AP CALCULUS Curriculum Module: Extrema

8/9/2019 AP CALCULUS Curriculum Module: Extrema

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Extrema:

A Curriculum Modulefor AP Calculus

2010Curriculum Module


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Extrema

Contents

AP Calculus Extrema .......................................................................................... 4

Worksheet 1: Classifying Critical Points, Part I...................................... 8

Worksheet 2: Classifying Critical Points, Part II ................................... 10

Teaching Examples for Extrema ................................................................... 13

Worksheet 3: Extrema in a Variety of Settings ...................................... 18

About the Author .................................................................................................. 23


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2010 Te College Board.

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AP Calculus Extrema

Dixie RossPfugerville High SchoolPfugerville, Texas

In 1995 when graphing calculators were frst allowed on the AP Calculus Exams, Iremember thinking, Well, there go all the good extrema problems. Te kids will just use

the max/min eature o their calculators and wont have to know any calculus at all. Tatthought seems ridiculous in retrospect, as extrema problems are alive and well and muchmore interesting than they have ever been beore. Te AP Calculus Exam uses multiplerepresentations o unctions to test students understanding o extrema in so manydierent ways now, some calculator active and some inactive.

I am convinced that students comprehend the topic at greater depth than everbeore, rather than simply applying a rote process (set derivative equal to zero andsolve) to problems similar to ones they have solved previously. Te creativity o the APCalculus Development Committee continues to amaze me as they write problems I wouldnever have considered and have certainly never seen in a textbook. As these problems

have evolved, I have tried to change my teaching in various ways to keep pace. I used tothink o the topic o extrema as a ew sections o my textbook. Now I see it as one o themajor themes o the course, permeating every topic that I teach. When I teach motion,I think, How can extrema questions be asked in this context? When I teach f, f , f relationships, I think, How can extrema questions be asked in this context? My thoughtsare the same with unctions defned by integrals, dierential equations, etc.

Because this topic is so important, it is requently discussed on the AP CalculusLISSERV (electronic discussion group, or EDG), or which you can sign up throughAP Central (http://apcentral.collegeboard.com/apc/public/homepage/7173.html). Somequestions come up repeatedly as new teachers struggle with the concepts surrounding

extrema. One such question is: Can an endpoint be considered a relative extreme point?Te answer depends on the defnition o relative extreme being used, which varies romtextbook to textbook. Tereore, the AP Calculus Development Committee careully crasthe questions or the AP Exam to avoid this issue entirely. Its oen a good idea to search thearchives o the EDG or answers to these requently discussed issues. Tere is also an excellentarticle on justifcation o extrema, On the Role o Sign Charts in AP Calculus Exams, whichcan be accessed through AP Central (http://apcentral.collegeboard.com/apc/members/courses/teachers_corner/36693.html). Every teacher o AP Calculus should read this article.



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Extrema

AP Exam Questions

An inspection o recently released AP Calculus Exam ree-response questions provides anexcellent overview o the rich variety o ways that students understanding o extrema can

be assessed. All o these questions, along with solutions and student samples, are availableon AP Central (AP Calculus AB items are ound at http://apcentral.collegeboard.com/apc/members/exam/exam_questions/1997.html; AP Calculus BC items are ound at http://apcentral.collegeboard.com/apc/members/exam/exam_questions/8031.html).

2008 Exam

3b and 3c (AB only): Accumulating rates o change, determine the location andvalue o the maximum volume o an oil slick on an interval o infnite length.

4a (Motion): Find the time that the particle is arthest le (minimum position),given velocity defned graphically.

5a (BC only): Given an analytic defnition o , determine the type o relativeextreme.

6b (AB only): Given analytic defnitions ofand , determine the location andtype o relative extreme.

2008 Form B Exam

4c (AB only): Determine the location and value o maximum value o compositiono two unctions, one defned by an integral.

5b: Given the graph o , fnd the absolute maximum value ogon a closedinterval.

