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 $$x_0=1$$


for d)






is that correct ???

  • $\begingroup$ I would like to add that Newton's method does not, in general, give the roots of a polynomial. It approximates it to any desired degree of accuracy (doubling in decimal accuracy roughly each iteration) but it won't find the EXACT solution. $\endgroup$ – Trogdor Jun 15 '15 at 7:03
  • $\begingroup$ The goal of (a) and (b) is to see that $f(0)<0$ and $f(1)>0$, so there is a root of the polynomial in the interval $[0,1]$. Once you know this, you can choose a point $x_0$ accordingly, (c). $\endgroup$ – hjhjhj57 Jun 15 '15 at 7:07

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.


a) You have to determine $f(0)$, that is $f(0) = 2(0)^3+5(0)-6=\cdots$.

b) You have to determine $f(1)$. See above.

c) You have to guess a number, say $x_{0}=1$, since it's close to the zero point.

d) You have to solve $f(x)=0$ using the Newton's method. Since $f'(x_{0})=6x^{2}+5$, you have $$x_{1} = x_0 -\frac{f(x_0 )}{f'(x_0 )} = 10/11$$ where $x_{0}=1$. Find also $x_{2}$, $x_{3}$ etc.


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