TITLE: avoiding calculus
QUESTION [16 upvotes]: Some people may carelessly say that you need calculus to find such a thing as a local maximum of $f(x) = x^3 - 20x^2 + 96x$.  Certainly calculus is sufficient, but whether it's necessary is another question.
There's a global maximum if you restrict the domain to $[0,8]$, and $f$ is $0$ at the endpoints and positive between them.  Say the maximum is at $x_0$.  One would have
$$
\frac{f(x)-f(x_0)}{x-x_0}\begin{cases} >0 & \text{if }x<x_0, \\  <0 & \text{if }x>x_0. \end{cases}
$$
This difference quotient is undefined when $x=x_0$, but mere algebra tells us that the numerator factors and we get
$$
\frac{(x-x_0)g(x)}{x-x_0} = g(x)
$$
where $g(x)$ is a polynomial whose coefficients depend on $x_0$.  Then of course one seeks its zeros since it should change signs at $x_0$.
Have we tacitly used the intermediate value theorem, or the extreme value theorem?  To what extent can those be avoided?  Must one say that if there is a maximum point, then it is at a zero of $g(x)$?  And can we say that without the intermediate value theorem?  (At least in the case of this function, I think we stop short of needing the so-called fundamental theorem of algebra to tell us some zeros of $g$ exist!)

REPLY [3 votes]: There is a kind of algebraic structure called a real closed field. It's a structure with constants $0$, $1$, and has operations $+$ and $\cdot$ and a relation $<$.
The theory of real closed fields is interesting, because any (first-order) logical statement you can make using just $0,1,+,\cdot,<$ is true in a single real closed field if and only if it is true in all real closed fields. Similarly, the ones that are true can be proven using just first-order logic and the axioms of real closed fields -- i.e. there is a purely algebraic proof of it.
For example, for any particular algebraic function, the statement that a particular value is a local maximum can be expressed in first-order logic. Thus, if you can find a local maximum with calculus, you can find it with pure algebra too.
Of course, the axioms of real closed field typically include some variant of the intermediate value theorem: e.g.

every polynomial of odd degree has a root

This is the IVT, since an odd polynomial must be positive somewhere and negative somewhere. Of course, this can also be seen as a variant of the fundamental theorem of algebra too. Ultimately, in the purely algebraic setting, there isn't much difference between the two.