Chapter 1
Functions and Graphs 4. A vertical line at x at x = = 10
Exercise Set 1.1 1. Graph y Graph y =
3 2 y
−4.
Note that y is constant and therefore any value of x we choose will yield the same value for y, which is 4. Thus, we will have a horizontal line at y at y = 4.
−
−
1 0
2
4
x 6
8
10 10
–1
4 –2
y 2
–4
–2
–3
0
2 x
4
5. Graph. Find the slope and the y the y-intercept -intercept of y y =
–2
First, we find some points that satisfy the equation, then we plot the ordered pairs and connect the plotted points to get the graph.
–4
When x When x = = 0, y 0, y =
2. Horizontal line at y at y =
−3(0) = 0, ordered pair (0, (0, 0) When x When x = = 1, y 1, y = −3(1) = −3, ordered pair (1, (1, −3) When x = −1, y = − 3(−1) = 3, ordered pair (−1, 3)
−3.5
4
4
y 2
y 2
–4
–2
0
2 x
4
–4
4
–4
Compare the equation y equation y = = 3x to the general linear equation form of y = mx + mx + b b to conclude the equation has a slope of m = m = 3 and a y a y -intercept of (0, (0, 0).
−
−4.5.
Note that x is constant and therefore any value of y we choose will yield the same value for x, which is 4. 4 .5. Thus, we will have a vertical line at x at x = = 4.5. 2
−
−
6. Slop Slopee
of m
−0.5
–2
0
2 –2 –1
–4 –2
and y-in -interc tercep eptt
y 2
–4 –2
= 4
1
–4 x
2 x
–2
–4
–6
0
–2
–2
3. Graph x Graph x = =
−3x.
2 x
4
of (0, (0, 0)
2
Chapter 1: Functions and Graphs 7. Graph. Find the slope and the y the y-intercept -intercept of y y = 0.5x. First, we find some points that satisfy the equation, then we plot the ordered pairs and connect the plotted points to get the graph. When x When x = = 0, y 0, y = 0.5(0) = 0, ordered pair (0, (0, 0)
Compare the equation y = 2x + 3 to the general linear equation form of y y = mx = mx + b to conclude the equation has a slope of m = m = 2 and a y a y -intercept of (0, (0, 3).
−
− 10. Slope of m = m = −1 and y and y-intercept -intercept of (0, (0, 4) 6
When x When x = = 6, y 6, y = 0.5(6) = 3, ordered pair (6, (6, 3) When x When x = =
−2, y 2, y = 0.5(−2) = −1, ordered pair (−2, −1)
4 y
4
2
y 2
–4
–2
0
–2
2
x
4
6
–2
2 x
4 –4
–2
11. Graph. Find the slope and the y the y-intercept -intercept of y y = –4
Compare the equation y equation y = = 0.5x to the general linear equation form of y = mx + mx + b b to conclude the equation has a slope of m m = 0.5 and a y a y-intercept -intercept of (0, (0, 0). 8. Slope of m = m = 3 and y and y-intercept -intercept of (0, (0, 0)
–4
–2
First, we find some points that satisfy the equation, then we plot the ordered pairs and connect the plotted points to get the graph. When x When x = = 0, y 0, y =
−(0) − 2 = −2, ordered pair (0, (0, −2) When x When x = = 3, y 3, y = −(3) − 2 = −5, ordered pair (3, (3, −5) When x When x = = −2, y 2, y = −(−2) − 2 = 0, ordered pair (−2, 0)
4
4
y 2
y 2
0
2 x
4
–4
–2
0
–2
–2
–4
–4
9. Graph. Find the slope and the y the y-intercept -intercept of y y =
−2x + 3.
First, we find some points that satisfy the equation, then we plot the ordered pairs and connect the plotted points to get the graph. When x When x = = 0, y 0, y =
−2(0) + 3 = 3, ordered pair (0, (0, 3) When x When x = = 2, y 2, y = −2(2) + 3 = −1, ordered pair (2, (2, −1) When x When x = = −2, y 2, y = −2(−2) + 3 = 7, ordered pair (−2, 7)
2 x
4
Compare the equation y = x 2 to the general linear equation form of y y = mx = mx + b to conclude the equation has a slope of m = m = 1 and a y a y -intercept of (0, (0, 2).
− −
− − 12. Slope of m = m = −3 and y and y-intercept -intercept of (0, (0, 2) 4 y 2
4 –4
y 2
–2
0
2 x
4
–2 –4
–2
0
2 x
4
–4
–2
13. Find the slope and y and y -intercept of 2x 2x + y –4
−x − 2.
− 2 = 0.
Exercise Set 1.1
3
Solve the equation for y for y.. 2x + y
22.
−2 y
= 0 = 2x + 2
y
−
Compare to y = mx + mx + b b to conclude the equation has a slope of m m = 2 and a y a y-intercept -intercept of (0, (0, 2).
−
y =
y + 2 3x + 13
−
− y 1
−
Solve the equation for y for y..
y
=
−
16. y = x = x + 2, slope of m = m = 1 and y and y-intercept -intercept of (0, (0, 2). 17. Find the slope and y and y -intercept of x = x = 2y + 8.
=
= 2(x 3) = 2x 6
− −
24. y
Compare to y = mx + mx + b b to conclude the equation has a slope of m m = 1 and a y a y-intercept -intercept of (0, (0, 52 ).
−
−0 y
− −5 −x − 52
−3(x 3(x − 5) −3x + 15
Plug the given given informati information on into into the equation y m(x x1 ) and solve for y for y
15. Find the slope and y and y -intercept of 2x 2x + 2y 2y + 5 = 0.
2x + 2y 2y + 5 = 0 2y = 2x
= =
23. Find the equation of line: with m = m = 2, containing (3, (3, 0)
14. y = 2x + 3, slope of m = m = 2 and y and y-intercept -intercept of (0, (0, 3)
y
− (−2)
−0
= =
y
−5(x 5(x − 5) −5x + 25
25. Find Find the equatio equation n of line: line: with with y-intercept (0, (0, 6) and 1 m = 2
−
Plug the given information into the equation y equation y = mx = mx + b
Solve the equation for y for y.. y
x = 2y + 8 8 = 2y
x
− 1 x−4 2
y y
= y
Compare to y = mx + mx + b b to conclude the equation has a slope of m m = 21 and a y a y -intercept of (0, (0, 4).
−
− 14 x + 43 , slope of m = m = − 14 and y and y -intercept of (0, (0, 34 ) 19. Find the equation equation of the line: with m = − 5, containing (1, (1, −5) Plug the given information into equation y −y1 = m = m((x−x1 ) 18. y =
− y1 y − (−5) y + 5 y y
= m(x x1 ) = 5(x 5(x 1) = 5x + 5 = 5x + 5 5 = 5x
− − − − − − −
−
−
26. y = 34 x + 7 27. Find the equation of line: with m with m = 0, containing (2, (2, 3) Plug the given given informati information on into into the equation y m(x x1 ) and solve for y for y
− y 1
−
y y
and solve for y for y
y
= mx + b 1 = x + ( 6) 2 1 = x 6 2
−3 −3
= 0(x = 0 = 3
− 2)
−8 −8
= 0(x = 0 = 8
− 4)
y
=
28. y y
y
20. y y
−7 −7 y
= 7(x 1) = 7x 7 = 7x
− −
−2, containing (2, (2, 3) Plug the given information into the equation y − y 1 = m(x − x1 ) and solve for y for y y − 3 = −2(x 2(x − 2) y − 3 = −2x + 4 y = −2x + 4 + 3 y = −2x + 7
29. Find the slope given ( 4, 2) and ( 2, 1)
− −
−
y Use the slope equation m = xy − −x . NOTE: It does not matter which point is chosen as (x (x1 , y1) and which is chosen as (x (x2 , y2 ) as long as the order the point coordinates are subtracted subtracted in the same order as illustrat illustrated ed below below 2
1
2
1
21. Find the equation of line: with m = m =
m = = =
1 ( 2) 2 ( 4) 1+2 2+4 3 2
−− − −− −
4
Chapter 1: Functions and Graphs 39. Find the slope given (x, (x, 2x + 3) and (x (x + h, 2(x 2(x + h) + 3)
−2 − 1) −4 − (−2) −3 −2
m = =
3 2
= 1 30. m = 6−3− (−2) =
2 8
=
1 4
31. Find the slope given ( 25 , 12 ) and ( 3, 45 )
− 4 − 21 5 −3 − 52 8 − 105 10 −15 − 10 5 5
m = =
= 5 6 3 4
3 16 1 4
=
3 16
4 1
3 4
3x+3h +3h 1 3x+1 h
−−
−
−
−
−1 y−1 y−1
=
y
=
y
=
= =
8 −2 34. m = −410 −(−4) = 0 This line has no slope
y
=
35. Find the slope given (2, (2, 3) and ( 1, 3)
y
=
1 2
y
−
36. m = −7−(−6) = −01 = 0 37. Find the slope given (x, (x, 3x) and (x (x + h, 3(x 3(x + h)) m = = = = 38. m =
4(x 4(x+h) 4x x+h x
− −
=
3(x 3(x + h) 3x x+h x 3x + 3h 3h 3x h 3h h 3
4x+4h +4h 4x h
−
− − −
=
4h h
=4
−−
=
1 (x 6) 4 1 6 x 4 4 1 3 x + 3 4 2 1 3 x+ 4 2
− − −
43. Find equation of line containing ( 25 , 12 ) and ( 3, 45 )
−
From Exercise 31, we know that the slope of the line is
− 343 and using the point (−3, 45 ) − 45 4 y− 5 4 y− 5 y
1 2
3 (x ( 2)) 2 3 (x + 2) 2 3 x+3 2 3 x+3+1 2 3 x+4 2
42. Using m Using m = 41 and the point (6, (6, 3)
This line has no slope
= 0
−
NOTE: You NOTE: You could use either of the given points and you would reach the final equation.
0
=
=3
From Exercise 29, we know that the slope of the line is 23 . Using the point( 2, 1) and the value of the slope in the point-slope formula y formula y y1 = m( m (x x1 ) and solving for y for y we get:
=
m =
3h h
− −
−3 y−3
− 3−3 −1 − 2 0 −3
=
41. Find equation of line containing ( 4, 2) and ( 2, 1)
y
− · = − 33. Find the slope given (3, (3, −7) and (3, (3, −9) −9 − (−7) m = 3−3 − 2 = undefined quantity 1 2
− − − = −
·
=
− − − −
[3(x [3(x+h) 1] (3x (3x 1) x+h x
3 5 10 17 15 170 3 34
=
3 16
40. m =
3 10 17 5
=
− − − 32. m = − − − = ( )
[2(x [2(x + h) + 3] (2x (2x + 3) x+h x 2x + 2h 2h + 3 2x 3 = h 2h = h = 2
m =
= = =
y
=
y
=
y
=
− 343 (x − (−3)) − 343 (x + 3) − 343 x − 349 − 343 x − 349 + 45 45 136 − 343 x − 170 + 170 3 91 − 34 x + 170
Exercise Set 1.1 44. Using m Using m = =
5
− 134 and the point − 34 , 58 y
−
y
−
5 8 5 8
=
−
=
− − − −
y
=
y
=
y
=
52. Using m Using m = 3 and the point (x, (x, 3x y y
13 3 x 4 4 13 39 x 4 16 13 39 5 x + 4 16 8 13 39 10 x + 4 16 16 13 29 x 4 16
− −
y
− − − −
53. Slope = of 8%.
0.4 5
−
−
58.
48. Since the line has a slope of m = 0, it is horizontal. horizontal. The 1 equation of the line is y is y = 2 49. Find equation of line containing (x, (x, 3x) and (x (x+h, 3(x 3(x+h))
a) F ( F ( 10) = 59 ( 10) + 32 = 18 + 32 = 14o F F (0) F (0) = 59 (0) + 32 = 0 + 32 = 32o F F (10) F (10) = 59 (10) + 32 = 18 + 32 = 50o F F (40) F (40) = 59 (40) + 32 = 72 + 32 = 104o F b) F (30) F (30) = 59 (30) + 32 = 54 + 32 = 86o F c) Same temperature in both means F means F ((x) = x = x.. So
−
· · · ·
· −
−
F ( F (x) = x
From Exercise 37, we found that the line containing (x, (x, 3x) and (x (x + h, 3(x 3(x + h)) had a slope of m = m = 3. Using the point (x, 3x) and the value of the slope in the point-slope formula
9 x + 32 = x 5 9 x x = 32 5 4 x = 32 5
−
3(x 3(x x) 3(0) =0 = 3x
−
x = x =
50. Using m Using m = = 4 and the point (x, (x, 4x) 59.
− 4x − 4x y
= 4(x 4(x = 0 = 4x
− x)
51. Find equation of line containing (x, (x, 2x+3) and (x (x+h, 2(x 2(x+ h) + 3) From Exerci Exercise se 37, we found found that that the line line contai containin ningg (x, 2x + 3) and and (x (x + h, 2(x 2(x + h) + 3) had had a slope slope of m = m = 2. Using the point (x, (x, 3x) and the value of the slope in the point-slope formula y y y
− (2x (2x + 3) − (2x (2x + 3) − (2x (2x + 3) y
= = = =
2(x x) 2(0) 0 2x + 3
−
≈
− −
−
−
= 0.916
Change in Life Life expectancy expectancy Change in Time 76 76..9 73 73..7 = 2000 1990 3.2 = 10 = 0.32 per year year
47. Find equation of line containing (2, (2, 3) and ( 1, 3) From Exercise 35, we found that the line containing (2, (2 , 3) and ( 1, 3) has a slope of m = 0. We notic noticee that the the y -coordinate -coordinate does not change regardless regardless of the x-value. Theref Therefore ore,, the line line in horizo horizont ntal al and has the equation equation y = 3.
