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Kunci Jawaban dan Pembahasan Soal || Buku Calculus 9th Purcell Chapter 0 - 0.2 Number 64 – 78

Pembahasan Soal Buku Calculus 9th Purcell Chapter 0 Section 0.2

◀ CHAPTER 0 PRELIMINARIES ▶

◀ SECTION 0.2 Inequalities And Absolute Values ▶


Problem Set 0.2, Number 64 – 78.
  1. Use the result of Problem 63 to show that
    2. 0 < a < b ⇒ √a < √b
    Answer:
    0 < a < b ⇒ a ⇒ (√a)2 and b ⇒ (√b)2, so (√a)2 < (√b)2, and, by Problem 63, |√a| < |√b| ⇒ √a < √b.

  2. Use the properties of the absolute value to show that each of the following is true.
    1. |a – b| ≤ |a| + |b|
    2. |a – b| ≥ |a| – |b|
    3. |a + b + c| ≤ |a| + |b| + |c|
    Answer:
    1. |a – b| = |a + (–b)| ≤ |a| + |–b| = |a| + |b|
    2. |a – b| ≥ ||a| – |b|| ≥ |a| – |b| Use Property 4 of absolute values.
    3. |a + b + c| = |(a + b) + c| ≤ |a + b| + |c|, so |a + b + c| ≤ |a| + |b| + |c|

    Calculus 9th Purcell Chapter 0 - 0.2 Number 64 - 78

  3. Use the Triangle Inequality and the fact that 0 < |a| < |b| ⇒ 1/|b| < 1/|a| to establish the following chain of inequalities.
    4.  
    1
    –
    1
     
    ≤
    1
    +
    1
    ≤
    1
    +
    1
    x2 + 3
    |x| + 2
    x2 + 3
    |x| + 2
    3
    2

    Answer:
     
    1
    –
    1
     
    =
     
    1
    +
    ⌈
    –
    1
    ⌉
    x2 + 3
    |x| + 2
    x2 + 3
    ⌊
    |x| + 2
    ⌋

     
     
     
     
     
    ≤
     
    1
     
    +
     
    –
    1
     
     
     
    x2 + 3
    |x| + 2
     

     
     
     
     
     
    =
     
    1
     
    +
     
    1
     
     
     
    x2 + 3
    |x| + 2
     

     
     
     
     
     
    =
    1
    +
    1
     
     
    x2 + 3
    |x| + 2

    by the Triangular Inequality, and since
    x2 + 3 > 0, |x + 2 > 0 ⇒ 1/x2 + 3 > 0, 1/|x| + 2 > 0.
    x2 + 3 ≥ 3 and |x| + 2 ≥ 2, so
    1/x2 + 3 ≤ 1/3 and 1/|x| + 2 ≤ 1/2
    1/x2 + 3 + 1/|x| + 2 ≤ 1/3 + 1/2

  4. Show that (see Problem 66)
    5.  
    x – 2
     
    ≤
    |x| + 2
    x2 + 9
    9
    Answer:
     
    x – 2
     
    =
     
    x + (–2)
     
    x2 + 9
    x2 + 9

     
    x – 2
     
    ≤
     
    x
     
    +
     
    –2
     
    x2 + 9
    x2 + 9
     
    x2 + 9
     

     
    x – 2
     
    ≤
     
    |x|
     
    +
     
    2
     
    x2 + 9
    x2 + 9
     
    x2 + 9
     

     
    x – 2
     
    =
     
    |x| + 2
    x2 + 9
    x2 + 9

    Since x2 + 9 ≥ 9,

     
    1
     
    ≤
     
    1
    x2 + 9
    9

     
    |x| + 2
     
    ≤
     
    |x| + 2
     
    x2 + 9
    9

     
    x – 2
     
    ≤
     
    |x| + 2
     
    x2 + 9
    9

  5. Show that
    6. |x| ≤ 2 ⇒
     
    x2 + 2x + 7
     
    ≤ 15
    x2 + 1
    Answer:
    |x| ≤ 2 ⇒ |x2 + 2x + 7|
    ≤
    |x2| + |2x| + 7
     
    ≤
    4 + 4 + 7 = 15

    and |x2 + 1| ≥ 1 so
    1
    ≤ 1.
    x2 + 1

    Thus,
     
    x2 + 2x + 7
     
    =
    |x2 + 2x + 7|
     
    1
     
    x2 + 1
    x2 + 1
     
     
     
     
    ≤
    15 · 1 = 15
     
     
     
     
     

  6. Show that
    7. |x| ≤ 1 ⇒ |x4 + 1/2 · x3 + 1/4 · x2 + 1/8 · x + 1/16| < 2
    Answer:
    |x4 + 1/2 · x3 + 1/4 · x2 + 1/8 · x + 1/16|
    ≤ |x4| + 1/2 · |x3| + 1/4 · |x2| + 1/8 · |x| + 1/16
    ≤ 1 + 1/2 + 1/4 + 1/8 + 1/16 since |x ≤ 1.
    So |x4 + 1/2 · x3 + 1/4 · x2 + 1/8 · x + 1/16| ≤ 1.9375 < 2.

