34.1 Perform the indicated operations.
6 + 2(8 − 3)
12(3+5)+(4−1)6
4 + 2 ÷ 2 • 6
SOLUTION
6 + 2(8 − 3) = 6 + 2(5) = 6 + 10 = 16
12(3+5)+(4−1)6=12(8)+36=96+36=996=332
4 + 2 ÷ 2 • 6 = 4 + 1 • 6 = 4 + 6 = 10
34.2 Evaluate the given expression when x = −1, y = 4, a = 2/3, b = −1/2
(x/y)2 ÷ (a/b)
a2x + by2
(a + xy)/(b − xy)
SOLUTION
(x/y)2 ÷ (a/b) = (−1/4)2 ÷ ([2/3]/[−1/2]) = (1/16) ÷ (6[2/3])/(6[−1/2]) = (1/16) ÷ (4/[−3]) = (1/16) • (−3/4) = −3/64
a2x + by2 = (2/3)2(−1) + (−1/2)42 = (4/9)(−1) + (−1/2)(16) = (−4/9) + (−8) = (−4/9) + (−72/9) = −76/9
(a + xy)/(b − xy) = [(2/3) + (−1)(4)]/[(−1/2) − (−1)(4)] = [(2/3) − 4]/[(−1/2) + 4] = [(2/3) − (12/3)]/[(−1/2) + 8/2] = [−10/3]/[7/2] = [−10/3] • [2/7] = −20/21
CHAPTER 2
CHAPTER 3
34.5 Which of the properties closure, identity, inverse, associativity, and commutativity are true for the set {−1, 0, 1} under addition?
SOLUTION
No closure since (−1) + (−1) = −2 and −2 is not in the set.
Has identity since (−1) + 0 = 0 + (−1) = −1, 0 + 0 = 0, and 1 + 0 = 0 + 1 = 1, so 0 is the identity.
Has inverses since 0 + 0 = 0 the identity with 0 being its own inverse and (−1) + 1 = 1 + (−1) = 0 the identity with 1 and −1 being the inverse of each other.
Has associativity under addition since −1, 0, and 1 are integers and the integers are associative under addition.
Has commutativity under addition since −1, 0, and 1 are integers and the integers are commutative under addition.
34.6 Which of the properties closure, identity, and inverse does the set {0, 1} have under addition?
SOLUTION
No closure since 1 + 1 = 2 and 2 is not in the set.
Has identity since 0 + 0 = 0 and 1 + 0 = 0 + 1 = 1, so 0 is the identity.
No inverse since 1 + 1 = 2 and 1 + 0 = 1, so 1 does not have an inverse.
CHAPTER 4
CHAPTER 5
CHAPTER 6
CHAPTER 7
CHAPTER 8
CHAPTER 9
CHAPTER 10
CHAPTER 11
34.21 If a : b = c : d, prove (a + 3b)/(a − 3b) = (c + 3d)/(c − 3d).
SOLUTION
a : b = c : d means a/b = c/d, so a/b + 3 = c/d + 3 and a/b − 3 = c/d − 3.
a/b+3=a/b+(3b)/b=(a+3b)/b and c/d+3=c/d+(3d)/d=(c+3d)/d
Therefore (a + 3b)/b = (c + 3d)/d and (a + 3b) = [b(c + 3d)]/d and finally (a + 3b)/(c + 3d) = b/d.
a/b−3=a/b−(3b)/b=(a−3b)/b and c/d−3=c/d−(3d)/d=(c−3d)/d
Therefore (a − 3b)/b = (c − 3d)/d and (a − 3b) = [b(c − 3d)]/d and finally (a − 3b)/(c − 3d) = b/d.
Since (a + 3b)/(c + 3d) = b/d and (a − 3b)/(c − 3d) = b/d, we can conclude that
(a+3b)/(c+3d)=(a−3b)(c−3d).
From (a + 3b)/(c + 3d) = (a − 3b)/(c − 3d) we see that (a + 3b) = [(c + 3d)(a − 3b)]/(c − 3d) and finally we have (a + 3b)/(a − 3b) = (c + 3d)/(c − 3d).
34.22 Solve for x. 3 : 5 = (x − 3) : (2x + 18).
SOLUTION
3 : 5 = (x − 3) : (2x + 18) means that 3/5 = (x − 3)/(2x + 18).
[5(2x + 18)](3/5) = [5(2x + 18)][(x − 3)/(2x + 18)]
(2x + 18)3 = 5(x − 3)
6x + 54 = 5x − 15
x + 54 = −15
x = −69
CHAPTER 12
34.23 If f (x) = x3 − 5x + 3, Find f(−4), f(−2), f(1), and f(2.5).
SOLUTION
f(x) = x3 − 5x + 3
f(−4) = (−4)3 − 5(−4) + 3 = −64 + 20 + 3 = −41
f(−2) = (−2)3 − 5(−2) + 3 = −8 + 10 + 3 = 5
f(1) = 13 − 5(1) + 3 = 1 − 5 + 3 = −1
f(2.5) = (2.5)3 − 5(2.5) + 3 = 15.625 − 12.5 + 3 = 6.125
34.24 Test x2 + x + y + y2 = 8 for symmetry.
SOLUTION
Substituting −x for x we get (−x)2 + (−x) + y + y2 = 8 and x2 − x + y + y2 = 8. Since the expression changed, the graph is NOT symmetric with respect to the y-axis.
Substituting −y for y we get x2 + x + (−y) + (−y)2 = 8 and x2 + x − y + y2 = 8. Since the expression changed, the graph is NOT symmetric with respect to the x-axis.
Substituting −x for x and −y for y we get (−x)2 + (−x) + (−y) + (−y)2 = 8 and x2 − x − y + y2 = 8. Since the expression changed, the graph is NOT symmetric with respect to the origin.
CHAPTER 13
34.25 If the radius of a circle is increased by 7 inches, the area is increased by 440 square inches. Find the radius of the original circle. Use π ≈ 22/7.
SOLUTION
Let r = the radius of the original circle in inches.
πr2 is the area of the original circle in square inches.
π(r + 7)2 is the area of the new circle in square inches.
π(r + 7)2 − πr2 = 440
πr2 + 14rπ + 49π + πr2 = 440
14rπ + 49π = 440
14r(22/7) + 49(22/7) = 440
2r(22) + 7(22) = 440
44r + 154 = 440
44r = 286
r = 6.5
The radius of the original circle is 6.5 inches.
34.26 Alice can complete a project in 15 days and her younger brother Jose can complete it in 25 days. After they have worked together for 3 days, Alice got called away. How much longer would it take Jose to complete the project working alone?
