Answer :
Let's provide a detailed, step-by-step solution to approximate the solution to the given equation by performing three iterations of successive approximation.
Given the equation:
[tex]$ x^2 - 5x + 1 = \frac{2}{x-1} $[/tex]
We will use an iteration method to approximate a solution. Let's begin with an initial guess from between [tex]\(x = 4\)[/tex] and [tex]\(x = 5\)[/tex] because they are close to each other. Let's use [tex]\(x = 4.5\)[/tex] as our starting point.
Step 1: Initial guess
Start with [tex]\(x = 4.5\)[/tex].
Step 2: Define the function [tex]\(f(x)\)[/tex]
We rearrange the original equation to define a function [tex]\(f(x)\)[/tex] that will be zero at the solution:
[tex]\[ f(x) = x^2 - 5x + 1 - \frac{2}{x-1} \][/tex]
Step 3: Compute the derivative [tex]\(f'(x)\)[/tex]
[tex]\[ f'(x) = \frac{d}{dx} \left( x^2 - 5x + 1 - \frac{2}{x-1} \right) \][/tex]
[tex]\[ f'(x) = 2x - 5 + \frac{2}{(x-1)^2} \][/tex]
Step 4: Perform iterations using Newton's method
Using the formula:
[tex]\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \][/tex]
Let’s calculate the next approximations:
Iteration 1:
For [tex]\(x = 4.5\)[/tex],
[tex]\[ f(4.5) = 4.5^2 - 5 \cdot 4.5 + 1 - \frac{2}{4.5 - 1} \][/tex]
[tex]\[ f(4.5) = 20.25 - 22.5 + 1 - \frac{2}{3.5} \][/tex]
[tex]\[ f(4.5) = -1.25 - 0.5714 \][/tex]
[tex]\[ f(4.5) = -1.8214 \][/tex]
[tex]\[ f'(4.5) = 2 \cdot 4.5 - 5 + \frac{2}{(4.5 - 1)^2} \][/tex]
[tex]\[ f'(4.5) = 9 - 5 + \frac{2}{3.5^2} \][/tex]
[tex]\[ f'(4.5) = 4 + \frac{2}{12.25} \][/tex]
[tex]\[ f'(4.5) = 4 + 0.1633 \][/tex]
[tex]\[ f'(4.5) = 4.1633 \][/tex]
Next approximation:
[tex]\[ x_1 = 4.5 - \frac{-1.8214}{4.1633} \][/tex]
[tex]\[ x_1 \approx 4.938 \][/tex]
Iteration 2:
For [tex]\(x \approx 4.938\)[/tex],
[tex]\[ f(4.938) = 4.938^2 - 5 \cdot 4.938 + 1 - \frac{2}{4.938 - 1} \][/tex]
[tex]\[ f(4.938) = 24.3785 - 24.69 + 1 - \frac{2}{3.938} \][/tex]
[tex]\[ f(4.938) = -0.3115 - 0.508 \][/tex]
[tex]\[ f(4.938) = -0.8195 \][/tex]
[tex]\[ f'(4.938) = 2 \cdot 4.938 - 5 + \frac{2}{(4.938 - 1)^2} \][/tex]
[tex]\[ f'(4.938) = 9.876 - 5 + \frac{2}{15.5016} \][/tex]
[tex]\[ f'(4.938) = 4.876 + 0.129 \][/tex]
[tex]\[ f'(4.938) = 5.005 \][/tex]
Next approximation:
[tex]\[ x_2 = 4.938 - \frac{-0.8195}{5.005} \][/tex]
[tex]\[ x_2 \approx 5.102 \][/tex]
Iteration 3:
For [tex]\(x \approx 5.102\)[/tex],
[tex]\[ f(5.102) = 5.102^2 - 5 \cdot 5.102 + 1 - \frac{2}{5.102 - 1} \][/tex]
[tex]\[ f(5.102) = 26.042 - 25.51 + 1 - \frac{2}{4.102} \][/tex]
[tex]\[ f(5.102) = 0.532 - 0.4877 \][/tex]
[tex]\[ f(5.102) = 0.0443 \][/tex]
[tex]\[ f'(5.102) = 2 \cdot 5.102 - 5 + \frac{2}{(5.102 - 1)^2} \][/tex]
[tex]\[ f'(5.102) = 10.204 - 5 + \frac{2}{16.816} \][/tex]
[tex]\[ f'(5.102) = 5.204 + 0.119 \][/tex]
[tex]\[ f'(5.102) = 5.323 \][/tex]
Next approximation:
[tex]\[ x_3 = 5.102 - \frac{0.0443}{5.323} \][/tex]
[tex]\[ x_3 \approx 5.0946 \][/tex]
Step 5: Compare to given options
We now compare [tex]\(x \approx 5.0946\)[/tex] to the given options in their decimal form:
A. [tex]\(\frac{33}{8} \approx 4.125\)[/tex]
B. [tex]\(\frac{65}{16} \approx 4.0625\)[/tex]
C. [tex]\(\frac{79}{16} \approx 4.9375\)[/tex]
D. [tex]\(\frac{39}{8} \approx 4.875\)[/tex]
After comparing, it is evident that the closest option is D.
