<s>[INST] Use Newton’s method to find the solution of $2x^3+5x-6=0$ by using the following steps Use Newton’s method to find the solution of  $$2x^3+5x-6=0$$
by using the following steps
a)  Find, $f(0)$  where $$f(x)=2x^3+5x-6$$
b)  Find, $f(1)$  where $$f(x)=2x^3+5x-6$$
c)  Choose a guess point  $$x_0$$
d)  Perform the iterations, until the values stabilize to the 8-digits after the decimal sign.
my work
for a)
newton formula
$$x_{n+1}  = x_n -\frac{f(x_n )}{f'(x_n )}   $$ ,      for $n = 0,1,2,3,...$
$$f' (x)=6x^2+5$$
$$x_1=x_0-\frac{-6}{5}  $$
$$ F(0)= x_0- \frac{-6}{5}  $$
for b)
$$x_2=x_1-\frac{-6}{5} $$
for c)
Let $$x_0=1$$
$$x_1=1-\frac{-6}{5}=\frac{11}{5}$$
for d)
$$x_1=x_0-\frac{-6}{5}$$
$$x_2=x_1-\frac{-6}{5}$$
$$x_3=x_2-\frac{-6}{5}$$
$$x_4=x_3-\frac{-6}{5}$$
$$x_5=x_4-\frac{-6}{5}$$
is that correct ???
 [/INST] As you wrote, Newton method updates the initial guess $x_0$ according to $$x_{n+1}  = x_n -\frac{f(x_n )}{f'(x_n )}$$ So, if $f(x)=2 x^3+5 x-6$, $f'(x)=6 x^2+5$, this gives, after simplification $$x_{n+1} =\frac{4 x_n^3+6}{6 x_n^2+5}$$ As AjmalW answered, choose $x_0=1$ and start repeating the iterations. You will find $x_1=\frac{10}{11}$, $x_2=\frac{11986}{13255}$ and so on until the convergence criteria is met.
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