Question

Obtain the root of `x^3-x-1=0` by Newton Raphson Method` (upto three decimal places). 

Answer 1

Equation :`x^3-2x-5=0`

∴ `f(x)=x^3-2x-5`

`f(0)=-5<0 and f(1)=-2<0 and f(2)=7> 0`  

Root of given eqn lies between 1 and 2.

`f,(x)=3x^2+2`

Let take `x_0=2` 

`x_1=x_0-f(x_0)/(f'(x_0))`

=`2-7/14=1.5`

Next iteration : 

∴` x_2=x_1-f(x_1)/(f'(x_1))`

= 1.343 

∴ `x_3=x^2-f(x_2)/(f'(x^2))=1.329`

For next iteration : 

∴ `x_4=x_3-f(x_3)/(f'(x^3))=1.329-f(1.329)/(f'(1.329))`

= 1.3283 

The root of eqn is `  x = 1.3283`