To prove it we take any real number note that the quantity under the square root is positive or zero.. Since is positive or zero, the To prove this fact algebraically we take any real n
Trang 1Chapter 1 Introduction to Models and Functions
1.3 Variables, Parameters, and Functions
1.3.1 The variables are the altitude and the wombat density, which we can call and
respectively The parameter is the rainfall, which we can call R
1.3.2 The variables are the altitude and the bandicoot density, which we can call and b,
respectively The parameter is the wombat density which we can call W
1.3.3 The graph is the horizontal line crossing the -axis at -4.2 is neither increasing
1.3.4 The graph is the line of slope 3 whose -intercept is -6 is increasing for all real
1.3.5 The graph is a hyperbola; we draw it by plotting points and joining them with a smooth
curve (in the next section we will learn how to transform the graph of to obtain
1.3.6 The graph is a cube root parabola; we draw it by plotting points and joining them with a
smooth curve The function is increasing for all real numbers
Trang 21.3.8 The graph is V-shaped; it is the graph of the absolute value function moved 5 units to the
Trang 31.3.10
0 5 10 15
Trang 41.3.16
the domain is = (−∞,∞)
equal to 4 The domain is x ∈ x ≠ 4{ }; or, we can say that it consists of two intervals,
never zero The domain of consists of all real numbers In symbols, the domain is = (−∞,∞)
Using interval notation, we write the domain as
1.3.24 There are no restrictions coming from the linear function The square root is
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1.3.25 There are no restrictions on the domain coming from the square root, since it is computed
of the numbers that are greater than or equal to 4 The only restriction is the denominator
1.3.26 There are no restrictions on the domain coming from the square root, since it is computed
of the numbers that are greater than or equal to 0 The remaining pieces are linear functions, which are always defined The domain is the set of all real numbers
1.3.27 Both pieces are quadratic functions, so the domain of is the set of all real numbers
The graph of is the parabola for positive values of , and its reflection across the -axis for negative values of and for
for short
1.3.29 The graph is the line of slope going through the origin Thus, the range is the set of
all real numbers To confirm this fact algebraically, we take any real number and find
Trang 61.3.30 Squaring a number we obtain a positive number, or zero Thus, we believe that the range
and (note that the quantity under the square root is positive or zero) 1.3.31 The graph suggests that the range is To prove it we take any real number
(note that the quantity under the square root is positive or zero) Thus, any number greater than or equal to 3 can be obtained by applying the function
1.3.32 The graph suggests that the range is Alternatively, we ask ourselves: what
numbers can we get as a result of subtracting from 3? Since is positive or zero, the
To prove this fact algebraically we take any real number and find a number such
the quantity under the square root is positive or zero) Thus, any number less than or
1.3.34 The range of the square root function consists of zero and all positive numbers, and that’s
our guess To prove it, we take any real number and find a number such that
Trang 7number larger than or equal to zero can be obtained by applying the function
1.3.35 Because and the cube root of a positive number is positive (and the cube root of
zero is zero), the range of consists of zero and positive numbers To confirm: we
1.3.36
1.3.37
0 20 40 60 80
Cooper’s Goshawk Sharp-shinned
Bird
.38 The cell volume is generally increasing but decreases during part of its cycle The cell
might get smaller when it gets ready to divide or during the night
.39 The fish population is steadily declining between 1950 and 1990
.40 Initially, the height is about 1 metre, then increases until about age 30 when the trees
reach the approximate height of 7 metres; after that, the height decreases
.41 The stock increases sharply, then crashes (falls to a bit below its original value), then
increases sharply again, crashes to an even lower value than before; around day 12 the stock increases again (bit less sharply than previously), crashes to about its initial value, and levels out
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initial
tiny intermediate
start tiny levels out
start plateau decline crash
day night day night
dawn noon evening midnight dawn
Time
plant even if there are no flowers
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0 10 20 30 40 50
Number of flowers
.47 If r = 0,c = 0; if r = 4,c = 0; if r = 6,c = 2; if It looks like these cells can
tolerate up to 5 rad before they begin to become cancerous
0 2 4 6
Trang 101.3.50 The function increases rapidly until , where its value raises above 15 Then it
decreases until , reaching the value slightly below 5 After remaining around 5 for some time, it starts decreasing sharply at It reaches its lowest value of about
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
−15
−10
−5 0 5 10 15
x f(x)
larger and larger, or smaller and smaller, the function approaches 5 Near the origin it drops sharply from values between 4 and 5 to below 0, and then rises sharply back to the values above 4 The graph is decreasing for negative and increasing for positive It seems to be symmetric with respect to the -axis
