7

Inverse Functions: Exponential, Logarithmic, and Inverse Trigonometric Functions 7.1 INVERSE FUNCTIONS SUGGESTED TIME AND EMPHASIS 1 class

Essential material

POINTS TO STRESS 1. The use of multiple representations (verbal, numeric, visual, algebraic) to understand inverse functions, always coming back to the central idea of reversing inputs and outputs. 2. Tests for one-to-one functions and techniques for graphing inverses. 3. Derivatives of inverse functions. QUIZ QUESTIONS • Text Question: Example 5 describes a technique to graph the inverse of a one-to-one function using the line y = x. Why does this technique work? Does it work for y = x 2 ? Answer: Reflecting across the line y = x is the same as reversing the roles of y and x in the function. This technique does not work for x 2 because it is not one-to-one. • Drill Question: If f (1) = 2, f (2) = 3, and f (3) = 1, then what are f ( f (1)), f −1 ( f (1)), and

f f −1 (1) ? Answer: 3, 1, 1 MATERIALS FOR LECTURE √ • Starting with f (x) = 3 x − 4 compute f −1 (−2) and f −1 (0). Then use algebra to find a formula for f −1 (x). Have the students try to repeat the process with g (x) = x 3 + x − 2. Note that facts such as g −1 (−2) = 0, g −1 (0) = 1, and and g −1 (8) = 2 can be found by looking at a table of values for g(x) but that the algebraic approach fails to give us a general formula for g −1 (x). Finally, draw graphs of f , f −1 , g, and g −1 . • Show that f (x) = 3x 2 + 2x + 1 is increasing for x ≥ − 13 by computing f  (x). Explicitly compute √ 3x − 2 − 1 −1 and discuss its domain and range. f (x) = 3 • Pose the question: If f is always increasing, is f −1 always increasing? Give the students time to try prove their answer. Answer: This is true. Proofs may involve diagrams and reflections about y = x, or you may try to get them to be more rigorous. This is an excellent opportunity to discuss concavity, noting that if f is concave up and increasing, then f −1 is concave down and increasing. • Have the students use graphing technology to check if f (x) = x 3 − 2x 2 − x + 2 is one-to-one, and then show how the domain can be restricted to get a one-to-one function. 345

CHAPTER 7 Inverse Functions: Exponential, Logarithmic, and Inverse Trigonometric Functions

WORKSHOP/DISCUSSION • Sketch a graph of the inverse of f (x) = 3x 3 + 2x by first sketching f (x) and then reflecting about y = x.

 Show how it would be difficult to find an algebraic formula for the inverse. Then compute f −1 (5). • Show why f (x) = sin x + cos x is one-to-one for − π4 ≤ x ≤ π4 . Graph the inverse and then compute −1  f (1). • Suppose f is one-to-one and we have the following information:

Show the students how to determine −1 

 f (7) = f −1 (13).

x 3 4 5 6

f (x) f  (x) 7 2 9 1 13 2 19 0.5



 f −1 (9) and f −1 (19).

See if they can explain why

GROUP WORK 1: Inverse Functions: Domains and Ranges While discussing the domains and ranges of inverse functions, this exercise will also foreshadow later excursions into the maximum and minimum values of functions. If a group finishes early, ask them this question: √ “Now consider the graph of f (x) = 2x − 3 + 2. What are the domain and range of f (x)? Try to figure out the domain and range of f −1 (x) by looking at the graph of f . In general, what information do you need to be able to compute the domain and range of f −1 (x) from the graph of a function f ?” Answers: 1. It is one-to-one, because the problem says it climbs steadily. 2. a −1 is the time in minutes at which it achieves a given altitude. 3. Reverse the data columns in the given table to get the table for the inverse function. The domain and range of a are 0 ≤ t ≤ 30 and 0 ≤ a ≤ 29,000, so the domain and range of a −1 are 0 ≤ x ≤ 29,000 and 0 ≤ a −1 ≤ 30. 4. After approximately 8.5 minutes 5. a is no longer 1-1, because heights are now achieved more than once. Bonus The domain of f −1 is the set of all y-values on the graph of f , and the range of f −1 is the set of all x-values on the graph of f . GROUP WORK 2: Functions in the Classroom Before starting this one, review the definition of “function”. Some of the problems can only be answered by polling the class after they are finished working. Don’t forget to take leap years into account for the eighth problem. For an advanced class, follow up by defining “one-to-one” and “bijection”, then determining which of the functions have these properties. Answers: Chairs: Function, one-to-one, bijection (if all chairs are occupied) 346

SECTION 7.1 INVERSE FUNCTIONS

Eye color: Function, not one-to-one Mom & Dad’s birthplace: Not a function; mom and dad could have been born in different places Molecules: Function, one-to-one (with nearly 100% probability); inverse assigns a number of molecules to the appropriate student. Spleens: Function, one-to-one, bijection. Inverse assigns each spleen to its owner. Pencils: Not a function; some people may have more than one or (horrors!) none. Student number: Function, one-to-one; inverse assigns each number to its owner. February birthday: Not a function; not defined for someone born on February 29. Birthday: Function, perhaps one-to-one. Cars: Not a function; some have none, some have more than one. Cash: Function, perhaps one-to-one. Middle names: Not a function; some have none, some have more than one. Identity: Function, one-to-one, bijection. Inverse is the same as the function. Calculus instructor: Function, not one-to-one. Number of hairs: Function, not one-to-one. There are more people in New York City than there are possible values for this function. Therefore, at least two New Yorkers have the same number of hairs on their heads, and so the function does not have an inverse. GROUP WORK 3: The Column of Liquid If the students need a hint, you can mention that the liquid in the mystery device was mercury. Answers: 1. The liquid is 1 cm high when the temperature is 32 ◦ F. 2. The liquid is 2 cm high when the temperature is 212 ◦ F 3. The inverse function takes a height in cm, and gives the temperature. So it is a device for measuring temperature. 4. A thermometer HOMEWORK PROBLEMS Core Exercises: 2, 6, 9, 20, 24, 33, 37, 38, 41 Sample Assignment: 2, 3, 6, 9, 15, 20, 21, 24, 30, 33, 34, 37, 38, 41 Exercise 2 3 6 9 15 20 21

D ×

× × ×

A

N ×

× × ×

G

Exercise 24 30 33 34 37 38 41

× ×

347

D

A × × × × × × ×

N

G × × ×

GROUP WORK 1, SECTION 7.1 Inverse Functions: Domains and Ranges Let a (t) be the altitude in feet of a plane that climbs steadily from takeoff until it reaches its cruising altitude after 30 minutes. We don’t have a formula for a, but extensive research has given us the following table of values: t a (t) 0.1 50 0.5 150 1 500 3 2000 7 8000 10 12,000 20 21,000 25 27,000 30 29,000 1. Is a (t) a one-to-one function? How do you know?

2. What does the function a −1 measure in real terms? Your answer should be descriptive, similar to the way a (t) was described above.

348

Inverse Functions: Domains and Ranges

3. We are interested in computing values of a −1 . Fill in the following table for as many values of x as you can. What quantity does x represent? a −1 (x)

x

What are the domain and range of a? What are the domain and range of a −1 ?

4. You are allowed to turn on electronic equipment after the plane has reached 10,000 feet. Approximately when can you expect to turn on your laptop computer after taking off?

5. Suppose we consider a (t) from the time of takeoff to the time of touchdown. Is a (t) still one-to-one?

349

GROUP WORK 2, SECTION 7.1 Functions in the Classroom Which of the following relations are functions? Domain

Function Values

Function

All the people in your classroom

Chairs

f (person) = his or her chair

All the people in your classroom

{blue, brown, green, hazel}

f (person) = his or her eye color

All the people in your classroom

Cities

f (person) = birthplace of his or her mom and dad

All the people in your classroom

R, the real numbers

f (person) = number of molecules in his or her body

All the people in your classroom

Spleens

f (person) = his or her spleen

All the people in your classroom

Pencils

f (person) = his or her pencil

All the students in your classroom

Integers from 0–99999999

f (person) = his or her student number

All the living people born in February

Days in February, 2007

f (person) = his or her birthday in February 2007

All the people in your classroom

Days of the year

f (person) = his or her birthday

All the people in your classroom

Cars

f (person) = his or her car

All the people in your classroom

R, the real numbers

f (person) = how much cash they have on them

All the people in your college

Names

f (person) = his or her middle name

All the people in your classroom

People

f (person) = theirself

All the students in your classroom

People

f (person) = his or her calculus instructor

All the people in New York City

W, the whole numbers

f (person) = the number of hairs on his or her head

350

GROUP WORK 3, SECTION 7.1 The Column of Liquid It is a fact that if you take a tube and fill it partway with liquid, the liquid will rise and fall based on the temperature. Assume that we have a tube of liquid, and we have a function h (T ), where h is the height of the liquid in cm at temperature T in ◦ F. 1. It is true that h (32) = 1. What does that mean in physical terms?

2. It is true that h (212) = 10. What does that mean in physical terms?

3. Describe the inverse function h −1 . What are its inputs? What are its outputs? What does it measure?

4. There is a device, currently available at your local drugstore, that measures the function h −1 . (It doesn’t use a column of liquid anymore, but its ancestors did.) What is the name of this device?

351

7.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES TRANSPARENCY AVAILABLE #13 (Figures 3, 4, 5, and 10) SUGGESTED TIME AND EMPHASIS 1 class

Essential material

POINTS TO STRESS 1. The graphs and properties of exponential functions, including growth rates. 2. Exponential functions as models for population growth and decay. 3. The basic differentiation formula for exponential functions, and how it is developed from the limit definition of the derivative. 4. The definition of e. 5. Growth rates of exponential functions as compared to growth rates of polynomials. QUIZ QUESTIONS

√ √ • Text Question: 33 = 3 · 3 · 3, 33/4 = 4 3 · 3 · 3. How does one make sense of 3 7 ? √ Answer: Answers will vary. One example: we can approximate 7 by a sequence of rational numbers, √ 7

and can thus similarly approximate 3

to any degree of precision. 3

• Drill Question: What is the derivative of f (x) = e x ? Answer: 3x 2 e x

3

MATERIALS FOR LECTURE • Start to draw a graph of 2x vs x, using the scale of one inch per unit on both axes. Point out that after one foot, the height would be over 100 yards (the length of a football field). After two feet, the height would be 264 miles, after three feet it would be 1,000,000 miles (four times the distance to the moon), after three and a half feet it would be in the heart of the sun. If the graph extended five feet to the right, x = 60, then y would be over one light year up. • Point out this contrast between exponential and linear functions: For equally spaced x-values, linear functions have constant differences in y-values, while pure exponential functions have constant ratios in y-values. Use this fact to show that the following table describes an exponential function, not a linear one. x y −6.2 0.62000 −2.4 0.65100 1.4 0.68355 5.2 0.71773 9.0 0.75361 12.8 0.79129 352

SECTION 7.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES

• In 1985 there were 15,948 diagnosed cases of AIDS in the United States. In 1990 there were 156,024. Scientists said that if there was no research done, the disease would grow exponentially. Compute the number of cases this model predicts for the year 2000. The actual number was 774,467. Discuss possible flaws in the model with the students, and point out the dangers of extrapolation. • Review the derivation of the formula f  (x) = f  (0) a x for an exponential function. Point out that for f (x) = 2x , f  (0) ≈ 0.693, and for g (x) = 3x , g  (0) ≈ 1.099.

