Question

Starting with the first approximation as \(x = 0.5\), the second approximation for the root of the following function by the Newton-Raphson method is (rounded off to two decimal places).

\[ f(x) = e^{-x} - x \]

Hint:

Apply \(x_1 = x_0 - f(x_0)/f'(x_0)\) with \(f(x)=e^{-x}-x\) and \(x_0=0.5\).

Answer 1

Correct Answer: 0.57

Step 1: Recall the Newton-Raphson update rule.
The Newton-Raphson method finds a root of \(f(x)=0\) by repeatedly moving from the current guess \(x_n\) to a better guess \(x_{n+1}\) using
\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]
Here \(f(x) = e^{-x} - x\), so we first need its derivative.

Step 2: Differentiate \(f(x)\).
\[ f'(x) = \frac{d}{dx}\left(e^{-x}\right) - \frac{d}{dx}(x) = -e^{-x} - 1 \]

Step 3: Evaluate \(f(x)\) and \(f'(x)\) at the starting guess \(x_0 = 0.5\).
\[ f(0.5) = e^{-0.5} - 0.5 = 0.60653 - 0.5 = 0.10653 \]
\[ f'(0.5) = -e^{-0.5} - 1 = -0.60653 - 1 = -1.60653 \]

Step 4: Apply the Newton-Raphson formula to get the second approximation.
\[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 0.5 - \frac{0.10653}{-1.60653} \]
\[ x_1 = 0.5 + \frac{0.10653}{1.60653} = 0.5 + 0.06631 = 0.56631 \]

Step 5: Round off to two decimal places.
\(0.56631\) rounds to \(0.57\), since the digit after the second decimal place is \(6\), which rounds the second decimal up from \(6\) to \(7\).

Final Answer:
The second approximation to the root is \(x_1 \approx 0.57\).
\[ \boxed{x_1 = 0.57} \]