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Kunci Jawaban dan Pembahasan Soal || Buku Calculus 9th Purcell Chapter 0 - 0.6 Number 1 – 21

Pembahasan Soal Buku Calculus 9th Purcell Chapter 0 Section 0.6

CHAPTER 0 PRELIMINARIES

SECTION 0.6 Operations on Functions


Calculus 9th Purcell Chapter 0 - 0.6

Problem Set 0.6, Number 1 – 21.
  1. For f(x) = x + 3 and g(x) = x2 find each value (if possible).
    2. (a)
    (f + g)(2)
    (b)
    (f · g)(0)
    (c)
    (g / f)(3)
    (d)
    (f ◦ g)(1)
    (e)
    (g ◦ f)(1)
    (f)
    (g ◦ f)(–8)
    💬 SOLUTION:
    (a)
    (f + g)(2) = (2 + 3) + 22 = 9
    (b)
    (f · g)(0) = (0 + 3)(02) = 0
    (c)
    (g / f)(3) = 32 / [3 + 3] = 9/6 = 3/2
    (d)
    (f ◦ g)(1) = f(12) = 1 + 3 = 4
    (e)
    (g ◦ f)(1) = g(1 + 3) = 42 = 16
    (f)
    (g ◦ f)(–8) = g(–8 + 3) = (–5)2 = 25

  2. For f(x) = x2 + x and g(x) = 2 / (x + 3) , find each value.
    3. (a)
    (f – g)(2)
    (b)
    (f / g)(1)
    (c)
    g2(3)
    (d)
    (f ◦ g)(1)
    (e)
    (g ◦ f)(1)
    (f)
    (g ◦ g)(3)
    💬 SOLUTION:
    (a)
    (f – g)(2) = (22 + 2) – [2 / (2 + 3)] = 6 – [2 / 5] = 28/5
    (b)
    (f / g)(1) = [12 + 1] / [2 / (1 + 3)] = [2] / [2 / 4] = 4
    (c)
    g2(3) = [(2 / (3 + 3))]2 = [1 / 3]2 = 1/9
    (d)
    (f ◦ g)(1) = f[(2) / (1 + 3)] = [1/2]2 + [1/2] = ¾
    (e)
    (g ◦ f)(1) = g(12 + 1) = (2) / (2 + 3) = 2/5
    (f)
    (g ◦ g)(3) = g[2 / (3 + 3)] = (2) / [(1/3) + 3] = (2) / [10/3] = 3/5

  3. For Φ(u) = u3 + 1 and ψ(v) = 1 / v, find each value.
    4. (a)
    (Φ + ψ)(t)
    (b)
    (Φ ◦ ψ)(r)
    (c)
    (ψ ◦ Φ)(r)
    (d)
    Φ3(z)
    (e)
    (Φ – ψ)(5t)
    (f)
    ((Φ – ψ) ◦ ψ)(t)
    💬 SOLUTION:
    (a)
    (Φ + ψ)(t) = t3 + 1 + (1/t)
    (b)
    (Φ ◦ ψ)(r) = Φ(1/r) = (1/r)3 + 1 = (1/r3) + 1
    (c)
    (ψ ◦ Φ)(r) = ψ(r3 + 1) = (1) / [r3 + 1]
    (d)
    Φ3(z) = (z3 + 1)3
    (e)
    (Φ – ψ)(5t) = [(5t)3 + 1] – [1/5t] = 125t3 + 1 – [1/5t]
    (f)
    ((Φ – ψ) ◦ ψ)(t) = (Φ – ψ)(1/t) = (1/t)3 + 1 – [1 / (1/t)] = (1/t3) + 1 – t

  4. If f(x) = √(x2 – 1) and g(x) = 2 / x, find formulas for the following and state their domains.
    5. (a)
    (f ∙ g)(x)
    (b)
    f4(x) + g4(x)
    (c)
    (f ◦ g)(x)
    (d)
    (g ◦ f)(x)
    💬 SOLUTION:
    (a)
    (f ∙ g)(x) = [2√(x2 – 1)] / x
     
    Domain: (–∞, –1] ∪ [1, ∞)
    (b)
    f4(x) + g4(x) = [√(x2 – 1)]4 + [2 / x]4
     
    = (x2 – 1)2 + [16 / x4]
     
    Domain: (–∞, 0) ∪ (0, ∞)
    (c)
    (f ◦ g)(x) = f(2 /x) = √[(2 / x)2 – 1] = √[(4 / x2) – 1]
     
    Domain: (–2, 0) ∪ (0, 2]
    (d)
    (g ◦ f)(x) = g[√(x2 – 1)] = 2 / √(x2 – 1)
     
    Domain: (–∞,–1) ∪ (1, ∞)

  5. If f(s) = √(s2 – 4) and g(w) = |1 + w|, find formulas for (f ◦ g)(x) and (g ◦ f)(x).
    6. 💬 SOLUTION:
    -
    (f ◦ g)(x) = f(|1 + x|) = √(|1 + x|2 – 4)
     
    √(x2 + 2x – 3)
    -
    (g ◦ f)(x) = g[√(x2 – 4)] = |1 + √(x2 – 4)|
     
    1 + √(x2 – 4)

