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Numerical Methods : Numerical Methods

Study concepts, example questions & explanations for Numerical Methods

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Example Questions

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Example Question #1 : Numerical Methods

Use Newton's method to determine $x_{4}$ for $x^{9} + x^{7} + x^{4} + 35$ if $x_{0} = 4$ .

Possible Answers:

3.54938489087712

2.47005214248551

-0.2492605855343779

-3.6230338271289018

Correct answer:

2.47005214248551

Explanation:

In order to solve this problem, we need to recall newton's method.

$x_{n+1} = x_{n} - \frac{f(x_{n)}}{f'(x_{n)}}$

So we need to run this equation 4 times

First let's find the derivative of $x^{9} + x^{7} + x^{4} + 35$

$f'(x) = 9 x^{8} + 7 x^{6} + 4 x^{3}$

Now Newton's Method looks like.

$x_{n+1} = x_{n} - \frac{x_{n}^{9} + x_{n}^{7} + x_{n}^{4} + 35}{9 x_{n}^{8} + 7 x_{n}^{6} + 4 x_{n}^{3}}$

Now plug in 4 for $x_{n}$

$x_{1} = 3.54938489087712$

Now we take this answer, and plug it back into the equation for $x_{n}$

We keep doing this until, we get to $x_{4}$

Below is the results of Newton's Method.

$\\x_{0} = 4 \\x_1 = 3.54938489087712 \\x_2 = 3.14795030253403 \\x_3 = 2.78995941644637 \\x_4 = 2.47005214248551$

So our final answer will be

x_4 = 2.47005214248551

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Example Question #1 : Numerical Methods

Use Newton's method to determine $x_{4}$ for $x^{7} + x^{4} + x + 64$ if $x_{0} = 4$ .

Possible Answers:

3.42244806249784

-4.043481696297789

2.07997258909272

1.6604321009452203

Correct answer:

2.07997258909272

Explanation:

In order to solve this problem, we need to recall newton's method.

$x_{n+1} = x_{n} - \frac{f(x_{n)}}{f'(x_{n)}}$

So we need to run this equation 4 times

First let's find the derivative of $x^{7} + x^{4} + x + 64$

$f'(x) = 7 x^{6} + 4 x^{3} + 1$

Now Newton's Method looks like.

$x_{n+1} = x_{n} - \frac{x_{n}^{7} + x_{n}^{4} + x_{n} + 64}{7 x_{n}^{6} + 4 x_{n}^{3} + 1}$

Now plug in 4 for $x_{n}$

$x_{1} = 3.42244806249784$

Now we take this answer, and plug it back into the equation for $x_{n}$

We keep doing this until, we get to $x_{4}$

Below is the results of Newton's Method.

$\\x_{0} = 4 \\x_1 = 3.42244806249784 \\x_2 = 2.9225078708163 \\x_3 = 2.48309675959336 \\x_4 = 2.07997258909272$

So our final answer will be

x_4 = 2.07997258909272

Report an Error

Example Question #1 : Function Evaluations, Real & Complex Zeros

Use Newton's method to determine $x_{3}$ for $x^{9} + x^{6} + x^{3} + 88$ if $x_{0} = 3$ .

Possible Answers:

-4.77716184119023

2.07704836290301

0.3818985329267486

2.66090131165956

Correct answer:

2.07704836290301

Explanation:

In order to solve this problem, we need to recall newton's method.

$x_{n+1} = x_{n} - \frac{f(x_{n)}}{f'(x_{n)}}$

So we need to run this equation 3 times

First let's find the derivative of $x^{9} + x^{6} + x^{3} + 88$

$f'(x) = 9 x^{8} + 6 x^{5} + 3 x^{2}$

Now Newton's Method looks like.

$x_{n+1} = x_{n} - \frac{x_{n}^{9} + x_{n}^{6} + x_{n}^{3} + 88}{9 x_{n}^{8} + 6 x_{n}^{5} + 3 x_{n}^{2}}$

Now plug in 3 for $x_{n}$

$x_{1} = 2.66090131165956$

Now we take this answer, and plug it back into the equation for $x_{n}$

We keep doing this until, we get to $x_{3}$

Below is the results of Newton's Method.

$\\x_{0} = 3 \\x_1 = 2.66090131165956 \\x_2 = 2.3559080299424 \\x_3 = 2.07704836290301$

So our final answer will be

x_3 = 2.07704836290301

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Example Question #4 : Numerical Methods

Use Newton's method to determine $x_{4}$ for $x^{7} + x^{4} + x + 92$ if $x_{0} = 4$ .

Possible Answers:

-5.093216906705919

3.4214801756023

-4.964409052627277

2.05430142585667

Correct answer:

2.05430142585667

Explanation:

In order to solve this problem, we need to recall newton's method.