  7. Show each of the following
    1. x < x2 for x < 0 or x > 1
    2. x2 < x
    Answer:
    1. x
      < 
      x2
      x – x2
      < 
      0
      x(1 – x)
      < 
      0
      x
      < 
      0 or x > 1

    2. x2
      < 
      x
      x2 – x
      < 
      0
      x(x – 1)
      < 
      0
      0
      < 
      x < 1

  8. Show that a ≠ 0 ⇒ a2 + 1/a2 ≥ 2. Hint: Consider (a – 1/a2).
    9. Answer:
    a ≠ 0 ⇒
    0 ≤ (a – 1/a)2 = a2 – 2 + 1/a2
    so, 2 ≤ a2 + 1/a2 or a2 + 1/a2 ≥ 2.

  9. The number ½(a + b) is called the average, or arithmetic mean, of a and b. Show that the arithmetic mean of two numbers is between the two numbers; that is, prove that
    10. a < b ⇒ a  <
    a + b
     < b
    2
    Answer:
    a < b
    a + a < a + b and a + b < b + b
    2a < a + b < 2b
    a < a + b/2 < b

  10. The number √ab is called the geometric mean of two positive numbers a and b. Prove that
    11. 0 < a < b ⇒ a < √ab < b
    Answer:
    0 < a < b
    a2 < ab and ab < b2
    a2 < ab < b2
    a < √ab < b

  11. For two positive numbers a and b, prove that
    12. √ab ≤ ½(a + b)
    This is the simplest version of a famous inequality called the geometric mean-arithmetic mean inequality.
    Answer:
    √ab ≤ ½ (a + b) ⇔ ab ≤ ¼ (a2 + 2ab + b2)
    ⇔ 0 ≤ ¼ · a2 – ½ · ab + ¼ · b2 = ¼ (a2 – 2ab + b2)
    ⇔ 0 ≤ ¼ (a – b)2 which is always true.

  12. Show that, among all rectangles with given perimeter p. the square has the largest area. Hint: If a and b denote the lengths of adjacent sides of a rectangle of perimeter p, then the area is ab, and for the square the area is a2 = [(a + b)/2]2. Now see Problem 74.
    13. Answer:
    For a rectangle the area is ab, while for a square the area is a2 = (a + b/2)2. From Problem 74, √ab ≤ ½ (a + b) ⇔ ab ≤ (a + b/2)2 so the square has the largest area.

  13. Solve 1 + x + x2 + x3 + ... + x99 ≤ 0
    14. Answer:
    1 + x + x2 + x3 + ... + x99 ≤ 0;
    (–∞, –1]

  14. The formula 1/R = 1/R1 + 1/R2 + 1/R3 gives the total resistance R in an electric circuit due to three resistances, R1, R2, and R3, connected in parallel. If 10 ≤ R1 ≤ 20, 20 ≤ R2 ≤ 30, and 30 ≤ R3 ≤ 40, find the range of values for R.
    15. Answer:
    1/R ≤ 1/10 + 1/20 + 1/30
    1/R ≤ 6 + 3 + 2/60
    1/R ≤ 11/60
    R ≥ 60/11
    1/R ≥ 1/20 + 1/30 + 1/40
    1/R ≥ 6 + 4 + 3/120
    R ≤ 120/13
    Thus, 60/11 ≤ R ≤ 120/13

  15. The radius of a sphere is measured to be about 10 inches. Determine a tolerance δ in this measurement that will ensure an error of less than 0.01 square inch in the calculated value of the surface area of the sphere.
    16. Answer:
    A = 4π · r2; A = 4π (10)2 = 400π
    |4π · r2 – 400π| < 0.01
    4π |r2 – 100| < 0.01
    |r2 – 100| < 0.01/4π
    – 0.01/4π < r2 – 100 < 0.01/4π
    √[100 – 0.01/4π] < r < √[100 + 0.01/4π]
    δ ≈ 0.00004 in

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