To solve the equation [tex]\( \frac{1}{2} x^3 + x - 7 = -3 \sqrt{x - 1} \)[/tex] using successive approximation, we will follow the steps required for this method:
### 1. Set Up the Function
First, we rewrite the given equation in a form suitable for successive approximation. This involves isolating [tex]\( x \)[/tex] on one side of the equation. We can rewrite the equation as:
[tex]\[
x = \left( \frac{7 - 3 \sqrt{x - 1} - \frac{1}{2} x^3}{3} \right)^{1/2}
\][/tex]
Let's denote this new function as:
[tex]\[
g(x) = \left( \frac{7 - 3 \sqrt{x - 1} - \frac{1}{2} x^3}{3} \right)^{1/2}
\][/tex]
### 2. Initial Guess
Using the graph provided, we choose an initial approximation. An initial guess that is around [tex]\( x = 3 \)[/tex] seems reasonable.
Let:
[tex]\[
x_0 = 3
\][/tex]
### 3. Perform Iterations
We perform three iterations of the successive approximation method.
First Iteration:
[tex]\[
x_1 = g(x_0) = \left( \frac{7 - 3 \sqrt{3 - 1} - \frac{1}{2} \cdot 3^3}{3} \right)^{1/2}
\][/tex]
Calculating this, we get:
[tex]\[
x_1 \approx (1.1587 \times 10^{-16} + 1.8923i)
\][/tex]
(Note: This value is complex since the square root of a negative number is involved.)
Second Iteration:
[tex]\[
x_2 = g(x_1)
\][/tex]
Substitute [tex]\( x_1 \)[/tex] into the function [tex]\( g \)[/tex]:
[tex]\[
x_2 = g\left(1.1587 \times 10^{-16} + 1.8923i\right) \approx 1.2572 - 0.0492i
\][/tex]
Third Iteration:
[tex]\[
x_3 = g(x_2)
\][/tex]
Substitute [tex]\( x_2 \)[/tex] into the function [tex]\( g \)[/tex]:
[tex]\[
x_3 = g\left(1.2572 - 0.0492i\right) \approx 1.2229 + 0.0356i
\][/tex]
### 4. Conclusion
After performing three iterations, the approximations we have obtained are:
[tex]\[
x_0 = 3
\][/tex]
[tex]\[
x_1 \approx 1.1587 \times 10^{-16} + 1.8923i
\][/tex]
[tex]\[
x_2 \approx 1.2572 - 0.0492i
\][/tex]
[tex]\[
x_3 \approx 1.2229 + 0.0356i
\][/tex]
Therefore, the successive approximations suggest that the solutions are complex numbers after the initial step. This could be due to the nature of the function and the iterative process undertaken.
