Hard A-Level Locating roots and iteration Questions

Challenging, exam-style A-Level Locating roots and iteration questions with worked solutions. Stretch yourself on the hardest iteration, fixed-point, interval-bisection, root-location problems.

iterationfixed-pointinterval-bisectionroot-locationrearrangementchange-of-sign
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
A fixed-point iteration is defined by xn+1=1x+1x_{n+1}=\frac{1}{x + 1} with fixed point α\alpha. Using the value of g′(α)g'(\alpha), which statement best describes the behaviour of the iteration and its staircase or cobweb diagram?
Show worked solution

Worked solution

  1. Locate the fixed point alpha

    α≈0.6180\alpha\approx 0.6180

    The fixed point satisfies x=g(x).

  2. Differentiate g and evaluate the derivative at alpha

    g′(x)=−1(x+1)2,g′(α)≈−0.3820g'(x)=- \frac{1}{\left(x + 1\right)^{2}},\quad g'(\alpha)\approx -0.3820

    The size and sign of g'(alpha) fix the behaviour of the iteration.

  3. Apply the iterative formula to find x_{1}

    x1=11.00000+1≈0.50000x_{1}=\frac{1}{1.00000 + 1}\approx 0.50000

    Substitute the previous estimate and evaluate to 5 decimal places.

  4. Apply the iterative formula to find x_{2}

    x2=10.50000+1≈0.66667x_{2}=\frac{1}{0.50000 + 1}\approx 0.66667

    Substitute the previous estimate and evaluate to 5 decimal places.

  5. Apply the iterative formula to find x_{3}

    x3=10.66667+1≈0.60000x_{3}=\frac{1}{0.66667 + 1}\approx 0.60000

    Substitute the previous estimate and evaluate to 5 decimal places.

  6. Apply the iterative formula to find x_{4}

    x4=10.60000+1≈0.62500x_{4}=\frac{1}{0.60000 + 1}\approx 0.62500

    Substitute the previous estimate and evaluate to 5 decimal places.

  7. Apply the iterative formula to find x_{5}

    x5=10.62500+1≈0.61538x_{5}=\frac{1}{0.62500 + 1}\approx 0.61538

    Substitute the previous estimate and evaluate to 5 decimal places.

  8. Apply the iterative formula to find x_{6}

    x6=10.61538+1≈0.61905x_{6}=\frac{1}{0.61538 + 1}\approx 0.61905

    Substitute the previous estimate and evaluate to 5 decimal places.

  9. Apply the iterative formula to find x_{7}

    x7=10.61905+1≈0.61765x_{7}=\frac{1}{0.61905 + 1}\approx 0.61765

    Substitute the previous estimate and evaluate to 5 decimal places.

  10. Apply the iterative formula to find x_{8}

    x8=10.61765+1≈0.61818x_{8}=\frac{1}{0.61765 + 1}\approx 0.61818

    Substitute the previous estimate and evaluate to 5 decimal places.

  11. Apply the iterative formula to find x_{9}

    x9=10.61818+1≈0.61798x_{9}=\frac{1}{0.61818 + 1}\approx 0.61798

    Substitute the previous estimate and evaluate to 5 decimal places.

  12. Apply the iterative formula to find x_{10}

    x10=10.61798+1≈0.61806x_{10}=\frac{1}{0.61798 + 1}\approx 0.61806

    Substitute the previous estimate and evaluate to 5 decimal places.

  13. Apply the iterative formula to find x_{11}

    x11=10.61806+1≈0.61803x_{11}=\frac{1}{0.61806 + 1}\approx 0.61803

    Substitute the previous estimate and evaluate to 5 decimal places.

  14. Apply the iterative formula to find x_{12}

    x12=10.61803+1≈0.61804x_{12}=\frac{1}{0.61803 + 1}\approx 0.61804

    Substitute the previous estimate and evaluate to 5 decimal places.

  15. State the behaviour of the iteration near alpha

    ∣g′(α)∣<1\left|g'(\alpha)\right|<1

    This determines whether the staircase/cobweb converges or diverges.

Answer
The iteration converges to the fixed point, and the diagram is a cobweb (because -1 < g'(alpha) < 0).
Question 2
8 markschallenging
Consider the function f(x)=(x−5)(x−2)f(x)=\left(x - 5\right) \left(x - 2\right) on the interval [1, 6][1,\,6]. Here f(1)f(1) and f(6)f(6) have the same sign, yet f(x)=0f(x)=0 has two solutions in [1, 6][1,\,6]. Which statement explains why the change-of-sign method does not behave as usual here?
Show worked solution

Worked solution

  1. Evaluate f at x=1.333

    f(1.333)≈2.444f(1.333)\approx 2.444

    Sampling f across the interval reveals its behaviour.

