7 Inverse Functions: Exponential, Logarithmic, and Inverse ...
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7 <strong>Inverse</strong> <strong>Functions</strong>: <strong>Exponential</strong>, <strong>Logarithmic</strong>,<br />
<strong>and</strong> <strong>Inverse</strong> Trigonometric <strong>Functions</strong><br />
7.1 INVERSE FUNCTIONS<br />
SUGGESTED TIME AND EMPHASIS<br />
1 class Essential material<br />
POINTS TO STRESS<br />
1. The use of multiple representations (verbal, numeric, visual, algebraic) to underst<strong>and</strong> inverse functions,<br />
always coming back to the central idea of reversing inputs <strong>and</strong> outputs.<br />
2. Tests for one-to-one functions <strong>and</strong> techniques for graphing inverses.<br />
3. Derivatives of inverse functions.<br />
QUIZ QUESTIONS<br />
• Text Question: Example 5 describes a technique to graph the inverse of a one-to-one function using the<br />
line y = x. Why does this technique work? Does it work for y = x 2 ?<br />
Answer: Reflecting across the line y = x is the same as reversing the roles of y <strong>and</strong> x in the function.<br />
This technique does not work for x 2 because it is not one-to-one.<br />
• Drill Question: If f (1) = 2, f (2) = 3, <strong>and</strong> f (3) = 1, then what are f ( f (1)), f −1 ( f (1)), <strong>and</strong><br />
f ( f −1 (1) ) ?<br />
Answer: 3, 1, 1<br />
MATERIALS FOR LECTURE<br />
• Starting with f (x) = 3√ x − 4compute f −1 (−2) <strong>and</strong> f −1 (0). Then use algebra to find a formula for<br />
f −1 (x). Have the students try to repeat the process with g (x) = x 3 + x − 2. Note that facts such as<br />
g −1 (−2) = 0, g −1 (0) = 1, <strong>and</strong> <strong>and</strong> g −1 (8) = 2 can be found by looking at a table of values for g(x) but<br />
that the algebraic approach fails to give us a general formula for g −1 (x). Finally, draw graphs of f , f −1 ,<br />
g,<strong>and</strong>g −1 .<br />
• Show that f (x) = 3x 2 + 2x + 1 is increasing for x ≥− 1 3 by computing f ′ (x). Explicitly compute<br />
√ 3x − 2 − 1<br />
f −1 (x) =<br />
<strong>and</strong> discuss its domain <strong>and</strong> range.<br />
3<br />
• Pose the question: If f is always increasing, is f −1 always increasing? Give the students time to try prove<br />
their answer.<br />
Answer: This is true. Proofs may involve diagrams <strong>and</strong> reflections about y = x, or you may try to get<br />
them to be more rigorous. This is an excellent opportunity to discuss concavity, noting that if f is concave<br />
up <strong>and</strong> increasing, then f −1 is concave down <strong>and</strong> increasing.<br />
• Have the students use graphing technology to check if f (x) = x 3 − 2x 2 − x + 2 is one-to-one, <strong>and</strong> then<br />
show how the domain can be restricted to get a one-to-one function.<br />
345
CHAPTER 7<br />
<strong>Inverse</strong> <strong>Functions</strong>: <strong>Exponential</strong>, <strong>Logarithmic</strong>, <strong>and</strong> <strong>Inverse</strong> Trigonometric <strong>Functions</strong><br />
WORKSHOP/DISCUSSION<br />
• Sketch a graph of the inverse of f (x) = 3x 3 +2x by first sketching f (x) <strong>and</strong> then reflecting about y = x.<br />
Show how it would be difficult to find an algebraic formula for the inverse. Then compute ( f −1) ′ (5).<br />
• Show why f (x) = sin x + cos x is one-to-one for − π 4 ≤ x ≤ π 4<br />
. Graph the inverse <strong>and</strong> then compute<br />
( f<br />
−1 ) ′ (1).<br />
• Suppose f is one-to-one <strong>and</strong> we have the following information:<br />
x f (x) f ′ (x)<br />
3 7 2<br />
4 9 1<br />
5 13 2<br />
6 19 0.5<br />
Show the students how to determine ( f −1) ′ (<br />
(9) <strong>and</strong> f<br />
−1 ) ′ (19).<br />
( f<br />
−1 ) ′ (<br />
(7) = f<br />
−1 ) ′ (13).<br />
See if they can explain why<br />
GROUP WORK 1: <strong>Inverse</strong> <strong>Functions</strong>: Domains <strong>and</strong> Ranges<br />
While discussing the domains <strong>and</strong> ranges of inverse functions, this exercise will also foreshadow later excursions<br />
into the maximum <strong>and</strong> minimum values of functions.<br />
If a group finishes early, ask them this question:<br />
“Now consider the graph of f (x) = √ 2x − 3 + 2. What are the domain <strong>and</strong> range of f (x)? Try to figure<br />
out the domain <strong>and</strong> range of f −1 (x) by looking at the graph of f . In general, what information do you need<br />
to be able to compute the domain <strong>and</strong> range of f −1 (x) from the graph of a function f ?”<br />
Answers:<br />
1. It is one-to-one, because the problem says it climbs steadily.<br />
2. a −1 is the time in minutes at which it achieves a given altitude.<br />
3. Reverse the data columns in the given table to get the table for the inverse function. The domain <strong>and</strong> range<br />
of a are 0 ≤ t ≤ 30 <strong>and</strong> 0 ≤ a ≤ 29,000, so the domain <strong>and</strong> range of a −1 are 0 ≤ x ≤ 29,000 <strong>and</strong><br />
0 ≤ a −1 ≤ 30.<br />
4. After approximately 8.5 minutes<br />
5. a is no longer 1-1, because heights are now achieved more than once.<br />
Bonus The domain of f −1 is the set of all y-values on the graph of f , <strong>and</strong> the range of f −1 is the set of all<br />
x-values on the graph of f .<br />
GROUP WORK 2: <strong>Functions</strong> in the Classroom<br />
Before starting this one, review the definition of “function”. Some of the problems can only be answered by<br />
polling the class after they are finished working. Don’t forget to take leap years into account for the eighth<br />
problem. For an advanced class, follow up by defining “one-to-one” <strong>and</strong> “bijection”, then determining which<br />
of the functions have these properties.<br />
Answers:<br />
Chairs: Function, one-to-one, bijection (if all chairs are occupied)<br />
346
SECTION 7.1<br />
INVERSE FUNCTIONS<br />
Eye color: Function, not one-to-one<br />
Mom & Dad’s birthplace: Not a function; mom <strong>and</strong> dad could have been born in different places<br />
Molecules: Function, one-to-one (with nearly 100% probability); inverse assigns a number of molecules to<br />
the appropriate student.<br />
Spleens: Function, one-to-one, bijection. <strong>Inverse</strong> assigns each spleen to its owner.<br />
Pencils: Not a function; some people may have more than one or (horrors!) none.<br />
Student number: Function, one-to-one; inverse assigns each number to its owner.<br />
February birthday: Not a function; not defined for someone born on February 29.<br />
Birthday: Function, perhaps one-to-one.<br />
Cars: Not a function; some have none, some have more than one.<br />
Cash: Function, perhaps one-to-one.<br />
Middle names: Not a function; some have none, some have more than one.<br />
Identity: Function, one-to-one, bijection. <strong>Inverse</strong> is the same as the function.<br />
Calculus instructor: Function, not one-to-one.<br />
Number of hairs: Function, not one-to-one. There are more people in New York City than there are possible<br />
values for this function. Therefore, at least two New Yorkers have the same number<br />
of hairs on their heads, <strong>and</strong> so the function does not have an inverse.<br />
GROUP WORK 3: The Column of Liquid<br />
If the students need a hint, you can mention that the liquid in the mystery device was mercury.<br />
Answers: 1. The liquid is 1 cm high when the temperature is 32 ◦ F. 2. The liquid is 2 cm high when the<br />
temperature is 212 ◦ F 3. The inverse function takes a height in cm, <strong>and</strong> gives the temperature. So it is a<br />
device for measuring temperature. 4. A thermometer<br />
HOMEWORK PROBLEMS<br />
Core Exercises: 2, 6, 9, 20, 24, 33, 37, 38, 41<br />
Sample Assignment: 2, 3, 6, 9, 15, 20, 21, 24, 30, 33, 34, 37, 38, 41<br />
Exercise D A N G<br />
2 ×<br />
3 ×<br />
6 ×<br />
9 ×<br />
15 ×<br />
20 × × ×<br />
21 × ×<br />
Exercise D A N G<br />
24 ×<br />
30 × ×<br />
33 × ×<br />
34 × ×<br />
37 ×<br />
38 ×<br />
41 ×<br />
347
GROUP WORK 1, SECTION 7.1<br />
<strong>Inverse</strong> <strong>Functions</strong>: Domains <strong>and</strong> Ranges<br />
Let a (t) be the altitude in feet of a plane that climbs steadily from takeoff until it reaches its cruising altitude<br />
after 30 minutes. We don’t have a formula for a, but extensive research has given us the following table of<br />
values:<br />
t a (t)<br />
0.1 50<br />
0.5 150<br />
1 500<br />
3 2000<br />
7 8000<br />
10 12,000<br />
20 21,000<br />
25 27,000<br />
30 29,000<br />
1. Is a (t) a one-to-one function? How do you know?<br />
2. What does the function a −1 measure in real terms? Your answer should be descriptive, similar to the way<br />
a (t) was described above.<br />
348
<strong>Inverse</strong> <strong>Functions</strong>: Domains <strong>and</strong> Ranges<br />
3. We are interested in computing values of a −1 . Fill in the following table for as many values of x as you<br />
can. What quantity does x represent?<br />
x<br />
a −1 (x)<br />
What are the domain <strong>and</strong> range of a? What are the domain <strong>and</strong> range of a −1 ?<br />
4. You are allowed to turn on electronic equipment after the plane has reached 10,000 feet. Approximately<br />
when can you expect to turn on your laptop computer after taking off?<br />
5. Suppose we consider a (t) from the time of takeoff to the time of touchdown. Is a (t) still one-to-one?<br />
349
Which of the following relations are functions?<br />
GROUP WORK 2, SECTION 7.1<br />
<strong>Functions</strong> in the Classroom<br />
Domain Function Values Function<br />
All the people in your classroom Chairs f (person) = his or her chair<br />
All the people in your classroom {blue, brown, green, hazel} f (person) = hisorhereyecolor<br />
All the people in your classroom Cities f (person) = birthplace of his or her mom <strong>and</strong> dad<br />
All the people in your classroom R, the real numbers f (person) = number of molecules in his or her body<br />
All the people in your classroom Spleens f (person) = hisorherspleen<br />
All the people in your classroom Pencils f (person) = hisorherpencil<br />
All the students in your classroom Integers from 0–99999999 f (person) = his or her student number<br />
All the living people born in February Days in February, 2007 f (person) = his or her birthday in February 2007<br />
All the people in your classroom Days of the year f (person) = his or her birthday<br />
All the people in your classroom Cars f (person) = his or her car<br />
All the people in your classroom R, the real numbers f (person) = howmuchcashtheyhaveonthem<br />
All the people in your college Names f (person) = his or her middle name<br />
All the people in your classroom People f (person) = theirself<br />
All the students in your classroom People f (person) = his or her calculus instructor<br />
All the people in New York City W, the whole numbers f (person) = the number of hairs on his or her head<br />
350
GROUP WORK 3, SECTION 7.1<br />
The Column of Liquid<br />
It is a fact that if you take a tube <strong>and</strong> fill it partway with liquid, the liquid will rise <strong>and</strong> fall based on the<br />
temperature. Assume that we have a tube of liquid, <strong>and</strong> we have a function h (T ),whereh is the height of the<br />
liquid in cm at temperature T in ◦ F.<br />
1. It is true that h (32) = 1. What does that mean in physical terms?<br />
2. It is true that h (212) = 10. What does that mean in physical terms?<br />
3. Describe the inverse function h −1 . What are its inputs? What are its outputs? What does it measure?<br />
4. There is a device, currently available at your local drugstore, that measures the function h −1 . (It doesn’t<br />
use a column of liquid anymore, but its ancestors did.) What is the name of this device?<br />
351
7.2 EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES<br />
TRANSPARENCY AVAILABLE<br />
#13 (Figures 3, 4, 5, <strong>and</strong> 10)<br />
SUGGESTED TIME AND EMPHASIS<br />
1 class Essential material<br />
POINTS TO STRESS<br />
