TITLE: Minimum of $x^2 + 3x - 1$ on $[0,1]$ and $[-2,2]$
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: The derivative of $f$ is $f´(x)=2x+3$.
That means that on $(-\infty,-3/2)$ the function $f$ is decreasing because of $f´(x)<0$ and that on $(-3/2,+\infty)$ the function $f$ is increasing because of $f`(x)>0$. So it is obvious that the minimum is attained at $0$ and is equal to $f(0)=-1$.
Practically the same applies to $g$ because $g´(-3/2)=0$ and if $g`(x_0)=0$ then $x_0$ could be minimum but it doesn´t necessarily needs to be but for this $g$ it is also the case that $g´´(-3/2)=3$ so the second derivative shows that it is really minimum. You only need basics of differential caluculus, call it analysis if you want.
If you want theorems which are used, as you say, then:
1) If $f$ is differentiable on $(a,b)$ and $f`(x)<0$ on $(a,b)$ then $f$ is decreasing on $(a,b)$
2) If $f$ is differentiable on $(a,b)$ and $f`(x)>0$ on $(a,b)$ then $f$ is increasing on $(a,b)$
3) If $f$ is differentiable at the point $x_0$ nad $f`(x_0)=0$ then $x_0$ could be minimum, or maximum, or saddle point. To deduce the character of the point $x_0$ you need second derivative or sometimes even higher derivatives.

REPLY [2 votes]: Critical Points
Suppose $f:[a,b]\rightarrow \mathbb{R}$ is continuous. Call $x\in[a,b]$ is a critical point for $f$ on $[a,b]$ if either

$x$ is an endpoint.
$x$ is a stationary point: $f'(x)=0$.
$f$ is not differentiable at $x$.

Proposition: Closed Interval Method
Let $f:[a,b]\rightarrow\mathbb{R}$ be continuous. If $f$ attains it absolute maximum or absolute minimum at some point $x\in[a,b]$, then $x$ is a critical point for $f$ on $[a,b]$.
Proof: Suppose $f$ attains an absolute extremum at $x\in [a,b]$. If $x$ is an endpoint then $x$ is a critical point. Therefore suppose $x\in(a,b)$ and suppose further that $f$ is differentiable at $x$. Then $f'(x)=0$ and so $x$ is a critical point. If $f$ is not differentiable at $x$ then $x$ is still a critical point $\bullet$
Remark
Therefore to find the absolute maximum/ minimum of a continuous function on a closed interval, find the critical points $\{x_1,x_2,\dots,x_n\}$ and evaluate the function at each. Then the absolute maximum/ minimum is the maximum/ minimum of $\{f(x_i):i=1,\dots,n\}$.