Hard A-Level Newton–Raphson Questions

Challenging, exam-style A-Level Newton–Raphson questions with worked solutions. Stretch yourself on the hardest newton-raphson, iteration, root, accuracy problems.

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A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
Use the Newton-Raphson method, starting from x0=1x_{0}=1, to find the root of x3+2x−2=0x^{3} + 2 x - 2=0 correct to 6 decimal places.
Show worked solution

Worked solution

  1. Set up the Newton-Raphson iteration

    xn+1=xn−xn3+2xn−23xn2+2x_{n+1}=x_{n}-\frac{x_{n}^{3} + 2 x_{n} - 2}{3 x_{n}^{2} + 2}

    Form the explicit formula from f and f'.

  2. Iterate from x_0 until the value stabilises

    x0=1 → x=0.770917x_{0}=1\ \to\ x=0.770917

    Repeated iteration converges to the root to the required accuracy.

  3. Confirm the root with a sign change

    f(0.7709165)=−0.000001880,f(0.7709175)=0.000001903f(0.7709165)=-0.000001880,\quad f(0.7709175)=0.000001903

    Opposite signs either side confirm the root rounds to this value.

  4. Write down the function

    f(x)=x3+2x−2f(x)=x^{3} + 2 x - 2

    This is the function whose root we are approximating.

  5. Differentiate the function

    f′(x)=3x2+2f'(x)=3 x^{2} + 2

    The derivative gives the slope of the tangent used at each step.

  6. Substitute f and f' into the formula

    xn+1=xn−xn3+2xn−23xn2+2x_{n+1}=x_{n}-\dfrac{x_{n}^{3} + 2 x_{n} - 2}{3 x_{n}^{2} + 2}

    This gives the explicit iterative formula for this function.

  7. Interpret the step geometrically

    y−f(xn)=f′(xn)(x−xn)y-f(x_{n})=f'(x_{n})\left(x-x_{n}\right)

    The next estimate is where the tangent at x_n meets the x-axis.

  8. Set y=0 on the tangent

    0−f(xn)=f′(xn)(x−xn)0-f(x_{n})=f'(x_{n})\left(x-x_{n}\right)

    The x-intercept of the tangent defines the next approximation.

  9. Rearrange the tangent equation for x

    x=xn−f(xn)f′(xn)x=x_{n}-\dfrac{f(x_{n})}{f'(x_{n})}

    This rearrangement recovers the Newton-Raphson formula.

  10. Recall the Newton-Raphson iteration

    xn+1=xn−f(xn)f′(xn)x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}

    Each new estimate is the previous one minus f divided by f'.

  11. Note a possible failure condition

    f′(xn)=0f'(x_{n})=0

    A stationary point gives a horizontal tangent, so the method breaks down.

  12. Comment on the starting value

    ∣x0−α∣ small\left|x_{0}-\alpha\right|\ \text{small}

    A poor starting point may diverge or converge to a different root.

  13. Confirm a root by a sign change

    f(a) f(b)<0f(a)\,f(b)<0

    Opposite signs across an interval show a root lies between a and b.

  14. State the required accuracy

    round only the final value\text{round only the final value}

    Full precision is kept while iterating and rounding is done at the end.

  15. State the root to the required accuracy

    x=0.770917x=0.770917

    This is the root to 6 decimal places.

Answer
0.7709170.770917
Question 2
8 markschallenging
The equation x3+2x−2=0x^{3} + 2 x - 2=0 is to be solved by the Newton-Raphson method with f(x)=x3+2x−2f(x)=x^{3} + 2 x - 2 and x0=1x_{0}=1. Find x3x_{3}, giving your answer to 8 decimal places.
Show worked solution

Worked solution

  1. Write the function and its derivative

    f(x)=x3+2x−2,f′(x)=3x2+2f(x)=x^{3} + 2 x - 2,\quad f'(x)=3 x^{2} + 2

    Both f and f' are needed for every iteration.

  2. State the starting value

    x0=1x_{0}=1

    The iteration begins from this value.

