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Kunci Jawaban dan Pembahasan Soal || Buku Calculus 9th Purcell Chapter 0 - 0.5 Number 1 – 30

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 1 – 30.
  1. For f(x) = 1 – x2 find each value.
    2. (a)
    f(1)
    (b)
    f(–2)
    (c)
    f(0)
    (d)
    f(k)
    (e)
    f(–5)
    (f)
    f1/4
    (g)
    f(1 + h)
    (h)
    f(1 + h) – f(1)
    (i)
    f(2 + h) – f(2)
    💬 SOLUTION:
    a.
    f(1) = 1 – 12 = 0
    b.
    f(–2) = 1 – (–2)2 = –3
    c.
    f(0) = 1 – 02 = 1
    d.
    f(k) = 1 – k2
    e.
    f(–5) = 1 – (–5)2 = –24
    f.
    f(1/4) = 1 – (1/4)2 = 1 – 1/16 = 15/16
    g.
    f(1 + h) = 1 – (1 + h)2 = –2h – h2
    h.
    f(1 + h) – f(1) = –2h – h2 – 0 = –2h – h2
    i.
    f(2 + h) – f(2) = 1 – (2 + h)2 + 3 = –4h – h2

  2. For F(x) = x3 + 3x , find each value.
    3. (a)
    F(1)
    (b)
    F(√(2))
    (c)
    F(1/4)
    (d)
    F(1 + h)
    (e)
    F(1 + h) – F(1)
    (f)
    F(2 + h) – F(2)
    💬 SOLUTION:
    a.
    F(1) = 13 + 3 × 1 = 4
    b.
    F(√(2)) = (√(2))3 + 3(√(2)) = 2√(2) + 3√(2) = 5√(2)
    c.
    F(1/4) = (1/4)3 + 3(1/4) = 1/64 + 3/4 = 49/64
    d.
    F(1 + h) = (1 + h)3 + 3(1 + h)
    F(1 + h) = 1 + 3h + 3h2 + h3 + 3 + 3h
    F(1 + h) = 4 + 6h + 3h2 + h3
    e.
    F(1 + h) – 1 = 3 + 6h + 3h2 + h3
    f.
    F(2 + h) – F(2)
    = (2 + h)3 + 3(2 + h) – [23 – 3(2)]
    = 8 + 12h + 6h2 + h3 + 6 + 3h – 14
    = 15h + 6h2 + h3

  3. For G(y) = 1/(y – 1) find each value.
    4. (a)
    G(0)
    (b)
    G(0.999)
    (c)
    G(1.01)
    (d)
    G(y2)
    (e)
    G(–x)
    (f)
    G(1/x2)
    💬 SOLUTION:
    a.
    G(0) = 1/(0 – 1) = –1
    b.
    G(0.999) = 1/(0.999 – 1) = –1000
    c.
    G(1.01) = 1/(1.01 – 1) = 100
    d.
    G(y2) = 1/(y2 – 1)
    e.
    G(–x) = 1/(–x – 1) = – [1/(x + 1)]
    f.
    G(1/(x2)) = 1/([1/(x2)] – 1) = (x2)/(1 – x2)

  4. For Φ(u) = (u + u2)/(√(u)), find each value. (Φ is the uppercase Greek letter phi.)
    5. (а)
    Φ(1)
    (b)
    Φ(–t)
    (c)
    Φ(1/2)
    (d)
    Φ(u + 1)
    (e)
    Φ(x2)
    (f)
    Φ(x2 + x)
    💬 SOLUTION:
    a.
    Φ(1) = (1 + 12)/(√(1)) = 2
    b.
    Φ(–t) = (–t + (–t)2)/(√(–t)) = (t2 – t)/(√(–t))
    c.
    Φ(1/2) = (1/2 + (1/2)2)/(√(1/2)) = 3/4/(√(1/2)) ≈ 1.06
    d.
    Φ(u + 1) = ((u + 1) + (u + 1)2)/(√(u + 1)) = (u2 + 3u + 2)/(√(u + 1))
    e.
    Φ(x2) = ((x2) + (x2)2)/(√(x2)) = (x2 + x4)/|x|
    f.
    Φ(x2 + x) = ((x2 + x) + (x2 + x)2)/(√(x2 + x)) = (x4 + 2x3 + 2x2 + x)/(√(x2 + x))

  5. For
    6. f(x)
    =
    1
    √(x – 3)
    find each value.
    a.
    f(0.25)
    b.
    f(π)
    c.
    f(3 + √(2))
    💬 SOLUTION:
    a.
    f(0.25)
    =
    1
    √(0.25 – 3)
    f(0.25)
    =
    1
    √(–2.75)
    is not defined
    b.
    f(x)
    =
    1
    √(π – 3)
    f(x)
    ≈
    2.658
    c.
    f(3 + √(2))
    =
    1
    √(3 + √(2) – 3)
    f(3 + √(2))
    =
    1
    √(√(2))
    f(3 + √(2))
    =
    2–0.25
    f(3 + √(2))
    ≈
    0.841

