Kunci Jawaban dan Pembahasan Soal || Buku Calculus 9th Purcell Chapter 0 - 0.2 Number 1 – 34

Pembahasan Soal Buku Calculus 9th Purcell Chapter 0 Section 0.2

◀ CHAPTER 0 PRELIMINARIES ▶

◀ SECTION 0.2 Inequalities And Absolute Values ▶

Concepts Review

The set {x: –1 ≤ x < 5} is written in interval notation as __________ and the set {x: x ≤ –2} is written as __________.

Answer:

[–1, 5); (–∞, –2]

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If a/b < 0, then either a < 0 and __________ or a > 0 and __________.

Answer:

b > 0; b < 0

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Which of the following are always true?

(a) |–x| = x

(b) |x|² = x²

(d) √(x²) = x

Answer:

(b) and (c)

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The inequality |x – 2| ≤ 3 is equivalent to __________ ≤ x ≤ __________.

Answer:

–1 ≤ x ≤ 5

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Problem Set 0.2, Number 1 - 42.

Show each of the following intervals on the real line.

(a) [–1,1]

(b) (–4, 1]

(d) [1, 4]

(e) [–1, ∞)

(f) (–∞, 0]

Answer:

(a)

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(b)

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(c)

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(d)

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(e)

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(f)

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Use the notation of Problem 1 to describe the gollowing intervals.

(a)

(b)

(c)

(d)

Answer:

(a) (2, 7)

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(b) [–3, 4)

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(d) [–1, 3]

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In each of Problems 3-26, express the solution set of the given inequality in interval natation and sketch its graph.

x – 7 < 2x – 5

Answer:

x – 7 < 2x – 5

–2 < x; (–2, ∞)

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3x – 5 < 4x – 6

Answer:

3x – 5 < 4x – 6

1 < x; (1, ∞)

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7x – 2 ≤ 9x + 3

Answer:

7x – 2 ≤ 9x + 3

–5 ≤ 2x

x ≥ –5/2; [–5/2 ,∞)

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5x – 3 > 6x – 4

Answer:

5x – 3 > 6x – 4

1 > x; (–∞,1)

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–4 < 3x + 2 < 5

Answer:

–4 < 3x + 2 < 5

–6 < 3x < 3

–2 < x < 1; (–2,–1)

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–3 < 4x – 9 < 11

Answer:

–3 < 4x – 9 < 11

6 < 4x < 20

3/2 < x < 5; (3/2 ,5)

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–3 < 1 – 6x ≤ 4

Answer:

–3 < 1 – 6x ≤ 4

–4 < –6x ≤ 3

2/3 > x ≥ –½; [–½, 2/3)

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4 < 5 – 3x < 7

Answer:

4 < 5 – 3x < 7

–1 < –3x < 2

1/3 > x < –2/3 ; (–2/3 , 1/3)

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x² + 2x – 12 < 0

Answer:

x² + 2x – 12 < 0

x = ^{–2±√[(2)² – 4(1)(–12)]}/₂₍₁₎

x = ^–2±√52/₂

x = –1±√13

[x–(–1+√13)][x–(–1-√13)] < 0;

(–1-√13,–1+√13)

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x² – 5x – 6 > 0

Answer:

x² – 5x – 6 > 0

(x + 1)(x – 6) > 0;

(–∞,–1) ∪ (6,∞)

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2x² + 5x – 3 > 0

Answer:

2x² + 5x – 3 > 0

(–∞,–3) ∪ (½,∞)

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4x² – 5x – 6 < 0

Answer:

4x² – 5x – 6 < 0

(4x + 3)(x – 2) < 0 ; (–3/4 ,2)

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^{x + 4}/_{x – 3} ≤ 0

Answer:

^{x + 4}/_{x – 3} ≤ 0; [–4, 3)

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^{3x – 2}/_{x – 1} ≥ 0

Answer:

^{3x – 2}/_{x – 1} ≥ 0; (–∞, 2/3] ∪ (1,∞)

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²/_x < 5

Answer:

²/_x < 5

²/_x – 5 < 0

^{2 – 5x}/_x < 0;

(–∞,0) ∪ (2/5 ,∞)

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⁷/_4x ≤ 7

Answer:

⁷/_4x ≤ 7

⁷/_4x – 7 ≤ 0

^{7 – 28x}/_4x ≤ 0

(–∞,0) ∪ [1/4 ,∞)

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¹/_{3x – 2} ≤ 4

Answer:

¹/_{3x – 2} ≤ 4

¹/_{3x – 2} – 4 ≤ 0

^{1 – 4(3x – 2)}/_{3x – 2} ≤ 0

^{9 – 12x}/_{3x – 2} ≤ 0; (–∞, 2/3) ∪ [3/4 ,∞)

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³/_{x + 5} > 2

Answer:

³/_{x + 5} > 2

³/_{x + 5} – 2 > 0

^{3 – 2(x + 5)}/_{x + 5} > 0

^{–2x – 7}/_{x + 5} > 0; (–5,–7/2)

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(x + 2)(x – 1)(x – 3) > 0

Answer:

(x + 2)(x – 1)(x – 3) > 0

(x + 2)(x – 1)(x – 3) > 0; (–2, 1) ∪ (3, 8)

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(2x + 3)(3x – 1)(x – 2) < 0

Answer:

(2x + 3)(3x – 1)(x – 2) < 0

(2x + 3)(3x – 1)(x – 2) < 0; (–∞, –3/2) ∪ (1/3, 2)

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(2x – 3)(x – 1)²(x – 3) ≥ 0

Answer:

