🔍 How to Use Newton-Raphson Method to Solve 𝑒 𝑥 − 𝑥 3 − cos ⁡ ( 25 𝑥 ) = 0 near to x=4.5 (correct to three decimal places )

🔍 Newton-Raphson Method—Find the Root of

ex - x3 - cos(25x) = 0

Given: Initial guess x₀ = 4.5
Method: Newton-Raphson
xₙ₊₁ = xₙ - f(xₙ) / f'(xₙ)

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🔢 Step-by-Step Iterations:
Let
f(x) = ex - x3 - cos(25x)
f'(x) = ex - 3x2 + 25sin(25x)

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✅ I – Iteration:
x₀ = 4.5
f(4.5) = e⁴⋅⁵ - 4.5³ - cos(25×4.5) = -1.9347
f'(4.5) = e⁴⋅⁵ - 3×4.5² + 25sin(25×4.5) = 15.2061
x₁ = 4.5 - (-1.9347 / 15.2061) 

x₁ = 4.6272

✅ II – Iteration:
f(4.6272) = 4.0024 ; f'(4.6272) = 51.2514
x₂ = 4.6272 - (4.0024 / 51.2514) 

x₂ = 4.5491

✅ III – Iteration:
f(4.5491) = -0.4011 ; f'(4.5491) = 47.1963
x₃ = 4.5491 - (-0.4011 / 47.1963) 

x₃ = 4.5576

✅ IV – Iteration:
f(4.5576) = 0.0197 ; f'(4.5576) = 51.6997
x₄ = 4.5576 - (0.0197 / 51.6997) 
x₄ = 4.5572

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🧠 Final Answer:
x = 4.557 (Correct to 3 decimal places)

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📘 Note: The Newton-Raphson method is powerful for approximating roots. Always check convergence by comparing iterations.