Chapter 12 Resource Masters

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1 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 2 _ 8 9 3 8 1 3 . Chapter 12 i nihet...... 26 Enrichment ...... 25 Word ProblemPractice ...... Practice Study GuideandIntervention 22 ...... Derivatives Lesson 12-4 Spreadsheet Activity 21 ...... 20 Enrichment ...... 19 Word ProblemPractice ...... Practice Study GuideandIntervention 16 ...... Tangent LinesandVelocity Lesson 12-3 15 Enrichment ...... 14 Word ProblemPractice ...... Practice Study GuideandIntervention 11 ...... Evaluating LimitsAlgebraically Lesson 12-2 10 Graphing CalculatorActivity ...... 9 Enrichment ...... 8 Word ProblemPractice ...... Practice Study GuideandIntervention 5 ...... Estimating LimitsGraphically Lesson 12-1 Anticipation Guide(Spanish) 4 ...... 3 Anticipation Guide(English) ...... Student-Built Glossary 1 ...... Chapter Resources Resource Masters Teacher’s GuidetoUsingtheChapter12 iv ...... n d d

S e c 2 : i i i ...... 24 ...... 18 ...... 13 ...... 7 ...... iii Contents tnadzdTs rcie...... 54 Standardized TestPractice ...... Chapter 12Extended-ResponseTest 53 ...... 51 Chapter 12Test,Form3 ...... 49 Chapter 12Test,Form2D ...... 47 Chapter 12Test,Form2C ...... Chapter 12Test,Form2B Chapter 12Test,Form2A 41 Chapter 12Test,Form1 ...... 40 Chapter 12VocabularyTest ...... Chapter 12Mid-ChapterTest 39 ...... 38 Chapter 12Quizzes3and4 ...... 37 Chapter 12Quizzes1and2 ...... Assessment 36 Enrichment ...... 35 Word ProblemPractice ...... Practice Study GuideandIntervention 32 ...... The FundamentalTheoremofCalculus Lesson 12-6 31 Enrichment ...... 30 Word ProblemPractice ...... Practice Study GuideandIntervention 27 ...... Area UnderaCurveandIntegration Lesson 12-5 Answers ...... 34 ...... 29 ...... A1–A26 ......

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Exercises and includes Glencoe Precalculus Pdf Pass This master These activities

These activities may extend Student Edition This master closely follows the This master closely Practice found in the types of problems section of the Use as an additional word problems. or as homework for practice option of the lesson. second-day teaching Word Problem Practice solving word includes additional practice in of the problems that apply to the concepts or as lesson. Use as an additional practice of homework for second-day teaching the lesson. Enrichment an historical the concepts of the lesson, offer or or multicultural look at the concepts, the widen students’ perspectives on They are mathematics they are learning. students. written for use with all levels of or Graphing Calculator, TI–Nspire, Spreadsheet Activities can be present ways in which technology lessons of used with the concepts in some approach this chapter. Use as an alternative part of to some concepts or as an integral your lesson presentation. Guided These includes the core materials needed for Chapter 12. These Chapter 12. These needed for the core materials includes (pages 1–2) These (pages 1–2) These as an instructional Teacher’s Guide to Using the to Using Guide Teacher’s (pages 3–4) This Chapter 12 Resource Masters 12 Resource Chapter . Encourage them to v i : 2 c e S

Lesson 12-1

d d Student Edition iv n i exercises to use as a reteaching . 3 1 8 3 9 Chapter 12 Resource Masters Chapter 8 _ 2 1 C M Study Guide and Intervention Study Guide and Intervention Lesson Resources masters provide vocabulary, key concepts, masters provide vocabulary, key and additional worked-out examples Chapter Resources Glossary Student-Built study tool that masters are a student of the key vocabulary presents up to twenty chapter. Students are to terms from the and/or examples for each record definitions suggest that students term. You may which they highlight or star the terms with before are not familiar. Give this to students beginning Practice activity. It can also be used in conjunction with the tool for students who have been absent. add these pages to their mathematics study add these pages to their mathematics the notebooks. Remind them to complete each lesson. appropriate words as they study Anticipation Guide and master, presented in both English beginning Spanish, is a survey used before may the chapter to pinpoint what students in the or may not know about the concepts survey chapter. Students will revisit this to see if after they complete the chapter their perceptions have changed. The these answers for options. The and assessment extensions, include worksheets, materials the back of this booklet. pages appear at Chapter 12 R C C P _ 4 0 0 _ i i 00ii_004_PCCRMC12_893813.indd Sec2:iv 00ii_004_PCCRMC12_893813.indd Sec2:v i i _ 0 0 4 _ P C C R M C

1 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 2 _ 8 9 3 8 1 3 . Chapter 12 i Mid-Chapter the chapter.Itparallelstimingof provides anoptiontoassessthefirsthalfof Mid-Chapter Test the chapter. assessment atappropriateintervalsin Quizzes assessment. assessment andsummative(final) assessment toolsforformative(monitoring) Resource Masters The assessmentmastersinthe Assessment Options chapter tests. used inconjunctionwithoneoftheleveled knowledge ofthosewords.Thiscanalsobe words andquestionstoassessstudents’ all students.Itincludesalistofvocabulary Vocabulary Test free-response questions. and includesbothmultiple-choice n d d

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1 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 2 _ 8 9 3 8 1 3 . Chapter 12 your PrecalculusStudyNotebooktoreviewvocabularyattheendofchapter. Remember toaddthepagenumberwhereyoufoundterm.Addthesepages As youstudythischapter,completeeachterm’sdefinitionordescription. This isanalphabeticallistofkeyvocabularytermsyouwilllearninChapter12. NAME i n indeterminate form indefinite integral Calculus Fundamental Theoremof direct substitution differentiation differential operator differential equation derivative definite integral antiderivative d d 12

S e c 1 : 1 Vocabulary Term 1 Student-Built Glossary on Page Found DT PERIOD DATE Defi nition/Description/Example

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nition/Description/Example Defi DATE DATE PERIOD Found on Page 2 : 1 c Student-Built Glossary Student-Built e S

d d 2 n i . 3 1 Vocabulary Term Vocabulary 8 3 9 8 _ 2 1 12 C upper limit tangent line two-sided limit right Riemann sum regular partition one-sided limit integration lower limit instantaneous velocity instantaneous rate of change instantaneous rate M Chapter 12 NAME NAME R C C P _ 4 0 0 _ i i 00ii_004_PCCRMC12_893813.indd Sec1:2 00ii_004_PCCRMC12_893813.indd Sec1:3 i i _ 0 0 4 _ P C C R M C

STEP 1

S Step 2 Step 1 write anexampleofwhy youdisagree. For thosestatementsthatyoumarkwithaD, useapieceofpaperto Did anyofyouropinionsaboutthestatements changefromthefirstcolumn? Reread eachstatementandcompletethelast column byenteringanAoraD. agree ordisagree,writeNS(NotSure). Write AorDinthefirstcolumnORifyouarenotsurewhether Decide whetheryouAgree(A)orDisagree(D)withthestatement. Read eachstatement. e c 1 : 3 3 Limits andDerivatives Anticipation Guide 10. 9. 7. 2. 1. 8. 6. 5. 4. 3.

f The function The processofevaluatinganintegraliscalledintegration. The derivativeofaconstantfunctionistheconstant. The processoffindingaderivativeiscalleddifferentiation. instantaneous rateofchange. The slopeofanonlineargraphatspecificpointisthe found bydirectsubstitution. Limits ofpolynomialandmanyrationalfunctionscanbe the point. The limitofaconstantfunctionatanypointisthe either theleft-handlimitorright-handexists. The limitofafunction on thevalueoffunctionatpoint The limitofafunction Fundamental TheoremofCalculus. antiderivatives issoimportantthatitcalledthe The connectionbetweendefiniteintegralsand ( x ) =

F ( x ). F ( x ) isanantiderivativeofthefunction f f ( ( Statement x x ) as ) as x x approaches approaches DT PERIOD DATE c . c c existsproviding doesnotdepend

Pdf Pass x -value of f ( x ) if Glencoe Precalculus STEP 2 A orD 33/17/09 11:36:02 AM / 1 7 / 0 9

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2 Chapter Resources

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), si x ( f . c se aproxima a se aproxima a x x FECHA FECHA PERÍODO ) a medida que ) a medida que x x ( ( f f Enunciado del punto. ) es la antiderivada de la función x x ( F ). x ( F

= ) x ( La relación entre las integrales definidas y las antiderivadas La relación entre las integrales definidas y las antiderivadas es tan importante que se llama teorema fundamental del cálculo. diferenciación. es igual a la constante. La derivada de una función constante integral se llama integración. El proceso de evaluación de una La función f El límite de una función por la derecha o un límite existe si y sólo si existe un límite por la izquierda. en un punto cualesquiera El límite de una función constante es el valor de y de las funciones El límite de las funciones polinomiales sustitución directa. racionales se puede calcular por lineal en un punto específico La pendiente de una gráfica no es igual a su tasa de cambio instantánea. derivada se llama El proceso de obtención de una El límite de una función en el punto no depende del valor de la función

7. 9. 2. 3. 4. 5. 6. 1. 4 : 8. 10.

1 c Ejercicios preparatorios Ejercicios y derivadas Límites e S

d d 4 n i . 3 1 8 3 9 8 en una hoja aparte un ejemplo de por qué no estás de acuerdo. Relee cada enunciado y escribe A o D en la última columna. sobre Compara la última columna con la primera. ¿Cambiaste de opinión alguno de los enunciados? escribe En los casos en que hayas estado en desacuerdo con el enunciado, Escribe A o D en la primera columna O si no estás seguro(a), escribe NS (no estoy estás seguro(a), escribe NS (no la primera columna O si no Escribe A o D en seguro(a)). Lee cada enunciado. Lee cada enunciado. (D) con el enunciado. de acuerdo (A) o en desacuerdo Decide si estás Paso 2 Paso 1 _ 2 PASO 1 1 12 A, D o NS C M Capítulo 12 • • • el Capítulo 12 Después de que termines • • • 12 Antes de que comiences el Capítulo NOMBRE NOMBRE R C C P _ 4 0 0 _ i i 00ii_004_PCCRMC12_893813.indd Sec1:4 0005_036_PCCRMC12_893813.indd 5 0 5 _ 0 3 6 _ P C C R M

Estimateeachone-sidedortwo-sidedlimit,ifitexists. Chapter 12 4. of Because theleft-andright-handlimits The graphof Estimate LimitsatFixedValues NAME 1. Estimate eachone-sidedortwo-sidedlimit,ifitexists. Exercises 12-1 x x x . lim left, then limits existandareequal.Thatis,if number If thevalueof The limitofafunction Existence ofaLimitatPoint i

lim lim n → → → d f d

Example x x

( lim

2 2 2

lim

5 x → →

- -

) as x

x x 0 0

+ (1

does notexist.

− = 5 ⎪ L

x x 3 x

lim 1 and 1 - approaches2arenotthesame,

1 x x →

as lim

⎥ → cos

2. Left-Hand Limit Estimating LimitsGraphically Study GuideandIntervention 2 f

+ ( c

x x - x f

) ( f

approaches x ( x = , x ) approachesaunique and ) )

= x x lim

→

x L

suggests that x

lim lim f c →

→ - . ( f

2 ) as 2

(

x c x ) fromthe =

x =

approaches

x 2. lim

→

x x

lim → → - lim c

3 f

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x -

− x

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right, then number If thevalueof -

27 c - L 9 existsifandonlybothone-sided 2 4 , then

6. ⎥

3. DT PERIOD DATE L

x lim

2 → as Right-Hand Limit

x c lim

→

f x f ( c approaches x (

+ x )

f ) approachesaunique = ( x

) L = 0 .

x x lim

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x − c

− fromthe ( f

x ( + x

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Glencoe Precalculus - x 2)

2 x - 2

112/7/09 12:32:15 PM 2 / 7 / 0 9

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5 Lesson 12-1

P M M P

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π x + Glencoe Precalculus

Pdf Pass x 0. y 0

2 =

(2 1 x

−

cos 2

∞

∞ ∞ (continued) 1 +

lim → → → - x lim lim

−

x x x

1 +

∞ x = → lim

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π x

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e −

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x lim lim → - → - → ( lim

1 +

x x x f

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∞ →

lim 0. x

y . 2 . ) approaches 0. This 1 L x

( Make a table of values, choosing L ) approaches a unique number ) approaches a unique ) approaches a unique number ) approaches a unique

The graph of f = x x ( ( = ) f f x ) ( 5. 8.

x )

f x (

2. x

approaches infinity f 5

∞ 1 Study Guide and Intervention Intervention Guide and Study Graphically Limits Estimating

- 6

+ ∞

2 x 10 100 1000 10,000 100,000 3

d sin x 0.08 0.01 0.001 0.0001 0.00001 lim d x → → - x lim

n x 2 i 3 2 . 6 − − x x

(

8 ∞ ∞ ∞ 3 9 8 ) increases, the height of the graph gets closer to 0. The limit indicates a increases, the height of the graph → → → _ x lim lim lim

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C M • • R 12-1 increasingly larger, supports our graphical analysis. Estimate 7. 4. Estimate each limit, if it exists. 1. Exercises Chapter 12 The pattern of outputs suggests that as The pattern of outputs suggests Estimate Limits at Infinity Limits at Estimate NAME NAME Analyze Graphically As horizontal asymptote at Support Numerically

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C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 _ 8 9 3 8 1 3 Chapter 12 12. 11. 9. 7. Estimate eachlimit,ifitexists. 5. 3. 1. Estimate eachone-sidedortwo-sidedlimit,ifitexists. NAME 12-1 . i n d d

base ofthepoleandbarn.Graph r the topofpolewillmovedownsidebarnatarate of thepoleispulledawayfrombarnatarate3feetpersecond, Find POLLUTANTS RATE OFCHANGE C the pollutantscreatedbyoneofitsmanufacturing processesisgivenby 7 x x x x x lim lim

lim ( → → - → - → →

lim lim x = ) 0 4 0

= +

−

100 7 1 ∞

− (4

−

sin 2 x

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−

lim

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- 400 100

Estimating LimitsGraphically Practice 16

x 4 4 + + x x

3 1 √

8 2 + -

,where

- x

. )

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x x x x x lim

lim lim lim DT PERIOD DATE → → - → → → lim

x 0 3 ∞ ∞

→ lim

−

− − 3 ⎪ 2

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x x

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+ 1 ≤ x 3

2 (

8 x

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100. 7 Pdf Pass

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of an object 8 Glencoe Precalculus Pdf 2nd m is the mass of 0 is the speed of ? 6 s Suppose a a Suppose m

m . Explain why th

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t projectile is thrown upward where its is thrown upward projectile height THEORY OF RELATIVITY Theoretically, the mass with velocity m ELECTRICITY PROJECTILE HEIGHT PROJECTILE the object at rest and light. What is the voltage in an electrical outlet in his home is modeled by the function V is determined by the function is determined by height of the projectile at table shows the its flight. various times during a. b. DATE DATE PERIOD

4. 5. 3.

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t t

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t + 0 d ( d v 1 n

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30,000(0.7) 3 is the time in hours. 1 t 8 = = 3 Interpret your result. Interpret your results. Graph Estimate Estimate Graph of the car after 10 years. Use the graph to estimate the value if necessary. Use the graph to estimate the number Use the graph to estimate the of grams of bacteria present after tenth, 8 hours. Round to the nearest 9 ) ) 8 t t _ ( ( 2 1 a car purchased for $30,000 is v a. are growing according to the function according are growing f c. CARS c. BACTERIA GROWTH BACTERIA a. is the weight of the bacteria in grams is the weight of and b. b. C

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C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 _ 8 9 3 8 1 3 Chapter 12 7. or doesnotexist infinite, ortheymaynotexist.Classifyeach limitas One specialfeatureofmathematicallimits isthattheymaybefinite, Fill inthechartbelow. guidelines, andstillothersresultinapenaltyiftheyareexceeded. absolute limits,inthattheycanneverbeexceeded.Othersarelike There aremanyexamplesoflimitsinourworld.Somethese A MatterofLimits NAME 10. 6. 5. 4. 3. 2. 1. 12-1 . i n d d

credit card creditlimitona craft accelerating space thespeedofan in acoolroom warm objectplaced temperatureofa on anairlineflight luggagelimit underpass on aroad heightlimit highway speedlimitona 9 x x lim lim

→ → 0 0

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1 + ⎪

x Enrichment 1 ⎥

8. . Ifthelimitisfinite,giveitsvalue. 11.

x x lim lim

→ → 0 1

−

− sin x x 2

+ x -

⎪ 1 1 x ⎥

12. DT PERIOD DATE finite, infinite,

Pdf Pass

x x lim lim

→ → Penalty orconsequence if thelimitisexceeded 0 2

− − - x x 2 4 x 1

- 2

Glencoe Precalculus - x 4 -

33/17/09 11:36:36 AM / 1 7 / 0 9

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Glencoe Precalculus ) Pdf Pass

and use ENTER

TRACE and use —

is close to 2. is close to 0.

2 x x x ÷

DATE DATE PERIOD TRACE

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), first graph the equation ), first graph the -coordinate should be close to -coordinate should 1

x y

( ENTER f — (

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lim ÷ x

ENTER

TRACE ) ZOOM 2 TRACE

— ZOOM as you like. The as you like. The

and

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[

1 x 1? - ln = ZOOM x

−

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6. Then press =

2 y

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x 0 Graphing Calculator Activity Calculator Graphing -

ZOOM 2

6. Then press d to examine the limit of the function when to examine the limit of the function to examine the limit of the function when to examine the limit of the function - x x d -

Y= Y= is when n 2 i x

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Will the graphing calculator give you the exact answer for every limit Will the graphing calculator give you the exact answer for every problem? Explain. If you graph

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R 12-1 4. 3. Chapter 12 2. 1. x limit. the value of the limit. Evaluate each Finding Limits Finding with less work than an calculator to find a limit You can use a graphing calculator. To find ordinary scientific NAME NAME

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C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 _ 8 9 3 8 1 3 direct substitutiontofindthelimit. Since thisisthelimitofapolynomialfunction,wecanapplymethod Usefactoringtoevaluate By directsubstitution,youobtain Chapter 12 Compute LimitsataPoint NAME 4. Userationalizingtoevaluate = 1. Evaluate eachlimit. Exercises of thefractionbeforefactoringanddividingcommonfactors. 12-2 x x x x . lim i

lim n lim lim → → - → → - d Example 3 Example 2 Example 1 d

x x

lim lim

4 16

1 → →

2 2

x −

(

−

2 4 3 √

- (2

- -

−

- √ 11 x 2

x 9 2 -

x x 16

x - x - x

4 4 -

- +

Evaluating LimitsAlgebraically Study GuideandIntervention 2

20 5 = = = = = 3

x x

Use directsubstitution,ifpossible,toevaluate

x x x x x

)

−

3 √

lim lim lim lim =

→ → → → = 3 -

16 5.

4 16 16 16 16

1 x x

lim lim

5 2.

+ →

→ −

(

x -

− − −

( ( √ √ x x

4 x x + 4 5,or 4

x x

(

- - or

3) ( − + - x x 16

16)( 16)( 3).

x x

4) 4 - -

− = = - - ( 8 1

5)( x

5) -

16 16

√ √

- - - 1 −

√ √ x x x

4) 32 2(

x x

- + +

− x x

- √

lim 16 + +

4) 4) 4) lim - → - → 16

2) 4 4 24

- 4 16

+ √

4 ( + x 3(

− 1 x

- + -

x 4

2) lim x

−

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3 .Rationalizethenumerator

3,or - 4

lim

5( DT PERIOD DATE x − → conjugate of Apply directsubstitutionandsimplify. Divide outthecommonfactorandsimplify. Factor. Apply directsubstitutionandsimplify. Simplify. Divide outthecommonfactor. Simplify. Multiply thenumeratoranddenominatorby 2 3.

- - 16 -

x 9 2)

43 −

- √ x x

4 x

6. - +

- 16 20 3

√ 4 x

. -

x x lim

Pdf Pass lim → → - 2

(

x − x 2

2 +

+ x Glencoe Precalculus 9

+ 5 x

2 x +

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+ 4,the 112/7/09 11:39:40 AM 2 / 7 / 0 9

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0 Lesson 12-2

A M M A

9 4 : 6 3 : 1 1

9 0 / 7 1

+ 7

1 3 Glencoe Precalculus + Pdf Pass

- x

x 0. x

6 (3 = 10 −

∞ n

1 x ∞ −

nity nity (continued)

lim → → -

∞ lim

x x

nity nity Limits of Reciprocal Limits of Reciprocal Functions at Infinity lim → ±

x For any positive integer For any positive

n n . x x 0 n n a a

∞ + ∞

1 → lim lim

→ - DATE DATE PERIOD 1 a

1 x

+ 3 = 2 = x Limits of Polynomials at Infi Scalar Multiple Property Limits of Power Functions at Infi Limits of Polynomials at Infi Limits of Power Functions at Infi x

)

) 2 x x … ( + (

+ p 3

p +

Infinity

x n 2 2 ∞

∞ 5 − x 5 x −

→

lim

a ∞ ∞

x lim

→ -

be a polynomial be a polynomial x = → →

p lim lim

)

x x

5 x ( x Limits of Polynomials at Limits of Polynomials

Then and Let function p

∞ ∞ 4 4

x x

lim 2

→ - ∞ ,

x -∞ ∞

n ∞

→

lim

· → x = =

lim

is odd. x

2 2 2. n is even.

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+ 2 )

5 2 x x if x 5. x

+ 6 5 .

∞ -∞ 6 5 3

1 2 ∞ - x + - 2 + Study Guide and Intervention Intervention Guide and Study Algebraically Limits Evaluating

= =

2 1

5 4 + -

= n n x 3 x 2 d - x x

( d x x x x n

( ( i (2 . 6 12 − x

∞ ∞

∞ ∞ 1

8 -∞ Limits of Power Limits of Power ∞ ∞ ∞ 3

9 8 lim lim lim lim lim → ∞ → → - → → - → → - → → - _

lim

lim lim

Functions at Infinity lim 2 x x x x x x x x x 1 Example

C M • • For any positive integer For any positive • R 12-2 4. Exercises Evaluate each limit. 1.

b. Chapter 12 Compute Limits at Infinity Limits at Compute NAME NAME a. Evaluate each limit. C C P _ 6 3 0 _ 5 0 0005_036_PCCRMC12_893813.indd 12 0005_036_PCCRMC12_893813.indd 13 0 5 _ 0 3 6 _ P C C R M

Find the of medicine her asthmamedicine.Thegraphshowstheamount MEDICINE BOOKS be worth?Thatis,findthe represented as 1 x x x x x lim lim lim 3

lim lim → → → → → 1 0 3 ∞ ∞

(

(2 −

(3 −

x 2 − x 2 x 2

+ 5 Supposethevalue 3

x + -

+ x Evaluating LimitsAlgebraically Practice

2 - d x 4

3 6 x 5 lim

- ) 4 → Eachday,Tamekatakes2milligrams of -

x x x m

x 2 x -

- + - leftinherbloodstreamafter + 9 x

7 2 m v 8) 5 1

(

t (

) d 3

) ) and = 4. 6. 10.

− 6 2. + 8.

d 35(0.2) 300 lim

→

t v

lim → 3

ofabookindollarsafter

∞

( v . Howmuchwillthebookeventually d ( t ). ). d days.

x x x x x lim lim

DT PERIOD DATE lim → → → - → - → - lim lim 3 4

6 ∞ ∞

− 2 √

(6 −

x +

− 4 x x 2 2

x 5

- x t

+ √ - 5 x - yearscanbe

7 36 2 + 6

8 -

- x x

x x 2 3

Tameka’s Asthma

)

Medicine (mg) 1 10

Pdf Pass 0 8 2 4 6

m 15 2 Days 3 Glencoe Precalculus 46 d 33/17/09 11:36:54 AM / 1 7 / 0 9

1 1 : 3 6 : 5

4 Lesson 12-2

A M M A

8 5 : 6 3 : 1 1

9 0 / 7 1

+ )

models the

. p x p ( 3000 ). p c −

of producing a certain x

= ( c y 0.3 ) ). The function The

x p ) by

( (

driving a car, While p x c f 2 200,000

( 15,000

x c lim → + lim

→ x

x p 2000

lim The cost The → 250 −

0.005

= = On Tuesday, the company produced On Tuesday, the company produced $21,000 worth of parts. How many parts did the company produce? by The average cost per part is found dividing Find the cost of producing 100 parts. Find the cost of producing 100 Find ) ) x x is the number of parts produced. ( microwaves for professional kitchens. ( it is important to maintain a safe to maintain it is important you and the car in distance between the function front of you. Suppose y average cost of a microwave manufactured by a company that makes x Find small engine part in dollars is given by small engine part in dollars is the equation b. c. MICROWAVES CAR SAFETY PARTS f speed in miles per hour, gives the speed in miles per distance, in yards, recommended safe and the one in front of between your car you. Find p a. DATE DATE PERIOD 6. 4. 5.