$x_{n+1} = x_{n} - \frac{f(x_{n)}}{f'(x_{n)}}$

So we need to run this equation 4 times

First let's find the derivative of $x^{7} + x^{4} + x + 92$

$f'(x) = 7 x^{6} + 4 x^{3} + 1$

Now Newton's Method looks like.

$x_{n+1} = x_{n} - \frac{x_{n}^{7} + x_{n}^{4} + x_{n} + 92}{7 x_{n}^{6} + 4 x_{n}^{3} + 1}$

Now plug in 4 for $x_{n}$

$x_{1} = 3.42148017560234$

Now we take this answer, and plug it back into the equation for $x_{n}$

We keep doing this until, we get to $x_{4}$

Below is the results of Newton's Method.

$\\x_{0} = 4 \\x_1 = 3.42148017560234 \\x_2 = 2.91920755930525 \\x_3 = 2.47383605844084 \\x_4 = 2.05430142585667$

So our final answer will be

x_4 = 2.05430142585667

Report an Error

Example Question #5 : Numerical Methods

Use Newton's method to determine $x_{2}$ for $2 x^{6} + x^{2} + 53$ if $x_{0} = 2$ .

Possible Answers:

1.52319587628866

-3.928159865648844

0.731597377649590

3.1305546016579893

Correct answer:

0.731597377649590

Explanation:

In order to solve this problem, we need to recall newton's method.

$x_{n+1} = x_{n} - \frac{f(x_{n)}}{f'(x_{n)}}$

So we need to run this equation 2 times

First let's find the derivative of $2 x^{6} + x^{2} + 53$

$f'(x) = 12 x^{5} + 2 x$

Now Newton's Method looks like.

$x_{n+1} = x_{n} - \frac{2 x_{n}^{6} + x_{n}^{2} + 53}{12 x_{n}^{5} + 2 x_{n}}$

Now plug in 2 for $x_{n}$

$x_{1} = 1.52319587628866$

Now we take this answer, and plug it back into the equation for $x_{n}$

We keep doing this until, we get to $x_{2}$

Below is the results of Newton's Method.

$\\x_{0} = 2 \\x_1 = 1.52319587628866 \\x_2 = 0.73159737764959$

So our final answer will be

x_2 = 0.73159737764959

Report an Error

Example Question #6 : Numerical Methods

Use Newton's method to determine $x_{3}$ for $x^{7} + x^{4} + x^{3} + 98$ if $x_{0} = 3$ .

Possible Answers:

1.64956709844858

1.6970343451644085

1.1077655881655772

2.54314623902253

Correct answer:

1.64956709844858

Explanation:

In order to solve this problem, we need to recall newton's method.

$x_{n+1} = x_{n} - \frac{f(x_{n)}}{f'(x_{n)}}$

So we need to run this equation 3 times

First let's find the derivative of $x^{7} + x^{4} + x^{3} + 98$

$f'(x) = 7 x^{6} + 4 x^{3} + 3 x^{2}$

Now Newton's Method looks like.

$x_{n+1} = x_{n} - \frac{x_{n}^{7} + x_{n}^{4} + x_{n}^{3} + 98}{7 x_{n}^{6} + 4 x_{n}^{3} + 3 x_{n}^{2}}$

Now plug in 3 for $x_{n}$

$x_{1} = 2.54314623902253$

Now we take this answer, and plug it back into the equation for $x_{n}$

We keep doing this until, we get to $x_{3}$

Below is the results of Newton's Method.

$\\x_{0} = 3 \\x_1 = 2.54314623902253 \\x_2 = 2.11651044448722 \\x_3 = 1.64956709844858$

So our final answer will be

x_3 = 1.64956709844858

Report an Error

Example Question #7 : Numerical Methods

Use Newton's method to determine $x_{2}$ for $x^{7} + x^{6} + x^{2} + 91$ if $x_{0} = 2$ .

Possible Answers:

0.726004314790471

1.55434782608696

3.5212421209971385

-6.867263981313135

Correct answer:

0.726004314790471

Explanation:

In order to solve this problem, we need to recall newton's method.

$x_{n+1} = x_{n} - \frac{f(x_{n)}}{f'(x_{n)}}$

So we need to run this equation 2 times

First let's find the derivative of $x^{7} + x^{6} + x^{2} + 91$

$f'(x) = 7 x^{6} + 6 x^{5} + 2 x$

Now Newton's Method looks like.

$x_{n+1} = x_{n} - \frac{x_{n}^{7} + x_{n}^{6} + x_{n}^{2} + 91}{7 x_{n}^{6} + 6 x_{n}^{5} + 2 x_{n}}$

Now plug in 2 for $x_{n}$

$x_{1} = 1.55434782608696$

Now we take this answer, and plug it back into the equation for $x_{n}$

We keep doing this until, we get to $x_{2}$

Below is the results of Newton's Method.

