Question

The equation \(f(x)\) is given as \(x^2-4=0\). Considering the initial approximation at \(x=6\) then the value of \(x_1\) is given as

Answer 1

The equation you've given is a quadratic equation, and it appears you're trying to solve it using the Newton-Raphson method, which is a root-finding algorithm that produces successively better approximations to the roots (or zeroes) of a real-valued function.

The Newton-Raphson formula is given by:

\(x_{n+1} = x_n - f(x_n) / f'(x_n)\)

where:

• \(x_{n+1}\) is the next guess
• \(x_n\) is the current guess
• \(f(x_n)\) is the value of the function at x_n
• \(f'(x_n)\) is the value of the derivative of the function at x_n

Given the function \(f(x) = x^2 - 4\), its derivative \(f'(x) = 2x\).

Given that the initial approximation \(x_0 = 6\), we can find the next approximation \(x_1\) using the Newton-Raphson formula:

\(x_1 = x_0 - f(x_0) / f'(x_0) \)

\(= 6 - (6^2 - 4) / (2*6) \)

\(= 6 - (36 - 4) / 12 \)

\(= 6 - 32 / 12 \)

\(= 6 - 8/3 \)

\(= 18/3 - 8/3 \)

\(= 10/3\)

So, the value of \(x_1\) is \(10/3\).