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Locating Roots and Iterative Methods

Use continuity and opposite signs to locate a root, rearrange an equation as x = g(x), and calculate successive estimates. Check a justified error bound or a sign-change bracket before reporting the required accuracy.

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What a learner can do afterwards

  • Use a sign change to show a root lies in a given interval
  • Rearrange an equation into an iterative formula
  • Run an iteration to a stated accuracy and justify stopping

1 · Read

A root is an input x where f(x) = 0. To evaluate f(a), replace every x in the rule with a, then calculate powers before multiplication and subtraction. A function is continuous on an interval if its graph has no break there. Opposite endpoint signs then guarantee at least one root strictly between the endpoints. If an endpoint value is zero, that endpoint is itself a root.

Try it together

For f(x) = x³ - x - 3, f(1) = 1 - 1 - 3 = -3 and f(2) = 8 - 2 - 3 = 3. This polynomial is continuous, so a root lies between 1 and 2. Rearrange x³ = x + 3 as x = the cube root of (x + 3). Starting at 1.5, repeatedly apply this rule to the latest value. Keep the full calculator value between updates; the next picture shows rounded displays.

x0 = 1.5x1 ≈ 1.651x2 ≈ 1.669x3 ≈ 1.671
These estimates suggest a location. A later sign check will certify the rounded answer.

Opposite signs do not prove there is only one root. Same signs do not rule out roots: (x - 3)² is positive at 2 and 4 but zero at 3. Continuity matters too. The function 1/(x - 2) changes sign between 1 and 3, but is undefined at 2 and never equals zero. When searching a table, compare signs at neighbouring inputs to find a bracket.

To form x = g(x), use equivalent algebra and keep any restrictions. For x³ + 2x = 9, subtract 2x and take cube roots: x = the cube root of (9 - 2x). For x² = 2x + 8, divide every term by x: x = 2 + 8/x. This rule requires a nonzero input. Check that each proposed starting value is allowed.

x² = 2x + 8divide every term by xx = 2 + 8/xrequires x ≠ 0

Write an update as x_next = g(x). The notation x0, also written x_0, names the starting value. Then x1 is the first output and x2 is the second. Always use the latest output as the next input. For x_next = (x + 4)/2, starting at x0 = 0 gives x1 = 2, x2 = 3 and x3 = 3.5. Keep calculator precision between updates and round only the reported answer.

x0 = 0x1 = (0 + 4)/2 = 2x2 = (2 + 4)/2 = 3x3 = (3 + 4)/2 = 3.5

A fixed point r satisfies r = g(r). For g(x) = (x + 4)/2, 2r = r + 4 gives r = 4. A valid rearrangement need not converge. For a linear rule g(x) = mx + b, subtract r = mr + b: g(x) - r = m(x - r). Each absolute error is multiplied by |m|. It shrinks when |m| < 1 and grows when |m| > 1, unless the start is r.

fixed point: 4inputs: 0, 2, 3, 3.5errors: 4, 2, 1, 0.5each error halves

Absolute error is |estimate - root|. If an initial error bound is B and each update multiplies errors by c, where 0 < c < 1, multiply the bound by c after each update. Stop when it is below the requested tolerance. Here 1-decimal accuracy means error below 0.05; 2-decimal accuracy means below 0.005. A sign bracket's midpoint is at most half its width from a root inside it. Correct rounding also needs a bracket within one rounding interval.

Suppose a root is bracketed between 4.314 and 4.316. The interval crosses 4.315, the boundary between rounding to 4.31 and rounding to 4.32. Values inside the bracket can round either way. Refine the bracket before reporting the root to two decimal places.

Root bracket. Ticks: 4.314, 4.315, 4.316. Marked values: 4.314, 4.316. 4.3144.3154.316Root bracket
Try it together

Return to f(x) = x³ - x - 3. Calculate f(1.671) = -0.005165289 and f(1.672) = 0.004216448. Continuity and opposite signs place a root between 1.671 and 1.672. Every value in this bracket rounds to 1.67, so the sign check certifies the two-decimal answer suggested by iteration.

f(1.671) < 0f(1.672) > 01.671 < root < 1.672root rounds to 1.67

Matching rounded iterates do not certify the root. For x_next = 0.9999x + 0.0001, the fixed point is 1. Starting at zero gives 0.0001 and then 0.00019999. Both round to 0.000 at three decimals, although both are far from 1. Use a justified error bound or a continuous-function sign bracket that certifies the requested accuracy.

x1 = 0.0001x2 = 0.00019999both round to 0.000fixed point = 1

A small function value is not an x-error bound either. For f(x) = 0.001(x - 10), the root is 10. At x = 9, f(9) = -0.001, but the estimate is still 1 away from the root. The function's scale affects its values. Check the distance using a justified bound or bracket.

f(9) = -0.001root = 10|9 - 10| = 1

Locate a root using continuity and opposite signs. Feed each output into x = g(x). Certify the required accuracy with a justified error bound or sign bracket.

This lesson, question bank and flashcards were written for this dataset.

These readings, additional questions and flashcards were written for this dataset. External records informed the mathematical review and question design; no source text is reproduced.

More to read

Check the interval before using its signs

For f(x) = x² - 11, the values f(3) = -2 and f(4) = 5 have opposite signs. A polynomial is continuous, so a root lies between these inputs. This calculation gives an interval containing a root, not its exact position.

The same conclusion does not follow for every rule with opposite endpoint signs. For 1/(x - 5), the values at 4 and 6 have opposite signs, but the function is undefined at 5. There is no zero of this function.

Equal signs have a different limitation. A continuous curve can touch zero inside an interval while both endpoint values are positive. It can also cross zero several times. Report what the endpoint evidence establishes, and check continuity throughout the interval before using the sign-change argument.

A usable rule may still fail to converge

There are several checks between rearranging an equation and finding a root. First check that the new equation is equivalent on the allowed inputs. Then check that the update is defined at the chosen starting value and at later values.

For x² = 4x + 9, division by nonzero x gives x = 4 + 9/x. Zero does not solve the original equation and cannot be used as an input to this update. A valid rearrangement does not make every starting value usable.

Even a rule defined at every update may fail to approach its fixed point. The equation x = 5 - x has fixed point 2.5. Starting at 1 gives 4, then 1, then 4 again. This repeating pair does not approach 2.5. Keep algebraic validity, allowed inputs and convergence as separate questions.

What certifies the answer's accuracy?

Suppose continuity and opposite endpoint signs place a root in a bracket. The midpoint is at most half the bracket's width from a root inside it. This gives a distance bound without knowing the exact root.

A rounded answer needs another check. A bracket from 6.724 to 6.726 crosses the rounding boundary 6.725. Possible roots within it can round to 6.72 or 6.73. Reducing the bracket's width is useful, but check where the whole bracket lies before claiming the rounded result.

Neither matching rounded updates nor a small function value supplies a distance bound by itself. For f(x) = 0.002(x - 7), the value f(6) = -0.002 is small, yet 6 is one unit from the root 7. Use evidence that bounds the distance to a root or fixes its rounding interval.

2 · Watch

En anglais

Take it off screen

Print a worksheetA4 with an answer key page for grown-ups. No screen, no internet.

Where it sits

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Locating Roots and Iterative Methods · Mathematics, ages 17 to 18 · LightMySky