Question

While solving the equation \(e^{-x} - x = 0\) for \(x\), using the Newton-Raphson method with initial guess \(x_0 = 2\), the solution after the first iteration (\(x_1\)) is ________ (Rounded off to two decimal places)

Hint:

Apply the Newton-Raphson update \(x_1 = x_0 - f(x_0)/f\prime(x_0)\) with \(f(x) = e^{-x} - x\).

Answer 1

Correct Answer: 0.36

Step 1: Write the Newton-Raphson formula.
For a function \(f(x)\), the update rule is \(x_1 = x_0 - \dfrac{f(x_0)}{f\prime(x_0)}\).
Here \(f(x) = e^{-x} - x\), so its derivative is \(f\prime(x) = -e^{-x} - 1\).

Step 2: Evaluate \(f(x_0)\) and \(f\prime(x_0)\) at \(x_0 = 2\).
\(e^{-2} = 0.13534\), so \(f(2) = 0.13534 - 2 = -1.86466\).
\(f\prime(2) = -0.13534 - 1 = -1.13534\).

Step 3: Substitute into the update formula.
\[ x_1 = 2 - \frac{-1.86466}{-1.13534} = 2 - 1.64239 \] Dividing gives \(1.86466 / 1.13534 = 1.64239\), and this ratio is positive since both terms are negative.
So \(x_1 = 2 - 1.64239 = 0.35761\).

Final Answer:
Rounded to two decimal places, the first iterate is \[ \boxed{0.36} \]