Find a root of an equation `f(x)=2x^3-2x-5` using Fixed Point Iteration method

Solution:
Method-1
Let `f(x) = 2x^3-2x-5`

Here `2x^3-2x-5=0`

`:.2x^3=2x+5`

`:.x^3=(2x+5)/2`

`:.x=root (3)((2x+5)/2)`

`:.phi(x)=root (3)((2x+5)/2)`

Here

`x`	0	1	2
`f(x)`	-5	-5	7

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

`:.` Root lies between `1` and `2`

`x_0 = (1 + 2)/2 = 1.5`


`x_1 = phi(x_0) = phi(1.5) = 1.5874`

`x_2 = phi(x_1) = phi(1.5874) = 1.59888`

`x_3 = phi(x_2) = phi(1.59888) = 1.60037`

`x_4 = phi(x_3) = phi(1.60037) = 1.60057`


Approximate root of the equation `2x^3-2x-5` using Iteration method is `1.60057` (After 4 iterations)

`n`	`x_0`	`x_1=phi(x_0)`	Update	Difference `|x_1-x_0|`
2	1.5	1.5874	`x_0 = x_1`	0.0874
3	1.5874	1.59888	`x_0 = x_1`	0.01148
4	1.59888	1.60037	`x_0 = x_1`	0.0015
5	1.60037	1.60057	`x_0 = x_1`	0.00019

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