1. The graphs <strong>and</strong> properties of exponential functions, including growth rates.<br />
2. <strong>Exponential</strong> functions as models for population growth <strong>and</strong> decay.<br />
3. The basic differentiation formula for exponential functions, <strong>and</strong> how it is developed from the limit<br />
definition of the derivative.<br />
4. The definition of e.<br />
5. Growth rates of exponential functions as compared to growth rates of polynomials.<br />
QUIZ QUESTIONS<br />
• Text Question: 3 3 = 3 · 3 · 3, 3 3/4 = 4√ 3 · 3 · 3. How does one make sense of 3 √7 ?<br />
Answer: Answers will vary. One example: we can approximate √ 7 by a sequence of rational numbers,<br />
<strong>and</strong> can thus similarly approximate 3 √7 to any degree of precision.<br />
• Drill Question: What is the derivative of f (x) = e x3 ?<br />
Answer: 3x 2 e x3<br />
MATERIALS FOR LECTURE<br />
• Starttodrawagraphof2 x vs x, using the scale of one inch per unit on both axes. Point out that after one<br />
foot, the height would be over 100 yards (the length of a football field). After two feet, the height would<br />
be 264 miles, after three feet it would be 1,000,000 miles (four times the distance to the moon), after three<br />
<strong>and</strong> a half feet it would be in the heart of the sun. If the graph extended five feet to the right, x = 60, then<br />
y would be over one light year up.<br />
• Point out this contrast between exponential <strong>and</strong> linear functions: For equally spaced x-values, linear<br />
functions have constant differences in y-values, while pure exponential functions have constant ratios in<br />
y-values. Use this fact to show that the following table describes an exponential function, not a linear one.<br />
x y<br />
−6.2 0.62000<br />
−2.4 0.65100<br />
1.4 0.68355<br />
5.2 0.71773<br />
9.0 0.75361<br />
12.8 0.79129<br />
352
SECTION 7.2<br />
EXPONENTIAL FUNCTIONS AND THEIR DERIVATIVES<br />
• In 1985 there were 15,948 diagnosed cases of AIDS in the United States. In 1990 there were 156,024.<br />
Scientists said that if there was no research done, the disease would grow exponentially. Compute the<br />
number of cases this model predicts for the year 2000. The actual number was 774,467. Discuss possible<br />
flaws in the model with the students, <strong>and</strong> point out the dangers of extrapolation.<br />
• Review the derivation of the formula f ′ (x) = f ′ (0) a x for an exponential function. Point out that for<br />
f (x) = 2 x , f ′ (0) ≈ 0.693, <strong>and</strong> for g (x) = 3 x , g ′ (0) ≈ 1.099.<br />
WORKSHOP/DISCUSSION<br />
• 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<br />
e x3 <strong>and</strong> then do a straightforward Product Rule problem such as finding the derivative of f (x) = e x sin x.<br />
e x<br />
Then have the students try to compute the derivatives of<br />
1 + e x , ( x 2) e ,<strong>and</strong>ex 3 + e x3 + xe 3 + x e3 + ( e 3) x .<br />
• Draw the graph of e −x from the graph of e x . Then use calculus to derive the properties of the graph of<br />
e x/(1−x) , <strong>and</strong> sketch the graph.<br />
• Estimate where 3 x > x 3 <strong>and</strong> where 2 x > x 8 using technology. Notice that exponential functions start by<br />
growing slower than polynomial functions, <strong>and</strong> then wind up growing much faster. For example, if one<br />
were to graph x 2 vs x using one inch per unit, then when x = 60, y would be only 100 yards, as opposed<br />
to a light year for y = 2 x . (The sun is only 8 light minutes from the earth.)<br />
• One way to measure the growth of the internet is to measure the number of hosts. The following data show<br />
the number of internet hosts over time. Try to determine with the students if this is exponential growth.<br />
(Note: Do not show the student the third column right away. Let them come up with the idea of finding<br />
growth rates between data points.)<br />
Month Hosts Growth<br />
Aug 1981 213 −<br />
May 1982 235 1.1400<br />
Aug 1983 562 2.0088<br />
Oct 1984 1024 1.6724<br />
Oct 1985 1961 1.9150<br />
Feb 1986 2308 1.6303<br />
Nov 1986 5089 2.8699<br />
Dec 1987 28,174 4.8534<br />
Jul 1988 33,000 1.3113<br />
Oct 1988 56,000 8.2927<br />
Jan 1989 80,000 4.1649<br />
Jul 1989 130,000 2.6406<br />
Oct 1989 159,000 2.2378<br />
Oct 1990 313,000 1.9686<br />
Jan 1991 376,000 2.0824<br />
Jul 1991 535,000 2.0246<br />
Oct 1991 617,000 1.7690<br />
Month Hosts Growth<br />
Jan 1992 727,000 1.9275<br />
Apr 1992 890,000 2.2461<br />
Jul 1992 992,000 1.5434<br />
Oct 1992 1,136,000 1.7198<br />
Jan 1993 1,313,000 1.7846<br />
Apr 1993 1,486,000 1.6407<br />
Jul 1993 1,776,000 2.0403<br />
Oct 1993 2,056,000 1.7961<br />
Jan 1994 2,217,000 1.3520<br />
Jul 1994 3,212,000 2.0990<br />
Oct 1994 3,864,000 2.0943<br />
Jan 1995 4,852,000 2.4862<br />
Jul 1995 6,642,000 1.8739<br />
Jan 1996 9,472,000 2.0337<br />
Jul 1996 12,881,000 1.8493<br />
Jan 1997 16,146,000 1.5712<br />
Jul 1997 19,540,000 1.4646<br />
353<br />
Month Hosts Growth<br />
Jan 1998 29,670,000 2.3056<br />
Jul 1998 36,739,000 1.5333<br />
Jan 1999 43,230,000 1.3846<br />
Jul 1999 56,218,000 1.6911<br />
Jan 2000 72,398,092 1.6585<br />
Jul 2000 93,047,785 1.6518<br />
Jan 2001 109,574,429 1.3868<br />
Jul 2001 125,888,197 1.3199<br />
Jan 2002 147,344,723 1.3699<br />
Jul 2002 162,128,493 1.2107<br />
Jan 2003 171,638,297 1.1208<br />
Jul 2003 233,101,481 1.8444<br />
Jan 2004 285,139,107 1.4963<br />
Jul 2004 317,646,084 1.2410<br />
Jan 2005 353,284,187 1.2370
CHAPTER 7<br />
<strong>Inverse</strong> <strong>Functions</strong>: <strong>Exponential</strong>, <strong>Logarithmic</strong>, <strong>and</strong> <strong>Inverse</strong> Trigonometric <strong>Functions</strong><br />
Answer: We can graph the data, <strong>and</strong> get a curve that looks like exponential growth. We can also graph<br />
growth rate <strong>and</strong> see (except for two spikes in the late 1980s) a relatively constant growth rate.<br />
H<br />
G<br />
360,000,000<br />
9<br />
300,000,000<br />
240,000,000<br />
180,000,000<br />
120,000,000<br />
60,000,000<br />
8<br />
7<br />
6<br />
5<br />
4<br />
3<br />
2<br />
1<br />
82 84 86 88 90 92 94 96 98 00 02 04 t<br />
82 84 86 88 90 92 94 96 98 00<br />
02<br />
04<br />
t<br />
GROUP WORK 1: I’ve Grown Accustomed to Your Growth<br />
Before h<strong>and</strong>ing out this activity, it may be prudent to review the rules of exponentiation. This exercise enables<br />
students to discover for themselves the equal ratio property of exponential functions.<br />
Answers: 1. Yes (m = 1), no, yes (m ≈ 2.08), yes (m ≈ 2.01) 2.Equally spaced changes in x-values result<br />
in equally spaced changes in y-values 3. Equally spaced changes in x values result in equally proportioned<br />
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<br />
when taking the ratio. However, the property is close to being true when A <strong>and</strong> b are large compared to C.<br />
GROUP WORK 2: Comparisons (Part 1)<br />
The purpose of this group work is to give the students a bit of “picture sense”. It is acceptable if they do this<br />
by looking at the graphs on their calculators, setting the windows appropriately.<br />
Answers: 1. 0 < x < 1.374 <strong>and</strong> x > 9.940 2. 0 < x < 1.052 <strong>and</strong> x > 95.7168 3. 0 < x < 1.18 <strong>and</strong><br />
x > 22.4 4. 0 < x < 1.34 <strong>and</strong> x > 10.9<br />
HOMEWORK PROBLEMS<br />
Core Exercises: 2, 10, 13, 16, 17, 24, 25, 35, 48, 59, 73, 88<br />
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,<br />
88<br />
Exercise D A N G<br />
1 × ×<br />
2 ×<br />
3 × ×<br />
5 × ×<br />
10 ×<br />
13 ×<br />
14 ×<br />
16 ×<br />
17 × ×<br />
Exercise D A N G<br />
20 × ×<br />
21 ×<br />
24 ×<br />
25 ×<br />
32 ×<br />
35 ×<br />
43 ×<br />
48 ×<br />
354<br />
Exercise D A N G<br />
56 ×<br />
59 × × ×<br />
64 ×<br />
65 ×<br />
73 ×<br />
79 ×<br />
86 ×<br />
88 × ×
GROUP WORK 1, SECTION 7.2<br />
I’ve Grown Accustomed to Your Growth<br />
1. Two or three of the following four tables of data have something in common: linear growth. Without<br />
trying to find complete equations of lines, determine which of them are linear growth, <strong>and</strong> determine their<br />
rate of change:<br />
x y<br />
1 2<br />
2 3<br />
3 4<br />
4 5<br />
x y<br />
21.5 4.32<br />
32.6 4.203<br />
43.7 4.090<br />
54.8 3.980<br />
x y<br />
−3 1.1<br />
−2.5 2.14<br />
−2 3.18<br />
−1.5 4.25<br />
x y<br />
1 −5.00<br />
3 −0.98<br />
6 5.05<br />
8 9.07<br />
2. In a sentence, describe a property of linear growth that can be determined from a table of values.<br />
355
I’ve Grown Accustomed to Your Growth<br />
3. The following four tables of data have something in common: exponential growth. <strong>Functions</strong> of the form<br />
y = Ab x (or Ae kx ) have a property in common analogous to the one you stated in Question 2. Find the<br />
property, <strong>and</strong> then find the value of b.<br />
x y<br />
1 5<br />
2 10<br />
3 20<br />
4 40<br />
x y<br />
21.5 4.32<br />
32.6 4.203<br />
43.7 4.090<br />
54.8 3.980<br />
x y<br />
−3 1.1<br />
−2.5 1.65<br />
−2 2.475<br />
−1.5 3.7125<br />
x y<br />
1 0.8<br />
3 7.2<br />
6 194.4<br />
8 1749.6<br />
4. Unfortunately, the above property does not hold for functions of the form y = Ab x + C. What goes<br />
wrong? For what kinds of values of A, b,<strong>and</strong>C does the property come close to being true?<br />
356
GROUP WORK 2, SECTION 7.2<br />
Comparisons (Part 1)<br />
You have learned that an exponential function grows faster than a polynomial function. Find the values of<br />
x > 0forwhich<br />
1. 2 x ≥ x 3 .<br />
2. (1.1) x ≥ x 2 .<br />
3. 2 x ≥ x 5 .<br />
4. 3 x ≥ x 5 .<br />
357
7.3 LOGARITHMIC FUNCTIONS<br />
TRANSPARENCY AVAILABLE<br />
#14 (Figures 2, 3, 5, <strong>and</strong> 6) can be used with this section.<br />
SUGGESTED TIME AND EMPHASIS<br />
1 class Essential material<br />
POINTS TO STRESS<br />
1. <strong>Logarithmic</strong> functions <strong>and</strong> their properties, including their geometric properties as inverses of<br />
exponentials.<br />
2. Graphs of logarithmic functions, including asymptotic behavior.<br />
QUIZ QUESTIONS<br />
• Text Question: Why does a log a x = x only hold for x > 0?<br />
Answer: log a x is defined only for x > 0.<br />
• Drill Question: What is the inverse function for y = log a x?<br />
Answer: a x<br />
MATERIALS FOR LECTURE<br />
√<br />
• Point out that log √ ln √ 11<br />
17 11 =<br />
ln √ ln 11<br />
= looks much simpler in the latter form.<br />
17 ln 17<br />
(<br />
• Sketch a graph of f (x) = log 2 (x + 3)<br />
2 ) ,sketch Answer: y y=f(x)<br />
the inverse function, <strong>and</strong> then find an algebraic<br />
formula for the inverse.<br />
y=f Ð!(x)<br />
2<br />
0<br />
2<br />
x<br />
f −1 (x) = 2 x/2 − 3<br />
• Sketch a graph of f (x) = 4ln(x + 2), starting with the graph of y = ln x. Then show that this is also a<br />
graph of the function written as ln (x + 2) 4 .<br />
• Discuss why<br />
ln (ln x) > μ.<br />
lim<br />
x→∞<br />
ln (ln x) = ∞. Point<br />
out that if ln x > eμ where μ is a positive number, then<br />
WORKSHOP/DISCUSSION<br />
( )<br />
• Discuss the graph of f (x) = 3 −x 2 x after writing it as f (x) = 2x 2 x<br />
3 x = . See if students can identify<br />
3<br />
( ) 2 x<br />
its inverse as g (x) = log 2/3 x. Ask them to solve = 1, <strong>and</strong> to explain why the answer is negative.<br />
3<br />
• 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<br />
log n x grows more slowly than x/n,forx sufficiently large.<br />
• Estimate where 5 x > 10 7 by the use of technology. Then have the students use logarithms to find an exact<br />
answer.<br />
358
SECTION 7.3<br />
LOGARITHMIC FUNCTIONS<br />
GROUP WORK 1: Comparisons (Part 2)<br />
The purpose of this group work is to give students a bit of “picture sense”. It is acceptable if they do this by<br />
looking at graphs on their calculators, setting the windows appropriately.<br />
Answers:<br />
1. 0 < x < 1.208 9, x > 18.6387 2. 0 < x < 1.0409, x > 125.3635<br />