WORKSHOP/DISCUSSION 3

3

• Point out that if f (x) = e x , then f  (x) is not equal to e x , as seen in the quiz. Compute the derivative of 3

e x and then do a straightforward Product Rule problem such as finding the derivative of f (x) = e x sin x. 2 e ex 3 + e x 3 + xe3 + x e3 + e3 x . , x , and ex Then have the students try to compute the derivatives of 1 + ex • Draw the graph of e−x from the graph of e x . Then use calculus to derive the properties of the graph of e x/(1−x) , and sketch the graph. • Estimate where 3x > x 3 and where 2x > x 8 using technology. Notice that exponential functions start by growing slower than polynomial functions, and then wind up growing much faster. For example, if one were to graph x 2 vs x using one inch per unit, then when x = 60, y would be only 100 yards, as opposed to a light year for y = 2x . (The sun is only 8 light minutes from the earth.) • One way to measure the growth of the internet is to measure the number of hosts. The following data show the number of internet hosts over time. Try to determine with the students if this is exponential growth. (Note: Do not show the student the third column right away. Let them come up with the idea of finding growth rates between data points.) Month

Hosts

Growth

Month

Hosts

Growth

Month

Hosts

Growth

Aug 1981 May 1982 Aug 1983 Oct 1984 Oct 1985 Feb 1986 Nov 1986 Dec 1987 Jul 1988 Oct 1988 Jan 1989 Jul 1989 Oct 1989 Oct 1990 Jan 1991 Jul 1991 Oct 1991

213 235 562 1024 1961 2308 5089 28,174 33,000 56,000 80,000 130,000 159,000 313,000 376,000 535,000 617,000

− 1.1400 2.0088 1.6724 1.9150 1.6303 2.8699 4.8534 1.3113 8.2927 4.1649 2.6406 2.2378 1.9686 2.0824 2.0246 1.7690

Jan 1992 Apr 1992 Jul 1992 Oct 1992 Jan 1993 Apr 1993 Jul 1993 Oct 1993 Jan 1994 Jul 1994 Oct 1994 Jan 1995 Jul 1995 Jan 1996 Jul 1996 Jan 1997 Jul 1997

727,000 890,000 992,000 1,136,000 1,313,000 1,486,000 1,776,000 2,056,000 2,217,000 3,212,000 3,864,000 4,852,000 6,642,000 9,472,000 12,881,000 16,146,000 19,540,000

1.9275 2.2461 1.5434 1.7198 1.7846 1.6407 2.0403 1.7961 1.3520 2.0990 2.0943 2.4862 1.8739 2.0337 1.8493 1.5712 1.4646

Jan 1998 Jul 1998 Jan 1999 Jul 1999 Jan 2000 Jul 2000 Jan 2001 Jul 2001 Jan 2002 Jul 2002 Jan 2003 Jul 2003 Jan 2004 Jul 2004 Jan 2005

29,670,000 36,739,000 43,230,000 56,218,000 72,398,092 93,047,785 109,574,429 125,888,197 147,344,723 162,128,493 171,638,297 233,101,481 285,139,107 317,646,084 353,284,187

2.3056 1.5333 1.3846 1.6911 1.6585 1.6518 1.3868 1.3199 1.3699 1.2107 1.1208 1.8444 1.4963 1.2410 1.2370

353

CHAPTER 7 Inverse Functions: Exponential, Logarithmic, and Inverse Trigonometric Functions

Answer: We can graph the data, and get a curve that looks like exponential growth. We can also graph growth rate and see (except for two spikes in the late 1980s) a relatively constant growth rate. H

G

360,000,000

9 8

300,000,000

7 240,000,000

6 5

180,000,000

4 120,000,000

3 2

60,000,000

1 82

84

86

88

90

92

94

96

98

00

02

04

t

82

84

86

88

90

92

94

96

98

00

02

04

t

GROUP WORK 1: I’ve Grown Accustomed to Your Growth Before handing out this activity, it may be prudent to review the rules of exponentiation. This exercise enables students to discover for themselves the equal ratio property of exponential functions. Answers: 1. Yes (m = 1), no, yes (m ≈ 2.08), yes (m ≈ 2.01) 2.Equally spaced changes in x-values result in equally spaced changes in y-values 3. Equally spaced changes in x values result in equally proportioned changes in y-values with the same ratio. b = 2, b = 0.9975, b = 2.25, b = 3 4. The “ + C” gets in the way when taking the ratio. However, the property is close to being true when A and b are large compared to C. GROUP WORK 2: Comparisons (Part 1) The purpose of this group work is to give the students a bit of “picture sense”. It is acceptable if they do this by looking at the graphs on their calculators, setting the windows appropriately. Answers: 1. 0 < x < 1.374 and x > 9.940 x > 22.4 4. 0 < x < 1.34 and x > 10.9

2. 0 < x < 1.052 and x > 95.7168

3. 0 < x < 1.18 and

HOMEWORK PROBLEMS Core Exercises: 2, 10, 13, 16, 17, 24, 25, 35, 48, 59, 73, 88 Sample Assignment: 1, 2, 3, 5, 10, 13, 14, 16, 17, 20, 21, 24, 25, 32, 35, 43, 48, 56, 59, 64, 65, 73, 79, 86, 88 Exercise 1 2 3 5 10 13 14 16 17

D × × ×

A ×

× × ×

N

G × × × × ×

Exercise 20 21 24 25 32 35 43 48

D ×

A × × × × × × ×

× 354

N

G ×

Exercise 56 59 64 65 73 79 86 88

D ×

×

A × × × × × × × ×

N

G ×

GROUP WORK 1, SECTION 7.2 I’ve Grown Accustomed to Your Growth 1. Two or three of the following four tables of data have something in common: linear growth. Without trying to find complete equations of lines, determine which of them are linear growth, and determine their rate of change: x 1 2 3 4

y 2 3 4 5

x 21.5 32.6 43.7 54.8

y 4.32 4.203 4.090 3.980

x −3 −2.5 −2 −1.5

y 1.1 2.14 3.18 4.25

x 1 3 6 8

y −5.00 −0.98 5.05 9.07

2. In a sentence, describe a property of linear growth that can be determined from a table of values.

355

I’ve Grown Accustomed to Your Growth

3. The following four tables of data have something in common: exponential growth. Functions of the form y = Ab x (or Aekx ) have a property in common analogous to the one you stated in Question 2. Find the property, and then find the value of b. x 1 2 3 4

y 5 10 20 40

x 21.5 32.6 43.7 54.8

y 4.32 4.203 4.090 3.980

x −3 −2.5 −2 −1.5

y 1.1 1.65 2.475 3.7125

x 1 3 6 8

y 0.8 7.2 194.4 1749.6

4. Unfortunately, the above property does not hold for functions of the form y = Ab x + C. What goes wrong? For what kinds of values of A, b, and C does the property come close to being true?

356

GROUP WORK 2, SECTION 7.2 Comparisons (Part 1) You have learned that an exponential function grows faster than a polynomial function. Find the values of x > 0 for which 1. 2x ≥ x 3 .

2. (1.1)x ≥ x 2 .

3. 2x ≥ x 5 .

4. 3x ≥ x 5 .

357

7.3 LOGARITHMIC FUNCTIONS TRANSPARENCY AVAILABLE #14 (Figures 2, 3, 5, and 6) can be used with this section. SUGGESTED TIME AND EMPHASIS 1 class

Essential material

POINTS TO STRESS 1. Logarithmic functions and their properties, including their geometric properties as inverses of exponentials. 2. Graphs of logarithmic functions, including asymptotic behavior. QUIZ QUESTIONS • Text Question: Why does a loga x = x only hold for x > 0? Answer: loga x is defined only for x > 0. • Drill Question: What is the inverse function for y = loga x? Answer: a x MATERIALS FOR LECTURE √ √ 11 ln 11 ln • Point out that log√17 11 = √ = looks much simpler in the latter form. ln 17 ln 17

Answer: • Sketch a graph of f (x) = log2 (x + 3)2 , sketch y

y=f(x)

the inverse function, and then find an algebraic formula for the inverse.

y=f Ð!(x) 2 2 0

x

f −1 (x) = 2x/2 − 3 • Sketch a graph of f (x) = 4 ln (x + 2), starting with the graph of y = ln x. Then show that this is also a graph of the function written as ln (x + 2)4 . • Discuss why lim ln (ln x) = ∞. Point out that if ln x > eμ where μ is a positive number, then x→∞

ln (ln x) > μ. WORKSHOP/DISCUSSION

x 2x 2 after writing it as f (x) = x = . See if students can identify • Discuss the graph of f (x) = 3 3 x 2 = 1, and to explain why the answer is negative. its inverse as g (x) = log2/3 x. Ask them to solve 3 • If x satisfies 3x > x 3 , take logs to base 3 on both sides to get x > 3 log3 x, or 13 x > log3 x. Point out that logn x grows more slowly than x/n, for x sufficiently large. • Estimate where 5x > 107 by the use of technology. Then have the students use logarithms to find an exact answer. 3−x 2x

358

SECTION 7.3 LOGARITHMIC FUNCTIONS

GROUP WORK 1: Comparisons (Part 2) The purpose of this group work is to give students a bit of “picture sense”. It is acceptable if they do this by looking at graphs on their calculators, setting the windows appropriately. Answers: 1. 0 < x < 1.208 9, x > 18.6387 3. 0 < x < 1.1772, x > 22.4400

2. 0 < x < 1.0409, x > 125.3635 4. 0 < x < 1.05138, x > 95.7168

GROUP WORK 2: Irrational, Impossible Relations Before starting this activity, review the definitions of rational and irrational numbers. The hint sheet should be given out only after the students have tried to show that log2 3 is irrational, or at least discussed it enough to understand what they are trying to show. If a group finishes early, have them show that log2 a is always irrational if a is an odd integer. ln 3 γ 2. −γ 3. Answers: 1. −γ 4. log2 3 = 5. The proof is outlined in the hint sheet. 2 ln 2 √ Answers (Hint Sheet): 1. 2a/b = 3 ⇒ b 2a = 3 ⇒ 2a = 3b 2. If a = b = 0, then log2 3 = 0/0, which is undefined. 3. 2a = 2 · 2 · · · · · 2, and 3b = 3 · 3 · · · · · 3, so these numbers can never be equal, because the left is always divisible by two, and the right never is (unless b = 0). 4. Irrational. HOMEWORK PROBLEMS Core Exercises: 2, 4, 9, 15, 23, 26, 40, 43, 47, 55 Sample Assignment: 2, 3, 4, 8, 9, 10, 15, 18, 21, 23, 26, 31, 37, 40, 43, 44, 47, 55, 66 Exercise 2 3 4 8 9 10 15 18 21 23

D ×

× × ×

A × × × × ×

N

G ×

Exercise 26 31 37 40 43 44 47 55 66

× ×

359

D

× × ×

A × × × × × × × × ×

N

G

GROUP WORK 1, SECTION 7.3 Comparisons (Part 2) You have been told that an exponential function grows more quickly than a polynomial function, and that logarithmic functions grow more slowly than linear functions. Find the values of x > 0 for which 1. 3x ≥ x 7 .

