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Applications of Derivatives · Applied Problems

Newton's Method

xn+1=xn−f(xn)f′(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

Newton's method iteratively approximates roots of f(x) = 0. Each step uses the tangent line to get a better approximation.

Conditions. f'(xₙ) ≠ 0 at each step. Convergence depends on the initial guess.
  • 3 variables
  • 2 worked examples
  • 15 in Applications of Derivatives

Variables

Variables used in the Newton's Method formula
SymbolNameUnit
xnCurrent approximation
Current x value		 -
fxnf(xₙ)
Function value at xₙ		 -
fpxnf'(xₙ)
Derivative at xₙ		 -

Worked examples

Use Newton's method with x₀ = 2 to approximate √5 (root of x²-5=0).
  1. f(x) = x²-5, f'(x) = 2x
  2. x₁ = 2 - (4-5)/(4) = 2 + 1/4 = 2.25
  3. x₂ = 2.25 - (5.0625-5)/(4.5) = 2.25 - 0.01389 ≈ 2.2361

Answer: x₂ ≈ 2.2361 (actual √5 ≈ 2.23607)

1 more worked examplePremium

Use Newton's method with x₀ = 1 to approximate ∛2 (root of x³ - 2 = 0).

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Common questions

What is the Newton's Method?

Newton's method iteratively approximates roots of f(x) = 0. Each step uses the tangent line to get a better approximation. It is one of the applications of derivatives formulas in the CalcRef reference.

When does the Newton's Method apply?

The Newton's Method holds under this condition: f'(xₙ) ≠ 0 at each step. Convergence depends on the initial guess.

How do you use the Newton's Method?

Worked example. Use Newton's method with x₀ = 2 to approximate √5 (root of x²-5=0). f(x) = x²-5, f'(x) = 2x. x₁ = 2 - (4-5)/(4) = 2 + 1/4 = 2.25. x₂ = 2.25 - (5.0625-5)/(4.5) = 2.25 - 0.01389 ≈ 2.2361. Answer: x₂ ≈ 2.2361 (actual √5 ≈ 2.23607).

Practice

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