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# Friday Notes 9/18/15 MA 16100 - 375

Purdue

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This 8 page Class Notes was uploaded by Rachel Mooar on Friday September 18, 2015. The Class Notes belongs to MA 16100 - 375 at Purdue University taught by Professor Yeung in Fall 2015. Since its upload, it has received 45 views. For similar materials see MA 161 in Mathematics (M) at Purdue University.

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Date Created: 09/18/15

MA 161 Exam 1 Studyguide Vertical Line Check Run your pencil along the graph If it hits only one point at a time then it s a function Composition of Functions f0 ClXfClX Function Not a Function JI u II J u I 4 abquotxaquotxbquotx Inverse Functions fx is onetoone 1 x2 which is equival nt to fx1 fx2 Horizontal Line Test The same as the vertical line test but horizontal This is to discover if a funcion is inverse NOTE The graph must be a function for this test to apply Properties of log and natural log i NOTE all examples here have base 39a E I ogxyogxogy i This is a function but it has no inverse aquotogxyXyaquotOgXaquotIOQW ogxyogxog ogxquotrrogx Finding the Inverse Function Example Solve equot2x37equot2x370 Lnequot2x3ln7 2X3n7 Example Find Inverse Function for YequotX17e Solve in terms of y L yeAxM12eAx 2 y12equotxequotx 3 yequotX2yequotx12yequotx 4 y12ye 5 Xny12y Tangent One point on a curve Slope of the tangent line is approached by slope of secant lines Point Q Stchange in L l l Stl Tangent Point fl quot k l l il l l l l tchange in t Slope of Secantmrmjin schange in t Slope of Tangel Change limchange in pchange in t O De nition of a Lir lt Lim fXL if fx gets arbitrarily close to L as X app oaches a X X gt a Fact lim fXL is equiveljgwt to limfxL Xgt a X 9 a limfxL X a a LiI Assume all have x approaching a lim fX plus or minus gxim fX plus or minus lim gX X 9 a X 9 a X 9 a 2 lim C fxC lim fX 3 lim fxgxim fX lim gX 4 lim fXgXim fXimgX if lim gXO 5 lim fXquotnim fXquotn 6HmCC 7 8 9 1 H imXa imXquotnquotblimXquotnquotfaquotn imquotnfXquot12aquot1n OimquotnfXquot12lim fXquot1n Example as X approaches 2 lim X1Xquot21im X1lim x1 RULE 6 im2lim1limXquot2im1 RULE 2 212A21141 RULES 7 8 9 Theorem 1 Suppose fX is less than or equal to gX for X near a but X DOES NOT equal a then as X approaches a lim fX is less than or equal to lim gX Theorem 2 Squeeze If fX is less than or equal to gX is less than or equal to hX for X near a but X DOES NOT equal a and as X approaches a lim fX lim hX L then lim gXL USE PATHAGOREAN THEOREM TO FIND THIRD SIDES X 1Xquot2quot12 X 1 Xquot21quot12 11Xquot2quot12 Sin Cos Tan Csc Sec Cot Find the csc of the sin triangle and you get the answer of Limits at In nity Horizontal asymptotes De nition Let fX be a function de ned on a in nity respectany on in nity a thiefnilirn fXL if values of f X can be made arbitrarily close to L as X is suf cientlylar As X approaches in nity lim 1X 0 As X approaches in nity lim 1X0 Continua ion of de nition If X approaches in nity fxL and X approaches in nity lim fXL we call yL a horizontal asymptote In the previous example yO is a i i ihoriLontallasymptote Derivatives and Rate of Change faT To nd slo loser and closer to a Recall that w rve at point p y taking a limit of slope of secant nnes ie slope of curve at p mpas Q approaches P lim as X approached a lim fXfaXa OR mpas h approaches 0 lim fahfah Note that hXa De nition f a as h approaches 0 lim fahfah This is called the derivative of f at a f a as X approaches a lim fXfaXa 7wwww Example Find the equation of the tangent line to yXquot2 a when X1 Solution a X1 DfXXquot21 slope at X1 is m as X approaches 1 lim fXf1X1 lim Xquot21X1 lim X1X1X1 lim X1 Sub 1 for X 2 The equation as it passes through 11 Use yymXX y12X1 In case that a moving object is at displacement sst at time t instantaneous velocity at ta is As t approaches a Iim sXsaXasa Similarly for any function yfx the derivative f a at a represents the instantaneous rate of change of fx