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Exam ReviewSupplemental InstructionIowa State UniversityLeader:Anna SteffensmeierCourse:Math 165Instructor:Kristopher LeeDate:September 15, 2016Section 2.1 Key Points:Average Rate of Change at two given points. fb-fab-a at the point (b,f(b)) and (a,(f(a)).Instantaneous Rate of Change or the Definition of the Derivative.lim?h→0 fx+h-f(x)hlim?z→x fz-f(x)z-xTangent Line- Equation of a line that runs through one particular point of a functionSecant Line- Equation of a line that runs through two different points of a functionSection 2.2 Key Points:Limit Definitionlimx→c fx=LEvaluating limits as h→0When you get 00….. Do more work!When you get c0 where c is some constant… Limit Does Not Exist! Congrats Mean Girls!When you get 0c where c is some constant… Limit =0The Sandwich Theoremg(x)≤f(x)≤h(x) for all x in some open interval including c except at x=c. Supposed that limx→cgx=limx→ch(x)=LSection 2.3 Key Points:One-Sided Limits:limx→c-f(x)ANDlimx→c+f(x)Limit Existence:limx→cfx=Llimx→c+fx=Llimx→c-fx=LAll of these statements must be equal in order to be true. If any of these limits are not the Same, then the limit does not exist at point climh→0sinhh=1Section 2.5 Key Points:Continuity at a Point:limx→c-fx=limx→c+fxf(c) has to existThe first two must be equal to each otherTypes of Discontinuity:Removable Discontinuity- the function has an open hole at a certain (x,y) with a filled in point at the same x coordinate, but different y coordinateJump Discontinuity- the function literally jumps from one y-value to another at a certain x coordinateInfinite Discontinuity- occurs where there is an asymptoteOscillating discontinuity- occurs when a function oscillates too much to have a limit as x→cIntermediate Value Theorem:Functions that are CONTINUOUS on an interval from [a,b] are said to have a point c that exists to give a point (c,f(c)).In order for the IVT to apply to a function or graph, it has to be 100% continuous over that interval.i.e. fx=1x [-2,2]f-2=1-2f2=12Yes, there is a sign change from -2 to 2, but the function is not continuous, therefore the IVT does not apply here.Section 2.6 Key Points:Infinite Limits:limx→∞fx=Llimx→-∞fx=LThis indicates a horizontal asymptotelimx→c+fx=∞limx→c-fx=∞Both of these indicate a vertical asymptotelimx→c+fx=∞limx→c-fx=-∞Horizontal Asymptotes:A horizontal asymptote exists if the limx→±∞fx=LVertical Asymptotes:A vertical asymptote exists when the denominator of a rational function is equal to 0 ................
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