  2. Evaluate f at x=1.667

    f(1.667)≈1.111f(1.667)\approx 1.111

    Sampling f across the interval reveals its behaviour.

  3. Evaluate f at x=2.000

    f(2.000)≈0.000f(2.000)\approx 0.000

    Sampling f across the interval reveals its behaviour.

  4. Evaluate f at x=2.333

    f(2.333)≈−0.889f(2.333)\approx -0.889

    Sampling f across the interval reveals its behaviour.

  5. Evaluate f at x=2.667

    f(2.667)≈−1.556f(2.667)\approx -1.556

    Sampling f across the interval reveals its behaviour.

  6. Evaluate f at x=3.000

    f(3.000)≈−2.000f(3.000)\approx -2.000

    Sampling f across the interval reveals its behaviour.

  7. Evaluate f at x=3.333

    f(3.333)≈−2.222f(3.333)\approx -2.222

    Sampling f across the interval reveals its behaviour.

  8. Evaluate f at x=3.667

    f(3.667)≈−2.222f(3.667)\approx -2.222

    Sampling f across the interval reveals its behaviour.

  9. Evaluate f at x=4.000

    f(4.000)≈−2.000f(4.000)\approx -2.000

    Sampling f across the interval reveals its behaviour.

  10. Evaluate f at x=4.333

    f(4.333)≈−1.556f(4.333)\approx -1.556

    Sampling f across the interval reveals its behaviour.

  11. Evaluate f at x=4.667

    f(4.667)≈−0.889f(4.667)\approx -0.889

    Sampling f across the interval reveals its behaviour.

  12. Evaluate f at x=5.000

    f(5.000)≈0.000f(5.000)\approx 0.000

    Sampling f across the interval reveals its behaviour.

  13. Evaluate f at x=5.333

    f(5.333)≈1.111f(5.333)\approx 1.111

    Sampling f across the interval reveals its behaviour.

  14. Evaluate f at x=5.667

    f(5.667)≈2.444f(5.667)\approx 2.444

    Sampling f across the interval reveals its behaviour.

  15. Explain why the change-of-sign test is unreliable here

    f(a) f(b)>0f(a)\,f(b)>0

    The interval contains an even number of roots (two), so the values of f at the two endpoints have the same sign.

Answer
The interval contains an even number of roots (two), so the values of f at the two endpoints have the same sign.
Question 3
8 markschallenging
Consider the function f(x)=1x−1f(x)=\frac{1}{x - 1} on the interval [0, 2][0,\,2]. Although f(0)f(0) and f(2)f(2) have opposite signs, the equation f(x)=0f(x)=0 has no root in [0, 2][0,\,2]. Which statement explains why the change-of-sign method does not behave as usual here?
Show worked solution

Worked solution

  1. Evaluate f at x=0.133

    f(0.133)≈−1.154f(0.133)\approx -1.154

    Sampling f across the interval reveals its behaviour.

  2. Evaluate f at x=0.267

    f(0.267)≈−1.364f(0.267)\approx -1.364

    Sampling f across the interval reveals its behaviour.

  3. Evaluate f at x=0.400

    f(0.400)≈−1.667f(0.400)\approx -1.667

    Sampling f across the interval reveals its behaviour.

  4. Evaluate f at x=0.533

    f(0.533)≈−2.143f(0.533)\approx -2.143

    Sampling f across the interval reveals its behaviour.

  5. Evaluate f at x=0.667

    f(0.667)≈−3.000f(0.667)\approx -3.000

    Sampling f across the interval reveals its behaviour.

  6. Evaluate f at x=0.800

    f(0.800)≈−5.000f(0.800)\approx -5.000

    Sampling f across the interval reveals its behaviour.

  7. Evaluate f at x=0.933

    f(0.933)≈−15.000f(0.933)\approx -15.000

    Sampling f across the interval reveals its behaviour.

  8. Evaluate f at x=1.067

    f(1.067)≈15.000f(1.067)\approx 15.000

    Sampling f across the interval reveals its behaviour.

  9. Evaluate f at x=1.200

    f(1.200)≈5.000f(1.200)\approx 5.000

    Sampling f across the interval reveals its behaviour.

  10. Evaluate f at x=1.333

    f(1.333)≈3.000f(1.333)\approx 3.000

    Sampling f across the interval reveals its behaviour.

  11. Evaluate f at x=1.467

    f(1.467)≈2.143f(1.467)\approx 2.143

    Sampling f across the interval reveals its behaviour.

