Application of Derivatives

APPLICATION DE DERIVATIVESTangents and Normalsy= = f(x)normaltangentR/24, yi)The Slope of the tangent at P = t'(x4)The slope of normal at P = -Yf((24)The equation of the tangent at P = 9-4/2-24The equation of the normal at P = - (2-24/g-y)y ==t(x)Q(24,y,)P(20, yo)I'm the tangent PQ , hipo = mo . Therefore,y, = ((x4) and yo - 4, = dy/no - 24 dxQThe tangent to a polynomial can befound by equating the least degreeterm to gero. (tangent at origin) DateRigehermaly-f(x)tangene((x,y)Q4CiTMNThelengthofthetangent=PTly/sinollyritcotto=The length of subtangent TM = ly/tano= ly dr/dylThe length of the normal = PN- ly/cosol= lyri+tanol= +The length of the subnormal MN = ly/cotol= 1y.dy/alAngle between CurvesacuteandP(20,90)0o = tan my m2-= tan('xx) - g'(xo)It mmI + ((20) (((20)Note that the angle between the werescan be different at different pointsof intersection. CurvesDistancebetweenShottestis theshortestdistanceThecommonlengthbotofthenormal.IntermediateValueTheorem(IMUT)[a,b]thenIffiscontinuousinforauvaluesofBE [f(a), fl6) 1 ,Thereexistsatleastonesuchf(x)=, bl .thatBxEThereexistsatleast onemf(b) + nf(a)thatf(c)=CE[a,b]suchm thm,n must beB lb, f(b))positive(x,f(x))(a,Has), AbCorouaryIffiscontin[a,6]andf(a)andf(b)areofoppsignsthenThegraphinterectsxaxisatleastoneThereisatleastoneCE[a,b]such thatf(c)0.f(x) = 0wasatleastoneroot. CLASSMATEDatePageRolle'sMeanValuefunctionTheorena(RT)Ify=t(x)isin6],differentiablecontinuousinf(b)(a,b)f (a)=thenandThereexistsatleastoneCEsuchthat(a,b)f'(c) = 0ThereexistsatLeastonerootoft'(x)=0in(a,b)Theslope off(x) iszero at leastoncein(a/b)..slope =slope = 0ABAB>aCbaGC2bNotethatRTfailsifanyoneconditionfails-althotheslopestillcanbezeroTheconveneofRTisalsonottrue.Tothatf'(x)hasatleastoneproverootin[a, b]f'(x) = f(x)andapplyRMVT on f(x)Iff(x)iscantanddiffsuchthatf(x)=Dhasmroots=Dm-1roots(alleast)(x)hashasrootsf'(x) = Dmhasatmostm+lroots.f(x)=0 Langrange's Mean Value continous TheornIfafunctionisininCa,[a,b]differentiableb) ,andthereexists atleastonethencsuchthatt'(c)=f(b) - fla)-b - aBslope WILLbe same atAleast once.aCLetn(x) =f'(z)-f(b) - f(a)dxb - a=f (x) -(f(b)-f(a))x=f(x)-kxb-aSo w(a) = f(a) - a ('f(b) - (las)b- a=bf(a) - af(b)b - aAndh(b) =bf(a) af(b)6 - aThereforebyRMVTh'(x) = Dat leastonce. DatePageApproximation and Errorsdy=sy=4f - 4idxDxf'(xi)xf - xiso=yfyi+f'(xi) (xx-xi)ThiscanbeusedtofindtheapproximatevalueofafunctionForexampletofind Sin (31)Letf(x)=sin (x')= 31 = T/6 + 0.0175f"2)=CDS(x')xi = n/6 E (rad)?so(T16)=4f-1/2V3+0.0175=T/62V3(0.0175) +I~0.515yf =22Therelativeerrorinyisdaywhile,errorinisdy=f'(x)dxtheyForexampletheradiusofaifsphereiscmI0.05cmfindthe10approxandrelativeerrorav = f'(x) drThe approx errordudwhere(unria= 4rr2drdrSodv4rr2dr0.05 chn=4n (10) (0.05)N20ndv/V=20n~0.015Theisrelative Vertical TangenthasaverticalThecurvey = f(x)x = noif thefirst+00or-tangent at-00principlederivativeequalsbut not bothlim f(26+h)- f(xo) = I 