  3. Apply the iteration to find x_{1}

    x1=1−1.000000005.00000000=0.80000000x_{1}=1-\frac{1.00000000}{5.00000000}=0.80000000

    Substitute the current estimate into the formula and round to 8 d.p.

  4. Apply the iteration to find x_{2}

    x2=0.80000000−0.112000003.92000000=0.77142857x_{2}=0.80000000-\frac{0.11200000}{3.92000000}=0.77142857

    Substitute the current estimate into the formula and round to 8 d.p.

  5. Apply the iteration to find x_{3}

    x3=0.77142857−0.001935863.78530612=0.77091716x_{3}=0.77142857-\frac{0.00193586}{3.78530612}=0.77091716

    Substitute the current estimate into the formula and round to 8 d.p.

  6. Write down the function

    f(x)=x3+2x−2f(x)=x^{3} + 2 x - 2

    This is the function whose root we are approximating.

  7. Differentiate the function

    f′(x)=3x2+2f'(x)=3 x^{2} + 2

    The derivative gives the slope of the tangent used at each step.

  8. Substitute f and f' into the formula

    xn+1=xn−xn3+2xn−23xn2+2x_{n+1}=x_{n}-\dfrac{x_{n}^{3} + 2 x_{n} - 2}{3 x_{n}^{2} + 2}

    This gives the explicit iterative formula for this function.

  9. Interpret the step geometrically

    y−f(xn)=f′(xn)(x−xn)y-f(x_{n})=f'(x_{n})\left(x-x_{n}\right)

    The next estimate is where the tangent at x_n meets the x-axis.

  10. Set y=0 on the tangent

    0−f(xn)=f′(xn)(x−xn)0-f(x_{n})=f'(x_{n})\left(x-x_{n}\right)

    The x-intercept of the tangent defines the next approximation.

  11. Rearrange the tangent equation for x

    x=xn−f(xn)f′(xn)x=x_{n}-\dfrac{f(x_{n})}{f'(x_{n})}

    This rearrangement recovers the Newton-Raphson formula.

  12. Recall the Newton-Raphson iteration

    xn+1=xn−f(xn)f′(xn)x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}

    Each new estimate is the previous one minus f divided by f'.

  13. Note a possible failure condition

    f′(xn)=0f'(x_{n})=0

    A stationary point gives a horizontal tangent, so the method breaks down.

  14. Comment on the starting value

    ∣x0−α∣ small\left|x_{0}-\alpha\right|\ \text{small}

    A poor starting point may diverge or converge to a different root.

  15. State the required approximation

    x3=0.77091716x_{3}=0.77091716

    This is x_3 to 8 decimal places.

Answer
0.770917160.77091716
Question 3
8 markschallenging
Which situation does NOT typically cause the Newton-Raphson method to fail?
Show worked solution

Worked solution

  1. State the Newton-Raphson formula

    xn+1=xn−f(xn)f′(xn)x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}

    The method is built on following the tangent at each estimate.

  2. Examine the role of the derivative

    step=f(xn)f′(xn)\text{step}=\dfrac{f(x_{n})}{f'(x_{n})}

    If f'(x_n) is zero or very small the step becomes undefined or huge.

  3. Write down the function

    f(x)=x3−2x−5f(x)=x^{3} - 2 x - 5

    This is the function whose root we are approximating.

  4. Differentiate the function

    f′(x)=3x2−2f'(x)=3 x^{2} - 2

    The derivative gives the slope of the tangent used at each step.

  5. Substitute f and f' into the formula

    xn+1=xn−xn3−2xn−53xn2−2x_{n+1}=x_{n}-\dfrac{x_{n}^{3} - 2 x_{n} - 5}{3 x_{n}^{2} - 2}

    This gives the explicit iterative formula for this function.

  6. Interpret the step geometrically

    y−f(xn)=f′(xn)(x−xn)y-f(x_{n})=f'(x_{n})\left(x-x_{n}\right)

    The next estimate is where the tangent at x_n meets the x-axis.

  7. Set y=0 on the tangent

    0−f(xn)=f′(xn)(x−xn)0-f(x_{n})=f'(x_{n})\left(x-x_{n}\right)

    The x-intercept of the tangent defines the next approximation.

