chapter 12 resource masters - anderson1.org · chapter 12 resource masters ... limits of polynomial...

Chapter 12 Resource Masters

Pdf Pass00i_PCCRMC12_893813.indd 100i_PCCRMC12_893813.indd 1 3/28/09 12:01:35 PM3/28/09 12:01:35 PM

Pdf Pass

Copyright © by The McGraw-Hill Companies, Inc. All rights reserved. Permission is granted to reproduce the material contained herein on the condition that such materials be reproduced only for classroom use; be provided to students, teachers, and families without charge; and be used solely in conjunction with the Glencoe Precalculus program. Any other reproduction, for sale or other use, is expressly prohibited.

Send all inquiries to:Glencoe/McGraw-Hill8787 Orion PlaceColumbus, OH 43240 - 4027

ISBN: 978-0-07-893813-9MHID: 0-07-893813-9

Printed in the United States of America.

2 3 4 5 6 7 8 9 10 079 18 17 16 15 14 13 12 11 10

StudentWorks PlusTM includes the entire Student Edition text along with the worksheets in this booklet.

TeacherWorks PlusTM includes all of the materials found in this booklet for viewing, printing, and editing.

Cover: Jason Reed/Photodisc/Getty Images

0ii_004_PCCRMC12_893813.indd Sec2:ii0ii_004_PCCRMC12_893813.indd Sec2:ii 12/5/09 4:41:12 PM12/5/09 4:41:12 PM

Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

Pdf Pass

Contents

Teacher’s Guide to Using the Chapter 12 Resource Masters ........................................... iv

Chapter ResourcesStudent-Built Glossary ....................................... 1Anticipation Guide (English) .............................. 3Anticipation Guide (Spanish) ............................. 4

Lesson 12-1Estimating Limits GraphicallyStudy Guide and Intervention ............................ 5Practice .............................................................. 7Word Problem Practice ..................................... 8Enrichment ........................................................ 9Graphing Calculator Activity ............................ 10

Lesson 12-2Evaluating Limits AlgebraicallyStudy Guide and Intervention .......................... 11Practice ............................................................ 13Word Problem Practice ................................... 14Enrichment ...................................................... 15

Lesson 12-3Tangent Lines and VelocityStudy Guide and Intervention .......................... 16Practice ............................................................ 18Word Problem Practice ................................... 19Enrichment ...................................................... 20Spreadsheet Activity ........................................ 21

Lesson 12-4DerivativesStudy Guide and Intervention .......................... 22Practice ............................................................ 24Word Problem Practice ................................... 25Enrichment ...................................................... 26

Lesson 12-5Area Under a Curve and IntegrationStudy Guide and Intervention .......................... 27Practice ............................................................ 29Word Problem Practice ................................... 30Enrichment ...................................................... 31

Lesson 12-6The Fundamental Theorem of CalculusStudy Guide and Intervention .......................... 32Practice ............................................................ 34Word Problem Practice ................................... 35Enrichment ...................................................... 36

AssessmentChapter 12 Quizzes 1 and 2 ........................... 37Chapter 12 Quizzes 3 and 4 ........................... 38Chapter 12 Mid-Chapter Test .......................... 39Chapter 12 Vocabulary Test ........................... 40Chapter 12 Test, Form 1 ................................. 41Chapter 12 Test, Form 2A ............................... 43Chapter 12 Test, Form 2B ............................... 45Chapter 12 Test, Form 2C .............................. 47Chapter 12 Test, Form 2D .............................. 49Chapter 12 Test, Form 3 ................................. 51Chapter 12 Extended-Response Test ............. 53Standardized Test Practice ............................. 54

Answers ........................................... A1–A26

Chapter 12 iii Glencoe Precalculus

0ii_004_PCCRMC12_893813.indd Sec2:iii0ii_004_PCCRMC12_893813.indd Sec2:iii 3/17/09 11:35:50 AM3/17/09 11:35:50 AM

Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

Pdf Pass

Teacher’s Guide to Using theChapter 12 Resource Masters

The Chapter 12 Resource Masters includes the core materials needed for Chapter 12. These materials include worksheets, extensions, and assessment options. The answers for these pages appear at the back of this booklet.

Chapter ResourcesStudent-Built Glossary (pages 1–2) These masters are a student study tool that presents up to twenty of the key vocabulary terms from the chapter. Students are to record definitions and/or examples for each term. You may suggest that students highlight or star the terms with which they are not familiar. Give this to students before beginning Lesson 12-1. Encourage them to add these pages to their mathematics study notebooks. Remind them to complete the appropriate words as they study each lesson.

Anticipation Guide (pages 3–4) This master, presented in both English and Spanish, is a survey used before beginning the chapter to pinpoint what students may or may not know about the concepts in the chapter. Students will revisit this survey after they complete the chapter to see if their perceptions have changed.

Lesson Resources Study Guide and Intervention These masters provide vocabulary, key concepts, additional worked-out examples and Guided Practice exercises to use as a reteaching activity. It can also be used in conjunction with the Student Edition as an instructional tool for students who have been absent.

Practice This master closely follows the types of problems found in the Exercises section of the Student Edition and includes word problems. Use as an additional practice option or as homework for second-day teaching of the lesson.

Word Problem Practice This master includes additional practice in solving word problems that apply to the concepts of the lesson. Use as an additional practice or as homework for second-day teaching of the lesson.

Enrichment These activities may extend the concepts of the lesson, offer an historical or multicultural look at the concepts, or widen students’ perspectives on the mathematics they are learning. They are written for use with all levels of students.

Graphing Calculator, TI–Nspire, or Spreadsheet Activities These activities present ways in which technology can be used with the concepts in some lessons of this chapter. Use as an alternative approach to some concepts or as an integral part of your lesson presentation.

Chapter 12 iv Glencoe Precalculus

0ii_004_PCCRMC12_893813.indd Sec2:iv0ii_004_PCCRMC12_893813.indd Sec2:iv 3/17/09 11:35:52 AM3/17/09 11:35:52 AM

Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

Pdf Pass

Assessment OptionsThe assessment masters in the Chapter 12 Resource Masters offer a wide range of assessment tools for formative (monitoring) assessment and summative (final) assessment.

Quizzes Four free-response quizzes offer assessment at appropriate intervals in the chapter.

Mid-Chapter Test This one-page test provides an option to assess the first half of the chapter. It parallels the timing of the Mid-Chapter Quiz in the Student Edition and includes both multiple-choice and free-response questions.

Vocabulary Test This test is suitable for all students. It includes a list of vocabulary words and questions to assess students’ knowledge of those words. This can also be used in conjunction with one of the leveled chapter tests.

Leveled Chapter Tests

• Form 1 contains multiple-choice questions and is intended for use with below grade level students.

• Forms 2A and 2B contain multiple-choice questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.

• Forms 2C and 2D contain free-response questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.

• Form 3 is a free-response test for use with above grade level students.

All of the above mentioned tests include a free-response Bonus question.

Extended-Response Test Performance assessment tasks are suitable for all students. Sample answers are included for evaluation.

Standardized Test Practice Thesethree pages are cumulative in nature. It includes two parts: multiple-choice questions with bubble-in answer format and short-answer free-response questions.

Answers• The answers for the Anticipation Guide

and Lesson Resources are provided as reduced pages.

• Full-size answer keys are provided for the assessment masters.

Chapter 12 v Glencoe Precalculus

0ii_004_PCCRMC12_893813.indd Sec2:v0ii_004_PCCRMC12_893813.indd Sec2:v 3/17/09 11:35:54 AM3/17/09 11:35:54 AM

Pdf Pass0ii_004_PCCRMC12_893813.indd Sec2:vi0ii_004_PCCRMC12_893813.indd Sec2:vi 3/17/09 11:35:56 AM3/17/09 11:35:56 AM

Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

NAME DATE PERIOD

Ch

apte

r R

eso

urc

es

Pdf Pass

(continued on the next page)

This is an alphabetical list of key vocabulary terms you will learn in Chapter 12. As you study this chapter, complete each term’s definition or description. Remember to add the page number where you found the term. Add these pages to your Precalculus Study Notebook to review vocabulary at the end of the chapter.

Vocabulary TermFound

on PageDefi nition/Description/Example

antiderivative

definite integral

derivative

differential equation

differential operator

differentiation

direct substitution

Fundamental Theorem of Calculus

indefinite integral

indeterminate form

Student-Built Glossary12

Chapter 12 1 Glencoe Precalculus

0ii_004_PCCRMC12_893813.indd Sec1:10ii_004_PCCRMC12_893813.indd Sec1:1 3/17/09 11:35:57 AM3/17/09 11:35:57 AM

Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NAME DATE PERIOD

Pdf Pass

Vocabulary TermFound

on PageDefi nition/Description/Example

instantaneous rate of change

instantaneous velocity

integration

lower limit

one-sided limit

regular partition

right Riemann sum

tangent line

two-sided limit

upper limit

Student-Built Glossary12



Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

NAME DATE PERIOD

Ch

apte

r R

eso

urc

es

Pdf Pass

12

Before you begin Chapter 12

• Read each statement.

• Decide whether you Agree (A) or Disagree (D) with the statement.

• Write A or D in the first column OR if you are not sure whether you agree or disagree, write NS (Not Sure).

After you complete Chapter 12

• Reread each statement and complete the last column by entering an A or a D.

• 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.

STEP 1 A, D, or NS

StatementSTEP 2 A or D

1. The limit of a function f(x) as x approaches c does not depend on the value of the function at point c.

2. The limit of a function f(x) as x approaches c exists providing either the left-hand limit or right-hand limit exists.

3. The limit of a constant function at any point is the x-value of the point.

4. Limits of polynomial and many rational functions can be found by direct substitution.

5. The slope of a nonlinear graph at a specific point is the instantaneous rate of change.

6. The process of finding a derivative is called differentiation.

7. The derivative of a constant function is the constant.

8. The process of evaluating an integral is called integration.

9. The function F(x) is an antiderivative of the function f(x) iff ′(x) = F(x).

10. The connection between definite integrals and antiderivatives is so important that it is called the Fundamental Theorem of Calculus.

Anticipation GuideLimits and Derivatives

Step 2

Step 1



Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NOMBRE FECHA PERÍODO

Pdf Pass

Antes de que comiences el Capítulo 12

• Lee cada enunciado.

• Decide si estás de acuerdo (A) o en desacuerdo (D) con el enunciado.

• Escribe A o D en la primera columna O si no estás seguro(a), escribe NS (no estoy seguro(a)).

Después de que termines el Capítulo 12

• Relee cada enunciado y escribe A o D en la última columna.

• Compara la última columna con la primera. ¿Cambiaste de opinión sobre alguno de los enunciados?

• En los casos en que hayas estado en desacuerdo con el enunciado, escribe en una hoja aparte un ejemplo de por qué no estás de acuerdo.

PASO 1 A, D o NS

EnunciadoPASO 2 A o D

1. El límite de una función f(x) a medida que x se aproxima a c, no depende del valor de la función en el punto c.

2. El límite de una función f(x) a medida que x se aproxima a c existe si y sólo si existe un límite por la derecha o un límite por la izquierda.

3. El límite de una función constante en un punto cualesquiera es el valor de x del punto.

4. El límite de las funciones polinomiales y de las funciones racionales se puede calcular por sustitución directa.

5. La pendiente de una gráfica no lineal en un punto específico es igual a su tasa de cambio instantánea.

6. El proceso de obtención de una derivada se llama diferenciación.

7. La derivada de una función constante es igual a la constante.

8. El proceso de evaluación de una integral se llama integración.

9. La función F(x) es la antiderivada de la función f(x), si f ′(x) = F(x).

10. La relación entre las integrales definidas y las antiderivadas es tan importante que se llama teorema fundamental del cálculo.

Ejercicios preparatoriosLímites y derivadas

12

Paso 2

Paso 1

Capítulo 12 4 Precálculo de Glencoe


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

NAME DATE PERIOD

Less

on

12-

1

Pdf Pass


12-1

Estimate Limits at Fixed ValuesLeft-Hand Limit

If the value of f(x) approaches a unique number L1 as x approaches c from the left, then lim x → c-

f (x) = L1.

Right-Hand LimitIf the value of f (x) approaches a unique number L2 as x approaches c from the right, then lim x → c+

f(x) = L2.

Existence of a Limit at a Point The limit of a function f (x) as x approaches c exists if and only if both one-sided limits exist and are equal. That is, if

lim x → c- f (x) = lim x → c+

f (x) = L, then lim x → c f (x) = L.

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

lim x → 2-

�x� , lim

x → 2+ �x� , and lim

x → 2 �x�

The graph of f(x) = �x� suggests that

lim x → 2-

�x� = 1 and lim

x → 2+ �x� = 2.

Because the left- and right-hand limits of f (x) as x approaches 2 are not the same,

lim x → 2

�x� does not exist.

Exercises


1. lim x → 0+

⎪3x⎥

− x 2. lim x → -2-

⎪x - 2⎥

− x2 - 4

3. lim x → 2

x2 + 3x - 10 −

x - 2

4. lim x → 0

(1 - cos2 x) 5. lim x → 3-

x

3 + 27 − x2 - 9

6. lim x → -2

1 −

(x + 2)2

Study Guide and InterventionEstimating Limits Graphically

Example

y

x

f (x) = � x �

005_036_PCCRMC12_893813.indd 5005_036_PCCRMC12_893813.indd 5 12/7/09 12:32:15 PM12/7/09 12:32:15 PM

Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.

NAME DATE PERIOD

Pdf Pass


12-1

Estimate Limits at Infinity• If the value of f (x) approaches a unique number L1 as x increases,

then lim x → ∞ f (x) = L1.

• If the value of f(x) approaches a unique number L2 as x decreases, then lim x → -∞

f (x) = L2.

Estimate lim x → ∞ 1 − x + 3

, if it exists.

Analyze Graphically The graph of f (x) = 1 − x + 3

suggests that lim x → ∞ 1 − x + 3

= 0.

As x increases, the height of the graph gets closer to 0. The limit indicates a horizontal asymptote at y = 0.

Support Numerically Make a table of values, choosing x-values that grow increasingly large. x approaches infinity

x 10 100 1000 10,000 100,000

f(x) 0.08 0.01 0.001 0.0001 0.00001

The pattern of outputs suggests that as x grows increasingly larger, f (x) approaches 0. This supports our graphical analysis.

Exercises

Estimate each limit, if it exists.

1. lim x → ∞ 2x + 1 − x 2. lim x → -∞

-3x + 1 −

x - 2 3. lim x → ∞

1 − x2

4. lim x → ∞ 2x2 - 5 − 3x3 + 2x

5. lim x → ∞ (ex sin 2xπ) 6. lim x → -∞

(2x + x)

7. lim x → ∞ (x sin x) 8. lim x → -∞

e2x 9. lim x → ∞

cos 2xπ

Study Guide and Intervention (continued)

Estimating Limits Graphically

Example

y

x

(x) =1

x + 3


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

NAME DATE PERIOD

Less

on

12-

1

Pdf Pass


12-1


1. lim x → 0+

(4 - √ � x ) 2. lim

x → 3+ 3 - x − ⎪x - 3⎥

3. lim x → 4

x2 - 16 − x - 4

4. lim x → -1-

x + 7 − x2 + 8x + 7

5. lim x → -1+

x + 7 − x2 + 8x + 7

6. lim x → 0

x2 + 1 −

x2

Estimate each limit, if it exists.

7. lim x → -∞ -4x2

− x2 + 1

8. lim x → ∞ 3x - 2 −

x - 1

9. lim x → 0

sin 2x − x 10. lim x → ∞

e3x + 2

11. RATE OF CHANGE A 20-foot pole is leaning against a barn. If the base of the pole is pulled away from the barn at a rate of 3 feet per second, the top of the pole will move down the side of the barn at a rate of r (x) = 3x −

√ �� 400 - x2 feet per second, where x is the distance between the

base of the pole and the barn. Graph r (x) to find lim x → 20-

r (x).

12. POLLUTANTS The cost in millions of dollars for a company to clean up the pollutants created by one of its manufacturing processes is given by C = 312x −

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

Find lim x → 100-

C.

PracticeEstimating Limits Graphically


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.

NAME DATE PERIOD

Pdf 2nd


12-1

1. BACTERIA GROWTH Bacteria in a dish are growing according to the function f (t) = 4 −

1 + 0.35e-0.2t , for t ≥ 0, where f (t)

is the weight of the bacteria in grams and t is the time in hours.

a. Graph f(t) for 0 ≤ t ≤ 20.

b. Use the graph to estimate the number of grams of bacteria present after 8 hours. Round to the nearest tenth, if necessary.

c. Estimate lim t → ∞ f (t), if it exists.

Interpret your result.

2. CARS After t years, the value of a car purchased for $30,000 is v(t) = 30,000(0.7)t.

a. Graph v(t) for 0 ≤ t ≤ 20.

b. Use the graph to estimate the value of the car after 10 years.

c. Estimate lim t → ∞ v(t), if it exists.

Interpret your results.

3. PROJECTILE HEIGHT Suppose a projectile is thrown upward where its height h in feet at any time t in seconds is determined by the function h(t). The table shows the height of the projectile at various times during its flight.

a. Graph the data and draw a curve through the data points to model the function h(t).

b. Use your graph to estimate lim t → 8-

h(t).

4. THEORY OF RELATIVITY Theoretically, the mass m of an object with velocity v is given by

m = m0 −

√ �� 1 - v

2 −

s2 , where m0 is the mass of

the object at rest and s is the speed of light. What is lim

v → s- m?

5. ELECTRICITY Ahmed determined that the voltage in an electrical outlet in his home is modeled by the function V(t) = 140 sin 120πt. Explain why

lim t → ∞ V(t) does not exist.

Word Problem PracticeEstimating Limits Graphically

t

4

6

8

2

124 8 16

f (t )

t

200

100

300

400

2 4 6 8

h (t )

t

v (t )

t h(t) t h(t)

0 256 4 384

1 336 5 336

2 384 6 256

3 400 7 144


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

NAME DATE PERIOD

Less

on

12-

1

Pdf Pass


12-1

A Matter of LimitsThere are many examples of limits in our world. Some of these are absolute limits, in that they can never be exceeded. Others are like guidelines, and still others result in a penalty if they are exceeded. Fill in the chart below.

Limit How is the limit set? Is the limit absolute?Penalty or consequence

if the limit is exceeded

1. speed limit on a highway

2. height limit on a road underpass

3. luggage limit on an airline flight

4. temperature of a warm object placed in a cool room

5. the speed of an accelerating space craft

6. credit limit on a credit card

One special feature of mathematical limits is that they may be finite, infinite, or they may not exist. Classify each limit as finite, infinite, or does not exist. If the limit is finite, give its value.

7. lim x → 0

1 −

x2 + 1 8. lim

x → 1

x + 1 − x2 - 1

9. lim x → 2

x2 - 4 −

x2 - x -2

10. lim x → 0

ln ⎪x⎥ 11. lim x → 0

sin ⎪x⎥

− x 12. lim x → 0

-1 − x4

Enrichment

005_036_PCCRMC12_893813.indd 9005_036_PCCRMC12_893813.indd 9 3/17/09 11:36:36 AM3/17/09 11:36:36 AM

Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NAME DATE PERIOD

Pdf Pass


12-1

Finding LimitsYou can use a graphing calculator to find a limit with less work than an ordinary scientific calculator. To find lim

x → a f (x), first graph the equation

y = f(x). Then use ZOOM and TRACE to locate a point on the graph whose x-coordinate is as close to a as you like. The y-coordinate should be close to the value of the limit.

Evaluate each limit.

1. lim x → 0

ex

- 1 − x

Press Y= ( 2nd [e] ) — 1 ) ÷ ENTER ZOOM 6. Then press ZOOM 2 ENTER . Press TRACE and use

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

2. lim x → 2

x2 - 4 −

x2 -3x + 2

Press Y= ( x2 — 4 ) ÷ ( x2 — 3 + 2 ) ENTER ZOOM 6. Then press ZOOM 2 ENTER . Press TRACE and use

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

3. If you graph y = ln x − x - 1

and use TRACE , why doesn’t the calculator tell you what y is when x = 1?

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

Graphing Calculator Activity


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

NAME DATE PERIOD

Less

on

12-

2

Pdf Pass


12-2 Study Guide and InterventionEvaluating Limits Algebraically

Example 1

Example 2

Example 3

Compute Limits at a Point

Use direct substitution, if possible, to evaluate lim

x → -2 (-2x4 + 3x3 - 5x + 3).

Since this is the limit of a polynomial function, we can apply the method of direct substitution to find the limit.

lim x → -2

(-2x4 + 3x3 - 5x + 3) = -2(-2)4 + 3(-2)3 - 5(-2) + 3 = -32 - 24 + 10 + 3, or -43

Use factoring to evaluate lim x → 4

x2 - 9x + 20

− x - 4

.

lim x → 4

x2 - 9x + 20 −

x - 4 = lim

x → 4 (x - 5)(x - 4)

− (x - 4)

Factor.

= lim x → 4

(x - 5) Divide out the common factor and simplify.

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

Use rationalizing to evaluate lim x → 16

√ � x - 4

− x - 16

.

By direct substitution, you obtain √ �� 16 - 4

− 16 - 16

or 0 − 0 . Rationalize the numerator

of the fraction before factoring and dividing common factors.

lim x → 16

√ � x - 4

− x - 16

= lim x → 16

√ � x - 4

− x - 16

· √ � x + 4

−

√ � x + 4 Multiply the numerator and denominator by √ � x + 4, the

conjugate of √ � x - 4.

= lim x → 16

x - 16 −

(x - 16)( √ � x + 4) Simplify.

= lim x → 16

x - 16 −

(x - 16)( √ � x + 4) Divide out the common factor.

= lim x → 16

1 −

( √ � x + 4) Simplify.

= 1 −

√ �� 16 + 4 or 1 −

8 Apply direct substitution and simplify.

Exercises


1. lim x → 3

(2x2 - 5x) 2. lim x → 5

√ �� x3 - 4 3. lim x → -2

x2 + 9x + 14 −

x + 2

4. lim x → 4

√ � x - 2

− x - 4

5. lim x → -4

( 1 − x + x) 6. lim x → 2

(-x2 + 5x - 1)


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NAME DATE PERIOD

Pdf Pass


Compute Limits at Infinity

Limits of Power Functions at Infinity

Limits of Polynomials at Infinity

Limits of Reciprocal Functions at Infinity

For any positive integer n,• lim x → ∞

xn = ∞.

• lim x → -∞ xn = ∞ if n is even.

• lim x → -∞ xn = -∞ if n is odd.

Let p be a polynomial functionp(x) = anx

n + … + a1x + a0.Then lim x → ∞

p(x) = lim x → ∞

anx

n

and lim x → -∞ p(x) = lim x → -∞

anx

n.

For any positive integer n,

lim x → ±∞ 1 − xn = 0.


a. lim x → -∞ (x5 - 6x + 1)

lim x → -∞ (x5 - 6x + 1) = lim x → -∞

x5 Limits of Polynomials at Infi nity

= -∞ Limits of Power Functions at Infi nity

b. lim x → ∞ (2x4 + 5x2)

lim x → ∞ (2x4 + 5x2) = lim x → ∞

2x4 Limits of Polynomials at Infi nity

= 2 lim x → ∞ x4 Scalar Multiple Property

= 2 · ∞ = ∞ Limits of Power Functions at Infi nity

Exercises


1. lim x → -∞ (-2x3 + 5x) 2. lim x → ∞

5 − x2

3. lim x → ∞ 6x - 1 − 10x + 7

4. lim x → ∞ 6x2 - 2x −

x3 + 1 5. lim x → ∞

5x4 + 2x3 - 1 −

2x3 + x2 - 1 6. lim x → -∞

(3x3 + 5x - 1)


Evaluating Limits Algebraically

12-2

Example


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

NAME DATE PERIOD

Less

on

12-

2

Pdf Pass



1. lim x → 3

(x2 + 3x - 8) 2. lim

x → -6 x2 - 36 − x + 6

3. lim x → 0

(3 + x)2 - 9

− x 4. lim x → 4

√ �� x2 - 2x + 1

5. lim x → 1

x2 - x −

2x2 + 5x - 7 6. lim

x → 3

x2 −

2 + √ �� x - 3

7. lim x → ∞ (2 - 6x + 5x3) 8. lim x → -∞

x

5 - 8x2 −

4x5 + 3x

9. lim x → ∞ 2x3 - 4x + 1 −

5x4 - 2x2 10. lim x → -∞

(6x7 - x2)

11. BOOKS Suppose the value v of a book in dollars after t years can be

represented as v(t) = 300 − 6 + 35(0.2)t

. How much will the book eventually

be worth? That is, find the lim t → ∞ v(t).

12. MEDICINE Each day, Tameka takes 2 milligrams of her asthma medicine. The graph shows the amount of medicine m left in her blood stream after d days. Find the lim

d → 3- m(d) and lim

d → 3+ m(d).

12-2 PracticeEvaluating Limits Algebraically

Tam

eka’

s As

thm

aM

edic

ine

(mg)

4

6

2

m

d

0

8

10

Days321 54 6


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NAME DATE PERIOD

Pdf Pass


1. POOLS A pool contains 75,000 liters of pure water. A mixture that contains 0.3 gram of chlorine per liter of water is pumped into the pool at a rate of 75 liters per minute. The concentration C of chlorine in grams per liter t minutes later in the pool is given by

C(t) = 0.3t − 1000 + t

. Find lim t → ∞ C(t).

2. POLLUTANTS As a by-product of one of its processes, a manufacturing company creates an airborne pollutant. The cost C of removing p% of the pollutant is

C = 60,000p

− 100 - p

, 0 ≤ p ≤ 100.

Find lim p → 100-

C.

