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Kunci Jawaban dan Pembahasan Soal || Buku Calculus 9th Purcell Chapter 0 - 0.6 Number 22 – 42

Pembahasan Soal Buku Calculus 9th Purcell Chapter 0 Section 0.6

CHAPTER 0 PRELIMINARIES

SECTION 0.6 Operations on Functions


Problem Set 0.6, Number 22 – 42.
  1. Sketch the graph of G(t) = t –〚t〛.
    2. 💬 SOLUTION:
    G(t) = t –〚t〛
    Calculus 9th Purcell Chapter 0 - 0.6

  2. State whether each of the following is an odd function, an even function, or neither. Prove your statements.
    3. (a)
    The sum of two even functions
    (b)
    The sum of two odd functions
    (c)
    The product of two even functions
    (d)
    The product of two odd functions
    (e)
    The product of an even function and an odd function
    💬 SOLUTION:
    (a)
    Even;
     
    (f + g)(–x) = f(–x) + g(–x) = f(x) + g(x)
     
    = (f + g)(x) if f and g are both even functions.
    (b)
    Odd;
     
    (f + g)(–x) = f(–x) + g(–x) = –f(x) – g(x)
     
    = –(f + g)(x) if f and g are both odd functions.
    (c)
    Even;
     
    (f ∙ g)(–x) = [f(–x)][g(–x)]
     
    = [f(x)][g(x)] = (f ∙ g)(x)
     
    if f and g are both even functions.
    (d)
    Even;
     
    (f ∙ g)(–x) = [f(–x)][g(–x)]
     
    = [–f(x)][–g(x)] = [f(x)][g(x)]
     
    = (f ∙ g)(x)
     
    if f and g are both odd functions.
    (e)
    Odd;
     
    (f ∙ g)(–x) = [f(–x)][g(–x)]
     
    = [f(x)][–g(x)] = –[f(x)][g(x)]
     
    = –(f ∙ g)(x)
     
    if f is an even function and g is an odd function.

  3. Let F be any function whose domain contains –x whenever it contains x. Prove each of the following.
    4. (a)
    F(x) – F(–x) is an odd function.
    (b)
    F(x) + F(–x) is an odd function.
    (c)
    F can always be expressed as the sum of an odd and an even function.
    💬 SOLUTION:
    (a)
    F(x) – F(–x) is odd because
     
    F(–x) – F(x) = –[F(x) – F(–x)]
    (b)
    F(x) + F(–x) is even because
     
    F(–x) + F(–(–x)) = F(–x) + F(x)]
     
    = F(x) + F(–x)
    (c)
    [F(x) – F(–x)] / 2 is odd an [F(x) + F(–x)] / 2 is even.
     
    [[F(x) – F(–x)] / 2 ] + [[F(x) + F(–x)] / 2]
     
    = [2F(x)] / 2
     
    = F(x)

  4. Is every polynomial of even degree an even function? Is every polynomial of odd degree an odd function? Explain.
    5. 💬 SOLUTION:
    Not every polynomial of even degree is an even function. For example f(x) = x2 + x is neither even nor odd. Not every polynomial of odd degree is an odd function. For example g(x) = x3 + x2 is neither even nor odd.

  5. Classify each of the following as a PF (polynomial function), RF (rational function but not a polynomial function), or neither.
    6. (a)
    f(x) = 3x1/2 + 1
    (b)
    f(x) = 3x2 + 2x–1
    (c)
    f(x) = 1 / (x + 1)
    (d)
    f(x) = 3
    (e)
    f(x) = πx3 – 3π
    (f)
    f(x) = (x + 1) / [√(x + 3)]
    💬 SOLUTION:
    (a)
    Neither
    (b)
    PF
    (c)
    RF
    (d)
    PF
    (e)
    RF
    (f)
    Neither

  6. The relationship between the unit price P (in cents) for a certain product and the demand D (in thousans of units) appears to satisfy
    7. P = √(29 – 3D + D2)
    On the other hand, the demand has risen over the t tears since 1970 according to D = 2 + √(t).
    (a)
    Express P as a function of t.
    (b)
    Evaluate P when t = 15.
    💬 SOLUTION:
    (a)
    P = √[29 – 3(2 + √(t)) + (2 + √(t))2]
     
    P = √[(t) + √(t) + 27]
    (b)
    When t = 15, P = √[(15) + √(15) + 27] ≈ 6.773

  7. After being in business for t years, a manufacturer of cars is making 120 + 2t + 3t2 units per year. The sales price in dollars per unit has risen according to the formula 6000 + 700t. Write a formula for the manufacturer's yearly revenue R(t) after t years.
    8. 💬 SOLUTION:
    P(t) = (120 + 2t + 3t2)(6000 + 700t)
    = 2100t3 + 19, 400t2 + 96, 000t + 720, 000

