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Kunci Jawaban dan Pembahasan Soal || Buku Calculus 9th Purcell Chapter 0 - 0.5 Number 31 – 50

Pembahasan Soal Buku Calculus 9th Purcell Chapter 0 Section 0.5

CHAPTER 0 PRELIMINARIES

SECTION 0.5 Functions and Their Graphs


Calculus 9th Purcell Chapter 0 - 0.5

Problem Set 0.5, Number 31 – 50.
  1. A plant has the capacity to produce from 0 to 100 computers per day. The daily overhead for the plant is $5000, and the direct cost (labor and materials) of producing one computer is $805. Write a formula for T(x) the total cost of producing x computers in one day, and also for the unit cost u(x) (average cost per computer). What are the domains of these functions?
    2. 💬 SOLUTION:
    T(x) = 5000 + 805x
    Domain : {x ∈ integers: 0 ≤ x ≤ 100}
    u(x)
    =
    T(x)
    =
    5000
    + 805
    x
    x
    Domain : {x ∈ integers: 0 ≤ x ≤ 100}

  2. It costs the ABC Company 400 + 5√(x(x - 4)) dollars to make x toy stoves that sell for $6 each.
    3. (a)
    Find a formula for P(x), the total profit in making x stoves.
    (b)
    Evaluate P(200) and P(1000).
    (c)
    How many stoves does ABC have to make to just break even?
    💬 SOLUTION:
    a.
    P(x) = 6x – (400 + √(x(x – 4)))
     
    = 6x – 400 – 5√(x(x – 4))
    b.
    P(200) ≈ –190 ; P(1000) ≈ 610
    c.
    ABC breaks even when P(x) = 0;
     
    6x – 400 – 5√(x(x – 4)) = 0; x ≈ 390

  3. Find the formula for the amount E(x) by which a number x exceeds its square. Plot a graph of E(x) for 0 ≤ x ≤ 1. Use the graph to estimate the positive number less than or equal to 1 that exceeds its square by the maximum amount.
    4. 💬 SOLUTION:
    E(x) = x – x2

    Number 33

    ½ exceeds its square by the maximum amount.

  4. Let p denote the perimeter of an equilateral triangle. Find a formula for A(p) the area of such a triangle.
    5. 💬 SOLUTION:
    Each side has length p/3. The height of the triangle is (√3p)/6.
    A(p)
    =
    1
    ×
    p
    ×
    √(3) . p
    =
    √(3) . (p)2
    2
    3
    6
    36

  5. A right triangle has a fixed hypotenuse of length h and one leg that has length x. Find a formula for the length L(x) of the other leg.
    6. 💬 SOLUTION:
    Let y denote the length of the other leg. Then
    x2 + y2
    =
    h2
    y2
    =
    h2 – x2
    y
    =
    √(h2 – x2)
    L(x)
    =
    √(h2 – x2)

  6. A right triangle has a fixed hypotenuse of length h and one leg that has length x. Find a formula for the area A(x) of the triangle.
    7. 💬 SOLUTION:
    The area is
    A(x) = ½ base × height
    A(x) = ½ x √(h2 – x2)

  7. The Acme Car Rental Agency charges $24 a day for the rental of a car plus $0.40 per mile.
    8. (a)
    Write a formula for the total rental expense E(x) for one day, where x is the number of miles driven.
    (b)
    If you rent a car for one day, how many miles can you drive for $120?
    💬 SOLUTION:
    a.
    E(x) = 24 + 0.40x
    b.
    120 = 24 + 0.40x
     
    0.40x = 96;
     
    x = 240 mi

Semoga Bermanfaat 😁

  1. A right circular cylinder of radius r is inscribed in a sphere of radius 2r. Find a formula for V(r), the volume of the cylinder. in terms of r.
    2. 💬 SOLUTION:
    The volume of the cylinder is πr2h, where h is the height of the cylinder. From the figure,
    r2 + (h/2)2 = (2r)2 ; h2/4 = 3r2 ;
    h = √(12r2) = 2r√(3).
    V(r) = πr2 (2r√(3)) = 2πr3√(3)
    Number 38

  2. A 1-mile track has parallel sides and equal semicircular ends. Find a formula for the area enclosed by the track, A(d), in terms of the diameter d of the semicircles. What is the natural domain for this function?
    3. 💬 SOLUTION:
    The area of the two semicircular ends is (πd2)/4 .
    The length of each parallel side is (1 – πd)/2 .
    A(d) = (πd2)/4 + d[(1 – πd)/2]
    A(d) = (πd2)/4 + [(d – πd2)/2]
    A(d) = (2d – πd2)/4
    Since the track is one mile long, πd < 1, so d < 1/π.
    Domain: {x ∈ ℝ : 0 < d < 1/π}.