3. 0 < x < 1.1772, x > 22.4400 4. 0 < x < 1.05138, x > 95.7168<br />
GROUP WORK 2: Irrational, Impossible Relations<br />
Before starting this activity, review the definitions of rational <strong>and</strong> irrational numbers. The hint sheet should<br />
be given out only after the students have tried to show that log 2 3 is irrational, or at least discussed it enough<br />
to underst<strong>and</strong> what they are trying to show. If a group finishes early, have them show that log 2 a is always<br />
irrational if a is an odd integer.<br />
Answers: 1. −γ 2. −γ 3. γ 4. log<br />
2<br />
2 3 = ln 3 5. The proof is outlined in the hint sheet.<br />
ln 2<br />
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,<br />
which is undefined. 3. 2 a = 2 · 2 ·····2, <strong>and</strong> 3 b = 3 · 3 ·····3, so these numbers can never be equal,<br />
because the left is always divisible by two, <strong>and</strong> the right never is (unless b = 0). 4. Irrational.<br />
HOMEWORK PROBLEMS<br />
Core Exercises: 2, 4, 9, 15, 23, 26, 40, 43, 47, 55<br />
Sample Assignment: 2, 3, 4, 8, 9, 10, 15, 18, 21, 23, 26, 31, 37, 40, 43, 44, 47, 55, 66<br />
Exercise D A N G<br />
2 × ×<br />
3 ×<br />
4 ×<br />
8 ×<br />
9 ×<br />
10 ×<br />
15 ×<br />
18 ×<br />
21 × ×<br />
23 ×<br />
Exercise D A N G<br />
26 ×<br />
31 ×<br />
37 ×<br />
40 × ×<br />
43 × ×<br />
44 × ×<br />
47 ×<br />
55 ×<br />
66 ×<br />
359
GROUP WORK 1, SECTION 7.3<br />
Comparisons (Part 2)<br />
You have been told that an exponential function grows more quickly than a polynomial function, <strong>and</strong> that<br />
logarithmic functions grow more slowly than linear functions. Find the values of x > 0forwhich<br />
1. 3 x ≥ x 7 .<br />
( ) 7 x<br />
2. ≥ x 4 .<br />
6<br />
3. x 5 ≥ log 2 x.<br />
4. x 2 ≥ log 1.1 x. 360
1. If log 2 x = γ, then what is log 1/2 x?<br />
GROUP WORK 2, SECTION 7.3<br />
Irrational, Impossible Relations<br />
2. If log b x = γ, then what is log 1/b x (assuming b > 1)?<br />
3. If log b x = γ, then what is log b 2 x?<br />
4. We are going to estimate log 2 3. In pre-calculus, you memorized that log 2 3 ≈ 1.584962501. Suppose you<br />
didn’t have this fact memorized. There is no log 2 3 button on your calculator! How would you compute<br />
it?<br />
5. Unfortunately, the calculator gives us only a finite number of digits. If log 2 3 were a rational number,<br />
we would be able to express it as a fraction, giving us perfect accuracy. Do you think it is rational or<br />
irrational? Try to prove your result.<br />
361
GROUP WORK 2, SECTION 7.3<br />
Irrational, Impossible Relations (Hint Sheet)<br />
So, you realize that it’s not easy to determine whether log 2 3 is rational!<br />
Onewaytoattempttoshowthatlog 2 3 is rational is to assume that it is, <strong>and</strong> try to find integers a <strong>and</strong> b such<br />
that log 2 3 = a b . If we can show that there are no such a <strong>and</strong> b,thenlog 2 3 cannot be rational.<br />
1. Assume that log 2 3 = a b . Show that a <strong>and</strong> b must then satisfy 2a = 3 b<br />
2. Notice that a = 0, b = 0 satisfies 2 a = 3 b . Show that this fact doesn’t help us.<br />
3. Find a ≠ 0<strong>and</strong>b ≠ 0 that satisfy 2 a = 3 b ,orshowthatnosuch{a, b} exists.<br />
4. Is log 2 3 rational or irrational? Why?<br />
362
7.4 DERIVATIVES OF LOGARITHMIC FUNCTIONS<br />
SUGGESTED TIME AND EMPHASIS<br />
1 class Essential material<br />
POINTS TO STRESS<br />
1. The basic logarithmic differentiation formula.<br />
2. The technique of logarithmic differentiation.<br />
3. The concept of e as a limit.<br />
QUIZ QUESTIONS<br />
• Text Question: Why is logarithmic differentiation necessary? For instance, could you do Example 15<br />
using the Product <strong>and</strong> Quotient Rules?<br />
Answer: It is a lot easier to do this problem with logarithmic differentiation. Also, something like x x is<br />
very hard to differentiate conventionally.<br />
• Drill Question: What is the instantaneous rate of change of the function f (x) = ln ( x 2 + 2x + 1 ) at<br />
x = 1?<br />
Answer: 1<br />
MATERIALS FOR LECTURE<br />
• Review logarithmic <strong>and</strong> exponential functions. Point out that there is no way to simplify ln (a + b) or ln a<br />
ln b .<br />
{ ln x + 2 if x > 0<br />
• Discuss Example 7 <strong>and</strong> extend it in the following way: Let f (x) =<br />
Graph<br />
ln (−x) − 1 if x < 0<br />
{<br />
f (x) <strong>and</strong> compute f ′ ln x + a if x > 0<br />
(x). Close with discussions of f (x) =<br />
from geometric <strong>and</strong><br />
ln (−x) + b if x < 0<br />
algebraic perspectives.<br />
• Derive the number e as a limit, as done in the text. Have them confirm Formulas 8 <strong>and</strong> 9 by plugging<br />
small numbers in for x <strong>and</strong> large numbers in for n. Show that we can also write e = lim (1 + n→∞ n)1/n by<br />
replacing x with 1/n in the text’s derivation.<br />
• Ask what the second derivative of f (x) = ln (√ x ) tells us about the shape of the graph of f .<br />
WORKSHOP/DISCUSSION<br />
• Compute derivatives of functions that involve logarithms <strong>and</strong> exponents, <strong>and</strong> that require the Product,<br />
Quotient, or Chain Rule. For example, use f (x) = ln (sin x) sin (ln x) or g (x) = ln (e x + ln x).<br />
• Compute d<br />
dx ln ( (x + 1) 5) , first by simplifying using properties of logarithms, <strong>and</strong> then using the Chain<br />
Rule without prior simplification.<br />
• Present the function f (x) = ln ln x. Before drawing the graph, have the students try to find the domain<br />
<strong>and</strong> range of f , as well as lim f (x). Then graph the function, find an equation of the line tangent to it at<br />
x→∞<br />
the point (e, 0), <strong>and</strong> graph the tangent line.<br />
363
CHAPTER 7<br />
<strong>Inverse</strong> <strong>Functions</strong>: <strong>Exponential</strong>, <strong>Logarithmic</strong>, <strong>and</strong> <strong>Inverse</strong> Trigonometric <strong>Functions</strong><br />
GROUP WORK 1: <strong>Logarithmic</strong> Differentiation<br />
In addition to its face value, the process of logarithmic differentiation is a good way of mastering computations<br />
involving logarithms. Don’t give a group the second page until they have finished with the first page. If a<br />
group finishes early, have them compute the derivative of ( x 2 + 1 ) sin x .<br />
Answers:<br />
1. The Power Rule requires that the exponent be a constant.<br />
2. The Exponent Rule requires that the base be a constant.<br />
3. -- 6.<br />
1 dy<br />
(x + 1) x dx<br />
y = (x + 1) x<br />
ln y = ln (x + 1) x<br />
ln y = x ln (x + 1)<br />
1 dy<br />
= ln (x + 1) +<br />
x<br />
y dx x + 1<br />
= ln (x + 1) +<br />
x<br />
x + 1<br />
[<br />
dy<br />
dx = (x + 1)x ln (x + 1) +<br />
x ]<br />
x + 1<br />
[<br />
7. They are identical. 8. e x ln(x+1) ln (x + 1) +<br />
x ]<br />
x + 1<br />
9. e x ln(x+1) [<br />
ln (x + 1) +<br />
x ]<br />
[<br />
= e ln((x+1)x ) ln (x + 1) +<br />
x + 1<br />
x ] [<br />
= (x + 1) x ln (x + 1) +<br />
x + 1<br />
x ]<br />
x + 1<br />
10. y ′ x sin y sin y + xy cos x sin x ln y<br />
=−y<br />
x ( x sin y y cos y ln x − y cos x cos x ) or y sin y + x sin x ln y<br />
x cos x − y ln x cos y<br />
364
SECTION 7.4<br />
DERIVATIVES OF LOGARITHMIC FUNCTIONS<br />
GROUP WORK 2: e as a Limit<br />
One can introduce this activity by deriving the expression in Problem 1 in terms of compound interest. Some<br />
students will require more guidance on what types of things to look for in Problem 1. Properties to notice<br />
include concavity, asymptotes, extrema, increase/decrease, <strong>and</strong> positivity/negativity.<br />
Answers:<br />
1.<br />
y<br />
3<br />
2<br />
1<br />
2. The horizontal asymptote of the graph<br />
(<br />
corresponds to lim 1 + 1 ) x<br />
.<br />
x→∞ x<br />
3. The limit is e ≈ 2.718.<br />
0<br />
1 2 3 4 5 6 7 8 9 10 x<br />
GROUP WORK 3: The Right Procedure at the Right Time<br />
This drill activity can be done as a group activity, or the students can work on it individually. The main idea<br />
is to make sure they underst<strong>and</strong> the fundamental idea.<br />
Answers:<br />
1. π 4 (sin x)π/4−1 cos x 2. ( sin π ) x ( ) ( √ ) x<br />
4 ln sin<br />
π<br />
4<br />
=−<br />
2<br />
√ (<br />
2<br />
ln 2 3.<br />
π4<br />
) sin x ( ) ln<br />
π<br />
4<br />
cos x<br />
4. sin ( π<br />
4<br />
) x sin(π/4)−1 =<br />
(<br />
√ √2−2 )<br />
2 /2<br />
2<br />
x<br />
5. 4πx 3 cos ( x 4)<br />
365
CHAPTER 7<br />
<strong>Inverse</strong> <strong>Functions</strong>: <strong>Exponential</strong>, <strong>Logarithmic</strong>, <strong>and</strong> <strong>Inverse</strong> Trigonometric <strong>Functions</strong><br />
GROUP WORK 4: Some Surprising Areas<br />
Have the groups work on the first three problems, <strong>and</strong> then h<strong>and</strong> the fourth out to groups that finish early.<br />
When closing the exercise, impart to the students how surprising the result is: that equal areas are created by<br />
any choice of m > 1. Note that the result of Problem 4 can be generalized for the area between x = m a <strong>and</strong><br />
x = m b (b > a).<br />
Answers:<br />
1. Each of the areas is ln 2 ≈ 0.693147. The students will probably not be able to simplify their answers to<br />
ln 2, <strong>and</strong> may leave them as ln (ln 1.0201) − ln (ln 1.01),ln(ln 16) − ln (ln 4),<strong>and</strong>ln(ln 100) − ln (ln 10).<br />
2. The second endpoint is the square of the first.<br />
3. The area between x = m <strong>and</strong> x = m 2 is the same, regardless of the value of m.<br />
∫ m 2 ∫<br />
dx ln m 2<br />
4.<br />
m x ln x = du<br />
ln m u = ln ( ( )<br />
ln m 2) 2lnm<br />
− ln (ln m) = ln = ln 2<br />
ln m<br />
HOMEWORK PROBLEMS<br />
Core Exercises: 2, 9, 16, 28, 38, 45, 57, 68, 74, 78<br />
Sample Assignment: 2, 9, 16, 28, 31, 35, 38, 39, 45, 46, 54, 57, 67, 68, 70, 74, 75, 78, 87, 91<br />
Exercise D A N G<br />
2 ×<br />
9 ×<br />
16 ×<br />
28 ×<br />
31 ×<br />
35 ×<br />
38 ×<br />
39 × ×<br />
45 ×<br />
46 ×<br />
Exercise D A N G<br />
54 ×<br />
57 × ×<br />
67 × × ×<br />
68 × × × ×<br />
70 ×<br />
74 ×<br />
75 ×<br />
78 ×<br />
87 ×<br />
91 ×<br />
366
GROUP WORK 1, SECTION 7.4<br />
<strong>Logarithmic</strong> Differentiation<br />
Let f (x) = (x + 1) x . As of now, none of the rules that we’ve learned seem to help us find f ′ (x).<br />
1. Why can’t you use the Power Rule to compute f ′ (x)?<br />
2. Why can’t you use a formula like d<br />
dx 3x = 3 x ln3tocompute f ′ (x)?<br />
3. Start with the equation y = (x + 1) x . Take the logarithm of both sides to get ln y = ln ( (x + 1) x) . Then<br />
use a property of logarithms to help you find the derivative of ln ( (x + 1) x) .<br />
4. Now use implicit differentiation to find d ln y.<br />
dx<br />
5. Since ln y = ln ( (x + 1) x) , you can equate your answers to Parts 3 <strong>and</strong> 4. Do so, <strong>and</strong> then replace all y<br />
terms by (x + 1) x . Why can you do this substitution?<br />
6. Perform some algebra to get dy<br />
dx<br />
alone on the left h<strong>and</strong> side, <strong>and</strong> you are done!<br />
367
<strong>Logarithmic</strong> Differentiation<br />
7. Let g (x) = e x ln(x+1) . What is the relationship between f (x) <strong>and</strong> g (x)?<br />
8. If g (x) = e x ln(x+1) ,computeg ′ (x). What techniques are you using in this case?<br />
9. Since you have done the same problem in two different ways, show that your answers to Problems 6 <strong>and</strong><br />
8 are identical.<br />
10. Let y = f (x) be implicitly defined by x sin y = y cos x . Compute y ′ in terms of x <strong>and</strong> y. (Hint: Can<br />
logarithms help you?)<br />
368
1. Graph the function f (x) =<br />
y<br />
3<br />
2<br />
1<br />
GROUP WORK 2, SECTION 7.4<br />
e as a Limit<br />
(<br />
1 + 1 ) x<br />
, <strong>and</strong> describe its properties.<br />
x<br />
0<br />
1 2 3 4 5 6 7 8 9 10 x<br />
2. What is the relationship between the graph you drew for Problem 1 <strong>and</strong> Formula 9 in the text?<br />
(<br />
3. Using Problems 1 <strong>and</strong> 2 as a guide, what do you think lim 1 + 1 ) x<br />