C . ). t t ( C

∞ → lim

24(0.98) 100. = ≤

) p t

( Flint bought a new ). v t ≤ . Find (

. As a by-product of one of t years can be represented v

% of the pollutant is

, 0

t ∞ - p p

4 Word Problem Practice Problem Word Algebraically Limits Evaluating 1 → 0.3 lim

100

d liters of contains 75,000 A pool d lim 1000 of his motorcycle, in thousands of of his motorcycle, in thousands

− → n

v . 14 p 3

100 60,000

− 1 =

8 Value Years 1 5 10 20 3 ) Find the nearest hundredth. Complete the table. Round answers to 9 t = 8

of chlorine in grams per liter of chlorine in grams ( _ 2 minutes later in the pool is given by minutes later in 1 motorcycle for $24,000. Suppose the motorcycle for $24,000. Suppose value dollars, after by the equation creates an airborne pollutant. The cost creates an airborne pollutant. of removing C its processes, a manufacturing company its processes, a manufacturing b. MOTORCYCLES POLLUTANTS C t C POOLS that contains A mixture pure water. per liter of water is 0.3 gram of chlorine pool at a rate of pumped into the The concentration 75 liters per minute. a. Find C M R 12-2 Chapter 12 3. 2. 1. NAME NAME C C P _ 6 3 0 _ 5 0 0005_036_PCCRMC12_893813.indd 14 0005_036_PCCRMC12_893813.indd 15 0 5 _ 0 3 6 _ P C C R M

C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 _ 8 9 3 8 1 3 First, notethat - as because the SqueezeTheoremwith this inequalityby Chapter 12 5. 3. 1. Use theSqueezeTheoremtofindeachlimit. Exercises you learnedthat In Lesson12-1,youlearnedthatthe The SqueezeTheorem NAME will helpyouanswerthisquestionalgebraically. Theorem, let 12-2 . at If The SqueezeTheorem i n 1 and1.Butwhataboutthe d d

x x x x h

lim lim lim

getscloserto0,thecorrespondingfunctionvaluesoscillatebetween 1 → → → ( itself,andif 5 x ) 0 0 0

≤ x x

− tan

x 15 2 4 lim

cos sin x f →

( x x 0

) Enrichment sin

−

− 1

x ≤ √ 3 ( 1

) g

− x 1 x ( 1

= lim x

x x lim

→ doesnotexist?AtheoremknownastheSqueezeTheorem

→ 2 ) forall 2. ≤

, obtaining - 4. 0

sin c

h 2 ( , ( x x f ) and

− ( 1 x ) x

= c x ) ≤

inanopenintervalcontaining = =

1forall L

0.Theresultisthat x

x lim x -

lim = →

2 → sin x

2 x 0 0

lim

≤ →

x g 2

− ( 1 x sin c lim

x x

→

, and , 2 g ) bothequal0.Youcannowapply , exceptfor sin ( 0 x

sin

− ) ,then 1 x

? Doesthislimitnotexistsimply

− 1 x g

(

− 1 ≤ x x 6.

doesnotexistbecause )

x = 2

x x x x . ToapplytheSqueeze lim lim lim lim

DT PERIOD DATE → → → x → x

2 lim

= 0 0 4 . FromLesson12-2, → c

⎪ ⎪ 0.Next,multiply − f sin 3 x x

0 (

⎥ ⎥

x x ) existsandisequalto sn sin

− 2

√ sin x x x

+ , exceptpossibly

− π + x 4

− 1 x 2

= 0. Pdf Pass Glencoe Precalculus L . 33/17/09 11:37:02 AM / 1 7 / 0 9

1 1 : 3 7 : 0

2 Lesson 12-2

A M M A

6 0 : 7 3 : 1 1

9 0 / 7 1

2 x )) is the slope

x 3 ( = f ) , x ( x f 1 and + .

2 x ) x h

6 7 +

x - 4 √ 3(

m −

. 4 ) at the point ( h = x ) = = ( DATE DATE PERIOD h

, provided the limit exists. , provided the limit

y y ) +

x x ( ( Reduce a Scalar Multiple, Sum Property, and Limit of Constant Function Property of Limits Instantaneous Rate of Change Formula f Expand and simplify. Simplify and factor. f

) h h

+ ]

1 x ( − f +

0 x

→

lim 1] h

- [

] 2 =

x 1

+ m [3

2 - 2.

) 3 x 1] h ( 4.

+ )

2 ) h

- )

hx h

) 6 h h

h + x + +

+ 2

x (2 x 6 Study Guide and Intervention Guide and Study Velocity Lines and Tangent x x 1 h (

−−

− [3( [ 3 f − 1

−− d d

1 at any point. 3(2

n + i

. 0 0 0 0 0 16

3 + 2 3 1

x 8 −

x 2 → → → → → 3

x lim lim lim lim lim 9 x h h h h h

8 6 = =

2 3 1 Example y y ======C

M the tangent line given by the tangent line Instantaneous Rate of Change Instantaneous of rate of change of the graph The instantaneous R 12-3 m 3. Exercises at Find an equation for the slope of the graph of each function any point. 1. m graph at any point is An equation for the slope of the m m m m Chapter 12 Tangent Lines Tangent NAME NAME y Find an equation for the slope of the graph of C C P _ 6 3 0 _ 5 0 0005_036_PCCRMC12_893813.indd 16 0005_036_PCCRMC12_893813.indd 17 0 5 _ 0 3 6 _ P C C R M

C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 _ 8 9 3 8 1 3 h of aravine.Theheighttherockafter Chapter 12 3. 1. object atthegivenvaluefor dropped isgivenby The distance Exercises The instantaneousvelocityoftherockat4secondsis128feetpersecond. = = = = = Instantaneous Velocity NAME v rock at4seconds. 12-3 . velocity If thedistanceanobjecttravelsisgivenasafunctionoftime Instantaneous Velocity i ( n ( t d t d ) d d Example

)

= ( ( 1 = t t 7 ) )

- 1500 h h h h h = = lim lim lim lim lim

→ → → → → 128 800 70 17 v 0 0 ( 0 0 0

(

t ) atatime

f − [ − − h - −−− - 1500

- ( - Tangent LinesandVelocity Study GuideandIntervention - ( t 128 - -

16(0)or 128 16 + 16 d 16 128

A rockisdroppedfrom1500feetabovethebase anobjectisabovetheground h h - h t h t )

h t

- 2 -

16(4

2 - - . Findtheinstantaneousvelocity

; ; 16 16 t

t 16 f

= ( = t h t d -

h h ) isgivenby 1 + 2

(

) )

128

4. ). Findtheinstantaneousvelocityof 2. h ) 2 h ]

[ t 1500 .

v (

t - ) 16(4) =

h lim

→ 2 ] t

secondsisgivenby

Factor. Multiply andsimplify. f Instantaneous VelocityFormula and IdentityFunctions Difference PropertyofLimitsandLimitConstant Divide by

f − ( t (

+ t

d d

+ h DT PERIOD DATE ( ( t )

t t secondsafteritis h = ) ) h )

1500 = = h

- andsimplify.

- -

v f ( 16 16 ( - t t ) 16(4 ) ofthe

t t , providedthelimitexists. 2 2

+ + f + (

90 1700; t

), thentheinstantaneous h ) 2 and t Pdf Pass

+ (continued) 10; f t ( t

) = = 5 1500 t

= Glencoe Precalculus 2 - 16(4) 2 112/5/09 5:03:13 PM 2 / 5 / 0 9

5 : 0 3 : 1

3 Lesson 12-3

P M M P

3 2 : 0 1 : 5

9 0 / 5 / 2

t Glencoe Precalculus Pdf Pass

1 700; 6. + +

+ t t

t 4 + 58

200 2 5) t + +

t 2 2 6 2 t t x 2 ) of the football. 1, - t

16 ( + 2 16 3 -

) if the path of an

of the football, in t

v - t t - - 3 is seconds passed −

( ; ( 5

3 v = x 5 = −

x = =

) ) ) ) t seconds after it is t = = . t t ( t ( DATE DATE PERIOD

( ( t h s y y s d , where time 2 t 16 - . in feet of a sky diver relative to the ground t h ) of the sky diver. t ( 18,000 6. 4. v = ). Find the instantaneous velocity of the ). Find the instantaneous velocity 2 t ) for any point in time ) ( t t 2. = ( (

d s t h ; 2 t 10.

8. 8. 2 t The position The an object is above the ground an object is above the ground 16 8 A quarterback throws a football with a velocity of 58 feet 2 d - + 1

; (3, 6) - 2 8 Practice Velocity Lines and Tangent x

+ t 1

x d - √

300 2 seconds after he throws it is defined as

17 n i 2 t . 18 3 = - x = = 1

8 How fast is the football traveling after 1.5 seconds? Find an expression for the instantaneous velocity ) ) ) 3 t = = t t 9 (

8 ( ( _ after the sky diver exited the plane. Find an expression for the after the sky diver exited the plane. Find an expression for the instantaneous velocity feet, per second toward a teammate. Suppose the height s s d y y b. FOOTBALL SKY DIVING a. can be defined by 2 1 C 9. M 7. 5. 3. 1. R 12-3 Chapter 12 12. 11. Find an equation for the instantaneous velocity Find an equation for the instantaneous object is defined as The distance dropped is given by object at the given value for Find an equation for the slope of the graph of each function for the slope of the graph Find an equation at any point. Find the slope of the line tangent to the graph of each function at each function to the graph of line tangent slope of the Find the point. the given NAME NAME C C P _ 6 3 0 _ 5 0 0005_036_PCCRMC12_893813.indd 18 0005_036_PCCRMC12_893813.indd 19 0 5 _ 0 3 6 _ P C C R M

C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 _ 8 9 3 8 1 3 Chapter 12 4. 3. 2. 1. NAME 12-3 . i n d d a. from atowerthatis800feethigh.The given by position oftheballafter a. exited theplane. seconds passedaftertheskydiver h the groundcanbedefinedby of afree-fallingskydiverrelativeto velocity Find anexpressionfortheinstantaneous the ballfallingafter2seconds? t 1200 feet.Thepositionoftherockafter FALLING OBJECT c. b. FREE FALLING BUNGEE JUMPING d. c. b. PROJECTILE t height

secondsisgivenby secondsisgivenby

( 1 t 9 ) 4 seconds? How fastistheballfallingafter of theskydiver. instantaneous velocity Find anexpressionforthe after 4seconds? What istheskydiver’s velocity after 2seconds? What istheskydiver’sheight it hitstheground? What isthevelocityofrockwhen instantaneous velocity Find anexpressionforthe When willtherockhitground? of therock. = 15,000 19 h v infeetrelativetotheground s ( ( t Tangent LinesandVelocity Word ProblemPractice t ) ofthejumper. ) Titodropsarockfrom = -

The position - 16 16 Mirandadropsaball t 2 t , where Abungeejumper’s 2

s h + ( ( t t 800.Howfastis ) ) t = = secondsis v v

900 ( ( - t t t ) ) is 16 h

infeet t - 2

+ 16 1200. t 2 . 5. 6. DT PERIOD DATE modeled by seconds aftertheballisthrown the height velocity of80feetpersecond.Suppose toward homeplatewithaninitial a. c. b. BASEBALL b. AREA d. a. of thesquareshownischanging. 5.4 inchesto5.6inches. instantaneous velocity Findanexpressionforthe reach itsmaximumheight? Forwhatvalueof after 0.5second? Howfastisthebaseballtraveling x of theareaatmoment Findtheinstantaneousrateofchange of thebaseball? Whatisthemaximumheight area Findtheaveragerateofchange the baseball.

= Supposethelength 5inches. a ( x h ) as Anoutfielderthrowsaball x h ofthebaseball,infeet,

( t ) x = a Pdf Pass changesfrom ( x

)= - x x 16 2 t willthebaseball t 2

+ Glencoe Precalculus v 80 x ( t ofeachside ) of t

+ 6.5. t

33/17/09 11:37:17 AM / 1 7 / 0 9

1 1 : 3 7 : 1

7 Lesson 12-3

A M M A

0 2 : 7 3 : 1 1

9 0 / 7 1

> 0 and . a x 4 9 x + -

x x 1 x 2 6 4 y - - - 0 1

2 2 2 x x x

, one where 1 2 − - c

= -

y = = + DATE DATE PERIOD

y y ? bx c

+ +

2 x bx ax

3 + = -

2 x y 4 2 ax +

2 4 y 0 x =

= y y 2. 4. -value into its respective equation to find the -value into its respective equation x -value of the vertex. x 0. Then draw the line tangent to the vertex of each line tangent to the vertex of 0. Then draw the <

a 2 +

x 0 Enrichment 2 4 25

d - - d

n i 2 2 -value of the vertex. . 20 3 x x Substitute each y Set each equation for the slope equal to zero and solve for Set each equation for the slope This gives the one where Find the slope of the tangent line to each graph at the vertex. Find the slope of the tangent line parabola and place the equation of each parabola below its graph. the equation of each parabola parabola and place Graph a parabola of the form Graph a parabola 1 pair. Write each vertex as an ordered

point. Find the equation for the slope of each graph at any

8 3 4 2 9 8 = = _

2 1 y y C M R 12-3 3. Chapter 12 Exercises Find the vertex of each parabola. 1. Step 6 Step 5 Step 4 Step 3 Tangents and Vertices Tangents at any point to find the equation for the slope of a function Can you use the of the form vertex of a parabola NAME NAME Step 2 Step 1 C C P _ 6 3 0 _ 5 0 0005_036_PCCRMC12_893813.indd 20 0005_036_PCCRMC12_893813.indd 21 0 5 _ 0 3 6 _ P C C R M

C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 _ 8 9 3 8 1 3 the valuesfor Chapter 12 3. 2. 1. Exercises The equationinpoint-slopeformofthetangentlineto method ofapproximatingafunction. chosen point.Setupaspreadsheetliketheoneshownbelowtostudythis approximation ofthefunctionforvalues in point-slopeform.Thetangentlineofafunctionisusuallygood point. Thisslopecanthenbeusedtowritetheequationoftangentline You havelearnedhowtofindtheslopeofatangentlinefunctionat Using theTangentLinetoApproximateaFunction NAME row 3areremarkablyclosetothevaluesof copied totheothercellsinrow3.Noticethatvaluesof approximating function, entered incellB2as y values of 12-3 . i

n = d d that occurs? Use thespreadsheettoapproximatevalues of Use thespreadsheettoapproximatevaluesof Use thespreadsheettoapproximatevaluesof for where that occurs? for where point, where about theequationoftangentlinetograph ofthefunctionat x

2 = 1 x 0.7, 0.8,0.9, 1,1.1,1.2,and1.3.Usethedatatomakeaconjecture x x

= = 21 1.Inthespreadsheet,function 1.7, 1.8,1.9,2,2.1,2.2,and2.3.Whatisthemaximumerror x x x -

near = = 0.3, 0, byusingtheappropriatetangentline.Comparevalues 2, byusingtheappropriatetangentline.Comparevalues Spreadsheet Activity x areenteredinrow1,columnsB–H.Thefunction x - x

3 2 1 =

0.2, = 1. 1byusing 2 x x = 204 .408 .114 1.69 1.44 1.21 1 0.81 0.64 0.49 ^2 * - x B1^2andcopiedtotheothercellsinrow2.The BCDEFGH A

- 0.1, 0,0.2,and0.3.Whatisthemaximum error g 1 ( x ) = . . . . . 1.6 1.4 1.3 1.2 1.2 1.1 1 0.8 1 0.6 0.9 0.4 0.8 0.7 2 the function x

- 1, isenteredincellB3as x f g nearthecoordinateof f ( ( x ( x x ) ) ) = = =

2 x

DT PERIOD DATE x 2 x inrow2. 2 isapproximatedfor ƒ ƒ f

( - ( ( x x x f 1.Forthisexample, ) ) ) ( x = = = )

g = x x √ ( 2 2

x nearthepoint, x nearthepoint, x

) = 2 for at(1,1)is =

2 f 2 * ( B1 x x Pdf Pass )

- = -

1 in 1and x 2 is Glencoe Precalculus 33/17/09 11:37:25 AM / 1 7 / 0 9

1 1 : 3 7 : 2

5 Lesson 12-3

A M M A

9 2 : 7 3 : 1 1

9 0 / 7 1

x 8 . 2 4 = x . ) - 2 x x Glencoe Precalculus ( 15 0.

Pdf Pass f 3 = = = ) ) x x )

( ( x

( f f 1 and 2 1 and

, then f ) is x 14 - = x

( 3 - x f

2; + 0.2 3 , then ; 3 2 2 x x , then 2, then - x x 3

x - x 5 3 3 4

x 5

+ + = = = =

+ ) ) ) ) 2 +

x x x x

x 3.4 3 ( ( ( ( 4 f f f f 4 x x x If If If - - - 3

= = = = ). If x ) ) ) ) ( is x x x x DATE DATE PERIOD h ( ( ( ( n

f f f f , provided the limit exists. , provided the limit ±

) ) x . x ( 1 ( f - g

n 8. = -

Original equation Sum and Difference Rules Constant, Constant Multiple of a Power, and Simplify. Original equation Distributive Property Difference Rules Constant Multiple of a Power, and Sum and Simplify. )

) Limits were used in Lesson 12-3 to used in Lesson Limits were cnx 6. h x

h (

f = + )

x 0. x (

( is a constant and = f − f

)

of a function. The derivative of of a function. The ), then

x 0 ( x

is a real number, then is a real number, then 4. 0 ( f n → h 3

2. + lim

. - 2 ±

1 1 , where 1 ) n - − ). Then evaluate the derivative for the -

and x = - , then n ( n 1 x ) cx c x 4 g ( x

nx x x f

( = = = = 1 = 4 · ) ) ) derivative ) f · x x x ) x ; 0 and ( ( ( ( 2 x f f f f 3 and

4 2 ( x 3

5 4 4

If If The derivative of a constant function is zero. The derivative of a If f a real number, then If = x + -

+ 5) + - 3 2 −

x 4

x 1 x 5) 1 x x x - x - x 20

- - 2 2 3 3

5 2 7 - 3 5; 3 2 x 3

x - x 6 - x

- - 2 Study Guide and Intervention Guide and Study Derivatives - - + -

(4 1 2 7

· − 7 2 2 2

(4 x 4 x 8 · d x x x x x 4 d x n x - 2 28 6 3

i . 22 4

3 4 6 4 4 3 1 ======8 = 3 ) ) ) ) ) ) ) 9 ) ) ) ) x x 8 x x x x x x x x x ( ( _ ( ( ( ( 2 ( (

( ( ( ), which is given by ), which is given 1 Example f f f f f f f f f f f x C ( M Constant Multiple of a Power Sum and Difference Constant Power Rule

R 12-4 There are several rules of derivatives that are useful when finding the rules of derivatives that are There are several that contain several terms. derivatives of functions 7. a. Find the derivative of each function. Chapter 12 Find the derivative of each function. 5. given values of 1. 3. Exercises Find the derivative of b.

Derivatives and the Basic Rules Basic Rules and the Derivatives NAME NAME determine the slope of a line tangent to the graph of a function at any point. function at any the graph of a a line tangent to the slope of determine the This limit is called f C C P _ 6 3 0 _ 5 0 0005_036_PCCRMC12_893813.indd 22 0005_036_PCCRMC12_893813.indd 23 0 5 _ 0 3 6 _ P C C R M

C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 _ 8 9 3 8 1 3 h h derivative oftheproductorquotienttwofunctions. Chapter 12 f f Product andQuotientRules NAME 3. 1. Find thederivativeofeachfunction. Exercises Findthederivativeof Findthederivativeof 12-4 f f .

i n

( ( ( ( d Example 2 Example 1 ( ( x x x x d x x d h

) ) ) )

) ) ( ( 2 = = = = x x 3 ======) ) 2 4 2 10 4 (2

x = = f

- 4 −− f 4 −− −− 2 x x x

Product Rule Quotient Rule x x

23 x ( ( ( 2

- x −

x ( x − 3 x 4 x 3 x

x - x

)( 2 + ) ) 2

- 2 +

g 2

+ 4 4 ( + - 4 ( 10 Derivatives Study GuideandIntervention x 1 x [ 3 x + ) 1) 3 g x ) 3

- (

2 + - x x x

+ (2 x

) - 2 2 f 4 2 ] x

2 f x (

- + (2 2

Constant RulesforDerivatives Sum RuleforLimits,Powerand Original equation Constant RulesforDerivatives Sum RuleforLimits,Powerand Original equation x x ) ( )(2 2 x ) x 3 10 6

x ) ) -

g 2 x x ( +

8 x

4 + ( +

) ( x +

x 4)2

) 3) x

5 2

− If If

dx - d x x f f

and and

2)(6

⎣ ⎢ ⎡ - 4.

− g f ( ( 12 x x ) ) g g

⎦ ⎤ aredifferentiableat aredifferentiableat x

x = 2 2

Use thefollowingrulestofind +

−− - f

( 5) x 10 ) g ( h h x) g ( (

- ( x x x

) ) ) f ( 2

x = = ) g (

( − (2 m k x x ( g g x x ) ( DT PERIOD DATE x g g 2

. x , then and ( x

Simplify. Distributive Property Substitution Product Rule Simplify. Distributive Property Substitution Quotient Rule

2 ( ( x ( ( 2 - )

x x x x

- ) + = ) ) ) ) 2)(2 = 1) 4) = = = = g

(3 −

− ( 3 dx 2 d 2

x 6 2

x x x

) .

3 2 2 x x x x x ≠

+ - ( 3

2 - 3 0,then x -

4 + + ) 1 + 1 g 1)(

( 5

x ) x x

= ). 2 Pdf Pass

+ f

Constant RulesforDerivatives Sum RuleforLimits,Powerand Original equation (continued) Constant RulesforDerivatives Sum RuleforLimits,Powerand Original equation ( x 5 ) g x ( ) x ) + Glencoe Precalculus

f ( x ) g

( x ). 33/17/09 11:37:33 AM / 1 7 / 0 9

1 1 : 3 7 : 3

3 Lesson 12-4

A M M P

0 5 : 1 1 : 5

9 0 / 5 / 2

) 2 x

1 and 1 1; + 3 and 0

- 3

+ =

= ) x x )(

2 5, where 3) x x x ; 6 + 2

- 3 2 x -

2 1; 2 + -

x 6

x 6 3 - 2

( 3

2 3 x - x − x

x 7

( x 2 x 2 - √ t - -

(3 4

3 = 3 ======) Acceleration is the = ) ) ) ) ) x ) x x x x ( ) x x ( ( ( DATE DATE PERIOD ( t ( ( ( h q f n h m t v Hint: 6. . 4. x 3 2. 2. - 1 and 3 -

12. =

2 2 and 2 and 1 x x = ; -

5 4

x x - =

2 3 x 7; x 8. - ; 2

+ ) x 6 3

x 14.

