◀ CHAPTER 0 PRELIMINARIES ▶
◀ SECTION 0.2 Real Numbers, Estimation, and Logic ▶
Problem Set 0.2, Number 35 - 63.
- |x – 2| ≥ 5
- |x + 2| < 1
- |4x + 5| ≤ 10
- |2x – 1| > 2
- |2x/7 – 5| ≥ 7
- |x/4 + 1| < 1
- |5x – 6| > 1
- |2x – 7| > 3
- |1/x – 3| > 6
- |2 + 5/x| > 1
- x2 – 3x – 4 ≥ 0
- 3x2 + 17x – 6 > 0
- x2 – 4x + 4 ≤ 0
- 14x2 + 11x – 15 ≤ 0
- |x – 3| < 0.5 ⇒ |5x – 15| < 2.5
- |x + 2| < 0.3 ⇒ |4x + 8| < 1.2
- |x – 2| < ɛ/6 ⇒ |6x – 12| < ɛ
- |x + 4| < ɛ/2 ⇒ |2x + 8| < ɛ
- |x – 5| < δ ⇒ |3x – 15| < ɛ
- |x – 2| < δ ⇒ |4x – 8| < ɛ
- |x + 6| < δ ⇒ |6x + 36| < ɛ
- |x + 5| < δ ⇒ |5x + 25| < ɛ
- On a lathe, you are to turn out a isk (this right circular cylinder) of curcumference 10 inches. This is done by continually measuring the diameter as you make the disk smaller. How closely must you measure the diameter if you can tolerate an error of most 0.02 inch in the circumference?
- Fahrenheit temperatures and Celsius temperatures are related by the formula C = 5/9 (F – 32). An experiment requires that a solution be kept at 50°C with an error of at most 3% (or 1.5°). You have only a Fahrenheit thermometer. What error are you allowed on it?
- |x – 1| < 2|x – 3|
- |2x – 1| ≥ |x + 1|
- 2|2x – 3| < |x + 10|
- |3x – 1| < 2|x + 6|
- Prove that |x| < |y| ⇔ x2 < y2 by giving a reason for each of these steps:
In Problems 35–44, find the solution sets of the given inequalities.
Answer:
|x – 2| ≥ 5
x – 2 ≤ –5 or x – 2 ≥ 5
x ≤ –3 or x ≥ 7
(–∞, –3] ∪ [7, ∞)
■
Answer:
|x + 2| < 1
–1 < x + 2 < 1
–1 < x < –3
(–3, –1)
■
Answer:
|4x + 5| ≤ 10
–10 ≤ 4x + 5 ≤ 10
–15 ≤ 4x ≤ 5
-15/4 ≤ x ≤ 5/4;
[-15/4, 5/4]
■
Answer:
|2x – 1| > 2
2x – 1 < 2 or 2x – 1 > 2
2x < –1 or 2x > 3;
x < –1/2 or x > 3/2, (–∞, –1/2) ∪ (3/2, ∞)
■
Answer:
|2x/7 – 5| ≥ 7
2x/7 – 5 ≤ –7 or 2x/7 – 5 ≥ 7
2x/7 ≤ –2 or 2x/7 ≥ 12
x ≤ –7 or x ≥ 42;
(–∞, –7] ∪ [42, ∞)
■
Answer:
|x/4 + 1| < 1
–1 < x/4 + 1 < 1
–2 < x/4 < 0;
–8 < x < 0; (–8, 0)
■
Answer:
|5x – 6| > 1
5x – 6 < –1 or 5x – 6 > 1
5x < 5 or 5x > 7
x < 1 or x > 7/5; (–∞, 1) ∪ (7/5, ∞)
■
Answer:
|2x – 7| > 3
2x – 7 < –3 or 2x – 7 > 3
2x < 4 or 2x > 10
x < 2 or x > 5; (–∞, 2) ∪ (5, ∞)
■
Answer:
|1/x – 3| > 6
1/x – 3 < –6 or 1/x – 3 > 6
1/x + 3 < 0 or 1/x – 9 > 0
1 + 3x/x < 0 or 1 – 9x/x > 0;
(–1/3, 0) ∪ (0, 1/9)
■
Answer:
|2 + 5/x| > 1
2 + 5/x < –1 or 2 + 5/x > 1
3 + 5/x < 0 or 1 + 5/x > 0
3x + 5/x < 0 or x + 5/x > 0;
(–∞, –5) ∪ (–5/3), 0) ∪ (0, ∞)
■
In Problems 45–48, solve the given quadratic inequality using the Quadratic Formula.
Answer:
x2 – 3x – 4 ≥ 0;
x = 3±√[(–3)2 – 4(1)(–4)]/2(1)
x = 3±√5/2
x = –1,4
(x + 1)(x – 4) = 0; (–∞, –1] ∪ [4, ∞)
■
Answer:
3x2 + 17x – 6 > 0
x = –17±√[(17)2 – 4(3)(–6)]/2(3)
x = –17±19/6
x = –6,1/3
(3x – 1)(x + 6) > 0; (–∞, –6] ∪ [1/3, ∞)
■
Answer:
x2 – 4x + 4 ≤ 0
x = 4±√[(–4)2 – 4(1)(4)]/2(1)
x = 2
(x – 2)(x – 2) ≤ 0; x = 2
■
Answer:
14x2 + 11x – 15 ≤ 0
x = –11±√[(11)2 – 4(14)(–15)]/2(14)
x = –11±31/28
x = –3/2,5/7
(x + 3/2)(x – 5/7) ≤ 0; [–3/2,5/7]
■
In Problem 49–52, show that the indicated implication is thue.
