does NOT exist. WHAT IF THE NUMBER X APPROACHES CANNOT BE PLUGGED INTO F(X)??????

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1 MATH 000 Miterm Review.3 Te it of a function f ( ) L Tis means tat in a given function, f(), as APPROACHES c, a constant, it will equal te value L. Tis is c only true if f( ) f( ) L. Tat means if te verticle line c is rawn, te function from te left of tat + c c line an te rigt, will approac te same number. Te following grap is a jump iscontinuity. If you raw a line at, te function approaces a ifferent value on 0 te left an rigt sie. Terefore f( ) oes NOT eist. 0 Te following grap as a iscontinuity base on an asymptote tat breaks te grap were te vertical line is rawn at, te function approaces ifferent values from bot sies, terefore f( ) oes NO eist. Te function of te following grap is f( ) Te it can be foun generally by te following f ( ) f ( c ). Tis means you can plug in te number tat X is c approacing into te function an fin te value. Eample: ( + 3 8) [(3) + 3(3) 8] 0 3 WHAT IF THE NUMBER X APPROACHES CANNOT BE PLUGGED INTO F(X)?????? I'm gla you aske Morpeus. Te scenarios you're trying to avoi are te following ) 0 a, CANNOT ivie by 0 ) a, CANNOT take te square root of a negative number

2 Well not entirely Keanu. It's a possibility, but not 00% true all te time. Tere are a couple tricks you can try to o to algebraically manipulate te function so you can plug in C. ) Multiply by a conjugate ) Epan out any part of te equation if it as some function raise to a constant eponent 3) Try to factor out polynomials in te numerator to cancel polynomials in te enominator Eample # (using a conjugate) ( + 4) (4) ( ) 0 0 ( ) ) Eample # (Epaning) ( + ) 4 ( ) Eample #3( Factoring) + 8 ( + 4)( ) Calculating Limits Tis formula soul be known for te eam: Te following is an eample of ow to use it. sin 0 Eample: sin 9 sin Wat I want te enominator to be is 9 so it can matc te numerator, if it matces, you can use te ientity above an equate it to one. Tis is a little algebra trick to swap in 9 for te enominator but still ave an equivalent equation.

3 Squeeze Teorem if f( ) g( ) ( ) wen is near a (ecept possibly at a) f( ) ( ) Lten tat implies g ( ) L a a a Wat tis means tat if you ave a function g() an you fin two functions, f() an (),to wege above an below it, you can say tat if te it approaces te same for f() an (), it must be te same for te function in between, g() For eample, let's take sin 0 I can't plug in 0 rigt away because I'll be iviing by 0, but I o know s sin can only take values between - an, so we can say sin, but ten multiply it by on bot sies to get sin an ten take te it as goes to 0 on bot sies to get sin 0 sin 0 an now since it's wege between 0 an 0, te it as goes to 0 of tis function must be 0..6 Limits involving infinity Tere are tree scenarios to look for wen taking its as goes to infinity c ) 0 ) c c c 3) b b Just wen given a function wit a numerator an enominator in te form of an te igest value of in g() an ignore te rest of te equation. f( ), take te igest value of in f() g ( ) Eample #(wen te enominator is bigger): cos 0 9 Eample #(Wen te numerator is bigger) Eample #3(Wen te numerator an enominator grow at a ratio)