2.

x 7 ≥ x 4. 6

3.

x ≥ log2 x. 5

4.

x ≥ log1.1 x. 2

360

GROUP WORK 2, SECTION 7.3 Irrational, Impossible Relations 1. If log2 x = γ , then what is log1/2 x?

2. If logb x = γ , then what is log1/b x (assuming b > 1)?

3. If logb x = γ , then what is logb2 x?

4. We are going to estimate log2 3. In pre-calculus, you memorized that log2 3 ≈ 1.584962501. Suppose you didn’t have this fact memorized. There is no log2 3 button on your calculator! How would you compute it?

5. Unfortunately, the calculator gives us only a finite number of digits. If log2 3 were a rational number, we would be able to express it as a fraction, giving us perfect accuracy. Do you think it is rational or irrational? Try to prove your result.

361

GROUP WORK 2, SECTION 7.3 Irrational, Impossible Relations (Hint Sheet) So, you realize that it’s not easy to determine whether log2 3 is rational! One way to attempt to show that log2 3 is rational is to assume that it is, and try to find integers a and b such a that log2 3 = . If we can show that there are no such a and b, then log2 3 cannot be rational. b a 1. Assume that log2 3 = . Show that a and b must then satisfy 2a = 3b b

2. Notice that a = 0, b = 0 satisfies 2a = 3b . Show that this fact doesn’t help us.

3. Find a = 0 and b = 0 that satisfy 2a = 3b , or show that no such {a, b} exists.

4. Is log2 3 rational or irrational? Why?

362

7.4 DERIVATIVES OF LOGARITHMIC FUNCTIONS SUGGESTED TIME AND EMPHASIS 1 class

Essential material

POINTS TO STRESS 1. The basic logarithmic differentiation formula. 2. The technique of logarithmic differentiation. 3. The concept of e as a limit. QUIZ QUESTIONS • Text Question: Why is logarithmic differentiation necessary? For instance, could you do Example 15 using the Product and Quotient Rules? Answer: It is a lot easier to do this problem with logarithmic differentiation. Also, something like x x is very hard to differentiate conventionally.

• Drill Question: What is the instantaneous rate of change of the function f (x) = ln x 2 + 2x + 1 at x = 1? Answer: 1 MATERIALS FOR LECTURE ln a • Review logarithmic and exponential functions. Point out that there is no way to simplify ln (a + b) or . ln b  ln x + 2 if x > 0 • Discuss Example 7 and extend it in the following way: Let f (x) = Graph ln (−x) − 1 if x < 0  ln x + a if x > 0  from geometric and f (x) and compute f (x). Close with discussions of f (x) = ln (−x) + b if x < 0 algebraic perspectives. • Derive the number e as a limit, as done in the text. Have them confirm Formulas 8 and 9 by plugging small numbers in for x and large numbers in for n. Show that we can also write e = lim (1 + n)1/n by n→∞

replacing x with 1/n in the text’s derivation. • Ask what the second derivative of f (x) = ln

√ x tells us about the shape of the graph of f .

WORKSHOP/DISCUSSION • Compute derivatives of functions that involve logarithms and exponents, and that require the Product, Quotient, or Chain Rule. For example, use f (x) = ln (sin x) sin (ln x) or g (x) = ln (e x + ln x).

d • Compute ln (x + 1)5 , first by simplifying using properties of logarithms, and then using the Chain dx Rule without prior simplification. • Present the function f (x) = ln ln x. Before drawing the graph, have the students try to find the domain and range of f , as well as lim f (x). Then graph the function, find an equation of the line tangent to it at x→∞

the point (e, 0), and graph the tangent line. 363

CHAPTER 7 Inverse Functions: Exponential, Logarithmic, and Inverse Trigonometric Functions

GROUP WORK 1: Logarithmic Differentiation

In addition to its face value, the process of logarithmic differentiation is a good way of mastering computations involving logarithms. Don’t give a group the second page until they have finished with the first page. If a

sin x . group finishes early, have them compute the derivative of x 2 + 1

Answers:

1. The Power Rule requires that the exponent be a constant.

2. The Exponent Rule requires that the base be a constant.

3. -- 6. y = (x + 1)x ln y = ln (x + 1)x ln y = x ln (x + 1) x 1 dy = ln (x + 1) + y dx x +1 1 x dy = ln (x + 1) + x x +1 (x + 1) dx   x dy x = (x + 1) ln (x + 1) + dx x +1

7. They are identical.

9.

e x ln(x+1)

 ln (x + 1) +

8.

e x ln(x+1)

 ln (x + 1) +

x x +1



     x x x x x ln (x+1) ( ) =e = (x + 1) ln (x + 1) + ln (x + 1) + x +1 x +1 x +1

y sin y + x sin x ln y x sin y sin y + x y cos x sin x ln y

or 10. y  = −y sin y cos x x cos x − y ln x cos y x x y cos y ln x − y cos x 364

SECTION 7.4 DERIVATIVES OF LOGARITHMIC FUNCTIONS

GROUP WORK 2: e as a Limit

One can introduce this activity by deriving the expression in Problem 1 in terms of compound interest. Some students will require more guidance on what types of things to look for in Problem 1. Properties to notice include concavity, asymptotes, extrema, increase/decrease, and positivity/negativity.

Answers:

1.

y

2

2. The horizontal asymptote of the graph

1 x . corresponds to lim 1 + x→∞ x

1

3. The limit is e ≈ 2.718.

3

0

1

2

3

4

5

6

7

8

9

10

x

GROUP WORK 3: The Right Procedure at the Right Time

This drill activity can be done as a group activity, or the students can work on it individually. The main idea is to make sure they understand the fundamental idea.

Answers:

1.

π 4

(sin x)π/4−1 cos x

4. sin

π 4

x sin(π/4)−1

=

 √ x √

x

2. sin π4 ln sin π4 = − 22 ln 2

 √ √ 2−2 /2 2 2 x

5. 4π x 3 cos x 4

365

3.

π sin x π ln 4 cos x 4

CHAPTER 7 Inverse Functions: Exponential, Logarithmic, and Inverse Trigonometric Functions

GROUP WORK 4: Some Surprising Areas Have the groups work on the first three problems, and then hand the fourth out to groups that finish early. When closing the exercise, impart to the students how surprising the result is: that equal areas are created by any choice of m > 1. Note that the result of Problem 4 can be generalized for the area between x = m a and x = m b (b > a). Answers: 1. Each of the areas is ln 2 ≈ 0.693147. The students will probably not be able to simplify their answers to ln 2, and may leave them as ln (ln 1.0201) − ln (ln 1.01), ln (ln 16) − ln (ln 4), and ln (ln 100) − ln (ln 10). 2. The second endpoint is the square of the first. 3. The area between x = m and x = m 2 is the same, regardless of the value of m.

 m2  ln m 2

dx du 2 ln m 2 4. = = ln ln m − ln (ln m) = ln = ln 2 u ln m m x ln x ln m HOMEWORK PROBLEMS Core Exercises: 2, 9, 16, 28, 38, 45, 57, 68, 74, 78 Sample Assignment: 2, 9, 16, 28, 31, 35, 38, 39, 45, 46, 54, 57, 67, 68, 70, 74, 75, 78, 87, 91 Exercise 2 9 16 28 31 35 38 39 45 46

D

A × × × × × × × × × ×

N

G

Exercise 54 57 67 68 70 74 75 78 87 91

×

366

D

× ×

A × × × × × × × × × ×

N

× ×

G × ×

GROUP WORK 1, SECTION 7.4 Logarithmic Differentiation Let f (x) = (x + 1)x . As of now, none of the rules that we’ve learned seem to help us find f  (x). 1. Why can’t you use the Power Rule to compute f  (x)?

2. Why can’t you use a formula like

d x 3 = 3x ln 3 to compute f  (x)? dx

3. Start with the equation y = (x + 1)x . Take the logarithm of both sides to get ln y = ln (x + 1)x . Then

use a property of logarithms to help you find the derivative of ln (x + 1)x .

4. Now use implicit differentiation to find

d ln y. dx

5. Since ln y = ln (x + 1)x , you can equate your answers to Parts 3 and 4. Do so, and then replace all y terms by (x + 1)x . Why can you do this substitution?

6. Perform some algebra to get

dy alone on the left hand side, and you are done! dx

367

Logarithmic Differentiation

7. Let g (x) = e x ln(x+1) . What is the relationship between f (x) and g (x)?

8. If g (x) = e x ln(x+1) , compute g  (x). What techniques are you using in this case?

9. Since you have done the same problem in two different ways, show that your answers to Problems 6 and 8 are identical.

10. Let y = f (x) be implicitly defined by x sin y = y cos x . Compute y  in terms of x and y. (Hint: Can logarithms help you?)

368

GROUP WORK 2, SECTION 7.4 e as a Limit



1 x 1. Graph the function f (x) = 1 + , and describe its properties. x

y 3 2 1 0

1

2

3

4

5

6

7

8

9

10 x

2. What is the relationship between the graph you drew for Problem 1 and Formula 9 in the text?



1 x 3. Using Problems 1 and 2 as a guide, what do you think lim 1 + is? x→∞ x

369

GROUP WORK 3, SECTION 7.4 The Right Procedure at the Right Time Compute the derivatives of the following functions: 1. (sin x)π/4

x 2. sin π4

3.

π sin x 4

4. x sin(π/4)

5. π sin x 4

370

GROUP WORK 4, SECTION 7.4 Some Surprising Areas Meet

1 : x ln x

1. Compute the area under f (x) =

1 from x = 1.01 to 1.0201, from 4 to 16, and from 10 to 100. x ln x

2. What do the three pairs of endpoints used in Problem 1 have in common?

3. The answers to Problems 1 and 2 should suggest an interesting property of the curve f (x) = What is it?

371

1 . x ln x

Some Surprising Areas

1 , the x-axis, the line x = m and the line x = m 2 is x ln x independent of your choice of m (as long as m > 1).

