Stationary Points - RadfordMathematics.com

Stationary points, aka critical points, of a curve are points at which its derivative is equal to zero, 0. Local maximum, minimum and horizontal points of ... RadfordMathematics.com OnlineMathematicsBook × SearchRadfordMathematics Close StationaryPoints-Part1 (Definition&HowtoFindStationaryPoints) Astationarypoint,orcriticalpoint,isapointatwhichthecurve'sgradientequalstozero.Consequentlyifacurvehasequation\(y=f(x)\)thenatastationarypointwe'llalwayshave: \[f'(x)=0\] whichcanalsobewritten: \[\frac{dy}{dx}=0\] Inotherwordsthederivativefunctionequalstozeroatastationarypoint. DifferentTypesofStationaryPoints Therearethreetypesofstationarypoints: local(orglobal)maximumpoints local(orglobal)minimumpoints horizontal(increasingordecreasing)pointsofinflexion. Itisworthpointingoutthatmaximumandminimumpointsareoftencalledturningpoints. TurningPoints Aturningpointisastationarypoint,whichiseither: alocal(orglobal)minimum alocal(orglobal)maximum eachofwhichareillustaredinthegraphsshownhere,wherethehorizontaltangentisshowninorange: HorizontalPointsofInflection Ahorizontalpointofinflectionisastationarypoint,whichiseither: aincreasinghorizontalpointofinflection adecreasinghorizontalpointofinflection eachofwhichareillustaredinthegraphsshownhere,wherethehorizontaltangentisshowninorange: Method:findingstationarypoints Givenafunction\(f(x)\)anditscurve\(y=f(x)\),tofindanystationarypoint(s)wefollowthreesteps: Step1:find\(f'(x)\) Step2:solvetheequation\(f'(x)=0\),thiswillgiveusthe\(x\)-coordinate(s)ofanystationarypoint(s). Step3(ifneeded/asked):calculatethe\(y\)-coordinate(s)ofthestationarypoint(s)bypluggingthe\(x\)valuesfoundinstep2into\(f(x)\). Tutorial1:HowtoFindStationaryPoints Inthefollowingtutorialweillustratehowtouseourthree-stepmethodtofindthecoordinatesofanystationarypoints,byfindingthestationarypoint(s)ofthecurves: \(y=x^2-4x+1\),and \(y=2x^3+3x^2-12x-1\) Example1 Giventhefunctiondefinedbytheequation: \[y=x^2-4x+5\] findthecoordinatesofanystationarypoint(s). SeeDetailedSolution Solution Followingourthree-stepmethod: Step1:find\(\frac{dy}{dx}\). For\(y=x^2-4x+5\),wefind: \[\frac{dy}{dx}=2x-4\] Step2:solvetheequation\(\frac{dy}{dx}=0\). That's: \[2x-4=0\] Solvingthisleadsto: \[2x=4\] Finally: \[x=2\] Atthisstagewecanstatethecurve\(y=x^2-4x+5\)hasonestationarypointwhose\(x\)-coordinateis\(x=2\). Step3:calculatethesationarypoint's\(y\)-coordinate. Todothiswereplace\(x\)by\(2\)inthecurve'sequation\(y=x^2-4x+5\). That's: \[\begin{aligned} y&=2^2-4\times2+5\\ &=4-8+5\\ &=-4+5\\ y&=1 \end{aligned}\] Sothestationarypointhas\(y\)-coordinate\(y=1\). Wecanthereforestatethatthecurve\(y=x^2-4x+5\)hasonestationarypointwithcoordinates\(\begin{pmatrix}2,1\end{pmatrix}\). Thisresultisconfirmed,usingourgraphicalcalculatorandlookingatthecurve\(y=x^2-4x+5\): Wecanseequiteclearlythatthecurvehasaglobalminimumpoint,whichisastationarypoint,at\(\begin{pmatrix}2,1\end{pmatrix}\). Example2 Findthecoordinatesofanystationarypoint(s)ofthefunctiondefinedby: \[y=2x^3+3x^2-12x+1\] SeeDetailedSolution Solution Followingourthree-stepmethod: Step1:find\(\frac{dy}{dx}\). Usingthepowerrulefordifferentiationwefind: \[\begin{aligned} \frac{dy}{dx}&=3\times2x^{3-1}+2\times3x^{2-1}-12\\ &=6x^2+6x^1-12\\ \frac{dy}{dx}&=6x^2+6x-12 \end{aligned}\] Step2:solvetheequation\(\frac{dy}{dx}=0\). Since\(\frac{dy}{dx}=6x^2+6x-12\)weneedtosolvethequadraticequation: \[6x^2+6x-12=0\] Doingthiseitherusingourgraphicalcalculator,thequadraticformula,orbyfactoring,wefindtwosolutions: \[x=-2\quad\text{and}\quadx=1\] Soatthisstagewecanstatethatthefunction\(y=2x^3+3x^2-12x+1\)hastwostationarypoints.Onewith\(x\)-coordinate\(x=-2\)andtheotherwith\(x\)-coordinate\(x=1\). Step3:calculatethesationarypoint's\(y\)-coordinate. Sincewefoundtwovaluesof\(x\),instep2,therearetwo\(y\)-coordinatestocalculate,oneforeachvalueof\(x\). when\(x=-2\): Replacing\(x\)by\(-2\)in\(y=2x^3+3x^2-12x+1\),wefind: \[\begin{aligned} y&=2\times(-2)^3+3\times(-2)^2-12\times(-2)+1\\ &=2\times(-8)+3\times4-(-24)+1\\ &=-16+12+24+1\\ y&=21 \end{aligned}\] Sothefunctionhasastationarypointat: \[\begin{pmatrix}-2,21\end{pmatrix}\] when\(x=1\): Replacing\(x\)by\(1\)in\(y=2x^3+3x^2-12x+1\),wefind: \[\begin{aligned} y&=2\times1^3+3\times1^2-12\times1+1\\ &=2\times1+3\times1-12+1\\ &=2+3-12+1\\ y&=-6 \end{aligned}\] Sothefunctionhasitssecondstationarypointat: \[\begin{pmatrix}1,-6\end{pmatrix}\] Wecanseebothofthesestationarypointsonthegraphshownbelow: Wecanseequiteclearlythatthestationarypointat\(\begin{pmatrix}-2,21\end{pmatrix}\)isalocalmaximumandthestationarypointat\(\begin{pmatrix}1,-6\end{pmatrix}\)isalocalminimum. Example3 Giventhefunctiondefinedby: \[y=x^3-6x^2+12x-12\] Findthecoordinatesofanystationarypoint(s)alongthisfunction'scurve'slength. SeeDetailedSolution Solution Followingourthree-stepmethod: Step1:find\(\frac{dy}{dx}\). Since\(y=x^3-6x^2+12x-12\),weusethepowerrulefordifferentiation,tofindthisfunction'sderivative: \[\begin{aligned} \frac{dy}{dx}&=3\timesx^{3-1}-2\times6x^{2-1}+12x^{1-1}+0\\ &=3x^2-12x^1+12x^0\\ \frac{dy}{dx}&=3x^2-12x+12 \end{aligned}\] Step2:solvetheequation\(\frac{dy}{dx}=0\). Since\(\frac{dy}{dx}=3x^2-12x+12\)wehavetosolvethequadraticequation: \[3x^2-12x+12=0\] Wecansolvethisusingthequadraticformulaorbyfactoringorevenusingourgraphicalcalculator. Indoingsowefindonesolution: \[x=2\] So,atthisstage,wecanstatethatthisfunctionhasonestationarypointwhose\(x\)-coordinateis\(x=2\). Step3:calculatethesationarypoint's\(y\)-coordinate. Sincewefoundthatthestationarypointhad\(x\)-coordinate\(x=2\),tofindits\(y\)-coordinatewereplace\(x\)by\(2\)inthefunction'sequation\(y=x^3-6x^2+12x-12\).That's: \[\begin{aligned} y&=2^3-6\times2^2+12\times2-12\\ &=8-6\times4+24-12\\ &=8-24+24-12\\ y&=-4 \end{aligned}\] Sothisfunctionhasastationarypointwithcoordinates: \[\begin{pmatrix}2,-4\end{pmatrix}\] Thisresultisconfirmedwhenwelookatthegraphof\(y=x^3-6x^2+12x-12\): Lookingatthisgraph,wecanseethatthiscurve'sstationarypointat\(\begin{pmatrix}2,-4\end{pmatrix}\)isanincreasinghorizontalpointofinflection. Exercise1 Findthecoordinatesofanystationarypoint(s)alongthelengthofeachofthefollowingcurves: \(y=x^2-2x-8\) \(y=-x^2-6x-8\) \(y=2x^3-12x^2-30x-10\) \(y=-2x^3+3x^2+36x-6\) \(y=x^3+3x^2+3x-2\) Answersw/outWorking Note:thisexercisecanbedownloadedasaworksheettopracticewith:worksheet AnswersWithoutWorking Wefindthederivativetobe\(\frac{dy}{dx}=2x-2\)andthiscurvehasonestationarypoint: \[\begin{pmatrix}1,-9\end{pmatrix}\] Wefindthederivativetobe\(\frac{dy}{dx}=-2x-6\)andthiscurvehasonestationarypoint: \[\begin{pmatrix}-3,1\end{pmatrix}\] Wefindthederivativetobe\(\frac{dy}{dx}=2x^3-12x^2-30x-10\)andthiscurvehastwostationarypoints: \[\begin{pmatrix}-1,6\end{pmatrix}\] and \[\begin{pmatrix}5,-210\end{pmatrix}\] Wefindthederivativetobe\(\frac{dy}{dx}=-2x^3+3x^2+36x-6\)andthiscurvehastwostationarypoints: \[\begin{pmatrix}-2,-50\end{pmatrix}\] and \[\begin{pmatrix}3,75\end{pmatrix}\] Wefindthederivativetobe\(\frac{dy}{dx}=x^3+3x^2+3x-2\)andthiscurvehasonestationarypoint: \[\begin{pmatrix}-1,-3\end{pmatrix}\] × AnswerswithWorking Selectthequestionnumberyou'dliketoseetheworkingfor: Qu.1 Qu.2 Qu.3 Qu.4 Qu.5 Close Tutorial2:HowtoFindStationaryPoints Inthefollowingtutorialweillustratehowtouseourthree-stepmethodtofindthecoordinatesofanystationarypoints,byfindingthestationarypoint(s)alongthecurve: \[y=x+\frac{9}{x}\] Example4 Giventhefunctiondefinedby: \[y=x+\frac{4}{x}\] findthecoordinatesofanystationarypointsalongthiscurve'slength. SeeDetailedSolution Solution Followingourthree-stepmethod: Step1:find\(\frac{dy}{dx}\). Wecanre-write\(y=x+\frac{4}{x}\),usingnegativeexponents: \[y=x+4.x^{-1}\] Wecannowusethepowerrulefordifferentiationtofindthederivative: \[\begin{aligned} \frac{dy}{dx}&=1+(-1).4.x^{-1-1}\\ &=1-4x^{-2}\\ \frac{dy}{dx}&=1-\frac{4}{x^2} \end{aligned}\] Step2:solvetheequation\(\frac{dy}{dx}=0\). Since\(\frac{dy}{dx}=1-\frac{4}{x^2}\)weneedtosolvetheequation: \[1-\frac{4}{x^2}=0\] Tosolvethisequationbyhandwestartbywritingtheentirelefthandsideover\(x^2\)usingthefactthat\(1=\frac{x^2}{x^2}\)sothat: \[1-\frac{4}{x^2}=0\] canbewritten: \[\frac{x^2}{x^2}-\frac{4}{x^2}=0\] andthereforewehavetosolve: \[\frac{x^2-4}{x^2}=0\] Thiswillequalto\(0\)ifandonlyifthenumerator,\(x^2-4\),equalsto\(0\). Soallweneedtosolveis: \[x^2-4=0\] That's: \[x^2=4\] whichleadstotwosolutions: \[x=-2\quad\text{and}\quadx=2\] Step3:calculatethesationarypoint's\(y\)-coordinate. Sincewefoundtwovaluesof\(x\),instep2,therearetwo\(y\)-coordinatestocalculate,oneforeachvalueof\(x\). when\(x=-2\): replacing\(x\)by\(-2\)in\(y=x+\frac{4}{x}\),wefind: \[\begin{aligned} y&=-2+\frac{4}{-2}\\ &=-2+(-2)\\ &=-2-2\\ y&=-4 \end{aligned}\] Sooneofthisfunction'sstationarypointsis: \[\begin{pmatrix}-2,-4\end{pmatrix}\] when\(x=2\): replacing\(x\)by\(2\)in\(y=x+\frac{4}{x}\),wefind: \[\begin{aligned} y&=2+\frac{4}{2}\\ &=2+2\\ y&=4 \end{aligned}\] Sothefunction'ssecondstationarypointhascoordinates: \[\begin{pmatrix}2,4\end{pmatrix}\] Wecanseebothofthesestationarypointsonthegraphshownbelow: Wecanseequiteclearlythatthestationarypointat\(\begin{pmatrix}-2,-4\end{pmatrix}\)isalocalmaximumandthestationarypointat\(\begin{pmatrix}2,4\end{pmatrix}\)isalocalminimum. Exercise2 Findthecoordinatesofanystationarypoint(s)alongthelengthofeachofthefollowingcurves: \(y=2x+\frac{8}{x}\) \(y=-x-\frac{1}{x}\) \(y=3x+\frac{27}{x}\) \(y=-2x-\frac{72}{x}\) \(y=x+\frac{25}{x}\) Answersw/outWorking Note:thisexercisecanbedownloadedasaworksheettopracticewith:worksheet AnswersWithoutWorking Wefindthederivativetobe\(\frac{dy}{dx}=2-\frac{8}{x^2}\)andthiscurvehastwostationarypoints: \[\begin{pmatrix}-2,-8\end{pmatrix}\] and \[\begin{pmatrix}2,8\end{pmatrix}\] Wefindthederivativetobe\(\frac{dy}{dx}=-1+\frac{1}{x^2}\)andthiscurvehastwostationarypoints: \[\begin{pmatrix}-1,2\end{pmatrix}\] and \[\begin{pmatrix}1,-2\end{pmatrix}\] Wefindthederivativetobe\(\frac{dy}{dx}=3-\frac{27}{x^2}\)andthiscurvehastwostationarypoints: \[\begin{pmatrix}-3,-18\end{pmatrix}\] and \[\begin{pmatrix}3,18\end{pmatrix}\] Wefindthederivativetobe\(\frac{dy}{dx}=-22+\frac{72}{x^2}\)andthiscurvehastwostationarypoints: \[\begin{pmatrix}-6,48\end{pmatrix}\] and \[\begin{pmatrix}6,-48\end{pmatrix}\] Wefindthederivativetobe\(\frac{dy}{dx}=1-\frac{25}{x^2}\)andthiscurvehastwostationarypoints: \[\begin{pmatrix}-5,-10\end{pmatrix}\] and \[\begin{pmatrix}5,10\end{pmatrix}\] × AnswerswithWorking Selectthequestionnumberyou'dliketoseetheworkingfor: Qu.1 Qu.2 Qu.3 Qu.4 Qu.5 Close