is?<br />
x→∞ x<br />
369
Compute the derivatives of the following functions:<br />
1. (sin x) π/4<br />
GROUP WORK 3, SECTION 7.4<br />
The Right Procedure at the Right Time<br />
2. ( sin π ) x<br />
4<br />
3. ( π<br />
4<br />
) sin x<br />
4. x sin(π/4)<br />
5. π sin ( x 4) 370
GROUP WORK 4, SECTION 7.4<br />
Some Surprising Areas<br />
Meet<br />
1<br />
x ln x :<br />
1. Compute the area under f (x) = 1<br />
x ln x<br />
from x = 1.01 to 1.0201, from 4 to 16, <strong>and</strong> from 10 to 100.<br />
2. What do the three pairs of endpoints used in Problem 1 have in common?<br />
3. The answers to Problems 1 <strong>and</strong> 2 should suggest an interesting property of the curve f (x) = 1<br />
x ln x .<br />
What is it?<br />
371
Some Surprising Areas<br />
4. Prove that the area enclosed by f (x) = 1<br />
x ln x ,thex-axis, the line x = m <strong>and</strong> the line x = m2 is<br />
independent of your choice of m (as long as m > 1).<br />
372
7.2* THE NATURAL LOGARITHMIC FUNCTION<br />
SUGGESTED TIME AND EMPHASIS<br />
1 class Essential material<br />
POINTS TO STRESS<br />
1. The definition <strong>and</strong> properties of the natural logarithmic function ln (x) =<br />
2. The basic differentiation <strong>and</strong> integration formulas for logarithms.<br />
3. The concept of e as the number x for which ln x = 1.<br />
4. The technique of logarithmic differentiation.<br />
∫ x<br />
1<br />
dt<br />
t , x > 0.<br />
QUIZ QUESTIONS<br />
• Text Question: Why is logarithmic differentiation necessary? For instance, could you do Example 14<br />
using the Product <strong>and</strong> Quotient Rules?<br />
Answer: It is a lot easier to do this problem with logarithmic differentiation. Also, something like x x is<br />
very hard to differentiate conventionally.<br />
∫ x<br />
dt<br />
• Drill Question: If we have ln x = , x > 0, why is ln x negative for 0 < x < 1?<br />
t<br />
Answer: Because if 0 < x < 1,<br />
is positive.<br />
MATERIALS FOR LECTURE<br />
• Discuss properties of ln x =<br />
∫ x<br />
1<br />
1<br />
∫ x<br />
1<br />
dt<br />
t<br />
=−<br />
∫ 1<br />
x<br />
dt<br />
t<br />
where ∫ 1<br />
x<br />
dt<br />
t , the area under the curve 1 t<br />
dt<br />
t , using the integral definition, <strong>and</strong> review why if ln x = ∫ x<br />
d<br />
dx ln x = 1 x , x > 0.<br />
• Derive the formula ln ( x 1/q) = 1 q ln x by rewriting the equation y = x1/q as y q<br />
1<br />
from x to 1,<br />
1<br />
t<br />
dt, then<br />
= x. Then use this<br />
equation to show that ln ( x p/q) = p q<br />
ln x.<br />
( x<br />
2<br />
)<br />
• Discuss the properties <strong>and</strong> graph of f (x) = ln<br />
x 2 . Point out that the two other forms of f (x),<br />
+ 1<br />
( )<br />
1<br />
namely f (x) = ln<br />
1 + 1/x 2 =−ln<br />
(1 + 1 )<br />
x 2 or f (x) = 2lnx − ln ( x 2 + 1 ) , could be helpful.<br />
WORKSHOP/DISCUSSION<br />
• Compute derivatives of functions that involve logarithms <strong>and</strong> that require the Product, Quotient, or Chain<br />
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<br />
use the absolute value in the formula d<br />
dx ln |x| = 1 x .<br />
• Present an example of logarithmic differentiation, such as finding the derivative of (sin x) x2 +3x .<br />
• Go over Exercise 73 with the students.<br />
373
CHAPTER 7<br />
<strong>Inverse</strong> <strong>Functions</strong>: <strong>Exponential</strong>, <strong>Logarithmic</strong>, <strong>and</strong> <strong>Inverse</strong> Trigonometric <strong>Functions</strong><br />
GROUP WORK 1: <strong>Logarithmic</strong> Differentiation<br />
In addition to its face value, the process of logarithmic differentiation is a good way of mastering computations<br />
involving logarithms. Don’t give a group the second page until they have finished with the first page. If a<br />
group finishes early, have them compute the derivative of ( x 2 + 1 ) sin x .<br />
Answers:<br />
1. The Power Rule requires that the exponent be a constant.<br />
2. The Exponent Rule requires that the base be a constant.<br />
3. -- 6.<br />
y = (x + 1) x<br />
1 dy<br />
(x + 1) x dx<br />
ln y = ln (x + 1) x<br />
ln y = x ln (x + 1)<br />
1 dy<br />
= ln (x + 1) +<br />
x<br />
y dx x + 1<br />
= ln (x + 1) +<br />
x<br />
7. They are identical. 8. e x ln(x+1) [<br />
ln (x + 1) +<br />
9. e x ln(x+1) [<br />
ln (x + 1) +<br />
x + 1<br />
[<br />
dy<br />
dx = (x + 1)x ln (x + 1) +<br />
x ]<br />
x + 1<br />
x ]<br />
[<br />
= e ln((x+1)x ) ln (x + 1) +<br />
x + 1<br />
10. y ′ x sin y sin y + xy cos x sin x ln y<br />
=−y<br />
x ( x sin y y cos y ln x − y cos x cos x ) or y sin y + x sin x ln y<br />
x cos x − y ln x cos y<br />
x ]<br />
x + 1<br />
x ] [<br />
= (x + 1) x ln (x + 1) +<br />
x + 1<br />
x ]<br />
x + 1<br />
GROUP WORK 2: Euler’s Constant<br />
This is a difficult problem <strong>and</strong> the students may need some help. Here are a few suggested hints:<br />
Problem 3(a): Each shaded section is part of a larger rectangle with opposite corners on the graph of f (t) = 1 t .<br />
The combined area of these rectangles is<br />
(<br />
1 − 1 ) ( 1<br />
+<br />
2 2 3)<br />
− 1 ( 1<br />
+ ... +<br />
n − 1 )<br />
n + 1<br />
Problem 3(b): Because f (t) = 1 t<br />
is concave upward, each shaded section is larger than the triangle that is<br />
one-half of the rectangle described in part a). The combined area of these triangles is<br />
[(<br />
1<br />
1 − 1 ) ( 1<br />
+<br />
2 2 2 3)<br />
− 1 ( 1<br />
+ ... +<br />
n − 1 )]<br />
n + 1<br />
374
SECTION 7.2*<br />
THE NATURAL LOGARITHMIC FUNCTION<br />
Answers:<br />
1. ln (n + 1) 2. γ n = (The area under the rectangles) − ∫ n+1 dt<br />
1<br />
t<br />
3. (a) Applying the hint gives an area of 1 − 1<br />
n + 1 .<br />
(b) The hint gives this result directly.<br />
(c) If n ≥ 9, 1 − 1<br />
n + 1 = 0.9.<br />
4. 25<br />
761<br />
1,145,993<br />
12<br />
− ln 5 ≈ 0.47389542,<br />
280<br />
− ln 9 ≈ 0.52063257,<br />
360,360<br />
− ln 14 ≈ 0.54107643<br />
GROUP WORK 3: How Slowly can ln x Grow?<br />
This activity leads the students to discover first that ln x grows more slowly than any root function x 1/N .<br />
Answers:<br />
1. 1 t < 1<br />
t (N−1)/N for t ≥ 1. ∫ x<br />
1<br />
dt<br />
t<br />
<<br />
∫ x<br />
1<br />
dt<br />
for x ≥ 1.<br />
t (N−1)/N<br />
2. Computing the integrals gives the result.<br />
3. Transitivity gives the result.<br />
GROUP WORK 4: Some Surprising Areas<br />
Have the groups work on the first three problems, <strong>and</strong> then h<strong>and</strong> the fourth out to groups that finish early.<br />
When closing the exercise, impart to the students how surprising the result is: that equal areas are created by<br />
any choice of m > 1. Note that the result of Problem 4 can be generalized for the area between x = m a <strong>and</strong><br />
x = m b (b > a).<br />
Answers:<br />
1. Each of the areas is ln 2 ≈ 0.693147. The students will probably not be able to simplify their answers to<br />
ln 2, <strong>and</strong> may leave them as ln (ln 1.0201) − ln (ln 1.01),ln(ln 16) − ln (ln 4),<strong>and</strong>ln(ln 100) − ln (ln 10).<br />
2. The second endpoint is the square of the first.<br />
3. The area between x = m <strong>and</strong> x = m 2 is the same, regardless of the value of m.<br />
4.<br />
∫ m 2<br />
m<br />
∫<br />
dx ln m 2<br />
x ln x = ln m<br />
du<br />
u = ln ( ln m 2) − ln (ln m) = ln<br />
375<br />
( ) 2lnm<br />
= ln 2<br />
ln m
CHAPTER 7<br />
<strong>Inverse</strong> <strong>Functions</strong>: <strong>Exponential</strong>, <strong>Logarithmic</strong>, <strong>and</strong> <strong>Inverse</strong> Trigonometric <strong>Functions</strong><br />
HOMEWORK PROBLEMS<br />
Core Exercises: 2, 4, 5, 10, 17, 25, 35, 42, 51, 68<br />
Sample Assignment: 2, 4, 5, 9, 10, 11, 17, 23, 25, 35, 38, 42, 45, 51, 59, 64, 68<br />
Exercise D A N G<br />
2 ×<br />
4 ×<br />
5 ×<br />
9 ×<br />
10 ×<br />
11 ×<br />
17 ×<br />
23 ×<br />
25 ×<br />
35 ×<br />
38 ×<br />
42 ×<br />
45 ×<br />
51 × ×<br />
59 ×<br />
64 ×<br />
68 ×<br />
376
GROUP WORK 1, SECTION 7.2*<br />
<strong>Logarithmic</strong> Differentiation<br />
Let f (x) = (x + 1) x . As of now, none of the rules that we’ve learned seem to help us find f ′ (x).<br />
1. Why can’t you use the Power Rule to compute f ′ (x)?<br />
2. Why can’t you use a formula like d<br />
dx 3x = 3 x ln3tocompute f ′ (x)?<br />
3. Start with the equation y = (x + 1) x . Take the logarithm of both sides to get ln y = ln ( (x + 1) x) . Then<br />
use a property of logarithms to help you find the derivative of ln ( (x + 1) x) .<br />
4. Now use implicit differentiation to find d ln y.<br />
dx<br />
5. Since ln y = ln ( (x + 1) x) , you can equate your answers to Parts 3 <strong>and</strong> 4. Do so, <strong>and</strong> then replace all y<br />
terms by (x + 1) x . Why can you do this substitution?<br />
6. Perform some algebra to get dy<br />
dx<br />
alone on the left h<strong>and</strong> side, <strong>and</strong> you are done!<br />
377
<strong>Logarithmic</strong> Differentiation<br />
7. Let g (x) = e x ln(x+1) . What is the relationship between f (x) <strong>and</strong> g (x)?<br />
8. If g (x) = e x ln(x+1) ,computeg ′ (x). What techniques are you using in this case?<br />
9. Since you have done the same problem in two different ways, show that your answers to Problems 6 <strong>and</strong><br />
8 are identical.<br />
10. Let y = f (x) be implicitly defined by x sin y = y cos x . Compute y ′ in terms of x <strong>and</strong> y. (Hint: Can<br />
logarithms help you?)<br />
378
GROUP WORK 2, SECTION 7.2*<br />
Euler’s Constant<br />
Consider the following picture:<br />
1. Find a formula for the left-h<strong>and</strong> sum of<br />
bound for the integral.<br />
∫ n+1<br />
1<br />
1<br />
t<br />
dt with n subintervals. Notice that this sum is an upper<br />
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 −<br />
ln (n + 1).<br />
3. It is a fact that lim<br />
n→∞ γ n = γ exists. We want to estimate γ.<br />
(a) Using a geometric argument, show that γ n ≤ 1 for all n.<br />
(b) Using a geometric argument, show that 1 2<br />
(<br />
1 − 1 )<br />
≤ γ<br />
n + 1 n for all n.<br />
(c) Conclude from parts (a) <strong>and</strong> (b) that 0.45 ≤ γ n ≤ 1ifn ≥ 9.<br />
4. Compute precise values for γ n for n = 4, 8, <strong>and</strong> 13. How do these values compare to the lower bound<br />
computed in Problem 3(c)?<br />
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<br />
whether γ is rational or irrational.<br />
379
1<br />
GROUP WORK 3, SECTION 7.2*<br />
How Slowly can ln x Grow?<br />
1. Let N be a positive integer. How do the functions 1<br />
∫<br />
t<br />
x ∫<br />
dt x<br />
1<br />
≤ dt for x ≥ 1.<br />
t t (N−1)/N<br />
1<br />
<strong>and</strong><br />
1<br />
compare for t ≥ 1? Show that<br />
t (N−1)/N<br />
2. Use Problem 1 to show that ln x ≤ N ( x 1/N − 1 ) for all x ≥ 1.<br />
3. Use Problem 2 to show that ln x ≤ Nx 1/N , for any positive integer N. Conclude that ln x grows “more<br />
slowly” than any root function x 1/N ,whereN is a positive integer.<br />
380
GROUP WORK 4, SECTION 7.2*<br />
Some Surprising Areas<br />
Meet<br />
1<br />
x ln x :<br />
1. Compute the area under f (x) = 1<br />
x ln x<br />
from x = 1.01 to 1.0201, from 4 to 16, <strong>and</strong> from 10 to 100.<br />
2. What do the three pairs of endpoints used in Problem 1 have in common?<br />
3. The answers to Problems 1 <strong>and</strong> 2 should suggest an interesting property of the curve f (x) = 1<br />
x ln x .<br />
What is it?<br />
381
4. Prove that the area enclosed by f (x) = 1<br />
x ln x ,thex-axis, the line x = m <strong>and</strong> the line x = m2 is<br />
independent of your choice of m (as long as m > 1).<br />
382
7.3* THE NATURAL EXPONENTIAL FUNCTION<br />
SUGGESTED TIME AND EMPHASIS<br />
1 class Essential material<br />
POINTS TO STRESS<br />
1. The definition of e x as the inverse of ln x.<br />
2. The graph <strong>and</strong> properties of e x .<br />
3. Derivatives <strong>and</strong> integrals involving e x .<br />
QUIZ QUESTIONS<br />