SOLUTION
1/15 = the part of the project Alice completes per day
1/25 = the part of the project Jose completes per day
3(1/15 + 1/25) = the part of the project they complete when working together for 3 days.
x = the number of days for Jose to complete the project after Alice leaves
x(1/25) = the part of the project Jose completes after Alice leaves
3(1/15 + 1/25) + x(1/25) = 1
3/15 + 3/25 + x/25 = 1
1/5 + 3/25 + x/25 = 1
25(1/5 + 3/25 + x/25) = 25(1)
5 + 3 + x = 25
8 + x = 25
x = 17
It would take Jose 17 days to complete the project after Alice left.
CHAPTER 14
34.27 Can the given three points, A, B, and C, be the vertices of a triangle ABC?
A = (3, 2), B = (−5, 6), C = (8, −4)
A = (−2, 42), B = (3, 69/2), C = (8, 27)
A = (−6, −4), B = (8, 3), C = (9, −2)
A = (3, 4), B = (4, 6), C = (3, 6)
SOLUTION
Recall that if any two of the three line segments formed by three points have the same slope, then the three points are collinear. In fact, all three such line segments will have the same slope when the three points are collinear.
A = (3, 2), B = (−5, 6), C = (8, −4)
Slope of line AB = (6 − 2)/(−5 − 3) = 4/(−8) = −1/2
Slope of line BC = (−4 − 6)/(8 − (−5)) = −10/13
Slope of line CA = (2 − (−4))/(3 − 8) = −6/5
The points A, B, and C are not collinear, so they can be the vertices of a triangle ABC.
A = (−2, 42), B = (3, 69/2), C = (8, 27)
Slope of line AB = (69/2 − 42)/(3 − (−2)) = (−15/2)/(5) = −3/2
Slope of line BC = (27 − 69/2)/(8 − 3) = (−15/2)/(5) = −3/2
Slope of line CA = (42 − 27)/(−2 − 8) = 15/(−10) = −3/2
The points A, B, and C are collinear, so they cannot be the vertices of a triangle ABC.
A = (−6, −4), B = (8, 3), C = (9, −2)
Slope of line AB = (3 − (−4))/(8 − (−6)) = 7/14 = 1/2
Slope of line BC = (−2 − 3)/(9 − 8) = −5/1 = −5
Slope of line CA = (−4 − (−2))/(−6 − 9) = −2/−15 = 2/15
The points A, B, and C are not collinear, so they can be the vertices of a triangle ABC.
A = (3, 4), B = (4, 6), C = (3, 6)
Slope of line AB = (6 − 4)/(4 − 3) = 2/1 = 2
Slope of line BC = (6 − 6)/(3 − 4) = 0/(−1) = 0
Slope of line CA = (4 − 6)/(3 − 3) = −2/0 slope is not defined
The points A, B, and C are not collinear, so they can be the vertices of a triangle ABC.
34.28 Do the given pair of lines intersect?
16x + 7 = 15y and 4x + 5y = 0
3x − 8y = 13 and x − 6y = 0
9x − 2y = 18 and 45x = 10y + 14
2x + 5y = 18 and x − 10y = 9
SOLUTION
16x+7=15yand4x+5y=0(16/15)x+7/15=yand5y=−4xslope=16/15andy=(−4/5)xslope=−4/5
Since the slopes of the lines are different, the lines will intersect.
3x−8y=13andx−6y=0−8y=−3x+13andx=6yy=(3/8)x−13/8and(1/6)x=yslope=3/8andslope=1/6
Since the slopes of the lines are different, the lines will intersect.
9x−2y=18and45x=10y+14−2y=−9x+18and45x−14=10yy=(9/2)x−9and(45/10)x−14/10=yslope=9/2and(9/2)x−7/5=yslope=9/2
Since the lines have the same slope, the lines are parallel and do not intersect.
2x+5y=18andx−10y=95y=−2x+18and−10y=−x+9y=(−2/5)x+18/5andy=(1/10)x−9/10slope=−2/5andslope=1/10
Since the slopes of the lines are different, the lines will intersect.
CHAPTER 15
34.29 A boat whose rate in still water is 12 miles per hour goes up a stream whose current has a rate of 2 miles per hour, and returns. The entire trip took 24 hours. Find how many hours are required for the trip upstream and how many hours for the return trip.
SOLUTION
Let x = hours for the trip upstream
Let y = hours for the return trip
x(12 − 2) = 10x = distance traveled upstream
y(12 + 2) = 14y = distance traveled for return trip
10x = 14y
x + y = 24
y = 24 − x
10x = 14(24 − x)
10x = 336 − 14x
24x = 336
x = 14
y = 24 − 14 = 10
The trip upstream took 14 hours and the return trip took 10 hours.
34.30 A chemist has the same acid in two strengths. Eight liters of one acid mixed 12 liters of the second acid gives a mixture that is 84% pure. Three liters of the first acid mixed with 2 liters of the second acid gives a mixture that is 86% pure. Find the purity of each acid.
SOLUTION
Let x = the purity of the first strength acid as a decimal
Let y = the purity of the second strength acid as a decimal
8x + 12y = 0.84(8 + 12) = 16.8
3x + 2y = 0.86(3 + 2) = 4.3
−6(3x + 2y = 4.3) yields − 18x − 12y = −25.8
−18x−12y=−25.88x+12y=16.8¯ −10x=−9
x = 0.9 (0.9 = 90%)
8(0.9) + 12y = 16.8
7.2 + 12y = 16.8
12y = 9.6
y = 0.8 (0.8 = 80%)
The purity of the first strength acid is 90% and the purity of the second strength acid is 80%.
CHAPTER 16
CHAPTER 17
34.33 Identify the type of conic section and write the equation in standard form for y2 − 5x − 6y − 26 = 0.
SOLUTION
y2 − 5x − 6y − 26 = 0 has only one squared term, so it is a parabola.
y2 − 5x − 6y − 26 = 0
y2 − 6y = 5x + 26
y2 − 6y + 9 = 5x + 26 + 9
(y − 3)2 = 5x + 35
(y − 3)2 = 5(x + 7)
It is a parabola with standard form (y − 3)2 = 5(x + 7).
34.34 Identify the type of conic section and write the equation in standard form for 25x2 − 144y2 + 450x − 1440y + 2025 = 0.
SOLUTION
25x2 − 144y2 + 450x − 1440y + 2025 = 0 has two squared terms with unequal coefficients having opposite signs, so it is a hyperbola.
25x2 − 144y2 + 450x − 1440y + 2025 = 0
25x2 − 144y2 + 450x − 1440y = −2025
25x2 + 450x − 144y2 − 1440y = −2025
25(x2 + 18x + ) − 144(y2 + 10y + ) = −2025
25(x2 + 18x + 81) − 144(y2 + 10y + 25) = −2025 + 25(81) − 144(25)
25(x + 9)2 − 144(y + 5)2 = −2025 + 2025 − 3600
25(x + 9)2 − 144(y + 5)2 = −3600
−25(x + 9)2 + 144(y + 5)2 = +3600
[−25(x + 9)2]/3600 + [144(y + 5)2]/3600 = + 3600/3600
[−(x + 9)2]/144 + [(y + 5)2]/25 = 1
(y + 5)2/25 − (x + 9)2/144 = 1
It is a hyperbola with standard form of (y + 5)2/25 − (x + 9)2/144 = 1.