8.25 9
Rate =
−
y y
= 0. 0 .035 = 3. 3.5%
57. The average rate of change of life expectancy at birth is computed by finding the slope of the line containing the two points (1990, (1990, 73 73..7) and (2000, (2000, 76 76..9), which is given by
46. Since the line has no slope, it is vertical. The equation of the line is x is x = = 4.
y
43. 43.33 1238
56. The stairs stairs have have a max maxim imum um grade grade of 0.9167 = 91. 91.67%
From Exercise 33, we found that the line containing (3, (3, 7) and (3, (3, 9) has no slope. We notice that the x the x-coordinate -coordinate does not change regardless of the y-value. Therefore, the line in vertical and has the equation x equation x = = 3.
−
−1
≈ 0.3171, or 31. 31.71%
55. The slope (or head) of the river is
−
− 3x = − 3x = − 3x =
= = 0 = 3x
= 0. 0 .08. This means the treadmill has a grade
54. The roof has a slope of 62..62
45. Find equation of line containing (3, (3, 7) and (3, (3, 9)
y y y
− (3x (3x − 1) − (3x (3x − 1)
− 1) 3(x − x)
− − −32 · 54 −40o
a) Since R Since R and and T T are are directly proportional we can write that R that R = = kT kT ,, where k where k is is a constant of proportionality. Using R Using R = = 12 12..51 when T when T = = 3 we can find k find k.. R 12 12..51 12 12..51 3 4.17
= kT = k(3) = k = k
Thus, we can write the equation of variation as R as R = 4.17 17T T b) This is the same as asking: find R when T T = 6. So, we use the variation equation R = 4.17 17T T = 4.17(6) = 25 25..02
6
Chapter 1: Functions and Graphs
60. We need to find t find t when D when D = 6.
64.
a)
D = 293t 6 = 293t 6 = t 293 0.0205 seconds seconds t
D(5) = D(10) =
≈
61.
D(20) =
a) Since B s directly proportional to W W we can write B = kW . kW .
D(50) =
b) When W When W = = 200 200 B B = 5 means that
D(65) =
B 5 5 200 0.025 2.5%
= kW = k(200)
= k = k 65.
a) M ( M (x) = 2.89 89x x + 70. 70.64 M (26) = 2.89(26) + 70. 70.64 = 75 75..14 + 70. 70.64 = 145.78
c) Find B Find B when W when W = = 120 B
· · · · ·
b) c) Since cars cannot have negative speed, and since the car will not need to stop if it has speed of 0 then the domain is any positive real number. NOTE: The domain will have an upper bound since cars have a top speed limit, depending on the make and model of the car.
= k
This means that the weight of the brain is 2. 2.5% the weight of the person.
11 0 + 5 5 = = 0. 0 .5 f t 10 10 11 10 + 5 115 = = 11. 11 .5 f t 10 10 11 20 + 5 225 = = 22. 22 .5 f t 10 10 11 50 + 5 555 = = 55. 55 .5 f t 10 10 11 65 + 5 720 = = 72 f t 10 10
= 0.025 025W W = 0.025(120 l 025(120 lbs bs)) = 3 lbs
The male was 145.78 cm tall. 62.
a)
b) F ( F (x) = 2.75 75x x + 71. 71.48 71.48 F ( F (26) = 2.75(26) + 71. = 71 71..5 + 71. 71.48 = 142.98
M = kW 80 = k(200) 0.4 = k Thus, the equation of variation is M is M = 0. 0 .4W b) k = 0.4 = 40% means that 40% of the body weight is the weight of muscles.
The female was 142.98 cm tall. 66.
c) M = 0.4(120) = 48 lb 63.
a) The equati equation on of varia variatio tion n is given given by N = P + 0.02 02P P = 1. 1 .02 02P P .. 1 .02(200000) = 204000 b) N = 1. c) 367200 = 1.02 02P P 367200 = P 1.02 3600 360000 00 = P
a) D(0) (0) D( 20) D(10) (10) D(32) (32)
−
= = = =
2(0) 2(0) + 11 1155 = 0 + 11 1155 f t 2( 20) + 115 = 40 + 115 = 75 f 75 f t 2(10 2(10)) + 11 1155 = 20 + 11 1155 = 13 1355 f t 2(32 2(32)) + 11 1155 = 64 + 11 1155 = 17 1799 f t
−
−
b) The stopping distance has to be a non-negative value. Therefore we need to solve the inequality 0 115 57 57..5
− −
≤ ≤ ≤
2F + + 115 2F F
The 32o limit comes from the fact that for any temperature above that there would be no ice. Thus, the domain of the function is restricted in the interval [ 57 57..5, 32].
−
67.
a) A(0) A(1) A(10) A(30) A(50)
= = = = =
0.08(0) + 19. 19.7 = 0 + 19. 19.7 = 19 19..7 0.08(1) + 19. 19.7 = 0. 0 .08 + 19. 19.7 = 19. 19.78 0.08(10) + 19. 19.7 = 0. 0 .8 + 19. 19.7 = 20. 20.5 0.08(30) + 19. 19.7 = 2. 2 .4 + 19. 19.7 = 22. 22.1 0.08(50) + 19. 19.7 = 4 + 19. 19.7 = 23 23..7
b) First we find the value of t4, t 4, which is 2003 53. So, we have to find A find A(53). (53).
− 1950 =
0.08(53) + 19. 19.7 = 4.24 + 19. 19.8 = 23. 23.94 A(53) = 0. The median age of women at first marriage in the year 2003 is 23.94 years.
Exercise Set 1.2
7 3. y = x = x 2 and y and y = (x
c) A(t) = 0.08 08tt + 19. 19.7
− 1)2
120 23
100 80
22
60 21
40 20
20 0
10
20
t
30
40
50
68. The use of the slope-intercept equation or the point-slope equation depends on the problem. If the problem gives the slope and the y-intercept then one should use the slopeinterce intercept pt equation. equation. If the problem problem gives the slope and a point that falls on the line, or two points that fall on the line then the point-slope point-slope equation equation should be used.
–10 –8
–6
–4
–2
4. y = x = x 2 and y and y = (x
1. y =
and y and y =
4 x 6
8
10
2
4 x 6
8
10
2
4 x 6
8
10
2
4 x 6
8
10
− 3)2
160 140 120 100 80
Exercise Set 1.2 1 2 2x
2
60
−
1 2 2x
40 20
40
–10 –8
–6
–4
–2 0
5. y = x = x 2 and y and y = (x + 1)2
20
120
–10 –8
–6
–4
–2 0
2
4 x 6
8
10
100
–20
80 60
–40
40
2. y =
1 2 4x
and y and y =
−
1 2 4x
20
20
–10 –8
–6
–4
–2
6. y = x = x 2 and y and y = (x + 3)2
10
160
–10 –8
–6
–4
–2 0
2
4 x 6
8
10
140 120
–10
100
–20
80 60 40 20 –10 –8
–6
–4
–2 0
8
Chapter 1: Functions and Graphs 7. y = x = x3 and y and y = x = x 3 + 1
13. y = x = x 2
− 4x + 3
10
14
8
12
6 y 4
10 y8 6
2 –4
0 –2
–2
2 x
4
4 2
–4 –8
0 –2
–10
–4
–6
8. y = x = x3 and y and y = x = x 3
–4
–2
14. y = x = x 2
−1
10
14
8
12
10
2
4 x 6
8
10
2 x
4
y8 6
0 –2
2 x
4
4 2
–4 –8
0 –2
–10
–4
–6
–4
9. Since the equation has the form ax form ax 2 + bx + c, with a with a = 0, the graph of the function is a parabola. The x The x-value -value of the vertex is given by
x =
15. y =
–2
−x2 + 2x 2x − 1 2
–4
–2
0
4 − 2ba = − 2(1) = −2
–2 –4
The y The y-value -value of the vertex is given by y
8
10
2 –2
4 x 6
− 6x + 5
6 y 4
–4
2
–6
= ( 2)2 + 4( 2) = 4 8 7 = 11
− − −7 − − − Therefore, Therefore, the vertex vertex is (−2, 11).
y
–8
–10 –12 –14 2
10. Since the equation is not in the form of ax + bx + bx + c c,, the graph of the function is not a parabola.
16. y =
−x2 − x + 6
11. Since the equation is not in the form of ax2 + bx + bx + c c,, the graph of the function is not a parabola.
8
12. Since the equation has the form ax form ax 2 + bx + c, with a with a = 0, the graph of the function is a parabola. The x The x-value -value of the vertex is given by
y4
6
x =
−
b = 2a
− −6
2(3)
–4
=1
The y The y-value -value of the vertex is given by 2
y = 3(1) = 3 6 = 3
− − Therefore, Therefore, the vertex vertex is (1, (1, −3).
2
− 6(1)
–2
0 –2 –4 –6 –8
2 x
4
Exercise Set 1.2 17. y = 2x2 + 4x 4x
9 x2 2x 2 x 2 = 0, then then use the quadra quadratic tic formula formula,, with with a = 1, b 1, b = 2, and c and c = 2, to solve for x for x..
−7
− −
10 8
x =
6 y 4
x =
2 –4
0 –2
–2
−
2 x
4
=
–4
=
–6 –8
=
–10
18. y = 3x2
8 6 y 4
22. x2
2 0 –2
2 x
4
–4 –8
=
−5 10 6 y 4 2
–6 x –4
0 –2
–2
y 2
y
–6 –10
+ 4x 4x
−2
= =
5 –2
=
=
10
–4
=
=
–8
20. y =
2
4
–4
1 2 3x
2 √ 2 ± 20
2
4
6
x8
10
12
14
=
0
± √ 5)
−b ± √ b2 − 4ac 2a
−8 ± (8)2 − 4(3)(2) 2(3) √ −8 ± 64 − 24 6 √ −8 ± 40 6 √ −8 ± 2 10
6 2( 4 10) 6 4 10 3
− ± √ − ± √ √
–5
The solutions are −4+3 24. 2 p2
y –10 –15
21. Solve x Solve x 2 2x = 2 Write the equation so that one side equals zero, that is
−
=
2(1
23. Solve 3y 3y 2 + 8y 8y + 2 = 0 Use the quadratic formula, with a = 3, b = 8, and c = 2, to solve for y for y .
8
–8
2
2 √ = 1± 5 √ √ The solutions are 1 + 5 and 1 − 5
–10
–10
2
± √ 12 2 √ 2±2 3 2
− 2x + 1 = 5 can be rewritten as x as x 2 − 2x − 4 = 0 − (−2) ± (−2)2 − 4(1)(−4) x = 2(1) √ 2 ± 4 + 16 =
–6
19. y = 21 x2 + 3x 3x
−(−2) ± (−2)2 − 4(1)(−2) 2(1) √ 2± 4+8
√ 3) ± = √ 2 = 1± 3 √ √ The solutions are 1 + 3 and 1 − 3
10
–2
2a
2(1
− 9x + 2
–4
− −b ± √ b2 − 4ac
10
√
and −4−3
10
− 5 p = p = 1 can be rewritten as 2 p2 − 5 p − 1 − (−5) ± (−5)2 − 4(2)(−1) p = 2(2) √ 5 ± 25 + 8 = =
4 √ 5 ± 33 4
10
Chapter 1: Functions and Graphs The solutions are
√
5+ 33 4
√
25. Solve x Solve x 2 2x + 10 = 0 Using the quadratic formula with a = 1, b = c = 10
−
= = = =
2
−2, and
=
x =
=
− 3i
2a
−6 ± (6)2 − 4(1)(10) 2(1) √ −6 ± 36 − 40
− −b ± √ b2 − 4ac
2a −6 ± (6) − 4(1)(−1) 2
x = = = = =
2a −4 ± (4) − 4(1)(8)
2(1) 36 + 4
−6 ± √ 2 −6 ± √ 40 2
−6 ± 2√ 10 2 √ 2(−3 ± 10)
2 √ = −3 ± 10 √ √ The solutions are −3 + 10 and −3 − 10
2(1) √ −4 ± 16 − 32
2 √ −4 ± −16 = 2 − 4 ± 4i = 2 4(1 ± i) = 2 = 2(1 ± i) = 2 ± 2i The solutions solutions are 2 + 2i 2i and 2 − 2i
−b ± √ b2 − 4ac
27. Solve x Solve x 2 + 6x 6x = 1 Write the equation so that one side equals zero, that is x2 + 6x 1 = 0, then then use the quadra quadratic tic formula formula,, with with a = 1, b 1, b = 6, and c and c = = 1, to solve for x for x.. x =
−b ± √ b2 − 4ac 2
x =
√ 2 − 6 ± −4 = 2 − 6 ± 2i = 2 2(−3 ± i) = 2 = −3 ± i The solutions are −3 + i and −3 − i −
=
2 √ = −2 ± 7 √ √ The solutions are −2 + 7 and −2 − 7
26.
x =
2
− ± √ 7)
2( 2
29. Solve x Solve x 2 + 4x 4x + 8 = 0 Using the quadratic formula with a with a = = 1, b 1, b = 4, and c and c = = 8
The solutions solutions are 1 + 3i 3i and 1
x =
2 √ −4 ± 28
=
−(−2) ± (−2)2 − 4(1)(10) 2(1) √ 2 ± 4 − 40
2 √ 2 ± −36 2 2 ± 6i 2 2(1 ± 3i) 2 1 ± 3i
2(1) √ −4 ± 16 + 12
=
2a
x =
−3= 0 −4 ± (4) − 4(1)(−3)
x =
−b ± √ b2 − 4ac
x =
=
28. x2 + 4x 4x = 3 can be rewritten as x 2 + 4x 4x
and 5−4 33
30. x = = = =
−10 ± (10)2 − 4(1)(27) 2(1) √ −10 ± 100 − 108 2
−10 ± √ −8 2 −10 ± 2i√ 2 2
√ − ± i 2) = 2√ = −5 ± i 2 √ √ The solutions are −5 + i 2 and −5 + i 2 2( 5
31. Solve 4x 4x2 = 4x 4x 1 Write the equation so that one side equals zero, that is 4x2 4x 1 = 0, then use the quadratic formula, with a = 4, b 4, b = 4, and c and c = 1, to solve for x for x..