3. MOTORCYCLES Flint bought a new motorcycle for $24,000. Suppose the value v of his motorcycle, in thousands of dollars, after t years can be represented by the equation v(t) = 24(0.98)t.

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

Years 1 5 10 20

Value

b. Find lim t → ∞ v(t).

4. CAR SAFETY While driving a car, it is important to maintain a safe distance between you and the car in front of you. Suppose the function y(x) = 0.005x2 + 0.3x + 3, where x is the speed in miles per hour, gives the recommended safe distance, in yards, between your car and the one in front of you. Find lim

x → 70 y(x).

5. PARTS The cost c of producing a certain small engine part in dollars is given by the equation c(p) = 3000 + 20p, where p is the number of parts produced.

a. Find the cost of producing 100 parts.

b. On Tuesday, the company produced $21,000 worth of parts. How many parts did the company produce?

c. The average cost per part is found by dividing c(p) by p.

Find lim p → 15,000

c(p)

− p .

6. MICROWAVES The function

f(x) = 250x + 200,000

− x models theaverage cost of a microwave f (x) manufactured by a company that makes x microwaves for professional kitchens. Find lim

x → 2000 f (x).

12-2 Word Problem PracticeEvaluating Limits Algebraically


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

NAME DATE PERIOD

Less

on

12-

2

Pdf Pass


The Squeeze Theorem

In Lesson 12-1, you learned that the lim x → 0

sin 1 − x does not exist because

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

-1 and 1. But what about the lim x → 0

x2 sin 1 − x ? Does this limit not exist simply

because lim x → 0

sin 1 − x does not exist? A theorem known as the Squeeze Theorem

will help you answer this question algebraically.

The Squeeze Theorem

If h(x) ≤ f (x) ≤ g(x) for all x in an open interval containing c, except possibly at c itself, and if lim x → c

h(x) = L = lim x → c g(x) , then lim x → c

f (x) exists and is equal to L.

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

this inequality by x2, obtaining -x2 ≤ x2 sin 1 − x ≤ x2. To apply the Squeeze

Theorem, let h(x) = -x2, f(x) = x2 sin 1 − x , and g(x) = x2. From Lesson 12-2,

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

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

x2 sin 1 − x = 0.

Exercises

Use the Squeeze Theorem to find each limit.

1. lim x → 0 x4 sin 1 − x 2. lim x → 4 ⎪x⎥ √ � x + 2

− x + 4

3. lim x → 0 x2 cos 1 −

3 √ � x 4. lim x → 0 ⎪x⎥ sin π − x

5. lim x → 0 tan x − x 6. lim x → 0

sin 3x − x

Enrichment12-2


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NAME DATE PERIOD

Pdf Pass


Tangent LinesInstantaneous Rate of Change

The instantaneous rate of change of the graph of f (x) at the point (x, f (x)) is the slope m of

the tangent line given by m = lim h → 0

f (x + h) - f (x)

− h , provided the limit exists.

Find an equation for the slope of the graph of y = 3x2 + 1 at any point.

m = lim h → 0

f (x + h) - f (x)

− h Instantaneous Rate of Change Formula

m = lim h → 0

[3(x + h)2 + 1] - [3 x 2 + 1]

−− h f(x + h) = 3(x + h)2 + 1 and f(x) = 3x2 + 1

m = lim h → 0

[3 x 2 + 6hx + 3 h 2 + 1] - [3 x 2 + 1]

−− h Expand and simplify.

m = lim h → 0

3h(2x + h)

− h Simplify and factor.

m = lim h → 0

3(2x + h) Reduce h.

m = 6x Scalar Multiple, Sum Property, and Limit of a

Constant Function Property of Limits

An equation for the slope of the graph at any point is m = 6x.

Exercises

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

1. y = x3 + 1 2. y = 4 - 7x

3. y = 3 − x2

4. y = 4 − √ � x

12-3 Study Guide and InterventionTangent Lines and Velocity

Example


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

NAME DATE PERIOD

Less

on

12-

3

Pdf Pass


Instantaneous VelocityInstantaneous Velocity

If the distance an object travels is given as a function of time f (t), then the instantaneous

velocity v(t) at a time t is given by v(t) = lim h → 0

f (t + h) - f (t)


A rock is dropped from 1500 feet above the base of a ravine. The height of the rock after t seconds is given by h(t) = 1500 - 16t2. Find the instantaneous velocity v(t) of the rock at 4 seconds.

v(t) = lim h → 0

f (t + h) - f (t)

− h Instantaneous Velocity Formula

= lim h → 0

[1500 - 16(4 + h ) 2 ] - [1500 - 16(4)2

] −−−

h f(t + h) = 1500 - 16(4 + h)2 and f(t) = 1500 - 16(4)2

= lim h → 0

-128h - 16 h 2 −

h Multiply and simplify.

= lim h → 0

h(-128 - 16h)

− h Factor.

= lim h → 0

(-128 - 16h) Divide by h and simplify.

= -128 - 16(0) or -128 Difference Property of Limits and Limit of Constant

and Identity Functions

The instantaneous velocity of the rock at 4 seconds is 128 feet per second.

Exercises

The distance d an object is above the ground t seconds after it is dropped is given by d(t). Find the instantaneous velocity of the object at the given value for t.

1. d(t) = 800 - 16t2; t = 3 2. d(t) = -16t2 + 1700; t = 5

3. d(t) = 70t - 16t2; t = 1 4. d(t) = -16t2 + 90t + 10; t = 2

12-3 Study Guide and Intervention (continued)

Tangent Lines and Velocity

Example


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.

NAME DATE PERIOD

Pdf Pass


Find the slope of the line tangent to the graph of each function at the given point.

1. y = x2 - x; (3, 6) 2. y = 5 − x ; (-1, -5)

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

3. y = -2x + 1 4. y = x3 - 2x2


5. d(t) = 300 - 16t2; t = 2 6. d(t) = -16t2 + 200t + 700; t = 3

Find an equation for the instantaneous velocity v(t) if the path of an object is defined as s(t) for any point in time t.

7. s(t) = 17t2 + 8 8. s(t) = 5t3 - 6t2 + 4t + 1

9. s(t) = √ � t - 2t2 10. s(t) = 3 − t + 2t

11. SKY DIVING The position h in feet of a sky diver relative to the ground can be defined by h(t) = 18,000 - 16t2, where time t is seconds passed after the sky diver exited the plane. Find an expression for the instantaneous velocity v(t) of the sky diver.

12. FOOTBALL A quarterback throws a football with a velocity of 58 feet per second toward a teammate. Suppose the height h of the football, in feet, t seconds after he throws it is defined as h(t) = -16t2 + 58t + 6.

a. Find an expression for the instantaneous velocity v(t) of the football.

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

12-3 PracticeTangent Lines and Velocity


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

NAME DATE PERIOD

Less

on

12-

3

Pdf Pass


1. FALLING OBJECT Miranda drops a ball from a tower that is 800 feet high. The position of the ball after t seconds is given by s(t) = -16t2 + 800. How fast is the ball falling after 2 seconds?

2. PROJECTILE Tito drops a rock from 1200 feet. The position of the rock after t seconds is given by s(t) = -16t2 + 1200.

a. How fast is the ball falling after 4 seconds?

b. When will the rock hit the ground?

c. Find an expression for the instantaneous velocity v(t)

of the rock.

d. What is the velocity of the rock when it hits the ground?

3. BUNGEE JUMPING A bungee jumper’s height h in feet relative to the ground in t seconds is given by h(t) = 900 - 16t2. Find an expression for the instantaneous velocity v(t) of the jumper.

4. FREE FALLING The position h in feet of a free-falling sky diver relative to the ground can be defined by h(t) = 15,000 - 16t2, where t is seconds passed after the sky diver exited the plane.

a. Find an expression for the instantaneous velocity v(t) of the sky diver.

b. What is the sky diver’s height after 2 seconds?

c. What is the sky diver’s velocity after 4 seconds?

5. BASEBALL An outfielder throws a ball toward home plate with an initial velocity of 80 feet per second. Suppose the height h of the baseball, in feet, t seconds after the ball is thrown is modeled by h(t) = -16t2 + 80t + 6.5.

a. Find an expression for the instantaneous velocity v(t) of the baseball.

b. How fast is the baseball traveling after 0.5 second?

c. For what value of t will the baseball reach its maximum height?

d. What is the maximum height of the baseball?

6. AREA Suppose the length x of each side of the square shown is changing.

a. Find the average rate of change of the area a(x) as x changes from 5.4 inches to 5.6 inches.

b. Find the instantaneous rate of change of the area at the momentx = 5 inches.

12-3 Word Problem PracticeTangent Lines and Velocity

x

x a(x ) = x 2


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NAME DATE PERIOD

Pdf Pass


Tangents and VerticesCan you use the equation for the slope of a function at any point to find the vertex of a parabola of the form y = ax2 + bx + c? Step 1 Graph a parabola of the form y = ax2 + bx + c, one where a > 0 and

one where a < 0. Then draw the line tangent to the vertex of each parabola and place the equation of each parabola below its graph.

Step 2 Find the slope of the tangent line to each graph at the vertex.

Step 3 Find the equation for the slope of each graph at any point.

Step 4 Set each equation for the slope equal to zero and solve for x. This gives the x-value of the vertex.

Step 5 Substitute each x-value into its respective equation to find the y-value of the vertex.

Step 6 Write each vertex as an ordered pair.

Exercises

Find the vertex of each parabola.

1. y = 2x2 - 4x + 2 2. y = -x2 - 6x - 9

3. y = 4x2 - 25 4. y = 1 − 2 x2 - 2x + 4

12-3 Enrichment

y

x

y = -x 2 - 4x

1

1

y

x

y = x 2 + 2x - 3

4

4


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

NAME DATE PERIOD

Less

on

12-

3

Pdf Pass


Using the Tangent Line to Approximate a FunctionYou have learned how to find the slope of a tangent line to a function at a point. This slope can then be used to write the equation of the tangent line in point-slope form. The tangent line of a function is usually a good approximation of the function for values of x near the coordinate of the chosen point. Set up a spreadsheet like the one shown below to study this method of approximating a function.

A B C D E F G H

1 x 0.7 0.8 0.9 1 1.1 1.2 1.3

2 x^2 0.49 0.64 0.81 1 1.21 1.44 1.69

3 2*x - 1 0.4 0.6 0.8 1 1.2 1.4 1.6

The equation in point-slope form of the tangent line to f (x) = x2 at (1, 1) is y = 2x - 1. In the spreadsheet, the function f (x) = x2 is approximated for values of x near x = 1 by using the function g(x) = 2x - 1. For this example, the values for x are entered in row 1, columns B–H. The function f (x) = x2 is entered in cell B2 as = B1^2 and copied to the other cells in row 2. The approximating function, g(x) = 2x - 1, is entered in cell B3 as =2*B1 - 1 and copied to the other cells in row 3. Notice that the values of g(x) = 2x - 1 in row 3 are remarkably close to the values of f (x) = x2 in row 2.

Exercises

1. Use the spreadsheet to approximate the values of f(x) = x2 near the point, where x = 2, by using the appropriate tangent line. Compare the values for x = 1.7, 1.8, 1.9, 2, 2.1, 2.2, and 2.3. What is the maximum error that occurs?

2. Use the spreadsheet to approximate the values of ƒ(x) = x2 near the point, where x = 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 that occurs?

3. Use the spreadsheet to approximate the values of ƒ(x) = √ � x for 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 point, where x = 1.

12-3 Spreadsheet Activity


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NAME DATE PERIOD

Pdf Pass


Derivatives and the Basic Rules Limits were used in Lesson 12-3 to determine the slope of a line tangent to the graph of a function at any point. This limit is called the derivative of a function. The derivative of f (x) is

f ′(x), which is given by f ′(x) = lim h → 0

f(x + h) - f(x)


There are several rules of derivatives that are useful when finding the derivatives of functions that contain several terms.

Power RuleIf f(x) = xn and n is a real number, then

f ′(x) = nxn − 1.

If f(x) = x3, then f ′(x) = 3x2.

ConstantThe derivative of a constant function is zero.

If f(x) = c, then f ′(x) = 0.

If f(x) = -2, then f ′(x) = 0.

Constant Multiple

of a Power

If f(x) = cxn, where c is a constant and n is

a real number, then f ′(x) = cnxn - 1.

If f(x) = 5x3, then f ′(x) = 15x2.

Sum and Difference If f(x) = g(x) ± h(x), then f ′(x) = g′(x) ± h′(x). If f(x) = 4x2 + 3x, then f ′(x) = 8x + 3.

Find the derivative of each function.

a. f (x) = 3x2 - 2x + 4

f (x) = 3x2 - 2x + 4 Original equation

f ′(x) = 2 · 3x2 - 1 - 2 · 1x1 - 1 + 0 Constant, Constant Multiple of a Power, and Sum and Difference Rules

= 6x - 2 Simplify.

b. f (x) = x4(4x3 - 5)

f (x) = x4(4x3 - 5) Original equation

f (x) = 4x7 - 5x4 Distributive Property

f ′(x)= 4 · 7x7 - 1 - 5 · 4x4 - 1 Constant Multiple of a Power, and Sum and Difference Rules

= 28x6 - 20x3 Simplify.

Exercises

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

1. f (x) = 4x2 - 5; x = 3 and -2 2. f (x) = -x3 + 5x2; x = 1 and -4

3. f (x) = -8 + 3x - x2; 0 and -3 4. f (x) = 3x4 + x5 -2; -1 and 2


5. f (x) = 6x2 - 3x + 4 6. f (x) = -x3.4 + 3x0.2

7. f (x) = 4 x 1 − 2 - 3 x

3 − 2 8. f (x) = -4x2 + 3x3 - 14

12-4 Study Guide and InterventionDerivatives

Example


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

NAME DATE PERIOD

Less

on

12-

4

Pdf Pass


Product and Quotient Rules Use the following rules to find the derivative of the product or quotient of two functions.

Product Rule If f and g are differentiable at x, then d − dx

[f (x)g(x)] = f ′(x)g(x) + f (x)g ′(x).

Quotient Rule

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

d − dx

⎡

⎢

⎣ f(x)

− g(x)

⎤

�

⎦ =

f ′(x)g(x) - f (x)g ′(x) −−

[g(x)] 2 .

Find the derivative of h(x) = (x2 - 2)(2x3 + 5x).

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

= (2x)( 2x3 + 5x) + (x2 - 2)( 6x2 + 5) Substitution

= 4x4 + 10x2 + 6x4 + 5x2 - 12x2 - 10 Distributive Property

= 10x4 + 3x2 - 10 Simplify.

Find the derivative of h(x) = (2x2 + 4) −

(x2 - 1) .

h ′(x) = f ′(x)g(x) - f (x)g ′(x)

−−

[g(x)]2 Quotient Rule

= 4x(x2 - 1) - (2x2 + 4)2x

−− (2x)2

Substitution

= 4x3 - 4x - 4x3 - 8x −− 4x2

Distributive Property

= - 3 − x Simplify.

Exercises


1. h(x) = (-4 + 2x2)(2x + 3) 2. m(x) = (3x - 1)(x2 + 5x)

3. d(x) = x2 + 3 −

x - 1 4. k(x) = 3x3 + 4 −

2x2 - 1


Derivatives

Example 1

f (x) = x2 - 2 Original equation

f ′(x) = 2x Sum Rule for Limits, Power and

Constant Rules for Derivatives

g(x) = 2x3 + 5x Original equation

g ′(x) = 6x2 + 5 Sum Rule for Limits, Power and


Example 2

f (x) = 2x2 + 4 Original equation

f ′(x) = 4x Sum Rule for Limits, Power and


g(x) = x2 - 1 Original equation

g ′(x) = 2x Sum Rule for Limits, Power and



Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.

NAME DATE PERIOD

Pdf Pass


Find the derivative of each function. Then evaluate the derivative of each function for the given values of x.

1. g(x) = 3x2 - 5x; x = -2 and 1 2. h(x) = 4x3 - x2; x = 3 and 0

3. f (x) = x2 - 4x + 7; x = 2 and -3 4. m(x) = -2x2 - 6x + 1; x = 0 and -3

5. q(x) = -1 + x3 - 2x4; x = -1 and 3 6. t(x) = 3x7 - 1; x = -1 and 1


7. f (x) = (x2 + 5x)2 8. f (x) = x2(x3 + 3x2)

9. f (x) = 5 √ � x6 10. h(x) = -

3 − x6

11. p(x) = -4x5 + 6x3 - 5x2 12. n(x) = (3x2 - 2x)(x3 + x2)

13. r(x) = 3x - 1 − x2 + 2

14. q(x) = √ � x (x2 - 3)

15. PHYSICS Acceleration is the rate at which the velocity of a moving object changes. The velocity in meters per second of a particle moving along a straight line is given by the function v(t) = 3t2 - 6t + 5, where t is the time in seconds. Find the acceleration of the particle in meters per second squared after 5 seconds. (Hint: Acceleration is the derivative of velocity.)

PracticeDerivatives

12-4


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

NAME DATE PERIOD

Less

on

12-

4

Pdf Pass


1. BIRDS The height h, in feet, of a flying bird can be defined by h(t) = -t3

− 3 + 7 −

2 t2 + 18 on the interval

[1, 10], where time t is given in seconds. Find the maximum and minimum height of the bird.

2. CLIFF DIVING At time t = 0, a diver jumps from a cliff 192 feet above the surface of the water. The height h of the diver is given by h(t) = -16t2 + 16t + 192, where h is measured in feet and time t is measured in seconds.

a. Find the equation for the velocity h'(t) of the diver at any time t.

b. Find the velocity of the diver after 1 second has passed.

c. Find the time when the diver hits the water.

d. What is the diver’s velocity when she hits the water?

3. GEOMETRY The formula to find the volume V of a cylinder in terms of its height h and radius r is V = πr2h. Consider a cylinder with a height of 10 inches and a changing radius when answering the following questions.a. Write a formula for the volume of the

cylinder in terms of its radius.

b. Find an equation for the instantaneous rate of change of the volume in terms its radius.

c. Find the value of V ′(r) when r = 3 inches.

4. VOLUME Suppose the length x of each side of the cube shown is changing.

a. Find the average rate of change of the volume V(x) as x changes from 3.2 inches to 3.4 inches.

b. Find the instantaneous rate of change of the volume V(x) at the moment x = 4 inches.

c. Explain the relationship between the volume formula and the derivative of the volume formula.

5. PROJECTILE Suppose a ball is hit straight upward from a height of 6 feet with an initial velocity of 80 feet per second. The height h of the ball in feet at any time t is given by the function h(t) = -16t2 + 80t + 6.

a. Find the equation for the velocity v(t) of the ball at any time t by finding the derivative of h(t).

b. Find the instantaneous velocity of the ball at t = 2 seconds.

Word Problem PracticeDerivatives

12-4

=


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NAME DATE PERIOD

Pdf Pass


Powerful DifferentiationIn Chapter 10, the series expansions of some transcendental functions were presented. In particular, the even function y = cos x, was shown to be a sum of even powers of x:

cos x = 1 - x2 −

2! + x

4 −

4! - x

6 −

6! +

x8 −

8! - …

and the sine function, being odd, was shown to be a sum of odd powers of x:

sin x = x - x3 −

3! + x

5 −

5! - x

7 −

7! + x

9 −

9! - … .

The power functions in these series expansions can be differentiated.

1. a. Find d(sin x) −

dx by differentiating the series expansion of sin x term by

term and simplifying the result.

b. What function does this new infinite series represent?

c. So, d(sin x) −

dx = .

2. a. What would you guess might be the derivative of cos x?

b. Find d(cos x) −

dx using the series expansion of cos x.

c. So, d(cos x) −

dx = .

3. a. The series expansion for ex, ex = 1 + x + x2 −

2! + x

3 −

3! +

x4 −

4! + …

was also discussed in Chapter 10. Differentiate the series expansion of ex term by term and simplify the result.

b. Thus, d(ex) −

dx = .

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

4. f (x) = xex 5. f (x) = sin x2 6. f (x) = (cos x)2

12-4 Enrichment

?

?

?


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

NAME DATE PERIOD

Less

on

12-

5

Pdf Pass


Area Under a Curve You can use the area of rectangles to find the area between the graph of a function f (x) and the x-axis on an interval [a, b] in the domain of f (x).

Approximate the area between the curve f (x) = 1 − 2 x2 and the x-axis

on the interval [0, 4] by first using the right endpoints and then by using the left endpoints of the rectangles. Use rectangles with a width of 1.Using right endpoints for the height of each rectangle produces four rectangles with a width of one unit (Figure A). Using left endpoints for the height of each rectangle produces four rectangles with a width of 1 unit (Figure B). However, the first rectangle has a height of f(0) or 0 and thus, has an area of 0 square units.

Area using right endpoints Area using left endpointsR1 = 1 · f (1) or 0.5 R1 = 1 · f (0) or 0R2 = 1 · f (2) or 2 R2 = 1 · f (1) or 0.5R3 = 1 · f (3) or 4.5 R3 = 1 · f (2) or 2R4 = 1 · f (4) or 8 R4 = 1 · f (3) or 4.5 total area = 15 total area = 7

The area using the right and left endpoints is 15 and 7 square units, respectively. We now have lower and upper estimates for the area of the region, 7 < area < 15. Averaging the two areas would give a better approximation of 11 square units.

Exercises

1. Approximate the area between the curve f (x) = 3x2 + 1 and the x-axis on the interval [0, 4] by first using the right endpoints and then by using the left endpoints. Use rectangles of width 1 unit. Then find the average for both approximations.

2. Approximate the area between the curve f (x) = -x2 + 5x + 6 and the x-axis on the interval [1, 5] by first using the right endpoints and then by using the left endpoints. Use rectangles of width 1 unit. Then find the average for both approximations.

12-5 Study Guide and InterventionArea Under a Curve and Integration

Example

2 4

8

4

x

Figure A Figure B

2 4

8

4

x


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.

NAME DATE PERIOD

Pdf Pass


Integration

Definite Integral

The area of a region under the graph of a function is

b

⌠

⌡

a

f (x) dx = lim n → ∞ ∑

i = 1

n

f (xi)Δx,

where a and b are the lower limits and upper limits, respectively, Δx = b - a − n and xi = a + iΔx.

Use limits to find the area of the region between the graph of

y = 4x2 and the x-axis on the interval [0, 5], or

5

⌠

⌡

0

4x2 dx.

5

⌠

⌡

0

4x2 dx = lim n → ∞ ∑

i = 1

n

f (xi)Δx Definition of definite integral

= lim n → ∞ ∑

i = 1

n

4xi2 Δx f(xi) = 4xi

2

= lim n → ∞ ∑

i = 1

n

4 ( 5i − n ) 2 5 − n xi = 5i − n and Δx = 5 − n

= lim n → ∞ 20 − n ( 25 −

n2 ∑

i = 1

n

i2) Expand and factor.

= lim n → ∞ 20 − n ( 25 −

n2 · n(n + 1)(2n + 1)

− 6 ) ∑

i = 1

n

i2 = n(n + 1)(2n + 1)

− 6

= lim n → ∞ 500 −

6 ( 2n2 + 3n + 1 −

n2 ) Simplify and expand.

= lim n → ∞ 500 −

6 (2 + 3 − n + 1 −

n2 ) Factor and divide each term by n2.

= ( lim n → ∞ 500 −

6 ) [ lim n → ∞

2 + ( lim n → ∞

3) ( lim n → ∞

1 − n ) + lim n → ∞

1 − n2

] Limit theorems

= 500 −

6 [2 + 3(0) + 0] or about 166.67 square units Simplify.

Exercise

Use limits to find the area between the graph of each function and the x-axis given by the definite integral.

1.

2

⌠

⌡

0

x3 dx 2.

4

⌠

⌡

2

(x2 + 3) dx

3.

6

⌠

⌡

4

(1 + x) dx 4.

3

⌠

⌡

1

4x3 dx


Area Under a Curve and Integration

Example


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

NAME DATE PERIOD

Less

on

12-

5

Pdf Pass


Approximate the area between the curve f(x) and the x-axis on the indicated interval using the indicated endpoints. Use rectangles with a width of 1.

1. f (x) = x + 3 2. f (x) = -x2 + 6x -4

[1, 5] [2, 5]

left endpoints right endpoints

3. g(x) = 3x3 4. p(x) = 1 + x2

[0, 4] [1, 6]

left endpoints right endpoints


5.

2

⌠

⌡

0

x2 dx 6. 6

⌠

⌡

1

6x2 dx

7. 3

⌠ ⌡

1

(x2 - x) dx 8. 1

⌠

⌡

-2

(-x2 -2x + 11) dx

9. Architecture and Design A designer is making a stained-glass window for a new building. The shape of the window can be modeled by the parabola y = 5 - 0.05x2. What is the area of the window?

12-5 PracticeArea Under a Curve and Integration

y

x−10

−10

10

−5−5 105


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NAME DATE PERIOD

Pdf Pass


1. DOG HOUSE Charlie is building a dog house for Fido. The entrance to the dog house is in the shape of the region shown. What is the area of the entrance to Fido’s dog house if x is given in feet?

2. MINING The entrance to a coal mine is in the shape of the region shown. What is the area of the entrance if x is given in meters?

3. DAMS The face of a dam is in the shape of the region shown. What is the area of the face of the dam if x is given in kilometers?

4. TRIANGLE AREA On a coordinate plane, draw the triangle formed by the x-axis and the lines x = 5 and y = x + 4.

a. Shade the interior of this triangle.

b. Find the height and length of the base of the triangle. Then calculate the area of the triangle using its height and base length.

c. Calculate the area of the triangle by

evaluating 5

⌠

⌡

-4

(x + 4) dx.

5. GRASS SEED Mr. Bower is seeding part 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

7

⌠

⌡

1

(-x2 + 8x - 7) dx, will he

have enough seed to complete the task? Explain.

