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Chapter 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
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
respectively. The parameter is the wombat density which we can call W.
1.3.3. The graph is the horizontal line crossing the
nor decreasing.

-axis at -4.2.

and

and b,

is neither increasing

y
0

x

-4.2

f (x) = -4.2

1.3.4. The graph is the line of slope 3 whose
numbers .

y

-intercept is -6.

is increasing for all real

f (x) = 3x-6
x

2

0

-6

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
this graph). The function
is decreasing on
and on
.
y

y = 2/x +4

6
4
01


x

1.3.6. The graph is a cube root parabola; we draw it by plotting points and joining them with a
is increasing for all real numbers .
smooth curve. The function

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y
3
f (x) = 2 x

2

-1

x

01

-2

1.3.7. The graph is the hyperbola
and decreasing on

.

shifted one unit upward.

is increasing on

y
f (x) = 1/x2 +1
2

-1

x

0 1

1.3.8. The graph is V-shaped; it is the graph of the absolute value function moved 5 units to the
right.
is increasing on
and decreasing on
.
y
f (x) = | x - 5 |
5
0

1.3.9.

2


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5

x

.

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Chapter 1

1.3.10.
15

y

10
5
0

0

1

2

x

3

4

1.3.11.
20

y

15
10
5
0

0

1

2
x

3

4

1.3.12.
30
25


y

20
15
10
5
0

0

1

2
x

3

4

1.3.13.
1.3.14.
⎛ c ⎞ 1 ⎛ 5⎞ c
1
1.3.15. h ⎜ ⎟ = ,h ⎜ ⎟ = ,h(c + 1) =
.
5c + 5
⎝ 5 ⎠ c ⎝ c ⎠ 25

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1.3.16.

1.3.17.

is a polynomial, so its domain consists of all real numbers. In symbols,
the domain is
= (−∞,∞) .

1.3.18. From

we get

{

. The domain consists of all real numbers

that are not

}


equal to 4. The domain is x ∈
x ≠ 4 ; or, we can say that it consists of two intervals,
and

.

1.3.19. The equation
has no real number solutions, and so the denominator of
is
never zero. The domain of
consists of all real numbers. In symbols, the domain is

= (−∞,∞) .
1.3.20. From

{

}

. The domain is x ∈
x ≥ 7 2 ; using intervals,

we get
.

1.3.21.

is defined for all

, and

is defined when


all non-zero real numbers. Using set notation,
and

. Thus, the domain consists of
; using interval notation,

.

1.3.22. From

we get

In short, we write

and

. Thus, the domain of

is

.

.

1.3.23. The requirements
(because of the fraction) and
(because of the square
root) imply
. Thus, the domain consists of all real numbers such that
.

Using interval notation, we write the domain as
.

1.3.24. There are no restrictions coming from the linear function
. The square root is
defined when
, and therefore the domain of
consists of all real numbers.
3

y

x

1
2x+1

4

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01

x

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Chapter 1

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
in
, and thus the domain is
, or, in short,
.
y 1/x
2
1/4

0

x

x

4

1/x

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.
y

2


x

2
0

x -1

x

-1
-2

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
.
y
x2
0

x

-x2

1.3.28. The graph of

is the horizontal line that crosses the


-axis at

range consists of the single real number 3. Using set notation, we write

. Thus, the
, or

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
an such that
. From
we get
.

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1.3.30. Squaring a number we obtain a positive number, or zero. Thus, we believe that the range
of
consists of all numbers
. To confirm this fact algebraically, we take any
real number

and find a number

and

such that

. From

we get

(note that the quantity under the square root is positive or zero).

1.3.31. The graph suggests that the range is
and find a number

such that

. To prove it we take any real number
. From

we get

and


(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
to two values of

and

.

f (x) = x2+3

y

3

,

0

x

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
expression
cannot be larger than 3.

is positive or zero, the


To prove this fact algebraically we take any real number
and find a number such
that
. From
we get that
and
(note that
the quantity under the square root is positive or zero). Thus, any number less than or
equal to 3 can be obtained by applying the function
to two values of ,
and

.

y
3

f (x) = -x2+3
x

1.3.33. Because the range of
take any

. From

is

, the range of
we compute


is the same. Formally:
.

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

6

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Chapter 1
. From
we get that
and
. Thus, any
number larger than or equal to zero can be obtained by applying the function
to

; in other words, the range of

is


.

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
take any real number
and find a number such that
. From
we get

and

.

