g(x) = e^x + x - 6 a) Show that the equation g(x) = 0 can be written as x = ln(6 - x) + 1, x < 6 The root of g(x) = 0 is a. The iterative formula x_{n+1} = ln(6 - x_n) + 1, x_0 = 2 is used to find an approximate value for a. b) Calculate the values of x_1, x_2, and x_3 to 4 decimal places. c) By choosing a suitable interval, show that a = 2.307 correct to 3 decimal places.

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g(x) = e^x + x - 6 a) Show that the equation g(x) = 0 can be written as x = ln(6 - x) + 1, x < 6 The root of g(x) = 0 is a - Edexcel - A-Level Maths Pure - Question 24 - 2013 - Paper 1

Question 24

g(x) = e^x + x - 6 a) Show that the equation g(x) = 0 can be written as x = ln(6 - x) + 1, x < 6 The root of g(x) = 0 is a. The iterative formula x_{n+1} = ln(... show full transcript

Worked Solution & Example Answer:g(x) = e^x + x - 6 a) Show that the equation g(x) = 0 can be written as x = ln(6 - x) + 1, x < 6 The root of g(x) = 0 is a - Edexcel - A-Level Maths Pure - Question 24 - 2013 - Paper 1

Step 1

Show that the equation g(x) = 0 can be written as

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Answer

To rewrite the equation g(x) = 0, we start with:

$e^x + x - 6 = 0$

Rearranging this gives:

$e^x = 6 - x$

Taking the natural logarithm of both sides:

$x = ext{ln}(6 - x)$

Adding 1 to both sides results in:

$x = ext{ln}(6 - x) + 1$

This shows that the equation can be expressed in the required form.

Step 2

Calculate the values of x_1, x_2, and x_3 to 4 decimal places.

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Answer

Using the iterative formula.

Start with x_0 = 2: $x_1 = ext{ln}(6 - x_0) + 1 = ext{ln}(6 - 2) + 1 = ext{ln}(4) + 1 \\ = 2.3863$
Now using x_1 to find x_2: $x_2 = ext{ln}(6 - x_1) + 1 = ext{ln}(6 - 2.3863) + 1 = ext{ln}(3.6137) + 1 \\ = 2.2847$
Finally, using x_2 to find x_3: $x_3 = ext{ln}(6 - x_2) + 1 = ext{ln}(6 - 2.2847) + 1 = ext{ln}(3.7153) + 1 \\ = 2.3125$

Thus, values obtained are:

x_1 = 2.3863
x_2 = 2.2847
x_3 = 2.3125

Step 3

By choosing a suitable interval, show that a = 2.307 correct to 3 decimal places.

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Answer

To show that a = 2.307, we can choose the interval [2.3065, 2.3075].

Calculating g(x) at the endpoints:

For x = 2.3065: $g(2.3065) = e^{2.3065} + 2.3065 - 6 \\ = 0.0002$
For x = 2.3075: $g(2.3075) = e^{2.3075} + 2.3075 - 6 \\ = -0.0044$

Since g(2.3065) is positive and g(2.3075) is negative, by the Intermediate Value Theorem, there exists a root in the interval [2.3065, 2.3075].

Thus, we conclude that a = 2.307 correct to 3 decimal places.

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