4. Prove that the area enclosed by f (x) =

372

7.2* THE NATURAL LOGARITHMIC FUNCTION SUGGESTED TIME AND EMPHASIS 1 class

Essential material

POINTS TO STRESS



1. The definition and properties of the natural logarithmic function ln (x) =

x dt 1

t

, x > 0.

2. The basic differentiation and integration formulas for logarithms. 3. The concept of e as the number x for which ln x = 1. 4. The technique of logarithmic differentiation. QUIZ QUESTIONS • Text Question: Why is logarithmic differentiation necessary? For instance, could you do Example 14 using the Product and Quotient Rules? Answer: It is a lot easier to do this problem with logarithmic differentiation. Also, something like x x is very hard to differentiate conventionally.  x dt • Drill Question: If we have ln x = , x > 0, why is ln x negative for 0 < x < 1? 1 t  1  x  1 dt 1 dt dt =− where x , the area under the curve from x to 1, Answer: Because if 0 < x < 1, t t 1 t x t is positive. MATERIALS FOR LECTURE  x  x dt 1 , using the integral definition, and review why if ln x = dt, then • Discuss properties of ln x = t 1 1 t 1 d ln x = , x > 0. dx x

• Derive the formula ln x 1/q = q1 ln x by rewriting the equation y = x 1/q as y q = x. Then use this

equation to show that ln x p/q = qp ln x.

x2 . Point out that the two other forms of f (x), • Discuss the properties and graph of f (x) = ln 2 x +1





1 1 namely f (x) = ln = − ln 1 + 2 or f (x) = 2 ln x − ln x 2 + 1 , could be helpful. 2 1 + 1/x x WORKSHOP/DISCUSSION • Compute derivatives of functions that involve logarithms and that require the Product, Quotient, or Chain Rule. For example, use f (x) = ln (sin x) sin (ln x) or f (x) = ln (x e + ln x). Include the reason that we 1 d ln |x| = . use the absolute value in the formula dx x • Present an example of logarithmic differentiation, such as finding the derivative of (sin x)x • Go over Exercise 73 with the students. 373

2 +3x

.

CHAPTER 7 Inverse Functions: Exponential, Logarithmic, and Inverse Trigonometric Functions

GROUP WORK 1: Logarithmic Differentiation In addition to its face value, the process of logarithmic differentiation is a good way of mastering computations involving logarithms. Don’t give a group the second page until they have finished with the first page. If a

sin x group finishes early, have them compute the derivative of x 2 + 1 . Answers: 1. The Power Rule requires that the exponent be a constant. 2. The Exponent Rule requires that the base be a constant. 3. -- 6. y = (x + 1)x ln y = ln (x + 1)x ln y = x ln (x + 1) x 1 dy = ln (x + 1) + y dx x +1 1 dy x = ln (x + 1) + x x +1 (x + 1) dx   dy x x = (x + 1) ln (x + 1) + dx x +1 7. They are identical. 9.

e x ln(x+1)

 ln (x + 1) +

8.

e x ln(x+1)

 ln (x + 1) +

x x +1



     x x x x x ln (x+1) ( ) =e ln (x + 1) + = (x + 1) ln (x + 1) + x +1 x +1 x +1

x sin y sin y + x y cos x sin x ln y y sin y + x sin x ln y

or 10. y  = −y sin y cos x x cos x − y ln x cos y x x y cos y ln x − y cos x GROUP WORK 2: Euler’s Constant This is a difficult problem and the students may need some help. Here are a few suggested hints: Problem 3(a): Each shaded section is part of a larger rectangle with opposite corners on the graph of f (t) = 1t . The combined area of these rectangles is





1 1 1 1 1 1− + − + ... + − 2 2 3 n n+1 Problem 3(b): Because f (t) =

1 t

is concave upward, each shaded section is larger than the triangle that is

one-half of the rectangle described in part a). The combined area of these triangles is 





 1 1 1 1 1 1 1− + − + ... + − 2 2 2 3 n n+1 374

SECTION 7.2* THE NATURAL LOGARITHMIC FUNCTION

Answers: 1. ln (n + 1)

2. γ n = (The area under the rectangles) −

3. (a) Applying the hint gives an area of 1 −

 n+1 dt 1

t

1 . n+1

(b) The hint gives this result directly. (c) If n ≥ 9, 1 − 4.

25 12

1 = 0.9. n+1

− ln 5 ≈ 0.47389542,

761 280

− ln 9 ≈ 0.52063257,

1,145,993 360,360

− ln 14 ≈ 0.54107643

GROUP WORK 3: How Slowly can ln x Grow? This activity leads the students to discover first that ln x grows more slowly than any root function x 1/N . Answers: 1 1 1. < (N −1)/N for t ≥ 1. t t

 1

x dt

t


1. Note that the result of Problem 4 can be generalized for the area between x = m a and x = m b (b > a). Answers: 1. Each of the areas is ln 2 ≈ 0.693147. The students will probably not be able to simplify their answers to ln 2, and may leave them as ln (ln 1.0201) − ln (ln 1.01), ln (ln 16) − ln (ln 4), and ln (ln 100) − ln (ln 10). 2. The second endpoint is the square of the first. 3. The area between x = m and x = m 2 is the same, regardless of the value of m. 

m2

4. m

dx = x ln x





2 ln m 2 = ln 2 = ln ln m − ln (ln m) = ln u ln m

ln m 2 du ln m

375

CHAPTER 7 Inverse Functions: Exponential, Logarithmic, and Inverse Trigonometric Functions

HOMEWORK PROBLEMS Core Exercises: 2, 4, 5, 10, 17, 25, 35, 42, 51, 68 Sample Assignment: 2, 4, 5, 9, 10, 11, 17, 23, 25, 35, 38, 42, 45, 51, 59, 64, 68 Exercise 2 4 5 9 10 11 17 23 25 35 38 42 45 51 59 64 68

D

A × × × × × × × × × × × × × × ×

376

N

G

× ×

×

GROUP WORK 1, SECTION 7.2* Logarithmic Differentiation Let f (x) = (x + 1)x . As of now, none of the rules that we’ve learned seem to help us find f  (x). 1. Why can’t you use the Power Rule to compute f  (x)?

2. Why can’t you use a formula like

d x 3 = 3x ln 3 to compute f  (x)? dx

3. Start with the equation y = (x + 1)x . Take the logarithm of both sides to get ln y = ln (x + 1)x . Then

use a property of logarithms to help you find the derivative of ln (x + 1)x .

4. Now use implicit differentiation to find

d ln y. dx



5. Since ln y = ln (x + 1)x , you can equate your answers to Parts 3 and 4. Do so, and then replace all y terms by (x + 1)x . Why can you do this substitution?

6. Perform some algebra to get

dy alone on the left hand side, and you are done! dx

377

Logarithmic Differentiation

7. Let g (x) = e x ln(x+1) . What is the relationship between f (x) and g (x)?

8. If g (x) = e x ln(x+1) , compute g  (x). What techniques are you using in this case?

9. Since you have done the same problem in two different ways, show that your answers to Problems 6 and 8 are identical.

10. Let y = f (x) be implicitly defined by x sin y = y cos x . Compute y  in terms of x and y. (Hint: Can logarithms help you?)

378

GROUP WORK 2, SECTION 7.2* Euler’s Constant Consider the following picture:

 1. Find a formula for the left-hand sum of

n+1 1 1

t

dt with n subintervals. Notice that this sum is an upper

bound for the integral.

2. Let γ n be the area of the shaded region in the diagram above. Show that γ n = 1 + ln (n + 1).

1 1 1 + + ··· + − 2 3 n

3. It is a fact that lim γ n = γ exists. We want to estimate γ . n→∞

(a) Using a geometric argument, show that γ n ≤ 1 for all n.



1 1 (b) Using a geometric argument, show that 1− ≤ γ n for all n. 2 n+1

(c) Conclude from parts (a) and (b) that 0.45 ≤ γ n ≤ 1 if n ≥ 9.

4. Compute precise values for γ n for n = 4, 8, and 13. How do these values compare to the lower bound computed in Problem 3(c)?

γ is known as Euler’s constant. It is a fact that γ = 0.577 to three decimal places. It is still not known today whether γ is rational or irrational. 379

GROUP WORK 3, SECTION 7.2* How Slowly can ln x Grow? 1 1 1. Let N be a positive integer. How do the functions and (N −1)/N compare for t ≥ 1? Show that t t  x  x dt 1 dt for x ≥ 1. ≤ t (N −1)/N 1 t 1



2. Use Problem 1 to show that ln x ≤ N x 1/N − 1 for all x ≥ 1.

3. Use Problem 2 to show that ln x ≤ N x 1/N , for any positive integer N . Conclude that ln x grows “more slowly” than any root function x 1/N , where N is a positive integer.

380

GROUP WORK 4, SECTION 7.2* Some Surprising Areas Meet

1 : x ln x

1. Compute the area under f (x) =

1 from x = 1.01 to 1.0201, from 4 to 16, and from 10 to 100. x ln x

2. What do the three pairs of endpoints used in Problem 1 have in common?

3. The answers to Problems 1 and 2 should suggest an interesting property of the curve f (x) = What is it?

381

1 . x ln x

1 , the x-axis, the line x = m and the line x = m 2 is x ln x independent of your choice of m (as long as m > 1).

4. Prove that the area enclosed by f (x) =

382

7.3* THE NATURAL EXPONENTIAL FUNCTION SUGGESTED TIME AND EMPHASIS 1 class

Essential material

POINTS TO STRESS 1. The definition of e x as the inverse of ln x. 2. The graph and properties of e x . 3. Derivatives and integrals involving e x . QUIZ QUESTIONS • Text Question: Why is the equation eln x = x only true for x > 0, but ln (e x ) = x holds for all x? Answer: Because ln x is defined only for x > 0. • Drill Question: What property of the graph of ln x ensures that lim e x = 0? x→−∞

Answer: lim ln x = −∞ x→0+

MATERIALS FOR LECTURE • Using properties of inverse functions, graph e x from ln x, and illustrate several laws of exponents. Point out using the graphs that since lim ln x = −∞, it follows that lim e x = 0. x→0+

• Compute

d dx

• Compute

d dx



 3

x→−∞

 3  3 3 d e x = e x . Show that e x = 3x 2 e x . (e x ) = e x and then point out that dx   2 

2 d e x +x (there are two possible ways to do this), ddx (sin (e x )) and dx 1 + e−x .

