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Calculate two iterations of Newton's Method to approximate a zero of the function using the given initial guess. (Round your answers to three decimal places.) f(x) = x9 − 9, x1 = 1.6

Answers

Answer 1

Answer:

Answer:

Iteration 1: $x_(2)=1.446$

Iteration 2: $x_(3)=1.337$

Step-by-step explanation:

Formula for Newton's method is,

$x_(n+1)=x_n-(f\left(x_n\right))/(f'\left(x_n\right))$

Given the initial guess as $x_(1)=1.6$ , therefore value of n = 1.

Also, $f\left(x\right)=x^(9)-9$ .

Differentiating with respect to x,

$(d)/(dx)\left(f\left(x\right)\right)=(d)/(dx)\left(x^9-9\right)$

Applying difference rule of derivative,

$(d)/(dx)\left(f\left(x\right)\right)=(d)/(dx)\left(x^9\right)-(d)/(dx)\left(9\right)$

Applying power rule and constant rule of derivative,

$(d)/(dx)\left(f\left(x\right)\right)=\left(9x^(9-1)\right)-0$

$(d)/(dx)\left(f\left(x\right)\right)=9x^(8)$

Substituting the value,

$x_(1+1)=x_1-(f\left(x_1\right))/(f'\left(x_1\right))$

$x_(2)=1.6-(f\left(1.6\right))/(f'\left(1.6\right))$

Calculating the value of $f\left(1.6\right)$ and $f'\left(1.6\right)$

Calculating $f\left(1.6\right)$

$f\left(1.6\right)=\left(1.6\right)^(9)-9$

$f\left(1.6\right)=59.71947674$

Calculating $f'\left(1.6\right)$ ,

$f'\left(1.6\right)=9\left(1.6\right)^(8)$

$f'\left(1.6\right)=386.5470566$

Substituting the value,

$x_(2)=1.6-(59.71947674)/(386.5470566)$

$x_(2)=1.446$

Therefore value after second iteration is $x_(2)=1.446$

Now use $x_(2)=1.446$ as the next value to calculate second iteration. Here n = 2

Therefore,

$x_(2+1)=x_2-(f\left(x_2\right))/(f'\left(x_2\right))$

$x_(3)=1.446-(f\left(1.446\right))/(f'\left(1.446\right))$

Calculating the value of $f\left(1.446\right)$ and $f'\left(1.446\right)$

Calculating $f\left(1.446\right)$

$f\left(1.446\right)=\left(1.446\right)^(9)-9$

$f\left(1.446\right)=18.63851065$

Calculating $f'\left(1.446\right)$ ,

$f\left(1.446\right)=9\left(1.446\right)^(8)$

$f\left(1.446\right)=172.0239252$

Substituting the value,

$x_(3)=1.446-(18.63851065)/(172.0239252)$

$x_(3)=1.337$

Therefore value after second iteration is $x_(3)=1.337$

Answer 2

Answer:

Final answer:

To calculate two iterations of Newton's Method, use the formula xn+1 = xn - f(xn)/f'(xn). Given an initial guess of x1 = 1.6 and the function f(x) = x9 - 9, calculate f(xn) and f'(xn) at x1 and then use the formula to find x2 and x3.

Explanation:

To calculate two iterations of Newton's Method, we need to use the formula:

xn+1 = xn - f(xn)/f'(xn)

Given an initial guess of x1 = 1.6 and the function f(x) = x9 - 9, we can proceed as follows:

Calculate f(xn) at x1: f(1.6) = (1.6)9 - 9 = 38.5432
Calculate f'(xn) at x1: f'(1.6) = 9(1.6)8 = 368.64
Calculate x2: x2 = 1.6 - f(1.6)/f'(1.6) = 1.6 - 38.5432/368.64 = 1.494
Repeat the process to find x3 using the updated x2 as the initial guess.

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Answers

Final answer:

To sketch the curve of intersection, we substitute the equation of the parabolic cylinder into the equation of the ellipsoid. We use the discriminant to determine the nature of the curve and find its parametric equations.

Explanation:

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Find two consecutive positive integers such that the square of the first decreased by 67 is equal to three times the second

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Answer:

The numbers are 10 and 11.

Step-by-step explanation:

Let $x$ and $x+1$ be consecutive positive integers.

When the problem says that the square of the first decreased by 67 this means $x^2-67$ and this is equal to three times the second $3(x+1)$ .

So $x^2-67=3(x+1)$ will be our equation.

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Solve by factoring

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Using the Zero Factor Principle: If ab = 0, then either a = 0 or b = 0, or both a and b are 0.

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Because the numbers need to be positive integers, we only take x = 10 as a valid solution.

The numbers are 10 and 11.

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How to determine the number distance it would fly

from the question, we have the following parameters that can be used in our computation:

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Evaluate

Distance = 9 miles

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Answers

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Step-by-step explanation:

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