$\\x_{0} = 2 \\x_1 = 1.55434782608696 \\x_2 = 0.726004314790471$

So our final answer will be

x_2 = 0.726004314790471

Report an Error

Example Question #1 : Numerical Methods

Use Newton's method to determine $x_{2}$ for $x^{7} + x^{4} + x^{2} + 22$ if $x_{0} = 2$ .

Possible Answers:

-3.17484823018874

1.64876033057851

-2.0120657408948617

1.24572412489887

Correct answer:

1.24572412489887

Explanation:

In order to solve this problem, we need to recall newton's method.

$x_{n+1} = x_{n} - \frac{f(x_{n)}}{f'(x_{n)}}$

So we need to run this equation 2 times

First let's find the derivative of $x^{7} + x^{4} + x^{2} + 22$

$f'(x) = 7 x^{6} + 4 x^{3} + 2 x$

Now Newton's Method looks like.

$x_{n+1} = x_{n} - \frac{x_{n}^{7} + x_{n}^{4} + x_{n}^{2} + 22}{7 x_{n}^{6} + 4 x_{n}^{3} + 2 x_{n}}$

Now plug in 2 for $x_{n}$

$x_{1} = 1.64876033057851$

Now we take this answer, and plug it back into the equation for $x_{n}$

We keep doing this until, we get to $x_{2}$

Below is the results of Newton's Method.

$\\x_{0} = 2 \\x_1 = 1.64876033057851 \\x_2 = 1.24572412489887$

So our final answer will be

x_2 = 1.24572412489887

Report an Error

Example Question #9 : Numerical Methods

Use Newton's method to determine $x_{2}$ for $2 x^{6} + x^{2} + 80$ if $x_{0} = 2$ .

Possible Answers:

-3.846704901790435

0.203630506439732

1.45360824742268

2.3264894836566743

Correct answer:

0.203630506439732

Explanation:

In order to solve this problem, we need to recall newton's method.

$x_{n+1} = x_{n} - \frac{f(x_{n)}}{f'(x_{n)}}$

So we need to run this equation 2 times

First let's find the derivative of $2 x^{6} + x^{2} + 80$

$f'(x) = 12 x^{5} + 2 x$

Now Newton's Method looks like.

$x_{n+1} = x_{n} - \frac{2 x_{n}^{6} + x_{n}^{2} + 80}{12 x_{n}^{5} + 2 x_{n}}$

Now plug in 2 for $x_{n}$

$x_{1} = 1.45360824742268$

Now we take this answer, and plug it back into the equation for $x_{n}$

We keep doing this until, we get to $x_{2}$

Below is the results of Newton's Method.

$\\x_{0} = 2 \\x_1 = 1.45360824742268 \\x_2 = 0.203630506439732$

So our final answer will be

x_2 = 0.203630506439732

Report an Error

Example Question #10 : Numerical Methods

Use Newton's method to determine $x_{4}$ for $2 x^{6} + x^{4} + 34$ if $x_{0} = 4$ .

Possible Answers:

0.5443159376072018

1.4075970071619803

1.83512388860283

3.32382015306122

Correct answer:

1.83512388860283

Explanation:

In order to solve this problem, we need to recall newton's method.

$x_{n+1} = x_{n} - \frac{f(x_{n)}}{f'(x_{n)}}$

So we need to run this equation 4 times

First let's find the derivative of $2 x^{6} + x^{4} + 34$

$f'(x) = 12 x^{5} + 4 x^{3}$

Now Newton's Method looks like.

$x_{n+1} = x_{n} - \frac{2 x_{n}^{6} + x_{n}^{4} + 34}{12 x_{n}^{5} + 4 x_{n}^{3}}$

Now plug in 4 for $x_{n}$

$x_{1} = 3.32382015306122$

Now we take this answer, and plug it back into the equation for $x_{n}$

We keep doing this until, we get to $x_{4}$

Below is the results of Newton's Method.

$\\x_{0} = 4 \\x_1 = 3.32382015306122 \\x_2 = 2.75495816879288 \\x_3 = 2.26903747948471 \\x_4 = 1.83512388860283$

So our final answer will be

x_4 = 1.83512388860283

Report an Error

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Numerical Methods : Numerical Methods

Study concepts, example questions & explanations for Numerical Methods

All Numerical Methods Resources

Example Questions

Example Question #1 : Numerical Methods

Example Question #1 : Numerical Methods

Example Question #1 : Function Evaluations, Real & Complex Zeros

Example Question #4 : Numerical Methods

Example Question #5 : Numerical Methods

Example Question #6 : Numerical Methods

Example Question #7 : Numerical Methods

Example Question #1 : Numerical Methods

Example Question #9 : Numerical Methods

Example Question #10 : Numerical Methods

All Numerical Methods Resources

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