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

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

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

`:.x^3+2x^2+x=1`

`:.x(x^2+2x+1)=1`

`:.x=(1)/(x^2+2x+1)`

`:.phi(x)=(1)/(x^2+2x+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_1 = phi(x_0) = phi(0.5) = 0.44444`

`x_2 = phi(x_1) = phi(0.44444) = 0.47929`

`x_3 = phi(x_2) = phi(0.47929) = 0.45698`

`x_4 = phi(x_3) = phi(0.45698) = 0.47108`

`x_5 = phi(x_4) = phi(0.47108) = 0.46209`

`x_6 = phi(x_5) = phi(0.46209) = 0.46779`

`x_7 = phi(x_6) = phi(0.46779) = 0.46416`

`x_8 = phi(x_7) = phi(0.46416) = 0.46647`

`x_9 = phi(x_8) = phi(0.46647) = 0.465`

`x_10 = phi(x_9) = phi(0.465) = 0.46593`

`x_11 = phi(x_10) = phi(0.46593) = 0.46534`

`x_12 = phi(x_11) = phi(0.46534) = 0.46572`


Approximate root of the equation `x^3+2x^2+x-1` using Iteration method is `0.46572` (After 12 iterations)

`n`	`x_0`	`x_1=phi(x_0)`	Update	Difference `|x_1-x_0|`
2	0.5	0.44444	`x_0 = x_1`	0.05556
3	0.44444	0.47929	`x_0 = x_1`	0.03485
4	0.47929	0.45698	`x_0 = x_1`	0.02231
5	0.45698	0.47108	`x_0 = x_1`	0.0141
6	0.47108	0.46209	`x_0 = x_1`	0.00899
7	0.46209	0.46779	`x_0 = x_1`	0.0057
8	0.46779	0.46416	`x_0 = x_1`	0.00363
9	0.46416	0.46647	`x_0 = x_1`	0.0023
10	0.46647	0.465	`x_0 = x_1`	0.00146
11	0.465	0.46593	`x_0 = x_1`	0.00093
12	0.46593	0.46534	`x_0 = x_1`	0.00059
13	0.46534	0.46572	`x_0 = x_1`	0.00038

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