• Draw the graph of e−x from the graph of e x . Then use calculus to derive properties of e x/(1−x) , and sketch its graph.     ex 3x x x dx, and sin xecos x dx. • Compute the following integrals: e dx, e cos e dx, 1 + ex WORKSHOP/DISCUSSION 2

• Have the students graph f (x) = e x and estimate the slope at x = 0 and x = 1. Ask why the slope at x = 0 is 0, while the slope of g (x) = e x at x = 0 is 1. e1/x , and sketch its graph. Point out that writing f (x) = e x+(1/x) e−x simplifies the process. Where is f  (x) = 0?  2 5 2 5 • Do some differentiations using e x , such as those of esin x , x 2 e x , e x + 1 and cos (x e + e x ). A nice x 3 3 summary problem is g (x) = ex 3 + e x + xe3 + x e + e3 .  2     e2x 2xeln x +1 ln (sin e x ) dx, and dx. • Compute dx, √ e x cos e x x2 + 1 1 + e4x

• Derive the properties of f (x) =

383

GROUP WORK: On and On with Exponential Inequalities This group work is essentially Exercises 87 and 88, using a slightly different initial approach. It might be useful to assign Exercise 89 after the students have completed this group work. Answers: 1. This can be done by using Property 7 of Integrals in Section 5.2. 2. Computation of the integrals in Problem 1 gives x ≤ e x − 1. 3. Integration of both sides of the inequality in Problem 2 gives the desired result. x3 x2 + 4. P (x) = 1 + x + 2 6 HOMEWORK PROBLEMS Core Exercises: 2, 4, 8, 18, 23, 28, 39, 50, 61, 77, 90 Sample Assignment: 1, 2, 4, 5, 8, 15, 18, 21, 23, 25, 28, 39, 45, 50, 61, 63, 72, 77, 78, 90 Exercise 1 2 4 5 8 15 18 21 23 25 28 39 45 50 61 63 72 77 78 90

D

A × × × × ×

×

×

384

× × × × × × × × × × × ×

N

G ×

×

× ×

GROUP WORK 1, SECTION 7.3* On and On Exponential Inequalities 1. Using the fact that the constant function f (x) ≡ 1 ≤ e x for x ≥ 0, why can we conclude that x u x 0 f (u) du ≤ 0 e du?

2. Show that 1 + x ≤ e x if x ≥ 0.

3. Using the same process as you used to derive the solution to Problem 2, show that 1 + x + x ≥ 0.

x2 ≤ e x if 2

4. Write the equation for the polynomial P (x) such that P (x) ≤ e x if x ≥ 0, if we apply the same process one more time to the answer of Problem 3.

385

7.4* GENERAL LOGARITHMIC AND EXPONENTIAL FUNCTIONS SUGGESTED TIME AND EMPHASIS 1 – 1 12 classes

Essential material

POINTS TO STRESS 1. Exponential graphs, from both symbolic and geometric perspectives, and their properties (derived from e x using a x = e x ln a ). 2. General logarithmic functions, their properties as inverses of exponentials, and their graphs (including asymptotic behavior). 3. The basic derivative formulas. 4. The concept of e as a limit QUIZ QUESTIONS • Text Question: Why does a loga x = x hold only for x > 0? Answer: loga x is defined only for x > 0. • Drill Question: What is the inverse function for y = loga x? Answer: a x MATERIALS FOR LECTURE • Start to draw a graph of 2x vs x, using the scale of one inch per unit on both axes. Point out that after one foot, the height would be over 100 yards (the length of a football field). After two feet, the height would be 264 miles, after three feet it would be 1,000,000 miles (four times the distance to the moon), after three and a half feet it would be in the heart of the sun. If the graph extended five feet to the right, x = 60, then y would be over one light year up. • Discuss the general definition of a x = e x ln a for a > 0 and x a real number, and discuss why this equation √  √ √ √ √3 3 ln 2 /2 3 ln 2 makes sense even for irrational numbers. [For example, 2 = e =e ]. Compare the x properties of a for a > 1 and 0 < a < 1. • Derive the general Power Rule (x n ) = nx n−1 for any real number n.

 • By differentiating the equation a loga x = x implicitly, derive the formula loga x =

1 . x ln a

• Estimate where 3x > x 3 and 2x > x 8 using technology. Notice that exponential functions start by growing slower than polynomial functions, and then wind up growing much faster. For example, if one were to graph x 2 vs x using the scale of one inch per unit on both axes, then when x = 60, y would be only 100 yards, as opposed to a light year for e x . (The sun is only 8 light seconds from the earth.) 386

SECTION 7.4* GENERAL LOGARITHMIC AND EXPONENTIAL FUNCTIONS

WORKSHOP/DISCUSSION

• Sketch a graph of f (x) = log2 (x + 3)2 , sketch

Answer:

y

y=f(x)

the inverse function, and then find an algebraic formula for the inverse.

y=f Ð!(x) 2 2 0

x

f −1 (x) = 2x/2 − 3 • Compute derivatives of functions that involve logarithms and exponents and that require the Product,

Quotient, or Chain Rule. For example, use f (x) = log2 (sin x) sin log2 x or f (x) = log3 e x + log2 x .  x • See if the students can identify the inverse of f (x) = 23 as g (x) = log2/3 x. Ask them to solve  x 2 = 1, and to explain why the answer is negative. 3

x sin x , and x sin(π/4) . Ideally, give the • Compute the derivatives of the functions (sin x)π/4 , sin π4 , π4 students time to work independently, and then discuss the answers as a class.   √ x 2x dx

2 dx and 2 . • Compute x + 1 ln 2

GROUP WORK 1: Comparisons The purpose of this group work is to give the students a bit of “picture sense”. It is acceptable if they do this by looking at the graphs on their calculators, setting the windows appropriately. Answers: 1. 0 < x < 1.374 and x > 9.940 x > 22.4 4. 0 < x < 1.34 and x > 10.9

2. 0 < x < 1.052 and x > 95.7168

3. 0 < x < 1.18 and

GROUP WORK 2: Irrational, Impossible Relations Before starting this activity, review the definitions of rational and irrational numbers. The hint sheet should be given out only after the students have tried to show that log2 3 is irrational, or at least discussed it enough to understand what they are trying to show. If a group finishes early, have them show that log2 a is always irrational if a is an odd integer. ln 3 γ 2. −γ 3. Answers: 1. −γ 4. log2 3 = 5. The proof is outlined in the hint sheet. 2 ln 2 √ Answers (Hint Sheet): 1. 2a/b = 3 ⇒ b 2a = 3 ⇒ 2a = 3b 2. If a = b = 0, then log2 3 = 0/0, which is undefined. 3. 2a = 2 · 2 · · · · · 2, and 3b = 3 · 3 · · · · · 3, so these numbers can never be equal, because the left is always divisible by two, and the right never is (unless b = 0). 4. Irrational. 387

GROUP WORK 3: e as a Limit One can introduce this activity by deriving the expression in Problem 1 in terms of compound interest. Some students will require more guidance on what types of things to look for in Problem 1. Properties to notice include concavity, asymptotes, extrema, increase/decrease, and positivity/negativity. Answers: 1.

2. The horizontal asymptote of the graph

1 x . corresponds to lim 1 + x→∞ x

y 3 2

3. The limit is e ≈ 2.718.

1 0

1

2

3

4

5

6

7

8

9

10

x

HOMEWORK PROBLEMS Core Exercises: 5, 7, 16, 25, 32, 44, 47, 51, 62 Sample Assignment: 1, 5, 7, 11, 16, 21, 25, 32, 37, 40, 44, 46, 47, 48, 51, 61, 62 Exercise 1 5 7 11 16 21 25 32 37 40 44 46 47 48 51 61 62

D × × × ×

× ×

388

A × × × × × × × × × × × × × × ×

N

G

× ×

×

× ×

×

GROUP WORK 1, SECTION 7.4* Comparisons You have learned that an exponential function grows faster than a polynomial function. Find the values of x > 0 for which 1. 2x ≥ x 3 .

2. (1.1)x ≥ x 2 .

3. 2x ≥ x 5 .

4. 3x ≥ x 5 .

389

GROUP WORK 2, SECTION 7.4* Irrational, Impossible Relations 1. If log2 x = γ , then what is log1/2 x?

2. If logb x = γ , then what is log1/b x (assuming b > 1)?

3. If logb x = γ , then what is logb2 x?

4. We are going to estimate log2 3. In pre-calculus, you memorized that log2 3 ≈ 1.584962501. Suppose you didn’t have this fact memorized. There is no log2 3 button on your calculator! How would you compute it?

5. Unfortunately, the calculator gives us only a finite number of digits. If log2 3 were a rational number, we would be able to express it as a fraction, giving us perfect accuracy. Do you think it is rational or irrational? Try to prove your result.

390

GROUP WORK 2, SECTION 7.4* Irrational, Impossible Relations (Hint Sheet) So, you realize that it’s not easy to determine whether log2 3 is rational! One way to attempt to show that log2 3 is rational is to assume that it is, and try to find integers a and b such a that log2 3 = . If we can show that there are no such a and b, then log2 3 cannot be rational. b a 1. Assume that log2 3 = . Show that a and b must then satisfy 2a = 3b b

2. Notice that a = 0, b = 0 satisfies 2a = 3b . Show that this fact doesn’t help us.

3. Find a = 0 and b = 0 that satisfy 2a = 3b , or show that no such {a, b} exists.

4. Is log2 3 rational or irrational? Why?

391

GROUP WORK 3, SECTION 7.4* e as a Limit



1 x 1. Graph the function f (x) = 1 + , and describe its properties. x

y 3 2 1 0

1

2

3

4

5

6

7

8

9

10 x

2. What is the relationship between the graph you drew for Problem 1 and Formula 9 in the text?



1 x 3. Using Problems 1 and 2 as a guide, what do you think lim 1 + is? x→∞ x

392

7.5 EXPONENTIAL GROWTH AND DECAY SUGGESTED TIME AND EMPHASIS 1 class

Recommended material

POINTS TO STRESS 1. The general growth/decay equation dy/dt = ky, y (0) = y0 and its associated solution curves. 2. Systems that exhibit exponential growth or decay. 3. The importance of the sign of the exponential growth/decay constant k. QUIZ QUESTIONS • Text Question: If dy/dt = ky, and k is a non-zero constant, then y could be (B) 2ekt (C) ekt + 3 (D) kt y + 5 (E) 12 ky 2 + 12 (A) 2ekt y Answer: (B) • Drill Question: A population grows according to the equation d P/dt = k P, where k is a constant and t is measured in years. If the population doubles every 10 years, then what is the value of k? Answer: k = 0.069 MATERIALS FOR LECTURE • Present a few real-world examples of exponential decay. For example, assume that a certain amount of pollution was dumped into a water reservoir, and that clean water was being added to the reservoir and taken from it at a constant rate. If the water is constantly being mixed, then the concentration of pollution will decay exponentially over time. • Derive Equation 2 by solving dy/dt = ky, y (0) = y0 using separation of variables. • Discuss exponential decay problems with an initial y value. Recall Newton’s Law of Cooling (Group Work 3, 100 Section 10.3). Show how the condition y(0) = y0 80 determines whether the solution is a growth or decay 60 function. Perhaps illustrate with a slope-field diagram. 40 20 0

2

4

6

8

10 x

WORKSHOP/DISCUSSION • Referring back to earlier material, show that the expression y = ekt can be written as y = a t and that ekt+c can be written as Aekt . • Do this variant on Exercise 11: We know that the half-life of carbon-14 is 5,730 years. The Shroud of Turin is an ancient artifact that many believe to be the burial shroud of Jesus Christ. In 1988, the Vatican consented to give a few fibers to scientists to carbon-date. They found that the fibers had 92% of the original 14 C left. Discuss what this implies about the age of the shroud. • Go over Exercise 3, using the doubling rate to write solutions in the form y = A2kt , k = 1/T , where T is the doubling time. 393

GROUP WORK 1: The Rule of 72 If a group finishes early, have them redo Problems 1–5 for annual compounding. Answers: 1. 13.86 years 2. The doubling time is not affected by changes in principal. Algebraically, the P0 drops out of the exponential growth equation. Intuitively, the doubling time is a property of the ratio of two numbers, not a property of the numbers themselves. 3. Estimate: 14.4 years, a 3.90% error. 4.

