Question

Use Newton's method to find 3x^2+2x-1=0, x0=1 Find right hand solution

Answer 1

\[\large f'(x) = 6x+2\]

Answer 2

newtons method:
\[\large x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\]
\begin{array}{ l |c | c | c | r }
n& x_n & f(x_n) & f'(x_n) & x_{n+1} = x_n -\frac{f(x_n)}{f'(x_n)}\\
\hline
0 & 1 & 4& 8 & 0.5 \\
1 & 0.5 & 0.75& 5& 0.35\\
2& 0.35&0.0675 & 4.1&0.33354\\
3& 0.33354&0.000826\\
\hline
\end{array}
how close to 0 you you need to be?

Answer 3

I need to find right hand solution

Answer 4

Ok. This quadratic has two solutions, one positive (right-hand sol) and a negative solution (left-hand sol). The above method is finding the positive solution. However, since it is a numerical method, you will need to specify how close to the actual solution you want to be.
For example at x = 0.33354, f(0.33354) = 0.000826 which is within 10^-3 of 0.

Answer 5

got it, thanks. can you help me on some other problems?