Minimum of $x^2 + 3x - 1$ on $[0,1]$ and $[-2,2]$ Consider the problem of finding the absolute minimum of the function $f : [0,1] \rightarrow \mathbb{R}$ that satisfies $f(x)=x^2 + 3x - 1$ everywhere.
Suppose we suspect, by graphical methods, that $f$ attains its absolute minimum at the point $0 \in [0,1]$. Thus its absolute minimum is believed to equal $f(0)=-1$.
How would you prove, in a rigorous and logically explicit manner, that $f$ really does achieve its absolute minimum at the point $0$ in its domain? Please, make your premises explicit and identify all the major theorems you use.
Consider also the function $g : [-2,2] \rightarrow \mathbb{R}$ with the same defining equation. Suppose you believe that $g$ attains its absolute minimum at $-3/2$ (which is a local minimum), thus its absolute minimum is $g(-3/2)=-13/4$. The same question applies: how would you confirm your suspicions?

For any $x\in [0,1]$, $f(x)=x^2+3x-1\geq 0+0-1=-1$.
For any $x\in [-2,2]$, $g(x)=x^2+3x-1=(x+\frac 32)^2-\frac{13}{4}\geq 0-\frac{13}{4}=-\frac{13}{4}$.
