Question

The root of the equation \( \sin x - 4x + 1 = 0 \) after its first iteration, using the Newton-Raphson method with an initial guess of \( x_0 = 0.2 \), is _____.\( \textit{[Round off to three decimal places.]}\)

Hint:

In Newton-Raphson method, the next approximation is found by using both the function value and its derivative at the current approximation.

Answer 1

Correct Answer: 0.25

The Newton-Raphson method is given by the formula: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] where \( f(x) = \sin x - 4x + 1 \) and its derivative is \( f'(x) = \cos x - 4 \). For the initial guess \( x_0 = 0.2 \), we compute: \[ f(x_0) = \sin(0.2) - 4(0.2) + 1 = 0.1987 - 0.8 + 1 = 0.3987 \] \[ f'(x_0) = \cos(0.2) - 4 = 0.9801 - 4 = -3.0199 \] Now, applying the Newton-Raphson formula: \[ x_1 = 0.2 - \frac{0.3987}{-3.0199} = 0.2 + 0.1325 = 0.3325 \] Thus, the root after the first iteration is approximately \( \boxed{0.333} \) (rounded to three decimal places).