12-5 Word Problem PracticeArea Under a Curve and Integration

x

y

y = -x3 + 4x

y = x4 - 5x2 + 4

y

x

y = -x3 + 2x2y

x

y

x


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

NAME DATE PERIOD

Less

on

12-

5

Pdf Pass


12-5 Enrichment

Reading MathematicsThere is a lot of special notation used in calculus that is not used in other branches of mathematics. In addition, there is often more than one notation for the same thing. You have already seen this in the case of the derivative.

1. Let f (x) = x2. What does lim h → 0

(x + h)2 - x2

− h find?

2. List several other ways of expressing this quantity.

Yet another notation for the derivative of a function y = f (x) is y. . This was the notation developed by Isaac Newton. Each of these notations also can be used to indicate higher-order derivatives.

For example, f �(x) d2y

− dx2

, and ÿ all indicate the second derivative of some function y = f (x).

3. What is the order of each derivative?

a. f �′(x) b. y. c. d4y

− dx4

d. y�

The Leibniz notation for the derivative dy

− dx

is usually read “dy dx,”or more formally, “the derivative of y with respect to x.” Note that

dy −

dx

is not a fraction of any kind. To indicate the value of the derivative at a specific value of x using the Leibniz notation, one might use the

following: dy

− dx

�

�

� x = 2

, read “dy dx evaluated at x = 2.”

Given f(x) = x3 + 3x2 - 4, find the value of each expression.

4. f ′(2) 5. dy

− dx

�

�

� x = -1

6. f �(0) 7. d3y

− dx3

�

�

�

x = 4


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NAME DATE PERIOD

Pdf Pass


Study Guide and InterventionThe Fundamental Theorem of Calculus

Antiderivatives and Indefinite Integrals

Given a function f (x), we say that F(x) is an antiderivative of f (x) if F ′(x) = f (x).

Rules for Antiderivatives

Power RuleIf f(x) = xn, where n is a rational number other than -1,

F(x) = xn + 1

− n + 1

+ C.

Constant

Multiple

of a Power

If f(x) = kxn, where n is a rational number other than -1 and k is a

constant, then F(x) = kxn + 1

− n + 1

+ C.

Sum and

Difference

If the antiderivatives of f(x) and g(x) are F(x) and G(x), respectively,

then the antiderivatives of f(x) ± g(x) are F(x) ± G(x).

Find all antiderivatives for each function.

a. f (x) = -3x5

f (x) = -3x5 Original equation

F(x) = -3x5 + 1

− 5 + 1

+ C Constant Multiple of a Power

= - 1 − 2 x6 + C Simplify.

b. f (x) = x3 + 4x2 - 2

f (x) = x3 + 4x2 - 2 Original equation

= x3 + 4x2 - 2x0 Rewrite the function so each term has a power of x.

F(x) = x3 + 1 −

3 + 1 + 4x2 + 1

− 2 + 1

- 2x0 + 1

− 0 + 1

Use all three rules.

= 1 − 4 x4 + 4 −

3 x3 - 2x + C Simplify.

Exercises


1. f (x) = 2x4 + 3x2 - 5 2. g(x) = 2 − x3

3. t(x) = 3 − 4 x6 -

1 − 2 x3 4. n(x) = 5 √ � x - 2

Example

12-6


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

NAME DATE PERIOD

Less

on

12-

6

Pdf Pass


The Fundamental Theorem of Calculus The indefinite integral of

f (x) is defined by ⌠ ⌡

f (x)dx = F(x) + C, where F(x) is an antiderivative of f (x)

and C is any constant.

Fundamental

Theorem of

Calculus

If F(x) is the antiderivative of the continuous function f (x), then

b

⌠

⌡

a

f(x) = F(b) - F(a).

The right side of this statement may also be written as F(x) ⎢

⎢

⎢ b

a .

Evaluate each integral.

a. ⌠ ⌡

(3x2 + 4x - 1) dx

⌠ ⌡

(3x2 + 4x - 1) dx = 3x2 + 1 −

2 + 1 + 4x1 + 1

− 1 + 1

- x0 + 1

− 0 + 1

+ C Constant Multiple of a Power

= 3x3

− 3 + 4x2

− 2 - x + C Simplify.

= x3 + 2x2 - x + C Simplify.

b.

4

⌠

⌡

2

(x3 - 1) dx

4

⌠ ⌡

2

(x3- 1) dx = ( x

4 −

4 - x)

⎢ ⎢

⎢ 4

2 Fundamental Theorem of Calculus

= ( 44

− 4 - 4) - ( 2

4

− 2 - 2) b = 4 and a = 2

= 60 - 6 or 54 Simplify.

Exercises


1. ⌠ ⌡

(3x7- x2) dx 2.

2

⌠ ⌡

1

(x2 + 1) dx

3. 2

⌠ ⌡

1

(x2- 1) dx 4.

1

⌠

⌡

-1

(x3 - 2x + 1) dx


The Fundamental Theorem of Calculus

Example

12-6


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NAME DATE PERIOD

Pdf Pass



1. f (x) = 4x3 2. f (x) = 2x + 3

3. f (x) = x(x2 - 3) 4. f (x) = 8x2 + 2x - 3


5. ⌠

⌡ 8 dx 6. ⌠

⌡

(2x3 + 6x) dx

7. ⌠

⌡ (-6x5 - 2x2 + 5x) dx 8.

5

⌠ ⌡

2

2x dx

9. -1

⌠ ⌡

-5

(-4x3 - 3x2) dx 10. 1

⌠ ⌡

-2

(1 - x)(x + 3) dx

11. PHYSICS The work in foot-pounds to compress a certain spring a distance of � feet from

its natural length is given by W = �

⌠ ⌡

0

2x dx. How much work is required to compress the

spring 6 inches from its natural length?

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

Suppose the number of hours needed to create p pieces is given by h = p

⌠ ⌡

0

(30 - 3x) dx.

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

PracticeThe Fundamental Theorem of Calculus

12-6


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

NAME DATE PERIOD

Less

on

12-

6

Pdf Pass


Word Problem PracticeThe Fundamental Theorem of Calculus

1. VERTICAL JUMP Lila tested her vertical jump in physical education class. The velocity of her jump can be defined as v(t) = -32t + 24, where t is given in seconds and the velocity is given in feet per second.

a. Find the position function s(t) for Lila’s jump. Assume that for t = 0, s(t) = 0.

b. After Lila jumps, how long does it take before she lands on the ground?

2. ADVERTISING New Wave’s business logo is in the shape of the region shown below. If the company intends to use it as part of its letterhead, how much space will the logo occupy at the top of each document for x between 0 inch and 1 inch?

3. SPRING STRETCHING The work, in joules, required to stretch a certain spring 36 inches beyond its natural length is

given by 3 ⌠

⌡ 0 80x dx. How much work

is required?

4. VOLUME In the figure below, find the volume of the solid formed by revolving the graph of f (x) = x2 over the interval [0, 3], if the volume of the solid is given

by 3 ⌠

⌡ 0π(x2)2 dx.

5. SPRING COMPRESSION A force of 800 pounds compresses a spring 2 inches from its natural length of 12 inches. The work, in inch-pounds, required to compress the spring another 2 inches is

given by 4 ⌠

⌡ 2 400x dx. How much work is

required to compress the spring another2 inches?

6. BILLBOARD The Squared and Linear Trucking Company has purchased a billboard to advertise the company. The central figure on the billboard, measured in feet, is shown in the diagram below. What is the area of this figure?

x

y

y = x4 - 2x2 + 1

x

y

2

4

6

8

10

2 4 6 8 10

y = x2

y = -3x + 18

12-6

2 4

f (x ) = x2

3

6

9

-3-6-9

y

x


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NAME DATE PERIOD

Pdf Pass


Derivatives of Exponential and Logarithmic Functions

Exponential Rule The derivative of y = ex is ex and the derivative of y = eu is eu du − dx

.

Find the derivative of y = e 3x.

Let u = 3x. Then dy

− dx

= eu · du − dx

.

Since du − dx

= 3, dy

− dx

= eu · 3 or 3eu.

The derivative of y = e3x is 3e3x.

Logarithmic Rule The derivative of y = ln x is 1 − x and the derivative of y = ln u is 1 − u · du −

dx

.

Find the derivative of y = ln (x2 + 3).

Let u = x2 + 3. Then dy

− dx

= 1 − u · du − dx

.

Since du − dx

= 2x, dy

− dx

= 1 − u · 2x, or 1 − x2 + 3

· 2x. Simplify to get 2x − x2 + 3

.

The derivative of y = ln (x2 + 3) is 2x − x2 + 3

.

Exercises


1. y = e-x 2. y = e √ � x − 2 3. y = e -

x − 4

4. y = e6x 5. y = 4ex 6. y = x2ex

7. y = ln (x3) 8. y = ln (2x + 5) 9. y = ln (sin x + 4)

10. y = ln ( 1 − x ) 11. y = x ln x 12. y = ln (2x3 + 4x)

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

Enrichment

Example 1

Example 2

12-6


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

Ass

essm

ent

NAME DATE PERIOD NAME DATE PERIOD

SCORE

NAME DATE PERIOD NAME DATE PERIOD

SCORE

Pdf Pass


12

12

Estimate each one-sided or two-sided limit for

f(x) =

⎧ ⎨

⎩ x2 + 1 if x < 2

x - 1 if x ≥ 2

, if it exists.

1. lim x → 2 -

f(x) 2. lim

x → 2 + f(x) 3. lim

x → 2 f(x)


4. lim x → 3

(-5x2 - 2x + 4)

5. lim x → -4

x2

- 16 − x + 4

6. lim x → - ∞ (3x2 - x)

A -∞ B -3 C 3 D ∞

Chapter 12 Quiz 1(Lessons 12-1 and 12-2)

1.

2.

3.

4.

5.

6.

Find the slope of the line tangent to the graph of each function at the given point.

1. y = x3 - 2x; (-1, 1)

2. y = 12 - 4x; (3, 0)

3. Find an equation for the slope of the graph of y = 2x3 + 5x2 - 2x at any point.

A m = 6x2 C m = 6x2 + 10x - 2

B m = 6x2 + 10x D m = 6


4. d(t) = 300 - 16t2; t = 2

5. d(t) = -16t2 + 90t; t = 4

Chapter 12 Quiz 2(Lesson 12-3)

1.

2.

3.

4.

5.

037-056_PCCRMC12_893813.indd 37037-056_PCCRMC12_893813.indd 37 12/7/09 12:09:58 PM12/7/09 12:09:58 PM

Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.

NAME DATE PERIOD

SCORE

NAME DATE PERIOD

SCORE

Pdf Pass


12

12


1. f(x) = 7x2 - 3x

A 14x - 3 B 14x C 7 D -3

2. f(x) = 4x2(x + 5)

3. f(x) = x2

− x3 - 1


4. 4

⌠ ⌡

1

x2 dx

5. 2

⌠ ⌡

0

(-x2 + 3x) dx

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


1. f(x) = 3x5

A x6

− 2 B x

6

− 2 + C C 3x4

− 4 D 3x4

− 4 + C

2. f(x) = 2x2 - 6x + 1

3. f(x) = -4x - x3


4. ⌠ ⌡

(5x2 - 1) dx

5. 1

⌠ ⌡

0

(x3 - 2x) dx

Chapter 12 Quiz 4(Lesson 12-6)

1.

2.

3.

4.

5.

1.

2.

3.

4.

5.


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

Ass

essm

ent

NAME DATE PERIOD

SCORE

Pdf Pass


12

Part I Write the letter for the correct answer in the blank at the right of each question.

Estimate each one-sided or two-sided limit.

1. lim x → 2 +

x2 - 7x + 10 − x - 2

A -∞ B -3 C 3 D ∞

2. lim x → -3+

2x2 + 6x −

x2 - 9

F -∞ G -1 H 1 J ∞


3. lim x → 1

x2 - 3x + 5 − x

A -∞ B -3 C 3 D ∞

4. lim x → ∞

5x3 - x − 6x4 - 2x2

F -∞ G 0 H 5 − 6 J ∞

5. Find the slope of the line tangent to the graph of y = x2 - 4x + 8 at any point.

6. Evaluate lim x → 0

3x2 + 4 −

8 - e 8 − x .

SLINGSHOT Jerry uses a slingshot to launch a rock into the air with an upward velocity of 40 feet per second. Suppose the height of the rock h, in feet, t seconds after it is launched is modeled by h(t) = -16t2 + 40t + 5.

7. Find an expression for the instantaneous velocity v(t) of the rock.

8. How fast is the rock traveling after 2 seconds?

9. For what value of t will the rock reach its maximum height?

10. What is the maximum height that the rock will reach?

Chapter 12 Mid-Chapter Test(Lessons 12-1 through 12-3)

1.

2.

3.

4.

Part 2

5.

6.

7.

8.

9.

10.


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NAME DATE PERIOD

SCORE

Pdf Pass


12

antiderivative

defi nite integral

derivative

difference quotient


differential operator

differentiation

direct substitution

indefi nite integral

indeterminate form


instantaneous velocity

integration

lower limit

one-sided limit

regular partition

right Riemann sum

tangent line

two-sided limit

upper limit

Choose the correct term in parentheses to complete each sentence correctly.

1. The function F(x) is said to be a(n) (antiderivative, derivative) of a function f(x) provided F ′(x) = f(x).

2. (One-sided limits, Two-sided limits) are used when we look at the value of a function f(x) as x approaches c from either the left side or right side of c.

3. The slope of a nonlinear graph at a specific point is called the (instantaneous rate of change, tangent line) of the graph at that point.

4. The (definite integral, indefinite integral) of a function

f(x) is defined by ⌠ ⌡

f(x) dx = F(x) + C, where F(x) is an

antiderivative of f(x) and C is any constant.

5. The result of finding a derivative of a function is called a (differential equation, differential operator).

Define each term in your own words.

6. indeterminate form

7. instantaneous velocity

Chapter 12 Vocabulary Test

1.

2.

3.

4.

5.


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

Ass

essm

ent

NAME DATE PERIOD

SCORE

Pdf Pass


12

Write the letter for the correct answer in the blank at the right of each question.

For Questions 1 and 2, use the graph of y = f(x) below to find each value.

1. lim x → 1

f(x)

A 0 C 2

B 1 D 3

2. lim x → 3 +

f(x)

F 3 G 2 H 1 J 0

3. MOTOR HOME After t years, the value v of a motor home purchased for $150,000 is v(t) = 150,000(0.92)t. Estimate lim t → ∞

v(t).

A $150,000 B $100,000 C $75,000 D $0


4. lim x → 4

√ � x - 2

− x - 4

F 1 − 4 G 1 −

2 H 1 J 0

5. lim x → ∞

3x2 - 2x − 5x3 + 7x2

A ∞ B 3 − 5 C 0 D -∞

6. lim x → - ∞ 2x3 - x2 + 3

F -∞ G 2 H 3 J ∞

7. Find the slope of the line tangent to the graph of y = x3 - 1 at the point (-2, -9).

A 12 B 9 C -9 D -12

8. Find an equation for the slope of the graph of y = -2x2 + 5x at any point.

F m = -4 G m = 5 H m = -4x J m = -4x + 5

9. FALLING OBJECTS Kyle drops a golf ball from a 1600-foot building. The position of the golf ball after t seconds is given by s(t) = -16t2 + 1600. How fast is the golf ball falling after 3 seconds?

A -32 ft/s B -96 ft/s C -144 ft/s D 1456 ft/s

10. Find an equation for the instantaneous velocity v(t) if the height of an object is defined as h(t) = 5 -6t + t2 for any point in time t.

F v(t) = 2t G v(t) = t2 H v(t) = -6 + 2t J v(t) = -6

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

y

x

f (x )

Chapter 12 Test, Form 1


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.

NAME DATE PERIOD

Pdf Pass


12


11. f(x) = x3 - x

A 3x2 - x B 3x - 1 C 3x2 D 3x2 - 1

12. f(x) = (4x - 5)2

F 4x - 5 G 8x - 10 H 32x J 32x - 40

13. HEIGHT The height of a ball in feet after t seconds is given by h(t) = 80t - 16t2 + 10 for 0 ≤ t ≤ 5. Find h′(2.5).

A 110 ft/s B 5 ft/s C 0 ft/s D -110 ft/s

Use the Quotient Rule to find the derivative of each function.

14. h(x) = 4x2 −

x - 4

F h′(x) = 8x H h′(x) = 4x2 - 32x − x - 4

G h′(x) = 12x2 - 32x − (x - 4)

2 J h′(x) = 4x2 - 32x −

(x - 4)2

15. g(x) = 4x + 3 − 3x - 2

A g ′(x) = 4 − 3 B g ′(x) = -17 −

(3x - 2)2 C g ′(x) = -17 −

3x - 2 D g ′(x) = 7x + 6 −

(3x - 2)2

16. Find all antiderivatives of f(x) = 8x3 - 3x2.

F 8x2 - 3 + C G 2x4 - x3 + C H 8x4 - 3x3 + C J 4x2 - 3x + C


17. ⌠ ⌡

(x3 - 2x) dx

A x4 - 2x2 + C B x4 - x2 + C C 1 − 4 x4 - x2 + C D 1 −

4 x4 + x2 + C

18. 2

⌠ ⌡

-2

5x2 dx

F 39 G 26 2 − 3 H 26 J 13 1 −

3

19. 3

⌠ ⌡

0

(3x2 - x3) dx

A 60.75 B 9 C 6.75 D 6

20. SPRINGS The work, in joules, required to stretch a spring one more foot, whose

natural length is one foot, is given by 1

⌠ ⌡

0

10x dx. How much work is required?

F 3 joules G 4 joules H 5 joules J 6 joules

Bonus Evaluate 2

⌠ ⌡

1

x-3 dx.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

Chapter 12 Test, Form 1 (continued)


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

Ass

essm

ent

NAME DATE PERIOD

SCORE

Pdf Pass


12



1. lim x → -1

f(x)

A -1 C 1

B 0 D 2

2. lim x → 1 +

f(x)

F 3 G 2 H 1 J 0

3. POP-UP CAMPER After t years, the value v of a pop-up camper purchased for $7000 is v(t) = 7000(0.89)t. Estimate lim t → ∞

v(t).

A $0 B $1000 C $5500 D $7000


4. lim x → 4

x2 - 16 − x - 4

F 0 G 1 H 4 J 8

5. lim x → ∞

-3x3 + 4x2

− 5x3 - 6x

A ∞ B - 3 − 5 C 0 D -∞

6. lim x → - ∞

1 − x3

F -∞ G 1 H 0 J ∞

7. Find the slope of the line tangent to the graph of y = 2 − x at the point (1, 2).

A 2 B 1 C -1 D -2

8. Find an equation for the slope of the graph of y = (x + 3)2 at any point.

F m = 2(x - 3) G m = x H m = 2x + 6 J m = x + 3

9. FALLING OBJECTS Tanesha drops a softball from a 1300-foot building. The position of the softball after t seconds is given by s(t) = -16t2 + 1300. How fast is the softball falling after 3 seconds?


10. Find an equation for the instantaneous velocity v(t) if the height of an object is defined as h(t) = √ � t + t2 for any point in time t.

F v(t) = 1 − 2 t + 2t G v(t) = 1 −

2 t

1 − 2 + 2 H v(t) = 1 −

2 t -

1 − 2 + 2t J v(t) = 1 −

2 t -

1 − 2 + 2

y

xf (x ) 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Chapter 12 Test, Form 2A


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NAME DATE PERIOD

Pdf Pass


12


11. f(x) = 3x2 + x

A x3 + x2 −

2 B 6x C 6x + 1 D x3 - x2

12. f(x) = 3(x - 2)2 + 5

F 6x - 12 G x - 2 H 6(x - 2) + 5 J 3x2 - 12x + 17

13. HEIGHT The height of a ball in feet h after t seconds is given by h(t) = -16t2 + 64t for 0 ≤ t ≤ 4. Find h′(2).

A -32 ft/s B 0 ft/s C 74 ft/s D 100 ft/s


14. h(x) = 3 - 2x − 3 + 2x

F h′(x) = -12 − 3 + 2x

G h′(x) = -12 − (3 + 2x)2

H h′(x) = -12 - 8x − (3 + 2x)2

J h′(x) = -12 − (3 - 2x)2

15. g(x) = x2 + 4 −

3 - x2

A g ′(x) = -2x − (3-x2)2

B g ′(x) = 14x − (3-x2)

C g ′(x) = 14x - 4x3 −

(3-x2)2 D g ′(x) = 14x −

(3-x2)2

16. Find all antiderivatives of f(x) = 12x5 + 9x2 - 4x.

F 12x6 + 9x3 - 4x2 + C H 60x4 + 18x2 - 4 + C

G 2x6 + 3x3 - 2x2 + C J 12x4 + 9x - 4 + C


17. ⌠ ⌡

x(x2 - 4) dx

A x4 - 4x2 + C B 1 − 4 x4 - 2x2 + C C 1 −

4 x4 + 2x2 + C D 1 −

4 x4 - 2x + C

18. 3

⌠ ⌡

0

0.8x3 dx

F 16.2 G 12.62 H 8.4 J 3

19. 3

⌠ ⌡

1

(4x3 - 3x) dx

A 68 B 95 C 135 D 202.5

20. SPRINGS The work, in joules, required to stretch a spring 2 inches from its

natural length of 3 inches is given by 2

⌠ ⌡

0



Bonus Evaluate 4

⌠ ⌡

0

x - 1 − 2 dx.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

Chapter 12 Test, Form 2A (continued)


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

Ass

essm

ent

NAME DATE PERIOD

SCORE

Pdf Pass


12



1. lim x → 0

f(x)

A -1 C 1

B 0 D 2

2. lim x → -2 -

f(x)

F -3 G -2 H -1 J 0

3. YACHT After t years, the value v of a yacht purchased for $200,000 is v(t) = 200,000(0.97)t. Estimate lim t → ∞

v(t).

A $0 B $100,000 C $150,000 D $200,000


4. lim x → ∞

-5x3 - 4x2 + x − 2x4 + x

F 5 − 2 G 1 H 0 J -5 −

2

5. lim x → 7

x

2 - 49 − x(x - 7)

A -∞ B -7 C 2 D 7

6. lim x → - ∞ x

5

F -∞ G -5 H 0 J ∞

7. Find the slope of the line tangent to the graph of y = -3x2 + 5x at the point (2, -2).

A m = -7 B m = -6 C m = -2 D m = 0

8. Find an equation for the slope of the graph of y = (x - 4)3 at any point.

F m = 3x2 - 24x + 48 H m = 2x - 12

G m = 2(x - 4)2 J m = x - 4

9. FALLING OBJECTS Tito drops a pebble from a 900-foot tower. The position of the pebble after t seconds is given by s(t) = -16t2 + 900. How fast is the pebble falling after 2 seconds?


10. Find an equation for the instantaneous velocity v(t) if the height of an object is defined as h(t) = 2 −

3 √ � t - t for any point in time t.

F v(t) = 1 − 3 t -

1 − 2 + 1 G v(t) = 4 −

3 t

1 − 2 - t2 H v(t) = 4 −

3 t -

1 − 2 - t J v(t) = 1 −

3 t -

1 − 2 - 1

y

xf (x ) 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Chapter 12 Test, Form 2B


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.

NAME DATE PERIOD

Pdf Pass


12


11. f(x) = -2x3 - x2

A - x4

− 2 -

x3 −

3 B -

x4 −

2 - x2 C -6x2 - x2 D -6x2 - 2x

12. f(x) = 7 - (x + 4)3

F -3(x + 4) H 7 - 3(x + 4)2

G -3x2 - 24x - 48 J x + 4

13. HEIGHT The height h of a ball in feet after t seconds is given by h(t) = -16t2 + 120t for 0 ≤ t ≤ 5. Find h′(3).

A -32 ft/s B -16 ft/s C 24 ft/s D 120 ft/s


14. h(x) = 4 - 3x − 4 + 3x

F h′(x) = -24 − 4 - 3x

H h′(x) = -24 - 18x − (4 + 3x)2

G h′(x) = -24 − 4 + 3x

J h′(x) = -24 − (4 + 3x)2

15. g(x) = x2 + 5

− 8 - x2

A g ′(x) = 26x − (x2 + 5)2

B g ′(x) = 26x − (8 - x2)2

C g ′(x) = 6x − (8 - x2)2

D g ′(x) = 26x - 4x3 −

(8 - x2)2

16. Find all antiderivatives of f(x) = (x + 1)2 - 1.

F (x + 1)3 + C G (x + 1)3

− 3 + C H (x + 1)3 - x + C J x

3 −

3 + x2

+ C


17. ⌠ ⌡

(1 - 3x2) dx

A x - 6x3 + C B 1 – x3 + C C -3x + C D x – x3 + C

18.

3

⌠ ⌡

0

1 − 4 x4 dx

F 0 G 8.1 H 12.15 J 20.25

19.

5

⌠ ⌡

2

(2x - 3x2) dx

A -8 B -21 C -65 D -96

20. SPRINGS The work, in joules, required to stretch a spring 5 inches from its

natural length of 4 inches is given by

5

⌠ ⌡

0



Bonus Evaluate

16

⌠ ⌡

0

x 1 − 4 dx.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

Chapter 12 Test, Form 2B (continued)


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

Ass

essm

ent

NAME DATE PERIOD

SCORE

Pdf Pass


12


1. lim x → -1

f(x)

2. lim x → 2 +

f(x) and f(2)

3. FISH TANK A fish tank’s population P in tens after t years

can be estimated by P(t) = 30t2 - 2t −

5t2 + 10

. What is the maximum

number of fish that can live in the tank?


