Locating Roots and Iterative Methods · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Trapping a root, then walking in

Mathematics · Calculus & Analysis · ages 17-18
Name ______________________   Date ____________

  1. For f(x) = x cubed minus 2x minus 5, what is f(2)?

    Answer: ______________

  2. For f(x) = (x - 5)², f(4) = f(6) = 1. Despite these matching endpoint signs, there is a root between 4 and 6.

    Circle one:   True   False

  3. A function has f(-4) = -7, f(-1) = 0 and f(2) = 6. Which of the listed x-values is itself a root?

    Answer: ______________

  4. A continuous f has f(1) = -2 and f(2) = 3. What follows?

    • At least one root lies exactly at x = 1
    • At least one root lies exactly at x = 2
    • At least one root lies between 1 and 2

  5. The iteration x_next = (x + 10)/3 converges to a fixed point r. What is r?

    Answer: ______________

  6. For x_next = (x + 8)/3, x0 = 1, x1 = 3 and x2 = 11/3. Rowan computes x3 = (1 + 8)/3 = 3. Which input should Rowan use instead?

    • Use 11/3, the latest output
    • Use 1, the original starting value
    • Use 3, the earlier output

  7. For x not equal to zero, divide x² = 3x + 6 by x. Which equivalent x = g(x) form results?

    • x = 3 + 6/x
    • x = (3 + 6)/x
    • x = 3x + 6/x
    • x = 3 + x/6

  8. Iterate x next = (x + 3)/2 from x0 = 1. What is x2?

    Answer: ______________

  9. Use x_next = (x + 11)/3 from x0 = 5. A justified bound is |x0 - r| ≤ 0.6, and each update divides this bound by 3. Continue until the bound first falls below 0.005. Which exact iterate and stopping justification are valid?

    • x4 = 445/81; its small step alone certifies error below 0.005.
    • x6 = 4009/729; the error bound first falls below 0.005 at update 6.
    • x5 = 1336/243; the error bound falls below 0.005 at update 5.

  10. A continuous f has opposite signs at 1.234 and 1.236, locating a root between them. Does this bracket determine the root rounded to two decimal places?

    • Yes, its small width proves every root rounds to 1.23
    • No, it crosses the rounding boundary 1.235
    • Yes, its small width proves every root rounds to 1.24
    • No, opposite signs exclude roots throughout this interval
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Answer key

For grown-ups. Fold this page away before handing over the rest.

Trapping a root, then walking in W1-mt_2COO6TWerM-s1

  1. -1 · Substituting gives 8 - 4 - 5 = -1.
  2. True · At x = 5, f(5) = (5 - 5)² = 0. Thus 5 is a root between the two positive endpoint values.
  3. -1 · A root is an input where the function value is zero. Since f(-1) = 0, the listed root is -1.
  4. At least one root lies between 1 and 2 · Opposite signs force the graph to cross the axis between 1 and 2.
  5. 5 · At a fixed point, r = (r + 10)/3. Thus 3r = r + 10, giving 2r = 10 and r = 5.
  6. Use 11/3, the latest output · The next update uses x2 = 11/3, giving x3 = (11/3 + 8)/3 = 35/9. Using 1 repeats the calculation for x1.
  7. x = 3 + 6/x · Dividing each term by x gives x²/x = 3x/x + 6/x, so x = 3 + 6/x.
  8. 2.5 · One step gives 2, and the next gives (2 + 3)/2 = 2.5.
  9. x5 = 1336/243; the error bound falls below 0.005 at update 5. · Repeated updates give x5 = 1336/243. The bound at update 4 is 0.6/81 > 0.005; at update 5 it is 0.6/243 < 0.005. Update 5 is the first certified stopping point.
  10. No, it crosses the rounding boundary 1.235 · The bracket contains values below and above 1.235. They can round to 1.23 or 1.24, so the bracket alone does not determine the rounded root.
Worksheet · LightMySky