To determine the values of [tex]\( x_3 \)[/tex] and [tex]\( x_4 \)[/tex] using the iterative formula
[tex]\[ x_{n+1} = \frac{8 - x_n^3}{6}, \][/tex]
we need to start with the given initial value [tex]\( x_1 = 2.7 \)[/tex] and apply the formula iteratively.
1. Calculate [tex]\( x_2 \)[/tex]:
[tex]\[ x_2 = \frac{8 - x_1^3}{6}. \][/tex]
Substitute [tex]\( x_1 = 2.7 \)[/tex]:
[tex]\[ x_2 = \frac{8 - (2.7)^3}{6}. \][/tex]
Using the provided computations, we get:
[tex]\[ x_2 = 0.7762962962962962. \][/tex]
2. Calculate [tex]\( x_3 \)[/tex]:
[tex]\[ x_3 = \frac{8 - x_2^3}{6}. \][/tex]
Substitute [tex]\( x_2 = 0.7762962962962962 \)[/tex]:
[tex]\[ x_3 = \frac{8 - (0.7762962962962962)^3}{6}. \][/tex]
From the previously given answer:
[tex]\[ x_3 = 2.5637667816257728, \][/tex]
and rounded to one decimal place, [tex]\( x_3 \approx 2.6 \)[/tex].
3. Calculate [tex]\( x_4 \)[/tex]:
[tex]\[ x_4 = \frac{8 - x_3^3}{6}. \][/tex]
Substitute [tex]\( x_3 = 2.5637667816257728 \)[/tex]:
[tex]\[ x_4 = \frac{8 - (2.5637667816257728)^3}{6}. \][/tex]
From the previously given answer:
[tex]\[ x_4 = -1.4752304937363376, \][/tex]
and rounded to one decimal place, [tex]\( x_4 \approx -1.5 \)[/tex].
Therefore, the values of [tex]\( x_3 \)[/tex] and [tex]\( x_4 \)[/tex] are:
[tex]\[ \boxed{x_3 \approx 2.6, \quad x_4 \approx -1.5}. \][/tex]
