Time (hours) Morphine sulfate (mg)

Transcription

1 Mat Xa Fall 2002 Review Notes Limits and Definition of Derivative Important Information: 1 According to te most recent information from te Registrar, te Xa final exam will be eld from 9:15 am to 12:15 pm on Monday, January 13 in Science Center Lecture Hall D 2 Te test will include twelve problems (eac wit multiple parts) 3 You will ave 3 ours to complete te test 4 You may use your calculator and one page (8 by 115 ) of notes on te test 5 I ave cosen tese problems because I tink tat tey are representative of many of te matematical concepts tat we ave studied Tere is no guarantee tat te problems tat appear on te test will resemble tese problems in any way watsoever 6 Remember: On exams, you will ave to supply evidence for your conclusions, and explain wy your answers are appropriate 7 Good sources of elp: Section leaders office ours (posted on Xa web site) Mat Question Center (during te reading period) Course-wide review on Friday 1/10 from 4:00-6:00 pm in Science Center E and Sunday 1/12 from 3:00-5:00 pm in Science Center A 1 Left Hand, Rigt Hand and Overall Limits 11 Te Idea of a Limiting Value Te manufacturers of te popular morpine solution Roxanol suggest a dose of 20ml initially, followed by 20ml every four ours Te grap sows te amount of morpine sulfate tat a patient receiving tis treatment would ave in teir bloodstream as a function of time Te letal dose of Roxanol for te average person is about 180 ml Is a patient wo receives te suggested dose of Roxanol in any danger? Solution To try to answer tis question, we can try to work out te amount of Roxanol present in te patient s body as time goes on Te numbers are sown in te following table Time (ours) Morpine sulfate (mg)

2 Some points to observe from tis situation are: As time gets larger and larger, te amount of morpine sulfate gets closer and closer to 40 Te amount of morpine sulfate never actually reaces 40, but if you keep going long enoug, you can get te amount of morpine sulfate to be as close to 40 as you like Te value of 40 is te limit or limiting value of te amount of morpine sulfate in te patient s blood stream Te limiting value is like te value of te dependent variable tat you d get if it were someow possible to keep going and going for an infinite amount of time (Tis is a limit or limiting value tat occurs wen te independent variable (in tis case, time) keeps going and going towards infinity) If time is given by t and te amount of Roxanol by M(t) ten te symbols tat are normally used to describe tis situation are: Lim t M(t) = Left Hand and Rigt Hand Limits Te morpine sulfate grap given in Section 11 was not (strictly speaking) complete as it did not specify exactly wat was appening in te places were te grap suddenly jumped up A more complete version of tat grap is sown below 1 1 I ave assumed tat te injection is always given exactly at te correct time if you know about te pysiology of drug absorption into te body you could probably successfully argue tat te pattern of open and filled in endpoints sould be reversed on te grap

3 Te left and limit is te eigt on te grap tat you expect to get to wen you flow along te grap, eading for te point tat you are interested in from te left side For example, if you flowed along te morpine sulfate grap towards t = 4 from te left side, you would expect to reac a eigt of 10 as you got closer and closer to t = 4 In matematical symbols tis left and limit is written: Lim t 4 M(t) = 10 Te - superscript tat appears on te 4 sows tat tis is a left and limit Te rigt and limit is te eigt on te grap tat you expect to reac wen you flow along te grap, approacing te point tat you are interested in from te rigt side For example, if you flowed along te morpine sulfate grap towards t = 4 from te rigt side, you would expect to reac a eigt of 30 as you got closer and closer to t = 4 In matematical symbols tis rigt and limit is written: Lim t 4 + M(t) = 30 Te + superscript tat appears on te 4 sows tat tis is a rigt and limit 13 Overall Limits In situations were you are looking at te left and rigt and limits of a function at a finite x-value, tere are essentially two tings tat can appen: Te left and limit equals te rigt and limit, or, Te left and limits does not equal te rigt and limit In te case were te left and limit and te rigt and limit bot exist, ten normally one simply says tat te limit of te function exists at te point in question, and tat te value of te limit is te same as te common value of te left and rigt and limits Te complete list of all situations tat could possibly occur wen looking at left and and rigt and limits (and trying to interpret tem) is given in te following table Left and Limit Rigt and Limit Interpretation for Limit L ± Does not exist ± L Does not exist L M Does not exist L L Limit exists and is equal to L + + Limit exists and is equal to + Limit exists and is equal to + Does not exist + Does not exist Were L and M are finite and L M

