Newtons method -x^2+4x-1 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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Newtons method -x^2+4x-1 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)

Mathematics
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  • phi
what are the options? perhaps you have a typo in the question
Sorry, 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
the 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.

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exactly what I got. I'm emailing her now. Thank you. This is the difference between an A and B on the exam
  • phi
The 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)
Email sent. Thank again for confirming. Thought I was going crazy

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