Therefore, the correct answer is:
D. [tex]\( x \approx \frac{39}{8} \)[/tex]
Given the equation:
[tex]$ x^2 - 5x + 1 = \frac{2}{x-1} $[/tex]
We will use an iteration method to approximate a solution. Let's begin with an initial guess from between [tex]\(x = 4\)[/tex] and [tex]\(x = 5\)[/tex] because they are close to each other. Let's use [tex]\(x = 4.5\)[/tex] as our starting point.
Step 1: Initial guess
Start with [tex]\(x = 4.5\)[/tex].
Step 2: Define the function [tex]\(f(x)\)[/tex]
We rearrange the original equation to define a function [tex]\(f(x)\)[/tex] that will be zero at the solution:
[tex]\[ f(x) = x^2 - 5x + 1 - \frac{2}{x-1} \][/tex]
Step 3: Compute the derivative [tex]\(f'(x)\)[/tex]
[tex]\[ f'(x) = \frac{d}{dx} \left( x^2 - 5x + 1 - \frac{2}{x-1} \right) \][/tex]
[tex]\[ f'(x) = 2x - 5 + \frac{2}{(x-1)^2} \][/tex]
Step 4: Perform iterations using Newton's method
Using the formula:
[tex]\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \][/tex]
Let’s calculate the next approximations:
Iteration 1:
For [tex]\(x = 4.5\)[/tex],
[tex]\[ f(4.5) = 4.5^2 - 5 \cdot 4.5 + 1 - \frac{2}{4.5 - 1} \][/tex]
[tex]\[ f(4.5) = 20.25 - 22.5 + 1 - \frac{2}{3.5} \][/tex]
[tex]\[ f(4.5) = -1.25 - 0.5714 \][/tex]
[tex]\[ f(4.5) = -1.8214 \][/tex]
[tex]\[ f'(4.5) = 2 \cdot 4.5 - 5 + \frac{2}{(4.5 - 1)^2} \][/tex]
[tex]\[ f'(4.5) = 9 - 5 + \frac{2}{3.5^2} \][/tex]
[tex]\[ f'(4.5) = 4 + \frac{2}{12.25} \][/tex]
[tex]\[ f'(4.5) = 4 + 0.1633 \][/tex]
[tex]\[ f'(4.5) = 4.1633 \][/tex]
Next approximation:
[tex]\[ x_1 = 4.5 - \frac{-1.8214}{4.1633} \][/tex]
[tex]\[ x_1 \approx 4.938 \][/tex]
Iteration 2:
For [tex]\(x \approx 4.938\)[/tex],
[tex]\[ f(4.938) = 4.938^2 - 5 \cdot 4.938 + 1 - \frac{2}{4.938 - 1} \][/tex]
[tex]\[ f(4.938) = 24.3785 - 24.69 + 1 - \frac{2}{3.938} \][/tex]
[tex]\[ f(4.938) = -0.3115 - 0.508 \][/tex]
[tex]\[ f(4.938) = -0.8195 \][/tex]
[tex]\[ f'(4.938) = 2 \cdot 4.938 - 5 + \frac{2}{(4.938 - 1)^2} \][/tex]
[tex]\[ f'(4.938) = 9.876 - 5 + \frac{2}{15.5016} \][/tex]
[tex]\[ f'(4.938) = 4.876 + 0.129 \][/tex]
[tex]\[ f'(4.938) = 5.005 \][/tex]
Next approximation:
[tex]\[ x_2 = 4.938 - \frac{-0.8195}{5.005} \][/tex]
[tex]\[ x_2 \approx 5.102 \][/tex]
Iteration 3:
For [tex]\(x \approx 5.102\)[/tex],
[tex]\[ f(5.102) = 5.102^2 - 5 \cdot 5.102 + 1 - \frac{2}{5.102 - 1} \][/tex]
[tex]\[ f(5.102) = 26.042 - 25.51 + 1 - \frac{2}{4.102} \][/tex]
[tex]\[ f(5.102) = 0.532 - 0.4877 \][/tex]
[tex]\[ f(5.102) = 0.0443 \][/tex]
[tex]\[ f'(5.102) = 2 \cdot 5.102 - 5 + \frac{2}{(5.102 - 1)^2} \][/tex]
[tex]\[ f'(5.102) = 10.204 - 5 + \frac{2}{16.816} \][/tex]
[tex]\[ f'(5.102) = 5.204 + 0.119 \][/tex]
[tex]\[ f'(5.102) = 5.323 \][/tex]
Next approximation:
[tex]\[ x_3 = 5.102 - \frac{0.0443}{5.323} \][/tex]
[tex]\[ x_3 \approx 5.0946 \][/tex]
Step 5: Compare to given options
We now compare [tex]\(x \approx 5.0946\)[/tex] to the given options in their decimal form:
A. [tex]\(\frac{33}{8} \approx 4.125\)[/tex]
B. [tex]\(\frac{65}{16} \approx 4.0625\)[/tex]
C. [tex]\(\frac{79}{16} \approx 4.9375\)[/tex]
D. [tex]\(\frac{39}{8} \approx 4.875\)[/tex]
After comparing, it is evident that the closest option is D.
Therefore, the correct answer is:
D. [tex]\( x \approx \frac{39}{8} \)[/tex]