    Semoga Bermanfaat 😁

  8. Starting at noon, airplane A flies due north at 400 miles per hour. Starting 1 hour later, airplane B flies due east at 300 miles per hour. Neglecting the curvature of the Earth and assuming that they fly at the same altitude, find a formula for D(t), the distance between the two airplanes t hours after noon. Hint: There will be two formulas for D(t), one if 0 < t < 1 and the other if t ≥ 1.
    9. 💬 SOLUTION:
    D(t) =
    {
    400t
    if 0 < t < 1
    √((400t)2 + [300(t – 1]2)
    If t ≥ 1
    D(t) =
    {
    400t
    If 0 < t < 1
    √(250, 00t2 – 180, 000t + 90, 000)
    If t ≥ 1

  9. Fin the distance between the airplanes of Problem 29 at 2:30 p.m.
    10. 💬 SOLUTION:
    D(2.5) ≈ 1097 mi

  10. Let f(x) = [ax + b] / [cx – a]. Show that f(f(x)) = x, provided a2 + bc ≠ 0 and x ≠ a/c.
    11. 💬 SOLUTION:
    f(f(x))
    =
    f(
    ax + b
    )
    cx – a
     
    =
    a(
    ax + b
    ) + b
    cx – a
    c(
    ax + b
    ) – a
    cx – a
    If a2 + bc = 0, f(f(x)) is undefined, while if x = a/c, f(x) is undefined.

  11. Let f(x) = [x – 3] / [x + 1]. Show that f(f(f(x))) = x, provided x ≠ ±1.
    12. 💬 SOLUTION:
    f(f(f(x)))
    =
    f(f(
    x – 3
    ))
    x + 1
     
    =
    f(
    x – 3
    – 3
    )
    x + 1
    x – 3
    + 1
    x + 1
     
    =
    f(
    x – 3 – 3x – 3
    )
    x – 3 + x + 1
     
    =
    f(
    –2x – 6
    )
    2x – 2
     
    =
    f(
    –x – 3
    )
    x – 1
     
    =
    –x – 3
    – 3
    x – 1
    –x – 3
    + 1
    x – 1
     
    =
    –x – 3 – 3x + 3
    –x – 3 + x – 1
     
    =
    –4x
    –4
     
    =
    x
    If x = –1, f(x) is undefined, while if x = 1, f(f(x)) is undefined.

  12. Let f(x) = [x] / [x – 1]. Find and simplify each value.
    13. (a)
    f(1/x)
    (b)
    f(f(x))
    (c)
    f(1/f(x))
    💬 SOLUTION:
    (a)
    f(
    1
    )
    =
    1/x
    =
    1
    x
    (1/x) – 1
    1 – x
    (b)
    f(f(x))
    =
    f(
    x
    )
    x – 1
     
     
    =
    x
    x – 1
    x
    – 1
    x – 1
     
     
    =
    x
    x – x + 1
     
     
    =
    x
    (c)
    f(
    1
    )
    =
    f(
    x
    )
    f(x)
    x – 1
     
     
    =
    x – 1
    x
    x – 1
    – 1
    x
     
     
    =
    x – 1
    x – 1 – x
     
     
    =
    1 – x

  13. Let f(x) = [x] / [√(x) – 1]. Find and simplify.
    14. (a)
    f(1/x)
    (b)
    f(f(x))
    💬 SOLUTION:
    (a)
    f(
    1
    )
    =
    1/x
    =
    1
    x
    √(1/x) – 1
    √(x) – x
    (b)
    f(f(x))
    =
    f(x/(√(x) – 1))
     
     
    =
    x/(√(x) – 1)
    √(
    x
    ) – 1
     
    √(x) – 1
     
     
    =
    x
    √[x(√(x) – 1)] + 1 – √(x)

  14. Prove that the operation of composition of functions is associative; that is, f1 ◦ (f2 ◦ f3) = (f1 ◦ f2) ◦ f3.
    15. 💬 SOLUTION:
    (f1 ◦ (f2 ◦ f3))(x)
    =
    f1 ◦ ((f2 ◦ f3)(x))
     
    =
    f1 (f2 (f3(x)))
    ((f1 ◦ f2) ◦ f3)(x)
    =
    (f1 ◦ f2) (f3(x))
     
    =
    f1 (f2 (f3(x)))
     