−
− − −
x = x = = =
− −b ± √ b2 − 4ac 2a
−(−4) ± (−4)2 − 4(4)(−1) 2(4) √ 4 ± 16 + 16 8 √ 4 ± 32 8
Exercise Set 1.2
11
± 4√ 2 8 √ 4(1 ± 2) 4
= =
√
1+ 2 2
√
and 1−2
2
− 4x2 = 4x 4 x − 1 can be rewritten as 0 = 4x 4x2 + 4x 4x − 1 −4 ± (4)2 − 4(4)(−1) x = 2(4) √ −4 ± 16 + 16 = =
8 √ −4 ± 32
=
−1 ± √ 2
8
=
35. Solve 50 = 9. 9.41 0.19 19x x + 0. 0.09 09x x2 . First, let us rewrite the equation as 0 = 40 40..59 0.19 19x x + 0. 0.09 09x x2 then we can use the quadratic formula to solve for x for x
− −
√ 4(−1 ± 4)
√ 2
and −1−2
8
=
f (7) = = = = f ( f (10) = = = = f ( f (12) = = = 34.
a) x = 2009
Therefore, the average price of a ticket will be $50 will happen during the, 1990 + 22. 22.3183 = 2012. 2012.3183 2012-13 season. NOTE: season. NOTE: We We could not choose the negative option of the quadratic formula since it would result in the result that is negative which corresponds to a year before 1990 and that does not make physical sense.
1 3 1 2 1 (7) + (7) + (7) 6 2 2 343 49 7 + + 6 2 2 343 147 21 + + 6 6 6 511 85 85..16 85 oranges 6 1 1 1 (10)3 + (10)2 + (10) 6 2 2 1000 + 50 + 5 6 500 150 15 + + 3 3 3 665 221..6 222 oranges 221 3 1 1 1 (12)3 + (12)2 + (12) 6 2 2 288 288 + 72 + 6 36 3666 oran orange gess
≈
≈
36.
≈
−
b) Solve 170 = 0. 0.0728 0728h h2 6.986 986h h + 289, which can be 2 written as 0. 0.0728 0728h h 6.986 986h h + 119 = 0
−
h =
≈
= =
37. f ( f (x) = x = x 3
b) Solve 100 = 4. 4.8565 + 0. 0.2841 2841x x + 0.1784 1784x x2 . First, First, let let us rewrite the equation as 0 = 95 95..1435+0 1435+0..2841 2841x x+ 0.1784 1784x x2 then we can use the quadratic formula to solve for x for x
−
−0.2841 ± 0.28412 − 4(0. 4(0.1784)(−95 95..1435) 2(0. 2(0.1784) √ 67..9751 √ −0.2841 ± 0.0807 + 67. 67.8944 −0.2841 ± 67 0.3568
=
0.3568
− −(−6.986) ± (−6.986)2 − 4(0. 4(0.0728)(119) 2(0. 2(0.0728) √ 6.986 ± 14 14..1514 0.1456 6.986 3.7618) 0.1456
±
−3.7618 = The The poss possib ible le two answ answer erss are are 6.986 0.1456 22 22..1440 in, in, which is out side of the domain of the 986+3.7618 function, and 6.986+3. = 73 73..8173 in, in, which is in 0.1456 the domain interval of the function w function w.. Therefore, the man is about 73. 73.8 inches tall.
− 1985 = 24
The average payroll for 2009-10 is $114. $114.4333 million
=
a) w (72) = 0.0728(72)2 6.986(72) + 289 = 163.4032 pounds
f (24) = 4.8565 + 0. 0.2841(24) + 0. 0.1784(24)2 = 4.8565 + 6. 6.8184 + 102. 102.7584 = 114.4333
x =
0.18
±
=
33. Find f Find f (7), (7), f f (10), (10), and f and f (12) (12)
=
0.18 0.19 3.8273 0.18 0.19 + 3. 3.8273 = 22. 22 .3183 0.18
=
√ 2
−
−(−0.19) ± (−0.19)2 − 4(0. 4(0.09)(−40 40..59) 2(0. 2(0.09) √ 14..6485 √ 0.19 ± 0.0361 + 14. 14.6124 0.19 ± 14
x =
2
The solutions are −1+2
0.3568
Therefore Therefore,, the average payroll payroll will be $100 million million is the, 1985 + 23. 23.3111 = 2007. 2007.3111, 2007-08 season. NOTE: NOTE: We could not choose the negative option of the quadratic formula since it would result in the result that is negative which corresponds to a year before 1985 and that does not make sense.
2
The solutions are 32.
=
8 √ 1± 2
=
−0.2841 ± 8.2447 0.3568 −0.2841 + 8.8.2447 = 22. 22 .3111
=
− x2
a) For large values of x, x3 would be larger than x2 . x3 = x x x and x and x 2 = x x so for very large values of x there is an extra factor of x x in x 3 which causes x causes x 3 2 to be larger than x than x . b) As x gets very large the values of x3 become much larger than those of x x 2 and therefore we can “ignore” 2 the effect of x in the expression x3 x2 . Thus, Thus, we we can approximate the function to look like x like x 3 for very large values of x. x .
· ·
·
−
12
Chapter 1: Functions and Graphs c) Below is a graph of x3 x2 and x3 for 100 x 200. It is hard to distinguish between the two graphs confirming the conclusion reached in part b).
−
0.01
≤ ≤
0.005
8e+06 7e+06
–0.01
6e+06
–0.006
–0.002
5e+06
0.002
0.006 x
0 .0 1
–0.005
4e+06 3e+06
–0.01
2e+06
40. f ( f (x) = x = x 3 + 2x 2x
1e+06 100
38. f ( f (x) = x = x 4
120
140
160
x
180
200
− 10 10x x3 + 3x 3x2 − 2x + 7
a) For large values of x, x , x 4 will be larger than 10 10x x3 + 2 3x 2x + 7 since the second term is a third degree polynomial (compared to a fourth degree polynomial) and has terms being subtracted.
−
|−
|
b) Since the values of x4 “dominate” the function for very large values of x the function will look like x4 for very large values of x. x .
a) For x or x values values very close to 0, 2x 2x is larger than x than x 3 since for x value valuess less less than than 1 the higher higher the degree degree the smaller the values of the term. b) For x values very close to 0, the function will looks like 2x 2x since the x the x 3 term may be “ignored”. c) Below is a graph of x x 3 + 2x 2x and 2x 2x for 0.01 x 0.01. It is very difficult difficult to distinguis distinguish h between between the two graphs confirming our conclusion in part b).
−
0.02
c) Below is a graph of x x 4 10 10x x3 + 3x2 2x +7 and x and x4 for 100 x 200. The graphs graphs are close to each other confirming our conclusion from part b).
−
≤ ≤
≤ ≤
−
0.01
1.6e+09
–0.01
–0.006
–0.002
0.002
0.006 x
1.4e+09
0 .0 1
–0.01
1.2e+09 1e+09
–0.02
8e+08 6e+08
41. f ( f (x) = x = x 3
4e+08
−x
2e+08 100
120
140
x
160
180
39. f ( f (x) = x = x 2 + x
x(x
a) For values very close to 0, x is larger than x2 since for values of x x less than 1 x 1 x 2 < x. x.
−
b) For values of x x very close to 0 f 0 f ((x) looks like x like x since 2 the x the x can be “ignored”. “ignored”. c) Below is a graph of x x2 + x and x and x for for 0.01 x 0.01. It is very hard to distinguish between the two graphs confirming our conclusion from part b).
−
≤ ≤
f (x) x x 2 x(x 1) 1)(x 1)(x + 1) x x x 3
200
− −
= = = = = = =
0 0 0 0 0 1
−1
42. x = 2.359 43. x =
−1.831, 831, x x = = −0.856, and x and x = = 3.188
44. x = 2.039, and x and x = = 3.594 45. x = 10 10..153, x = 1.871, x = 0.821, x = x = 0.098, 098, x x = = 0.535, x 535, x = = 1.219, and x and x = = 3.297
−
46. y = 8.254 254x x 47. y =
−
− 5.457
−0.279 279x x + 4. 4.036
48. y = 1.004 004x x2 + 1. 1.904 904x x
− 0.601
−
−0.303,
Exercise Set 1.3
13
49. y = 0.942 942x x2
− 2.651 651x x − 27 27..943 50. y = 0.218 218x x3 + 0. 0.188 188x x2 − 29 29..643 643x x + 57. 57.067 51. y = 0.237 237x x4 − 0.885 885x x3 − 29 29..224 224x x2 + 165. 165.166 166x x − 210 210..135
√ x and y = √ x − 2
5 4 3
Exercise Set 1.3
y
1. y = x and y = x + 3
| |
4. y =
|
|
2
10
1
8 0
2
6
5. y =
y
4
x
6
8
10
2 x
4
6 2
y –10 –8
–6
–4
–2 0
2
2. y = x and y = x + 1
| |
|
4 x 6
8
2
10
–6
|
4
–4
0
–2
2
x4
6
2
x4
6
2
x4
6
–2
10
–4 8
–6 6
6. y =
y
3 x
4
6 2
y –10 –8
3. y =
–6
–4
–2 0
2
4 x 6
8
2
10
√ x and y = √ x + 1
4
–6
–4
0
–2
–2
5
–4 4
–6 3
7. y = −x2
y 2
6 1
y 0
2
4
x
6
8
4 2
10
–6
–4
–2
0 –2 –4 –6
14
Chapter 1: Functions and Graphs
8. y = −x3
12. y = |x1 | 6 y
10
4
y 5
2 –6
–4
0
–2
2
x4
6
–10 –8
–6
–4
–2
2
–2
8
10
–5
–4 –6
9. y =
4 x 6
–10
1 x2
2
−9 . It is important to note here that x 13. y = xx+3 that x = = in the domain of the plotted function.
10
−3 is not
10
8
8 6 y 4
6 y
2
4 –10 –8
–6
–4
2
–2 0 –2
2
4 x 6
8
10
–4 –6
–4
–2
2 x
4
–8 –10
1 10. y = x− 1
2
14. y = xx−−24 . Note: x = 2 is not in the domain of the plotted function.
10 8
10
6 y 4
8 6 y 4
2 –4
0 –2
–2
2 x
4
2 –10 –8
–4
–6
–4
–6
–2 0 –2
2
4 x 6
8
10
–4
–8
–6
–10
–8
√ 11. y = x
–10
3
1
15. y = xx−−11 . It is important important to note here that x = 1 is not in the domain of the plotted function.
4 y 2
–10 –8
–6
–4
–2 0
10 8
2
4 x 6
8
6 y 4
10
2
–2 –10 –8
–4
–6
–4
–2 0 –2 –4 –6 –8 –10
2
4 x 6
8
10
Exercise Set 1.3
15
2
−25 . No 16. y = xx+5 Note te:: x = plotted function.
− 5 is not in the domain of the
41.
10 8
42.