+ 5

5 10. 1 5 2 4

- +

6 + - the velocity of a moving Acceleration is the rate at which + 4 Practice Derivatives 2 -

4 x 1 2

2 x x

x 5 x √ d - - 3 −

( n i . 24 = = = 3 = = = = 1 ) ) ) 8 ) ) ) ) 3 x x x x x x x 9 ( ( ( ( ( ( 8 ( is the time in seconds. Find the acceleration of the particle in is the time in seconds. Find the acceleration of the particle in _ f q f object changes. The velocity in meters per second of a particle moving object changes. The velocity in meters per second of a particle along a straight line is given by the function r p f g PHYSICS derivative of velocity.) t meters per second squared after 5 seconds. ( 2 1 C 9. 5. 3. M 1. R 12-4 Chapter 12 15. 13. 11. Find the derivative of each function. Find the derivative of each 7. Find the derivative of each function. Then evaluate the derivative of each the derivative Then evaluate each function. derivative of Find the values of for the given function NAME NAME C C P _ 6 3 0 _ 5 0 0005_036_PCCRMC12_893813.indd 24 0005_036_PCCRMC12_893813.indd 25 0 5 _ 0 3 6 _ P C C R M

C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 _ 8 9 3 8 1 3 Chapter 12 3. 2. 1. NAME 12-4 . i n d d h the diverisgivenby surface ofthewater.Theheight jumps fromacliff192feetabovethe a. measured inseconds. is measuredinfeetandtime h flying birdcanbedefinedby a. answering thefollowingquestions. 10 inchesandachangingradiuswhen Consider acylinderwithheightof height ofthebird. Find themaximumandminimum [1, 10],wheretime BIRDS c. b. GEOMETRY d. c. b. CLIFF DIVING height volume

( ( 2 t t 5 ) ) of thediveratanytime Find theequationforvelocity cylinder intermsofitsradius. Write aformulaforthevolumeof r Find thevalueof the volumeintermsitsradius. instantaneous rateofchange Find anequationforthe she hitsthewater? What isthediver’svelocitywhen hits thewater. Find thetimewhendiver 1 secondhaspassed. Find thevelocityofdiverafter

= = =

25 3inches. The height -

− - h 3 V 16 t andradius

3 ofacylinderintermsits

Derivatives Word ProblemPractice + t 2

Theformulatofindthe +

− 7 2

At time

16 t 2

+ t

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) when V =

= 0,adiver t .

π t is r 2 h h h . of

h ' ( t ) 4. 5. DT PERIOD DATE a. h at anytime second. Theheight with aninitialvelocityof80feetper straight upwardfromaheightof6feet c. b. VOLUME b. PROJECTILE a. side ofthecubeshownischanging. ( t ) 3.2 inchesto3.4inches. derivative of of theballatanytime Findtheequationforvelocity Explaintherelationshipbetween x of thevolume Findtheinstantaneousrateofchange ball at Findtheinstantaneousvelocityof the volumeformula. volume formulaandthederivativeof volume Findtheaveragerateofchange = =

4inches. - 16 Supposethelength t t

V Y 2 =

t ( Supposeaballishit

+ isgivenbythefunction x 2seconds. ) as 80 7 h Pdf Pass

Y V ( t

t =

x ( ). + x changesfrom h Y Y ) atthemoment 6. oftheballinfeet Y Glencoe Precalculus t byfindingthe x ofeach v ( t 112/7/09 1:28:31 PM ) 2 / 7 / 0 9

1 : 2 8 : 3

1 Lesson 12-4

P M M A

5 4 : 7 3 : 1 1

9 0 / 7 1

(cos = ) x ( f term by x … ? x +

! 4 4 x − + + .

x ! 3 , was shown to be a sum , was shown to 3 x −

DATE DATE PERIOD x

! 2 2 cos x −

2 = x

y x

sin + = 1 ) x = (

x f e , x e … … . … . . . - -

! 8 ! 9 8 x 9 − x −

? ? ? + + +

! 6 ! 7 6 x 7 − x −

term by term and simplify the result.

: e -

- using the series expansion of cos by differentiating the series expansion of sin

x ) ! ) 4 ! = 5 = = x x 4 x

5 − x

−

)

) )

x x x e +

dx + dx

(

6 Enrichment ! (cos 2 (sin ! 3 2 d −

dx dx 2 x 5.

3 d x −

d −

− −

d (cos (sin d d d n − −

- i - xe . 26

3 x 1 1

8 = 3 Find was also discussed in Chapter 10. Differentiate the series expansion of The series expansion for term and simplifying the result. term and simplifying the result. 9 ) = = 8

Thus, Thus, Find Find What function does this new infinite series represent?

What would you guess might be the derivative of cos So, So, So, So, x _ ( x x 2 1 f C M R 12-4 1. a. The power functions in these series expansions can be differentiated. in these series expansions The power functions b. of each function. 3 to find the derivative Use the results of Exercises 1– 4. c. 3. a. 2. a. b. c. b. Chapter 12 Powerful Differentiation Powerful transcendental functions were the series expansions of some In Chapter 10, the even function presented. In particular, NAME NAME and the sine function, being odd, was shown to be a sum of odd powers of being odd, was shown to and the sine function, sin cos of even powers of of even powers C C P _ 6 3 0 _ 5 0 0005_036_PCCRMC12_893813.indd 26 0005_036_PCCRMC12_893813.indd 27 0 5 _ 0 3 6 _ P C C R M

C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 _ 8 9 3 8 1 3 areas wouldgiveabetterapproximationof11squareunits. have lowerandupperestimatesfortheareaofregion,7 The areausingtherightandleftendpointsis157squareunits,respectively.Wenow total R area R R Approximatetheareabetweencurve graph ofafunction 2. 1. Exercises Chapter 12 Area UnderaCurve NAME rauigrgtedons Areausingleftendpoints R Area usingrightendpoints f rectangles withawidthof1unit(FigureB).However,thefirstrectanglehasheight of oneunit(FigureA).Usingleftendpointsfortheheighteachrectangleproducesfour Using rightendpointsfortheheightofeachrectangleproducesfourrectangleswithawidth endpoints oftherectangles.Userectangleswithawidth1. on theinterval[0,4]byfirstusingrightendpointsandthenleft 12-5 . (0) or0andthus,hasanareaofsquareunits. i n 4 3 2 1 d

d Approximate theareabetweencurve Approximate theareabetweencurve Use rectanglesofwidth1unit.Thenfindthe average forbothapproximations. interval [1,5]byfirstusingtherightendpoints andthenbyusingtheleftendpoints. Use rectanglesofwidth1unit.Thenfindthe average forbothapproximations. interval [0,4]byfirstusingtherightendpoints andthenbyusingtheleftendpoints. Example = = = =

2 1 1 1 7 1 4 8 · · · ·

0 f f f f 27 4 r8 R R R R (4) or8 (3) or4.5 (2) or2 (1) or0.5 Figure A 24 Area UnderaCurveandIntegration Study GuideandIntervention = x 1 totalarea 15 f ( x ) andthe You canusetheareaofrectanglestofindbetween x -axis onaninterval[ f f ( ( x x ) ) = = 3 - DT PERIOD DATE x x 2 2

+ + 4 3 2 1 ,

1andthe = = = = b 5 4 8 ] inthedomainof 1 1 1 1 x

0 + < · · · ·

Figure B 6andthe f f f f 24 area f (3) or4.5 (2) or2 (1) or0.5 (0) or0

( x ) x Pdf Pass = -axis onthe < x

= 15.Averagingthetwo

− 2 1 7

x x 2 -axis onthe andthe f Glencoe Precalculus ( x ). x -axis 33/17/09 11:37:49 AM / 1 7 / 0 9

1 1 : 3 7 : 4

9 Lesson 12-5

A M M P

6 4 : 1 5 : 3

9 0 / 8 / 2

x Δ 1) ) i + x

(

5 n −

Glencoe Precalculus 6

Pdf Pass

1 n = - 1)(2 n

x ∑ i + b

−

(

= − 2 n

→ ∞ x lim x

and Δ (continued) = n

4 i

Δ n 2 5 −

i =

)

i =

x n i = (

Expand and factor. ∑ i Simplify and expand. Factor and divide each term by Limit theorems Simplify. Definition of definite integral f x dx

) x

( f

]

dx 2 ⌠ ⌡ b a 1 n −

3) ∞ 2 dx + x

→

3 lim 4 2

n x

⌠ ⌡ 5 0 4 (

DATE DATE PERIOD

+ ⌠ ⌡ ⌠ ⌡ 3 4

1 2

)

1 n −

∞ →

lim n

( )

)

3 1)

∞ +

→

lim n ) n

(

) 6

2 1 1)(2 n + + −

) + are the lower limits and upper limits, respectively, are the lower limits

n 2

2 i + b i

5 n n

x −

n 3 ∞

(

+ 3 n 2 −

− 0] or about 166.67 square units n + 1 Δ

) n → Δ a )

i 2 lim

i and + i ∑ · + 2 n n x

a i 5

= −

( [

− 2 2 2 i x (

f x 2. ) n n

4 4 ( ( 25 25 − −

( (

-axis on the interval [0, 5], or -axis on the interval

3(0) 6 n

1 1 = 6 6 x where The area of a region under the graph of a function is under the graph of a function is The area of a region and

n n 500 −

i ∑ = = n n +

500 20 20 500

− i i − − −

∑ ∑

∞ ∞

∞ ∞ ∞ ∞ ∞ ∞ [2

Study Guide and Intervention Intervention Guide and Study and Integration a Curve Area Under → →

lim dx lim → → → → → → 8 6

n lim lim lim lim lim lim n 2

)

n n n n n n 500

( −

d x

28 = n i

======. and the + dx 3 2 1

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x (1 9

x 2 8

4 _ x ⌡ ⌡ ⌠ ⌠ 6 2 Example 2

0 4

1 4 =

Definite Integral M ⌡ ⌠ 12-5 5 1. 3. Exercise and Use limits to find the area between the graph of each function the

y between the graph of Use limits to find the area of the region Chapter 12 0 Integration NAME NAME

R C C P _ 6 3 0 _ 5 0 0005_036_PCCRMC12_893813.indd 28 0005_036_PCCRMC12_893813.indd 29 0 5 _ 0 3 6 _ P C C R M

C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 _ 8 9 3 8 1 3 5. x Use limitstofindtheareabetweengraphofeachfunctionand 3. Chapter 12 9. 7. 1. with awidthof1. indicated intervalusingtheendpoints.Userectangles Approximate theareabetweencurve NAME 12-5 . i n -axis givenbythedefiniteintegral. d d g left endpoints [0, 4] left endpoints [1, 5] f Architecture andDesign 1 0 y of thewindowcanbemodeledbyparabola stained-glass windowforanewbuilding.The shape 3 2 ⌡ ⌠ ⌡ ⌠

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29 3 x

x 0.05

x + )

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6. 4. 2. 8. Adesignerismakinga f ( x ) andthe right endpoints right endpoints [1, 6] [2, 5] f - 1 6 p

1 ⌡ ⌠ ⌡ ⌠

(

( DT PERIOD DATE 2

x 6 x ( ) - ) x = 2 = x

dx 2

- -

x + 2 2 x

x +

x + -axis onthe 2 6

11)

− - 10 Pdf Pass 4 dx −

5 − − 10 10 5 y Glencoe Precalculus 5 10 x 112/5/09 5:19:04 PM 2 / 5 / 0 9

5 : 1 9 : 0

4 Lesson 12-5

P M M A

2 0 : 8 3 : 1 1

9 0 / 7 1

dx x x

. = 7) Glencoe Precalculus Pdf Pass - dx

y x

8 4) + + y

coordinate plane, On a 2 5 and x x ( 0

4 =

⌠ ⌡ 5 - (

Mr. Bower is seeding part x

⌠ ⌡ 7 1 area of the triangle using its height area of the triangle using its height and base length. by Calculate the area of the triangle evaluating Find the height and length of the base Find the height and length of the the of the triangle. Then calculate Shade the interior of this triangle. Shade the interior of his lawn, but he has only enough seed to cover 35 square yards. If the area in square yards that he needs to seed can be found by c. GRASS SEED TRIANGLE AREA TRIANGLE b. draw the triangle formed by the triangle formed draw the and the lines have enough seed to complete the task? Explain. a. DATE DATE PERIOD 5. 4. is given is given x x 4 x 4 + 2 2 x + x x is given in feet? 5 2 3 x x - + 4 3 x x x =- x y = y =- y 0 y y 0 a dog is building Charlie y 0 0 Word Problem Practice Problem Word and Integration a Curve Area Under 3

The entrance to a coal mine is d d The face of a dam is in the n i . 30 3 1 8 3 9 8 _ 2 1 house for Fido. The entrance to the dog to the dog Fido. The entrance house for shape of the region house is in the the area of the entrance shown. What is if to Fido’s dog house DAMS MINING DOG HOUSE area of the face of the dam if in kilometers? shape of the region shown. What is the in the shape of the region shown. What in the shape of the region shown. is the area of the entrance if in meters? C M R 12-5 Chapter 12 3. 2. 1. NAME NAME C C P _ 6 3 0 _ 5 0 0005_036_PCCRMC12_893813.indd 30 0005_036_PCCRMC12_893813.indd 31 0 5 _ 0 3 6 _ P C C R M

C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 _ 8 9 3 8 1 3 4. Given Chapter 12 1. for thesamething.Youhavealreadyseenthisincaseofderivative. branches ofmathematics.Inaddition,thereisoftenmorethanonenotation There isalotofspecialnotationusedincalculusthatnotother Reading Mathematics NAME following: at aspecificvalueof is 3. function Yet anothernotationforthederivativeofafunction 2. 6. The Leibniznotationforthederivative For example, used toindicatehigher-orderderivatives. the notationdevelopedbyIsaacNewton.Eachofthesenotationsalsocanbe or moreformally,“thederivativeof 12-5 . i n d not d f Let What istheorderofeachderivative? List severalotherwaysofexpressingthisquantity. f a.

3 (2) (0) 1

afractionofanykind.Toindicatethevaluederivative f

f f ( ( ( x 31 x y x ) 7. )

)

− = dx dy = = 5.

Enrichment x

( 2 x

f x . Whatdoes x

3 ).

( = + x ) 2

3 −

dx d , read“ 2 x y x 2 2

usingtheLeibniznotation,onemightuse ,and

- b. 4,findthevalueofeachexpression.

y . dy dx

ÿ lim

→ allindicatethesecondderivativeofsome 0

evaluatedat ( − x

+ y

h with respectto h )

− dx dy -

isusuallyread“ x 2

c. find?

x −

dx d

4 = y 4

− −

2.”

dx d dx dy DT PERIOD DATE 3 y 3

.” Notethat x = x

= -

= f ( 4 x dy dx

) is 1

y . d.

. Thiswas ,”

y −

dx dy Pdf Pass

Glencoe Precalculus 33/17/09 11:38:09 AM / 1 7 / 0 9

1 1 : 3 8 : 0

9 Lesson 12-5

A M M A

2 1 : 8 3 : 1 1

9 0 / 7 1

. = x

) x ( is a

k F ) if 1 and x 1, ), respectively, ( - 2 ). x f - ( x ( - G

3 ± 5 2 √ x

−

)

x ) and ( x = = ( F ) ) F x x ( DATE DATE PERIOD ( n g ) are x ) are ( x ( g

g C. ±

) + x

( f 1

) and 1 + x n ( + f

n kx − Constant Multiple of a Power Simplify. Original equation of Rewrite the function so each term has a power Use all three rules. Simplify. Original equation ) is an antiderivative of ) is an antiderivative is a rational number other than is a rational number x is a rational number other than is a rational number = = n ( n )

x F

(

F 1 1 Rules for Antiderivatives + +

+ 0

, where 1 x

1 n 0 , where + − 2 C n

+ n

x kx x

n −

+ - = =

= ) )

x x ) 0 1 1 ( ( x f f 2. 2 + x ( + 2 C then the antiderivatives of If constant, then If the antiderivatives of If the antiderivatives If F 2 5 - 2 2 x

), we say that

2 − 4 3

- x - C - - 4. +

(

2 f 2 2

+ 3 x 4 x x

−

+ x x 1

1 1 2 − + 6

3 4 5 4 4 5 1 5 x + + 1

x - - x x 2 Study Guide and Intervention Guide and Study of Calculus Theorem The Fundamental + 4 + + +

+ + 3 1 2

5 −

3 6

4 −

3 3 3 x d 3 x − x

- 4 1

d x - - x - − x Power Rule Sum and Difference Constant Multiple of a Power

3 4 i

− . 32 2

3 = = 1 ======8 = 3 ) ) ) ) ) ) 9 ) ) x x x x 8 x x x ( ( ( ( _ x ( 2 ( ( (

f f 1 F F Example

f t f f C M R 12-6

1. 3. Exercises Find all antiderivatives for each function.

b. a. Find all antiderivatives for each function. Chapter 12 Antiderivatives and Indefinite Integrals and Antiderivatives Given a function NAME NAME C C P _ 6 3 0 _ 5 0 0005_036_PCCRMC12_893813.indd 32 0005_036_PCCRMC12_893813.indd 33 0 5 _ 0 3 6 _ P C C R M

1. Evaluate eachintegral. Exercises

b. a. Chapter 12 The FundamentalTheoremofCalculus NAME f 12-6 . Calculus Theorem of Fundamental i ( n x

d 2 ⌡ ⌠ d

4 2 4

) isdefinedby 1 Example ⌠ ⌠ ⌡ ⌠ ⌡ ⌡ ⌠ ⌡

⌡ ⌠

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( C

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3 x x x isanyconstant. 3 x x 3 x 2 -

- 7 2 2 - -

33 + 1) + 1)

4 x 4 2 dx dx ) x The FundamentalTheoremofCalculus Study GuideandIntervention x dx

If The rightsideofthisstatementmayalsobewrittenas dx

= = - = - F 1) 60 ( Evaluate eachintegral. 1)

( x (

⌡ ⌠ − ) istheantiderivativeofcontinuousfunction 4 4 − x 4 4

f 4 4.

( - - x 6or54 ) = = = 4

dx x

) )

x − 3 3

−

⎢ ⎢ ⎢ 2 - 3 = 3 x x 2 4

2 3 + +

( F + 1 1 2

− 2 2 (

4 x

− x

+ ) - 2 2 x

2. 4

−

- 1 C

x ) x 1 +

, where

x + + 1 1

C -

− 0 x F 0 + + ( 1 x 1

) isanantiderivativeof

+ ⌡ ⌠ - 1 2 ⌡ ⌠

DT PERIOD DATE The indefiniteintegralof

(

C ( x x 2

+ f - ( x Simplify. b Fundamental TheoremofCalculus 1) F Constant MultipleofaPower Simplify. Simplify. 2 ), then

( = x x 4and ) dx

⎢ ⎢ ⎢ + a b

. 1) b a ⌡ ⌠

f ( = dx x ) (continued) 2 Pdf Pass

F ( b f ) ( x - )

F Glencoe Precalculus ( a ). 33/17/09 11:53:38 PM / 1 7 / 0 9

1 1 : 5 3 : 3

8 Lesson 12-6

P M M A

1 2 : 8 3 : 1 1

9 0 / 7 1

(30

Glencoe Precalculus

p Pdf Pass ⌠ ⌡ 0

h 3 dx -

3) x dx + 2

3 ) x + x

+ )( 2

6 x x x

+ 2 8

- 3 = = x x dx (1 ) ) pieces is given by

x x

2 DATE DATE PERIOD p ( (

1 5 ⌠ ⌡ ⌠ ⌡ ⌡ ⌠ 2 -

f f

2. 4. . How much work is required to compress the hours to create one piece of furniture. h x dx

⌠ ⌡

0 =

10. A craftsman works ) x dx 5 )

+ x

2 3) 3

x - -

2 compress a certain spring a distance of The work in foot-pounds to 4 Practice of Calculus Theorem The Fundamental 3 3 3 x

x (

d 5 4 x 6. d

4 x n i . - 34 6

3 = = ( dx 1

) ) - 5 3 1 ( 8 x x

9 ⌠ ⌡ -

8 ( ( - _ ⌠ ⌠ ⌡ ⌡ Suppose the number of hours needed to create f its natural length is given by

f WOODWORKING PHYSICS How many hours does it take the craftsman to make 6 pieces? spring 6 inches from its natural length? 2

1 C M 3. 1. R 12-6 Chapter 12 12. 11. 9. 7. Evaluate each integral. 5.

Find all antiderivatives for each function. antiderivatives Find all NAME NAME C C P _ 6 3 0 _ 5 0 0005_036_PCCRMC12_893813.indd 34 0005_036_PCCRMC12_893813.indd 35 0 5 _ 0 3 6 _ P C C R M

C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 _ 8 9 3 8 1 3 Chapter 12 3. 2. 1. NAME 12-6 . i n d d is required? jump inphysicaleducationclass.The a. feet persecond. seconds andthevelocityisgivenin v velocity ofherjumpcanbedefinedas given by 36 inchesbeyonditsnaturallengthis joules, requiredtostretchacertainspring VERTICAL JUMP SPRING STRETCHING ADVERTISING b. for logo occupyatthetopofeachdocument its letterhead,howmuchspacewillthe If thecompanyintendstouseitaspartof is intheshapeofregionshownbelow.

( 3 t

5 ) jump. Assumethatfor Find thepositionfunction before shelandsontheground? After Lilajumps,howlongdoesittake x =- between0inchand1inch? 35 32 0 ⌡ ⌠ 3 The FundamentalTheoremofCalculus Word ProblemPractice t 80 + 24,where x dx. 0 y NewWave’sbusinesslogo y Lilatestedhervertical = How muchwork x 4 - Thework,in 2 x t 2 x isgivenin + t = 1 0, s ( t ) forLila’s s ( t ) = 0. 4. 6. 5. 2 inches? required tocompressthespringanother given by compress thespringanother2inchesis The work,ininch-pounds,requiredto from itsnaturallengthof12inches. 800 poundscompressesaspring2inches by [0, 3],ifthevolumeofsolidisgiven VOLUME BILLBOARD SPRING COMPRESSION the graphof volume ofthesolidformedbyrevolving What istheareaofthisfigure? in feet,isshownthediagrambelow. central figureonthebillboard,measured billboard toadvertisethecompany.The Trucking Companyhaspurchaseda DT PERIOD DATE 0 ⌡ ⌠ 3 π ( x 2 2 ⌡ ⌠ 4 - - - )

Inthefigurebelow,find 2 400 3 6 9 9 6 3 10 dx. 2 4 6 8 TheSquaredandLinear 0 f y

f ( 0 ( y x dx x x y ) ) = = 246810 Pdf Pass = x 2 x . Howmuchworkis 2 2 x 2 overtheinterval y =- 4 A force of Glencoe Precalculus 3 x x + 18 x 33/17/09 11:38:24 AM / 1 7 / 0 9

1 1 : 3 8 : 2

4 Lesson 12-6

A M M A

1 3 : 8 3 : 1 1

9 0 / 7 1

x +

3 Glencoe Precalculus x Pdf Pass

x x 4 − e . 2

x e

ln (sin ln (2 dx du

− · = = = =

1 u

− y y y y dx du − u is e . u

is 3

u ln x e +

2 =

2 = y x −

y 3). +

DATE DATE PERIOD 2 9. x . x 5) 3 + e 12.

ln (

x x = =

3. 6.

x x

y y .

and the derivative of √ e

. Simplify to get

2 and the derivative of and the derivative of x 1 e x −

−

x x 3

4 ln (2

e 2 x + · is = = = = 2

2 x is x x y y y y −

3 e

ln + 1 .

