Answer:
Step-by-step explanation:
Given the initial value of X0 = -1, we can determine the solution of the equation x³ + 5x +2 = 0 using the Newton's method. According to newton's approximation formula;
[tex]y = f(x_0) + f'(x_0)(x-x_0)[/tex]
[tex]x_n = x_n_-_1 - \frac{f(x_n_-_1 )}{f'(x_n_-_1 )}[/tex]
If [tex]x_0 = 1\\[/tex]
We will iterate using the formula;
[tex]x_1 = x_0 - \frac{f(x_0 )}{f'(x_0 )}[/tex]
Given f(x) = x³ + 5x +2
f(x0) = f(-1) = (-1)³ + 5(-1) +2
f(-1) = -1 -5 +2
f(-1) = -4
f'(x) = 3x²+5
f'(-1) = 3(-1)²+5
f'(-1) = 8
[tex]x_1 = -1+4/8\\x_1 = -1+0.5\\x_1 = -0.5\\\\x_2 = x_1 - \frac{f(x_1)}{f'(x_1)}\\x_2 = -0.5 - \frac{f(-0.5)}{f'(-0.5)}[/tex]
f(-0.5) = (-0.5)³ + 5(-0.5) +2
f(-0.5) = -0.125-2.5+2
f(-0.5) = -0.625
f'(-0.5) = 3(-0.5)²+5
f'(-0.5) = 3(0.25)+5
f'(-0.5) = 0.75+5
f'(-0.5) = 5.75
[tex]x_2 = -0.5 - \frac{(-0.625)}{5.75}\\x_2 = -0.5 + \frac{(0.625)}{5.75}\\x_2 = -0.5 + 0.1086957\\x_2 = -0.3913[/tex]
The estimated solution is -0.3913 (to 4dp)
