Newton-Raphson Method: Solve e^x − 2 = 0 in One Iteration

The Newton–Raphson method is used to solve the nonlinear equation
e^x − 2 = 0. Starting from the initial guess
x₀ = 1, one iteration gives the approximate root
x₁ ≈ 0.74 (rounded to two decimal places).

Newton-Raphson Method Formula

The general Newton–Raphson iteration formula is:

x_n+1 = x_n −
f(x_n) / f′(x_n)

For this problem:

• f(x) = e^x − 2
• f′(x) = e^x

This method exhibits quadratic convergence and is well suited for
exponential equations.

Step-by-Step Iteration

1. Initial guess:

   x₀ = 1

2. Evaluate the function:

   
   f(1) = e¹ − 2 ≈ 2.71828 − 2 = 0.71828
   

3. Evaluate the derivative:

   
   f′(1) = e¹ ≈ 2.71828
   

4. Apply the Newton–Raphson formula:

   
   x₁ = 1 − (0.71828 / 2.71828)
   

5. Simplify:

   
   x₁ = 1 − (1 − 2/e) = 2/e ≈ 0.73576
   

6. Rounded to two decimal places:

   x₁ ≈ 0.74

Correct Answer

The approximate root after one Newton–Raphson iteration is:

x ≈ 0.74

Common Mistakes Explained

• Forgetting the derivative:
  Taking f′(x) = 1 incorrectly gives
  x₁ ≈ 0.28, which is far from the correct value.
• Misrounding e:
  Using e ≈ 2.7 still gives ~0.74, but precise calculation confirms accuracy.
• Solving the wrong equation:
  Treating e^x = 1 leads to x = 0, which is not the given problem.
• Confusing with two iterations:
  A second iteration gives x₂ ≈ 0.85, but the question asks for
  only one iteration.

This result matches GATE-style numerical answers rounded to
two decimal places.