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f(x) = 3x^3 - 2x - 6 (a) Show that f(x) = 0 has a root, α, between x = 1.4 and x = 1.45 - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 6
Question 2
f(x) = 3x^3 - 2x - 6 (a) Show that f(x) = 0 has a root, α, between x = 1.4 and x = 1.45. (b) Show that the equation f(x) = 0 can be written as x = \sqrt{\frac{2}{... show full transcript
Worked Solution & Example Answer:f(x) = 3x^3 - 2x - 6 (a) Show that f(x) = 0 has a root, α, between x = 1.4 and x = 1.45 - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 6
Step 1
Show that f(x) = 0 has a root, α, between x = 1.4 and x = 1.45.
Answer
To show that f(x) = 0 has a root between x = 1.4 and x = 1.45, we first evaluate the function at these points:
- Calculate f(1.4):
f(1.4) = 3(1.4)^3 - 2(1.4) - 6 = 3(2.744) - 2.8 - 6 = 8.232 - 2.8 - 6 = -0.568
- Calculate f(1.45):
f(1.45) = 3(1.45)^3 - 2(1.45) - 6 = 3(3.066125) - 2.9 - 6 = 9.198375 - 2.9 - 6 = 0.245
Since f(1.4) < 0 and f(1.45) > 0, there is a change of sign in the interval (1.4, 1.45). By the Intermediate Value Theorem, there exists at least one root α in this interval.
Step 2
Show that the equation f(x) = 0 can be written as x = \sqrt{\frac{2}{x} + \frac{2}{3}}.
Answer
Starting from the equation f(x) = 3x^3 - 2x - 6 = 0:
Rearranging gives:
to go to x = \sqrt{\frac{2}{x} + \frac{2}{3}}.
Dividing the equation by 3:
x^3 = \frac{2x + 6}{3}
This implies:
x = \sqrt{\frac{2}{x} + \frac{2}{3}} as stated. This formula avoids division by zero by ensuring that x ≠ 0.
Step 3
Starting with x₀ = 1.43, use the iteration x₁ = \sqrt{\frac{2}{x₀} + \frac{2}{3}} to calculate the values of x₁, x₂, and x₃, giving your answers to 4 decimal places.
Answer
Applying the iteration:
-
For x₀ = 1.43:
- x₁ = \sqrt{\frac{2}{1.43} + \frac{2}{3}} = \sqrt{1.396 + 0.6667} = \sqrt{2.0627} ≈ 1.4371
-
For x₁ ≈ 1.4371:
- x₂ = \sqrt{\frac{2}{1.4371} + \frac{2}{3}} ≈ \sqrt{1.392 + 0.6667} ≈ \sqrt{2.0587} ≈ 1.4366
-
For x₂ ≈ 1.4366:
- x₃ = \sqrt{\frac{2}{1.4366} + \frac{2}{3}} ≈ \sqrt{1.392 + 0.6667} ≈ \sqrt{2.0587} ≈ 1.4364
Thus:
- x₁ ≈ 1.4371
- x₂ ≈ 1.4366
- x₃ ≈ 1.4364
Step 4
By choosing a suitable interval, show that α = 1.435 is correct to 3 decimal places.
Answer
Choose the interval (1.4345, 1.4355):
- Calculate f(1.4345):
f(1.4345) ≈ -0.01
- Calculate f(1.4355):
f(1.4355) ≈ 0.003
Since f(1.4345) < 0 and f(1.4355) > 0, we have a change of sign.
Thus, by the Intermediate Value Theorem, α = 1.435 is confirmed to be correct to 3 decimal places.