Find `sqrt(12)` using Newton Raphson method

Solution:
Let `x=sqrt(12)`

`:.x^2=12`

`:.x^2-12=0`

i.e. `f(x)=x^2-12`

`d/(dx)(x^2-12)=2x`

`d/(dx)(x^2-12)`

`=d/(dx)(x^2)-d/(dx)(12)`

`=2x-0`

`=2x`

`:. f'(x) = 2x`

Here

`x`	0	1	2	3	4
`f(x)`	-12	-11	-8	-3	4

Here `f(3) = -3 < 0 and f(4) = 4 > 0`

`:.` Root lies between `3` and `4`

`x_0 = (3 + 4)/2 = 3.5`

`x_0 = 3.5`


`1^(st)` iteration :

`f(x_0)=f(3.5)=3.5^2-12=0.25`

`f'(x_0)=f'(3.5)=2*3.5=7`

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

`x_1=3.5 - (0.25)/(7)`

`x_1=3.4643`


`2^(nd)` iteration :

`f(x_1)=f(3.4643)=3.4643^2-12=0.0013`

`f'(x_1)=f'(3.4643)=2*3.4643=6.9286`

`x_2 = x_1 - f(x_1)/(f'(x_1))`

`x_2=3.4643 - (0.0013)/(6.9286)`

`x_2=3.4641`


`3^(rd)` iteration :

`f(x_2)=f(3.4641)=3.4641^2-12=0`

`f'(x_2)=f'(3.4641)=2*3.4641=6.9282`

`x_3 = x_2 - f(x_2)/(f'(x_2))`

`x_3=3.4641 - (0)/(6.9282)`

`x_3=3.4641`


Approximate root of the equation `x^2-12=0` using Newton Raphson method is `3.4641` (After 3 iterations)

`n`	`x_0`	`f(x_0)`	`f'(x_0)`	`x_1`	Update
1	3.5	0.25	7	3.4643	`x_0 = x_1`
2	3.4643	0.0013	6.9286	3.4641	`x_0 = x_1`
3	3.4641	0	6.9282	3.4641	`x_0 = x_1`

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