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f(x) = ln(x + 2) - x + 1, \, x > -2, \, x \in \mathbb{R}.\n\n(a) Show that there is a root of \( f(x) = 0 \) in the interval \( 2 < x < 3 \).\n\n(b) Use the iterative formula \( x_{n+1} = ln(x_n + 2) + 1 \)\n to calculate the values of \( x_1, x_2, \) and \( x_3 \), giving your answers to 5 decimal places.\n\n(c) Show that \( x = 2.505 \) is a root of \( f(x) = 0 \) correct to 3 decimal places. - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 6
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f(x) = ln(x + 2) - x + 1, \, x > -2, \, x \in \mathbb{R}.\n\n(a) Show that there is a root of \( f(x) = 0 \) in the interval \( 2 < x < 3 \).\n\n(b) Use the iterativ... show full transcript
Worked Solution & Example Answer:f(x) = ln(x + 2) - x + 1, \, x > -2, \, x \in \mathbb{R}.\n\n(a) Show that there is a root of \( f(x) = 0 \) in the interval \( 2 < x < 3 \).\n\n(b) Use the iterative formula \( x_{n+1} = ln(x_n + 2) + 1 \)\n to calculate the values of \( x_1, x_2, \) and \( x_3 \), giving your answers to 5 decimal places.\n\n(c) Show that \( x = 2.505 \) is a root of \( f(x) = 0 \) correct to 3 decimal places. - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 6
Step 1
Show that there is a root of f(x) = 0 in the interval 2 < x < 3
Answer
To show that there is a root in the interval ( 2 < x < 3 ), we evaluate the function at the endpoints.\n\nCalculating ( f(2) = ln(2 + 2) - 2 + 1 = ln(4) - 2 \approx 0.3863 - 2 = -1.6137 ) (approximately).\n\nNext, for ( f(3) = ln(3 + 2) - 3 + 1 = ln(5) - 2 \approx 1.6094 - 2 = -0.3906 ) (approximately).\n\nThus, ( f(2) > 0 ) and ( f(3) < 0 ), demonstrating a change of sign, which by the Intermediate Value Theorem indicates that there is at least one root in the interval ( (2, 3) ).
Step 2
Use the iterative formula x_{n+1} = ln(x_n + 2) + 1 to calculate x_1, x_2, and x_3, giving your answers to 5 decimal places.
Answer
Starting with ( x_0 = 2.5 ):\n1. Calculate ( x_1 ): \n [ x_1 = ln(2.5 + 2) + 1 = ln(4.5) + 1 \approx 1.5041 + 1 = 2.5041 ]\n\n2. Calculate ( x_2 ): \n [ x_2 = ln(2.5041 + 2) + 1 \approx ln(4.5041) + 1 \approx 1.5046 + 1 = 2.5046 ]\n\n3. Calculate ( x_3 ): \n [ x_3 = ln(2.5046 + 2) + 1 \approx ln(4.5046) + 1 \approx 1.5047 + 1 = 2.5047 ]\n\nThus, we have: ( x_1 \approx 2.50410, x_2 \approx 2.5046, x_3 \approx 2.5047 ).
Step 3
Show that x = 2.505 is a root of f(x) = 0 correct to 3 decimal places.
Answer
To verify that ( x = 2.505 ) is a root to three decimal places, we calculate ( f(2.505) ):\n\n[ f(2.505) = ln(2.505 + 2) - 2.505 + 1 = ln(4.505) - 2.505 \approx 1.5049 - 2.505 = -0.0001 ] \n\nNext, calculate ( f(2.5045) ) and ( f(2.5055) ):\n[ f(2.5045) = ln(4.5045) - 2.5045 \approx -0.00006 ] \n[ f(2.5055) = ln(4.5055) - 2.5055 \approx 0.00007 ] \n\nBoth calculations show a change of sign between ( [2.5045, 2.5055] ), confirming that there is a root at ( x = 2.505 ).