  6. For
    7. f(x)
    =
    √(x2 + 9)
    x – √(3)
    find each value.
    a.
    f(0.79)
    b.
    f(12.26)
    c.
    f(√(3))
    💬 SOLUTION:
    a.
    f(0.79)
    =
    √((0.79)2 + 9)
    0.79 – √(3)
     
    ≈
    –3.293
    b.
    f(12.26)
    =
    √((12.26)2 + 9)
    12.26 – √(3)
     
    ≈
    1.199
    c.
    f(√(3))
    =
    √((√(3))2 + 9)
    √(3) – √(3)
     
    ;
    undefined

  7. Which of the following determine a function f with formula y = f(x)? For those that do, find f(x). Hint: Solve for y in terms of x and note that the definition of a function requires a single y for each x.
    8. a.
    x2 + y2 = 1
    b.
    xy + y + x = 1, x ≠ –1
    c.
    x = √(2y + 1)
    d.
    x = y/(y + 1)
    💬 SOLUTION:
    a.
    x2 + y2
    =
    1
    y2
    =
    1 – x2
    y
    =
    ±√(1 – x2)
     
    ;
    not a function
    b.
    xy + y + x
    =
    1
    y(x + 1)
    =
    1 – x
    y =
    1 – x
    x + 1
    ; f(x) =
    1 – x
    x + 1
    c.
    x
    =
    √(2y + 1)
    x2
    =
    2y + 1
    y =
    x2 – 1
    2
    ; f(x) =
    x2 – 1
    2
    d.
    x  =
    y
    y + 1
    xy + x
    =
    2y + 1
    x
    =
    y – xy
    x
    =
    y(1 – x)
    y  =
    x
    1 – x
    ; f(x)  =
    x
    1 – x

  8. Which of the graphs in Figure 12 are graphs of functions?
    9. Figure 12
    This problem suggests a rule: For a graph to be the graph of a function, each vertical line must meet the graph in at most one point.
    💬 SOLUTION:
    The graphs on the left are not graphs of functions, the graphs on the right are graphs of functions.

  9. For f(x) = 2x2 – 1, find and simplify [f(a + h) – f(a)]/h.
    10. 💬 SOLUTION:
    f(a + h) – f(a)
    =
    [2(a + h)2 – 1] – (2a2 – 1)
    h
    h
     
    =
    4ah + 2h2
     
    h
     
    =
    4a + 2h

  10. For F(t) = 2t3, find and simplify [F(a + h) – f(a)]/h.
    11. 💬 SOLUTION:
    F(a + h) – F(a)
    =
    4(a + h)3 – 4a3
    h
    h
     
    =
    4a3 + 12a2h + 12ah2 + 4h3 – 4a3
    h
     
    =
    12a2h + 12ah2 + 4h3
    h
     
    =
    12a2 + 12ah + 4h2

Semoga Bermanfaat 😁

  1. For g(u) = 3/(u – 2), find and simplify [g(x + h) – f(x)]/h.
    2. 💬 SOLUTION:
    g(x + h) – g(x)
     
    3
    –
    3
    =
    x + h – 2
    x – 2
    h
    h
     
     
    3x – 6 – 3x – 3h + 6
    =
    x2 – 4x + hx – 2h + 4
    h
     
    =
    –3h
    h(x2 – 4x + hx – 2h + 4)
     
    =
    –
    3
    x2 – 4x + hx – 2h + 4

  2. For G(t) = t/(t + 4), find and simplify [G(a + h) – f(a)]/h.
    3. 💬 SOLUTION:
    G(a + h) – G(x)
     
    a + h
    –
    a
    =
    a + h + 4
    a + 4
    h
    h
     
     
    a2 + 4a + ah + 4h – a2 – ah – 4a
    =
    a2 + 8a + ah + 4h + 16
    h
     
    =
    4h
    h(a2 + 8a + ah + 4h + 16)
     