  12. Evaluate f at x=1.600

    f(1.600)≈1.667f(1.600)\approx 1.667

    Sampling f across the interval reveals its behaviour.

  13. Evaluate f at x=1.733

    f(1.733)≈1.364f(1.733)\approx 1.364

    Sampling f across the interval reveals its behaviour.

  14. Evaluate f at x=1.867

    f(1.867)≈1.154f(1.867)\approx 1.154

    Sampling f across the interval reveals its behaviour.

  15. Explain why the change-of-sign test is unreliable here

    f(a) f(b)<0f(a)\,f(b)<0

    The function is discontinuous at x=1 (a vertical asymptote), so a change of sign does not guarantee a root.

Answer
The function is discontinuous at x=1 (a vertical asymptote), so a change of sign does not guarantee a root.
Question 4
8 markschallenging
The equation x3−8x−5=0x^{3} - 8 x - 5=0 is to be solved by a fixed-point iteration of the form xn+1=g(xn)x_{n+1}=g(x_{n}). Which of the following is a valid rearrangement x=g(x)x=g(x) of the equation?
Show worked solution

Worked solution

  1. Write the equation to be solved

    x3−8x−5=0x^{3} - 8 x - 5=0

    We look for a rearrangement into the form x=g(x).

  2. Rearrange the equation into the form x=g(x)

    x=8x+53x=\sqrt[3]{8 x + 5}

    Adding 8 x+5 and taking the cube root isolates x validly.

  3. Apply the iterative formula to find x_{1}

    x1=8⋅2.00000+53≈2.75892x_{1}=\sqrt[3]{8 \cdot 2.00000 + 5}\approx 2.75892

    Substitute the previous estimate and evaluate to 5 decimal places.

  4. Apply the iterative formula to find x_{2}

    x2=8⋅2.75892+53≈3.00264x_{2}=\sqrt[3]{8 \cdot 2.75892 + 5}\approx 3.00264

    Substitute the previous estimate and evaluate to 5 decimal places.

  5. Apply the iterative formula to find x_{3}

    x3=8⋅3.00264+53≈3.07306x_{3}=\sqrt[3]{8 \cdot 3.00264 + 5}\approx 3.07306

    Substitute the previous estimate and evaluate to 5 decimal places.

  6. Apply the iterative formula to find x_{4}

    x4=8⋅3.07306+53≈3.09282x_{4}=\sqrt[3]{8 \cdot 3.07306 + 5}\approx 3.09282

    Substitute the previous estimate and evaluate to 5 decimal places.

  7. Apply the iterative formula to find x_{5}

    x5=8⋅3.09282+53≈3.09832x_{5}=\sqrt[3]{8 \cdot 3.09282 + 5}\approx 3.09832

    Substitute the previous estimate and evaluate to 5 decimal places.

  8. Apply the iterative formula to find x_{6}

    x6=8⋅3.09832+53≈3.09985x_{6}=\sqrt[3]{8 \cdot 3.09832 + 5}\approx 3.09985

    Substitute the previous estimate and evaluate to 5 decimal places.

  9. Apply the iterative formula to find x_{7}

    x7=8⋅3.09985+53≈3.10027x_{7}=\sqrt[3]{8 \cdot 3.09985 + 5}\approx 3.10027

    Substitute the previous estimate and evaluate to 5 decimal places.

  10. Apply the iterative formula to find x_{8}

    x8=8⋅3.10027+53≈3.10039x_{8}=\sqrt[3]{8 \cdot 3.10027 + 5}\approx 3.10039

    Substitute the previous estimate and evaluate to 5 decimal places.

  11. Apply the iterative formula to find x_{9}

    x9=8⋅3.10039+53≈3.10042x_{9}=\sqrt[3]{8 \cdot 3.10039 + 5}\approx 3.10042

    Substitute the previous estimate and evaluate to 5 decimal places.

  12. Apply the iterative formula to find x_{10}

    x10=8⋅3.10042+53≈3.10043x_{10}=\sqrt[3]{8 \cdot 3.10042 + 5}\approx 3.10043

    Substitute the previous estimate and evaluate to 5 decimal places.

  13. Apply the iterative formula to find x_{11}

    x11=8⋅3.10043+53≈3.10043x_{11}=\sqrt[3]{8 \cdot 3.10043 + 5}\approx 3.10043

    Substitute the previous estimate and evaluate to 5 decimal places.

  14. Apply the iterative formula to find x_{12}

    x12=8⋅3.10043+53≈3.10043x_{12}=\sqrt[3]{8 \cdot 3.10043 + 5}\approx 3.10043

    Substitute the previous estimate and evaluate to 5 decimal places.