4. lim x → 3

x2 + 2x - 15 −

x - 3

5. lim x → 16-

x − x - 16

6. lim x → - ∞ (x3 + 1000)

7. Find the slope of the line tangent to the graph of y = x(5x2 - 3x + 4) at the point (-1, -12).

8. Find an equation for the slope of the graph of

y = 1 − 3 x3 - 4x2 - 1 −

2 at any point.

9. FIREWORKS Fireworks are launched with an upward velocity of 160 feet per second. Suppose the height h of one firework in feet t seconds after it is fired is modeled by h(t) = -16t2 + 160t + 25. For what value of t will the firework reach its maximum height? What is the maximum height?

10. Find an equation for the instantaneous velocity v(t) if the height of an object is defined as h(t) = 3 √ � t + √ � t

y

xf (x )

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

Chapter 12 Test, Form 2C


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NAME DATE PERIOD

Pdf Pass


12


11. f(x) = -4x3 + x2 - 2x + 7

12. f(x) = (3x + 4)2

13. ARCHERY An arrow is shot upward with a velocity of 40 feet per second. Suppose the height h of the arrow in feet t seconds after it is shot is defined as h(t) = -16t2 + 40t + 6. How fast is the arrow traveling after 1 second?


14. h(x) = 4x2 - 2 − x2 - 4

15. g(x) = x4 + 1 −

2 - x3

16. Find all antiderivatives of f(x) = x2(x + 1)2.


17. ⌠ ⌡

(3x2 - 1)2 dx

18. 2

⌠ ⌡

0

(0.2x2 + 1) dx

19. -2

⌠ ⌡

-4

(x2 - 3x) dx

20. TREE HOUSE Leroy is in his tree house 35 feet above the ground when he drops his binoculars. The instantaneous velocity of his binoculars can be defined as v(t) = -32t, where time t is given in seconds and velocity v is measured in feet per second. Find the position function s(t) of the dropped binoculars.

Bonus Evaluate 2x

⌠ ⌡

1

(6x2 - 4x) dx.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

Chapter 12 Test, Form 2C (continued)


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

Ass

essm

ent

NAME DATE PERIOD

SCORE

Pdf Pass


12


1. lim x → -1

f(x)

2. lim x → 1 +

f(x) and f(1)

3. DEER A park’s deer population P in hundreds after t years

can be estimated by P(t) = 75t3 - 3t2 −

15t3 + 2t . What is the maximum

number of deer that can live in the park?


4. lim x → -4

x

2 - 3x - 28 − x + 4

5. lim x → 2 +

3x − x - 2

6. lim x → - ∞ (x5 + x3

)

7. Find the slope of the line tangent to the graph of y = x2(x - 3) at the point (-1, -4).

8. Find an equation for the slope of the graph of y = 1 −

5 x5 - 4x3

+ 1 at any point.

9. ROCKET A model rocket is launched with an upward velocity of 176 feet per second. Suppose the height h of the rocket in feet t seconds after it is fired is modeled by h(t) = -16t2 + 176t + 10. For what value of t will the rocket reach its maximum height? What is the maximum height?

10. Find an equation for the instantaneous velocity v(t) if the height of an object is defined as h(t) = 2 4 √ � t - 5 √ � t for any point in time t.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

y

x

f (x )

Chapter 12 Test, Form 2D


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NAME DATE PERIOD

Pdf Pass


12


11. f(x) = 3x3 - 5x2 - 8

12. f(x) = x2(2x - 5)(x + 1)

13. PROJECTILE A projectile is shot with a velocity of 175 feet per second toward a target. Suppose the height h of the projectile in feet t seconds after it is shot is defined as h(t) = -16t2 + 175t + 5. How fast is the projectile traveling after 2 seconds?


14. h(x) = 3x2 - 4 − x2 - 2

15. g(x) = x4 + 2 −

4 - x3

16. Find all antiderivatives of f(x) = x(x - 2)(x + 7).


17. ⌠ ⌡

(2x + 3)2 dx

18.

2

⌠ ⌡

0

(0.7x2 + 2) dx

19. -1

⌠ ⌡

-3

(x2 - 7x) dx

20. ROOFTOP Michelle is on the roof of her house 20 feet above the ground when she drops her hammer. The instantaneous velocity of her hammer can be defined as v(t) = -32t, where time t is given in seconds and velocity v is measured in feet per second. Find the position function s(t) of the dropped hammer.

Bonus Evaluate x2

⌠ ⌡

2

(4x3 - 6x) dx.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

Chapter 12 Test, Form 2D (continued)


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

Ass

essm

ent

NAME DATE PERIOD

SCORE

Pdf Pass


12


1. lim x → 2 -

f(x) and f(-2)

2. lim x → 1

f(x)

3. QUAIL A park’s quail population P in thousands after

t years can be estimated by P(t) = 18t4 - 2t2 −

9t4 + 3t2 . What is the

maximum number of quail that can live in the park?


4. lim x → 2

x3 - 8 − x - 2

5. lim x → 0

3 sin x − x

6. lim x → - ∞ (x3 - x2

)

7. Find the slope of the line tangent to the graph of y = x2(x - 2)2 at the point (2, 0).

8. Find an equation for the slope of the graph of y = (x + 1)3 at any point.

9. ROCKET A rocket is launched into the air with an upward velocity of 256 feet per second. Suppose the height h of the rocket, in feet, t seconds after it is fired is modeled by h(t) = -16t2 + 256t. For what value of t will the rocket reach its maximum height? What is the maximum height?

10. Find an equation for the instantaneous velocity v(t) if the height of an object is defined as h(t) = 0.2t1.7 + 1 −

3 3 √ � t for any point in time t.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

x

y

f (x )

Chapter 12 Test, Form 3


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NAME DATE PERIOD

Pdf Pass


12


11. f(x) = 1 − 8 x7 - 0.4x3 + 1 −

2

12. f(x) = √ � x (x + 1)

13. SKI JUMPING A jumper leaves the incline with a velocity of 55 feet per second. Suppose the height h of the jumper, in feet, t seconds after she jumps is defined as h(t) = -16t2 + 55t + 60. How fast is the jumper traveling after 0.5 second?


14. h(x) = √ � x + 1 −

-x + 1

15. g(x) = x2 - 4x + 1 −

x3- 3x

16. Find all antiderivatives of f(x) = (2x + 1)2(x - 3).


17. ⌠ ⌡

(x + 1)3 dx

18. 3

⌠ ⌡

1

( 1 − x2

+ 1) dx

19. 0

⌠ ⌡

-2

[x(x + 8) + 12] dx

20. FARMS Neil is in the loft of his barn 12 feet above the ground when he drops his pitchfork. The instantaneous velocity of his pitchfork can be defined as v(t) = -32t, where time t is given in seconds and velocity v is measured in feet per second. Find the position function s(t) of the dropped pitchfork.

Bonus Evaluate 2x

⌠ ⌡

x

(3x2 - 2x) dx.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

Chapter 12 Test, Form 3 (continued)


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

Ass

essm

ent

NAME DATE PERIOD

SCORE

Pdf Pass


12

Demonstrate your knowledge by giving a clear, concise solution to each problem. Be sure to include all relevant drawings and justify your answers. You may show your solution in more than one way or investigate beyond the requirements of the problem.

1. Let f(x) = x2 + 5x −

x2 - 6x .

a. Make a table of values for f(x). Then draw the graph of the function.

b. Using the graph in part a, explain how to find the limit of f(x) as x approaches 3. Then explain how to find the limit as x approaches 0.

c. Find the limit of the numerator, x2 + 5x, as x approaches 3. Justify your answer.

d. Find the limit of the denominator, x2 - 6x, as x approaches 3. Justify your answer.

e. Find the limit as x approaches 3 of x2 + 5x −

x2 - 6x algebraically.

f. Find the limit as x approaches 0 of x2 + 5x −

x2 - 6x algebraically.

2. Find two different functions, f and g, such that lim x → 0

f(x) = lim x → 0

g(x).

3. The speed of an object is given by s = 4t - t2, where time t is in seconds.

a. Graph the function.

b. Use two different methods to find the area or approximate area bounded by the curve and the t-axis from t = 0 to t = 4. Which method is more accurate? Why?

c. The area of a rectangle is given by a = � · w. In this case, � = s (speed) and w = t (time). What do you think the area under the curve represents? Why?

Extended-Response Test


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NAME DATE PERIOD

SCORE

Pdf Pass


12

1. Given f(x) = 2x - 1 and g(x) = x2 + 2, find (f � g)(x).

A (f � g)(x) = 2x3 - x2 + 4x - 2 C (f � g)(x) = 2x2 + 3

B (f � g)(x) = 4x2 - 4x + 3 D (f � g)(x) = x2 + 2x + 1 1.

2. Find the slope of the line tangent to the graph of y = x2 - 5x + 1 at the point (2, -5).

F -15 G -5 H -1 J 0 2.

3. Name a different pair of polar coordinates for the point (3, 120°).

A (3, 300°) B (-3, 480°) C (-3, 300°) D (-3, -240°) 3.

4. Find the coefficient of the sixth term in the expansion of (a + b)8.

F 28 G 56 H 6720 J 20,160 4.

5. Use the discriminant to identify the conic section x2 - 9y2 + 36y - 72 = 0.

A hyperbola B circle C ellipse D parabola 5.

6. Suppose θ is the measure of the angle that a loading dock ramp makes

with the ground and that tan θ = 3 − 4 . Find sin θ.

F 3 − 5 G 4 −

5 H 4 −

3 J 5 −

3 6.

7. Let A = ⎡ ⎢

⎣ -2

0 4 1

⎤ �

⎦ and B =

⎡ ⎢

⎣ 5 -2

1 0 ⎤ �

⎦ . Find AB.

A ⎡ ⎢

⎣ -10

4 21

-8 ⎤ �

⎦ B

⎡

⎢

⎣ 3 -2

5 1 ⎤ �

⎦ C

⎡ ⎢

⎣ -7

2 3 1 ⎤ �

⎦ D

⎡ ⎢

⎣ -18

-2 -2

0 ⎤ �

⎦ 7.

Part 1: Multiple Choice

Instructions: Fill in the appropriate circle for the best answer.

Standardized Test Practice(Chapters 1–12)

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J

A B C D


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

Ass

essm

ent

NAME DATE PERIOD

Pdf Pass


12

8. Find the exact distance between the points at A(-2, 4, 1) and B(0, 3, -2).

F 2 √ �� 37 G 2 √ �� 14 H 3 √ � 6 J √ �� 14 8.

9. Find the absolute value of the complex number z = 7 + 24i. A 25 B 49 C 576 D 625 9.

10. List all possible rational zeros of p(x) = 3x3 - 5x2 + 7.

F ±1, ±3, ± 1 − 7 , ± 3 −

7 H ±1, ±3

G ±1, ±7, ± 1 − 3 , ± 7 −

3 J ±1, ±7 10.

11. Find the exact value of cos 5π −

6 .

A √ � 3

− 2 B -

1 − 2 C -

√ � 2 −

2 D -

√ � 3 −

2 11.

12. The vector u has a magnitude of 4.3 centimeters and a direction of 45°. Find the magnitude of its vertical component.

F 2.45 cm G 3.04 cm H 3.25 cm J 6.53 cm 12.

13. Evaluate ⌠ ⌡

(x3 - 4x) dx.

A x4 - 4x2 + C C 1 − 4 x4 + 2x2 + C

B 1 − 4 x4 - x2 + C D 1 −

4 x4 - 2x2 + C 13.


x −

(x + 2)2 - 4 or state that the limit does not exist.

F 1 − 4 G 0 H 4 J does not exist 14.

15. Express the series 11 + 18 + 27 + . . . + 171 using sigma notation.

A ∑

n = 1

11

(n2 + n) B ∑

n = 3

13

(n2 + n) C ∑

n = 3

13

(n2 + 2) D ∑

n = 2

9

2n2 15.

Standardized Test Practice (continued)

(Chapters 1–12)

F G H J

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J

A B C D


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.

NAME DATE PERIOD

Pdf Pass


12

16. Solve 5x + 1 = 25x - 2.

17. Find the magnitude of vector a = ⟨-8, 15⟩.

18. Classify the random variable X as discrete or continuous. Explain your reasoning. X represents the weight of rice in a 20-ounce box of rice chosen randomly off a grocery store shelf.


x2 - 8x + 12 −

x - 2 .

20. Write the pair of parametric equations in rectangular form. Identify the related conic.

x = 4 cos t y = 3 sin t

21. Write an explicit formula for finding the nth term of a geometric sequence with a first term of a1 = -4 and a common ratio of r = 1 −

2 .

22. Find the partial fraction decomposition of 4x − x2 - 9

.

23. Find z if X = 49, μ = 42, and σ = 1.9.

24. Find all antiderivatives for ⌠ ⌡

(x4 - 3x2) dx.

25. BOUNCING BALL The velocity of a ball that was bounced off a sidewalk can be defined as v(t) = -32t + 80, where time t is given in seconds and velocity v is given in feet per second.

a. Find the position function s(t) for the ball after it bounces off the sidewalk. Assume that for t = 0, s(t) = 0.

b. What is the height of the ball after 2 seconds?

c. How long does it take for the ball to return to the ground?

Part 2: Short Response

Instructions: Write your answers in the space provided.

16.

17.

18.

19.

20.

21.

22.

23.

24.

Standardized Test Practice (continued)

(Chapters 1–12)

25a.

25b.

25c.


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.

Chapter 12 A1 Glencoe Precalculus

An

swer

s

Answers (Anticipation Guide and Lesson 12-1)

Pdf Pass

Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.

NA

ME

DA

TE

PE

RIO

D

Lesson 12-1

Ch

ap

ter

12

5

Gle

ncoe

Pre

calc

ulus

12-1

Esti

mat

e Li

mit

s at

Fix

ed V

alue

sL

eft-

Han

d L

imit

If t

he v

alue

of f

(x) a

ppro

ache

s a

uniq

ue

num

ber

L 1 as

x ap

proa

ches

c fr

om t

he

left

, the

n lim

x

→ c

- f

(x) =

L1.

Rig

ht-H

and

Lim

itIf

the

val

ue o

f f (x

) app

roac

hes

a un

ique

nu

mbe

r L 2 a

s x

appr

oach

es c

from

the

ri

ght,

then

lim

x

→ c

+ f(

x) =

L2.

Exi

sten

ce o

f a

Lim

it a

t a

Poi

nt

The

limit

of a

func

tion

f (x

) as

x ap

proa

ches

c e

xist

s if

and

only

if b

oth

one-

side

d lim

its

exis

t an

d ar

e eq

ual.

That

is, i

f

lim

x

→ c

- f

(x) =

lim

x

→ c

+ f

(x) =

L, t

hen

lim

x

→ c f

(x) =

L.

E

stim

ate

each

one

-sid

ed o

r tw

o-si

ded

lim

it, i

f it

exi

sts.

lim

x

→ 2

- � x

� , lim

x

→ 2

+ � x

� , a

nd lim

x

→ 2

� x�

The

grap

h of

f(x)

= �

x� s

ugge

sts

that

lim

x

→ 2

- � x

� =

1

and

lim

x

→ 2

+ � x

� =

2.

Bec

ause

the

left

- and

rig

ht-h

and

limit

s of

f (x

) as

x ap

proa

ches

2 a

re n

ot t

he s

ame,

lim

x

→ 2

� x� d

oes

not

exis

t.

Exer

cise

s

Est

imat

e ea

ch o

ne-s

ided

or

two-

side

d li

mit

, if

it e

xist

s.

1. lim

x

→ 0

+ ⎪ 3

x⎥

−

x

2.

lim

x →

-2-

⎪ x -

2⎥

−

x2 - 4

3.

lim

x

→ 2

x2 + 3

x -

10

−

x

- 2

3

∞

7

4. lim

x

→ 0

(1 -

cos

2 x)

5. lim

x

→ 3

- x3 +

27

−

x2 - 9

6.

lim

x

→ -

2 1

−

(x +

2)2

0

-∞

∞

Stud

y Gu

ide

and

Inte

rven

tion

Est

imati

ng

Lim

its

Gra

ph

ically

Exam

ple

y

x

f(x)

= �

x�

005_

036_

PC

CR

MC

12_8

9381

3.in

dd5

12/7

/09

12:3

2:15

PM


NA

ME

DA

TE

PE

RIO

D

Chapter Resources

DA D AA A D A D A

12

B

efor

e yo

u be

gin

Cha

pter

12

•

Rea

d ea

ch s

tate

men

t.

•

Dec

ide

whe

ther

you

Agr

ee (A

) or

Dis

agre

e (D

) wit

h th

e st

atem

ent.

•

Wri

te A

or

D in

the

firs

t co

lum

n O

R if

you

are

not

sur

e w

heth

er y

ou

agre

e or

dis

agre

e, w

rite

NS

(Not

Sur

e).

A

fter

you

com

plet

e C

hapt

er 1

2

•

Rer

ead

each

sta

tem

ent

and

com

plet

e th

e la

st c

olum

n by

ent

erin

g an

A o

r a

D.

•

Did

any

of y

our

opin

ions

abo

ut t

he s

tate

men

ts c

hang

e fr

om t

he fi

rst

colu

mn?

•

For

thos

e st

atem

ents

tha

t yo

u m

ark

wit

h a

D, u

se a

pie

ce o

f pap

er t

o w

rite

an

exam

ple

of w

hy y

ou d

isag

ree.

ST

EP

1

A,

D,

or

NS

Sta

tem

en

tS

TE

P 2

A

or

D

1.

The

limit

of a

func

tion

f(x)

as

x ap

proa

ches

c d

oes

not

depe

nd

on t

he v

alue

of t

he fu

ncti

on a

t po

int

c.

2.

The

limit

of a

func

tion

f(x)

as

x ap

proa

ches

c e

xist

s pr

ovid

ing

eith

er t

he le

ft-h

and

limit

or

righ

t-ha

nd li

mit

exi

sts.

3.

The

limit

of a

con

stan

t fu

ncti

on a

t an

y po

int

is t

he x

-val

ue o

f th

e po

int.

4.

Lim

its

of p

olyn

omia

l and

man

y ra

tion

al fu

ncti

ons

can

be

foun

d by

dir

ect

subs

titu

tion

.

5.

The

slop

e of

a n

onlin

ear

grap

h at

a s

peci

fic p

oint

is t

he

inst

anta

neou

s ra

te o

f cha

nge.

6.

The

proc

ess

of fi

ndin

g a

deri

vati

ve is

cal

led

diffe

rent

iati

on.

7.

The

deri

vati

ve o

f a c

onst

ant

func

tion

is t

he c

onst

ant.

8.

The

proc

ess

of e

valu

atin

g an

inte

gral

is c

alle

d in

tegr

atio

n.

9.

The

func

tion

F(x

) is

an a

ntid

eriv

ativ

e of

the

func

tion

f(x)

iff ′(

x) =

F(x

).

10.

The

conn

ecti

on b

etw

een

defin

ite

inte

gral

s an

d an

tide

riva

tive

s is

so

impo

rtan

t th

at it

is c

alle

d th

e Fu

ndam

enta

l The

orem

of C

alcu

lus.

Antic

ipat

ion

Guid

eLim

its

an

d D

eri

vati

ves

Step

2

Step

1

Ch

ap

ter

12

3

Gle

ncoe

Pre

calc

ulus

0ii_

004_

PC

CR

MC

12_8

9381

3.in

ddS

ec1:

33/

17/0

911

:36:

02A

M

A01_A17_PCCRMC12_893813.indd 1A01_A17_PCCRMC12_893813.indd 1 12/7/09 1:35:41 PM12/7/09 1:35:41 PM

Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.


Answers (Lesson 12-1)

Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

12

6

Gle

ncoe

Pre

calc

ulus

12-1

Esti

mat

e Li

mit

s at

Infi

nity

• If

the

val

ue o

f f (x

) app

roac

hes

a un

ique

num

ber

L 1 as

x in

crea

ses,

th

en lim

x

→ ∞

f (x

) = L

1.

• If

the

val

ue o

f f(x

) app

roac

hes

a un

ique

num

ber

L 2 as

x de

crea

ses,

th

en

lim

x →

-∞

f (x

) = L

2.

E

stim

ate

lim

x

→ ∞

1

−

x +

3 ,

if it

exi

sts.

Ana

lyze

Gra

phic

ally

The

gra

ph o

f f (x

) =

1 −

x

+ 3

sug

gest

s th

at lim

x

→ ∞

1

−

x +

3 =

0.

As

x in

crea

ses,

the

hei

ght

of t

he g

raph

get

s cl

oser

to

0. T

he li

mit

indi

cate

s a

hori

zont

al a

sym

ptot

e at

y =

0.

Supp

ort N

umer

ical

ly M

ake

a ta

ble

of v

alue

s, c

hoos

ing

x-va

lues

tha

t gr

ow

incr

easi

ngly

larg

e.

x ap

proa

ches

infin

ity

x1

01

00

10

00

10

,00

01

00

,00

0

f(x)

0.0

80

.01

0.0

01

0.0

00

10

.00

00

1

The

patt

ern

of o

utpu

ts s

ugge

sts

that

as

x gr

ows

incr

easi

ngly

larg

er, f

(x) a

ppro

ache

s 0.

Thi

s su

ppor

ts o

ur g

raph

ical

ana

lysi

s.

Exer

cise

s

Est

imat

e ea

ch li

mit

, if

it e

xist

s.

1. lim

x

→ ∞

2x +

1

−

x

2.

lim

x →

-∞

-3x

+ 1

−

x -

2

3.

lim

x

→ ∞

1 −

x2

2

-3

0

4. lim

x

→ ∞

2x2 -

5

−

3x3 +

2x

5. lim

x

→ ∞

(ex s

in 2

xπ)

6.

lim

x →

-∞

(2x +

x)

0

do

es

no

t e

xis

t

-∞

7. lim

x

→ ∞

(x s

in x

) 8.

lim

x →

-∞

e2x

9.

lim

x

→ ∞

cos

2xπ

d

oe

s n

ot

ex

ist

0

do

es

no

t e

xis

t

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

Est

imati

ng

Lim

its

Gra

ph

ically

Exam

ple

y

x

(x)=

1x

+ 3

005_

036_

PC

CR

MC

12_8

9381

3.in

dd6

12/5

/09

4:49

:57

PM


NA

ME

DA

TE

PE

RIO

D

Lesson 12-1

Ch

ap

ter

12

7

Gle

ncoe

Pre

calc

ulus

12-1

Est

imat

e ea

ch o

ne-s

ided

or

two-

side

d li

mit

, if

it e

xist

s.

1.

lim

x →

0+ (4

- √

� x )

4

2.

lim

x →

3+

3 -

x

−

⎪ x -

3⎥

-1

3.

lim

x

→ 4

x2 - 1

6 −

x

- 4

8

4.

lim

x →

-1-

x

+ 7

−

x2 + 8

x +

7

-∞

5.

lim

x →

-1+

x

+ 7

−

x2 + 8

x +

7

∞

6.

lim

x →

0 x2 +

1

−

x2

∞

Est

imat

e ea

ch li

mit

, if

it e

xist

s.

7.

lim

x →

-∞

-4x

2 −

x2 +

1

-4

8.

lim

x →

∞ 3x

- 2

−

x -

1

3

9.

lim

x

→ 0

sin

2x

−

x

2

10.

lim

x →

∞ e

3x +

2

∞

11. R

ATE

OF

CHA

NG

E A

20-

foot

pol

e is

lean

ing

agai

nst

a ba

rn. I

f the

bas

e of

the

pol

e is

pul

led

away

from

the

bar

n at

a r

ate

of 3

feet

per

sec

ond,

th

e to

p of

the

pol

e w

ill m

ove

dow

n th

e si

de o

f the

bar

n at

a r

ate

of

r (x)

=

3x

−

√ �

��

�

400

- x

2 f

eet

per

seco

nd, w

here

x is

the

dis

tanc

e be

twee

n th

e

base

of t

he p

ole

and

the

barn

. Gra

ph r

(x) t

o fin

d

lim

x →

20-

r (x)

.

∞

12. P

OLL

UTA

NTS

The

cos

t in

mill

ions

of d

olla

rs fo

r a

com

pany

to

clea

n up

th

e po

lluta

nts

crea

ted

by o

ne o

f its

man

ufac

turi

ng p

roce

sses

is g

iven

by

C =

31

2x

−

100

- x

, w

here

x is

the

num

ber

of p

ollu

tant

s an

d 0

≤ x

≤ 1

00.

Find

lim

x →

100

- C

. ∞

Prac

tice

Est

imati

ng

Lim

its

Gra

ph

ically

005_

036_

PC

CR

MC

12_8

9381

3.in

dd7

12/5

/09

4:50

:51

PM

A01_A17_PCCRMC12_893813.indd 2A01_A17_PCCRMC12_893813.indd 2 12/7/09 10:17:59 AM12/7/09 10:17:59 AM

Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.



An

swer

s

Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

12

8

Gle

ncoe

Pre

calc

ulus

12-1

1. B

ACT

ERIA

GRO

WTH

Bac

teri

a in

a d

ish

are

grow

ing

acco

rdin

g to

the

func

tion

f (

t) =

4

−

1 +

0.3

5e-

0.2t ,

for

t ≥ 0

, whe

re f

(t)

is t

he w

eigh

t of

the

bac

teri

a in

gra

ms

and

t is

the

tim

e in

hou

rs.

a. G

raph

f(t)

for

0 ≤

t ≤

20.

b. U

se t

he g

raph

to

esti

mat

e th

e nu

mbe

r of

gra

ms

of b

acte

ria

pres

ent

afte

r 8

hour

s. R

ound

to

the

near

est

tent

h,

if ne

cess

ary.

3

.7 g

c. E

stim

ate

lim

t →

∞ f

(t),

if it

exi

sts.