• Text Question: Why is the equation e ln x = x only true for x > 0, but ln (e x ) = x holds for all x?<br />
Answer: Because ln x is defined only for x > 0.<br />
• Drill Question: What property of the graph of ln x ensures that lim<br />
x→−∞ ex = 0?<br />
Answer: lim ln x =−∞<br />
x→0 +<br />
MATERIALS FOR LECTURE<br />
• Using properties of inverse functions, graph e x from ln x, <strong>and</strong> illustrate several laws of exponents. Point<br />
out using the graphs that since lim<br />
x→0<br />
• Compute d<br />
• Compute d<br />
dx<br />
d<br />
dx<br />
ln x =−∞, it follows that lim<br />
+ x→−∞ ex = 0.<br />
(<br />
e x3) (<br />
≠ e x3 . Show that<br />
dx (ex ) = e x <strong>and</strong> then point out that<br />
( ) e x2 +x<br />
(there are two possible ways to do this),<br />
d<br />
dx (sin (ex )) <strong>and</strong><br />
dx<br />
d<br />
e x3) ′<br />
= 3x 2 e x3 .<br />
√<br />
1 + ( e −x) 2 .<br />
• Draw the graph of e −x from the graph of e x . Then use calculus to derive properties of e x/(1−x) , <strong>and</strong> sketch<br />
its graph.<br />
∫ ∫<br />
∫<br />
• Compute the following integrals: e 3x dx, e x cos e x e x ∫<br />
dx,<br />
1 + e x dx,<strong>and</strong> sin xe cos x dx.<br />
WORKSHOP/DISCUSSION<br />
• Have the students graph f (x) = e x2 <strong>and</strong> estimate the slope at x = 0<strong>and</strong>x = 1. Ask why the slope at<br />
x = 0 is 0, while the slope of g (x) = e x at x = 0is1.<br />
• Derive the properties of f (x) = e1/x<br />
, <strong>and</strong> sketch its graph. Point out that writing f (x) = ex+(1/x)<br />
e−x simplifies the process. Where is f ′ (x) = 0?<br />
( 5<br />
• Do some differentiations using e x , such as those of e sin2 x , x 2 e x5 , e x2 + 1)<br />
<strong>and</strong> cos (x e + e x ).Anice<br />
summary problem is g (x) = ex 3 + e x3 + xe 3 + x e3 + ( e 3) x .<br />
∫ 2xe<br />
ln ( x 2 +1 ) ∫ ln (sin e x ∫<br />
)<br />
• Compute<br />
x 2 dx,<br />
+ 1 e x cos e x dx,<strong>and</strong> e 2x<br />
√ dx.<br />
1 + e 4x<br />
383
GROUP WORK: On <strong>and</strong> On with <strong>Exponential</strong> Inequalities<br />
This group work is essentially Exercises 87 <strong>and</strong> 88, using a slightly different initial approach. It might be<br />
useful to assign Exercise 89 after the students have completed this group work.<br />
Answers:<br />
1. This can be done by using Property 7 of Integrals in Section 5.2.<br />
2. Computation of the integrals in Problem 1 gives x ≤ e x − 1.<br />
3. Integration of both sides of the inequality in Problem 2 gives the desired result.<br />
4. P (x) = 1 + x + x2<br />
2 + x3<br />
6<br />
HOMEWORK PROBLEMS<br />
Core Exercises: 2, 4, 8, 18, 23, 28, 39, 50, 61, 77, 90<br />
Sample Assignment: 1, 2, 4, 5, 8, 15, 18, 21, 23, 25, 28, 39, 45, 50, 61, 63, 72, 77, 78, 90<br />
Exercise D A N G<br />
1 ×<br />
2 ×<br />
4 ×<br />
5 ×<br />
8 ×<br />
15 ×<br />
18 ×<br />
21 ×<br />
23 ×<br />
25 ×<br />
28 ×<br />
39 ×<br />
45 ×<br />
50 ×<br />
61 × × ×<br />
63 ×<br />
72 ×<br />
77 ×<br />
78 ×<br />
90 × ×<br />
384
GROUP WORK 1, SECTION 7.3*<br />
On <strong>and</strong> On <strong>Exponential</strong> Inequalities<br />
1. Using the fact that the constant function f (x) ≡ 1 ≤ e x for x ≥ 0, why can we conclude that<br />
∫ x<br />
0 f (u) du ≤ ∫ x<br />
0 eu du?<br />
2. Show that 1 + x ≤ e x if x ≥ 0.<br />
3. Using the same process as you used to derive the solution to Problem 2, show that 1 + x + x2<br />
2 ≤ ex if<br />
x ≥ 0.<br />
4. Write the equation for the polynomial P (x) such that P (x) ≤ e x if x ≥ 0, if we apply the same process<br />
one more time to the answer of Problem 3.<br />
385
7.4* GENERAL LOGARITHMIC AND EXPONENTIAL FUNCTIONS<br />
SUGGESTED TIME AND EMPHASIS<br />
1–1 1 2<br />
classes Essential material<br />
POINTS TO STRESS<br />
1. <strong>Exponential</strong> graphs, from both symbolic <strong>and</strong> geometric perspectives, <strong>and</strong> their properties (derived from<br />
e x using a x = e x ln a ).<br />
2. General logarithmic functions, their properties as inverses of exponentials, <strong>and</strong> their graphs (including<br />
asymptotic behavior).<br />
3. The basic derivative formulas.<br />
4. The concept of e as a limit<br />
QUIZ QUESTIONS<br />
• Text Question: Why does a log a x = x hold only for x > 0?<br />
Answer: log a x is defined only for x > 0.<br />
• Drill Question: What is the inverse function for y = log a x?<br />
Answer: a x<br />
MATERIALS FOR LECTURE<br />
• Starttodrawagraphof2 x vs x, using the scale of one inch per unit on both axes. Point out that after one<br />
foot, the height would be over 100 yards (the length of a football field). After two feet, the height would<br />
be 264 miles, after three feet it would be 1,000,000 miles (four times the distance to the moon), after three<br />
<strong>and</strong> a half feet it would be in the heart of the sun. If the graph extended five feet to the right, x = 60, then<br />
y would be over one light year up.<br />
• Discuss the general definition of a x = e x ln a for a > 0<strong>and</strong>x a real number, <strong>and</strong> discuss why this equation<br />
makes sense even for irrational numbers. [For example, √ 3<br />
√ √<br />
2√<br />
= e<br />
3ln 2<br />
= e<br />
properties of a x for a > 1<strong>and</strong>0< a < 1.<br />
( √3ln2 )<br />
/2 ]. Compare the<br />
• Derive the general Power Rule (x n ) ′ = nx n−1 for any real number n.<br />
• By differentiating the equation a log a x = x implicitly, derive the formula ( log a x ) ′ 1 =<br />
x ln a .<br />
• Estimate where 3 x > x 3 <strong>and</strong> 2 x > x 8 using technology. Notice that exponential functions start by growing<br />
slower than polynomial functions, <strong>and</strong> then wind up growing much faster. For example, if one were to<br />
graph x 2 vs x using the scale of one inch per unit on both axes, then when x = 60, y would be only<br />
100 yards, as opposed to a light year for e x . (The sun is only 8 light seconds from the earth.)<br />
386
SECTION 7.4*<br />
GENERAL LOGARITHMIC AND EXPONENTIAL FUNCTIONS<br />
WORKSHOP/DISCUSSION<br />
(<br />
• Sketch a graph of f (x) = log 2 (x + 3)<br />
2 ) ,sketch<br />
the inverse function, <strong>and</strong> then find an algebraic<br />
formula for the inverse.<br />
Answer:<br />
y<br />
y=f(x)<br />
y=f Ð!(x)<br />
2<br />
0<br />
2<br />
x<br />
f −1 (x) = 2 x/2 − 3<br />
• Compute derivatives of functions that involve logarithms <strong>and</strong> exponents <strong>and</strong> that require the Product,<br />
Quotient, or Chain Rule. For example, use f (x) = log 2 (sin x) sin ( log 2 x ) (<br />
or f (x) = log 3 e x + log 2 x ) .<br />
• See if the students can identify the inverse of f (x) =<br />
(<br />
23<br />
) x<br />
as g (x) = log2/3 x. Ask them to solve<br />
(<br />
23<br />
) x<br />
= 1, <strong>and</strong> to explain why the answer is negative.<br />
• Compute the derivatives of the functions (sin x) π/4 , ( sin ( ))<br />
π x, ( π4<br />
) sin x,<br />
4<br />
<strong>and</strong>x sin(π/4) . Ideally, give the<br />
students time to work independently, <strong>and</strong> then discuss the answers as a class.<br />
∫ ∫<br />
√2 x 2xdx<br />
• Compute dx <strong>and</strong> ( x 2 + 1 ) ln 2 .<br />
GROUP WORK 1: Comparisons<br />
The purpose of this group work is to give the students a bit of “picture sense”. It is acceptable if they do this<br />
by looking at the graphs on their calculators, setting the windows appropriately.<br />
Answers: 1. 0 < x < 1.374 <strong>and</strong> x > 9.940 2. 0 < x < 1.052 <strong>and</strong> x > 95.7168 3. 0 < x < 1.18 <strong>and</strong><br />
x > 22.4 4. 0 < x < 1.34 <strong>and</strong> x > 10.9<br />
GROUP WORK 2: Irrational, Impossible Relations<br />
Before starting this activity, review the definitions of rational <strong>and</strong> irrational numbers. The hint sheet should<br />
be given out only after the students have tried to show that log 2 3 is irrational, or at least discussed it enough<br />
to underst<strong>and</strong> what they are trying to show. If a group finishes early, have them show that log 2 a is always<br />
irrational if a is an odd integer.<br />
Answers: 1. −γ 2. −γ 3. γ 4. log<br />
2<br />
2 3 = ln 3 5. The proof is outlined in the hint sheet.<br />
ln 2<br />
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,<br />
which is undefined. 3. 2 a = 2 · 2 ·····2, <strong>and</strong> 3 b = 3 · 3 ·····3, so these numbers can never be equal,<br />
because the left is always divisible by two, <strong>and</strong> the right never is (unless b = 0). 4. Irrational.<br />
387
GROUP WORK 3: e as a Limit<br />
One can introduce this activity by deriving the expression in Problem 1 in terms of compound interest. Some<br />
students will require more guidance on what types of things to look for in Problem 1. Properties to notice<br />
include concavity, asymptotes, extrema, increase/decrease, <strong>and</strong> positivity/negativity.<br />
Answers:<br />
1. y<br />
2. The horizontal asymptote of the graph<br />
(<br />
3<br />
corresponds to lim 1 + 1 ) x<br />
.<br />
x→∞ x<br />
2<br />
3. The limit is e ≈ 2.718.<br />
1<br />
0<br />
1 2 3 4 5 6 7 8 9 10 x<br />
HOMEWORK PROBLEMS<br />
Core Exercises: 5, 7, 16, 25, 32, 44, 47, 51, 62<br />
Sample Assignment: 1, 5, 7, 11, 16, 21, 25, 32, 37, 40, 44, 46, 47, 48, 51, 61, 62<br />
Exercise D A N G<br />
1 × ×<br />
5 × ×<br />
7 ×<br />
11 × ×<br />
16 × ×<br />
21 ×<br />
25 ×<br />
32 ×<br />
37 ×<br />
40 ×<br />
44 × ×<br />
46 ×<br />
47 ×<br />
48 ×<br />
51 ×<br />
61 × × ×<br />
62 × × × ×<br />
388
GROUP WORK 1, SECTION 7.4*<br />
Comparisons<br />
You have learned that an exponential function grows faster than a polynomial function. Find the values of<br />
x > 0forwhich<br />
1. 2 x ≥ x 3 .<br />
2. (1.1) x ≥ x 2 .<br />
3. 2 x ≥ x 5 .<br />
4. 3 x ≥ x 5 .<br />
389
1. If log 2 x = γ, then what is log 1/2 x?<br />
GROUP WORK 2, SECTION 7.4*<br />
Irrational, Impossible Relations<br />
2. If log b x = γ, then what is log 1/b x (assuming b > 1)?<br />
3. If log b x = γ, then what is log b 2 x?<br />
4. We are going to estimate log 2 3. In pre-calculus, you memorized that log 2 3 ≈ 1.584962501. Suppose you<br />
didn’t have this fact memorized. There is no log 2 3 button on your calculator! How would you compute<br />
it?<br />
5. Unfortunately, the calculator gives us only a finite number of digits. If log 2 3 were a rational number,<br />
we would be able to express it as a fraction, giving us perfect accuracy. Do you think it is rational or<br />
irrational? Try to prove your result.<br />
390
GROUP WORK 2, SECTION 7.4*<br />
Irrational, Impossible Relations (Hint Sheet)<br />
So, you realize that it’s not easy to determine whether log 2 3 is rational!<br />
Onewaytoattempttoshowthatlog 2 3 is rational is to assume that it is, <strong>and</strong> try to find integers a <strong>and</strong> b such<br />
that log 2 3 = a b . If we can show that there are no such a <strong>and</strong> b,thenlog 2 3 cannot be rational.<br />
1. Assume that log 2 3 = a b . Show that a <strong>and</strong> b must then satisfy 2a = 3 b<br />
2. Notice that a = 0, b = 0 satisfies 2 a = 3 b . Show that this fact doesn’t help us.<br />
3. Find a ≠ 0<strong>and</strong>b ≠ 0 that satisfy 2 a = 3 b ,orshowthatnosuch{a, b} exists.<br />
4. Is log 2 3 rational or irrational? Why?<br />
391
GROUP WORK 3, SECTION 7.4*<br />
e as a Limit<br />
(<br />
1. Graph the function f (x) = 1 + 1 ) x<br />
, <strong>and</strong> describe its properties.<br />
x<br />
y<br />
3<br />
2<br />
1<br />
0<br />
1 2 3 4 5 6 7 8 9 10 x<br />
2. What is the relationship between the graph you drew for Problem 1 <strong>and</strong> Formula 9 in the text?<br />
(<br />
3. Using Problems 1 <strong>and</strong> 2 as a guide, what do you think lim 1 + 1 ) x<br />
is?<br />
x→∞ x<br />
392
7.5 EXPONENTIAL GROWTH AND DECAY<br />
SUGGESTED TIME AND EMPHASIS<br />
1 class Recommended material<br />
POINTS TO STRESS<br />
1. The general growth/decay equation dy/dt = ky, y (0) = y 0 <strong>and</strong> its associated solution curves.<br />
2. Systems that exhibit exponential growth or decay.<br />