CHAPTER 18
34.35 Solve for all points (x, y) that satisfy the system of equations xy + 3y2 = 6 and x2 + y2 = 10.
SOLUTION
We can combine the two equations to obtain an equation in which the constant term is 0.
5(xy + 3y2 = 6) yields 5xy + 15y2 = 30
3(x2 + y2 = 10) yields 3x2 + 3y2 = 30
3x2 + 3y2 = 5xy + 15y2
3x2 − 5xy − 12y2 = 0
(3x + 4y)(x − 3y) = 0
3x+4y=0orx−3y=03x=−4yorx=3y
x = ( 4/3)y
Substituting x = (−4/3)y into x2 + y2 = 10, we get [(−4/3)y]2 + y2 = 10.
(16/9)y2+y2=1016y2+9y2=9025y2=90y2=90/25y=±(90/25)y=±(3/5)10x=(−4/3)y=(−4/3)[±(3/5)10]=−(4/5)10 or+(4/5)10([−4/5]10,[3/5]10) and ([4/5]10,[−3/5]10)
Substituting x = 3y into x2 + y2 = 10, we get (3y)2 + y2 =10.
9y2 + y2 = 10
10y2 = 10
y2 = 1
y = ±1
x = 3y = 3(±1) = ±3
(3, 1) and (−3, −1)
The solutions for (x, y) are (3, 1), (−3, −1), ([−4/5]10,[3/5]10), and ([4/5]10,[−3/5]10).
34.36 Solve for all points (x, y) that satisfy the system of equation x + y = 7 and xy = 6.
SOLUTION
We square x + y = 7 to get x2 + 2xy + y2 = 49 and we multiply xy = 6 by 4 to get 4xy = 24.
Now (x2 + 2xy + y2 = 49) − (4xy = 24) = x2 − 2xy + y2 = 25. So (x − y)2 = 25 and x − y = ±5. Finally we have x = y ± 5 and x = y + 5 or x = y − 5.
Substituting x = y + 5 into x + y = 7. We get y + 5 + y = 7. Then 2y = 2 and y = 1. So x = y + 5 = 1 + 5 = 6 and (x, y) = (6, 1).
Substituting x = y − 5 into x + y = 7. We get y − 5 + y = 7. Then 2y = 12 and y = 6. So x = y − 5 = 6 − 5 = 1 and (x, y) = (1, 6).
The solutions for (x, y) are (6, 1) and (1, 6).
CHAPTER 19
34.37 Solve the inequality (6x2 − 11x + 3)/(x + 4) ≤ 0.
SOLUTION
(6x2 − 11x + 3)/(x + 4) ≤ 0
[(3x − 1)(2x − 3)]/(x + 4) ≤ 0
The critical values, values that make a factor of the problem zero, are at x = 1/3, x = 3/2, and at x = −4. Since x = −4 would make the problem undefined, it is not to be in the solution. We will separate our number line into parts by using a “) (” to show the critical point is not in the solution and a “] [” to show the critical point is included in the solution.
Since we want the problem to be ≤ 0, we have the solution when the problem is negative or zero. Thus, the solution is x ≤ −4 or 1/3 ≤ x ≤ 3/2. (See Fig. 34-1.)
34.38 Solve the inequality |5 − 3x | ≤ 2.
SOLUTION
|5 − 3x | ≤ 2
Since we want the absolute value of the quantity to be less than or equal to 2, the quantity has to be between −2 and 2 and including both.
The solution is 1 ≤ x ≤ 7/3.
CHAPTER 20
34.39 Find all of the real zeros of P(x).
P(x) = x4 − 5x3 − 5x2 + 45x − 36 = 0
P(x) = x4 − 12x3 + 44x2 − 40x − 25 = 0
SOLUTION
P(x) = x4 − 5x3 − 5x2 + 45x − 36 = 0
Since the coefficient of the highest degree term, x4, is 1, the only rational zeros of P(x) are integral factors of 36: ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36.
P(1) = 14 − 5(1)3 − 5(1)2 + 45(1) − 36 = 1 − 5 − 5 + 45 − 36 = 0. Since 1 is a zero of P(x) and we use synthetic division to get the depressed equation of one degree less than P(x), so degree 3.
P(x) = (x − 1)(x3 − 4x2 − 9x + 36) = (x − 1)P′(x)
We try x = 1 in the depressed equation, P′(x), to see if 1 is a double zero of P(x).
P′(1) = 13 − 4(1)2 − 9(1) + 36 = 1 − 4 − 9 + 36 = 24. Thus, 1 is not a double zero of P(x).
P′(2) = 23 − 4(2)2 − 9(2) + 36 = 8 − 16 − 18 + 36 = 10
P′(3) = 33 − 4(3)2 − 9(3) + 36 = 27 − 36 − 27 + 36 = 0 Thus, 3 is a zero of P′(x) and therefore a zero of P(x).
We will use synthetic division to get the depressed equation for P′(x).
P′(x) = (x − 3)(x2 − x − 12) = (x − 3)(x − 4)(x + 3)
P(x) = (x − 1)(x − 3)(x − 4)(x + 3) = 0
The zeros of P(x) = x4 − 5x3 − 5x2 + 45x − 36 are 1, 3, 4, −3.
P(x) = x4 − 12x3 + 44x2 − 40x − 25 = 0
Since the coefficient of the highest degree term is one, all the rational zeros are factors of the constant term 25. The possible rational zeros of P(x) are ±1, ±5, ±25.
P(1) = 14 − 12(1)3 + 44(1)2 − 40(1) − 25 = 1 − 12 + 44 − 40 − 25 = −32
P(5) = 54 − 12(5)3 + 44(5)2 − 40(5) − 25 = 625 − 1500 + 1100 − 200 − 25 = 0 So 5 is a zero of P(x).
Now we find the depressed equation by synthetic division.
P(x) = (x − 5)(x3 − 7x2 + 9x + 5) = (x − 5)P′(x) = 0
Now we check to see if 5 is a double zero of P(x) by Finding P′(5).
P′(5) = 53 − 7(5)2 + 9(5) + 5 = 125 − 175 + 45 + 5 = 0. So 5 is a zero of P′(x) and a double zero of P(x).
Now we use synthetic division to find the depressed equation for P′(x).
P(x) = (x − 5)(x − 5)(x2 − 2x − 1) = (x − 5)2P″(x) = 0.
Solving the quadratic equation by completing the square method.
x2−2x−1=0x2−2x=1x2−2x+1=1+1(x−1)2=2x−1=±2x=1±2
The zeros of P(x) = x4 − 12x3 + 44x2 − 40x − 25 = 0 are 5, 5, 1+2, 1−2.
