Find a root of an equation `f(x)=2x^3-2x-5` using False Position method (regula falsi method)

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

Let `f(x) = 2x^3-2x-5`

Here

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

`1^(st)` iteration :

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

`:.` Now, Root lies between `x_0 = 1` and `x_1 = 2`

`x_2 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_2 = 1 - (-5) * (2 - 1)/(7 - (-5))`

`x_2 = 1.41667`

`f(x_2)=f(1.41667)=2*1.41667^(3)-2*1.41667-5=-2.14699 < 0`


`2^(nd)` iteration :

Here `f(1.41667) = -2.14699 < 0` and `f(2) = 7 > 0`

`:.` Now, Root lies between `x_0 = 1.41667` and `x_1 = 2`

`x_3 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_3 = 1.41667 - (-2.14699) * (2 - 1.41667)/(7 - (-2.14699))`

`x_3 = 1.55359`

`f(x_3)=f(1.55359)=2*1.55359^(3)-2*1.55359-5=-0.60759 < 0`


`3^(rd)` iteration :

Here `f(1.55359) = -0.60759 < 0` and `f(2) = 7 > 0`

`:.` Now, Root lies between `x_0 = 1.55359` and `x_1 = 2`

`x_4 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_4 = 1.55359 - (-0.60759) * (2 - 1.55359)/(7 - (-0.60759))`

`x_4 = 1.58924`

`f(x_4)=f(1.58924)=2*1.58924^(3)-2*1.58924-5=-0.15063 < 0`


`4^(th)` iteration :

Here `f(1.58924) = -0.15063 < 0` and `f(2) = 7 > 0`

`:.` Now, Root lies between `x_0 = 1.58924` and `x_1 = 2`

`x_5 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_5 = 1.58924 - (-0.15063) * (2 - 1.58924)/(7 - (-0.15063))`

`x_5 = 1.59789`

`f(x_5)=f(1.59789)=2*1.59789^(3)-2*1.59789-5=-0.0361 < 0`


`5^(th)` iteration :

Here `f(1.59789) = -0.0361 < 0` and `f(2) = 7 > 0`

`:.` Now, Root lies between `x_0 = 1.59789` and `x_1 = 2`

`x_6 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_6 = 1.59789 - (-0.0361) * (2 - 1.59789)/(7 - (-0.0361))`

`x_6 = 1.59996`

`f(x_6)=f(1.59996)=2*1.59996^(3)-2*1.59996-5=-0.00858 < 0`


`6^(th)` iteration :

Here `f(1.59996) = -0.00858 < 0` and `f(2) = 7 > 0`

`:.` Now, Root lies between `x_0 = 1.59996` and `x_1 = 2`

`x_7 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_7 = 1.59996 - (-0.00858) * (2 - 1.59996)/(7 - (-0.00858))`

`x_7 = 1.60045`

`f(x_7)=f(1.60045)=2*1.60045^(3)-2*1.60045-5=-0.00203 < 0`


`7^(th)` iteration :

Here `f(1.60045) = -0.00203 < 0` and `f(2) = 7 > 0`

`:.` Now, Root lies between `x_0 = 1.60045` and `x_1 = 2`

`x_8 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_8 = 1.60045 - (-0.00203) * (2 - 1.60045)/(7 - (-0.00203))`

`x_8 = 1.60056`

`f(x_8)=f(1.60056)=2*1.60056^(3)-2*1.60056-5=-0.00048 < 0`


Approximate root of the equation `2x^3-2x-5=0` using False Position method is `1.60056`

`n`	`x_0`	`f(x_0)`	`x_1`	`f(x_1)`	`x_2`	`f(x_2)`	Update
1	1	-5	2	7	1.41667	-2.14699	`x_0 = x_2`
2	1.41667	-2.14699	2	7	1.55359	-0.60759	`x_0 = x_2`
3	1.55359	-0.60759	2	7	1.58924	-0.15063	`x_0 = x_2`
4	1.58924	-0.15063	2	7	1.59789	-0.0361	`x_0 = x_2`
5	1.59789	-0.0361	2	7	1.59996	-0.00858	`x_0 = x_2`
6	1.59996	-0.00858	2	7	1.60045	-0.00203	`x_0 = x_2`
7	1.60045	-0.00203	2	7	1.60056	-0.00048	`x_0 = x_2`

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