7 Inverse Functions: Exponential, Logarithmic, and Inverse ...

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 − 4compute 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,andg −1 .
• Show that f (x) = 3x 2 + 2x + 1 is increasing for x ≥− 1 3 by computing f ′ (x). Explicitly compute
√ 3x − 2 − 1
f −1 (x) =
and discuss its domain and range.
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 − π 4 ≤ x ≤ π 4
. Graph the inverse and then compute
( f
−1 ) ′ (1).
• Suppose f is one-to-one and we have the following information:
x f (x) f ′ (x)
3 7 2
4 9 1
5 13 2
6 19 0.5
Show the students how to determine ( f −1) ′ (
(9) and f
−1 ) ′ (19).
( f
−1 ) ′ (
(7) = f
−1 ) ′ (13).
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 D A N G
2 ×
3 ×
6 ×
9 ×
15 ×
20 × × ×
21 × ×
Exercise D A N G
24 ×
30 × ×
33 × ×
34 × ×
37 ×
38 ×
41 ×
347

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?
x
a −1 (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

Which of the following relations are functions?
GROUP WORK 2, SECTION 7.1
Functions in the Classroom
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) = hisorhereyecolor
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) = hisorherspleen
All the people in your classroom Pencils f (person) = hisorherpencil
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) = howmuchcashtheyhaveonthem
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 ),whereh 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: 3 3 = 3 · 3 · 3, 3 3/4 = 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,
and can thus similarly approximate 3 √7 to any degree of precision.
• Drill Question: What is the derivative of f (x) = e x3 ?
Answer: 3x 2 e x3
MATERIALS FOR LECTURE
• Starttodrawagraphof2 x 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) = 2 x , f ′ (0) ≈ 0.693, and for g (x) = 3 x , g ′ (0) ≈ 1.099.
WORKSHOP/DISCUSSION
• Point out that if f (x) = e x3 , then f ′ (x) is not equal to e x3 , as seen in the quiz. Compute the derivative of
e x3 and then do a straightforward Product Rule problem such as finding the derivative of f (x) = e x sin x.
e x
Then have the students try to compute the derivatives of
1 + e x , ( x 2) e ,andex 3 + e x3 + xe 3 + x e3 + ( e 3) x .
• 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 3 x > x 3 and where 2 x > 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 = 2 x . (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
Aug 1981 213 −
May 1982 235 1.1400
Aug 1983 562 2.0088
Oct 1984 1024 1.6724
Oct 1985 1961 1.9150
Feb 1986 2308 1.6303
Nov 1986 5089 2.8699
Dec 1987 28,174 4.8534
Jul 1988 33,000 1.3113
Oct 1988 56,000 8.2927
Jan 1989 80,000 4.1649
Jul 1989 130,000 2.6406
Oct 1989 159,000 2.2378
Oct 1990 313,000 1.9686
Jan 1991 376,000 2.0824
Jul 1991 535,000 2.0246
Oct 1991 617,000 1.7690
Month Hosts Growth
Jan 1992 727,000 1.9275
Apr 1992 890,000 2.2461
Jul 1992 992,000 1.5434
Oct 1992 1,136,000 1.7198
Jan 1993 1,313,000 1.7846
Apr 1993 1,486,000 1.6407
Jul 1993 1,776,000 2.0403
Oct 1993 2,056,000 1.7961
Jan 1994 2,217,000 1.3520
Jul 1994 3,212,000 2.0990
Oct 1994 3,864,000 2.0943
Jan 1995 4,852,000 2.4862
Jul 1995 6,642,000 1.8739
Jan 1996 9,472,000 2.0337
Jul 1996 12,881,000 1.8493
Jan 1997 16,146,000 1.5712
Jul 1997 19,540,000 1.4646
353
Month Hosts Growth
Jan 1998 29,670,000 2.3056
Jul 1998 36,739,000 1.5333
Jan 1999 43,230,000 1.3846
Jul 1999 56,218,000 1.6911
Jan 2000 72,398,092 1.6585
Jul 2000 93,047,785 1.6518
Jan 2001 109,574,429 1.3868
Jul 2001 125,888,197 1.3199
Jan 2002 147,344,723 1.3699
Jul 2002 162,128,493 1.2107
Jan 2003 171,638,297 1.1208
Jul 2003 233,101,481 1.8444
Jan 2004 285,139,107 1.4963
Jul 2004 317,646,084 1.2410
Jan 2005 353,284,187 1.2370

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
300,000,000
240,000,000
180,000,000
120,000,000
60,000,000
8
7
6
5
4
3
2
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 2. 0 < x < 1.052 and x > 95.7168 3. 0 < x < 1.18 and
x > 22.4 4. 0 < x < 1.34 and x > 10.9
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 D A N G
1 × ×
2 ×
3 × ×
5 × ×
10 ×
13 ×
14 ×
16 ×
17 × ×
Exercise D A N G
20 × ×
21 ×
24 ×
25 ×
32 ×
35 ×
43 ×
48 ×
354
Exercise D A N G
56 ×
59 × × ×
64 ×
65 ×
73 ×
79 ×
86 ×
88 × ×