34.40 Use synthetic division to show that Q(x) = 5x + 2 is a divisor of P(x) = 60x3 − 41x2 − 11x + 6.
SOLUTION 1
P(x)/Q(x) = [60x3 − 41x2 − 11x + 6]/[5x + 2] = (1/5)[60x3 − 41x2 − 11x + 6]/{(1/5)[5x + 2]} = [12x3 − (41/5)x2 − (11/5)x + (6/5)]/[x + (2/5)]
−2/512−41/5−11/5+6/5−24/5+26/5−6/5¯ 12−65/5+15/5+ 0
Since the remainder is 0, Q(x) is a divisor of P(x).
SOLUTION 2
Since Q(x) = 5x + 2 = 5(x + 2/5), we can divide P(x) by (x + 2/5).
−2/560−41−11+6−24+26−6¯ 60−65+15+0
P′(x) = 60x2 − 65x + 15
(x+2/5)P′(x)=(x+2/5)(60x2−65x+15)=(x+2/5)(5)(12x2−13x+3)=(5x+2)(12x2−13x+3)=60x3−41x2−11x+6=P(x)
So Q(x) is a divisor of P(x) since (5x + 2)(12x2 − 13x + 3) = 60x3 − 41x2 − 11x + 6 and Q(x)(12x2 − 13x + 3) = P(x).
CHAPTER 21
34.41 Determine the zeros, the vertical asymptotes, the horizontal asymptotes, and holes for each rational function R(x).
R(x) = (x2 + 4x − 5)/(x3 − x2)
R(x) = (x3 − x2 − 14x + 24)/(x + 4)
SOLUTION
R(x) = (x2 + 4x − 5)/(x3 − x2) = [(x + 5)(x − 1)]/[x2(x − 1)]
Since −5 makes the numerator zero but not the denominator, there is a zero at x = −5.
Since 1 makes the numerator zero and makes the denominator zero, there is a hole at x = 1.
Since 0 makes the denominator zero but not the numerator, there is a vertical asymptote at x = 0.
Since the degree of the denominator is greater than the degree of the numerator, there is a horizontal asymptote at y = 0.
R(x) = (x3 − x2 − 14x + 24)/(x + 4) = [(x + 4)(x2 − 5x + 6)]/(x + 4) = [(x + 4)(x − 3)(x − 2)]/(x + 4)
Since both 2 and 3 make the numerator zero but not the denominator, there are zeros at x = 2 and x = 3.
Since −4 makes the numerator zero and makes the denominator zero, there is a hole at x = −4.
Since there is no value for x that makes only the denominator zero, there is no vertical asymptote.
Since the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.
34.42 Determine the domain of each rational function R(x).
R(x) = (x + 3)/(x2 + 2x + 1)
R(x) = (x2 + x − 6)/(x2 + 7x + 12)
SOLUTION
R(x) = (x + 3)/(x2 + 2x + 1) = (x + 3)/(x + 1)2
Since x = −1 makes R(x) undefined, the domain of R(x) is all real numbers ≠ −1.
R(x) = (x2 + x − 6)/(x2 + 7x + 12) = [(x − 2)(x + 3)]/[(x + 3)(x + 4)]
Since both −3 and −4 make R(x) undefined, the domain of R(x) is all real numbers except −3 and −4.
CHAPTER 22
34.43 Identify each sequence as arithmetic, geometric, harmonic, or none of these.
1, 2, 3, 4, 5, ...
2, 4, 6, 8, ...
9, 8, 7, 6, ...
−1. −3, −5, −7, ...
12, 32, 52, ...
13,23,33, …
2, 4, 8, 16, ...
1, 1/2, 1/4, 1/8, ...
1, 1/3, 1/5, 1/7, ...
1,1+2,1+22,1+32,…
SOLUTION
1, 2, 3, 4, 5, ...
2 − 1 = 1, 3 − 2 = 1, 4 − 3 = 1, 5 − 4 = 1 Since there is a common difference between terms, the sequence is arithmetic.
2, 4, 6, 8, ...
4 − 2 = 2, 6 − 4 = 2, 8 − 6 = 2 Since there is a common difference between terms, the sequence is arithmetic.
9, 8, 7, 6, ...
8 − 9 = −1, 7 − 8 = −1, 6 − 7 = −1 Since there is a common difference between terms, the sequence is arithmetic.
−1, −3, −5, −7, ...
−3 − (−1) = −2, −5 − (−3) = −2, −7 − (−5) = −2 Since there is a common difference between terms, the sequence is arithmetic.
12, 32, 52, ...
9 − 1 = 8, 25 − 9 = 16 Since the differences are different, the sequence is not arithmetic.
9/1 = 9, 25/9 ≠ 9 Since the ratios are different, the sequence is not geometric.
We now check the sequence of the reciprocals of the terms in the given sequence to see if it is harmonic. 1/1, 1/9, 1/25, ... 1/9 − 1 = −8/9, 1/25 − 1/9 = −16/225 Since the sequence of the reciprocals of the terms in the given do not have a common difference, they do not form an arithmetic sequence and the given sequence is not harmonic.
Thus, the sequence is none of these three types of sequences: arithmetic, geometric, or harmonic.
13,23,33, …
23−1≠33−23, so the sequence is not arithmetic.
2/13≠33/23, so the sequence is not geometric.
1/23−1/1≠1/33−1/23, so the sequence is not harmonic.
Thus, the sequence is none of these.
2, 4, 8, 16, ...
4 − 2 ≠ 8 − 4, so the sequence is not arithmetic.
4/2 = 2, 8/4 = 2, 16/8 = 2 Since the sequence has a common ratio, the sequence is geometric.
1, 1/2, 1/4, 1/8, ...
1/2 − 1 ≠ 1/4 − 1/2, so the sequence is not arithmetic.
(1/2)/1 = 1/2, (1/4)/(1/2) = 1/2, (1/8)/(1/4) = 1/2. Since there is a common ratio, the sequence is geometric.
1, 1/3, 1/5, 1/7, ...
1/3 − 1 ≠ 1/5 − 1/3, so the sequence is not arithmetic.
(1/3)/1 ≠ (1/5)/(1/3), so the sequence is not geometric.
1/1, 1/(1/3), 1/(1/5), 1/(1/7) ... = 1, 3, 5, 7, ... 3 − 1 = 2, 5 − 3 = 2, 7 − 5 = 2 Since the sequence of the reciprocals of the given sequence forms an arithmetic sequence, the original sequence is harmonic.
1, 1+2, 1+22, 1+32,…
(1+2)−1=2,(1+22)−(1+2)=2, (1+32)−(1+22)=2 Since there is a common difference, the sequence is arithmetic.
34.44 Find the requested terms.
Insert five arithmetic means between (7x − 3y) and (13x + 9y).
Insert three geometric means between x and y.
