Analyzing Functions with Derivatives: Monotonicity & Extrema
From Fermat's Theorem to Applied Optimization — A Complete Treatment
The Bridge Theorems
Three classical results form a logical chain that connects the local behaviour of a derivative to the global behaviour of a function. Each theorem builds on the one before it.
1.1 Fermat's Theorem (Interior Extremum Theorem)
Statement. Let $f$ be defined on an open interval containing $c$. If $f$ has a local extremum at $c$ and $f$ is differentiable at $c$, then
$$f'(c) = 0.$$
Proof
Suppose $f$ has a local maximum at $c$. Then there exists $\delta > 0$ such that $f(c) \ge f(x)$ for all $x$ with $|x - c| < \delta$. Consider the difference quotient:
- For $h > 0$: $\;\displaystyle\frac{f(c+h)-f(c)}{h} \le 0$ (numerator $\le 0$, denominator $> 0$).
- For $h < 0$: $\;\displaystyle\frac{f(c+h)-f(c)}{h} \ge 0$ (numerator $\le 0$, denominator $< 0$).
Taking the limit as $h \to 0$:
$$f'(c) = \lim_{h\to 0^+}\frac{f(c+h)-f(c)}{h} \le 0 \quad\text{and}\quad f'(c) = \lim_{h\to 0^-}\frac{f(c+h)-f(c)}{h} \ge 0.$$Both inequalities force $f'(c) = 0$. The argument for a local minimum is analogous (inequalities reverse). $\square$
Geometric Meaning
At a local extremum of a smooth curve, the tangent line is horizontal. The slope of the tangent equals zero.
⚠ Converse is FALSE. $f'(c)=0$ does not guarantee an extremum. The classic counterexample is $f(x)=x^3$ at $x=0$: we have $f'(0)=0$, yet $f$ is strictly increasing through $0$.
1.2 Rolle's Theorem
Statement. If $f$ is
- continuous on $[a,b]$,
- differentiable on $(a,b)$, and
- $f(a) = f(b)$,