    =
    (f1 ◦ (f2 ◦ f3))(x)

    Semoga Bermanfaat 😁

  15. Let f1(x) = x, f2(x) = 1/x, f3(x) = 1 – x, f4(x) = 1 / (1 – x), f5(x) = (x – 1) / x, and f6(x) = x / (x – 1) . Note that f3(f4(x)) = f3(1 / (1 – x)) = 1 – 1/(1 – x) = x / (x – 1) = f6(x) that is, f3 ◦ f4 = f4. In fact, the composition of any two of these functions is another one in the list. Fill in the composition table in Figure 11.
    16. Calculus 9th Purcell Chapter 0 - 0.6
    Figure 11
    Then use this table to find each of the following. From Problem 35, you know that the associative law holds.
    (a)
    f3 ◦ f3 ◦ f3 ◦ f3 ◦ f3
    (b)
    f1 ◦ f2 ◦ f3 ◦ f4 ◦ f5 ◦ f6
    (c)
    F if F ◦ f6 = f1
    (d)
    G if G ◦ f3 ◦ f6 = f1
    (e)
    G if f2 ◦ f5 ◦ H = f5
    💬 SOLUTION:
    f1(f1(x))
    =
    x
    ;
    f1(f2(x))
    =
    1
    ;
    x
    f1(f3(x))
    =
    1 – x
    ;
    f1(f4(x))
    =
    1
    ;
    1 – x
    f1(f5(x))
    =
    x – 1
    ;
    x
    f1(f6(x))
    =
    x
    ;
    x – 1

    f2(f1(x))
    =
    1
    ;
    x
    f2(f2(x))
    =
    1
    =
    x
    ;
    1/x
    f2(f3(x))
    =
    1
    ;
    1 – x
    f2(f4(x))
    =
    1
    =
    1 – x
    ;
    1/(1 – x)
    f2(f5(x))
    =
    1
    =
    x
    ;
    (1 – x)/x
    x – 1
    f2(f6(x))
    =
    1
    =
    x – 1
    ;
    x/(x – 1)
    x

    f3(f1(x))
    =
    1 – x
    ;
    f3(f2(x))
    =
    1 –
    1
    =
    x – 1
    ;
    x
    x
    f3(f3(x))
    =
    1 – (1 – x)
    =
    x
    ;
    f3(f4(x))
    =
    1 –
    1
    =
    x
    ;
    1 – x
    x – 1
    f3(f5(x))
    =
    1 –
    x – 1
    =
    1
    ;
    x
    x
    f3(f6(x))
    =
    1 –
    x
    =
    1
    ;
    x – 1
    1 – x

    f4(f1(x))
    =
    1
    ;
    1 – x
    f4(f2(x))
    =
    1
    =
    x
    ;
    1 – (1/x)
    x – 1
    f4(f3(x))
    =
    1
    =
    1
    ;
    1 – (1 – x)
    x
    f4(f4(x))
    =
    1
    =
    1 – x
    =
    x – 1
    ;
    1 – [1/(1 – x)]
    1 – x – 1
    x
    f4(f5(x))
    =
    1
    =
    x
    =
    x
    ;
    1 – [(x – 1)/x]
    x – (x – 1)
    f4(f6(x))
    =
    1
    =
    x – 1
    =
    1 – x
    ;
    1 – [x/(x – 1)]
    x – 1 – x

    f5(f1(x))
    =
    x – 1
    ;
    x
    f5(f2(x))
    =
    (1/x) – 1
    =
    1 – x
    ;
    1/x
    f5(f3(x))
    =
    1 – x – 1
    =
    x
    ;
    1 – x
    x – 1
    f5(f4(x))
    =
    [1/(1 – x)] – 1
    =
    1 – (1 – x)
    =
    x
    ;
    1/(1 – x)
    1
    f5(f5(x))
    =
    [(x – 1)/x] – 1
    =
    x – 1 – x
    =
    1
    ;
    (x – 1)/x
    x – 1
    1 – x
    f5(f6(x))
    =
    [x/(x – 1)] – 1
    =
    x – (x – 1)
    =
    1
    ;
    x/(x – 1)
    x
    x

    f6(f1(x))
    =
    x
    ;
    x – 1
    f6(f2(x))
    =
    1/x
    =
    1
    ;
    (1/x) – 1
    1 – x
    f6(f3(x))
    =
    1 – x
    =
    x – 1
    ;
    1 – x – 1
    x
    f6(f4(x))
    =
    [1/(1 – x)]
    =
    1
    =
    1
    ;
    [1/(1 – x)] – 1
    1 – (1 – x)
    x
    f6(f5(x))
    =
    [(x – 1)/x]
    =
    x – 1
    =
    1 – x
    ;
    [(x – 1)/x] – 1
    x – 1 – x
    f6(f6(x))
    =
    x/(x – 1)
    =
    x
    =
    x
    [x/(x – 1)] – 1
    x – (x – 1)