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 y
1 2
2 3
3 4
4 5
x y
21.5 4.32
32.6 4.203
43.7 4.090
54.8 3.980
x y
−3 1.1
−2.5 2.14
−2 3.18
−1.5 4.25
x y
1 −5.00
3 −0.98
6 5.05
8 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 Ae kx ) 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 y
1 5
2 10
3 20
4 40
x y
21.5 4.32
32.6 4.203
43.7 4.090
54.8 3.980
x y
−3 1.1
−2.5 1.65
−2 2.475
−1.5 3.7125
x y
1 0.8
3 7.2
6 194.4
8 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,andC 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 > 0forwhich
1. 2 x ≥ x 3 .
2. (1.1) x ≥ x 2 .
3. 2 x ≥ x 5 .
4. 3 x ≥ 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 log a x = x only hold for x > 0?
Answer: log a x is defined only for x > 0.
• Drill Question: What is the inverse function for y = log a x?
Answer: a x
MATERIALS FOR LECTURE
√
• Point out that log √ ln √ 11
17 11 =
ln √ ln 11
= looks much simpler in the latter form.
17 ln 17
(
• Sketch a graph of f (x) = log 2 (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
0
2
x
f −1 (x) = 2 x/2 − 3
• Sketch a graph of f (x) = 4ln(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
ln (ln x) > μ.
lim
x→∞
ln (ln x) = ∞. Point
out that if ln x > eμ where μ is a positive number, then
WORKSHOP/DISCUSSION
( )
• Discuss the graph of f (x) = 3 −x 2 x after writing it as f (x) = 2x 2 x
3 x = . See if students can identify
3
( ) 2 x
its inverse as g (x) = log 2/3 x. Ask them to solve = 1, and to explain why the answer is negative.
3
• If x satisfies 3 x > x 3 , take logs to base 3 on both sides to get x > 3log 3 x,or 1 3 x > log 3 x. Point out that
log n x grows more slowly than x/n,forx sufficiently large.
• Estimate where 5 x > 10 7 by the use of technology. Then have the students use logarithms to find an exact
answer.
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 2. 0 < x < 1.0409, x > 125.3635
3. 0 < x < 1.1772, x > 22.4400 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 log 2 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 log 2 a is always
irrational if a is an odd integer.
Answers: 1. −γ 2. −γ 3. γ 4. log
2
2 3 = ln 3 5. The proof is outlined in the hint sheet.
ln 2
Answers (Hint Sheet): 1. 2 a/b = 3 ⇒ b√ 2 a = 3 ⇒ 2 a = 3 b 2. If a = b = 0, then log 2 3 = 0/0,
which is undefined. 3. 2 a = 2 · 2 ·····2, and 3 b = 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 D A N G
2 × ×
3 ×
4 ×
8 ×
9 ×
10 ×
15 ×
18 ×
21 × ×
23 ×
Exercise D A N G
26 ×
31 ×
37 ×
40 × ×
43 × ×
44 × ×
47 ×
55 ×
66 ×
359

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 > 0forwhich
1. 3 x ≥ x 7 .
( ) 7 x
2. ≥ x 4 .
6
3. x 5 ≥ log 2 x.
4. x 2 ≥ log 1.1 x. 360

1. If log 2 x = γ, then what is log 1/2 x?
GROUP WORK 2, SECTION 7.3
Irrational, Impossible Relations
2. If log b x = γ, then what is log 1/b x (assuming b > 1)?
3. If log b x = γ, then what is log b 2 x?
4. We are going to estimate log 2 3. In pre-calculus, you memorized that log 2 3 ≈ 1.584962501. Suppose you
didn’t have this fact memorized. There is no log 2 3 button on your calculator! How would you compute
it?
5. Unfortunately, the calculator gives us only a finite number of digits. If log 2 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 log 2 3 is rational!
Onewaytoattempttoshowthatlog 2 3 is rational is to assume that it is, and try to find integers a and b such
that log 2 3 = a b . If we can show that there are no such a and b,thenlog 2 3 cannot be rational.
1. Assume that log 2 3 = a b . Show that a and b must then satisfy 2a = 3 b
2. Notice that a = 0, b = 0 satisfies 2 a = 3 b . Show that this fact doesn’t help us.
3. Find a ≠ 0andb ≠ 0 that satisfy 2 a = 3 b ,orshowthatnosuch{a, b} exists.
4. Is log 2 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
• Review logarithmic and exponential functions. Point out that there is no way to simplify ln (a + b) or ln a
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
{
f (x) and compute f ′ ln x + a if x > 0
(x). Close with discussions of f (x) =
from geometric and
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→∞ n)1/n by
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).
• Compute d
dx ln ( (x + 1) 5) , first by simplifying using properties of logarithms, and then using the Chain
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
group finishes early, have them compute the derivative of ( x 2 + 1 ) sin x .
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.
1 dy
(x + 1) x dx
y = (x + 1) x
ln y = ln (x + 1) x
ln y = x ln (x + 1)
1 dy
= ln (x + 1) +
x
y dx x + 1
= ln (x + 1) +
x
x + 1
[
dy
dx = (x + 1)x ln (x + 1) +
x ]
x + 1
[
7. They are identical. 8. e x ln(x+1) ln (x + 1) +
x ]
x + 1
9. e x ln(x+1) [
ln (x + 1) +
x ]
[
= e ln((x+1)x ) ln (x + 1) +
x + 1
x ] [
= (x + 1) x ln (x + 1) +
x + 1
x ]
x + 1
10. y ′ x sin y sin y + xy cos x sin x ln y
=−y
x ( x sin y y cos y ln x − y cos x cos x ) or y sin y + x sin x ln y
x cos x − y ln x cos y
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
3
2
1
2. The horizontal asymptote of the graph
(
corresponds to lim 1 + 1 ) x
.
x→∞ x
3. The limit is e ≈ 2.718.
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 2. ( sin π ) x ( ) ( √ ) x
4 ln sin
π
4
=−
2
√ (
2
ln 2 3.
π4
) sin x ( ) ln
π
4
cos x
4. sin ( π
4
) x sin(π/4)−1 =
(
√ √2−2 )
2 /2
2
x
5. 4πx 3 cos ( x 4)
365

