Math Problem Statement

x1 2 and 3 of newtons method starting at x0 = -1.45 for the equation x^3+2x+5

Solution

To apply Newton's Method for solving the equation f(x)=x3+2x+5f(x) = x^3 + 2x + 5, starting at x0=1.45x_0 = -1.45, follow these steps:

Newton's Method Formula:

xn+1=xnf(xn)f(xn)x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} where:

  • f(x)=x3+2x+5f(x) = x^3 + 2x + 5
  • f(x)=3x2+2f'(x) = 3x^2 + 2

Step 1: Calculate f(x)f(x) and f(x)f'(x)

f(x)=x3+2x+5f(x) = x^3 + 2x + 5 f(x)=3x2+2f'(x) = 3x^2 + 2

Step 2: Perform Iterations

Start with x0=1.45x_0 = -1.45.


Iteration 1 (x1x_1):

f(1.45)=(1.45)3+2(1.45)+5=3.051125f(-1.45) = (-1.45)^3 + 2(-1.45) + 5 = -3.051125 f(1.45)=3(1.45)2+2=8.3025f'(-1.45) = 3(-1.45)^2 + 2 = 8.3025 x1=1.453.0511258.3025=1.082725x_1 = -1.45 - \frac{-3.051125}{8.3025} = -1.082725


Iteration 2 (x2x_2):

f(1.082725)=(1.082725)3+2(1.082725)+5=0.679437f(-1.082725) = (-1.082725)^3 + 2(-1.082725) + 5 = -0.679437 f(1.082725)=3(1.082725)2+2=5.514036f'(-1.082725) = 3(-1.082725)^2 + 2 = 5.514036 x2=1.0827250.6794375.514036=0.959274x_2 = -1.082725 - \frac{-0.679437}{5.514036} = -0.959274


Iteration 3 (x3x_3):

f(0.959274)=(0.959274)3+2(0.959274)+5=0.049126f(-0.959274) = (-0.959274)^3 + 2(-0.959274) + 5 = -0.049126 f(0.959274)=3(0.959274)2+2=4.759453f'(-0.959274) = 3(-0.959274)^2 + 2 = 4.759453 x3=0.9592740.0491264.759453=0.948923x_3 = -0.959274 - \frac{-0.049126}{4.759453} = -0.948923


Results:

  • x1=1.082725x_1 = -1.082725
  • x2=0.959274x_2 = -0.959274
  • x3=0.948923x_3 = -0.948923

Would you like me to expand on the method, verify calculations, or explain further?


Relative Questions:

  1. How does Newton's method converge, and what are its limitations?
  2. Why is the derivative important in Newton's method?
  3. Can Newton's method fail to converge? If so, under what conditions?
  4. How can initial guesses affect the convergence of Newton's method?
  5. What are the differences between Newton's method and other root-finding techniques?

Tip:

Always check f(x)0f'(x) \neq 0 at each step to avoid division by zero.

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Math Problem Analysis

Mathematical Concepts

Numerical Methods
Newton's Method
Polynomials

Formulas

Newton's Method formula: x_(n+1) = x_n - f(x_n)/f'(x_n)
Cubic polynomial: f(x) = x^3 + 2x + 5
Derivative: f'(x) = 3x^2 + 2

Theorems

Newton-Raphson Method

Suitable Grade Level

Grades 11-12 and Undergraduate