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 $x_{n+1} = \sqrt{\frac{1}{3 - x_n}}$ to calculate the values of $x_2$, $x_3$ and $x_4$, giving all your answers to 4 decimal places. (c) Show that $x = 0.653$ is a root of f(x) = 0 correct to 3 decimal places.

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f(x) = -x³ + 3x² - 1 - Edexcel - A-Level Maths Pure - Question 6 - 2007 - Paper 5

Question 6

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 6 - 2007 - Paper 5

Step 1

Show that the equation f(x) = 0 can be rewritten as

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Answer

To show that the equation f(x) = 0 can be rewritten as $x = \sqrt{\frac{1}{3 - x}}$ , we start with:

$-x^3 + 3x^2 - 1 = 0$

Rearranging this gives:

$x^3 - 3x^2 + 1 = 0$

We can express this as:

$x^3 = 3x^2 - 1$

Taking the cube root on both sides and isolating $x$ , we have:

$x = \sqrt[3]{3x^2 - 1}$

Then we divide both sides by $(3 - x)$ to arrive at the required equation:

$x = \sqrt{\frac{1}{3 - x}}$

Step 2

Starting with x_1 = 0.6, use the iteration x_{n+1} = √(1/(3 - x_n)) to calculate the values of x_2, x_3 and x_4, giving all your answers to 4 decimal places.

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Answer

We will perform the iteration step:

Start with $x_1 = 0.6$ : $x_2 = \sqrt{\frac{1}{3 - 0.6}} = \sqrt{\frac{1}{2.4}} \approx 0.6455$
Now use $x_2$ to find $x_3$ : $x_3 = \sqrt{\frac{1}{3 - 0.6455}} = \sqrt{\frac{1}{2.3545}} \approx 0.6526$
Finally, use $x_3$ to compute $x_4$ : $x_4 = \sqrt{\frac{1}{3 - 0.6526}} = \sqrt{\frac{1}{2.3474}} \approx 0.6531$

Thus, we have:

$x_2 \approx 0.6455$
$x_3 \approx 0.6526$
$x_4 \approx 0.6531$

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 show that $x = 0.653$ is a root correct to three decimal places, we will evaluate f(x) at this point:

Compute: $f(0.653) = - (0.653)^3 + 3(0.653)^2 - 1$
Calculating each term:
- $-(0.653)^3 \approx -0.2805$
- $3(0.653)^2 \approx 1.2812$
Now substitute back into f(x): $f(0.653) \approx -0.2805 + 1.2812 - 1 = -0.0003$

Since this value is close to 0, we can conclude:

Change of sign between $0.652$ and $0.654$ indicates $0.653$ is indeed a root of f(x) = 0 correct to 3 decimal places.

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