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.

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.

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

It's supposedly power of 3 !

anonymous · Answer

= 5.519924

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

anonymous · Answer

You missed "2" in 2x^7

anonymous · Answer

Manually result is -.003644806

anonymous · Answer

Yep, correct result is -.003644806