



 
Hi Nancy, You have an error in the iteration step for Newton's Method, it should be
or as I want to write it \[x_{n +1} = x_{n} \frac{f(x_n)}{f '(x_n)}\] With your function this becomes \[x_{n +1} = x_{n} \frac{x_n^5 +2x_n^2 +3}{5x_n^4 +4x_n}\] To get started with Newton's Method you need to select an initial value $x_0.$ Newton's Method works best if the starting value is close to the root you seeking. You might just think, why not just start with $x_0 = 0.$ Unfortunately $f'(0) = 0$ and $f'(x_0)$ is in the denominator so that won't work. Let's be a little more careful. Looking at the function $f(x) = x^5 +2x^2 +3$ you can see that if $x$ is positive so is $f(x).$ $f(0)$ is also positive so any root must be negative. $f(1)$ is also positive but $f(2)$ is negative so there is a root between $1$ and $2.$ Hence it seems like starting with $x_0 = 1$ or $x_0 = 2$ would be good choices but again $f'(1) = 0$ so the remaining choice is $x_0 = 2.$ The iterative step with $x_0 = 2$ gives \begin{eqnarray*} Now find $x_2$ using the iterative step \[x_{2} = x_{1} \frac{x_1^5 +2x_1^2 +3}{5x_1^4 +4x_1}\] Penny  


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