  8. Rearrange the tangent equation for x

    x=xn−f(xn)f′(xn)x=x_{n}-\dfrac{f(x_{n})}{f'(x_{n})}

    This rearrangement recovers the Newton-Raphson formula.

  9. Recall the Newton-Raphson iteration

    xn+1=xn−f(xn)f′(xn)x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}

    Each new estimate is the previous one minus f divided by f'.

  10. Note a possible failure condition

    f′(xn)=0f'(x_{n})=0

    A stationary point gives a horizontal tangent, so the method breaks down.

  11. Comment on the starting value

    ∣x0−α∣ small\left|x_{0}-\alpha\right|\ \text{small}

    A poor starting point may diverge or converge to a different root.

  12. Confirm a root by a sign change

    f(a) f(b)<0f(a)\,f(b)<0

    Opposite signs across an interval show a root lies between a and b.

  13. State the required accuracy

    round only the final value\text{round only the final value}

    Full precision is kept while iterating and rounding is done at the end.

  14. Describe convergence

    xn+1≈xnx_{n+1}\approx x_{n}

    Iteration stops once successive estimates agree to the required accuracy.

  15. Select the correct statement

    Starting close to a simple root where f'(x) is not zero.

    This correctly describes the behaviour of the method.

Answer
Starting close to a simple root where f'(x) is not zero.
Question 4
8 markschallenging
If the tangent at xnx_{n} is nearly horizontal, what happens to xn+1x_{n+1}?
Show worked solution

Worked solution

  1. State the Newton-Raphson formula

    xn+1=xn−f(xn)f′(xn)x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}

    The method is built on following the tangent at each estimate.

  2. Examine the role of the derivative

    step=f(xn)f′(xn)\text{step}=\dfrac{f(x_{n})}{f'(x_{n})}

    If f'(x_n) is zero or very small the step becomes undefined or huge.

  3. Write down the function

    f(x)=x3−2x−5f(x)=x^{3} - 2 x - 5

    This is the function whose root we are approximating.

  4. Differentiate the function

    f′(x)=3x2−2f'(x)=3 x^{2} - 2

    The derivative gives the slope of the tangent used at each step.

  5. Substitute f and f' into the formula

    xn+1=xn−xn3−2xn−53xn2−2x_{n+1}=x_{n}-\dfrac{x_{n}^{3} - 2 x_{n} - 5}{3 x_{n}^{2} - 2}

    This gives the explicit iterative formula for this function.

  6. Interpret the step geometrically

    y−f(xn)=f′(xn)(x−xn)y-f(x_{n})=f'(x_{n})\left(x-x_{n}\right)

    The next estimate is where the tangent at x_n meets the x-axis.

  7. Set y=0 on the tangent

    0−f(xn)=f′(xn)(x−xn)0-f(x_{n})=f'(x_{n})\left(x-x_{n}\right)

    The x-intercept of the tangent defines the next approximation.

  8. Rearrange the tangent equation for x

    x=xn−f(xn)f′(xn)x=x_{n}-\dfrac{f(x_{n})}{f'(x_{n})}

    This rearrangement recovers the Newton-Raphson formula.

  9. Recall the Newton-Raphson iteration

    xn+1=xn−f(xn)f′(xn)x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}

    Each new estimate is the previous one minus f divided by f'.

  10. Note a possible failure condition

    f′(xn)=0f'(x_{n})=0

    A stationary point gives a horizontal tangent, so the method breaks down.

  11. Comment on the starting value

    ∣x0−α∣ small\left|x_{0}-\alpha\right|\ \text{small}

    A poor starting point may diverge or converge to a different root.

  12. Confirm a root by a sign change

    f(a) f(b)<0f(a)\,f(b)<0

    Opposite signs across an interval show a root lies between a and b.

  13. State the required accuracy

    round only the final value\text{round only the final value}

    Full precision is kept while iterating and rounding is done at the end.

  14. Describe convergence

    xn+1≈xnx_{n+1}\approx x_{n}

    Iteration stops once successive estimates agree to the required accuracy.