Interest Rate Actual Doubling Time Estimated Doubling Time 3% 23.1 years 24 years 8% 8.66 years 9 years 12% 5.78 years 6 years 18% 3.85 years 4 years The Rule of 72 is accurate to within 4% for the given range of interest rates.

Error 3.90% 3.92% 3.81% 3.90%

5. If r is the interest rate (as a number, not a percentage), then the doubling time is D =

ln 2 . r

6. 100 ln 2 ≈ 69. This estimate gives a doubling time of 13.8, for an error of 0.4%. 7. If the compounding is assumed to be annual, 72 works best for values of r between 3% and 15%. For example, 10% compounded annually doubles every 7.2735 years. The rule of 72 gives 7.2, a good approximation, and the rule of 69 gives 6.9, a bad one.

GROUP WORK 2: More of Homer’s Blood Although this is a thematic sequel to the Group Work in Section 9.4, it can also stand on its own. Answers: 1. It is nearly constant. 2. 30 minutes: 3.

dC = kC dt

dC dC dC ≈ −0.1165; 60 minutes: ≈ −0.0259; 80 minutes: ≈ −0.0096 dt dt dt 5. y = 10e−0.052t

4. k ≈ −0.052

7. Any time after 3:52 P. M .

6. After approximately 13.33 minutes

GROUP WORK 3: Find the Error Answer: When we divide by y in the first step, we assume that y = 0. We would have to check y = 0 as a solution separately. The integration gives ln |y| = kx + c. The derivation assumed y > 0. If we assumed y < 0 we would get y = −Aekx , where A > 0. 394

SECTION 7.5 EXPONENTIAL GROWTH AND DECAY

HOMEWORK PROBLEMS Core Exercises: 2, 5, 8, 18 Sample Assignment: 2, 3, 5, 8, 11, 13, 16, 18, 20 Exercise 2 3 5 8 11 13 16 18 20

D × × × × ×

395

A × × × × × × × × ×

N

×

G

×

×

GROUP WORK 1, SECTION 7.5 The Rule of 72 In this exercise, we attempt to answer the question asked by many investors: “How long is it going to take for me to double my money?” 1. Consider an investment of $100 invested at 5%, compounded continuously. How long would it take for the investor to have $200?

2. What would the doubling time be if the initial investment were $1,000? $10,000? What effect does changing the principal have on the doubling time, and why?

One of the first things that is taught in an economics class is the Rule of 72. It can be summarized thusly: “The number of years it takes an investment to double is equal to 72 divided by the annual percentage interest rate.” 3. What would the Rule of 72 say the doubling time of a 5% investment is? Is it a good estimate?

4. Repeat Problems 1 and 3 for investments of 3%, 8%, 12% and 18%. What can you say about the accuracy of the Rule of 72?

5. Derive a precise formula for the time T to double an initial investment.

6. There is an integer that gives a more accurate answer for continuous or nearly continuous compounding than the Rule of 72. What is this number? Check your answer by using it to estimate the doubling time of a 5% investment.

7. It turns out that there is a reason that we use the number 72 in the Rule. It has to do with one of the assumptions we made. Why do economists use the Rule of 72?

396

GROUP WORK 2, SECTION 7.5 More of Homer’s Blood In a previous exercise, we discussed Homer’s blood pressure. Well, it turns out that Homer has to have an operation. When he is brought into the hospital, he is given a sedative to help him sleep. The doctor wants to operate, but cannot safely do so until the concentration of sedative in his body is less than 0.03 milligrams/liter. In this exercise, we will determine how many hours the doctor must wait until he can operate. The following table of data was obtained by monitoring the levels of the sedative in Homer’s blood. Samples were taken every ten minutes, and the concentration of the drug was determined and reported in milligrams per liter. Time (minutes)

Concentration (mg/L)

0 10 20 30 40 50 60 70 80 90

10.0 6.07 3.68 2.23 1.35 0.820 0.498 0.302 0.183 0.111

1. Is the rate of change of concentration with respect to time a constant?

2. Estimate the rate of change of concentration with respect to time at 30 minutes, 60 minutes, and 80 minutes.

3. Show that the rate of change is (roughly) proportional to the concentration C. Write this relationship as a differential equation for dC/dt.

4. Find the constant of proportionality.

397

More of Homer’s Blood

5. Solve the differential equation you wrote in Problem 3, using the constant of proportionality from Problem 4.

6. When is the drug’s concentration halved?

7. If Homer received the sedative at 2:00 P. M ., when can the doctor start the operation?

398

GROUP WORK 3, SECTION 7.5 Find the Error It is a beautiful spring afternoon, you are sitting in the park, and you are happy. Your favorite volleyball team is going to the Olympics, but that isn’t what is making you happy. Your sweetheart has just told you that you are getting a nice surprise for May Day, but that isn’t what is making you happy. A magazine company has just informed you that you are entered in a million-dollar sweepstakes, but that isn’t what is making you happy. You are happy because you have learned a new theorem: If

dy = ky, then y = Aekx . dx

The nice thing about this theorem, you think, is that the converse is also true. In other words: dy = ky. dx You hear some tinkly bells, and there is the ice cream cart! You wave, and it comes towards you. But there is something wrong with the ice cream vendor. He is wearing large dark glasses and an obviously fake beard and mustache. You buy the ice cream anyway (one of those ice cream bars on a stick), but after he has gone away, you notice this written on the wrapper. If y = Aekx , then

There are two unusual things going on here. First of all, this message appeared on your wrapper in the first place. Secondly, you are most of the way through a “Find the error” problem, and there isn’t an error here! In fact, you might have seen this very proof in your calculus class. You eat your ice cream, and then notice this written on the stick: Since A = ec , then A can’t be zero, and A can’t be negative. dy = ky, no matter what the value of A. But if y = Aekx , then dx So why didn’t your teacher’s ‘proof’ catch all the solutions? Why not, indeed? Find the error.

399

7.6 INVERSE TRIGONOMETRIC FUNCTIONS SUGGESTED TIME AND EMPHASIS 1 2

– 1 class

Recommended material (especially the inverse sine, inverse cosine and inverse tangent func-

tions) POINTS TO STRESS 1. Definitions, and domains and ranges of inverse trigonometric functions. 2. Derivatives and integrals of inverse trigonometric functions. 3. Uses of inverse trigonometric functions in techniques of integration.

QUIZ QUESTIONS

• Text Question: Why is it true that sin−1 (sin x) = x only when − π2 ≤ x ≤ π2 and sin sin−1 x = x only when −1 ≤ x ≤ 1? Answer: The range of sin−1 x is limited to − π2 ≤ sin−1 x ≤ π2 , and its domain is −1 ≤ x ≤ 1. d 2 −1 √ • Drill Question: What is x sin x ? dx   √ √ √ x 4 1 − x sin−1 x + x Answer: √ 1−x

MATERIALS FOR LECTURE • Ask students to guess the largest domain for cos x containing 0 on which cos x is one-to-one. Then define cos−1 x. Answer: 0 ≤ x ≤ π −1

d cos x = − ddx sin−1 x , cos−1 x + sin−1 x is a constant, and show that • Point out that since dx cos−1 x + sin−1 x =

π 2.

This provides an easy way to remember the graph of cos−1 x.

• Solve some right angle identities involving inverse trigonometric functions, such as sin tan−1 x ,



tan sin−1 x and cos sin−1 x .  • Solve some integration problems using inverse trigonometric functions, such as  2x dx . √ 1 − x4 400

0

2

1 dx and 1 + 4x 2

SECTION 7.6 INVERSE TRIGONOMETRIC FUNCTIONS

WORKSHOP/DISCUSSION • Set up a problem that requires some additional skills or information. A good example is Exercise 47: Where should the point P be chosen on the line segment AB in the figure at right so as to maximize the angle θ? 5

Œ

A

¬ P

2 º B

3

Answer: We can maximize θ by minimizing the sum S of the

S

complementary angles α and β. Let x be the distance from A to



2 5 + arctan , with 0 < x < 3. P. Then S = arctan x 3−x

2.5 2.4 2.3 2.2

2 −5 2.1 + S = 2 . The critical points turn out to be 0 2 2 3 x 1 x + 25 (3 − x) + 4 √ √ the roots of x 2 − 10x + 5, that is, 5 ± 2 5. So we check x = 0, x = 5 − 2 5, and x = 3 to find that √ x = 5 − 2 5 ≈ 0.528 gives a minimum. The result is verified by the graph of S.

• Define and graph tan−1 x, noting that tan tan−1 x = x for −∞ < x < ∞, while tan−1 (tan x) = x for − π2 < x < π2 . • Sketch the graph of f (x) = tan−1



1+x . 1 + x2



• Differentiate y = tan−1 x 2 + e x , y 2 = sin eθ + cos−1 θ and x 4 + x = cos (x y).

GROUP WORK 1: Inverse Trickery This activity helps the students build a skill that is useful when doing inverse trigonometric substitutions. If they have never seen anything like this before, you may want to do an example on the board with the students before handing out the activity. Answers:

1. √

x x2 + 4

√ 2.

x2 + 9 3

3.

√ 1 − x4 401

x 4. √ 16 − x 2

5.