SOLUTION
Since we want five arithmetic means in addition to the two given terms, (7x − 3y) is the first term and (13x + 9y) is the seventh term of the sequence. The nth term formula for an arithmetic sequence is l = a + (n − 1)d where l is the nth term, a is the first term, n is the number of terms, and d is the common difference between terms.
l = 13x + 9y, a = 7x + 3y, n = 7, d = ? So we substitute into the nth term formula and solve for d.
13x + 9y = 7x − 3y + (7 − 1)d
6x + 12y = 6d
x + 2y = d
We add this value for d to the first term to get the second, add d to the second term to get the third term and so on.
2nd term = (7x − 3y) + (x + 2y) = 8x − y, 3rd term = (8x − y) + (x + 2y) = 9x + y, 4th term = (9x + y) + (x + 2y) = 10x + 3y, 5th term = (10x + 3y) + (x + 2y) = 11x + 5y, 6th term = (11x + 5y) + (x + 2y) = 12x + 7y, 7th term = (12x + 7y) + (x + 2y) = 13x + 9y, which is our given last term.
The five arithmetic means between (7x − 3y) and (13x + 9y) are 8x − y, 9x + y, 10x + 3y, 11x + 5y, and 12x + 7y.
Since we want to insert three geometric means between x and y there will be five terms in the sequence. The first term is x and the fifth term is y. The nth term formula for a geometric sequence is
l = arn − 1 where l is the nth term, a is the first term, r is the common ratio, and n is the number of terms.
l = y, a = x, r = ?, n = 5 So we need to substitute into the formula and solve for r.
l=arn−1y=xr5−1y=xr4y/x=r4±(y/x)4=r±(x2y)4x=r
Using the positive value first, the second term is x((x3y)4/x)=(x3y)4. The third term is [(x3y)4][(x3y)4/x]=[(x6y2)4]/x=[x(x2y2)4]/x=(xy). The fourth term is [(xy)][(x3y)4/x]=[(x2y2)4][(x3y)4/x]=[(x5y3)4/x]=[x(xy3)4/x]=(xy3)4. The fifth term is [(xy3)4][(x3y)4/x]=[(x4y4)4/x]=(xy)/x=y. Using the negative value we get the same results but the opposite signs for the second and fourth terms.
The three geometric means are (x3y)4, (xy), (xy3)4 or −(x3y)4, (xy), −(xy3)4.
CHAPTER 23
34.45 Simplify each expression using the laws of logarithms.
log27 81 − log3 27
log25 125 + log5 25 − log125 5
log3 (1/3) − log9 (1/27) + log27 9
log (25/4) + log (36/50) − log (3/4)
SOLUTION
We will change from the logarithmic form of the expression to its exponential form and then use the laws of exponents. Recall that if two powers of the same base are equal, then the exponents are equal.
log27 81 − log3 27
Let a = log27 81, so 27a = 81. (33)a = 34, 33a = 34, 3a = 4, a = 4/3
Let b = log3 27, so 3b = 27. 3b = 33, b = 3
log27 81 − log3 27 = a − b = 4/3 − 3 = 4/3 − 9/3 = −5/3
Therefore, log27 81 − log3 27 = −5/3.
log25 125 + log5 25 − log125 5
Let a = log25 125, so 25a = 125. (52)a = 53, 52a = 53, 2a = 3, a = 3/2
Let b = log5 25, so 5b = 25. 5b = 52, b = 2
Let c = log125 5, so 125c = 5. (53)c = 51, 53c = 51, 3c = 1, c = 1/3
log25 125 + log5 25 − log125 5 = a + b − c = 3/2 + 2 − 1/3 = 9/6 + 12/6 − 2/6 = 19/6
Therefore, log25 125 + log5 25 − log125 5 = 19/6.
log3 (1/3) − log9 (1/27) + log27 9
Let a = log3 (1/3), so 3a = 1/3. 3a = 3−1, a = −1
Let b = log9 (1/27), so 9b = 1/27. (32)b = 3−3, 32b = 3−3, 2b = −3, b = −3/2
Let c = log27 9, so 27c = 9. (33)c = 32, 33c = 32, 3c = 2, c = 2/3
log3 (1/3) − log9 (1/27) + log27 9 = a − b + c = −1 − (−3/2) + 2/3 = −1 + 3/2 + 2/3 = −6/6 + 9/6 + 4/6 = 7/6
Therefore, log3 (1/3) − log9 (1/27) + log27 9 = 7/6.
log (25/4)+log (36/50)−log (3/4)log (25/4)+log (36/50)−log (3/4)=log[(25/4)×(36/50)÷(3/4)]=log[(25/4)×(36/50)×(4/3)]=log [3600/600]=log 6
Therefore, log (25/4) + log (36/50) − log (3/4) = log 6.
34.46 Solve for x. a, b, c are nonzero constants.
ex = e−x
ax + 1 = b2x/cx−1
a4x + 8a2x = 6a3x
a5x + a4x = 6a4x − 6a3x
SOLUTION
ex = e −x
log ex = log e−x
x(log e) = −x(log e)
x = −x
2x = 0
x = 0
ax + 1 = b2x/cx − 1
log(ax + 1) = log(b2x/cx − 1)
(x + 1)(log a) = log(b2x) − log(cx − 1)
(x + 1)(log a) = 2x(log b) − (x − 1)(log c)
x(log a) + log a = 2x(log b) − x(log c) + log c
x(log a) − 2x(log b) + x(log c) = log c − log a
x[log a − 2(log b) + log c] = log c − log a
x = [log c − log b]/[log a − 2(log b) + log c]
a4x + 8a2x = 6a3x
a4x + 8a2x − 6a3x = 0
a4x − 6a3x + 8a2x = 0
a2x[a2x − 6ax + 8] = 0
a2x = 0 or a2x − 6ax + 8 = 0
a2x = 0 or (ax − 4)(ax − 2) = 0
a2x = 0 or (ax − 4) = 0 or (ax − 2) = 0
a2x = 0 no solution, since no power of a nonzero number is zero.
ax−4=0orax−2=0ax=4orax=2log ax=log 4orlog ax=log 2x(log a)=log 4orx(loga)=log 2x=(log 4)/(log a)orx=(log 2)/(log a)
The solutions are x = (log 4)/(log a) or x = (log 2)/(log a).
a5x + a4x = 6a4x − 6a3x
a5x − 5a4x + 6a3x = 0
a3x(a2x − 5ax + 6) = 0
a3x = 0 or a2x − 5ax + 6 = 0
a3x = 0 or (ax − 3)(ax − 2) = 0
a3x = 0 no solution, since no power of a nonzero number is zero.
ax−3=0orax−2=0ax=3orax=2log ax=log 3orlog ax=log 2x(log a)=log 3orx(loga)=log 2x=(log 3)/(log a)orx=(log 2)/(log a)
The solutions are x = (log 3)/(log a) or x = (log 2)/(log a).
