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f(x) = ln(x + 2) - x + 1, \, x > -2, \, x \in \mathbb{R} - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 6
Question 4
f(x) = ln(x + 2) - x + 1, \, x > -2, \, x \in \mathbb{R}. (a) Show that there is a root of \( f(x) = 0 \) in the interval \( 2 < x < 3 \). (b) Use the iterative fo... show full transcript
Worked Solution & Example Answer:f(x) = ln(x + 2) - x + 1, \, x > -2, \, x \in \mathbb{R} - 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 first calculate the function values at the endpoints:
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Calculate ( f(2) ): [ f(2) = ln(2 + 2) - 2 + 1 = ln(4) - 2 + 1 = ln(4) - 1 \approx 0.3863 \approx 0.38 ]
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Calculate ( f(3) ): [ f(3) = ln(3 + 2) - 3 + 1 = ln(5) - 3 + 1 = ln(5) - 2 \approx 0.6094 - 2 \approx -1.3906 \approx -1.39 ]
Since ( f(2) ) is positive and ( f(3) ) is negative, by the Intermediate Value Theorem, 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, x_0 = 2.5 to calculate the values of x_1, x_2, and x_3
Answer
Starting with ( x_0 = 2.5 ):
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Calculate ( x_1 ): [ x_1 = ln(2.5 + 2) + 1 = ln(4.5) + 1 \approx 1.5041 + 1 = 2.5041 ]
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Calculate ( x_2 ): [ x_2 = ln(2.5041 + 2) + 1 = ln(4.5041) + 1 \approx 1.5049 + 1 = 2.5049 ]
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Calculate ( x_3 ): [ x_3 = ln(2.5049 + 2) + 1 = ln(4.5049) + 1 \approx 1.5048 + 1 = 2.5048 ]
The values are approximately:\n- ( x_1 \approx 2.5041 )\n- ( x_2 \approx 2.5049 )\n- ( x_3 \approx 2.5048 )
Step 3
Show that x = 2.505 is a root of f(x) = 0 correct to 3 decimal places
Answer
To confirm that ( x = 2.505 ) is a root to 3 decimal places, evaluate ( f(2.505) ):
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Calculate ( f(2.505) ): [ f(2.505) = ln(2.505 + 2) - 2.505 + 1 = ln(4.505) - 2.505 + 1 \approx 1.5050 - 2.505 \approx -0.0001 \approx 0 ]
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Also, calculate the function at the bounds, ( f(2.5045) ) and ( f(2.5055) ):
- ( f(2.5045) \approx 6 \times 10^{-4} )
- ( f(2.5055) \approx -2 \times 10^{-4} )
Since ( f(2.5045) ) is positive and ( f(2.5055) ) is negative, and ( f(2.505) ) is approximately zero, by the Intermediate Value Theorem, we can confirm that ( x = 2.505 ) is a root correct to 3 decimal places.