  15. Confirm the valid iterative formula

    x3−8x−5=0  ⇔  x=8x+53x^{3} - 8 x - 5=0\;\Leftrightarrow\;x=\sqrt[3]{8 x + 5}

    This rearrangement is equivalent to the original equation.

Answer
x=8x+53x=\sqrt[3]{8 x + 5}
Question 5
8 markschallenging
The equation x3−4x−3=0x^{3} - 4 x - 3=0 is to be solved by a fixed-point iteration of the form xn+1=g(xn)x_{n+1}=g(x_{n}). Which of the following is a valid rearrangement x=g(x)x=g(x) of the equation?
Show worked solution

Worked solution

  1. Write the equation to be solved

    x3−4x−3=0x^{3} - 4 x - 3=0

    We look for a rearrangement into the form x=g(x).

  2. Rearrange the equation into the form x=g(x)

    x=4x+33x=\sqrt[3]{4 x + 3}

    Adding 4 x+3 and taking the cube root isolates x validly.

  3. Apply the iterative formula to find x_{1}

    x1=4⋅2.00000+33≈2.22398x_{1}=\sqrt[3]{4 \cdot 2.00000 + 3}\approx 2.22398

    Substitute the previous estimate and evaluate to 5 decimal places.

  4. Apply the iterative formula to find x_{2}

    x2=4⋅2.22398+33≈2.28279x_{2}=\sqrt[3]{4 \cdot 2.22398 + 3}\approx 2.28279

    Substitute the previous estimate and evaluate to 5 decimal places.

  5. Apply the iterative formula to find x_{3}

    x3=4⋅2.28279+33≈2.29774x_{3}=\sqrt[3]{4 \cdot 2.28279 + 3}\approx 2.29774

    Substitute the previous estimate and evaluate to 5 decimal places.

  6. Apply the iterative formula to find x_{4}

    x4=4⋅2.29774+33≈2.30151x_{4}=\sqrt[3]{4 \cdot 2.29774 + 3}\approx 2.30151

    Substitute the previous estimate and evaluate to 5 decimal places.

  7. Apply the iterative formula to find x_{5}

    x5=4⋅2.30151+33≈2.30246x_{5}=\sqrt[3]{4 \cdot 2.30151 + 3}\approx 2.30246

    Substitute the previous estimate and evaluate to 5 decimal places.

  8. Apply the iterative formula to find x_{6}

    x6=4⋅2.30246+33≈2.30270x_{6}=\sqrt[3]{4 \cdot 2.30246 + 3}\approx 2.30270

    Substitute the previous estimate and evaluate to 5 decimal places.

  9. Apply the iterative formula to find x_{7}

    x7=4⋅2.30270+33≈2.30276x_{7}=\sqrt[3]{4 \cdot 2.30270 + 3}\approx 2.30276

    Substitute the previous estimate and evaluate to 5 decimal places.

  10. Apply the iterative formula to find x_{8}

    x8=4⋅2.30276+33≈2.30277x_{8}=\sqrt[3]{4 \cdot 2.30276 + 3}\approx 2.30277

    Substitute the previous estimate and evaluate to 5 decimal places.

  11. Apply the iterative formula to find x_{9}

    x9=4⋅2.30277+33≈2.30277x_{9}=\sqrt[3]{4 \cdot 2.30277 + 3}\approx 2.30277

    Substitute the previous estimate and evaluate to 5 decimal places.

  12. Apply the iterative formula to find x_{10}

    x10=4⋅2.30277+33≈2.30278x_{10}=\sqrt[3]{4 \cdot 2.30277 + 3}\approx 2.30278

    Substitute the previous estimate and evaluate to 5 decimal places.

  13. Apply the iterative formula to find x_{11}

    x11=4⋅2.30278+33≈2.30278x_{11}=\sqrt[3]{4 \cdot 2.30278 + 3}\approx 2.30278

    Substitute the previous estimate and evaluate to 5 decimal places.

  14. Apply the iterative formula to find x_{12}

    x12=4⋅2.30278+33≈2.30278x_{12}=\sqrt[3]{4 \cdot 2.30278 + 3}\approx 2.30278

    Substitute the previous estimate and evaluate to 5 decimal places.

  15. Confirm the valid iterative formula

    x3−4x−3=0  ⇔  x=4x+33x^{3} - 4 x - 3=0\;\Leftrightarrow\;x=\sqrt[3]{4 x + 3}

    This rearrangement is equivalent to the original equation.

Answer
x=4x+33x=\sqrt[3]{4 x + 3}

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