1. Find a root of an equation `f(x)=x^3-x-1` using False Position method

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

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

Here

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

`1^(st)` iteration :

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

`x_2 = 1.16667`

`f(x_2) = f(1.16667) = -0.5787 < 0`


`2^(nd)` iteration :

Here `f(1.16667) = -0.5787 < 0` and `f(2) = 5 > 0`

`:.` Now, Root lies between `x_0 = 1.16667` and `x_1 = 2`

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

`x_3 = 1.16667 - (-0.5787) * (2 - 1.16667)/(5 - (-0.5787))`

`x_3 = 1.25311`

`f(x_3) = f(1.25311) = -0.28536 < 0`


`3^(rd)` iteration :

Here `f(1.25311) = -0.28536 < 0` and `f(2) = 5 > 0`

`:.` Now, Root lies between `x_0 = 1.25311` and `x_1 = 2`

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

`x_4 = 1.25311 - (-0.28536) * (2 - 1.25311)/(5 - (-0.28536))`

`x_4 = 1.29344`

`f(x_4) = f(1.29344) = -0.12954 < 0`


`4^(th)` iteration :

Here `f(1.29344) = -0.12954 < 0` and `f(2) = 5 > 0`

`:.` Now, Root lies between `x_0 = 1.29344` and `x_1 = 2`

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

`x_5 = 1.29344 - (-0.12954) * (2 - 1.29344)/(5 - (-0.12954))`

`x_5 = 1.31128`

`f(x_5) = f(1.31128) = -0.05659 < 0`


`5^(th)` iteration :

Here `f(1.31128) = -0.05659 < 0` and `f(2) = 5 > 0`

`:.` Now, Root lies between `x_0 = 1.31128` and `x_1 = 2`

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

`x_6 = 1.31128 - (-0.05659) * (2 - 1.31128)/(5 - (-0.05659))`

`x_6 = 1.31899`

`f(x_6) = f(1.31899) = -0.0243 < 0`


`6^(th)` iteration :

Here `f(1.31899) = -0.0243 < 0` and `f(2) = 5 > 0`

`:.` Now, Root lies between `x_0 = 1.31899` and `x_1 = 2`

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

`x_7 = 1.31899 - (-0.0243) * (2 - 1.31899)/(5 - (-0.0243))`

`x_7 = 1.32228`

`f(x_7) = f(1.32228) = -0.01036 < 0`


`7^(th)` iteration :

Here `f(1.32228) = -0.01036 < 0` and `f(2) = 5 > 0`

`:.` Now, Root lies between `x_0 = 1.32228` and `x_1 = 2`

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

`x_8 = 1.32228 - (-0.01036) * (2 - 1.32228)/(5 - (-0.01036))`

`x_8 = 1.32368`

`f(x_8) = f(1.32368) = -0.0044 < 0`


`8^(th)` iteration :

Here `f(1.32368) = -0.0044 < 0` and `f(2) = 5 > 0`

`:.` Now, Root lies between `x_0 = 1.32368` and `x_1 = 2`

`x_9 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_9 = 1.32368 - (-0.0044) * (2 - 1.32368)/(5 - (-0.0044))`

`x_9 = 1.32428`

`f(x_9) = f(1.32428) = -0.00187 < 0`


`9^(th)` iteration :

Here `f(1.32428) = -0.00187 < 0` and `f(2) = 5 > 0`

`:.` Now, Root lies between `x_0 = 1.32428` and `x_1 = 2`

`x_10 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_10 = 1.32428 - (-0.00187) * (2 - 1.32428)/(5 - (-0.00187))`

`x_10 = 1.32453`

`f(x_10) = f(1.32453) = -0.00079 < 0`


`10^(th)` iteration :

Here `f(1.32453) = -0.00079 < 0` and `f(2) = 5 > 0`

`:.` Now, Root lies between `x_0 = 1.32453` and `x_1 = 2`

`x_11 = x_0 - f(x_0) * (x_1 - x_0)/(f(x_1) - f(x_0))`

`x_11 = 1.32453 - (-0.00079) * (2 - 1.32453)/(5 - (-0.00079))`

`x_11 = 1.32464`

`f(x_11) = f(1.32464) = -0.00034 < 0`


Approximate root of the equation `x^3-x-1=0` using False Position method is `1.32464`

`n`	`x_0`	`f(x_0)`	`x_1`	`f(x_1)`	`x_2`	`f(x_2)`	Update
1	1	-1	2	5	1.16667	-0.5787	`x_0 = x_2`
2	1.16667	-0.5787	2	5	1.25311	-0.28536	`x_0 = x_2`
3	1.25311	-0.28536	2	5	1.29344	-0.12954	`x_0 = x_2`
4	1.29344	-0.12954	2	5	1.31128	-0.05659	`x_0 = x_2`
5	1.31128	-0.05659	2	5	1.31899	-0.0243	`x_0 = x_2`
6	1.31899	-0.0243	2	5	1.32228	-0.01036	`x_0 = x_2`
7	1.32228	-0.01036	2	5	1.32368	-0.0044	`x_0 = x_2`
8	1.32368	-0.0044	2	5	1.32428	-0.00187	`x_0 = x_2`
9	1.32428	-0.00187	2	5	1.32453	-0.00079	`x_0 = x_2`
10	1.32453	-0.00079	2	5	1.32464	-0.00034	`x_0 = x_2`