`f(x) =` `x^2+sinx`

`f'(x) = 2x+cosx`

At absolute minimum/maximum f'(x) = 0

Let `f'(x) = p(x)`

Then;

`p'(x) = 2-sinx`

Using Newton's method for the first approximation `x_0` we will get a better approximation `x_1` by;

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

Let us say `x_0 = 30`

Then we can get `x_1` as;

`x_1 = 30-(2xx30+cos30)/(2-sin30) = -10.5774`

Now the second approximation `x_2` can be given by;

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

`x_2 = -10.5774-(2xx-10.5774+cos(-10.5774))/(2-sin(-10.5774)) `

`x_2 = -1.3394`

Similarly we can get;

`x_3 = -0.0509`

`x_4 = -0.5000`

`x_5 = -0.4999`

`x_6 = -0.4999`

So at `x = -0.4999` we have a stationary point.

`f'(x) = 2x+cosx`

`f''(x) = 2-sinx`

`f''(x)_((x = -0.4999)) = 2.0087 > 0`

So f(x) has a minimum at x = -0.4999

So the absolute minimum of f(x) occurs at x = -0.4999