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Chapter 12 Resource Masters
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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
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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
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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
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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
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(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
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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
Chapter 12 2 Glencoe Precalculus
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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
Chapter 12 3 Glencoe Precalculus
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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
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Chapter 12 5 Glencoe Precalculus
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
Estimate each one-sided or two-sided limit, if it exists.
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 �
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Chapter 12 6 Glencoe Precalculus
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
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Chapter 12 7 Glencoe Precalculus
12-1
Estimate each one-sided or two-sided limit, if it exists.
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
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Chapter 12 8 Glencoe Precalculus
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
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Chapter 12 9 Glencoe Precalculus
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
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Chapter 12 10 Glencoe Precalculus
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
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Chapter 12 11 Glencoe Precalculus
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
Evaluate each limit.
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)
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Chapter 12 12 Glencoe Precalculus
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.
Evaluate each limit.
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
Evaluate each limit.
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)
Study Guide and Intervention (continued)
Evaluating Limits Algebraically
12-2
Example
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Chapter 12 13 Glencoe Precalculus
Evaluate each limit.
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
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aM
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ine
(mg)
4
6
2
m
d
0
8
10
Days321 54 6
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Chapter 12 14 Glencoe Precalculus
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
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Chapter 12 15 Glencoe Precalculus
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
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Chapter 12 16 Glencoe Precalculus
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
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Chapter 12 17 Glencoe Precalculus
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)
− h , provided the limit exists.
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
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Chapter 12 18 Glencoe Precalculus
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
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.
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
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Chapter 12 19 Glencoe Precalculus
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
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Chapter 12 20 Glencoe Precalculus
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
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Chapter 12 21 Glencoe Precalculus
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
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Chapter 12 22 Glencoe Precalculus
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)
− h , provided the limit exists.
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
Find the derivative of each function.
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
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Chapter 12 23 Glencoe Precalculus
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
Find the derivative of each function.
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
12-4 Study Guide and Intervention (continued)
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
Constant Rules for Derivatives
Example 2
f (x) = 2x2 + 4 Original equation
f ′(x) = 4x Sum Rule for Limits, Power and
Constant Rules for Derivatives
g(x) = x2 - 1 Original equation
g ′(x) = 2x Sum Rule for Limits, Power and
Constant Rules for Derivatives
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Chapter 12 24 Glencoe Precalculus
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
Find the derivative of each function.
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
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Chapter 12 25 Glencoe Precalculus
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
=
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Chapter 12 26 Glencoe Precalculus
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
?
?
?
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Chapter 12 27 Glencoe Precalculus
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
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Chapter 12 28 Glencoe Precalculus
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
12-5 Study Guide and Intervention (continued)
Area Under a Curve and Integration
Example
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Chapter 12 29 Glencoe Precalculus
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
Use limits to find the area between the graph of each function and the x-axis given by the definite integral.
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
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Chapter 12 30 Glencoe Precalculus
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
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Chapter 12 31 Glencoe Precalculus
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
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Chapter 12 32 Glencoe Precalculus
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
Find all antiderivatives for each function.
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
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Chapter 12 33 Glencoe Precalculus
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
Evaluate each integral.
1. ⌠ ⌡
(3x7- x2) dx 2.
2
⌠ ⌡
1
(x2 + 1) dx
3. 2
⌠ ⌡
1
(x2- 1) dx 4.
1
⌠
⌡
-1
(x3 - 2x + 1) dx
Study Guide and Intervention (continued)
The Fundamental Theorem of Calculus
Example
12-6
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Chapter 12 34 Glencoe Precalculus
Find all antiderivatives for each function.
1. f (x) = 4x3 2. f (x) = 2x + 3
3. f (x) = x(x2 - 3) 4. f (x) = 8x2 + 2x - 3
Evaluate each integral.
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
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Chapter 12 35 Glencoe Precalculus
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
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Chapter 12 36 Glencoe Precalculus
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
Find the derivative of each function.
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
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Chapter 12 37 Glencoe Precalculus
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)
Evaluate each limit.
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
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.
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.
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Chapter 12 38 Glencoe Precalculus
12
12
Find the derivative of each function.
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
Use limits to find the area between the graph of each function and the x-axis given by the definite integral.
4. 4
⌠ ⌡
1
x2 dx
5. 2
⌠ ⌡
0
(-x2 + 3x) dx
Chapter 12 Quiz 3(Lessons 12-4 and 12-5)
Find all antiderivatives for each function.
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
Evaluate each integral.
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.
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Chapter 12 39 Glencoe Precalculus
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 ∞
Evaluate each limit.
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.
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Chapter 12 40 Glencoe Precalculus
12
antiderivative
defi nite integral
derivative
difference quotient
differential equation
differential operator
differentiation
direct substitution
indefi nite integral
indeterminate form
instantaneous rate of change
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.
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Chapter 12 41 Glencoe Precalculus
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
Evaluate each limit.
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
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Chapter 12 42 Glencoe Precalculus
12
Find the derivative of each function.
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
Evaluate each integral.
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)
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Chapter 12 43 Glencoe Precalculus
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 -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
Evaluate each limit.
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?
