anonymous
  • anonymous
What is the third approximation of a root of 2x^7+3x^4+3=0, if the second approximation is .6922. Use Newton's Method.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
second approximation is 0.6922 third is \[0.6922-\frac{f(.6922)}{f'(.6922)}\] i would use a calculator
anonymous
  • anonymous
\[f'(x)=14x^6+12x^2\] \[f'(.6922)=14(.6922)^6+12(.6922)^2\] who knows?
anonymous
  • anonymous
I come out with -0.0035, but that is not the right answer...

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anonymous
  • anonymous
f' (.6922)=14(.6922)^ 6+12(.6922)^ 3 Did you plug into 12(.6922)^ 3 ?
anonymous
  • anonymous
It's supposedly power of 3 !
anonymous
  • anonymous
= 5.519924
anonymous
  • anonymous
you are right it should be \[f'(.6922)=14(.6922)^6+12(.6922)^3\]
anonymous
  • anonymous
Of course, it's just a tiny typo :)
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=.6922-%28%28.6922%29^7%2B3%28.6922%29^3%2B3%29%2F%2814%28.6922%29^6%2B12%28.6922%29^3%29
anonymous
  • anonymous
You missed "2" in 2x^7
anonymous
  • anonymous
Manually result is -.003644806
anonymous
  • anonymous
Yep, correct result is -.003644806

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