2007 Exam

2c: Accumulating rates o change, determine the location and value o maximumon a closed interval.

4a (Motion): Find the time at which the particle is arthest le (minimumposition).

6b (Analytic): For a given critical point, determine the type o relative extremeusing second derivative test.

2007 Form B Exam:

2d (Motion): Find the time that the particle is arthest right (maximum position),given velocity defned analytically and graphically.


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4a and 4d: Given the graph o , determinex-value o the relative maximum andminimum values of(x).

5c (Dierential equation): For a given point, determine the type o relative extremeusing the second derivative test.

2006 Exam

3b: Function defned by integral, determine the type o relative extreme o aparticularx-value.

2006 Form B Exam

2b: Given the graph of9, determine thex-value o absolute maximum.

4b: Determine the time at which fis increasing at its greatest rate.

2005 Exam

2d: Accumulating rates o change, determine the time and value when the amounto sand is at its minimum.

4a and 4c: Given the table o inormation on f, f , f , determine the locationsand types o relative extrema; given a unction defned by integral, do the same.

2005 Form B Exam

2c: Accumulating rates o change, determine the time at which the amount o

water is at an absolute minimum.

2004 Exam

3c: Given velocity defned analytically, determine the maximum position.

4c: Given an implicitly defned unction, determine the type o relative extremeusing the second derivative test.

5b and 5c: Given a unction defned by integral, determinex-values o relativemaximum and absolute minimum values o the unction.

2004 Form B Exam

2d: Given a rate o change, determine the maximum number o mosquitoes.

4b: Given a graph o , determinex-values o absolute minimum and maximum.

6c: Determine the value on that will maximize the area o region S.



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Extrema

Student Worksheets and Teaching Examples

On the ollowing pages, the frst two student worksheets can be used any time aerstudents have learned the closed interval test (or candidates test), the frst derivative test

and the second derivative test or extrema. I would not use them back-to-back, but wouldspace them apart by a ew weeks. Te teaching examples and third student worksheet canbe used either as each o the relevant topics is covered or aer all relevant topics have beencovered and the students are reviewing or the AP Exam.


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Worksheet 1: Classifying Critical Points, Part I

Complete each statement by choosing one o the our phrases rom the box below. Phrasesmay be used more than once. Unless otherwise specifed, assume each unction is defnedand continuous or all real numbers.

the absolute (global) maximum the absolute (global) minimum

a local (relative) maximum a local (relative) minimum

1. A unctionfdefned and continuous on the interval22# x#5 has critical points(or critical numbers) only atx521 andx52. Te unctionfhas values as givenin the table below.

x f(x)

22 1

21 2

2 0

5 2

Te valuex52 locates ___________________ value o the unction. Te value

f(x)52 is ____________________ value o the unction.

2. Ix52 is the only critical point o a unctionfand , thenx52 locates

__________________ value o the unction.

3. I and changes rom negative to positive atx52, thenx52locates

____________________ value o the unctionf.

4. I and , thenx52 locates _____________________ value othe unctionf.

5. Ix52 is a critical point o the unctionf, and decreases throughx52,thenx52 locates _________________________ value o the unction.

6. I a continuous unctionfincreases throughout a closed interval, then the leendpoint o the graph ofon the interval is _________________________ pointo the unction.

7. A student ound the critical points o a unction fto bex52 andx54, and



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Extrema

produced the chart below.

Interval x < 2 2 4

positive negative positive

Te valuex52 locates ___________________________ value o the unction.

8. Ix52 is the only critical point o a unctionfand , thenx52 locates a____________________ value o the unction.

Worksheet 1: Solutions

Note that critical points also are reerred to in some texts as critical numbers or criticalvalues. Tey are values ox at which a unctionfsatisfes or at which is notdefned (does not exist).

1. the absolute (global) minimum, the absolute (global) maximumOne also can say, Te valuex52 locates, or is thex-coordinate o, the absolute(global) minimum point o the graph o the unction. Note that the graph ofhastwo absolute (global) maximum points, and , but just one absolute(global) maximum value, .