Inte

rpre

t yo

ur r

esul

t. 4

g;

Ov

er

tim

e,

the

we

igh

t o

f th

e b

ac

teri

a i

n t

he

dis

h w

ill

ap

pro

ac

h a

ma

xim

um

of

4 g

.

2. C

ARS

Aft

er t

year

s, t

he v

alue

of

a ca

r pu

rcha

sed

for

$30,

000

is

v(t)

= 3

0,00

0(0.

7)t .

a. G

raph

v(t

) for

0 ≤

t ≤

20.

b. U

se t

he g

raph

to

esti

mat

e th

e va

lue

of t

he c

ar a

fter

10

year

s.

$8

47

c. E

stim

ate lim

t →

∞ v

(t),

if it

exi

sts.

Inte

rpre

t yo

ur r

esul

ts.

$0

; O

ve

r ti

me

, th

e v

alu

e o

f th

e c

ar

wil

l re

ac

h a

min

imu

m o

f $

0.

3. P

ROJE

CTIL

E H

EIG

HT

Supp

ose

a pr

ojec

tile

is t

hrow

n up

war

d w

here

its

heig

ht h

in fe

et a

t an

y ti

me

t in

seco

nds

is d

eter

min

ed b

y th

e fu

ncti

on h

(t).

The

tabl

e sh

ows

the

heig

ht o

f the

pro

ject

ile a

t va

riou

s ti

mes

dur

ing

its

fligh

t.

a.

Gra

ph t

he d

ata

and

draw

a c

urve

th

roug

h th

e da

ta p

oint

s to

mod

el t

he

func

tion

h(t

).

b. U

se y

our

grap

h to

est

imat

e lim

t →

8- h

(t).

0 f

t

4. T

HEO

RY O

F RE

LATI

VIT

Y

Theo

reti

cally

, the

mas

s m

of a

n ob

ject

w

ith

velo

city

v is

giv

en b

y

m =

m

0 −

√ �

��

1 -

v2 −

s2 ,

whe

re m

0 is

the

mas

s of

the

obje

ct a

t re

st a

nd s

is t

he s

peed

of

light

. Wha

t is

lim

v

→ s

- m

? ∞

5. E

LECT

RICI

TY A

hmed

det

erm

ined

tha

t th

e vo

ltag

e in

an

elec

tric

al o

utle

t in

his

ho

me

is m

odel

ed b

y th

e fu

ncti

on

V(t

) = 1

40 s

in 1

20π

t. E

xpla

in w

hy

lim

t →

∞ V

(t) d

oes

not

exis

t.

A

s t

in

cre

as

es

, th

e g

rap

h

os

cil

late

s b

etw

ee

n 1

40

an

d -

14

0.

Wor

d Pr

oble

m P

ract

ice

Est

imati

ng

Lim

its

Gra

ph

ically

t

468 2

124

816

f(t)

t

200

100

300

400

24

68

h(t)

t

v(t)

th

(t)

t

h(t

)

02

56

438

4

13

36

53

36

23

84

625

6

34

00

714

4

005_

036_

PC

CR

MC

12_8

9381

3.in

dd8

12/7

/09

12:4

7:58

PM


NA

ME

DA

TE

PE

RIO

D

Lesson 12-1

Ch

ap

ter

12

9

Gle

ncoe

Pre

calc

ulus

12-1

A M

att

er

of

Lim

its

The

re a

re m

any

exam

ples

of

lim

its

in o

ur w

orld

. So

me

of t

hese

are

ab

solu

te li

mit

s, in

tha

t th

ey c

an n

ever

be

exce

eded

. Oth

ers

are

like

gu

idel

ines

, and

sti

ll o

ther

s re

sult

in a

pen

alty

if t

hey

are

exce

eded

. F

ill i

n th

e ch

art

belo

w.

Lim

itH

ow

is

th

e l

imit

se

t?Is

th

e l

imit

ab

so

lute

?P

en

alt

y o

r c

on

se

qu

en

ce

if t

he

lim

it i

s e

xc

ee

de

d

1.

spee

d lim

it o

n a

high

way

Th

e g

ov

ern

me

nt

se

ts t

he

sp

ee

d

lim

its

.n

os

pe

ed

ing

tic

ke

t,

fi n

e,

an

d s

o o

n

2.

heig

ht li

mit

on

a r

oad

unde

rpas

sH

eig

ht

lim

it i

s s

et

for

sa

fe c

lea

ran

ce

.y

es

Da

ma

ge

is

do

ne

to

v

eh

icle

an

d/o

r s

tru

ctu

re.

3.

lugg

age

limit

on

an

airl

ine

fligh

tA

irli

ne

se

ts l

imit

o

n a

mo

un

t o

f b

ag

ga

ge

all

ow

ed

.n

o

Pa

ss

en

ge

r m

us

t p

ay

fo

r a

ny

lu

gg

ag

e

be

yo

nd

lim

it.

4.

tem

pera

ture

of a

w

arm

obj

ect

plac

ed

in a

coo

l roo

m

ca

n o

nly

ge

t a

s

co

ol

as

its

s

urr

ou

nd

ing

sy

es

no

t p

os

sib

le

5.

the

spee

d of

an

acce

lera

ting

spa

ce

craf

tp

hy

sic

al

co

ns

tan

t,

the

sp

ee

d o

f li

gh

ty

es

no

t p

os

sib

le

6.

cred

it li

mit

on

a cr

edit

car

ds

et

by

ba

nk

is

su

ing

ca

rdn

o

fi n

an

cia

l p

en

alt

y

se

t b

y b

an

k,

po

ss

ibly

ma

y l

os

e

ca

rd

One

spe

cial

fea

ture

of

mat

hem

atic

al li

mit

s is

tha

t th

ey m

ay b

e fi

nite

, in

fini

te, o

r th

ey m

ay n

ot e

xist

. Cla

ssif

y ea

ch li

mit

as

fini

te, i

nfin

ite,

or

doe

s no

t ex

ist.

If t

he li

mit

is f

init

e, g

ive

its

valu

e.

7.

lim

x

→ 0

1

−

x2 + 1

8.

lim

x

→ 1

x +

1

−

x2 - 1

9.

lim

x

→ 2

x2 -

4

−

x2 - x

-2

fi n

ite

; 1

do

es

no

t e

xis

t

fi n

ite

; 4

−

3

10. lim

x

→ 0

ln ⎪

x⎥

11.

lim

x

→ 0

sin

⎪ x⎥

−

x

12.

lim

x

→ 0

-1

−

x4

infi

nit

e

d

oe

s n

ot

ex

ist

in

fi n

ite

Enri

chm

ent

005_

036_

PC

CR

MC

12_8

9381

3.in

dd9

3/17

/09

11:3

6:36

AM


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.


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

Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

12

10

Gle

ncoe

Pre

calc

ulus

12-1

Fin

din

g L

imit

sY

ou c

an u

se a

gra

phin

g ca

lcul

ator

to

find

a lim

it w

ith

less

wor

k th

an a

n or

dina

ry s

cien

tific

cal

cula

tor.

To

find

lim

x

→ a

f (x

), fir

st g

raph

the

equ

atio

n

y =

f(x)

. The

n us

e Z

OO

M a

nd

TR

AC

E t

o lo

cate

a p

oint

on

the

grap

h w

hose

x-

coor

dina

te is

as

clos

e to

a a

s yo

u lik

e. T

he y

-coo

rdin

ate

shou

ld b

e cl

ose

to

the

valu

e of

the

lim

it.

Eva

luat

e ea

ch li

mit

.

1.

lim

x

→ 0

ex -

1

−

x

Pr

ess

Y=

(

2

nd

[e]

) —

1

) ÷

E

NT

ER

Z

OO

M 6

. The

n pr

ess

ZO

OM

2 E

NT

ER

. Pre

ss

TR

AC

E a

nd u

se

an

d t

o ex

amin

e th

e lim

it o

f the

func

tion

whe

n x

is c

lose

to

0.

1

2. lim

x

→ 2

x2 -

4

−

x2 -3x

+ 2

Pr

ess

Y=

(

x2 —

4

) ÷

(

x2 —

3

+

2

)

EN

TE

R

ZO

OM

6. T

hen

pres

s Z

OO

M 2

EN

TE

R. P

ress

T

RA

CE

and

use

and

to

exam

ine

the

limit

of t

he fu

ncti

on w

hen

x is

clo

se t

o 2.

4

3. I

f you

gra

ph y

= ln

x

−

x -

1 a

nd u

se

TR

AC

E, w

hy d

oesn

’t th

e ca

lcul

ator

tel

l you

w

hat

y is

whe

n x

= 1

?

S

am

ple

an

sw

er:

Th

e f

un

cti

on

is

un

de

fi n

ed

at

x =

1.

4. W

ill t

he g

raph

ing

calc

ulat

or g

ive

you

the

exac

t an

swer

for

ever

y lim

it

prob

lem

? E

xpla

in.

N

o;

Sa

mp

le a

ns

we

r: T

he

tra

ce

fu

nc

tio

n d

oe

s n

ot

hig

hli

gh

t e

ve

ry p

os

sib

le n

um

be

r.

Grap

hing

Cal

cula

tor

Activ

ity

005_

036_

PC

CR

MC

12_8

9381

3.in

dd10

3/17

/09

11:3

6:40

AM


NA

ME

DA

TE

PE

RIO

D

Lesson 12-2

Ch

ap

ter

12

11

Gle

ncoe

Pre

calc

ulus

12-2

Stud

y Gu

ide

and

Inte

rven

tion

Eva

luati

ng

Lim

its

Alg

eb

raic

ally

Exam

ple

1

Exam

ple

2

Exam

ple

3

Com

pute

Lim

its

at a

Poi

nt

Use

dir

ect

subs

titu

tion

, if

poss

ible

, to

eval

uate

li

m

x →

-2 (

-2x

4 + 3

x3 - 5

x +

3).

Sinc

e th

is is

the

lim

it o

f a p

olyn

omia

l fun

ctio

n, w

e ca

n ap

ply

the

met

hod

of

dire

ct s

ubst

itut

ion

to fi

nd t

he li

mit

.

lim

x →

-2 (-

2x4 +

3x3 -

5x

+ 3

) = -

2(-

2)4 +

3(-

2)3 -

5(-

2) +

3

= -

32 -

24

+ 1

0 +

3, o

r -

43

U

se f

acto

ring

to

eval

uate

lim

x

→ 4

x2 - 9

x +

20

−

x

- 4

.

lim

x

→ 4

x2 - 9

x +

20

−

x

- 4

=

lim

x

→ 4

(x -

5)(x

- 4

)

−

(x -

4)

F

act

or.

=

lim

x

→ 4

(x -

5)

Div

ide

ou

t th

e c

om

mo

n f

act

or

an

d s

imp

lify.

=

4 -

5, o

r -

1 A

pp

ly d

ire

ct s

ub

stitu

tion

an

d s

imp

lify.

U

se r

atio

nali

zing

to

eval

uate

lim

x

→ 1

6 √ �

x -

4

−

x -

16 .

By

dire

ct s

ubst

itut

ion,

you

obt

ain

√ �

�

16 -

4

−

16 -

16

or 0 −

0 . R

atio

naliz

e th

e nu

mer

ator

of t

he fr

acti

on b

efor

e fa

ctor

ing

and

divi

ding

com

mon

fact

ors.

lim

x

→ 1

6 √ �

x -

4

−

x -

16

= lim

x

→ 1

6 √ �

x -

4

−

x -

16 ·

√ �

x +

4

−

√ �

x +

4

Multi

ply

th

e n

um

era

tor

an

d d

en

om

ina

tor

by

√ �

x +

4,

the

con

jug

ate

of

√ �

x -

4.

=

lim

x

→ 1

6 x

- 1

6 −

(x -

16)

( √ �

x +

4)

Sim

plif

y.

=

lim

x

→ 1

6 x

- 1

6 −

(x -

16)

( √ �

x +

4)

Div

ide

ou

t th

e c

om

mo

n f

act

or.

=

lim

x

→ 1

6 1

−

( √ �

x +

4)

Sim

plif

y.

=

1

−

√ �

�

16 +

4 o

r 1 −

8 A

pp

ly d

ire

ct s

ub

stitu

tion

an

d s

imp

lify.

Exer

cise

s

Eva

luat

e ea

ch li

mit

.

1. lim

x

→ 3

(2x2 -

5x)

3

2.

lim

x

→ 5

√ �

��

x3 - 4

1

1

3. lim

x

→ -

2 x2 + 9

x +

14

−

x

+ 2

5

4. lim

x

→ 4

√ �

x -

2

−

x -

4

1

−

4

5. lim

x

→ -

4 ( 1 −

x + x

)

-4

.25

6.

lim

x

→ 2

(-x2 +

5x

- 1

) 5

005_

036_

PC

CR

MC

12_8

9381

3.in

dd11

12/5

/09

4:54

:56

PM


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s


Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

12

12

Gle

ncoe

Pre

calc

ulus

Com

pute

Lim

its

at In

fini

ty

Lim

its

of P

ower

F

unct

ions

at

Infi

nity

Lim

its

of P

olyn

omia

ls a

t In

fini

tyL

imit

s of

Rec

ipro

cal

Fun

ctio

ns a

t In

fini

ty

For

any

posi

tive

inte

ger

n,•

lim

x

→ ∞

xn =

∞.

•

lim

x →

-∞

xn =

∞ if

n is

eve

n.

•

lim

x →

-∞

xn =

-∞

if n

is o

dd.

Let

p be

a p

olyn

omia

l fu

ncti

onp(

x) =

anx

n + …

+ a

1x +

a0.

Then

lim

x

→ ∞

p(x

) = lim

x

→ ∞

anx

n

and

lim

x →

-∞

p(x

) =

lim

x →

-∞

anx

n .

For

any

posi

tive

inte

ger

n,

lim

x →

±∞

1 −

xn

= 0

.

E

valu

ate

each

lim

it.

a.

lim

x →

-∞

(x5 -

6x

+ 1

)

lim

x →

-∞

(x5 -

6x

+ 1

) =

lim

x →

-∞

x5

Lim

its o

f P

oly

no

mia

ls a

t In

fi nity

=

-∞

L

imits

of

Po

we

r F

un

ctio

ns

at

Infi n

ity

b. li

m

x

→ ∞

(2x

4 + 5

x2 )

lim

x

→ ∞

(2x4 +

5x2 )

= lim

x

→ ∞

2x4

Lim

its o

f P

oly

no

mia

ls a

t In

fi nity

=

2 lim

x

→ ∞

x4

Sca

lar

Mu

ltip

le P

rop

ert

y

=

2 ·

∞ =

∞

Lim

its o

f P

ow

er

Fu

nct

ion

s a

t In

fi nity

Exer

cise

s

Eva

luat

e ea

ch li

mit

.

1.

lim

x →

-∞

(-2x

3 + 5

x)

2. lim

x

→ ∞

5 −

x2

3. lim

x

→ ∞

6x -

1

−

10x

+ 7

∞

0

3

−

5

4. lim

x

→ ∞

6x2 -

2x

−

x3 + 1

5.

lim

x

→ ∞

5x4 +

2x3 -

1

−

2x

3 + x

2 - 1

6.

lim

x →

-∞

(3x3 +

5x

- 1

)

0

∞

-

∞

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

Eva

luati

ng

Lim

its

Alg

eb

raic

ally

12-2

Exam

ple

005_

036_

PC

CR

MC

12_8

9381

3.in

dd12

3/17

/09

11:3

6:49

AM


NA

ME

DA

TE

PE

RIO

D

Lesson 12-2

Ch

ap

ter

12

13

Gle

ncoe

Pre

calc

ulus

Eva

luat

e ea

ch li

mit

.

1.

lim

x

→ 3

(x2 +

3x

- 8

) 1

0

2.

lim

x →

-6 x2 -

36

−

x +

6

-1

2

3.

lim

x

→ 0

(3 +

x)2 -

9

−

x

6

4.

lim

x →

4 √

��

��

�

x2 - 2

x +

1

3

5.

lim

x

→ 1

x2 -

x

−

2x2 +

5x

- 7

1

−

9

6.

lim

x →

3

x2 −

2 +

√ �

��

x -

3

9

−

2

7.

lim

x →

∞ (2

- 6

x +

5x3 )

∞

8.

lim

x →

-∞

x5 - 8

x2 −

4x

5 + 3

x 1

−

4

9.

lim

x →

∞ 2x

3 - 4

x +

1

−

5x

4 - 2

x2

0

10.

lim

x →

-∞

(6x7 -

x2 )

-∞

11. B

OO

KS

Supp

ose

the

valu

e v

of a

boo

k in

dol

lars

aft

er t

year

s ca

n be

repr

esen

ted

as v

(t) =

30

0 −

6

+ 3

5(0.

2)t .

How

muc

h w

ill t

he b

ook

even

tual

ly

be w

orth

? Th

at is

, fin

d th

e lim

t →

∞ v

(t).

$5

0

12. M

EDIC

INE

Eac

h da

y, T

amek

a ta

kes

2 m

illig

ram

s of

he

r as

thm

a m

edic

ine.

The

gra

ph s

how

s th

e am

ount

of

med

icin

e m

left

in h

er b

lood

str

eam

aft

er d

day

s.

Find

the

lim

d

→ 3

- m

(d) a

nd lim

d

→ 3

+ m

(d).

2 m

g,

4 m

g

12-2

Prac

tice

Eva

luati

ng

Lim

its

Alg

eb

raic

ally

Tameka’s AsthmaMedicine (mg)

46 2

m

d

0810

Days

32

15

46

005_

036_

PC

CR

MC

12_8

9381

3.in

dd13

3/17

/09

11:3

6:54

AM


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.



Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

12

14

Gle

ncoe

Pre

calc

ulus

1. P

OO

LS A

poo

l con

tain

s 75

,000

lite

rs o

f pu

re w

ater

. A m

ixtu

re t

hat

cont

ains

0.

3 gr

am o

f chl

orin

e pe

r lit

er o

f wat

er is

pu

mpe

d in

to t

he p

ool a

t a

rate

of

75 li

ters

per

min

ute.

The

con

cent

rati

on

C o

f chl

orin

e in

gra

ms

per

liter

t m

inut

es la

ter

in t

he p

ool i

s gi

ven

by

C(t

) =

0.3t

−

1000

+ t .

Find

lim

t →

∞ C

(t).

0

.3 g

/L

2. P

OLL

UTA

NTS

As

a by

-pro

duct

of o

ne o

f it

s pr

oces

ses,

a m

anuf

actu

ring

com

pany

cr

eate

s an

air

born

e po

lluta

nt. T

he c

ost

C

of r

emov

ing

p% o

f the

pol

luta

nt is

C =

60,0

00p

−

100

- p

, 0

≤ p

≤ 1

00.

Find

lim

p →

100

- C

.

∞

3. M

OTO

RCY

CLES

Flin

t bo

ught

a n

ew

mot

orcy

cle

for

$24,

000.

Sup

pose

the

va

lue

v of

his

mot

orcy

cle,

in t

hous

ands

of

dolla

rs, a

fter

t ye

ars

can

be r

epre

sent

ed

by t

he e

quat

ion

v(t)

= 2

4(0.

98)t .

a. C

ompl

ete

the

tabl

e. R

ound

ans

wer

s to

th

e ne

ares

t hu

ndre

dth.

Ye

ars

15

10

20

Va

lue

23

.52

21

.69

19

.61

16

.02

b. F

ind

lim

t →

∞ v

(t).

0

4. C

AR

SAFE

TY W

hile

dri

ving

a c

ar,

it is

impo

rtan

t to

mai

ntai

n a

safe

di

stan

ce b

etw

een

you

and

the

car

in

fron

t of

you

. Sup

pose

the

func

tion

y(

x) =

0.0

05x2 +

0.3

x +

3, w

here

x is

the

sp

eed

in m

iles

per

hour

, giv

es t

he

reco

mm

ende

d sa

fe d

ista

nce,

in y

ards

, be

twee

n yo

ur c

ar a

nd t

he o

ne in

fron

t of

yo

u. F

ind

lim

x

→ 7

0 y(x

).

4

8.5

yd

5. P

ART

S Th

e co

st c

of p

rodu

cing

a c

erta

in

smal

l eng

ine

part

in d

olla

rs is

giv

en b

y th

e eq

uati

on c

(p) =

300

0 +

20p

, whe

re

p is

the

num

ber

of p

arts

pro

duce

d.

a. F

ind

the

cost

of p

rodu

cing

100

par

ts.

$5

00

0

b. O

n Tu

esda

y, t

he c

ompa

ny p

rodu

ced

$21,

000

wor

th o

f par

ts. H

ow m

any

part

s di

d th

e co

mpa

ny p

rodu

ce?

90

0

c. T

he a

vera

ge c

ost

per

part

is fo

und

by

divi

ding

c(p

) by

p.

Find

lim

p →

15,

000 c(

p)

−

p .

$2

0.2

0

6. M

ICRO

WA

VES

The

func

tion

f(x) =

250x

+ 2

00,0

00

−

x

mod

els

the

aver

age

cost

of a

mic

row

ave

f (x)

m

anuf

actu

red

by a

com

pany

tha

t m

akes

x

mic

row

aves

for

prof

essi

onal

kit

chen

s.

Find

lim

x →

200

0 f (x

).

3

50

12-2

Wor

d Pr

oble

m P

ract

ice

Eva

luati

ng

Lim

its

Alg

eb

raic

ally

005_

036_

PC

CR

MC

12_8

9381

3.in

dd14

3/17

/09

11:3

6:58

AM


NA

ME

DA

TE

PE

RIO

D

Lesson 12-2

Ch

ap

ter

12

15

Gle

ncoe

Pre

calc

ulus

Th

e S

qu

eeze

Th

eo

rem

In L

esso

n 12

-1, y

ou le

arne

d th

at t

he lim

x

→ 0

sin

1 −

x doe

s no

t ex

ist

beca

use

as x

get

s cl

oser

to

0, t

he c

orre

spon

ding

func

tion

val

ues

osci

llate

bet

wee

n

-1

and

1. B

ut w

hat

abou

t th

e lim

x

→ 0

x2 s

in 1 −

x ? D

oes

this

lim

it n

ot e

xist

sim

ply

beca

use

lim

x

→ 0

sin

1 −

x doe

s no

t ex

ist?

A t

heor

em k

now

n as

the

Squ

eeze

The

orem

will

hel

p yo

u an

swer

thi

s qu

esti

on a

lgeb

raic

ally

.

The

Squ

eeze

The

orem

If h

(x) ≤

f (x

) ≤ g

(x) f

or a

ll x

in a

n op

en in

terv

al c

onta

inin

g c,

exc

ept

poss

ibly

at

c it

self,

and

if lim

x

→ c h

(x) =

L =

lim

x

→ c g

(x) ,

the

n lim

x

→ c f (

x) e

xist

s an

d is

equ

al t

o L.

Firs

t, no

te t

hat

-1

≤ s

in 1 −

x ≤ 1

for

all x

, exc

ept

for

x =

0. N

ext,

mul

tipl

y

this

ineq

ualit

y by

x2 ,

obta

inin

g -

x2 ≤ x

2 sin

1 −

x ≤ x

2 . To

app

ly t

he S

quee

ze

Theo

rem

, let

h(x

) = -

x2 , f(x

) = x

2 sin

1 −

x , an

d g(

x) =

x2 .

From

Les

son

12-2

,

you

lear

ned

that

lim

x

→ 0

h(x

) and

lim

x

→ 0

g(x

) bot

h eq

ual 0

. You

can

now

app

ly

the

Sque

eze

Theo

rem

wit

h c

= 0

. The

res

ult

is t

hat

lim

x

→ 0

x2 s

in 1 −

x = 0

.

Exer

cise

s

Use

the

Squ

eeze

The

orem

to

find

eac

h li

mit

.

1. lim

x

→ 0

x4 s

in 1 −

x 0

2.

lim

x →

4 ⎪

x⎥ √

x +

2

−

x +

4

2

3. lim

x

→ 0

x2 c

os

1 −

3 √

x

0

4. lim

x

→ 0

⎪x⎥

sin

π

−

x 0

5. lim

x

→ 0

tan

x −

x

1

6.

lim

x

→ 0

sin

3x

−

x

3

Enri

chm

ent

12-2

005_

036_

PC

CR

MC

12_8

9381

3.in

dd15

3/17

/09

11:3

7:02

AM


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s


Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

12

16

Gle

ncoe

Pre

calc

ulus

Tang

ent

Line

sIn

stan

tane

ous

Rat

e of

Cha

nge

The

inst

anta

neou

s ra

te o

f cha

nge

of t

he g

raph

of f

(x) a

t th

e po

int

(x, f

(x))

is t

he s

lope

m o

f

the

tang

ent

line

give

n by

m =

lim

h

→ 0

f (x

+ h

) - f

(x)

−

h

, pr

ovid

ed t

he li

mit

exi

sts.

F

ind

an e

quat

ion

for

the

slop

e of

the

gra

ph o

f y

= 3

x2 + 1

at

any

poin

t.

m =

lim

h

→ 0

f (

x +

h) -

f (x

)

−

h

Inst

an

tan

eo

us

Ra

te o

f C

ha

ng

e F

orm

ula

m =

lim

h

→ 0

[3

(x +

h)2 +

1] -

[3 x 2 +

1]

−−

h

f(x +

h)

= 3

(x +

h)2

+ 1

an

d f(

x) =

3x2

+ 1

m =

lim

h

→ 0

[3

x 2 + 6

hx +

3 h 2 +

1] -

[3 x

2 + 1

]

−

−

h

E

xpa

nd

an

d s

imp

lify.

m =

lim

h

→ 0

3h

(2x

+ h

) −

h

S

imp

lify

an

d f

act

or.

m =

lim

h

→ 0

3(2

x +

h)

Re

du

ce h

.

m =

6x

Sca

lar

Mu

ltip

le,

Su

m P

rop

ert

y, a

nd

Lim

it o

f a

Co

nst

an

t F

un

ctio

n P

rop

ert

y o

f L

imits

An

equa

tion

for

the

slop

e of

the

gra

ph a

t an

y po

int

is m

= 6

x.