= =

, 1). y x

− y e dx . du −

. u x

3) is

· 3 e

e + 1 u dx −

−

, or

2 · x x

= u 2 is 3

x 3 or 3 · 3 e

· e dy dx

− 1 u ln (

−

= u

= = e

dy dx The derivative of

−

y y = The derivative of

Find the derivative of Find the derivative of Find the derivative dy dx −

8. dy dx −

, 11. ) x ) 6 Enrichment

3. Then 3 3

x 1

through the point ( − x

2. 2 3, + d ( . Then

x 5.

d x x x = = 2 n i - 6 .

x 3 e e 3

ln ( ln 8 dx dx du du 3

− − ======9

8 _ u u y Find an equation for a line that is tangent to the graph of y y y y 2 1 C Example 2 Example 1 M 7. Exponential Rule Logarithmic Rule R 12-6 Since Chapter 12 1. The derivative of Exercises function. Find the derivative of each

The derivative of The derivative Derivatives of Exponential and Logarithmic Functions and Logarithmic of Exponential Derivatives NAME NAME 13. 10. 4. Since

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Chapter 12 5. 1. function atthegivenpoint. Find theslopeoflinetangenttographeach f Estimate eachone-sidedortwo-sidedlimitfor NAME 4. Evaluate eachlimit. NAME DATE NAME DATE PERIOD PERIOD 1. . i ( n d x)

12 12 d B A A y Find anequationfortheslopeofgraph d y d y

x x x x

lim

3 ( ( lim -∞ → → - → →

lim = = = = 7 t t ) ) -∞ m m

2 12

3 2 ⎧ ⎨ ⎩

x = =

(

= = - x

300

37 3

(3 − - - x x

(

- x x 6 6 5 2

16 2 + x ) - - x 4 + x x 2 5 2

(Lessons 12-1and12-2) Chapter 12Quiz1 (Lesson 12-3) Chapter 12Quiz2

2 2

16 t 1 if 1 if x - - 4

- ; ( x 2 + ; (3,0)

d 2 16

- 2 10 - anobjectisabovetheground )

90 x 1, 1) x x 2 t

+ 2 x < ≥ ; x t

t ; atanypoint. 4)

2 2 t =

B , ifitexists. = 2 2. 4 -

x 3 lim → 2

f d ( x ( t ) ). Findtheinstantaneous D C C 3 m m

= = 6 6

t . x DT PERIOD DATE 2 t 3.

seconds + 10 x lim → D x 2

- ∞ f ( x 2 )

6. 5. 4. 3. 2. 1. 5. 4. 3. 2. 1. Pdf Pass SCORE SCORE Glencoe Precalculus 112/7/09 12:09:58 PM 2 / 7 / 0 9

1 2 : 0 9 : 5

8 Assessment

P M M P

7 4 : 5 2 : 5

SCORE SCORE 5. 1. 2. 3. 4. 1. 2. 3. 4. 5. C +

3 x 4 3 -

−

D D DATE PERIOD

x 4 3 7

−

C C

+ x

6 2 x

14 −

B B -axis given by the definite integral. 1 x +

3 x x 5) x

dx + 6 3

- 1 dx ) dx

Chapter 12 Quiz 4 (Lesson 12-6) Chapter 12 Quiz 3 12 Quiz Chapter 12-4 and 12-5) (Lessons

8 x - - 2 3 x

(

3 ) -

5 2 2 2 x

1) 3 4 x d 3 x x x x - d 2 x

− n -

i 38 - + . 4 3 7

2 x 3

2 2

1 - 6 ======8 x x dx 2 x 3 3

14 −

9 ) ) ) ) ) ) 2 x - (5

( x x x x x x

( x _

( ( ( ( ( ( 2

1 4 2 ⌡ ⌠ ⌠ ⌡ f

f f f f f 0

1 A A ⌠ ⌡ ⌠ ⌡ 0 1 12 12 C M R 2. 5. function and the 4. Use limits to find the area between the graph of each the area between the graph Use limits to find NAME DATE PERIOD DATE NAME 4. Evaluate each integral. Chapter 12 1. Find all antiderivatives for each function. Find all antiderivatives for Find the derivative of each function. derivative of Find the 1. 5. 3. 3. 2.

NAME C C P _ 6 5 0 - 7 3 0037-056_PCCRMC12_893813.indd 38 0037-056_PCCRMC12_893813.indd 39 3 7 - 0 5 6 _ P C C R M

C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 _ 8 9 3 8 1 3 6. 7. 1. Estimate eachone-sidedortwo-sidedlimit. Chapter 12 NAME SLINGSHOT 9. 8. 10. h rock an upwardvelocityof40feetpersecond.Supposetheheight 5. 4. 3. Evaluate eachlimit. 2. . Part 2 i Part I n ( d 12 t d )

A A 3 Evaluate Find anexpressionfortheinstantaneousvelocity

What isthemaximum height thattherockwillreach? For whatvalueof How fastistherocktravelingafter2seconds? y Find theslopeoflinetangenttograph

F F x x x x = 9 lim h lim lim lim

→ ∞ → → - → =

, infeet, - -∞ -∞ -∞ -∞

1 2

39 3 Write theletterforcorrectanswerinblankatrightofeachquestion.

x −

x − +

− 2

6 -

2 t

− 2 5

- 2 x -

x x 4 x 4 + (Lessons 12-1through12-3) Chapter 12Mid-ChapterTest 3 x

2 3 2 7

x Jerry usesaslingshottolaunchrockintotheairwith

x lim + - - 40 - x

x → t

+ secondsafteritislaunchedmodeledby 6

+ 9 x + 2 x 0

8atanypoint. t

− 3 + 8

x 5.

+ t

will therockreachitsmaximumheight? e 4 G G B B −

8 x

. - 0 - - 3 1 3 H H C C 3 1 3

− 6 5

DT PERIOD DATE v ( t ) oftherock.

D D J J

Pdf Pass ∞ ∞ ∞ ∞ SCORE 10. 9. 8. 7. 6. 5. Glencoe Precalculus 4. 3. 2. 1.

33/17/09 12:43:30 PM / 1 7 / 0 9

1 2 : 4 3 : 3

0 Assessment

P M M P

4 3 : 3 4 : 2 1

9 0 / 7 1

SCORE 3. 4. 5. 1. 2. one-sided limit one-sided regular partition right Riemann sum tangent line two-sided limit upper limit ) is an ). DATE DATE PERIOD x x from either ( ( c f F

= ) x (

F , where C

approaches + nite integral x ) x ( indeterminate form change instantaneous rate of instantaneous velocity integration lower limit direct substitution indefi F . ) as

c x ) provided is any constant. = ( x

f C ( f dx ) x ( ) and f

x ( ⌡ ⌠

f ) is said to be a(n) (antiderivative, ) is said to be a(n) x ( F Chapter 12 Vocabulary Test 12 Vocabulary Chapter 0 4

d d n i 40 . 3 1 8 3 nite integral 9 ) is defined by 8 x _ ( 2 derivative) of a function The function antiderivative of (One-sided limits, Two-sided limits) are used when we look (One-sided limits, Two-sided limits) at the value of a function instantaneous velocity The result of finding a derivative of a function is called a The result of finding a derivative operator). (differential equation, differential indeterminate form the left side or right side of The (definite integral, indefinite integral) of a function The (definite integral, indefinite f The slope of a nonlinear graph at a specific point is called The slope of a nonlinear graph tangent line) of the the (instantaneous rate of change, graph at that point. 1 differential operator differentiation antiderivative defi derivative difference quotient differential equation 12 C M R Chapter 12 Choose the correct term in parentheses to complete each sentence correctly. term in parentheses to Choose the correct 1. NAME NAME 2. 7. 5. Define each term in your own words. 6. 4. 3. C C P _ 6 5 0 - 7 3 0037-056_PCCRMC12_893813.indd 40 0037-056_PCCRMC12_893813.indd 41 3 7 - 0 5 6 _ P C C R M

5. 4. Evaluate eachlimit. 7.

Chapter 12 For Questions1and2,usethegraphof Write theletterforcorrectanswerinblankatrightofeachquestion. NAME

3. 6. 10.

8. 2. . i n d 12 d

Find theslopeoflinetangenttograph

Findanequationfortheslopeofgraph

Find anequationfortheinstantaneous velocity

B A A A A A 4

y object isdefinedas How fastisthegolfballfallingafter3seconds? FALLING OBJECTS MOTOR HOME The positionofthegolfballafter for $150,000is

F F F F F x x x x 1

lim x lim lim

lim → → - ∞ → ∞ →

12 $150,000

0 3 1 lim = → m -∞ - ∞ − v 4 1

4 3

x 1 (

41 32 ft/s

f = ) f −

−

5 √ ( 3 2

( -

= x x x x x

) - 3 1atthepoint( ) x 2 -

Chapter 12Test,Form1

- + 3 - 4

t 4 -

2 7

x x 2

v + After ( t 3 ) = Kyledropsa golfballfroma1600-footbuilding. h G G G G G D B B B B C ( 150,000(0.92) t

9 $100,000 2 2 3 ) t = m v - − −

- years,thevalue 5 3 2 1 (

5 96 ft/s 2, t

= )

= - - 5

6 9). t 2 t

+ t

secondsisgivenby t t 2 . Estimate foranypointintime H H H H H f C C C C y ( v x

)

$75,000 0 1

1 ofamotorhomepurchased = m - v -

( f 9 144 ft/s t

( y = )

0 x t DT PERIOD DATE

lim = → ∞ = ) belowtofindeachvalue.

v y -

(

- t - 4

) iftheheightofan

x 2

x + ( 2 t s

2 ). ( + x t ) t 5

t = . x

atanypoint. D D D D - J J J J J

$0 0

0 1456 ft/s 16 Pdf Pass m ∞ v -∞ - ( t 12 t 2

) SCORE +

- 1600.

- 4 6 x Glencoe Precalculus

+ 5 10. 9. 8. 7. 6. 5. 4. 3. 2. 1.

33/17/09 12:43:39 PM / 1 7 / 0 9

1 2 : 4 3 : 3

9 Assessment

P M M P

9 2 : 3 3 : 5

9 0 / 5 / 2

20. B: 15. 16. 17. 18. 19. 11. 12. 13. 14.

6 2) C C +

Glencoe Precalculus x x Pdf Pass + +

2 (3 x −

3 1 = -

- -

)

1 3

− 4 x 2 2 x 110 ft/s x (

x x 1 4

− - g 3

13 6 joules 6 4 32 6 J J J J

D D D D D

x x 2

)

32 32 4

C 2

C - - - -

(continued)

- 17 + 2 2 x

+ x x x - ( 3

2 4 4 3 − −

−

3 = = = - DATE DATE PERIOD - (2.5). ) )

)

4 x x x x 4 2 h ( ( ( x . How much work is required?

x x 1 4 g − h

h seconds is given seconds is given t . 26 5 joules 5 3 32 ft/s 0 8 6.75 2 x dx x J C C C C C H H H H H

1 ⌠ ⌡ 0

- 5. Find

2) 3 ≤

17 C - t

- 8 x

+ ≤

= (3 + −

) x 2

x = x 1 10 (

- f )

- - 2 3 − - x

( x x x 4 . g x 10 for 0 dx 4 joules 4 2 3 8 ft/s 5 26

3 + B B B B B G G G G x

- 2 2 x t )

32 2 4 ⌠ ⌡ 1

- 16

2 C C -

x x

( t

x +

dx x 12 + dx 2 −

4 3

−

2 ) - 2 8 )

- 3 +

The work, in joules, required to stretch a spring one more foot, whose x x Chapter 12 Test, Form 1 Form 12 Test, Chapter 80 x 2

x 3

- a ball in feet after The height of x x 4 = =

x 5 4 x

3 2

= x 2 - - 3 ) ) 4 −

) − d

- d - ) dx

(4 x x - x

2 2 n

- t

( ( 42 ( . 2

2 x x x ( 4 3 = = = 3 = x 1 x x h g h ) Evaluate Evaluate h )

8 x ) ) 5 x 3 (

3 joules 3 4 ft/s 110 39 60.75 3 (3 x x 9 ( 2 (

8 ( ( 3 2 ⌡ ⌠ ⌠ ⌡ ⌠ ⌡ _ F F F F F f

g HEIGHT SPRINGS

h f natural length is one foot, is given by natural length is one foot, is given by by 0 - A A A A A G 2 Find all antiderivatives of

1 12 C M R

16.

Chapter 12 18. Find the derivative of each function. derivative of Find the 11. NAME NAME Bonus

15. 17. Evaluate each integral. Use the Quotient Rule to find the derivative of each function. Use the Quotient Rule to find 14. 12. 13. 19. 20.

C C P _ 6 5 0 - 7 3 0037-056_PCCRMC12_893813.indd 42 0037-056_PCCRMC12_893813.indd 43 3 7 - 0 5 6 _ P C C R M

Chapter 12 For Questions1and2,usethegraphof Write theletterforcorrectanswerinblankatrightofeachquestion. NAME

3. 6. 10.

. i n 1. 2. d 12 d

Find theslopeoflinetangenttograph

Findanequationfortheslopeofgraph

Find anequationfortheinstantaneous velocity

B A A A A A 4

purchased for$7000is object isdefinedas How fastisthesoftballfallingafter3seconds? FALLING OBJECTS POP-UP CAMPER The positionofthesoftballafter

F F F F F x x x x x 3 lim lim lim lim → → - ∞ → ∞ → → -

2 0

3 0 lim v - -∞ m ∞ - 4 1

(

43 1 1332 ft/s 1

)

− x

2 − - ( = x ( 2(

x -

− x

x 5 1 3 -

) Chapter 12Test,Form2A ) 3

− x 16

x 2 1

4 3 3

- - +

3) 6 4 2 x x

2 t

After Taneshadropsasoftballfrom1300-foot building. h G G G G G D B B B B C ( t

1 $1000

1 1 1 2 2 ) v = m - - v ( ( t t

1300 ft/s t

− ) years,thevalue 5 3

√ = )

= t

7000(0.89) +

−

2 1

t t 2 − t 2 1 foranypointintime

secondsisgivenby

+ 2 f t ( H H H H H . Estimate x C C C C y )

0 $5500

4 0

1 = v m v - - ofapop-upcamper

( f 0 1 96 ft/s t

( y = ) x DT PERIOD DATE

y = = ) belowtofindeachvalue. 2 v ( y ( x t − 2 1

) iftheheightofan x

+ t t

lim - + → ∞

− − 2 x 2 1 s 3)

( atthepoint(1,2). + t t

)

2 v atanypoint. = ( t

- ). D D D D J J J J J 16

$7000 8

0 32 ft/s Pdf Pass m ∞ -∞ v - t ( 2 2 t

) = + SCORE = 1300. x

−

2 1 +

t 3 Glencoe Precalculus - − 2 1

+ 2 10. 9. 8. 7. 6. 5. 4. 3. 2. 1.

33/17/09 12:43:53 PM / 1 7 / 0 9

1 2 : 4 3 : 5

3 Assessment

P M M P

9 5 : 3 4 : 2 1

9 0 / 7 1

20. B: 15. 16. 17. 18. 19. 11. 12. 13. 14.

)

x C )

2 17 x 2 x 12 + + -

Glencoe Precalculus - - 14

Pdf Pass x x (3 −

(3 −

2 2

x =

- =

) 4 ) - - x

x x

2 ( 3 ( 1 4 x x

− g h C 100 ft/s 100 3 40 joules 40

202.5 3

J J J C J D D D D D +

3 2 +

x ) 4 2 8 ) x 4 (continued) C 2 - 4

- 2 - 2 - 5 + - + x

x 2 12 + x (3 x (3 14 − -

−

18 9

2 2) + + 1 . = =

+ x 4 4 -

DATE DATE PERIOD ) ) +

4 x x

x x x 4 x

(

x ( seconds is given seconds is given . How much work is required? t 1 4

− - g h

(2). 6( 2 dx 74 ft/s 74 8.4 32 joules 32 6 12 135 60

x h J x C C C C H H H H H C

2 2 ⌡ ⌠ after

+ )

x 5 C )

2 x 2 x x 12 +

+ - - 4. Find 2 14 12 x (3 −

≤ (3

− =

2 t )

= = x - ( ) 2

≤ ) f 4 x

x - ( x

(

. 1 4

x − g

h C 0 ft/s 0 12.62 28 joules 28 95 6

for 0 C

B B B B B t G G G G

2 1 − +

- - + 2

x x 64 2 5

4 x ⌠ ⌡ 0 2 + 4 +

x )

2 2 x 2 2 - C 2 x t 2

- 3

+ x -

x x dx - 2 - 2)

+ x 3 16 dx

x 4 3 (3 )

− 2

− x 2

- +

9 -

x The work, in joules, required to stretch a spring 2 inches from its

x 2 + Chapter 12 Test, Form 2A Form 12 Test, Chapter

+ 4 - - 2

3 2 a ball in feet The height of x 4 x = = +

− 2 4)

12 3

= 3 3 3 − dx x

− +

) 6 ) d

d - ) 3 3( 3 x - + x x

6 n

t i

( 32 ft/s 44 ( 2 . x 3 = x x ( = 3 4 3 = = x x 1 - x h g x ) Evaluate Evaluate h ( ) 8 ) ) x 3 x

16.2 14 joules 14 6 12 68 2

(4 0.8 x x 9 ( (

8 ( ( 3 3 ⌡ ⌠ ⌠ ⌡ ⌠ ⌡ _ F F F F F h natural length of 3 inches is given by natural length of 3 inches is given by g f f HEIGHT SPRINGS

by 1 0 A A A A A G 2

Find all antiderivatives of

1 12 C M R Use the Quotient Rule to find the derivative of each function. Use the Quotient Rule to find 14.

15. Chapter 12 19. 12. Find the derivative of each function. derivative of Find the 11. NAME NAME Bonus

16. 13. 18.

20.

17. Evaluate each integral. C C P _ 6 5 0 - 7 3 0037-056_PCCRMC12_893813.indd 44 0037-056_PCCRMC12_893813.indd 45 3 7 - 0 5 6 _ P C C R M

9. 6. Chapter 12 For Questions1and2,usethegraphof Write theletterforcorrectanswerinblankatrightofeachquestion. NAME 5. 4. Evaluate eachlimit. 10.

7. 2. 3. . i n d 12 d

Findanequationfortheslopeofgraph

Find theslopeoflinetangenttograph

Find anequationfortheinstantaneous velocity

G B A A A A A 4

object isdefinedas pebble fallingafter2seconds? at thepoint(2,

FALLING OBJECTS YACHT of thepebbleafter is

F F F F F x x x x x 5 lim lim lim → → - ∞ → → ∞ → -

0 lim lim v m - m m -∞ -∞ − - - 2 5 ( 7 0

t (

45 2 836 ft/s 1 3 )

= = = )

f − = x

x - −

(

= ( 2(

x f After 2 x 200,000(0.97) - x ( 5

) - x Chapter 12Test,Form2B 5 x

− x - 1 3 7 x ) 3 2 2 49

x 7) - - - - 4 −

4 1 2

24 4)

+ t x years,thevalue - x 2 2

1 +

2).

+ t x Titodropsapebble froma900-foottower.Theposition h secondsisgivenby 48 48

G G G G D B B B C B ( t t

$100,000

2 ) . Estimate = m - - - v - (

64 ft/s 7 5 2 t

− 3 2 = )

√

- t

− 4 3 6

t v

− 1 2

ofayachtpurchasedfor$200,000 t - lim → ∞ for anypointintime t 2

( v x s H H H H H ) C C C C ( y J ( t t

2 $150,000 0

) = = m m - v m -

32 ft/s 1 0 ( ( y - t = - = = x DT PERIOD DATE )

y = 16 ) belowtofindeachvalue. 2

v = x ( y (

x t t 2 -

− ) iftheheightofan 4 3 x

2 = -

+ t -

- 12 - 900.Howfastisthe 4) − 1 2 x

- t x 3 . 2 atanypoint. t

D D D D 5 J J J J

$200,000

836 ft/s 0 x Pdf Pass

− v ∞ - 2 ( 5 t

) =

SCORE = 0

− 1 3

t Glencoe Precalculus - − 2 1

- 1 10. 9. 8. 7. 6. 5. 4. 3. 2. 1.

33/17/09 12:44:04 PM / 1 7 / 0 9

1 2 : 4 4 : 0

4 Assessment

P M M P

3 5 : 5 3 : 5

9 0 / 5 / 2

20. B: 15. 16. 17. 18. 19. 11. 12. 13. 14.

3 2 x ) 2 4

- - Glencoe Precalculus C x Pdf Pass x

C (8

2 26 −

2 +

3 =

2 ) x + x

96 6

( 3 3 x - − - x – x

g 120 ft/s 120

20.25 125 joules 125

J J J

D D D D D

x 2

)

2 C 2 x )

) 18

x (continued) 2 3 + 2

x -

x 24 + x 6

+ - 4)

- 24 2

(4 - + x - (4

(8 −

−

C 3

x - = = =

+ 1)

2 DATE DATE PERIOD 3( ) ) ) 4 x x . How much work is required? x x + x - seconds is given by seconds is given

( ( + 65 6 3 (

t x - h h - - g x 1. x dx 24 ft/s 24

7 12.15 75 joules 75

( (3). -

J J C C C C C 2 H H H H H

after

⌡ ⌠ 0

2 1) . ) 2

+ x

x C

26 (

C 2 5. Find +

= (8 −

3 ) ≤ +

x = t 3 - (

) 4 3 2 + x ≤ x −

16 ft/s 21 ( – x x . ( − - - - - g

of a ball in feet

8.1

50 joules 50 1 h dx

B B B B B G G G 1 4 for 0 − t

3 2 ⌠ ⌡ 48 0

x x

2 -

5) 4) x 120

x 3 3 C

x C +

24 24

dx - +

x 2 x

dx 2 - - -

+ x x

3 ) + x t 4 4

− − ( 3 ) 3 3

− 4)

2 3 24

x 3 (

− 2 3 The work, in joules, required to stretch a spring 5 inches from its x x Chapter 12 Test, Form 2B Form 12 Test, Chapter + + 6 - x - + - x

The height The 16 4 - = = 2 =

- - 2

1) 3

4 2

4 − 8

x x 3 − ) )

) 6

- d 4 -

d dx

7 x x + 2 - x x n

− - i

( ( - 32 ft/s 3( 8 3 46 ( . = 4 =

= 3 = = x x x 1

h - h - - - x g - - ) Evaluate Evaluate ) ) 8 ) ) (1 1 4 x t 3

−

0 joules 0

( x x 9

( ( (2

(

8 ( (

5 3 ⌡ ⌠ _ F F F F F h

natural length of 4 inches is given by f g HEIGHT f SPRINGS

h A A A A A ⌠ ⌡ ⌠ ⌡ G 2 0 2

Find all antiderivatives of

1 12 C M R Use the Quotient Rule to find the derivative of each function. Use the Quotient Rule to find 14. 20. 19.

16. Chapter 12 G Find the derivative of each function derivative of Find the 11. NAME NAME Bonus

15. 12.

