To solve the equation, we will use successive approximations. We are looking to solve the equation:
[tex]\[
\frac{2}{x+4} = 3^x + 1
\][/tex]
### Step-by-Step Solution
1. Starting with an Initial Guess
Let's start with an initial guess for [tex]\( x \)[/tex]. Here, we take:
[tex]\[
x = \frac{-39}{6}
\][/tex]
Simplifying this:
[tex]\[
x = -6.5
\][/tex]
2. Defining the Function
Define the function [tex]\( f(x) \)[/tex] based on the original equation:
[tex]\[
f(x) = \frac{2}{x + 4} - 3^x - 1
\][/tex]
3. Defining the Derivative
We also need the derivative [tex]\( f'(x) \)[/tex]. This derivative incorporates the parts of the function:
[tex]\[
f'(x) = -\frac{2}{(x + 4)^2} - 3^x \ln(3)
\][/tex]
4. Applying Successive Approximations
We use the Newton-Raphson method for successive approximations. The Newton-Raphson method formula is:
[tex]\[
x_{\text{new}} = x - \frac{f(x)}{f'(x)}
\][/tex]
5. Iterations
We'll proceed with three iterations as indicated:
1. First iteration:
Given our initial guess [tex]\( x = -6.5 \)[/tex]:
[tex]\[
f(x) = \frac{2}{-6.5 + 4} - 3^{-6.5} - 1
\][/tex]
and
[tex]\[
f'(x) = -\frac{2}{(-6.5 + 4)^2} - 3^{-6.5} \ln(3)
\][/tex]
Calculate [tex]\( f(x) \)[/tex] and [tex]\( f'(x) \)[/tex].
Update [tex]\( x \)[/tex] using Newton-Raphson:
[tex]\[
x_{\text{new}} = -6.5 - \frac{f(-6.5)}{f'(-6.5)}
\][/tex]
2. Second iteration:
With the updated [tex]\( x \)[/tex]:
[tex]\[
f(x) = \frac{2}{x_{\text{new}} + 4} - 3^{x_{\text{new}}} - 1
\][/tex]
and
[tex]\[
f'(x) = -\frac{2}{(x_{\text{new}} + 4)^2} - 3^{x_{\text{new}}} \ln(3)
\][/tex]
Update [tex]\( x \)[/tex] again using the new values.
3. Third iteration:
With the further updated [tex]\( x \)[/tex]:
[tex]\[
f(x) = \frac{2}{x_{\text{new}} + 4} - 3^{x_{\text{new}}} - 1
\][/tex]
and
[tex]\[
f'(x) = -\frac{2}{(x_{\text{new}} + 4)^2} - 3^{x_{\text{new}}} \ln(3)
\][/tex]
Update [tex]\( x \)[/tex] once more.
6. Final Approximation
After three iterations, we find that [tex]\( x \approx -1308.935921746094 \)[/tex].
Thus, the approximate solution to the equation [tex]\(\frac{2}{x + 4} = 3^x + 1\)[/tex] after three iterations of successive approximations is [tex]\( x \approx -1308.935921746094 \)[/tex].