A -1332 ft/s B -1300 ft/s C -96 ft/s D 32 ft/s
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
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Chapter 12 44 Glencoe Precalculus
12
Find the derivative of each function.
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
Use the Quotient Rule to find the derivative of each function.
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
Evaluate each integral.
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
7x dx. How much work is required?
F 14 joules G 28 joules H 32 joules J 40 joules
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)
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Chapter 12 45 Glencoe Precalculus
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 → 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
Evaluate each limit.
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?
A -836 ft/s B -64 ft/s C -32 ft/s D 836 ft/s
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
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Chapter 12 46 Glencoe Precalculus
12
Find the derivative of each function.
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
Use the Quotient Rule to find the derivative of each function.
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
Evaluate each integral.
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
10x dx. How much work is required?
F 0 joules G 50 joules H 75 joules J 125 joules
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)
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Chapter 12 47 Glencoe Precalculus
12
For Questions 1 and 2, use the graph of y = f(x) below to find each value.
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?
Evaluate each limit.
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
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Chapter 12 48 Glencoe Precalculus
12
Find the derivative of each function.
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?
Use the Quotient Rule to find the derivative of each function.
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.
Evaluate each integral.
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)
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Chapter 12 49 Glencoe Precalculus
12
For Questions 1 and 2, use the graph of y = f(x) below to find each value.
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?
Evaluate each limit.
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
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Chapter 12 50 Glencoe Precalculus
12
Find the derivative of each function.
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?
Use the Quotient Rule to find the derivative of each function.
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).
Evaluate each integral.
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)
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Chapter 12 51 Glencoe Precalculus
12
For Questions 1 and 2, use the graph of y = f(x) below to find each value.
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?
Evaluate each limit.
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
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Chapter 12 52 Glencoe Precalculus
12
Find the derivative of each function.
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?
Use the Quotient Rule to find the derivative of each function.
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).
Evaluate each integral.
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)
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Chapter 12 53 Glencoe Precalculus
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
037-056_PCCRMC12_893813.indd 53037-056_PCCRMC12_893813.indd 53 3/17/09 12:44:43 PM3/17/09 12:44:43 PM
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Chapter 12 54 Glencoe Precalculus
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
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Chapter 12 55 Glencoe Precalculus
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.
14. Evaluate lim x → 0
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
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Chapter 12 56 Glencoe Precalculus
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.
19. Evaluate lim x → 2
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.
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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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
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ec1:
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17/0
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:36:
02A
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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.
Chapter 12 A2 Glencoe Precalculus
Answers (Lesson 12-1)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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_
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/09
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:57
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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
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_
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/09
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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.
Chapter 12 A3 Glencoe Precalculus
Answers (Lesson 12-1)
An
swer
s
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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_
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/09
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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
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_
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6:36
AM
A01_A17_PCCRMC12_893813.indd 3A01_A17_PCCRMC12_893813.indd 3 12/7/09 1:49:56 PM12/7/09 1:49:56 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.
Chapter 12 A4 Glencoe Precalculus
Answers (Lesson 12-1 and Lesson 12-2)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
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12_8
9381
3.in
dd10
3/17
/09
11:3
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AM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
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dd11
12/5
/09
4:54
:56
PM
A01_A17_PCCRMC12_893813.indd 4A01_A17_PCCRMC12_893813.indd 4 12/7/09 1:50:31 PM12/7/09 1:50:31 PM
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 A5 Glencoe Precalculus
An
swer
s
Answers (Lesson 12-2)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
A01_A17_PCCRMC12_893813.indd 5A01_A17_PCCRMC12_893813.indd 5 12/7/09 10:21:39 AM12/7/09 10:21:39 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.
Chapter 12 A6 Glencoe Precalculus
Answers (Lesson 12-2)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
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/09
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AM
A01_A17_PCCRMC12_893813.indd 6A01_A17_PCCRMC12_893813.indd 6 3/17/09 8:19:52 PM3/17/09 8:19:52 PM
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 A7 Glencoe Precalculus
An
swer
s
Answers (Lesson 12-3)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
A01_A17_PCCRMC12_893813.indd 7A01_A17_PCCRMC12_893813.indd 7 12/7/09 10:28:35 AM12/7/09 10:28:35 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 A8 Glencoe Precalculus
Answers (Lesson 12-3)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
A01_A17_PCCRMC12_893813.indd 8A01_A17_PCCRMC12_893813.indd 8 12/7/09 10:29:57 AM12/7/09 10:29:57 AM
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 A9 Glencoe Precalculus
An
swer
s
Answers (Lesson 12-3)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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_
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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_
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A01_A17_PCCRMC12_893813.indd 9A01_A17_PCCRMC12_893813.indd 9 12/7/09 10:30:54 AM12/7/09 10:30:54 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.
Chapter 12 A10 Glencoe Precalculus
Answers (Lesson 12-4)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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_
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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_
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AM
A01_A17_PCCRMC12_893813.indd 10A01_A17_PCCRMC12_893813.indd 10 3/17/09 8:20:08 PM3/17/09 8:20:08 PM
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 A11 Glencoe Precalculus
An
swer
s
Answers (Lesson 12-4)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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_
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12_8
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/09
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:50
PM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
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12_8
9381
3.in
dd25
12/7
/09
9:58
:06
AM
A01_A17_PCCRMC12_893813.indd 11A01_A17_PCCRMC12_893813.indd 11 12/7/09 10:34:23 AM12/7/09 10:34:23 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.