−1 0 1 2 3 4 5
x g(x)
1.3.52 The graph of the function consists of four curves (although they might look like it, they
are not straight lines) The function starts at 3 when , then decreases until ,
reaching its initial value of 3 again when The graph seems to be symmetric with respect to the vertical line
1.8 2 2.2 2.4 2.6 2.8 3 3.2
x h(x)
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1.3.53 The function is decreasing from until just before , where it reaches its lowest
value of 0 Then it shows a small bump, reaching about 0.6 before it falls back to 0 at a small positive value of Afterward, it increases first sharply, and then slows down, reaching the value of 5 as grows larger As well, the function reaches 5 as grows larger and larger negative The graph seems to be symmetric with respect to the -axis
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
x
|g(x)|
1.3.54 The function decreases (initially) quickly until , then increases until
, decreases rapidly until , reaching its lowest value of approximately After it starts a rapid increase
x k(x)
1.3.55 The function decreases very rapidly until , when it hits 0 The graph remains
constant at 0 until , except that it experiences a small bump (reaching the value of
increasing
0 0.5 1 1.5 2 2.5 3 3.5 4
x k(x)+|k(x)|
Trang 121.4 Working with Functions
1.4.1
–12–11
–1–2
–7–8
1–1
–2–5
30
3–2
51
81
72
−15
−10
−5 0 5 10 15
x
f(x) g(x)
72
Trang 134–1
5–2
20
2–1
21
10
42
21
83
52
−2 0 2 4 6 8 10 12
x
F(x) G(x)
sum
1.4.4
83
5–2
42
2–1
21
10
20
21
4–1
52
−2 0 2 4 6 8 10 12
–1–2
–8–8
1–1
–15–5
30
–10–2
51
71
72
Trang 14f(x) g(x)
product
1.4.6
6–6
–1–2
–9–9
1–1
–36–12
30
–75–15
51
–126–18
72
x
f(x) h(x)
product
1.4.7
–5–1
5–2
00
2–1
11
10
42
21
153
52
−5 0 5 10 15
x
F(x) G(x)
product
Trang 15153
5–2
42
2–1
11
10
00
21
–5–1
52
−1 0 1 2 3 4 5 6 7 8
is again a polynomial, so its domain is = (−∞,∞) The quotient
is defined for all such that 1.4.10 The domain of is = (−∞,∞) and the domain of is The product
Trang 16of the quotient
(2x) + 3⋅4(2x)2− (2x)3 = 64 − 96x + 48x2− 8x3
and
( )g f (x) = g f (x)( )= g x( )3 = 4 − 2x31.4.14 We compute
(f g)(x) = f g(x)( )= f 4( )= 12 − 42 = −4 and
(g f)(x) = g f (x)( )= g 12 − x( 2)= 4 1.4.15 We compute
( )f g (x) = f g(x)( )= f x − 3( )= 1
x− 3and
x− 3
Trang 181.4.21 We find
( f g)(x) = f (3x − 5) = 2⋅(3x − 5) + 3 = 6x − 7
(g f )(x) = g(2x + 3) = 3⋅(2x + 3) − 5 = 6x + 4
These don’t match, so the functions do not commute
1.4.22 We find
( f h)(x) = f (−3x −12) = 2⋅(−3x −12) + 3 = −6x − 21
1.4.27 The function fails the horizontal line test because, for example,
Therefore, it has no inverse
1.4.28 The function now passes the horizontal line test because the values get larger and
Trang 19−5 −4 −3 −2 −1 0 1 2 3 4 5 6
y
The inverse the point (5, 1)
1.4.31 This function doesn’t have an inverse because it fails the horizontal line test From the
graph, we couldn’t tell whether F –1(2) is 1 or –1
0 5 10 15 20 25 30
(1, 2)
1.4.32
0 5 10 15 20 25 30
y
The function
0 0.5 1 1.5 2 2.5
x
The inverse
Trang 20the graph we see that both the domain and the range consist of all real numbers (To prove the claim for the range algebraically, we proceed as in Exercises 28-35 in Section 1.3) is an increasing function, and therefore it satisfies the horizontal line test; in
prove the claim for the range algebraically, we proceed as in Exercises 28-35 in Section 1.3) satisfies the horizontal line test because it is an increasing function Thus, it
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range is The function does not pass the horizontal line test – thus; it does not have an inverse function
prove the claim for the range algebraically, we proceed as in Exercises 28 to 35 in Section 1.3) The function does not pass the horizontal line test – thus, it does not have an inverse function
1.4.37 To sketch the graph of , start with the graph of , shift it to the right for 3
units (thus getting the graph of ), and then expand vertically by a factor of 13
function passes the horizontal line test, so it has an inverse To find it, we solve for :
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0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6
−4 −3 −2 −1 0 1 2 3 4 5
x
g(x) 4g(x)
1.4.47.The vertical axis is shifted by a value less than 0, moving it down
−0.4
−0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4
− 5 − 4 − 3 − 2 − 1 0 1 2 3 4 5
x
g(x) g(x/3)
1.4.49.The horizontal axis is shifted by a value greater than 0
0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4
− 5 − 4 − 3 − 2 − 1 0 1 2 3 4 5
x g(x) g(x+1)
Trang 261.4.50 The graph of is the reflection of the graph of with respect to the
we shift the graph of to the right units
shift the graph of to the right units
shift the graph of to the right units
Trang 27Copyright © 2015 Nelson Education Ltd 27
to the right 2 units (to obtain the graph of ) then reflect that graph across the
1.4.56 Start with the graph of and reflect it across the -axis (thus getting the graph of
) Move the resulting graph 1 unit up
Trang 28Thus, to graph , we shift the graph of ½ units to the left and then
1.4.58 Completing the square, we get
the right and then units down
1.4.59 Start with the graph of and move it 2 units to the right (thus getting the graph of
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0 2 4 6 8 10
Age Length as a function of age
1.4.61
0 0.2 0.4 0.6 0.8 1
Age Tail length as a function of age
1.4.62.
