GCSE Iteration Practice Questions

Free GCSE Iteration practice questions with full step-by-step worked solutions. Covers substitution, evaluating f(x), sign of f(x), iterative formula. Practise exam-style problems and check your method.

substitutionevaluating f(x)sign of f(x)iterative formularoundingsign change
GCSE Higher70 questionsStep-by-step solutions
Question 1
1 markeasy
f(x)=x3−2x−5f(x) = x^3 - 2x - 5. Work out the value of f(2)f(2).
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Worked solution

  1. Substitute the value into the function

    f(2)=23−2×2−5f(2) = 2^{3} - 2 \times 2 - 5

    Replace every xx in f(x)f(x) with 22.

  2. Work out each term

    f(2)=8−4−5f(2) = 8 - 4 - 5

    Deal with the powers first, then the multiplications, then add and subtract.

  3. State the value

    f(2)=−1f(2) = -1

    So f(2)f(2) is −1-1, which is negative.

Answer
f(2)=−1f(2) = -1
Question 2
2 markseasy
The equation x2−4x−3=0x^2 - 4x - 3 = 0 is to be solved by iteration. Which of these iterative formulae is a correct rearrangement of the equation?
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Worked solution

  1. Rearrange the equation

    x2=4x+3x^2 = 4x + 3

    Move 4x+34x + 3 to the right-hand side.

  2. Undo the power

    x=4+3xx = 4 + \frac{3}{x}

    Divide every term by xx.

  3. Write it as an iterative formula

    xn+1=4+3xnx_{n+1} = 4 + \frac{3}{x_n}

    The new value xn+1x_{n+1} comes from putting the old value xnx_n into the right-hand side.

Answer
xn+1=4+3xnx_{n+1} = 4 + \frac{3}{x_n}
Question 3
2 marksintermediate
The iterative formula xn+1=4xn+1x_{n+1} = \frac{4}{x_n + 1} converges to a number xx. This number is a root of which equation?
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Worked solution

  1. Write down what the limit satisfies

    x=4x+1x = \frac{4}{x + 1}

    If the iterates settle on a value xx, then feeding xx in gives xx back out.

  2. Clear the root or the fraction

    x(x+1)=4x(x + 1) = 4

    Multiply both sides by (x+1)(x + 1).

  3. Rearrange into the form f(x)=0f(x) = 0

    x2+x−4=0x^2 + x - 4 = 0

    Expand the bracket and take the 44 across.

  4. State the equation

    x2+x−4=0x^2 + x - 4 = 0

    This is the equation the iteration is solving.

  5. Check numerically

    x5=1.617021…x_5 = 1.617021\ldots

    Running the iteration from x0=1x_0 = 1 gives x5≈1.617021x_5 \approx 1.617021.

  6. Substitute the limit back in

    f(1.617021)≈0.231779f(1.617021) \approx 0.231779

    The value makes f(x)f(x) essentially zero, confirming it is a root of this equation and not of any of the others.

Answer
x2+x−4=0x^2 + x - 4 = 0
Question 4
3 markshard
The equation x2−5=0x^2 - 5 = 0 can be rearranged to give the iterative formula xn+1=5xnx_{n+1} = \frac{5}{x_n}. Starting with x0=2x_0 = 2, describe what happens to the sequence of iterates.
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Worked solution

  1. Write down the formula and the starting value

    xn+1=5xn,x0=2x_{n+1} = \frac{5}{x_n}, \qquad x_0 = 2

    The rearrangement is valid algebra, but that alone does not make the iteration work.

  2. Work out x1x_{1}

    x1=2.5000x_{1} = 2.5000

    Substitute the previous iterate into the right-hand side, keeping full calculator accuracy.

  3. Work out x2x_{2}

    x2=2.0000x_{2} = 2.0000

    Substitute the previous iterate into the right-hand side, keeping full calculator accuracy.

  4. Work out x3x_{3}

    x3=2.5000x_{3} = 2.5000

    Substitute the previous iterate into the right-hand side, keeping full calculator accuracy.

  5. Work out x4x_{4}

    x4=2.0000x_{4} = 2.0000

    Substitute the previous iterate into the right-hand side, keeping full calculator accuracy.

  6. Work out x5x_{5}

    x5=2.5000x_{5} = 2.5000

    Substitute the previous iterate into the right-hand side, keeping full calculator accuracy.

  7. Work out x6x_{6}

    x6=2.0000x_{6} = 2.0000

    Substitute the previous iterate into the right-hand side, keeping full calculator accuracy.