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),andln(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.
∫ m 2 ∫
dx ln m 2
4.
m x ln x = du
ln m u = ln ( ( )
ln m 2) 2lnm
− ln (ln m) = ln = ln 2
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 D A N G
2 ×
9 ×
16 ×
28 ×
31 ×
35 ×
38 ×
39 × ×
45 ×
46 ×
Exercise D A N G
54 ×
57 × ×
67 × × ×
68 × × × ×
70 ×
74 ×
75 ×
78 ×
87 ×
91 ×
366

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
dx 3x = 3 x ln3tocompute f ′ (x)?
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
dx
alone on the left hand side, and you are done!
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) ,computeg ′ (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

1. Graph the function f (x) =
y
3
2
1
GROUP WORK 2, SECTION 7.4
e as a Limit
(
1 + 1 ) x
, and describe its properties.
x
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?
(
3. Using Problems 1 and 2 as a guide, what do you think lim 1 + 1 ) x
is?
x→∞ x
369

Compute the derivatives of the following functions:
1. (sin x) π/4
GROUP WORK 3, SECTION 7.4
The Right Procedure at the Right Time
2. ( sin π ) x
4
3. ( π
4
) sin x
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
x ln x
from x = 1.01 to 1.0201, from 4 to 16, and from 10 to 100.
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) = 1
x ln x .
What is it?
371

Some Surprising Areas
4. Prove that the area enclosed by f (x) = 1
x ln x ,thex-axis, the line x = m and the line x = m2 is
independent of your choice of m (as long as m > 1).
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) =
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.
∫ x
1
dt
t , x > 0.
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?
t
Answer: Because if 0 < x < 1,
is positive.
MATERIALS FOR LECTURE
• Discuss properties of ln x =
∫ x
1
1
∫ x
1
dt
t
=−
∫ 1
x
dt
t
where ∫ 1
x
dt
t , the area under the curve 1 t
dt
t , using the integral definition, and review why if ln x = ∫ x
d
dx ln x = 1 x , x > 0.
• Derive the formula ln ( x 1/q) = 1 q ln x by rewriting the equation y = x1/q as y q
1
from x to 1,
1
t
dt, then
= x. Then use this
equation to show that ln ( x p/q) = p q
ln x.
( x
2
)
• Discuss the properties and graph of f (x) = ln
x 2 . Point out that the two other forms of f (x),
+ 1
( )
1
namely f (x) = ln
1 + 1/x 2 =−ln
(1 + 1 )
x 2 or f (x) = 2lnx − ln ( x 2 + 1 ) , could be helpful.
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
use the absolute value in the formula d
dx ln |x| = 1 x .
• Present an example of logarithmic differentiation, such as finding the derivative of (sin x) x2 +3x .
• Go over Exercise 73 with the students.
373

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
group finishes early, have them compute the derivative of ( x 2 + 1 ) sin x .
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
1 dy
(x + 1) x dx
ln y = ln (x + 1) x
ln y = x ln (x + 1)
1 dy
= ln (x + 1) +
x
y dx x + 1
= ln (x + 1) +
x
7. They are identical. 8. e x ln(x+1) [
ln (x + 1) +
9. e x ln(x+1) [
ln (x + 1) +
x + 1
[
dy
dx = (x + 1)x ln (x + 1) +
x ]
x + 1
x ]
[
= e ln((x+1)x ) ln (x + 1) +
x + 1
10. y ′ x sin y sin y + xy cos x sin x ln y
=−y
x ( x sin y y cos y ln x − y cos x cos x ) or y sin y + x sin x ln y
x cos x − y ln x cos y
x ]
x + 1
x ] [
= (x + 1) x ln (x + 1) +
x + 1
x ]
x + 1
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) = 1 t .
The combined area of these rectangles is
(
1 − 1 ) ( 1
+
2 2 3)
− 1 ( 1
+ ... +
n − 1 )
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
+
2 2 2 3)
− 1 ( 1
+ ... +
n − 1 )]
n + 1
374