4 2 Calculating Limits Using te Algebraic Structure of a Function Generally speaking, wen te denominator of a function approaces zero, te function as no value and te grap of te function as a vertical asymptote Let s ceck te grap of te function f( x) = x3 x 2 + x 1 x 1 Near te point x = 1 If you grap tis on your calculator, te following window settings will display te part of te grap tat we are interested in xmin = 0 xmax = 2 ymin = 1 ymax = 3 If you enter te formula for f(x) correctly into your calculator, you sould see someting tat looks like a parabola wit a point missing in te center Tis is quite a bit different from te vertical asymptote tat you migt ave expected to see Te function f(x) is undefined at x = 1 but tere was no vertical asymptote on te grap of y = f(x) (Te reason te point is missing in te center is because x = 1 is not in te domain of te function f so te calculator does not evaluate te function tere) Wat is it about te function f( x) = x3 x 2 + x 1 x 1 tat allows it to be undefined at x = 1 and allows it to avoid aving a vertical asymptote at x = 1? Observe tat: x 3 x 2 + x 1 = (x 1) (x 2 + 1) You migt be tempted to go aead and write someting along te following lines: f( x)= x 3 x 2 + x 1 ( = x 1 ) x 2 +1 = x 2 +1, x 1 x 1 and from tis conclude (incorrectly) tat f(1) = 2 Tis simplification is valid at all point except x = 1, wic (unfortunately) is te point we are interested in So, wen x is close to but not equal to one, we ave tat f(x) = x Tis means tat wen x is close to one, f(x) will be close to = 2 Te main ting tat saves te grap of f(x) from aving a vertical asymptote at x = 1 is te fact tat te denominator, x 1, wic approaces zero as x 1, is perfectly balanced by a twin factor of (x 1) in te numerator wic approaces zero in exactly te same way as te denominator does wen x 1 If tere were no perfect twin factor on te top line of te function to balance te x 1 in te denominator ten te grap of f(x) would ave ad a vertical asymptote at x = 1 21 Calculating Limits for Functions Defined by Equations In addition to finding te values of limits by examining graps and performing numerical calculations, you can often deduce te value of a limit by analyzing te symbolic representations of te function We will concentrate on te two kinds of limits: limits as x ± and limits at finite values of x Te reasoning processes for analyzing te beavior of te function in tese two kinds of limit situation are demonstrated troug te following two examples:

5 211 Example: A Limit as x ± Calculate te limit of f(x) = 1 (05) x as x ± Solution Limit of f(x) = 1 (05) x as x + Te exponential function: y = (05) x is a decreasing function As x gets really, really big tis exponential gets closer and closer to zero So, f(x) = 1 (05) x gets closer and closer to 1-0 = 1 Limit of f(x) = 1 (05) x as x - Te exponential function: y = (05) x is a decreasing function Terefore, as x gets larger, (05) x gets smaller Conversely, wen x gets more and more negative, (05) x gets larger and larger So, as x -, (05) x will get very large and f(x) = 1 (05) x will resemble 1 - (very large number), so te limit will be Example: Limits at Finite x-values Calculate te limit of g( x) = x +1 x 2 1 as: (I) x 1 + (II) x 1 - (III) x 1 Solution Te limit of g( x) = x +1 x 2 1 as x 1+ Wen x is close to 1, te numerator of g(x) is basically equal to 2, and te denominator is very close to zero If you are approacing from te rigt, ten x > 1, so te numerator x 2 1 > 0 Tis means tat wen x is sligtly larger tan 1, gx Te limit of g( x) = x +1 x = very_l arg e _ positive_ number very _ small_ positive _ number as x 1- Wen x is close to 1, te numerator of g(x) is basically equal to 2, and te denominator is very close to zero If you are approacing from te left, ten x < 1, so te numerator x 2 1 < 0 Tis means tat wen x is sligtly less tan 1, gx 2 very _ small_ negative_ number = very_ larg e _ negative_ number