  3. Let A(c) denote the area of the region bounded from above by the line y = x + 1, from the left by the y-axis, from below by the x-axis, and from the right by the line x = c Such a function is called an accumulation function. (See Figure 13.) Find
    4. (a)
    A(1)
    (b)
    A(2)
    (c)
    A(0)
    (d)
    A(c)
    (e)
    Sketch the graph of A(c).
    (f)
    What are the domain and range of A?
    💬 SOLUTION:
    a.
    A(1) = 1(1) + ½ (1)(2 – 1) = 3/2
    Number 40a
    b.
    A(2) = 2(1) + ½ (2)(3 – 1) = 4
    c.
    A(0) = 0
    d.
    A(c) = c(1) + ½ (c)(c + 1 – 1) = ½ c2 + c
    e.
     
    Number 40e
    f.
    Domain: {c ∈ ℝ : c ≥ 0}
     
    Range: {y ∈ ℝ : y ≥ 0}

  4. Let B(c) denote the area of the region bounded from above by the graph of the curve y = x(1 – x) from below by the x-axis, and from the right by the line x = c. The domain of B is the interval [0, 1]. (See Figure 14.) Given that B(1) = 1/6,
    5. (a)
    Find B(0).
    (b)
    Find B(1/2).
    (c)
    As best you can, sketch a graph of B(c).
    💬 SOLUTION:
    a.
    B(0) = 0
    b.
    B(½) = ½ B(1) = ½ × 1/6 = 1/12
    c.
     
    Number 41c

  5. Which of the following functions satisfies f(x + y) = f(x) + f(y) for all real numbers x and y?
    6. (a)
    f(t) = 2t
    (b)
    f(t) = t2
    (c)
    f(t) = 2t + 1
    (d)
    f(t) = –3t
    💬 SOLUTION:
    a.
    f(x + y) = 2(x + y) = 2x + 2y = f(x) + f(y)
    b.
    f(x + y) = (x + y)2 = x2 + 2xy + y2 ≠ f(x) + f(y)
    c.
    f(x + y) = 2(x + y) + 1 = 2x + 2y + 1 ≠ f(x) + f(y)
    d.
    f(x + y) = –3(x + y) = –3x – 3y = f(x) + f(y)

  6. Let f(x + y) = f(x) + f(y) for all x and y. Prove that there is a number m such that f(t) = mt for all rational numbers t. Hint: First decide what m has to be. Then proceed in steps, starting with f(0) = 0, f(p) = mp for a natural number p, f(1/p) = m/p and so on.
    7. 💬 SOLUTION:
    For any x, x + 0 = x, so f(x) = f(x + 0) = f(x) + f(0), hence f(0) = 0.
    Let m be the value of f(1).
    For p in ℕ, p = p · 1 = 1 + 1 + … + 1, so
    f(p) = f(1 + 1 + … + 1) = f(1) + f(1) + … + f(1) = pf(1) = pm.
    1 = p(1/p) = 1/p + 1/p + … + 1/p, so
    m = f(1) = f (1/p + 1/p + … + 1/p)
    = f(1/p) + f(1/p) + … + f(1/p) = pf(1/p),
    hence f(1/p) = m/p.
    Any rational number can be written as p/q with p, q in ℕ.
    p/q = p(1/q) = 1/q + 1/q + … + 1/q,
    so f(p/q) = f(1/q + 1/q + … + 1/q)
    = f(1/q) + f(1/q) + … + f(1/q)
    = pf(1/q) = p(m/q) = m(p/q)

  7. A baseball diamond is a square with sides of 90 feet. A player, after hitting a home run, loped around the diamond at 10 feet per second. Let s represent the player's distance from home plate after t seconds.
    8. (a)
    Express s as a function of t by means of a four-part formula.
    (b)
    Express s as a function of t by means of a three-part formula.
    💬 SOLUTION:
    The player has run 10t feet after t seconds.
    He reaches first base when t = 9, second base when t = 18, third base when t = 27, and home plate when t = 36.
    The player is 10t – 90 feet from first base when 9 ≤ t ≤ 18, hence √[902 + (10t – 90)2] feet from home plate.
    The player is 10t – 180 feet from second base when 18 ≤ t ≤ 27, thus he is 90 – (10t – 180) = 270 – 10t feet from third base and √[902 + (270 – 10t)2] feet from home plate.
    The player is 10t – 270 feet from third base when 27 ≤ t ≤ 36, thus he is 90 – (10t – 270) = 360 – 10t feet from home plate.
    a.
    s = {
    10t
    if 0 ≤ t ≤ 9
     
    √[902 + (10t – 90)2]
    if 9 < t ≤ 18
     
    √[902 + (270 – 10t)2]
    if 18 < t ≤ 27
     
    360 – 10t
    if 27 < t ≤ 36
     
     
     
     
    b.
    s = {
    180 – |180 – 10t|
    if 0 ≤ t ≤ 9
     
     
    or 27 < t ≤ 36
     
    √[902 + (10t – 90)2]
    if 9 < t ≤ 18
     
    √[902 + (270 – 10t)2]
    if 18 < t ≤ 27

Semoga Bermanfaat 😁

    To use technology effectively, you need to discover its capabilities, its strengths, and its weaknesses. We urge you to practice graphing functions of various types using your own computer package or calculator. Problems 45–50 are designed for this purpose.
  1. Let f(x) = (x3 + 3x – 5)/(x2 + 4).
    2. (a)
    Evaluate f(1.38) and f(4.12).
    (b)
    Construct a table of values for this function corresponding to – = –4, –3 , … , 3, 4.
    💬 SOLUTION:
    a.
    f(1.38) ≈ 0.2994
     
    f(4.12) ≈ 3.6852
    b.
     
