Take for example: $f(x) = \frac{x^4}{4} - 2x^3 + \frac{11x^2}{2}-6x + 2$, for all $x \in R$
The critical points are 1,2, and 3.
1 and 3 are local minimums and 2 is a local maximum from 2nd derivative test.
I can draw a graph and look at the intervals in between and on the sides of the critical points to deduce the graph shape: I found out that x = 2 is a global min and that there is no global max.
The problem with this method is that the more critical points there are the more I have to crunch numbers in my calculator to see how the graph is shaped around critical points.
Is there another way to reach the same conclusion that avoids this graphical way of doing things? For instance could I take the limit and deduce something from that, or maybe a more rigorous way?
Since$$\lim_{x\to\pm\infty}f(x)=\infty,\tag1$$$f$ cannot have a global maximum. But $(1)$ also implies that $f$ has a global minimum: Just take $R>0$ such that $|x|>R\implies|f(x)|>|f(0)|$. Then the minimum of $f|_{[-R,R]}$ has to be a global minimum.