A-Level Edexcel Maths Pure Exponential & Logarithms: f(x) = ln(x + 2)

f(x) = ln(x + 2) - x + 1, x > -2, x e R - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 6

Question 5

f(x) = ln(x + 2) - x + 1, x > -2, x e R. (a) Show that there is a root of f(x) = 0 in the interval 2 < x < 3. (b) Use the iterative formula x_{n+1} = ln(x_n... show full transcript

Worked Solution & Example Answer:f(x) = ln(x + 2) - x + 1, x > -2, x e R - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 6

Step 1

Show that there is a root of f(x) = 0 in the interval 2 < x < 3.

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Answer

To show that there is a root in the interval (2, 3), we can evaluate the function at the endpoints:

Calculate f(2):

$f(2) = ln(2 + 2) - 2 + 1 = ln(4) - 1 \approx 0.3869 > 0$
Calculate f(3):

$f(3) = ln(3 + 2) - 3 + 1 = ln(5) - 2 \approx -0.3863 < 0$

Since f(2) > 0 and f(3) < 0, by the Intermediate Value Theorem, there is a root in the interval (2, 3).

Step 2

Use the iterative formula x_{n+1} = ln(x_n + 2) + 1, x_0 = 2.5 to calculate the values of x_1, x_2, and x_3, giving your answers to 5 decimal places.

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Answer

For x_0 = 2.5:

$x_1 = ln(2.5 + 2) + 1 = ln(4.5) + 1 \approx 2.50408$
For x_1 = 2.50408:

$x_2 = ln(2.50408 + 2) + 1 = ln(4.50408) + 1 \approx 2.50498$
For x_2 = 2.50498:

$x_3 = ln(2.50498 + 2) + 1 = ln(4.50498) + 1 \approx 2.50518$

Thus, to 5 decimal places:

x_1 = 2.50408
x_2 = 2.50498
x_3 = 2.50518

Step 3

Show that x = 2.505 is a root of f(x) = 0 correct to 3 decimal places.

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Answer

To verify that x = 2.505 is a root correct to 3 decimal places, we evaluate f(2.505):

$f(2.505) = ln(2.505 + 2) - 2.505 + 1 = ln(4.505) - 2.505.$

Calculating this, we find that:

Estimate f(2.505):

$f(2.505) \approx -0.00022$ (using a calculator)

Next, check f(2.504) and f(2.506):

For f(2.504): $f(2.504) \approx 0.00002$
For f(2.506): $f(2.506) \approx -0.00006$

Since there is a sign change between f(2.504) and f(2.505), and f(2.505) is very close to 0, we can conclude that x = 2.505 is indeed a root correct to 3 decimal places.

f(x) = ln(x + 2) - x + 1, x > -2, x e R - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 6

Worked Solution & Example Answer:f(x) = ln(x + 2) - x + 1, x > -2, x e R - Edexcel - A-Level Maths Pure - Question 5 - 2008 - Paper 6

Show that there is a root of f(x) = 0 in the interval 2 < x < 3.

Use the iterative formula x_{n+1} = ln(x_n + 2) + 1, x_0 = 2.5 to calculate the values of x_1, x_2, and x_3, giving your answers to 5 decimal places.

Show that x = 2.505 is a root of f(x) = 0 correct to 3 decimal places.

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