Math Problem Statement

Write the formula for​ Newton's method and use the given initial approximation to compute the approximations x 1 and x 2. f left parenthesis x right parenthesis equals x squared plus 2 x minus 8​, x 0 equals 3 Question content area bottom Part 1 Write the formula for​ Newton's method for the given function. Choose the correct answer below. A.x Subscript n plus 1 Baseline equals x Subscript n Baseline plus StartFraction x Subscript n Superscript 2 Baseline plus 2 x Subscript n Baseline minus 8 Over 2 x Subscript n Baseline plus 2 EndFraction x Subscript n plus 1 Baseline equals x Subscript n Baseline plus StartFraction x Subscript n Superscript 2 Baseline plus 2 x Subscript n Baseline minus 8 Over 2 x Subscript n Baseline plus 2 EndFraction B.x Subscript n plus 1 Baseline equals x Subscript n Baseline minus StartFraction x Subscript n Superscript 2 Baseline plus 2 x Subscript n Baseline minus 8 Over x Subscript n Superscript 2 Baseline minus 8 EndFraction x Subscript n plus 1 Baseline equals x Subscript n Baseline minus StartFraction x Subscript n Superscript 2 Baseline plus 2 x Subscript n Baseline minus 8 Over x Subscript n Superscript 2 Baseline minus 8 EndFraction C.x Subscript n plus 1 Baseline equals x Subscript n Baseline minus StartFraction x Subscript n Superscript 2 Baseline plus 2 x Subscript n Baseline minus 8 Over 2 x Subscript n Baseline plus 2 EndFraction x Subscript n plus 1 Baseline equals x Subscript n Baseline minus StartFraction x Subscript n Superscript 2 Baseline plus 2 x Subscript n Baseline minus 8 Over 2 x Subscript n Baseline plus 2 EndFraction Your answer is correct.D.x Subscript n plus 1 Baseline equals x Subscript n Baseline minus StartFraction 2 x Subscript n Baseline plus 2 Over x Subscript n Superscript 2 Baseline plus 2 x Subscript n Baseline minus 8 EndFraction x Subscript n plus 1 Baseline equals x Subscript n Baseline minus StartFraction 2 x Subscript n Baseline plus 2 Over x Subscript n Superscript 2 Baseline plus 2 x Subscript n Baseline minus 8 EndFraction Part 2 Use the given initial approximation to compute the approximations x 1 and x 2. x 1equals    2.125 ​(Do not round until the final answer. Then round to three decimal places as​ needed.) Part 3 x 2equals    enter your response here ​(Do not round until the final answer. Then round to three decimal places as​ needed.)

Solution

Step 1: Newton's Method Formula

Newton's Method formula is:

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

For the given function:

f(x)=x2+2x−8f(x) = x^2 + 2x - 8

First, compute its derivative:

f′(x)=2x+2f'(x) = 2x + 2

Substituting f(x)f(x) and f′(x)f'(x) into Newton's formula:

xn+1=xn−xn2+2xn−82xn+2x_{n+1} = x_n - \frac{x_n^2 + 2x_n - 8}{2x_n + 2}

Thus, the correct answer is Option C.

Step 2: Compute x1x_1

Given x0=3x_0 = 3:

x1=3−32+2(3)−82(3)+2x_1 = 3 - \frac{3^2 + 2(3) - 8}{2(3) + 2}

=3−9+6−86+2= 3 - \frac{9 + 6 - 8}{6 + 2}

=3−78= 3 - \frac{7}{8}

=3−0.875= 3 - 0.875

=2.125= 2.125

So, x1=2.125x_1 = 2.125.

Step 3: Compute x2x_2

Using x1=2.125x_1 = 2.125:

x2=2.125−(2.125)2+2(2.125)−82(2.125)+2x_2 = 2.125 - \frac{(2.125)^2 + 2(2.125) - 8}{2(2.125) + 2}

First, compute the numerator:

(2.125)2=4.515625,2(2.125)=4.25(2.125)^2 = 4.515625, \quad 2(2.125) = 4.25

4.515625+4.25−8=0.7656254.515625 + 4.25 - 8 = 0.765625

Next, compute the denominator:

2(2.125)+2=6.252(2.125) + 2 = 6.25

x2=2.125−0.7656256.25x_2 = 2.125 - \frac{0.765625}{6.25}

=2.125−0.1225= 2.125 - 0.1225

=2.0025= 2.0025

Rounding to three decimal places:

x2≈2.003x_2 \approx 2.003

So, x2=2.003x_2 = 2.003.

Final Answers:

  • x1=2.125x_1 = 2.125
  • x2=2.003x_2 = 2.003

Would you like a further breakdown of the calculations?

Related Questions:

  1. How does Newton's Method work geometrically?
  2. What are the conditions for Newton's Method to converge?
  3. What happens if the derivative f′(x)f'(x) is very small?
  4. Can Newton's Method fail? Under what circumstances?
  5. How does Newton's Method compare to other root-finding methods?

Tip:

Newton's Method converges quadratically, meaning it can be very fast if the initial guess is close to the root!

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Math Problem Analysis

Mathematical Concepts

Numerical Methods
Newton's Method
Derivatives
Root Approximation

Formulas

Newton's Method: x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}
Given function: f(x) = x^2 + 2x - 8
Derivative: f'(x) = 2x + 2

Theorems

Newton's Method for Root Finding

Suitable Grade Level

Grades 10-12, College

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