17. Evaluate each integral. 18. 13. C C P _ 6 5 0 - 7 3 0037-056_PCCRMC12_893813.indd 46 0037-056_PCCRMC12_893813.indd 47 3 7 - 0 5 6 _ P C C R M

C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 _ 8 9 3 8 1 3 10. 1. 8. 3. 2. Chapter 12 9. For Questions1and2,usethegraphof NAME 7. 4. Evaluate eachlimit. 6. 5. . i n d 12 d

Findanequationfortheslope ofthegraph

Findtheslopeoflinetangent tothegraphof

Find anequation fortheinstantaneousvelocity

4 height ofanobjectisdefined as reach itsmaximumheight?Whatisthe height?

y h firework infeet velocity of160feetpersecond.Supposetheheight

FIREWORKS FISH TANK can beestimatedby y number offishthatcanliveinthetank?

x x x x x 7 lim lim

( lim -∞ → → → → → - lim lim = = t )

16 3 2

− x

3 1

47 1

x − (5

- (

3 f

− (

x x x

+ ( x - 16 x 2 3 x - )

Chapter 12Test,Form2C

) x 2

- and 4 + -

t x 16

2 3

3 x Afishtank’spopulation

1000 - + 2

Fireworksarelaunchedwithanupward x

- 15 160

f + (2)

−

4)atthepoint( 2 1 ) secondsafteritisfiredmodeledby

atanypoint. t

+ P 25.Forwhatvalueof ( t ) =

− 30 5 t 2 t + 2 - h 10 - 2 ( t t 1,

)

.Whatisthemaximum = - P 0

12).

intensafter √ 3 y

f t

y ( x

+ ) = t

will the firework willthefirework

√ f ( t

x DT PERIOD DATE x

) belowtofindeachvalue. v ( t ) ifthe h ofone t years

3. 2. 1. 10. 9. 8. 7. 6. 5. 4. Pdf Pass SCORE Glencoe Precalculus 112/5/09 5:39:03 PM 2 / 5 / 0 9

5 : 3 9 : 0

3 Assessment

P M M P

0 2 : 4 4 : 2 1

9 0 / 7 1

B: 20. 15. 16. 17. 18. 19. 11. 12. 13. 14. (continued) , where t 32 -

DATE DATE PERIOD = ) t ( . v 2 of the arrow h is measured in feet per 1) v +

x ) of the dropped binoculars. ( t 2 ( s x

= . ) x dx ( f ) x 4 7 - +

6. How fast is the arrow traveling 6. How fast is the 2 x x +

t (6

x -

⌠ ⌡ 1 2 40 2 x

feet above the Leroy is in his tree house 35

dx +

dx 2 3 4) + 4 2

x dx t 1 of arrow is shot upward with a velocity An 2 - 3

+ - 1)

)

x 2 + Chapter 12 Test, Form 2C Form 12 Test, Chapter 8 - 2

x 16 x 4 x + 4 1) 4

seconds after it is shot is defined as seconds after it x

4 t 2 − x

−

2 - d 3 - -

d is given in seconds and velocity x

n t i 2 48 - . = = =

3 = = x 2 1 ) Evaluate Evaluate ) ) 8 ) )

x t x 3 (3 x

(

2 x x

( ( (0.2

(

4 8 ( (

- 2 ⌠ ⌡ _

g f TREE HOUSE ARCHERY Suppose the height 40 feet per second. in feet h ground when he drops his binoculars. The instantaneous velocity of his binoculars can be defined as time second. Find the position function f h after 1 second? ⌠ ⌡ ⌠ ⌡ 0

- 2

Find all antiderivatives of

1 12 C M R 17. Evaluate each integral. 15. Chapter 12 18. 13. Find the derivative of each function. derivative of Find the 11. NAME NAME 20. Bonus 16. 12. 19. 14. Use the Quotient Rule to find the derivative Use the Quotient Rule to find of each function. C C P _ 6 5 0 - 7 3 0037-056_PCCRMC12_893813.indd 48 0037-056_PCCRMC12_893813.indd 49 3 7 - 0 5 6 _ P C C R M

C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 _ 8 9 3 8 1 3 1. 8. 10. 3. 6. 4. Evaluate eachlimit. 2. Chapter 12 7. For Questions1and2,usethegraphof NAME 9. 5. . i n d 12 d

Findanequationfortheslope ofthegraph

Findtheslopeoflinetangent tothegraphof

Find anequation fortheinstantaneousvelocity

4 height ofanobjectisdefined as

y reach itsmaximumheight?Whatisthe height? can beestimatedby number ofdeerthatcanliveinthepark? point intime ROCKET DEER y h rocket infeet velocity of176feetpersecond.Supposetheheight x x x x x 9 lim lim

( lim lim -∞ → → → - → → - lim = = t )

2 1

− x = 5 1

+ +

49 4 1

Apark’sdeerpopulation

(

− x f

5 - x

f ( (

( x − 3

x - - x x 16 - 2 x Amodelrocket islaunchedwithanupward ) and Chapter 12Test,Form2D 5

) 2

3)atthepoint( 4 + t x

3 x 2

3 + x + x

+ t 3 t

4 . ) - 176 secondsafteritisfiredmodeledby f 1atanypoint. (1)

28 t

+ P 10.Forwhatvalueof ( t ) = -

− 75 15 1, t t 3 - 3

- + P h 4). 3 2 inhundredsafter ( t t t ) 2

= . Whatisthemaximum 2 f (

√ x 4 y )

t f

willtherocket (

0 √ 5 x DT PERIOD DATE

) belowtofindeachvalue. y v forany ( t ) ifthe h ofthe t years x

3. 2. 1. 10. 9. 8. 7. 6. 5. 4. Pdf Pass SCORE Glencoe Precalculus 33/17/09 12:44:24 PM / 1 7 / 0 9

1 2 : 4 4 : 2

4 Assessment

P M M P

9 2 : 4 4 : 2 1

9 0 / 7 1

B: 20. 15. 16. 17. 18. 19. 11. 12. 13. 14. (continued) , where t 32 of the - h

DATE DATE PERIOD = 7). ) t + (

v x is measured in feet per 2)( v - ) of the dropped hammer.

t x ( ( s x

= . ) x dx ( f ) x 6 5. How fast is the projectile 5. How fast is the -

3 + 1)

x t +

seconds after it is shot is defined as seconds after it 8

(4 2 t x

x - ⌠ ⌡ 2

2 175

x 5)(

+ 175 feet is shot with a velocity of A projectile

4 3 - 5 2 2 dx

dx x Michelle is on the roof of her house 20 feet above t 2

x - - 2

)

2) 2 + Chapter 12 Test, Form 2D Form 12 Test, Chapter 0 - 2 3

x 16 5 x (2 4

x + 2 3)

3 4 − x

−

- d 7 x

d is given in seconds and velocity 3

+ n t i x

50 - . = = =

3 = = x 2 1 ) Evaluate Evaluate ) ) 8 ) )

x t x 3 (2

(

x x 9

( ( (0.7

(

3 8 ( (

- 2 ⌡ ⌠ _ g

f ROOFTOP

PROJECTILE per second toward a target. Suppose the height a target. Suppose the height per second toward projectile in feet h the ground when she drops her hammer. The instantaneous velocity of her hammer can be defined as time second. Find the position function f h traveling after 2 seconds?

⌠ ⌡ ⌡ ⌠ 0

- 2

Find all antiderivatives of

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1 2 : 4 4 : 3

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9 3 : 4 4 : 2 1

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B: 20. 15. 16. 17. 18. 19. 11. 12. 13. 14. (continued) , t 32 - ) of the

t DATE DATE PERIOD is measured ( = 3). s v ) - t of the jumper,

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Find all antiderivatives of

1 12 C M R 19. 15. 13. 18. Chapter 12 20. Find the derivative of each function. derivative of Find the 11. 12. NAME NAME 16. Evaluate each integral. 17. 14. Use the Quotient Rule to find the derivative Use the Quotient Rule to find of each function. Bonus C C P _ 6 5 0 - 7 3 0037-056_PCCRMC12_893813.indd 52 0037-056_PCCRMC12_893813.indd 53 3 7 - 0 5 6 _ P C C R M

C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 _ 8 9 3 8 1 3 3. Chapter 12 2. 1. investigate beyondtherequirementsofproblem. your answers.Youmayshowsolutioninmorethanonewayor each problem.Besuretoincludeallrelevantdrawingsandjustify Demonstrate yourknowledgebygivingaclear,concisesolutionto NAME . i n d 12 d c. c. b. The speedofanobjectisgivenby f. e. d. Findtwodifferentfunctions, Let a. b. a.

x 5 lim 3 → Findthelimitofnumerator, under thecurverepresents? Why? Theareaofarectangle isgivenby method ismoreaccurate?Why? approaches 3.Thenexplainhowtofindthelimitas Usingthegraphinpart Find thelimitas Findthelimitas Justify youranswer. Findthelimitofdenominator, Justify youranswer. Make atableofvaluesfor bounded bythecurveand Usetwodifferentmethodstofindtheareaor approximate area Graph thefunction.

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Glencoe Precalculus 33/17/09 12:44:43 PM / 1 7 / 0 9

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1 1 : 5 1 : 3

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P M M P

6 5 : 4 4 : 2 1

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20. 21. 22. 23. 24. 16. 17. 18. 19. 25a. 25b. 25c. (continued) .

DATE DATE PERIOD 4 and 9 = 0.

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1 2 −

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Write your answers in the space provided. Write your answers

42, and x - = = =

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bounced The velocity of a ball that was 2 - μ

- 2

x − x

2 49, 25 = → =

lim

x Instructions:

1 X Standardized Test Practice Test Practice Standardized (Chapters 1–12) (Chapters 6 + 5

d if d z n i 56 . 3 1 Find the position function off the sidewalk. Assume that for 8 3 . What is the height of the ball after 2 seconds? 9 . . How long does it take for the ball to return to the ground? is given in seconds and velocity 8 _ Find the magnitude of vector Find the magnitude Solve 5 Write an explicit formula for finding the Write an explicit formula for finding term of geometric sequence with a first Write the pair of parametric equations in rectangular Write the pair of parametric equations form. Identify the related conic. BOUNCING BALL Find all antiderivatives for Find the partial fraction decomposition of Find the partial fraction decomposition Find t a off a sidewalk can be defined as a common ratio of Explain your reasoning. Explain your reasoning. randomly off a grocery in a 20-ounce box of rice chosen store shelf. Evaluate Classify the random variable Classify the random c b 2 1 12 C M R 17. Chapter 12 16. NAME NAME 21. 20. 24. 22. 23. 25. 19. 18. C C P _ 6 5 0 - 7 3 0037-056_PCCRMC12_893813.indd 56 AA01_A17_PCCRMC12_893813.indd 1 0 1 _ A 1 7 _ P C C R M

NAME DATE PERIOD NAME DATE PERIOD 1 12 Anticipation Guide 12-1 Study Guide and Intervention A1 Limits and Derivatives Estimating Limits Graphically Estimate Limits at Fixed Values Step 1 Before you begin Chapter 12 Left-Hand Limit Right-Hand Limit

• Read each statement. If the value off( x) approaches a unique If the value off (x) approaches a unique Answers numberL as x approachesc from the numberL as x approachesc from the • Decide whether you Agree (A) or Disagree (D) with the statement. 1 2 left, then lim f (x) = L1. right, then lim f(x) = L2. - + x → c x → c Chapter Resources • Write A or D in the first column OR if you are not sure whether you Existence of a Limit at a Point agree or disagree, write NS (Not Sure). The limit of a functionf (x ) as x approachesc exists if and only if both one-sided limits exist and are equal. That is, if

STEP 1 STEP 2 lim f (x) = lim f (x) = L, then lim f (x) = L. (AnticipationGuideandLesson12-1) - + Statement x → c x → c x → c A, D, or NS A or D 1. The limit of a functionf(x ) as x approachesc does not depend on the value of the function at pointc. A Lesson 12-1 Example 2. The limit of a functionf(x ) as x approachesc exists providing Estimate each one-sided or two-sided limit, if it exists. either the left-hand limit or right-hand limit exists. D lim x , lim x , and lim x

- + 3. The limit of a constant function at any point isx -valuethe of x → 2 x → 2 x → 2 D the point. y The graph off( x) x suggests that f (x) x 4. Limits of polynomial and many rational functions can be = = A lim x = 1 and lim x = 2. found by direct substitution. - + x → 2 x → 2 5. The slope of a nonlinear graph at a specific point is the instantaneous rate of change. A Because the left- and right-hand limits 0 x 6. The process of finding a derivative is called differentiation. A of f (x) as x approaches 2 are not the same, lim x does not exist.

7. The derivative of a constant function is the constant. D x → 2 8. The process of evaluating an integral is called integration. A 9. The functionF (x) is an antiderivative of the functionf(x) if Exercises f (x) = F(x). D Estimate each one-sided or two-sided limit, if it exists. 10. The connection between definite integrals and ⎪3x⎥ ⎪x - 2⎥ x2 + 3x - 10 antiderivatives is so important that it is called the A 1. lim 2. lim 3.lim −x 2 + - x 2 −x - 4 - Fundamental Theorem of Calculus. x → 0 x → -2 x → 2 − Pdf Pass

3 ∞ 7 Step 2 After you complete Chapter 12 Glencoe Precalculus 3 • Reread each statement and complete the last column by entering an A or a D. 2 x + 27 1 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 4. lim (1 - cos x) 5. lim 6. lim 2 2 - x 9 (x 2) − - − + x → 0 x → 3 x → -2 • Did any of your opinions about the statements change from the first column?

• For those statements that you mark with a D, use a piece of paper to write an example of why you disagree. 0 -∞ ∞

Chapter 12 3 Glencoe Precalculus Chapter 12 5 Glencoe Precalculus 112/7/09 1:35:41 PM 2 / 7 / 0 9

1 : 3

5 0ii_004_PCCRMC12_893813.indd Sec1:3 3/17/09 11:36:02 AM 005_036_PCCRMC12_893813.indd 5 12/7/09 12:32:15 PM : 4

1 Answers

P M M A

9 5 : 7 1 : 0 1

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112/7/09 10:17:59 AM

250 :05 PM 4:50:51 12/5/09 7 005_036_PCCRMC12_893813.indd 0_3_CRC2831.nd6 005_036_PCCRMC12_893813.indd 250 :95 PM 4:49:57 12/5/09

7 7 6 6 Chapter 12 Chapter Chapter 12 Chapter Glencoe Precalculus Glencoe Glencoe Precalculus Glencoe

does not exist 0 does not exist not does 0 exist not does

→

100 x

∞

lim . . C Find Find

−

x 100

∞ ∞ ∞

→ → → → - → x x x

= ≤ ≤ C 100. 100. x 0 and pollutants of number the is x where ,

π x 312 lim lim lim x ( x 2 cos e ) x sin 8. 8. 7. 7. 9. 9.

x 2

the pollutants created by one of its manufacturing processes is given by by given is processes manufacturing its of one by created pollutants the

POLLUTANTS The cost in millions of dollars for a company to clean up up clean to company a for dollars of millions in cost The 12. 12.

∞ - de o xs exist not does 0

Glencoe Precalculus

Pdf Pass

−

∞ ∞ ∞

→ → → - → → x x x

3 x 2 x

+ π lim lim lim

(2 e ( ) x ) x 2 sin 4. 4. 6. 6. 5.

∞

x x

2 5 x

2 → →

20 x

lim ). x ( r ) to find find to ) x ( r Graph barn. the and pole the of base

−

- x 400

√

2 - 2 2 3 0 3

= =

is the distance between the the between distance the is x where second, per feet ) x ( r

x 3

the top of the pole will move down the side of the barn at a rate of of rate a at barn the of side the down move will pole the of top the

of the pole is pulled away from the barn at a rate of 3 feet per second, second, per feet 3 of rate a at barn the from away pulled is pole the of

−

∞ ∞

∞

→ → → → → - → x x x

x −

2 x

RATE OF CHANGE OF RATE

A 20-foot pole is leaning against a barn. If the base base the If barn. a against leaning is pole 20-foot A 2 11. 11.

−

1. 1. 3. 3. 2. lim lim lim

1 + - + 2 1 x 3 1 x

Estimate each limit, if it exists. it if limit, each Estimate

Exercises

→ → 0 x

∞

→ → x

−

∞

2 lim lim e

10. 10. 9.

+ 2 x 3 x 2 sin

supports our graphical analysis. graphical our supports

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∞ ∞

→ → - → x x

− + 1 x

1 x −

) ( .800 .0 .010.00001 0.0001 0.001 0.01 0.08 x f

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lim lim

8. 8. 7. 7.

e =

(x)

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1 2 01010 000100,000 10,000 1000 100 10

(Lesson 12-1) (Lesson x

Estimate each limit, if it exists. it if limit, each Estimate

approaches infinity approaches x y

increasingly large. increasingly

Support Numerically Support -values that grow grow that -values x choosing values, of table a Make

= horizontal asymptote at at asymptote horizontal 0. y

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Lesson 12-1

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2 2

increases, the height of the graph gets closer to 0. The limit indicates a a indicates limit The 0. to closer gets graph the of height the increases, x As

∞ ∞

lim lim

5. 5. 6. 6.

+ +

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lim

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→ → x

3 x

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Estimate Estimate , if it exists. it if , Example

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→ → - →

4 x 1 x

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4. 4. 3.

- 16 x + 7 x

∞

→ - → x

= lim L ) x ( f . then then

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Estimate Limits at Infinity at Limits Estimate Estimate each one-sided or two-sided limit, if it exists. it if limit, two-sided or one-sided each Estimate

Estimating Limits Graphically Limits Estimating Estimating Limits Graphically Limits Estimating d d

3 Practice Study Guide and Intervention Intervention and Guide Study 12-1 12-1 (continued) 1 8 3 9 8

_ NAME NAME DATE PERIOD PERIOD DATE NAME NAME DATE PERIOD PERIOD DATE 2 1 C M R Chapter 12 C C P _ 7 1 A _ 1 0 AA01_A17_PCCRMC12_893813.indd 2 AA01_A17_PCCRMC12_893813.indd 3 0 1 _ A 1 7 _ P C C R M

NAME DATE PERIOD NAME DATE PERIOD 3 12-1 Word Problem Practice 12-1 Enrichment A3 Estimating Limits Graphically 1. BACTERIA GROWTH Bacteria in a dish 3. PROJECTILE HEIGHT Suppose a A Matter of Limits are growing according to the function projectile is thrown upward where its 4 There are many examples of limits in our world. Some of these are f (t) = , for t ≥ 0, wheref (t) heighth in feet at any timet in seconds -0.2t absolute limits, in that they can never be exceeded. Others are like −1 + 0.35e is determined by the functionh(t ). The is the weight of the bacteria in grams table shows the height of the projectile at guidelines, and still others result in a penalty if they are exceeded. and t is the time in hours. various times during its flight. Fill in the chart below. a. Graphf (t) for 0 ≤ t ≤ 20. Penalty or consequence Limit How is the limit set? Is the limit absolute? f(t) th(t) th(t) if the limit is exceeded 8 02564384 1. speed limit on a 6 The government 1 336 5 336 highway speeding ticket, sets the speed no 4 23846256 fine, and so on limits. 2 34007144 Answers 2. height limit 0 4 81612 t a. Graph the data and draw a curve

Damage is done to Lesson 12-1 through the data points to model the on a road Height limit is set b. Use the graph to estimate the number yes vehicle and/or functionh (t). underpass for safe clearance. of grams of bacteria present after structure. 8 hours. Round to the nearest tenth, h(t) if necessary. 3.7 g 400 3. luggage limit Passenger must on an airline flight Airline sets limit 300 pay for any on amount of no luggage (Lesson12-1) c. Estimate lim f (t), if it exists. 200 baggage allowed. t → ∞ beyond limit. Interpret your result. 100 4. temperature of a 4 g; Over time, the weight of can only get as 2 4 6 8t warm object placed the bacteria in the dish will 0 Copyright ©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc. in a cool room cool as its yes not possible approach a maximum of 4 g. surroundings b. Use your graph to estimatelim h(t).

2. CARS Aftert years, the value of - 0 ft t → 8 5. the speed of an a car purchased for $30,000 is accelerating space physical constant, v(t) = 30,000(0.7)t. yes not possible 4. THEORY OF RELATIVITY craft the speed of light a. Graphv (t) for 0 ≤ t ≤ 20. Theoretically, the massm of an object with velocityv is given by v(t) m 6. credit limit on a financial penalty m = 0 , wherem is the mass of 2 0 credit card −v set by bank set by bank, 1 - − no √ s2 issuing card possibly may lose the object at rest ands is the speed of Pdf Pass card light. What is lim m? ∞ - v → s One special feature of mathematical limits is that they may be finite, 0 t infinite, or they may not exist. Classify each limitfinite, as infinite, 5. ELECTRICITY Ahmed determined that or does not exist. If the limit is finite, give its value.

the voltage in an electrical outlet in his 2 Glencoe Precalculus x + 1 x 4 b. Use the graph to estimate the value 7. lim 1 8.lim 9.lim - home is modeled by the function 2 2 2 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. − − x + 1 x - 1 x - x - 2 of the car after 10 years.$847 x 0 x 1 x 2 → → → − V(t) = 140 sin 120πt. Explain why 4 lim V(t) does not exist. finite; 1 does not exist nite; fi− 3 c. Estimatelim v(t), if it exists. t → ∞

t → ∞ sin ⎪ x⎥ Interpret your results. ⎪ ⎥ -1 $0; Over As t increases, the graph 10. lim ln x 11. lim 12.lim x −4 − x time, the value of the car will oscillates between 140 and-140. x → 0 x → 0 x → 0 reach a minimum of $0. infinite does not exist nite infi

Chapter 12 8 Glencoe Precalculus Chapter 12 9 Glencoe Precalculus 112/7/09 1:49:56 PM 2 / 7 / 0 9

1 : 4

9 005_036_PCCRMC12_893813.indd 8 12/7/09 12:47:58 PM 005_036_PCCRMC12_893813.indd 9 3/17/09 11:36:36 AM : 5

6 Answers

P M M P

1 3 : 0 5 : 1

9 0 / 7 / 2

250 :45 PM 4:54:56 12/5/09 11 005_036_PCCRMC12_893813.indd 0_3_CRC2831.nd10 005_036_PCCRMC12_893813.indd /70 13:0AM 11:36:40 3/17/09

11 11 Chapter 12 Chapter Glencoe Precalculus Glencoe 10 10 Chapter 12 Chapter Glencoe Precalculus Glencoe

→ → - → → 2 x 4 x 4 x

−

every possible number. possible every

4 x

( ) − - 5 4.25

- - + + − lim lim lim x ( 1) 1) x 5 x 6. 6. 5. 4.

1 1 No; Sample answer: The trace function does not highlight highlight not does function trace The answer: Sample No;

- 2 x √

−

- → → → 2 x 5 x 3 x

problem? Explain. problem?

+ 2 x

5 11 3 - -

lim lim lim x (2 4 x ) x 5

3. 3. 2. 1.

√

Will the graphing calculator give you the exact answer for every limit limit every for answer exact the you give calculator graphing the Will 4. 4.

3 2 + + 14 x 9 x 2

Glencoe Precalculus

Pdf Pass

Exercises

Sample answer: The function is undefi undefi is function The answer: Sample 1. at ned x

− + 4 16 √

= − or

Apply direct substitution and simplify. and substitution direct Apply

1 1

= =

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− →

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x ln = lim

Simplify.

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Divide out the common factor. common the out Divide

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Simplify.

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→ → → 16 x 16 x

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is close to 2. to close is x when function the of limit the examine to and and

· = lim lim

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0 16 16

− . Rationalize the numerator numerator the Rationalize . or obtain you substitution, direct By

4 16 −

√

→ →

2 x

+ -

x 2 x 3

lim

2. 2.

→

16 x

- 4 x

−

16 x

Lesson 12-2 2

lim

. evaluate to rationalizing Use

Example 3 Example

- 4 x √

1 - - = 1 1 or 5, 4 Apply direct substitution and simplify. and substitution direct Apply

→ → 4 x

- = lim 5) 5) x (

Divide out the common factor and simplify. and factor common the out Divide

−

→ →

4 x 4 x

is close to 0. to close is x when function the of limit the examine to and and

4) x (

- 4 x

lim lim

Factor.

+ - 20 x 9 x

- -

4) x 5)( x (

use and Press . 2 press Then 6.

TRACE ENTER ZOOM ZOOM 2

−

→ → 4 x

1 ] e [ Press Press

2nd Y= ( -

4 x ENTER

— ) )

lim

. Use factoring to evaluate evaluate to factoring Use

Example 2 Example

+ - 20 x 9 x

(Lesson 12-1 and Lesson 12-2) Lesson 12-1 and (Lesson 2

→ → 0 x

−

lim

1. 1.