4 Left an Rigt an its c + c f( ) left an it f( ) rigt an it Lets say a rational function is unefine at some value at c, ten tere is a vertical asymptope at c. If you want to evaluate te left or rigt an it at tat asymptope just keep in min tat it will eiter go to positive or negative infinity. Just test values very close to tat value an see if you get a positive or negative value. If you get a positive value, it will go to positive infinity an if you get a negative value it will go to negative infinity. Eample Evaluate te following it 3 + Since it s a left an it, I m going to pick a value sligtly less tan. I will pick.9 an test it into te function 3 (.9) + ( + ) ( ) All I m concerne about is weter or not it is a positive or negative number because (.9) ( ) at te vertical asymptote it will eiter go up to positive infinity or own to negative infinity. Since I got a positive number in te numerator an a negative in te enominator, te result will be negative wen I ivie te two. Terefore 3 +. Derivatives an rates of cange Te picture to te left illustrates ow to form te it efinition of te erivative. Te erivative elps us fin te slope at one specific point, were in te past te slope formula requires two points. Now wit te erivative, we can fin instantaneous rate of cange or slope of a tangent line to a curve. Please try to restrain yourself from jumping out of your cair wit joy from now iscovering tat you can fin te slope of a tangent line to a curve. Now tat you are reseate, tis is te it efinition of te erivative. ) 0 f( + ) f( ) You migt ask yourself, ow i tis formula come about? Well gla you aske. Tis formula is base off te original efinition of te slope of a line from previous mat courses. Usually, te y y formula tat was use is m. Now we ave our picture about, so we use f(+) for y an f() for y because tose are our Y values of te function above, wic is a very general function tat as a curve. We ten use te +, wic is a coorinate, for an te first coorinate,, for. Wen substituting tese values in for te slope

5 formula, we get f( + ) f( ), wic ten simplifies to ( + ) f( + ) f( ) because te value cancels out. However wat if te istance between te two points goes to 0? Te istance is given by so as it approaces 0, we fin instantaneous slope at one point, ence te formula above saying LIMIT as 0. Wen computing its, we normally just plug in values of watever te it is approacing into tat variable, owever wit tis formula given above, we cannot plug in 0 for. Wy? Well because we woul get a number ivie by zero. Ten te worl will imploe an all is lost. Not really, but we can t ivie by zero. Te reason wy is tat if I ave ¼, I know tat its object ivie into 4 parts. But if I ave 0, I are you to take one object an ivie it into 0 parts. It oesn t make logical sense because ow can you make object separate into 0 parts? So te wole point is we simplify te numerator, f( + ) f( ), so we can etract an H to cancel wit te H in te enominator. Eample: f( ) fin te it efinition of te erivative 0 f( + ) f( ) ( + ) ) 0 ( + ) ( + ) ( ) ) + 0 ( + ) ( + ) 3) 0 ( + + ) ( + ) 4) 0 4 ( + ) 5) 0 4 ( + ) 6) 0 ( 4 ) ( + ) 7) 0 ( 4 ) 8) 0 ( ) + ( 4 ) 9) 0 ( ) + ( 4 (0)) 0) 0 ( (0)) + 4 ) ( ) 4 ) 4 4 3) 3 Steps ) Plug in te formula into te it efinition ) Fin te LCD an combine fractions in te numerator 3) Epress te numerator as a single fraction 4) Epan te equation in parentesis 5) Distribute te - to te function in parentesis 6) Cancel terms (a lot soul always cancel at tis point) 7) Factore a H out of te numerator 8) Rewrite te comple fraction as a multiplying by te reciprocal 9) Cancel te H (te wole goal of tis process) 0) Plug in 0 for H now tat it s remove from te enominator ) Combine te in te enominator ) Write as a single fraction 3) Reuce te fraction by reucing te eponents of

6 Fining te average rate of cange To fin te average rate of cange of a function, f() between two points, A an B, te following formula must be use. f() b f() a. b a Eample: Fin te average rate of cange for te following function over te interval to 3, f( ) f(3) f() Basic ifferentiation formulas tis just means to ifferentiate. ) kf ( ) k f ( ) kf '( ) ) [ f( ) ± g( )] f( ) ± g( ) n n 3) n ) Constant multiple rule- Constants can be factore from an equation an ten be multiplie at te en after ifferentiating. E. ( ) 4 ) Sum rule- If two functions are being ae, tey can be ifferentiate separately an ae at te en. E ) General power rule- Wen aving a variable raise to a constant power, multiply te variable by te power, ten subtract one from te power E. 6 4(6 ) 4 also note tat if f()cos(), f'() -sin() If f()sin(), f'()cos().4 Te prouct an quotient rule Here are some oter rules for ifferentiation Quotient rule: f( ) f '( ) g( ) f( ) g'( ) g( ) [ g( )] Use wen tere are fractions