CHAPTER 24
34.47 A family borrowed $6000 to repair their home, agreeing to pay $50 monthly until the principal and interest at 6% is paid. Find the number of full payments required.
SOLUTION
Let x=the number of monthly payments required to pay off the loan. A=P+Prt P=the amount loaned, r=the rate per unit of time, t=the number of time units A=the amount paid
50x=6000+(6000)(0.06/12)(x)50x=6000+30x20x=6000x=300
It would require 300 monthly payments to pay off the loan at $50 per month.
34.48 What is the present value of an annual pension of $1200 at 4% if the pension is to run 18 years and the present value, V, is found by the formula V = (A/r)[1 − 1/(1 + r)n].
SOLUTION
V=(A/r)[1−1/(1+r)n]V=(1200/0.04)[1−1/(1+0.04)18]V=(30,000)[1−1/(1.04)18]V=15191.15637
The present value of the pension is $15,191.16.
CHAPTER 25
34.49 From an alphabet containing 21 consonants and 5 vowels, six-letter words are to be formed. How many of the words will contain 4 consonants and 2 vowels with no letters repeated?
SOLUTION
First we must select 4 consonants and 2 vowels to get the groups of 6 letters. Then since words have their letters in a specific order, we need to count the ordering for each group of 6 letters.
The number of ways to select 4 consonants from the 21 consonants is the number of combinations of selecting 4 objects from 21 objects, C(21, 4).
The number of ways to select 2 vowels from the 5 vowels is the number of combinations of selecting 2 objects from 5 objects, C(5, 2).
The number of ways 6 letters can be ordered to form six letter words is the number of permutations of selected 6 from 6 objects, P(6, 6).
Thus, the number of six-letter words that can be formed having 4 consonants and 2 vowels is
C(21,4)×C(5,2)×P(6,6)=5985×10×720=43,092,000.
34.50 In how many ways can a committee of 5 senators be appointed from the U.S. Senate, if the majority leader is always a member of the committee?
SOLUTION
There are 100 senators in the US Senate. Since the majority leader is on every committee, we are actually choosing 4 people from a group of 99 people. The order of selection is not important, so we want to find the combinations for selecting 4 objects from a group of 99 objects, C(99, 4).
C(99,4)=3,764,376
There are 3,764,376 possible committees with the majority leader and four other senators.
CHAPTER 26
34.51 Write the term which contains x5 in the expansion of (x2 + 1/x)10.
SOLUTION
The rth term formula for (a + b)n is rth term = C(n, n − 1)an − r + 1br − 1.
At first we are only interested in the variable part with a = x2, b = 1/x, n = 10, r = ?.
(x2)n − r + 1(1/x)r − 1 = (x2)10 − r + 1(x− 1)r − 1 = (x2)11 − r(x)1 − r = (x)22 − 2r(x)1 − r = x22 − 2r + 1 − r = x23 − 3r.
Now we set x5 = x23 − 3r and solve for r, using the fact that when two powers of the same base are equal the exponents are equal.
5 = 23 − 3r
3r + 5 = 23
3r = 18
r = 6
Thus we need to find the sixth term of (x2 + 1/x)10.
The sixth term is C(6, 5) · (x2)5 · (1/x)5 = 252(x10)x−5 = 252x5.
34.52 Expand the binomial (x/2 + 2/x2)5.
SOLUTION
(x/2+2/x2)5=(x/2)5+C(5,1) ⋅ (x/2)4 ⋅ (2/x2)1+C(5,2) ⋅ (x/2)3 ⋅ (2/x2)2+C(5,3) ⋅ (x/2)2 ⋅ (2/x2)3+C(5,4) ⋅ (x/2)1 ⋅ (2/x2)4+(2/x2)5=x5/32+5(x4/16)(2/x2)+10(x3/8)(4/x4)+10(x2/4)(8/x6)+5(x/2)(16/x8)+32/x10(x/2+2/x2)5=x5/32+5x2/8+5/x+20/x4+40/x7+32/x10
CHAPTER 27
34.53 A man has in his pocket 3 pennies, 2 nickels, and a dime. If he draws two coins from his pocket at random, what is the probability that the amount drawn exceeds 6 cents?
SOLUTION
To total more than 6 cents, he must select a penny and a dime, a nickel and a dime, or two nickels.
The ways to select a dime and a penny is C(1, 1) · C(3, 1) = 1(3) = 3 ways.
The ways to select a dime and a nickel is C(1, 1) · C(2, 1) = 1(2) = 2 ways.
The ways to select two nickels is C(2, 2) = 1 way.
Thus, there are (3 + 2 + 1) = 6 ways for the value of the two coins to exceed six cents.
The ways to select 2 coins from 6 coins is C(6, 2) = 15 ways.
The probability that the value of the coins selected exceeds six cents is 6/15 = 2/5.
The probability that the man will select two coins whose value exceeds six cents is 2/5 or 40%.
34.54 A club has a lottery which offers two prizes. Twenty tickets are sold. If one person bought 4 of the tickets, what is the probability that they will win at least one prize?
SOLUTION
The tickets the person bought may contain 2 winners and 2 losers, 1 winner and 3 losers, or 4 losers.
The ways the tickets can contain 2 winners and 2 losers is C(2, 2) · C(18, 2) = 1(153) = 153 ways.
The ways the tickets can contain 1 winner and 3 losers is C(2, 1) · C(18, 3) = 2(816) = 1632 ways.
Thus, the ways the 4 tickets can have at least 1 winner is 153 + 1632 = 1785 ways.
The total ways of drawing 4 tickets out 20 is C(20, 4) = 4845 ways.
Therefore, the probability of winning at least one prize with 4 tickets is 1785/4845 = 7/19 or about 37%.
CHAPTER 28
34.55 Solve each system of equations using determinants. Write the solutions in form (x, y, z).
3x+y=14−zx+z=1+2yx+y=15−2z
x + y = 13 + 2z
x + 7 = 3y
x + 4z = −14
SOLUTION
3x+y=14−zx+z=1+2yx+y=15−2z3x+y+z=14x−2y+z=1x+y+2z=15D=|3111−21112|=−13Dx=|14111−211512|=−26Dy=|31411111152|=−39Dz=|31141−211115|=−65x=Dx/D=−26/(−13)=2,y=Dy/D=−39/(−13)=3,z=Dz/D=−65/(−13)=5
Thus, the solution (x, y, z) of the system is (2, 3, 5).
x+y=13+2zx+7=3yx+4z=−14x+y−2z=13x−3y+0z=−7x+0y+14z=−14D=|11−21−30104|=−22Dx=|131−2−7−30−1404|=−44Dy=|113−21−701−144|=−66Dz=|11131−3−710−14|=−88x=Dx/D=−44/(−22)=2,y=Dy/D=−66/(−22)=3,z=Dz/D=88/(−22)=−4
The solution (x, y, z) for the system is (2, 3, −4).
