habeebe
New member
- Joined
- Nov 2, 2011
- Messages
- 2
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?
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?