- 1 e - + + - - =

43 or 3, 10 24 32

x x

- → 2 x

- - + + + - - - + - - = + lim x 2 ( x 3 3 2) 5( 2) 3( 2) 2( 3) x 5

Evaluate each limit. each Evaluate

3 4 3 4

direct substitution to find the limit. the find to substitution direct

Since this is the limit of a polynomial function, we can apply the method of of method the apply can we function, polynomial a of limit the is this Since

the value of the limit. the of value the

as you like. The The like. you as a to close as is -coordinate x -coordinate should be close to to close be should -coordinate y

Answers → - → 2 x

- - + + = lim x 2 ( x 3 3). x 5 ). Then use use Then ). x ( f y and and to locate a point on the graph whose whose graph the on point a locate to ZOOM

TRACE

3 4

→ → a x Use direct substitution, if possible, to evaluate evaluate to possible, if substitution, direct Use

Example 1 Example

lim ), first graph the equation equation the graph first ), x ( f ordinary scientific calculator. To find find To calculator. scientific ordinary

You can use a graphing calculator to find a limit with less work than an an than work less with limit a find to calculator graphing a use can You

Finding Limits Finding Compute Limits at a Point a at Limits Compute 4

Evaluating Limits Algebraically Limits Evaluating d d

3 Study Guide and Intervention and Guide Study 12-2 Graphing Calculator Activity Calculator Graphing 12-1 1 8 3 9 8

NAME NAME DATE PERIOD PERIOD DATE NAME NAME DATE PERIOD PERIOD DATE 2 1 C M R Chapter 12 C C P _ 7 1 A _ 1 0 AA01_A17_PCCRMC12_893813.indd 4 AA01_A17_PCCRMC12_893813.indd 5 0 1 _ A 1 7 _ P C C R M

NAME DATE PERIOD NAME DATE PERIOD 5

12-2 Study Guide and Intervention (continued) 12-2 Practice A5 Evaluating Limits Algebraically Evaluating Limits Algebraically Compute Limits at Infinity Evaluate each limit. x2 - 36 Limits of Power Limits of Polynomials at Limits of Reciprocal 1. lim (x2 + 3x - 8) 10 2.lim -12 −x + 6 Functions at Infinity Infinity Functions at Infinity x → 3 x → -6 For any positive integern, Let p be a polynomial For any positive integern, n function 1 • lim x = ∞. lim = 0. −xn x → ∞ p(x) = a xn + … + a x + a . x → ±∞ n 1 0 2 (3 + x) - 9 2 n n • lim x = ∞ if n is even. Then lim p(x) = lim a x 3. lim x 6 4.lim √x - 2x + 1 3 n x → -∞ x → ∞ x → ∞ x → 0 − x → 4 n and lim p(x) = lim a x . n n • lim x = -∞ if n is odd. x → -∞ x → -∞

x → -∞ Answers

x2 - x 1 x2 9 Example Evaluate each limit. 5. lim − 6.lim − 2 2x + 5x - 7 9 2 + √x - 3 2 x → 1 − x → 3 − lim 5 a. (x - 6x + 1) x → -∞ lim (x5 - 6x + 1) = lim x5 Limits of Polynomials at Infinity

x → -∞ x → -∞ Limits of Power Functions at nityInfi 5 2 = -∞ x - 8x 1 (Lesson12-2) 7. lim (2 - 6x + 5x3) ∞ 8. lim − 4x5 + 3x x → ∞ x → -∞ − 4

4 2

b. lim (2x + 5x ) Lesson 12-2

x → ∞ x → ∞ 4 3 2x - 4x + 1 = 2 lim x Scalar Multiple Property 9. lim 0 10. lim (6x7 - x2) -∞ x → ∞ 4 2 5x - 2x x → ∞ − x → -∞ = 2 · ∞ = ∞ Limits of Power Functions at nityInfi

Exercises

Evaluate each limit. 11. BOOKS Suppose the valuev of a book in dollars aftert years can be

300 represented asv (t) = . How much will the book eventually 3 5 6x - 1 6 + 35(0.2) t 1. lim (-2x + 5x) 2. lim − 3.lim 2 − 10x + 7 x x → -∞ x → ∞ x → ∞ − Pdf Pass be worth? That is, find thelim v(t). t → ∞ 3 $50 ∞ 0 − 5

2 4 3 m Glencoe Precalculus 6x - 2x 5x + 2x - 1 12. MEDICINE Each day, Tameka takes 2 milligrams of 4. 5. 6. 3 lim 3 lim 3 2 lim (3x + 5x - 1) 10

Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. x + 1 2x + x - 1 x x x her asthma medicine. The graph shows the amount → ∞ − → ∞ → -∞ − 8 of medicinem left in her blood stream afterd days. 6 Find the lim m(d) and lim m(d). 4 - + d → 3 d → 3

0 ∞ -∞ Medicine (mg) 2 d Tameka’s Asthma 2 mg, 4 mg 0 152 3 46 Days

112/7/09 10:21:39 AM Chapter 12 12 Glencoe Precalculus Chapter 12 13 Glencoe Precalculus 2 / 7 / 0 9

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9 Answers

A M M P

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/70 13:2AM 11:37:02 3/17/09 15 005_036_PCCRMC12_893813.indd 0_3_CRC2831.nd14 005_036_PCCRMC12_893813.indd /70 13:8AM 11:36:58 3/17/09

15 15 14 14 Chapter 12 Chapter Chapter 12 Chapter Glencoe Precalculus Glencoe Glencoe Precalculus Glencoe

350

∞

→ → t

→ → →

0 x 0 x

Find Find ). ). t ( v b. b.

−

x x

→ 2000 x

lim

3 1 lim lim

6. 6. 5.

x 3 sin x tan

lim ). x ( f Find Find

microwaves for professional kitchens. kitchens. professional for microwaves x

Glencoe Precalculus

Pdf Pass manufactured by a company that makes makes that company a by manufactured

) ) x ( f microwave a of cost − average

35 16 96 16.02 19.61 21.69 23.52

Value

= = ) x ( f models the models

→ → →

0 x 0 x

x x √

200,000 x 250

3 −

− 0 0

lim lim sin x cos x 4. 4. 3. er 020 10 5 1 Years ⎥ ⎪

2 π 1 MICROWAVES The function function The 6. 6.

the nearest hundredth. nearest the

Complete the table. Round answers to to answers Round table. the Complete a. a.

$20.20

. by the equation equation the by 24(0.98) ) t ( v

→ 15,000 p

− → → 4 x 0 x

4 x

−

lim

Find Find .

2 0

− lim lim x sin x 2. 2. 1. 1. dollars, after after dollars, years can be represented represented be can years t

⎥ ⎪

4 1

+ + ) p ( c

2 x √

of his motorcycle, in thousands of of thousands in motorcycle, his of v value

. . p by ) p ( c dividing

Use the Squeeze Theorem to find each limit. each find to Theorem Squeeze the Use motorcycle for $24,000. Suppose the the Suppose $24,000. for motorcycle

The average cost per part is found by by found is part per cost average The c. c.

MOTORCYCLES Flint bought a new new a bought Flint 3. 3.

Exercises

900

→ → 0 x

parts did the company produce? company the did parts

= = − lim 0. The result is that that is result The 0. c with Theorem Squeeze the x 0.

∞

→ → → 0 x 0 x

On Tuesday, the company produced produced company the Tuesday, On b. b.

lim lim ) both equal 0. You can now apply apply now can You 0. equal both ) x ( g and ) x ( h you learned that that learned you

→

p Lesson 12-2 100

lim . . C Find Find x

- = = = −

x ) x ( h let Theorem, . From Lesson 12-2, 12-2, Lesson From . x ) x ( g and , sin x ) x ( f ,

2 2 2

1 −

- 100 100 p $5000

x ≤ ≤ = =

100. 100. p 0 , C

≤ ≤ - − this inequality by x by inequality this . To apply the Squeeze Squeeze the apply To . x sin x x

(Lesson 12-2) (Lesson obtaining ,

60,000 p

2 2 2 2 1

of removing removing of % of the pollutant is is pollutant the of % p

≤ - = ≤ − sin sin 1 that note First, 0. Next, multiply multiply Next, 0. x for except , x all for 1

Find the cost of producing 100 parts. 100 producing of cost the Find a. a.

creates an airborne pollutant. The cost cost The pollutant. airborne an creates C

its processes, a manufacturing company company manufacturing a processes, its

is the number of parts produced. parts of number the is p

POLLUTANTS

As a by-product of one of of one of by-product a As 2. 2.

+ = , where where , p 20 3000 ) p ( c equation the

→ → → → c x c x c x

= = = lim lim lim f then , ) x ( g L ) x ( h ) exists and is equal to . L to equal is and exists ) x ( itself, and if if and itself, c at by given is dollars in part engine small

≤ ≤ , except possibly possibly except , c containing interval open an in x all for ) x ( g ) x ( f ) x ( h If PARTS of producing a certain certain a producing of c cost The 5. 5.

Answers

The Squeeze Theorem Squeeze The

0.3 g/L 0.3

48.5 yd 48.5

→ → 70 x

will help you answer this question algebraically. question this answer you help will

−

lim ). x ( y you. Find Find you.

∞

→ → t

+ t 1000

→ → 0 x

= =

lim ). t ( C . Find Find . ) t ( C

between your car and the one in front of of front in one the and car your between

t 0.3

− lim sin sin because because does not exist? A theorem known as the Squeeze Theorem Theorem Squeeze the as known theorem A exist? not does

recommended safe distance, in yards, yards, in distance, safe recommended minutes later in the pool is given by by given is pool the in later minutes t

→ → 0 x

− lim x 1 and 1. But what about the the about what But 1. and 1 ? Does this limit not exist simply simply exist not limit this Does ? sin speed in miles per hour, gives the the gives hour, per miles in speed of chlorine in grams per liter liter per grams in chlorine of C

2 1

+ + =

is the the is x where 3, x 0.3 75 liters per minute. The concentration concentration The minute. per liters 75 x 0.005 ) x ( y

gets closer to 0, the corresponding function values oscillate between between oscillate values function corresponding the 0, to closer gets x as

of rate a at pool the into pumped front of you. Suppose the function function the Suppose you. of front

→ → 0 x

0.3 gram of chlorine per liter of water is is water of liter per chlorine of gram 0.3 distance between you and the car in in car the and you between distance

− lim sin sin In Lesson 12-1, you learned that the the that learned you 12-1, Lesson In does not exist because because exist not does

pure water. A mixture that contains contains that mixture A water. pure it is important to maintain a safe safe a maintain to important is it

The Squeeze Theorem Squeeze The

POOLS CAR SAFETY CAR A pool contains 75,000 liters of of liters 75,000 contains pool A While driving a car, car, a driving While 1. 1. 4. 4.

Evaluating Limits Algebraically Limits Evaluating d d

3 Enrichment Word Problem Practice Problem Word 12-2 12-2 1 8 3 9 8

NAME NAME DATE PERIOD PERIOD DATE NAME NAME DATE PERIOD PERIOD DATE 2 1 C M R Chapter 12 C C P _ 7 1 A _ 1 0 AA01_A17_PCCRMC12_893813.indd 6 AA01_A17_PCCRMC12_893813.indd 7 0 1 _ A 1 7 _ P C C R M

NAME DATE PERIOD NAME DATE PERIOD 7

12-3 Study Guide and Intervention 12-3 Study Guide and Intervention (continued) A7 Tangent Lines and Velocity Tangent Lines and Velocity Tangent Lines Instantaneous Velocity

Instantaneous Rate of Change Instantaneous Velocity

The instantaneous rate of change of the graphf (x of) at the pointx (, f (x)) is the slopem of If the distance an object travels is given as a function off (timet), then the instantaneous f (x + h) - f (x) f (t + h) - f (t) the tangent line given bym lim , provided the limit exists. velocityv (t) at a timet is given byv (t) lim , provided the limit exists. = = h h h → 0 − h → 0 −

Example Find an equation for the slope of the graph of Example A rock is dropped from 1500 feet above the base y = 3x2 + 1 at any point. of a ravine. The height of the rock aftert seconds is given by h(t) = 1500 - 16t2. Find the instantaneous velocityv(t) of the f (x + h) - f (x) m Instantaneous Rate of Change Formula rock at 4 seconds. = lim Answers h h → 0 − f (t + h) - f (t) v(t) = lim Instantaneous Velocity Formula 2 2 [3(x + h) + 1] - [3x + 1] h 2 2 m = lim f(x + h) = 3(x + h) + 1 andf (x) = 3x + 1 h → 0 − 2 2 h [1500 - 16(4 + h) ] - [1500 - 16(4)] h → 0 −− = lim f(t + h) = 1500 - 16(4 + h)2 and f(t) = 1500 - 16(4)2

2 2 2 h h 0 [3x + 6hx + 3h + 1] - [3x + 1] → −−− 2 m = lim Expand and simplify. -128h - 16h h = lim Multiply and simplify. h → 0

h −− h → 0 − 3h(2x + h) h(-128 - 16h) m lim Simplify and factor. (Lesson12-3) = = lim Factor. h h h → 0 − h → 0 − = lim (-128 - 16h) Divide byh and simplify. m = lim 3(2x + h) Reduceh . h → 0 h → 0 Copyright ©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc. = -128 - 16(0) or- 128 Difference Property of Limits and Limit of Constant m = 6x Scalar Multiple, Sum Property, and Limit of a and Identity Functions Constant Function Property of Limits The instantaneous velocity of the rock at 4 seconds is 128 feet per second. An equation for the slope of the graph at any pointm =is 6 x. Exercises Exercises The distanced an object is above the groundt seconds after it is Lesson 12-3 Find an equation for the slope of the graph of each function at dropped is given byd( t). Find the instantaneous velocity of the any point. object at the given value fort.

2 1. y = x3 + 1 m = 3x 2.y = 4 - 7x m = -7 1. d(t) = 800 - 16t2; t = 3 2.d( t) = -16t2 + 1700;t = 5 Pdf Pass

v(3) = -96 ft/s v(5) = -160 ft/s

6 Glencoe Precalculus 3 4 2 3. y = − m = −− 4.y = m = − 2 3 − √x − x x x √x Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 3. d(t) = 70t - 16t2; t = 1 4.d( t) = -16t2 + 90t + 10; t = 2

v(1) = 38 ft/s v(2) = 26

112/7/09 10:28:35 AM Chapter 12 16 Glencoe Precalculus Chapter 12 17 Glencoe Precalculus 2 / 7 / 0 9

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5 Answers

A M M A

7 5 : 9 2 : 0 1

9 0 / 7 /

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0_3_CRC2831.nd18 005_036_PCCRMC12_893813.indd 250 :02 PM 5:10:23 12/5/09 /70 13:7AM 11:37:17 3/17/09 19 005_036_PCCRMC12_893813.indd

18 18 19 19 Chapter 12 Chapter Glencoe Precalculus Glencoe Chapter 12 Chapter Glencoe Precalculus Glencoe

10 ft/s 10

10 sq in. per in. per in. sq 10

128 ft/s 128 after 4 seconds? seconds? 4 after

How fast is the football traveling after 1.5 seconds? 1.5 after traveling football the is fast How b. b.

What is the sky diver’s velocity velocity diver’s sky the is What c. c.

5 inches. 5 x

+ - = ( 58 32 ) t v t

of the area at the moment the at area the of 14,936 ft 14,936 after 2 seconds? seconds? 2 after

Find the instantaneous rate of change change of rate instantaneous the Find b. b. Find an expression for the instantaneous velocity velocity instantaneous the for expression an Find ) of the football. the of ) t ( v a. a. What is the sky diver’s height height diver’s sky the is What b. b.

+ + - = 6. t 58 t feet, t 16 ) t ( as h defined is it throws he after seconds

= - = 32 ) ( t t v of the sky diver. diver. sky the of

2 Glencoe Precalculus

per second toward a teammate. Suppose the height height the Suppose teammate. a toward second per of the football, in in football, the of h Pdf Pass ) ) t ( v velocity instantaneous

5.4 inches to 5.6 inches. 5.6 to inches 5.4

FOOTBALL A quarterback throws a football with a velocity of 58 feet feet 58 of velocity a with football a throws quarterback A 12. 12. Find an expression for the the for expression an Find a. a.

changes from from changes x as ) x ( a area

Find the average rate of change of the the of change of rate average the Find a. a. = - = ) ( 32 t v t exited the plane. the exited

seconds passed after the sky diver diver sky the after passed seconds

- = is is t where , t 16 15,000 ) t ( h

instantaneous velocity velocity instantaneous ) of the sky diver. sky the of ) t ( v 2

the ground can be defined by by defined be can ground the

after the sky diver exited the plane. Find an expression for the the for expression an Find plane. the exited diver sky the after

x x )= x ( a

of a free-falling sky diver relative to to relative diver sky free-falling a of

2 - = is seconds passed passed seconds is t time where , can be defined by by defined be can t 16 18,000 ) t ( h

FREE FALLING FREE in feet feet in h position The 4. 4.

SKY DIVING SKY in feet of a sky diver relative to the ground ground the to relative diver sky a of feet in h position The 11. 11.

Lesson 12-3

= - = 32 ) ( t t v

) of the jumper. jumper. the of ) t ( v velocity

2 t

−

of the square shown is changing. is shown square the of

= + - = - − ) ) ( 2 ) ( 4 t v t v t

Find an expression for the instantaneous instantaneous the for expression an Find

t t

√ AREA of each side side each of x length the Suppose 6. 6.

- = . . t 16 900 ) t ( h by given is seconds t

= + = - − ) t ( s t 2 ) t ( s t 2 t

9. 9. 10. 10. √

in feet relative to the ground in in ground the to relative feet in h height

2 3

106.5 ft 106.5

BUNGEE JUMPING BUNGEE A bungee jumper’s jumper’s bungee A 3. 3.

= + - = ) ( 4 12 15 ) ( 34 t v t t t v t

of the baseball? the of

What is the maximum height height maximum the is What d. d.

+ + - = + = 1 t 4 t 6 t 5 ) t ( s 8 17 ) t ( s t 8. 8. 7. 7.

2 3 2 it hits the ground? the hits it

2.5 s 2.5

What is the velocity of the rock when when rock the of velocity the is What d. d.

object is defined as as defined is object ) for any point in time . t time in point any for ) t ( s

Find an equation for the instantaneous velocity velocity instantaneous the for equation an Find ) if the path of an an of path the if ) t (

(Lesson 12-3) (Lesson v = − = 32 ) ( t t v of the rock. rock. the of

reach its maximum height? maximum its reach

) t ( v velocity instantaneous

will the baseball baseball the will t of value what For c. c.

= - = = (2) (2) 104 ft/s 104 (3) ft/s 64 v v Find an expression for the the for expression an Find c. c.

64 ft/s 64

s 3 5 √

= + + - = = = - 3 t 700; t 200 t 16 ) t ( d 2 t ; ) ) t ( d t 16 300 6. 6. 5. 5.

2 2

When will the rock hit the ground? the hit rock the will When b. b.

after 0.5 second? 0.5 after object at the given value for . t for value given the at object

How fast is the baseball traveling traveling baseball the is fast How b. b.

128 ft/s 128 4 seconds? seconds? 4 ). Find the instantaneous velocity of the the of velocity instantaneous the Find ). t ( d

Answers by given is dropped

How fast is the ball falling after after falling ball the is fast How a. a. an object is above the ground ground the above is object an d distance The seconds after it is is it after seconds t + - = 80 32 ) ( t t v

+ - = 1200. 1200. seconds is given by by given is seconds t t 16 ) t ( s

the baseball. the

= - = - = 2 4 3 m x x m - = + - =

x 2 x y 1 1 x 2 y 4. 4. 3. 3.

1200 feet. The position of the rock after after rock the of position The feet. 1200 2 3

instantaneous velocity velocity instantaneous ) of of ) t ( v

PROJECTILE Tito drops a rock from from rock a drops Tito 2. 2.

Find an expression for the the for expression an Find a. a. at any point. any at

Find an equation for the slope of the graph of each function function each of graph the of slope the for equation an Find 64 ft/s 64

+ + - = 6.5. t 80 t 16 ) t ( h by modeled

the ball falling after 2 seconds? 2 after falling ball the seconds after the ball is thrown is is thrown is ball the after seconds

+ - =

800. How fast is is fast How 800. 16 ) t ( s by given of the baseball, in feet, feet, in baseball, the of h height the t t - 5 5

- - - = = − ; (3, 6) 6) (3, ; x 5) 5) 1, ( ; y x y 2. 2. 1. 1.

2 2

position of the ball after after ball the of position velocity of 80 feet per second. Suppose Suppose second. per feet 80 of velocity is seconds t

from a tower that is 800 feet high. The The high. feet 800 is that tower a from toward home plate with an initial initial an with plate home toward the given point. given the

FALLING OBJECT FALLING BASEBALL An outfielder throws a ball ball a throws outfielder An ball a drops Miranda 1. 1. 5. 5. Find the slope of the line tangent to the graph of each function at at function each of graph the to tangent line the of slope the Find

Tangent Lines and Velocity and Lines Tangent Tangent Lines and Velocity and Lines Tangent d d

3 Practice 12-3 Word Problem Practice Problem Word 12-3 1 8 3 9 8

NAME NAME DATE PERIOD PERIOD DATE NAME NAME DATE PERIOD PERIOD DATE 2 1 C M R Chapter 12 C C P _ 7 1 A _ 1 0 AA01_A17_PCCRMC12_893813.indd 8 AA01_A17_PCCRMC12_893813.indd 9 0 1 _ A 1 7 _ P C C R M

NAME DATE PERIOD NAME DATE PERIOD 9 12-3 Enrichment 12-3 Spreadsheet Activity A9

Tangents and Vertices Using the Tangent Line to Approximate a Function Can you use the equation for the slope of a function at any point to find the You have learned how to find the slope of a tangent line to a function at a vertex of a parabola of the formy = ax2 + bx + c? point. This slope can then be used to write the equation of the tangent line Step 1 Graph a parabola of the formy = ax2 + bx + c, one wherea > 0 and in point-slope form. The tangent line of a function is usually a good one wherea < 0. Then draw the line tangent to the vertex of each approximation of the function for valuesx nearof the coordinate of the parabola and place the equation of each parabola below its graph. chosen point. Set up a spreadsheet like the one shown below to study this method of approximating a function. y y A BCDEFGH 4 1 x 0.7 0.8 0.9 1 1.1 1.2 1.3 0 4 x 1 2 x^2 0.49 0.64 0.81 1 1.21 1.44 1.69 0 1 x 3 2*x - 1 0.4 0.6 0.8 1 1.2 1.4 1.6 Answers

The equation in point-slope form of the tangent linef (x )to = x2 at (1, 1) is 2 2 y =x +2x - 3 y = -x -4x y = 2x - 1. In the spreadsheet, the functionf (x) = x2 is approximated for values ofx near x = 1 by using the functiong(x ) = 2x - 1. For this example, Step 2 Find the slope of the tangent line to each graph at the vertex. the values forx are entered in row 1, columns B–H. The functionf (x) = x2 is entered in cell B2 as= B1^2 and copied to the other cells in row 2. The The slope of each tangent line is 0. approximating function,g(x ) 2x 1, is entered in cell B3 as2 B1 1 and

= - = * - (Lesson12-3) Step 3 Find the equation for the slope of each graph at any point. copied to the other cells in row 3. Notice that the valuesg(x) of = 2x - 1 in row 3 are remarkably close to the valuesf ( xof) = x2 in row 2. For y = x2 + 2x - 3, m = 2x + 2. Fory = -x2 - 4x, m = -2x - 4. Step 4 Set each equation for the slope equal to zero and solvex. for Exercises This gives thex -value of the vertex. Copyright ©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.

2 2x + 2 = 0 -2x - 4 = 0 1. Use the spreadsheet to approximate the valuesf(x )of = x near the point, wherex = 2, by using the appropriate tangent line. Compare the values 2x = -2 -2x = 4 for x = 1.7, 1.8, 1.9, 2, 2.1, 2.2, and 2.3. What is the maximum error x = -1 x = -2 that occurs? Step 5 Substitute eachx -value into its respective equation to find the The values of ƒ(x) = x2 are 2.89, 3.24, 3.61, 4, 4.41, 4.84, and 5.29. The y-value of the vertex. values ofg (x) = 4x - 4 are 2.8, 3.2, 3.6, 4, 4.4, 4.8, and 5.2. The maximum Lesson 12-3

2 2 error is 0.09 at the points wherex = 1.7 and 2.3.

y = x + 2x - 3; x = -1 y = -x - 4x; x = -2 2 2 y = (-1) + 2(-1) - 3 y = -(-2) - 4(-2) 2. Use the spreadsheet to approximate the valuesƒ(x of) = x2 near the point,

Pdf Pass y = -4 y = 4 wherex = 0, by using the appropriate tangent line. Compare the values for x = -0.3, -0.2, -0.1, 0, 0.1, 0.2, and 0.3. What is the maximum error Step 6 Write each vertex as an ordered pair. that occurs? For y = x2 + 2x - 3, (-1, -4). Fory = -x2 - 4x, (-2, 4). The values of ƒ(x) = x2 are 0.09, 0.04, 0.01, 0, 0.01, 0.04, and 0.09. The values ofg (x) = 0 are 0, 0, 0, 0, 0, 0, and 0. The maximum error is 0.09 at the points wherex = -0.3 and 0.3. Glencoe Precalculus Exercises Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 3. Use the spreadsheet to approximate the valuesƒ(x of) = √x for Find the vertex of each parabola. x = 0.7, 0.8, 0.9, 1, 1.1, 1.2, and 1.3. Use the data to make a conjecture about the equation of the tangent line to the graph of the function at the 1. y = 2x2 - 4x + 2 (1, 0) 2.y = -x2 - 6x - 9 (−3, 0) point, wherex = 1.

1 The values of ƒ(x) = √x are approximately 0.837, 0.894, 0.949, 1, 1.049, 3. y = 4x2 - 25 (0, −25) 4.y = − x2 - 2x + 4 (2, 2) 2 1.095, and 1.140. The exact equation of the tangent liney = is0.5 x + 0.5.

112/7/09 10:30:54 AM Chapter 12 20 Glencoe Precalculus Chapter 12 21 Glencoe Precalculus 2 / 7 / 0 9

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4 Answers

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23 23 22 22 Chapter 12 Chapter Chapter 12 Chapter

Glencoe Precalculus Glencoe Glencoe Precalculus Glencoe

x − −

- = + - =

) ( 9 8 ) ( x f x x x f + - = - = -

x 3 x 4 ) x ( f x 3 x 4 ) x ( f 14 14 8. 8. 7.