SECTION 7.2*
THE NATURAL LOGARITHMIC FUNCTION
Answers:
1. ln (n + 1) 2. γ n = (The area under the rectangles) − ∫ n+1 dt
1
t
3. (a) Applying the hint gives an area of 1 − 1
n + 1 .
(b) The hint gives this result directly.
(c) If n ≥ 9, 1 − 1
n + 1 = 0.9.
4. 25
761
1,145,993
12
− ln 5 ≈ 0.47389542,
280
− ln 9 ≈ 0.52063257,
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 t < 1
t (N−1)/N for t ≥ 1. ∫ x
1
dt
t
<
∫ x
1
dt
for x ≥ 1.
t (N−1)/N
2. Computing the integrals gives the result.
3. Transitivity gives the result.
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),andln(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.
4.
∫ m 2
m
∫
dx ln m 2
x ln x = ln m
du
u = ln ( ln m 2) − ln (ln m) = ln
375
( ) 2lnm
= ln 2
ln m

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 D A N G
2 ×
4 ×
5 ×
9 ×
10 ×
11 ×
17 ×
23 ×
25 ×
35 ×
38 ×
42 ×
45 ×
51 × ×
59 ×
64 ×
68 ×
376

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
dx 3x = 3 x ln3tocompute f ′ (x)?
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
dx
alone on the left hand side, and you are done!
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) ,computeg ′ (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
bound for the integral.
∫ n+1
1
1
t
dt with n subintervals. Notice that this sum is an upper
2. Let γ n be the area of the shaded region in the diagram above. Show that γ n = 1 + 1 2 + 1 3 +···+ 1 n −
ln (n + 1).
3. It is a fact that lim
n→∞ γ n = γ exists. We want to estimate γ.
(a) Using a geometric argument, show that γ n ≤ 1 for all n.
(b) Using a geometric argument, show that 1 2
(
1 − 1 )
≤ γ
n + 1 n for all n.
(c) Conclude from parts (a) and (b) that 0.45 ≤ γ n ≤ 1ifn ≥ 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

1
GROUP WORK 3, SECTION 7.2*
How Slowly can ln x Grow?
1. Let N be a positive integer. How do the functions 1
∫
t
x ∫
dt x
1
≤ dt for x ≥ 1.
t t (N−1)/N
1
and
1
compare for t ≥ 1? Show that
t (N−1)/N
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 ≤ Nx 1/N , for any positive integer N. Conclude that ln x grows “more
slowly” than any root function x 1/N ,whereN 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
x ln x
from x = 1.01 to 1.0201, from 4 to 16, and from 10 to 100.
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) = 1
x ln x .
What is it?
381

4. Prove that the area enclosed by f (x) = 1
x ln x ,thex-axis, the line x = m and the line x = m2 is
independent of your choice of m (as long as m > 1).
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 e ln 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
x→−∞ ex = 0?
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
x→0
• Compute d
• Compute d
dx
d
dx
ln x =−∞, it follows that lim
+ x→−∞ ex = 0.
(
e x3) (
≠ e x3 . Show that
dx (ex ) = e x and then point out that
( ) e x2 +x
(there are two possible ways to do this),
d
dx (sin (ex )) and
dx
d
e x3) ′
= 3x 2 e x3 .
√
1 + ( e −x) 2 .
• 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.
∫ ∫
∫
• Compute the following integrals: e 3x dx, e x cos e x e x ∫
dx,
1 + e x dx,and sin xe cos x dx.
WORKSHOP/DISCUSSION
• Have the students graph f (x) = e x2 and estimate the slope at x = 0andx = 1. Ask why the slope at
x = 0 is 0, while the slope of g (x) = e x at x = 0is1.
• Derive the properties of f (x) = e1/x
, and sketch its graph. Point out that writing f (x) = ex+(1/x)
e−x simplifies the process. Where is f ′ (x) = 0?
( 5
• Do some differentiations using e x , such as those of e sin2 x , x 2 e x5 , e x2 + 1)
and cos (x e + e x ).Anice
summary problem is g (x) = ex 3 + e x3 + xe 3 + x e3 + ( e 3) x .
∫ 2xe
ln ( x 2 +1 ) ∫ ln (sin e x ∫
)
• Compute
x 2 dx,
+ 1 e x cos e x dx,and e 2x
√ dx.
1 + e 4x
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.
4. P (x) = 1 + x + x2
2 + x3
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 D A N G
1 ×
2 ×
4 ×
5 ×
8 ×
15 ×
18 ×
21 ×
23 ×
25 ×
28 ×
39 ×
45 ×
50 ×
61 × × ×
63 ×
72 ×
77 ×
78 ×
90 × ×
384