6 y 4
–6
–2 0 –2
–4
1 t2 3
=
4
1 1 (y +7) 4
= √ y1+7 4
= √ 1 3
1
w
−1
−4 5
t2 4 5
= w =
√ 4 5
w
√ √ 44. 165/2 = ( 16)5 = (4)5 = 1024 √ 45. 642/3 = ( 64)3 = (4)2 = 16 √ 46. 82/3 = ( 8)2 = (2)2 = 4 √ 47. 163/4 = ( 16)3 = (2)3 = 8 √ 48. 255/2 = ( 25)5 = (5)5 = 3125 43. 93/2 = ( 9)3 = (3)3 = 27
2 –10 –8
40. (y + 7)
2
4 x 6
8
10
3
–4
3
–6 –8
4
–10
√ 17. x3 = x ( ) √ 18. x5 = x ( ) √ 19. a3 = a( ) √ 20. b2 = b( ) = b ( √ = t( ) 21. t = t √ = c( ) 22. c = c 3 2
49. The domain consists of all x all x-values -values such that the denominator does not equal 0, that is x is x 5 = 0, which leads to x = 5. Therefore, the domain is x x = 5
5 2
5
2 4
4
23. √ 1 = 3
1 t( ) 4 3
4
t
24. √ 1 = 5
1
t( ) 6 5
t6
1 25. √ = t
1
m(
27. √ x1 +7 =
2 3
2 5
32. t = −2
34. y 35. b 36. b 37. e
1 2
)
1 2
)
(x3 +4)( 2 ) 1
1 2
= (x ( x3 + 4)−( ) 1 2
−2 3
−1 3
−1 5
1
5
t2
=
1 1 b3
= √ 1
1
= √ 1
−19
1
=
− 3)
17 6
54. The domain is the solution to 2x 2x 2x
b2
b2
=
6
e
6
m19
1 1 (x 3) 2
−6 ≥ 2x ≥ x ≥
− 6 ≥ 0. 0 6 3
55. To complete the table we will plug the given W given W values values into the equation
17
= √ 1
−
≥ 0 ≥ −4 ≥ −54
2
= √ 1
1 19 m6
−1 2
3
5
1 e
√ 1y 3
b5
=
53. The domain of a square root function is restricted by the value value where the radicant is positive. positive. Thus, Thus, the domain domain of f ( f (x) = 5x + 4 can be found by finding the solution to the inequality 5x 5x + 4 0.
x
=
=
−
= √ 1
2 5
1 2 y3
6
6
t
t
−
5x + 4 5x
2
=
−17
39. (x
5
52. Solving x Solving x 2 + 6x 6x + 5 = 0 leads to (x (x + 3)(x 3)(x + 2) = 0 which means the domain consists of all real numbers such that x = 3 and x and x = 2
≥
√ 2
=
5
38. m
3
√
5
y
= (x ( x2 + 7)−( )
7
31. y =
Which means that the domain is the set of all x -values such that x that x = 3 or x or x = 2
= m−( )
√ x √ = t
29. x = 1 7
1 2
= 0 = 0
So x = 3 an and x = 2
6 5
1
3
1 5
= t −( )
(x2 +7)(
28. √ x1 +4 =
− 5x + 6 (x − 3)(x 3)(x − 2)
4 3
1
2
x2
= t −( )
1 2
1 2
Theref Therefore ore,, the domain domain is
51. Solving for the values of the x in the denominator that make it 0.
= t −( )
1
t( )
26. √ 1m =
33. t
− 2. −∞ − ∪ − ∞
1 8
8
30. t
50. x + 2 = 0 leads to x = ( , 2) ( 2, ).
)
1 2
1 7
7
− { | }
3 5
= √ x1−3
T ( T (20) = (20)1.31 = 50. 50 .623 51 1.31 T ( T (30) = (30) = 86. 86 .105 86 1.31 T ( T (40) = (40) = 125. 125.516 126
≈ ≈ ≈
16
Chapter 1: Functions and Graphs T ( T (50) = (50)1.31 = 168. 168.132 168 1.31 T (1 T (100 00)) = (100 (100)) = 416. 416.869 417 1.31 T (1 T (150 50)) = (150 (150)) = 709. 709.054 709
b) y(2. (2.7) = 0. 0.73(7)3.63
≈ ≈ ≈
c)
5000 = 0.73(x 73(x)3.63 5000 = x3.63 0.73
Therefore the table is given by W T
0 0
10 20
20 51
30 86
40 126
50 168
100 417
≈ 853 853..156 156 k kgg.
5000
150 709
1 3.63
= x
0.73 11 11..393 393 m m
Now the graph
≈
59. Let V Let V be be the velocity of the blood, and let A be the cross sectional area of the blood vessel. Then
700 600
V =
500 y400
k A
Using V Using V = = 30 when A when A = = 3 we can find k find k..
300
k 3 (30) (30)(3 (3)) = k 90 = k 30
200 100 0
20
40
60
80 x
100
1 20
140
=
Now we can write the proportial equation
56. First find the constant of the variation. variation. Let N N represent the number of cities with a population greater than S. N =
k S
V =
90 A
we need to find A find A when V when V = 0. 0 .026
k 350000 (48)(3 (48)(3500 50000) 00) = k 16 1680 8000 0000 00 = k
90 A 0.026 026A A = 90 90 A = 0.026 = 3461.538 538 m m 2 0.026 =
48 =
So the variation equation is N = 16800000 . Now, we have have S to find N find N when S when S = = 200000.
60. Let V Let V be be the velocity of the blood, and let A be the cross sectional area of the blood vessel. Then
16800000 N = 200000 = 84 57.
x
V =
a) f (180) = 0.144(180)1/2 = 0.144(13. 144(13.41640786) 2 1.932 932 m m .
≈
b) f (170) = 0.144(170)1/2 = 0.144(13. 144(13.03840481) 1.878 878 m m 2 .
≈
c) The graph 5
k A
Using V Using V = = 28 when A when A = = 2.8 we can find k find k.. k 2.8 (28)(2. (28)(2.8) = k 78 78..4 = k 28 =
Now we can write the proportial equation 4
V = 3 y
we need to find A find A when V when V = 0. 0 .025
2
78 78..4 A 0.025 025A A = 78 78..4 78 78..4 A = 0.025 = 3136 m 2 0.025 =
1
0
58.
78 78..4 A
50
100 x
a) y(2. (2.7) = 0. 0.73(2. 73(2.7)3.63
150
≈ 26 26..864 864 k kgg.
20 0
Exercise Set 1.4
17 67. x = 2.6458 and x and x = =
61. 9 = 0 x 9 x(x + 7 + ) = x(0) x x2 + 7x 7x + 9 = 0 7 x =
68. x =
x+7+
= x = and x =
−2.6458
−2 and x and x = = 3
69. The function has no zeros 70. x = 1 and x and x = = 2
− ± 49 − 4(1)(9) 2 −7 ± √ 13
Exercise Set 1.4
2
2π 3
rad 1. (120o )( π180 o )=
−7 − √ 13
120
2
rad
1
−7 + √ 13
0.5
2
62. –1
1 1 1 = w w2 2 w w = 1 2 w w 1 = 0 1 w =
− − − −
=
0
–0.5
1
–0.5
± √ 1 + 4 √ 2 1± 5
–1 5π 6
rad 2. (150o )( π180 o )=
2
rad
1
63. P = P = 1000 1000tt5/4 + 14000
150
0.5
a) t = 37, P 37, P = = 1000(37)5/4 + 14000 = 105254. 105254.0514. 5/4 t = 40, P 40, P = = 1000(40) + 14000 = 114594. 114594.6744 50, P = = 1000(50)5/4 + 14000 = 146957. 146957.3974 t = 50, P b) Below is the graph of P P for 0
0.5
–1
≤ t ≤ 50.
0
–0.5
0.5
1
–0.5
140000 120000
–1
100000 80000
rad 3. (240o )( π180 o )=
60000
1
4π 3
rad
40000
0.5
20000 0
10
20
t
30
40
50
64. At most a function of degree n can have n y-intercepts. A polynomial of degree n can be factored into at most n linear terms and each of those linear terms leads to a yintercept. This is sometimes called the Fundamental Theorem of Algebra 65. A rational function is a function given by the quotient of two polynomial functions while a polynomial function is a function that has the form an xn + a n−1 xn−1 + + a1 x + a0 . Since every polynomial function can be written as a quotient of two other polynomial function then every polynomial function is a rational function.
·· · · ·
66. x = 1.5 and x and x = = 9.5
–1
–0.5
0
–0.5
240
–1
0.5
1
18
Chapter 1: Functions and Graphs
rad 4. (300o )( π180 o )=
–1
5π 3
o
o 8. ( 76π )( π180 rad ) = 210
rad
1
1
0.5
0.5
0
–0.5
0.5
1
–1
–0.5
300 o
o 9. ( 32π )( π180 rad ) = 270
1
1
0.5
0.5 –0.5
–1 –1
1
–1
rad 5. (540o )( π180 o ) = 3π rad
540
0.5
–0.5
210 –1
0
–0.5
–0.5
0.5
0.5
1
1 –0.5
–0.5 –1 –1
270 o
− 5π rad rad 6. ( 450o )( π180 o )= 2
o (3π)( π180 10. (3π rad ) = 540
−
1
1
0.5 –1
0.5
–0.5
0.5
1 540
–1
–0.5
0.5
1
–0.5 –0.5 –1 –1
–450 o
o 11. ( −3π )( π180 rad ) =
o 7. ( 34π )( π180 rad ) = 135
−60o
1
1
0.5
0.5
135
–1
–0.5
0
0.5
1
–1
–0.5
0
–0.5
–0.5
–1
–1
0.5
–60
1
Exercise Set 1.4
19
o 11π 11π 12. ( −15 )( π180 rad ) =
−132o
17. We need to solve θ1 = θ 2 + 2π 2 π(k) for k for k . If the solution is an integer then the angles are coterminal otherwise they are not coterminal.
1
π 2 π π 2π 1 2
0.5
–1
–0.5
0
0.5
− − −
1
–0.5 –132
3π + 2π 2π(k ) 2 = 2π(k ) =
= k = k
Since k is not an integer, we conclude that not coterminal.
–1
13. We need to solve θ solve θ 1 = θ 2 + 360(k 360(k) for k for k.. If the solution is an integer then the angles are coterminal otherwise they are not coterminal. 395 395 380 380 360 1.05
7π 6 2π 2π 2π 1
−
= k = k
−135o are
15. We need to solve θ solve θ 1 = θ 2 + 360(k 360(k) for k for k.. If the solution is an integer then the angles are coterminal otherwise they are not coterminal.
3π 4 π π 2π 1 2
= k = k o
16.
− −
= k = k
= 440 440 + 360( 360(k k) = 360(k ) = k = k
Since k Since k is not an integer, we conclude that 140o and 440o are not coterminal.
and −23π are
π 2
=
−5π + 2π 2π(k)
6 = 2π(k ) = k = k
7π 6
=
and −65π are
−π + 2π 2π(k )
4 = 2π(k) = k = k
Since k Since k is not an integer, we conclude that not coterminal. 21. sin 34 sin 34 o = 0. 0 .5592
140 140 300 300 360 1.61
= 2π(k )
20. We need to solve θ1 = θ 2 + 2π 2 π(k) for k for k . If the solution is an integer then the angles are coterminal otherwise they are not coterminal.
−
Since k is not an integer, we conclude that 15 and 395 are not coterminal.
are
− 32π + 2π 2π(k)
Since k is an integer, we conclude that coterminal.
= 107 + 360(k 360(k) = 360(k )
o
3π 2
19. We need to solve θ1 = θ 2 + 2π 2 π(k) for k for k . If the solution is an integer then the angles are coterminal otherwise they are not coterminal.
= 135 + 360(k 360(k) = 360(k)
Since k Since k is is an integer, we conclude that 225o and coterminal.
107 214 214 360 0.594
=
Since k is an integer integer,, we conclude conclude that coterminal.
14. 225 360 360 360 1
π 2 2π 2π 2π 1
= k
Since k is not an integer, we conclude that 15o and 395o are not coterminal.
and
18.
= 15 + 360( 360(k k) = 360(k )
= k
π 2
22. sin 82 sin 82 o = 0. 0 .9903 23. cos 12 cos 12 o = 0. 0 .9781 24. cos 41 cos 41 o = 0. 0 .7547 25. tan 5 tan 5 o = 0. 0 .0875 26. tan 68 tan 68 o = 2. 2 .4751
3π 4
and −4π are
20
Chapter 1: Functions and Graphs
27. cot 34 cot 34 o =
1 tan 34o
= 1. 1 .4826
28. cot 56 cot 56 o =
1 tan 56o
= 0. 0 .6745
29. sec 23 sec 23 o =
1 cos 23o
= 1. 1 .0864
30. csc 72 csc 72 o =
1 sin 72o
= 1. 1 .0515
54. 1.4 x 1.4 x = sin 25 sin 25 o x = 3.3127
sin 25 sin 25 o
31. sin( sin( π5 ) = 0.5878 32.
cos( cos( 25π )
33.
tan( tan( π7 ) =
34.
1 cot( cot( 311π ) = tan( π ) tan( 311
36.
55.
= 0. 0 .3090
35. sec( sec( 38π ) =
=
cos t =
0.4816
40 ) 60 t = 48 48..1897o t = cos−1 (
= 0. 0 .8665
1 cos( cos( 38π )
= 2. 2 .6131
1 csc( csc( 413π ) = sin( π ) sin( 413
= 1. 1 .2151
40 60
56. tan t =
37. sin(2 sin(2..3) = 0. 0.7457
20 25
20 ) 25 t = 38 38..6598o t = tan−1 (
38. cos(0 cos(0..81) = 0. 0.6895 39. t = sin = sin−1 (0. (0.45) = 26. 26.7437o 57.
