Global min/max verification

Question

Take $x^3-3x$ for example, would $x=1$ be a local min as well as a global min if we consider the range of the domain to be all of $x \in \mathbb{R}$ as min/max must be at a finite $x$?

Answer 1

For $f(x) = x^3 - 3x$ the point $x = 1$ is a local minimizer, but not a global minimizer. The reason for this is that in a small region around $x = 1$, $f(1) =-2$ is the smallest value, meaning it is a local min. The reason it is not a global min is because there are points where the $y$-value is smaller than $-2$. For example, $f(-3) = -18 < -2$, which means that $f(1) = -2$ cannot be the global min. In this case, the function has no global min or global max, because we have $$\lim_{x\to-\infty}f(x) = -\infty \quad \text{and}\quad \lim_{x\to\infty}f(x) = \infty.$$