Find a root of an equation `f(x)=2x^3-2x-5` using Newton Raphson method

Solution:
Here `2x^3-2x-5=0`

Let `f(x) = 2x^3-2x-5`

`:. f'(x) = 6x^2-2`

Here

`x`	0	1	2
`f(x)`	-5	-5	7

Here `f(1) = -5 < 0 and f(2) = 7 > 0`

`:.` Root lies between `1` and `2`

`x_0 = (1 + 2)/2 = 1.5`


`1^(st)` iteration :

`f(x_0)=f(1.5)=2*1.5^(3)-2*1.5-5=-1.25`

`f'(x_0)=f'(1.5)=6*1.5^(2)-2=11.5`

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

`x_1 = 1.5 - (-1.25)/(11.5)`

`x_1 = 1.6087`


`2^(nd)` iteration :

`f(x_1)=f(1.6087)=2*1.6087^(3)-2*1.6087-5=0.1089`

`f'(x_1)=f'(1.6087)=6*1.6087^(2)-2=13.52741`

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

`x_2 = 1.6087 - (0.1089)/(13.52741)`

`x_2 = 1.60065`


`3^(rd)` iteration :

`f(x_2)=f(1.60065)=2*1.60065^(3)-2*1.60065-5=0.00062`

`f'(x_2)=f'(1.60065)=6*1.60065^(2)-2=13.37239`

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

`x_3 = 1.60065 - (0.00062)/(13.37239)`

`x_3 = 1.6006`


`4^(th)` iteration :

`f(x_3)=f(1.6006)=2*1.6006^(3)-2*1.6006-5=0`

`f'(x_3)=f'(1.6006)=6*1.6006^(2)-2=13.37149`

`x_4 = x_3 - f(x_3)/(f'(x_3))`

`x_4 = 1.6006 - (0)/(13.37149)`

`x_4 = 1.6006`


Approximate root of the equation `2x^3-2x-5=0` using Newton Raphson method is `1.6006`

`n`	`x_0`	`f(x_0)`	`f'(x_0)`	`x_1`	Update
1	1.5	-1.25	11.5	1.6087	`x_0 = x_1`
2	1.6087	0.1089	13.52741	1.60065	`x_0 = x_1`
3	1.60065	0.00062	13.37239	1.6006	`x_0 = x_1`
4	1.6006	0	13.37149	1.6006	`x_0 = x_1`

---

---