Theorem 15.1.1. Fermat’s Theorem.
If \(f(a)\) is a local extremum (either a maximum or a minimum) of \(f(x)\) at \(x=a\text{,}\) and \(f(x)\) is differentiable at \(x=a\) then \(f^\prime(a)=0\text{.}\)
Section 15.1 Fermat’s Theorem
It is by logic that we prove, but by intuition that we discover.―Henri Poincaré (1854–1912)
It is actually a little surprising how much effort it takes to prove the First Derivative Test. We will start by proving some preliminary results that will make it a little easier to follow the logic behind the proof of the First Derivative Test. We’ll begin with (Fermat’s Theorem 9.1.1) which says that if \(f\) attains a maximum (or minimum) at \(x=a\) then \(f^\prime(a)=0.\) More formally:
Recall that Fermat’gs Theorem does not say that if \(f^\prime(a)=0\) then \(f(a)\) is an extremum. In fact, we know that this is not true. Rather, it states the converse: If we know that \(f(a)\) is an extremum and \(f^\prime(a)\) exists, then \(f^\prime(a) =0\text{.}\)
It is very rare that we can develop a proof of a theorem by directly writing down the logical steps in order. Usually the process takes a lot of trying, backtracking, trying again, and so on much as we described in our analogy in Chapter 2 about finding your way out of a forest.
Of course, in a textbook it is not practical to list all of the bad ideas we might have just to see that they are, in fact, bad ideas. So we will use the following Scrapwork construct when we are just “thinking about” a problem. The scrapwork is not the proof. The purpose of scrapwork is to engage our intuition and to begin organizing our intuitive understanding so that a rigorous proof will emerge. So, not every statement we make inside a scrapwork construct will necessarily be fully rigorous. If you see a gap in the logic inside a scrapwork construct watch to see how it gets filled in the proof.
Scrapwork 15.1.
Notice that in Definition 13.2.3 the quantity \(\frac{f(a+h)-f(a)}{h}\) is the slope of a particular secant line, as in the sketch below.
If \((a,f(a))\) is a (local) maximum then the slope of the secant line in our diagram, \(\frac{f(a+h)-f(a)}{h}\text{,}\) must be negative when \(h\gt 0\text{.}\) Since the slope is less than zero it follows that
\begin{equation*}
f^\prime(a)=\limit{h}{0}{\frac{f(a+h)-f(a)}{h}}\le0.
\end{equation*}
Problem 15.1.2.
Draw a similar diagram to convince yourself that \(f^\prime(a)\) must also be greater than or equal to zero when \(h\lt 0\text{.}\)
We can now rigorously prove Fermat’s Theorem.
Proof of Fermat’s Theorem.
We will only prove the case when \(f(a)\) is a local maximum. The case of a local minimum is very similar. Since \(f(a)\) is a local maximum there is an interval containing \(a\) such that for any \(h\) (sufficiently small that \(a+h\) is also in the interval), \(f(a+h)\leq f(a)\text{.}\) Thus \(f(a+h)-f(a)\leq0\) as seen in the sketch above. If \(h\lt0\) then \(\frac{f(a+h)-f(a)}{h}\geq 0\) and so
\begin{equation}
f^\prime(a)=\limit{h}{0}{\frac{f(a+h)-f(a)}{h}}\ge0.\tag{15.1}
\end{equation}
However if \(h\gt0\) then
\begin{equation}
f^\prime(a)=\limit{h}{0}{\frac{f(a+h)-f(a)}{h}}\le0.\tag{15.2}
\end{equation}
The only way that inequalities in Formulas (15.1) and (15.2)) can be both true is if \(f^\prime(a)=0.\)
Problem 15.1.3.
Use our proof of the maximum case (Proof 15.1.1) of Fermat’s Theorem 15.1.1 as a guide to constructing a proof of Fermat’s Theorem when \(f(a)\) is a local minimum.