8h-soForexample,f(x) = sgn(x) wasa verticaltangent -I0-1t'(o+) = f'(o-)=+00soitdoeshaveaverticaltangentatx=D .,However,f(x)=x243doesnothaveaverticaltangent -f (lo+) = +00f'(o-) =00LUD # PHD classmateDatePageConcavitywhilef'(x)tellsusthethegraphoff(x),slopeoff"(x)tellsthecurvatureusofthegraphf'(x)always thealways vemonotonicallymonotonicallyincreasingdecreasingf"(x)always thealways -veconcave upwardconcave downwardI(smiley curve)(sad cureIft"(x) = 0 , , the graph hasnocurvature . arevarious cases, the graphs-Ingraph (f(x))f (x)t"/x)+++-+--NotethatiffisInvertibleandf is concave upwards, f-' will beconcave downwards andiceversa InequalitiesusingConcavitiesInaconcavefunctionf(x)mf(22)+nf(x4)>m+nf(mx2 + ux)m+ noutput atm: nBoutput at MPAMxmxz + hmX2m+nThisinequalitywillbeHippedoftheshapeisconcavedownwardTheinequalityholdsforallX4andX2as+longasnx,MonotonicityAfunctionissaidtobemanotonicifitalwaysincreasingorisalwaysdecreasingoralwaysconst.neitherM IMIMI(check wbhd)like LLL 4 P212 In its domain , = f(x)tobe -Increasing (NON decreasing) f(x) < f(x2)- If &24X2f'(x) > 0-Decreasing (Non increasing)-If 24 < x,((x) > (Xx)t((x)Strictly Increasing (MI)-If(xx) < f(x2)240can be = 0 only at somepoints, not in an intervalStrictly Decreasing (MD)- If x, V x2f(x4)>f(x2)t'(x)< 0noteThegraphoff(x)=ax+biscx+da rectangular hyperbola havingasympbotesx=-d/yC&y=a/cPoint of InflectionIt isapointatwhich-f(x)changes its concavity (f" changes S146f(x)hasa tangent line(vertical tangentItalsooccursatpointswhere f"(x)=0or does not existAt a POI thecurvecrossesitstangent DatePageCriticalandStationaryPointsCriticalpointsarepointsy=f(x)wheref'(x)=0orf'(x)doesnotexist.Notethatendpointsarenotconsideredcriticalpointsstationarypointsarepointswheref'(x)=0.ExtremumLocal MaximaLocal Minimicfla=h) f(a)a-h a atha-h a atht'(x)changes sign(wavyf'(x) changes signcurvefrom -hethefromthe-vef'(a) = Df"(a) <0f'(a)=0&f"(a)> 0doesn't work itf(a) =0 f f"(a) oL-max andLainarethe greatest/leastvalues intheirneighbour woodofLmax&LevinareThe possiblepointscriticalandendpaintsf'(x)=0butf'(x)doesn'tAtpointswhereitssignaPOI exists .changeIf f'(a) DNEf"(a)alsoDNE The global maxima is Gmax = { au Hand allisGmir = (f(a),whileglobalminimahim of ffbinterval,NotethatinanopenIn (a, b)Gmin = t(c)Gmax = flbnot an exactvalue so DAFaCbRemembertocheckextremumandcriticalpointstoseeiftheyarewithinthegivenrangesInequalitiesUsingMonotonicityToprovethatHx) > g(x) inso meinterval, define n(x) = f(x)-g(x)andproveh(x)>0inthatintervalusingmonotonicityForexampletoproveSinxletforx0,h(x) = x- - sinxh'(x)=I-wsxh (x) >0MIh(o) 0 &(2) is MI0 < h (x)sinn < xNotethattanxisMI&sinx is MDxx DatePageCubic Polynomial AnalysisFor a f(x) = x3 + ax2 + bx+c,->f'(x) = 3x2 + 2ax+ bD = 4(a2-36)If D < .00 I real & 2 imaginary & t'(x)>0If D > 0, 3 real roots & f((x) = 3(x-24)(2-X2)POI24=X2>xx x2x3x3x3(img)If D = 0, f'(x) = 3(x-x4)2 & f(x) = (x-x4) + k24= X2 XX3tok = Ck # O