6 Te limit of g( x) = x +1 x 2 1 as x -1 Wen x is close to 1, bot numerator and denominator are very close to zero Tis is a very confusing situation, and some care and finesse are required to make sense of it Note tat te polynomial in te denominator can be factored to give: gx = x +1 ( x +1) ( x 1) = x + 1 x x 1 x +1 x wen x is very close to 1 It is te presence of te factors of (x + 1) in bot te numerator and denominator tat prevent g(x) from aving a vertical asymptote at x 1, as te very tiny value in te denominator is precisely balanced by and equal, tiny value in te numerator Wen tese two tiny values ave balanced, all tat remains is -1/2 3 Te Limit Definition of te Derivative 31 Te Tangent Line as te Limit of Secant Lines Te instantaneous rate of cange of a function is te same as te slope of te tangent line to te grap of te function at watever point you are interested in If you find secant lines over smaller and smaller intervals, ten wen te intervals become very sort, te secant line starts to resemble a tangent line Tangent line Secant lines Te slope of te secant line joining te points (a, f(a)) and (b, f(b)) is given by Slope = fb f a b a

7 Wat we are doing as we bring te secant line closer and closer to te tangent line is keeping te point (a, f(a)) fixed and bringing te oter point (b, f(b)) closer in As you can see from te diagram, as te point (b, f(b)) is brougt closer and closer to (a, f(a)) te slope of te secant line gets closer and closer to te slope of te tangent line To obtain te slope of te tangent line, you can calculate te limit of: Slope = fb f a b a as b a Te numerical value of tis limit is called te derivative of te function f(x) at te point x = a 32 A Numerical Example of Calculating te Derivative Te objective of tis example will be to calculate te derivative of te function f(x) = x 2 at te point x = 2 Te number tat we get will be te slope of te tangent line tat just touces te grap of y = x 2 at te point were x = 2 Tangent line Solution First of all, let s try just doing some average rates of cange over very small intervals near x = 2 to see if we can guess te slope of te tangent line from te entries in te table Two points for secant line Average rate of cange of f(x) = x 2 X = 2 to x = X = 2 to x = X = 2 to x = X = 2 to x =

8 Te average rate of cange seems to get closer and closer to a value of 4 as we make te interval sorter and sorter Terefore, te limit of te average rate of cange (as te distance between te two points tat form te secant line srinks down to zero) will be equal to 4 Te slope of te tangent line to te function f(x) = x 2 at te point x = 2 is, terefore, 4 You can also reac tis conclusion by setting up a difference quotient for te slope of te secant line and reading te algebra Te slope of Tangent Line at x = 2 is given by: Lim 0 f ( 2 + ) f 2 () Let s write out wat te difference quotient is, and try to simplify it algebraically as muc as possible Difference Quotient = f ( 2 + ) f 2 () ( = 2 + ) = = As 0, te limit of te difference quotient will be: = ( 4 + ) 4 So, te slope of te tangent line to te function f(x) = x 2 at te point x = 2 is equal to 4 33 Calculating a Formula for te Derivative Using Limits Te derivative tat is, te slope of te tangent line of a function f(x) at a general point x is given by: f ( x)= Lim 0 Calculate a formula for te derivative of te function f(x) = x 2 f( x+ ) f x Solution Te difference quotient for te function f(x) = x 2 at a general point x will be given by: f( x+ ) f x Te limit of tis as 0 will be: ( = x + ) 2 x 2 = x x + 2 x 2 = 2 x x + 2 = ( 2 x + ) 2 x So, a formula for te derivative of f(x) is: f ( x)= 2 x

