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f(x) = 2x^3 - x - 4 - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 5
Question 7
f(x) = 2x^3 - x - 4. (a) Show that the equation f(x) = 0 can be written as x = \sqrt{\frac{2}{x + 2}}. The equation 2x^3 - x - 4 = 0 has a root between 1.35 and ... show full transcript
Worked Solution & Example Answer:f(x) = 2x^3 - x - 4 - Edexcel - A-Level Maths Pure - Question 7 - 2006 - Paper 5
Step 1
Show that the equation f(x) = 0 can be written as x = \sqrt{\frac{2}{x + 2}}
Answer
To express the equation 2x^3 - x - 4 = 0 in the desired form, we start by isolating x:
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Add x and 4 to both sides: [ 2x^3 = x + 4 ]
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Divide both sides by 2: [ x^3 = \frac{x + 4}{2} ]
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Take the cube root on both sides: [ x = \sqrt[3]{\frac{x + 4}{2}} ]
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Alternatively, we can rearrange to show: [ x = \sqrt{\frac{2}{x + 2}}. ]
Step 2
Use the iteration formula with x_0 = 1.35, to find, to 2 decimal places, the value of x_1, x_2 and x_3.
Answer
Using the iteration formula: [ x_{n+1} = \frac{2}{x_n + 2} ]
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For n = 0 (x_0 = 1.35): [ x_1 = \frac{2}{1.35 + 2} = \frac{2}{3.35} \approx 0.597 \
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For n = 1 (x_1 = 0.597): [ x_2 = \frac{2}{0.597 + 2} = \frac{2}{2.597} \approx 0.771 ]
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For n = 2 (x_2 = 0.771): [ x_3 = \frac{2}{0.771 + 2} = \frac{2}{2.771} \approx 0.722. ]
Step 3
By choosing a suitable interval, prove that \alpha = 1.392, to three decimal places.
Answer
To prove \alpha = 1.392, we can choose the interval (1.391, 1.3925):
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Evaluate f(1.391) and f(1.3925): [ f(1.391) = 2(1.391)^3 - (1.391) - 4 \approx -0.002, ] [ f(1.3925) = 2(1.3925)^3 - (1.3925) - 4 \approx 0.008. ]
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Since f(1.391) < 0 and f(1.3925) > 0, by the Intermediate Value Theorem:
- There is at least one root in (1.391, 1.3925).
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This confirms that \alpha = 1.392 to three decimal places.