Iteration worksheet
9 GCSE practice questions with answers, free to print. Every sheet is generated, so you can make a fresh one whenever you need it.
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2 marks
\(f(x) = x^3 + 7x - 38\)
Show that the equation \(f(x) = 0\) has a solution in the interval \(2 < x < 3\).
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2 marks
Rearrange \(x^3 + 3x - 44 = 0\) into the form \(x = \sqrt[3]{p + qx}\), where \(p\) and \(q\) are integers.
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2 marks
\(f(x) = x^3 - {a:coef}x - {b}\)
Show that the equation \(f(x) = 0\) has a solution between \(x = {v:1dp}\) and \(x = {v2:1dp}\).
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2 marks
The iteration formula
\(x_{n+1} = \sqrt[3]{ 4x_n + 54 }\)
The values it produces get closer and closer to one number.
Write down the equation, in the form \(x^3 + px + q = 0\), that this number is a solution of.
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3 marks
\(x_{n+1} = \sqrt[3]{ 9x_n + 33 }\)
Use the iteration formula with \(x_0 = 3\) to work out \(x_1\), \(x_2\) and \(x_3\).
Give your answers correct to 4 decimal places.
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2 marks
\(f(x) = x^3 + 7x - 108\)
Show that the equation \(f(x) = 0\) has a root between \(x = 4\) and \(x = 5\).
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2 marks
Rearrange the equation \(x^3 + 4x - 38 = 0\) into the form \(x = \sqrt[3]{p + qx}\), where \(p\) and \(q\) are integers.
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2 marks
\(f(x) = x^3 - {a:coef}x - {b}\)
Show that the equation \(f(x) = 0\) has a solution between \(x = {v:1dp}\) and \(x = {v2:1dp}\).
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2 marks
A student uses the iteration formula
\(x_{n+1} = \sqrt[3]{ 2x_n + 37 }\)
The values it produces get closer and closer to one number.
Write down the equation, in the form \(x^3 + px + q = 0\), that this number is a solution of.
Answers
- \(f(2) = -16\) and \(f(3) = 10\). One is negative and the other is positive, so \(f(x) = 0\) has a solution between \(x = 2\) and \(x = 3\).
- \(x = \sqrt[3]{ 44 - 3x }\)
- \(f({v:1dp}) = {fv:3dp}\) and \(f({v2:1dp}) = {fv2:3dp}\). The sign changes between them, so \(f(x) = 0\) has a solution between \(x = {v:1dp}\) and \(x = {v2:1dp}\).
- \(x^3 - 4x - 54 = 0\)
- \(x_1 \approx 3.9149\), \(x_2 \approx 4.0863\), \(x_3 \approx 4.1169\)
- \(f(4) = -16\) and \(f(5) = 52\). One is negative and the other is positive, so \(f(x) = 0\) has a solution between \(x = 4\) and \(x = 5\).
- \(x = \sqrt[3]{ 38 - 4x }\)
- \(f({v:1dp}) = {fv:3dp}\) and \(f({v2:1dp}) = {fv2:3dp}\). The sign changes between them, so \(f(x) = 0\) has a solution between \(x = {v:1dp}\) and \(x = {v2:1dp}\).
- \(x^3 - 2x - 37 = 0\)