34.56 Find the solution for each system of equations using determinants.
x + y + z = 6
2x + 3y + 4z = 16
2x + 3y = 7
x + 5y = 9
8x + 5y = 17
SOLUTION
x + y + z = 6
2x + 3y + 4z = 16
In this system, we have more variables than equations, so we consider one variable as a free variable and determine a solution for the other two variables in terms of the third variable. We can choose any of the variables, say z, to be the free variable. We treat the free variable as a constant in solving the system, so we use our solution methods for systems of two equations in two variables.
x+y=6−z2x+3y=16−4zD=|1123|=1Dx=|6−z116−4z3|=18−3z−16+4z=2+zDy=|16−z216−4z|=16−4z−12+2z=4−2zx=Dx/D=(2+z)/1=2+zy=Dy/D=(4−2z)/1=4−2z
The solution of the system is (x, y, z) = (2 + z, 4 − 2z, z) where z is any real number.
Sample values in the solution are found by making a choice for z and finding the specific point.
z = −2 yields (2 + (−2), 4 − 2(−2), −2) = (0, 8, −2)
z = 0 yields (2 + 0, 4 − 2(0), 0) = (2, 4, 0)
z = 5 yields (2 + 5, 4 − 2(5), 5) = (7, −6, 5)
Note: The number of free variables we use in the solution is based on having the same number of non-free variables as the number of given equations.
2x + 3y = 7
x + 5y = 9
8x + 5y = 17
In the system, we have more equations than variables, so we select a subset of the given equations that has the same number of equations as we have variables, and we solve that system of equations. We use the solution of that system, if any, to substitute into each of the remaining equations. If that solution satisfies each of the remaining equations, then the system has a solution; otherwise the system has no solution.
Choosing the first two equations 2x + 3y = 7 and x + 5y = 9 and then solving this system,
D=|2315|=7, Dx=|7395|=8, Dy=|2719|=11
x = Dx/D = 8/7, y = Dy/D = 11/7, We get (x, y) = (8/7, 11/7) as our trial solution.
We now substitute our trial solution into the remaining equation 8x + 5y = 17. to see if the trial solution is a solution for the system or if there is no solution for the system.
8x + 5y = 8(8/7) + 5(11/7) = 64/7 + 55/7 = 119/7 = 17, which is the value in the equation. The trial solution satisfies the remaining equation.
Since (x, y) = (8/7, 11/7) satisfies each equation in the given system, it is the solution for the system.
Note: If D = 0 when you use the reduced number equations, you need to determine if the equations are dependent or inconsistent. If they are inconsistent, then the system has no solution. If they are dependent then eliminate one of the equations from the system and find the solution for this set of equations. Any solution for this set of equations is a solution for the original system of equations.
CHAPTER 29
34.57 Determine if the system of equations has a solution other than (0, 0, 0) for 3x + y + 9z = 0, x − y − z = 0, and 2x − 3y − 5z = 0. Use matrices to solve the system.
SOLUTION
3x + y + 9z = 0
x − y − z = 0
2x − 3y − 5z = 0
We will transform the system to one where we treat z as a free variable and solve the system for x and y in terms of z.
3x + y = −9z
x − y = z
2x − 3y = 5z
We now select any two of these three equations and solve that system and then verify that the solution satisfies the third equation also.
3x + y = −9z
x − y = z
A=[31|−91−1|1] [31|−91−1|1]∼R2R1[1−1|131|−9]∼R2−3R1[1−1|104|−12]∼(1/4)R2[1−1|101|−3]∼R1+R2[10|−201|−3]
x = −2z, y = −3z. Verifying that they satisfy the third equation, 2x − 5y = 2(−2z) − 3(−3z) = −4z + 9z = 5z.
So the solution (x, y, z) = (−2z, −3z, z) where z is any real number is a solution of the given system of equations.
34.58 Determine if the system of equations has any solutions other than (0, 0, 0) for 3x + 2y + 3z = 0, 5x − y + 2z = 0, and 6x + y − 3z = 0. Use matrices to solve the system.
SOLUTION
3x + 2y + 3z = 0
5x − y + 2z = 0
6x + y − 3z = 0
Let z be a free variable and transform the equations.
3x + 2y = −3z
5x − y = −2z
6x + y = 3z
Solving 3x + 2y = −3z and 5x − y = −2z in terms of z.
A=[32|−35−1|−2]~(1/3)R1[12/3|−15−1|−2]~R2−5R1[12/3|−10−13/3|3]~(−3/13)R2[12/3|−101|−9/13]~ R1−(2/3)R2 [10|−7/1301|−9/13] x=(−7/13)z, y=(−9/13)z
We need to verify x = (−7/13)z and y = (−9/13)z in 6x + y = 3z.
6x + y = 6[(−7/13)z] + [(−9/13)z] = (−42/13)z + (−9/13)z = (−51/13)z ≠ 3z
Thus, x = (−7/13)z and y = (−9/13)z does not satisfy the third equation and ([−7/13]z, [−9/13]z, z) is not a solution of the given system of equations. The only solution of the system is (0, 0, 0).
CHAPTER 30
34.59 Prove by mathematical induction that 32n − 1 is divisible by 8 for all natural numbers n.
SOLUTION
Note: The set of natural numbers is the same as the set of positive integers and the names can be used interchangeably.
34.60 Prove by mathematical induction that 1 · 2 + 2 · 3 + 3 · 4 + ⋯ + n(n + 1) = (n/3)(n + 1)(n + 2) where n is a natural number.
SOLUTION
CHAPTER 31
34.61 Resolve (2x2 − 6x − 2)/(x2 − 4x + 3) into partial fractions.
SOLUTION
Since the degree of the numerator is not less than the degree of the denominator, we need to divide before we can use our decomposition procedure.
(2x2 − 6x − 2)/(x2 − 4x + 3) = 2 + (2x − 8)/(x2 − 4x + 3) from polynomial division as shown on page 15.
We use our decomposition procedure on the the remainder (2x − 8)/(x2 − 4x + 3) and later add this to 2.
(2x−8)/(x2−4x+3)=(2x−8)/[(x−3)(x−1)]=A/(x−3)+B(x−1)=[A(x−1)+B(x−2)]/[(x−3)(x−1)]
Thus, 2x − 8 = A(x − 1) + B(x − 3)
We need to solve for 2 variables and we see that 1 and 3 will make factors zero, we will use these two substitutions to get the equations to solve for A and B.
Let x=1, 2(1)−8=A(1−1)+B(1−3)−6=A(0)+B(−2)−6=−2B3=B
Let x=3, 2(3)−8=A(3−1)+B(3−3)−2=A(2)+B(0)−2=2A−1=A
Thus, (2x − 8)/(x2 − 4x + 3) = −1/(x − 3) + 3/(x − 1)
The decomposition of (2x2 − 6x − 2)/(x2 − 4x + 3) = 2 − 1/(x − 3) + 3/(x − 1).
