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: Think of Newton-Raphson as following the tangent line to the axis.
At any guess $x_0$, draw the tangent line to the curve $y=f(x)$ at the point $(x_0, f(x_0))$. Its slope is $f'(x_0)$, so the tangent line is
\[ y - f(x_0) = f'(x_0)\,(x - x_0) \]
The next guess $x_1$ is taken as the point where this tangent line crosses the x-axis, that is, where $y=0$:
\[ 0 - f(x_0) = f'(x_0)\,(x_1 - x_0) \]
\[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} \]
This is the same update rule as before, derived here from the geometry of a tangent line rather than quoted directly.

Step 2: Compute the derivative of $f(x) = e^{-x} - x$.
Using $\frac{d}{dx}e^{-x} = -e^{-x}$ and $\frac{d}{dx}x = 1$:
\[ f'(x) = -e^{-x} - 1 \]

Step 3: Evaluate the function and slope at $x_0 = 0.5$.
$e^{-0.5} \approx 0.60653$.
\[ f(0.5) = 0.60653 - 0.5 = 0.10653 \]
\[ f'(0.5) = -0.60653 - 1 = -1.60653 \]
Since $f(0.5) > 0$ and the slope $f'(0.5)$ is negative, the tangent line drawn at $x_0=0.5$ slopes downward and crosses the x-axis to the right of $x_0$, so we expect $x_1 > x_0$.

Step 4: Find where the tangent line meets the x-axis.
\[ x_1 = 0.5 - \frac{0.10653}{-1.60653} = 0.5 + \frac{0.10653}{1.60653} \]
Dividing, $\frac{0.10653}{1.60653} \approx 0.06631$, so
\[ x_1 \approx 0.5 + 0.06631 = 0.56631 \]
This matches the direction predicted in Step 3: $x_1$ came out bigger than $x_0=0.5$, consistent with the sign of the slope.

Step 5: Round to the required precision.
Rounded to two decimal places, $x_1 \approx 0.57$.

Final Answer:
\[ \boxed{x_1 = 0.57} \]