9 Note tat if you plug x = 2 into tis formula, you get tat te derivative wen x = 2 is four (just as in te previous example) 34 Summary of te Procedure for Calculating a Formula for a Derivative Using Limits Te calculation of te derivative f ( x) can be broken down into four distinct stages Stage 1: Formulate f(x + ) Stage 2: Formulate te difference quotient: f( x + ) f x Stage 3: Simplify te difference quotient as muc as possible Stage 4: Take te limit as 0 Te result tat you get will be a formula for te derivative of te function f(x): f ( x) = Lim 0 f( x+ ) f( x) Note tat te only variable in your final derivative formula sould be x (So, tere sould be no s wen you are finised) Particular trouble spots include: Formulating f(x + ) correctly Sometimes people forget tat te + is part of te input tat goes into te function and incorrectly write down an algebraic expression tat puts te + on te outside of te function, ie tey incorrectly write down f(x) + Expanding brackets Sometimes people forget to multiply brackets by FOILing and incorrectly write tings like: (x + a) 2 = x 2 + a 2 Combining fractions Remember to cross-multiply Dividing by correctly Remember tat you can only divide by wen you ave factored out of everyting tat is in te numerator of te difference quotient So pay lots of attention to wat you are doing wenever you ave to do any of tese operations 35 Anoter Example: Calculating a Formula for te Derivative Using Limits Use te limit definition of te derivative to find an equation for te derivative of: f( x)= 1 x

10 Solution Step 1: Formulate f(x + ) f( x+ )= 1 x + Step 2: Formulate te difference quotient: f( x + ) f x f( x+ ) f x = 1 x + 1 x Step 3: Simplify te difference quotient until you can cancel out te tat is present in te denominator 1 x + 1 x = x x + x x + = x x x ( x + ) = x ( x + ) Step 4: Take te limit as 0 f ( x)= Lim 0 x ( x + ) = Lim 0 1 x ( x + ) = 1 x 2 4 Sketcing te Grap of te Derivative Function Te derivative of a function is also a function in its own rigt, and so it is possible to draw te grap of te derivative function Te guiding principle for ow to produce te grap of te derivative is as follows: Te eigts on te derivative grap are te slopes of tangent lines to te original function grap Some oter observations tat can elp to guide your efforts to sketc a grap of te derivative are listed below 1 Were will te slope of te tangent line be zero on te original function grap? Tese points will be te x-intercepts of te derivative grap

11 2 Were will te slope of te tangent line be positive on te original function grap? Tese are te intervals were te eigt of te derivative grap will be positive 3 Were will te slope of te tangent line be negative on te original function grap? Tese are te intervals were te eigt of te derivative grap will be negative Tree oter points tat are wort noting are: Wen te grap of te original function is concave up, te derivative grap is increasing (and vice versa) Wen te grap of te original function is concave down, te derivative grap is decreasing (and vice versa) (As always!) Te eigts on te derivative grap are given by te slopes of tangent lines to te original function grap 41 Example: Sketcing Graps of a Function and Its Derivative Sketc a grap of te function: f (x) = 1 3 x 3 x + 2 and use tis original function grap to sketc a grap of te derivative Solution Te function and its derivative are sown on te next page Note tat te formula for te derivative of f(x) is: f (x) = x 2 1 Tis is te equation for a quadratic function (wic sould ave a parabola for a grap) Te derivative grap sown on te next page does indeed resemble a parabola, wic provides some additional confirmation tat te derivative grap as been drawn correctly

12 ORIGINAL FUNCTION y = f(x) DERIVATIVE y = f (x) 5 A Practical Interpretation of te Derivative One interpretation tat can be attaced to te numerical value of te derivative at x = a, tat is to te numerical value of f ( a), is tis: is te approximate amount tat te value Te numerical value of f a of te function, f(a), canges by wen x is increased by exactly one unit from x = a to x = a + 1 For example, if x was te distance from te source of te Mississippi river (in units of miles) and f(x) was te eigt above sea level (in feet), ten te meaning of: f 1000 = 200 would be tat wen you are 1000 miles from te course of te Mississippi, your eigt above sea level is 200 feet

