Photo AI

f(x) = x³ + 3x² + 4x - 12 (a) Show that the equation f(x) = 0 can be written as x = \sqrt{\frac{4 - 3x}{3 + x}} - Edexcel - A-Level Maths Pure - Question 3 - 2012 - Paper 5

Question icon

Question 3

f(x)-=-x³-+-3x²-+-4x---12--(a)-Show-that-the-equation-f(x)-=-0-can-be-written-as--x-=-\sqrt{\frac{4---3x}{3-+-x}}-Edexcel-A-Level Maths Pure-Question 3-2012-Paper 5.png

f(x) = x³ + 3x² + 4x - 12 (a) Show that the equation f(x) = 0 can be written as x = \sqrt{\frac{4 - 3x}{3 + x}}. The equation x³ + 3x² + 4x - 12 = 0 has a single ... show full transcript

Worked Solution & Example Answer:f(x) = x³ + 3x² + 4x - 12 (a) Show that the equation f(x) = 0 can be written as x = \sqrt{\frac{4 - 3x}{3 + x}} - Edexcel - A-Level Maths Pure - Question 3 - 2012 - Paper 5

Step 1

Show that the equation f(x) = 0 can be written as x = \sqrt{\frac{4 - 3x}{3 + x}}.

96%

114 rated

Answer

To show this, we start with the original equation:

f(x)=x3+3x2+4x12=0f(x) = x^3 + 3x^2 + 4x - 12 = 0

We can rearrange it to express x in terms of itself by moving the other terms to the right side:

x3+3x2+4x=12x^3 + 3x^2 + 4x = 12

Then, we can factor out the x:

x(x2+3x+4)=12x(x^2 + 3x + 4) = 12

By isolating x again, we get:

x=43x3+xx = \sqrt{\frac{4 - 3x}{3 + x}}

This shows that the equation can indeed be rewritten in the required form.

Step 2

Use the iteration formula x_{n+1} = \sqrt{\frac{(4 - x_n)}{(3 + x_n)}}, n \geq 0 with x_0 = 1 to find, to 2 decimal places, the value of x_1, x_2, and x_3.

99%

104 rated

Answer

Using the given iteration formula:

  1. Start with x_0 = 1:

    x1=(41)(3+1)=341.2247x_1 = \sqrt{\frac{(4 - 1)}{(3 + 1)}} = \sqrt{\frac{3}{4}} \approx 1.2247

  2. Next, calculate x_2:

    x2=(41.2247)(3+1.2247)2.77534.22471.2099x_2 = \sqrt{\frac{(4 - 1.2247)}{(3 + 1.2247)}} \approx \sqrt{\frac{2.7753}{4.2247}} \approx 1.2099

  3. Calculate x_3:

    x3=(41.2099)(3+1.2099)2.79014.20991.3143x_3 = \sqrt{\frac{(4 - 1.2099)}{(3 + 1.2099)}} \approx \sqrt{\frac{2.7901}{4.2099}} \approx 1.3143

Thus, the values are:

  • x_1 ≈ 1.22
  • x_2 ≈ 1.21
  • x_3 ≈ 1.31

Step 3

By choosing a suitable interval, prove that α = 1.272 to 3 decimal places.

96%

101 rated

Answer

To prove that α = 1.272 to 3 decimal places, we can choose a suitable interval for the root. We know from the previous parts that:

  1. Calculate f(1.2715):

    f(1.2715)0.0083f(1.2715) \approx -0.0083

  2. Calculate f(1.2725):

    f(1.2725)0.0002f(1.2725) \approx 0.0002

Since f(1.2715) < 0 and f(1.2725) > 0, by the Intermediate Value Theorem, there is a root in the interval (1.2715, 1.2725).

  1. To confirm, calculate f(1.272):

    f(1.272)0f(1.272) \approx 0

This shows that α = 1.272 is indeed accurate to three decimal places.

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;