Section 3.1 Exponential and Logistic Functions 133 ■ Section 3.1 Exponential and Logistic Functions Exploration 1 1. The point (0, 1) is common to all four graphs, and all four functions can be described as follows: Domain: Range: Continuous Always increasing Not symmetric No local extrema Bounded below by , which is also the only asymptote 2. The point (0, 1) is common to all four graphs, and all four functions can be described as follows: Domain: Range: Continuous Always decreasing Not symmetric No local extrema Bounded below by , which is also the only asymptote [–2, 2] by [–1, 6] [–2, 2] by [–1, 6] [–2, 2] by [–1, 6] [–2, 2] by [–1, 6] lim xSq g 1 x 2 = 0, lim xS–q g 1 x 2 = q y = 0 1 0, q 2 1 -q, q 2 [–2, 2] by [–1, 6] [–2, 2] by [–1, 6] [–2, 2] by [–1, 6] [–2, 2] by [–1, 6] lim xSq f 1 x 2 = q, lim xS–q f 1 x 2 = 0 y = 0 1 0, q 2 1 -q, q 2 Chapter 3 Exponential, Logistic, and Logarithmic Functions y 1 =2 x y 2 =3 x y 3 =4 x y 4 =5 x y 1 = a 1 2 b x y 2 = a 1 3 b x y 3 = a 1 4 b x y 4 = a 1 5 b x
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Section 3.1 Exponential and Logistic Functions 133
■ Section 3.1 Exponential and LogisticFunctions
Exploration 1
1. The point (0, 1) is common to all four graphs, and all fourfunctions can be described as follows:
Domain:Range:ContinuousAlways increasingNot symmetricNo local extremaBounded below by , which is also the only
asymptote
2. The point (0, 1) is common to all four graphs, and all fourfunctions can be described as follows:
Domain:Range:ContinuousAlways decreasingNot symmetricNo local extremaBounded below by , which is also the only
asymptote
[–2, 2] by [–1, 6]
[–2, 2] by [–1, 6]
[–2, 2] by [–1, 6]
[–2, 2] by [–1, 6]
limxSq
g 1x 2 = 0, limxS–q
g 1x 2 = q
y = 0
10, q 21- q, q 2
[–2, 2] by [–1, 6]
[–2, 2] by [–1, 6]
[–2, 2] by [–1, 6]
[–2, 2] by [–1, 6]
limxSq
f 1x 2 = q, limxS–q
f 1x 2 = 0
y = 0
10, q 21- q, q 2
Chapter 3Exponential, Logistic, and Logarithmic Functions
y1=2x
y2=3x
y3=4x
y4=5x
y1 = a 12b x
y2 = a 13b x
y3 = a 14b x
y4 = a 15b x
134 Chapter 3 Exponential, Logistic, and Logarithmic Functions
Exploration 2
1.
2.
most closely matches the graph of f(x).
3.
Quick Review 3.1
1.
2.
3. 272/3=(33)2/3=32=9
4. 45/2=(22)5/2=25=32
5.
6.
7.
8.
9. –1.4 since (–1.4)5=–5.37824
10. 3.1, since (3.1)4=92.3521
Section 3.1 Exercises1. Not an exponential function because the base is variable
and the exponent is constant. It is a power function.
2. Exponential function, with an initial value of 1 and base of 3.
3. Exponential function, with an initial value of 1 and baseof 5.
4. Not an exponential function because the exponent is con-stant. It is a constant function.
5. Not an exponential function because the base is variable.
6. Not an exponential function because the base is variable.It is a power function.
7.
8.
9.
10.
11.
12.
13.
14.
15. Translate by 3 units to the right. Alternatively,
so it can be
obtained from f(x) using a vertical shrink by a factor of .
16. Translate f(x)=3x by 4 units to the left. Alternatively,g(x)=3x±4=34 3x=81 3x=81 f(x), so it can beobtained by vertically stretching f(x) by a factor of 81.
