Let [tex]f(x) = x^3 + 3x + \sin(x) - 5[/tex]. Using Newton's method to approximate a solution to [tex]f(x) = 0[/tex], we consider the recurrence
[tex]\begin{cases} x_1 = 1 \\ x_{n + 1} = x_n - \frac{f(x_n)}{f'(x_n)} & \text{for } n \ge 1 \end{cases}[/tex]
Differentiating [tex]f(x)[/tex] gives
[tex]f'(x) = 3x^2 + 3 + \cos(x)[/tex]
Then
[tex]x_2 = 1 - \dfrac{f(1)}{f'(1)} = 1 + \dfrac{1 - \sin(1)}{6 + \cos(1)} \approx 1.024238790[/tex]
[tex]x_3 = x_2 - \dfrac{f(x_2)}{f'(x_2)} \approx 1.024009549[/tex]
[tex]x_4 = x_3 - \dfrac{f(x_3)}{f'(x_3)} \approx \boxed{1.024009528}[/tex]
which agrees numerically with the actual root of [tex]f(x)[/tex] up to at least 9 digits after the decimal point.
