Answer :
To find the solution to the equation [tex]\( f(x) = g(x) \)[/tex] given [tex]\( f(x) = \frac{1}{x + 3} + 1 \)[/tex] and [tex]\( g(x) = 2 \log(x) \)[/tex], we will use successive approximation starting with a value obtained from the graph.
### Step-by-Step Solution:
1. Identify the given functions:
[tex]\[ f(x) = \frac{1}{x + 3} + 1 \quad \text{and} \quad g(x) = 2 \log(x) \][/tex]
2. Determine the initial guess from the graph:
Based on the initial options, assume the initial guess [tex]\( x_0 \approx \frac{61}{16} \)[/tex].
3. Define the process of successive approximation:
Successive approximation often uses a method like Newton's method for iterative solving, which involves the following steps:
[tex]\[ x_{n+1} = x_n - \frac{f(x_n) - g(x_n)}{f'(x_n) - g'(x_n)} \][/tex]
4. Compute the derivatives of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex]:
[tex]\[ f'(x) = -\frac{1}{(x + 3)^2} \][/tex]
[tex]\[ g'(x) = \frac{2}{x} \][/tex]
5. Iterate using the initial guess [tex]\( x_0 \approx 3.8125 \)[/tex] (which is [tex]\( \frac{61}{16} \)[/tex]) and the Newton's method iteration:
[tex]\[ x_{n+1} = x_n - \frac{\frac{1}{x_n + 3} + 1 - 2 \log(x_n)}{-\frac{1}{(x_n + 3)^2} - \frac{2}{x_n}} \][/tex]
6. Perform successive iterations until the result converges:
This iterative process continues by updating [tex]\( x_n \)[/tex] until the change between iterations is smaller than a given tolerance, typically [tex]\( 10^{-6} \)[/tex].
### Final Result:
Upon performing the successive approximation method with the initial guess and iterating a sufficient number of times, the approximate solution is:
[tex]\[ x \approx 1.8285983038394122 \][/tex]
Therefore, the closest value to this result among the given options:
- [tex]\( \frac{61}{16} \approx 3.8125 \)[/tex]
- [tex]\( \frac{31}{8} \approx 3.875 \)[/tex]
- [tex]\( \frac{55}{16} \approx 3.4375 \)[/tex]
does not match the found result closely. Hence,
The approximate solution to the equation [tex]\( f(x) = g(x) \)[/tex] after using successive approximation is:
[tex]\[ x \approx 1.8285983038394122 \][/tex]
This value is the converged point where [tex]\( f(x) \)[/tex] equals [tex]\( g(x) \)[/tex] according to the method used.
### Step-by-Step Solution:
1. Identify the given functions:
[tex]\[ f(x) = \frac{1}{x + 3} + 1 \quad \text{and} \quad g(x) = 2 \log(x) \][/tex]
2. Determine the initial guess from the graph:
Based on the initial options, assume the initial guess [tex]\( x_0 \approx \frac{61}{16} \)[/tex].
3. Define the process of successive approximation:
Successive approximation often uses a method like Newton's method for iterative solving, which involves the following steps:
[tex]\[ x_{n+1} = x_n - \frac{f(x_n) - g(x_n)}{f'(x_n) - g'(x_n)} \][/tex]
4. Compute the derivatives of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex]:
[tex]\[ f'(x) = -\frac{1}{(x + 3)^2} \][/tex]
[tex]\[ g'(x) = \frac{2}{x} \][/tex]
5. Iterate using the initial guess [tex]\( x_0 \approx 3.8125 \)[/tex] (which is [tex]\( \frac{61}{16} \)[/tex]) and the Newton's method iteration:
[tex]\[ x_{n+1} = x_n - \frac{\frac{1}{x_n + 3} + 1 - 2 \log(x_n)}{-\frac{1}{(x_n + 3)^2} - \frac{2}{x_n}} \][/tex]
6. Perform successive iterations until the result converges:
This iterative process continues by updating [tex]\( x_n \)[/tex] until the change between iterations is smaller than a given tolerance, typically [tex]\( 10^{-6} \)[/tex].
### Final Result:
Upon performing the successive approximation method with the initial guess and iterating a sufficient number of times, the approximate solution is:
[tex]\[ x \approx 1.8285983038394122 \][/tex]
Therefore, the closest value to this result among the given options:
- [tex]\( \frac{61}{16} \approx 3.8125 \)[/tex]
- [tex]\( \frac{31}{8} \approx 3.875 \)[/tex]
- [tex]\( \frac{55}{16} \approx 3.4375 \)[/tex]
does not match the found result closely. Hence,
The approximate solution to the equation [tex]\( f(x) = g(x) \)[/tex] after using successive approximation is:
[tex]\[ x \approx 1.8285983038394122 \][/tex]
This value is the converged point where [tex]\( f(x) \)[/tex] equals [tex]\( g(x) \)[/tex] according to the method used.