Question

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.

Answer 1

second approximation is 0.6922 third is
\[0.6922-\frac{f(.6922)}{f'(.6922)}\] i would use a calculator

Answer 2

\[f'(x)=14x^6+12x^2\]
\[f'(.6922)=14(.6922)^6+12(.6922)^2\] who knows?

Answer 3

I come out with -0.0035, but that is not the right answer...

Answer 4

f' (.6922)=14(.6922)^ 6+12(.6922)^ 3
Did you plug into 12(.6922)^ 3 ?

Answer 5

It's supposedly power of 3 !

Answer 6

= 5.519924

Answer 7

you are right it should be
\[f'(.6922)=14(.6922)^6+12(.6922)^3\]

Answer 8

Of course, it's just a tiny typo :)

Answer 9

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

Answer 10

You missed "2" in 2x^7

Answer 11

Manually result is -.003644806

Answer 12

Yep, correct result is -.003644806