📂Calculus

Fermat's Last Theorem Proof

Theorem¹

If the function $f(x)$ is either a maximum or a minimum at $x=c$ and $f ' (c)$ exists, then $f ' (c) = 0$

Explanation

While high school textbooks generally only introduce Rolle’s Theorem up to Rolle’s Theorem, to rigorously prove Rolle’s Theorem, one must be able to show why the derivative at a critical point is $0$, and Fermat’s Theorem guarantees that.

Proof

Strategy: Divide the proof into two cases: maxima and minima.

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• Case 1. $f(x)$ is a maximum at $x=c$

  For a sufficiently small positive number $h$, $f(c) \ge f(c \pm h)$, therefore

  $$ \lim _{h \to 0^+} {{f(c+h)-f(c)} \over h} \le 0 \quad \text{and} \quad \lim _{h \to 0^-} {{f(c+h)-f(c)} \over h} \ge 0 $$

  Since $\displaystyle f '(c) = \lim _{n \to 0} {{f(c+h)-f(c)} \over h}$ exists by assumption, $0 \le f '(c) \le 0$, and simplifying gives

  $$ f '(c)=0 $$

• Case 2. $f(x)$ is a minimum at $x=c$

  For a sufficiently small positive number $h$, $f(c) \le f(c \pm h)$, therefore

  $$ \lim _{h \to 0^+} {{f(c+h)-f(c)} \over h} \ge 0 \quad \text{and} \quad \lim _{h \to 0^-} {{f(c+h)-f(c)} \over h} \le 0 $$

  Since $\displaystyle f '(c) = \lim _{n \to 0} {{f(c+h)-f(c)} \over h}$ exists by assumption, $0 \le f '(c) \le 0$ and simplifying gives

  $$f ' (c)=0$$

Therefore, in either case, if $c$ is a critical point and $f ' (c)$ exists, then $f ' (c) = 0$ must be true.

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