A-Level Newton–Raphson Practice Questions

Free A-Level Newton–Raphson practice questions with full step-by-step worked solutions. Covers newton-raphson, differentiation, iterative-formula, iteration. Practise exam-style problems and check your method.

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A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Given f(x)=x3−2x−5f(x)=x^{3} - 2 x - 5, find f′(x)f'(x) (the derivative needed for the Newton-Raphson method).
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Worked solution

  1. Write down the function

    f(x)=x3−2x−5f(x)=x^{3} - 2 x - 5

    Identify the function that is to be differentiated.

  2. Differentiate with respect to x

    3x2−23 x^{2} - 2

    Apply the standard rules of differentiation term by term.

  3. State the derivative

    f′(x)=3x2−2f'(x)=3 x^{2} - 2

    This derivative is used in the Newton-Raphson formula.

Answer
3x2−23 x^{2} - 2
Question 2
2 markseasy
For f(x)=x3−x−1f(x)=x^{3} - x - 1, which of the following is the correct Newton-Raphson iterative formula?
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Worked solution

  1. Differentiate the function

    f′(x)=3x2−1f'(x)=3 x^{2} - 1

    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. Select the correct iterative formula

    xn+1=xn−xn3−xn−13xn2−1x_{n+1}=x_{n}-\frac{x_{n}^{3} - x_{n} - 1}{3 x_{n}^{2} - 1}

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

Answer
xn+1=xn−xn3−xn−13xn2−1x_{n+1}=x_{n}-\frac{x_{n}^{3} - x_{n} - 1}{3 x_{n}^{2} - 1}
Question 3
3 marksintermediate
The equation x3−2x2−1=0x^{3} - 2 x^{2} - 1=0 is to be solved by the Newton-Raphson method with f(x)=x3−2x2−1f(x)=x^{3} - 2 x^{2} - 1 and x0=2x_{0}=2. Find x3x_{3}, giving your answer to 4 decimal places.
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Worked solution

  1. Write the function and its derivative

    f(x)=x3−2x2−1,f′(x)=3x2−4xf(x)=x^{3} - 2 x^{2} - 1,\quad f'(x)=3 x^{2} - 4 x

    Both f and f' are needed for every iteration.

  2. State the starting value

    x0=2x_{0}=2

    The iteration begins from this value.

  3. Apply the iteration to find x_{1}

    x1=2−−1.00004.0000=2.2500x_{1}=2-\frac{-1.0000}{4.0000}=2.2500

    Substitute the current estimate into the formula and round to 4 d.p.

  4. Apply the iteration to find x_{2}

    x2=2.2500−0.26566.1875=2.2071x_{2}=2.2500-\frac{0.2656}{6.1875}=2.2071

    Substitute the current estimate into the formula and round to 4 d.p.

  5. Apply the iteration to find x_{3}

    x3=2.2071−0.00875.7852=2.2056x_{3}=2.2071-\frac{0.0087}{5.7852}=2.2056

    Substitute the current estimate into the formula and round to 4 d.p.

  6. State the required approximation

    x3=2.2056x_{3}=2.2056

    This is x_3 to 4 decimal places.

Answer
2.20562.2056
Question 4
5 markshard
Use the Newton-Raphson method, starting from x0=0.75x_{0}=0.75, to find the root of x−cos⁡(x)=0x - \cos{\left(x \right)}=0 correct to 5 decimal places.
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Worked solution

  1. Set up the Newton-Raphson iteration

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

    Form the explicit formula from f and f'.

  2. Iterate from x_0 until the value stabilises

    x0=0.75 → x=0.73909x_{0}=0.75\ \to\ x=0.73909

    Repeated iteration converges to the root to the required accuracy.

  3. Confirm the root with a sign change

    f(0.739085)=−0.00000022,f(0.739095)=0.00001651f(0.739085)=-0.00000022,\quad f(0.739095)=0.00001651

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

  4. Write down the function

    f(x)=x−cos⁡(x)f(x)=x - \cos{\left(x \right)}

    This is the function whose root we are approximating.

  5. Differentiate the function

    f′(x)=sin⁡(x)+1f'(x)=\sin{\left(x \right)} + 1

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

  6. Substitute f and f' into the formula

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

    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. State the root to the required accuracy

    x=0.73909x=0.73909

    This is the root to 5 decimal places.

Answer
0.739090.73909
Question 5
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.
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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

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