[–7, 3] by [–2, 8]
###
[–3, 7] by [–2, 8]
18
g 1x 2 = 2x-3 = 2- 3 # 2x =18
# 2x =18
# f 1x 2 ,f 1x 2 = 2x
g 1x 2 = 2 # a 1eb x
= 2e-x
f 1x 2 = 3 # 112 2x = 3 # 2x>2g 1x 2 = 12 # a 1
3b x
f 1x 2 =32
# a 12b x
f a -32b = 8 # 4-3>2 =
8
122 2 3>2 =823 =
88
= 1
f a 13b = -2 # 31>3 = -2 313
f 1-2 2 = 6 # 3-2 =69
=23
f 10 2 = 3 # 50 = 3 # 1 = 3
b15
1
a6
138
1
212
3B1258
=52
since 53 = 125 and 23 = 8
32-216 = -6 since 1-6 2 3 = -216
k L 0.693
k = 0.7[–4, 4] by [–2, 8]
[–4, 4] by [–2, 8]
[–4, 4] by [–2, 8]
[–4, 4] by [–2, 8]
[–4, 4] by [–2, 8]
[–4, 4] by [–2, 8]
g 1x 2 = e0.4xf 1x 2 = 2x
g 1x 2 = e0.5xf 1x 2 = 2x
g 1x 2 = e0.6xf 1x 2 = 2x
g 1x 2 = e0.7xf 1x 2 = 2x
g 1x 2 = e0.8xf 1x 2 = 2x
f 1x 2 = 2x
Section 3.1 Exponential and Logistic Functions 135
17. Reflect over the y-axis.
18. Reflect over the y-axis and then shift by 5 unitsto the right.
19. Vertically stretch by a factor of 3 and thenshift 4 units up.
20. Vertically stretch by a factor of 2 and thenhorizontally shrink by a factor of 3.
21. Reflect across the y-axis and horizontallyshrink by a factor of 2.
22. Reflect across the x-axis and y-axis. Then,horizontally shrink by a factor of 3.
23. Reflect across the y-axis, horizontally shrink bya factor of 3, translate 1 unit to the right, and verticallystretch by a factor of 2.
24. Horizontally shrink by a factor of 2, verticallystretch by a factor of 3 and shift down one unit.
25. Graph (a) is the only graph shaped and positioned likethe graph of
26. Graph (d) is the reflection of across the y-axis.
27. Graph (c) is the reflection of across the x-axis.
28. Graph (e) is the reflection of across the x-axis.
29. Graph (b) is the graph of translated down 2 units.
30. Graph (f) is the graph of translated down 2 units.
31. Exponential decay;
32. Exponential decay;
33. Exponential decay:
34. Exponential growth:
35.
36.
[–0.25, 0.25] by [0.5, 1.5]
x 7 0
[–2, 2] by [–0.2, 3]
x 6 0
limxSq
f 1x 2 = q; limxS–q
f 1x 2 = 0
limxSq
f 1x 2 = 0; limxS–q
f 1x 2 = q
limxSq
f 1x 2 = 0; limxS–q
f 1x 2 = q
limxSq
f 1x 2 = 0; limxS–q
f 1x 2 = qy = 1.5x
y = 3-x
y = 0.5x
y = 2x
y = 2x
y = bx, b 7 1.
[–3, 3] by [–2, 8]
f 1x 2 = ex
[–2, 3] by [–1, 4]
f 1x 2 = ex
[–3, 3] by [–5, 5]
f 1x 2 = ex
[–2, 2] by [–1, 5]
f 1x 2 = ex
[–2, 3] by [–1, 4]
f 1x 2 = 0.6x
[–5, 5] by [–2, 18]
f 1x 2 = 0.5x
[–3, 7] by [–5, 45]
f 1x 2 = 2x
[–2, 2] by [–1, 9]
f 1x 2 = 4x
136 Chapter 3 Exponential, Logistic, and Logarithmic Functions
Section 3.1 Exponential and Logistic Functions 137
48.
Domain:Range:ContinuousAlways decreasingNot symmetricBounded below by , which is also the only asymptoteNo local extrema
49.