(2x – 3)(x – 1)²(x – 3) ≥ 0

(2x – 3)(x – 1)²(x – 3) ≥ 0; (–∞, –3/2] ∪ [3,∞)

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(2x – 3)(x – 1)²(x – 3) > 0

Answer:

(2x – 3)(x – 1)²(x – 3) > 0;

(–∞,1) ∪ (1, 3/2) ∪ (3,∞)

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x³ – 5x² – 6x < 0

Answer:

x³ – 5x² – 6x < 0

x(x² – 5x – 6) < 0

x(x + 1)(x – 6) < 0

(–∞,–1) ∪ (0, 6)

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x³ – x² – x + 1 > 0

Answer:

x³ – x² – x + 1 > 0

(x² – 1)(x – 1) > 0

(x + 1)(x – 1)² > 0;

(–1, 1) ∪ (1,∞)

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Tell whether each of the following is true or false.

(a) –3 < –7

(b) –1 > –17

Answer:

(a) False.

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(b) True.

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Tell whether each of the following is true or false.

(a) –5 > –√26

(b) ⁶/₇ < ³⁴/₃₉

Answer:

(a) True.

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(b) True.

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Assume that a > 0, b > 0. Prove each statement. Hint: Each part requires two proofs: one for ⇒ and one for ⇐.

(a) a < b ⇔ a² < b²

(b) a < b ⇔ ¹/_a > ¹/_b

Answer:

(a)

⇒ Let a < b, so ab < b². Also, b² < ab. Thus, a² < ab < b² and a² < b². ⇐ Let a² < b², so a ≠ b Then

0 < (a – b)² = a² – 2ab + b²

< b² – 2ab + b² = 2b(b – a)

Since b > 0, we can divide by 2b to get b – a > 0.

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(b)

We can divide or multiply an inequality by any positive number.

a < b ⇔ ^a/_b < 1 ⇔ ¹/_b < ¹/_a.

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Which of the following are true if a ≤ b?

(a) a² ≤ ab

(b) a – 3 ≤ a – 3

(d) –a ≤ –b

Answer:

(b) and (c) are true.

(a) is false: Take a = –1, b = 1.

(d) is false: if a ≤ b, then –a ≥ –b

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Find all values of x that satisfy both inequalities simultaneously.

(a) 3x + 7 > 1 and 2x + 1 < 3

(b) 3x + 7 > 1 and 2x + 1 > –4

Answer:

(a)

3x + 7 > 1 and 2x + 1 < 3

3x > –6 and 2x < 2

x > –2 and x < 1; (–2, 1)

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(b)

3x + 7 > 1 and 2x + 1 > –4

3x > –6 and 2x > –5

x > –2 and x > -5/2; (-2, ∞)

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(c)

3x + 7 > 1 and 2x + 1 < –4

x > –2 and –5/2; ∅

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Find all the values of x that satisfy at least one of the two inequalities.

(a) 2x – 7 > 1 or 2x + 1 < 3

(b) 2x – 7 ≤ 1 or 2x + 1 < 3

Answer:

(a)

2x – 7 > 1 or 2x + 1 < 3

2x > 8 or 2x < 2

x > 4 or x < 1

(–∞, 1) ∪ (4, ∞)

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(b)

2x – 7 ≤ 1 or 2x + 1 < 3

2x ≤ 8 or 2x < 2

x ≤ 4 or x < 1

(–∞, 4]

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(c)

2x – 7 ≤ 1 or 2x + 1 > 3

2x ≤ 8 or 2x > 2

x ≤ 4 or x > 1

(–∞, ∞)

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Solve for x, expressing your answer in interval notation.

(a) (x + 1)(x² + 2x – 7) ≥ x² – 1

(b) x⁴ – x² ≥ 8

Answer:

(a)

(x + 1)(x² + 2x – 7) ≥ x² – 1

x³ + 3x² – 5x – 7 ≥ x² – 1

x³ + 2x² – 5x – 6 ≥ 0

(x + 3)(x + 1)(x – 2) ≥ 0

[–3, –1] ∪ [2, ∞)

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(b)

x⁴ – x² ≥ 8

x⁴ – x² – 8 ≥ 0

(x² – 4)(x² + 2) ≥ 0

(x² + 2)(x + 2)(x – 2) ≥ 0

(–∞, –2] ∪ [2, ∞)

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(c)

(x² + 1)² – 7 (x² + 1) + 10 < 0

[(x² + 1) – 5] [(x² + 1) – 2] < 0

(x² – 4)(x² – 1) < 0

(x + 2)(x + 1)(x – 1)(x – 2) < 0

(–2, –1) ∪ (1, 2)

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Solve each inequality. Express your solution in interval notation.

(a) 1.99 < ¹/_x < 2.01

(b) 2.99 < ¹/_{x + 2} < 3.01

Answer:

(a)

1.99 < ¹/_x < 2.01

1.99x < 1 < 2.01x

1.99x < 1 and 1 < 2.01x

x < ¹/_1.99 and x > ¹/_2.01

¹/_2.01 < x < ¹/_1.99

(¹/_2.01 , x < ¹/_1.99)

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(b)

2.99 < ¹/_{x + 2} < 3.01

2.99(x + 2) < 1 < 3.01(x + 2)

2.99x + 5.98 < 1 and 1 < 3.01x + 6.02

x < ^–4.98/_2.99 and x > ^–5.02/_3.01

–^5.02/_3.01 < x < –^4.98/_2.99

(–^5.02/_3.01 , –^4.98/_2.99)

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