40. t = sin = sin−1 (0. (0.87) = 60. 60.4586o
tan t =
41. t = cos = cos−1 (0. (0.34) = 70. 70.1231o
18 ) 9.3 t = 62 62..6761o t = tan−1 (
42. t = cos = cos−1 (0. (0.72) = 43. 43.9455o 43. t = tan = tan−1 (2. (2.34) = 66. 66.8605o 44. t = tan = tan−1 (0. (0.84) = 40. 40.0302o
58.
45. t = sin = sin−1 (0. (0.59) = 0. 0.6311
sin t =
= sin−1 (0. (0.26) = 0. 0.2630 46. t = sin
30 ) 50 36..8699o t = 36
48. t = cos = cos−1 (0. (0.78) = 0. 0.6761
59. We can rewrite 75o = 30 o + 45o then use a sum identity
49. t = tan = tan−1 (0. (0.11) = 0. 0.1096 50. t = tan = tan−1 (1. (1.26) = 0. 0.8999
cos( cos(A + B ) = cos Acos B sin Asin B cos 75 cos 75 o = cos(30 cos(30o + 45o ) = cos 30 cos 30 o cos 45 cos 45 o sin 30 sin 30 o sin 45 sin 45 o 3 1 1 1 = 2 2 2 2
−
51. x 40 x = 40 40sin sin 57 57 o x = 33 33..5468 =
√
= =
52. tan 20 tan 20
o
=
15 x
15 x = tan 20 tan 20 o x = 41 41..2122 53. cos 50 cos 50
o
30 50
t = sin−1
47. t = cos = cos−1 (0. (0.60) = 0. 0.9273
sin 57 sin 57 o
18 9.3
=
15 x
15 cos 50 cos 50 o x = 23 23..3359 x =
− · √ − · √
√ 3 1 √ − √ 2 2 2 2 −1 +√ √ 3 2 2
60. The x The x coordinate can be found as follows x cos 20 cos 20 o = 200 x = 200c0s 20 o = 187.939 The y The y coordinate y 200 y = 200sin 20 sin 20 o = 68 68..404
sin 20 sin 20 o
=
Exercise Set 1.4
21
61. Five miles is the same as 5 5280 f t = 26400 f t. The difference in elevation, y elevation, y,, is
b)
·
62. First a grade of 5% means that the ratio of the y the y coordinate coordinate to the x coordinate is 0.05 since since tan t = xy . This This means means that x that x = = 0.y05 = 20y 20 y The distance from the base to the top is 6 5280 5280 f f t = 31680 f 31680 f t. Using the pythagorean theorem
h(1
·
x2 + y 2 (20y (20y )2 + y 2 401yy 2 401
= 3168 3168002 = 10 1003 0362 6224 2400 00 = 10 1003 0362 6224 2400 00
y
63.
1003622400
=
−
−
= 555.567 f t 67.
401 = 1582.02 02 f f t
a) When we consider the two triangles we have a new triangle that has three equal angles which is the definition of an equilateral triangle. b) The short leg of each triangle is given by 2sin 2sin(30) (30) = 1 2( 2 ) = 1
a)
c) The long leg (L) is given by x cos 40 cos 40 o = 150 x = 150cos 40 cos 40 o = 114.907
22 = L2 + 12 4 1 = L2 3 = L
−√
d) By considering all possible ratios between the long, short and hypotenuse of small triangles we obtain the trigonometric functions of π6 = 30 o and π3 = 60 o
b) y 150 y = 150sin 40 sin 40 o = 96 96..4181
sin 40 sin 40 o
=
68.
c)
a) Since the triangle has two angles equal in magnitude it should have two sides that are equal as well. (It is an isosceles triangle) b)
z2
(x + 180)2 + y 2 (114.907 + 180)2 + (96. (96.4181)2 9626 962666.58866 96266..58866 96266 310.268
= = = = =
z2 z
h2 = 11 + 11 h2 = 2 h = 2
√
√
c) Since the hypotenuse is known then we can use the figure to find the trigonometric functions of π4 = 45o using the ratios of the sides of the triangle.
64. 77000 200 sec 60 sec 60 o 5000000 15400000 = 5000000cos 5000000 cos 60 60 o = 6.16 16 cm/sec cm/sec
v
=
·
·
65. 77000 100 sec 65 sec 65 4000000 7700000 = 4000000cos 4000000 cos 65 65 o = 4.55494 55494 cm/sec
v
66.
h 1012 + x h = tan(24 tan(24o )(1012 + x) h = (101 (1012) 2)tan tan(24 (24o ) + x tan(24 tan(24o ) h h = (101 (1012) 2)tan tan(24 (24o ) + tan(24 tan(24o ) tan(67 tan(67o ) tan(24 tan (24o ) = (101 (1012) 2)tan tan(24 (24o ) tan(67 tan(67o ) (1012)tan (1012)tan(24 (24o ) h = tan(24 tan(24o ) 1 tan(67 tan(67o ) tan(24 tan(24o ) =
y sin 4 sin 4 = 26400 y = 2640 264000sin 4 sin 4 o = 1841.57 57 f f t o
a) tan(67 tan(67o ) =
=
h x
so, x so, x = =
·
h tan(67 tan(67o )
·
o
69.
a) The tangent of an angle is equal to the ratio of the opposite side to the adjacent side (of a right triangle), and for the small triangle that ratio is 75 . b) For the large right triangle, the opposite side is 10 and the adjacent side is 7+7 = 14. Thus the tangent 10 is 14 c) Because the trigonometric functions depend on the ratios ratios of the sides and not the size of triangle. triangle. Note that the answer in part b) is equivalent to that in part a) even though the triangle in part b) was larger that that used in part a)
70. Let (x, (x, y ) be a non-origin point that defines the terminal side of an angle, t, and let r let r = x2 + y2 be the distance from the origin to the point (x, (x, y ). Then the trigonometric function are defined as follows:
22
Chapter 1: Functions and Graphs sin t = t = yr and csc and csc t = t = yr (y = 0) x cos t = t = r and sec and sec t = t = xr (x = 0) and y tan t = t = x (x = 0) and cot and cot t = t = xy (y = 0) From the above definitions and recalling that the reciprocal of a non zero number x number x is given by x1 we show that 1 sin t = t = csc t
cos t = t =
1 sec t
tan t = t =
1 cot t
x) e) sin( sin(s + t) = (w+ = w + x. Using the results we have 1 obtained from previous parts we can conclude sin( sin(s + t) = w = w + + x = sin = sin((s)cos( cos(t) + cos( cos(s)sin( sin(t)
73.
a) sin( sin(t) =
= = = =
− − −
y/r x/r y x r r y r r x y x tan t
= = = =
d) sin( sin(r) = uz , which means z means z = sin = sin((r)u. Using results from part a) and part b) we get sin get sin((r) = sin( sin(s) and u and u = = sin sin((t), therefore z therefore z = sin = sin((s)sin( sin(t) z) e) cos( cos(s + t) = (y− = y z. Replacing ur results for y for y 1 and z and z we get cos get cos((s + t) = cos = cos((s)cos( cos(t) sin( sin(s)sin( sin(t)
−
74.
x/r y/r x y r r x r r y x y cot t
a) sin( sin(t) =
u 1
− t) = u1 = u = u = sin sin((t) π v b) sin( sin( 2 − t) = 1 = v = v = cos cos((t)
75. Use cos Use cos 2 t + sin2 t = 1 as follows cos2 t + sin2 t = 1 cos2 t sin 2 t 1 + = 2 2 cos t cos t cos2 t t 1 + tan = sec2 t
÷ ·
76. cos2 t + sin2 t = 1 cos2 t sin 2 t 1 + = sin2 t sin2 sin2 t cot2 t + 1 = csc2 t
= u
b) Consider the triangle made by the sides v , w, and y . The angl anglee vw has a value of 90 r (completes a straight straight angle). angle). The sum of angles angles in any triangle triangle is 180. Therefore
−
s + 90 + (90 r) = s + 180 r = s r = s =
− − −
−
a) cos( cos( π2
Thus cos t = cot t sin t 72.
180 180 0 r
means y = cos = cos((s)v . c) cos( cos(s) = vy , which means y But But from from part part a) v = cos( cos(t), theref therefore ore y = cos( cos(s)cos( cos(t)
÷ ·
and =
= v/ v /
s + 90 + (90 r) = s + 180 r = s r = s =
Thus sin t = tan t cos t
cos t sin t
v 1
−
71. =
= u, u, and cos and cos((t) =
b) Consider the triangle made by the sides v, w, and y. The angl anglee vw has a value of 90 r (completes a straight straight angle). angle). The sum of angles angles in any triangle triangle is 180. Therefore
and
sin t cos t
u 1
180 180 0 r
77. Let 2t 2t = t = t + t sin( sin(a + b) = sin( sin(a)cos( cos(b) + cos( cos(a)sin( sin(b) sin(2 sin(2tt) = sin( sin(t + t) = sin( sin(t)cos( cos(t) + cos( cos(t)sin( sin(t) = 2sin( sin(t)cos( cos(t) 78.
a) cos(2 cos(2tt) = cos( cos(t + t) = cos( cos(t)cos( cos(t) sin( sin(t)sin( sin(t) 2 2 = cos (t) sin (t)
c) sin( sin(s) = wv which means that w that w = sin( sin(s)v . cos( cos(t) = v = v 1 Thus, w Thus, w = sin = sin((s)v = sin = sin((s) cos( cos (t) d) sin( sin(t) = u1 = u and cos( cos(r) = ux which means that x = u = u cos( cos(r). In part b) we showed that r = s therefore cos( cos(r) = cos( cos(s). So, x So, x = = u u cos( cos(r) = sin = sin((t)cos( cos(s)
−
−
b) cos(2 cos(2tt) = cos2 (t) sin2 (t) = cos2 (t) (1 cos2 (t)) = 2cos2 (t) 1
− − − −
Exercise Set 1.5
23
c)
Exercise Set 1.5
cos(2 cos(2tt) = cos2 (t) sin2 (t) = (1 sin2 (t)) sin2 (t) = 1 2sin2 (t)
−
− −
−
1. 5π/4 π/4 1
79. Using the result from Exercise 78 part (c) cos(2 cos(2tt) cos(2 cos(2tt) 1 cos(2 cos(2tt) 1 2 1 cos(2 cos(2tt) 2
−
−
− −
= 1 2sin2 (t) = 2sin2 (t)
− −
0.5
= sin2 (t) –1
5/4*Pi
−
82.
1
–0.5
cos(2 cos(2tt) = 2cos2 1 cos(2 cos(2tt) + 1 = 2cos2 (t) cos(2 cos(2tt) + 1 = cos2 (t) 2 81.
0.5
= sin (t)
80.
p
0
–0.5
2
q
r
–1
2.
− 5π/6 π/6 1
s
a) V (0) V (0) = sin = sin (0)sin (0)sin (0)sin (0)sin (0)sin (0)sin (0) = 0 V (1) V (1) = sin = sin p( π2 )sinq ( π2 )sinr ( π2 )sins ( π2 ) = 1 b) When h When h = = 0 the volume of the tree is zero since there is no height and therefore the proportion of volume under under that heigh heightt is zero. zero. While While at the top of the tree, h tree, h = 1, the proportion of volume under the tree is 1 since the entire tree volume falls below its height.
0.5
–1
a)
–0.5
√
√
×
√
3
4
1
–0.5
–5/6*Pi
π π 48.646 V (0 V (0..5) = sin−3.728 ( )sin48. ( ) 4 2 2 π π 123.208 86.629 sin−123. ( )sin86. ( ) 2 2 2 2 = 0.8208
0. 5
–1
3.
− π
1
b) V ( V (h) 0.5
1 0.8
-Pi
–1
–0.5
0.5
1
0.6
–0.5 0.4
–1
0.2
4. 2π 0
0.2
0.4
h
0.6
0.8
1
1
c) The result from part b) agrees with the definition of V ( V (h) since the values of V ( V (h) are limited between 0 and 1.
0.5
83. 1 π π 74.831 V ( V ( ) = sin−5.621 ( )sin74. ( ) 2 4 2 2 π π 195.644 138.959 sin−195. ( )sin138. ( ) 2 2 2 2 = 0.8219
√
×
√ 3
√
–1
–0.5
0.5 –0.5
4
–1
1
–2*Pi
24
Chapter 1: Functions and Graphs 26. t = sin = sin −1 ( 1) =
5. 13 13π/ π/66
−
3π 2
27. 2t = sin = sin −1 (0) = nπ = nπ so t so t = =
1
0.5
1
−√
–0.5
− − −
–1
6.
√ − −√
π ) = 3 3 π 3 sin( sin(t + ) = 3 2 π 3 t+ = sin−1 ( ) 3 2 π π t = + 2nπ 2nπ 3 3 2π = + 2nπ 2nπ 3 and π 4π 4 π t = + + 2nπ 2nπ 3 3 = π + 2nπ 2nπ
2sin( sin(t + –0.5
nπ 2
28.
