Starting values for Newton's Method

[habeebe]

Use Newton's method to find all the roots of the equation correct to six decimal places. \(\displaystyle \frac{1}{x}=1+x^3\)

In the problem I had prior to this, my resulting function was factorable into the form \(\displaystyle a*b=0\), so I knew I had two roots, and could easily find starting values with the intermediate value theorem. With this problem, I worked my way up to \(\displaystyle f(x)=0=x^4+x-1=x(x^3+1)-1\). It looks like there are two roots, but the \(\displaystyle -1\) is throwing me off. I am not sure how to get my two functions for the two seperate roots, and because of that I can't find my \(\displaystyle x_1\)'s for either root.

How do I get to the next step?

 

[habeebe]

Nevermind. I got there through a circuitous route involving finding the minimum and taking the the limit as \(\displaystyle x\to\pm\infty\) to prove that I had two roots.

 

[fatemeh8989]

Hi;
I don't know about the newton's method but as a simple way you can have this:
1/x = 1+x^3
x^4 + x -1 =0
y(1)= x^4
y(2)=1-x
by plotting y(1) and y(2) these two functions will cross eachother in just two points which are the roots of first equation.