$f: \mathbb{R}\rightarrow \mathbb{R}$. Then determining if each of the solutions is a global maximum or a global minimum. $$f(x) = \frac{x^{4}}{4} - 2x^{3} + \frac{11}{2}x^{2} - 6x + 2$$ for all $x \in \mathbb{R}$.
I have used the synthetic division to find this equation: $$\left ( x^{2} - 5x + 6 \right )\left ( x - 1 \right ) = 0$$
So the critical points are $x^{*} = 1,2,3$.
I know that if $x^{*} = 1$, then ${f}''\left ( 1 \right ) = 2 > 0$, so this is minimum.
If $x^{*} = 2$, then ${f}''\left ( 2 \right ) = 1 < 0$, so this is maximum.
If $x^{*} = 3$, then ${f}''\left ( 3 \right ) = 2 > 0$, so this is minimum.
How do I determine whether these solutions are local max/min and global max/min?
Are all of them global solutions as $x \in \mathbb{R}$?
EDIT: I have plugged the values of the $x's$ in the main function.
For $x=1$ and $x=3$, $f(1)= -0.25 = f(3)$, and for $x=2$, $f(2)= 0$.
So is $x=1,3$ a local minimum and $x=2$ a global maximum?
From the second derivative test the extremum points that you have found are all local. Note that $\lim_{x\to \pm\infty}f(x)=+\infty$, so $x=1$ is not a global maximum point. On the other hand, since $f(1)=f(3)=-1/4$, it follows that $x=1$ and $x=3$ are global minimum points.