2 1 + x2

GROUP WORK 2: Where to Hang That Picture? Students will probably need some help getting started on this activity. In Problem 1, they should write θ as the sum of the angles above and below the horizontal (see diagram below). Each of these angles can then be written in terms of the position of the painting using the inverse tangent. θ =α+β x tan α = d h−x tan β = d

d is the (fixed) distance from the observer to the painting. In Problem 2, let α be the angle between the bottom of the painting and the horizontal. Write tan θ as tan x − tan y tan (θ + α − α) and use the identity tan (x − y) = . 1 + tan x tan y Note that Problem 2 of this activity is identical to Exercise 48. Answers:

dα

dβ 1 1 dθ = and sec2 β = − . We set = 0: 1. Starting as indicated above, we get sec2 α dx d dx d dx x h−x h cos2 α cos2 β − =0 ⇔ α=β ⇔ = ⇔ x= . d d d d 2 2. Starting as indicated above, and calling the distance from the observer to the wall x we get tan θ = h+d d − hx x

x = 2 . We can maximize θ by maximizing tan θ, and we find that h+d d x + dh + d 2 1+ x x √ hx is maximized when x = d (h + d). 2 2 x + dh + d

HOMEWORK PROBLEMS Core Exercises: 1, 3, 13, 16, 18, 19, 24, 39, 43, 57, 64 Sample Assignment: 1, 3, 6, 13, 14, 16, 18, 19, 22, 24, 33, 37, 39, 41, 43, 49, 57, 63, 64, 72 Exercise 1 3 6 13 14 16 18 19 22 24

D

×

A × × × × × × × × ×

N

G

Exercise 33 37 39 41 43 49 57 63 64 72

×

402

D

×

A × × × × × × × × × ×

N

G

×

GROUP WORK 1, SECTION 7.6 Inverse Trickery Rewrite the following expressions in terms of x without any trigonometric or inverse trigonometric functions.

1. sin tan−1 2x

2. sec tan−1 3x

3. cos sin−1 x 2

4. tan sin−1 4x

5. sin 2 tan−1 x

403

GROUP WORK 2, SECTION 7.6 Where to Hang That Picture? 1. Suppose we want to hang a rectangular painting of height h on the wall so that an observer standing in front of the painting has the best view. In other words, we want to maximize the angle θ (see diagram) subtended by the painting. Prove that θ is maximized when the center of the painting is at the level of the observer’s eyes.

2. Now suppose the painting has already been hung at a distance d above the eye of the observer (see diagram). How far from the wall should the observer stand to get the best view?

404

APPLIED PROJECT

Where to Sit at the Movies

It is hard to imagine an undergraduate student who has not had an argument with his or her friends concerning the best seat in the movie theater. “Up front” connoisseurs, “Middle center” partisans and “Balcony” fetishists have nearly come to blows over this decision. In this exercise, the students use the mathematics they’ve learned to settle the question for a model movie theater. Problems 3 and 4 can be done with a scientific calculator if a very messy, but still within the students’ abilities.)

CAS

is not available. (The derivative will be

If a more in-depth experience is desired, students could be assigned to perform measurements at a local theater, and thus determine the actual “best seat in the house”.

405

7.7 HYPERBOLIC FUNCTIONS SUGGESTED TIME AND EMPHASIS 1 class

Optional material

POINTS TO STRESS 1. The reason why hyperbolic functions are called hyperbolic (cosh2 x − sinh2 x = 1). 2. The basic hyperbolic differentiation formulas. 3. Inverse hyperbolic functions as logarithms. QUIZ QUESTIONS • Text Question: What is the relationship among cosh x, sinh x and the hyperbola x 2 − y 2 = 1? Answer: The point (cosh x, sinh x) is on that hyperbola. d x • Drill Question: Compute (e cosh 2x). dx d x Answer: (e cosh 2x) = e x (cosh 2x + 2 sinh 2x) dx MATERIALS FOR LECTURE • Present definitions of cosh x and sinh x, and the identity cosh2 x − sinh2 x = 1. Discuss the geometric connection between this identity and the hyperbola x 2 − y 2 = 1, similar to the link between the trigonometric functions cos x and sin x and the unit circle x 2 + y 2 = 1. y (cosh u, sinh u)

x@-y@=1 0

x

x 2 − y 2 = cosh2 u − sinh2 u = 1 • Prove the differentiation rules for sinh x, cosh x and discuss the general pattern. • Present the graphs of sinh x and cosh x, emphasizing that y = cosh x [or y = a cosh x] is not a parabola. x  • Show how c + a cosh models the shape of a hanging chain. a WORKSHOP/DISCUSSION • Develop the graphs of sinh−1 x and cosh−1 x.

• Compute the derivatives of y = sinh (e x ), y = cos (cosh x), y = cosh x 2 , and y = ecosh x . • Show that (cosh x + sinh x)2 = cosh 2x + sinh 2x. In fact, (cosh x + sinh x)n = cosh nx + sinh nx for any integer n. 406

SECTION 7.7 HYPERBOLIC FUNCTIONS

GROUP WORK 1: A Tangent Line to a Hyperbolic Curve Answers: 2. y = 2, y = y0 + (x − x0 ) sinh

y

1.

10

x 2 e + ex − 1 = 0 ex =

0



(multiply by e x )

√ −1+ 5 2

x = ln

x

5

1 2 x0

e x − e−x = −1

3.

5

_5





(the other root is extraneous) √ 

−1+ 5 2

GROUP WORK 2: A Hyperbolic Paradox After the students have completed Problem 4, ask them how we can have 0 · ∞ = 1. Draw an analogy:

1 1 · x = 1, even though lim = 0 and lim x = ∞. lim x→∞ x x→∞ x x→∞ Answers: 1.

2. cosh x − sinh x = e−x , and so the limit is 0.

y 10

3. This can be done graphically (as in Problem 1) or algebraically (as in Problem 2).

5

0

4. The students may need to be reminded that cosh2 x − sinh2 x = 1. 1

2

3 x

The limit of the difference appears to be 0. HOMEWORK PROBLEMS Core Exercises: 2, 8, 22, 23, 32, 51, 54, 58, 67 Sample Assignment: 2, 5, 8, 9, 14, 22, 23, 28, 32, 35, 43, 48, 51, 54, 58, 63, 65, 67 Exercise 2 5 8 9 14 22 23 28 32

D

×

A × × × × × × × ×

N

G

Exercise 35 43 48 51 54 58 63 65 67

× ×

407

D

A × × × × × × × × ×

N

G

×

GROUP WORK 1, SECTION 7.7 A Tangent Line to a Hyperbolic Curve x 1. Graph the curve y = 2 cosh . 2 y

1 0

1

x

2. Find the equation for the tangent line at the point (0, 2), then find the equation for the tangent line at the point (x0 , y0 ).

3. At what point is the slope of the tangent line equal to −1?

408

GROUP WORK 2, SECTION 7.7 A Hyperbolic Paradox 1. Graph y = sinh x and y = cosh x on one graph. What does lim (cosh x − sinh x) appear to be? x→∞

y

1 0

1

x

2. Find an equation for cosh x − sinh x and justify your guess in Problem 1 analytically.

3. Show that lim (cosh x + sinh x) = ∞. x→∞

4. What is lim (cosh x − sinh x) (cosh x + sinh x)? x→∞

409

7.8 INDETERMINATE FORMS AND L’HOSPITAL’S RULE SUGGESTED TIME AND EMPHASIS 1 class

Essential material

POINTS TO STRESS 1. The various types of indeterminate forms. 2. The use of l’Hospital’s Rule to determine the behavior of indeterminate forms. 3. The geometric plausibility of l’Hospital’s Rule, as discussed in the text. QUIZ QUESTIONS • Text Question: In Example 2, why wasn’t the Quotient Rule used when taking the derivatives? ex Answer: We are not taking the derivative of 2 . We are using l’Hospital’s Rule to evaluate x f (x) f  (x) = lim  lim . x→∞ g (x) x→∞ g (x) 2x + tan 3x • Drill Question: Find lim . x→0 sin x Answer: 5 MATERIALS FOR LECTURE ∞ is called ”indeterminate” by inviting the students to come up with different pairs of • Demonstrate why ∞ f (x) = 1, 0, 23 , and ∞. Urge functions f and g such that lim f (x) = ∞ and lim g (x) = ∞ but lim x→∞ x→∞ x→∞ g (x) them to come up with simple examples [such as f (x) = g (x) = x, f (x) = x, g (x) = x 2 , and so forth]. If there is student interest, continue by asking for similar functions with lim f (x) = lim g (x) = ∞. x→5

x→5

Finally, remind the students of what happened when they tried to find lim ln (ln x) graphically and x→∞

numerically (putting in very large values of x) to show that it is possible to be led astray by estimations of infinite behavior.

• Show that we can compute lim x a 1/x − 1 = ln a first by using l’Hospital’s Rule and then without x→∞

a 1/x

−1 az − 1 = lim = (a z ) at z = 0]. x→∞ z→0 1/x z 0 ∞ • Show how to manipulate other indeterminate forms into or . [Good examples are lim sin x ln x, 0 ∞ x→0+ 1/x 1/x lim (1 + x) and lim (1 + x) .] l’Hospital’s Rule [ lim

x→∞

x→0+

• Give a graphical explanation for l’Hospital’s Rule as in Figure 1. If the class is interested in a more formal proof, state the Cauchy Mean Value Theorem and outline the formal proof given in the text. Point out that the conclusion of the Cauchy Mean Value Theorem is not always geometrically evident. For example, if f (x) = x 3 , g (x) = e x , a = 1 and b = 2, the conclusion states that for some c with 1 < c < 2, 7 3c2 . = c e e (e − 1) 410

SECTION 7.8 INDETERMINATE FORMS AND L’HOSPITAL’S RULE

WORKSHOP/DISCUSSION • Approximate a few of the limits discussed in the text using numerical methods. For example, look at ln x f (x) = for x = 0.9, 0.99, 0.999, 1.01, 1.001, etc. x −1 1 − cos x 6 first graphically, then numerically, and finally compute it using x→0 x 12

• Have the students estimate lim l’Hospital’s Rule.

x −3 x −3 and lim 3 . Have them try 2 x→2 x − 10x 2 + 31x − 30 − 10x + 31x − 30 to figure out why one exists and the other doesn’t. (This can be done algebraically, or by looking at a graph. The students should be led to try both.)

• Have the students compute lim

x→3 x 3



1/x and lim • Do some challenging limit problems, such as lim sin π2 + x x→0+

x→1



1 1 − . ln x x −1

GROUP WORK 1: Find the Error Most students, once they understand l’Hospital’s Rule, believe the conclusion to be stronger than it is. This activity helps them to clarify what it does and does not say. Answers:

sin x x 2 + sin x =1 = lim 1 + 2 1. lim x→∞ x→∞ x2 x f  (a) exists or is ±∞. If not, then l’Hospital’s Rule does not apply. x→a g  (a) Have the students reread the rule carefully, particularly the last clause.