0 0.2 0.4 0.6 0.8 1
Length Tail length as a function of length
1.4.63.
0 10 20 30 40 50
Mass Length as a function of mass
The graph of length as a function of mass looks like the graph of mass as a function of length turned on its side
Trang 30The population is increasing, whereas decreases to 0 at time 1 and then increases;
the sum decreases slightly and then increases
2.5 30.625 37.5 68.125
3 33.5 37 70.5
Trang 31Copyright © 2015 Nelson Education Ltd 31
20 30 40 50 60 70
increases, decreases, and the total increases
1.4.70.Because the mass is the same at ages 2.5 days and 3 days, the function relating a and M
has no inverse Knowing the mass does not give enough information to estimate the age
0 1 2 3 4 5 6
Age
1.4.71.Volume never has the same value twice, and therefore V has an inverse that contains
sufficient information to find the age
0 2 4 6 8 10
Age
1.4.72.Glucose production is 8.2 mg at ages 2 days and 3 days We cannot figure out the age
from this measurement
0 2 4 6 8 10
Trang 32because it fails the vertical line test Furthermore, we couldn’t tell if was 5.1 or 5.6
0 2 4 6 8 10
The population increases, the mass per individual decreases, and the total mass increases and then decreases
1.4.75.Denote the total mass by Then
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0 1
Year
0 20 40 60 80 100 120
Mass per individual (kilograms) Year
0 40 80 120 160
Year
The population decreases, the mass per individual increases, and the total mass decreases
1.4.76.Denote the total mass by Then
Mass per individual (kilograms) Year
Mass per individual
150 200 250 300 350 400
Year Total mass
The population decreases, the mass per individual increases, and the total mass increases
1.4.77.We denote the total mass by Then
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2 2.5 3 3.5
30 35 40 50 60 70 75
120 130 140 150 160 170 180 190 200
vertically by a factor of 2 (to obtain ), then reflect across the -axis (to obtain
), and finally move up 2 units The final graph is shown below
0 0.5 1 1.5 2
1.4.79 Using long division
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Alternatively, we add and subtract in the numerator:
by a factor of (to obtain ), then reflect across the -axis (to obtain
), and finally move up units See below
c
r A
0
horizontal line test, so have no inverse functions
Trang 36
, and so on (we used Maple) In the case of a linear function, Maple returns the correct inverse function For the functions involving quadratic and cubic terms, Maple returns two and three solutions respectively, ignoring the fact that does not have an inverse and has a (unique) inverse Maple is unable to solve for the inverses
1.4.81 The function increases rapidly until just after , where it reaches its highest value;
then it starts decreasing, and levels off
1.4.82 The function oscillates, with the amplitude of its oscillations decreasing First, it climbs
to a maximum value of about 2.4, then decreases to a bit above 0.8, then climbs again to about 2, and so on
1.4.83 The function consists of numerous oscillations, with a rapid interchange of highs and
lows The complicated oscillating pattern is actually periodic, with period of about 7
Trang 37
Copyright © 2015 Nelson Education Ltd 37
the function , as can be seen well in the initial behaviour: the first peak of
is higher than in Exercise 1.4.82
1.4.85 The function increases rapidly until it reaches a bit about 1.2 and then decreases rapidly
to just after ; after further slower decrease, it levels off at zero
1.4.86 The function seems to be periodic, with the period of a bit over 3 units It consists of a
number of sharp peaks, some with high values and some with low values
Trang 38
1.1 FALSE
real number have inverse functions
the function is equal to )
1.9 TRUE, by the definition of the inverse function
Chapter Summary and Review: Supplementary Problems
1.1 Divide both the numerator and the denominator by , to get