Question

The Newton-Raphson method is used to find the roots of the equation \(x = e^{-2x}\). If the initial guess \(x_0 = 0\), the estimate of the root after the first iteration, \(x_1 =\)

• A. 1
• B. 1/2
• C. 1/3
• D. 1/4

Hint:

Use \(x_1 = x_0 - f(x_0)/f'(x_0)\) with \(f(x)=x-e^{-2x}\); at \(x_0=0\), \(f=-1\) and \(f'=3\).

Answer 1

The Correct Option is C

Instead of just plugging numbers into the Newton-Raphson formula, we can think of it geometrically: draw the tangent line to the curve $y = f(x)$ at the current guess, and the next guess is where that tangent line crosses the x-axis.

1. Write $f(x)$: rearrange $x = e^{-2x}$ into $f(x) = x - e^{-2x} = 0$, since a root of this $f$ is exactly a solution of the original equation.
2. Value and slope at $x_0 = 0$: $f(0) = 0 - e^{0} = -1$, and the slope is $f'(x) = 1 + 2e^{-2x}$, so $f'(0) = 1 + 2(1) = 3$.
3. Equation of the tangent line at $x_0$: a line through the point $(0, -1)$ with slope $3$ is $y - (-1) = 3(x - 0)$, or $y = 3x - 1$.
4. Find where this tangent line crosses $y = 0$: set $0 = 3x - 1$, which gives $x = 1/3$.

This x-intercept of the tangent line is exactly $x_1$, the next Newton-Raphson estimate. So $x_1 = 1/3$, the same result reached by plugging numbers directly into the update formula.

Let's summarize:

• Newton-Raphson replaces the curve near $x_0$ with its tangent line.
• The next guess is simply where that tangent line meets the x-axis.
• Here that gives $x_1 = 1/3$, matching option (C).

So the correct estimate after one iteration is $x_1 = 1/3$.