The statement is True.

Let's break this down:

The Newton-Raphson method iteratively refines guesses for the root of the equation $f(x) = 0$. The method uses the formula:

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

Where:

• $f(x)$ is the function.
• $f'(x)$ is the derivative of the function.
• $x_n$ is the current guess, and $x_{n+1}$ is the next guess.

For the Newton-Raphson method to work, the derivative $f'(x)$ should not be zero or very close to zero, as this would cause division by zero or result in a very large update that could lead to divergence.

Given the function:

$f(x) = e^{-0.5x}(4 - x) - 2$

The derivative $f'(x)$ will involve both the product rule and the chain rule.

At $x_0 = 6$, if the derivative $f'(6)$ is negative, the Newton-Raphson method could still work mathematically (since it is defined as long as the derivative is not zero), but a negative value of $f'(x_0)$ might lead to undesirable behavior or poor convergence depending on the function's behavior.

To confirm this specifically, we'd need to compute $f'(x)$ at $x_0 = 6$ and check whether the derivative is negative. However, the general idea is that the Newton-Raphson method may not work well for certain initial guesses if the derivative is negative or close to zero, as the sign of the update can cause erratic behavior.

Would you like to see a step-by-step calculation to further explore this?