Use Newton's method with the specified initial approximation x 1 to find x 3, the third approximation to the root of the given equation. (Round your answer to four decimal places).

x^3 + 4x − 3 = 0, x 1 = 1

Question

Use Newton's method with the specified initial approximation x 1 to find x 3, the third approximation to the root of the given equation. (Round your answer to four decimal places).

x^3 + 4x − 3 = 0, x 1 = 1

zale101 · Answer

Newton's method: 

$x_2=x_1-\Large\frac{f(x_1)}{f'(x_1)}$

$x_3=x_2-\Large\frac{f(x_2)}{f'(x_2)}$
 and so on.

Makes sense? 
So you have x^3 + 4x − 3 = 0, x 1 = 1

anonymous · Answer

okay I understand that part

zale101 · Answer

to find $x_2$, you must take whatever the number of $x_1$, which is 1, subtract it by the f(1), divided by the derivative f'(1).

zale101 · Answer

So $f(x)=x^3 + 4x − 3 $  and  $x_1=1$

$x_2=1-\Large\frac{f(1)}{f'(1)}$

dan815 · Answer

Hi would you like a more detailed explanation of newtons method