Domain:Range: (0, 5)ContinuousAlways increasingSymmetric about (0.69, 2.5)Bounded below by and above by ; both are
asymptotesNo local extrema
50.
Domain:Range: (0, 6)ContinuousAlways increasingSymmetric about (0.69, 3)Bounded below by and above by ; both
are asymptotesNo local extrema
For #51–52, refer to Example 7 on page 285 in the text.
51. Let P(t) be Austin’s population t years after 1990.Then withexponential growth, where .From Table 3.7, . So,
Solving graphically, we find that the curveintersects the line y=800,000 at
. Austin’s population will pass 800,000 in 2006.
52. Let P(t) be the Columbus’s population t years after 1990.Then with exponential growth, whereP0=632,910. From Table 3.7,P(10)=632,910 b10=711,470. So,
.
Solving graphically, we find that the curveintersects the line y=800,000 at
. Columbus’s population will pass 800,000 in2010.
53. Using the results from Exercises 51 and 52, we representAustin’s population as y=465,622(1.0350)t andColumbus’s population as y=632,910(1.0118)t. Solvinggraphically, we find that the curves intersect at .The two populations will be equal, at 741,862, in 2003.
54. From the results in Exercise 53, the populations are equalat 741,862. Austin has the faster growth after that, becauseb is bigger (1.0350>1.0118). So Austin will reach 1 mil-lion first. Solving graphically, we find that the curvey=465,622(1.0350)t intersects the line y=1,000,000 at
. Austin’s population will reach 1 million in 2012.
55. Solving graphically, we find that the curve
intersects the line y=10 when
. Ohio’s population stood at 10 million in 1969.
56. (a)
or 1,794,558 people
(b) or
19,161,673 people
(c)
57. (a) When
(b) When .
58. (a) When
(b) When . After about 5700.22years, 10 grams remain.
59. False. If a>0 and 0<b<1, or if a<0 and b>1, thenis decreasing.
60. True. For the horizontal asymptotes
are y=0 and y=c, where c is the limit of growth.
61. Only 8x has the form with a nonzero and b positivebut not equal to 1. The answer is E.
62. For b>0, f(0)=b0=1. The answer is C.
63. The growth factor of is the base b. Theanswer is A.
64. With x>0, ax>bx requires a>b (regardless ofwhether x<1 or x>1). The answer is B.
f 1x 2 = a # bx
a # bx
f 1x 2 =c
1 + a # bx
f 1x 2 = a # bx
C L 5.647t = 10,400,
t = 0, C = 20 grams.
t = 6, B L 6394
t = 0, B = 100.
limxSq
P 1 t 2 = 19.875 or 19,875,000 people.
P 1210 2 =19.875
1 + 57.993e-0.03500512102 L 19.161673
P 150 2 =19.875
1 + 57.993e-0.0350051502 L 1.794558
t L 69.67
y = 12.79
11 + 2.402e-0.0309x 2
t L 22.22
t L 13.54
t L 20.02y = 632,910 11.0118 2 tb = 10B
711,470632,910
L 1.0118
P 1 t 2 = P0 bt
t L 15.75y = 465,622 11.0350 2 tb =
10
B656,562465,622
L 1.0350.
b10 = 656,562P 110 2 = 465,622P0 = 465,622P 1 t 2 = P0 b
t
limxSq
f 1x 2 = 6, limxS –q
f 1x 2 = 0
y = 6y = 0
1- q, q 2[–3, 7] by [–2, 8]
limxSq
f 1x 2 = 5, limxS–q
f 1x 2 = 0
y = 5y = 0
1- q, q 2[–3, 4] by [–1, 7]
limxSq
f 1x 2 = 0, limxS–q
f 1x 2 = q
y = 0
10, q 21- q, q 2[–2, 2] by [–1, 9]
138 Chapter 3 Exponential, Logistic, and Logarithmic Functions
65. (a)
Domain:
Range:
Intercept: (0, 0)
Decreasing on : Increasing on
Bounded below by
Local minimum at
Asymptote .