    =
    4
    a2 + 8a + ah + 4h + 16

  3. Find the natural domain for each of the following.
    4. a.
    F(z) = √(2z + 3)
    b.
    G(v) = 1/(4v – 1)
    c.
    Ψ(x) = √(x2 – 9)
    d.
    H(y) = – √(625 – y4)
    💬 SOLUTION:
    a. F(z) = √(2z + 3)
    2z + 3 ≥ 0; z ≥ –(3/2)
    Domain: {z ∈ ℝ : z ≥ –(3/2)}
    b. g(v) = 1/(4v – 1)
    4v – 1 = 0; v = ¼
    Domain: {v ∈ ℝ : v ≠ ¼}
    c. Ψ(x) = √(x2 – 9)
    x2 – 9 ≥ 0; x2 ≥ 9; |x| ≥ 3
    Domain: {x ∈ ℝ : |x| ≥ 3}
    d. H(y) = –√(625 – y4)
    625 – y4 ≥ 0; 625 ≥ y4; |y| ≤ 5
    Domain: {y ∈ ℝ : |y| ≤ 5}

  4. Find the natural domain in each case.
    5. a.
    f(x) = (4 – x2)/(x2 – x – 6)
    b.
    G(y) = √(y + 1)–1
    c.
    Φ(u) = |2u + 3|
    d.
    F(t) = t2/3 – 4
    💬 SOLUTION:
    a. f(x) = (4 – x2)/(x2 – x – 6) = (4 – x2)/[(x – 3)(x + 2)]
    Domain: {x ∈ ℝ : x ≠ –2, 3}
    b. G(y) = √[(y + 1)–1]
    1/(y + 1) ≥ 0; y > –1
    Domain: {y ∈ ℝ : y > –1}
    c. Φ(u) = |2u + 3|
    Domain: ℝ (all real numbers)
    d. F(t) = t2/3 – 4
    Domain: ℝ (all real numbers)

    In Problems 15–30, specify whether the given function is even, odd, or neither, and then sketch its graph.
  5. f(x) = – 4
    6. 💬 SOLUTION:
    f(x) = –4; f(–x) = –4; even function

    Number 15

  6. f(x) = 3x
    7. 💬 SOLUTION:
    f(x) = 3x; f(–x) = –3x; odd function

    Number 16

  7. F(x) = 2x + 1
    8. 💬 SOLUTION:
    F(x) = 2x + 1; F(–x) = –2x + 1; neither

    Number 17

  8. F(x) = 3x – √2
    9. 💬 SOLUTION:
    F(x) = 3x – √2; F(–x) = –3x – √2; neither

    Number 18

  9. g(x) = 3x2 + 2x – 1
    10. 💬 SOLUTION:
    g(x) = 3x2 + 2x – 1; g(–x) = 3x2 – 2x – 1; neither

    Number 19

  10. g(u) = u3/8
    11. 💬 SOLUTION:
    g(u) = u3/8; g(–u) = –(u3/8); odd function

    Number 20

Semoga Bermanfaat 😁

  1. g(x) = x/(x2 – 1)
    2. 💬 SOLUTION:
    g(x) = x/(x2 – 1); g(–x) = (–x)/(x2 – 1); odd

    Number 21

  2. Φ(z) = (2z + 1)/(z – 1)
    3. 💬 SOLUTION:
    Φ(z) = (2z + 1)/(z – 1); Φ(–z) = (–2z + 1)/(–z – 1); neither

    Number 22

  3. f(w) = √(w – 1)
    4. 💬 SOLUTION:
    f(w) = √(w – 1); f(–w) = √(–w – 1); neither

    Number 23

  4. h(x) = √(x2 + 4)
    5. 💬 SOLUTION:
    h(x) = √(x2 + 4); h(–x) = √(x2 + 4); even function

    Number 24

  5. f(x) = |2x|
    6. 💬 SOLUTION:
    f(x) = |2x|; f(–x) = |–2x| = |2x|; even function

    Number 25

  6. F(t) = –|t + 3|
    7. 💬 SOLUTION:
    F(t) = –|t + 3|; F(–t) = –|–t + 3|; neither

    Number 26

  7. g(x) = 〚x/2〛
    8. 💬 SOLUTION:
    g(x) = 〚x/2〛; g(–x) = 〚–x/2〛; neither

    Number 27

  8. G(x) = 〚2x – 1〛
    9. 💬 SOLUTION:
    G(x) = 〚2x – 1〛; G(–x) = –〚2x – 1〛; neither

    Number 28

    10. g(t)
    =
    {
    1
    if t ≤ 0
    t + 1
    if 0 < t < 2
    t2 – 1
    if t ≥ 2
    💬 SOLUTION:
    g(t)
    =
    {
    1
    if t ≤ 0
    t + 1
    if 0 < t < 2
    t2 – 1
    if t ≥ 2
    ; neither

    Number 29

    11. h(x)
    =
    {
    –x2 + 4
    if x ≤ 1
    3x
    if x > 1
    💬 SOLUTION:
    h(x)
    =
    {
    –x2 + 4
    if x ≤ 1
    3x
    if x > 1
    ; neither

    Number 30

Semoga Bermanfaat 😁

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