Find a root of an equation `f(x)=x^3+2x^2+x-1` using False Position method (regula falsi method)

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

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

Here

`x`	0	1
`f(x)`	-1	3

`1^(st)` iteration :

Here `f(0) = -1 < 0` and `f(1) = 3 > 0`

`:.` Now, Root lies between `x_0 = 0` and `x_1 = 1`

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

`x_2=0 - (-1) * (1 - 0)/(3 - (-1))`

`x_2=0.25`

`f(x_2)=f(0.25)=0.25^3+2*0.25^2+0.25-1=-0.6094 < 0`


`2^(nd)` iteration :

Here `f(0.25) = -0.6094 < 0` and `f(1) = 3 > 0`

`:.` Now, Root lies between `x_0 = 0.25` and `x_1 = 1`

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

`x_3=0.25 - (-0.6094) * (1 - 0.25)/(3 - (-0.6094))`

`x_3=0.3766`

`f(x_3)=f(0.3766)=0.3766^3+2*0.3766^2+0.3766-1=-0.2863 < 0`


`3^(rd)` iteration :

Here `f(0.3766) = -0.2863 < 0` and `f(1) = 3 > 0`

`:.` Now, Root lies between `x_0 = 0.3766` and `x_1 = 1`

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

`x_4=0.3766 - (-0.2863) * (1 - 0.3766)/(3 - (-0.2863))`

`x_4=0.4309`

`f(x_4)=f(0.4309)=0.4309^3+2*0.4309^2+0.4309-1=-0.1177 < 0`


`4^(th)` iteration :

Here `f(0.4309) = -0.1177 < 0` and `f(1) = 3 > 0`

`:.` Now, Root lies between `x_0 = 0.4309` and `x_1 = 1`

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

`x_5=0.4309 - (-0.1177) * (1 - 0.4309)/(3 - (-0.1177))`

`x_5=0.4524`

`f(x_5)=f(0.4524)=0.4524^3+2*0.4524^2+0.4524-1=-0.0457 < 0`


`5^(th)` iteration :

Here `f(0.4524) = -0.0457 < 0` and `f(1) = 3 > 0`

`:.` Now, Root lies between `x_0 = 0.4524` and `x_1 = 1`

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

`x_6=0.4524 - (-0.0457) * (1 - 0.4524)/(3 - (-0.0457))`

`x_6=0.4606`

`f(x_6)=f(0.4606)=0.4606^3+2*0.4606^2+0.4606-1=-0.0173 < 0`


`6^(th)` iteration :

Here `f(0.4606) = -0.0173 < 0` and `f(1) = 3 > 0`

`:.` Now, Root lies between `x_0 = 0.4606` and `x_1 = 1`

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

`x_7=0.4606 - (-0.0173) * (1 - 0.4606)/(3 - (-0.0173))`

`x_7=0.4637`

`f(x_7)=f(0.4637)=0.4637^3+2*0.4637^2+0.4637-1=-0.0065 < 0`


`7^(th)` iteration :

Here `f(0.4637) = -0.0065 < 0` and `f(1) = 3 > 0`

`:.` Now, Root lies between `x_0 = 0.4637` and `x_1 = 1`

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

`x_8=0.4637 - (-0.0065) * (1 - 0.4637)/(3 - (-0.0065))`

`x_8=0.4649`

`f(x_8)=f(0.4649)=0.4649^3+2*0.4649^2+0.4649-1=-0.0024 < 0`


`8^(th)` iteration :

Here `f(0.4649) = -0.0024 < 0` and `f(1) = 3 > 0`

`:.` Now, Root lies between `x_0 = 0.4649` and `x_1 = 1`

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

`x_9=0.4649 - (-0.0024) * (1 - 0.4649)/(3 - (-0.0024))`

`x_9=0.4653`

`f(x_9)=f(0.4653)=0.4653^3+2*0.4653^2+0.4653-1=-0.0009 < 0`


`9^(th)` iteration :

Here `f(0.4653) = -0.0009 < 0` and `f(1) = 3 > 0`

`:.` Now, Root lies between `x_0 = 0.4653` and `x_1 = 1`

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

`x_10=0.4653 - (-0.0009) * (1 - 0.4653)/(3 - (-0.0009))`

`x_10=0.4655`

`f(x_10)=f(0.4655)=0.4655^3+2*0.4655^2+0.4655-1=-0.0003 < 0`


Approximate root of the equation `x^3+2x^2+x-1=0` using False Position method is `0.4655` (After 9 iterations)

`n`	`x_0`	`f(x_0)`	`x_1`	`f(x_1)`	`x_2`	`f(x_2)`	Update
1	0	-1	1	3	0.25	-0.6094	`x_0 = x_2`
2	0.25	-0.6094	1	3	0.3766	-0.2863	`x_0 = x_2`
3	0.3766	-0.2863	1	3	0.4309	-0.1177	`x_0 = x_2`
4	0.4309	-0.1177	1	3	0.4524	-0.0457	`x_0 = x_2`
5	0.4524	-0.0457	1	3	0.4606	-0.0173	`x_0 = x_2`
6	0.4606	-0.0173	1	3	0.4637	-0.0065	`x_0 = x_2`
7	0.4637	-0.0065	1	3	0.4649	-0.0024	`x_0 = x_2`
8	0.4649	-0.0024	1	3	0.4653	-0.0009	`x_0 = x_2`
9	0.4653	-0.0009	1	3	0.4655	-0.0003	`x_0 = x_2`

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