Find `root(3)(48)` using Bisection method

Solution:
Let `x=48^(1/3)`

`:.x^3=48`

`:.x^3-48=0`

i.e. `f(x)=x^3-48`

Here

`x`	0	1	2	3	4
`f(x)`	-48	-47	-40	-21	16

`1^(st)` iteration :

Here `f(3) = -21 < 0` and `f(4) = 16 > 0`

`:.` Now, Root lies between `3` and `4`

`x_0=(3+4)/2=3.5`

`f(x_0)=f(3.5)=3.5^3-48=-5.125 < 0`


`2^(nd)` iteration :

Here `f(3.5) = -5.125 < 0` and `f(4) = 16 > 0`

`:.` Now, Root lies between `3.5` and `4`

`x_1=(3.5+4)/2=3.75`

`f(x_1)=f(3.75)=3.75^3-48=4.7344 > 0`


`3^(rd)` iteration :

Here `f(3.5) = -5.125 < 0` and `f(3.75) = 4.7344 > 0`

`:.` Now, Root lies between `3.5` and `3.75`

`x_2=(3.5+3.75)/2=3.625`

`f(x_2)=f(3.625)=3.625^3-48=-0.3652 < 0`


`4^(th)` iteration :

Here `f(3.625) = -0.3652 < 0` and `f(3.75) = 4.7344 > 0`

`:.` Now, Root lies between `3.625` and `3.75`

`x_3=(3.625+3.75)/2=3.6875`

`f(x_3)=f(3.6875)=3.6875^3-48=2.1414 > 0`


`5^(th)` iteration :

Here `f(3.625) = -0.3652 < 0` and `f(3.6875) = 2.1414 > 0`

`:.` Now, Root lies between `3.625` and `3.6875`

`x_4=(3.625+3.6875)/2=3.6562`

`f(x_4)=f(3.6562)=3.6562^3-48=0.8773 > 0`


`6^(th)` iteration :

Here `f(3.625) = -0.3652 < 0` and `f(3.6562) = 0.8773 > 0`

`:.` Now, Root lies between `3.625` and `3.6562`

`x_5=(3.625+3.6562)/2=3.6406`

`f(x_5)=f(3.6406)=3.6406^3-48=0.2534 > 0`


`7^(th)` iteration :

Here `f(3.625) = -0.3652 < 0` and `f(3.6406) = 0.2534 > 0`

`:.` Now, Root lies between `3.625` and `3.6406`

`x_6=(3.625+3.6406)/2=3.6328`

`f(x_6)=f(3.6328)=3.6328^3-48=-0.0566 < 0`


`8^(th)` iteration :

Here `f(3.6328) = -0.0566 < 0` and `f(3.6406) = 0.2534 > 0`

`:.` Now, Root lies between `3.6328` and `3.6406`

`x_7=(3.6328+3.6406)/2=3.6367`

`f(x_7)=f(3.6367)=3.6367^3-48=0.0982 > 0`


`9^(th)` iteration :

Here `f(3.6328) = -0.0566 < 0` and `f(3.6367) = 0.0982 > 0`

`:.` Now, Root lies between `3.6328` and `3.6367`

`x_8=(3.6328+3.6367)/2=3.6348`

`f(x_8)=f(3.6348)=3.6348^3-48=0.0208 > 0`


`10^(th)` iteration :

Here `f(3.6328) = -0.0566 < 0` and `f(3.6348) = 0.0208 > 0`

`:.` Now, Root lies between `3.6328` and `3.6348`

`x_9=(3.6328+3.6348)/2=3.6338`

`f(x_9)=f(3.6338)=3.6338^3-48=-0.0179 < 0`


`11^(th)` iteration :

Here `f(3.6338) = -0.0179 < 0` and `f(3.6348) = 0.0208 > 0`

`:.` Now, Root lies between `3.6338` and `3.6348`

`x_10=(3.6338+3.6348)/2=3.6343`

`f(x_10)=f(3.6343)=3.6343^3-48=0.0014 > 0`


`12^(th)` iteration :

Here `f(3.6338) = -0.0179 < 0` and `f(3.6343) = 0.0014 > 0`

`:.` Now, Root lies between `3.6338` and `3.6343`

`x_11=(3.6338+3.6343)/2=3.634`

`f(x_11)=f(3.634)=3.634^3-48=-0.0082 < 0`


`13^(th)` iteration :

Here `f(3.634) = -0.0082 < 0` and `f(3.6343) = 0.0014 > 0`

`:.` Now, Root lies between `3.634` and `3.6343`

`x_12=(3.634+3.6343)/2=3.6342`

`f(x_12)=f(3.6342)=3.6342^3-48=-0.0034 < 0`


`14^(th)` iteration :

Here `f(3.6342) = -0.0034 < 0` and `f(3.6343) = 0.0014 > 0`

`:.` Now, Root lies between `3.6342` and `3.6343`

`x_13=(3.6342+3.6343)/2=3.6342`

`f(x_13)=f(3.6342)=3.6342^3-48=-0.001 < 0`


`15^(th)` iteration :

Here `f(3.6342) = -0.001 < 0` and `f(3.6343) = 0.0014 > 0`

`:.` Now, Root lies between `3.6342` and `3.6343`

`x_14=(3.6342+3.6343)/2=3.6342`

`f(x_14)=f(3.6342)=3.6342^3-48=0.0002 > 0`


Approximate root of the equation `x^3-48=0` using Bisection method is `3.6342` (After 15 iterations)

`n`	`a`	`f(a)`	`b`	`f(b)`	`c=(a + b)/2`	`f(c)`	Update
1	3	-21	4	16	3.5	-5.125	`a = c`
2	3.5	-5.125	4	16	3.75	4.7344	`b = c`
3	3.5	-5.125	3.75	4.7344	3.625	-0.3652	`a = c`
4	3.625	-0.3652	3.75	4.7344	3.6875	2.1414	`b = c`
5	3.625	-0.3652	3.6875	2.1414	3.6562	0.8773	`b = c`
6	3.625	-0.3652	3.6562	0.8773	3.6406	0.2534	`b = c`
7	3.625	-0.3652	3.6406	0.2534	3.6328	-0.0566	`a = c`
8	3.6328	-0.0566	3.6406	0.2534	3.6367	0.0982	`b = c`
9	3.6328	-0.0566	3.6367	0.0982	3.6348	0.0208	`b = c`
10	3.6328	-0.0566	3.6348	0.0208	3.6338	-0.0179	`a = c`
11	3.6338	-0.0179	3.6348	0.0208	3.6343	0.0014	`b = c`
12	3.6338	-0.0179	3.6343	0.0014	3.634	-0.0082	`a = c`
13	3.634	-0.0082	3.6343	0.0014	3.6342	-0.0034	`a = c`
14	3.6342	-0.0034	3.6343	0.0014	3.6342	-0.001	`a = c`
15	3.6342	-0.001	3.6343	0.0014	3.6342	0.0002	`b = c`

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