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 α - 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 α.
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 α - Edexcel - A-Level Maths Pure - Question 24 - 2013 - Paper 1
Step 1
Show that the equation g(x) = 0 can be written as x = ln(6 - x) + 1
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Answer
To prove this, we start by setting g(x) = 0:
ex+x−6=0
Rearranging gives:
ex=6−x
Taking the natural log of both sides:
x=extln(6−x)
Adding 1 to both sides:
x=extln(6−x)+1
This confirms the required format. We note that this is valid for x < 6.
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:
For x_0 = 2:
x_1 \\ ext{approx} = 2.3863 $$
For x_1 = 2.3863:
x_2 \\ ext{approx} = 2.2847 $$
For x_2 = 2.2847:
x_3 \\ ext{approx} = 2.3125 $$
Step 3
By choosing a suitable interval, show that α = 2.307 correct to 3 decimal places.
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Answer
To establish that α lies within the interval [2.3065, 2.3075], we evaluate g(x):
Thus, we find one positive and one negative evaluation in the interval, suggesting that a root exists there. Since α lies within this interval, we conclude that α = 2.307 correct to 3 decimal places.
