Lec3 - Calculus I Lecture 3 Problem set 3 Section Section Section Section Section 4.1 4.2 4.3 4.4 4.7 51 55 59 26 27 33 75-78 54 60 44 45 76 Outline

Lec3 - Calculus I Lecture 3 Problem set 3 Section...

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Calculus ILecture 3December 15, 2017
Problem set 3Section 4.1: 51, 55, 59Section 4.2: 26, 27, 33Section 4.3: 75-78Section 4.4: 54, 60Section 4.7: 44, 45, 76
Outline
Minimum and maximumfattainslocal minimumatciff(x)f(c)forxnearc.fattainslocal maximumatciff(x)f(c)forxnearc.Examples:Ix3-3x2+6 atc=0 andc=2.Icosxatc=0 andc=πfattainsabsolute minimum (maximum)atciff(x)f(c)for allxin the domain under consideration.E.g.:x3-3x2+6 has absolute minimum atc=2 if the domain is[1,5]. It does not attain absolute minimum atc=2 if the domain isR.Extreme Value Theorem:Iffis continuous on a bounded interval[a,b]. Then there existsc,d[a,b]such thatf(c)f(x)f(d)forallx[a,b].Both assumptions of continuity and closed bounded domain areessential.
Fermat theoremFermat’s theorem: Iffattains local maximum/minimum atctheneitherf0(c)does not exist orf0(c) =0.The pointscwheref0(c)does not exists orf0(c) =0 are calledcritical points(orcritical numbers) off.Examples:|x|,x2-3x+2 ,x1/3(1-x).The converse of Fermat’s theorem is not true, i.e. not all criticalpoints are local maximum/minimum. E.g.x3atx=0.Fermat theorem provides a way to find absolute max and min of acontinuous functionfon a bounded closed interval[a,b]:IFind all critical points offon(a,b).ICompare the values offat these points and at the end pointsaandbto find the absolute max and min.
Mean value theoremMean value theorem: Iffis continuous on[a,b]and differentiableon(a,b)then there existsc(a,b)such thatf0(c) =f(b)-f(a)b-a.Rolle’s theorem:Iff(a) =f(b), then there existsc(a,b)suchthatf0(c) =0.Corollary: Iff0(x) =0 for allx(a,b)thenfis constant on[a,b].Iff0(x) =g0(x)on(a,b)then there existscRsuch thatf(x) =g(x) +con[a,b].Examples: prove|sina-sinb| ≤ |a-b|for alla,bR. In particular,|sina| ≤ |a|.
Derivative testsFrom the Mean Value Theorem, we see that iff0(x)>0 on(a,b)thenfis increasing on[a,b]. Similarly iff0(x)<0 thenfis decreasing.Examples:x3-3x,x-sinx.The First Derivative Test: Suppose thatcis a critical point off.IIff0changes from positive to negative atc, thenfhas a localmaximum atc.IIff0changes from negative to positive atc, thenfhas a localminimum atc.IIff0does not change sign atc, thenfhas no local maximum orminimum atc.The Second Derivative Test: Supposef00is continuous nearc.IIff0(c) =0 andf00(c)>0 thenfhas a local minimum atc.IIff0(c) =0 andf00(c)<0 thenfhas a local maximum atc.
Concavityfisconcave upwardonI= (a,b)if the graph offlies above all thetangents on that interval. Similarly, it isconcave downwardonIif itsgraph lies below all the tangents.Concavity test:IIff00(x)>0 for allx(a,b)thenfis concave upward on(a,b).IIff00(x)<0 for allx(a,b)thenfis concave downward on(a,b).Examples:I14x4+x3-x22-3x+1Ie1/x,e-x22.

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