Question

If \(x_0=1.2\) is the initial guess of the solution of \(x^3+2x-1=0\), then the \(1^{\text{st}}\) iteration solution \(x_1=\)

• A. \(0.705\)
• B. \(0.675\)
• C. \(0.815\)
• D. \(0.795\)

Hint:

Newton-Raphson iteration is \[ \boxed{ x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)}. } \] It provides quadratic convergence when the initial guess is sufficiently close to the root.

Answer 1

The Correct Option is A

Step 1: Use the Newton-Raphson formula. For \[ f(x)=x^3+2x-1, \] Newton-Raphson iteration is \[ \boxed{ x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)}. } \] Differentiate: \[ f'(x)=3x^2+2. \]

Step 2: Evaluate \(f(x_0)\) and \(f'(x_0)\). Given, \[ x_0=1.2. \] Then, \[ f(1.2) = (1.2)^3+2(1.2)-1 = 1.728+2.4-1 = 3.128. \] Also, \[ f'(1.2) = 3(1.2)^2+2 = 3(1.44)+2 = 6.32. \]

Step 3: Compute the first iteration. \[ x_1 = 1.2-\frac{3.128}{6.32} = 1.2-0.4949 \approx0.705. \] Hence, \[ \boxed{x_1\approx0.705.} \] Therefore, \[ \boxed{(A)} \] is the correct answer.