f(x) = -x³ + 3x² - 1.
(a) Show that the equation f(x) = 0 can be rewritten as
$x = \sqrt{\frac{1}{3 - x}}$.
(b) Starting with $x_1 = 0.6$, use the iteration
$x_{... show full transcript
Worked Solution & Example Answer:f(x) = -x³ + 3x² - 1 - Edexcel - A-Level Maths Pure - Question 5 - 2007 - Paper 5
Step 1
Show that the equation f(x) = 0 can be rewritten as $x = \sqrt{\frac{1}{3 - x}}$
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Answer
We start with the equation:
−x3+3x2−1=0.
Rearranging this gives: x3−3x2+1=0.
Next, we isolate the terms involving x:
x3=3x2−1.
Taking the cube root leads us to the iteration form:
x=3−x1.
This completes part (a).
Step 2
Starting with $x_1 = 0.6$, use the iteration $x_{n+1} = \sqrt{\frac{1}{3 - x_n}}$ to calculate the values of $x_2, x_3$ and $x_4$
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Answer
We begin our iteration from x1=0.6:
For x1=0.6:
x2=3−0.61=2.41≈0.6455.
For x2=0.6455:
x3=3−0.64551=2.35451≈0.6526.
For x3=0.6526:
x4=3−0.65261=2.34741≈0.6530.
Thus, the values calculated are approximately:
x2≈0.6455
x3≈0.6526
x4≈0.6530.
Step 3
Show that $x = 0.653$ is a root of f(x) = 0 correct to 3 decimal places.
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Answer
To verify that x=0.653 is a root, we evaluate:
f(0.653)=−(0.653)3+3(0.653)2−1.
Calculating this:
First, compute (0.653)3≈0.2804.
Next, compute 3(0.653)2≈3⋅0.4276≈1.2828.
Then, combine the results:
f(0.653)≈−0.2804+1.2828−1=0.0024.
Since f(0.653)≈0.0024, we find that this value is very close to zero, confirming that 0.653 is indeed a root of f(x)=0 to three decimal places.
