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anonymous
 one year ago
Newtons method
x^2+4x1 find x2 x0=0 No matter what I do I do not get one of the options.
I know how to do this x1=x0 f(x)/f'(x)
anonymous
 one year ago
Newtons method x^2+4x1 find x2 x0=0 No matter what I do I do not get one of the options. I know how to do this x1=x0 f(x)/f'(x)

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phi
 one year ago
Best ResponseYou've already chosen the best response.1what are the options? perhaps you have a typo in the question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, left hand. It was on a final so its a moot point now, but .25, .23, .33 and .08 i went with .23 which I know is wrong, but its the only thing I could come to and I had to add f(x)/f'(x) in order to get it. Its just bothering me now.

phi
 one year ago
Best ResponseYou've already chosen the best response.1the first iteration gives \(x_1= 0.25 \) the second iteration gives \[ 0.25  \frac{0.25^2 + 1 1 }{0.5+4} =0.25  \frac{0.0625}{3.5} \\ = 0.25+ 0.017857 = 0.267857 \] I would talk to your teacher, because the question has a problem.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0exactly what I got. I'm emailing her now. Thank you. This is the difference between an A and B on the exam

phi
 one year ago
Best ResponseYou've already chosen the best response.1The exact root is \( 2\sqrt{3} \approx 0.267949 \) and we should see the 2nd iteration close to that value (especially when rounded to two digits)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Email sent. Thank again for confirming. Thought I was going crazy
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