7 Prouct rule: f ( ) g ( ) f '( ) g ( ) + f ( ) g '( ) Use wen tere are multiple functions multiplying eac oter Eample (Quotient rule) + K ( ) 6 5 First write own wat f() is an g() an ifferentiate tem. f ( ) + 5 f '( ) 4 g ( ) 5 g'( ) ()(6 5) (+ )(6) k'( ) (6 5) ( 0) (+ 6) k'( ) (6 5) 0 6 k'( ) (6 5) 6 k'( ) (6 5) Eample (Prouct rule) ( + 5)( 5) First write own wat f() is an g() an ifferentiate tem f ( ) + 5 f '( ) 4 g ( ) 5 g'( ) k ( ) (4 )( 5) + ( + 5)() k ( ) (4 0 ) + ( + 5) k ( ) k ( ) Differentiating square roots an ealing wit negative eponents f( ) f() to ifferentiate tis, cange te raical to a fractional eponent Now use te power rule ( f '( ) ) Wen given f( ) f '( ) ( )( ) ( ). Now it is in a format tat te power rule can be use.

8 Derivatives of trigonometric functions sin cos cos sin tan sec.5 Cain rule csc csc cot sec sec tan cot csc Cain rule: f ( g ( )) f '( g ( ))( g '( )) Cain rule is use wen you ave a function embee in te oter. For eample, ( ) 6 6 k ( ) + can be split into f( ), g( ) +. Now wen aske to fin f(g()), te original formula will be foun. Tis illustrates tat te function above can be split into two separate functions. How to interpret tis is treating te insie function as a variable an ifferentiating wit te power rule, but ten multiplying but te insie erivative. 6 5 Eample: k ( ) ( + ) Treat ( + ) as a variable, so wen ifferentiating, te result is 6( + ). But, at tis point you ave to multiply by te erivative of te insie of te function, ( + ). Tis results in 5 k'( ) 6( + ) ( )..6 Implicit ifferentiation Implicit ifferentiation is use wen te X an Y variables are tangle togeter an not easily separate. For eample, 3 y + y 5. If we are aske to fin y, we ave to ifferentiate every variable wit respect to itself, only wen Y is being ifferentiate, we put y or y. For simplicity purposes, I ll put y. Eample: 3 y + y 5 We ave to ifferentiate te wole function, an treat X an Y as teir own variables an ifferentiate wit respect to itself. 3 )[()( y ) + ( )( y)( y')] + [(6 )( y) + ( )()( y')] 5 )( )( ') ( )( ') y y + y y y 3) '( ) y y+ y y 5 y 6 y 4) y ' 3 y + ) Differentiate everyting using prouct rule were necessary ) Everyting wit y keep on te left, anyting else move to te rigt sie by subtraction 3) Factor out y from te left sie 4) Divie te left over parentesis to get y by itself

9 .7 Relate rates Tese are usually wor problems were we are given a rate of cange of a variable an aske to fin te rate of cange of anoter variable. For eample: Eample Air is being pumpe into a sperical balloon at a rate of 5 cm 3 /min. Determine te rate at wic te raius of te balloon is increasing wen te iameter of te balloon is 0 cm. Solution Te first ting tat we ll nee to o ere is to ientify wat information tat we ve been given an wat we want to fin. Before we o tat let s notice tat bot te volume of te balloon an te raius of te balloon will vary wit time an so are really functions of time, i.e. v(t) an r(t). We know tat air is being pumpe into te balloon at a rate of 5 cm 3 /min. Tis is te rate at wic te volume is increasing. Recall tat rates of cange are noting more tan erivatives an so we know tat, We want to etermine te rate at wic te raius is canging. Again, rates are erivatives an so it looks like we want to etermine, Note tat we neee to convert te iameter to a raius. Now tat we ve ientifie wat we ave been given an wat we want to fin we nee to relate tese two quantities to eac oter. In tis case we can relate te volume an te raius wit te formula for te volume of a spere. As in te previous section wen we looke at implicit ifferentiation, we will typically not use te part of tings in te formulas, but since tis is te first time troug one of tese we will o tat to remin ourselves tat tey are really functions of t. Now we on t really want a relationsip between te volume an te raius. Wat we really want is a relationsip between teir erivatives. We can o tis by ifferentiating bot sies wit respect to t. In oter wors, we will nee to o implicit ifferentiation on te above formula. Doing tis gives, Note tat at tis point we went aea an roppe te from eac of te terms. Now all tat we nee to o is plug in wat we know an solve for wat we want to fin.