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
0 f (u) du ≤ ∫ x
0 eu 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 + x2
2 ≤ ex if
x ≥ 0.
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 1 2
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 log a x = x hold only for x > 0?
Answer: log a x is defined only for x > 0.
• Drill Question: What is the inverse function for y = log a x?
Answer: a x
MATERIALS FOR LECTURE
• Starttodrawagraphof2 x 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 > 0andx a real number, and discuss why this equation
makes sense even for irrational numbers. [For example, √ 3
√ √
2√
= e
3ln 2
= e
properties of a x for a > 1and0< a < 1.
( √3ln2 )
/2 ]. Compare the
• Derive the general Power Rule (x n ) ′ = nx n−1 for any real number n.
• By differentiating the equation a log a x = x implicitly, derive the formula ( log a x ) ′ 1 =
x ln a .
• Estimate where 3 x > x 3 and 2 x > 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) = log 2 (x + 3)
2 ) ,sketch
the inverse function, and then find an algebraic
formula for the inverse.
Answer:
y
y=f(x)
y=f Ð!(x)
2
0
2
x
f −1 (x) = 2 x/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) = log 2 (sin x) sin ( log 2 x ) (
or f (x) = log 3 e x + log 2 x ) .
• See if the students can identify the inverse of f (x) =
(
23
) x
as g (x) = log2/3 x. Ask them to solve
(
23
) x
= 1, and to explain why the answer is negative.
• Compute the derivatives of the functions (sin x) π/4 , ( sin ( ))
π x, ( π4
) sin x,
4
andx sin(π/4) . Ideally, give the
students time to work independently, and then discuss the answers as a class.
∫ ∫
√2 x 2xdx
• Compute dx and ( x 2 + 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 2. 0 < x < 1.052 and x > 95.7168 3. 0 < x < 1.18 and
x > 22.4 4. 0 < x < 1.34 and x > 10.9
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 log 2 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 log 2 a is always
irrational if a is an odd integer.
Answers: 1. −γ 2. −γ 3. γ 4. log
2
2 3 = ln 3 5. The proof is outlined in the hint sheet.
ln 2
Answers (Hint Sheet): 1. 2 a/b = 3 ⇒ b√ 2 a = 3 ⇒ 2 a = 3 b 2. If a = b = 0, then log 2 3 = 0/0,
which is undefined. 3. 2 a = 2 · 2 ·····2, and 3 b = 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. y
2. The horizontal asymptote of the graph
(
3
corresponds to lim 1 + 1 ) x
.
x→∞ x
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 D A N G
1 × ×
5 × ×
7 ×
11 × ×
16 × ×
21 ×
25 ×
32 ×
37 ×
40 ×
44 × ×
46 ×
47 ×
48 ×
51 ×
61 × × ×
62 × × × ×
388

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 > 0forwhich
1. 2 x ≥ x 3 .
2. (1.1) x ≥ x 2 .
3. 2 x ≥ x 5 .
4. 3 x ≥ x 5 .
389

1. If log 2 x = γ, then what is log 1/2 x?
GROUP WORK 2, SECTION 7.4*
Irrational, Impossible Relations
2. If log b x = γ, then what is log 1/b x (assuming b > 1)?
3. If log b x = γ, then what is log b 2 x?
4. We are going to estimate log 2 3. In pre-calculus, you memorized that log 2 3 ≈ 1.584962501. Suppose you
didn’t have this fact memorized. There is no log 2 3 button on your calculator! How would you compute
it?
5. Unfortunately, the calculator gives us only a finite number of digits. If log 2 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 log 2 3 is rational!
Onewaytoattempttoshowthatlog 2 3 is rational is to assume that it is, and try to find integers a and b such
that log 2 3 = a b . If we can show that there are no such a and b,thenlog 2 3 cannot be rational.
1. Assume that log 2 3 = a b . Show that a and b must then satisfy 2a = 3 b
2. Notice that a = 0, b = 0 satisfies 2 a = 3 b . Show that this fact doesn’t help us.
3. Find a ≠ 0andb ≠ 0 that satisfy 2 a = 3 b ,orshowthatnosuch{a, b} exists.
4. Is log 2 3 rational or irrational? Why?
391

GROUP WORK 3, SECTION 7.4*
e as a Limit
(
1. Graph the function f (x) = 1 + 1 ) x
, 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?
(
3. Using Problems 1 and 2 as a guide, what do you think lim 1 + 1 ) x
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) = y 0 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,andk is a non-zero constant, then y could be
(A)2e kty (B)2e kt (C) e kt + 3 (D) kty + 5 (E) 1 2 ky2 + 1 2
Answer: (B)
• Drill Question: A population grows according to the equation dP/dt = kP,wherek 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) = y 0 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) = y 0
80
determines whether the solution is a growth or decay
function. Perhaps illustrate with a slope-field diagram.
60
40
20
0
2 4 6 8 10 x
WORKSHOP/DISCUSSION
• Referring back to earlier material, show that the expression y = e kt can be written as y = a t and that
e kt+c can be written as Ae kt .
• 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 = A2 kt , 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 P 0 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 Error
3% 23.1 years 24 years 3.90%
8% 8.66 years 9 years 3.92%
12% 5.78 years 6 years 3.81%
18% 3.85 years 4 years 3.90%
The Rule of 72 is accurate to within 4% for the given range of interest rates.
5. If r is the interest rate (as a number, not a percentage), then the doubling time is D = ln 2
r .
6. 100ln2≈ 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: dC
dt
3. dC
dt
≈−0.1165; 60 minutes: dC
dt
≈−0.0259; 80 minutes: dC
dt
≈−0.0096
= kC 4. k ≈−0.052 5. y = 10e −0.052t
6. After approximately 13.33 minutes 7. Any time after 3:52 P.M.
GROUP WORK 3: Find the Error
Answer: When we divide by y in the first step, we assume that y ≠ 0. Wewouldhavetochecky = 0as
a solution separately. The integration gives ln |y| = kx + c. The derivation assumed y > 0. If we assumed
y < 0 we would get y =−Ae kx ,whereA > 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 D A N G
2 × ×
3 × ×
5 × × ×
8 × ×
11 × ×
13 × ×
16 ×
18 × ×
20 ×
395