2. l’Hospital’s Rule applies only if lim

GROUP WORK 2: The Sector Ratio The students may not remember the formula for the area of a sector of the unit circle. Before handing out the exercise, give enough of a trigonometry review so that they will be able to work the problem. Students may need to be prompted to use sin (θ/2) = x to write θ as a function of x. Make sure to cover Part 4 with the class at the end. It is important that they understand that taking limits isn’t some sort of symbolic game. These limits actually do make physical sense when looked at geometrically. If the students need to see this exercise another way, have them graph S (x) /T (x), and see what happens for large x. 411

Answers:

√ θ S (x) =1 3. lim , so S (x) = arcsin x. 2. T (x) = x 1 − x 2 x→0 T (x) 2 4. When x gets very small, the isosceles triangle and the circular sector have virtually the same area, as the arc subtended by θ starts to look more and more like a straight line. 1. S (θ) =

HOMEWORK PROBLEMS Core Exercises: 1, 6, 31, 51, 66, 77, 85, 91 Sample Assignment: 1, 6, 11, 24, 31, 46, 51, 59, 66, 67, 77, 81, 84, 85, 89, 91, 92, 97 Exercise 1 6 11 24 31 46 51 59 66

D

A × × × × × × × × ×

N

G

Exercise 67 77 81 84 85 89 91 92 97

×

412

D

× × ×

A × × × × × × × × ×

N

G × ×

×

GROUP WORK 1, SECTION 7.8 Find the Error It is a beautiful Spring day. You are again jolly, for you have a backpack filled with schoolbooks and cornbread, and you are ready to have your lunch and start working on your calculus homework. You lay out your food and notebook on a picnic table, and suddenly a familiar scent catches your attention. “L’Hospital’s Rule, eh?” comments the wild-eyed stranger, who has sat down next to you. “Interesting stuff. . . if you don’t mind reading LIES!” “But l’Hospital’s Rule is useful! It is a true and wonderful technique of computing limits,” you say. “Useless!” he bellows. “You show me a steak and baked potato with a side of l’Hospital’s Rule and I’ll show you a steak and baked potato, by cracky!” “What are you talking about?” “Look at this, and see the lies exposed!” The stranger writes this in your notebook: ∞ x 2 + sin x 2x + cos x  = lim both indeterminate forms of type x→∞ x→∞ 2x ∞ x2 2 − sin x which does not exist! = lim x→∞ 2 “Why, yes, that is a very good use of l’Hospital’s Rule, and it proves my point,” you say. “You started with a hard limit in one of the standard indeterminate forms, used l’Hospital twice, correctly I might add, and arrived x 2 + sin x at the answer, that lim does not exist.” x→∞ x2 At that point the stranger laughs and laughs, and winds up saying between chortles, “Graph it, Sonny.” lim

Fearlessly, you graph

x 2 + sin x , and come up with this: x2 y

1 0

x

1

“Looks like the limit is 1 to me, boy. And last thing I heard was that 1 exists. Enjoy your johnnycake!” You start to lose your appetite. Was l’Hospital lying all this time? Is the limit 1? Or doesn’t it exist? If 1 doesn’t exist, then don’t things start looking grim for 2, 3 and 4? Does it still “take 1 to know 1”? Find the error. x 2 + sin x without using l’Hospital’s Rule. x→∞ x2

1. Compute lim

2. What is wrong with the stranger’s argument? 413

GROUP WORK 2, SECTION 7.8 The Sector Ratio Consider the following circular sector: x

1

1 ¬ _ 2

Let S (x) be the total area of the sector. Let T (x) be the total area of the isosceles triangle formed by the radii and the chord. 1. Find a formula for S (x).

2. Find a formula for T (x).

3. Compute lim

x→0

S (x) . T (x)

4. Explain why this result makes sense geometrically.

414

WRITING PROJECT

The Origins of l’Hospital’s Rule

This is a nice historical project on the mathematical origins of l’Hospital’s Rule. It would be good as a longterm project for a liberal arts course, or as a shorter project in which just the historical aspects are emphasized. Details about directions for investigation are given in the text. Plenty of information about the historical figures can be found on the World Wide Web, but urge the students to use caution — information from the Web may not be meticulously researched, and may therefore be unreliable. This project could also be done by a pair of motivated students, as opposed to a larger group.

415

7

SAMPLE EXAM

Problems marked with an asterisk (*) are particularly challenging and should be given careful consideration. 1. Let f be a one-to-one function whose inverse function is given by the formula f −1 (x) = x 5 + 2x 3 + 3x + 1 (a) Compute f −1 (1) and f (1).

(b) Compute the value of x0 such that f (x0 ) = 1.

(c) Compute the value of y0 such that f −1 (y0 ) = 1.

(d) Below is a graph of f −1 . Draw an approximate graph of f . y fÐ!

0

416

x

CHAPTER 7 SAMPLE EXAM

2. Find constants A, B, and k such that the equation f (x) = A2kx + B satisfies the following three conditions: • f (x) is always decreasing. • f (x) has a horizontal asymptote at y = 1, and • f (x) goes through the point (0, 4).

3. Find constants A and k such that the equation f (x) = A2kx fits the following data as closely as possible: x y 1 1.8000 2 2.7000 4 6.0750 5 9.1125 7 20.503

4. Compute derivatives of the following functions. (a) f (x) = e2π x

(b) g (x) = x 2πe

(c) h (x) = (eπ )2x

(d) l (x) = π



e2x



5. Consider the function h (x) = (1 + sin π x)g(x) . Suppose g (1) = 2 and g  (1) = −1. Find h  (1).

417

2

6. (a) Find the point on the curve y = e−x where the slope of the tangent line is 2/e.

(b) Find the x-intercept c of the line tangent to the curve at that point. y

1

_2

_1

0

1

c

2 x

7. Let f (x) = ln 1 + x 2 . (a) What is lim f (x)? x→∞

(b) What is lim f  (x)? x→∞

(c) Using parts (a) and (b), explain the behavior of the function as x gets large.

8. Draw a graph of f (x) = ln ln x (a) Over the range [2, 10]. (b) Over the range [2, 100]. (c) What is limx→∞ ln ln x? 418

CHAPTER 7 SAMPLE EXAM

9. (a) Below is a graph of f (x) = A ln x + B. What are A and B? y 3 2 (1, 1)

1 0

1

2

3

4

5

x

_1

(b) Below is a graph of f (x) = ln kx. What is k? y 2 1

0

1

2

3

x

_1

 10. Let g (x) = (a) Compute

5x dt

t 1  g (x).

(b) Using part (a), find an equation for g (x) in terms of ln x.  11. Using the equation ln x = (a) The expression eln x

x1

dt, x > 0, justify the following statements: t = x is true only for positive values of x. 1

(b) ln x is increasing everywhere.

(c) ln x 1/2 = 12 ln x. 12. Let g (x) =

 x2 0

 ln

1 2

+

√ t dt for all x. Compute the critical points of g.

13. Consider the functions f (x) = e x − 1 and g (x) = 2 ln (x + 1). (a) Draw a graph of these functions for − 12 ≤ x ≤ 2. (b) Write an equation for the area between these two functions for − 12 ≤ x ≤ 2. 419

7

SAMPLE EXAM SOLUTIONS

1. f −1 (x) = x 5 + 2x 3 + 3x + 1 (a) f −1 (1) = 7, f (1) = 0 (b) The value x0 such that f (x0 ) = 1 is f −1 (1) = 7. (c) The value y0 such that f −1 (y0 ) = 1 is f (1) = 0. (d) The graph of f (x) is the graph of f −1 (x) reflected about the line y = x. y

fÐ! f

0

x

2. Let f (x) = 3 (2)−x + 1. Then f (x) is always decreasing, has a horizontal asymptote at y = 1, and f (0) = 4. 3. f (x) = 1.2 (2)0.585x 4. (a) f  (x) = 2π e2π x (b) g  (x) = 2π e · x 2πe−1 (c) h  (x) = 2 ln eπ · (eπ)2x = 2 (1 + ln π ) (eπ)2x (d) l  (x) = (ln π ) π



e2x



· 2e2x   π cos π x g(x)   5. h (x) = (1 + sin π x) g (x) ln (1 + sin π x) + g (x) ; 1 + sin π x   2π  2 h (1) = 1 · −1 · 0 + (−1) = −2π 1  2  2 2 2 6. (a) The slope is e−x = −2xe−x . If −2xe−x = , then −x 2 = −1 and −2x = 2. So x = −1 and e

1 . the point is −1, e

1 1 2 (b) The tangent line at −1, is y = + (x + 1). Set y = 0 and solve for x to get x = − 32 . e e e 7. (a) lim f (x) = ∞ x→∞

2x 1 2x · 2x = . lim f  (x) = lim =0 2 2 x→∞ 1 + x 2 1+x 1 + x x→∞ (c) The function increases at a slower and slower rate, but still goes to infinity.

(b) f  (x) =

420

CHAPTER 7 SAMPLE EXAM SOLUTIONS

8. (a)

(b)

y 1

(c) lim ln ln x = ∞

y

x→∞

2

1

0

10

x 0

100

x

_1

_1

9. (a) Use f (1) = 1 and f (3) = 0 to get A = −

1 and B = 1. ln 3

(b) Use f (0.2) = 0 to get K = 5.

1 1  = 10. (a) g (x) = 5 5x x  (b) g (x) must differ from ln x by a constant, so g (x) = ln x + c. Using g (1) = c = ln 5, so g (x) = ln x + ln 5.

1

5 dt

t

= ln 5, we get

11. (a) eln x = x is true only for positive values of x because ln x is defined only for positive values of x. This is because we cannot integrate across the discontinuity at t = 0. 1 is positive for t > 0, so ln x is increasing everywhere. t  x 1/2 1/2 1 (c) ln x = dt. Let u = t 2 . Then u 1/2 = t so 12 u −1/2 du = dt and t 1  x  x 1/2

1 1 dt = du = 12 ln x. So ln x 1/2 = 12 ln x. t 1 1 2u   12. g  (x) = 2x ln 12 + |x| = 0 where x = 0, x = 12 , or x = − 12 . g (x) has critical points at x = 0 and at (b)

x = ± 12 . 13. (a) The points of intersections are at x = 0 and x = c, where c ≈ 0.752. y 8 6 4 2 _1

(b)

1 _ _2

0

c

1

2

3 x

  x e − 1 − 2 ln (x + 1) dx −1/2 0  x  c

 = −1/2 e − 1 − 2 ln (x + 1) dx + 0 2 ln (x + 1) − e x − 1 dx 2  + c e x − 1 − 2 ln (x + 1) dx

A =

2

421

where c ≈ 0.752.

422