Find a root of an equation `f(x)=x^3+2x^2+x-1` using Newton Raphson method

Solution:
Here `x^3+2x^2+x-1=0`

Let `f(x) = x^3+2x^2+x-1`

`d/(dx)(x^3+2x^2+x-1)=3x^2+4x+1`

`d/(dx)(x^3+2x^2+x-1)`

`=d/(dx)(x^3)+d/(dx)(2x^2)+d/(dx)(x)-d/(dx)(1)`

`=3x^2+4x+1-0`

`=3x^2+4x+1`

`:. f'(x) = 3x^2+4x+1`

Here

`x`	0	1
`f(x)`	-1	3

Here `f(0) = -1 < 0 and f(1) = 3 > 0`

`:.` Root lies between `0` and `1`

`x_0 = (0 + 1)/2 = 0.5`

`x_0 = 0.5`


`1^(st)` iteration :

`f(x_0)=f(0.5)=0.5^3+2*0.5^2+0.5-1=0.125`

`f'(x_0)=f'(0.5)=3*0.5^2+4*0.5+1=3.75`

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

`x_1=0.5 - (0.125)/(3.75)`

`x_1=0.4667`


`2^(nd)` iteration :

`f(x_1)=f(0.4667)=0.4667^3+2*0.4667^2+0.4667-1=0.0039`

`f'(x_1)=f'(0.4667)=3*0.4667^2+4*0.4667+1=3.52`

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

`x_2=0.4667 - (0.0039)/(3.52)`

`x_2=0.4656`


`3^(rd)` iteration :

`f(x_2)=f(0.4656)=0.4656^3+2*0.4656^2+0.4656-1=0`

`f'(x_2)=f'(0.4656)=3*0.4656^2+4*0.4656+1=3.5126`

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

`x_3=0.4656 - (0)/(3.5126)`

`x_3=0.4656`


Approximate root of the equation `x^3+2x^2+x-1=0` using Newton Raphson method is `0.4656` (After 3 iterations)

`n`	`x_0`	`f(x_0)`	`f'(x_0)`	`x_1`	Update
1	0.5	0.125	3.75	0.4667	`x_0 = x_1`
2	0.4667	0.0039	3.52	0.4656	`x_0 = x_1`
3	0.4656	0	3.5126	0.4656	`x_0 = x_1`

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