GROUP WORK 1, SECTION 7.5
TheRuleof72
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 10.0
10 6.07
20 3.68
30 2.23
40 1.35
50 0.820
60 0.498
70 0.302
80 0.183
90 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
dx = ky,theny = Aekx .
The nice thing about this theorem, you think, is that the converse is also true. In other words:
If y = Ae kx , then dy
dx = ky.
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.
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 = e c ,thenA can’t be zero, and A can’t be negative.
But if y = Ae kx ,then dy
dx = ky,nomatterwhatthevalueofA.
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 functions)
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: Whyisittruethatsin −1 (sin x) = x only when − π 2
≤ x ≤ π 2 and sin ( sin −1 x ) = x
only when −1 ≤ x ≤ 1?
Answer: The range of sin −1 x is limited to − π 2 ≤ sin−1 x ≤ π 2
, and its domain is −1 ≤ x ≤ 1.
• Drill Question: What is d ( x 2 sin −1 √ x ) ?
dx
Answer: x [ 4 √ 1 − x sin −1 √ x + √ x ]
√ 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
• Point out that since
dx
d ( cos −1 x ) = −dx
d ( sin −1 x ) ,cos −1 x + sin −1 x is a constant, and show that
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
∫
2xdx
√ . 1 − x 4
400
∫ 2
0
1
dx and
1 + 4x2

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
Answer: We can maximize θ by minimizing the sum S of the
complementary angles α and β. Let x be the distance from A to
( ) ( )
5 2
P. Then S = arctan + arctan , with 0 < x < 3.
x 3 − x
S ′ =
−5
x 2 + 25 + 2
(3 − x) 2 + 4 . The critical points turn out to be 2.1
0 1
S
2.5
2.4
2.3
2.2
A
3
Œ ¬ º
P
2 3 x
the roots of x 2 − 10x + 5, that is, 5 ± 2 √ 5. So we check x = 0, x = 5 − 2 √ 5, and x = 3tofindthat
x = 5 − 2 √ 5 ≈ 0.528 gives a minimum. The result is verified by the graph of S.
2
B
• Define and graph tan −1 x, noting that tan ( tan −1 x ) = x for −∞ < x < ∞, while tan −1 (tan x) = x for
− π 2 < x < π 2 .
( ) 1 + x
• Sketch the graph of f (x) = tan −1 1 + x 2 .
• Differentiate y = tan −1 ( x 2 + e x) , y 2 = sin ( e θ + cos −1 θ ) and x 4 + x = cos (xy).
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
√
x 2 + 4
2.
√
x 2 + 9
3
3. √ 1 − x 4 4.
401
x
√
16 − x 2
5.
2
1 + x 2

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
tan α = x d
tan β = h − x
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:
1. Starting as indicated above, we get ( sec 2 α ) dα
dx
= 1 d and ( sec 2 β ) dβ
dx = −1 dθ
. We set
d dx = 0:
cos 2 α
− cos2 β
= 0 ⇔ α = β ⇔ x d d
d = h − x ⇔ x = h 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 x x
hx
( )( ) =
h + d d x
1 +
2 + dh + d 2 . We can maximize θ by maximizing tan θ, and we find that
x x
hx
x 2 + dh + d 2 is maximized when x = √ d (h + 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 D A N G
1 ×
3 ×
6 ×
13 ×
14 ×
16 × ×
18 ×
19 ×
22 ×
24 ×
Exercise D A N G
33 ×
37 ×
39 ×
41 × ×
43 ×
49 × ×
57 ×
63 ×
64 ×
72 ×
402

