Answer :
To solve the given equation [tex]\(-\left(\frac{3}{2}\right)^x + 12 = 2x - 3\)[/tex] using three iterations of successive approximation, follow these steps:
1. Initial Guess:
Start with an initial guess for [tex]\( x \)[/tex]. From the options provided, a reasonable starting value might be around [tex]\( x = 2 \)[/tex].
2. Rearrange the Equation:
Modify the given equation to be suitable for successive approximation. Rearranging the equation to isolate a new form [tex]\( x_{n+1} \)[/tex], we have:
[tex]\[ \left(\frac{3}{2}\right)^x = 15 - 2x \][/tex]
This yields our successive approximation function:
[tex]\[ g(x) = \left( \frac{3}{2} \right)^x - 15 + 2x \][/tex]
3. Define the Iteration:
Use the iterative formula:
[tex]\[ x_{n+1} = g(x_n) = \left( \frac{3}{2} \right)^{x_n} - 15 + 2x_n \][/tex]
4. Perform Iterations:
Using the initial guess [tex]\( x_0 = 2 \)[/tex], apply three iterations of the function [tex]\( g \)[/tex]:
- Iteration 1:
[tex]\[ x_1 = g(x_0) = \left( \frac{3}{2} \right)^2 - 15 + 2 \times 2 = 2.25 - 15 + 4 \approx -8.75 \][/tex]
- Iteration 2:
[tex]\[ x_2 = g(x_1) = \left( \frac{3}{2} \right)^{-8.75} - 15 + 2 \times -8.75 = (very small number) - 15 - 17.5 \approx -32.5 \quad \text{(since the small number is negligible)} \][/tex]
- Iteration 3:
[tex]\[ x_3 = g(x_2) = \left( \frac{3}{2} \right)^{-32.5} - 15 + 2 \times -32.5 = (even smaller number) - 15 - 65 \approx -80 \][/tex]
5. Final Result:
After three iterations, the successive approximation yields an approximate solution:
[tex]\[ x \approx -80 \][/tex]
6. Match the Result with the Options:
Comparing the result to the provided options, we see that [tex]\( -80 \)[/tex] does not match any given option directly. But it confirms that the true root is quite far from these options, suggesting deeper investigation or re-evaluation of the initial conditions might be needed in a precise context.
Since the options do not seem to line up well with the successive approximation result, the given values might highlight potential approximations closer to the initial guess, indicating further detailed reevaluation if constrained to the multiple choice.
1. Initial Guess:
Start with an initial guess for [tex]\( x \)[/tex]. From the options provided, a reasonable starting value might be around [tex]\( x = 2 \)[/tex].
2. Rearrange the Equation:
Modify the given equation to be suitable for successive approximation. Rearranging the equation to isolate a new form [tex]\( x_{n+1} \)[/tex], we have:
[tex]\[ \left(\frac{3}{2}\right)^x = 15 - 2x \][/tex]
This yields our successive approximation function:
[tex]\[ g(x) = \left( \frac{3}{2} \right)^x - 15 + 2x \][/tex]
3. Define the Iteration:
Use the iterative formula:
[tex]\[ x_{n+1} = g(x_n) = \left( \frac{3}{2} \right)^{x_n} - 15 + 2x_n \][/tex]
4. Perform Iterations:
Using the initial guess [tex]\( x_0 = 2 \)[/tex], apply three iterations of the function [tex]\( g \)[/tex]:
- Iteration 1:
[tex]\[ x_1 = g(x_0) = \left( \frac{3}{2} \right)^2 - 15 + 2 \times 2 = 2.25 - 15 + 4 \approx -8.75 \][/tex]
- Iteration 2:
[tex]\[ x_2 = g(x_1) = \left( \frac{3}{2} \right)^{-8.75} - 15 + 2 \times -8.75 = (very small number) - 15 - 17.5 \approx -32.5 \quad \text{(since the small number is negligible)} \][/tex]
- Iteration 3:
[tex]\[ x_3 = g(x_2) = \left( \frac{3}{2} \right)^{-32.5} - 15 + 2 \times -32.5 = (even smaller number) - 15 - 65 \approx -80 \][/tex]
5. Final Result:
After three iterations, the successive approximation yields an approximate solution:
[tex]\[ x \approx -80 \][/tex]
6. Match the Result with the Options:
Comparing the result to the provided options, we see that [tex]\( -80 \)[/tex] does not match any given option directly. But it confirms that the true root is quite far from these options, suggesting deeper investigation or re-evaluation of the initial conditions might be needed in a precise context.
Since the options do not seem to line up well with the successive approximation result, the given values might highlight potential approximations closer to the initial guess, indicating further detailed reevaluation if constrained to the multiple choice.