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)

Question

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)

phi · Answer

what are the options?  perhaps you have a typo in the question

anonymous · Answer

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 · Answer

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.

anonymous · Answer

exactly what I got.  I'm emailing her now.  Thank you.  This is the difference between an A and B on the exam

phi · Answer

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)

anonymous · Answer

Email sent.  Thank again for confirming.  Thought I was going crazy