2. the absolute (global) maximum

3. a local (relative) minimum

4. a local (relative) minimum

5. a local (relative) maximum

6. the absolute (global) minimumOne also can say, Te le endpoint o the interval locates the absolute (global)minimum value o the unction, or Te le endpoint o the interval locates, or isthex-coordinate o, the absolute minimum point o the graph o the unction.

7. a local (relative) maximum

8. the absolute (global) minimum


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Worksheet 2: Classifying Critical Points, Part II

Determine i each o the ollowing statements is true or alse. I you decide a statementis alse, provide a counterexample to show why it is alse and then rewrite the statementin order to make it true. Unless otherwise specifed, assume each unction is defned and

continuous or all real numbers.

1. A critical point (or critical number) o a unction fo a variable x is thex-coordinate o a relative maximum or minimum value o the unction.

2. A continuous unction on a closed interval can have only one maximum value.

3. I is always positive, then the unction fmust have a relative minimumvalue.

4. I a unction fhas a local minimum value at x 5 c, then .

5. I and , then x52 locates a relative maximum value of.

6. I , then x5c is a point o inection or the unction fand cannot be thex-coordinate o a maximum or minimum point on the graph of.

7. I a unction fis defned on a closed interval and or all x in the interval,then the absolute maximum value o the unction will occur at the right endpointo the interval.

8. Te absolute minimum value o a continuous unction on a closed interval canoccur at only one point.

9. Ix52 is the only critical point o a unction fand , then is the

minimum value o the unction.

10. o locate the absolute extrema o a continuous unction on a closed interval, youneed only compare they-values o all critical points.

11. I and decreases through x 5c, then x 5c locates a localminimum value or the unction.

12. Absolute extrema o a continuous unction on a closed interval can occur only atendpoints or critical points.



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Extrema

Worksheet 2: Solutions and Teacher Notes

Note to instructors: Tis is a great worksheet to have students work on in pairs. Tediscussions can get quite heated! Tere is certainly more than one counterexample to eachalse statement and oen more than one way to rewrite the alse statements; the solutions

below provide you with one idea or each. Students might have other ideas that make orgood discussion items. o make the work go aster, I oen require students only to providea counterexample or to rewrite the alse statement so that it is true.

1. False

Counterexample: Note that x 50 is a critical point or the unction ,but that x50 corresponds to neither a relative maximum nor a relative minimumvalue of.

Rewrite: A critical point is a POSSIBLE location or a relative maximum orminimum value o a unction.

2. True. Note that the maximum value can occur at more than onex-value but thatthe maximum value itsel is unique. See Problem 8 or an example.

3. False

Counterexample: For , is always positive, but the unctionhas no relative extrema.

Rewrite: I and is always positive, then the unction must have arelative minimum at x5c.

4. False

Counterexample: Te unction has a local minimum at x 50, butis not defned.

Rewrite: I a DIFFERENIABLE unction has a local minimum value at x 5c,then .

5. True

6. False

Counterexample: For the unction , , but x50 is not a point

o inection. Note thatx5

0

does correspond to a relative and absolute minimumvalue of.

Rewrite: I or a unctionf, then x5c may or may not be an inectionpoint or fand x5c may or may not correspond to a relative minimum ormaximum value of.


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7. True

8. False

Counterexample: Te unction takes on its minimum value o21 at

the points in the interval 0, x , 6.

Rewrite: Tere is exactly one absolute minimum value o a continuous unctionon a closed interval, but this minimum value can occur at more than one point inthe interval. See Problem 2.

9. True

10. False

Counterexample: Te unction on the interval 2# x# 6 has itsabsolute minimum value at x52 and its absolute maximum value at x5 6.

Neither x5

2 norx5

6 is a critical point o the unction. Rewrite: o locate the absolute extrema o a continuous unction on a closed

interval, you must compare they-values o all critical points AND ENDPOINS.

11. False

Counterexample: For , , decreases throughx50, and fhas a local (and global) maximum value at x50.