1.3.36.

1.3.37.
80

Length

60
40
20
0

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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.42.

Population

initial






intermediate

tiny
start




tiny levels out
Time




.43.

DNA

maximum

initial
0

start

plateau

decline crash

Time




.44.
Temperature

maximum

average temp

minimum
day

night
day
Time

night



.45.

Wetness

wet

dry
dawn

noon

evening midnight dawn


Time



.46. If
if
if
plant even if there are no flowers.

8

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Perhaps one bee will check out the

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Chapter 1

Number of bees

50
40
30
20
10

0

0

5

10
15
20
Number of flowers

25

Number of cancerous cells



.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.
6
4
2
0
0

2

4
6
Radiation


8

10



.48. If
if
if
if
The insect
develops most quickly (in the shortest time) at the highest temperatures.

Development time

40
30
20
10
0

10

20
30
Temperature

40




.49. If a = 0,h = 0 ; if a = 100,h = 50 ; if a = 500,h = 83.3 ; if a = 1000,h = 90.9 ; if
It looks like it would reach 100 m.
100

Height

80
60
40
20
0

0

200

400
600
Age

800

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1.3.50. 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
when
and then starts increasing rapidly.
f (x)

15

10
5

0

−5
−10

−15
0

0.5


1

1.5

2

2.5

3

3.5

4

4.5

x

5

1.3.51. The function reaches its lowest value of approximately
when
. As grows
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.
g (x)


5

4

3

2

1

0

−1

−6

−4

−2

0

2

x

4

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 lowest point
. Between
and
the function is increasing,
reaching its initial value of 3 again when
. The graph seems to be symmetric with
respect to the vertical line
.
h(x)

3.2

3

2.8

2.6

2.4

2.2

2

1.8

10


Full file at .

0

1

2

3

4

5

6

x

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Chapter 1

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.
|g(x)|

5
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
−6

−4

−2

0

2

x

4


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.
k(x)

.

4
3
2
1
0
−1
−2
−3
−4
−1.5

−1

−0.5

0

0.5


1

1.5

2

2.5

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
approximately 0.75) between
and
. Around
it seems to start
increasing.
k(x)+|k(x)|

4
3.5
3
2.5
2
1.5
1
0.5
0

−1.5

−1

−0.5

0

0.5

1

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1.5

2

2.5

x

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1.4 Working with Functions
1.4.1.
–2
–1
0
1
2

–1
1
3
5
7

–11
–8
–5
–2
1

–12
–7
–2
3
8

15
sum

f(x)

10
5

g(x)

0
−5
−10
−15

−2

−1

0
x

1

2

1.4.2.
–2
1
0
1
2


–1
1
3
5
7

–6
–9
–12
–15
–18

10

–7
–8
–9
–10
–11

f(x)

5
0
−5
−10

sum

−15

−20

12

Full file at .

h(x)
−1.5

−0.5

x

0.5

1.5

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Chapter 1

1.4.3.
–2
–1
0
1

2

5
2
1
2
5

–1
0
1
2
3

12

4
2
2
4
8
sum

10
8

F(x)

6
4


G(x)

2
0
−2

−2

−1

0
x

1

2

1.4.4.
–2
–1
0
1
2

5
2
1
2
5


3
2
1
0
–1

8
4
2
2
4

12
10
8

F(x)
sum

6
4
2
0
−2

H(x)
−2

−1


0
x

1

2

1.4.5.
–2
–1
0
1
2

–1
1
3
5
7

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–11
–8
–5
–2
1


11
–8
–15
–10
7

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15
10

f (x)
product
g(x)

5
0
−5
−10
−15
−20
−2

−1


0
x

1

2

1.4.6.
–2
–1
0
1
2
20
0
−20
−40
−60
−80
−100
−120
−140
−2

–1
1
3
5
7


–6
–9
–12
–15
–18

6
–9
–36
–75
–126

f(x)
h(x)

product
−1

0
x

1

2

1.4.7.
–2
–1
0

1
2

5
2
1
2
5

–1
0
1
2
3

–5
0
1
4
15

15

product

10
F(x)
G(x)

5

0
−5
−2

14

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−1

0
x

1

2

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Chapter 1

1.4.8.
–2
–1
0
1

2

5
2
1
2
5

8
7
6
5
4
3
2
1
0
−1
−2

−1

1.4.9. Both

and

3
2
1
0

–1

15
4
1
0
–5

F(x)
product

0
x

H(x)
2

1

are polynomials, so their domains are all real numbers. We compute

is again a polynomial, so its domain is
= (−∞,∞) . The quotient

is defined for all
1.4.10. The domain of

such that

is
= (−∞,∞) and the domain of


is defined whenever

is

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is

. The product

. The domain of the quotient

is the common domain of
1.4.11. The domain of

.

