Find `root(3)(48)` using Newton Raphson method

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

`:.x^3=48`

`:.x^3-48=0`

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

`d/(dx)(x^3-48)=3x^2`

`d/(dx)(x^3-48)`

`=d/(dx)(x^3)-d/(dx)(48)`

`=3x^2-0`

`=3x^2`

`:. f'(x) = 3x^2`

Here

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

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

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

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

`x_0 = 3.5`


`1^(st)` iteration :

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

`f'(x_0)=f'(3.5)=3*3.5^2=36.75`

`x_1 = x_0 - f(x_0)/(f'(x_0))`

`x_1=3.5 - (-5.125)/(36.75)`

`x_1=3.6395`


`2^(nd)` iteration :

`f(x_1)=f(3.6395)=3.6395^3-48=0.2069`

`f'(x_1)=f'(3.6395)=3*3.6395^2=39.7369`

`x_2 = x_1 - f(x_1)/(f'(x_1))`

`x_2=3.6395 - (0.2069)/(39.7369)`

`x_2=3.6342`


`3^(rd)` iteration :

`f(x_2)=f(3.6342)=3.6342^3-48=0.0003`

`f'(x_2)=f'(3.6342)=3*3.6342^2=39.6233`

`x_3 = x_2 - f(x_2)/(f'(x_2))`

`x_3=3.6342 - (0.0003)/(39.6233)`

`x_3=3.6342`


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

`n`	`x_0`	`f(x_0)`	`f'(x_0)`	`x_1`	Update
1	3.5	-5.125	36.75	3.6395	`x_0 = x_1`
2	3.6395	0.2069	39.7369	3.6342	`x_0 = x_1`
3	3.6342	0.0003	39.6233	3.6342	`x_0 = x_1`

---

---