Question:
The formula to solve simple non-linear equation f(x) = 0 by Newton’s Raphson Method is given by :
The formula to solve simple non-linear equation f(x) = 0 by Newton’s Raphson Method is given by :
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The Newton-Raphson method has quadratic convergence ($p = 2$), making it one of the fastest iterative methods for root finding, provided the initial guess is close to the root and $f'(x) \neq 0$.
Updated On: Jul 24, 2026
- $x_{n+1} = x_n - \frac{f'(x_n)}{f(x_n)}$
- $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$
- $x_{n+1} = f(x_n) - \frac{f''(x_n)}{f'(x_n)}$
- $x_{n+1} = f(x_n) - \frac{f'(x_n)}{f''(x_n)}$
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
The Newton-Raphson method is an iterative root-finding algorithm used to find successively better approximations to the roots of a real-valued function $f(x) = 0$.
Detailed Explanation:
Let us derive the formula geometrically:
1. Let $x_n$ be an approximate root of $f(x) = 0$.
2. Consider the tangent line to the curve $y = f(x)$ at the point $(x_n, f(x_n))$. The slope of this tangent line is $f'(x_n)$.
3. The equation of this tangent line is: \[ y - f(x_n) = f'(x_n)(x - x_n) \] 4. To find the next approximation $x_{n+1}$, we find where this tangent line intersects the $x$-axis (set $y = 0$): \[ 0 - f(x_n) = f'(x_n)(x_{n+1} - x_n) \] \[ x_{n+1} - x_n = -\frac{f(x_n)}{f'(x_n)} \] \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] This is the standard Newton-Raphson iteration formula.
Step 2: Final Answer:
The iteration formula matches Option (B).
The Newton-Raphson method is an iterative root-finding algorithm used to find successively better approximations to the roots of a real-valued function $f(x) = 0$.
Detailed Explanation:
Let us derive the formula geometrically:
1. Let $x_n$ be an approximate root of $f(x) = 0$.
2. Consider the tangent line to the curve $y = f(x)$ at the point $(x_n, f(x_n))$. The slope of this tangent line is $f'(x_n)$.
3. The equation of this tangent line is: \[ y - f(x_n) = f'(x_n)(x - x_n) \] 4. To find the next approximation $x_{n+1}$, we find where this tangent line intersects the $x$-axis (set $y = 0$): \[ 0 - f(x_n) = f'(x_n)(x_{n+1} - x_n) \] \[ x_{n+1} - x_n = -\frac{f(x_n)}{f'(x_n)} \] \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] This is the standard Newton-Raphson iteration formula.
Step 2: Final Answer:
The iteration formula matches Option (B).
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