Numerical Methods - Function Evaluations, Real & Complex Zeros

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

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

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

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

Use Newton's method to determine $x_{4}$ for $x^{9} + x^{6} + x^{4} + 73$ if $x_{0} = 4$ .

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 $x^{9} + x^{6} + x^{4} + 73$

$f'(x) = 9 x^{8} + 6 x^{5} + 4 x^{3}$

Now Newton's Method looks like.

$x_{n+1} = x_{n} - \frac{x_{n}^{9} + x_{n}^{6} + x_{n}^{4} + 73}{9 x_{n}^{8} + 6 x_{n}^{5} + 4 x_{n}^{3}}$

Now plug in 4 for $x_{n}$

$x_{1} = 3.55290461303134$

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.

$\x_{0} = 4 \x_1 = 3.55290461303134 \x_2 = 3.15455262559778 \x_3 = 2.79898902956631 \x_4 = 2.48035572128041$

So our final answer will be

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Function Evaluations, Real & Complex Zeros

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