    ◦
    f1
    f2
    f3
    f4
    f5
    f6
    f1
    f1
    f2
    f3
    f4
    f5
    f6
    f2
    f2
    f1
    f4
    f3
    f6
    f5
    f3
    f3
    f5
    f1
    f6
    f2
    f4
    f4
    f4
    f6
    f2
    f5
    f1
    f3
    f5
    f5
    f3
    f6
    f1
    f4
    f2
    f6
    f6
    f4
    f5
    f2
    f3
    f1

    (a)
    f3 ◦ f3 ◦ f3 ◦ f3 ◦ f3
     
    = (((( f3 ◦ f3 ) ◦ f3 ) ◦ f3 ) ◦ f3 )
     
    = ((( f1 ◦ f3 ) ◦ f3 ) ◦ f3 )
     
    = (( f3 ◦ f3 ) ◦ f3 )
     
    = f1 ◦ f3
     
    = f3
    (b)
    f1 ◦ f2 ◦ f3 ◦ f4 ◦ f5 ◦ f6
     
    = ((((( f1 ◦ f2 ) ◦ f3 ) ◦ f4 ) ◦ f5 ) ◦ f6 )
     
    = (((( f2 ◦ f3 ) ◦ f4 ) ◦ f5 ) ◦ f6 )
     
    = ( f4 ◦ f4 ) ◦ ( f5 ◦ f6 )
     
    = f5 ◦ f2
     
    = f3
    (c)
    If F ◦ f6 = f1, then F = f6.
    (d)
    If G ◦ f3 ◦ f6 = f1, then G ◦ f4 = f1 so
     
    G = f5.
    (e)
    If f2 ◦ f5 ◦ H = f5, then f6 ◦ H = f5 so
     
    H = f3.

    Use a computer or a graphing calculator in Problems 37 – 40
  16. Let f(x) = x2 – 3x. Using the same axes, draw the graphs of y = f(x), y = f(x – 0.5) – 0.6, and y = f(1.5x), all on the domain [–2, 5].
    17. 💬 SOLUTION:
    Calculus 9th Purcell Chapter 0 - 0.6

  17. Let f(x) = |x3|. Using the same axes, draw the graphs of y = f(x), y = f(3x), and y = f(3(x – 0.8)), all on the domain [–3, 3].
    18. 💬 SOLUTION:
    Calculus 9th Purcell Chapter 0 - 0.6

  18. Let f(x) = 2√(x) – 2x + 0.25x2. Using the same axes, draw the graphs of y = f(x), y = f(1.5x), and y = f(x – 1) + 0.5, all on the domain [0, 5].
    19. 💬 SOLUTION:
    Calculus 9th Purcell Chapter 0 - 0.6

  19. Let f(x) = 1/(x2 + 1). Using the same axes, draw the graphs of y = f(x), y = f(2x), and y = f(x – 2) + 0.6, all on the domain [–4, 4].
    20. 💬 SOLUTION:
    Calculus 9th Purcell Chapter 0 - 0.6

  20. Your computer algebra system (CAS) may allow the use of parameters in defining functions. In each case, draw the graph of y = f(x) for the specified values of the parameter k, using the same axes and –5 ≤ x ≤ 5.
    21. (a)
    f(x) = |kx|0.7 for k = 1, 2, 0.5, and 0.2.
    (b)
    f(x) = |x – k|0.7 for k = 0, 2, –0.5, and –3.
    (c)
    f(x) = |x|k for k = 0.4, 0.7, 1, and 1.7.
    💬 SOLUTION:
    (a)
     
    Calculus 9th Purcell Chapter 0 - 0.6
    (b)
     
    Calculus 9th Purcell Chapter 0 - 0.6
    (c)
     
    Calculus 9th Purcell Chapter 0 - 0.6

  21. Using the same axes, draw the graph of f(x) = |k(x – c)|n for the following of parameters.
    22. (a)
    c = –1, k = 1.4, n = 0.7
    (b)
    c = 2, k = 1.4, n = 1
    (c)
    c = 0, k = 0.9, n = 0.6
    💬 SOLUTION:
    Calculus 9th Purcell Chapter 0 - 0.6

Semoga Bermanfaat 😁

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