Rewrite: I and decreases through x5c, then x5c locates alocal MAXIMUM value or the unction. Or, i and INCREASESthrough x5c, then x5c locates a local minimum value or the unction.

12. True



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Extrema

Teaching Examples for Extrema

Example 1 (noncalculator): Given the unctionf(x)52 x 2 e12x on the closed interval[0,3], convince me in three dierent ways that the maximum value of(x) occurs at x 51.

Solution and explanation:

Method 1: Closed Interval (or Candidates) TestSincef9(x) 5211e12x only or x51, then x51 is the only critical point (or criticalnumber) orf.

x f(x)

0 2e

1 22

3 2321/e2

Since fis continuous on the interval [0,3] andf(1)522 is greater than the values of(x) at either endpoint, the absolute maximum occurs at x51. Note that the absolutemaximum value isf(1)522 and the absolute maximum point is (1,22).

Method 2: First Derivative Test

Since = + f x e x( ) 1 1 only or x 51, then x 51 is the only critical point orf.

Interval 0 x, 1 1,x 3

positive negativef(x) increasing decreasing

Since x51 is the only critical point in the interval, fis continuous at x51, andgoes rom positive to negative at x51, then the unction fmust go rom increasing todecreasing there, thus establishing an absolute maximum at x 51.

Note that the First Derivative est (normally a test or relative, or local, extrema) works herebecause x51 is the only critical point in the interval, so the unction fcan change romincreasing to decreasing or rom decreasing to increasing at most once in the interval. Wealso could argue that since is positive or 0 x,1 and negative or 1, x 3, thenf(x) increases or 0, x,1 and decreases or 1, x#3. Since fis continuous at x51, this

behavior guarantees that fhas an absolute maximum value at x51.Method 3: Second Derivative Test

Since = + f x e x( ) 1 1 =0 only or x 51, then x 51 is the only critical point orf. Since

= f x e x( ) 1 , then . Since x 51 is the only critical point, fis continuous

near x51, and is negative, we have established an absolute maximum atx51.

Note that the Second Derivative est (normally a test or relative, or local, extrema) works


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here because x51 is the only critical point in the interval, so the unction fcan changerom increasing to decreasing or rom decreasing to increasing at most once in the interval.We also could argue that since is negative or 0# x#3 and , then the graphof(x) is concave down throughout the interval with a horizontal tangent at the point (1,22). Tis behavior guarantees that fhas an absolute maximum value o22 at x51.

Note to instructors: By doing the same problem using three dierent methods, students candiscuss the relative merits o each method and the importance o particular phrasing, suchas speciying in the frst and second derivative tests that x51 is the ONLY critical point.

Example 2 (graphical, noncalculator): Te graph o , shown below on the interval[0,10], consists o a semicircle o radius 2 and a line segment meeting at the point ( 4,0).Te line segment has right endpoint (10,2).

Graph o

a. For what value ox, 0, x ,10, does fhave a local minimum? Justiy your answer.

b. For what value ox, 0# x#10, does fhave an absolute maximum? Justiy youranswer.


a. fhas a local minimum at x54 since goes rom negative (graph o belowthex-axis) to positive (graph o above thex-axis) at that point, indicating thatfgoes rom decreasing to increasing at x54, thus establishing a local minimumthere.

b. Tere are no values ox or which goes rom positive to negative. Tis meansthere are no internal candidates or the absolute maximum, hence the maximummust occur at one o the endpoints. Since the amount o decrease (representedby the area below thex-axis or 2) is greater than the amount o increase

(represented by the area above thex-axis or 6), the absolute maximum value ofmust be at the le endpoint x 50. More specifcally, supposef(0)5y

0. Ten

f(10)5y022 + 6,y

0. Hence fhas its absolute maximum value at x50. Note

also thatf(4)5y022. Sincef(4)5y

022 is smaller thanf(0)5y

0and smaller

thanf(10)5y022 + 6, then fhas its absolute minimum value at x54.



