Find a root of an equation `f(x)=root(3)(48)` using Fixed Point Iteration method

Solution:
`x=root(3)(48)`

`:.x^3=48`

`:.x^3-48=0`

Here

`x`	0	1	2	3	4
`f(x)`	48	47	40	21	-16

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

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


`:.48-x^3=0`

Adding `10x` in both the sides, we get

`10x=48-x^3+10x`

`:.x=(48-x^3+10x)/10`

`:.phi(x)=(48-x^3+10x)/10`

`x_1 = phi(x_0) = phi(-0.5) = 4.3125`

`x_2 = phi(x_1) = phi(4.3125) = 1.09226`

`x_3 = phi(x_2) = phi(1.09226) = 5.76195`

`x_4 = phi(x_3) = phi(5.76195) = -8.56777`

`x_5 = phi(x_4) = phi(-8.56777) = 59.12531`

`x_6 = phi(x_5) = phi(59.12531) = -20605.1162`


The method lead us away from the solution, so it is divergent.

`n`	`x_0`	`x_1=phi(x_0)`	Update	Difference `|x_1-x_0|`
2	-0.5	4.3125	`x_0 = x_1`	4.8125
3	4.3125	1.09226	`x_0 = x_1`	3.22024
4	1.09226	5.76195	`x_0 = x_1`	4.66969
5	5.76195	-8.56777	`x_0 = x_1`	14.32972
6	-8.56777	59.12531	`x_0 = x_1`	67.69308
7	59.12531	-20605.1162	`x_0 = x_1`	20664.24151

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