and

, i.e.,

and the domain of

.
is

. The product


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is defined whenever
of the quotient

is in the domain of both

and

, i.e., when

is the common domain of and , with the extra requirement that
Thus, the domain of
is
.
1.4.12. The domain of

is

is defined whenever
of the quotient

and the domain of

and


, i.e.,

.

. The product

is in the domain of both

is the common domain of
1.4.13. We compute

is

. The domain

and

, i.e., when

. The domain

, i.e., x ≠ 0 .

( f
g ) (x) = f ( g(x)) = f ( 4 − 2x ) = ( 4 − 2x )

3

= 43 − 3⋅ 42 (2x) + 3⋅ 4(2x)2 − (2x)3
= 64 − 96x + 48x 2 − 8x 3

and

( g
f ) (x) = g ( f (x)) = g ( x ) = 4 − 2x
3

1.4.14. We compute
and

( f
g ) (x) = f ( g(x)) = f ( 4) = 12 − 4

2

3

= −4

( g
f ) (x) = g ( f (x)) = g (12 − x ) = 4
2

1.4.15. We compute

( f
g ) (x) = f ( g(x)) = f ( x − 3) = x 1− 3
and

( g
f ) (x) = g ( f (x)) = g ⎛⎜⎝ 1x ⎞⎟⎠ = 1x − 3
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Chapter 1

1.4.16. We compute

and
x −1 2 − x +1
⎛ x − 1⎞ 1− 2
( g
f ) (x) = g ( f (x)) = g ⎜⎝ 2 ⎟⎠ = 2 = 22 = −x4+ 3

1.4.17. We compute
1
1− x
−
1
⎛ ⎞
x
( f
g ) (x) = f ( g(x)) = f ⎜⎝ 1x ⎟⎠ = 1x = 1+x x = 1−
1+ x
+1
x
x

and

( g
f ) (x) = g ( f (x)) = g ⎛⎜⎝ xx −+ 11⎞⎟⎠ = x 1− 1 = xx +− 11

x +1

1.4.18. We compute

and

( f
g ) (x) = f ( g(x)) = f (

2− x =

)

( g
f ) (x) = g ( f (x)) = g (

x−2 = 2− x−2

( f
g ) (x) = f ( g(x)) = f (

x =

2− x −2

)

1.4.19. We compute

)

x


1+

( x)

=

2

x
1+ x

and
⎛

( g
f ) (x) = g ( f (x)) = g ⎜⎝ 1+xx
1.4.20. We compute
and

( f
g ) (x) = f ( g(x)) = f ( x

3

2

⎞
x
⎟⎠ = 1+ x 2

) (


)

− 1 = x3 − 1 − 2 = x3 − 3

( g
f ) (x) = g ( f (x)) = g ( x − 2 ) = ( x − 2 )

3

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−1= x − 2 −1

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1.4.21. We find

( f
g)(x) = f (3x − 5) = 2 ⋅(3x − 5) + 3 = 6x − 7

and


(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
(h
f )(x) = h(2x + 3) = −3⋅(2x + 3) − 12 = −6x − 21.
These match, so the functions do commute.

1.4.23. We find

These do not match, so the functions do not commute.
1.4.24. We find

These do not match, so the functions do not commute.
1.4.25. If we write

we can solve for with the steps

. Also, f (1) = 5 and

Therefore,
1.4.26. If we write

we can solve for

Therefore,

as


Also,

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

larger for positive

If we write

Also,

18

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, we can solve for as

or

and

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Chapter 1

1.4.29.
The function

The inverse

15
10
5

x

y

the point (1, 5)

0
−5
−10
−5 −4 −3 −2 −1

0
x

1

2


3

4

5

2
1
0
−1
−2
−3
−4
−5

the point (5, 1)

−5 −4 −3 −2 −1

0 1 2 3 4 5 6
y

1.4.30.
The function
10
5
0
the point (1, −2)

x


y

−5
−10
−15
−20
−5 −4 −3 −2 −1

0
x

1

2

3

4

5

The inverse
4
3.5
3
2.5
2
1.5
1

the point (−2, 1)
0.5
0
−0.5
−6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6
y

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.
The function

What the inverse would look like

30
25
20
15
10

the point
(1, 2)

5
0
−5 −4 −3 −2 −1

1.4.32.

