Find `root(3)(48)` using Secant 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 :

`x_0 = 3` and `x_1 = 4`

`f(x_0) = f(3) = -21` and `f(x_1) = f(4) = 16`

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

`x_2 = 3 - (-21) * (4 - 3)/(16 - (-21))`

`x_2 = 3.5676`

`:. f(x_2)=f(3.5676)=3.5676^3-48=-2.5936`


`2^(nd)` iteration :

`x_1 = 4` and `x_2 = 3.5676`

`f(x_1) = f(4) = 16` and `f(x_2) = f(3.5676) = -2.5936`

`:. x_3 = x_1 - f(x_1) * (x_2 - x_1)/(f(x_2) - f(x_1))`

`x_3 = 4 - 16 * (3.5676 - 4)/(-2.5936 - 16)`

`x_3 = 3.6279`

`:. f(x_3)=f(3.6279)=3.6279^3-48=-0.2513`


`3^(rd)` iteration :

`x_2 = 3.5676` and `x_3 = 3.6279`

`f(x_2) = f(3.5676) = -2.5936` and `f(x_3) = f(3.6279) = -0.2513`

`:. x_4 = x_2 - f(x_2) * (x_3 - x_2)/(f(x_3) - f(x_2))`

`x_4 = 3.5676 - (-2.5936) * (3.6279 - 3.5676)/(-0.2513 - (-2.5936))`

`x_4 = 3.6344`

`:. f(x_4)=f(3.6344)=3.6344^3-48=0.0047`


`4^(th)` iteration :

`x_3 = 3.6279` and `x_4 = 3.6344`

`f(x_3) = f(3.6279) = -0.2513` and `f(x_4) = f(3.6344) = 0.0047`

`:. x_5 = x_3 - f(x_3) * (x_4 - x_3)/(f(x_4) - f(x_3))`

`x_5 = 3.6279 - (-0.2513) * (3.6344 - 3.6279)/(0.0047 - (-0.2513))`

`x_5 = 3.6342`

`:. f(x_5)=f(3.6342)=3.6342^3-48=0`


Approximate root of the equation `x^3-48=0` using Secant method is `3.6342` (After 4 iterations)

`n`	`x_0`	`f(x_0)`	`x_1`	`f(x_1)`	`x_2`	`f(x_2)`	Update
1	3	-21	4	16	3.5676	-2.5936	`x_0 = x_1` `x_1 = x_2`
2	4	16	3.5676	-2.5936	3.6279	-0.2513	`x_0 = x_1` `x_1 = x_2`
3	3.5676	-2.5936	3.6279	-0.2513	3.6344	0.0047	`x_0 = x_1` `x_1 = x_2`
4	3.6279	-0.2513	3.6344	0.0047	3.6342	0	`x_0 = x_1` `x_1 = x_2`

---

---