \end{equation*} Nowdifferentiate: \begin{equation*} f'(x)=-\frac{\cos(x)}{\sin^2(x)}\text{.} \end{equation*} Overtheinterval\([0,\pi]\text{,}\)\(f'(x)\)isundefinedat\(x=0\)and\(x=\pi\text{.}\)Sincetheseareendpoints,wedonotclassifythemascriticalpoints.Furthermore, \begin{equation*} f'(x)=0\implies\cosx=0\impliesx=\frac{\pi}{2}+n\pi,n\in\mathbb{Z}\text{,} \end{equation*} andso\(f(x)\)hasonecriticalpointat\(x=\frac{\pi}{2}\text{.}\)Wenowcompare: \begin{equation*} \lim_{x\to0}f(x)=\infty,\f(\pi/2)=1,\\lim_{x\to\pi}f(x)=\infty\text{.} \end{equation*} Hence,\(f(x)\)hasanabsoluteminimumat\(\left(\frac{\pi}{2},1\right)\)andnoabsolutemaxima. \(f(x)=\ln(x)/x^2\)on\([1,4]\) AnswerAbsolutemaximum\((\sqrt{e},1/(2e))\text{;}\)Absoluteminimum\((1,0)\text{.}\) Solution Wecalculate \begin{equation*} f'(x)=\diff{}{x}\left(\frac{\lnx}{x^2}\right)=\frac{1-2\lnx}{x^3}\text{,} \end{equation*} andsolvefor\(f'(x)=0\text{.}\) \begin{equation*} \begin{split}f'(x)\amp=0\\\frac{1-2\lnx}{x^3}\amp=0\\\lnx^2\amp=1\\(x>0)\\x\amp=\sqrt{e}\approx1.65.\end{split} \end{equation*} Since\(f'\)isundefinedonlyat\(x=0\text{,}\)whichisoutsidetheinterval\([1,4]\text{,}\)\(x=\sqrt{e}\)givesouronlyinteriorcriticalpoint.Comparing \begin{equation*} f(1)=0,\\f(4)\approx0.0867,\\f(\sqrt{e})=1/(2e)\approx0.18 \end{equation*} wefindtheabsolutemaximumof\(f\)on\([1,4]\)occursat\(x=\sqrt{e}\)andtheabsoluteminimumoccursat\(x=1\text{.}\) \(f(x)=x\sqrt{1-x^2}\)on\([-1,1]\) AnswerAbsolutemaximum\(\left(\frac{1}{\sqrt{2}},\frac{1}{2}\right)\);Absoluteminimum\(\left(-\frac{1}{\sqrt{2}},-\frac{1}{2}\right)\) Solution Welookforrelativeextrema: \begin{equation*} f'(x)=\frac{1-2x^2}{\sqrt{1-x^2}}\text{.} \end{equation*} Hence,\(f'(x)\)isundefinedat\(x=-1\)and\(x=1\text{.}\)Sincethesepointsareendpoints,wedonotconsiderthemtobeendpoints.Next,welookforrootsof\(f'(x)\text{:}\) \begin{equation*} f'(x)=\frac{1-2x^2}{\sqrt{1-x^2}}=0\impliesx^2=\frac{1}{2}\impliesx=\pm\frac{1}{\sqrt{2}}\text{,} \end{equation*} andso\(f\)hastwocriticalpoints.Nowcompare: \begin{equation*} f(-1)=0,\f\left(-\frac{1}{\sqrt{2}}\right)=-\frac{1}{2},\f\left(\frac{1}{\sqrt{2}}\right)=\frac{1}{2},\f(1)=0\text{.} \end{equation*} Therefore,\(f(x)\)hasanabsolutemaximumat\(\left(\frac{1}{\sqrt{2}},\frac{1}{2}\right)\)andanabsoluteminimumat\(\left(-\frac{1}{\sqrt{2}},-\frac{1}{2}\right)\text{.}\) \(f(x)=xe^{-x^2/32}\)on\([0,2]\) AnswerAbsoluteminimum\((0,0)\text{;}\)Absolutemaximum\((2,2e^{1/8})\) Solution Wefirstlookforanyrelativeextrema: \begin{equation*} \begin{split}f'(x)\amp=e^{-x^2/32}+x\left(\frac{-2x}{32}\right)e^{-x^2/32}\\\amp=\left(1-16x^2\right)e^{-x^2/32}\end{split} \end{equation*} So\(f'(x)=0\)when \begin{equation*} 0=1-16x^2\impliesx=\pm4\text{.