13 = 3 would be tat wen you increase your distance from te source of te f 1000 Mississippi from 1000 miles to 1001 miles, your eigt above sea level will drop by approximately 3 feet Tis practical interpretation for te value of te derivative can be justified grapically Heigt above sea level Rise of tangent line Exact cange in value of function wen you increase by one, from 1000 to Tangent line Distance from source of river Since te run is equal to one, te rise of te tangent line is equal to te value of te derivative at x = 1000 Te rise of te tangent line between x = 1000 and x = 1001 is approximately equal to te exact cange of te value of te function between x = 1000 to x = 1001 Hence, te value of te derivative at x = 1000 is approximately equal to te cange in te value of te function wen you increase x by one unit from x = 1000 to x = Te Second Derivative Te second derivative is a matematical tool tat enables us to easily determine te concavity of te original function witout aving to grap it If f(x) is te function, ten is te second derivative f x is te derivative and Te second derivative of a function f(x) is te derivative of te derivative f ( x) 61 Examples: Calculating Derivatives and Second Derivatives (a) f(x) = x 3 + x 2 FUNCTION Ten: f ( x) = 3x 2 + 2x DERIVATIVE And: f x = 6x + 2 2nd DERIVATIVE f x

14 (b) f( x)= 4 7 x + x 4 + ln( x) FUNCTION Ten: f ( x)= ln( 7) 4 7 x + 4 x x And: f ( x)= ln( 7) ln( 7) 4 7 x +12 x x 2 DERIVATIVE 2nd DERIVATIVE 7 Te Connections Between te Original Function, First Derivative and Second Derivative 71 Relationsip between a function and its derivative 711 Maximum and minimum values of te original function Tese usually occur wen te derivative is equal to zero At a maximum (te top of a ill) te grap must go from an increasing grap to a decreasing grap At te point were te cange-over appens, te tangent line must be orizontal and te derivative equal to zero At a minimum (te bottom of a valley) te grap must go from a decreasing grap to an increasing grap At te point were te cange-over appens, te tangent line must be orizontal and te derivative equal to zero 712 Places were te original function is increasing and decreasing Beavior of original function Sign of derivative Increasing + Decreasing

15 713 Places were te original function is concave up and concave down Beavior of original function Concave up Concave down Beavior of derivative Increasing (may be + or, all tat is important is tat te derivative is increasing) Decreasing (may be + or, all tat is important is tat te derivative is decreasing) 72 Using te First Derivative to Classify Critical Points as Maximums or Minimums A critical point is a point at wic eiter te derivative is equal to zero, or at wic te derivative is undefined (for example, at a point were te grap of te original function as a sarp corner) To ceck te type (maximum or minimum) of critical point, you can evaluate te derivative on te left and on te rigt of te critical point Sign of derivative just to te left of te critical point Sign of derivative just to te rigt of te critical point Type of critical point + Maximum + Minimum Te reason tat examining te signs of te derivative on eiter side of a critical point can tell you about te nature of te point (ie weter it is a maximum or minimum) is indicated below MAXIMUM + Function is increasing Derivative is positive Function is decreasing Derivative is negative Function is decreasing Derivative is negative Function is increasing Derivative is positive + MINIMUM

16 73 Te Sign of te Second Derivative and te Concavity of te Original Function Te sign of te 2nd derivative tells you about te concavity of te original function Since te second derivative is te derivative of te derivative, wen te original function is concave down, te derivative is decreasing, and wen te derivative is decreasing, te derivative of te derivative is negative Overall, wen te original function is concave down te second derivative is negative On te oter and, wen te original function is concave up, te derivative is increasing Wen te derivative is increasing, te derivative of te derivative is positive Overall, wen te original function is concave up, te second derivative is positive Second Derivative is Derivative is Original function is Positive Increasing Concave up Negative Decreasing Concave down Zero Neiter increasing nor decreasing Possibly at a point of inflection If te second derivative is equal to zero at a point, ten te original function often as a point of inflection tere - but not always (NOTE: g(x) = x 4 is a good counter-example) 74 Te Second Derivative and Classifying Critical Points Near a local maximum, te original function is concave down So, near a local maximum te second derivative is negative Near a local minimum, te original function is concave up So, near a local minimum te second derivative is positive Type of Critical Point Concavity of Original Sign of Second Derivative Function Local maximum Concave down Local minimum Concave up +