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Extrema

Example 3 (accumulating rates o change, calculator active): Te rate, in gallonsper hour, at which water ows into a tank over the time interval 0#t#3, where tismeasured in hours, is given byR(t)5t2 sin(t) and the rate, in gallons per hour, at which

water ows out o the tank over the same time interval is given by . Te tank

initially holds 10 gallons o water.

a. When is the rate at which water ows into the tank the greatest? Justiy youranswer.

b. Write an expression that represents the amountA(t) o water in the tank at anygiven time t.

c. Determine when there is the least amount o water in the tank. How much water isin the tank at this time? Show the analysis that leads to your conclusion.


a. o fnd the value otor which the rate at which water ows into the tankis greatest, we seek the maximum value oR(t). Start by setting :

. Solving on our calculator, we see that when t50 or when t52.289. We can now choose among three methods o justifcation(closed interval test, frst derivative test or second derivative test). You can havedierent groups o students use each method and then have them display theirwork or the rest o the class to discuss and critique. I choose to use the secondderivative test since it gives me the opportunity to review the keystrokes orNDeriv. Since t52.289 is the only critical point in the interior o the interval,

, and , which is negative, we have established anabsolute maximum orR at t52.289.

b. , 0#t#3.

c. Te least amount o water means the absolute minimum value oA(t). Start byusing the Second Fundamental Teorem o Calculus to fnd

. Set and solve using the calculator to fnd

t50.867 hours. Again, we have three possible methods o justifcation. Sincet50.867 is our only critical point and goes rom negative to positive there,

we have established an absolute minimum at t50.867 hours. Finally, evaluateA(0.867) using the FnInt eature o the calculator to fnd that there were 9.201gallons o water in the tank at that time.


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Example 4 (motion, noncalculator): Te velocity o a particle moving on thex-axis isgiven byv(t)5t32 6t2 or the time interval 0 t 10.

a. When is the particle arthest to the le?

b. When is the velocity o the particle increasing the astest?Solution and explanation:

a. Te phrase arthest to the le tells us we are looking or the minimum position.o fnd minimum position, we start by seeing where the derivative o position(which is velocity) equals zero or ails to exist: or t50 ort5 6.

ime interval 0 ,t, 6 6,t# 10

Sign o velocity negative positive

Direction o movement le right

Since the particle moves to the le or 0,t, 6 (indicated by negative velocity)and moves to the right or 6 ,t 10, it is arthest to the le at t5 6.

b. o fnd where velocity is increasing the astest, we really are looking orthe maximum acceleration. Acceleration is given bya(t)53t2212t. Since

or t52 only, the only critical point is at t52. Compare thevalues oa(t) at the critical point and endpoints:

t a(t)

0 0

2 212

10 180

Tereore, velocity is increasing the astest at t510.

Or, students may note that a(t) is a quadratic unction or which the graph is aparabola opening upward. Since the vertex o the parabola has t-coordinate t52

in the interval [0,10], then the minimum acceleration occurs at t52 and themaximum acceleration occurs at one o the endpoints o the interval. Comparingvalues as above, we see that the maximum acceleration occurs at t510.



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Extrema

Example 5 (diferential equation, noncalculator): Let .

a. How do we know that (1,22) is a critical point or the unctiony5f(x)?

b. Is the point (1,2

2) a relative maximum point, relative minimum point, orneither? Justiy your answer.


Tis is a great problem or which to have students draw a slope feld, so they can see thatthe particular solution passing through (1,22) looks as i it has a relative minimum atthis point. In calculus, though, seeing is not enough to make us believe. We need analysisto veriy our answer.

a. We know that (1,22) is a critical point since the value o the derivative

at x 51,y522 is zero .

b. Find . Since we already know the frst derivative at (1,22) is zero, we

now know that the second derivative at this point is positive

. Tereore, (1,22) is a relative minimum point.

You might discuss with students why a closed interval test or frst derivative test

would be impossible or this problem.