Chapter 12 A12 Glencoe Precalculus
Answers (Lesson 12-4 and Lesson 12-5)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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_
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MC
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3.in
dd26
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/09
11:3
7:45
AM
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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_
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3/17
/09
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7:49
AM
A01_A17_PCCRMC12_893813.indd 12A01_A17_PCCRMC12_893813.indd 12 3/17/09 8:20:16 PM3/17/09 8:20:16 PM
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 A13 Glencoe Precalculus
An
swer
s
Answers (Lesson 12-5)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
A01_A17_PCCRMC12_893813.indd 13A01_A17_PCCRMC12_893813.indd 13 12/7/09 10:36:01 AM12/7/09 10:36:01 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 A14 Glencoe Precalculus
Answers (Lesson 12-5)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
A01_A17_PCCRMC12_893813.indd 14A01_A17_PCCRMC12_893813.indd 14 12/7/09 2:04:13 PM12/7/09 2:04:13 PM
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 A15 Glencoe Precalculus
An
swer
s
Answers (Lesson 12-6)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
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3/17
/09
11:5
3:38
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A01_A17_PCCRMC12_893813.indd 15A01_A17_PCCRMC12_893813.indd 15 12/7/09 10:38:12 AM12/7/09 10:38:12 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.
Chapter 12 A16 Glencoe Precalculus
Answers (Lesson 12-6)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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_
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MC
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/09
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AM
A01_A17_PCCRMC12_893813.indd 16A01_A17_PCCRMC12_893813.indd 16 3/17/09 8:31:11 PM3/17/09 8:31:11 PM
Copyright
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lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
Chapter 12 A17 Glencoe Precalculus
An
swer
s
Answers (Lesson 12-6)
Pdf Pass
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
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/09
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AM
A01_A17_PCCRMC12_893813.indd 17A01_A17_PCCRMC12_893813.indd 17 12/7/09 10:38:42 AM12/7/09 10:38:42 AM
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pyrig
ht ©
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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
Chapter 12 A18 Glencoe Precalculus
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
A18_A26_PCCRMC12_893813.indd 18A18_A26_PCCRMC12_893813.indd 18 12/5/09 5:44:47 PM12/5/09 5:44:47 PM
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cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
Chapter 12 Assessment Answer Key
An
swer
s
Pdf Pass
Chapter 12 A19 Glencoe Precalculus
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
instantaneous rate of change
indefi nite integral
differential equation
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
A18_A26_PCCRMC12_893813.indd 19A18_A26_PCCRMC12_893813.indd 19 3/17/09 11:42:48 AM3/17/09 11:42:48 AM
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pyrig
ht ©
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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.
Chapter 12 Assessment Answer Key
Pdf Pass
Chapter 12 A20 Glencoe Precalculus
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
A18_A26_PCCRMC12_893813.indd 20A18_A26_PCCRMC12_893813.indd 20 3/17/09 11:42:52 AM3/17/09 11:42:52 AM
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© G
lencoe/M
cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
Chapter 12 Assessment Answer Key
An
swer
s
Pdf Pass
Chapter 12 A21 Glencoe Precalculus
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
A18_A26_PCCRMC12_893813.indd 21A18_A26_PCCRMC12_893813.indd 21 3/17/09 11:42:54 AM3/17/09 11:42:54 AM
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ht ©
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e/M
cG
raw
-Hill, a
div
isio
n o
f Th
e M
cG
raw
-Hill C
om
pa
nie
s, In
c.
Chapter 12 Assessment Answer Key
Pdf Pass
Chapter 12 A22 Glencoe Precalculus
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
A18_A26_PCCRMC12_893813.indd 22A18_A26_PCCRMC12_893813.indd 22 3/17/09 11:42:58 AM3/17/09 11:42:58 AM
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cG
raw
-Hill
, a
div
isio
n o
f T
he
McG
raw
-Hill
Co
mp
an
ies,
Inc.
Chapter 12 Assessment Answer Key
An
swer
s
Pdf Pass
Chapter 12 A23 Glencoe Precalculus
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
A18_A26_PCCRMC12_893813.indd 23A18_A26_PCCRMC12_893813.indd 23 3/17/09 11:43:01 AM3/17/09 11:43:01 AM
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raw
-Hill, a
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-Hill C
om
pan
ies, In
c.
Chapter 12 Assessment Answer Key
Pdf Pass
Chapter 12 A24 Glencoe Precalculus
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
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Chapter 12 Assessment Answer Key
An
swer
s
Pdf Pass
Chapter 12 A25 Glencoe Precalculus
Standardized Test PracticePage 54 Page 55
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Chapter 12 Assessment Answer Key
Pdf Pass
Chapter 12 A26 Glencoe Precalculus
Standardized Test Practice (continued)Page 56
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25a.
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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
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