34.62 Resolve (3x3 − 12x2 + 9x − 21)/[(x2 + 5)(x − 2)2] into partial fractions.
SOLUTION
(3x3 − 12x2 + 9x − 21)/[(x2 + 5)(x − 2)2] = (Ax + B)/(x2 + 5) + C/(x − 2)2 + D/(x − 2)
(3x3 − 12x2 + 9x − 21)/[(x2 + 5)(x − 2)2] = [(Ax + B)(x − 2)2 + C(x2 + 5) + D(x2 + 5)(x − 2)]/[(x2 + 5)(x − 2)2]
Thus, (3x3 − 12x2 + 9x − 21) = [(Ax + B)(x − 2)2 + C(x2 + 5) + D(x2 + 5)(x − 2)]
We need to solve for 4 variables and we see that 2 will make a factor zero, we will use it and three other numbers for the four substitutions to get the equations to solve for A, B, C and D. Any numbers can be used, but we use 0, 1 and −1 to make the other three substitutions.
Let x = 2, 3(2)3 − 12(2)2 + 9(2) − 21 = (A(2) + B)(2 − 2)2 + C(22 + 5) + D(22 + 5)(2 − 2)
24−48+18−21=(2A+B)(0)+C(9)+D(9)(0)−27=9C−3=C
Let x=0,3(0)3−12(0)2+9(0)−21=(A(0)+B)(0−2)2+C(02+5)+D(02+5)(0−2)−21=B(4)+(−3)(5)+D(5)(−2)−21=4B−15−10D−6=4B−10D−3=2B−5D
Let x=1, 3(1)3−12(1)2+9(1)−21=(A(1)+B)(1−2)2+C(12+5)+D(12+5)(1−2)−21=(A+B)(−1)2+(−3)(6)+D(6)(−1)−21=A+B−18−6D−3=A+B−6D
Let x=−1, 3(−1)3−12(−1)2+9(−1)−21=(A(−1)+B)(−1−2)2+C((−1)2+5)+D((−1)2+5)((−1)−2)]−45=(−A+B)(9)+(−3)(6)+D(6)(−3)−45=−9A+9B−18−18D−27=−9A+9B−18D−3=−A+B−2D
Now we have a system of three equations to solve for A, B, and D,
2B−5D=−3equation 1A+B−6D=−3equation 2−A+B−2D=−3equation 3
We eliminate A in equations 2 and 3.
A+B−6D=−3−A+B−2D=−3¯ 2B−8D=−6equation 4
Now we solve equations 1 and 4 for B and D.
Using equation 2, B = 1, and D = 1, we get A + 1 − 6(1) = −3 and A = 2.
We now have A = 2, B = 1, C = −3, and D = 1.
(3x3 − 12x2 + 9x − 21)/[(x2 + 5)(x − 2)2] = (Ax + B)/(x2 + 5) + C/(x − 2)2 + D/(x − 2)
(3x3 − 12x2 + 9x − 21)/[(x2 + 5)(x − 2)2] = (2x + 1)/(x2 + 5) − 3/(x − 2)2 + 1/(x − 2)
The decomposition of (3x3 − 12x2 + 9x − 21)/[(x2 + 5)(x − 2)2] = (2x + 1)/(x2 + 5) − 3/(x − 2)2 + 1/(x − 2).
CHAPTER 32
34.63 Use the iteration method to find the smallest positive real zero of P(x) = x3 − 9x + 5 to the nearest hundredth.
SOLUTION
P(x) = x3 − 9x + 5
P(0) = 5, P(1) = 1 − 9 + 5 = −3 Since P(0) and P(1) have opposite signs, the smallest positive real zero of P(x) is between 0 and 1.
x3 − 9x + 5 = 0
x3 − 9x = −5
x(x2 − 9) = −5
x = −5/(x2 − 9)
The iteration function we will use is g(x) = −5/(x2 − 9).
Let x0 = 0 g(0) = −5/(02 − 9) = 0.556
Let x1 = 0.556 g(0.556) = −5/([0.556]2 − 9) = 0.575
Let x2 = 0.575 g(0.575) = −5/([0.575]2 − 9) = 0.577
Since 0.575 rounds to 0.58 and 0.577 round to 0.58, to the nearest hundredth the smallest positive real zero of P(x) = x3 − 9x + 5 is 0.58.
34.64 Use the bisection method to find the smallest positive real zero of P(x) = x3 − 6x2 + 5x − 3 to the nearest hundredth.
SOLUTION
P(x) = x3 − 6 x2 + 5x − 3
P(0) = 03 − 6(0)2 + 5(0) − 3 = −3, P(1) = 13 − 6(1)2 + 5(1) − 3 = −3, P(2) = 23 − 6(2)2 + 5(2) − 3 = −9,
P(3) = 33 − 6(3)2 + 5(3) − 3 = −15, P(4) = 43 − 6(4)2 + 5(4) − 3 = −15, P(5) = 53 − 6(5)2 + 5(5) − 3 = −3,
P(6) = 63 − 6(6)2 + 5(6) − 3 = 27 Since P(5) and P(6) have opposite signs, there is a zero between 5 and 6. The zero interval is (5, 6).
The midpoint of the interval is M = (5 + 6)/2 = 5.5. If P(5.5) is negative we would replace the 5 in the zero interval since P(5) is negative, and if P(5.5) is positive we would replace the 6 in the zero interval with 5.5 since P(6) is positive.
Since both endpoints of the zero interval (5.137, 5.141) round to 5.14 to the nearest hundredth, the smallest positive real zero of P(x) = x3 − 6x2 + 5x − 3 is 5.14.
CHAPTER 33
34.65 Find the derivative of f (x) = 2x2 + x + 6 using the definition of the derivative.
SOLUTION
f(x)=2x2+x+6f′(x)=lim h→0 [f(x+h)−f(x)]/h=lim h→0 {[2(x+h)2+(x+h)+6]−[2x2+x+6]}/h=lim h→0 {2x2+4xh+2h2+x+h+6−2x2−x−6}/h=lim h→0 {4xh+2h2+h}/h=lim h→0 {h[4x+2h+1]}/h=lim h→0 {4x+2h+1}f′(x)=4x+1
The derivative of f (x) = 2x2 + x + 6 is f ′(x) = 4x + 1.
34.66 What number exceeds its square by the greatest amount?
SOLUTION
Let x be the number.
f(x)=x−x2
We want the maximum value for the function f (x) = x − x2.
Since f (x) = −x2 + x has a graph that is a parabola opening downward, we want the vertex of the parabola. (See Section 17.4)
f(x)=−x2+xy=−x2+xy=−(x2−x)y−1/4=−(x2−x+1/4)y−1/4=−(x−1/2)2
So the vertex of the parabola is (1/2, 1/4).
f(x)=x−x2 has its maximum at x=1/2.
Thus, 1/2 is the number that exceeds its square by the greatest amount.