} \end{equation*} Thus,\(f\)hasnorelativeextremaintheinterval\([0,2]\text{.}\)Since \begin{equation*} f(0)=0,\text{and}f(2)=2e^{-1/8}\text{,} \end{equation*} weseethat\(f\)hasanabsoluteminimumat\(0\)andanabsolutemaximumat\(2\)overtheinterval\([0,2]\text{.}\) \(f(x)=x-\tan^{-1}(2x)\)on\([0,2]\) AnswerAbsoluteminimum\((1/2,\frac{2-\pi}{4})\text{;}\)Absolutemaximum\((2,2-\tan^{-1}(4))\) Solution Wefirstdifferentiate: \begin{equation*} f'(x)=1-\frac{2}{4x^2+1}\text{.} \end{equation*} Andso\(f'(x)\)isdefinedoverthedomainof\(f\text{.}\)Wenowcompute: \begin{equation*} f'(x)=0\implies\frac{2}{4x^2+1}=1\impliesx=\pm\frac{1}{2}\text{.} \end{equation*} Werejectthenegativesolutionsincethedomainof\(f\)is\([0,2]\text{.}\)Hence,\(f(x)\)hasoncecriticalpointat\(x=\frac{1}{2}\text{.}\)Nowcompare: \begin{equation*} f(0)=0\f(1/2)\approx-0.285398,\f(2)\approx0.674182\text{.} \end{equation*} Therefore,\(f(x)\)hasanabsoluteminimumat\(\left(\frac{1}{2},-0.285398\right)\)andanabsolutemaximumat\(\left(2,0.674182\right)\text{.}\) \(f(x)=\frac{x}{x^2+1}\) AnswerAbsolutemaximum\((1,1/2)\text{;}\)Absoluteminimum\((-1,-1/2)\) Solution Wearelookingforabsolutemaximaandminimaof\(f\)over\((-\infty,\infty)\text{.}\)Welookforanyrelativeextrema: \begin{equation*} \begin{split}f'(x)\amp=\frac{(x^2+1)-x(2x)}{(x^2+1)^2}\\\amp=\frac{1-x^2}{(x^2+1)^2}\end{split} \end{equation*} Noticethat\(x^2+1=0\)hasnorealsolutions,andso\(f\)isdefinedover\(\mathbb{R}\text{.}\)Hence,\(f'(x)=0\)when \begin{equation*} 1-x^2=0\impliesx=\pm1\text{.} \end{equation*} Wenowtakethefollowinglimits: \begin{equation*} \lim_{x\to\infty}\frac{x}{x^2+1}\Heq\lim_{x\to\infty}\frac{1}{2x}=0 \end{equation*} \begin{equation*} \lim_{x\to-\infty}\frac{x}{x^2+1}\Heq\lim_{x\to-\infty}\frac{1}{2x}=0 \end{equation*} Since\(f(-1)=-1/2\text{,}\)\(f(1)=1/2\)and\(f\to0\)as\(x\to\pm\infty\text{,}\)wededucethat\(f\)hasanabsoluteminimumatthepoint\((-1,-1/2)\)andanabsolutemaximumatthepoint\((1,1/2)\text{.}\) Exercise5.5.9.Foreachofthefollowing,sketchapotentialgraphofacontinuousfunctionontheclosedinterval\([0,4]\)withthegivenproperties. Absoluteminimumat0,absolutemaximumat2,relativeminimumat3.SolutionWesketchafunctionon\([0,4]\)whichhasanabsoluteminimumat\(x=0\text{,}\)anabsolutemaximumat\(x=2\text{,}\)andarelativeminimumat\(x=3\text{:}\) Absolutemaximumat1,absoluteminimumat2,relativemaximumat3.SolutionWesketchafunctionon\([0,4]\)whichhasanabsolutemaximumat1,anabsoluteminimumat2,andarelativemaximumat3: Absoluteminimumat4,absolutemaximumat1,relativeminimumat2,relativemaximaat1and3.SolutionWesketchafunctionon\([0,4]\)whichhasanabsoluteminimumat\(x=4\text{,}\)anabsolutemaximumat\(x=1\text{,}\)arelativeminimumat\(x=2\)andrelativemaximaat\(x=1\)and\(x=3\text{:}\) login