  15. Select the correct statement

    Itisthrownfarfromxnbecausedividingbythesmallslopef′(xn)givesalargestep.It is thrown far from x_n because dividing by the small slope f'(x_n) gives a large step.

    This correctly describes the behaviour of the method.

Answer
Itisthrownfarfromxnbecausedividingbythesmallslopef′(xn)givesalargestep.It is thrown far from x_n because dividing by the small slope f'(x_n) gives a large step.
Question 5
8 markschallenging
For f(x)=−x2+sin⁡(x)f(x)=- \frac{x}{2} + \sin{\left(x \right)}, which of the following is the correct Newton-Raphson iterative formula?
Show worked solution

Worked solution

  1. Differentiate the function

    f′(x)=cos⁡(x)−12f'(x)=\cos{\left(x \right)} - \frac{1}{2}

    The derivative appears in the denominator of the formula.

  2. Recall the Newton-Raphson formula

    xn+1=xn−f(xn)f′(xn)x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}

    The correct formula subtracts f divided by f'.

  3. Write down the function

    f(x)=−x2+sin⁡(x)f(x)=- \frac{x}{2} + \sin{\left(x \right)}

    This is the function whose root we are approximating.

  4. Differentiate the function

    f′(x)=cos⁡(x)−12f'(x)=\cos{\left(x \right)} - \frac{1}{2}

    The derivative gives the slope of the tangent used at each step.

  5. Substitute f and f' into the formula

    xn+1=xn−−xn2+sin⁡(xn)cos⁡(xn)−12x_{n+1}=x_{n}-\dfrac{- \frac{x_{n}}{2} + \sin{\left(x_{n} \right)}}{\cos{\left(x_{n} \right)} - \frac{1}{2}}

    This gives the explicit iterative formula for this function.

  6. Interpret the step geometrically

    y−f(xn)=f′(xn)(x−xn)y-f(x_{n})=f'(x_{n})\left(x-x_{n}\right)

    The next estimate is where the tangent at x_n meets the x-axis.

  7. Set y=0 on the tangent

    0−f(xn)=f′(xn)(x−xn)0-f(x_{n})=f'(x_{n})\left(x-x_{n}\right)

    The x-intercept of the tangent defines the next approximation.

  8. Rearrange the tangent equation for x

    x=xn−f(xn)f′(xn)x=x_{n}-\dfrac{f(x_{n})}{f'(x_{n})}

    This rearrangement recovers the Newton-Raphson formula.

  9. Recall the Newton-Raphson iteration

    xn+1=xn−f(xn)f′(xn)x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}

    Each new estimate is the previous one minus f divided by f'.

  10. Note a possible failure condition

    f′(xn)=0f'(x_{n})=0

    A stationary point gives a horizontal tangent, so the method breaks down.

  11. Comment on the starting value

    ∣x0−α∣ small\left|x_{0}-\alpha\right|\ \text{small}

    A poor starting point may diverge or converge to a different root.

  12. Confirm a root by a sign change

    f(a) f(b)<0f(a)\,f(b)<0

    Opposite signs across an interval show a root lies between a and b.

  13. State the required accuracy

    round only the final value\text{round only the final value}

    Full precision is kept while iterating and rounding is done at the end.

  14. Describe convergence

    xn+1≈xnx_{n+1}\approx x_{n}

    Iteration stops once successive estimates agree to the required accuracy.

  15. Select the correct iterative formula

    xn+1=xn−−xn2+sin⁡(xn)cos⁡(xn)−12x_{n+1}=x_{n}-\frac{- \frac{x_{n}}{2} + \sin{\left(x_{n} \right)}}{\cos{\left(x_{n} \right)} - \frac{1}{2}}

    This matches x_n minus f(x_n) over f'(x_n).

Answer
xn+1=xn−−xn2+sin⁡(xn)cos⁡(xn)−12x_{n+1}=x_{n}-\frac{- \frac{x_{n}}{2} + \sin{\left(x_{n} \right)}}{\cos{\left(x_{n} \right)} - \frac{1}{2}}

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