Exer

cise

s

Fin

d an

equ

atio

n fo

r th

e sl

ope

of t

he g

raph

of

each

fun

ctio

n at

an

y po

int.

1. y

= x

3 + 1

m

= 3

x2

2. y

= 4

- 7

x m

= -

7

3. y

= 3 −

x2

m =

− 6

−

x3

4.

y =

4

−

√ �

x m

= −

2

−

x √

�

x

12-3

Stud

y Gu

ide

and

Inte

rven

tion

Tan

gen

t Lin

es

an

d V

elo

cit

y

Exam

ple

005_

036_

PC

CR

MC

12_8

9381

3.in

dd16

3/17

/09

11:3

7:06

AM


NA

ME

DA

TE

PE

RIO

D

Lesson 12-3

Ch

ap

ter

12

17

Gle

ncoe

Pre

calc

ulus

Inst

anta

neou

s V

eloc

ity

Inst

anta

neou

s V

eloc

ity

If t

he d

ista

nce

an o

bjec

t tr

avel

s is

giv

en a

s a

func

tion

of t

ime

f (t)

, the

n th

e in

stan

tane

ous

velo

city

v(t

) at

a ti

me

t is

give

n by

v(t

) = lim

h

→ 0

f (t +

h) -

f (t

)

−

h ,

prov

ided

the

lim

it e

xist

s.

A r

ock

is d

ropp

ed f

rom

150

0 fe

et a

bove

the

bas

e of

a r

avin

e. T

he h

eigh

t of

the

roc

k af

ter

t se

cond

s is

giv

en b

y h(

t) =

150

0 -

16t

2 . F

ind

the

inst

anta

neou

s ve

loci

ty v

(t)

of t

he

rock

at

4 se

cond

s.

v(t)

= lim

h

→ 0

f (t +

h) -

f (t

)

−

h

Inst

an

tan

eo

us

Ve

loci

ty F

orm

ula

= lim

h

→ 0

[150

0 -

16(

4 +

h ) 2 ] -

[150

0 -

16(

4)2 ]

−−

−

h

f(t

+ h

) =

15

00

- 1

6(4

+ h

)2 a

nd

f(t)

= 1

50

0 -

16

(4)2

= lim

h

→ 0

-12

8h -

16 h

2

−

h

Mu

ltip

ly a

nd

sim

plif

y.

= lim

h

→ 0

h(-

128

- 1

6h)

−

h

F

act

or.

= lim

h

→ 0

(-12

8 -

16h

) D

ivid

e b

y h

an

d s

imp

lify.

= -

128

- 1

6(0)

or

-12

8 Di

ffe

ren

ce P

rop

ert

y o

f L

imits

an

d L

imit

of

Co

nst

an

t

an

d I

de

ntit

y F

un

ctio

ns

The

inst

anta

neou

s ve

loci

ty o

f the

roc

k at

4 s

econ

ds is

128

feet

per

sec

ond.

Exer

cise

s

The

dis

tanc

e d

an o

bjec

t is

abo

ve t

he g

roun

d t

seco

nds

afte

r it

is

drop

ped

is g

iven

by

d(t)

. Fin

d th

e in

stan

tane

ous

velo

city

of

the

obje

ct a

t th

e gi

ven

valu

e fo

r t.

1. d

(t) =

800

- 1

6t2 ;

t = 3

2.

d(t

) = -

16t2 +

170

0; t

= 5

v(

3)

= -

96

ft/

s

v(

5)

= -

16

0 f

t/s

3. d

(t) =

70t

- 1

6t2 ;

t = 1

4.

d(t

) = -

16t2 +

90t

+ 1

0; t

= 2

v(

1)

= 3

8 f

t/s

v(2

) =

26

12-3

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

Tan

gen

t Lin

es

an

d V

elo

cit

y

Exam

ple

005_

036_

PC

CR

MC

12_8

9381

3.in

dd17

12/5

/09

5:03

:13

PM


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.



Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

12

18

Gle

ncoe

Pre

calc

ulus

Fin

d th

e sl

ope

of t

he li

ne t

ange

nt t

o th

e gr

aph

of e

ach

func

tion

at

the

give

n po

int.

1. y

= x

2 - x

; (3,

6)

5

2. y

= 5 −

x ; (-

1, -

5)

-5

Fin

d an

equ

atio

n fo

r th

e sl

ope

of t

he g

raph

of

each

fun

ctio

n at

any

poi

nt.

3. y

= -

2x +

1

m =

-2

4.

y =

x3 -

2x2

m =

3x2

- 4

x

The

dis

tanc

e d

an o

bjec

t is

abo

ve t

he g

roun

d t

seco

nds

afte

r it

is

drop

ped

is g

iven

by

d(t)

. Fin

d th

e in

stan

tane

ous

velo

city

of

the

obje

ct a

t th

e gi

ven

valu

e fo

r t.

5. d

(t) =

300

- 1

6t2 ;

t = 2

6.

d(t

) = -

16t2 +

200

t + 7

00; t

= 3

v(2

) =

-6

4 f

t/s

v(

3)

= 1

04

ft/

s

Fin

d an

equ

atio

n fo

r th

e in

stan

tane

ous

velo

city

v(t

) if

the

pat

h of

an

obje

ct is

def

ined

as

s(t)

for

any

poi

nt in

tim

e t.

7. s

(t) =

17t

2 + 8

8.

s(t

) = 5

t3 - 6

t2 + 4

t + 1

v(t)

= 3

4t

v(

t) =

15

t2 -

12

t +

4

9. s

(t) =

√ � t -

2t2

10.

s(t)

= 3 −

t + 2

t

v(t)

= √

� t −

2t

- 4

t v(

t) =

- 3

−

t2

+ 2

11. S

KY

DIV

ING

The

pos

itio

n h

in fe

et o

f a s

ky d

iver

rel

ativ

e to

the

gro

und

can

be d

efin

ed b

y h(

t) =

18,

000

- 1

6t2 ,

whe

re t

ime

t is

seco

nds

pass

ed

afte

r th

e sk

y di

ver

exit

ed t

he p

lane

. Fin

d an

exp

ress

ion

for

the

inst

anta

neou

s ve

loci

ty v

(t) o

f the

sky

div

er.

v(t)

= -

32

t

12. F

OO

TBA

LL A

qua

rter

back

thr

ows

a fo

otba

ll w

ith

a ve

loci

ty o

f 58

feet

pe

r se

cond

tow

ard

a te

amm

ate.

Sup

pose

the

hei

ght

h of

the

foot

ball,

in

feet

, t s

econ

ds a

fter

he

thro

ws

it is

def

ined

as

h(t)

= -

16t2 +

58t

+ 6

.

a. F

ind

an e

xpre

ssio

n fo

r th

e in

stan

tane

ous

velo

city

v(t

) of t

he fo

otba

ll.

v(t)

= -

32

t +

58

b. H

ow fa

st is

the

foot

ball

trav

elin

g af

ter

1.5

seco

nds?

10

ft/

s

12-3

Prac

tice

Tan

gen

t Lin

es

an

d V

elo

cit

y

005_

036_

PC

CR

MC

12_8

9381

3.in

dd18

12/5

/09

5:10

:23

PM


NA

ME

DA

TE

PE

RIO

D

Lesson 12-3

Ch

ap

ter

12

19

Gle

ncoe

Pre

calc

ulus

1. F

ALL

ING

OBJ

ECT

Mir

anda

dro

ps a

bal

l fr

om a

tow

er t

hat

is 8

00 fe

et h

igh.

The

po

siti

on o

f the

bal

l aft

er t

seco

nds

is

give

n by

s(t

) = -

16t2 +

800

. How

fast

is

the

ball

falli

ng a

fter

2 s

econ

ds?

6

4 f

t/s

2. P

ROJE

CTIL

E Ti

to d

rops

a r

ock

from

12

00 fe

et. T

he p

osit

ion

of t

he r

ock

afte

r t s

econ

ds is

giv

en b

y s(

t) =

-16

t2 + 1

200.

a. H

ow fa

st is

the

bal

l fal

ling

afte

r 4

seco

nds?

1

28

ft/

s

b. W

hen

will

the

roc

k hi

t th

e gr

ound

?

5 √

�

3 s

c. F

ind

an e

xpre

ssio

n fo

r th

e in

stan

tane

ous

velo

city

v(t

)

of t

he r

ock.

v(

t) =

−3

2t

d. W

hat

is t

he v

eloc

ity

of t

he r

ock

whe

n it

hit

s th

e gr

ound

?

16

0 √

�

3 f

t/s

3. B

UN

GEE

JU

MPI

NG

A b

unge

e ju

mpe

r’s

heig

ht h

in fe

et r

elat

ive

to t

he g

roun

d in

t s

econ

ds is

giv

en b

y h(

t) =

900

- 1

6t2 .

Find

an

expr

essi

on fo

r th

e in

stan

tane

ous

velo

city

v(t

) of t

he ju

mpe

r.

v(t)

= -

32t

4. F

REE

FALL

ING

The

pos

itio

n h

in fe

et

of a

free

-falli

ng s

ky d

iver

rel

ativ

e to

th

e gr

ound

can

be

defin

ed b

y h(

t) =

15,

000

- 1

6t2 ,

whe

re t

is

seco

nds

pass

ed a

fter

the

sky

div

er

exit

ed t

he p

lane

.

a. F

ind

an e

xpre

ssio

n fo

r th

e in

stan

tane

ous

velo

city

v(t

) of

the

sky

div

er.

v(t)

= -

32

t

b. W

hat

is t

he s

ky d

iver

’s he

ight

af

ter

2 se

cond

s?

14

,93

6 f

t

c. W

hat

is t

he s

ky d

iver

’s ve

loci

ty

afte

r 4

seco

nds?

1

28

ft/

s

5. B

ASE

BALL

An

outf

ield

er t

hrow

s a

ball

tow

ard

hom

e pl

ate

wit

h an

init

ial

velo

city

of 8

0 fe

et p

er s

econ

d. S

uppo

se

the

heig

ht h

of t

he b

aseb

all,

in fe

et, t

se

cond

s af

ter

the

ball

is t

hrow

n is

m

odel

ed b

y h(

t) =

-16

t2 + 8

0t +

6.5

.

a. F

ind

an e

xpre

ssio

n fo

r th

e in

stan

tane

ous

velo

city

v(t

) of

the

base

ball.

v(t)

= -

32

t +

80

b. H

ow fa

st is

the

bas

ebal

l tra

velin

g af

ter

0.5

seco

nd?

64

ft/

s

c. F

or w

hat

valu

e of

t w

ill t

he b

aseb

all

reac

h it

s m

axim

um h

eigh

t?

2.5

s

d. W

hat

is t

he m

axim

um h

eigh

t of

the

bas

ebal

l?

10

6.5

ft

6. A

REA

Sup

pose

the

leng

th x

of e

ach

side

of

the

squ

are

show

n is

cha

ngin

g.

a. F

ind

the

aver

age

rate

of c

hang

e of

the

ar

ea a

(x) a

s x

chan

ges

from

5.

4 in

ches

to

5.6

inch

es.

11

sq

in

. p

er

in.

b. F

ind

the

inst

anta

neou

s ra

te o

f cha

nge

of t

he a

rea

at t

he m

omen

tx

= 5

inch

es.

10

sq

in

. p

er

in.

12-3

Wor

d Pr

oble

m P

ract

ice

Tan

gen

t Lin

es

an

d V

elo

cit

y

x

xa(

x)=

x 2

005_

036_

PC

CR

MC

12_8

9381

3.in

dd19

3/17

/09

11:3

7:17

AM


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s


Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

12

20

Gle

ncoe

Pre

calc

ulus

Tan

gen

ts a

nd

Vert

ices

Can

you

use

the

equ

atio

n fo

r th

e sl

ope

of a

func

tion

at

any

poin

t to

find

the

ve

rtex

of a

par

abol

a of

the

form

y =

ax2 +

bx

+ c

? St

ep 1

Gra

ph a

par

abol

a of

the

form

y =

ax2 +

bx

+ c

, one

whe

re a

> 0

and

on

e w

here

a <

0. T

hen

draw

the

line

tan

gent

to

the

vert

ex o

f eac

h pa

rabo

la a

nd p

lace

the

equ

atio

n of

eac

h pa

rabo

la b

elow

its

grap

h.

Step

2 F

ind

the

slop

e of

the

tan

gent

line

to

each

gra

ph a

t th

e ve

rtex

.

Th

e s

lop

e o

f e

ac

h t

an

ge

nt

lin

e i

s 0

.

Step

3 F

ind

the

equa

tion

for

the

slop

e of

eac

h gr

aph

at a

ny p

oint

.

Fo

r y

= x

2 +

2x

- 3

, m

= 2

x +

2.

Fo

r y

= -

x2 -

4x,

m =

-2x

- 4

.

Step

4 S

et e

ach

equa

tion

for

the

slop

e eq

ual t

o ze

ro a

nd s

olve

for

x.

This

giv

es t

he x

-val

ue o

f the

ver

tex.

2x

+ 2

= 0

-2

x -

4 =

0

2x

= -

2

-

2x

= 4

x =

-1

x =

-2

Step

5 S

ubst

itut

e ea

ch x

-val

ue in

to it

s re

spec

tive

equ

atio

n to

find

the

y-

valu

e of

the

ver

tex.

y =

x2 +

2x

- 3

; x

= -

1

y =

-x2

- 4

x; x

= -

2

y =

(-

1)2

+ 2

(-1

) -

3

y =

-(-

2)2

- 4

(-2

)

y =

-4

y =

4

Step

6 W

rite

eac

h ve

rtex

as

an o

rder

ed p

air.

Fo

r y

= x

2 +

2x

- 3

, (-

1,

-4

). F

or

y =

-x2

- 4

x, (

-2

, 4

).

Exer

cise

s

Fin

d th

e ve

rtex

of

each

par

abol

a.

1. y

= 2

x2 - 4

x +

2

(1,

0)

2. y

= -

x2 - 6

x -

9

(−3

, 0

)

3. y

= 4

x2 - 2

5 (0

, −

25

) 4.

y =

1 −

2 x2 -

2x

+ 4

(2

, 2

)

12-3

Enri

chm

ent

y

x

y= -x2

-4x

1

1

y

x

y=x2

+2x

- 3

4

4

005_

036_

PC

CR

MC

12_8

9381

3.in

dd20

3/17

/09

11:3

7:20

AM


NA

ME

DA

TE

PE

RIO

D

Lesson 12-3

Ch

ap

ter

12

21

Gle

ncoe

Pre

calc

ulus

Usi

ng

th

e T

an

gen

t Lin

e t

o A

pp

roxi

mate

a F

un

cti

on

You

hav

e le

arne

d ho

w t

o fin

d th

e sl

ope

of a

tan

gent

line

to

a fu

ncti

on a

t a

poin

t. Th

is s

lope

can

the

n be

use

d to

wri

te t

he e

quat

ion

of t

he t

ange

nt li

ne

in p

oint

-slo

pe fo

rm. T

he t

ange

nt li

ne o

f a fu

ncti

on is

usu

ally

a g

ood

appr

oxim

atio

n of

the

func

tion

for

valu

es o

f x n

ear

the

coor

dina

te o

f the

ch

osen

poi

nt. S

et u

p a

spre

adsh

eet

like

the

one

show

n be

low

to

stud

y th

is

met

hod

of a

ppro

xim

atin

g a

func

tion

.

AB

CD

EF

GH

1x

0.7

0.8

0.9

11

.11

.21

.3

2x^

20

.49

0.6

40

.81

11

.21

1.4

41

.69

32

*x

- 1

0.4

0.6

0.8

11

.21

.41

.6

The

equa

tion

in p

oint

-slo

pe fo

rm o

f the

tan

gent

line

to

f (x)

= x

2 at

(1, 1

) is

y =

2x

- 1

. In

the

spre

adsh

eet,

the

func

tion

f (x

) = x

2 is

appr

oxim

ated

for

valu

es o

f x n

ear

x =

1 b

y us

ing

the

func

tion

g(x

) = 2

x -

1. F

or t

his

exam

ple,

th

e va

lues

for

x ar

e en

tere

d in

row

1, c

olum

ns B

–H. T

he fu

ncti

on f

(x) =

x2 i

s en

tere

d in

cel

l B2

as =

B1^

2 an

d co

pied

to

the

othe

r ce

lls in

row

2. T

he

appr

oxim

atin

g fu

ncti

on, g

(x) =

2x

- 1

, is

ente

red

in c

ell B

3 as

=2*

B1

- 1

and

co

pied

to

the

othe

r ce

lls in

row

3. N

otic

e th

at t

he v

alue

s of

g(x

) = 2

x -

1 in

ro

w 3

are

rem

arka

bly

clos

e to

the

val

ues

of f

(x) =

x2 i

n ro

w 2

.

Exer

cise

s

1. U

se t

he s

prea

dshe

et t

o ap

prox

imat

e th

e va

lues

of f

(x) =

x2 n

ear

the

poin

t, w

here

x =

2, b

y us

ing

the

appr

opri

ate

tang

ent

line.

Com

pare

the

val

ues

for

x =

1.7

, 1.8

, 1.9

, 2, 2

.1, 2

.2, a

nd 2

.3. W

hat

is t

he m

axim

um e

rror

th

at o

ccur

s?

T

he

va

lue

s o

f ƒ

(x)

= x

2 a

re 2

.89

, 3

.24

, 3

.61

, 4

, 4

.41

, 4

.84

, a

nd

5.2

9.

Th

e

va

lue

s o

f g

(x)

= 4

x -

4 a

re 2

.8,

3.2

, 3

.6,

4,

4.4

, 4

.8,

an

d 5

.2.

Th

e m

ax

imu

m

err

or

is 0

.09

at

the

po

ints

wh

ere

x =

1.7

an

d 2

.3.

2. U

se t

he s

prea

dshe

et t

o ap

prox

imat

e th

e va

lues

of ƒ

(x) =

x2 n

ear

the

poin

t, w

here

x =

0, b

y us

ing

the

appr

opri

ate

tang

ent

line.

Com

pare

the

val

ues

for

x =

-0.

3, -

0.2,

-0.

1, 0

, 0.1

, 0.2

, and

0.3

. Wha

t is

the

max

imum

err

or

that

occ

urs?

T

he

va

lue

s o

f ƒ

(x)

= x

2 a

re 0

.09

, 0

.04

, 0

.01

, 0

, 0

.01

, 0

.04

, a

nd

0.0

9.

Th

e

va

lue

s o

f g

(x)

= 0

are

0,

0,

0,

0,

0,

0,

an

d 0

. T

he

ma

xim

um

err

or

is 0

.09

at

the

po

ints

wh

ere

x =

-0

.3 a

nd

0.3

.

3. U

se t

he s

prea

dshe

et t

o ap

prox

imat

e th

e va

lues

of ƒ

(x) =

√ �

x fo

r x

= 0

.7, 0

.8, 0

.9, 1

, 1.1

, 1.2

, and

1.3

. Use

the

dat

a to

mak

e a

conj

ectu

re

abou

t th

e eq

uati

on o

f the

tan

gent

line

to

the

grap

h of

the

func

tion

at

the

poin

t, w

here

x =

1.

T

he

va

lue

s o

f ƒ

(x)

= √

�

x a

re a

pp

rox

ima

tely

0.8

37

, 0

.89

4,

0.9

49

, 1

, 1

.04

9,

1.0

95

, a

nd

1.1

40

. T

he

ex

ac

t e

qu

ati

on

of

the

ta

ng

en

t li

ne

is

y =

0.5

x +

0.5

.

12-3

Spre

adsh

eet A

ctiv

ity

005_

036_

PC

CR

MC

12_8

9381

3.in

dd21

3/17

/09

11:3

7:25

AM


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.



Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

12

22

Gle

ncoe

Pre

calc

ulus

Der

ivat

ives

and

the

Bas

ic R

ules

Lim

its

wer

e us

ed in

Les

son

12-3

to

dete

rmin

e th

e sl

ope

of a

line

tan

gent

to

the

grap

h of

a fu

ncti

on a

t an

y po

int.

This

lim

it is

cal

led

the

deri

vati

ve o

f a fu

ncti

on. T

he d

eriv

ativ

e of

f (x

) is

f ′(x)

, whi

ch is

giv

en b

y f ′(

x) =

lim

h

→ 0

f(x +

h) -

f(x)

−

h

, pr

ovid

ed t

he li

mit

exi

sts.

Ther

e ar

e se

vera

l rul

es o

f der

ivat

ives

tha

t ar

e us

eful

whe

n fin

ding

the

de

riva

tive

s of

func

tion

s th

at c

onta

in s

ever

al t

erm

s.

Po

we

r R

ule

If f(

x) =

xn

an

d n

is a

re

al n

um

be

r, t

he

n

f ′(x)

= n

xn −

1.

If f(

x) =

x3,

the

n f

′(x)

= 3

x2.

Co

ns

tan

tT

he

de

riva

tive

of

a c

on

sta

nt

fun

ctio

n is

ze

ro.

If f(

x) =

c,

the

n f

′(x)

= 0

.

If f(

x) =

-2

, th

en

f ′(x

) =

0.

Co

ns

tan

t M

ult

iple

of

a P

ow

er

If f(

x) =

cxn ,

wh

ere

c is

a c

on

sta

nt

an

d n

is

a r

ea

l nu

mb

er,

th

en

f ′(x

) =

cnx

n -

1.

If f(

x) =

5x3

, th

en

f ′(x

) =

15

x2.

Su

m a

nd

Dif

fere

nc

eIf f(

x) =

g(x

) ±

h(x

), t

hen f

′(x)

= g

′(x)

± h

′(x).

If f(

x) =

4x2

+ 3

x, t

he

n f

′(x)

= 8

x +

3.

F

ind

the

deri

vati

ve o

f ea

ch f

unct

ion.

a. f

(x)

= 3

x2 - 2

x +

4

f (x)

= 3

x2 - 2

x +

4

Orig

ina

l eq

ua

tion

f ′(x)

= 2

· 3x

2 -

1 -

2 ·

1x1

- 1 +

0

Co

nst

an

t, C

on

sta

nt

Mu

ltip

le o

f a

Po

we

r, a

nd

Su

m a

nd

Diff

ere

nce

Ru

les

=

6x

- 2

S

imp

lify.

b. f

(x)

= x

4 (4x

3 - 5

)

f (x)

= x

4 (4x3 -

5)

Orig

ina

l eq

ua

tion

f (x)

= 4

x7 - 5

x4 D

istr

ibu

tive

Pro

pe

rty

f ′(x)

= 4

· 7x

7 -

1 -

5 ·

4x4

- 1

Co

nst

an

t M

ulti

ple

of

a P

ow

er,

an

d S

um

an

d D

iffe

ren

ce R

ule

s

=

28x

6 - 2

0x3

Sim

plif

y.

Exer

cise

s

Fin

d th

e de

riva

tive

of

f(x)

. The

n ev

alua

te t

he d

eriv

ativ

e fo

r th

e gi

ven

valu

es o

f x.

1. f

(x) =

4x2 -

5; x

= 3

and

-2

2. f

(x) =

-x3 +

5x2 ;

x =

1 a

nd -

4

f ′

(x)

= 8

x; 2

4,

-1

6

f ′

(x)

= -

3x2

+ 1

0x;

7,

-8

8

3. f

(x) =

-8

+ 3

x -

x2 ;

0 an

d -

3 4.

f (x

) = 3

x4 + x

5 -2;

-1

and

2

f ′

(x)

= 3

- 2

x; 3

, 9

f ′(x

) =

12

x3 +

5x4

; -

7,

17

6

Fin

d th

e de

riva

tive

of

each

fun

ctio

n.

5. f

(x) =

6x2 -

3x

+ 4

f ′

(x)

= 1

2x

- 3

6.

f (x

) = -

x3.4 +

3x0.

2

7. f

(x) =

4 x 1

− 2 -

3 x 3

− 2

f ′(x

) =

2 √

�

x

−

x -

9 √

�

x

−

2

8.

f (x

) = -

4x2 +

3x3 -

14

f ′(x

) =

-8x

+ 9

x2

12-4

Stud

y Gu

ide

and

Inte

rven

tion

Deri

vati

ves

Exam

ple

f ′(x

) =

-3.4

x2.4 +

0.6

x-0.8

005_

036_

PC

CR

MC

12_8

9381

3.in

dd22

3/17

/09

11:3

7:29

AM


NA

ME

DA

TE

PE

RIO

D

Lesson 12-4

Ch

ap

ter

12

23

Gle

ncoe

Pre

calc

ulus

Prod

uct

and

Quo

tien

t Ru

les

Use

the

follo

win

g ru

les

to fi

nd t

he

deri

vati

ve o

f the

pro

duct

or

quot

ient

of t

wo

func

tion

s.

Pro

du

ct

Ru

leIf

f a

nd

g a

re d

iffe

ren

tiab

le a

t x,

th

en

d −

dx

[f (x

)g(x

) ] =

f ′(x

)g(x

) +

f (x

)g ′(

x).

Qu

oti

en

t R

ule

If f

an

d g

are

diff

ere

ntia

ble

at

x a

nd

g(x

) ≠

0,

the

n

d −

dx

⎡

⎢ ⎣ f(x

) −

g(

x) ⎤

� ⎦ =

f ′(x)

g(x)

- f

(x)g

′(x)

−

−

[g(x

) ] 2

.