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 2
x )
2. sec ( tan −1 3
x )
3. cos ( sin −1 x 2)
4. tan ( sin −1 4
x )
5. sin ( 2tan −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 onthewallsothatanobserverstandingin
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
WheretoSitattheMovies
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 CAS is not available. (The derivative will be
very messy, but still within the students’ abilities.)
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 (cosh 2 x − sinh 2 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,sinhx and the hyperbola x 2 − y 2 = 1?
Answer: The point (cosh x, sinh x) is on that hyperbola.
• Drill Question: Compute d
dx (ex cosh 2x).
d
Answer:
dx (ex cosh 2x) = e x (cosh 2x + 2sinh2x)
MATERIALS FOR LECTURE
• Present definitions of cosh x and sinh x, and the identity cosh 2 x − sinh 2 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
x@-y@=1
(cosh u, sinh u)
0 x
x 2 − y 2 = cosh 2 u − sinh 2 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) ,andy = e cosh 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:
1.
y
10
( )
2. y = 2, y = y 0 + (x − x 0 ) sinh 12
x 0
3. e x − e −x = −1
( e
x ) 2 + e x − 1 = 0 (multiply by e x )
5
_5 0
5 x
e x = −1+√ 5
2
(the other root is extraneous)
)
x = ln
(
−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
lim
x→∞ x · x 1
= 1, even though lim = 0 and lim
x→∞ x x =∞.
x→∞
Answers:
1.
2. cosh x − sinh x = e −x , and so the limit is 0.
10
3. This can be done graphically (as in Problem 1) or algebraically (as
in Problem 2).
5
0 y
1 2 3 x
The limit of the difference
appears to be 0.
HOMEWORK PROBLEMS
4. The students may need to be reminded that cosh 2 x − sinh 2 x = 1.
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 D A N G
2 ×
5 ×
8 ×
9 ×
14 ×
22 ×
23 ×
28 × × ×
32 ×
Exercise D A N G
35 ×
43 ×
48 × ×
51 ×
54 ×
58 ×
63 ×
65 ×
67 ×
407