3. The importance of the sign of the exponential growth/decay constant k.<br />
QUIZ QUESTIONS<br />
• Text Question: If dy/dt = ky,<strong>and</strong>k is a non-zero constant, then y could be<br />
(A)2e kty (B)2e kt (C) e kt + 3 (D) kty + 5 (E) 1 2 ky2 + 1 2<br />
Answer: (B)<br />
• Drill Question: A population grows according to the equation dP/dt = kP,wherek is a constant <strong>and</strong> t<br />
is measured in years. If the population doubles every 10 years, then what is the value of k?<br />
Answer: k = 0.069<br />
MATERIALS FOR LECTURE<br />
• Present a few real-world examples of exponential decay. For example, assume that a certain amount of<br />
pollution was dumped into a water reservoir, <strong>and</strong> that clean water was being added to the reservoir <strong>and</strong><br />
taken from it at a constant rate. If the water is constantly being mixed, then the concentration of pollution<br />
will decay exponentially over time.<br />
• Derive Equation 2 by solving dy/dt = ky, y (0) = y 0 using separation of variables.<br />
• Discuss exponential decay problems with an initial y<br />
value. Recall Newton’s Law of Cooling (Group Work 3, 100<br />
Section 10.3). Show how the condition y(0) = y 0<br />
80<br />
determines whether the solution is a growth or decay<br />
function. Perhaps illustrate with a slope-field diagram.<br />
60<br />
40<br />
20<br />
0<br />
2 4 6 8 10 x<br />
WORKSHOP/DISCUSSION<br />
• Referring back to earlier material, show that the expression y = e kt can be written as y = a t <strong>and</strong> that<br />
e kt+c can be written as Ae kt .<br />
• Do this variant on Exercise 11: We know that the half-life of carbon-14 is 5,730 years. The Shroud of<br />
Turin is an ancient artifact that many believe to be the burial shroud of Jesus Christ. In 1988, the Vatican<br />
consented to give a few fibers to scientists to carbon-date. They found that the fibers had 92% of the<br />
original 14 C left. Discuss what this implies about the age of the shroud.<br />
• Go over Exercise 3, using the doubling rate to write solutions in the form y = A2 kt , k = 1/T , where T is<br />
the doubling time.<br />
393
GROUP WORK 1: The Rule of 72<br />
If a group finishes early, have them redo Problems 1–5 for annual compounding.<br />
Answers:<br />
1. 13.86 years<br />
2. The doubling time is not affected by changes in principal. Algebraically, the P 0 drops out of the<br />
exponential growth equation. Intuitively, the doubling time is a property of the ratio of two numbers,<br />
not a property of the numbers themselves.<br />
3. Estimate: 14.4 years, a 3.90% error.<br />
4. Interest Rate Actual Doubling Time Estimated Doubling Time Error<br />
3% 23.1 years 24 years 3.90%<br />
8% 8.66 years 9 years 3.92%<br />
12% 5.78 years 6 years 3.81%<br />
18% 3.85 years 4 years 3.90%<br />
The Rule of 72 is accurate to within 4% for the given range of interest rates.<br />
5. If r is the interest rate (as a number, not a percentage), then the doubling time is D = ln 2<br />
r .<br />
6. 100ln2≈ 69. This estimate gives a doubling time of 13.8, for an error of 0.4%.<br />
7. If the compounding is assumed to be annual, 72 works best for values of r between 3% <strong>and</strong> 15%. For<br />
example, 10% compounded annually doubles every 7.2735 years. The rule of 72 gives 7.2, a good<br />
approximation, <strong>and</strong> the rule of 69 gives 6.9, a bad one.<br />
GROUP WORK 2: More of Homer’s Blood<br />
Although this is a thematic sequel to the Group Work in Section 9.4, it can also st<strong>and</strong> on its own.<br />
Answers:<br />
1. It is nearly constant.<br />
2. 30 minutes: dC<br />
dt<br />
3. dC<br />
dt<br />
≈−0.1165; 60 minutes: dC<br />
dt<br />
≈−0.0259; 80 minutes: dC<br />
dt<br />
≈−0.0096<br />
= kC 4. k ≈−0.052 5. y = 10e −0.052t<br />
6. After approximately 13.33 minutes 7. Any time after 3:52 P.M.<br />
GROUP WORK 3: Find the Error<br />
Answer: When we divide by y in the first step, we assume that y ≠ 0. Wewouldhavetochecky = 0as<br />
a solution separately. The integration gives ln |y| = kx + c. The derivation assumed y > 0. If we assumed<br />
y < 0 we would get y =−Ae kx ,whereA > 0.<br />
394
SECTION 7.5<br />
EXPONENTIAL GROWTH AND DECAY<br />
HOMEWORK PROBLEMS<br />
Core Exercises: 2, 5, 8, 18<br />
Sample Assignment: 2, 3, 5, 8, 11, 13, 16, 18, 20<br />
Exercise D A N G<br />
2 × ×<br />
3 × ×<br />
5 × × ×<br />
8 × ×<br />
11 × ×<br />
13 × ×<br />
16 ×<br />
18 × ×<br />
20 ×<br />
395
GROUP WORK 1, SECTION 7.5<br />
TheRuleof72<br />
In this exercise, we attempt to answer the question asked by many investors: “How long is it going to take<br />
for me to double my money?”<br />
1. Consider an investment of $100 invested at 5%, compounded continuously. How long would it take for<br />
the investor to have $200?<br />
2. What would the doubling time be if the initial investment were $1,000? $10,000? What effect does<br />
changing the principal have on the doubling time, <strong>and</strong> why?<br />
One of the first things that is taught in an economics class is the Rule of 72. It can be summarized thusly:<br />
“The number of years it takes an investment to double<br />
is equal to 72 divided by the annual percentage interest rate.”<br />
3. What would the Rule of 72 say the doubling time of a 5% investment is? Is it a good estimate?<br />
4. Repeat Problems 1 <strong>and</strong> 3 for investments of 3%, 8%, 12% <strong>and</strong> 18%. What can you say about the accuracy<br />
of the Rule of 72?<br />
5. Derive a precise formula for the time T to double an initial investment.<br />
6. There is an integer that gives a more accurate answer for continuous or nearly continuous compounding<br />
than the Rule of 72. What is this number? Check your answer by using it to estimate the doubling time<br />
of a 5% investment.<br />
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<br />
assumptions we made. Why do economists use the Rule of 72?<br />
396
GROUP WORK 2, SECTION 7.5<br />
More of Homer’s Blood<br />
In a previous exercise, we discussed Homer’s blood pressure. Well, it turns out that Homer has to have an operation.<br />
When he is brought into the hospital, he is given a sedative to help him sleep. The doctor wants to operate,<br />
but cannot safely do so until the concentration of sedative in his body is less than 0.03 milligrams/liter.<br />
In this exercise, we will determine how many hours the doctor must wait until he can operate.<br />
The following table of data was obtained by monitoring the levels of the sedative in Homer’s blood. Samples<br />
were taken every ten minutes, <strong>and</strong> the concentration of the drug was determined <strong>and</strong> reported in milligrams<br />
per liter.<br />
Time (minutes) Concentration (mg/L)<br />
0 10.0<br />
10 6.07<br />
20 3.68<br />
30 2.23<br />
40 1.35<br />
50 0.820<br />
60 0.498<br />
70 0.302<br />
80 0.183<br />
90 0.111<br />
1. Is the rate of change of concentration with respect to time a constant?<br />
2. Estimate the rate of change of concentration with respect to time at 30 minutes, 60 minutes, <strong>and</strong><br />
80 minutes.<br />
3. Show that the rate of change is (roughly) proportional to the concentration C. Write this relationship as a<br />
differential equation for dC/dt.<br />
4. Find the constant of proportionality.<br />
397
More of Homer’s Blood<br />
5. Solve the differential equation you wrote in Problem 3, using the constant of proportionality from<br />
Problem 4.<br />
6. When is the drug’s concentration halved?<br />
7. If Homer received the sedative at 2:00 P.M., when can the doctor start the operation?<br />
398
GROUP WORK 3, SECTION 7.5<br />
Find the Error<br />
It is a beautiful spring afternoon, you are sitting in the park, <strong>and</strong> you are happy. Your favorite volleyball team<br />
is going to the Olympics, but that isn’t what is making you happy. Your sweetheart has just told you that<br />
you are getting a nice surprise for May Day, but that isn’t what is making you happy. A magazine company<br />
has just informed you that you are entered in a million-dollar sweepstakes, but that isn’t what is making you<br />
happy. You are happy because you have learned a new theorem:<br />
If dy<br />
dx = ky,theny = Aekx .<br />
The nice thing about this theorem, you think, is that the converse is also true. In other words:<br />
If y = Ae kx , then dy<br />
dx = ky.<br />
You hear some tinkly bells, <strong>and</strong> there is the ice cream cart! You wave, <strong>and</strong> it comes towards you. But there<br />
is something wrong with the ice cream vendor. He is wearing large dark glasses <strong>and</strong> an obviously fake beard<br />
<strong>and</strong> mustache. You buy the ice cream anyway (one of those ice cream bars on a stick), but after he has gone<br />
away, you notice this written on the wrapper.<br />
There are two unusual things going on here. First of all, this message appeared on your wrapper in the first<br />
place. Secondly, you are most of the way through a “Find the error” problem, <strong>and</strong> there isn’t an error here! In<br />
fact, you might have seen this very proof in your calculus class. You eat your ice cream, <strong>and</strong> then notice this<br />
written on the stick:<br />
Since A = e c ,thenA can’t be zero, <strong>and</strong> A can’t be negative.<br />
But if y = Ae kx ,then dy<br />
dx = ky,nomatterwhatthevalueofA.<br />
So why didn’t your teacher’s ‘proof’ catch all the solutions?<br />
Why not, indeed? Find the error.<br />
399
7.6 INVERSE TRIGONOMETRIC FUNCTIONS<br />
SUGGESTED TIME AND EMPHASIS<br />
1<br />
2<br />
– 1 class Recommended material (especially the inverse sine, inverse cosine <strong>and</strong> inverse tangent functions)<br />
POINTS TO STRESS<br />
1. Definitions, <strong>and</strong> domains <strong>and</strong> ranges of inverse trigonometric functions.<br />
2. Derivatives <strong>and</strong> integrals of inverse trigonometric functions.<br />
3. Uses of inverse trigonometric functions in techniques of integration.<br />
QUIZ QUESTIONS<br />
• Text Question: Whyisittruethatsin −1 (sin x) = x only when − π 2<br />
≤ x ≤ π 2 <strong>and</strong> sin ( sin −1 x ) = x<br />
only when −1 ≤ x ≤ 1?<br />
Answer: The range of sin −1 x is limited to − π 2 ≤ sin−1 x ≤ π 2<br />
, <strong>and</strong> its domain is −1 ≤ x ≤ 1.<br />
• Drill Question: What is d ( x 2 sin −1 √ x ) ?<br />
dx<br />
Answer: x [ 4 √ 1 − x sin −1 √ x + √ x ]<br />
√ 1 − x<br />
MATERIALS FOR LECTURE<br />
• Ask students to guess the largest domain for cos x containing 0 on which cos x is one-to-one. Then define<br />
cos −1 x.<br />
Answer: 0 ≤ x ≤ π<br />
• Point out that since<br />
dx<br />
d ( cos −1 x ) = −dx<br />
d ( sin −1 x ) ,cos −1 x + sin −1 x is a constant, <strong>and</strong> show that<br />
cos −1 x + sin −1 x = π 2 . This provides an easy way to remember the graph of cos−1 x.<br />
• Solve some right angle identities involving inverse trigonometric functions, such as sin ( tan −1 x ) ,<br />
tan ( sin −1 x ) <strong>and</strong> cos ( sin −1 x ) .<br />
• Solve some integration problems using inverse trigonometric functions, such as<br />
∫<br />
2xdx<br />
√ . 1 − x 4<br />
400<br />
∫ 2<br />
0<br />
1<br />
dx <strong>and</strong><br />
1 + 4x2
SECTION 7.6<br />
INVERSE TRIGONOMETRIC FUNCTIONS<br />
WORKSHOP/DISCUSSION<br />
• Set up a problem that requires some additional skills or information. A<br />
good example is Exercise 47: Where should the point P be chosen on the<br />
line segment AB in the figure at right so as to maximize the angle θ?<br />
5<br />
Answer: We can maximize θ by minimizing the sum S of the<br />
complementary angles α <strong>and</strong> β. Let x be the distance from A to<br />
( ) ( )<br />
5 2<br />
P. Then S = arctan + arctan , with 0 < x < 3.<br />
x 3 − x<br />
S ′ =<br />
−5<br />
x 2 + 25 + 2<br />
(3 − x) 2 + 4 . The critical points turn out to be 2.1<br />
0 1<br />
S<br />
2.5<br />
2.4<br />
2.3<br />
2.2<br />
A<br />
3<br />
Œ ¬ º<br />
P<br />
2 3 x<br />