F

ind

the

deri

vati

ve o

f h(

x) =

(x2 -

2)(

2x3 +

5x)

.

h ′(x

) = f

′(x)g

(x) +

f (x

)g ′(x

) P

rod

uct

Ru

le

=

(2x)

( 2x3 +

5x)

+ (x

2 - 2

)( 6x

2 + 5

) S

ub

stitu

tion

=

4x4 +

10x

2 + 6

x4 + 5

x2 - 1

2x2 -

10

Dis

trib

utiv

e P

rop

ert

y

=

10x

4 + 3

x2 - 1

0 S

imp

lify.

F

ind

the

deri

vati

ve o

f h(

x) =

(2x2 +

4)

−

(x2 -

1) .

h ′(x

) = f ′(

x)g(

x) -

f (x

)g ′(x

)

−−

[g(x

) ]2

Qu

otie

nt

Ru

le

=

4x(x

2 - 1

) - (2

x2 + 4

)2x

−

−

(2

x)2

S

ub

stitu

tion

=

4x3 -

4x

- 4

x3 - 8

x

−−

4x

2

Dis

trib

utiv

e P

rop

ert

y

=

- 3 −

x S

imp

lify.

Exer

cise

s

Fin

d th

e de

riva

tive

of

each

fun

ctio

n.

1. h

(x) =

(-4

+ 2

x2 )(2x

+ 3

) 12

x2 +

12x

- 8

2.

m(x

) = (3

x -

1)(x

2 + 5

x)

9x2

+ 2

8x

- 5

3. d

(x) =

x2 + 3

−

x -

1

x2 -

2x

- 3

−

(x -

1)2

4. k

(x) =

3x3 +

4

−

2x2 -

1

6x4

- 9

x2 -

16x

−

(2

x2 -

1)2

12-4

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

Deri

vati

ves

Exam

ple

1

f (x)

= x

2 - 2

O

rig

ina

l eq

ua

tion

f ′(x)

= 2

x Su

m R

ule

fo

r L

imits

, P

ow

er

an

d

Co

nst

an

t R

ule

s fo

r D

eriva

tive

s

g(x)

= 2

x3 + 5

x O

rig

ina

l eq

ua

tion

g ′(x

) = 6

x2 + 5

S

um

Ru

le f

or

Lim

its,

Po

we

r a

nd

Co

nst

an

t R

ule

s fo

r D

eriva

tive

s

Exam

ple

2

f (x)

= 2

x2 + 4

O

rig

ina

l eq

ua

tion

f ′(x)

= 4

x S

um

Ru

le f

or

Lim

its,

Po

we

r a

nd

Co

nst

an

t R

ule

s fo

r D

eriva

tive

s

g(x)

= x

2 - 1

O

rig

ina

l eq

ua

tion

g ′(x

) = 2

x S

um

Ru

le f

or

Lim

its,

Po

we

r a

nd

Co

nst

an

t R

ule

s fo

r D

eriva

tive

s

005_

036_

PC

CR

MC

12_8

9381

3.in

dd23

3/17

/09

11:3

7:33

AM


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s


Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

12

24

Gle

ncoe

Pre

calc

ulus

Fin

d th

e de

riva

tive

of

each

fun

ctio

n. T

hen

eval

uate

the

der

ivat

ive

of e

ach

func

tion

for

the

giv

en v

alue

s of

x.

1. g

(x) =

3x2 -

5x;

x =

-2

and

1 2.

h(x

) = 4

x3 - x

2 ; x

= 3

and

0

6x

- 5

; -

17

, 1

12x2

- 2

x; 1

02

, 0

3. f

(x) =

x2 -

4x

+ 7

; x =

2 a

nd -

3 4.

m(x

) = -

2x2 -

6x

+ 1

; x =

0 a

nd -

3

2x

- 4

; 0

, -

10

-4x

- 6

; -

6,

6

5. q

(x) =

-1

+ x

3 - 2

x4 ; x

= -

1 an

d 3

6. t

(x) =

3x7 -

1; x

= -

1 an

d 1

3x2

- 8

x3;

11

, -

18

9

2

1x6

; 2

1,

21

Fin

d th

e de

riva

tive

of

each

fun

ctio

n.

7.

f (x)

= (x

2 + 5

x)2

8. f

(x) =

x2 (x

3 + 3

x2 )

f'(x

) =

4x3

+ 3

0x2

+ 5

0x

f'

(x)

= 5

x4 +

12

x3

9. f

(x) =

5 √

�

x6

10.

h(x)

= -

3 −

x6

6

−

5 5

√ �

x

18

−

x7

11. p

(x) =

-4x

5 + 6

x3 - 5

x2 12

. n(

x) =

(3x2 -

2x)

(x3 +

x2 )

-2

0x4

+ 1

8x2

- 1

0x

1

5x4

+ 4

x3 -

6x2

13. r

(x) =

3x -

1

−

x2 + 2

14

. q(

x) =

√ �

x (x

2 - 3

)

-3x2

+ 2

x +

6

−

(x

2 +

2)2

5

−

2 x

3

−

2 -

3

−

2 x

- 1

−

2

15. P

HY

SICS

Acc

eler

atio

n is

the

rat

e at

whi

ch t

he v

eloc

ity

of a

mov

ing

obje

ct c

hang

es. T

he v

eloc

ity

in m

eter

s pe

r se

cond

of a

par

ticl

e m

ovin

g al

ong

a st

raig

ht li

ne is

giv

en b

y th

e fu

ncti

on v

(t) =

3t2 -

6t +

5, w

here

t i

s th

e ti

me

in s

econ

ds. F

ind

the

acce

lera

tion

of t

he p

arti

cle

in

met

ers

per

seco

nd s

quar

ed a

fter

5 s

econ

ds. (

Hin

t: A

ccel

erat

ion

is t

he

deri

vati

ve o

f vel

ocit

y.)

24

m/s

2

Prac

tice

Deri

vati

ves

12-4

005_

036_

PC

CR

MC

12_8

9381

3.in

dd24

12/5

/09

5:11

:50

PM


NA

ME

DA

TE

PE

RIO

D

Lesson 12-4

Ch

ap

ter

12

25

Gle

ncoe

Pre

calc

ulus

1. B

IRD

S Th

e he

ight

h, i

n fe

et, o

f a

flyin

g bi

rd c

an b

e de

fined

by

h(t)

= -

t3 −

3

+ 7 −

2 t2 + 1

8 on

the

inte

rval

[1

, 10]

, whe

re t

ime

t is

give

n in

sec

onds

. Fi

nd t

he m

axim

um a

nd m

inim

um

heig

ht o

f the

bir

d.

maxim

um

: 75 1

−

6 f

t, m

inim

um

: 21 1

−

6 f

t

2. C

LIFF

DIV

ING

At

tim

e t =

0, a

div

er

jum

ps fr

om a

clif

f 192

feet

abo

ve t

he

surf

ace

of t

he w

ater

. The

hei

ght

h of

th

e di

ver

is g

iven

by

h(t)

= -

16t2 +

16t

+ 1

92, w

here

h

is m

easu

red

in fe

et a

nd t

ime

t is

mea

sure

d in

sec

onds

.

a. F

ind

the

equa

tion

for

the

velo

city

h'(t

) of

the

div

er a

t an

y ti

me

t.

h'(

t) =

-3

2t

+ 1

6

b. F

ind

the

velo

city

of t

he d

iver

aft

er

1 se

cond

has

pas

sed.

h'(

1)

= -

16

ft/

s

c. F

ind

the

tim

e w

hen

the

dive

r hi

ts t

he w

ater

. t

= 4

s

d. W

hat

is t

he d

iver

’s ve

loci

ty w

hen

she

hits

the

wat

er?

h'(

4)

= -

11

2 f

t/s

3. G

EOM

ETRY

The

form

ula

to fi

nd t

he

volu

me

V o

f a c

ylin

der

in t

erm

s of

its

heig

ht h

and

rad

ius

r is

V =

πr2 h

. C

onsi

der

a cy

linde

r w

ith

a he

ight

of

10 in

ches

and

a c

hang

ing

radi

us w

hen

answ

erin

g th

e fo

llow

ing

ques

tion

s.a.

Wri

te a

form

ula

for

the

volu

me

of t

he

cylin

der

in t

erm

s of

its

radi

us.

V(r

) =

10

πr2

b. F

ind

an e

quat

ion

for

the

inst

anta

neou

s ra

te o

f cha

nge

of

the

volu

me

in t

erm

s it

s ra

dius

.

V′(

r) =

20

πr

c. F

ind

the

valu

e of

V ′(r

) whe

n r

= 3

inch

es.

60

π c

u i

n.

pe

r in

.

4. V

OLU

ME

Supp

ose

the

leng

th x

of e

ach

side

of t

he c

ube

show

n is

cha

ngin

g.

a. F

ind

the

aver

age

rate

of c

hang

e of

the

vo

lum

e V

(x) a

s x

chan

ges

from

3.

2 in

ches

to

3.4

inch

es.

32

.7 c

u i

n.

pe

r in

.

b. F

ind

the

inst

anta

neou

s ra

te o

f cha

nge

of t

he v

olum

e V

(x) a

t th

e m

omen

t x

= 4

inch

es.

48

cu

in

. p

er

in.

c. E

xpla

in t

he r

elat

ions

hip

betw

een

the

volu

me

form

ula

and

the

deri

vati

ve o

f th

e vo

lum

e fo

rmul

a.

Sam

ple

an

sw

er:

Th

e d

eri

vati

ve

o

f th

e f

orm

ula

fo

r th

e v

olu

me is

th

e f

orm

ula

fo

r o

ne h

alf

th

e

su

rface a

rea o

f th

e c

ub

e, o

r 3x2

. If

th

e v

olu

me o

f a c

ub

e is w

ritt

en

in

term

s o

f th

e a

po

them

s o

f it

s

faces (

8a

3),

th

e d

eri

vati

ve is t

he

fo

rmu

la f

or

the s

urf

ace a

rea o

f th

e c

ub

e (

24a

2).

5. P

ROJE

CTIL

E Su

ppos

e a

ball

is h

it

stra

ight

upw

ard

from

a h

eigh

t of

6 fe

et

wit

h an

init

ial v

eloc

ity

of 8

0 fe

et p

er

seco

nd. T

he h

eigh

t h

of t

he b

all i

n fe

et

at a

ny t

ime

t is

give

n by

the

func

tion

h(

t) =

-16

t2 + 8

0t +

6.

a. F

ind

the

equa

tion

for

the

velo

city

v(t

) of

the

bal

l at

any

tim

e t b

y fin

ding

the

de

riva

tive

of h

(t).

v(t)

= -

32

t +

80

b. F

ind

the

inst

anta

neou

s ve

loci

ty o

f the

ba

ll at

t =

2 s

econ

ds.

16

ft/

s

Wor

d Pr

oble

m P

ract

ice

Deri

vati

ves

12-4

=

005_

036_

PC

CR

MC

12_8

9381

3.in

dd25

12/7

/09

9:58

:06

AM


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.


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

Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

12

26

Gle

ncoe

Pre

calc

ulus

Po

werf

ul

Dif

fere

nti

ati

on

In C

hapt

er 1

0, t

he s

erie

s ex

pans

ions

of s

ome

tran

scen

dent

al fu

ncti

ons

wer

e pr

esen

ted.

In

part

icul

ar, t

he e

ven

func

tion

y =

cos

x, w

as s

how

n to

be

a su

m

of e

ven

pow

ers

of x

:

cos

x =

1 -

x2 −

2!

+ x4 −

4!

- x6 −

6!

+ x8 −

8!

- …

and

the

sine

func

tion

, bei

ng o

dd, w

as s

how

n to

be

a su

m o

f odd

pow

ers

of x

:

sin

x =

x -

x3 −

3!

+ x5 −

5!

- x7 −

7!

+ x9 −

9!

- …

.

The

pow

er fu

ncti

ons

in t

hese

ser

ies

expa

nsio

ns c

an b

e di

ffere

ntia

ted.

1. a

. Fi

nd d(

sin

x)

−

dx

by

diffe

rent

iati

ng t

he s

erie

s ex

pans

ion

of s

in x

ter

m b

y

term

and

sim

plify

ing

the

resu

lt.

1 -

x2

−

2

! + x4

−

4

! - x6

−

6

! + x8

−

8

! - …

b.

Wha

t fu

ncti

on d

oes

this

new

infin

ite

seri

es r

epre

sent

? c

os

x

c.

So,

d(si

n x)

−

dx

=

. c

os

x

2. a

. W

hat

wou

ld y

ou g

uess

mig

ht b

e th

e de

riva

tive

of c

os x

?

An

sw

ers

ma

y v

ary

.

b.

Fin

d d(

cos

x)

−

dx

usi

ng t

he s

erie

s ex

pans

ion

of c

os x

.

-x

+ x3

−

3

! - x5

−

5

! + x7

−

7

! - x9

−

9

! + …

c.

So,

d(co

s x)

−

dx

=

. -

sin

x

3. a

. Th

e se

ries

exp

ansi

on fo

r ex ,

ex = 1

+ x

+ x2 −

2!

+ x3 −

3!

+ x4 − 4!

+

…w

as a

lso

disc

usse

d in

Cha

pter

10.

Diff

eren

tiat

e th

e se

ries

ex

pans

ion

of e

x ter

m b

y te

rm a

nd s

impl

ify t

he r

esul

t.

1 +

x +

x2

−

2

! + x3

−

3

! + x4

−

4

! + …

b.

Thu

s, d(

ex ) −

dx

=

.

ex

Use

the

res

ults

of

Exe

rcis

es 1

– 3 t

o fi

nd t

he d

eriv

ativ

e of

eac

h fu

ncti

on.

4. f

(x) =

xex

5. f

(x) =

sin

x2

6. f

(x) =

(cos

x)2

xe

x +

ex

2x

co

s x

2

-

2 c

os

x s

in x

12-4

Enri

chm

ent

? ??

005_

036_

PC

CR

MC

12_8

9381

3.in

dd26

3/17

/09

11:3

7:45

AM


NA

ME

DA

TE

PE

RIO

D

Lesson 12-5

Ch

ap

ter

12

27

Gle

ncoe

Pre

calc

ulus

Are

a U

nder

a C

urve

You

can

use

the

are

a of

rec

tang

les

to fi

nd t

he a

rea

betw

een

the

grap

h of

a fu

ncti

on f

(x) a

nd t

he x

-axi

s on

an

inte

rval

[a, b

] in

the

dom

ain

of f

(x).

A

ppro

xim

ate

the

area

bet

wee

n th

e cu

rve

f (x)

= 1 −

2 x2 a

nd t

he x

-axi

s

on t

he in

terv

al [

0, 4

] by

fir

st u

sing

the

rig

ht e

ndpo

ints

and

the

n by

usi

ng t

he le

ft

endp

oint

s of

the

rec

tang

les.

Use

rec

tang

les

wit

h a

wid

th o

f 1.

Usi

ng r

ight

end

poin

ts fo

r th

e he

ight

of e

ach

rect

angl

e pr

oduc

es fo

ur r

ecta

ngle

s w

ith

a w

idth

of

one

uni

t (F

igur

e A

). U

sing

left

end

poin

ts fo

r th

e he

ight

of e

ach

rect

angl

e pr

oduc

es fo

ur

rect

angl

es w

ith

a w

idth

of 1

uni

t (F

igur

e B

). H

owev

er, t

he fi

rst

rect

angl

e ha

s a

heig

ht o

f f(0

) or

0 an

d th

us, h

as a

n ar

ea o

f 0 s

quar

e un

its.

Are

a us

ing

righ

t en

dpoi

nts

Are

a us

ing

left

end

poin

tsR

1 = 1

· f (

1) o

r 0.

5 R

1 = 1

· f (

0) o

r 0

R2 =

1 ·

f (2)

or

2 R

2 = 1

· f (

1) o

r 0.

5R

3 = 1

· f (

3) o

r 4.

5 R

3 = 1

· f (

2) o

r 2

R4 =

1 ·

f (4)

or

8 R

4 = 1

· f (

3) o

r 4.

5

tota

l are

a =

15

tota

l are

a =

7

The

area

usi

ng t

he r

ight

and

left

end

poin

ts is

15

and

7 sq

uare

uni

ts, r

espe

ctiv

ely.

We

now

ha

ve lo

wer

and

upp

er e

stim

ates

for

the

area

of t

he r

egio

n, 7

< a

rea

< 1

5. A

vera

ging

the

tw

o ar

eas

wou

ld g

ive

a be

tter

app

roxi

mat

ion

of 1

1 sq

uare

uni

ts.

Exer

cise

s

1. A

ppro

xim

ate

the

area

bet

wee

n th

e cu

rve

f (x)

= 3

x2 + 1

and

the

x-a

xis

on t

he

inte

rval

[0, 4

] by

first

usi

ng t

he r

ight

end

poin

ts a

nd t

hen

by u

sing

the

left

end

poin

ts.

Use

rec

tang

les

of w

idth

1 u

nit.

Then

find

the

ave

rage

for

both

app

roxi

mat

ions

.

9

4 u

nit

s2,

45

un

its

2;

69

.5 u

nit

s2

2. A

ppro

xim

ate

the

area

bet

wee

n th

e cu

rve

f (x)

= -

x2 + 5

x +

6 a

nd t

he x

-axi

s on

the

in

terv

al [1

, 5] b

y fir

st u

sing

the

rig

ht e

ndpo

ints

and

the

n by

usi

ng t

he le

ft e

ndpo

ints

. U

se r

ecta

ngle

s of

wid

th 1

uni

t. Th

en fi

nd t

he a

vera

ge fo

r bo

th a

ppro

xim

atio

ns.

4

0 u

nit

s2,

44

un

its

2;

42

un

its

2

12-5

Stud

y Gu

ide

and

Inte

rven

tion

Are

a U

nd

er

a C

urv

e a

nd

In

teg

rati

on

Exam

ple 2

4

8 4

x

Figu

re A

Figu

re B

24

8 4

x

005_

036_

PC

CR

MC

12_8

9381

3.in

dd27

3/17

/09

11:3

7:49

AM


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s


Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

12

28

Gle

ncoe

Pre

calc

ulus

Inte

grat

ion

De

fin

ite

In

teg

ral

Th

e a

rea

of

a r

eg

ion

un

de

r th

e g

rap

h o

f a

fu

nct

ion

is b

⌠

⌡

a f (x

) dx

=

lim

n

→ ∞

∑

i = 1

n f

(xi)Δ

x,

wh

ere

a a

nd

b a

re t

he

low

er

limits

an

d u

pp

er

limits

, re

spe

ctiv

ely

, Δ

x =

b -

a

−

n

an

d x

i = a

+ iΔ

x.

U

se li

mit

s to

fin

d th

e ar

ea o

f th

e re

gion

bet

wee

n th

e gr

aph

of y

=

4x2 a

nd t

he x

-axi

s on

the

inte

rval

[0,

5],

or 5

⌠

⌡

0 4

x2 dx.

5

⌠

⌡

0 4x2 d

x =

lim

n

→ ∞

∑

i = 1

n

f (x

i)Δx

De

finiti

on

of

de

finite

inte

gra

l

=

lim

n

→ ∞

∑

i = 1

n

4x i2 Δ

x f(x

i) =

4x i2

=

lim

n

→ ∞

∑

i = 1

n

4 ( 5i

−

n )

2 5 −

n

x i = 5

i −

n a

nd

Δx

= 5

−

n

=

lim

n

→ ∞

20

−

n ( 25

−

n2 ∑

i = 1

n

i2 )

Exp

an

d a

nd

fa

cto

r.

=

lim

n

→ ∞

20

−

n ( 25

−

n2

· n(

n +

1)(2

n +

1)

−

6

) ∑

i = 1

n

i2 =

n(n

+ 1

)(2n

+ 1

)

−

6

=

lim

n

→ ∞

500

−

6 ( 2n

2 + 3

n +

1

−

n2

) S

imp

lify

an

d e

xpa

nd

.

=

lim

n

→ ∞

500

−

6 (2

+ 3 −

n +

1 −

n2 )

F

act

or

an

d d

ivid

e e

ach

te

rm b

y n2

.

=

( lim

n

→ ∞

500

−

6 ) [ lim

n

→ ∞

2 +

( lim

n

→ ∞

3) (

lim

n

→ ∞

1 −

n ) +

lim

n

→ ∞

1 −

n2 ]

Lim

it th

eo

rem

s

=

500

−

6 [2

+ 3

(0) +

0] o

r ab

out

166.

67 s

quar

e un

its

Sim

plif

y.

Exer

cise

Use

lim

its

to f

ind

the

area

bet

wee

n th

e gr

aph

of e

ach

func

tion

and

th

e x-

axis

giv

en b

y th

e de

fini

te in

tegr

al.

1. 2

⌠

⌡

0 x3 d

x 4

un

its

2

2. 4

⌠

⌡

2 (x2 +

3) d

x 7

4

−

3

un

its

2

3. 6

⌠

⌡

4 (1

+ x

) dx

12

un

its

2

4. 3

⌠

⌡

1 4x3 d

x 8

0 u

nit

s2

12-5

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

Are

a U

nd

er

a C

urv

e a

nd

In

teg

rati

on

Exam

ple

005_

036_

PC

CR

MC

12_8

9381

3.in

dd28

12/5

/09

5:15

:42

PM


NA

ME

DA

TE

PE

RIO

D

Lesson 12-5

Ch

ap

ter

12

29

Gle

ncoe

Pre

calc

ulus

App

roxi

mat

e th

e ar

ea b

etw

een

the

curv

e f(

x) a

nd t

he x

-axi

s on

the

in

dica

ted

inte

rval

usi

ng t

he in

dica

ted

endp

oint

s. U

se r

ecta

ngle

s w

ith

a w

idth

of

1.

1. f

(x) =

x +

3

22

un

its

2

2. f

(x) =

-x2 +

6x

-4

10

un

its

2

[1

, 5]

[2

, 5]

le

ft e

ndpo

ints

righ

t en

dpoi

nts

3. g

(x) =

3x3

10

8 u

nit

s2

4. p

(x) =

1 +

x2

95

un

its

2

[0

, 4]

[1

, 6]

le

ft e

ndpo

ints

righ

t en

dpoi

nts

Use

lim

its

to f

ind

the

area

bet

wee

n th

e gr

aph

of e

ach

func

tion

and

the

x-

axis

giv

en b

y th

e de

fini

te in

tegr

al.

5. 2

⌠

⌡

0 x2 d

x 8

−

3 u

nit

s2

6. 6

⌠

⌡

1 6

x2 dx

43

0 u

nit

s2

7. 3 ⌠

⌡

1 (

x2 - x

) dx

14

−

3

un

its

2

8. 1

⌠

⌡

-

2 (-x2 -

2x +

11)

dx

33

un

its

2

9. A

rchi

tect

ure

and

Des

ign

A d

esig

ner

is m

akin

g a

stai

ned-

glas

s w

indo

w fo

r a

new

bui

ldin

g. T

he s

hape

of

the

win

dow

can

be

mod

eled

by

the

para

bola

y

= 5

- 0

.05x

2 . W

hat

is t

he a

rea

of t

he w

indo

w?

a

bo

ut

66

.67

un

its

2

12-5

Prac

tice

Are

a U

nd

er

a C

urv

e a

nd

In

teg

rati

on

y

x−

10

−1010 −5

−5

105

005_

036_

PC

CR

MC

12_8

9381

3.in

dd29

12/5

/09

5:19

:04

PM


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.



Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

12

30

Gle

ncoe

Pre

calc

ulus

1. D

OG

HO

USE

Cha

rlie

is b

uild

ing

a do

g ho

use

for

Fido

. The

ent

ranc

e to

the

dog

ho

use

is in

the

sha

pe o

f the

reg

ion

show

n. W

hat

is t

he a

rea

of t

he e

ntra

nce

to F

ido’

s do

g ho

use

if x

is g

iven

in fe

et?

4

ft2

2. M

ININ

G T

he e

ntra

nce

to a

coa

l min

e is

in

the

sha

pe o

f the

reg

ion

show

n. W

hat

is t

he a

rea

of t

he e

ntra

nce

if x

is g

iven

in

met

ers?

5

1

−

15 m

2

3. D

AM

S Th

e fa

ce o

f a d

am is

in t

he

shap

e of

the

reg

ion

show

n. W

hat

is t

he

area

of t

he fa

ce o

f the

dam

if x

is g

iven

in

kilo

met

ers?

1

1

−

3 k

m2

4. T

RIA

NG

LE A

REA

On

a co

ordi

nate

pla

ne,

draw

the

tri

angl

e fo

rmed

by

the

x-ax

is

and

the

lines

x =

5 a

nd y

= x

+ 4

.

a. S

hade

the

inte

rior

of t

his

tria

ngle

.

Sa

mp

le a

ns

we

r:

b. F

ind

the

heig

ht a

nd le

ngth

of t

he b

ase

of t

he t

rian

gle.

The

n ca

lcul

ate

the

area

of t

he t

rian

gle

usin

g it

s he

ight

an

d ba

se le

ngth

.

9 u

nit

s,

9 u

nit

s;

40

.5 u

nit

s2

c. C

alcu

late

the

are

a of

the

tri

angl

e by

eval

uati

ng 5

⌠

⌡

-4 (x

+ 4

) dx.

40

.5 u

nit

s2

5. G

RASS

SEE

D M

r. B

ower

is s

eedi

ng p

art

of h

is la

wn,

but

he

has

only

eno

ugh

seed

to

cov

er 3

5 sq

uare

yar

ds. I

f the

are

a in

sq

uare

yar

ds t

hat

he n

eeds

to

seed

can

be fo

und

by 7

⌠

⌡

1 (-

x2 + 8

x -

7) d

x, w

ill h

e

have

eno

ugh

seed

to

com

plet

e th

e ta

sk?

Exp

lain

.