GROUP WORK 1, SECTION 7.7
A Tangent Line to a Hyperbolic Curve
1. Graph the curve y = 2cosh x 2 . x
y
1
0
1
2. Find the equation for the tangent line at the point (0, 2), then find the equation for the tangent line at the
point (x 0 , y 0 ).
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
x→∞ (cosh x − sinh x)(cosh x + sinh 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?
Answer:
lim
x→∞
We are not taking the derivative of ex
x 2 .
f ′ (x)
g ′ (x) .
f (x)
g (x) = lim x→∞
2x + tan 3x
• Drill Question: Find lim
.
x→0 sin x
Answer: 5
We are using l’Hospital’s Rule to evaluate
MATERIALS FOR LECTURE
• Demonstrate why ∞ is called ”indeterminate” by inviting the students to come up with different pairs of
∞
f (x)
functions f and g such that lim f (x) =∞and lim g (x) =∞but lim
x→∞ x→∞ x→∞ g (x) = 1, 0, 2 3
,and∞. Urge
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
l’Hospital’s Rule [ 1/x − 1 a z − 1
lim = lim = (a z ) ′ at z = 0].
x→∞ 1/x z→0 z
• Show how to manipulate other indeterminate forms into 0 0 or ∞ . [Good examples are lim sin x ln x,
∞ x→0 +
lim (1 + x→∞ x)1/x and lim (1 +
x→0 x)1/x .]
+
• 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 = 1andb = 2, the conclusion states that for some c with 1 < c < 2,
3c 2 7
e c =
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
f (x) =
ln x for x = 0.9, 0.99, 0.999, 1.01, 1.001, etc.
x − 1
1 − cos x 6
• Have the students estimate lim
x→0 x 12
l’Hospital’s Rule.
first graphically, then numerically, and finally compute it using
x − 3
x − 3
• Have the students compute lim
x→3 x 3 − 10x 2 and lim
+ 31x − 30 x→2 x 3 − 10x 2 . Have them try
+ 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.)
[ (
• Do some challenging limit problems, such as lim sin π2
+ x )] (
1/x 1
and lim
x→0 + x→1 ln x − 1 )
.
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:
x 2 + sin x
1. lim
x→∞ x 2
(
= lim 1 + sin x )
x→∞ x 2 = 1
f ′ (a)
2. l’Hospital’s Rule applies only if lim x→a g ′ exists or is ±∞. If not, then l’Hospital’s Rule does not apply.
(a)
Have the students reread the rule carefully, particularly the last clause.
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.
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Answers:
1. S (θ) = θ 2 ,soS (x) = arcsin x. 2. T (x) = x√ 1 − x 2 S (x)
3. lim
x→0 T (x) = 1
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.
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 D A N G
1 ×
6 ×
11 ×
24 ×
31 ×
46 ×
51 ×
59 ×
66 × ×
Exercise D A N G
67 × ×
77 × ×
81 × ×
84 × ×
85 ×
89 ×
91 × × ×
92 ×
97 ×
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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
x→∞ x 2 = lim
both indeterminate forms of type ∞ )
x→∞ 2x
∞
2 − sin x
= lim
which does not exist!
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
x→∞ x 2 does not exist.”
At that point the stranger laughs and laughs, and winds up saying between chortles, “Graph it, Sonny.”
Fearlessly, you graph x2 + sin x
x 2 , and come up with this:
y
1
0
1
x
“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
1. Compute lim
x→∞ x 2 without using l’Hospital’s Rule.
2. What is wrong with the stranger’s argument?
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Consider the following circular sector:
GROUP WORK 2, SECTION 7.8
TheSectorRatio
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.
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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.
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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 x 0 such that f (x 0 ) = 1.
(c) Compute the value of y 0 such that f −1 (y 0 ) = 1.
(d) Below is a graph of f −1 . Draw an approximate graph of f .
y
fÐ!
0
x
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CHAPTER 7
SAMPLE EXAM
2. Find constants A, B, andk such that the equation f (x) = A2 kx + 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) = A2 kx 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) = e 2πx
(b) g (x) = x 2πe
(c) h (x) = (eπ) 2x
(d) l (x) = π ( e 2x )
5. Consider the function h (x) = (1 + sin πx) g(x) . Suppose g (1) = 2andg ′ (1) =−1. Find h ′ (1).
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6. (a) Find the point on the curve y = e −x2 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
x→∞ f ′ (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 lim x→∞ ln ln x?
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CHAPTER 7
SAMPLE EXAM
9. (a) Below is a graph of f (x) = A ln x + B. WhatareA and B?
y
3
2
1
(1, 1)
_1
0
1 2 3 4 5 x
(b) Below is a graph of f (x) = ln kx.Whatisk?
y
2
1
0
1 2 3 x
_1
∫ 5x
dt
10. Let g (x) =
1 t
(a) Compute g ′ (x).
(b) Using part (a), find an equation for g (x) in terms of ln x.
∫ x
1
11. Using the equation ln x = dt, x > 0, justify the following statements:
1 t
(a) The expression e ln x = x is true only for positive values of x.
(b) ln x is increasing everywhere.
(c) ln ( x 1/2) = 1 2
ln x.
12. Let g (x) = ∫ (
x 2
0
ln 12
+ √ )
t dt for all x. Compute the critical points of g.
13. Consider the functions f (x) = e x − 1andg (x) = 2ln(x + 1).
(a) Draw a graph of these functions for − 1 2 ≤ x ≤ 2.
(b) Write an equation for the area between these two functions for − 1 2 ≤ x ≤ 2.
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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 x 0 such that f (x 0 ) = 1is f −1 (1) = 7.
(c) The value y 0 such that f −1 (y 0 ) = 1is 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) = 2πe 2πx
(b) g ′ (x) = 2πe · x 2πe−1
(c) h ′ (x) = 2lneπ · (eπ) 2x = 2 (1 + ln π)(eπ) 2x
(d) l ′ (x) = (ln π) π ( e 2x ) · 2e 2x
5. h ′ (x) = (1 + sin πx) g(x) [
g ′ (x) ln (1 + sin πx) + g (x)
h ′ (1) = 1 2 ·
[−1 · 0 + 2π ]
1 (−1) =−2π
]
π cos πx
;
1 + sin πx
(
6. (a) The slope is e −x2) ′
=−2xe
−x 2 . If −2xe −x2 = 2 e ,then−x2 =−1and−2x = 2. So x =−1and
(
the point is −1, 1 )
.
e
(
(b) The tangent line at −1, 1 )
is y = 1 e e + 2 e (x + 1). Sety = 0 and solve for x to get x =−3 2 .
7. (a) lim f (x) =∞
x→∞
(b) f ′ (x) = 1
2x · 2x =
1 + x2 1 + x 2 . lim f ′ 2x
(x) = lim
x→∞ x→∞ 1 + x 2 = 0
(c) The function increases at a slower and slower rate, but still goes to infinity.
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CHAPTER 7
SAMPLE EXAM SOLUTIONS
8. (a)
y
1
(b)
y
2
(c) lim ln ln x =∞
x→∞
1
0
10 x
0
100 x
_1
9. (a) Use f (1) = 1and f (3) = 0togetA =− 1 and B = 1.
ln 3
(b) Use f (0.2) = 0togetK = 5.
( ) 1
10. (a) g ′ (x) = 5 = 1 5x x
(b) g (x) must differ from ln x by a constant, so g (x) = ln x + c. Usingg (1) =
c = ln 5, so g (x) = ln x + ln 5.
_1
∫ 5
1
dt
t
= ln 5, we get
11. (a) e ln 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.
(b) 1 t
is positive for t > 0, so ln x is increasing everywhere.
(c) ln ( x 1/2) =
∫ x 1/2
∫ x 1/2
∫
1 x
1 t dt = 1
( )
12. g ′ (x) = 2x ln 12
+ |x|
1
1
t dt. Letu = t2 .Thenu 1/2 = t so 1 2 u−1/2 du = dt and
1
2u du = 1 2 ln x. Soln( x 1/2) = 1 2
ln x.
= 0wherex = 0, x = 1 2 ,orx =−1 2
. g (x) has critical points at x = 0andat
x =± 1 2 .
13. (a) The points of intersections are at x = 0andx = c,wherec ≈ 0.752.
y
8
6
4
2
_1
_ 1 _ 2
0
c
1 2 3 x
(b) A = ∫ 2
−1/2
= ∫ 0
−1/2
∣ e x − 1 − 2ln(x + 1) ∣ dx
[ e x − 1 − 2ln(x + 1) ] dx + ∫ c [ (
0 2ln(x + 1) − e x − 1 )] dx
421
+ ∫ 2 [
c e x − 1 − 2ln(x + 1) ] dx

where c ≈ 0.752.
422