Let's solve the equation [tex]\(3(-x+2) - 1 = 5x - 4\)[/tex] using successive approximation method.
Given the starting values:
[tex]\[
\begin{array}{|c|c|c|c|c|c|}
\hline
x & 0 & 1 & 2 & 3 & 4 \\
\hline
3^{(-x+2)}-1 & 8 & 2 & 0 & -\frac{2}{3} & -\frac{8}{9} \\
\hline
5x-4 & -4 & 1 & 6 & 11 & 16 \\
\hline
\end{array}
\][/tex]
### Step-by-Step Solution
1. Identify Intervals Between Values:
- Notice that [tex]\(3^{(-x+2)}-1\)[/tex] and [tex]\(5x-4\)[/tex] are both equal when [tex]\(x\)[/tex] is between 1 and 2 since:
[tex]\[
3^{(-1+2)}-1 = 2, \quad 5 \cdot 1 - 4 = 1
\][/tex]
[tex]\[
3^{(-2+2)}-1 = 0, \quad 5 \cdot 2 - 4 = 6
\][/tex]
2. First Iteration:
- Estimate the midpoint between [tex]\(x = 1\)[/tex] and [tex]\(x = 2\)[/tex]:
[tex]\[
x_1 = \frac{1 + 2}{2} = 1.5
\][/tex]
- Evaluate the function at [tex]\(x_1\)[/tex]:
[tex]\[
3^{(-1.5+2)} - 1 = 3^{0.5} - 1 \approx 1.732 - 1 = 0.732
\][/tex]
[tex]\[
5 \cdot 1.5 - 4 = 7.5 - 4 = 3.5
\][/tex]
- Since [tex]\(0.732 < 3.5\)[/tex], the solution lies between [tex]\(x = 1\)[/tex] and [tex]\(x = 1.5\)[/tex].
3. Second Iteration:
- Estimate the midpoint between [tex]\(x = 1\)[/tex] and [tex]\(x = 1.5\)[/tex]:
[tex]\[
x_2 = \frac{1 + 1.5}{2} = 1.25
\][/tex]
- Evaluate the function at [tex]\(x_2\)[/tex]:
[tex]\[
3^{(-1.25+2)} - 1 = 3^{0.75} - 1 \approx 2.279 - 1 = 1.279
\][/tex]
[tex]\[
5 \cdot 1.25 - 4 = 6.25 - 4 = 2.25
\][/tex]
- Since [tex]\(1.279 < 2.25\)[/tex], the solution lies between [tex]\(x = 1\)[/tex] and [tex]\(x = 1.25\)[/tex].
4. Third Iteration:
- Estimate the midpoint between [tex]\(x = 1\)[/tex] and [tex]\(x = 1.25\)[/tex]:
[tex]\[
x_3 = \frac{1 + 1.25}{2} = 1.125
\][/tex]
- Evaluate the function at [tex]\(x_3\)[/tex]:
[tex]\[
3^{(-1.125+2)} - 1 = 3^{0.875} - 1 \approx 2.574 - 1 = 1.574
\][/tex]
[tex]\[
5 \cdot 1.125 - 4 = 5.625 - 4 = 1.625
\][/tex]
- Since [tex]\(1.574 < 1.625\)[/tex], the solution lies between [tex]\(x = 1.125\)[/tex] and [tex]\(x = 1.25\)[/tex].
Among the given options, we notice that [tex]\(x \approx \frac{17}{16}\)[/tex] which equals approximately [tex]\(1.0625\)[/tex], lies within the final range and is the closest approximation after a few iterations.
So, the correct answer is:
B. [tex]\(x \approx \frac{17}{16}\)[/tex]