(b)
Domain:Range:No interceptsIncreasing on ;Decreasing on Not boundedLocal maxima at Asymptotes: x=0, .
66. (a)
(b)
(c)
(d)
67. (a) —f(x) decreases less rapidly as x decreases.
(b) —as x increases, g(x) decreases ever more rapidly.
68. : to the graph of apply a vertical stretch by ,since .
69. .
70. .
71. .
72.
73. Since 0<b<1, and
. Thus, and
.
■ Section 3.2 Exponential and LogisticModeling
Quick Review 3.2
1. 0.15
2. 4%
3. (1.07)(23)
4. (0.96)(52)
5. .
6. .
7. b= ≠1.01
8. b= ≠1.41
9. b= ≠0.61
10. b= ≠0.89
Section 3.2 ExercisesFor #1–20, use the model .
1. r=0.09, so P(t) is an exponential growth function of 9%.
2. r=0.018, so P(t) is an exponential growth function of 1.8%.
3. r=–0.032, so f(x) is an exponential decay functionof 3.2%.
4. r=–0.0032, so f(x) is an exponential decay functionof 0.32%.
5. r=1, so g(t) is an exponential growth function of 100%.
6. r=–0.95, so g(t) is an exponential decay function of 95%.
11 + a # bx 2 = qa 7 0 and 0 6 b 6 1, or a 6 0 and b 7 1.
a 7 0 and b 7 1, or a 6 0 and 0 6 b 6 1
a 6 0, c = 1
a Z 0, c = 2
f 1ax + b 2 = 2ax + b = 2ax2b = 12b 2 12a 2x 2b12a 2xc = 2a
y3
y1
9x = 3x + 1, 132 2x = 3x + 1, 2x = x + 1, x = 1
3x = 4x + 4, x = -4
8x>2 = 4x + 1, 122 2x>2 = 122 2x + 1 # 3x
2= 2x + 2,
3x = 33, so x = 3
2x = 122 2 2 = 24, so x = 4
limxSq
g 1x 2 = 0, limxS-q
g 1x 2 = - qy = 0
1-1, -e 23-1, 0 2 ´ 10, q 21- q, -1 4
1- q, -e 4 ´ 10, q 21- q, 0 2 ´ 10, q 2[–3, 3] by [–7, 5]
limxSq
f 1x 2 = q, limxS-q
f 1x 2 = 0y = 0
a -1, -1eb
y = -1e
3-1, q 21- q, -1 4
B-1e
, q b1- q, q 2
[–5, 5] by [–2, 5]
Section 3.2 Exponential and Logistic Modeling 139
20.
g(x)=
For #21–22, use
21. Since
bfi= b=
22. Since
b›= b=
For #23–28, use the model .
23. c=40, a=3, so f(1)=
60b=20, b= , thus f(x)= .
24. c=60, a=4, so f(1)=
96b=36, b= , thus f(x)= .
25. c=128, a=7, so f(5)= ,
128=32+224bfi, 224bfi=96, bfi= ,
b= , thus .
26. c=30, a=5, so f(3)= ,
75b‹=15, b‹= b= ,
thus .
27. c=20, a=3, so f(2)=
30b¤=10, b¤= b= ,
thus f(x)= .
28. c=60, a=3, so f(8)=
90b°=30, b°= b= ,
thus f(x)= .
29. ; P(t)=1,000,000 whent≠20.73 years, or the year 2020.
30. ; P(t)=1,000,000 whent≠12.12 years, or the year 2012.
31. The model is .
(a) In 1915: about P(25)≠12,315. In 1940: aboutP(50)≠24,265.
(b) P(t)=50,000 when t≠76.65 years after 1890 — in1966.
32. The model is .
(a) In 1930: about P(20)≠6554. In 1945: aboutP(35)≠9151.
(b) P(t)=20,000 when t≠70.14 years after 1910 —about 1980.
33. (a) , where t is time in days.
