`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