10 Eample A 5 foot laer is resting against te wall. Te bottom is initially 0 feet away from te wall an is being puse towars te wall at a rate of ft/sec. How fast is te top of te laer moving up te wall secons after we start pusing? Solution Te first ting to o in tis case is to sketc picture tat sows us wat is going on. We ve efine te istance of te bottom of te laer from te wall to be an te istance of te top of te laer from te floor to be y. Note as well tat tese are canging wit time an so we really soul write an te. However, as is often te case wit relate rates/implicit ifferentiation problems we on t write part just try to remember tis in our eas as we procee wit te problem. Net we nee to ientify wat we know an wat we want to fin. We know tat te rate at wic te bottom of te laer is moving towars te wall. Tis is, Note as well tat te rate is negative since te istance from te wall,, is ecreasing. We always nee to be careful wit signs wit tese problems. We want to fin te rate at wic te top of te laer is moving away from te floor. Tis is tat tis quantity soul be positive since y will be increasing.. Note as well As wit te first eample we first nee a relationsip between an y. We can get tis using Pytagorean teorem. All tat we nee to o at tis point is to ifferentiate bot sies wit respect to t, remembering tat an y are really functions of t an so we ll nee to o implicit ifferentiation. Doing tis gives an equation tat sows te relationsip between te erivatives. Net, let s see wic of te various parts of tis equation tat we know an wat we nee to fin. We know an are being aske to etermine so it s okay tat we on t know tat. However, we still nee to etermine an y.

11 Determining an y is actually fairly simple. We know tat initially an te en is being puse in towars te wall at a rate of ft/sec an tat we are intereste in wat as appene after secons. We know tat, So, te en of te laer as been puse in 3 feet an so after secons we must ave coul ave compute tis in one step as follows,. Note tat we To fin y (after secons) all tat we nee to o is reuse te Pytagorean Teorem wit te values of tat we just foun above. Now all tat we nee to o is plug into our previous ifferentiate equation an solve for. Notice tat we got te correct sign for. If we gotten a negative ten we ave known tat we a mae a mistake an we coul go back an look for it..8 Linear approimation an Linearization Linear approimation elps us estimate values of functions tat cannot easily be computere. For eample, know 5 5, so 5. soul be a little more tan 5, but by ow muc? 5.. We f( A) f( X) + f '( X)( A X) In tis formula, te function is te operation being one to te value we cannot fin a value for. X is te value we want it to be an A is te value tat we actually ave to eal wit. Eample: Approimate te value of 5. Let f() because te operation we re ealing wit on 5. is taking te square root. Ten we can also conclue tat f '( ) Let A 5 because tat s te value we want it to be to make it easier

12 Let X 5. because tat s te value we ave to actually eal wit. Now time to plug in to te equation for linear approimation. f(5.) f(5) + f '(5)(5. 5) (.) (5) Now lets say we want to generalize a formula to approimate all values aroun a certain point so we won t ave to keep using te formula LX ( ) f( A) + f'( A)( X A), were X stays X, an A is te value you want te function to be. Eample: If f( ) + an A, use linearization to approimate 3.8, 3.9, 4 So A because tat will be te value we want to plug into te equation to make it can eal wit comfortably. If f( ) + ten f '( ) + 4 so ten it ll equal, a value we LX ( ) f() + f'()( X ) LX ( ) + + ( X ) + LX ( ) + ( X ) 4 LX ( ) + X 4 3 LX ( ) + X 4 3 So now we ave te function LX ( ) + X were we can plug in a value base on te A value. So if A, we re 4 using tis approimation base on 4, so if we calculate 3.8, tat will be.8 away from te A value, so we plug in 3.8 into X an L (.8) + (.8).95, but an I mean ey, tat s a really close estimate! 4 Now if you wante to try 3.9, it will be.9 away from te A value. So plug in.9 into X 3 L (.9) + (.9).975, but Last but not least, try 4, wic soul be, but lets ceck. 3 It is away from your A value, so let L () + ( ) SUCCESS!!!! It matces! 4