379847921 Solution Manual for Calculus for the Life Sciences 2nd Edition by Adler pdf PDF

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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 2 x

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 smooth curve. The function is increasing for all real numbers .

Instructor’s Solutions Manual to accompany Calculus for the Life Sciences, Second Canadian Edition

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

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) = . ⎝ 5 ⎠ c ⎝ c ⎠ 25 5c + 5

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Instructor’s Solutions Manual to accompany Calculus for the Life Sciences, Second Canadian Edition

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

{

we get

}

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

.

1.3.21.

is defined for all

, and

is defined when

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

1.3.22. From

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

.

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 0 1

4

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

x

1/ 4 0

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 -2

x -1

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

-x 2

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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Instructor’s Solutions Manual to accompany Calculus for the Life Sciences, Second Canadian Edition

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

,

.

f (x) = x 2+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 , . y 3 f (x) = -x2+3 x

1.3.33. Because the range of take any

. From

, 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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Instructor’s Solutions Manual to accompany Calculus for the Life Sciences, Second Canadian Edition

.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

noon

evening midnight dawn

 .45.

Wetness

wet

dry dawn

Time

 .46. If if if plant even if there are no flowers.

8

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 Temp erature

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 Ag e

800

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1000

9

Instructor’s Solutions Manual to accompany Calculus for the Life Sciences, Second Canadian Edition

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

x

−1 −6

−4

−2

0

2

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 . 3.2

h (x) 3

2.8

2.6

2.4

2.2

2

1.8 0

10

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

4

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. 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

11

Instructor’s Solutions Manual to accompany Calculus for the Life Sciences, Second Canadian Edition

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

–7 –8 –9 –10 –11

10 f (x)

5 0 −5 −10

sum

−15 −20

12

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

4 2 2 4 8

12 sum

10 8

F(x)

6 4 2

G(x)

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

13

Instructor’s Solutions Manual to accompany Calculus for the Life Sciences, Second Canadian Edition

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 5

F(x) G(x)

0 −5 −2

14

−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

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

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. is

. The product

15

Instructor’s Solutions Manual to accompany Calculus for the Life Sciences, Second Canadian Edition

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

is

and

, i.e.,

.

. The product

is in the domain of both

is the common domain of

. The domain

and

, i.e., when

. The domain

, i.e., x ≠ 0 .

1.4.13. We compute

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

2

2

= 4 − 3⋅ 4 (2x) + 3⋅ 4(2x) − (2x) 2

= 64 − 96 x + 48x − 8x

3

3

3

and

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

3

1.4.14. We compute

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

2

= −4

and

( 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 ⎜⎝ x1 ⎟⎠ = 16

1 −3 x

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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 −1 1− x ( f  g )(x) = f (g(x) ) = f ⎜⎝ x ⎟⎠ = 1 = 1+x x = 1+ x +1 x x

and ⎛

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

1 x +1 = x −1 x −1 x +1

1.4.18. We compute

( 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...