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Worksheet 3: Extrema in a Variety of Settings

1. (noncalculator) Te minimum value of(x)5x424x3 + kis 7. Determine thevalue ok.

2. (noncalculator) Suppose thatP(t) measures the proportion o the normal oxygenlevel in a pond, withP(t)51 corresponding to the normal (unpolluted) level and0 P(t) 1. Te time tis measured in weeks with t 0. At time t50, organicwaste is dumped into the pond and, as the waste material oxidizes, the proportion

o the normal oxygen level in the pond is given by , t > 0.

a. At what time tis the proportion o the normal oxygen level in the pond the least?

b. At what time tis the proportion o the normal oxygen level in the pondincreasing most rapidly?

3. (noncalculator) Te curve with derivative hasy523 as a tangent line.

At what point is the line tangent to the curve? Determine i the point that youound is a relative maximum point, relative minimum point or neither or thecurve. Justiy your answer.

4. (calculator active) Te rate o changeR, in kilometers per hour, o the altitudeo a hot air balloon is given byR(t)5t324t2 + 6 or time 0 t 4, where tismeasured in hours. Assume the balloon is initially at ground level.

a. What is the maximum altitude o the balloon during the interval 0#t#4?

b. At what time is the altitude o the balloon increasing most rapidly?

5. (noncalculator) Let fbe a unction that is continuous on the interval [21, 4]. Teunction fis twice dierentiable except at x51, and fand its derivatives have theproperties indicated in the table below, where DNE indicates that the derivativesofdo not exist at x 51.

x 21 21,x,0 0 0,x,1 1 1,x,2 2 2,x,4 4

1 positive 0 negative 21 negative 0 positive 3 26 negative 21 negative DNE positive 21 positive 8 3 positive 0 negative DNE negative 0 positive 4



19

Extrema

a. For21,x,4, fnd all values ox at which fhas a relative extreme. For eacho thesex-values, determine whether fhas a relative maximum or minimum.Justiy your answer.

b. For21 x 4, fnd the maximum value of. Justiy your answer.

c. On the axes provided, sketch the graph o a unction that has all thecharacteristics of.

3 421

-1

-2

3

f(x)

2

1

-1-2

d. Let h be the unction defned by on the interval [21, 4].

For21,x,4, fnd all values ox at which h has a relative extreme. For each o

thesex-values, determine whether h has a relative maximum or minimum. Justiyyour answers.

6. (calculator active) Te velocity o a particle moving along thex-axis is given byor time 0#t# 6.5. When t5 0, the particle is at x521.

a. Write an expression or the position o the particle at any time t.

b. Determine when the particle is arthest rom the origin.

c. When the particle is arthest rom the origin, is the velocity increasing ordecreasing?


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Worksheet 3: Solutions and Teacher Notes

1. First, set the derivative equal to zero to fnd critical numbers. Tat is, solve theequation to obtain the critical numbers x5 0 and x53.A frst derivative test reveals that the minimum must occur at x 53.

x, 0 0 , x , 3 x > 3

negative negative positive decreasing decreasing increasing

Substituting x53 into the original unction and setting the result equal to 7 yieldsk534. Tat is, the equation has solution k534.

2. a. Since we are looking or the minimum oP, we compute

and solve to obtain t51 and t521. Since t521 is not in ourdomain, our only critical point is t51. Since t51 is the only critical point ort$ 0 and is negative or 0#t,1 and positive or t> 1, we have establishedthat the minimum proportion o the normal oxygen level in the pond occurs att51 week.

b. Since we are looking or the maximum o , we need to compute the

derivative o , . We then solve to obtain

t5 0 and . Since is the only critical point in the interior o our

domain and changes rom positive to negative at (which means

goes rom increasing to decreasing), then the proportion o the normaloxygen level in the pond is increasing most rapidly at .