the roots of x 2 − 10x + 5, that is, 5 ± 2 √ 5. So we check x = 0, x = 5 − 2 √ 5, <strong>and</strong> x = 3tofindthat<br />
x = 5 − 2 √ 5 ≈ 0.528 gives a minimum. The result is verified by the graph of S.<br />
2<br />
B<br />
• Define <strong>and</strong> graph tan −1 x, noting that tan ( tan −1 x ) = x for −∞ < x < ∞, while tan −1 (tan x) = x for<br />
− π 2 < x < π 2 .<br />
( ) 1 + x<br />
• Sketch the graph of f (x) = tan −1 1 + x 2 .<br />
• Differentiate y = tan −1 ( x 2 + e x) , y 2 = sin ( e θ + cos −1 θ ) <strong>and</strong> x 4 + x = cos (xy).<br />
GROUP WORK 1: <strong>Inverse</strong> Trickery<br />
This activity helps the students build a skill that is useful when doing inverse trigonometric substitutions. If<br />
they have never seen anything like this before, you may want to do an example on the board with the students<br />
before h<strong>and</strong>ing out the activity.<br />
Answers:<br />
1.<br />
x<br />
√<br />
x 2 + 4<br />
2.<br />
√<br />
x 2 + 9<br />
3<br />
3. √ 1 − x 4 4.<br />
401<br />
x<br />
√<br />
16 − x 2<br />
5.<br />
2<br />
1 + x 2
GROUP WORK 2: Where to Hang That Picture?<br />
Students will probably need some help getting started on this activity. In Problem 1, they should write θ as<br />
the sum of the angles above <strong>and</strong> below the horizontal (see diagram below). Each of these angles can then be<br />
written in terms of the position of the painting using the inverse tangent.<br />
θ = α + β<br />
tan α = x d<br />
tan β = h − x<br />
d<br />
d is the (fixed) distance from the observer to the painting.<br />
In Problem 2, let α be the angle between the bottom of the painting <strong>and</strong> the horizontal. Write tan θ as<br />
tan x − tan y<br />
tan (θ + α − α) <strong>and</strong> use the identity tan (x − y) =<br />
1 + tan x tan y .<br />
Note that Problem 2 of this activity is identical to Exercise 48.<br />
Answers:<br />
1. Starting as indicated above, we get ( sec 2 α ) dα<br />
dx<br />
= 1 d <strong>and</strong> ( sec 2 β ) dβ<br />
dx = −1 dθ<br />
. We set<br />
d dx = 0:<br />
cos 2 α<br />
− cos2 β<br />
= 0 ⇔ α = β ⇔ x d d<br />
d = h − x ⇔ x = h d<br />
2 .<br />
2. Starting as indicated above, <strong>and</strong> calling the distance from the observer to the wall x we get tan θ =<br />
h + d<br />
− d x x<br />
hx<br />
( )( ) =<br />
h + d d x<br />
1 +<br />
2 + dh + d 2 . We can maximize θ by maximizing tan θ, <strong>and</strong> we find that<br />
x x<br />
hx<br />
x 2 + dh + d 2 is maximized when x = √ d (h + d).<br />
HOMEWORK PROBLEMS<br />
Core Exercises: 1, 3, 13, 16, 18, 19, 24, 39, 43, 57, 64<br />
Sample Assignment: 1, 3, 6, 13, 14, 16, 18, 19, 22, 24, 33, 37, 39, 41, 43, 49, 57, 63, 64, 72<br />
Exercise D A N G<br />
1 ×<br />
3 ×<br />
6 ×<br />
13 ×<br />
14 ×<br />
16 × ×<br />
18 ×<br />
19 ×<br />
22 ×<br />
24 ×<br />
Exercise D A N G<br />
33 ×<br />
37 ×<br />
39 ×<br />
41 × ×<br />
43 ×<br />
49 × ×<br />
57 ×<br />
63 ×<br />
64 ×<br />
72 ×<br />
402
GROUP WORK 1, SECTION 7.6<br />
<strong>Inverse</strong> Trickery<br />
Rewrite the following expressions in terms of x without any trigonometric or inverse trigonometric functions.<br />
1. sin ( tan −1 2<br />
x )<br />
2. sec ( tan −1 3<br />
x )<br />
3. cos ( sin −1 x 2)<br />
4. tan ( sin −1 4<br />
x )<br />
5. sin ( 2tan −1 x ) 403
GROUP WORK 2, SECTION 7.6<br />
Where to Hang That Picture?<br />
1. Suppose we want to hang a rectangular painting of height h onthewallsothatanobserverst<strong>and</strong>ingin<br />
front of the painting has the best view. In other words, we want to maximize the angle θ (see diagram)<br />
subtended by the painting. Prove that θ is maximized when the center of the painting is at the level of the<br />
observer’s eyes.<br />
2. Now suppose the painting has already been hung at a distance d above the eye of the observer (see<br />
diagram). How far from the wall should the observer st<strong>and</strong> to get the best view?<br />
404
APPLIED PROJECT<br />
WheretoSitattheMovies<br />
It is hard to imagine an undergraduate student who has not had an argument with his or her friends concerning<br />
the best seat in the movie theater. “Up front” connoisseurs, “Middle center” partisans <strong>and</strong> “Balcony” fetishists<br />
have nearly come to blows over this decision. In this exercise, the students use the mathematics they’ve learned<br />
to settle the question for a model movie theater.<br />
Problems 3 <strong>and</strong> 4 can be done with a scientific calculator if a CAS is not available. (The derivative will be<br />
very messy, but still within the students’ abilities.)<br />
If a more in-depth experience is desired, students could be assigned to perform measurements at a local<br />
theater, <strong>and</strong> thus determine the actual “best seat in the house”.<br />
405
7.7 HYPERBOLIC FUNCTIONS<br />
SUGGESTED TIME AND EMPHASIS<br />
1 class Optional material<br />
POINTS TO STRESS<br />
1. The reason why hyperbolic functions are called hyperbolic (cosh 2 x − sinh 2 x = 1).<br />
2. The basic hyperbolic differentiation formulas.<br />
3. <strong>Inverse</strong> hyperbolic functions as logarithms.<br />
QUIZ QUESTIONS<br />
• Text Question: What is the relationship among cosh x,sinhx <strong>and</strong> the hyperbola x 2 − y 2 = 1?<br />
Answer: The point (cosh x, sinh x) is on that hyperbola.<br />
• Drill Question: Compute d<br />
dx (ex cosh 2x).<br />
d<br />
Answer:<br />
dx (ex cosh 2x) = e x (cosh 2x + 2sinh2x)<br />
MATERIALS FOR LECTURE<br />
• Present definitions of cosh x <strong>and</strong> sinh x, <strong>and</strong> the identity cosh 2 x − sinh 2 x = 1. Discuss the geometric<br />
connection between this identity <strong>and</strong> the hyperbola x 2 − y 2 = 1, similar to the link between the<br />
trigonometric functions cos x <strong>and</strong> sin x <strong>and</strong> the unit circle x 2 + y 2 = 1.<br />
y<br />
x@-y@=1<br />
(cosh u, sinh u)<br />
0 x<br />
x 2 − y 2 = cosh 2 u − sinh 2 u = 1<br />
• Prove the differentiation rules for sinh x, cosh x <strong>and</strong> discuss the general pattern.<br />
• Present the graphs of sinh x <strong>and</strong> cosh x, emphasizing that y = cosh x [or y = a cosh x] is not a parabola.<br />
( x )<br />
• Show how c + a cosh models the shape of a hanging chain.<br />
a<br />
WORKSHOP/DISCUSSION<br />
• Develop the graphs of sinh −1 x <strong>and</strong> cosh −1 x.<br />
• Compute the derivatives of y = sinh (e x ), y = cos (cosh x), y = cosh ( x 2) ,<strong>and</strong>y = e cosh x .<br />
• Show that (cosh x + sinh x) 2 = cosh 2x + sinh 2x. In fact, (cosh x + sinh x) n = cosh nx + sinh nx for<br />
any integer n.<br />
406
SECTION 7.7<br />
HYPERBOLIC FUNCTIONS<br />
GROUP WORK 1: A Tangent Line to a Hyperbolic Curve<br />
Answers:<br />
1.<br />
y<br />
10<br />
( )<br />
2. y = 2, y = y 0 + (x − x 0 ) sinh 12<br />
x 0<br />
3. e x − e −x = −1<br />
( e<br />
x ) 2 + e x − 1 = 0 (multiply by e x )<br />
5<br />
_5 0<br />
5 x<br />
e x = −1+√ 5<br />
2<br />
(the other root is extraneous)<br />
)<br />
x = ln<br />
(<br />
−1+<br />
√<br />
5<br />
2<br />
GROUP WORK 2: A Hyperbolic Paradox<br />
After the students have completed Problem 4, ask them how we can have 0 ·∞ = 1. Draw an analogy:<br />
( ) 1<br />
lim<br />
x→∞ x · x 1<br />
= 1, even though lim = 0 <strong>and</strong> lim<br />
x→∞ x x =∞.<br />
x→∞<br />
Answers:<br />
1.<br />
2. cosh x − sinh x = e −x , <strong>and</strong> so the limit is 0.<br />
10<br />
3. This can be done graphically (as in Problem 1) or algebraically (as<br />
in Problem 2).<br />
5<br />
0 y<br />
1 2 3 x<br />
The limit of the difference<br />
appears to be 0.<br />
HOMEWORK PROBLEMS<br />
4. The students may need to be reminded that cosh 2 x − sinh 2 x = 1.<br />
Core Exercises: 2, 8, 22, 23, 32, 51, 54, 58, 67<br />
Sample Assignment: 2, 5, 8, 9, 14, 22, 23, 28, 32, 35, 43, 48, 51, 54, 58, 63, 65, 67<br />
Exercise D A N G<br />
2 ×<br />
5 ×<br />
8 ×<br />
9 ×<br />
14 ×<br />
22 ×<br />
23 ×<br />
28 × × ×<br />
32 ×<br />
Exercise D A N G<br />
35 ×<br />
43 ×<br />
48 × ×<br />
51 ×<br />
54 ×<br />
58 ×<br />
63 ×<br />
65 ×<br />
67 ×<br />
407
GROUP WORK 1, SECTION 7.7<br />
A Tangent Line to a Hyperbolic Curve<br />
1. Graph the curve y = 2cosh x 2 . x<br />
y<br />
1<br />
0<br />
1<br />
2. Find the equation for the tangent line at the point (0, 2), then find the equation for the tangent line at the<br />
point (x 0 , y 0 ).<br />
3. At what point is the slope of the tangent line equal to −1?<br />
408
GROUP WORK 2, SECTION 7.7<br />
A Hyperbolic Paradox<br />
1. Graph y = sinh x <strong>and</strong> y = cosh x on one graph. What does lim (cosh x − sinh x) appear to be?<br />
x→∞<br />
y<br />
1<br />
0<br />
1<br />
x<br />
2. Find an equation for cosh x − sinh x <strong>and</strong> justify your guess in Problem 1 analytically.<br />
3. Show that lim (cosh x + sinh x) =∞.<br />
x→∞<br />
4. What is lim<br />
x→∞ (cosh x − sinh x)(cosh x + sinh x)? 409
7.8 INDETERMINATE FORMS AND L’HOSPITAL’S RULE<br />
SUGGESTED TIME AND EMPHASIS<br />
1 class Essential material<br />
POINTS TO STRESS<br />
1. The various types of indeterminate forms.<br />
2. The use of l’Hospital’s Rule to determine the behavior of indeterminate forms.<br />
3. The geometric plausibility of l’Hospital’s Rule, as discussed in the text.<br />
QUIZ QUESTIONS<br />
• Text Question: In Example 2, why wasn’t the Quotient Rule used when taking the derivatives?<br />
Answer:<br />
lim<br />
x→∞<br />
We are not taking the derivative of ex<br />
x 2 .<br />
f ′ (x)<br />
g ′ (x) .<br />
f (x)<br />
g (x) = lim x→∞<br />
2x + tan 3x<br />
• Drill Question: Find lim<br />
.<br />
x→0 sin x<br />
Answer: 5<br />
We are using l’Hospital’s Rule to evaluate<br />
MATERIALS FOR LECTURE<br />
• Demonstrate why ∞ is called ”indeterminate” by inviting the students to come up with different pairs of<br />
∞<br />
f (x)<br />
functions f <strong>and</strong> g such that lim f (x) =∞<strong>and</strong> lim g (x) =∞but lim<br />
x→∞ x→∞ x→∞ g (x) = 1, 0, 2 3<br />
,<strong>and</strong>∞. Urge<br />
them to come up with simple examples [such as f (x) = g (x) = x, f (x) = x, g (x) = x 2 , <strong>and</strong> so forth].<br />
If there is student interest, continue by asking for similar functions with lim f (x) = lim g (x) =∞.<br />
x→5 x→5<br />
Finally, remind the students of what happened when they tried to find lim ln (ln x) graphically <strong>and</strong><br />
x→∞<br />
numerically (putting in very large values of x) to show that it is possible to be led astray by estimations of<br />
infinite behavior.<br />
• Show that we can compute lim x ( a 1/x − 1 ) = ln a first by using l’Hospital’s Rule <strong>and</strong> then without<br />
x→∞<br />
a<br />
l’Hospital’s Rule [ 1/x − 1 a z − 1<br />
lim = lim = (a z ) ′ at z = 0].<br />
x→∞ 1/x z→0 z<br />
• Show how to manipulate other indeterminate forms into 0 0 or ∞ . [Good examples are lim sin x ln x,<br />
∞ x→0 +<br />
lim (1 + x→∞ x)1/x <strong>and</strong> lim (1 +<br />
x→0 x)1/x .]<br />
+<br />
• Give a graphical explanation for l’Hospital’s Rule as in Figure 1. If the class is interested in a more formal<br />
proof, state the Cauchy Mean Value Theorem <strong>and</strong> outline the formal proof given in the text. Point out<br />
that the conclusion of the Cauchy Mean Value Theorem is not always geometrically evident. For example,<br />
if f (x) = x 3 , g (x) = e x , a = 1<strong>and</strong>b = 2, the conclusion states that for some c with 1 < c < 2,<br />
3c 2 7<br />
e c =<br />
e (e − 1) . 410
SECTION 7.8<br />
INDETERMINATE FORMS AND L’HOSPITAL’S RULE<br />
WORKSHOP/DISCUSSION<br />
• Approximate a few of the limits discussed in the text using numerical methods. For example, look at<br />
f (x) =<br />
ln x for x = 0.9, 0.99, 0.999, 1.01, 1.001, etc.<br />
x − 1<br />
1 − cos x 6<br />
• Have the students estimate lim<br />
x→0 x 12<br />
l’Hospital’s Rule.<br />
first graphically, then numerically, <strong>and</strong> finally compute it using<br />
x − 3<br />
x − 3<br />
• Have the students compute lim<br />
x→3 x 3 − 10x 2 <strong>and</strong> lim<br />
+ 31x − 30 x→2 x 3 − 10x 2 . Have them try<br />
+ 31x − 30<br />
