Question

The equation $x^4 - x - 10 = 0$ is to be solved using the Newton-Raphson method. If $x = 2$ is taken as the initial approximation of the equation, then the next approximation of the solution is

• A. \( \frac{38}{31} \)
• B. \( \frac{48}{31} \)
• C. \( \frac{58}{31} \)
• D. \( \frac{28}{31} \)

Hint:

The Newton-Raphson method is an iterative numerical method for finding successively better approximations to the roots (or zeroes) of a real-valued function. It is based on approximating the function by its tangent line at a given point. The convergence of the Newton-Raphson method is quadratic, meaning that the number of correct decimal places roughly doubles with each iteration, provided the initial guess is sufficiently close to the root. However, it requires the derivative of the function to be calculated, and it might fail if $f'(x_n) = 0$ at any iteration.

Answer 1

The Correct Option is C

The Newton-Raphson method formula for finding the root of an equation $f(x)=0$ is given by: $$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} $$ Given the equation $x^4 - x - 10 = 0$. So, $f(x) = x^4 - x - 10$.
Step 1: Find the derivative of $f(x)$. $$ f'(x) = \frac{d}{dx}(x^4 - x - 10) = 4x^3 - 1 $$
Step 2: Identify the initial approximation. The initial approximation is given as $x_n = x_0 = 2$.
Step 3: Calculate $f(x_0)$. Substitute $x_0 = 2$ into $f(x)$: $$ f(2) = (2)^4 - (2) - 10 = 16 - 2 - 10 = 14 - 10 = 4 $$
Step 4: Calculate $f'(x_0)$. Substitute $x_0 = 2$ into $f'(x)$: $$ f'(2) = 4(2)^3 - 1 = 4(8) - 1 = 32 - 1 = 31 $$
Step 5: Apply the Newton-Raphson formula to find the next approximation $x_1$. $$ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} $$ $$ x_1 = 2 - \frac{4}{31} $$
Step 6: Simplify the expression for $x_1$. $$ x_1 = \frac{2 \cdot 31 - 4}{31} = \frac{62 - 4}{31} = \frac{58}{31} $$ The next approximation of the solution is $\frac{58}{31}$.