Question

Find the positive root of x3-6x+4=0 by Newton Raphson method

Answer 1

Answer:

The correct answer is o.7301

Explanation:

f(x)=$x^{3} -6x+4$

By newton Raphson method,we have to find value of xn so that f(xn) = 0.

$df(x)/dx=3x^{2} -6$

f(0)=+4 and $df(0)/dx=-6$

the further value of x will be 0- 4/-6 =2/3

$f(2/3)=0.296$

the next value x is x2

$f(2/3)=0.296 and df(2/3)/dx=-14/3$

therefore,x2=$2/3 - 0.296/-14/3$

               x2=0.7301

f(0.730)=0.00840≅0

therefore,the answer is0.7301