2 2

3 2

− −

9 9 2 x x x

√ √

1 3

−

1) ( x - 1) (2 x

−

1 x 2

1 x −

2 2

+ - = - = 0.6 3.4 ) ( x x x f 3 12 ) ( x x f + - = + - = 2

x 3 x ) x ( f 4 x 3 x 6 ) x ( f 6. 6. 5. 5.

= =

) x ( k ) x ( d

4. 4. 3.

0.2 0.2 3.4 2

0.8 2.4

- - - - + + 16 9 6 x x x 3 2 x x 4 x 3 3 x

3 2 2 4 2 Find the derivative of each function. each of derivative the Find

4 3

- + - + 5 28 9 8 12 12 + - = + + - = x x x x ) ) x 5 x 1)( x (3 ) x ( m 3) x )(2 x 2 4 ( ) x ( h 2. 2. 1. 1.

Glencoe Precalculus 2 2 2 2

Pdf Pass

- = - - + = - - + 8 8 ) x ( f 1 and 2 and 1 2; x x 3 ) x ( f 3 and 0 ; x x 3 3. 3. 4. 4.

5 4 2

Lesson 12-4 function. each of derivative the Find

= - + - = - 8 ) ( 88 7, ; 10 3 ) ( 16 24, ; x f x x x f

x 2

Exercises

- = + - = - = - = 4 and 1 x ; x 5 x ) x ( f 2 and 3 x 5; x 4 ) x ( f

2. 2. 1. 1.

2 3 2

- = −

Simplify.

given values of of values given . x 3

−−

of derivative the Find ). Then evaluate the derivative for the the for derivative the evaluate Then ). x ( f

x 4

Distributive Property Distributive

- - - x 8 x 4 x 4 x 4

3 3 −−

Exercises

) x (2

= =

Substitution

+ - - x 4)2 x (2 1) x ( x 4

2 2

−−

] [

= - 28 x 20 x Simplify.

) x ( g

3 6 2

) x ( h Quotient Rule Quotient

) x ( g ) x ( f ) x ( g ) x ( f · = · - 4 4 ) x ( f x 4 5 x 7

Constant Multiple of a Power, and Sum and Difference Rules Difference and Sum and Power, a of Multiple Constant

- - - 1 4 1 7

= - x 4 ) x ( f x 5 Distributive Property Distributive

4 7 Constant Rules for Derivatives for Rules Constant Derivatives for Rules Constant

= - x ) x ( f (4 5) 5) x

Original equation Original

= = x 2 ) x ( g x 4 ) x ( f

Sum Rule for Limits, Power and and Power Limits, for Rule Sum and Power Limits, for Rule Sum 4 3

- = + = 1 1 x ) x ( g 4 x 2 ) x ( f Original equation Original equation Original

= - x ) x ( f b. 5) x (4

2 2 4 3

−

1) x (

= -

x 6 2 2

Simplify.

2 = . ) x ( of h derivative the Find

Example 2 Example

+ 4) x (2

= + · - · 2 2 ) x ( f 0 0 x 1 2 x 3

Constant, Constant Multiple of a Power, and Sum and Difference Rules Difference and Sum and Power, a of Multiple Constant Constant,

- - 1 1 1 2

= + -

x 3 ) x ( f 4 4 x 2 Original equation Original

- + = 10 10 x 3 x 10 Simplify.

2 (Lesson 12-4) (Lesson 4

+ - = 4 x 2 x 3 ) x ( f a.

- - + + + = 10 10 x 12 x 5 x 6 x 10 x 4

Distributive Property Distributive

2 2 4 2 4

+ - + + =

5) 5) x 6 2)( x ( ) x 5 x 2 )( x (2

Substitution

2 2 3 Find the derivative of each function. function. each of derivative the Find Example

+ = ) ) x ( g ) x ( f ) x ( g ) x ( f ) x ( h Product Rule Product

Constant Rules for Derivatives for Rules Constant Derivatives for Rules Constant

+ = + = ± = ± = 3. 8 ) ( then , 3 4 ) ( If ). ( ) ( ) ( then ), ( ) ( ) ( If x x f x x x f x h x g x f x h x g x f Sum and Difference and Sum

+ = = 5 5 x 6 ) x ( g x 2 ) x ( f

Sum Rule for Limits, Power and and Power Limits, for Rule Sum and Power Limits, for Rule Sum

2 = . ) ) ( then number, real a

x f cnx of a Power a of

1 n

- = + = 2 2 x ) x ( f x 5 x 2 ) x ( g

Original equation Original Original equation Original

= = = 5 ) ( If is a constant and and constant a is where , . 15 ) ( then , is ) ( If x x f c x x f n cx x f

Answers Multiple Constant

2 3 2 3 n

= = 0. ) ( then , ) ( If x f c x f

Constant

+ - = Find the derivative of of derivative the Find ). x 5 x 2)(2 x ( ) x ( h Example 1 Example

= - = 0. ) ( then 2, ) ( If x f x f The derivative of a constant function is zero. zero. is function constant a of derivative The 3 2

. ) ( nx x f

− 1 n Power Rule Power

= = = ) ( If ) ( If is a real number, then then number, real a is and . 3 ) ( then , x x f x x f n x x f

−− 2 3 n

) (

x g

⎦ ⎣

dx ) (

x g

−

= = .

⎢

Quotient Rule Quotient - ⎤ ⎡

) ( ) ( ( ) ( x g x f x) g x f ) ( x f

derivatives of functions that contain several terms. several contain that functions of derivatives

≠ 0, then 0, ) ( and at differentiable are and If x g x g f

There are several rules of derivatives that are useful when finding the the finding when useful are that derivatives of rules several are There

− → 0 h

−

= = lim

) ) x ( f by given is which ), x ( f , provided the limit exists. exists. limit the provided ,

+ = ) ) ( ) ( ) ( ) ( ). ( ) ( , then then , at differentiable are and If x g x f x g x f x g x f x g f Product Rule Product

d + - x ( f ) x ( f ) h

of a function. The derivative of of derivative The function. a of the called is limit This ) is is ) x ( f derivative

derivative of the product or quotient of two functions. two of quotient or product the of derivative determine the slope of a line tangent to the graph of a function at any point. point. any at function a of graph the to tangent line a of slope the determine

Product and Quotient Rules Rules Quotient and Product Derivatives and the Basic Rules Rules Basic the and Derivatives Use the following rules to find the the find to rules following the Use Limits were used in Lesson 12-3 to to 12-3 Lesson in used were Limits 0

Derivatives Derivatives d d

A10

Study Guide and Intervention Intervention and Guide Study Study Guide and Intervention and Guide Study 12-4 12-4 (continued) 1 8 3 9 8

NAME NAME DATE PERIOD PERIOD DATE NAME NAME DATE PERIOD PERIOD DATE 2 1 C M R Chapter 12 C C P _ 7 1 A _ 1 0 AA01_A17_PCCRMC12_893813.indd 10 AA01_A17_PCCRMC12_893813.indd 11 0 1 _ A 1 7 _ P C C R M

NAME DATE PERIOD

1 NAME DATE PERIOD 1 12-4 Practice 12-4 Word Problem Practice A11 Derivatives Derivatives Find the derivative of each function. Then evaluate the derivative of each 1. BIRDS The heighth , in feet, of a 4. VOLUME Suppose the lengthx of each function for the given valuesx of. flying bird can be defined by side of the cube shown is changing. -t3 7 h(t) = + − t2 + 18 on the interval Y 1. g(x) = 3x2 - 5x; x = -2 and 1 2.h (x) = 4x3 - x2; x = 3 and 0 −3 2 [1, 10], where timet is given in seconds. Find the maximum and minimum 2 6x - 5; -17, 1 x 12 - 2x; 102, 0 height of the bird. Y 7 Y = Y 1 1 maximum: 75 − ft, minimum: 21 − ft 6 6 Y 3. f (x) = x2 - 4x + 7; x = 2 and- 3 4.m (x) = -2x2 - 6x + 1; x = 0 and- 3 2. CLIFF DIVING At timet = 0, a diver a. Find the average rate of change of the 2x - 4; 0, -10 - 4x - 6; -6, 6 jumps from a cliff 192 feet above the volumeV (x) as x changes from surface of the water. The heighth of 3.2 inches to 3.4 inches.

the diver is given by Answers h(t) = -16t2 + 16t + 192, whereh 32.7 cu in. per in. 5. q(x) = -1 + x3 - 2x4; x = -1 and 3 6.t (x) = 3x7 - 1; x = -1 and 1 is measured in feet and timet is b. Find the instantaneous rate of change measured in seconds. of the volumeV (x) at the moment 3x2 - 8x3; 11, -189 x216; 21, 21 a. Find the equation for the velocityh'(t ) x = 4 inches. of the diver at any timet. 48 cu in. per in. Find the derivative of each function. h'(t) = -32t + 16 c. Explain the relationship between the (Lesson12-4) 7. f (x) = (x2 + 5x)2 f8. (x ) = x2(x3 + 3x2) b. Find the velocity of the diver after volume formula and the derivative of 1 second has passed. the volume formula. f'(x) = 4x3 + 30x2 + 50x f '(x) = 5x4 + 12x3 h'(1) = -16 ft/s Sample answer: The derivative

Copyright ©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc. of the formula for the volume is c. Find the time when the diver the formula for one half the 5 3 hits the water. t = 4 s 9. f (x) = √ x 6 10.h(x ) = - − 2 x6 surface area of the cube, orx .3 d. What is the diver’s velocity when If the volume of a cube is written 6 5 18 she hits the water? − √ x − in terms of the apothems of its 5 x7 h'(4) = -112 ft/s faces (8a3), the derivative is the formula for the surface area of 5 3 2 2 3 2 11. p(x) = -4x + 6x - 5x 12.n( x) = (3x - 2x)(x + x ) 3. GEOMETRY The formula to find the the cube (24a2). volumeV of a cylinder in terms of its

-20x4 + 18x2 - 10x 15x4 + 4x3 - 6x2 heighth and radiusr is V = πr2h. 5. PROJECTILE Suppose a ball is hit Consider a cylinder with a height of straight upward from a height of 6 feet Pdf Pass 10 inches and a changing radius when with an initial velocity of 80 feet per 3x - 1 13. r(x) = 14.q(x ) = √x (x2 - 3) answering the following questions. second. The heighth of the ball in feet 2 −x + 2 at any timet is given by the function a. Write a formula for the volume of the 2 3 1 2 -3x + 2x + 6 5 − 3 - − h(t) = -16t + 80t + 6. 2 2 − x - − x cylinder in terms of its radius. Lesson 12-4 2 2 2 2 (x + 2) 2 − a. Find the equation for the velocityv(t) V(r) = 10πr Glencoe Precalculus of the ball at any timet by finding the

Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. b. Find an equation for the 15. PHYSICS Acceleration is the rate at which the velocity of a moving derivative ofh (t). object changes. The velocity in meters per second of a particle moving instantaneous rate of change of v(t) = -32t + 80 along a straight line is given by the functionv(t) = 3t2 - 6t + 5, where the volume in terms its radius. t is the time in seconds. Find the acceleration of the particle in V(r) = 20πr b. Find the instantaneous velocity of the meters per second squared after 5 seconds.Hint: ( Acceleration is the ball att = 2 seconds. derivative of velocity.)24 m/s2 c. Find the value ofV (r) when r = 3 inches. 60π cu in. per in. 16 ft/s

112/7/09 10:34:23 AM Chapter 12 24 Glencoe Precalculus Chapter 12 25 Glencoe Precalculus 2 / 7 / 0 9

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27 27 26 26 Chapter 12 Chapter Chapter 12 Chapter Glencoe Precalculus Glencoe Glencoe Precalculus Glencoe

- + cos 2 sin sin cos 2 x x e xe x x

2 x x ; 42 units 42 ; units 44 , units 40

2 2 Lesson 12-5 2

= = = xe ) x ( f ) x (cos ) x ( f x sin ) x ( f 6. 6. 5. 4.

Use rectangles of width 1 unit. Then find the average for both approximations. both for average the find Then unit. 1 width of rectangles Use x 2 2

interval [1, 5] by first using the right endpoints and then by using the left endpoints. endpoints. left the using by then and endpoints right the using first by 5] [1, interval

Use the results of Exercises 1– 3 to find the derivative of each function. each of derivative the find to 3 1– Exercises of results the Use

+ + - = -axis on the the on -axis x the and 6 x 5 x ) x ( f curve the between area the Approximate 2. 2. 2

−

e = Thus, Thus, . .

b. Glencoe Precalculus

; 69.5 units 69.5 ; units 45 , units 94 ) e (

d Pdf Pass

x 2 2 2

! ! !

4 3 2

− − −

+ + + + +

1 … x Use rectangles of width 1 unit. Then find the average for both approximations. both for average the find Then unit. 1 width of rectangles Use

x x x

4 3 2

interval [0, 4] by first using the right endpoints and then by using the left endpoints. endpoints. left the using by then and endpoints right the using first by 4] [0, interval

term by term and simplify the result. the simplify and term by term expansion of of expansion e

+ = -axis on the the on -axis x the and 1 x 3 ) x ( f curve the between area the Approximate 1. 1.

2 was also discussed in Chapter 10. Differentiate the series series the Differentiate 10. Chapter in discussed also was

! ! ! 4 3 2

− − −

Exercises + + + + + = The series expansion for for expansion series The … x 1 e , e 3. a. 3.

x x x x x

4 3 2

areas would give a better approximation of 11 square units. square 11 of approximation better a give would areas

< <

15. Averaging the two two the Averaging 15. 7 area region, the of area the for estimates upper and lower have −

sin sin x = So, So, . .

The area using the right and left endpoints is 15 and 7 square units, respectively. We now now We respectively. units, square 7 and 15 is endpoints left and right the using area The (cos (cos d ) x

! ! ! !

9 3 7 5

= = 15 total area 7 area total 15 area total

− − − −

+ - + - + -

… x

x x x x

4 4 9 7 5 3 · = · = (3) or 4.5 or (3) f 1 R 8 or (4) f 1 R

−

3 3

· = · = R (2) or 2 or (2) f 1 R 4.5 or (3) f 1 Find Find using the series expansion of cos . x cos of expansion series the using b.

) x (cos d

2 2 · = · = (1) or 0.5 or (1) f 1 R 2 or (2) f 1 R

Answers may vary. may Answers

1 1

· = · = (0) or 0 or (0) f 1 R 0.5 or (1) f 1

What would you guess might be the derivative of cos cos of derivative the be might guess you would What ? x 2. a. 2. Area using right endpoints Area using left endpoints left using Area endpoints right using Area

Figure A Figure Figure B Figure −

cos cos x = So, So, . . c.

? (sin (sin d ) x

0 0 x x

cos cos x What function does this new infinite series represent? represent? series infinite new this does function What b.

4 4

8 8

! ! ! !

8 6 4 2

− − − −

- + - + - … 1 term and simplifying the result. result. the simplifying and term

x x x x

−

8 6 4 2

(Lesson 12-4 and Lesson 12-5) Lesson 12-4 and (Lesson dx

id Find by differentiating the series expansion of sin sin of expansion series the differentiating by term by by term x 1. a. 1.

) x (sin d

(0) or 0 and thus, has an area of 0 square units. square 0 of area an has thus, and 0 or (0) f

The power functions in these series expansions can be differentiated. be can expansions series these in functions power The

rectangles with a width of 1 unit (Figure B). However, the first rectangle has a height of of height a has rectangle first the However, B). (Figure unit 1 of width a with rectangles

of one unit (Figure A). Using left endpoints for the height of each rectangle produces four four produces rectangle each of height the for endpoints left Using A). (Figure unit one of

! ! ! ! 9 7 5 3

− − − −

- = - + - + x x sin … . … Using right endpoints for the height of each rectangle produces four rectangles with a width width a with rectangles four produces rectangle each of height the for endpoints right Using

x x x x

9 7 5 3

endpoints of the rectangles. Use rectangles with a width of 1. of width a with rectangles Use rectangles. the of endpoints and the sine function, being odd, was shown to be a sum of odd powers of of powers odd of sum a be to shown was odd, being function, sine the and : : x

on the interval [0, 4] by first using the right endpoints and then by using the left left the using by then and endpoints right the using first by 4] [0, interval the on

! ! ! ! 8 6 4

Answers 2

− − − −

- + - + - = … 1 x cos

x x x x

8 6 4 2

Approximate the area between the curve curve the between area the Approximate ) x ( f and the the and x -axis -axis x −

Example

2 1 : : x of powers even of

= presented. In particular, the even function function even the particular, In presented. , was shown to be a sum sum a be to shown was , x cos y

In Chapter 10, the series expansions of some transcendental functions were were functions transcendental some of expansions series the 10, Chapter In

graph of a function ( f function a of graph ] in the domain of f of domain the in ] b [ , a interval an on -axis x the and ) x ). x (

Powerful Differentiation Powerful Area Under a Curve Curve a Under Area You can use the area of rectangles to find the area between the the between area the find to rectangles of area the use can You 2

Area Under a Curve and Integration and Curve a Under Area d d

A12

3 Study Guide and Intervention and Guide Study Enrichment 12-5 12-4 1 8 3 9 8

NAME NAME DATE PERIOD PERIOD DATE NAME NAME DATE PERIOD PERIOD DATE 2 1 C M R Chapter 12 C C P _ 7 1 A _ 1 0 AA01_A17_PCCRMC12_893813.indd 12 AA01_A17_PCCRMC12_893813.indd 13 0 1 _ A 1 7 _ P C C R M

NAME DATE PERIOD

1 NAME DATE PERIOD 3

12-5 Study Guide and Intervention (continued) 12-5 Practice A13 Area Under a Curve and Integration Area Under a Curve and Integration Integration Approximate the area between the curvef(x) and thex -axis on the indicated interval using the indicated endpoints. Use rectangles b n with a width of 1. ⌠ The area of a region under the graph of a function fis (x ) dx = lim ∑ f (x)Δx, i ⌡ n → ∞ i = 1 2 2 2 Definite Integral a 1. f (x) = x + 3 22 units 2.f ( x) = -x + 6x -4 10 units b - a where a and b are the lower limits and upper limits, respectively,Δx = − n [1, 5] [2, 5] and x = a + iΔx. i left endpoints right endpoints

Example Use limits to find the area of the region between the graphy = of 5 ⌠ 4x2 and thex -axis on the interval [0, 5], or 4 x 2 dx.

Answers ⌡ 5 n 0 ⌠ 3 2 2 2 4x2 dx = ∑ f (x)Δx Definition of definite integral 3. g(x) = 3x 108 units 4.p( x) = 1 + x 95 units lim i ⌡ n → ∞ i = 1 0 [0, 4] [1, 6] n = lim ∑ 4 x 2 Δx f(x) = 4x2 left endpoints right endpoints i i i n → ∞ i = 1 n 2 5i 5 5i 5 x = and Δx = − = lim ∑ 4 (− ) − i −n n n n (Lesson12-5) n → ∞ i = 1 n 20 25 = lim − − ∑ i 2 Expand and factor. n (n2 ) n → ∞ i = 1 n Use limits to find the area between the graph of each function and the 20 25 n(n + 1)(2n + 1) n(n + 1)(2n + 1) = lim · ∑ i 2 = Copyright ©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc. −n (−2 ) 6 x-axis given by the definite integral. n 6 i = 1 n → ∞ − −

500 2n2 + 3n + 1 2 6 = lim Simplify and expand. − ( 2 ) ⌠ 2 8 2 ⌠ 2 2 6 n n → ∞ 5. x dx − units 6. 6 x dx 430 units − ⌡ ⌡ 3 0 1 500 3 1 = lim 2 + − + − Factor and divide each term nby2. − ( n 2) n → ∞ 6 n 500 1 1 = lim lim 2 + lim 3 lim − + lim − Limit theorems ( − ) ( )( n) 2 n → ∞ 6 [n → ∞ n → ∞ n → ∞ n → ∞ n ]

500 3 1 Simplify. = − [2 + 3(0) + 0] or about 166.67 square units ⌠ 2 14 2 ⌠ 2 2 6 7. (x - x) dx − units 8. (-x -2x + 11) dx 33 units ⌡ ⌡ Pdf Pass 3 1 -2 Exercise

Use limits to find the area between the graph of each function and the x-axis given by the definite integral. Glencoe Precalculus y 2 4 9. Architecture and Design A designer is making a 10 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. ⌠ 2 ⌠ 74 2 1. x3 dx 4 units 2. ( x2 + 3) dx − units stained-glass window for a new building. The shape ⌡ ⌡ 3 of the window can be modeled by the parabola 0 2 x y = 5 - 0.05x2. What is the area of the window? 10 5 5 10 6 3 − − 2 −5 ⌠ 2 ⌠ 3 2 about 66.67 units 3. (1 + x) dx 12 units 4. 4x dx 80 units ⌡ ⌡ −10 Lesson 12-5 4 1

112/7/09 10:36:01 AM Chapter 12 28 Glencoe Precalculus Chapter 12 29 Glencoe Precalculus 2 / 7 / 0 9

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31 31 30 30 Chapter 12 Chapter Glencoe Precalculus Glencoe Chapter 12 Chapter Glencoe Precalculus Glencoe

Lesson 12-5 3

km 1 −

2 1

4 x

to cover 35 square yards. square 35 cover to

− 3

6 6

(0) f

7. 7. 6.

y d

yards but only has enough seed seed enough has only but yards 3

enough seed to cover 36 square square 36 cover to seed enough

Glencoe Precalculus

= - =

1 x

dx Pdf Pass

−

3 24

(2) (2) f

4. 4. 5. 5.

+ =-

Explain.

x 2 x y

2 3

have enough seed to complete the task? task? the complete to seed enough have

- + = 4, find the value of each expression. each of value the find 4, x 3 x ) x ( f Given

2 3

⌡

- + -

, will he he will , dx 7) x 8 x ( be found by by found be

2 in kilometers? in ⌠

7 area of the face of the dam if if dam the of face the of area is given given is x

square yards that he needs to seed can can seed to needs he that yards square

2 x

shape of the region shown. What is the the is What shown. region the of shape

−

2.” x at evaluated dx dy “ read , following:

to cover 35 square yards. If the area in in area the If yards. square 35 cover to

DAMS The face of a dam is in the the in is dam a of face The 3. 3.

of his lawn, but he has only enough seed seed enough only has he but lawn, his of

using the Leibniz notation, one might use the use might one notation, Leibniz the using x of value specific a at

GRASS SEED GRASS Mr. Bower is seeding part part seeding is Bower Mr. 5. 5.

a fraction of any kind. To indicate the value of the derivative derivative the of value the indicate To kind. any of fraction a not is

−

m 5

−

.” Note that that Note .” x to respect with of y derivative “the formally, more or

40.5 units 40.5

−

,” dx “ dy read usually is derivative the for notation Leibniz The

- 4

⌡

+ . dx 4) x ( evaluating evaluating

⌠

dx 0

− 4

second fourth first third

y y ) ) x ( f d. c. b. a.

Calculate the area of the triangle by by triangle the of area the Calculate c. c.

. y d

What is the order of each derivative? each of order the is What 3. 3.

+ - =

4 x 5 x y

2 4

= ). x ( f y function

and base length. length. base and

− 2

y all indicate the second derivative of some some of derivative second the indicate all ÿ and , ) ) x ( f example, For

y d

area of the triangle using its height height its using triangle the of area 2

used to indicate higher-order derivatives. higher-order indicate to used of the triangle. Then calculate the the calculate Then triangle. the of

(Lesson 12-5) (Lesson meters? in

the notation developed by Isaac Newton. Each of these notations also can be be can also notations these of Each Newton. Isaac by developed notation the Find the height and length of the base base the of length and height the Find b. b. is the area of the entrance if if entrance the of area the is is given given is x

= . This was was This . y is ) x ( f y function a of derivative the for notation another Yet in the shape of the region shown. What What shown. region the of shape the in

MINING The entrance to a coal mine is is mine coal a to entrance The 2. 2.

dx dx

−

0 −

x , , D , ), ( , answers: Sample y y x f

) ( x d dy

4 ft 4 2 2

Answers

List several other ways of expressing this quantity. quantity. this expressing of ways other several List 2. 2.