Answer:
|x – 3| < 0.5 ⇒ 5|x – 3| < 5(0.5) < ⇒ |5x – 15| < 2.5
■
Answer:
|x + 2| < 0.3 ⇒ 4|x + 2| < 4(0.3) ⇒ |4x + 8| < 1.2
■
Answer:
|x – 2| < ɛ/6 ⇒ 6|x – 2| < ɛ ⇒ |6x – 12| < ɛ
■
Answer:
|x + 4| < ɛ/2 ⇒ 2|x + 4| < ɛ ⇒ |2x + 8| < ɛ
■
In Problems 53–56, find δ (depending on ɛ) so that the given implication is true.
Answer:
|3x – 15| < ɛ ⇒ |3(x – 5)| < ɛ
⇒ 3|x – 5| < ɛ
⇒ |x – 5| < ɛ/3; δ = ɛ/3
■
Answer:
|4x – 8| < ɛ ⇒ |4(x – 2)| < ɛ
⇒ 4|x – 2| < ɛ
⇒ |x – 2| < ɛ/4; δ = ɛ/4
■
Answer:
|6x + 36| < ɛ ⇒ |6(x + 6)| < ɛ
⇒ 6|x + 6| < ɛ
⇒ |x + 6| < ɛ/6; δ = ɛ/6
■
Answer:
|5x + 25| < ɛ ⇒ |5(x + 5)| < ɛ
⇒ 5|x + 5| < ɛ
⇒ |x + 5| < ɛ/5; δ = ɛ/5
■
Answer:
C = πd
|C – 10| ≤ 0.02
|πd – 10| ≤ 0.02
|π(d – 10/π)| ≤ 0.02
|d – 10/π| ≤ 0.02/π ≈ 0.0064
We must measure the diameter to an accuracy of 0.0064 in.
■
Answer:
|C – 50| ≤ 1.5, |5/9(F – 32)– 50| ≤ 1.5;
|5/9(F – 32)– 90| ≤ 1.5
|F – 122| ≤ 2.7
We are allowed an error of 2.7° F.
■
In Problems 59–62, solve the inequalities.
Answer:
|
|x – 1|
|
<
|
2|x
– 3|
|
|
|x – 1|
|
<
|
|2x
– 6|
|
|
(x – 1)2
|
<
|
(2x
– 6)2
|
|
x2
– 2x + 1
|
<
|
4x2
– 24x + 36
|
|
3x2
– 22x + 35
|
>
|
0
|
|
(3x – 7) (x – 5)
|
>
|
0
;
|
|
( –∞, 7/3 )
|
⋃
|
(5,
∞)
|
■
Answer:
|
|2x – 1|
|
≥
|
|x
+ 1|
|
|
(2x – 1)2
|
≥
|
(x
+ 1)2
|
|
4x2 – 4x + 1
|
≥
|
x2
+ 2x + 1
|
|
3x2 – 6x
|
≥
|
0
|
|
3x (x – 2)
|
≥
|
0
|
|
( –∞, 0]
|
⋃
|
[2,
∞)
|
■
Answer:
|
2|2x – 3|
|
<
|
|x
+ 10|
|
|
|4x – 6|
|
<
|
|x
+ 10|
|
|
(4x – 6)2
|
<
|
(x
+ 10)2
|
|
16x2 – 48x + 36
|
<
|
x2
+ 20x + 100
|
|
15x2 – 68x – 64
|
<
|
0
|
|
(5x + 4) (3x – 16)
|
<
|
0;
|
|
(–4/5, 16/3)
|
|
|
■
Answer:
|
|3x – 1|
|
<
|
2|x
+ 6|
|
|
|3x – 1|
|
<
|
|2x
+ 12|
|
|
(3x – 1)2
|
<
|
(2x
+ 12)2
|
|
9x2 – 6x + 1
|
<
|
4x2
+ 48x + 144
|
|
15x2 – 54x – 143
|
<
|
0
|
|
(5x + 11) (x – 13)
|
<
|
0;
|
|
(–11/5, 13)
|
|
|
■
|
|x| < |y|
|
⇒
|
|x||x|
≤ |x||y| and |x||y| < |y||y|
|
|
|
⇒
|
|x|2
< |y|2
|
|
|
⇒
|
x2
< y2
|
Conversely,
|
x2
< y2
|
⇒
|
|x|2
< |y|2
|
|
|
⇒
|
|x|2
– |y|2 < 0
|
|
|
⇒
|
(|x|
– |y|)(|x| + |y|) < 0
|
|
|
⇒
|
|x|
– |y| < 0
|
|
|
⇒
|
|x|
< |y|
|
Answer:
|
|x| < |y|
|
⇒
|
|x||x|
≤ |x||y| and |x||y| < |y||y|
|
|
|
|
|
Order property: x < y ⇔ xz < yz
when z is positive.
|
|
|
|
⇒
|
|x|2
< |y|2
|
|
|
|
|
Transitivity
|
|
|
|
⇒
|
x2
< y2
|
|
|
|
|
(|x|2 = x2)
|
|
|
Conversely,
|
|
||
|
x2
< y2
|
⇒
|
|x|2
< |y|2
|
|
|
|
|
(x2 = |x|2)
|
|
|
|
⇒
|
|x|2
– |y|2 < 0
|
|
|
|
|
Subtract |y|2 from each side.
|
|
|
|
⇒
|
(|x|
– |y|)(|x| + |y|) < 0
|
|
|
|
|
Factor the difference of two squares.
|
|
|
|
⇒
|
|x|
– |y| < 0
|
|
|
|
|
This is the only factor that can be negative.
|
|
|
|
⇒
|
|x|
< |y|
|
|
|
|
|
Add |y| to each side.
|
|
■
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