     
    x
    f(x)
     
    –4
    –4.05
     
    –3
    –3.1538
     
    –2
    –2.375
     
    –1
    –1.8
     
    0
    –1.25
     
    1
    –0.2
     
    2
    1.125
     
    3
    2.3846
     
    4
    3.55

  2. Follow the instructions in Problem 45 for f(x) = (sin2 x – 3 tan x)/(cos x).
    3. 💬 SOLUTION:
    a.
    f(1.38) ≈ –76.8204
     
    f(4.12) ≈ 6.7508
    b.
     
     
    x
    f(x)
     
    –4
    –6.1902
     
    –3
    0.4118
     
    –2
    13.7651
     
    –1
    9.9579
     
    0
    0
     
    1
    –7.3369
     
    2
    –17.7388
     
    3
    –0.4521
     
    4
    4.4378

  3. Draw the graph of f(x) = x3 – 5x2 + x + 8 on the domain [–2, 5].
    4. (a)
    Determine the range of f.
    (b)
    Where on this domain is f(x) ≥ 0?
    💬 SOLUTION:
    Number 47
    a.
    Range: {y ∈ ℝ : –22 ≤ y ≤ 13}
    b.
    f(x) = 0 when x ≈ –1.1, 1.7, 4.3
    f(x) ≥ 0 on [–1.1, 1.7] ∪ [4.3, 5]

  4. Superimpose the graph of g(x) = 2x2 – 8x – 1 with domain [–2, 5] on the graph of f(x) of Problem 47.
    5. (a)
    Estimate the x-values where f(x) = g(x).
    (b)
    Where on [–2,5] is f(x) ≥ g(x)?
    (c)
    Estimate the largest value of |f(x) – g(x)| on [–2, 5].
    💬 SOLUTION:
    Number 48
    a.
    f(x) = g(x) at x ≈ –0.6, 3.0, 4.6
    b.
    f(x) ≥ g(x) on [–0.6, 3.0] ∪ [4.6, 5]
    c.
    | f(x) – g(x)|
    = |x3 – 5x2 + x + 8 – 2x2 + 8x + 1|
    = | x3 – 7x2 + 9x + 9|
    Largest value |f(–2) – g(–2)| = 45

  5. Graph f(x) = (3x – 4)/(x2 + x – 6) on the domain [–6, 6].
    6. (a)
    Determine the x- and y-intercepts.
    (b)
    Determine the range of f for the given domain.
    (c)
    Determine the vertical asymptotes of the graph.
    (d)
    Determine the horizontal asymptote for the graph when the domain is enlarged to the natural domain.
    💬 SOLUTION:
    Number 49
    a.
    x-intercept: 3x – 4 = 0; x = 4/3
    y-intercept: (3 ∙ 0 – 4)/(02 + 0 – 6) = 2/3
    b.
    ℝ
    c.
    x2 + x – 6 = 0; (x + 3)(x – 2) = 0
    Vertical asymptotes at x = –3, x = 2
    d.
    Horizontal asymptote at y = 0

  6. Follow the directions in Problem 49 for the function g(x) = (3x2 – 4)/(x2 + x – 6).
    7. 💬 SOLUTION:
    Number 50
    a.
    x-intercept:
    3x2 – 4 = 0; x = ±√(4/3) = ±[(2√(3))/3]
    y-intercept: 2/3
    b.
    On [–6, –3) , g increases from g(–6) = 13/3 ≈ 4.3333 to ∞.
    On (2.6] , g decreased from ∞ to 26/9 ≈ 2.8889.
    On (–3, 2) the maximum occurs around x = 0.1451 with value 0.6748.
    Thus, the range is (–∞, 0.6748] ∪ [2.8889, ∞).
    c.
    x2 + x – 6 = 0; (x + 3)(x – 2) = 0
    Vertical asymptotes at x = –3, x = 2
    d.
    Horizontal asymptote at y = 3

Semoga Bermanfaat 😁

Demikian soal serta penjelasan untuk Pembahasan Soal Buku Calculus 9th Purcell Chapter 0 - 0.5 Number 31 – 50. Silahkan untuk berkunjung kembali dikarenakan akan selalu ada update terbaru 😊😄🙏. Silahkan juga untuk memilih dan mendiskusikan di tempat postingan ini di kolom komentar ya supaya semakin bagus diskusi pada setiap postingan. Diperbolehkan request di kolom komentar pada postingan ini tentang rangkuman atau catatan atau soal dan yang lain atau bagian hal yang lainnya, yang sekiranya belum ada di website ini. Terima kasih banyak sebelumnya 👍. Semoga bermanfaat dan berkah untuk kita semua. Amiiinnn 👐👐👐
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