Find the third approximation x3 to the root of the equation f(x) = x3 | Mathematics I 2077

Mathematics I20775 marksNumerical: Newton's method approximationNewton's method

Find the third approximation $x_3$ to the root of the equation $f(x) = x^3 - 2x - 7$ , setting $x_1 = 2$ . [5]

Find the third approximation $x_3$ to the root of the equation $f(x) = x^3 - 2x - 7$, setting $x_1 = 2$. [5]

$f(x) = x^3 - 2x - 7$ - Initial guess: $x1 = 2$ - Required: third approximation $x3$ $x{n+1} = xn - \frac{f(xn)}{f'(xn)}, \qquad f'(x) = 3x^2 - 2$ --- $f(2) = 8 - 4 - 7 = -3$ $f'(2) = 3(4) - 2 = 10$ $x2 = 2 - \frac{-3}{10} = 2 + 0.3 = 2.3$ --- $$f(2.3) = (2.3)^3 - 2(2.3) - 7 = 12.167 - ...

$f(x) = x^3 - 2x - 7$ - Initial guess: $x1 = 2$ - Required: third approximation $x3$ $$x{n+1} = xn - \frac{f(xn)}{f'(xn)}, \qquad f'(x) = 3x^2 - 2$$ --- $$f(2) = 8 - 4 - 7 = -3$$ $$f'(2) = 3(4) - 2 = 10$$ $$x2 = 2 - \frac{-3}{10} = 2 + 0.3 = 2.3$$ --- $$f(2.3) = (2.3)^3 - 2(2.3) - 7 = 12.167 - ...

Find the third approximation x3 to the root of the equation f(x) = x3 - 2x - 7, setting x1 = 2.