to figure out why one exists <strong>and</strong> the other doesn’t. (This can be done algebraically, or by looking at a<br />
graph. The students should be led to try both.)<br />
[ (<br />
• Do some challenging limit problems, such as lim sin π2<br />
+ x )] (<br />
1/x 1<br />
<strong>and</strong> lim<br />
x→0 + x→1 ln x − 1 )<br />
.<br />
x − 1<br />
GROUP WORK 1: Find the Error<br />
Most students, once they underst<strong>and</strong> l’Hospital’s Rule, believe the conclusion to be stronger than it is. This<br />
activity helps them to clarify what it does <strong>and</strong> does not say.<br />
Answers:<br />
x 2 + sin x<br />
1. lim<br />
x→∞ x 2<br />
(<br />
= lim 1 + sin x )<br />
x→∞ x 2 = 1<br />
f ′ (a)<br />
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.<br />
(a)<br />
Have the students reread the rule carefully, particularly the last clause.<br />
GROUP WORK 2: The Sector Ratio<br />
The students may not remember the formula for the area of a sector of the unit circle. Before h<strong>and</strong>ing out the<br />
exercise, give enough of a trigonometry review so that they will be able to work the problem. Students may<br />
need to be prompted to use sin (θ/2) = x to write θ as a function of x.<br />
Make sure to cover Part 4 with the class at the end. It is important that they underst<strong>and</strong> that taking limits isn’t<br />
some sort of symbolic game. These limits actually do make physical sense when looked at geometrically. If<br />
the students need to see this exercise another way, have them graph S (x) /T (x), <strong>and</strong> see what happens for<br />
large x.<br />
411
Answers:<br />
1. S (θ) = θ 2 ,soS (x) = arcsin x. 2. T (x) = x√ 1 − x 2 S (x)<br />
3. lim<br />
x→0 T (x) = 1<br />
4. When x gets very small, the isosceles triangle <strong>and</strong> the circular sector have virtually the same area, as the<br />
arc subtended by θ starts to look more <strong>and</strong> more like a straight line.<br />
HOMEWORK PROBLEMS<br />
Core Exercises: 1, 6, 31, 51, 66, 77, 85, 91<br />
Sample Assignment: 1, 6, 11, 24, 31, 46, 51, 59, 66, 67, 77, 81, 84, 85, 89, 91, 92, 97<br />
Exercise D A N G<br />
1 ×<br />
6 ×<br />
11 ×<br />
24 ×<br />
31 ×<br />
46 ×<br />
51 ×<br />
59 ×<br />
66 × ×<br />
Exercise D A N G<br />
67 × ×<br />
77 × ×<br />
81 × ×<br />
84 × ×<br />
85 ×<br />
89 ×<br />
91 × × ×<br />
92 ×<br />
97 ×<br />
412
GROUP WORK 1, SECTION 7.8<br />
Find the Error<br />
It is a beautiful Spring day. You are again jolly, for you have a backpack filled with schoolbooks <strong>and</strong> cornbread,<br />
<strong>and</strong> you are ready to have your lunch <strong>and</strong> start working on your calculus homework. You lay out your<br />
food <strong>and</strong> notebook on a picnic table, <strong>and</strong> suddenly a familiar scent catches your attention. “L’Hospital’s Rule,<br />
eh?” comments the wild-eyed stranger, who has sat down next to you. “Interesting stuff. . . if you don’t mind<br />
reading LIES!”<br />
“But l’Hospital’s Rule is useful! It is a true <strong>and</strong> wonderful technique of computing limits,” you say.<br />
“Useless!” he bellows. “You show me a steak <strong>and</strong> baked potato with a side of l’Hospital’s Rule <strong>and</strong> I’ll show<br />
you a steak <strong>and</strong> baked potato, by cracky!”<br />
“What are you talking about?”<br />
“Look at this, <strong>and</strong> see the lies exposed!”<br />
The stranger writes this in your notebook:<br />
x 2 + sin x 2x + cos x (<br />
lim<br />
x→∞ x 2 = lim<br />
both indeterminate forms of type ∞ )<br />
x→∞ 2x<br />
∞<br />
2 − sin x<br />
= lim<br />
which does not exist!<br />
x→∞ 2<br />
“Why, yes, that is a very good use of l’Hospital’s Rule, <strong>and</strong> it proves my point,” you say. “You started with a<br />
hard limit in one of the st<strong>and</strong>ard indeterminate forms, used l’Hospital twice, correctly I might add, <strong>and</strong> arrived<br />
x 2 + sin x<br />
at the answer, that lim<br />
x→∞ x 2 does not exist.”<br />
At that point the stranger laughs <strong>and</strong> laughs, <strong>and</strong> winds up saying between chortles, “Graph it, Sonny.”<br />
Fearlessly, you graph x2 + sin x<br />
x 2 , <strong>and</strong> come up with this:<br />
y<br />
1<br />
0<br />
1<br />
x<br />
“Looks like the limit is 1 to me, boy. And last thing I heard was that 1 exists. Enjoy your johnnycake!”<br />
You start to lose your appetite. Was l’Hospital lying all this time? Is the limit 1? Or doesn’t it exist? If 1<br />
doesn’t exist, then don’t things start looking grim for 2, 3 <strong>and</strong> 4? Does it still “take 1 to know 1”?<br />
Find the error.<br />
x 2 + sin x<br />
1. Compute lim<br />
x→∞ x 2 without using l’Hospital’s Rule.<br />
2. What is wrong with the stranger’s argument?<br />
413
Consider the following circular sector:<br />
GROUP WORK 2, SECTION 7.8<br />
TheSectorRatio<br />
x<br />
1<br />
1<br />
¬<br />
_ 2<br />
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<br />
<strong>and</strong> the chord.<br />
1. Find a formula for S (x).<br />
2. Find a formula for T (x).<br />
3. Compute lim<br />
x→0<br />
S (x)<br />
T (x) .<br />
4. Explain why this result makes sense geometrically.<br />
414
WRITING PROJECT<br />
The Origins of l’Hospital’s Rule<br />
This is a nice historical project on the mathematical origins of l’Hospital’s Rule. It would be good as a longterm<br />
project for a liberal arts course, or as a shorter project in which just the historical aspects are emphasized.<br />
Details about directions for investigation are given in the text.<br />
Plenty of information about the historical figures can be found on the World Wide Web, but urge the students<br />
to use caution — information from the Web may not be meticulously researched, <strong>and</strong> may therefore be<br />
unreliable.<br />
This project could also be done by a pair of motivated students, as opposed to a larger group.<br />
415
7 SAMPLE EXAM<br />
Problems marked with an asterisk (*) are particularly challenging <strong>and</strong> should be given careful consideration.<br />
1. Let f be a one-to-one function whose inverse function is given by the formula<br />
f −1 (x) = x 5 + 2x 3 + 3x + 1<br />
(a) Compute f −1 (1) <strong>and</strong> f (1).<br />
(b) Compute the value of x 0 such that f (x 0 ) = 1.<br />
(c) Compute the value of y 0 such that f −1 (y 0 ) = 1.<br />
(d) Below is a graph of f −1 . Draw an approximate graph of f .<br />
y<br />
fÐ!<br />
0<br />
x<br />
416
CHAPTER 7<br />
SAMPLE EXAM<br />
2. Find constants A, B, <strong>and</strong>k such that the equation f (x) = A2 kx + B satisfies the following three<br />
conditions:<br />
• f (x) is always decreasing.<br />
• f (x) has a horizontal asymptote at y = 1, <strong>and</strong><br />
• f (x) goes through the point (0, 4).<br />
3. Find constants A <strong>and</strong> k such that the equation f (x) = A2 kx fits the following data as closely as possible:<br />
x y<br />
1 1.8000<br />
2 2.7000<br />
4 6.0750<br />
5 9.1125<br />
7 20.503<br />
4. Compute derivatives of the following functions.<br />
(a) f (x) = e 2πx<br />
(b) g (x) = x 2πe<br />
(c) h (x) = (eπ) 2x<br />
(d) l (x) = π ( e 2x )<br />
5. Consider the function h (x) = (1 + sin πx) g(x) . Suppose g (1) = 2<strong>and</strong>g ′ (1) =−1. Find h ′ (1).<br />
417
6. (a) Find the point on the curve y = e −x2 where the slope of the tangent line is 2/e.<br />
(b) Find the x-intercept c of the line tangent to the curve at that point.<br />
y<br />
1<br />
_2<br />
_1<br />
0<br />
1<br />
c 2<br />
x<br />
7. Let f (x) = ln ( 1 + x 2) .<br />
(a) What is lim f (x)?<br />
x→∞<br />
(b) What is lim<br />
x→∞ f ′ (x)?<br />
(c) Using parts (a) <strong>and</strong> (b), explain the behavior of the function as x gets large.<br />
8. Draw a graph of f (x) = ln ln x<br />
(a) Over the range [2, 10].<br />
(b) Over the range [2, 100].<br />
(c) What is lim x→∞ ln ln x?<br />
418
CHAPTER 7<br />
SAMPLE EXAM<br />
9. (a) Below is a graph of f (x) = A ln x + B. WhatareA <strong>and</strong> B?<br />
y<br />
3<br />
2<br />
1<br />
(1, 1)<br />
_1<br />
0<br />
1 2 3 4 5 x<br />
(b) Below is a graph of f (x) = ln kx.Whatisk?<br />
y<br />
2<br />
1<br />
0<br />
1 2 3 x<br />
_1<br />
∫ 5x<br />
dt<br />
10. Let g (x) =<br />
1 t<br />
(a) Compute g ′ (x).<br />
(b) Using part (a), find an equation for g (x) in terms of ln x.<br />
∫ x<br />
1<br />
11. Using the equation ln x = dt, x > 0, justify the following statements:<br />
1 t<br />
(a) The expression e ln x = x is true only for positive values of x.<br />
(b) ln x is increasing everywhere.<br />
(c) ln ( x 1/2) = 1 2<br />
ln x.<br />
12. Let g (x) = ∫ (<br />
x 2<br />
0<br />
ln 12<br />
+ √ )<br />
t dt for all x. Compute the critical points of g.<br />
13. Consider the functions f (x) = e x − 1<strong>and</strong>g (x) = 2ln(x + 1).<br />
(a) Draw a graph of these functions for − 1 2 ≤ x ≤ 2.<br />
(b) Write an equation for the area between these two functions for − 1 2 ≤ x ≤ 2.<br />
419
7 SAMPLE EXAM SOLUTIONS<br />
1. f −1 (x) = x 5 + 2x 3 + 3x + 1<br />
(a) f −1 (1) = 7, f (1) = 0<br />
(b) The value x 0 such that f (x 0 ) = 1is f −1 (1) = 7.<br />
(c) The value y 0 such that f −1 (y 0 ) = 1is f (1) = 0.<br />
(d) The graph of f (x) is the graph of f −1 (x) reflected about the line y = x.<br />
y<br />
fÐ!<br />
f<br />
0<br />
x<br />
2. Let f (x) = 3 (2) −x + 1. Then f (x) is always decreasing, has a horizontal asymptote at y = 1, <strong>and</strong><br />
f (0) = 4.<br />
3. f (x) = 1.2 (2) 0.585x<br />
4. (a) f ′ (x) = 2πe 2πx<br />
(b) g ′ (x) = 2πe · x 2πe−1<br />
(c) h ′ (x) = 2lneπ · (eπ) 2x = 2 (1 + ln π)(eπ) 2x<br />
(d) l ′ (x) = (ln π) π ( e 2x ) · 2e 2x<br />
5. h ′ (x) = (1 + sin πx) g(x) [<br />
g ′ (x) ln (1 + sin πx) + g (x)<br />
h ′ (1) = 1 2 ·<br />
[−1 · 0 + 2π ]<br />
1 (−1) =−2π<br />
]<br />
π cos πx<br />
;<br />
1 + sin πx<br />
(<br />
6. (a) The slope is e −x2) ′<br />
=−2xe<br />
−x 2 . If −2xe −x2 = 2 e ,then−x2 =−1<strong>and</strong>−2x = 2. So x =−1<strong>and</strong><br />
(<br />
the point is −1, 1 )<br />
.<br />
e<br />
(<br />
(b) The tangent line at −1, 1 )<br />
is y = 1 e e + 2 e (x + 1). Sety = 0 <strong>and</strong> solve for x to get x =−3 2 .<br />
7. (a) lim f (x) =∞<br />
x→∞<br />
(b) f ′ (x) = 1<br />
2x · 2x =<br />
1 + x2 1 + x 2 . lim f ′ 2x<br />
(x) = lim<br />
x→∞ x→∞ 1 + x 2 = 0<br />
(c) The function increases at a slower <strong>and</strong> slower rate, but still goes to infinity.<br />
420
CHAPTER 7<br />
SAMPLE EXAM SOLUTIONS<br />
8. (a)<br />
y<br />
1<br />
(b)<br />
y<br />
2<br />
(c) lim ln ln x =∞<br />
x→∞<br />
1<br />
0<br />
10 x<br />
0<br />
100 x<br />
_1<br />
9. (a) Use f (1) = 1<strong>and</strong> f (3) = 0togetA =− 1 <strong>and</strong> B = 1.<br />
ln 3<br />
(b) Use f (0.2) = 0togetK = 5.<br />
( ) 1<br />
10. (a) g ′ (x) = 5 = 1 5x x<br />
(b) g (x) must differ from ln x by a constant, so g (x) = ln x + c. Usingg (1) =<br />
c = ln 5, so g (x) = ln x + ln 5.<br />
_1<br />
∫ 5<br />
1<br />
dt<br />
t<br />
= ln 5, we get<br />
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.<br />
This is because we cannot integrate across the discontinuity at t = 0.<br />
(b) 1 t<br />
is positive for t > 0, so ln x is increasing everywhere.<br />
(c) ln ( x 1/2) =<br />
∫ x 1/2<br />
∫ x 1/2<br />
∫<br />
1 x<br />
1 t dt = 1<br />
( )<br />
12. g ′ (x) = 2x ln 12<br />
+ |x|<br />
1<br />
1<br />
t dt. Letu = t2 .Thenu 1/2 = t so 1 2 u−1/2 du = dt <strong>and</strong><br />
1<br />
2u du = 1 2 ln x. Soln( x 1/2) = 1 2<br />
ln x.<br />
= 0wherex = 0, x = 1 2 ,orx =−1 2<br />
. g (x) has critical points at x = 0<strong>and</strong>at<br />
x =± 1 2 .<br />
13. (a) The points of intersections are at x = 0<strong>and</strong>x = c,wherec ≈ 0.752.<br />
y<br />
8<br />
6<br />
4<br />
2<br />
_1<br />
_ 1 _ 2<br />
0<br />
c<br />
1 2 3 x<br />
(b) A = ∫ 2<br />
−1/2<br />
= ∫ 0<br />
−1/2<br />
∣ e x − 1 − 2ln(x + 1) ∣ dx<br />
[ e x − 1 − 2ln(x + 1) ] dx + ∫ c [ (<br />
0 2ln(x + 1) − e x − 1 )] dx<br />
421<br />
+ ∫ 2 [<br />
c e x − 1 − 2ln(x + 1) ] dx
where c ≈ 0.752.<br />
422