  6. If g(x) = x2 + 1, find formulas for g3(x) and (g ◦ g ◦ g)(x).
    7. 💬 SOLUTION:
    -
    g3(x) = (x2 + 1)3 = (x4 + 2x2 + 1)(x2 + 1)
     
    = x6 + 3x4 + 3x2 + 1
    -
    (g ◦ g ◦ g)(x) = (g ◦ g)(x2 + 1)
     
    = g[(x2 + 1)2 + 1] = g(x4 + 2x2 + 2)
     
    = (x4 + 2x2 + 2)2 + 1
     
    = x8 + 4x6 + 8x4 + 8x2 + 5

  7. Calculate g(3.141) if g(u)
    =
    √(u3 + 2u)
    .
    2 + u
    8. 💬 SOLUTION:
    g(3.141) ≈ 1.188

    Semoga Bermanfaat 😁

  8. Calculate g(2.03) if g(x)
    =
    [√x – ∛x]4
    .
    1 – x + x2
    9. 💬 SOLUTION:
    g(2.03) ≈ 0.000205

  9. Calculate [g2(π) – g(π)]1/3 if g(v) = |11 – 7v|.
    10. 💬 SOLUTION:
    [g2(π) – g(π)]1/3
    = [(11 – 7π)2 – |11 – 7π|]1/3
    ≈ 4.789

  10. Calculate [g3(π) – g(π)]1/3 if g(x) = |6x – 11|.
    11. 💬 SOLUTION:
    [g3(π) – g(π)]1/3
    = [(6π – 11)3 – (6π – 11)]1/3
    ≈ 7.807

  11. Find f and g so that F = g ◦ f. (See Example 3.)
    12. (a)
    F(x) = √(x + 7)
    (b)
    F(x) = (x2 + x)15
    💬 SOLUTION:
    (a)
    g(x) = √x, f(x) = x + 7
    (b)
    g(x) = x15, f(x) = x2 + x

  12. Find f and g so that p = f ◦ g.
    13. (a)
    p(x)
    =
    2
    (x2 + x + 1)3
    (b)
    p(x)
    =
    1
     
    x3 + 3x
     
    💬 SOLUTION:
    (a)
    f(x) = 2 / x3, g(x) = x2 + x + 1
    (b)
    f(x) = 1 / x, g(x) = x3 + 3x

  13. Write p(x) = 1 / (√(x2 + 1)) as a composite of three functions in two different ways.
    14. 💬 SOLUTION:
    p = f ◦ g ◦ h if f(x) = 1 / x, g(x) = √(x), h(x) = x2 + 1
    p = f ◦ g ◦ h if f(x) = 1 / √(x), g(x) = x + 1, h(x) = x2

  14. Write p(x) = 1 / (√(x2 + 1)) as a composite of four functions.
    15. 💬 SOLUTION:
    p = f ◦ g ◦ h ◦ l if f(x) = 1 / x, g(x) = √(x), h(x) = x + 1, l(x) = x2

    Semoga Bermanfaat 😁

  15. Sketch the graph of f(x) = √(x – 2) – 3 by first sketching g(x) = √(x) and then translating. (See Example 4.)
    16. 💬 SOLUTION:
    Translate the graph of g(x) = √(x) to the right 2 units and down 3 units.
    Calculus 9th Purcell Chapter 0 - 0.6
    Calculus 9th Purcell Chapter 0 - 0.6

  16. Sketch the graph of g(x) = |x + 3| – 4 by first sketching h(x) = |x| and then translating.
    17. 💬 SOLUTION:
    Translate the graph of h(x) = |x| to the left 3 units and down 4 units.
    Calculus 9th Purcell Chapter 0 - 0.6
    Calculus 9th Purcell Chapter 0 - 0.6

  17. Sketch the graph of f(x) = (x – 2)2 – 4 using translations.
    18. 💬 SOLUTION:
    Translate the graph of y = x2 to the right 2 units and down 4 units.
    Calculus 9th Purcell Chapter 0 - 0.6

  18. Sketch the graph of g(x) = (x + 1)3 – 3 using translations.
    19. 💬 SOLUTION:
    Translate the graph of y = x3 to the left 1 unit and down 3 units.
    Calculus 9th Purcell Chapter 0 - 0.6

  19. Sketch the graphs of f(x) = (x – 3) / 2 and g(x) = √(x) using the same coordinate axes. Then sketch f + g by adding y-coordinates.
    20. 💬 SOLUTION:
    (f + g)(x) = [(x – 3) / 2] + √(x)
    Calculus 9th Purcell Chapter 0 - 0.6

  20. Follow the directions of Problem 19 for f(x) = x and g(x) = |x|.
    21. 💬 SOLUTION:
    (f + g)(x) = x + |x|
    Calculus 9th Purcell Chapter 0 - 0.6

  21. Sketch the graph of F(t)
    =
    |t| – t
    .
    t
    22. 💬 SOLUTION:
    F(t)
    =
    |t| – t
    .
    t
    Calculus 9th Purcell Chapter 0 - 0.6

Semoga Bermanfaat 😁

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