0
y


1

2

3

4

5

The function

30

the point (2, −1)

0

1

2

3
x

4

5


6

5

6

The inverse

2

20

1.5

15

1

10

the point (1, 2)

5
0

1

2

the point (2, 1)


0.5

y

3

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the point (2, 1)

2.5

25

0

2.5
2
1.5
1
0.5
0
−0.5
−1
−1.5
−2
−2.5


4

5

0

0

1

2

3
x

4

19


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Full file Ins.aanual
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1.4.33. To obtain the graph of
, move the graph of
upward for 4 units. From
the 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
other words, it has an inverse. To find the inverse we write

Thus,

and solve for

:

.
y
f (x) =

3

x +4

4

0 1

x

1.4.34. To obtain the graph of

, move the graph of

upward for 4 units. From


the graph we see that the domain of
is
and the range is
. (To
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
has an inverse. To find the inverse we write

Thus,
y

and solve for

:

.
g(x) =

x +4

4
0

20

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x


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Solution Manual for Calculus for the Life Sciences 2nd Edition by Adler
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Chapter 1

1.4.35. Since division by zero is not allowed, the domain of is the interval
from which
has been removed, i.e., the domain consists of two sets,
and
. The
. The function does not pass the horizontal line test – thus; it does not
range is
have an inverse function.
y

y = 1/x2
1

-1

0

x

1

1.4.36. The domain is given to be

. From the graph, we read off the range: it is
. (To
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.
f (x) = | x - 2 |

y

-6

0

2

10 x

1.4.37. To sketch the graph of
units (thus getting the graph of

, start with the graph of

, shift it to the right for 3

), and then expand vertically by a factor of 13.

. The range is
. The
The domain is the set of all real numbers such that

function passes the horizontal line test, so it has an inverse. To find it, we solve for

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:

21


Solution Manual for Calculus for the Life Sciences 2nd Edition by Adler
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nkDtoiraccompany
ect.eu/ Calculus for the Life Sciences, Second Canadian Edition
Thus,

.
y

y=13 / (x -3)

0

x

3

1.4.38. The graph of


is the line with slope
and the -intercept
. Both the domain and the range
consist of all real numbers. The function satisfies the horizontal line test because it is
an increasing function. Thus, it has an inverse. To find the inverse we solve for :

Thus,

.
y

y = (3x -5)/7

0 5/3

-5/7

x

1.4.39. We compute

22

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Solution Manual for Calculus for the Life Sciences 2nd Edition by Adler
Full file at .


Chapter 1

Thus,
1.4.40. We compute

Thus,

1.4.41. We compute

Thus,
1.4.42. We compute

Thus,

1.4.43. Solving

for

we get

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Solution Manual for Calculus for the Life Sciences 2nd Edition by Adler
Full file Instructor’s
at />Solutions Manual to accompany Calculus for the Life Sciences, Second Canadian Edition


Thus,

1.4.44. Solving

for

we find

Thus,

1.4.45. Solving

for

we find

Thus,

24

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Solution Manual for Calculus for the Life Sciences 2nd Edition by Adler
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Chapter 1

1.4.46.
The vertical axis is scaled by a value greater than 1.

6
5.5
5
4.5
4
3.5
3
2.5
2
1.5
1
0.5

4g(x)

g(x)
−4 −3 −2 −1

0
x

1

2

3

4

5


1.4.47.
The vertical axis is shifted by a value less than 0, moving it down.
1.4
1.2
1
0.8
0.6
0.4
0.2
0
−0.2
−0.4
−5 −4 −3 −2 −1

g(x)

g(x)−1
0
x

1

2

3

4

5


1.4.48.
The horizontal axis is scaled by a value less than 1.
1.4
1.3
1.2
1.1
1
0.9
0.8
0.7
0.6
−5 −4 −3 −2 −1

g(x/3)
g(x)

0
x

1

2

3

4

5

1.4.49.
The horizontal axis is shifted by a value greater than 0.
1.4

1.3
1.2
1.1
g(x+1)
1
0.9
0.8
0.7
0.6
−5 −4 −3 −2 −1

g(x)

0
x

1

2

3

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4

5

25