Given the equation:
[tex]\[
\frac{1}{2} x^3 + x - 7 = -3 \sqrt{x - 1}
\][/tex]
We want to approximate the solution using three iterations of successive approximations. Let's treat the right-hand side as a new function [tex]\( f(x) \)[/tex], such that:
[tex]\[
f(x) = \frac{1}{2} x^3 + x - 7 + 3 \sqrt{x - 1}
\][/tex]
We seek the fixed point where [tex]\( x = f(x) \)[/tex]. Let's start with an initial guess. Observing the graph, let's assume a reasonable initial guess [tex]\(x_0 = 1.5\)[/tex].
### Iteration 1
Plugging [tex]\( x_0 = 1.5 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[
f(1.5) = \frac{1}{2} (1.5)^3 + 1.5 - 7 + 3 \sqrt{1.5 - 1}
\][/tex]
[tex]\[
= \frac{1}{2} (3.375) + 1.5 - 7 + 3 \sqrt{0.5}
\][/tex]
[tex]\[
= 1.6875 + 1.5 - 7 + 3 \cdot 0.7071
\][/tex]
[tex]\[
= 1.6875 + 1.5 - 7 + 2.1213
\][/tex]
[tex]\[
= -1.6912
\][/tex]
Updating [tex]\( x_1 \approx -1.6912 \)[/tex], but this result is not feasible because [tex]\( x \)[/tex] needs to be greater than or equal to 1 (since we have [tex]\(\sqrt{x-1}\)[/tex]).
We must re-evaluate with a valid initial guess. Let's try [tex]\( x_0 = 2 \)[/tex].
### Iteration 2
Re-calculating with [tex]\( x_0 = 2 \)[/tex]:
[tex]\[
f(2) = \frac{1}{2} (2)^3 + 2 - 7 + 3 \sqrt{2 - 1}
\][/tex]
[tex]\[
= \frac{1}{2} (8) + 2 - 7 + 3 \sqrt{1}
\][/tex]
[tex]\[
= 4 + 2 - 7 + 3
\][/tex]
[tex]\[
= 2
\][/tex]
Updating [tex]\( x_1 = 2 \)[/tex].
### Iteration 3
Continuing with [tex]\( x_1 = 2 \)[/tex]:
[tex]\[
f(2) = \frac{1}{2} (2)^3 + 2 - 7 + 3 \sqrt{2 - 1}
\][/tex]
[tex]\[
= 4 + 2 - 7 + 3
\][/tex]
[tex]\[
= 2
\][/tex]
We see that the iteration has stabilized at [tex]\( x = 2 \)[/tex].
For validation, let's compare this with the given options:
[tex]\[
\frac{25}{16} \approx 1.5625
\][/tex]
[tex]\[
\frac{27}{16} \approx 1.6875
\][/tex]
[tex]\[
\frac{13}{8} \approx 1.625
\][/tex]
[tex]\[
\frac{15}{8} \approx 1.875
\][/tex]
None of the given options exactly matches [tex]\( x = 2 \)[/tex]. However, since we've noticed a stabilization around [tex]\( x = 2 \)[/tex], the closest match from among the options provided is:
[tex]\[
x \approx \frac{15}{8} \approx 1.875
\][/tex]
Thus, the correct answer is:
[tex]\[
\boxed{\frac{15}{8}}
\][/tex]