(b) After 38.11 days.
34. (a) , where t is time in days.
(b) After 117.48 days.
35. One possible answer: Exponential and linear functionsare similar in that they are always increasing or alwaysdecreasing. However, the two functions vary in howquickly they increase or decrease. While a linear functionwill increase or decrease at a steady rate over a giveninterval, the rate at which exponential functions increaseor decrease over a given interval will vary.
36. One possible answer: Exponential functions and logisticfunctions are similar in the sense that they are alwaysincreasing or always decreasing. They differ, however, inthe sense that logistic functions have both an upper andlower limit to their growth (or decay), while exponentialfunctions generally have only a lower limit. (Exponentialfunctions just keep growing.)
37. One possible answer: From the graph we see that thedoubling time for this model is 4 years. This is the timerequired to double from 50,000 to 100,000, from 100,000 to200,000, or from any population size to twice that size.Regardless of the population size, it takes 4 years for it todouble.
38. One possible answer: The number of atoms of a radioac-tive substance that change to a nonradioactive state in agiven time is a fixed percentage of the number of radioac-tive atoms initially present. So the time it takes for half ofthe atoms to change state (the half-life) does not dependon the initial amount.
39. When t=1, B≠200—the population doubles every hour.
40. The half-life is about 5700 years.
For #41–42, use the formula P(h)=14.7 0.5h/3.6, where h ismiles above sea level.
41. P(10)=14.7 0.510/3.6=2.14 lb/in2
42. when h≠9.20miles above sea level.
[–1, 19] by [–1, 9]
P 1h 2 = 14.7 # 0.5h>3.6 intersects y = 2.5
#
#
y = 3.5 a 12b t>65
y = 6.6 a 12b t>14
P 1 t 2 = 4200 11.0225 2 t
P 1 t 2 = 6250 11.0275 2 t
P 1 t 2 = 478,000 11.0628 2 tP 1 t 2 = 736,000 11.0149 2 t
601 + 3 # 0.87x
8B13
L 0.8713
,
601 + 3b8 = 30, 60 = 30 + 90b8,
201 + 3 # 0.58x
B13
L 0.5813
,
201 + 3b2 = 10, 20 = 10 + 30b2,
f 1x 2 L30
1 + 5 # 0.585x
3B15
L 0.5851575
=15
,
301 + 5b3 = 15, 30 = 15 + 75b3
f 1x 2 L128
1 + 7 # 0.844x5B
96224
L 0.844
96224
128
1 + 7b5 = 32
60
1 + 4 a 38b x
38
601 + 4b
= 24, 60 = 24 + 96b,
40
1 + 3 # a 13b x
13
401 + 3b
= 20, 20 + 60b = 40,
f 1x 2 =c
1 + a # bx
4B1.49
3L 0.84. f 1x 2 L 3 # 0.84x1.49
3 ,
f 14 2 = 3 # b4 = 1.49f0 = 3, so f 1x 2 = 3 # bx.
5B8.05
4L 1.15. f 1x 2 L 4 # 1.15x8.05
4,
f 15 2 = 4 # b5 = 8.05,f0 = 4, so f 1x 2 = 4 # bx.
f 1x 2 = f0# bx
-5.8 # 10.8 2x 1Decay Model 2g0 = -5.8,
-4.64-5.8
= 0.8 = r + 1, so
140 Chapter 3 Exponential, Logistic, and Logarithmic Functions
43. The exponential regression model iswhere is measured in
thousands of people and t is years since 1900. The predictedpopulation for Los Angeles for 2003 is or3,981,000 people. This is an overestimate of 161,000 people,
an error of .
44. The exponential regression model using 1950–2000 data iswhere is measured in
thousands of people and t is years since 1900. The predictedpopulation for Phoenix for 2003 is or1,874,000 people. This is an overestimate of 486,000 people,
an error of .
The exponential regression model using 1960–2000 data iswhere is measured in
thousands of people and t is years since 1900. The predict-ed population for Phoenix for 2003 is or1,432,000 people. This is an overestimate of 44,000 people,
an error of .