13/6*Pi
0.5
–1
+ 2nπ 2nπ
− 7π/4 π/4
−
1 –7/4*Pi
0.5
29. –1
–0.5
0.5
π 1 ) = 4 2 π 1 3t + = cos−1 ( ) 4 2 π 2π 2 π 3t = + + 2nπ 2nπ 4 3 5π 3t = 2nπ 12 5π 2 t = nπ 36 3 and π 4π 4 π 3t = + + 2nπ 2nπ 4 3 13 13π π 3t = + 2nπ 2nπ 12 13 2 t = π + nπ 36 3
cos(3 cos(3tt +
1
−
–0.5
−
–1
7. cos(9 cos(9π/ π/2) 2) = 0
√ 12 8. sin(5 sin(5π/ π/4) 4) = − 9. sin( sin( 5π/6) π/6) = −21
− −1 10. cos( cos(−5π/4) π/4) = √ 2 11. cos(5 cos(5π π) = −1
−
12. sin(6 sin(6π π) = 0
− −√ 3 √ 14. tan( tan(−7π/3) π/3) = − 3 15. cos 125 cos 125 o = −0.5736 13. tan( tan( 4π/3) π/3) =
30. cos(2 cos(2tt) = 0 2t = cos−1 (0) π 2t = + 2nπ 2nπ 2 π t = + nπ 4 and 3π 2t = + 2nπ 2nπ 2 3π t = + nπ 4
16. sin 164 sin 164 o = 0. 0 .2756 17. tan( tan( 220o ) =
− −0.8391 18. cos( cos(−253o ) = −0.2924 19. sec 286 sec 286 o =
1 cos 286o
= 3. 3 .62796
20. csc 312 csc 312 o =
1 sin 312o
=
−1.34563
21. sin(1 sin(1..2π) =
−0.587785 22. tan( tan(−2.3π) = −1.37638 23. cos( cos(−1.91) = −0.332736 24. sin( sin(−2.04) = −0.891929 25. t = sin = sin−1 (1/ (1/2) =
−
π 6
+ 2nπ 2nπ and
31.
5π 6
+ 2nπ 2nπ
cos(3 cos(3tt) = 1 3t = cos−1 (1) 3t = 2nπ 2 t = nπ 3
Exercise Set 1.5
25
32.
36. t 2cos( cos( ) 2 t cos( cos( ) 2 t 2 t 2
= =
√ − 3 −√ 3
sin2 t 2sin t 3 (sin t 3)(sin 3)(sin t + 1) The only solution solution comes from sin t + 1 t
−
2
√
− 3) = cos−1 ( =
t = and t = 2 t =
2 5π + 2nπ 2nπ 6 5π + 4nπ 4nπ 3
− −
− −
39. y = 5cos( cos(t/2) t/2) + 1 amplitude = 5, period = 2π = 4π 4 π, mid-line y mid-line y = 1 maximum maximum = 1 + 5 = 6, minimum minimum = 1 5 = 4 1 2
−
−
40. y = 3sin( sin(t/3) t/3) + 2 amplitude = 3, period = 2π = 6, mid-line y mid-line y = 2 maximum maximum = 2 + 3 = 5, minimum minimum = 2 3 = 1 1 3
− 12
1 t = sin−1 ( ) 2 7π t = + 2nπ 2nπ 6 and 11 11π π t = + 2nπ 2nπ 6
−
34.
−
−
− −
− −
− −
= 0 = sin−1 ( 1) 3π = + 2nπ 2nπ 2
38. y = 3cos 2 cos 2tt 3 amplitude = 3, period = 22π = π, π , mid-line y mid-line y = 3 maximum = 3 + 3 = 0, minimum minimum = 3 3 = 6
2sin2 t 5sin t 3 = 0 (2sin (2sin t + 1)(sin 1)(sin t 3) = 0 The only solution comes from (2sin (2sin t + 1) = 0
cos2 x + 5cos 5cos x 2 cos x + 5cos 5cos x 6 (cos x + 6)(cos 6)(cos x 1) The only solution comes from cos x 1 x x
= 0 = 0
−
7π + 2nπ 2nπ 6 7π + 4nπ 4nπ 3
sin t =
−
37. y = 2sin 2 sin 2tt + 4 amplitude = 2, period = 22π = π, π , mid-line y mid-line y = 4 maximum maximum = 4 + 2 = 6, minimum minimum = 4 2 = 2
33.
−
−
−
−
41. y = 21 sin(3 sin(3tt) 3 amplitude = 21 , period = 23π , mid-line y mid-line y = maximum 3 + 21 = −25 , minimum = 3
−
−3 − − 21 = −27
−
42. y = 21 cos(4 cos(4tt) + 2 amplitude = 21 , period = 24π = π2 , mid-line y mid-line y = 2 maximum = 2 + 21 = 25 , minimum = 2 21 = 23
−
43. y = 4sin( sin(πt) πt) + 2 amplitude = 4, period = 2ππ = 2, mid-line y mid-line y = 2 maximum maximum = 2 + 4 = 6, minimum minimum = 2 4 = 2
−
−
= 6 = 0 = 0
44. y = 3cos(3 cos(3πt πt)) 2 amplitude = 3, period = 32ππ = 32 , mid-line y mid-line y = 2 maximum = 2 + 3 = 1, minimum minimum = 2 3 = 5
= 0 = cos−1 (1) = 2nπ
45. The maximum is 10 and the minimum is -4 so the amplitude is 10−2(−4) = 7. The mid-line mid-line is y = 10 7 = 3, and the period is 2π 2π (the distance from one peak to the next one) which means that b that b = 22ππ = 1. From the information above, and the graph, we conclude that the function is
35.
− −
− − − − −
y = 7sin t + 3 cos2 x + 5cos 5cos x = 6 2 cos x + 5cos 5cos x + 6 = 0 5 cos x =
−
= =
± 25 − 4(1)(6)
− 2 −5 ± 1 2 −5 − 1 = −3 2
and =
−5 + 1 = −2 2
Since both values are larger than one, then the equation has no solutions. solutions.
46. The maximum is 4 and the minimum is -1 so the amplitude is 4−(2−1) = 25 . The midmid-li line ne is y = 4 25 = 23 , and the period is 4π 4π which means b = π2 . From the information information above, and the graph, we conclude that the function is
−
y =
5 3 cos( cos(t/2) t/2) + 2 2
47. The maximum is 1 and the minimum is -3 so the amplitude is 1−(2−3) = 2. The mid-l mid-line ine is y = 1 2 = 1, and the period is 4π 4π which means b = π2 . From the information information above, and the graph, we conclude that the function is
−
y = 2cos( cos(t/2) t/2)
−1
−
26
Chapter 1: Functions and Graphs
48. The maximum is -0.5 and the minimum is -1.5 so the amplitude amplitude is −0.5−2(−1.5) = 21 . The The midmid-li line ne is is y = 0.5 −21 = 1, and the period is 1 which means that b = 21π = 2π 2 π. From the information above, and the graph, we conclude conclude that the function function is
− −
−
y =
1 sin(2 sin(2πt πt)) 2
−1
59. The amplitude is given as 5.3. b = f 2π where f f is the frequency, b = 0.172 2π = 1.08071, k = 143. Therefore, Therefore, the function is
· ·
·
p( p(t) = 5.3cos(1 cos(1..08071 08071tt) + 143 60. p(t) = 6.7cos(0 cos(0..496372 496372tt) + 137 61. x = cos = cos(140 (140o ), y = sin = sin(140 (140o ), ( 0.76604 76604,, 0.64279)
−
49. 0.808 808cos 40 o cos 30 R = 0.339 + 0. cos 40 cos 30 o 0.196 196sin sin 40 40 o sin 30 sin 30 o 0.482 482cos cos 0 0 o cos 30 cos 30 o = 0.571045 megajoules/m2
−
−
62. ( 0.17365 17365,, 0.98481)
−
−
= cos(( 95π ), y = sin = sin(( 95π ), (0. (0.80902 80902,, 0.58779) 63. x = cos
−
64. ( 0.22252 22252,, 0.97493)
−
−
65. Rewrite 105o = 45 o + 60o and use a sum identity.
50. o
sin 105 sin 105 o
o
R = 0.339 + 0. 0.808 808cos cos 30 30 cos 20 cos 20 o 0.196 196sin sin 30 30 sin 20 sin 20 o 0.482 482cos cos 180 180 o cos 20 cos 20 o = 1.12788 megajoules/m2
−
−
= sin(45 sin(45o + 60o ) = sin 45 sin 45 o cos 60 cos 60 o + cos 45 cos 45 o sin 60 sin 60 o 1 1 1 3 = + 2 2 2 2 1 3 = + 2 2 2 2 1+ 3 = 2 2
√ ·
51.
√
R = 0.339 + 0. 0.808 808cos cos 50 50 o cos 55 cos 55 o 0.196 196sin sin 50 50 o sin 55 sin 55 o 0.482 482cos cos 45 45 o cos 55 cos 55 o = 0.234721 megajoules/m2
−
√ √
−
66.
52.
cos 165 cos 165 o
−
53. Period is 5 so b = 25π , k = 2500, a = 250. Therefore, Therefore, the function is 2πt V ( V (t) = 250cos 250cos + 2500 5 54. Period is 2 sp b = 22π = π, π , a = 3400 2 = 1700, k = 1700 + 1100 = 2800. Therefore Therefore,, the function function is V ( V (t) = 1700cos 1700cos πt + 2800 55. Since our lungs increase and decrease as we breathe then there is a maximum and minimum volume for the air capacity pacity in our lungs. We have a regular period of time at which which we breathe breathe (inhale and exhale). exhale). These facotrs facotrs are reasons why the cosine model is appropriate for describing lung capacity. capacity.
67.
a) From the graph we can see that the point with angle t has an opposite x opposite x and y coordinate than the point with angle t angle t + π. Since the x the x coordinate corresponds to the cos the cos of of the angle which the point makes and the y coordinate corresponds to the sin the sin of of the angle which the point makes it follows that sin that sin((t + π + π)) = sin( sin(t) and cos and cos((t + π) = cos( cos(t). b)
−
π + 36 36..1 12
57. The frequency frequency is the reciprocal reciprocal of the period. Therefore, Therefore, b 880π 880π f = 2π = 2π = 440 440 H H z 58. f =
440π 440π 2π =
220 H 220 H z
−
−
sin( sin(t + π) = = = and cos( cos(t + π) = = =
56. The minimum is 35. 35 .33 and the maximum is 36. 36.87 so the 36.87−35. 35.33 amplitude is 36. = 0 . 77. The period pe riod is is 24 so b = 2 2π π = , k = 36 36. . 87 0 . 77 = 36 36. . 1. Thus, the function is 24 12 T ( T (t) = 0.77 77cos cos
= cos(120 cos(120o + 45o ) = cos 120 cos 120 o cos 45 cos 45 o sin 120 sin 120 o sin 45 sin 45 o 1 1 3 1 = 2 2 2 2 1 3 = 2 2 2 2 1+ 3 = 2 2
− √ − · √ − · √ √ − √ − √ √ − √
R = 0.339 + 0. 0.808 808cos cos 50 50 o cos 0 cos 0 o 0.196 196sin sin 50 50 o sin 0 sin 0 o 0.482 482cos cos 0 0 o cos 0 cos 0 o = 0.858372 megajoules/m2
−
√ · √ √
√
sin tcos π + cos tsin π sin t 1 + cos t 0 sin t
−
·−
·
cos tcos π sin tsin π cos t 1 sin t 0 cos t
−
− ·− −
c) tan( tan(t + π) =
sin( sin(t + π) cos( cos(t + π)
·
Exercise Set 1.5
27
−sin t −cos t
=
n/12−1 2n/12
n/12−1 2n/12 Comparing exponents we can conclude that n 1 =1 12 n = 2 12 n = 24
sin t cos t = tan t =
68.
a) The amplitude could be thought of as half the difference ference between between the maximum maximum and minimum minimum,, a = max−min , whic w hich h impli i mplies es that 2 = a max min. min. k is 2 the average mean of the maximum and the minimum, max+min k = max+min , which which implies implies that 2k = max + max + min min.. 2 Solving the system of equations above for max and min gives min gives the desired results.
−
−
There are 24 notes above A above middle C. 76. n/12 880(2n/12 )π 2π
b) The average mean of the maximum and minimum, using the results from part a), implies that the mid)+(k−a) line equation is y is y = (k+a)+(k = 22k = k. k . 2
n/12−1 2n/12
c) Half the difference between the maximum and minimum, using the results from part a), implies that the amplitude is (k+a)−2 (k−a) = 22a = a. a . 69.
(
n 12
−
n/12−1 2n/12 1)ln 1)ln(2) (2) = ln = ln(1 (1..5)
n 12
n/12 880(2n/12 )π 2π
asin( asin(bt) bt) + k
n/12−1 2n/12
g(t) (
n 12
−
n/12−1 2n/12 1)ln 1)ln(2) (2) = ln = ln(2 (2..5)
n 12
71. Since at the base, L base, L is is small, T small, T is is large, and d and d is is large, then the basilar membrane is affected mostly by high frequency sounds.
·
440
78.
a) Left to the student
− 21
−
79.
a) Left to the student b) y = 21 cos(2 cos(2tt) +
= 1760
c) We use the double angle identity obtained in Exercise 79 of Section 1.4 and solve for sin2 (t) to obtain the model in part b).
n/12 for 880(22π )π
n/12 880(2n/12 )π 2π
=
There are 28 notes above A above middle C
= 261 261..626
75.
