Use Newton's method with initial approximation x1 = −1 to find x2, the second approximation to the root of the equation x3 + x + 8 = 0. (Round your answer to four decimal places.) x2 =

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Answer:

The second approximation to the root of the equation [tex]x^{3}+x+8 = 0[/tex] is -1.5000.

Step-by-step explanation:

The Newton's method is a numerical method by approximation that help find roots of a equation of the form [tex]f(x) = 0[/tex] with the help of the equation itself and its first derivative. The Newton's formula is:

[tex]x_{i+1} = x_{i} - \frac{f(x_{i})}{f'(x_{i})}[/tex]

Where:

[tex]x_{i}[/tex] - i-th approximation, dimensionless.

[tex]x_{i+1}[/tex] - (i+1)-th approximation, dimensionless.

[tex]f(x_{i})[/tex] - Function evaluated at the i-th approximation, dimensionless.

[tex]f'(x_{i})[/tex] - First derivative of the function evaluated at the i-th approximation, dimensionless.

The function and its first derivative are [tex]f(x) = x^{3}+x+8[/tex] and [tex]f'(x) = 3\cdot x^{2}+1[/tex], respectively. Now, the Newton's formula is expanded:

[tex]x_{i+1} = x_{i}-\frac{x_{i}^{3}+x_{i}+8}{3\cdot x_{i}^{2}+1}[/tex]

If [tex]x_{1} = -1[/tex], the value of [tex]x_{2}[/tex] is:

[tex]x_{2} = -1 - \frac{(-1)^{3}+(-1)+8}{3\cdot (-1)^{2}+1}[/tex]

[tex]x_{2} = -1.5000[/tex]

The second approximation to the root of the equation [tex]x^{3}+x+8 = 0[/tex] is -1.5000.

Answer:

-2.5000

Step-by-step explanation:

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