0 x

the derivative of of derivative the ) ( x x f

+ =-

x 4 x y

− 3

→ →

0 h

lim . What does does What . find?

x ) x ( f Let 1. 1.

Sample answer: Sample

- + x ) h x (

2 2

is given in feet? in given is if x house dog Fido’s to

Shade the interior of this triangle. this of interior the Shade a. a.

for the same thing. You have already seen this in the case of the derivative. the of case the in this seen already have You thing. same the for

shown. What is the area of the entrance entrance the of area the is What shown.

branches of mathematics. In addition, there is often more than one notation notation one than more often is there addition, In mathematics. of branches

+ = = house is in the shape of the region region the of shape the in is house 4. x y and 5 x lines the and

There is a lot of special notation used in calculus that is not used in other other in used not is that calculus in used notation special of lot a is There

house for Fido. The entrance to the dog dog the to entrance The Fido. for house -axis -axis x the by formed triangle the draw

Reading Mathematics Reading

DOG HOUSE DOG TRIANGLE AREA TRIANGLE Charlie is building a dog dog a building is Charlie On a coordinate plane, plane, coordinate a On 1. 1. 4. 4. 4

Area Under a Curve and Integration and Curve a Under Area d d

A14

3 Enrichment 12-5 Word Problem Practice Problem Word 12-5 1 8 3 9 8

NAME NAME DATE PERIOD PERIOD DATE NAME NAME DATE PERIOD PERIOD DATE 2 1 C M R Chapter 12 C C P _ 7 1 A _ 1 0 AA01_A17_PCCRMC12_893813.indd 14 AA01_A17_PCCRMC12_893813.indd 15 0 1 _ A 1 7 _ P C C R M

NAME DATE PERIOD NAME DATE PERIOD 1 5

12-6 Study Guide and Intervention 12-6 Study Guide and Intervention(continued) A15 The Fundamental Theorem of Calculus The Fundamental Theorem of Calculus

Antiderivatives and Indefinite Integrals The Fundamental Theorem of CalculusThe indefinite integral of ⌠ f (x) is defined by f (x)dx = F(x) + C, whereF (x) is an antiderivative off (x ) Given a functionf ( x), we say thatF (x) is an antiderivative off (x ) if F (x) = f (x). ⌡ and C is any constant. Rules for Antiderivatives If f(x) = xn, wheren is a rational number other than-1, b Power Rule xn + 1 F(x) = + C. ⌠ −n + 1 Fundamental If F(x) is the antiderivative of the continuous functionf (x), then f(x) = F(b) - F(a). ⌡ Constant n If f(x) = kx , wheren is a rational number other than-1 and k is a Theorem of a Multiple kxn + 1 Calculus ⎢b of a Power constant, thenF (x) = − + C. The right side of this statement may also be writtenF( xas) ⎢ . n + 1 ⎢a Sum and If the antiderivatives off(x ) and g(x) are F(x) and G(x), respectively,

Difference then the antiderivatives fof(x ) ± g(x) are F(x) ± G(x). Answers Example Evaluate each integral. Lesson 12-6 Example Find all antiderivatives for each function. ⌠ a. (3 x2 + 4x - 1) dx ⌡

5 2 + 1 1 + 1 0 + 1 a. f (x) = -3x ⌠ 2 3x 4x x (3 x + 4x - 1) dx = + - + C Constant Multiple of a Power ⌡ − − − 2 + 1 1 + 1 0 + 1 5 f (x) = -3x Original equation 3x3 4x2 (Lesson12-6) = − + − - x + C Simplify. -3x5 + 1 3 2 F(x) = + C Constant Multiple of a Power −5 1 + = x3 + 2x2 - x + C Simplify.

1 6 = - − x + C Simplify. Copyright ©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc. 2 4 ⌠ b. (x3 - 1) dx 3 2 b. f (x) = x + 4x - 2 ⌡ 2 3 2 4 Original equation f (x) = x + 4x - 2 4 ⎢4 ⌠ 3 x (x - 1) dx = ( − - x) ⎢ Fundamental Theorem of Calculus ⌡ 4 3 2 0 ⎢2 = x + 4x - 2x Rewrite the function so each term has a powerx. of 2 44 24 = ( − - 4) - ( − - 2) b = 4 anda = 2 x3 + 1 4x2 + 1 2x0 + 1 4 2 F(x) = + - Use all three rules. − − − 3 + 1 2 + 1 0 + 1 = 60 - 6 or 54 Simplify.

1 4 4 3

Pdf Pass = − x + − x - 2x + C Simplify. 4 3 Exercises

Exercises Evaluate each integral. 2 ⌠ 3x8 x3 ⌠ 10 Find all antiderivatives for each function. 1. (3 x7- x2) dx − - − + C 2. (x2 + 1) dx − ⌡ 8 3 ⌡ 3 4 2 2 Glencoe Precalculus 1. f (x) = 2x + 3x - 5 2.g(x ) = − 1 x3 2 1 Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. − x5 + x3 - 5x + C - − + C 5 x2 2 1 3 1 5 ⌠ 4 ⌠ 3. t(x) − 6 − 3 n4.(x ) √ x 2 3. (x2 1) dx − 4. (x3 2x 1) dx = x - x = - - - + 2 4 2 ⌡ 3 ⌡ 6 1 3 1 5 − -1 − x7 - − x4 + C − x 5 - 2x + C 28 8 6

112/7/09 10:38:12 AM Chapter 12 32 Glencoe Precalculus Chapter 12 33 Glencoe Precalculus 2 / 7 / 0 9

1 0 : 3

8 005_036_PCCRMC12_893813.indd 32 3/17/09 11:38:12 AM 005_036_PCCRMC12_893813.indd 33 3/17/09 11:53:38 PM : 1

2 Answers

A M M P

1 1 : 1 3 : 8

9 0 / 7 1 /

/70 13:4AM 11:38:24 3/17/09 35 005_036_PCCRMC12_893813.indd 0_3_CRC2831.nd34 005_036_PCCRMC12_893813.indd /70 13:1AM 11:38:21 3/17/09

35 35 34 34 Chapter 12 Chapter Chapter 12 Chapter Glencoe Precalculus Glencoe Glencoe Precalculus Glencoe

126 h 126 246810

0 x

How many hours does it take the craftsman to make 6 pieces? 6 make to craftsman the take it does hours many How

360 joules 360

2 0

⌡

- = = . . dx ) x 3 (30 Suppose the number of hours needed to create create to needed hours of number the Suppose h by given is pieces p

⌠

+ =-

is required? is 18 x 3 y

0 WOODWORKING hours to create one piece of furniture. furniture. of piece one create to hours h works craftsman A 12. 12.

⌡ 8

How much work work much How dx. x 80

given by by given Glencoe Precalculus ⌠

22.5 ft 22.5 Pdf Pass =

x y 2 10 2

0.25 ft-lb 0.25 y

36 inches beyond its natural length is is length natural its beyond inches 36

What is the area of this figure? this of area the is What

spring 6 inches from its natural length? natural its from inches 6 spring joules, required to stretch a certain spring spring certain a stretch to required joules,

in feet, is shown in the diagram below. below. diagram the in shown is feet, in

SPRING STRETCHING SPRING The work, in in work, The 3. 3.

central figure on the billboard, measured measured billboard, the on figure central

⌡

= =

its natural length is given by by given is length natural its W . How much work is required to compress the the compress to required is work much How . dx x 2

billboard to advertise the company. The The company. the advertise to billboard

⌠

Trucking Company has purchased a a purchased has Company Trucking 15

−

PHYSICS feet from from of feet distance a spring certain a compress to foot-pounds in work The 11. 11. BILLBOARD The Squared and Linear Linear and Squared The 6. 6. 2 8

500 9 500 2400 inch-pounds 2400

0 x

2 inches? 2 - - 5 2

⌡ ⌡

required to compress the spring another spring the compress to required

- + - -

4 ( dx 3) x )( x (1 dx ) x 3 x 9. 9. 10. 10.

2 3

⌠ ⌠ 2

⌡

- 1 1

. How much work is work much How . dx x 400 given by by given

+ - =

1 x 2 x y ⌠

2 4

compress the spring another 2 inches is is inches 2 another spring the compress

- - + + −

x 21

C x x −

between 0 inch and 1 inch? 1 and inch 0 between x for

The work, in inch-pounds, required to to required inch-pounds, in work, The 6 2 3 2 5

logo occupy at the top of each document document each of top the at occupy logo

from its natural length of 12 inches. inches. 12 of length natural its from 2

⌡ ⌡ its letterhead, how much space will the the will space much how letterhead, its 800 pounds compresses a spring 2 inches inches 2 spring a compresses pounds 800

+ - -

x dx x 2 dx ) x 5 x 2 x 6 ( 7. 7. 8. 8.

5 2

⌠ ⌠ If the company intends to use it as part of of part as it use to intends company the If SPRING COMPRESSION SPRING A force of of force A 5. 5.

(Lesson 12-6) (Lesson 5 is in the shape of the region shown below. below. shown region the of shape the in is

−

ADVERTISING New Wave’s business logo logo business Wave’s New 2. 2. units

3 π 243

+ + + − 8 3 C x C x x

2 4 1

1.5 s 1.5

⌡ ⌡

dx ) x 6 dx 8 x (2 5. 5.

Lesson 12-6 6.

- 6

3 ⌠ ⌠ before she lands on the ground? the on lands she before

4 2

After Lila jumps, how long does it take take it does long how jumps, Lila After b. 0

Answers x

Evaluate each integral. each Evaluate

+ - = 16 ) ( 24 t t s t

2 4 3

x ) x ( f

+ - = + - + =

− − − ) ( 3 ) ( C x x x F C x x x x F

2 4 2 3 9

1 3 8

= = jump. Assume that for for that Assume jump. 0. ) t ( s 0, t

Find the position function function position the Find ) for Lila’s Lila’s for ) t ( s

- = - + = 3) 3) x ( x ) x ( f 3 x 2 x 8 ) x ( f 3. 4.

2 2 ⌡

π dx. ) x ( by

2 2 ⌠ feet per second. per feet

seconds and the velocity is given in in given is velocity the and seconds

+ = + + =

) ( 3 ) ( C x x F C x x x F

[0, 3], if the volume of the solid is given given is solid the of volume the if 3], [0, 4 2 + =- 24, where where 24, t 32 ) t ( v is given in in given is t

= over the interval interval the over x ) x ( f of graph the

velocity of her jump can be defined as as defined be can jump her of velocity

2 + = =

3 x 2 ) x ( f x 4 ) x ( f 2. 1.

volume of the solid formed by revolving revolving by formed solid the of volume jump in physical education class. The The class. education physical in jump

VERTICAL JUMP VERTICAL VOLUME Lila tested her vertical vertical her tested Lila In the figure below, find the the find below, figure the In 1. 1. 4. 4. 6

Find all antiderivatives for each function. each for antiderivatives all Find 1

The Fundamental Theorem of Calculus of Theorem Fundamental The The Fundamental Theorem of Calculus of Theorem Fundamental The d d

A16

3 Word Problem Practice Problem Word Practice 12-6 12-6 1 8 3 9 8

NAME NAME DATE PERIOD PERIOD DATE NAME NAME DATE PERIOD PERIOD DATE 2 1 C M R Chapter 12 C C P _ 7 1 A _ 1 0 AA01_A17_PCCRMC12_893813.indd 16 AA01_A17_PCCRMC12_893813.indd 17 0 1 _ A 1 7 _ P C C R M

NAME DATE PERIOD 1 7 12-6 Enrichment A17

Derivatives of Exponential and Logarithmic Functions

Exponential Rule The derivative ofy = ex is ex and the derivative ofy = eu is eu du . −dx

Example 1 Find the derivative ofy = e3 x. dy du Let u = 3x. Then− = eu · − . dx dx du dy Since − = 3, − = eu · 3 or 3eu. dx dx

The derivative ofy = e3x is 3e3x. Answers 1 1 du Logarithmic RuleThe derivative ofy = ln x is − and the derivative ofy = ln u is − · − . x u dx

Example 2 Find the derivative ofy = ln (x2 + 3). dy 1 du Let u = x2 + 3. Then− = − · − . dx u dx

du dy 1 1 2x (Lesson12-6) Since = 2x, − = − · 2x, or · 2x. Simplify to get . − u −2 −2 dx dx x + 3 x + 3

Exercises

Find the derivative of each function.

√x x e - − 1. y = e-x 2.y = 3.y = e 4 −2 x -− -x 1 √x 1 -e e - − e 4 − 4 4 √x 4. y = e6x 5.y = 4ex 6.y = x2ex

6x x 2 x x Pdf Pass 6e 4e x e + 2xe

7. y = ln (x3) 8.y = ln (2x + 5) 9.y = ln (sinx + 4) 3 2 cos x − x − 2x 5 x + −sin + 4 1 3 10. y = ln ( − x ) 11.y = x ln x 12.y = ln (2x + 4x) Glencoe Precalculus 1 6x2 + 4 - − +1 ln x x −2x3 + 4x

13. Find an equation for a line that is tangent to the graph of y = ln x through the pointe ,( 1). 1 x y - 1 = − (x - e) or y = − e e

112/7/09 10:38:42 AM Chapter 12 36 Glencoe Precalculus 2 / 7 / 0 9

1 0 : 3

8 005_036_PCCRMC12_893813.indd 36 3/17/09 11:38:31 AM : 4

2 Answers

A M M P

7 4 : 4 4 : 5

9 0 / 5 / 2

1.25 s = = no limit -

) H G B C t ( m v

Mid-Chapter Test Mid-Chapter 1. 9. 10. 5. 6. 7. 8. 3. 2. 4.

x C

2 x + + )

+ 2

1 40

x x 4 − 3 4 x C − -

B 3 + -

10 A −

21 4 3 - - - - - 2 + (Lessons 12-4 and 12-5) (Lessons x

2 (Lesson 12-6) - x 3 (

x - 3 −

x 3 x 2

5 12 −

2 3 −

- Quiz 4 Page 38 4. 5. 1. 2. 3. 5. 1. 2. 3. 4. Quiz 3 Quiz Chapter 12 Assessment Answer Key Answer 12 Assessment Chapter 8 8 4 47 5 1 1 D

1 C

- - d d 64 ft/s 38 ft/s n i . no limit A18 (Lessons 12-1 and 12-2) (Lessons 3 - - (Lesson 12-3) 1 8 3 9 8 _ 2 1 C M R Quiz 2 Page 37 5. Chapter 12 4. 3. 1. 2. 6. 5. 3. 4. 1. 2. Quiz 1 Quiz 1 Page 37 Page 38 Page 39 C C P _ 6 2 A _ 8 1 AA18_A26_PCCRMC12_893813.indd 18 AA18_A26_PCCRMC12_893813.indd 19 1 8 _ A 2 6 _ P C C R M

C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 _ 8 9 3 8 1 Chapter 12 7. 6. 5. 4. 3. 2. 1. ae4 Pg 1 Page42 Page41 Form1 Page 40 Vocabulary Test 3 . i n d d One-sided limits in time. at anyspecifi velocity ofanobject velocity isthe Instantaneous limit ofafunction. attempting tofi substitution when quotient ruleordirect result ofapplyingthe fraction is whenthe Indeterminate form equation differential indefi of change instantaneous rate antiderivative

1 9 A19 nite integral −

0 0

isthe c point Chapter 12AssessmentAnswerKey nd the nd the 4. 3. 2. 1. 10. 9. 8. 7. 6. 5.

H B J A F C F D G B 14. 13. 12. B: 20. 19. 18. 17. 16. 15. 11.

0.375 Pdf Pass H C G C G B J C J D Glencoe Precalculus 33/17/09 11:42:48 AM / 1 7 / 0 9

1 1 : 4 2 : 4

8 Answers

A M M A

2 5 : 2 4 : 1 1

9 0 / 7 1

20. B: 15. 16. 17. 18. 19. 11. 12. 13. 14. C H A H C F A F B J

10. 5. 6. 7. 8. 9. 2. 3. 4. 1. 4 A F C F B G D G B F

19. 20. B: 17. 18. 13. 14. 15. 16. 11. 12. Chapter 12 Assessment Answer Key Answer 12 Assessment Chapter 0 2

d d n i . A20 3 1 8 3 9 H C H D H B J A H A 8 _ 2 1

C M R 10. Chapter 12 9. 8. 6. 7. 5. 1. Form 2A Form 2A Page 43 Page 44 Page 45 2B Form Page 46 4. 3. 2. C C P _ 6 2 A _ 8 1 AA18_A26_PCCRMC12_893813.indd 20 AA18_A26_PCCRMC12_893813.indd 21 1 8 _ A 2 6 _ P C C R M

2 1 ( t )

m = A21 5 s;425ft

− 3 1 =

t - - 0, 2

- 25 60 x 8 2 ∞ ∞

− 3 2 2

+ - Chapter 12AssessmentAnswerKey

− 2 1 x

t -

− 1 2

14. 13. 12. 11. 20. 19. 18. 17. 16. 15. B: g

− − 5 9 5 1 s ( h -

( x x x t 5 ) 5 ) ( 12

x + = 16 - = 18 )

x 2

− −

2 1 = - x 8 ft/s 2 - x

x 2 3 x x +

4 − 16 −

6 ( 3 15 (2 + - 8 + x + 2 −

+ 3 2

- 2

- t

− 24 3 1 28 - x 2 x x

x x x Pdf Pass + 3 4)

- 3 2

x 3 +

+ )

+ 2

35 2

C x

Glencoe Precalculus 33/17/09 11:42:54 AM / 1 7 / 0 9

1 1 : 4 2 : 5

4 Answers

A M M A

8 5 : 2 4 : 1 1

9 0 / 7 1

x C

6 C 2

2 20 + +

)

3 3 + 2 x 2) 10 x

x + 4 x x 4

- x 7

2 - -

- 16

4 x 2 3

− 10 + 2 - x ( (4 − + 13 15

− 6 16 3 x - 2 x 3 x

5 36 x - 2

= = - 5 3

− − - x

111 ft/s

) 6 - 8 = 9

x + = = 3 x + (

)

x 4 ) t 3 ( h x x 8

( s 1 4 − 4 3 −

g B: 20. 15. 16. 17. 18. 19. 11. 12. 13. 14.

4 5 −

- 2 t

x 1 5 −

12 Chapter 12 Assessment Answer Key Answer 12 Assessment Chapter -

- 2 3 4 −

∞

11 9 4 - ∞ - 2 500 2, 1 t - x -

1 2 − d

d = n i

. A22 = 3 5.5 s; 494 ft 1 8 m ) 3 t 9 ( 8 _ v 2 1 C M R 10. Chapter 12 9. 8. 7. 5. 6. 4. 3. 2. 1. Form 2D Page 49 50 Page C C P _ 6 2 A _ 8 1 AA18_A26_PCCRMC12_893813.indd 22 AA18_A26_PCCRMC12_893813.indd 23 1 8 _ A 2 6 _ P C C R M

(

2 t m 3 )

= = 8 s;1024ft A23

0.34 3 no limit x 2000 - 2, 1 2 12

3 0 t ∞ + 0.7 6 Chapter 12AssessmentAnswerKey

+ x

− 9 1

t 3 -

− 3 2

14. 13. 12. 11. 20. 19. 18. 17. 16. 15. B: h

− 4 1 x ( (

s 4 x x x

( 4 ) - )

t = + )

= −

− 2 3 8 7 − 3 8

= 7

x -

x x −− x x 3 39 ft/s

6 3 x − 2 1 − 2 - x

3 +

10 -

√ 2 + - + 16 ( − 2 3 − − 8

x 1.2 11 x 3 2

2 −

1 3

3 2

− x 2 1

( x

- t -

3 2

x x 2

x 3 - +

+ 2 x - x Pdf Pass 2 +

6 x

- 2

− 2 ) 2 1 + x 2 x

3 2 12

+ √ 1) +

3 C 2

Glencoe Precalculus C 33/17/09 11:43:01 AM / 1 7 / 0 9

1 1 : 4 3 : 0

1 Answers

A M M P

7 4 : 7 4 : 5

9 0 / 5 / 2

= 8 3 x

−

= - (4) s ) x · ( g , or

)

x x ) ) +

4 0 6) 5) ) ) ⎢ ⎢ 6

5 6 5

+ Glencoe Precalculus and - 6 - 2 5

t + -

. 2 Pdf Pass (3)

5 6

−

)

x 2 x s - + x

( x x x x (

3 x 1 3

−

( ( ( ( 64 - · (

− x

x − g (

(

= −

3 -

- 1 )

units

or 2

→ 0 x

→ 0 → → t +

lim ( x → 0 6 lim x lim 3 lim → 0 → 0

f x x x lim 32 5

−

lim

− lim

2 x - 32

−

(2) = = = =

s = = = =

x x x x 0 d 0 or 10 units 5 ) 5 6 6 1 2

= + t S + - - 0 + ) +

2 2 2 2 x 3 x x . −

x 4 3 2 1 2 ( - x −

st t (1) +

2235 3 s 4

4 ( → 0

= ·

→ → 0 lim

x ⌠ ⌡ lim lim

0 x x

The area under the curve represents the distance between the object and its starting point after 4 seconds because d Method 2: Estimate using rectangles. 1 Method 1 is more accurate because integrating yields the exact area under the curve. Method 2 yields an approximation. Method 1: Integrate.

Sample answer:

1e.

3c.

3a. 3b. 1f. = 2. 1, -

-axis 1. x, x x,

x 3 3 0. The → → 3, a value of lim lim = x x

) is

· x x x · → ( f 5 x 6 Page 53, Extended-Response Test Extended-Response Page 53, + 1. - . As

8 3

−

3 3 -coordinates )

- x y 4 7 6 5 7 4 8 3 9 2 3 8

− − − − − − → ( →

3. Therefore, 10 lim lim 3. For the limit as 12 x x

------

- =

- x = = 51015 undefined undefined

Chapter 12 Assessment Answer Key Answer 12 Assessment Chapter ) ) 0 is approximately (3), or f x x

) is about 6 → 5 x 4 = ( 2 )

x f

- + x )

( 2

2 d f

3. The intersection of this line and x d 0 x x ( n i f . 2 0 xf A24 2 3 4 4 3 2 1 (

( 0, follow the same procedure as 3

= 9 4 5 6 7 1 2 3 1 0 →

- 3 3 3 - - - lim 8 x

x - 3 9 → → → → 8 lim lim lim _ x x x

the curve is the limit as approximately x described above, except at limit as Draw a vertical line through the at or or 24 approaches 0, the approach a value of approximately so The graph indicates that continuous at

2 1 C M R 1c.

1d. 1a. Sample Answers Sample Chapter 12

1b. C C P _ 6 2 A _ 8 1 AA18_A26_PCCRMC12_893813.indd 24 AA18_A26_PCCRMC12_893813.indd 25 1 8 _ A 2 6 _ P C C R M

C Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 1 2 _ 8 9 3 8 1 Chapter 12 Page55 Page 54 Standardized TestPractice 7. 6. 5. 4. 3. 2. 1. 3 . i n d d ABC ABCD AB AB FGHJ F FG

2 5 HJ GH HJ CD CD A25 D Chapter 12AssessmentAnswerKey 15. 14. 13. 12. 11. 10. 9. 8.

AB FGHJ ABCD ABC ABC F FGH F HJ GH HJ GH CD D D J

Pdf Pass Glencoe Precalculus 33/17/09 11:43:11 AM / 1 7 / 0 9

1 1 : 4 3 : 1

1 Answers

A M M A

3 1 : 3 4 : 1 1

9 0 / 7 1

(continued) t

1 80 - C 3

2 + ) 2

+ t Chapter 12 Assessment Answer Key Answer 12 Assessment Chapter =

1 2 x − −

(

2 4 x 9 16 5 s y 5 4

−

+ + 96 ft 17

3.68 6

- 2 -

≈

+ 3

ellipse

5 d

n = 2 i - = 2 . x

A26

3 n ) x 1 16 5 1 x −

−

t 8 a 3 ( 0 and 20 ounces 9 8 s answer: the weight answer: the weight _ of the rice could be of the rice could any weight between continuous; Sample continuous; Sample 2 1 C M R 25c. Chapter 12 25a. 25b. 24. 22. 23. 21. 20. 18. 19. 17. 16. Standardized Test Practice Test Standardized Page 56 Page 56 C C P _ 6 2 A _ 8 1 AA18_A26_PCCRMC12_893813.indd 26