N

o;

Sa

mp

le a

ns

we

r: h

e n

ee

ds

e

no

ug

h s

ee

d t

o c

ov

er

36

sq

ua

re

ya

rds

bu

t o

nly

ha

s e

no

ug

h s

ee

d

to c

ov

er

35

sq

ua

re y

ard

s.

12-5

Wor

d Pr

oble

m P

ract

ice

Are

a U

nd

er

a C

urv

e a

nd

In

teg

rati

on

x

y

y=-

x3+

4x

y=x4

-5x

2+

4

y

x

y=-

x3+

2x2

y

x

y

x

005_

036_

PC

CR

MC

12_8

9381

3.in

dd30

3/17

/09

11:3

8:02

AM


NA

ME

DA

TE

PE

RIO

D

Lesson 12-5

Ch

ap

ter

12

31

Gle

ncoe

Pre

calc

ulus

12-5

Enri

chm

ent

Read

ing

Math

em

ati

cs

Ther

e is

a lo

t of

spe

cial

not

atio

n us

ed in

cal

culu

s th

at is

not

use

d in

oth

er

bran

ches

of m

athe

mat

ics.

In

addi

tion

, the

re is

oft

en m

ore

than

one

not

atio

n fo

r th

e sa

me

thin

g. Y

ou h

ave

alre

ady

seen

thi

s in

the

cas

e of

the

der

ivat

ive.

1. L

et f

(x) =

x2 .

Wha

t do

es lim

h

→ 0

(x +

h)2 -

x2

−

h

fin

d?

th

e d

eri

va

tiv

e o

f f(

x) =

x2

2. L

ist

seve

ral o

ther

way

s of

exp

ress

ing

this

qua

ntit

y.

S

am

ple

an

sw

ers

: d

(x2)

−

dx

, f′

(x),

dy

−

dx ,

Dxy

, y′

Yet

ano

ther

not

atio

n fo

r th

e de

riva

tive

of a

func

tion

y =

f (x

) is

y. . Thi

s w

as

the

nota

tion

dev

elop

ed b

y Is

aac

New

ton.

Eac

h of

the

se n

otat

ions

als

o ca

n be

us

ed t

o in

dica

te h

ighe

r-or

der

deri

vati

ves.

For

exam

ple,

f �(

x) d2 y

−

dx

2 , an

d ÿ

all i

ndic

ate

the

seco

nd d

eriv

ativ

e of

som

e fu

ncti

on y

= f

(x).

3. W

hat

is t

he o

rder

of e

ach

deri

vati

ve?

a.

f �′

(x)

thir

d

b. y.

firs

t c.

d4 y

−

dx4

fou

rth

d.

y�

se

co

nd

The

Leib

niz

nota

tion

for

the

deri

vati

ve dy

−

dx

is

usua

lly r

ead

“dy

dx,”

or m

ore

form

ally

, “th

e de

riva

tive

of y

wit

h re

spec

t to

x.”

Not

e th

at dy

−

dx

is

not

a fr

acti

on o

f any

kin

d. T

o in

dica

te t

he v

alue

of t

he d

eriv

ativ

e at

a s

peci

fic v

alue

of x

usi

ng t

he L

eibn

iz n

otat

ion,

one

mig

ht u

se t

he

follo

win

g: dy

−

dx

�

� � x

= 2

, rea

d “d

y dx

eva

luat

ed a

t x

= 2

.”

Giv

en f

(x)

= x

3 + 3

x2 - 4

, fin

d th

e va

lue

of e

ach

expr

essi

on.

4. f

′(2)

24

5.

dy

−

dx �

� � x

= -

1 -

3

6. f

�(0)

6

7. d3 y

−

dx

3 �

� �

x =

4

6

005_

036_

PC

CR

MC

12_8

9381

3.in

dd31

12/7

/09

12:0

5:06

PM


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s


Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

12

32

Gle

ncoe

Pre

calc

ulus

Stud

y Gu

ide

and

Inte

rven

tion

Th

e F

un

dam

en

tal

Th

eo

rem

of

Calc

ulu

s

Ant

ider

ivat

ives

and

Ind

efin

ite

Inte

gral

s

Giv

en a

func

tion

f (x

), w

e sa

y th

at F

(x) i

s an

ant

ider

ivat

ive

of f

(x) i

f F ′(x

) = f

(x).

Ru

les

fo

r A

nti

de

riv

ati

ve

s

Po

we

r R

ule

If f(

x) =

xn ,

wh

ere

n is

a r

atio

na

l nu

mb

er

oth

er

tha

n -

1,

F(x)

=

xn +

1

−

n

+ 1

+ C

.

Co

ns

tan

t

Mu

ltip

le

of

a P

ow

er

If f(

x) =

kxn ,

wh

ere

n is

a r

atio

na

l nu

mb

er

oth

er

tha

n -

1 a

nd

k is

a

con

sta

nt,

th

en

F(x

) =

kx

n +

1

−

n

+ 1

+ C

.

Su

m a

nd

Dif

fere

nc

e

If t

he

an

tide

riva

tive

s o

f f(x

) a

nd

g(x

) a

re F

(x)

an

d G

(x),

re

spe

ctiv

ely

,

the

n t

he

an

tide

riva

tive

s o

f f(x

) ±

g(x

) a

re F

(x)

± G

(x).

F

ind

all a

ntid

eriv

ativ

es f

or e

ach

func

tion

.

a. f

(x)

= -

3x5

f

(x) =

-3x

5 O

rig

ina

l eq

ua

tion

F

(x) =

-3x

5 +

1

−

5

+ 1

+ C

C

on

sta

nt

Mu

ltip

le o

f a

Po

we

r

= -

1 −

2 x6 +

C

Sim

plif

y.

b. f

(x)

= x

3 + 4

x2 - 2

f (

x) =

x3 +

4x2 -

2

Orig

ina

l eq

ua

tion

=

x3 +

4x2 -

2x0

Re

write

th

e f

un

ctio

n s

o e

ach

te

rm h

as

a p

ow

er

of

x.

F(

x) =

x3 +

1

−

3 +

1 +

4x2

+ 1

−

2

+ 1

- 2x

0 +

1

−

0

+ 1

U

se a

ll th

ree

ru

les.

=

1 −

4 x4

+ 4 −

3 x

3 - 2

x +

C

Sim

plif

y.

Exer

cise

s

Fin

d al

l ant

ider

ivat

ives

for

eac

h fu

ncti

on.

1. f

(x) =

2x4 +

3x2 -

5

2. g

(x) =

2 −

x3

2

−

5 x

5 +

x3 -

5x

+ C

- 1

−

x2

+ C

3. t

(x) =

3 −

4 x6 -

1 −

2 x3

4. n

(x) =

5 √

� x -

2

3

−

28 x

7 -

1

−

8 x

4 +

C

5

−

6 x

6

−

5 -

2x

+ C

Exam

ple

12-6

005_

036_

PC

CR

MC

12_8

9381

3.in

dd32

3/17

/09

11:3

8:12

AM


NA

ME

DA

TE

PE

RIO

D

Lesson 12-6

Ch

ap

ter

12

33

Gle

ncoe

Pre

calc

ulus

The

Fund

amen

tal T

heor

em o

f Ca

lcul

us T

he in

defin

ite

inte

gral

of

f (x)

is d

efin

ed b

y ⌠

⌡

f (x)

dx =

F(x

) + C

, whe

re F

(x) i

s an

ant

ider

ivat

ive

of f

(x)

and

C is

any

con

stan

t.

Fu

nd

am

en

tal

Th

eo

rem

of

Ca

lcu

lus

If F

(x)

is t

he

an

tide

riva

tive

of

the

co

ntin

uo

us

fun

ctio

n f

(x),

th

en

b

⌠

⌡

a f(x)

= F

(b)

- F

(a).

Th

e r

igh

t si

de

of

this

sta

tem

en

t m

ay

als

o b

e w

ritt

en

as

F(x)

⎢

⎢ ⎢ b

a .

E

valu

ate

each

inte

gral

.

a. ⌠

⌡

(3x2 +

4x

- 1

) dx

⌠

⌡

(3x2 +

4x

- 1

) dx

= 3x

2 +

1

−

2 +

1 +

4x1

+ 1

−

1

+ 1

- x0

+ 1

−

0

+ 1

+ C

C

on

sta

nt

Mu

ltip

le o

f a

Po

we

r

= 3x

3 −

3 +

4x2

−

2 -

x +

C

Sim

plif

y.

= x

3 + 2

x2 - x

+ C

S

imp

lify.

b. 4

⌠

⌡

2 (x3 -

1)

dx

4

⌠

⌡

2 (x3 -

1) d

x =

( x4 −

4 -

x) ⎢

⎢

⎢ 4 2

Fu

nd

am

en

tal T

he

ore

m o

f C

alc

ulu

s

= (

44 −

4 -

4)

- (

24 −

2 -

2)

b =

4 a

nd

a =

2

= 6

0 -

6 o

r 54

S

imp

lify.

Exer

cise

s

Eva

luat

e ea

ch in

tegr

al.

1. ⌠

⌡

(3x7 -

x2 )

dx

3x8

−

8 -

x3

−

3 +

C

2.

2

⌠

⌡

1 (x2 +

1) d

x 1

0

−

3

3.

2

⌠

⌡

1 (x2 -

1) d

x 4

−

3

4.

1

⌠

⌡

-1 (x

3 - 2

x +

1) d

x 2

Stud

y Gu

ide

and

Inte

rven

tion

(con

tinu

ed)

Th

e F

un

dam

en

tal

Th

eo

rem

of

Calc

ulu

s

Exam

ple

12-6

005_

036_

PC

CR

MC

12_8

9381

3.in

dd33

3/17

/09

11:5

3:38

PM


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.



Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

12

34

Gle

ncoe

Pre

calc

ulus

Fin

d al

l ant

ider

ivat

ives

for

eac

h fu

ncti

on.

1. f

(x) =

4x3

2. f

(x) =

2x

+ 3

F

(x)

= x

4 +

C

F

(x)

= x

2 +

3x

+ C

3. f

(x) =

x(x

2 - 3

) 4.

f (x

) = 8

x2 + 2

x -

3

F

(x)

= 1

−

4 x

4 -

3

−

2 x

2 +

C

F(x

) =

8

−

3 x

3 +

x2 -

3x

+ C

Eva

luat

e ea

ch in

tegr

al.

5.

⌠

⌡ 8

dx

6. ⌠

⌡

(2x3 +

6x)

dx

8

x +

C

1

−

2 x

4 +

3x2

+ C

7.

⌠

⌡ (-

6x5 -

2x2 +

5x)

dx

8.

5

⌠

⌡

2 2x

dx

-x

6 -

2

−

3 x

3 +

5

−

2 x

2 +

C

2

1

9.

-1

⌠

⌡

-5 (-

4x3 -

3x2 )

dx

10.

1

⌠

⌡

-2 (1

- x

)(x +

3) d

x

50

0

9

11. P

HY

SICS

The

wor

k in

foot

-pou

nds

to c

ompr

ess

a ce

rtai

n sp

ring

a d

ista

nce

of �

feet

from

its

natu

ral l

engt

h is

giv

en b

y W

= �

⌠

⌡

0 2x

dx. H

ow m

uch

wor

k is

req

uire

d to

com

pres

s th

e

spri

ng 6

inch

es fr

om it

s na

tura

l len

gth?

0.2

5 f

t-lb

12. W

OO

DW

ORK

ING

A c

raft

sman

wor

ks h

hou

rs t

o cr

eate

one

pie

ce o

f fur

nitu

re.

Supp

ose

the

num

ber

of h

ours

nee

ded

to c

reat

e p

piec

es is

giv

en b

y h

= p

⌠

⌡

0 (30 -

3x)

dx.

How

man

y ho

urs

does

it t

ake

the

craf

tsm

an t

o m

ake

6 pi

eces

?

12

6 h

Prac

tice

Th

e F

un

dam

en

tal

Th

eo

rem

of

Calc

ulu

s

12-6

005_

036_

PC

CR

MC

12_8

9381

3.in

dd34

3/17

/09

11:3

8:21

AM


NA

ME

DA

TE

PE

RIO

D

Lesson 12-6

Ch

ap

ter

12

35

G

lenc

oe P

reca

lcul

us

Wor

d Pr

oble

m P

ract

ice

Th

e F

un

dam

en

tal

Th

eo

rem

of

Calc

ulu

s

1. V

ERTI

CAL

JUM

P Li

la t

este

d he

r ve

rtic

al

jum

p in

phy

sica

l edu

cati

on c

lass

. The

ve

loci

ty o

f her

jum

p ca

n be

def

ined

as

v(t)

=-

32t+

24,

whe

re t

is g

iven

in

seco

nds

and

the

velo

city

is g

iven

in

feet

per

sec

ond.

a. F

ind

the

posi

tion

func

tion

s(t

) for

Lila

’s ju

mp.

Ass

ume

that

for

t= 0

, s(t

) = 0

.

s(t)

= -

16

t2+

24

t

b. A

fter

Lila

jum

ps, h

ow lo

ng d

oes

it t

ake

befo

re s

he la

nds

on t

he g

roun

d?

1.5

s

2. A

DV

ERTI

SIN

G N

ew W

ave’

s bu

sine

ss lo

go

is in

the

sha

pe o

f the

reg

ion

show

n be

low

. If

the

com

pany

inte

nds

to u

se it

as

part

of

its

lett

erhe

ad, h

ow m

uch

spac

e w

ill t

he

logo

occ

upy

at t

he t

op o

f eac

h do

cum

ent

for

x be

twee

n 0

inch

and

1 in

ch?

8 − 15

in

2

3. S

PRIN

G S

TRET

CHIN

G T

he w

ork,

in

joul

es, r

equi

red

to s

tret

ch a

cer

tain

spr

ing

36 in

ches

bey

ond

its

natu

ral l

engt

h is

give

n by

3 ⌠ ⌡ 0 8

0x d

x. H

ow m

uch

wor

k

is r

equi

red?

36

0 j

ou

les

4. V

OLU

ME

In t

he fi

gure

bel

ow, f

ind

the

volu

me

of t

he s

olid

form

ed b

y re

volv

ing

the

grap

h of

f (x

) =x2 o

ver

the

inte

rval

[0

, 3],

if th

e vo

lum

e of

the

sol

id is

giv

en

by 3 ⌠ ⌡ 0

π(x

2 )2dx

.

243

π−

5 u

nit

s3

5. S

PRIN

G C

OM

PRES

SIO

N A

forc

e of

80

0 po

unds

com

pres

ses

a sp

ring

2 in

ches

fr

om it

s na

tura

l len

gth

of 1

2 in

ches

. Th

e w

ork,

in in

ch-p

ound

s, r

equi

red

to

com

pres

s th

e sp

ring

ano

ther

2 in

ches

is

give

n by

4 ⌠ ⌡ 2 400

x dx

. How

muc

h w

ork

is

requ

ired

to

com

pres

s th

e sp

ring

ano

ther

2 in

ches

?

24

00

in

ch

-po

un

ds

6. B

ILLB

OA

RD T

he S

quar

ed a

nd L

inea

r Tr

ucki

ng C

ompa

ny h

as p

urch

ased

a

billb

oard

to

adve

rtis

e th

e co

mpa

ny. T

he

cent

ral f

igur

e on

the

bill

boar

d, m

easu

red

in fe

et, i

s sh

own

in t

he d

iagr

am b

elow

. W

hat

is t

he a

rea

of t

his

figur

e?

22

.5 f

t2

x

y

y=x4

-2x

2+

1

x

y

246810

24

68

10

y=x2

y=-

3x+

18

12-6

24

f(x)

=x2

369

-3

-6

-9

y

x

005_

036_

PC

CR

MC

12_8

9381

3.in

dd35

3/17

/09

11:3

8:24

AM


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s


Pdf Pass


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

12

36

Gle

ncoe

Pre

calc

ulus

Deri

vati

ves

of

Exp

on

en

tial

an

d L

og

ari

thm

ic F

un

cti

on

s

Ex

po

ne

nti

al

Ru

leT

he

de

riva

tive

of

y =

ex

is e

x a

nd

th

e d

eriva

tive

of

y =

eu

is e

u du

−

dx

.

Fin

d th

e de

riva

tive

of

y =

e 3x

.

Let

u =

3x.

The

n dy

−

dx

= e

u · du

−

dx

.

Sinc

e du

−

dx

= 3

, dy

−

dx

= e

u · 3

or

3eu .

The

deri

vati

ve o

f y =

e3x

is 3

e3x.

Lo

ga

rith

mic

Ru

leT

he

de

riva

tive

of

y =

ln

x is

1

−

x an

d t

he

de

riva

tive

of

y =

ln

u is

1

−

u · d

u

−

dx

.

F

ind

the

deri

vati

ve o

f y

= ln

(x2 +

3).

Let

u =

x2 +

3. T

hen

dy

−

dx =

1 −

u ·

du

−

dx .

Sinc

e du

−

dx

= 2

x, dy

−

dx

= 1 −

u ·

2x,

or

1

−

x2 + 3

· 2

x. S

impl

ify t

o ge

t

2x

−

x2 + 3

.

The

deri

vati

ve o

f y =

ln (x

2 + 3

) is

2x

−

x2 + 3

.

Exer

cise

s

Fin

d th

e de

riva

tive

of

each

fun

ctio

n.

1.

y =

e-

x 2.

y =

e √ �

x −

2

3. y

= e

- x −

4

-e

-x

1

−

4 √

�

x e

√ �

x

- 1

−

4 e

- x −

4

4.

y =

e6x

5.

y =

4ex

6. y

= x

2 ex

6e

6x

4e

x

x2e

x +

2xe

x

7.

y =

ln (x

3 )

8. y

= ln

(2x

+ 5

) 9.

y =

ln (s

in x

+ 4

)

3

−

x

2

−

2x

+ 5

co

s x

−

sin

x +

4

10. y

= ln

( 1 −

x ) 11

. y

= x

ln x

12

. y

= ln

(2x3 +

4x)

- 1

−

x

1

+ l

n x

6x2

+ 4

−

2x3

+ 4

x

13. F

ind

an e

quat

ion

for

a lin

e th

at is

tan

gent

to

the

grap

h of

y

= ln

x t

hrou

gh t

he p

oint

(e, 1

).

Enri

chm

ent

Exam

ple

1

Exam

ple

2

12-6 y

- 1

= 1

−

e (x

- e

) o

r y

= x −

e

005_

036_

PC

CR

MC

12_8

9381

3.in

dd36

3/17

/09

11:3

8:31

AM


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.

Chapter 12 Assessment Answer Key

Pdf Pass


Quiz 1 (Lessons 12-1 and 12-2) Quiz 3 (Lessons 12-4 and 12-5) Mid-Chapter TestPage 37 Page 38 Page 39

Quiz 2 (Lesson 12-3)

Page 37Quiz 4 (Lesson 12-6)

Page 38

1.

2.

3.

4.

5.

6.

1.

2.

3.

4.

5.

1.

2.

3.

4.

5.

1.

2.

3.

4.

5.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

5

1

no limit

-47

-8

D

1

-4

C

-64 ft/s

-38 ft/s

A

12x2 + 40x

-x4 - 2x

− (x3 - 1)2

21

10

− 3

B

2 − 3 x3 - 3x2 + x

+ C

-2x2 - x − 4 + C

5x3

− 3 - x + C

- 3 − 4

B

H

C

G

m = 2x - 4

no limit

v(t) = -32t + 40

-24 ft/s

1.25 s

30 ft


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s

Pdf Pass


Vocabulary Test Form 1Page 40 Page 41 Page 42

1.

2.

3.

4.

5.

6.

7.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

antiderivative

One-sided limits


indefi nite integral


Indeterminate form is when the

fraction 0 −

0 is the

result of applying the quotient rule or direct substitution when attempting to fi nd the limit of a function.

Instantaneous velocity is the velocity of an object at any specifi c point in time.

B

G

D

F

C

F

A

J

B

H

D

J

C

J

B

G

C

G

H

C

0.375


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.


Pdf Pass


Form 2A Form 2BPage 43 Page 44 Page 45 Page 46

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

A

H

A

J

B

H

D

H

C

H

C

F

B

G

D

G

B

F

F

A

4

C

H

A

H

C

F

A

F

B

J

D

G

C

J

B

J

D

H

J

D

25.6


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s

Pdf Pass


Form 2CPage 47 Page 48

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

2

0, 2

60

8

-∞

-∞

25

m = x2 - 8x

5 s; 425 ft

v(t) =

1 − 3 t

- 2 − 3 + 1 −

2 t

- 1 − 2

-12x2 + 2x - 2

18x + 24

8 ft/s

h′(x) = -28x −

(x2 - 4)2

g′(x) = -x6 + 8x3 + 3x2

− (2 - x3)2

1 − 5 x5 +

1 −

2 x4 +

1 −

3 x3 + C

9 − 5 x5 - 2x3 + x + C

2 8 −

15

36 2 − 3

s(t) = -16t2 + 35

16x3 - 8x2


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.


Pdf Pass


Form 2DPage 49 Page 50

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

-2

2, 1

500

-11

∞

-∞

9

m = x4 - 12x2

5.5 s; 494 ft

v(t) = 1 − 2 t

- 3 − 4 - 1 −

5 t

- 4 − 5

9x2 - 10x

8x3 - 9x2 - 10x

111 ft/s

h′(x) = -4x −

(x2 - 2)2

g′(x) = -x6 + 16x3 + 6x2

− (4 - x3)2

1 − 4 x4 +

5 −

3 x3 - 7x2 + C

4 − 3 x3 + 6x2 + 9x + C

5 13

− 15

36 2 − 3

s(t) = -16t2 + 20

x8 - 3x4- 4


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s

Pdf Pass


Form 3Page 51 Page 52

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

2, 1

no limit

2000

12

3

-∞

0

m = 3x2 + 6x + 3

8 s; 1024 ft

v(t) = 0.34 t 0.7 + 1 −

9 t

- 2 − 3

7 − 8 x6 - 1.2x2

3 −

2 x

1 − 2 + 1 −

2 x

- 1 − 2

39 ft/s

h′(x) = x + 1 + 2 √ � x

− 2 √ � x (-x + 1)2

g′(x) = -x4 + 8x3 - 6x2 + 3

−− (x3 - 3x)2

1 − 4 x4 + x3 + 3 −

2 x2 + x+ C

x4 - 8 −

3 x3 - 11

− 2 x2 - 3x + C

2 2 − 3

10 2 − 3

s(t) = -16t2 + 12

7x3 - 3x2


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pan

ies, In

c.


Pdf Pass


Page 53, Extended-Response Test Sample Answers

1a. x f(x)

-2 - 3 − 8

1 - 4 − 7

0 undefined

1 - 6 − 5

2 - 7 − 4

3 - 8 − 3

4 - 9 − 2

5 -10

6 undefined

7 12

1b. The graph indicates that f(x) is continuous at x = 3. Therefore,

lim x → 3

f(x) = f(3), or - 8 − 3 . As x

approaches 0, the y-coordinates approach a value of approximately -1, so lim

x → 0 f(x) is about -1.

Draw a vertical line through the x-axis at x = 3. The intersection of this line and the curve is the limit as x → 3, a value of approximately -3. For the limit as x → 0, follow the same procedure as described above, except at x = 0. The limit as x → 0 is approximately -1.

1c. lim x → 3

( x 2 + 5x) = lim x → 3

x+ 5 · lim x → 3

x,

or 24

1d. lim x → 3

( x 2 - 6x) = lim

x → 3 x- 6 · lim

x → 3 x,

or -9

1e. lim x → 3

x 2 + 5x − x2 - 6x

= lim x → 3

(x2+ 5x)

−

lim x → 3

(x2 - 6x)

, or - 8 − 3

1f. lim x → 0

x 2 + 5x − x 2 - 6x

= lim x → 0

x(x + 5)

− x(x - 6)

= lim x → 0

(x + 5)

− (x - 6)

= lim x → 0

(x + 5)

−

lim x → 0

(x - 6)

= 5 − -6

or - 5 − 6

2. Sample answer: f(x) = x 2 and g(x) = x

lim x → 0

f(x) = 0 = lim x → 0

g(x)

3a.

3b. Method 1: Integrate.

4

⌠ ⌡

0

(4t - t 2 ) dt = 2 t 2 - 1 − 3 t 3 ⎢

⎢

4

0

= 32 - 64 − 3

= 32 − 3 units2

Method 2: Estimate using rectangles. 1 · s(1) + 1 · s(2) + 1 · s(3) + 1 · s(4) = 3 + 4 + 3 + 0 or 10 units2.

Method 1 is more accurate because integrating yields the exact area under the curve. Method 2 yields an approximation.

3c. The area under the curve represents the distance between the object and its starting point after 4 seconds because d = st.

x

f(x)

1234

5 10 15-2-3-4 x

S

1234

-2-2 2 3 5


Copyright

© G

lencoe/M

cG

raw

-Hill

, a

div

isio

n o

f T

he

McG

raw

-Hill

Co

mp

an

ies,

Inc.


An

swer

s

Pdf Pass


Standardized Test PracticePage 54 Page 55

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J

A B C D

F G H J

A B C D


Co

pyrig

ht ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f Th

e M

cG

raw

-Hill C

om

pa

nie

s, In

c.


Pdf Pass


Standardized Test Practice (continued)Page 56

16.

17.

18.

19.

20.

21.

22.

23.

24.

25a.

25b.

25c.

5

17

continuous; Sample answer: the weight of the rice could be any weight between

0 and 20 ounces

-4

x2

− 16

+ y2

− 9 = 1;

ellipse

an= -4 ( 1 −

2 ) n - 1

2 − x - 3

+ 2 −

x + 3

≈3.68

1 − 5 x5 + x3 + C

s(t) = -16t2 + 80t

96 ft

5 s


chapter 12 resource masters - anderson1.org · chapter 12 resource masters ... limits of polynomial...

Documents