  8. Look at what is happening

    x0=2.0000,  x1=2.5000,  x2=2.0000,  x3=2.5000x_0 = 2.0000,\; x_1 = 2.5000,\; x_2 = 2.0000,\; x_3 = 2.5000

    The terms simply repeat: x2=x0x_2 = x_0 and x3=x1x_3 = x_1.

  9. Explain why

    xn+2=5  5xn  =xnx_{n+2} = \frac{5}{\;\frac{5}{x_n}\;} = x_n

    Applying the formula twice returns you to where you started, so the sequence is locked into a two-value cycle and can never close in on the root 2.23612.2361.

  10. Conclude

    xn alternates: 2.0000,  2.5000,  2.0000,  2.5000,…x_n \text{ alternates: } 2.0000,\; 2.5000,\; 2.0000,\; 2.5000,\ldots

    The sequence never converges, so this iteration cannot be used to find the root.

Answer
The terms just flip back and forth between 22 and 2.52.5 for ever, so the sequence never converges and the iteration never gives the root.
Question 5
5 markschallenging
The equation x3−2x−5=0x^3 - 2x - 5 = 0 can be rearranged in more than one way. Iteration A: xn+1=xn3−52x_{n+1} = \frac{x_n^3 - 5}{2} with x0=2x_0 = 2. Iteration B: xn+1=2xn+53x_{n+1} = \sqrt[3]{2x_n + 5} with x0=2x_0 = 2. Which statement correctly describes what the two iterations do?
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Worked solution

  1. Both rearrangements are valid

    x3−2x−5=0  ⇒  x=x3−52andx=2x+53x^3 - 2x - 5 = 0 \;\Rightarrow\; x = \frac{x^3 - 5}{2} \quad\text{and}\quad x = \sqrt[3]{2x + 5}

    Each can be reversed to give the original equation, so both have the root as a fixed point. Validity is not the issue — behaviour is.

  2. Start iteration A

    xn+1=xn3−52,x0=2x_{n+1} = \frac{x_n^3 - 5}{2}, \qquad x_0 = 2

    Apply the formula and watch the terms.

  3. Iteration A: x1x_{1}

    x1=1.5000x_{1} = 1.5000

    Substitute the previous term into the right-hand side.

  4. Iteration A: x2x_{2}

    x2=−0.8125x_{2} = -0.8125

    Substitute the previous term into the right-hand side.

  5. Iteration A: x3x_{3}

    x3=−2.7682x_{3} = -2.7682

    Substitute the previous term into the right-hand side.

  6. Iteration A: x4x_{4}

    x4=−13.1061x_{4} = -13.1061

    Substitute the previous term into the right-hand side.

  7. Iteration A: x5x_{5}

    x5=−1128.1244x_{5} = -1128.1244

    Substitute the previous term into the right-hand side.

  8. Iteration A is diverging

    ∣x1∣=1.500→∣x5∣=1128.12|x_1| = 1.500 \to |x_5| = 1128.12

    The terms are flying away from the root, not towards it. Iteration A is useless for finding the root.

  9. Now start iteration B

    xn+1=2xn+53,x0=2x_{n+1} = \sqrt[3]{2x_n + 5}, \qquad x_0 = 2

    The same equation, a different rearrangement.

  10. Iteration B: x1x_{1}

    x1=2.080083…x_{1} = 2.080083\ldots

    Keep the full calculator display between steps.

  11. Iteration B: x2x_{2}

    x2=2.092350…x_{2} = 2.092350\ldots

    Keep the full calculator display between steps.

  12. Iteration B: x3x_{3}

    x3=2.094216…x_{3} = 2.094216\ldots

    Keep the full calculator display between steps.

  13. Iteration B: x4x_{4}

    x4=2.094500…x_{4} = 2.094500\ldots

    Keep the full calculator display between steps.

  14. Iteration B converges

    x4=2.09450  ⇒  x=2.095 (3 d.p.)x_4 = 2.09450 \;\Rightarrow\; x = 2.095 \text{ (3 d.p.)}

    The terms settle down on the root x=2.095x = 2.095 to 33 decimal places.

  15. Conclude

    A diverges,B→2.095\text{A diverges},\qquad \text{B} \to 2.095

    Two valid rearrangements of the same equation can behave completely differently; you must check that an iteration actually converges before trusting it.

Answer
A diverges; B converges to x=2.095 (3 d.p.)\text{A diverges; B converges to } x = 2.095 \text{ (3 d.p.)}

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