2200 880 = 2.5 =
b) y = 21 cos(2 cos(2tt)
74. From the equation, n equation, n has to be 12 in order to equal 880. There are 12 notes above A above middle C.
= 2200
ln(2 ln(2..5) ln(2) ln(2) n ln(2 ln(2..5) = +1 12 ln(2) ln(2) ln(2 ln(2..5) n = 12 +1 ln(2) ln(2) n = 27 27..86314
− 1
70. Since at the apex, L apex, L is is large, T large, T is is small, and d and d is is small, then the basilar membrane is affected mostly by low frequency sounds.
880 2−9/12 π 2π
77.
73. f =
=
There are 19 notes above A above middle C.
asin[ asin[b(t + 2π/b 2π/b)] )] + k asin( asin(bt + 2π 2π ) + k
c) Since the function evaluated at t at t + 2π/b has π/b has the same value as the function evaluated at t and 2π/b 2π/b = 0 then t + 2π/b is evaluated after t. Since Since we have have a periodic function in g in g((t) it follows that the period of the function is implied to be 2π/b 2 π/b..
880π 880π 2π =
1320 880 = 1.5 =
b)
72. f =
= 1320
ln(1 ln(1..5) ln(2) ln(2) n ln(1 ln(1..5) = +1 12 ln(2) ln(2) ln(1 ln(1..5) n = 12 +1 ln(2) ln(2) n = 19 19..01955
− 1
a) Since the radius of a unit circle is 1, the circumference of the unit circle is 2π 2 π. Therefore Therefore any point point t + 2π will have exactly the same terminal side as the point t point t,, that is to say that the points t points t and and t t + 2π are coterminal on the unit circle. Therefore, sin Therefore, sin t = t = sin sin((t + 2π 2π ) for all numbers t numbers t..
g (t + 2π/b 2π/b)) = = from part a) = by definition g (t + 2π/b 2π/b)) =
1760 880 = 2 =
1 2
c) We use the double angle identity obtained in Exercise 79 of Section 1.4 and solve for cos2 (t) to obtain the model in part b).
28 80.
Chapter 1: Functions and Graphs 9. y = 3x3
a) Left to the student b) y = sin = sin(2 (2tt)
− 6x + 1 300
c) We use the double angle identity for sin for sin(2 (2tt) obtained in Exercise 77 of Section 1.4 to obtain the model in part b). 81.
200 100
a) Left to the student b) Left to the student
–4
0
–2
c) The horizontal shift moves every point of the original graph π4 units to the right. 82.
2 x
4
–100 –200
a) Left to the student
–300
b) Left to the student c) The horizontal shift moves every point of the original graph π3 units to the left.
10. y = x + 1
|
| 6
83. Left to the student
5
84. Left to the student
4
85. Left to the student
3
Chapter Review Exercises 2
1.
a) 100 live births per 1000 women
1
b) 20 years old and 30 years old 2. f ( f ( 2) = 2( 2)2
− − (−2) + 3 = 13
−
–4
3.
11. f ( f (x) = (x f (1 f (1 + h) = = = =
2(1 + h)2 (1 + h) + 3 2(1 2(1 + 2h + h2 ) 1 h + 3 2 + 4h + 2h 2h2 1 h + 3 2h2 + 3h 3h + 4
−
− 2)2
4
2
6. = (1 = (h = h2
− (2 − h))2 − 1)2 − 2h + 1
0
–1
1
2
–4
2 x
–2
0 –2
20
–4
15
–6
10 –8
5
–1
0
13. 1
x
2
5
4
−
25
–2
3
2
− 4)2 = ( −3)2 = 9 8. f ( f (x) = 2x2 + 3x 3x − 1
–3
x
16 12. f ( f (x) = xx− importantt to note note that x = 4 does +4 . It is importan not belong to the domain of the plotted function.
7. f (4) f (4) = (1
–4
4
6
− (0) + 3 = 3 5. f ( f (−5) = (1 − (−5))2 = (1 + 5)2 = 6 2 = 36 − h)
2 x
8
− − − −
4. f (0) f (0) = 2(0)2
f (2 (2
0
–2
3
a) f (2) (2) = 1. 1.2 b) x =
−3
4
Chapter Review Exercises 14. x =
−2
29 20. x2 7x + 12 (x 3)(x 3)(x 4) x 3 x Or x 4 x
2
1
–4
= = = =
− − − −
–2
2 x
4
= 0 = 4
−
–1
0 0 0 3
21. x2 + 2x 2x = 8 x2 + 2x 2x 8 (x + 4)(x 4)(x 2) x+4 x Or x 2 x
15. y = 4
− 2x 10 8
4
x2 + 6x 6x
0
–1
1
2
x
− 20
= 0
x =
3
–2
=
− 7 2−5 16. m = 4− −(−7) = 11
=
y
− y1 y − (−2) y y
= m(x x1 ) 7 = (x (x 4) 11 7x 28 = + 11 11 7x 6 = + 11 11
− − − − − 2 −
y
− y1 y − 11 y y
−
− x1) 1 8(x − ) 2 8x − 4 + 11
x + 5x 5x + 4 (x + 1)(x 1)(x + 4) x+1 x Or x+4 x
√
29
= = = = = =
0 0 0 0 0
−3
= 0 = 1 = 0 = 1
−
24.
19. 2
− − − −
−
= = 8x + 7
− 16 , y-intercept y -intercept (0, (0, 3)
2
−3 ±
x3 + 3x 3x2 x 3 x2 (x + 3) (x + 3) (x + 3)(x 3)(x2 1) (x + 3)(x 3)(x 1)(x 1)(x + 1) x+3 x Or x 1 x Or x+1 x
= m(x =
−6 ± √ 36 + 80 2 √ −6 ± 2 29
23.
17. Use the slope-point equation
18. Slope =
= 0 = 2
22. x2 + 6x 6x = 20
2 –2
−
−
6
–3
= 0 = 0 = 0 = 4
− −
–2
= 0 = 0 = 0 = 1
−
= 0 = 4
−
x4 + 2x 2x3 x 2 x3 (x + 2) (x + 2) (x + 2)(x 2)(x3 1) x x
−
− − −
= 0 = 0 = 0 = 2 = 1
− −
25. Using the points (one could use any two points on the line) (0, (0, 50, and (4, (4, 350) the rate of change is 350 50 300 = = 75 pages pages per per day day 4 0 4
− −
30
Chapter 1: Functions and Graphs
26. The rate of change is
36. tan(7 tan(7π/ π/4) 4) =
20 100 80 = = 12 0 12
− −
−
−20 meters per second second
−1
37. x 127 x = 127 sin(70 sin (70o ) = 119.341
3
sin(70 sin(70o ) =
27. The variati variation on equation equation is M = kW , kW , with with k constant. When W When W = = 150, M 150, M = = 60 means 60 = k(150) 60 = k 150 2 = k 5
38. t = sin = sin −1 (1) =
3
40. 2t = cos = cos−1 (2), No solution 41.
− π4 ) π cos2 (2t (2t − ) 4 π cos2 (2t (2t − ) 4 π cos(2 cos(2tt − ) 4
12 12cos cos2 (2t (2t
2 M = W 5 2 = (210) 5 = 84 84 l lbs bs
−x−7 =0 =
30.
10
√ y
=
2t =
x = x3/20
√
t =
3
− −
4
2
–2
2
x 4
6
–2
a) m =
− = −
18 14
G
=
(2 sin (2 sin((t)
9 7
− 74
=
G = G = b) G(18) = 79 (18) + G(25) = 79 (25) +
35. cos( cos( π) =
3/4 cos− ( 3/4)
±
1
π + 2nπ 2nπ 6 π π + + 2nπ 2nπ 6 4 5π + 2nπ 2nπ 12 5 pi + nπ 24
−
− −
42.
92 74 23 9
34. sin(2 sin(2π/ π/3) 3) =
9 12 3 4
and π 2t = cos−1 ( 3/4) 4 π π 2t = + 2nπ 2nπ 4 6 π π 2t = + + 2nπ 2nπ 6 4 π 2t = + 2nπ 12 pi t = + nπ 24
32.
−
=
6
31. 272/3 = ( 27)2 = 3 2 = 9
33.
=
− −
√ 3
20
= 9
Two solutions π 2t = 4 π 2t = 4 2t =
± √ 1 + 140 √ 10 1 ± 141 1
x =
29. y1/6 =
+ 2nπ 2nπ
√ 39. t = tan = tan −1 ( 3) = π + nπ
Find M Find M when W when W = = 210
28. 5x2
π 2
√ 3 2
−1
437 7 437 7
− 1)(sin 1)(sin((t) + 4)
9 (x 9) 7 9 81 x + 74 7 7 9 437 x+ 7 7
− −
= 0 sin( sin(t) + 4 = 0 sin( sin(t) = 4 No solu solutio tion n 2 sin( sin (t) 1 = 0 1 sin( sin(t) = 2 t = sin−1 (1/ (1/2) π t = + 2nπ 2nπ 6 5π t = + 2nπ 2nπ 6
−
= 85. 85 .6 = 94. 94 .6
−
43. y = 2 sin( sin (t/3) t/3) 4 2π amplitude = 2, period = (1/ 6π (1/3) = 6π mid-line y mid-line y = 4, max = 4 + 2 = 2, min =
− −
−
−4 − 2 = −6
Chapter 1 Test
31
44. y = 21 cos(2 cos(2πt πt)) + 3 amplitude = 21 , period = 22ππ = 1 mid-line y mid-line y = 3, max = 3 + 21 = 27 , min = 3
6. m = 10−−3(−−25) = 3 7. Use the points (0, (0, 30) and (3, (3, 9) − 21 Average rate of change = 93−−30 0 = 3 = 7 The computer loses $700 of its value each year.
− 21 = 25
−
5 1 2
45. Amplitude = − = 22,, period period = π = π,, mid-line value = 3 y = 2sin(2 sin(2tt) + 3
46. Amplitude = 1−(2−5) = 3, period = 2, mid-line value = 2 y = 3cos( cos(πt) πt) 2
−
−
47.
0 8. Rate of change = 63− −0 =
9. Variation equation F equation F = kW . kW . Use F Use F = = 120 when W when W = = 180 to find k find k
a) Amplitude = 1352−1 = 67, 67, period period = 1/ 1/2 means that 2π b = (1/ = 4π 4 π , mid-line midline value val ue = 1 + 67 = 68 (1/2) Since the heel begins on the top of the eye we will use a cosine model h(t) = 67 cos 67 cos(4 (4πt πt)) + 68 b) h(10) = 67 cos 67 cos(40 (40π π) + 68 = 101. 101.5 m
1 48. (645/3 )−1/2 = 64 −5/6 = ( √ 64) = 6
5
1 32
120 = k(180) 120 = k 180 2 = k 3 The equation of variation is F is F = 32 W a) f (1) (1) = 4 b) x = 3 and x and x = = 3
10.
−2, and x and x = = 2 √ √ 50. x = ± 10 and x and x = = ±2 2 51. (−1.8981 8981,, 0.7541), (−0.2737 2737,, 1.0743), and (2. (2.0793 0793,, 0.6723)
−
49. x = 0, x 0, x = =
52.
53.
a) w(h) = 0.003968 003968x x + 3. 3.269048 269048x x b)
− 76 76..428571
w(67) = 0.003968(67)2 + 3. 3.269048(67) = 160.415 lbs
11. x2 + 4x 4x
−2
= = = 12.
–10 –8
–6
–4
–2 0
2
4 x 6
8
10
–10
–20
a) f ( 3) = ( 3)2 + 2 = 11 b) f (x + h) = (x + h)2 + 2 = x 2 + 2xh 2xh + h2 + 2
−
√ 13. 1/ t = 1/t1/2 = t−1/2 √ 3 −3/5 3/5 14. t
2
3. f ( f (x) = 2x + 3
5
= 1/t 1 /t
= 1/ 1/ t
2
2
a) f ( 2) = 2( 2) + 3 = 8 + 3 = 11 b) f (x + h + h)) = 2(x 2(x + h + h))2 + 3 = 2(x 2(x2 + 2xh 2xh + + h h2 ) + 3 = 2 2 2x + 4xh 4xh + 2h 2h + 3 4. Slope =
2 −4 ± 2√ 6 2√ −2 ± 6
y10
2. f ( f (x) = x = x 2 + 2
−
−4 ± √ 16 + 8 2 −4 ± √ 24
20
a) Approximately 1150 minutes per month b) About 62 years old
−
= 0
x =
− 76 76..428571
Chapter 1 Test 1.
−
a) G(x) = 0.6255 6255x x + 75. 75.4766 b) G(18) = 0. 0.6255(18) + 75. 75.4766 = 86. 86.7356 G(25) = 0. 0.6255(25) + 75. 75.4766 = 91. 91.1141 c) In Exercise 33, G 33, G(18) (18) = 85. 85.6 and G and G(25) (25) = 94. 94.6. The results obtained with the regression line are close to those obtained in Exercise 33. 2
1 2
−
−1 . It is important important to note that x = 15. f ( f (x) = xx+1 in the domain of the plotted function 4
−3, y 3, y -intercept (0, (0, 2)
2
5.
