Use Newton's method to find all roots of the equation correct to six decimal places? Square root (x+1)=x^2-x

aruv | Student

Let us define a function


and find its derivative with respsct to x

`f'(x)=2x-1-(1/2)1/sqrt(x+1)` ,

from above expression we conclude that if x=-1 ,then f'(x) is not defined.

The Newton method is

`x_(k+1)=x_k-f(x_k)/(f'(x_k))`    , prvided `f'(x_k)!=0`  for any `x_k.`

Now draw the graph of f(x) to estimate the initial approximation

From graph we observe that roots of the given equation lies in       (-1,0) and  (1,2).

1. let approximation of first root be -.5 i.e `x_0=-.5`






`` `f'(-.4841553)=2(-.4841553)-1-.5/sqrt(-.4841553+1)`



Simlarly second root can be calculated which is near t the point x=2.