Find a root of an equation `f(x)=x^3+2x^2+x-1` using Bisection 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 `0` and `1`

`x_0=(0+1)/2=0.5`

`f(x_0)=f(0.5)=0.5^3+2*0.5^2+0.5-1=0.125 > 0`


`2^(nd)` iteration :

Here `f(0) = -1 < 0` and `f(0.5) = 0.125 > 0`

`:.` Now, Root lies between `0` and `0.5`

`x_1=(0+0.5)/2=0.25`

`f(x_1)=f(0.25)=0.25^3+2*0.25^2+0.25-1=-0.6094 < 0`


`3^(rd)` iteration :

Here `f(0.25) = -0.6094 < 0` and `f(0.5) = 0.125 > 0`

`:.` Now, Root lies between `0.25` and `0.5`

`x_2=(0.25+0.5)/2=0.375`

`f(x_2)=f(0.375)=0.375^3+2*0.375^2+0.375-1=-0.291 < 0`


`4^(th)` iteration :

Here `f(0.375) = -0.291 < 0` and `f(0.5) = 0.125 > 0`

`:.` Now, Root lies between `0.375` and `0.5`

`x_3=(0.375+0.5)/2=0.4375`

`f(x_3)=f(0.4375)=0.4375^3+2*0.4375^2+0.4375-1=-0.0959 < 0`


`5^(th)` iteration :

Here `f(0.4375) = -0.0959 < 0` and `f(0.5) = 0.125 > 0`

`:.` Now, Root lies between `0.4375` and `0.5`

`x_4=(0.4375+0.5)/2=0.4688`

`f(x_4)=f(0.4688)=0.4688^3+2*0.4688^2+0.4688-1=0.0112 > 0`


`6^(th)` iteration :

Here `f(0.4375) = -0.0959 < 0` and `f(0.4688) = 0.0112 > 0`

`:.` Now, Root lies between `0.4375` and `0.4688`

`x_5=(0.4375+0.4688)/2=0.4531`

`f(x_5)=f(0.4531)=0.4531^3+2*0.4531^2+0.4531-1=-0.0432 < 0`


`7^(th)` iteration :

Here `f(0.4531) = -0.0432 < 0` and `f(0.4688) = 0.0112 > 0`

`:.` Now, Root lies between `0.4531` and `0.4688`

`x_6=(0.4531+0.4688)/2=0.4609`

`f(x_6)=f(0.4609)=0.4609^3+2*0.4609^2+0.4609-1=-0.0162 < 0`


`8^(th)` iteration :

Here `f(0.4609) = -0.0162 < 0` and `f(0.4688) = 0.0112 > 0`

`:.` Now, Root lies between `0.4609` and `0.4688`

`x_7=(0.4609+0.4688)/2=0.4648`

`f(x_7)=f(0.4648)=0.4648^3+2*0.4648^2+0.4648-1=-0.0026 < 0`


`9^(th)` iteration :

Here `f(0.4648) = -0.0026 < 0` and `f(0.4688) = 0.0112 > 0`

`:.` Now, Root lies between `0.4648` and `0.4688`

`x_8=(0.4648+0.4688)/2=0.4668`

`f(x_8)=f(0.4668)=0.4668^3+2*0.4668^2+0.4668-1=0.0043 > 0`


`10^(th)` iteration :

Here `f(0.4648) = -0.0026 < 0` and `f(0.4668) = 0.0043 > 0`

`:.` Now, Root lies between `0.4648` and `0.4668`

`x_9=(0.4648+0.4668)/2=0.4658`

`f(x_9)=f(0.4658)=0.4658^3+2*0.4658^2+0.4658-1=0.0009 > 0`


`11^(th)` iteration :

Here `f(0.4648) = -0.0026 < 0` and `f(0.4658) = 0.0009 > 0`

`:.` Now, Root lies between `0.4648` and `0.4658`

`x_10=(0.4648+0.4658)/2=0.4653`

`f(x_10)=f(0.4653)=0.4653^3+2*0.4653^2+0.4653-1=-0.0008 < 0`


`12^(th)` iteration :

Here `f(0.4653) = -0.0008 < 0` and `f(0.4658) = 0.0009 > 0`

`:.` Now, Root lies between `0.4653` and `0.4658`

`x_11=(0.4653+0.4658)/2=0.4656`

`f(x_11)=f(0.4656)=0.4656^3+2*0.4656^2+0.4656-1=0 > 0`


Approximate root of the equation `x^3+2x^2+x-1=0` using Bisection method is `0.4656` (After 12 iterations)

`n`	`a`	`f(a)`	`b`	`f(b)`	`c=(a + b)/2`	`f(c)`	Update
1	0	-1	1	3	0.5	0.125	`b = c`
2	0	-1	0.5	0.125	0.25	-0.6094	`a = c`
3	0.25	-0.6094	0.5	0.125	0.375	-0.291	`a = c`
4	0.375	-0.291	0.5	0.125	0.4375	-0.0959	`a = c`
5	0.4375	-0.0959	0.5	0.125	0.4688	0.0112	`b = c`
6	0.4375	-0.0959	0.4688	0.0112	0.4531	-0.0432	`a = c`
7	0.4531	-0.0432	0.4688	0.0112	0.4609	-0.0162	`a = c`
8	0.4609	-0.0162	0.4688	0.0112	0.4648	-0.0026	`a = c`
9	0.4648	-0.0026	0.4688	0.0112	0.4668	0.0043	`b = c`
10	0.4648	-0.0026	0.4668	0.0043	0.4658	0.0009	`b = c`
11	0.4648	-0.0026	0.4658	0.0009	0.4653	-0.0008	`a = c`
12	0.4653	-0.0008	0.4658	0.0009	0.4656	0	`b = c`

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