The equations in #45–46 can be solved either algebraically orgraphically; the latter approach is generally faster.
45. (a) P(0)=16 students.(b) P(t)=200 when t≠13.97 — about 14 days.(c) P(t)=300 when t≠16.90 — about 17 days.
46. (a) P(0)=11.(b) P(t)=600 when t≠24.51 — after 24 or 25 years.(c) As , —the population never rises
above this level.47. The logistic regression model is
, where x is the num-
ber of years since 1900 and is measured in millionsof people. In the year 2010, so the model predictsa population of
people.
48. which is the same model as
the solution in Example 8 of Section 3.1. Note that trepresents the number of years since 1900.
49. where x is the number of
years after 1800 and P is measured in millions. Our modelis the same as the model in Exercise 56 of Section 3.1.
50. , where x is the number of
years since 1800 and P is measured in millions.
As or nearly 16 million, which issignificantly less than New York’s population limit of20 million. The population of Arizona, according to ourmodels, will not surpass the population of New York. Ourgraph confirms this.
51. False. This is true for logistic growth, not for exponentialgrowth.
52. False. When r<0, the base of the function, 1+r, ismerely less than 1.
53. The base is 1.049=1+0.049, so the constant percentagegrowth rate is 0.049=4.9%. The answer is C.
54. The base is 0.834=1-0.166, so the constant percentagedecay rate is 0.166=16.6%. The answer is B.
55. The growth can be modeled as P(t)=1 2t/4. SolveP(t)=1000 to find t≠39.86. The answer is D.
56. Check S(0), S(2), S(4), S(6), and S(8). The answer is E.
57. (a) where x is the number of
years since 1900 and P is measured in millions.P(100)≠277.9, or 277,900,000 people.
(b) The logistic model underestimates the 2000 popula-tion by about 3.5 million, an error of around 1.2%.
(c) The logistic model predicted a value closer to theactual value than the exponential model, perhapsindicating a better fit.
58. (a) Using the exponential growth model and the datafrom 1900–2050, Mexico’s population can be repre-sented by M(x)≠ where x is the num-ber of years since 1900 and M is measured in millions.Using 1900–2000 data for the U.S., and the exponen-tial growth model, the population of the United Statescan be represented by P(x)≠ where xis the number of years since 1900 and P is measuredin millions. Since Mexico’s rate of growth outpaces theUnited States’ rate of growth, the model predicts that
Domain: (0, q)Range: (–q, q)ContinuousIncreasing on its domainNo symmetryNot boundedNo local extremaAsymptote at x=0
58.
Domain: (–q, 2)Range: (–q, q)ContinuousDecreasing on its domainNo symmetryNot boundedNo local extremaAsymptote at x=2.
59. (a) ı= =10 dB
(b) ı= =70 dB
(c) ı= =150 dB
60. I=12 10–0.0705≠10.2019 lumens.
61. The logarithmic regression model isln x, where x is the
year and y is the population. Graph the function and useTRACE to find that when The population of San Antonio will reach 1,500,000 peo-ple in the year 2023.
62. The logarithmic regression model isln x, where x is the year
and y is the population. Graph the function and useTRACE to find that when Thepopulation of Milwaukee will reach 500,000 people in theyear 2024.
63. True, by the definition of a logarithmic function.
64. True, by the definition of common logarithm.
65. log 2≠0.30103. The answer is C.
66. log 5≠0.699 but 2.5 log 2≠0.753. The answer is A.
67. The graph of f(x)=ln x lies entirely to the right of theorigin. The answer is B.
(d) log r=(–0.30) log (450)+2.36≠1.58, r≠37.69,very close
(e) One possible answer: Consider the power functiony=a xb then:
log y=log (a xb)=log a+log xb
=log a+b log x=b(log x)+log a
which is clearly a linear function of the formf(t)=mt+c where m=b, c=log a, f(t)=log yand t=log x. As a result, there is a linear relation-ship between log y and log x.