Fermat's stationary-point theorem

1. Suppose \(f\) has a local maximum at \(c \in (a,b)\) and \(f'(c)\) exists.Proof: by definition of local maximum, there is \(\delta > 0\) with \(f(x) \le f(c)\) for all \(x \in (c-\delta, c+\delta) \subseteq (a,b)\).

2. The left-hand derivative is \(\ge 0\).Proof: for \(h < 0\) with \(|h| < \delta\), \(\frac{f(c+h) - f(c)}{h} \ge 0\) (numerator \(\le 0\), denominator \(< 0\)). Hence \(f'(c) = \lim_{h\to 0^-}\frac{f(c+h)-f(c)}{h} \ge 0\).

3. The right-hand derivative is \(\le 0\).Proof: for \(h > 0\) with \(h < \delta\), \(\frac{f(c+h) - f(c)}{h} \le 0\) (numerator \(\le 0\), denominator \(> 0\)). Hence \(f'(c) = \lim_{h\to 0^+}\frac{f(c+h)-f(c)}{h} \le 0\).

4. Conclude.Proof: \(f'(c) \ge 0\) and \(f'(c) \le 0\), so \(f'(c) = 0\).(For a local minimum, apply the argument to \(-f\).)

5. Q.E.D.