Example The displacement of a particle at time t is given by in meters t in seconds Find the velocity and speed at t2 Solution Velocity at t2 into a f 2 as X approaches 0 Iim f2hf2h im Sub 2h into the t of the equation im 1h13h13 Sub 2 in the t of the equation h does not equal 0 in the limit im 1h 33h3h im 1h h33h im h33h velocity speed speed is the magnitude of velocity so it ll always be positive MEMORIZE THE DERIVATIVE EQUATIONS Derivatives Derviative of function f of a f as h approaches 0 Iim fahfah Replace a by x to get f X as h approaches 0 Iim fXhfXh quotderivative of fquot delta y fXhfX delta X XhX 1 Equivalent to f x as delta x approaches 0 Iim delta ydelta X It can also be written as lim dydx or lim dfdx De nition f is differentiable at a if f a exists as a real number ie f a as h approaches 0 lim fahfah Theorem If f is differentiable at a then f is continuous at a Reason Continuity at a means as x approaches a Iim fxfaDDlim fafaxa Suppose f is differentiable at a then f a xa Iim fxfaxa It ll be a real number Write Xa lim fXfaXa The function is differentiable if it is continuous andisn t a differentiable if it isn t continuous 7 Situations where f is NOT Differentiable 8 1 Discontinuity from the theorem 2 VerticalTangent 3 hle Iim fxhfXh 1 Slgpe in does NOT eq ual I h0 Iim fxhfXh 2 75 a Slobe In Example fx the absolute value abs of x l fx x greater than or equal to O l X X lt O h0 at X 0 Iim fxhfXh h0 Iim fhf0h Iim hOh h0 Iim fxhfXh Iim fhf0h Iim hOh 1 because hljo hOim fhfOh DOES NOT equal hDO Iim fhf0h Therefore the Derivative does not exist Domain what x cannot be 1xquot2xquot24quot12 Because the denominator can t be zero xquot2 4quot12 cannot equal zero 0 Xquot2 4quot12 OXquot2 4 4Xquot2 X cannot equal 2 Also square roots cannot have negatives and be a real number so we have to make sure there s nothing else that would make the denominator negative under the square root 3A2 9 and 94 3 so that s okay 1A2 1 and 14 3 so in nity 3 U 3 in nity Range What y can be It s easiest to have a graph to look at After looking at the graph for 1Xquot2Xquot24quot12 you can see that the graph covers from in nity to in nity meaning the range is all real numbers Shortcuts for the Derivative 1 FxC Because the line is constantly going horizontal the secant slope is 0 Therefore This is a commonly used rue yC 2 yfxx Because the line is going in a diagonal from the origin at a slope of yX l 11 because of the equation yx Therefore This is a commonly used rule 3 In general dxquotndxnxquotn1 n is all real numbers dxquotndxh0 lim xhquotnXquotnh nxquotn1 This can only be used when n is a positive integer This equation WILL save your life in calculus 4 De ne e to be the real umber satisfying h0 lim equoth1h1 NOTE DO NOT USE THESE IN THE WRITTEN PORTION FOR THIS TEST UNLESS SPECIFICALLY SPECIFIED OR IT DOESN T SAY AT ALL IF IT SAYS SHOW WORK DO THE CALCULATIONS SO YOU DON T LOSE POINT Rules of Derivatives a dCdXClellCddx fx b dcFl3 tal gXdfvx or dgdX Example dxquotw 2 xquot3dquotx dXquot2dX 2dXquot3dx 2X23Xquot2 This uses rule b and shortcut 3 Example dxquot5 xquot 12 equotXdx dXquot5dX dXquot 12dx deAXdx 5Xquot4 12Xquot 121 equotx This uses rule b and shortcuts 3 and 4 Example Find the point on the graph of yxquot3 Xquot2 X 1 with horizontal tangent Solution horizontal tangent at X fX H In outcome fXquot3 Xquot2 X 1 O f X3Xquot2 2X 1 O expand the binomial 3X 1 X 1 DD X13 OR 1 This uses shortcut 3 Example Find the equation of lines that are parallel to the line 12X y1 and tangential to the curve y1 Xquot3 What we re er Onelooking for y 1 XA xx xxxxiiiiilciig 1 2X y 1 Solution Slope of 12X y 1 is equivalent to y 12X 1 Therefore its slope is 12 because of the general equation ymXb Find X such that the slope of y1 Xquot3 at X is 12 Le f X12 therefore 12f X ddX 1 Xquot3 divide both sides by 12 Xquot24 OR 3Xquot2 X2 a 2 At X2 of tangent is fX 1 2A3 1 8 8 Yf212X2 y 12X 24 9 At X2 of tangent is yf2 12X 2 712X 24 f 2 l 1 2quot3 y 12X 17

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