–4
y
− y1 y − (−5) y y
= m(x x1 ) 1 = (x 8) 4 1 = x 2 5 4 1 = x 7 4
− − − − −
2 x
–2
–2 –4 –6
4
−1 is not
32
Chapter 1: Functions and Graphs
16. sin(11 sin(11π/ π/6) 6) = 17. cos( cos(
− 12
27.
25
sqrt2 3π/4) π/4) = sqrt2 2
−
20
18. tan( tan(π) = 0
15
19. 3.28 x 3.28 x = tan(40 tan(40o ) = 3.909
10
tan(40 tan(40o ) =
5
–4
0
–2
28.
20. tan( tan(t) =
±
√
3
4
4
√
2
t = tan−1 ( 3) π t = + 2nπ 2nπ 3 and t = tan−1 ( 3) π t = + 2nπ 2nπ 3
−
2 x
–4
√
–2
2 x
4
–2
−
–4
21. cos2 (t) = 2
29.
√
cos( cos(t) = 2 cos( cos(t) = 1.414
±
−170 a) Find the slope m slope m = = 176 80−50 = Use slope-point equation
1 5
M
− − −
− − M 1 − 170 M −
No solution, cos solution, cos((t) cannot have values larger than 1. 22. 2sin3 (2t (2t) 3sin2 (2t (2t) 2sin(2 sin(2tt) = 0 sin(2 sin(2tt)(2sin )(2sin(2 (2tt) 1)(sin 1)(sin(2 (2tt) + 2) = 0 nπ t = 2 Or π t = + nπ 12 Or 7π t = + nπ 12
−
−
−
−
= m(r r1 ) 1 = (r 50) 5 1 M = r 10 + 170 5 1 M = r + 160 5
b) M (62) (62) = 51 (62) + 160 = 172. 172.4 M (75) (75) = 51 (75) + 160 = 175 30. 3x + 3x2
23. Amplitude = 4, period = 22π = π, π , mid-line y mid-line y = 4 max = 4 + 4 = 8, min = 4 4 = 0
−
8 1 = 0 x x+8 = 0 1 x =
−
−
2π (1/ (1/3)
24. Amplitude = 6, period = = 6π 6 π, mid-line y mid-line y = max = 10 + 6 = 4, min = 10 6 = 16
−
−
− −
−
25. Amplitude = − 0.5−2(−1.5) = 21 , period = 23π , 2π b = (2π/ (2π/3) 3) = 3, mid-line value is -1 Thus, equation of the line is y is y = 21 cos(3 cos(3tt) 1
−10
= 31. x =
1 6
±
6 i 95 6
√
− − 1.2543
32. There are no real zeros for this function.
−
1 3 26. Amplitude = 4− 2 = 2 , period = 1, b = 21π = 2π 2 π, mid-line value is 2. 2 .5 Thus, equation of the line is y is y = 23 sin(2 sin(2πt πt)) +
=
± √ 1 − 96 6 √ 1 ± i 95
33. ( 1.21034 21034,, 2.36346)
−
5 2
34.
a) M (r) = 0.2r + 160 b) M (62) (62) = 172. 172.4 M (75) (75) = 174 c) The results from the regression model are exactly the same as the result obtained in Exercise 29.
Technology Connection 35.
33
a) Linear Model: y = 37 37..57614 57614x x + 294. 294.47744 Quadratic Quadratic Model: y = 0.59246 59246x x2 + 74 74..60681 60681x x 117..72472 117 Cubic Cubic Model: Model: y = 0.02203 02203x x3 2.60421 60421x x2 + 125..71434 125 71434x x 439 439..64751 Quartic Quartic Model: y = 0.00284 00284x x4 0.32399 32399x x3 + 2 11 11..45714 45714x x 88 88..51211 51211x x + 507. 507.83874
−
b)
−
− −
− −
2000 1500 1000 500
0
10
20
x
30
40
50
c) By consider consider the graph in part b) and the scatter scatter plot of the data points, it seems like quartic model best fits the data. data. The reason reason for this conclus conclusion ion is because the scatter plot and the quartic model have the least amount of deviation (sometimes called residue) between them compared to the other models. d) Left to the student (answers vary).
Technology Connection Left to the student
• Page 7: 1. The line will look like a vertical line. 2. The line will look like a horizontal line. 3. The line will look like a vertical line. 4. The line will look like a horizontal line.
• Page 10: 1. Graphs are parallel 2. The The functi function on values alues differ differ by the consta constant nt value value added. 3. Graphs are parallel
• Page 19: 1. f (−5) = 6, f ( f (−4.7) = 3. 3.99, f (11) (11) = 150, f (2/ (2/3) = −1.556 2. f (−5) = −21 21..3, f (−4.7) = −12 12..3, f (11) f (11) = −117 117..3, f (2/ (2/3) = 3. 3.2556
3. f ( 5) = 75, f ( f ( 4.7) = f (2/ (2/3) = 1. 1.6889
−
−
−
−45 45..6,
f (1 f (11) 1) = 420 420.6,
-2 -15
-1 -5
1 -1
2 -4
3 -5
4 -4
5 -1
6 4
7 11
1 3
2 1
3 -5
4 -15
5 - 29
1. x = 4.4149 2. x =
−0.618034 and x and x = = 1.618034
• Page 27: 1. y = −0.37393 37393x x + 1. 1.02464 2. y = 0.46786 46786x x2 − 3.36786 36786x x + 5. 5.26429 3. y = 0.975 975x x3 − 6.031 031x x2 + 8. 8.625 625x x − 3.055 • Page 28: 1. x = 2 and x and x = = −5 2. x = −4 and x and x = = 6 3. x = −2 and x and x = = 1 4. x = −1.414214, 414214, x x = = 0, and x and x == == 1.414214 5. x = 0 and x and x = = 700 6. x =
−2.079356, 079356, x x = = 0.46295543, and x and x = = 3.1164004 7. x = −3.095574, 095574, x x = = −0.6460838, x 6460838, x = = 0.6460838, and x = 3.095574
8. x =
−1 and x and x = = 1 9. x = −2, x 2, x = = 1.414214, 414214, x x = = 1, and x and x = = 1.414214 10. x = −3, x 3, x = = −1, x 1, x = = 2, and x and x = = 3 11. x = −0.3874259 and x and x = = 1.7207592 • Page 37: 1. [0, [0, ∞) 2. [−2, ∞) 3. (−∞, ∞) 4. (−∞, ∞) 5. [1, [1, ∞) 6. (−∞, ∞) 7. [−3, ∞) 8. (−∞, ∞) 9. (−∞, ∞) 10. (−∞, ∞)
11. Not correct 12. Correct
• Page 46: 1. t = 6.89210o 2. t = 46 46..88639o 3. No solution 5. t = 0.66874
0 4
0 1
• Page 23:
4. t = 1.01599
• Page 21: -1 11
-3 -3 -29
12. x = 6.1329332
• Page 5:
1.
2.
6. 0.46677
34
Chapter 1: Functions and Graphs
• Page 56:
b)
Number Number 2 equation: equation: shifts shifts the cos the cos((πx) πx) graph up by 1 unit Number Number 3 equation: equation: shifts shifts the cos the cos((πx) πx) graph up by 1 unit and shrinks the period by a factor of 2 Number 4 equation: shifts the cos the cos((πx) πx) graph up by 1 unit, shrinks the period by a factor of 2, and increases the amplitude by a factor of 3 Number 5 equation: shifts the cos the cos((πx) πx) graph up by 1 unit, shrinks the period by a factor of 2, increases the amplitude by a factor of 3, and shifts the graph to the right by 0. 0.5 units
370 360 350 340 330 320 0
20 x
30
40
to t = = 31 c) January 1990 corresponds to t y = 0.0122(31)2 + 0. 0.8300(31) + 314. 314.8104 = 352. 352.29 January 2000 corresponds to t to t = = 41 y = 0.0122(41)2 + 0. 0.8300(41) + 314. 314.8104 = 369. 369.39 The estimates seem to be reasonable when compared to the data.
Extended Life Science Connection 1.
10
a) y = 1.343450619 343450619x x + 311. 311.3019556 b) 360
d) January 2010 corresponds to t to t = = 51 y = 0.0122(51)2 + 0. 0.8300(51) + 314. 314.8104 = 388. 388.94 January 2050 corresponds to t to t = = 91 y = 0.0122(91)2 + 0. 0.8300(91) + 314. 314.8104 = 491. 491.57
350 340 330
e) Find x Find x when y when y = 500
320
0
10
20 x
30
500 = 0.0122 0122x x2 + 0. 0.8300 8300x x+ 314..8104 314 0 = 0.0122 0122x x2 + 0. 0.8300 8300x x 185..1896 185 0.8300 x = + 2(0. 2(0.0122)
40
c) January January 1990 corresponds corresponds to t to t = = 31 y = 1.343450619(31) + 311. 311.301955 3019556 6 = 352 352..95 January January 2000 corresponds to t to t = = 41 y = 1.343450619(41) + 311. 311.301955 3019556 6 = 366 366..38 The estimates seem to be reasonable when compared to the data. d) January January 2010 corresponds corresponds to t to t = = 51 y = 1.343450619(51) + 311. 311.301955 3019556 6 = 379 379..82 January January 2050 corresponds to t to t = = 91 y = 1.343450619(91) + 311. 311.301955 3019556 6 = 433 433..56 The estimates seem to be reasonable when compared to the data. e) Find x Find x when y when y = 500 y 500 500 311 311..30196 500 311 311..30196 1.34345 140..5 140
− −
= 1.34345 34345x x + 311. 311.30196 = 1.34345x 34345x + 311. 311.30196 = 1.34345 34345x x = x
≈
x
The carbon dioxide concentration will reach 500 parts per million sometime in the year 2099. 2.
a) y = 0.0122244281 0122244281x x2 + 0. 0.8300246407 8300246407x x + 314. 314.8103665
−
(0.(0.8300) − 4(0. 4(0.0122)(−185 185..1896) 2
2(0. 2(0.0122)
x
≈
161..63 161
The carbon dioxide concentration will reach 500 parts per million sometime in the year 2120. 3.
a) y = 0.000307 000307x x3 + 0.031536 031536x x2 + 0.509387 509387x x + 315..8660781 315
−
b) 370 360 350 340 330 320 0
10
20 x
30
40
to t = = 31 c) January 1990 corresponds to t y = 0.000307 000307x x3 + 0.031536 031536x x2 + 0.509387 509387x x + 315..866078 315 8660781 1 = 352 352..82 January 2000 corresponds to t to t = = 41 y = 0.000307 000307x x3 + 0.031536 031536x x2 + 0.509387 509387x x + 315..866078 315 8660781 1 = 368 368..60
− −
Extended Life Science Connection
35
The estimates seem to be reasonable when compared to the data d) January January 2010 corresponds corresponds to t to t = = 51 3 y = 0.000307 000307x x + 0.031536 031536x x2 + 0.509387 509387x x + 315..866078 315 8660781 1 = 383 383..15 January January 2050 corresponds to t to t = = 91 3 y = 0.000307 000307x x + 0.031536 031536x x2 + 0.509387 509387x x + 315..866078 315 8660781 1 = 392 392..02
− −
e) Find x when y = 500 500.. The maxim maximum um of the cubic cubic function does not intersect the line y line y = 500 therefore under under this model the carbon dioxide concentrati concentration on will never reach 500 parts per million. 4.
a)
380 370 360
• QUADRATIC QUADRATIC MODEL: MODEL: This model also resem-
bles the original data set’s scatter plot. At relatively small values of t it allows for a longer time for the increase in the concentration of carbon dioxide since it is a parabola. parabola. As time increases increases though though the level at which the concentration of carbon dioxide will increase will be quicker than the linear model.
• CUBIC MODEL: MODEL: This modelalso modelalso resembled resembled the
original data set’s scattor plot indicates that there is a level after which the concentration of carbon dioxide will will not increa increase. se. It is the only only model model that did not allow the concentration level of carbon dioxide to reach 500 parts per million. This model suggests that as time increased the concentration of carbon dioxide will begin to decrease indefinitely.
MODEL: This model, as the other, modeled the data set to a very good degree of accuracy. racy. It was the only model that allowed allowed for oscillating behavior in the future, which is more likely to happen than what the other models models suggested. suggested.
• PERIODIC PERIODIC
350 340 330 320 0
10
20
30
t
40
50
b) The graph represents a steady increase in the concentration of carbon dioxide. c) 0.4 0.2 0
46
47
t
48
49
50
–0.2 –0.4
d) The graph shows an oscillating behavior for the concentration of carbon dioxide. e) 378 376 374 372 370 45
46
47
t
48
49
50
f ) The graph behavior shows that there is a periodic fluctuation in the concentration of carbon dioxide. 5.
the scatter scatter plot of the original original data sets. Under this model, the concentration levels of carbon dioxide will increase with time indefinitely.
• LINEAR MODEL: This MODEL: This model is the easiest math-
ematical ematically ly to compute compute and explain. explain. It does resembl resemblee
