Get the maximum of a function with Newton Raphson

Question

I am having troubles with this exercise.

I have to get the maximum of this function on [-1,2] doing Newton Raphson:

$f(x) = 12 - 3x^4 - 2x^6$

So the derivatives are:

$f'(x) = -12x^3 - 12x^5$

$f''(x) = -36x^2 - 60x^4$

I don't know how to apply the hessian matrix and how to solve this problem. Could you help me?

Answer 1

$f'(x)=-12 x^3 (1 + x^2)=0$ at $x=0$

The first non zero derivative has an even order

$f^{(4)}(x)=-720 x^2-72$

Which is negative at $x=0$, so $x=0$ is a point of maximum and

$f(0)=12$ is the maximum attained by the function

Answer 2

solving the equation $$-12x^3-12x^5=0$$ we have $$-12x^3(1+x^2)=0$$ the only real solution is $$x=0$$