3. Te key to understanding this problem is in realizing that the tangent line to the

curve is horizontal, which means that the slope, , must equal zero. Note that

only at x5

3, . We conclude that the liney5

2

3 is tangent

to the curve at the point (3,23). o determine the type o point we are workingwith, we must use the second derivative test. Using the quotient rule to fnd

and evaluating at the point (3,23), we fnd the value o

the second derivative to be positive , which means the point (3,23) must bea relative (local) minimum point.



21

Extrema

4. a. By examining the graph oR on the graphing calculator, you can see thatR(t) ispositive or 0,t,1.572, negative or 1.572,t,3.514, and positive or3.514,t,4. Tis means that the balloon rises, alls and then rises. Temaximum altitude is either at t51.572 or t54. Since the area below thex-axisor 1.572,t,3.514 is larger than the area above thex-axis or 3.514,t,4,the decrease in altitude is greater than the increase in altitude during the lasttime interval (3.514,t,4), so the maximum altitude occurs at t51.572.

Indeed, we can check that the altitude at t51.572 hours is

kilometers, while the altitude at t54 hours is km.

b. We are looking or the time whenR(t) is a maximum. Tereore, we need to setthe derivative oR(t), , equal to zero: when t5 0 ort58/3. Since t58/3 is the only critical point in the interior o our domainandR0(t)5 6tis positive at t58/3,R has a local minimum there. (In act,R(8/3)523.481 km/hr, showing that the altitude o the balloon is decreasing att58/3.) Tereore, we should check the values oR(t) at the two endpoints t5 0and t54. SinceR(0)5R(4)5 6 kilometers per hour, the altitude o the balloonis increasing most rapidly at t5 0 and t54.

I like to include problems like this because students always assume the maximumor minimum will occur at the critical point they have ound (and many textbooksreinorce this idea because thats how ALL o the problems turn out). Its good tooccasionally have problems with endpoint answers.

5. a. fhas a relative minimum at x 51 since the sign o changes romnegative to positive at that point, which indicates that the unction changes

rom decreasing to increasing at x 51, giving us a relative minimum.

b. Since there are no points where changes rom positive to negative thereare no interior candidates or our maximum value. Tereore, the maximummust occur at an endpoint. Sincef(4)53 > 15f(21), the maximum value othe unction isf(4)53.


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c.

3 421

1

3

f(x)

2

1

1

d. Using the Fundamental Teorem o Calculus, . Since

changes rom positive to negative at x5

0, there is a relative maximum at x5

0.Since changes rom negative to positive at x52, there is relativeminimum at x 52.

6. a. . Integrating the velocity gives the

net change in position and, i we add the initial position to the change inposition, we get the new position.

b. We compare values o the position unction x at the endpoints and atall critical points. Critical points or the position unction occur whenthe derivative o the position, which is the velocity, equals zero. Since

or t5 0, t5 /2, and t53/2, the critical pointsare t5 0, t5 /2 and t53/2. We use our solution rom part a to fnd thepositionx(t) at the critical points and endpoints.

t 0 /2 3/2 6.5

x(t) 21 20.296 23.819 20.887

Te particle is arthest rom the origin at t53/2.

c. o determine i velocity is increasing or decreasing, we should check the sign

o . Using the NDeriv eature o the calculator, , which

indicates velocity is increasing because its derivative is positive.

Tis part o the problem has very little to do with extrema, but I try to fndevery opportunity I can to reinorce students calculator skills. Tey seem toget lots o practice evaluating defnite integrals on the calculator but much lesspractice evaluating derivatives.


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Extrema

About the Author

Dixie Ross is a classroom teacher at Pugerville High School in Pugerville, exas. Shestarted the Advanced Placement program at aylor High School in aylor, exas, in

1989, and began working as a College Board consultant and workshop leader in 1994. Sheassisted with the development o the AP Mathematics Vertical eams oolkit, the Settingthe Cornerstones workshop, the Building Success in Mathematics workshop, and theAPVertical Teams Guide for Mathematics and Statistics. Ross has served as a Reader or theAP Calculus Exam.

AP CALCULUS Curriculum Module: Extrema

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