Question

If the initial guess of the solution of \[ 1+x-x^3=0 \] is \(x_0=1\), then the \(2^{\text{nd}}\) approximate root of \[ 1+x-x^3=0 \] using Newton-Raphson method is

• A. \(2\)
• B. \(\dfrac{3}{2}\)
• C. \(\dfrac{31}{23}\)
• D. \(\dfrac{15}{8}\)

Hint:

For Newton-Raphson problems, carefully compute \[ f(x_n)\quad\text{and}\quad f'(x_n) \] before substituting into \[ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}. \] Arithmetic mistakes are the most common source of errors.

Answer 1

The Correct Option is C

Concept: The Newton-Raphson method is an iterative method for finding the roots of a nonlinear equation \[ f(x)=0. \] The iteration formula is \[ x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)}. \]

Step 1: Find \(f(x)\) and \(f'(x)\). Given, \[ f(x)=1+x-x^3. \] Differentiating, \[ f'(x)=1-3x^2. \]

Step 2: Find the first approximation. Given, \[ x_0=1. \] Now, \[ f(1)=1+1-1=1, \] and \[ f'(1)=1-3=-2. \] Therefore, \[ x_1 = 1-\frac{1}{-2} = 1+\frac12 = \frac32. \]

Step 3: Find the second approximation. Using \[ x_1=\frac32, \] we have \[ f\!\left(\frac32\right) = 1+\frac32-\left(\frac32\right)^3 = \frac52-\frac{27}{8} = -\frac78. \] Also, \[ f'\!\left(\frac32\right) = 1-3\left(\frac32\right)^2 = 1-\frac{27}{4} = -\frac{23}{4}. \] Hence, \[ x_2 = \frac32 - \frac{-\frac78}{-\frac{23}{4}} = \frac32-\frac{7}{46} = \frac{69-7}{46} = \frac{62}{46} = \frac{31}{23}. \] Therefore, \[ \boxed{x_2=\frac{31}{23}.} \] Hence, the correct option is \[ \boxed{\left(C\right)\;\frac{31}{23}.} \]