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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
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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 $10/3$.

by Platinum (93,169 points)

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